A compound with a molecular formula C7H14O produced the following high resolution 1H NMR spectrum.
\nA compound with a molecular formula C7H14O produced the following high resolution 1H NMR spectrum.
\nDeduce what information can be obtained from the 1H NMR spectrum.
\nIdentify the functional group that shows stretching at 1710 cm–1 in the infrared spectrum of this compound using section 26 of the data booklet and the 1H NMR.
\nSuggest the structural formula of this compound.
\nBromine was added to hexane, hex-1-ene and benzene. Identify the compound(s) which will react with bromine in a well-lit laboratory.
\nDeduce the structural formula of the main organic product when hex-1-ene reacts with hydrogen bromide.
\nState the reagents and the name of the mechanism for the nitration of benzene.
\nOutline, in terms of the bonding present, why the reaction conditions of halogenation are different for alkanes and benzene.
\nBelow are two isomers, A and B, with the molecular formula C4H9Br.
\nExplain the mechanism of the nucleophilic substitution reaction with NaOH(aq) for the isomer that reacts almost exclusively by an SN2 mechanism using curly arrows to represent the movement of electron pairs.
\nNumber of hydrogen environments: 3
\nRatio of hydrogen environments: 2:3:9
\nSplitting patterns: «all» singlets
\n\n
Accept any equivalent ratios such as 9:3:2.
\nAccept “no splitting”.
\n[3 marks]
\ncarbonyl
OR
C=O
\n
Accept “ketone” but not “aldehyde”.
\n[1 mark]
\nAccept (CH3)3CCH2COCH3.
\nAward [1] for any aldehyde or ketone with C7H14O structural formula.
\n[2 marks]
\nhexane AND hex-1-ene
\n\n
Accept “benzene AND hexane AND hex-1-ene”.
\n[1 mark]
\nCH3CH2CH2CH2CHBrCH3
\n\n
Accept displayed formula but not molecular formula.
\n[1 mark]
\nReagents: «concentrated» sulfuric acid AND «concentrated» nitric acid
\nName of mechanism: electrophilic substitution
\n[2 marks]
\nbenzene has «delocalized» bonds «that are susceptible to electrophile attack» AND alkanes do not
\n\n
Do not accept “benzene has single and double bonds”.
\n[1 mark]
\ncurly arrow going from lone pair/negative charge on O in –OH to C
\ncurly arrow showing Br leaving
\nrepresentation of transition state showing negative charge, square brackets and partial bonds
\n\n
\n
Accept OH– with or without the lone pair.
\nDo not allow curly arrows originating on H in OH–.
\nAccept curly arrows in the transition state.
\nDo not penalize if HO and Br are not at 180°.
\nDo not award M3 if OH–C bond is represented.
\nAward [2 max] if wrong isomer is used.
\n[3 marks]
\nEDTA is produced by reacting ethane-1,2-diamine with chloroethanoic acid, ClCH2COOH.
\nIdentify the other product formed.
\nExplain why EDTA, a chelating agent, is more effective in removing heavy metal ions from solution than monodentate ligands.
\nHCl/hydrogen chloride
\n\n
Accept “hydrochloric acid”.
\n[1 mark]
\nforms four/six/several/multiple coordinate/coordination bonds «to a central metal ion»
OR
is a polydentate/tetradentate/hexadentate ligand
forms more stable complex/stronger bonds with central metal ion
OR
increases entropy/S by releasing smaller «monodentate ligand» molecules previously complexed
complex ions are much larger «and can be removed easily due to large size of chelate complexes»
OR
heavy metal ions trapped inside the ligand/become «biologically» inactive/nontoxic/harmless
\n
Accept “dative «covalent»” for “coordinate/coordination”.
\nDo not accept just “chelates”.
\n[3 marks]
\nRhodium and palladium are often used together in catalytic converters. Rhodium is a good reduction catalyst whereas palladium is a good oxidation catalyst.
\nNickel(II) ions are least soluble at pH 10.5. Calculate the molar solubility of nickel(II) hydroxide at this pH. KspNi(OH)2 = 5.48 × 10–16.
\nRhodium is paramagnetic with an electron configuration of [Kr] 5s14d8.
\nExplain, in terms of electron spin pairing, why paramagnetic substances are attracted to a magnetic field and diamagnetic substances are not.
\nRhodium is a type 1 superconductor.
\nSketch graphs of resistance against temperature for a conductor and superconductor.
\n![\"M17/4/CHEMI/HP3/ENG/TZ2/05.c.ii\"](\"../assets/uploads/tinymce_asset/asset/3701/Schermafbeelding_2017-09-21_om_13.24.04.png\"/)
Contrast type 1 and type 2 superconductors by referring to three differences between them.
\nKsp= [Ni2+][OH−]2
OR
5.48 x 10−16 = [Ni2+][10−3.5]2
«[Ni2+] =» 5.48 x 10−9 «mol dm−3»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\nparamagnetic materials have unpaired electrons
OR
diamagnetic materials have all electrons «spin-»paired
unpaired electrons align with an external magnetic field
OR
paired electrons are not influenced by magnetic field
\n
Accept “diamagnetic materials have no unpaired electrons\" for M1.
\n[2 marks]
\n![\"M17/4/CHEMI/HP3/ENG/TZ2/05.c.ii_01/M\"](\"../assets/uploads/tinymce_asset/asset/3702/Schermafbeelding_2017-09-21_om_13.25.44.png\"/)
![\"M17/4/CHEMI/HP3/ENG/TZ2/05.c.ii_02/M\"](\"../assets/uploads/tinymce_asset/asset/3703/Schermafbeelding_2017-09-21_om_13.27.18.png\"/)
\n
Conductor:
Accept any concave upwards curve or line showing resistance increasing with temperature. There should be a
y-axis intercept. Do not accept x-axis intercept for conductor.
Superconductor:
Sharp transition with vertical line to x-axis. Greater than Tc, accept any concave upwards curve or line showing resistance increasing with temperature.
[2 marks]
\nAny three of:
\ntype 1 have lower critical temperature/Tc «than type 2»
OR
type 2 can superconduct at higher temperatures «than type 1»
type 1 are «elemental» metals AND type 2 can be alloys/composites/metal oxide ceramics/perovskites
\ntype 1 have sharp transition to superconductivity AND type 2 have more gradual transition
\ntype 1 have all «magnetic» flux expelled to normal state AND type 2 have partial penetration of flux in mixed state
\ntype 1 typically work via Cooper pairs AND type 2 may not necessarily use this mechanism
\nmagnetic fields can penetrate type 2 in the mixed state «in a type of Vortex» AND type 1 has no mixed state
\ntype 1 have one critical magnetic field/Bc AND type 2 have two/Bc1 and Bc2
\n\n
Award [1 max] if three correct pieces of information are given for one type only without contrasting with the other type.
\nMarks may also be awarded from suitable sketch(es).
\nAccept “H” for “B”.
\n[3 marks]
\nSoluble acids and bases ionize in water.
\nA solution containing 0.510 g of an unknown monoprotic acid, HA, was titrated with 0.100 mol dm–3 NaOH(aq). 25.0 cm3 was required to reach the equivalence point.
\nThe following curve was obtained using a pH probe.
\nState, giving a reason, the strength of the acid.
\nState a technique other than a pH titration that can be used to detect the equivalence point.
\nDeduce the pKa for this acid.
\nThe pKa of an anthocyanin is 4.35. Determine the pH of a 1.60 × 10–3 mol dm–3 solution to two decimal places.
\nweak AND pH at equivalence greater than 7
OR
weak acid AND forms a buffer region
[1 mark]
\ncalorimetry
OR
measurement of heat/temperature
OR
conductivity measurement
\n
Accept “indicator” but not “universal indicator”.
\n[1 mark]
\n«pKa = pH at half-equivalence =» 5.0
\n[1 mark]
\nKa =
\n[H3O+] = «mol dm–3»
\npH = «» 3.57
\n\n
Award [3] for correct final answer to two decimal places.
\nIf quadratic equation used, then: [H3O+] = 2.459 × 10–4 «mol dm–3» and pH = 3.61
\n[3 marks]
\nThe Bombardier beetle sprays a mixture of hydroquinone and hydrogen peroxide to fight off predators. The reaction equation to produce the spray can be written as:
\n| C6H4(OH)2(aq) + H2O2(aq) | \n→ | \nC6H4O2(aq) + 2H2O(l) | \n
| hydroquinone | \n\n | quinone | \n
Hydrogenation of propene produces propane. Calculate the standard entropy change, ΔS θ, for the hydrogenation of propene.
\nThe standard enthalpy change, ΔH θ, for the hydrogenation of propene is –124.4 kJ mol–1. Predict the temperature above which the hydrogenation reaction is not spontaneous.
\n«ΔS θ =» 270 «J K–1 mol–1» – 267 «J K–1 mol–1» – 131 «J K–1 mol–1»
\n«ΔS θ =» –128 «J K–1 mol–1»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\n«non spontaneous if» ΔG θ = ΔH θ – TΔS θ > 0
OR
ΔH θ > TΔS θ
«T above» » 972 «K»
\n\n
Award [2] for correct final answer.
\nAccept 699 °C.
\nDo not award M2 for any negative T value.
\n[2 marks]
\nThe structures of the amino acids cysteine, glutamine and lysine are given in section 33 of the data booklet.
\nAn aqueous buffer solution contains both the zwitterion and the anionic forms of alanine. Draw the zwitterion of alanine.
\nCalculate the pH of a buffer solution which contains 0.700 mol dm–3 of the zwitterion and 0.500 mol dm–3 of the anionic form of alanine.
\nAlanine pKa = 9.87.
\n\n
Penalize incorrect bond linkages or missing hydrogens once only in 8 (a) and 8 (c) (i).
\n[1 mark]
\n«pH = 9.87 + log»
\n«= 9.87 – 0.146»
\n= 9.72
\n\n
pH can be deduced by an alternative method.
\n[1 mark]
\nThere is a link between world energy consumption and carbon dioxide production.
\nClimate induced changes in the ocean can be studied using measurements such as the Atmospheric Potential Oxygen (APO). Trends in APO concentration from two stations, one in each hemisphere, are shown below.
\nTrends in atmospheric potential oxygen (APO) based on monthly averages between 1990 and 2010.
\n[Source: www.ioos.noaa.gov]
\nThe following graph represents world energy consumption by type for the years 1988–2013.
\nEstimate the percentage of energy consumption which did not directly produce CO2 in 2013.
\nO2 is consumed in producing CO2 for electricity generation. The graph shows the relationship between the world’s electricity generation and CO2 production between 1994 and 2013.
\nCalculate the mass, in million tonnes, of oxygen gas ultimately found in CO2 which is consumed in generating 18000 terawatts of electricity using the equation given for the best fit line. Give your answer to 2 significant figures.
\nAssume coal is the only energy source.
\nThe equilibrium expression for O2 exchange between the atmosphere and ocean is O2(g) O2(aq). Identify one factor which shifts the equilibrium to the right.
\nFactors such as photosynthesis and respiration are excluded so that APO is influenced by oceanic changes only. Suggest why the seasonal cycles from Alert station and Cape Grim observatory are different.
\nThe change in APO O2/N2 ratio, per meg, is measured relative to an O2/N2 reference.
\n\n
Calculate the APO Δ(O2/N2) value for an oxygen concentration of 209400 ppm assuming that any change in N2 concentration is negligible. Reference values for O2 and N2 are 209 460 and 790 190 ppm respectively.
\nSuggest a reason for the general negative gradient of the APO curve given in (c).
\n«»
\n«» 13 «%»
\n\n
Accept range of “11–16%”.
\n[1 mark]
\n«18000 = 0.54x – 2000»
x = 37037 «million tonnes of CO2»
« x 37037 = 26930»
27000/2.7 x 104 «million tonnes of O2»
\n
Accept “37000 «million tonnes of CO2»” for M1.
\nAward [2] for correct final answer with two significant figures.
\nAward [1] for non rounded answers in range 26903–26936 «million tonnes of O2».
\n[2 marks]
\nincrease in «atmospheric» pressure
OR
increase in [O2(g)]/concentration of O2(g)
OR
decrease in [O2(aq)]/concentration of O2(aq)
OR
decrease in temperature
\n
Accept “increase in volume of oceans «due to polar ice cap melting»” OR “consumption of O2 in oceans/O2(aq)
«by living organisms»”.
State symbols required for oxygen concentration.
\n[1 mark]
\nsummer in one station while winter in other
OR
stations are at different latitudes
oxygen dissolves better in colder water
\n\n
Accept “opposite seasons «in each hemisphere»”.
\nDo not accept “different locations with different temperatures” OR “stations are in different hemispheres”.
\n[2 marks]
\n« =» − 286.5 «per meg»
\n\n
The nitrogen cancels so is not needed in the calculation.
\nNegative sign required for mark.
\n[1 mark]
\ndecrease in [O2]/concentration of O2
OR
increasing combustion of fossil fuels «consumes more O2 so [O2]/concentration of O2 decreases»
OR
warmer oceans/seas/water «as oxygen dissolves better in colder water»
OR
deforestation
\n
Accept “decrease in level of O2”.
\nAccept \"increasing CO2 production «consumes more O2 so [O2]/concentration of O2 decreases»\".
\nDo not accept “decrease in amount of O2” OR “increase in greenhouse gases”.
\n[1 mark]
\nRetinal, one of the many forms of vitamin A, reacts with opsin to produce rhodopsin. Refer to section 35 of the data booklet for one structure of vitamin A.
\nIdentify the structural feature which enables rhodopsin to absorb visible light.
\nOutline the change that occurs in the retinal residue during the absorption of visible light.
\n«extensive system of» conjugation/alternating single and double «carbon to carbon» bonds
OR
delocalized electrons «over much of the molecule»
\n
Accept “delocalization”.
\n[1 mark]
\ncis«-retinal» converts to trans«-retinal»
OR
one of the C=C «fragments in retinal» changes «its configuration» from cis to trans
[1 mark]
\nThe graph of the rate of an enzyme-catalyzed reaction is shown below.
\n![\"M17/4/CHEMI/HP3/ENG/TZ2/13\"](\"../assets/uploads/tinymce_asset/asset/3704/Schermafbeelding_2017-09-21_om_13.52.27.png\"/)
Determine the value of the Michaelis constant, Km, including units, from the graph.
\nSketch a second graph on the same axes to show how the reaction rate varies when a competitive inhibitor is present.
\nOutline the significance of the value of Km.
\n«Km = [substrate] at Vmax»
\n4.2 x 10–3
\nmol dm–3
\n\n
Accept answers in the range of 3.5 x 10–3 to 5.0 x 10–3 mol dm–3.
\nM2 can be scored independently.
\n[2 marks]
\ngraph to right of curve AND finish at same Vmax
\n\n
Do not penalize if curve does not finish exactly at same Vmax as long as it is close to it (since drawn curve does not
flatten out completely at Vmax = 0.50).
[1 mark]
\nKm is inverse measure of affinity of enzyme for a substrate / Km is inversely proportional to enzyme activity
OR
high value of Km indicates higher substrate concentration needed for enzyme saturation
OR
low value of Km means reaction is fast at low substrate concentration
\n
Idea of inverse relationship must be conveyed.
\nAccept “high value of Km indicates low affinity of enzyme for substrate/less stable ES complex/lower enzyme activity”.
\nAccept “low value of Km indicates high affinity of enzyme for substrate/stable ES complex/greater enzyme activity”.
\n[1 mark]
\nDisposable plastic lighters contain butane gas. In order to determine the molar mass of butane, the gas can be collected over water as illustrated below:
\nList the data the student would need to collect in this experiment.
\nExplain why this experiment might give a low result for the molar mass of butane.
\nSuggest one improvement to the investigation.
\nmass/m of lighter before AND after the experiment
\nvolume of gas/Vgas «collected in the cylinder»
\n«ambient» pressure/P «of the room»
\ntemperature/T
\n\n
Accept “change in mass of lighter”.
\nAccept “weight” for “mass”.
\nDo not accept just “mass of lighter/gas”.
\nAccept “volume of water displaced”.
\nDo not accept “amount” for “volume” or “mass”.
\n[4 marks]
\nAny two of:
\npressure of gas not equalized with atmospheric/room pressure
\ntoo large a recorded volume «of gas produces a lower value for molar mass of butane»
OR
cylinder tilted
difficult to dry lighter «after experiment»
OR
higher mass of lighter due to moisture
OR
smaller change in mass but same volume «produces lower value for molar mass of butane»
using degrees Celcius/°C instead of Kelvin/K for temperature
\n\n
Accept “vapour pressure of water not accounted for” OR “incorrect vapour pressure of water used” OR “air bubbles trapped in cylinder”. Do not accept “gas/bubbles escaping «the cylinder»” or other results leading to a larger molar mass.
\nAccept “lighter might contain mixture of propane and butane”.
\nDo not accept only “human errors” OR “faulty equipment” (without a clear explanation given for each) or “mistakes in calculations”.
\n[2 marks]
\nrecord vapour pressure of water «at that temperature»
OR
equalize pressure of gas in cylinder with atmospheric/room pressure
OR
tap cylinder before experiment «to dislodge trapped air»
OR
collect gas using a «gas» syringe/eudiometer/narrower/more precise graduated tube
OR
collect gas through tubing «so lighter does not get wet»
OR
dry lighter «before and after experiment»
OR
hold «measuring» cylinder vertical
OR
commence experiment with cylinder filled with water
\n
Accept “adjust cylinder «up or down» to ensure water level inside cylinder matches level outside”.
\nAccept “repeat experiment/readings «to eliminate random errors»”.
\nAccept “use pure butane gas”.
\n[1 mark]
\nA hemoglobin-oxygen saturation curve does not follow the same model as enzyme-substrate reactions.
\n![\"M17/4/CHEMI/HP3/ENG/TZ2/14\"](\"../assets/uploads/tinymce_asset/asset/3705/Schermafbeelding_2017-09-21_om_14.26.13.png\"/)
Explain the shape of the curve from 0 to X kPa.
\nExplain why carbon monoxide is toxic to humans.
\noxygen binds to first active site «of deoxygenated heme» AND alters shape of other active sites
OR
cooperative binding
affinity of partially oxygenated hemoglobin for oxygen increases
\n[2 marks]
\nCO is a competitive inhibitor «of oxygen binding to hemoglobin»
OR
CO has greater affinity for hemoglobin «than oxygen»
less oxygen is transported
OR
uptake of oxygen decreases
OR
causes hypoxia
\n
Do not penalize “CO binds irreversibly” if included in answer.
\n[2 marks]
\nDNA is a complex molecule.
\nOutline how its structure allows it to be negatively charged in the body.
\nDeduce the nucleotide sequence of a complementary strand of a fragment of DNA with the nucleotide sequence –GACGGATCA–.
\nphosphate groups «in nucleotides fragments are almost completely» ionized
\n\n
Do not accept just “phosphate «groups»”.
\n[1 mark]
\n–CTGCCTAGT–
\n[1 mark]
\nPolymer nanocomposites often have better structural performance than conventional materials. Lithographic etching and metal coordination are two methods of assembling these nanocomposites.
\nNanoparticles anchor plasticizers in PVC so that they cannot escape from the polymer as easily.
\nState the two distinct phases of a composite.
\nIdentify the methods of assembling nanocomposites by completing the table.
\nExplain how the structure of plasticizers enables them to soften PVC.
\nSuggest a reason why nanoparticles can better anchor plasticizers in the polymer.
\nreinforcing «phase»
\n«embedded in» matrix «phase»
\n[2 marks]
\n\n
Award [2] for all 4, [1] for 2 or 3 correct.
\n[2 marks]
\nAny three of:
\ncontain a polar group «which locks into the polymer»
\na non-polar group «which weakens the forces between chains»
\nembedded between chains of polymers
\n«plasticizer molecules» fit between chains
\n«plasticizer molecules» prevent chains from forming crystalline regions
\n«plasticizer molecules» keeps strands/chains/molecules separated
\n«plasticizer molecules» increase space/volume between chains
\nweakens intermolecular/dipole-dipole/London/dispersion/instantaneous induced dipole-induced dipole/van der Waals/vdW forces
\n\n
Do not accept “«plasticizer molecules» “lower density” or “softer”.
\n[3 marks]
\nmore places «for plasticizers» to bond
OR
increased surface area
[1 mark]
\nThe sun is the main source of energy used on earth.
\nOne fusion reaction occurring in the sun is the fusion of deuterium, , with tritium, , to form helium, . State a nuclear equation for this reaction.
\nExplain why this fusion reaction releases energy by using section 36 of the data booklet.
\nState the technique used to show that the sun is mainly composed of hydrogen and helium.
\nColoured molecules absorb sunlight. Identify the bonding characteristics of such molecules.
\n+ → +
\n\n
Accept \"n\" for \"\"
\nAccept \"2H + 3H → 4He + 1n/n\".
\n[1 mark]
\nhigher binding energy/BE «per nucleon» for helium/products
OR
nucleons in products more tightly bound
mass defect/lost matter converted to energy
\n\n
Accept converse statement in M1.
\nAccept “mass deficit” for “mass defect”.
\n[2 marks]
\nspectrometry
\n\n
Accept “spectroscopy” for “spectrometry” OR more specific techniques such as “atomic absorption spectrometry/AAS”, “astrophotometry” etc. Do not award mark for incorrect specific spectrometric techniques.
\nDo not accept “spectrum”.
\n[1 mark]
\n«extensive system of» conjugation/alternating single and double «carbon to carbon» bonds
OR
delocalized electrons «over much of the molecule»
[1 mark]
\nThe sun is the main source of energy used on earth.
\nCalculate the energy released, in MeV, in this reaction, using section 36 of the data booklet.
\nΔBE = BE(4He) − (BE(2H) + BE(3H))
OR
ΔBE = 4 x 7.1 − (2 x 1.1 + 3 x 2.8)
= 17.8 «MeV»
\n
Accept answers in range 17.3 to 18.1 «MeV».
\nAward [1 max] for final answers in range of 3.0 to 3.4 «MeV».
\nAward [2] for correct final answer.
\n[2 marks]
\nInfrared (IR) spectra can be used to distinguish between various types of plastics. Some simplified IR spectra are given here.
\nExplain, with a reference to molecular structure, which two of the plastics can not be distinguished by IR spectroscopy.
\n\n
HDPE AND LDPE «have similar IR»
\nboth are polyethene/polyethylene
OR
only branching differs
OR
same bonds
OR
same bending/stretching/vibrations
\n
Accept “water bottle AND water bottle cap” for M1.
\n[2 marks]
\nRhodium and palladium are often used together in catalytic converters. Rhodium is a good reduction catalyst whereas palladium is a good oxidation catalyst.
\nIn a catalytic converter, carbon monoxide is converted to carbon dioxide. Outline the process for this conversion referring to the metal used.
\nNickel is also used as a catalyst. It is processed from an ore until nickel(II) chloride solution is obtained. Identify one metal, using sections 24 and 25 of the data booklet, which will not react with water and can be used to extract nickel from the solution.
\nDeduce the redox equation for the reaction of nickel(II) chloride solution with the metal identified in (b)(i).
\nAnother method of obtaining nickel is by electrolysis of a nickel(II) chloride solution. Calculate the mass of nickel, in g, obtained by passing a current of 2.50 A through the solution for exactly 1 hour. Charge (Q) = current (I) × time (t).
\ncarbon monoxide/CO adsorbs onto palladium/Pd
\nbonds stretched/weakened/broken
OR
«new» bonds formed
OR
activation energy/Ea «barrier» lowered «in both forward and reverse reactions»
products/CO2 desorb «from catalyst surface»
\n[3 marks]
\nFe/iron
OR
Zn/zinc
OR
Co/cobalt
OR
Cd/cadmium
OR
Cr/chromium
\n
Accept “Mn/manganese”.
\n[1 mark]
\n\n
Ni2+(aq) + Fe(s) → Ni(s) + Fe2+(aq)
OR
Ni2+(aq) + Zn(s) → Ni(s) + Zn2+(aq)
OR
Ni2+(aq) + Co(s) → Ni(s) + Co2+(aq)
OR
Ni2+(aq) + Cd(s) → Ni(s) + Cd2+(aq)
OR
Ni2+(aq) + Cr(s) → Ni(s) + Cr2+(aq)
\n
Accept “3Ni2+(aq) + 2Cr(s) → 3Ni(s) + 2Cr3+(aq)”.
\nDo not penalize similar equations involving formation of Fe3+(aq), Mn2+(aq) OR Co3+(aq).
\nIgnore Cl− ions.
\nAccept correctly balanced non-ionic equations eg, “NiCl2(aq) + Zn(s) → Ni(s) + ZnCl2(aq)” etc.
\nDo not allow ECF from (b)(i).
\n[2 mark]
\n «» = 0.09326 «mol»
OR
«» = 0.04663 «mol»
«= 0.04663 mol x 58.69 gmol–1» = 2.74 «g»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\nThere are many sources of energy available.
\nMethanol fuel cells provide a portable energy source. The process can be represented by the overall equation CH3OH(aq) + O2(g) → CO2(g) + 2H2O(g).
\nDeduce the half-cell equations occurring at each electrode during discharge.
\nOutline the function of the proton-exchange membrane (PEM) in the fuel cell.
\nExplain how the flow of ions allows for the operation of the fuel cell.
\nAnode (negative electrode):
CH3OH(aq) + H2O(l) → 6H+(aq) + 6e− + CO2(g)
Cathode (positive electrode):
O2(g) + 6H+(aq) + 6e− → 3H2O(l)
\n
Award [1 max] for correct equations at wrong electrode.
\nAccept “e” for “e–”.
\nAccept “O2(g) + 4H+(aq) + 4e− → 2H2O(l)”.
\n[2 marks]
\nallows H+/ions pass through/diffuse/move «from anode to cathode but not electrons or small molecules»
\n\n
Accept “acts as a salt bridge”.
\n[1 mark]
\nH+/ions pass through/diffuse/move from anode/negative electrode «through membrane» to cathode/positive electrode
\nH+/ions used to reduce oxygen at cathode/positive electrode
\n\n
Oxygen must be mentioned for M2.
\n[2 marks]
\nLiquid Crystal on Silicon, LCoS, uses liquid crystals to control pixel brightness. The degree of rotation of plane polarized light is controlled by the voltage received from the silicon chip.
\nTwo important properties of a liquid crystal molecule are being a polar molecule and having a long alkyl chain. Explain why these are essential components of a liquid crystal molecule.
\nMetal impurities during the production of LCoS can be analysed using ICP-MS. Each metal has a detection limit below which the uncertainty of data is too high to be valid. Suggest one factor which might influence a detection limit in ICP-MS/ICP-OES.
\nPolar molecule:
«orientation of molecule» influenced by electric field/«applied» voltage/«applied» potential «difference»/«applied» current
OR
can be switched on and off
Long alkyl chain:
prevent close packing of molecules
OR
molecules can align
OR
reduces the melting point of the liquid crystal/LC «phase making liquid at room temperature»
\n
Accept “makes molecule rod-shaped” for M2.
\n[2 marks]
\ninability to replicate calibrations below certain levels
OR
variation in methodology
OR
variation between machines calibrated with the same samples
OR
variation in plasma torches
OR
different detection limits for MS AND OES
OR
interference from solvents/other chemicals
OR
inability to produce pure standards
OR
chance that low signal AND blank are same
[1 mark]
\nThe combustion of fossil fuels produces large amounts of CO2, a greenhouse gas.
\nThe diagram below illustrates a range of wavelengths in the electromagnetic spectrum.
\nThe structures of 11-cis-retinal and β-carotene are given in section 35 of the data booklet. Suggest a possible wavelength of light absorbed by each molecule using section 3 of the data booklet.
\nboth between 400–700 «nm»
\nβ-carotene at higher wavelength than retinal
\n\n
Accept any wavelength within the 400-700 nm visible region range for M1 and any higher wavelength for β-carotene
within the same region for M2.
[2 marks]
\nThe structures of the amino acids cysteine, glutamine and lysine are given in section 33 of the data booklet.
\nDeduce the structural formula of the dipeptide Cys-Lys.
\nIdentify the type of bond between two cysteine residues in the tertiary structure of a protein.
\nDeduce the structural formula of the predominant form of cysteine at pH 1.0.
\nA mixture of the three amino acids, cysteine, glutamine and lysine, was placed in the centre of a square plate covered in polyacrylamide gel. The gel was saturated with a buffer solution of pH 6.0. Electrodes were connected to opposite sides of the gel and a potential difference was applied.
\nSketch lines on the diagram to show the relative positions of the three amino acids after electrophoresis.
\n![\"M17/4/CHEMI/SP3/ENG/TZ2/02.d\"](\"../assets/uploads/tinymce_asset/asset/3730/Schermafbeelding_2017-09-25_om_18.00.55.png\"/)
\n
correct order
\namide link
\n\n
Accept CO–NH but not CO–HN for amide link.
\nPenalize incorrect bond linkages or missing hydrogens once only in 7 (a) and 7 (c).
\n[2 marks]
\ncovalent
\n\n
Accept “S-S/disulfide”.
\n[1 mark]
\n\n
Penalize incorrect bond linkages or missing hydrogens once only in 7 (a) and 7 (c).
\n[1 mark]
\nCys and Gln move to positive electrode AND Lys to negative electrode
\nCys further to positive electrode than Gln
\n\n
Do not penalize if lines are omitted or if different markings are given (eg, spots etc.), as long as relative positions are correctly indicated.
\nAccept Gln on original position indicated.
\nAward [1 max] for reverse order of amino acids.
\n[2 marks]
\nA Grätzel dye-sensitized solar cell (DSSC) and a silicon based photovoltaic cell both convert solar energy into electrical energy by producing a charge separation.
\nContrast how absorption of photons and charge separation occur in each device.
\nSuggest one advantage a DSSC has over a silicon based photovoltaic cell.
\n\n
Accept “existence of holes AND electrons at p-n junction” for M2.
\n[4 marks]
\nAny of:
\ncheaper
OR
ease of fabrication
OR
use light of lower energy/lower frequency/longer wavelength
OR
plentiful and renewable resources «to construct DSSC cells»
OR
operate at lower «internal» temperatures/better at radiating heat away «since constructed with thin front layer of conductive plastic compared to glass box in photovoltaic cell»
OR
use of nanoparticles provides large surface area exposure to sunlight/sun/light
OR
can absorb better under cloudy conditions
OR
better conductivity
OR
more flexible
\n
Accept “lower mass/lighter «so greater flexibility to integrate into windows etc.»” OR “greater power-conversion efficiency «with latest DSSC models»”.
\n[1 mark]
\nWhich statements are correct for a voltaic cell?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nWhat is the order of increasing boiling point?
\nA. C4H10 < CH3COOH < CH3CH2CHO < CH3CH2CH2OH
\nB. C4H10 < CH3CH2CHO < CH3CH2CH2OH < CH3COOH
\nC. CH3COOH < CH3CH2CH2OH< CH3CH2CHO < C4H10
\nD. C4H10 < CH3CH2CH2OH < CH3CH2CHO < CH3COOH
\nB
\nWhat are the functional groups in the aspirin molecule?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nAspirin is one of the most widely used drugs in the world.
\nAspirin was synthesized from 2.65 g of salicylic acid (2-hydroxybenzoic acid) (Mr = 138.13) and 2.51 g of ethanoic anhydride (Mr = 102.10).
\nSuggest two absorbances, other than the absorbances due to the ring structure and C–H bonds, that would be present in the infrared (IR) spectrum of aspirin.
\nState two techniques, other than IR spectroscopy, which could be used to confirm the identity of aspirin.
\nAny two of:
\n2500–3000 «cm–1» / «absorbance» due to O–H in carboxyl
\n1700–1750 «cm–1» / «absorbance» due to C=O in carboxyl/ethanoate
\n1050–1410 «cm–1» / «absorbance» due to C–O bond in carboxyl/ethanoate
\n\n
Accept “carboxylic acid” for “carboxyl”, “acetate/ester” for “ethanoate”.
\nAccept specific wavenumber once within indicated range.
\nDo not award mark if reference is made to an alcohol/ether.
\n[2 marks]
\nAny two of:
\nmelting point
\nmass spectrometry/MS
\nhigh-performance liquid chromatography/HPLC
\nNMR/nuclear magnetic resonance
\nX-ray crystallography
\nelemental analysis
\n\n
Accept “spectroscopy” instead of “spectrometry” where mentioned but not “spectrum”.
\nAccept “ultraviolet «-visible» spectroscopy/UV/UV-Vis”.
\nDo not accept “gas chromatography/GC”.
\nAccept “thin-layer chromatography/TLC” as an alternative to “HPLC”.
\n[2 marks]
\nWhat is the name of the compound with this molecular structure applying IUPAC rules?
\nA. 1-methylpropanoic acid
\nB. 2-methylpropanoic acid
\nC. 2-methylbutanoic acid
\nD. 3-methylbutanoic acid
\nC
\nThe structures of morphine, diamorphine and codeine are given in section 37 of the data booklet.
\nMethadone is used to treat heroin addiction. 1H NMR spectroscopy can be used to study its structure.
\nPredict the number of different hydrogen environments in the molecule ignoring the benzene rings.
\nPredict the chemical shift and the splitting pattern seen for the hydrogens on the carbon atom circled in the diagram. Use section 27 of the data booklet.
\n6
\n[1 mark]
\nChemical shift:
2.2–2.7 «ppm»
Splitting pattern:
quartet/q
[2 marks]
\nWhich molecule has a tertiary nitrogen?
\nA. (CH3)2NH
\nB. (C2H5)4N+I−
\nC. C3H7N(CH3)2
\nD. C6H5NH2
\nC
\nWhat can be determined about a molecule from the number of signals in its 1HNMR spectrum?
\nA. Bonds present
\nB. Molecular formula
\nC. Molecular mass
\nD. Number of hydrogen environments
\nD
\nA number of drugs have been developed to treat excess acidity in the stomach.
\nOutline how ranitidine (Zantac) functions to reduce stomach acidity.
\nBlocks/binds H2-histamine receptors «in cells of stomach lining»
OR
prevents histamine molecules binding to H2-histamine receptors «and triggering acid secretion»
\n
Accept “H2 receptor antagonist”
\n[1 mark]
\nTaxol is produced using a chiral auxiliary. Describe how the chiral auxiliary functions to produce the desired product.
\nchiral molecule/auxiliary/optically active species added/connected/attached «to non-chiral starting molecule to force reaction to follow a certain path»
\none enantiomer produced
OR
chiral auxiliary creates stereochemical condition «necessary to follow a certain pathway»
OR
stereochemical induction
OR
existing chiral centre affects configuration of new chiral centres
«after new chiral centre created» chiral auxiliary removed «to obtain desired product»
\n[3 marks]
\nWhat is the density, in gcm−3, of a 34.79 g sample with a volume of 12.5 cm3?
\nA. 0.359
\nB. 0.36
\nC. 2.783
\nD. 2.78
\nD
\nWhat is the Index of Hydrogen Deficiency (IHD) for 1,3,5-hexatriene (C6H8)?
\nA. 1
\nB. 3
\nC. 5
\nD. 6
\nB
\nTargeted Alpha Therapy (TAT) is a technique that involves using alpha-radiation to treat leukemia and other dispersed cancers.
\nYttrium-90 and lutetium-177 are used in radiotherapy.
\nExplain why alpha-radiation is particularly suitable for this treatment.
\nOutline how the alpha-radiation in TAT is directed to cancer cells.
\nIdentify the type of radiation emitted by these two radioisotopes.
\nState an equation for the one-step decay of yttrium-90.
\nThe half-life of lutetium-177 is 6.75 days. Calculate the percentage remaining after 27 days.
\nmore damaging than other radiation types
OR
very damaging to «cancer» cells
OR
high ionizing density «of alpha particles»
absorbed within a very short range of emission
OR
causes little damage to surrounding tissues
\n
Accept “high ionizing power «of alpha particles»” for M1.
\nAccept “low penetrating power «of alpha particles»” for M2.
\n[2 marks]
\n«radioactive isotope/radionuclide/alpha-emitter» administered using carrier drug/protein/antibodies
\n[1 mark]
\nbeta/β «radiation»
\n[1 mark]
\n\n
\n
Accept \"\" OR \"\"
\nAccept ECF from (b) (i) if incorrect radiation identified, eg,
\n[1 mark]
\nALTERNATIVE 1:
«4 half-lives»
6.25 «%»
ALTERNATIVE 2:
«Nt = N0» 6.25 «%»
[1 mark]
\nWhich statement explains one of the decreases in first ionization energy (I.E.) across period 3?
\n
A. The nuclear charge of element Al is greater than element Mg.
\nB. The electron-electron repulsion is greater, for the electron with the opposite spin, in element S than in element P.
\nC. A new sub-level is being filled at element S.
\nD. The p orbital being filled in element Al is at a lower energy than the s orbital in element Mg.
\n\n
B
\nWhat is the order of decreasing ionic radius?
\nA. S2− > Cl− > Al3+ > Mg2+
\nB. Cl− > S2− > Al3+ > Mg2+
\nC. S2− > Cl− > Mg2+ > Al3+
\nD. Mg2+ > Al3+ > Cl− > S2−
\nC
\nWhat is the charge on the iron(III) complex ion in [Fe(OH)2(H2O)4]Br?
\nA. 0
\nB. 1+
\nC. 2+
\nD. 3+
\nB
\nWhich combination describes the PH4+ ion?
\nA
\nWhich of the following will form a buffer solution if combined in appropriate molar ratios?
\nA. HCl and NaCl
\nB. NaOH and HCOONa
\nC. NH4Cl and HCl
\nD. HCl and NH3
\nD
\nWhich indicator is appropriate for the acid-base titration shown below?
\nA. Thymol blue (pKa = 1.5)
B. Methyl orange (pKa = 3.7)
C. Bromophenol blue (pKa = 4.2)
D. Phenolphthalein (pKa = 9.6)
D
\nWhich combination describes the bonding and structure in benzoic acid, C6H5COOH?
\nB
\nThe rate of the acid-catalysed iodination of propanone can be followed by measuring how the concentration of iodine changes with time.
\nI2(aq) + CH3COCH3(aq) → CH3COCH2I(aq) + H+(aq) + I−(aq)
\nThe general form of the rate equation is:
\nRate = [H3CCOCH3(aq)]m × [I2(aq)]n × [H+(aq)]p
\nThe reaction is first order with respect to propanone.
\nSuggest how the change of iodine concentration could be followed.
\nA student produced these results with . Propanone and acid were in excess and iodine was the limiting reagent. Determine the relative rate of reaction when .
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/01.a.ii\"](\"../assets/uploads/tinymce_asset/asset/3648/Schermafbeelding_2017-09-19_om_17.58.35.png\"/)
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
The student then carried out the experiment at other acid concentrations with all other conditions remaining unchanged.
\nDetermine the relationship between the rate of reaction and the concentration of acid and the order of reaction with respect to hydrogen ions.
\nWhen the concentration of iodine is varied, while keeping the concentrations of acid and propanone constant, the following graphs are obtained.
\nDeduce, giving your reason, the order of reaction with respect to iodine.
\nWhen the reaction is carried out in the absence of acid the following graph is obtained.
\nDiscuss the shape of the graph between A and B.
\nuse a colorimeter/monitor the change in colour
\nOR
\ntake samples AND quench AND titrate «with thiosulfate»
\n\n
Accept change in pH.
\nAccept change in conductivity.
\nAccept other suitable methods.
\nMethod must imply “change”.
\n[1 mark]
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/01.a.ii/M\"](\"../assets/uploads/tinymce_asset/asset/3649/Schermafbeelding_2017-09-19_om_18.04.57.png\"/)
best fit line
\nrelative rate of reaction
\n\n
Best fit line required for M1.
\n\n
M2 is independent of M1.
\nAccept range from 0.0070 to 0.0080.
\n[2 marks]
\nRelationship:
rate of reaction is «directly» proportional to [H+]
OR
rate of reaction [H+]
Order of reaction with respect to [H+]:
first
\n
Accept \"doubling the concentration doubles the rate\".
\nDo not accept “rate increases as concentration increases”.
\n[2 marks]
\nzero order
\nrate of reaction is the same for all concentrations of iodine
\n\n
Accept “all graphs have same/similar gradient”.
\n[2 marks]
\nslow rate of reaction which gradually increases
\nas H+ ions are produced «to catalyse the reaction»
OR
reaction is autocatalytic
\n
M1 should mention “rate of reaction”.
\n[2 marks]
\nConsider the following half-equations:
\nI2 (s) + 2e– 2I– (aq) Eθ = +0.54 V
(brown) (colourless)
MnO4– (aq) + 8H+ (aq) + 5e– Mn2+ (aq) + 4H2O (l) Eθ = +1.51 V
(purple) (colourless)
Which statement is correct for the reaction between KMnO4 (aq) and KI (aq) in acidic conditions?
\nA. MnO4– reduces I– to I2.
\nB. I– reduces MnO4– to Mn2+.
\nC. The colour changes from brown to purple.
\nD. MnO4– is oxidized to Mn2+.
\nB
\nWhich species have resonance structures?
\nI. Ozone, O3
II. Carbon dioxide, CO2
III. Benzene, C6H6
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nTitanium and vanadium are consecutive elements in the first transition metal series.
\nreacts with water and the resulting titanium(IV) oxide can be used as a smoke screen.
\nDescribe the bonding in metals.
\nTitanium exists as several isotopes. The mass spectrum of a sample of titanium gave the following data:
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/02.b\"](\"../assets/uploads/tinymce_asset/asset/3658/Schermafbeelding_2017-09-20_om_08.37.43.png\"/)
Calculate the relative atomic mass of titanium to two decimal places.
\nState the number of protons, neutrons and electrons in the atom.
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/02.c\"](\"../assets/uploads/tinymce_asset/asset/3659/Schermafbeelding_2017-09-20_om_08.43.58.png\"/)
State the full electron configuration of the ion.
\nSuggest why the melting point of vanadium is higher than that of titanium.
\nSketch a graph of the first six successive ionization energies of vanadium on the axes provided.
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/02.d.iii\"](\"../assets/uploads/tinymce_asset/asset/3660/Schermafbeelding_2017-09-20_om_09.09.57.png\"/)
Explain why an aluminium-titanium alloy is harder than pure aluminium.
\nDescribe, in terms of the electrons involved, how the bond between a ligand and a central metal ion is formed.
\nOutline why transition metals form coloured compounds.
\nState the type of bonding in potassium chloride which melts at 1043 K.
\nA chloride of titanium, , melts at 248 K. Suggest why the melting point is so much lower than that of KCl.
\nFormulate an equation for this reaction.
\nSuggest one disadvantage of using this smoke in an enclosed space.
\nelectrostatic attraction
\nbetween «a lattice of» metal/positive ions/cations AND «a sea of» delocalized electrons
\n\n
Accept “mobile electrons”.
\nDo not accept “metal atoms/nuclei”.
\n[2 marks]
\n\n
\n
Answer must have two decimal places with a value from 47.90 to 48.00.
\nAward [2] for correct final answer.
\nAward [0] for 47.87 (data booklet value).
\n[2 marks]
\nProtons: 22 AND Neutrons: 26 AND Electrons: 22
\n[1 mark]
\n\n
[1 mark]
\nvanadium has smaller ionic radius «leading to stronger metallic bonding»
\n\n
Accept vanadium has «one» more valence electron«s» «leading to stronger metallic bonding».
\nAccept “atomic” for “ionic”.
\n[1 mark]
\nregular increase for first five AND sharp increase to the 6th
\n\n
A log graph is acceptable.
\nAccept log plot on given axes (without amendment of y-axis).
\nAward mark if gradient of 5 to 6 is greater than “best fit line” of 1 to 5.
\n[1 mark]
\ntitanium atoms/ions distort the regular arrangement of atoms/ions
\nOR
\ntitanium atoms/ions are a different size to aluminium «atoms/ions»
\nprevent layers sliding over each other
\n\n
Accept diagram showing different sizes of atoms/ions.
\n[2 marks]
\npair of electrons provided by the ligand
\n\n
Do not accept “dative” or “coordinate bonding” alone.
\n[1 mark]
\npartially filled d-orbitals
\n«ligands cause» d-orbitals «to» split
\nlight is absorbed as electrons transit to a higher energy level «in d–d transitions»
OR
light is absorbed as electrons are promoted
energy gap corresponds to light in the visible region of the spectrum
\ncolour observed is the complementary colour
\n[4 marks]
\nionic
\nOR
\n«electrostatic» attraction between oppositely charged ions
\n[1 mark]
\n«simple» molecular structure
\nOR
\nweak«er» intermolecular bonds
\nOR
\nweak«er» bonds between molecules
\n\n
Accept specific examples of weak bonds such as London/dispersion and van der Waals.
\nDo not accept “covalent”.
\n[1 mark]
\n correct products
correct balancing
\n
Accept ionic equation.
\nAward M2 if products are HCl and a compound of Ti and O.
\n[2 marks]
\nHCl causes breathing/respiratory problems
\nOR
\nHCl is an irritant
\nOR
\nHCl is toxic
\nOR
\nHCl has acidic vapour
\nOR
\nHCl is corrosive
\n\n
Accept TiO2 causes breathing
\nproblems/is an irritant.
\nAccept “harmful” for both HCl and TiO2.
\nAccept “smoke is asphyxiant”.
\n[1 mark]
\nWhich equation represents enthalpy of hydration?
\nA. Na(g) → Na+(aq) + e−
\nB. Na+(g) → Na+(aq)
\nC. NaCl(s) → Na+(g) + Cl−(g)
\nD. NaCl(s) → Na+(aq) + Cl−(aq)
\nB
\nWhich combination of ΔH θ and ΔS θ will result in a non-spontaneous reaction at all temperatures?
\nA
\nWhat are the products when an aqueous solution of copper(II) sulfate is electrolysed using inert graphite electrodes?
\nC
\nVanadium has a number of different oxidation states.
\nElectrode potentials for the reactions of vanadium and other species are shown below.
\nDetermine the oxidation state of vanadium in each of the following species.
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/03.a\"](\"../assets/uploads/tinymce_asset/asset/3663/Schermafbeelding_2017-09-20_om_09.58.14.png\"/)
Identify, from the table, a non-vanadium species that can reduce VO2+(aq) to V3+(aq) but no further.
\nIdentify, from the table, a non-vanadium species that could convert to V2+(aq).
\nFormulate an equation for the reaction between VO2+(aq) and V2+(aq) in acidic solution to form V3+(aq).
\nComment on the spontaneity of this reaction by calculating a value for using the data given in (b) and in section 1 of the data booklet.
\n\n
Do not penalize incorrect notation twice.
\n[2 marks]
\nH2SO3(aq)
OR
Pb(s)
[1 mark]
\nZn(s)
\n[1 mark]
\n\n
\n
Accept equilibrium sign.
\n[1 mark]
\n\n
\n
spontaneous as is negative
\n\n
Do not award M3 as a stand-alone answer.
\nAccept “spontaneous” for M3 if answer given for M2 is negative.
\nAccept “spontaneous as is positive” for M3.
\n[3 marks]
\nThe table gives rate data for the reaction in a suitable solvent.
\nC4H9Br + OH− → C4H9OH + Br−
\nWhich statement is correct?
\nA. The rate expression is rate = k [C4H9Br] [OH− ].
\nB. The rate increases by a factor of 4 when the [OH− ] is doubled.
\nC. C4H9Br is a primary halogenoalkane.
\nD. The reaction occurs via SN1 mechanism.
\nD
\nPropene reacts separately with H2O/H+ and H2/Ni to give products X and Z respectively.
\nWhat are the major products of the reactions?
\nA
\nWhat are the units for the rate constant, k, in the expression?
\nRate = k [X]2[Y]
\nA. mol2 dm−6 s−1
\nB. mol−1 dm3 s−1
\nC. mol dm−3 s−1
\nD. mol−2 dm6 s−1
\nD
\nTwo hydrides of nitrogen are ammonia and hydrazine, . One derivative of ammonia is methanamine whose molecular structure is shown below.
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/05\"](\"../assets/uploads/tinymce_asset/asset/3669/Schermafbeelding_2017-09-20_om_11.35.47.png\"/)
Hydrazine is used to remove oxygen from water used to generate steam or hot water.
\n\n
The concentration of dissolved oxygen in a sample of water is .
\nEstimate the H−N−H bond angle in methanamine using VSEPR theory.
\nState the electron domain geometry around the nitrogen atom and its hybridization in methanamine.
\n\n
Ammonia reacts reversibly with water.
Explain the effect of adding ions on the position of the equilibrium.
Hydrazine reacts with water in a similar way to ammonia. (The association of a molecule of hydrazine with a second H+ is so small it can be neglected.)
\n\n
\n
Calculate the pH of a solution of hydrazine.
\nSuggest a suitable indicator for the titration of hydrazine solution with dilute sulfuric acid using section 22 of the data booklet.
\nOutline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride.
\nDetermine the enthalpy change of reaction, , in kJ, when 1.00 mol of gaseous hydrazine decomposes to its elements. Use bond enthalpy values in section 11 of the data booklet.
\n\n
The standard enthalpy of formation of is . Calculate the enthalpy of vaporization, , of hydrazine in . (If you did not get an answer to (f), use but this is not the correct answer.)
\nCalculate, showing your working, the mass of hydrazine needed to remove all the dissolved oxygen from of the sample.
\nCalculate the volume, in , of nitrogen formed under SATP conditions. (The volume of 1 mol of gas = at SATP.)
\n107°
\n\n
Accept 100° to < 109.5°.
\nLiterature value = 105.8°
\n[1 mark]
\ntetrahedral
\nsp3
\n\n
\n
No ECF allowed.
\n[2 marks]
\nremoves/reacts with
\nmoves to the right/products «to replace ions»
\n\n
Accept ionic equation for M1.
\n[2 marks]
\nKb = 10–5.77 / 1.698 x 10–6
OR
[OH–]2 «= 1.698 × 10–6 × 0.0100» = 1.698 × 10–8
\nOR
\n[OH–] «» = 1.303 × 10–4 «mol dm–3»
\npH «» = 10.1
\n\n
Award [3] for correct final answer.
\nGive appropriate credit for other methods containing errors that do not yield correct final answer.
\n[3 marks]
\nmethyl red
\nOR
\nbromocresol green
\nOR
\nbromophenol blue
\nOR
\nmethyl orange
\n[1 mark]
\nbubbles
\nOR
\ngas
\nOR
\nmagnesium disappears
\n\n
\n
Do not accept “hydrogen” without reference to observed changes.
\nAccept \"smell of ammonia\".
\nAccept 2H+(aq) + Mg(s) Mg2+(aq) + H2(g)
\nEquation must be ionic.
\n[2 marks]
\nbonds broken:
\nE(N–N) + 4E(N–H)
\nOR
\n\n
bonds formed:
\nE(NN) + 2E(H–H)
\nOR
\n\n
\n
\n
Award [3] for correct final answer.
\nAward [2 max] for +95 «kJ».
\n[3 marks]
\n\n
![\"M17/4/CHEMI/HP2/ENG/TZ1/05.g/M\"](\"../assets/uploads/tinymce_asset/asset/3671/Schermafbeelding_2017-09-20_om_14.03.34.png\"/)
OR
\n\n
\n
\n
Award [2] for correct final answer. Award [1 max] for –44 «kJ mol–1».
\nAward [2] for:
\nΔHvap = –50.6 kJ mol–1 – (–85 J mol–1) = ＋34 «kJ mol–1».
\nAward [1 max] for –34 «kJ mol–1».
\n[2 marks]
\ntotal mass of oxygen
\n\n
OR
\n\n
\n
\n
Award [3] for correct final answer.
\n[3 marks]
\n\n
\n
\n
Award [1] for correct final answer.
\n[1 mark]
\nWhat is the product of the reaction between pentan-2-one and sodium borohydride, NaBH4?
\nA. Pentan-1-ol
\nB. Pentan-2-ol
\nC. Pentanoic acid
\nD. Pentanal
\nB
\nThe graph shows values of ΔG for a reaction at different temperatures.
\nWhich statement is correct?
\nA. The standard entropy change of the reaction is negative.
\nB. The standard enthalpy change of the reaction is positive.
\nC. At higher temperatures, the reaction becomes less spontaneous.
\nD. The standard enthalpy change of the reaction is negative.
\nB
\nThis question is about carbon and chlorine compounds.
\nEthane, , reacts with chlorine in sunlight. State the type of this reaction and the name of the mechanism by which it occurs.
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/06.a\"](\"../assets/uploads/tinymce_asset/asset/3674/Schermafbeelding_2017-09-20_om_15.22.26.png\"/)
Formulate equations for the two propagation steps and one termination step in the formation of chloroethane from ethane.
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/06.bi\"](\"../assets/uploads/tinymce_asset/asset/3672/Schermafbeelding_2017-09-20_om_14.32.42.png\"/)
Deduce the splitting patterns in the 1H NMR spectrum of C2H5Cl.
\nExplain why tetramethylsilane (TMS) is often used as a reference standard in 1H NMR.
\nOne possible product, X, of the reaction of ethane with chlorine has the following composition by mass:
\ncarbon: 24.27%, hydrogen: 4.08%, chlorine: 71.65%
\nDetermine the empirical formula of the product.
\nThe mass and 1H NMR spectra of product X are shown below. Deduce, giving your reasons, its structural formula and hence the name of the compound.
\nWhen the product X is reacted with NaOH in a hot alcoholic solution, C2H3Cl is formed. State the role of the reactant NaOH other than as a nucleophile.
\nChloroethene, , can undergo polymerization. Draw a section of the polymer with three repeating units.
\nsubstitution AND «free-»radical
\nOR
\nsubstitution AND chain
\n\n
Award [1] for “«free-»radical substitution” or “SR” written anywhere in the answer.
\n[1 mark]
\nTwo propagation steps:
\n\n
\n
One termination step:
\n\n
OR
\n\n
OR
\n\n
\n
Accept radical without if consistent throughout.
\nAllow ECF for incorrect radicals produced in propagation step for M3.
\n[3 marks]
\ntriplet AND quartet
\n[1 mark]
\nchemical shift/signal outside range of common chemical shift/signal
\nstrong signal/12/all H atoms in same environment
OR
singlet/no splitting of the signal
volatile/easily separated/easily removed
OR
inert/stabl
contains three common NMR nuclei/1H and 13C and 29Si
\n\n
Do not accept chemical shift = 0.
\n[2 marks]
\nAND AND
\n«hence» CH2Cl
\n\n
Accept : :
\nDo not accept C2H4Cl2.
\nAward [2] for correct final answer.
\n[2 marks]
\nmolecular ion peak(s) «about» m/z 100 AND «so» C2H4Cl2 «isotopes of Cl»
\ntwo signals «in 1H NMR spectrum» AND «so» CH3CHCl2
OR
«signals in» 3:1 ratio «in 1H NMR spectrum» AND «so» CH3CHCl2
OR
one doublet and one quartet «in 1H NMR spectrum» AND «so» CH3CHCl2
1,1-dichloroethane
\n\n
Accept “peaks” for “signals”.
\nAllow ECF for a correct name for M3 if an incorrect chlorohydrocarbon is identified.
\n[3 marks]
\nbase
OR
proton acceptor
[1 mark]
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/06.d/M\"](\"../assets/uploads/tinymce_asset/asset/3675/Schermafbeelding_2017-09-20_om_15.46.25.png\"/)
\n
Continuation bonds must be shown.
\nIgnore square brackets and “n”.
\nAccept ![\"M17/4/CHEMI/HP2/ENG/TZ1/06.d_2/M\"](\"../assets/uploads/tinymce_asset/asset/3676/Schermafbeelding_2017-09-20_om_15.47.42.png\"/)
.
Accept other versions of the polymer, such as head to head and head to tail.
\nAccept condensed structure provided all C to C bonds are shown (as single).
\n[1 mark]
\nWhich species produced by the successive dissociations of phosphoric acid, H3PO4, are amphiprotic?
\nA. HPO42− and PO43−
\nB. H2PO4− and HPO42−
\nC. H2PO4− and PO43−
\nD. HPO42− only
\nB
\nWhat is the number of optical isomers of isoleucine?
\nA. 0
\nB. 2
\nC. 4
\nD. 8
\nC
\nBenzene is an aromatic hydrocarbon.
\nDiscuss the physical evidence for the structure of benzene.
\nState the typical reactions that benzene and cyclohexene undergo with bromine.
\n![\"M17/4/CHEMI/HP2/ENG/TZ1/07.b\"](\"../assets/uploads/tinymce_asset/asset/3678/Schermafbeelding_2017-09-20_om_16.35.41.png\"/)
State the reagents used to convert benzene to nitrobenzene and the formula of the electrophile formed.
\nExplain the mechanism for the nitration of benzene, using curly arrows to show the movement of electron pairs.
\nState the reagents used in the two-stage conversion of nitrobenzene to aniline.
\nAny two of:
planar «X-ray»
C to C bond lengths all equal
OR
C to C bonds intermediate in length between C–C and C=C
all C–C–C bond angles equal
\n[2 marks]
\nbenzene: «electrophilic» substitution/SE
AND
cyclohexene: «electrophilic» addition/AE
\n
Accept correct equations.
\n[1 mark]
\n«concentrated» nitric AND sulfuric acids
\n+NO2
\n\n
Accept NO2+.
\n[2 marks]
\ncurly arrow going from benzene ring to N of +NO2/NO2+
\ncarbocation with correct formula and positive charge on ring
\ncurly arrow going from C–H bond to benzene ring of cation
\nformation of organic product AND H+
\n\n
Accept mechanism with corresponding Kekulé structures.
\nDo not accept a circle in M2 or M3.
\nAccept first arrow starting either inside the circle or on the circle.
\nM2 may be awarded from correct diagram for M3.
\nM4: Accept C6H5NO2 + H2SO4 if HSO4– used in M3.
\nFe/Zn/Sn AND HCl/H2SO4/CH3COOH
\nNaOH/KOH
\n\n
Accept other suitable metals and acids.
\nAccept other suitable bases.
\nAward [1 max] for single-step reducing agents (such as H2/Pt, Na2S etc.).
\nAccept formulas or names.
\n[2 marks]
\nWhich functional group is responsible for the pKb of 4.1 in this compound?
\nA. Amido
\nB. Amino
\nC. Chloro
\nD. Ether
\nB
\nWhat is the pH of 1.0 × 10−3 mol dm−3 sodium hydroxide, NaOH(aq)?
\nKw = 1.0 × 10−14
\nA. 3
\nB. 4
\nC. 10
\nD. 11
\nD
\nWhich compound gives this 1H NMR spectrum?
\nA. CH3CH2OCH2CH3
\nB. CH3CH2OH
\nC. CH3CH2CH3
\nD. CH3CH2CH2OH
\nD
\nHow many atoms of nitrogen are there in 0.50 mol of (NH4)2CO3?
\nA. 1
\nB. 2
\nC. 3.01 × 1023
\nD. 6.02 × 1023
\nD
\nWhich species acts as a Lewis and Brønsted–Lowry base?
\nA. [Al(H2O)6]3+
\nB. BF3
\nC. NH4+
\nD. OH−
\nD
\nA student titrated an ethanoic acid solution, CH3COOH (aq), against 50.0 cm3 of 0.995 mol dm–3 sodium hydroxide, NaOH (aq), to determine its concentration.
\nThe temperature of the reaction mixture was measured after each acid addition and plotted against the volume of acid.
\nCurves X and Y were obtained when a metal carbonate reacted with the same volume of ethanoic acid under two different conditions.
\nUsing the graph, estimate the initial temperature of the solution.
\nDetermine the maximum temperature reached in the experiment by analysing the graph.
\nCalculate the concentration of ethanoic acid, CH3COOH, in mol dm–3.
\nDetermine the heat change, q, in kJ, for the neutralization reaction between ethanoic acid and sodium hydroxide.
\nAssume the specific heat capacities of the solutions and their densities are those of water.
\nCalculate the enthalpy change, ΔH, in kJ mol–1, for the reaction between ethanoic acid and sodium hydroxide.
\nExplain the shape of curve X in terms of the collision theory.
\nSuggest one possible reason for the differences between curves X and Y.
\n21.4 °C
\nAccept values in the range of 21.2 to 21.6 °C.
\n29.0 «°C»
\nAccept range 28.8 to 29.2 °C.
\n\n
ALTERNATIVE 1
\n«volume CH3COOH =» 26.0 «cm3»
\n«[CH3COOH] = 0.995 mol dm–3 \\( \\times \\frac{{50.0\\,{\\text{cm3}}}}{{26.0\\,{\\text{cm3}}}} = \\)» 1.91 «mol dm−3»
\nALTERNATIVE 2
\n«n(NaOH) =0.995 mol dm–3 x 0.0500 dm3 =» 0.04975 «mol»
\n«[CH3COOH] = dm3 =» 1.91 «mol dm–3»
\nAccept values of volume in range 25.5 to 26.5 cm3.
\nAward [2] for correct final answer.
\n«total volume = 50.0 + 26.0 =» 76.0 cm3 AND «temperature change 29.0 – 21.4 =» 7.6 «°C»
\n«q = 0.0760 kg x 4.18 kJ kg–1 K–1 x 7.6 K =» 2.4 «kJ»
\nAward [2] for correct final answer.
\n«n(NaOH) = 0.995 mol dm–3 x 0.0500 dm3 =» 0.04975 «mol»
\nOR
\n«n(CH3COOH) = 1.91 mol dm–3 x 0.0260 dm3 =» 0.04966 «mol»
\n«ΔH = » –48 / –49 «kJ mol–1»
\nAward [2] for correct final answer.
\nNegative sign is required for M2.
\n«initially steep because» greatest concentration/number of particles at start
\nOR
\n«slope decreases because» concentration/number of particles decreases
\nvolume produced per unit of time depends on frequency of collisions
\nOR
\nrate depends on frequency of collisions
\nmass/amount/concentration of metal carbonate more in X
\nOR
\nconcentration/amount of CH3COOH more in X
\nWhat is the value of x when 32.2 g of Na2SO4•xH2O are heated leaving 14.2 g of anhydrous Na2SO4? Mr(H2O) = 18; Mr(Na2SO4) = 142.
\nNa2SO4•xH2O (s) → Na2SO4 (s) + xH2O (g)
\nA. 0.1
\nB. 1
\nC. 5
\nD. 10
\nD
\nA buffer is produced by mixing 20.0 cm3 of 0.10 mol dm−3 ethanoic acid, CH3COOH(aq), with 0.10 mol dm−3 sodium hydroxide, NaOH(aq).
\nWhat is the volume of NaOH required and the pH of the buffer?
\nD
\nHow many grams of sodium azide, NaN3, are needed to produce 68.1 dm3 of N2 (g) at STP?
\nMolar volume at STP = 22.7 dm3 mol–1; Mr(NaN3) = 65.0
\n2NaN3 (s) → 3N2 (g) + 2Na (s)
\nA. 32.5
\nB. 65.0
\nC. 130.0
\nD. 195.0
\nC
\nWhich change represents oxidation?
\nA. HClO4 to HClO3
\nB. N2 to NH3
\nC. N2O to NO
\nD. SO42− to SO32−
\nC
\nWhat is the sum of the coefficients when the following equation is balanced using the smallest whole numbers?
\n__C6H12O6 (aq) → __C2H5OH (aq) + __CO2 (g)
\nA. 4
\nB. 5
\nC. 9
\nD. 10
\nB
\nA reaction takes place when a rechargeable battery is used:
\nPb(s) + PbO2(s) + 4H+(aq) + 2SO42−(aq) → 2PbSO4(s) + 2H2O(l)
\nWhich statements are correct?
\nI. H+ is reduced
II. The oxidation state of Pb metal changes from 0 to +2
III. PbO2 is the oxidising agent
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nWhat is the number of protons and the number of neutrons in 131I?
\nA
\nWhich statement is correct for the overall reaction in a voltaic cell?
\n2AgNO3(aq) + Ni(s) → 2Ag(s) + Ni(NO3)2(aq) E θ= +1.06 V
\nA. Electrons flow from Ag electrode to Ni electrode.
\nB. Ni is oxidized to Ni2+ at the cathode (negative electrode).
\nC. Ag+ is reduced to Ag at the anode (positive electrode).
\nD. Ag has a more positive standard electrode potential value than Ni.
\nD
\nWhich is the electron configuration of a chromium atom in the ground state?
\nA. [Ne]3s23p64s13d4
\nB. [Ar]3d3
\nC. 1s22s22p63s23p64s23d4
\nD. [Ar]4s13d5
\nD
\nIn the electrolysis of aqueous potassium nitrate, KNO3(aq), using inert electrodes, 0.1 mol of a gas was formed at the cathode (negative electrode).
\nWhich is correct?
\nB
\nWhich trends are correct across period 3 (from Na to Cl)?
\nI. Atomic radius decreases
II. Melting point increases
III. First ionization energy increases
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nWhich oxide dissolves in water to give a solution with a pH below 7?
\nA. MgO
\nB. Li2O
\nC. CaO
\nD. P4O10
\nD
\nThe electronegativity values of four elements are given.
\nWhat is the order of increasing polarity of the bonds in the following compounds?
\nA. CO < OF2 < NO < CF4
\nB. CF4 < CO < OF2 < NO
\nC. NO < OF2 < CO < CF4
\nD. CF4 < NO < OF2 < CO
\nC
\nTrends in physical and chemical properties are useful to chemists.
\nThe Activity series lists the metal in order of reactivity.
\nExplain the general increasing trend in the first ionization energies of the period 3 elements, Na to Ar.
\nExplain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
\nState an equation for the reaction of phosphorus (V) oxide, P4O10 (s), with water.
\nDescribe the emission spectrum of hydrogen.
\nIdentify the strongest reducing agent in the given list.
\nA voltaic cell is made up of a Mn2+/Mn half-cell and a Ni2+/Ni half-cell.
\nDeduce the equation for the cell reaction.
\nThe voltaic cell stated in part (ii) is partially shown below.
\nDraw and label the connections needed to show the direction of electron movement and ion flow between the two half-cells.
\nincreasing number of protons
\nOR
\nincreasing nuclear charge
\n«atomic» radius/size decreases
\nOR
\nsame number of shells
\nOR
\nsimilar shielding «by inner electrons»
\n«greater energy needed to overcome increased attraction between nucleus and electrons»
\natomic/ionic radius increases
\nsmaller charge density
\nOR
\nforce of attraction between metal ions and delocalised electrons decreases
\nDo not accept discussion of attraction between valence electrons and
nucleus for M2.
Accept “weaker metallic bonds” for M2.
\nP4O10 (s) + 6H2O (l) → 4H3PO4 (aq)
\nAccept “P4O10 (s) + 2H2O (l) → 4HPO3 (aq)” (initial reaction).
\n«series of» lines
\nOR
\nonly certain frequencies/wavelengths
\nconvergence at high«er» frequency/energy/short«er» wavelength
\nM1 and/or M2 may be shown on a diagram.
\nMn
\nMn (s) + Ni2+ (aq) → Ni (s) + Mn2+ (aq)
\nwire between electrodes AND labelled salt bridge in contact with both electrolytes
\nanions to right (salt bridge)
OR
cations to left (salt bridge)
OR
arrow from Mn to Ni (on wire or next to it)
Electrodes can be connected directly or through voltmeter/ammeter/light bulb, but not a battery/power supply.
\nAccept ions or a specific salt as the label of the salt bridge.
\nWhich compound has the shortest C–N bond?
\nA. CH3NH2
\nB. (CH3)3CNH2
\nC. CH3CN
\nD. CH3CHNH
\nC
\nWhich of the following series shows increasing hydrogen bonding with water?
\nA. Propane < propanal < propanol < propanoic acid
\nB. Propane < propanol < propanal < propanoic acid
\nC. Propanal < propane < propanoic acid < propanol
\nD. Propanoic acid < propanol < propanal < propane
\nA
\nWhich statements are correct for metals?
\nI. They conduct electricity because they have free moving ions.
II. They consist of a close-packed lattice of positive ions with delocalized electrons.
III. They are malleable because the layers of positive ions can slide over each other.
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nWhich statement is correct for this reaction?
\nFe2O3 (s) + 3CO (g) → 2Fe (s) + 3CO2 (g) ΔH = −26.6 kJ
\nA. 13.3 kJ are released for every mole of Fe produced.
\nB. 26.6 kJ are absorbed for every mole of Fe produced.
\nC. 53.2 kJ are released for every mole of Fe produced.
\nD. 26.6 kJ are released for every mole of Fe produced.
\nA
\nThe enthalpy changes for two reactions are given.
\nBr2 (l) + F2 (g) → 2BrF (g) ΔH = x kJ
Br2 (l) + 3F2 (g) → 2BrF3 (g) ΔH = y kJ
What is the enthalpy change for the following reaction?
\nBrF (g) + F2 (g) → BrF3 (g)
\nA. x – y
\nB. –x + y
\nC. (–x + y)
\nD. (x – y)
\nC
\nWhat is the enthalpy change, in kJ, of the following reaction?
\n3H2 (g) + N2 (g) 2NH3 (g)
\nA. (6 × 391) − [(3 × 436) + 945]
\nB. (3 × 391) − (436 + 945)
\nC. −[(3 × 436) + 945] + (3 × 391)
\nD. −(6 × 391) + [(3 × 436) + 945]
\nD
\nA student set up a simple voltaic cell consisting of a copper electrode and a zinc electrode dipped in sodium chloride solution.
\nThe student gradually increased the distance, d, between the electrodes to study the effect on the initial current, I, passing through the light bulb.
\nThe student hypothesized that the initial current would be inversely proportional to the distance between the electrodes.
\nThe following data was collected over five trials.
\nThe data did not support the student’s hypothesis. He investigated other possible relationships by plotting a graph of the average current against the distance between the electrodes. He obtained the following best-fit line with a correlation coefficient (r) of −0.9999.
\nSketch a graph that would support the student’s hypothesis.
\nSuggest what the correlation coefficient of −0.9999 indicates.
\nState the equation of the straight line obtained using the data.
\nOutline how current flows in the sodium chloride solution.
\nOR
\nOR
\nCorrect labels of axes required for mark.
Accept d–1 instead of .
Accept I–1 instead of .
Plot of I vs d should not be linear.
\nnegative correlation
\nOR
\nmodel/prediction matches results
\nOR
\n99% of variance accounted for
\nI = – 0.001631 d + 0.09939
\nOR
\ny = – 0.001631 x + 0.09939
\nAccept correctly rounded values for m and b in equation.
\nDo not accept “y = mx + b”.
\nions move «across electrolyte»
\nAntacids react with hydrochloric acid in the stomach to relieve indigestion. A student investigated different brands of antacid to see which caused the largest increase in pH in a given time. She added the antacids to hydrochloric acid, and recorded the change in pH over five minutes.
\nState an equation for the reaction of magnesium hydroxide with hydrochloric acid.
\nSuggest two variables, besides the time of reaction, which the student should have controlled in the experiment to ensure a fair comparison of the antacids.
\nCalculate the uncertainty in the change in pH.
\nThe student concluded that antacid B was the most effective, followed by A then C and finally D. Discuss two arguments that reduce the validity of the conclusion.
\nMg(OH)2 (s) + 2HCl (aq) → MgCl2 (aq) + 2H2O (l)
\nAccept full or net ionic equation.
\nAny two from:
\nvolume «of HCl»
\nconcentration «of HCl»/[HCl]
\ntemperature «of HCl»
\nmass of antacid/tablets
\nsize of antacid particles/tablets
\nOR
\nsurface area of antacid «particles»/tablets
\nAccept “number of tablets/different doses”.
\nDo not accept “same pH meter” OR “initial pH” OR “concentration of antacid/[antacid]”.
\nA variable must be given so do not accept answers such as “stirring”, “whether tablets are whole or crushed” etc.
\n[Max 2 Marks]
\n(±) 0.04
\nOR
\n(±) 0.03
\nAny two of:
\nuncertainty «(±)0.04/(±)0.03» means A and C cannot be distinguished
\neach measurement was conducted once
\nstomach pH should not be raised a lot «so antacid B is not necessarily effective»
\nmass/number of tablets/dose «of antacid» used was not controlled
\nactual environment in stomach is different
\nAccept “amount of tablets” for “dose”.
\nDo not accept “nature/composition of tablets differs”.
\nAccept an answer such as “time frame is too short since some antacids could be long-acting drugs if they contain a gelatinisation/delaying agent” but not just “time frame is too short since some antacids could be long-acting drugs”.
\n[Max 2 Marks]
\nSodium chloride, NaCl, can be spread on icy roads to lower the freezing point of water.
\nThe diagram shows the effects of temperature and percentage by mass of NaCl on the composition of a mixture of NaCl and H2O.
\nEstimate the lowest freezing point of water that can be reached by adding sodium chloride.
\nEstimate the percentage by mass of NaCl dissolved in a saturated sodium chloride solution at +10 ºC.
\nCalculate the percentage of water by mass in the NaCl•2H2O crystals. Use the data from section 6 of the data booklet and give your answer to two decimal places.
\nSuggest a concern about spreading sodium chloride on roads.
\n–21 «ºC»
\n28 «%»
\nAccept any specific answer in the range 27 to 29 «%».
\nMr = 94.48
\n«» 38.15 «%»
\nAward M2 only if answer is to 2 decimal places.
\nAward [2] for correct final answer.
\nAward [1 max] for 38.10 %.
\nrust/corrosion «of cars and bridges»
OR
waste of important raw material
OR
soil/water salination/pollution «from run off»
OR
erosion of/damage to the road surface
OR
specific example of damage to the ecosystem
OR
«outdoor» temperatures may go below effective levels for NaCl «to lower freezing point» so NaCl could be wasted
OR
roads can refreeze causing hazards
Do not accept “tyre damage”.
\nDo not accept “economic issues” OR “environmental issues” unless specified (eg accept “increase in costs for local councils road budgets” but not “cost” alone).
\nDo not accept “makes roads more slippery”.
\nIbuprofen and paracetamol are mild analgesics. One of the IR spectra below belongs to ibuprofen and the other to paracetamol. The structures of both compounds are given in section 37 of the data booklet.
\nBoth spectra show a peak at wavenumber 1700 cm–1. Identify the bond responsible for this peak.
\nDeduce which spectrum belongs to paracetamol, giving two reasons for your choice. Use section 26 of the data booklet.
\nDescribe how mild analgesics function.
\nC=O
\nAccept “carbonyl”.
\n\n
X (must be identified) AND
\nAny two of:
\nFor X:
\nN–H «absorption» AND at 3300 – 3500 «cm–1» ✔
\nO–H «absorption» in phenol AND at 3200 – 3600 «cm–1» ✔
\nabsence of OH «absorption» in carboxylic acid AND 2500 – 3000 «cm–1»
\nAccept any specific wavenumber in the range 3300–3380 «cm–1» for M1.
\nAccept any specific wavenumber in the range 3100–3200 «cm–1».
\nAward [1 max] if Y is incorrectly identified for paracetamol but if a correct reason/reasons is/are given for the bond absorption(s).
\n[Max 2 Marks]
\nprevents/interferes with the production of prostaglandins
\nOR
\nprevents/interferes with the production of substances responsible for inflammation/pain/fever at the site of injury/source of pain
\nSome drugs are extracted from natural sources and others are synthetic.
\nExplain the role of the chiral auxiliary in the synthesis of Taxol.
\nAny three of:
\nchiral auxiliary is optically active
\nis attached to non-optically active/non-chiral substrate
\ncreates stereochemical condition necessary to follow a certain pathway
\nallows only the required enantiomer to form «so avoids need to separate a racemic mixture»
\n[Max 3 Marks]
\nEthanol slows down the reaction time of a driver leading to traffic accidents. Explain how the concentration of ethanol in a sample of breath can be determined using a fuel cell breathalyser.
\nethanol is oxidized «to ethanoic acid»
\nOR
\nelectrons are released
\ncurrent/potential proportional to concentration «of ethanol»
\nOR
\ncurrent compared to a reference «to determine concentration»
\nAccept “ethanol reacts with oxygen” for M1.
\nAccept “voltage” for “potential”.
\nIt is wise to fill dental cavities before irreversible tooth decay sets in. An amalgam (alloy of mercury, silver, and other metals) is often used although many prefer a white composite material.
\nOutline the composition of an alloy and a composite.
\nOutline why an alloy is usually harder than its components by referring to its structure.
\nAt present, composite fillings are more expensive than amalgam fillings.
\nSuggest why a patient might choose a composite filling.
\nExplain how Inductively Coupled Plasma (ICP) Spectroscopy could be used to determine the concentration of mercury in a sample of dental filling.
\nAlloy:
mixture of metal with other metals/non-metals
OR
\nmixture of elements that retains the properties of a metal
\nComposite:
reinforcing phase embedded in matrix phase
Award [1 max] for implying “composites only have heterogeneous/nonhomogeneous compositions”.
\n[Max 2 Marks]
\ndifference in ionic/atomic radius prevents layers sliding over each other
\nAccept “difference in diameter/packing of cations prevents layers sliding over each other”.
\nconcern about Hg poisoning
OR
«composite» is white «so looks more like tooth»
OR
galvanic response potential exists
OR
local allergic potential
OR
less damage/destruction of healthy tooth tissue
OR
long term corrosion requires replacement
OR
gradual darkening of tooth
Accept other correct responses.
\nAny three of:
\nsample injected into argon «plasma»
\natoms «of sample» are excited/ionised
\nOR
\nelectrons are promoted
\nelectrons drop back/recombine with ions AND emit photons of characteristic energies/wavelengths/frequencies
\ntotal number of photons is proportional to concentration of element
\nactual concentration found from calibration/standard curve
\nAccept \"graph/plot\" for “curve”.
\n[Max 3 Marks]
\nWhich transition on the diagram corresponds to the ionization of hydrogen in the ground state?
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/05\"](\"../assets/uploads/tinymce_asset/asset/4453/Schermafbeelding_2018-08-07_om_09.24.08.png\")
A
\nWhich complex has the greatest d orbital splitting?
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/08\"](\"../assets/uploads/tinymce_asset/asset/4455/Schermafbeelding_2018-08-07_om_09.28.41.png\"/)
B
\nWhich metal has the strongest metallic bonding?
\nA. Na
\nB. Mg
\nC. Al
\nD. Ca
\nC
\nWhich molecules have at least one sp2 hybridized atom?
\nI. CH3COOH
\nII. CH3COCH3
\nIII. CH2CHCH2OH
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nCatalysts can take many forms and are used in many industrial processes.
\nSuggest two reasons why it might be worth using a more expensive catalyst to increase the rate of a reaction.
\nAny two of:
\ngreater selectivity
\nhigher efficiency
\nlonger life expectancy
\nOR
\nnot easily poisoned
\neasier to recover
\nlow«er» environmental impact
\nlarge range of conditions/temperatures/pressures supported
\nlower energy costs
\nincrease in yield «per unit time» offsets cost of catalyst
\n[Max 2 Marks]
\n\n
Nanotechnology has many applications.
\nState equations for the formation of iron nanoparticles and carbon atoms from Fe(CO)5 in the HIPCO process.
\nOutline why the iron nanoparticle catalysts produced by the HIPCO process are more efficient than solid iron catalysts.
\nDiscuss one possible risk associated with the use of nanotechnology.
\nFe(CO)5 (g) → Fe (s) + 5CO (g)
\n2CO (g) → C (s) + CO2 (g)
\nlarge surface area «on which carbon nanotubes form»
\nunknown health effects
OR
unknown effect on immune systems
OR
unknown environmental effects
OR
greater inflammatory response
OR
lung damage/toxicity
OR
hazardous effect on biodiversity
OR
risk of explosion
Do not accept vague responses.
\nWhich can be represented with only one Lewis structure?
\nA. CH2O
\nB. C6H6
\nC. O3
\nD. NO3−
\nA
\nWhat is the enthalpy of solution of MgF2(s) in kJ mol−1?
\nLattice enthalpy of MgF2(s) = 2926 kJ mol−1
\nHydration enthalpy of Mg2+(g) = −1963 kJ mol−1
\nHydration enthalpy of F−(g) = −504 kJ mol−1
\nA. 2926 − 1963 + 2(−504)
\nB. 2926 − 1963 − 504
\nC. −2926 − (−1963) − (−504)
\nD. −2926 − (−1963) − 2(−504)
\nA
\nWhich statement is correct?
\nA. If ΔH < 0, reaction is always spontaneous.
\nB. If ΔH > 0, reaction is never spontaneous.
\nC. If ΔS < 0, reaction can be spontaneous if temperature is low enough.
\nD. If ΔS < 0, reaction can be spontaneous if temperature is high enough.
\nC
\nWhat are correct labels for the Maxwell−Boltzmann energy distribution curves?
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/19\"](\"../assets/uploads/tinymce_asset/asset/4456/Schermafbeelding_2018-08-07_om_09.44.40.png\"/)
D
\nThe development of materials with unique properties is critical to advances in industry.
\nLow density polyethene (LDPE) and high density polyethene (HDPE) are both addition polymers.
\nOutline two properties a substance should have to be used as liquid-crystal in a liquid-crystal display.
\nDescribe how the structures of LDPE and HDPE affect one mechanical property of the plastics.
\nOne of the two infrared (IR) spectra is that of polyethene and the other of polytetrafluoroethene (PTFE).
\nDeduce, with a reason, which spectrum is that of PTFE. Infrared data is given in section 26 of the data booklet.
\nMany plastics used to be incinerated. Deduce an equation for the complete combustion of two repeating units of PVC, (–C2H3Cl–)2.
\nAny two of:
\nability to form a LC phase
\nchemically stable
\n«LC phase that is» stable over suitable temperature range
\npolar
\nOR
\nbeing able to change orientation with applied electric field
\nrapid switching speed «responds to changes of voltage quickly»
\nAccept “ability of molecules to transmit light under certain conditions” OR “rodshaped molecules” OR “stable to light/not light sensitive”.
\n[Max 2 Marks]
\nbranching in LDPE prevents close packing «of chains»
\nLDPE is more flexible/less rigid
\nOR
\nLDPE has lower «tensile» strength
\nDo not accept “difference in density”.
\nAward [1 max] for stating “branching in LDPE AND little/no branching in HDPE”.
\nB AND absence «of absorption of» C–H at 2850–3090 «cm–1»
\nOR
\nB AND presence of «absorption of» C–F at 1000–1400 «cm–1»
\n(–C2H3Cl–)2 (s) + 5O2 (g) → 4CO2 (g) + 2H2O (l) + 2HCl (g)
\ncorrect species in reactants and products
\nbalanced
\nAccept “(–C2H3Cl–)2 (s) + 5.5O2 (g) → 4CO2 (g) + 3H2O (l) + Cl2 (g)”.
\nAward M2 only if M1 correct.
\nThe reaction between NO2 and F2 gives the following rate data at a certain temperature.
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/20\"](\"../assets/uploads/tinymce_asset/asset/4458/Schermafbeelding_2018-08-07_om_09.47.13.png\")
What is the overall order of reaction?
\nA. 3
\nB. 2
\nC. 1
\nD. 0
\nA
\nWhat is the effect of increasing temperature on the rate constant, k?
\nA. The rate constant does not change.
\nB. The rate constant decreases linearly.
\nC. The rate constant increases exponentially.
\nD. The rate constant increases proportionally with temperature.
\nC
\nWhat is the effect of increasing the temperature in this reaction?
\nCO2(g) + H2O(l) H+(aq) + HCO3−(aq) ΔH < 0
\nA. The pH will decrease.
\nB. The pH will increase.
\nC. CO2 pressure will decrease.
\nD. The equilibrium position will shift to the right.
\nB
\nConsider the following lipid and carbohydrate.
\nIn order to determine the number of carbon-carbon double bonds in a molecule of linoleic acid, 1.24 g of the lipid were dissolved in 10.0 cm3 of non-polar solvent.
\nThe solution was titrated with a 0.300 mol dm–3 solution of iodine, I2.
\nDetermine the empirical formula of linoleic acid.
\nThe empirical formula of fructose is CH2O. Suggest why linoleic acid releases more energy per gram than fructose.
\nState the type of reaction occurring during the titration.
\nCalculate the volume of iodine solution used to reach the end-point.
\nOutline the importance of linoleic acid for human health.
\nC9H16O
\nratio of oxygen to carbon in linoleic acid lower
\nOR
\nlinoleic acid less oxidized
\nOR
\nlinoleic acid more reduced
\nAccept “«average» oxidation state of carbon in linoleic acid is lower”.
\n«electrophilic» addition/AE
\nOR
\noxidation–reduction/redox
\n« =» 0.00442 «mol»
\n0.00884 mol of C=C
\nOR
\nratio of linoleic acid : iodine = 1:2
\n«volume of I2 solution = =» 0.0295 «dm3» / 29.5 «cm3»
\nAward [3] for correct final answer.
\nAny two of:
\nincreases «ratio of» HDL «to LDL» cholesterol
\nOR
\ndecreases LDL cholesterol «level»
\nremoves plaque from/unblocks arteries
\nOR
\ndecreases risk of heart disease
\ndecreases risk of stroke «in the brain»
\nAccept \"essential fatty acid\".
\nDo not accept “bad cholesterol” for “LDL cholesterol” OR “good cholesterol” for “HDL cholesterol”.
\nDo not accept general answers such as “source of energy” OR “forms triglycerides” OR “regulates permeability of cell membranes” etc.
\n[Max 2 Marks]
\n1.0 mol of N2(g), 1.0 mol of H2(g) and 1.0 mol of NH3(g) are placed in a 1.0 dm3 sealed flask and left to reach equilibrium. At equilibrium the concentration of N2(g) is 0.8 mol dm−3.
\nN2(g) + 3H2(g) 2NH3(g)
\nWhat are the equilibrium concentration of H2(g) and NH3(g) in mol dm−3?
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/23\"](\"../assets/uploads/tinymce_asset/asset/4459/Schermafbeelding_2018-08-07_om_09.56.26.png\"/)
B
\nUrea, (H2N)2CO, is excreted by mammals and can be used as a fertilizer.
\nUrea can also be made by the direct combination of ammonia and carbon dioxide gases.
\n2NH3(g) + CO2(g) (H2N)2CO(g) + H2O(g) ΔH < 0
\nCalculate the percentage by mass of nitrogen in urea to two decimal places using section 6 of the data booklet.
\nSuggest how the percentage of nitrogen affects the cost of transport of fertilizers giving a reason.
\nThe structural formula of urea is shown.
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/01.b_01\"](\"../assets/uploads/tinymce_asset/asset/4469/Schermafbeelding_2018-08-07_om_11.43.42.png\")
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/01.b_02\"](\"../assets/uploads/tinymce_asset/asset/4470/Schermafbeelding_2018-08-07_om_11.45.16.png\"/)
\n
Urea can be made by reacting potassium cyanate, KNCO, with ammonium chloride, NH4Cl.
\nKNCO(aq) + NH4Cl(aq) → (H2N)2CO(aq) + KCl(aq)
\nDetermine the maximum mass of urea that could be formed from 50.0 cm3 of 0.100 mol dm−3 potassium cyanate solution.
\nState the equilibrium constant expression, Kc.
\nPredict, with a reason, the effect on the equilibrium constant, Kc, when the temperature is increased.
\nDetermine an approximate order of magnitude for Kc, using sections 1 and 2 of the data booklet. Assume ΔGΘ for the forward reaction is approximately +50 kJ at 298 K.
\nSuggest one reason why urea is a solid and ammonia a gas at room temperature.
\nSketch two different hydrogen bonding interactions between ammonia and water.
\nThe combustion of urea produces water, carbon dioxide and nitrogen.
\nFormulate a balanced equation for the reaction.
\nCalculate the maximum volume of CO2, in cm3, produced at STP by the combustion of 0.600 g of urea, using sections 2 and 6 of the data booklet.
\nDescribe the bond formation when urea acts as a ligand in a transition metal complex ion.
\nThe C–N bonds in urea are shorter than might be expected for a single C–N bond. Suggest, in terms of electrons, how this could occur.
\nThe mass spectrum of urea is shown below.
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/01.j_01\"](\"../assets/uploads/tinymce_asset/asset/4475/Schermafbeelding_2018-08-07_om_13.00.41.png\")
Identify the species responsible for the peaks at m/z = 60 and 44.
\nThe IR spectrum of urea is shown below.
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/01.k_01\"](\"../assets/uploads/tinymce_asset/asset/4476/Schermafbeelding_2018-08-07_om_13.07.17.png\")
Identify the bonds causing the absorptions at 3450 cm−1 and 1700 cm−1 using section 26 of the data booklet.
\nPredict the number of signals in the 1H NMR spectrum of urea.
\nPredict the splitting pattern of the 1H NMR spectrum of urea.
\nOutline why TMS (tetramethylsilane) may be added to the sample to carry out 1H NMR spectroscopy and why it is particularly suited to this role.
\nmolar mass of urea «4 1.01 + 2 14.01 + 12.01 + 16.00» = 60.07 «g mol-1»
\n«% nitrogen = 100 =» 46.65 «%»
\n\n
Award [2] for correct final answer.
\nAward [1 max] for final answer not to two decimal places.
\n[2 marks]
\n«cost» increases AND lower N% «means higher cost of transportation per unit of nitrogen»
\nOR
\n«cost» increases AND inefficient/too much/about half mass not nitrogen
\n\n
Accept other reasonable explanations.
\nDo not accept answers referring to safety/explosions.
\n[1 mark]
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/01.b/M\"](\"../assets/uploads/tinymce_asset/asset/4471/Schermafbeelding_2018-08-07_om_11.46.41.png\"/)
\n
Note: Urea’s structure is more complex than that predicted from VSEPR theory.
\n[3 marks]
\nn(KNCO) «= 0.0500 dm3 0.100 mol dm–3» = 5.00 10–3 «mol»
\n«mass of urea = 5.00 10–3 mol 60.07 g mol–1» = 0.300 «g»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\n\n
[1 mark]
\n«Kc» decreases AND reaction is exothermic
\nOR
\n«Kc» decreases AND ΔH is negative
\nOR
\n«Kc» decreases AND reverse/endothermic reaction is favoured
\n\n
[1 mark]
\nln K « = » = –20
\n\n
«Kc =» 2 10–9
\nOR
\n1.69 10–9
\nOR
\n10–9
\n\n
Accept range of 20-20.2 for M1.
\nAward [2] for correct final answer.
\n[2 marks]
\nAny one of:
\nurea has greater molar mass
\nurea has greater electron density/greater London/dispersion
\nurea has more hydrogen bonding
\nurea is more polar/has greater dipole moment
\n\n
Accept “urea has larger size/greater van der Waals forces”.
\nDo not accept “urea has greater intermolecular forces/IMF”.
\n\n
[1 mark]
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/01.e.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4472/Schermafbeelding_2018-08-07_om_12.44.01.png\"/)
Award [1] for each correct interaction.
\n\n
If lone pairs are shown on N or O, then the lone pair on N or one of the lone pairs on O MUST be involved in the H-bond.
\nPenalize solid line to represent H-bonding only once.
\n[2 marks]
\n2(H2N)2CO(s) + 3O2(g) → 4H2O(l) + 2CO2(g) + 2N2(g)
\ncorrect coefficients on LHS
\ncorrect coefficients on RHS
\n\n
Accept (H2N)2CO(s) + O2(g) → 2H2O(l) + CO2(g) + N2(g).
\nAccept any correct ratio.
\n[2 marks]
\n«V = 22700 cm3 mol–1 =» 227 «cm3»
\n[1 mark]
\nlone/non-bonding electron pairs «on nitrogen/oxygen/ligand» given to/shared with metal ion
\nco-ordinate/dative/covalent bonds
\n[2 marks]
\nlone pairs on nitrogen atoms can be donated to/shared with C–N bond
\nOR
\nC–N bond partial double bond character
\nOR
\ndelocalization «of electrons occurs across molecule»
\nOR
\nslight positive charge on C due to C=O polarity reduces C–N bond length
\n[1 mark]
\n60: CON2H4+
\n44: CONH2+
\n\n
Accept “molecular ion”.
\n\n
\n
[2 marks]
3450 cm–1: N–H
\n1700 cm–1: C=O
\n\n
Do not accept “O–H” for 3450 cm–1.
\n[2 marks]
\n1
\n[2 marks]
\nsinglet
\n\n
Accept “no splitting”.
\n[1 mark]
\nacts as internal standard
\nOR
\nacts as reference point
\n\n
one strong signal
\nOR
\n12 H atoms in same environment
\nOR
\nsignal is well away from other absorptions
\n\n
Accept “inert” or “readily removed” or “non-toxic” for M1.
\n[2 marks]
\nWhat describes HPO42−?
\nA. Amphiprotic but not amphoteric
\nB. Amphoteric but not amphiprotic
\nC. Amphiprotic and amphoteric
\nD. Neither amphiprotic nor amphoteric
\nC
\nWhat is the pH of a solution in which the hydroxide ion concentration is 1 × 10−11 mol dm−3 at 298 K?
\nKw = 1 × 10−14 at 298 K
\nA. 3
\nB. 7
\nC. 11
\nD. 14
\nA
\nWhich statements are correct?
\nI. Lewis bases can act as nucleophiles.
\nII. Electrophiles are Lewis acids.
\nIII. Lewis acids are electron pair acceptors.
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nWhich combination of acid and base is most likely to have a pH of 8.5 at the equivalence point in a titration?
\nA. Hydrochloric acid and sodium hydroxide
\nB. Hydrochloric acid and ammonia
\nC. Nitric acid and ammonia
\nD. Methanoic acid and sodium hydroxide
\nD
\nWhich combination would electroplate an object with copper?
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/30_01\"](\"../assets/uploads/tinymce_asset/asset/4460/Schermafbeelding_2018-08-07_om_10.07.27.png\")
![\"M18/4/CHEMI/HPM/ENG/TZ1/30_02\"](\"../assets/uploads/tinymce_asset/asset/4461/Schermafbeelding_2018-08-07_om_10.08.26.png\"/)
B
\nLactose is a disaccharide formed by the condensation reaction of the monosaccharides galactose and glucose.
\nDescribe what is meant by a condensation reaction.
\nDraw the structure of galactose on the skeleton provided.
\nExplain how the inclusion of carbohydrates in plastics makes them biodegradable.
\n«reaction in which» two reactants/molecules/functional groups bond/react «to form a larger molecule/single main product»
\nsmall/tiny molecule
\nOR
\nH2O formed
\nAccept formula or name of a specified small molecule other than water such as ammonia, ethanoic/acetic acid,
ethanol, hydrogen sulfide etc. for M2.
Do not accept just “molecule formed”.
\nAward [1 max] for an example giving an equation of a condensation reaction such as the formation of a disaccharide.
\nAccept “alpha” or “beta” form of galactose.
\nAny two of:
\nmakes the plastic more hydrophilic/water soluble
\ncarbohydrates are broken down/hydrolysed by bacteria/microorganisms
\nmakes plastic more accessible to bacteria as holes/channels are created
\nOR
\nplastic of lower density is more permeable/susceptible to water/oxygen/heat/pressure
\nweakens intermolecular/London/dispersion/instantaneous induced dipole-induced dipole forces «between polymer chains in the plastic»
\nAccept “van der Waals/vdW” for “London” forces.
\n[Max 2 Marks]
\nWhat does not affect the mass of products formed in electrolysis of an aqueous solution?
\nA. Current
\nB. Duration of electrolysis
\nC. Initial mass of cathode
\nD. Charge on the ions
\nC
\nVitamins can be water-soluble or fat-soluble.
\nExplain, at the molecular level, why vitamin D is soluble in fats. Use section 35 of the data booklet.
\nState one function of vitamin D in the body.
\n«mainly» hydrocarbon/non-polar «structure»
\nforms London/dispersion/instantaneous induced dipole-induced dipole forces «with fats»
\nAccept “forms van der Waals’/vdW forces”.
\nAward [1 max] for “contains only one OH/hydroxyl AND cannot form «enough» H-bonds”.
\nhelps absorb calcium
OR
helps build bones
OR
helps keep bones healthy
OR
helps block the release of parathyroid hormone
OR
helps in muscle function
OR
helps immune system function
OR
cell growth
OR
reduction of inflammation
OR
protection from osteoporosis
OR
prevents rickets
Accept helps prevent colon/breast/prostate cancer.
\nAccept treat/prevent diabetes/heart disease/high blood pressure/multiple sclerosis.
\nAccept other correct answers.
\nWhich monomer could create this polymer?
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/33\"](\"../assets/uploads/tinymce_asset/asset/4462/Schermafbeelding_2018-08-07_om_10.11.19.png\")
A. But-2-ene
\nB. But-1-ene
\nC. Propene
\nD. 2-Methylprop-1-ene
\nA
\nWhich is a secondary alcohol?
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/34\"](\"../assets/uploads/tinymce_asset/asset/4463/Schermafbeelding_2018-08-07_om_10.13.28.png\"/)
B
\nWhat is name of this compound applying IUPAC rules?
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/X35\"](\"../assets/uploads/tinymce_asset/asset/4464/Schermafbeelding_2018-08-07_om_10.15.14.png\"/)
A. E 1-bromo-1-chlorobut-1-ene
\nB. Z 1-bromo-1-chlorobut-1-ene
\nC. E 1-bromo-1-chloro-2-ethylethene
\nD. Z 1-bromo-1-chloro-2-ethylethene
\nB
\nWhich molecule contains a chiral carbon?
\nA. CH3CH2CHBrCH2CH3
\nB. CH3CH2CHBrCH3
\nC. CH2BrCH(CH3)CH2Br
\nD. CH3CH2CH2CH2CH2Br
\nB
\nWhich reagents are needed to convert nitrobenzene to phenylamine in 2 steps?
\n![\"M18/4/CHEMI/HPM/ENG/TZ1/37\"](\"../assets/uploads/tinymce_asset/asset/4465/Schermafbeelding_2018-08-07_om_10.20.08.png\"/)
D
\nWhich would be the most effective method to distinguish between liquid propan-1-ol and propan-2-ol?
\nA. Observation of colour change when warmed with acidified potassium dichromate
\nB. Determination of m/z value of molecular ion in the mass spectrum
\nC. Determination of percentage composition
\nD. 1H NMR spectroscopy
\nD
\nThe emission spectrum of an element can be used to identify it.
\nElements show trends in their physical properties across the periodic table.
\nDraw the first four energy levels of a hydrogen atom on the axis, labelling n = 1, 2, 3 and 4.
\nDraw the lines, on your diagram, that represent the electron transitions to n = 2 in the emission spectrum.
\nOutline why atomic radius decreases across period 3, sodium to chlorine.
\nOutline why the ionic radius of K+ is smaller than that of Cl−.
\nCopper is widely used as an electrical conductor.
\nDraw arrows in the boxes to represent the electronic configuration of copper in the 4s and 3d orbitals.
\nImpure copper can be purified by electrolysis. In the electrolytic cell, impure copper is the anode (positive electrode), pure copper is the cathode (negative electrode) and the electrolyte is copper(II) sulfate solution.
\nFormulate the half-equation at each electrode.
\n\n
Outline where and in which direction the electrons flow during electrolysis.
\n![\"M18/4/CHEMI/SP2/ENG/TZ2/03.a.i/M\"](\"../assets/uploads/tinymce_asset/asset/4616/Schermafbeelding_2018-08-10_om_09.17.43.png\"/)
4 levels showing convergence at higher energy
\n[1 mark]
\n![\"M18/4/CHEMI/SP2/ENG/TZ2/03.a.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4617/Schermafbeelding_2018-08-10_om_09.19.58.png\"/)
arrows (pointing down) from n = 3 to n = 2 AND n = 4 to n = 2
\n[1 mark]
\nsame number of shells/«outer» energy level/shielding AND nuclear charge/number of protons/Zeff increases «causing a stronger pull on the outer electrons»
\n[1 mark]
\nK+ 19 protons AND Cl– 17 protons
\nOR
\nK+ has «two» more protons
\nsame number of electrons/isoelectronic «thus pulled closer together»
\n[2 marks]
![\"M18/4/CHEMI/SP2/ENG/TZ2/03.c.i/M\"](\"../assets/uploads/tinymce_asset/asset/4619/Schermafbeelding_2018-08-10_om_09.27.32.png\"/)
[1 mark]
\nAnode (positive electrode):
\nCu(s) → Cu2+(aq) + 2e–
\nCathode (negative electrode):
\nCu2+(aq) + 2e– → Cu(s)
\n\n
Accept Cu(s) – 2e– → Cu2+(aq).
\nAccept for →
\nAward [1 max] if the equations are at the wrong electrodes.
\n[2 marks]
\n«external» circuit/wire AND from positive/anode to negative/cathode electrode
\n\n
Accept “through power supply/battery” instead of “circuit”.
\n[1 mark]
\nEnthalpy changes depend on the number and type of bonds broken and formed.
\nThe table lists the standard enthalpies of formation, , for some of the species in the reaction above.
\n![\"M18/4/CHEMI/SP2/ENG/TZ2/04.b\"](\"../assets/uploads/tinymce_asset/asset/4612/Schermafbeelding_2018-08-10_om_08.21.04.png\"/)
Hydrogen gas can be formed industrially by the reaction of natural gas with steam.
\nCH4(g) + H2O(g) → 3H2(g) + CO(g)
\nDetermine the enthalpy change, ΔH, for the reaction, in kJ, using section 11 of the data booklet.
\nBond enthalpy for C≡O: 1077 kJ mol−1
\nOutline why no value is listed for H2(g).
\nDetermine the value of ΔHΘ, in kJ, for the reaction using the values in the table.
\nOutline why the value of enthalpy of reaction calculated from bond enthalpies is less accurate.
\nbonds broken: 4(C–H) + 2(H–O)/4(414) + 2(463)/2582 «kJ»
\nbonds made: 3(H–H) + C≡O/3(436) + 1077/2385 «kJ»
\nΔH «= ΣBE(bonds broken) – ΣBE(bonds made) = 2582 – 2385» = «+» 197 «kJ»
\n\n
Award [3] for correct final answer.
\nAward [2 max] for –197 «kJ».
\n[3 marks]
\nfor any element = 0 «by definition»
\nOR
\nno energy required to form an element «in its stable form» from itself
\n[1 mark]
\nΔHΘ « = (products) – (reactants) = –111 + 0 – [–74.0 + (–242)]»
\n= «+» 205 «kJ»
\n[1 mark]
\n«bond enthalpies» averaged values «over similar compounds»
\nOR
\n«bond enthalpies» are not specific to these compounds
\n[1 mark]
\nA mixture of 1.00 mol SO2(g), 2.00 mol O2(g) and 1.00 mol SO3(g) is placed in a 1.00 dm3 container and allowed to reach equilibrium.
\n2SO2(g) + O2(g) 2SO3(g)
\nDistinguish between the terms reaction quotient, Q, and equilibrium constant, Kc.
\nThe equilibrium constant, Kc, is 0.282 at temperature T.
\nDeduce, showing your work, the direction of the initial reaction.
\nQ: non-equilibrium concentrations AND Kc: equilibrium concentrations
\nOR
\nQ: «measured» at any time AND Kc: «measured» at equilibrium
\n[1 mark]
\n«Q = » = 0.500
\nreverse reaction favoured/reaction proceeds to the left AND Q > Kc/0.500 > 0.282
\n\n
Do not award M2 without M1.
\n[2 marks]
\nThe structure of an organic molecule can help predict the type of reaction it can undergo.
\nImprovements in instrumentation have made identification of organic compounds routine.
\nThe empirical formula of an unknown compound containing a phenyl group was found to be C4H4O. The molecular ion peak in its mass spectrum appears at m/z = 136.
\nThe Kekulé structure of benzene suggests it should readily undergo addition reactions.
\n ![\"M18/4/CHEMI/SP2/ENG/TZ2/07.a_01\"](\"../assets/uploads/tinymce_asset/asset/4625/Schermafbeelding_2018-08-10_om_10.35.21.png\"/)
Discuss two pieces of evidence, one physical and one chemical, which suggest this is not the structure of benzene.
\n\n
Formulate the ionic equation for the oxidation of propan-1-ol to the corresponding aldehyde by acidified dichromate(VI) ions. Use section 24 of the data booklet.
\nThe aldehyde can be further oxidized to a carboxylic acid.
\nOutline how the experimental procedures differ for the synthesis of the aldehyde and the carboxylic acid.
\n\n
Deduce the molecular formula of the compound.
\nIdentify the bonds causing peaks A and B in the IR spectrum of the unknown compound using section 26 of the data booklet.
\n![\"M18/4/CHEMI/SP2/ENG/TZ2/07.c.ii_01\"](\"../assets/uploads/tinymce_asset/asset/4630/Schermafbeelding_2018-08-10_om_10.50.17.png\"/)
Deduce full structural formulas of two possible isomers of the unknown compound, both of which are esters.
\nDeduce the formula of the unknown compound based on its 1H NMR spectrum using section 27 of the data booklet.
\n![\"M18/4/CHEMI/SP2/ENG/TZ2/07.c.iv\"](\"../assets/uploads/tinymce_asset/asset/4635/Schermafbeelding_2018-08-10_om_10.59.18.png\"/)
Physical evidence:
\nequal C–C bond «lengths/strengths»
\nOR
\nregular hexagon
\nOR
\n«all» C–C have bond order of 1.5
\nOR
\n«all» C–C intermediate between single and double bonds
\n\n
Chemical evidence:
\nundergoes substitution reaction «more readily than addition»
\nOR
\ndoes not discolour/react with bromine water
\nOR
\nsubstitution forms only one isomer for 1,2-disubstitution «presence of alternate double bonds would form two isomers»
\nOR
\nmore stable than expected «compared to hypothetical molecule cyclohexa-1,3,5-triene»
\nOR
\nenthalpy change of hydrogenation/combustion is less exothermic than predicted «for cyclohexa-1,3,5-triene»
\n\n
M1:
\nAccept “all C–C–C bond angles are equal”.
\n[2 marks]
\n3CH3CH2CH2OH(l) + Cr2O72–(aq) + 8H+(aq) → 3CH3CH2CHO(aq) + 2Cr3+(aq) + 7H2O(l)
\ncorrect reactants and products
\nbalanced equation
\n[2 marks]
\nAldehyde:
\nby distillation «removed from reaction mixture as soon as formed»
\nCarboxylic acid:
\n«heat mixture under» reflux «to achieve complete oxidation to –COOH»
\n\n
Accept clear diagrams or descriptions of the processes.
\n[2 marks]
\n«»
\nC8H8O2
\n[1 mark]
\nA: C–H «in alkanes, alkenes, arenes»
\nAND
\nB: C=O «in aldehydes, ketones, carboxylic acids and esters»
\n[1 mark]
\nAny two of:
\n OR C6H5COOCH3
OR CH3COOC6H5
OR HCOOCH2C6H5
\n
Do not penalize use of Kekule structures for the phenyl group.
\nAccept the following structures:
\n![\"M18/4/CHEMI/SP2/ENG/TZ2/07.c.iii_02/M\"](\"../assets/uploads/tinymce_asset/asset/4634/Schermafbeelding_2018-08-10_om_10.55.29.png\"/)
Award [1 max] for two correct aliphatic/linear esters with the molecular formula C8H8O2.
\n[2 marks]
\nC6H5COOCH3 «signal at 4 ppm (3.7 – 4.8 range in data table) due to alkyl group on ester
\n[1 mark]
\nSome physical properties of molecular substances result from the different types of forces between their molecules.
\nExplain why the hydrides of group 16 elements (H2O, H2S, H2Se and H2Te) are polar molecules.
\nThe graph shows the boiling points of the hydrides of group 16 elements.
\n![\"M18/4/CHEMI/SP2/ENG/TZ2/06.a.ii\"](\"../assets/uploads/tinymce_asset/asset/4622/Schermafbeelding_2018-08-10_om_10.27.26.png\"/)
Explain the increase in the boiling point from H2S to H2Te.
\nLewis structures show electron domains and are used to predict molecular geometry.
\nDeduce the electron domain geometry and the molecular geometry for the NH2− ion.
\n\n
polar bonds «between H and group 16 element»
\nOR
\ndifference in electronegativities «between H and group 16 element»
\n\n
uneven distribution of charge/electron cloud
\nOR
\nnon-linear/bent/V-shaped/angular shape «due to lone pairs»
\nOR
\npolar bonds/dipoles do not cancel out
\n\n
M2:
\nDo not accept “net/overall dipole moment” without further explanation.
\nAccept “non-symmetrical «shape/distribution of charge»”.
\n[2 marks]
\nnumber of electrons increases
\nLondon/dispersion/instantaneous induced dipole-induced dipole forces increase
\n\n
M1: Accept “Mr/Ar increases” or “molecules become larger in size/mass/surface area”.
\n[2 marks]
\nElectron domain geometry:
\ntetrahedral
\nMolecular geometry:
\nbent/V-shaped/angular
\n\n
Both marks can be awarded for clear diagrams. Electron domain geometry requires a 3-D diagram showing the tetrahedral arrangement.
\n[2 marks]
\nThe table summarizes some properties of graphite and graphene.
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/01\"](\"../assets/uploads/tinymce_asset/asset/4636/Schermafbeelding_2018-08-10_om_11.17.29.png\"/)
Graphene is two-dimensional, rather than three-dimensional, material.
\nJustify this by using the structure of graphene and information from the table.
\nShow that graphene is over 1600 times stronger than graphite.
\nIdentify a value from the table which can be used to support the information about graphene given below.
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/01.a.iii_01\"](\"../assets/uploads/tinymce_asset/asset/4642/Schermafbeelding_2018-08-10_om_12.38.15.png\"/)
Electrons in a solid are restricted to certain ranges, or bands, of energy (vertical axis). In an insulator or semiconductor, an electron bound to an atom can break free only if it gets enough energy from heat or a passing photon to jump the “band gap”, but in graphene the gap is infinitely small.
\n ![\"M18/4/CHEMI/SP3/ENG/TZ2/01.a.iii_02\"](\"../assets/uploads/tinymce_asset/asset/4643/Schermafbeelding_2018-08-10_om_12.39.44.png\"/)
Diamond, graphene, and graphite are all network solids.
\nSuggest, giving a reason, the electron mobility of diamond compared to graphene.
\nThe melting point of diamond at 1 × 106 kPa is 4200 K (in the absence of oxygen).
\nSuggest, based on molecular structure, why graphene has a higher melting point under these conditions.
\nconsists of single/one sheet/layer «of carbon atoms»
\n\n
graphene has no density measurement
\nOR
\ngraphene has no distance between layers data
\nOR
\ngraphene has large specific surface area «compared to graphite»
\n\n
Do not accept “sp2” alone without reference to single/one sheet/layer.
\nAccept “thickness of one atom” OR “consists of a plane” for M1.
\n[2 marks]
\nAny one of these alternatives:
\nALTERNATIVE 1
\n«»
\n1.7 × 103/1711
\n\n
ALTERNATIVE 2
\n1600 × 76 × 106 = 1.2 × 1011 «is less than tensile strength of graphene»
\n\n
ALTERNATIVE 3
\n= 8.1 × 107 «is greater than upper end of tensile strength for graphite»
\n\n
Accept any value in the range 1700–27 083. Answer may be expressed in scientific notation or otherwise.
\nAccept any value calculated which is less than the graphene tensile strength based on a value chosen from within the 4.8–76 × 106 range.
\n[1 mark]
\n«graphene has a high electron mobility of» 15 000–200 000 «cm2 V–1 s–1»
\n\n
A specific value or range of values must be given.
\nAccept any value in the 15 000–200 000 «cm2 V–1 s–1» range.
\n[1 mark]
\nsmaller/zero
\n\n
no delocalized electrons/electrons are bound/electrons not free to move/electrons not free to roam
\nOR
\nlocalized electrons «in sigma bonds»
\nOR
\nlarge band gap
\n\n
Accept “diamond is a dielectric” OR “diamond does not conduct electricity” for M2.
\nAward [1 max] for just “immobile/less mobile”.
\nAward [2] for “electrons immobile «in diamond» due to the large band gap” OR \"electrons «in diamond» immobile
\nsince electrons are localized «in the sigma bonds»”.
\n[2 marks]
\nshorter bonds in graphene
\nOR
\nbonds in graphene intermediate between single and double
\nOR
\nbond order in graphene is 1.33
\nOR
\ndelocalization creates stronger bonds
\nOR
\nshorter bonds are stronger
\n\n
stronger/shorter bonds require higher temperature/faster thermal motion to be altered
\nOR
\nstronger/shorter bonds require greater energy to be broken
\n\n
[2 marks]
\nThe process of converting heat to electricity is limited by its thermal (Carnot) efficiency.
\n\n
Calculate the thermal efficiency of a steam turbine supplied with steam at 540°C and using a river as the choice of sink at 23 °C.
\nPower plants generating electricity by burning coal to boil water operate at approximately 35% efficiency.
\nState what this means and suggest why it is lower than the thermal efficiency.
\n« × 100» = 64 «%»
\n[1 mark]
\n35% of chemical/potential energy available in coal is transformed to electricity/electrical energy
\n\n
not all chemical energy from burning fuel transferred into heating water
\nOR
\nenergy dispersed elsewhere/energy lost due to friction of moving parts
\nOR
\nheat loss to the surroundings
\n\n
Accept “stored energy” for “potential energy”.
\n[2 marks]
\nNuclear power is another source of energy.
\nCompare and contrast the process of nuclear fusion with nuclear fission.
\n\n
Dubnium-261 has a half-life of 27 seconds and rutherfordium-261 has a half-life of 81 seconds.
\nEstimate what fraction of the dubnium-261 isotope remains in the same amount of time that of rutherfordium-261 decays.
\nAward [1] for one similarity:
\nboth increase binding energy/energy yield «per nucleon»
\nOR
\nmass loss/defect in both «nuclear» reactions/mass converted to energy «from E = mc2»
\nOR
\nboth produce ionizing radiation
\n\n
Award [2 max] for any two differences:
\nin fusion, light nuclei combine to form heavier ones AND in fission, heavier nuclei split into lighter ones
\nfission produces radioactive/nuclear waste AND fusion does not
\n\n
fission is caused by bombarding with a neutron «or by spontaneous fission» AND fusion does not
\nOR
\nfission can initiate a chain reaction AND fusion does not
\n\n
fusion releases more energy per unit mass of fuel than fission
\nfuel is easier to obtain/cheaper for fusion reactions
\nfission reactions can be controlled in a power plant AND fusion cannot «yet»
\nfusion reactor less likely to cause a large-scale technological disaster compared to fission
\nfusion less dangerous than fission as radioactive isotopes produced have short half-lives so only cause a threat for a relatively short period of time
\nfusion is in experimental development AND fission used commercially
\n\n
Accept “small nuclei” OR “smaller atomic masses of nuclei” for “light nuclei” AND “large nuclei” OR “greater atomic masses of nuclei” for “heavier nuclei”.
\nDo not accept “no/less waste produced for fusion”.
\nAccept “higher specific energy for fusion”.
\n[3 marks]
\n//0.016
\n\n
Accept “1.6%”.
\n[1 mark]
\nOne method of producing biodiesel is by a transesterification process.
\nDeduce the equation for the transesterification reaction of pentyl octanoate, C7H15COOC5H11, with methanol.
\nOutline why the ester product of this reaction is a better diesel fuel than pentyl octanoate.
\nC7H15COOC5H11(l) + CH3OH(l) → C7H15COOCH3(l) + C5H11OH(l)
\nOR
\nC13H26O2(l) + CH4O(l) → C9H18O2(l) + C5H12O(l)
\nOR
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/14.a/M\"](\"../assets/uploads/tinymce_asset/asset/4660/Schermafbeelding_2018-08-10_om_14.49.59.png\"/)
\n
Accept correct equation in any format eg, skeletal, condensed structural formula, etc.
\nAccept equations with equilibrium arrow.
\n[1 mark]
\n\n
less viscous «and so does not need to be heated to flow»
\nOR
\nless likely to undergo incomplete combustion
\nOR
\nfewer intermolecular/London/dispersion forces
\nOR
\nvaporizes easier
\n\n
Ignore equation and products in 14a.
\nAccept “van der Waals’/vdW” for “London”.
\n[1 mark]
\nDrug testing is necessary to determine safe and effective doses.
\nDistinguish between the lethal dose (LD50) and the toxic dose (TD50).
\nLD50: amount/dose that kills 50% of the population
\nTD50: amount/dose that negatively affects/produces toxic effects in 50% of the population
\n\n
Award [1 max] for “LD50 used in animal trials AND TD50 used in human studies”.
\n[2 marks]
\nPenicillins and aspirin are important medicines.
\nDescribe how penicillin combats bacterial infections.
\nState how penicillins may be modified to increase their effectiveness.
\nState the type of reaction used to synthesize aspirin from salicylic acid.
\nExplain why aspirin is not stored in a hot, humid location.
\n«irreversibly» binds/bonds to enzyme/transpeptidase
\nOR
\ninhibits enzyme/transpeptidase «in bacteria» that produces cell walls
\nOR
\nprevents cross-linking of bacterial cell walls
\n\n
cells absorb water AND burst
\nOR
\ncells cannot reproduce
\n\n
[2 marks]
\nmodify side chain
\n[1 mark]
\ncondensation
\nOR
\nesterification
\nOR
\nnucleophilic substitution/nucleophilic displacement/SN2
\n\n
Do not accept just “substitution/displacement”.
\n[1 mark]
\nwater causes hydrolysis
\nOR
\naspirin reacts with water
\n\n
heat increases the rate of hydrolysis
\nOR
\nheat increases the rate of the reaction with water
\n\n
Accept “aspirin will convert into salicylic/ethanoic acid”.
\nDo not accept “aspirin dissolves in water” OR “aspirin absorbs water/is hygroscopic”.
\n[2 marks]
\nMorphine and diamorphine (heroin) are both opioids.
\nExplain why diamorphine is more potent than morphine using section 37 of the data booklet.
\nmorphine has hydroxyl/OH groups/is more polar AND diamorphine has ester/ethanoate/acetate groups/is less polar/is lipid soluble
\ncrossing blood brain barrier is easier for non-polar/less polar compounds/for lipid soluble compounds
\n\n
Accept “alcohol/hydroxy” for “hydroxyl” but not “hydroxide”.
\nAccept \"fats\" for \"lipid\".
\n[2 marks]
\nExcess acid in the stomach is often treated with calcium carbonate.
\nFormulate a chemical equation for the neutralization of stomach acid with calcium carbonate.
\nCalculate the amount, in mol, of stomach acid neutralized by an antacid tablet containing 0.750 g calcium carbonate.
\nExplain how omeprazole (Prilosec) regulates pH in the stomach.
\n2HCl(aq) + CaCO3(s) → H2O(l) + CO2(g) + CaCl2(aq)
\n\n
Accept ionic equation:
\n2H+(aq) + CO32–(aq) → CO2(g) + H2O(l)
\n[1 mark]
\n« =» 0.0150 «mol HCl»
\n[1 mark]
\ninhibits the secretion of stomach acid/H+
\n«active metabolites» bind «irreversibly» to «receptors of the» proton pump
\n\n
Do not accept “hydrogen/H/H2” for “H+”.
\nAccept “PPI/proton pump inhibitor” for M2.
\nAccept “H+/K+ ATPase” for “proton pump”.
\n[2 marks]
\nAntiviral medications such as zanamivir (Relenza) are commonly available for consumer use.
\nIdentify the names of two functional groups present in zanamivir using section 37 of the data booklet.
\nDistinguish between bacteria and viruses.
\nAny two of:
\nhydroxyl
\ncarboxyl/carbonyl
\nether
\namido/carbonyl
\n\n
Accept “alcohol/hydroxy” for “hydroxyl”, “carboxylic acid” for “carboxyl” and “amide/carboxamide” for “amido”.
\nAccept “amino/amine” OR “imine/imino” but these are not correct as they are part of the guanidino group.
\nAccept “alkenyl/alkene/carbon to carbon double bond” but not “C=C” OR “carbon double bond”.
\nAccept “carbonyl” only once.
\nAccept “heterocyclic ring” for “ether”.
\n[2 marks]
\nAny two of:
\nbacteria perform living functions «on their own» AND viruses do not «without host cell»
\n\n
bacteria have cell walls AND viruses do not
\nOR
\nbacteria do not have a capsid AND viruses do
\n\n
bacteria larger than viruses
\nbacteria reproduce by fission/budding AND viruses reproduce within a living host cell
\n\n
Accept examples of living functionsexcretion, reproduction etc for M1.
\nAccept “bacteria have flagella/cytoplasm/ribosome AND virus can have head/protein tail/double stranded RNA/single stranded DNA”.
\nAccept other specific structural differences for M2.
\nAccept “asexual reproduction for bacteria” for M4.
\n[2 marks]
\nDrug synthesis often involves solvents.
\nIdentify a common hazardous solvent and a Green solvent that could replace it.
\n\n
Hazardous solvent:
\nAny one of:
\nmethanal/formaldehyde
\nmethanol
\nchlorinated solvent/carbon tetrachloride/methylene chloride/dichloromethane
\ndiethyl ether/ethoxyethane
\n\n
benzene
\nOR
\nmethyl benzene/toluene
\nOR
\n«1,2/1,3/1,4» dimethylbenzene/«ortho/o-/meta/m-/para/p-» xylene
\n\n
Green solvent:
\nAny one of:
\nwater
\n«supercritical/liquid» carbon dioxide/supercritical fluids
\nethanol «only if replacing a hazardous solvent»
\npropan-2-ol/2-propanol/isopropanol «only if replacing a hazardous solvent»
\npropanone/acetone «only if replacing a hazardous solvent»
\nethyl ethanoate/ethyl acetate «only if replacing a hazardous solvent»
\norganic carbonates/dimethyl carbonate/diethyl carbonate/ethylene
\ncarbonate/propylene carbonate
\nionic liquids
\nfluorous solvents
\n\n
Accept correct names (either IUPAC or generic) or formulas.
\nDo not accept inorganic acids such as HCl, H2SO4, etc.
\nAccept any specific chlorinated solvent.
\nAccept other hazardous solvents.
\nDo not accept any solvent given as both hazardous and green.
\nAward [2] for combination “Hazardous solvent: dimethylformamide/DMF/N,N-dimethylmethanamide” AND “Green solvent: methanol «only if replacing a hazardous solvent»”.
\nAccept other green solvents but not “solvents from biomass/food waste”.
\n[2 marks]
\nWhat are the products of electrolysis when concentrated calcium bromide solution is electrolysed using graphite electrodes?
\nA
\nEnthalpy changes depend on the number and type of bonds broken and formed.
\nEnthalpy changes depend on the number and type of bonds broken and formed.
\nThe table lists the standard enthalpies of formation, , for some of the species in the reaction above.
\n
Hydrogen gas can be formed industrially by the reaction of natural gas with steam.
\nCH4(g) + H2O(g) → 3H2(g) + CO(g)
\nDetermine the enthalpy change, ΔH, for the reaction, in kJ, using section 11 of the data booklet.
\nBond enthalpy for C≡O: 1077 kJ mol−1
\nOutline why no value is listed for H2(g).
\nDetermine the value of ΔHΘ, in kJ, for the reaction using the values in the table.
\nThe table lists standard entropy, SΘ, values.
\n![\"M18/4/CHEMI/HP2/ENG/TZ2/05.c\"](\"../assets/uploads/tinymce_asset/asset/4522/Schermafbeelding_2018-08-08_om_15.07.35.png\"/)
Calculate the standard entropy change for the reaction, ΔSΘ, in J K−1.
\nCH4(g) + H2O(g) → 3H2(g) + CO(g)
\nCalculate the standard free energy change, ΔGΘ, in kJ, for the reaction at 298 K using your answer to (b)(ii).
\nDetermine the temperature, in K, above which the reaction becomes spontaneous.
\nbonds broken: 4(C–H) + 2(H–O)/4(414) + 2(463)/2582 «kJ»
\nbonds made: 3(H–H) + C≡O/3(436) + 1077/2385 «kJ»
\nΔH «= ΣBE(bonds broken) – ΣBE(bonds made) = 2582 – 2385» = «+» 197 «kJ»
\n\n
Award [3] for correct final answer.
\nAward [2 max] for –197 «kJ».
\n[3 marks]
\nfor any element = 0 «by definition»
\nOR
\nno energy required to form an element «in its stable form» from itself
\n[1 mark]
\nΔHΘ « = (products) – (reactants) = –111 + 0 – [–74.0 + (–242)]»
\n= «+» 205 «kJ»
\n[1 mark]
\n«ΔSΘ = ΣSΘproducts – ΣSΘreactants = 198 + 3 × 131 – (186 + 189) =» «+» 216 «J K–1»
\n[1 mark]
\n«ΔGΘ = ΔHΘ – TΔSΘ = 205 kJ – 298 K × kJ K–1 =» «+» 141 «kJ»
\n[1 mark]
\n«ΔHΘ = TΔSΘ»
\n«»
\n«T =» 949 «K»
\n\n
Do not award a mark for negative value of T.
\n[1 mark]
\nAmino acids are the building blocks of proteins.
\nDraw the structures of the main form of glycine in buffer solutions of pH 1.0 and 6.0.
\nThe pKa of glycine is 2.34.
\nCalculate the pH of a buffer system with a concentration of 1.25 × 10−3 mol dm−3 carbonic acid and 2.50 × 10−2 mol dm−3 sodium hydrogen carbonate. Use section 1 of the data booklet.
\npKa (carbonic acid) = 6.36
\nSketch the wedge and dash (3-D) representations of alanine enantiomers.
\nUV-Vis spectroscopy can be used to determine the unknown concentration of a substance in a solution.
\nCalculate the concentration of an unknown sample of pepsin with an absorbance of 0.725 using section 1 of the data booklet.
\nCell length = 1.00 cm
\nMolar absorptivity (extinction coefficient) of the sample = 49650 dm3 cm−1 mol−1
\nA different series of pepsin samples is used to develop a calibration curve.
\n
Estimate the concentration of an unknown sample of pepsin with an absorbance of 0.30 from the graph.
\n![\"M18/4/CHEMI/HP3/ENG/TZ2/08.c/M\"](\"../assets/uploads/tinymce_asset/asset/4542/Schermafbeelding_2018-08-08_om_18.45.38.png\"/)
\n
Penalize charge on incorrect atom once only.
\nPenalize missing hydrogens or incorrect bond connectivities once only in Option B.
\nAccept condensed structural formulas.
\nAccept skeletal structures.
\n[2 marks]
\nALTERNATIVE 1
\n«pH = 6.36 + log =»
\n7.66
\nALTERNATIVE 2
\n«Ka = 4.4 × 10–7 = [H+] , [H+] = 2.2 × 10–8 mol dm–3»
\n«pH =» 7.66
\n\n
Do not accept “«pH =» 8”.
\n[1 mark]
\nPenalize missing hydrogens or incorrect bond connectivities once only in Option B.
\nWedges AND dashes must be used.
\n[1 mark]
\n«» 1.46 × 10−5 «mol dm−3»
\n[1 mark]
\n\n
0.65 «μg cm–3»
\nAccept any value in the range 0.60–0.70 «μg cm–3»
\n[1 mark]
\nHow many moles of FeS2 are required to produce 32 g of SO2? (Ar: S = 32, O = 16)
\n4FeS2 (s) + 11O2 (g) → 2Fe2O3 (s) + 8SO2 (g)
\n\n
A. 0.25
\nB. 0.50
\nC. 1.0
\nD. 2.0
\nA
\nThe volume of a sample of gas measured at 27 °C is 10.0 dm3. What is the temperature when the volume is reduced to 9.0 dm3 at the same pressure?
\n\n
A. −3.0 °C
\nB. 24.3 °C
\nC. 29.7 °C
\nD. 57.0 °C
\nA
\n16 g of bromine react with 5.2 g of metal, M, to form MBr2. What is the relative atomic mass of the metal M? (Ar : Br = 80)
\n\n
A. 13
\nB. 26
\nC. 52
\nD. 104
\nC
\nAn antacid tablet containing 0.50 g of NaHCO3 (Mr = 84) is dissolved in water to give a volume of 250 cm3. What is the concentration, in mol dm−3, of HCO3− in this solution?
\n\n
A.
\nB.
\nC.
\nD.
\nB
\nWhich statements are correct for the emission spectrum of hydrogen?
\nI. The lines converge at higher frequencies.
\nII. Electron transitions to n = 2 are responsible for lines in the visible region.
\nIII. Lines are produced when electrons move from lower to higher energy levels.
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nWhich statement about 56Fe3+ and 54Fe2+ is correct?
\n\n
A. Both have the same numbers of protons and electrons.
\nB. Both have the same number of protons.
\nC. Both have the same number of neutrons.
\nD. Both have the same numbers of protons and neutrons.
\nB
\nWhich oxides produce an acidic solution when added to water?
\nI. Al2O3 and SiO2
\nII. P4O6 and P4O10
\nIII. NO2 and SO2
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nWhich species will require the least energy for the removal of one electron?
\n\n
A. Na+
\nB. Mg+
\nC. Al2+
\nD. C3+
\nB
\nWhich species has the same molecular geometry as SO32−?
\n\n
A. BF3
\nB. SO3
\nC. PF3
\nD. CO32−
\nC
\nHow many lone pairs and bonding pairs of electrons surround the central chlorine atom in ClF2+?
\n\n
\n
D
\nWhich compound has the highest boiling point?
\n\n
A. CH3CHO
\nB. CH3CH2F
\nC. CH3OCH3
\nD. CH3CH2NH2
\nD
\nWhich molecule is polar?
\nA. BeCl2
\nB. BCl3
\nC. NCl3
\nD. CCl4
\nC
\nConsider the following reactions:
\nFe2O3 (s) + CO (g) → 2FeO (s) + CO2 (g) ΔHΘ = −3 kJ
\nFe (s) + CO2 (g) → FeO (s) + CO (g) ΔHΘ = +11 kJ
\nWhat is the ΔHΘ value, in kJ, for the following reaction?
\nFe2O3 (s) + 3CO (g) → 2Fe (s) + 3CO2 (g)
\n\n
A. −25
\nB. −14
\nC. +8
\nD. +19
\nA
\nWhich is correct when Ba(OH)2 reacts with NH4Cl?
\nBa(OH)2 (s) + 2NH4Cl (s) → BaCl2 (aq) + 2NH3 (g) + 2H2O (l) ΔHΘ = +164 kJ mol−1
\n\n
C
\nConsider the following reaction:
\nN2 (g) + 3H2 (g) 2NH3 (g)
\n
Which calculation gives ΔHΘ, in kJ, for the forward reaction?
\nA. 2z − y − 3x
\nB. y + 3x − 2z
\nC. y + 3x − 6z
\nD. 6z − y − 3x
\n\n
C
\nSamples of sodium carbonate powder were reacted with separate samples of excess hydrochloric acid.
\nNa2CO3 (s) + 2HCl (aq) → CO2 (g) + 2NaCl (aq) + H2O (l)
\nReaction I: 1.0 g Na2CO3 (s) added to 0.50 mol dm−3 HCl (aq)
\nReaction II: 1.0 g Na2CO3 (s) added to 2.0 mol dm−3 HCl (aq)
\nWhat is the same for reactions I and II?
\nA. Initial rate of reaction
\nB. Total mass of CO2 produced
\nC. Total reaction time
\nD. Average rate of production of CO2
\nB
\nWhat decreases the activation energy of a reaction?
\nA. Increasing the temperature
\nB. Adding a catalyst
\nC. Adding more reactants
\nD. Increasing collision frequency of reactants
\nB
\nConsider the reaction:
\n2N2O (g) 2N2 (g) + O2 (g)
\nThe values of Kc at different temperatures are:
\nWhich statement is correct at higher temperature?
\nA. The forward reaction is favoured.
\nB. The reverse reaction is favoured.
\nC. The rate of the reverse reaction is greater than the rate of the forward reaction.
\nD. The concentration of both reactants and products increase.
\nA
\nWhich two species act as Brønsted–Lowry acids in the reaction?
\nH2PO4− (aq) + OH− (aq) HPO42− (aq) + H2O (l)
\n\n
A. HPO42− (aq) and OH− (aq)
\nB. H2PO4− (aq) and HPO42− (aq)
\nC. HPO42− (aq) and H2O (l)
\nD. H2PO4− (aq) and H2O (l)
\nD
\nWhich is correct for the reaction?
\nP4 (s) + 3H2O (l) + 3OH− (aq) → PH3 (g) + 3H2PO2− (aq)
\n\n
D
\nWhich describes the flow of electrons in a voltaic cell?
\n\n
A. From the cathode (positive electrode) to the anode (negative electrode) through the external circuit
\nB. From the anode (negative electrode) to the cathode (positive electrode) through the external circuit
\nC. From the oxidizing agent to the reducing agent through the salt bridge
\nD. From the reducing agent to the oxidizing agent through the salt bridge
\nB
\nThis question is about the decomposition of hydrogen peroxide.
\nHydrogen peroxide decomposes to water and oxygen when a catalyst such as potassium iodide, KI, is added.
\n
2H2O2 (aq) O2 (g) + 2H2O (l)
Suggest why many chemicals, including hydrogen peroxide, are kept in brown bottles instead of clear colourless bottles.
\nIn a laboratory experiment solutions of potassium iodide and hydrogen peroxide were mixed and the volume of oxygen generated was recorded. The volume was adjusted to 0 at t = 0.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5958/4bi_1.PNG\")
The data for the first trial is given below.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5959/4bi_2.PNG\")
Plot a graph on the axes below and from it determine the average rate of
formation of oxygen gas in cm3 O2 (g) s−1.
![\"\"](\"../assets/uploads/tinymce_asset/asset/5960/4bi_3.PNG\")
Average rate of reaction:
\nTwo more trials (2 and 3) were carried out. The results are given below.
\nDetermine the rate equation for the reaction and its overall order, using your answer from (b)(i).
\nRate equation:
\nOverall order:
\nAdditional experiments were carried out at an elevated temperature. On the axes below, sketch Maxwell–Boltzmann energy distribution curves at two temperatures T1 and T2, where T2 > T1.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5963/4biii.PNG\")
Apart from a greater frequency of collisions, explain, by annotating your graphs in (b)(iii), why an increased temperature causes the rate of reaction to increase.
\nMnO2 is another possible catalyst for the reaction. State the IUPAC name for MnO2.
\nComment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.
\nH2O2 (aq) + CH3COOH (aq) ⇌ CH3COOOH (aq) + H2O (l)
\nSodium percarbonate, 2Na2CO3•3H2O2, is an adduct of sodium carbonate and hydrogen peroxide and is used as a cleaning agent.
\nMr (2Na2CO3•3H2O2) = 314.04
\nCalculate the percentage by mass of hydrogen peroxide in sodium percarbonate, giving your answer to two decimal places.
\ndecomposes in light [✔]
\n\n
Note: Accept “sensitive to light”.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5961/4bi_m.PNG\")
points correctly plotted [✔]
\nbest fit line AND extended through (to) the origin [✔]
\nAverage rate of reaction:
«slope (gradient) of line =» 0.022 «cm3 O2 (g) s−1» [✔]
\n
Note: Accept range 0.020–0.024cm3 O2 (g) s−1.
\nRate equation:
Rate = k[H2O2] × [KI] [✔]
Overall order:
2 [✔]
\n
Note: Rate constant must be included.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5964/4biii_m.PNG\")
peak of T2 to right of AND lower than T1 [✔]
\nlines begin at origin AND T2 must finish above T1 [✔]
\nEa marked on graph [✔]
\nexplanation in terms of more “particles” with E ≥ Ea
\nOR
\ngreater area under curve to the right of Ea in T2 [✔]
\nmanganese(IV) oxide
\nOR
\nmanganese dioxide [✔]
\n\n
Note: Accept “manganese(IV) dioxide”.
\nmoves «position of» equilibrium to right/products [✔]
\n\n
Note: Accept “reactants are always present as the reaction is in equilibrium”.
\nM( H2O2) «= 2 × 1.01 + 2 × 16.00» = 34.02 «g» [✔]
\n«% H2O2 = 3 × × 100 =» 32.50 «%» [✔]
\n\n
Note: Award [2] for correct final answer.
\nThere were a couple of comments claiming that this NOS question on “why to store hydrogen peroxide in brown bottles” is not the syllabus. Most candidates were quite capable of reasoning this out.
\nMost candidates could plot a best fit line and find the slope to calculate an average rate of reaction.
\nGood performance but with answers that either typically included only [H2O2] with first or second order equation or even suggesting zero order rate equation.
\nFair performance; errors including not starting the two curves at the origin, drawing peak for T2 above T1, T2 finishing below T1 or curves crossing the x-axis.
\nThe majority of candidates earned at least one mark, many both marks. Errors included not annotating the graph with Ea and referring to increase of kinetic energy as reason for higher rate at T2.
\nA well answered question. Very few candidates had problem with nomenclature.
\nOne teacher suggested that “stored” would have been better than “sold” for this question. There were a lot of irrelevant answers with many believing the back reaction was an acid dissociation.
\nIt is recommended that candidates use the relative atomic masses given in the periodic table.
\nWhat is the order of increasing pH for the following solutions of the same concentration?
\n\n
A. HCl (aq) < NH3 (aq) < NaOH (aq) < CH3COOH (aq)
\nB. CH3COOH (aq) < HCl (aq) < NH3 (aq) < NaOH (aq)
\nC. HCl (aq) < CH3COOH (aq) < NH3 (aq) < NaOH (aq)
\nD. NaOH (aq) < NH3 (aq) < CH3COOH (aq) < HCl (aq)
\nC
\nBoth vinegar (a dilute aqueous solution of ethanoic acid) and bleach are used as cleaning agents.
\nBleach reacts with ammonia, also used as a cleaning agent, to produce the poisonous compound chloramine, NH2Cl.
\nOutline why ethanoic acid is classified as a weak acid.
\nA solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.
\nCl2 (g) + 2NaOH (aq) ⇌ NaOCl (aq) + NaCl (aq) + H2O (l)
\nSuggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.
\nDraw a Lewis (electron dot) structure of chloramine.
\nState the hybridization of the nitrogen atom in chloramine.
\nDeduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
\nMolecular geometry:
\nH–N–H bond angle:
\nState the type of bond formed when chloramine is protonated.
\nSketch a graph of pH against volume of hydrochloric acid added to ammonia solution, showing how you would determine the pKa of the ammonium ion.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5967/5di.PNG\")
\n
Suggest a suitable indicator for the titration, using section 22 of the data booklet.
\nExplain, using two equations, how an equimolar solution of ammonia and ammonium ions acts as a buffer solution when small amounts of acid or base are added.
\npartial dissociation «in aqueous solution» [✔]
\nethanoic acid/vinegar reacts with NaOH [✔]
\nmoves equilibrium to left/reactant side [✔]
\nreleases Cl2 (g)/chlorine gas
OR
Cl2 (g)/chlorine gas is toxic [✔]
\n
Note: Accept “ethanoic acid produces H+ ions”
\nAccept “ethanoic acid/vinegar reacts with NaOCl”.
\nDo not accept “2CH3COOH + NaOCl + NaCl → 2CH3COONa + Cl2 + H2O” as it does not refer to equilibrium.
\nAccept suitable molecular or ionic equations for M1 and M3.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5965/5ci.PNG\")
[✔]
\n
Note: Accept any combination of dots/crosses or lines to represent electron pairs.
\nsp3 [✔]
\nMolecular geometry:
«trigonal» pyramidal [✔]
H–N–H bond angle:
107° [✔]
\n
Note: Accept angles in the range of 100–109.
\ncovalent/dative/coordinate [✔]
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5968/5di_m.PNG\")
correct shape of graph AND vertical drop at Vn [✔]
\npKa = pH at /half neutralization/half equivalence [✔]
\n\n
Note: M1: must show buffer region at pH > 7 and equivalence point at pH < 7. Graph must start below pH = 14.
\nmethyl orange
OR
bromophenol blue
OR
bromocresol green
OR
methyl red [✔]
NH3 (aq) + H+ (aq) → NH4 + (aq) [✔]
\nNH4 + (aq) + OH− (aq) → NH3 (aq) + H2O(l) [✔]
\n\n
Note: Accept reaction arrows or equilibrium signs in both equations.
\nAward [1 max], based on two correct reverse equations but not clearly showing reacting with acid or base but rather dissociation.
\nMajority of candidates understood weak acids do not fully dissociate.
\nThe average score was 1 out 3. Many could not suggest why it is dangerous to mix chlorine with vinegar. Most students gained at least one mark for stating that “chlorine gas will be produced” but couldn’t link it to equilibrium ideas.
\nMost candidates correctly drew the Lewis structure of chloramine. Some left off lone pair electrons.
\nMostly correct with a surprising number stating sp or sp2 hybridization.
\nGenerally well done with some candidates misinterpreting the bond angle from the stated geometry.
\n“Ionic bond”, “hydrogen bond” and “intermolecular forces” were some common answers.
\nQuite poorly done with many candidates not indicating a vertical drop but rather a weak acid/weak base curve. Some did not have the correct location for the equivalence point.
\nGenerally well done although a number of candidates chose bromothymol blue as a suitable indicator for weak base with a strong acid.
\nNearly 30 % of candidates did not attempt to answer this question about buffer equations. It was also poorly answered because equations were not used to explain buffer action or the dissociation equations for the base and acid were given rather than their reactions with H+ or OH- .
\nThis question is about iron.
\nDeduce the full electron configuration of Fe2+.
\nExplain why, when ligands bond to the iron ion causing the d-orbitals to split, the complex is coloured.
\nState the nuclear symbol notation, , for iron-54.
\nMass spectrometry analysis of a sample of iron gave the following results:
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5969/6d.PNG\")
Calculate the relative atomic mass, Ar, of this sample of iron to two decimal places.
\nAn iron nail and a copper nail are inserted into a lemon.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5971/6e.PNG\")
Explain why a potential is detected when the nails are connected through a voltmeter.
\nCalculate the standard electrode potential, in V, when the Fe2+ (aq) | Fe (s) and Cu2+ (aq) | Cu (s) standard half-cells are connected at 298 K. Use section 24 of the data booklet.
\nCalculate ΔGθ, in kJ, for the spontaneous reaction in (f)(i), using sections 1 and 2 of the data booklet.
\nCalculate a value for the equilibrium constant, Kc, at 298 K, giving your answer to two significant figures. Use your answer to (f)(ii) and section 1 of the data booklet.
\n(If you did not obtain an answer to (f)(ii), use −140 kJ mol−1, but this is not the correct value.)
\n1s2 2s2 2p6 3s2 3p6 3d6 [✔]
\n«frequency/wavelength of visible» light absorbed by electrons moving between d levels/orbitals [✔]
\ncolour due to remaining frequencies
OR
complementary colour transmitted [✔]
[✔]
\n«Ar =» 54 × 0.0584 + 56 × 0.9168 + 57 × 0.0217 + 58 × 0.0031
\nOR
\n«Ar =» 55.9111 [✔]
\n«Ar =» 55.91 [✔]
\n\n
Note: Award [2] for correct final answer
\n
Do not accept data booklet value (55.85).
lemon juice is the electrolyte
OR
lemon juice allows flow of ions
OR
each nail/metal forms a half-cell with the lemon juice [✔]
\n
Any one of:
iron is higher than copper in the activity series
OR
each half-cell/metal has a different redox/electrode potential [✔]
\n
iron is oxidized
OR
Fe → Fe2+ + 2e−
OR
Fe → Fe3+ + 3e−
OR
iron is anode/negative electrode of cell [✔]
\n
copper is cathode/positive electrode of cell
OR
reduction occurs at the cathode
OR
2H+ + 2e− → H2 [✔]
electrons flow from iron to copper [✔]
«Eθ = +0.34 V −(−0.45 V) = +»0.79 «V» [✔]
\n«ΔGθ = −nFEθ = −2mol × 96 500 C mol−1 × =» −152 «kJ» [✔]
\n\n
Note: Accept range 150−153 kJ.
\n«lnKc = =» 61.38 [✔]
\nK = 4.5 × 1026 [✔]
\n\n
Note: Accept answers in range 2.0 × 1026 to 5.5 × 1026.
\nDo not award M2 if answer not given to two significant figures.
\nIf −140 kJmol−1 used, answer is 3.6 × 1024.
\nDone fairly well with common mistakes leaving in the 4s2 electrons as part of Fe2+ electron configuration, or writing 4s1 3d5
\nThis was poorly answered and showed a clear misconception and misunderstanding of the concepts. Most of the candidates failed to explain why the complex is coloured and based their answers on the emission of light energy when electrons fall back to ground state and not on light absorption by electrons moving between the split d-orbitals and complementary colour transmitted of certain frequencies.
\nMany candidates wrote the nuclear notation for iron as Z over A.
\nThis question on average atomic mass was the best answered question on the exam. A few candidates did not write the answer to two decimal places as per instructions.
\nVery few candidates scored M1 regarding the lemon juice role as electrolyte. Some earned M2 but a lot of answers were too vague, such as ‘electrons move through the circuit’, etc.
\nOnly 50 % of candidates earned this relatively easy mark on calculate EMF from 2 half-cell electrode potentials.
\nAverage performance; typical errors were using the incorrect value for n, the number of electrons, or not using consistent units and making a factor of 1000 error in the final answer.
\nThis question was left blank by quite a few candidates. Common errors included not using correct units, or more often, calculation error in converting ln Kc into Kc value.
\nAn aqueous solution of silver nitrate, AgNO3 (aq), can be electrolysed using platinum electrodes.
\nFormulate the half-equations for the reaction at each electrode during electrolysis.
\nCathode (negative electrode):
\nAnode (positive electrode):
\nCathode (negative electrode):
Ag+ (aq) + e− → Ag (s) [✔]
Anode (positive electrode):
2H2O(l) → O2 (g) + 4H+ (aq) + 4e− [✔]
\n
Note: Accept 4OH− (aq) → O2 (g) + 2H2O(l) + 4e−
\nAccept multiple or fractional coefficients in both half-equations.
\nVery few answers were correct, even for stronger candidates. Many failed to formulate the correct half equation for the reaction at the anode and used the nitrate ion instead of oxidation of H2O. Some candidates lost one of the marks for using equilibrium arrows in an electrolysis equation.
\nEthyne, C2H2, reacts with oxygen in welding torches.
\nEthyne reacts with steam.
\nC2H2 (g) + H2O (g) → C2H4O (g)
\nTwo possible products are:
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5980/4c.PNG\")
Product B, CH3CHO, can also be synthesized from ethanol.
\nWrite an equation for the complete combustion of ethyne.
\nDeduce the Lewis (electron dot) structure of ethyne.
\nCompare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
\nIdentify the type of interaction that must be overcome when liquid ethyne vaporizes.
\nState the name of product B, applying IUPAC rules.
\nDetermine the enthalpy change for the reaction, in kJ, to produce A using section 11 of the data booklet.
\nThe enthalpy change for the reaction to produce B is −213 kJ.
\nPredict, giving a reason, which product is the most stable.
\nThe IR spectrum and low resolution 1H NMR spectrum of the actual product formed are shown.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5981/1civ.PNG\")
Deduce whether the product is A or B, using evidence from these spectra together with sections 26 and 27 of the data booklet.
\nIdentity of product:
\nOne piece of evidence from IR:
\nOne piece of evidence from 1H NMR:
\n\n
Deduce the splitting pattern you would expect for the signals in a high resolution 1H NMR spectrum.
\n2.3 ppm:
\n9.8 ppm:
\nSuggest the reagents and conditions required to ensure a good yield of product B.
\nReagents:
\nConditions:
\nDeduce the average oxidation state of carbon in product B.
\nExplain why product B is water soluble.
\nC2H2 (g) + 2.5O2 (g) → 2CO2 (g) + H2O (l)
\nOR
\n2C2H2 (g) + 5O2 (g) → 4CO2 (g) + 2H2O (l) [✔]
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5979/1bi.PNG\")
[✔]
\n
Note: Accept any valid combination of lines, dots and crosses.
\n«ethyne» shorter AND a greater number of shared/bonding electrons
\nOR
\n«ethyne» shorter AND stronger bond [✔]
\nLondon/dispersion/instantaneous dipole-induced dipole forces [✔]
\nethanal [✔]
\n«sum of bond enthalpies of reactants =» 2(C—H)+C ≡ C + 2(O—H)
OR
2 × 414 «kJ mol-1» + 839 «kJ mol-1» + 2 × 463 «kJ mol-1»
OR
2593 «kJ» [✔]
\n
«sum of bond enthalpies of A =» 3(C—H) + C=C + C—O + O—H
OR
3 × 414 «kJ mol-1» + 614 «kJ mol-1» + 358 «kJ mol-1» + 463 «kJ mol-1»
OR
2677 «kJ» [✔]
«enthalpy of reaction = 2593 kJ – 2677 kJ» = –84 «kJ» [✔]
\n
Note: Award [3] for correct final answer.
\nB AND it has a more negative/lower enthalpy/«potential» energy
\nOR
\nB AND more exothermic «enthalpy of reaction from same starting point» [✔]
\nIdentity of product: «B»
\nIR spectrum:
1700–1750 «cm–1 band» AND carbonyl/CO group present
OR
no «band at» 1620–1680 «cm–1» AND absence of double bond/C=C
OR
no «broad band at» 3200–3600 «cm–1 » AND absence of hydroxyl/OH group [✔]
1H NMR spectrum:
«only» two signals AND A would have three
OR
«signal at» 9.4–10.0 «ppm» AND «H atom/proton of» aldehyde/–CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of alkyl/CH next to» aldehyde/CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of» RCOCH2- present
OR
no «signal at» 4.5–6.0 «ppm» AND absence of «H atom/proton next to» double bond/C=C ✔
\n
Note: Accept a specific value or range of wavenumbers and chemical shifts.
\nAccept “two signals with areas 1:3”.
\n2.3 ppm: doublet [✔]
\n9.8 ppm: quartet [✔]
\nReagents:
acidified/H+ AND «potassium» dichromate«(VI)»/K2Cr2O7/Cr2O72- [✔]
Conditions:
distil «the product before further oxidation» [✔]
\n
Note: Accept “«acidified potassium» manganate(VII)/KMnO4/MnO4-/permanganate”.
\nAccept “H2SO4” or “H3PO4” for “H+”.
\nAccept “more dilute dichromate(VI)/manganate(VII)” or “excess ethanol”.
\nAward M1 if correct reagents given under “Conditions”.
\n–1 [✔]
\nAny three of:
\nhas an oxygen/O atom with a lone pair [✔]
\nthat can form hydrogen bonds/H-bonds «with water molecules» [✔]
\nhydrocarbon chain is short «so does not disrupt many H-bonds with water molecules» [✔]
\n«large permanent» dipole-dipole interactions with water [✔]
\nAll candidates were able to write the correct reactants/products for combustion of ethyne, but a few failed to balance correctly.
\nMost drew correct Lewis structures for ethyne, though some drew ethene.
\nSurprisingly very few explained the difference in bond length/strength looking at electrons shared and just gave the shorter/triple or longer/single bond answer.
\nGood to see that most candidates identified the specific IMF correctly.
\nMost candidates gave the correct IUPAC name.
\nCandidates were able to calculate the ΔH of the given reaction correctly; a few inverted the calculations or made mathematical errors.
\nGenerally well done, most common error was stating that the enthalpy change was “larger” without the indication that it was an exothermic change or the sign.
\nInterpretation of spectra was very good and the few candidates that lost marks with 1H NMR data rather than IR, for example simply mentioning two signals for B. However, most candidates that attempted this question got full marks.
\nThe stronger candidates were able to predict the splitting pattern correctly, others inverted the answer, but many others repeated the information for protons with the given chemical shift, which is unexpected since wording was straightforward.
\nCandidates seemed to be confused by the prompts, reagent and conditions, so often included the acid among conditions. Careless errors were common such as the wrong charge on the dichromate ion. Few candidates suggest permanganate as an option.
\nMost candidates were able to calculate oxidation state of carbon in B.
\nCandidates did not understand that they must mention the IMF responsible for the solubility. Most candidates explained the polarity of the aldehyde and water but did not mention that this results in permanent dipole-dipole interactions; many did mention H-bonding. The mention of the lone pair on O atom and short hydrocarbon chain were very rare.
\nXylene is a derivative of benzene. One isomer is 1,4-dimethylbenzene.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5942/1.PNG\"/)
Xylene, like benzene, can be nitrated.
\nBromine reacts with alkanes.
\nState the number of 1H NMR signals for this isomer of xylene and the ratio in which they appear.
\n\n
Draw the structure of one other isomer of xylene which retains the benzene ring.
\nWrite the equation for the production of the active nitrating agent from concentrated sulfuric and nitric acids.
\nExplain the mechanism for the nitration of benzene, using curly arrows to indicate the movement of electron pairs.
\nIdentify the initiation step of the reaction and its conditions.
\n1,4-dimethylbenzene reacts as a substituted alkane. Draw the structures of the two products of the overall reaction when one molecule of bromine reacts with one molecule of 1,4-dimethylbenzene.
\nThe organic product is not optically active. Discuss whether or not the organic product is a racemic mixture.
\nNumber of signals: 2 [✔]
\nRatio:
\n3 : 2
\nOR
\n6 : 4 [✔]
\nNote: Accept any correct integer or fractional ratio. Accept ratios in reverse order.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5943/1a_m.PNG\")
[✔]
2H2SO4 + HNO3 ⇌ NO2+ + 2HSO4− + H3O+ [✔]
\nNote: Accept a single arrow instead of an equilibrium sign.
Accept “H2SO4 + HNO3 ⇌ NO2+ + HSO4− + H2O”.
Accept “H2SO4 + HNO3 ⇌ H2NO3+ + HSO4−”.
Accept equivalent two step reactions in which sulfuric acid first behaves as a strong acid and protonates the nitric acid, before behaving as a dehydrating agent removing water from it.
![\"\"](\"../assets/uploads/tinymce_asset/asset/5945/1cii.PNG\")
curly arrow going from benzene ring to N «of +NO2/NO2+» [✔]
carbocation with correct formula and positive charge on ring [✔]
curly arrow going from C–H bond to benzene ring of cation [✔]
formation of organic product nitrobenzene AND H+ [✔]
\n
Note: Accept mechanism with corresponding Kekulé structures.
Do not accept a circle in M2 or M3.
Accept first arrow starting either inside the circle or on the circle.
If Kekulé structure used, first arrow must start on the double bond.
M2 may be awarded from correct diagram for M3.
M4: Accept “C6H5NO2 + H2SO4” if HSO4− used in M3.
Br2 2Br• [✔]
«sun»light/UV/hv
OR
high temperature [✔]
Note: Do not penalize missing radical symbol on Br.
Accept “homolytic fission of bromine” for M1.
![\"\"](\"../assets/uploads/tinymce_asset/asset/5946/1dii_m.PNG\")
[✔]
HBr [✔]
\n\n
Note: Accept condensed formulae, such as CH3C6H4CH2Br.
\nno AND there is no chiral carbon
\nOR
\nno AND there is no carbon with four different substituents/groups [✔]
\n\n
Note: Accept “no AND no asymmetric carbon
atom”.
Many identified two correct peaks but quite a few less the correct ratio.
\nGenerally well done, although some candidates repeated the formula of the 1,4-isomer structure or drew the wrong bond, e.g. benzene ring to H rather than C on CH3.
\nThe production of NO3− was a common answer.
\nPerformance was fairly good by schools covering the topic while others had no idea. There were many careless steps, such as omission or misplacement of + sign.
\nVery well done, with a few making reference to a catalyst.
\nSome candidates lost one mark for the bond originated from H in CH3 instead of C. Some teachers thought the use of the word “substituted alkane” made the question more difficult than it should have been.
\nOne of the most poorly answered questions on the exam with only 10 % of candidates earning this mark. Some candidates just answered ‘yes’ or ‘no’ on whether the organic product is a racemic mix and very few mentioned the absence of a chiral carbon. One teacher though the use of benzene in this question made it unnecessarily tough, stating “the optical activity of benzene has not been covered due to the limited chemistry of benzene included in the specification. An aliphatic compound here would test the understanding of enantiomers without the confusion of adding benzene”. Candidates should recognize that carbon in benzene cannot be the centre of optical activity and look for chiral carbons in the substitution chains.
\nBenzoic acid, C6H5COOH, is another derivative of benzene.
\nIdentify the wavenumber of one peak in the IR spectrum of benzoic acid, using section 26 of the data booklet.
\nIdentify the spectroscopic technique that is used to measure the bond lengths in solid benzoic acid.
\nOutline one piece of physical evidence for the structure of the benzene ring.
\nDraw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
\nOutline why both C to O bonds in the conjugate base are the same length and suggest a value for them. Use section 10 of the data booklet.
\nThe pH of an aqueous solution of benzoic acid at 298 K is 2.95. Determine the concentration of hydroxide ions in the solution, using section 2 of the data booklet.
\nFormulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.
\nThe combustion reaction in (f)(ii) can also be classed as redox. Identify the atom that is oxidized and the atom that is reduced.
\nSuggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
\nState the reagent used to convert benzoic acid to phenylmethanol (benzyl alcohol), C6H5CH2OH.
\nAny wavenumber in the following ranges:
2500−3000 «cm−1» [✔]
1700−1750 «cm−1» [✔]
2850−3090 «cm−1» [✔]
X-ray «crystallography/spectroscopy» [✔]
\nAny one of:
\n«regular» hexagon
\nOR
\nall «H–C–C/C-C-C» angles equal/120º [✔]
all C–C bond lengths equal/intermediate between double and single
OR
\nbond order 1.5 [✔]
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5947/2d.PNG\")
[✔]
\n
Note: Accept Kekulé structures.
Negative sign must be shown in correct position.
electrons delocalized «across the O–C–O system»
\nOR
\nresonance occurs [✔]
\n122 «pm» < C–O < 143 «pm» [✔]
\n\n
Note: Accept “delocalized π-bond”.
Accept “bond intermediate between single and double bond” or “bond order 1.5” for M1.
Accept any answer in range 123 to 142 pm.
ALTERNATIVE 1:
[H+] «= 10−2.95» = 1.122 × 10−3 «mol dm−3» [✔]
«[OH−] = =» 8.91 × 10−12 «mol dm−3» [✔]
\n
ALTERNATIVE 2:
pOH = «14 − 2.95 =» 11.05 [✔]
«[OH−] = 10−11.05 =» 8.91 × 10−12 «mol dm−3» [✔]
\n
Note: Award [2] for correct final answer.
Accept other methods.
2C6H5COOH (s) + 15O2 (g) → 14CO2 (g) + 6H2O (l)
correct products [✔]
correct balancing [✔]
Oxidized:
\nC/carbon «in C6H5COOH»
\nAND
\nReduced:
\nO/oxygen «in O2» [✔]
\n«intermolecular» hydrogen bonding [✔]
\n\n
Note: Accept diagram showing hydrogen bonding.
\nlithium aluminium hydride/LiAlH4 [✔]
\nMost candidates could identify a wavenumber or range of wavenumbers in the IR spectrum of benzoic acid.
\nLess than half the candidates identified x-ray crystallography as a technique used to measure bond lengths. There were many stating IR spectroscopy and quite a few random guesses.
\nAgain less than half the candidates could accurately give a physical piece of evidence for the structure of benzene. Many missed the mark by not being specific, stating ‘all bonds in benzene with same length’ rather than ‘all C-C bonds in benzene have the same length’.
\nVery poorly answered with only 1 in 5 getting this question correct. Many did not show all the bonds and all the atoms or either forgot or misplaced the negative sign on the conjugate base.
\nThis question was a challenge. Candidates were not able to explain the intermediate bond length and the majority suggested the value of either the bond length of C to O single bond or double bond.
\nGenerally well done with a few calculating the pOH rather than the concentration of hydroxide ion asked for.
\nMost earned at least one mark by correctly stating the products of the reaction.
\nAnother question where not reading correctly was a concern. Instead of identifying the atom that is oxidized and the atom that is reduced, answers included formulas of molecules or the atoms were reversed for the redox processes.
\nThe other question where only 10 % of the candidates earned a mark. Few identified hydrogen bonding as the reason for carboxylic acids forming dimers. There were many G2 forms stating that the use of the word “dimer” is not in the syllabus, however the candidates were given that a dimer has double the molar mass and the majority seemed to understand that the two molecules joined together somehow but could not identify hydrogen bonding as the cause.
\nVery few candidates answered this part correctly and scored the mark. Common answers were H2SO4, HCl & Sn, H2O2. In general, strongest candidates gained the mark.
\nThis question is about sodium and its compounds.
\nThe Born-Haber cycle for sodium oxide is shown (not to scale).
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5949/3d.PNG\")
Sodium peroxide is used in diving apparatus to produce oxygen from carbon dioxide.
\n2Na2O2 (s) + 2CO2 (g) → 2Na2CO3 (s) + O2 (g)
\nPlot the relative values of the first four ionization energies of sodium.
\n\n
\n
Outline why the alkali metals (group 1) have similar chemical properties.
\nDescribe the structure and bonding in solid sodium oxide.
\nCalculate values for the following changes using section 8 of the data booklet.
\n
ΔHatomisation (Na) = 107 kJ mol−1
ΔHatomisation (O) = 249 kJ mol−1
O2(g) → O2- (g):
\nNa (s) → Na+ (g):
\n\n
\n
The standard enthalpy of formation of sodium oxide is −414 kJ mol−1. Determine the lattice enthalpy of sodium oxide, in kJ mol−1, using section 8 of the data booklet and your answers to (d)(i).
\n
(If you did not get answers to (d)(i), use +850 kJ mol−1 and +600 kJ mol−1 respectively, but these are not the correct answers.)
Justify why K2O has a lower lattice enthalpy (absolute value) than Na2O.
\nWrite equations for the separate reactions of solid sodium oxide and solid phosphorus(V) oxide with excess water and differentiate between the solutions formed.
\nSodium oxide, Na2O:
\nPhosphorus(V) oxide, P4O10:
\nDifferentiation:
\n\n
\n
Sodium peroxide, Na2O2, is formed by the reaction of sodium oxide with oxygen.
\n2Na2O (s) + O2 (g) → 2Na2O2 (s)
\nCalculate the percentage yield of sodium peroxide if 5.00g of sodium oxide produces 5.50g of sodium peroxide.
\nDetermine the enthalpy change, ΔH, in kJ, for this reaction using data from the table and section 12 of the data booklet.
\nOutline why bond enthalpy values are not valid in calculations such as that in (g)(i).
\nAn allotrope of molecular oxygen is ozone. Compare, giving a reason, the bond enthalpies of the O to O bonds in O2 and O3.
\nOutline why a real gas differs from ideal behaviour at low temperature and high pressure.
\nThe reaction of sodium peroxide with excess water produces hydrogen peroxide and one other sodium compound. Suggest the formula of this compound.
\nState the oxidation number of carbon in sodium carbonate, Na2CO3.
\n [✔]
Notes: Accept curve showing general trend.
Award mark only if the energy difference between the first two points is larger than that between points 2/3 and 3/4.
\n
same number of electrons in outer shell
\nOR
\nall are s1 [✔]
\n«3-D/giant» regularly repeating arrangement «of ions»
OR
lattice «of ions» [✔]
\n
electrostatic attraction between oppositely charged ions
OR
electrostatic attraction between Na+ and O2− ions [✔]
\n
Note: Do not accept “ionic” without description.
\nO2(g) → O2- (g)
\n«ΔHatomisation (O) + 1st EA + 2nd EA = 249 k Jmol−1 − 141 kJmol−1 + 753 kJmol−1 =» «+»861 «kJmol−1» [✔]
\n\n
Na (s) → Na+ (g)
\n«ΔHatomisation (Na) + 1st IE = 107 kJmol−1 + 496 kJmol−1 =» «+»603 «kJmol−1» [✔]
\nlattice enthalpy = 861 «kJ mol−1» + 2 × 603 «kJ mol−1» −(−414 «kJ mol−1») [✔]
\n«= +» 2481 «kJ mol−1» [✔]
\n\n
Note: Award [2] for correct final answer.
\nIf given values are used:
M1: lattice enthalpy = 850 «kJ mol−1» +
2 × 600 «kJ mol−1» −(−414 «kJ mol−1»)
M2: «= +» 2464 «kJ mol−1»
K+ ion is larger than Na+
\nOR
\nsmaller attractive force because of greater distance between ion «centres» [✔]
\nSodium oxide:
Na2O(s) + H2O(l) → 2NaOH (aq) [✔]
Phosphorus(V) oxide:
P4O10 (s) + 6H2O(l) → 4H3PO4 (aq) [✔]
Differentiation:
NaOH/product of Na2O is alkaline/basic/pH > 7 AND H3PO4/product of P4O10 is acidic/pH < 7 [✔]
n(Na2O2) theoretical yield «= » = 0.0807/8.07 × 10−2 «mol»
\nOR
mass of Na2O2 theoretical yield «= × 77.98 gmol−1» = 6.291 «g» [✔]
\n% yield «= × 100» OR « × 100» = 87.4 «%» [✔]
\n\n
Note: Award [2] for correct final answer.
\n∑ΔHf products = 2 × (−1130.7) / −2261.4 «kJ» [✔]
\n∑ΔHf reactants = 2 × (−510.9) + 2 × (−393.5) / −1808.8 «kJ» [✔]
\nΔH = «∑ΔHf products − ∑ΔHf reactants = −2261.4 −(−1808.8) =» −452.6 «kJ» [✔]
\n\n
Note: Award [3] for correct final answer.
\nAward [2 max] for “+ 452.6 «kJ»”.
\nonly valid for covalent bonds
\nOR
\nonly valid in gaseous state [✔]
\nbond in O3 has lower enthalpy AND bond order is 1.5 «not 2» [✔]
\n\n
Note: Accept “bond in ozone is longer”.
\nAny one of:
\nfinite volume of particles «requires adjustment to volume of gas» [✔]
\nshort-range attractive forces «overcomes low kinetic energy» [✔]
\nNaOH [✔]
\nIV [✔]
\nGenerally well done with a correct plot of ionization energies.
\nThe majority answered correctly stating same number of valence electrons as the reason. Some candidates stated same size or similar ionization energy but the majority scored well.
\nMany candidates lost one or two marks for missing “electrostatic forces” between “oppositely charged ions”, or “lattice”. Some candidates’ answers referred to covalent bonds and shapes of molecules.
\nGood performance with typical error being in the calculation for the first equation, ½O2 (g) → O2− (g), where the value for the first electron affinity of oxygen was left out.
\nMany candidates earned some credit for ECF based on (d)(i).
\nAverage performance with answers using atomic size rather than ionic size or making reference to electronegativities of K and Na.
\nAn average of 1.1 out of 3 earned here. Many candidates could write a balanced equation for the reaction of sodium oxide with water but not phosphorus(V) oxide. Mediocre performance in identifying the acid/base nature of the solutions formed.
\nThe majority earned one or two marks in finding a % yield.
\nThe average was 2.2 out 3 for this question on enthalpy of formation. Enthalpy calculations were generally well done.
\nThe majority of candidates referred to “bond enthalpy values are average”, rather than not valid for solids or only used for gases.
\nSome candidates recognized that ozone had a resonance structure but then only compared bond length between ozone and oxygen rather than bond enthalpy.
\nFew candidates could distinguish the cause for difference in behaviour between real and ideal gases at low temperature or high pressure. Many answers were based on increase in number of collisions or faster rate or movement of gas particles.
\nNa2O was a common formula in many candidates’ answers for the product of the reaction of sodium peroxide with water.
\nThe vast majority of candidates could correctly state the oxidation number of carbon in sodium carbonate.
\nBromine and methanoic acid react in aqueous solution.
\nBr2 (aq) + HCOOH (aq) → 2Br− (aq) + 2H+ (aq) + CO2 (g)
\nThe reaction was monitored by measuring the volume of carbon dioxide produced as time
progressed.
Determine from the graph the rate of reaction at 20 s, in cm3 s−1, showing your working.
\nOutline, with a reason, another property that could be monitored to measure the rate of this reaction.
\nDescribe one systematic error associated with the use of the gas syringe, and how the error affects the calculated rate.
\nIdentify one error associated with the use of an accurate stopwatch.
\ntangent drawn to curve at t = 20 s [✔]
\nslope/gradient calculation [✔]
\n0.35 «cm3 s–1» [✔]
\n\n
Note: Accept values in the range 0.32–0.42 «cm3 s–1»
\nALTERNATIVE 1
colour [✔]
Br2 /reactant is coloured «Br– (aq) is not» [✔]
\n
ALTERNATIVE 2
conductivity [✔]
greater/increased concentration of ions in products [✔]
Note: Do not accept “changes in temperature” or “number of bubbles”.
\n\n
ALTERNATIVE 3
mass/pressure [✔]
gas is evolved/produced [✔]
Note: Do not accept “mass of products is less than mass of reactants”.
\n\n
ALTERNATIVE 4
pH [✔]
methanoic acid is weak AND HBr is strong
OR
increase in [H+] [✔]
ALTERNATIVE 1
gas may leak/be lost/escape
OR
plunger may stick/friction «so pressure is greater than atmospheric pressure»
OR
syringe may be tilted «up» so plunger moves less «with gravity acting on plunger»
OR
CO2 dissolved in water [✔]
calculated rate lower [✔]
ALTERNATIVE 2
syringe may be tilted «down» so plunger moves more «with gravity acting on plunger»
OR
syringe is held in hand so gets warmer and gas expands [✔]
calculated rate higher [✔]
\n
Note: Calculated rate is lower or higher must be stated for M2.
\nDo not accept “scale on syringe is inaccurate”, “errors in reading syringe”, or “bubbles in syringe”.
\nhuman reaction time/delay «starting/stopping the stopwatch» [✔]
\nThis part proved to be challenging for some candidates whereas other candidates were able to draw a tangent at 20 sec and then calculate the rate. A significant number of candidates calculated the average rate and achieved one mark.
\nMajority of the candidates stated another property, which could be monitored correctly. The most common error was changes in temperature, which was stated by some candidates.
\nThis part was about the systematic error and was answered correctly, but many candidates failed to state how the error affected the calculated rate. Many candidates confused this with the concept of a random error and identified the uncertainty of reading the syringe, which is incorrect.
\nThis part was well answered by most candidates although some candidates did not read the question clearly and commented on the stopwatch not working properly or not being accurate.
\nDescribe the characteristics of the nematic liquid crystal phase.
\nShape of molecules:
\nDistribution:
\nShape of molecules:
linear
OR
rod «shaped» [✔]
Distribution:
no positional order AND «some» directional order [✔]
Note: Accept “partly ordered”.
\nMost candidates were able to obtain at least one mark on this question but struggled with the distribution of the nematic liquid crystal phase.
\nMetals are extracted from their ores by several methods, including electrolysis and reduction with carbon.
\nDetermine the mass of aluminium, in g, that could be extracted from an appropriate solution by a charge of 48 250 C. Use sections 2 and 6 of the data booklet.
\nOnce extracted, the purity of the metal can be assessed using ICP-MS. Suggest two advantages of using plasma technology rather than regular mass spectrometry.
\nExplain the action of metals as heterogeneous catalysts.
\nOutline how alloys conduct electricity and why they are often harder than pure metals.
\nConduct electricity:
\nHarder than pure metals:
\nCarbon nanotubes are added to metals to increase tensile strength.
\nWrite an equation for the formation of carbon nanotubes from carbon monoxide.
\nmoles of electrons «= » = 0.5000 «mol» [✔]
\nmoles of aluminium «= » = 0.1667 «mol» [✔]
\nmass of aluminium «= 26.98 g mol–1 × 0.1667 mol» = 4.50 «g» [✔]
\n\n
Note: Award [3] for correct final answer.
\nAny two of:
larger linear calibration [✔]
«accurate» detection of multiple elements/metals [✔]
\n«accurate» detection of elements in low concentration [✔]
\ntemperature around 10 000 K atomises/ionises every material [✔]
\nAny two of:
reactant(s) adsorb onto active sites/surface [✔]
bonds weakened/broken/stretched «in adsorbed reactants»
OR
activation energy lowered [✔]
products desorbed [✔]
\n\n
Note: Accept “products released” for M3.
\nConduct electricity:
«delocalized/valence» electrons free to move «under potential difference» [✔]
Harder than pure metals:
atoms/ions of different sizes prevent layers «of atoms/ions» from sliding over one another [✔]
2CO (g) → C (s) + CO2 (g) [✔]
\nMany candidates did reasonably well in this question but some struggled with the number of electrons required.
\nMost candidates did not seem to understand any advantages of using plasma technology rather than regular mass spectrometry.
\nThis question was reasonably answered with many candidates receiving a mark for the action of a catalyst. The terms adsorbed and desorbed were often missing.
\nMost candidates were awarded one mark for how alloys conduct electricity. Some struggled with describing why they are harder than pure metals.
\nCarbon nanotubes proved to be difficult for the majority of the candidates. Hardly any candidates stated an equation for the formation of carbon nanotubes from carbon monoxide.
\nPolymers have a wide variety of uses but their disposal can be problematic.
\nDraw a section of isotactic polychloroethene (polyvinylchloride, PVC) showing all the atoms and all the bonds of four monomer units.
\nThe infrared (IR) spectrum of polyethene is given.
\nSuggest how the IR spectrum of polychloroethene would diff er, using section 26 of the data booklet.
\nExplain how plasticizers affect the properties of plastics.
\nSuggest why the addition of plasticizers is controversial.
\nOutline, giving a reason, how addition and condensation polymerization compare with regard to green chemistry.
\nDraw the full structural formula of the organic functional group formed during the polymerization of the two reactants below.
\ncorrect bonding [✔]
\nCl atoms all on same side and alternate [✔]
\n\n
Note: Continuation bonds must be shown.
\nAward [1 max] if less than or more than four units shown.
\nAccept a stereo formula with all atoms and bonds shown.
\n«strong additional» absorption at 600–800 «cm–1» [✔]
\nAny two of:
embedded/fit between chains of polymers [✔]
prevent chains from forming crystalline regions [✔]
\nkeep polymer strands/chains/molecules separated/apart [✔]
\nincrease space/volume between chains [✔]
\nweaken intermolecular/dipole-dipole/London/dispersion/instantaneous dipoleinduced dipole/van der Waals/vdW forces «between chains» [✔]
\nincrease flexibility/durability/softness [✔]
\nmake polymers less brittle [✔]
\n\n
Note: Accept “lowers density/melting point”.
\nleach into foodstuffs/environment
OR
«unknown» health/environmental consequences [✔]
\n
Note: Accept “plasticizers cannot be recycled”.
\naddition produces only the polymer «AND more green»
OR
addition has no by/side-product/condensation produces by-product/small molecules/HCl/NH3 «AND less green»
OR
addition has high atom economy/condensation has lower atom economy «AND less green»
OR
condensation polymers «often» more biodegradable than addition polymers «AND more green» [✔]
\n
Note: Accept “if water produced by condensation «AND condensation and addition equally green»”.
\nAccept for addition “all of reactants change into products”.
\n [✔]
\n
Note: Continuation bonds must be shown.
\nDo not accept condensed formula.
\nFew candidates scored at least one mark although most either scored both or none for this polymer structure. Some did not read that only four monomer units are required.
\nAlmost all candidates received the mark for identifying the correct absorption band for polychloroethene.
\nThis was a well-answered question; with most candidates identifying at least one method plasticizers affect the properties of plastic.
\nSeveral candidates wrote vague answers as to why the addition of plasticizers is controversial.
\nCandidates seemed to have difficulty in comparing addition and condensation polymerisation with regard to green chemistry.
\nSeveral candidates struggled to draw the full structural formula of the peptide linkage formed during the polymerisation of the two reactants.
\nCalcium has a face-centred cubic (cubic close packing) arrangement of atoms.
\nState the number of atoms in the unit cell.
\nDetermine the density of calcium, in g cm−3, using section 2 of the data booklet.
\nAr = 40.08; metallic radius (r) = 1.97 × 10−10 m
\n« » 4 [✔]
\na = « =» 5.572 × 10–10 «m»
OR
volume of unit cell = «(5.572 × 10–10 m)3 × 106 =» 1.73 × 10–22 «cm3» [✔]
mass of unit cell =« =» 2.66 × 10–22 «g» [✔]
\ndensity = « » 1.54 «g cm–3» [✔]
\n\n
Note: Award [3] for correct final answer.
\nThe number of atoms in the unit cell was correctly calculated by most of the candidates.
\nMajority of the candidates managed to get three marks in determining the density of the calcium.
\nWhich change of state is exothermic?
\nA. CO2(s) → CO2(g)
B. H2O(l) → H2O(g)
C. NH3(g) → NH3(l)
D. Fe(s) → Fe(l)
\n
C
\nWhich volume, in cm3, of 0.20 mol dm-3 NaOH (aq) is needed to neutralize 0.050 mol of H2S(g)?
\nH2S(g) + 2NaOH(aq) → Na2S(aq) + 2H2O(l)
\nA. 0.25
B. 0.50
C. 250
D. 500
D
\nThe complete combustion of 15.0cm3 of a gaseous hydrocarbon X produces 60.0 cm3 of carbon dioxide gas and 75.0 cm3 of water vapour. What is the molecular formula of X? (All volumes are measured at the same temperature and pressure.)
\nA. C4H6
B. C4H8
C. C4H10
D. C6H10
C
\n5.0mol of Fe2O3(s) and 6.0mol of CO(g) react according to the equation below. What is the limiting reactant and how many moles of the excess reactant remain unreacted?
\nFe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
\nB
\nWhich is correct for the line emission spectrum for hydrogen?
\nA. Line M has a higher energy than line N.
B. Line N has a lower frequency than line M.
C. Line M has a longer wavelength than line N.
D. Lines converge at lower energy.
C
\nWhat is the condensed electron configuration of the Fe2+ ion?
\nA. [Ar]3d6
B. [Ar]3d44s2
C. [Ar]3d54s1
D. [Ar]3d64s2
A
\nWhich equation represents the first electron affinity of chlorine?
\nA. Cl(g)+e-→ Cl-(g)
B. Cl2(g) + e- → Cl-(g)
C. Cl+(g) + e- → Cl(g)
D. Cl(g) → Cl+(g) + e-
A
\nWhich solution forms when phosphorus(V) oxide, P4O10, reacts with water?
\nC
\nSunflower oil contains stearic, oleic and linoleic fatty acids. The structural formulas of these acids are given in section 34 of the data booklet.
\nExplain which one of these fatty acids has the highest boiling point.
\n10.0 g of sunflower oil reacts completely with 123 cm3 of 0.500 moldm–3 iodine solution. Calculate the iodine number of sunflower oil to the nearest whole number.
\nstearic acid AND chain has no kinks/more regular structure
OR
stearic acid AND it has straight chain
OR
stearic acid AND no C=C/carbon to carbon double bonds
OR
stearic acid AND saturated
OR
stearic acid AND chains pack more closely together
stronger London/dispersion/instantaneous induced dipole-induced dipole forces «between molecules»
\n\n
Accept “stearic acid AND greater surface area/electron density”.
\nM2 can only be scored if stearic acid is correctly identified.
\nAccept “stronger intermolecular/van der Waals’/vdW forces”.
\n[2 marks]
\n«n(I2) = 0.123 dm3 x 0.500 moldm–3 =» 0.0615 «mol»
\n«m(I2) = 0.0615 mol x 253.8 gmol–1 =» 15.6 «g»
\n«iodine number » = 156
\n\n
Award [3] for correct final answer.
\nIodine number must be a whole number.
\nAward [2 max] for 78.
\n[3 marks]
\nHeavy metals are toxic even in very low concentrations.
\nOutline why heavy metals are toxic.
\nDetermine the maximum concentration of lead(II) ions at 298 K in a solution in which the concentration of carbonate ions is maintained at 1.10 × 10−4 mol dm−3. Use section 32 of the data booklet.
\nState a method, other than precipitation, of removing heavy metal ions from solution.
\nAny one of:
disrupt endocrine system
OR
compete for active sites of enzymes/cellular receptors
OR
form complexes with/inhibit enzymes
OR
denature proteins
OR
change shape of active site [✔]
participate in redox reactions
OR
disturb normal redox balance «in cells» [✔]
initiate «free» radical reactions «in electron transfer» [✔]
\n«Ksp = 7.40 × 10–14»
Ksp = [Pb2+][CO32–] [✔]
[Pb2+] «= » = 6.73 × 10–10 «mol dm–3» [✔]
\n\n
Note: Award [2] for correct final answer.
\nAny one of:
chelation «by EDTA/polydentate ligand anchored» [✔]
ion exchange systems [✔]
\nadsorption by «water» plants [✔]
\n\n
Note: Accept “use of zeolites”.
\nThe candidates seemed to have difficulty in outlining why heavy metals are toxic.
\nMajority of the candidates managed to get two marks in determining the maximum concentration of lead(II) ions using solubility product constant.
\nThis was not well answered by most of the candidates.
\nA chemical reaction occurs when a phospholipid is heated with excess sodium hydroxide.
\nGlycerol is one product of the reaction. Identify the two other organic products.
\nIdentify the type of reaction which occurs.
\nC17H31COONa
\n[(CH3)3NCH2CH2OH]OH
\n\n
Accept “NaC17H31COO”.
\nAccept “(CH3)3N+CH2CH2OH OR [(CH3)3NCH2CH2OH]+” if positive charge is shown.
\nAccept suitable names (eg, sodium linoleate, choline hydroxide etc.) OR correct molecular formulas.
\n[2 marks]
\nhydrolysis
\n\n
Accept “nucleophilic substitution/displacement / SN/SN2 /saponification”.
\nDo not accept “acid hydrolysis”.
\n[1 mark]
\nSuperconductivity has many applications.
\nState what is meant by a superconductor.
\nOutline the difference in behaviour of Type 1 and Type 2 superconductors when the temperature is lowered.
\n«material with» no electrical resistance [✔]
\nType 1 has sharper transition to superconductivity [✔]
\n\n
Note: Accept annotated plot of electrical resistance against temperature.
\nThe question about superconductor was well answered by the candidates.
\nSome candidates struggled to outline the difference in the behaviour of Type 1 and Type 2 superconductors when the temperature is lowered.
\nProteins have structural or enzyme functions.
\nOil spills are a major environmental problem.
\nSome proteins form an α-helix. State the name of another secondary protein structure.
\nCompare and contrast the bonding responsible for the two secondary structures.
\nOne similarity:
\nOne difference:
\nExplain why an increase in temperature reduces the rate of an enzyme-catalyzed reaction.
\nState and explain how a competitive inhibitor affects the maximum rate, Vmax, of an enzyme-catalyzed reaction.
\nSuggest two reasons why oil decomposes faster at the surface of the ocean than at greater depth.
\nOil spills can be treated with an enzyme mixture to speed up decomposition.
\nOutline one factor to be considered when assessing the greenness of an enzyme mixture.
\nβ/beta pleated/sheet [✔]
\nOne similarity:
hydrogen bonding
OR
attractions between C=O and N–H «on main chain» [✔]
One difference:
α-helix has hydrogen bonds between amino acid residues that are closer than β-pleated sheet
OR
H-bonds in α-helix parallel to helix axis AND perpendicular to sheet in β-pleated sheet
OR
α-helix has one strand AND β-pleated sheet has two «or more» strands
OR
α-helix is more elastic «since H-bonds can be broken easily» AND β-pleated sheet is less elastic «since H-bonds are difficult to break» [✔]
\n
Note: Accept a diagram which shows hydrogen bonding between O of C=O and H of NH groups for M1.
Accept “between carbonyl/amido/amide/carboxamide” but not “between amino/amine” for M1.
\nenzyme denatured/ loss of 3-D structure/conformational change
OR
«interactions responsible for» for tertiary/quaternary structures altered [✔]
shape of active site changes
OR
fewer substrate molecules fit into active sites [✔]
Vmax unchanged [✔]
\nat high substrate concentration substrate outcompetes inhibitor/need a higher
substrate concentration to reach Vmax [✔]
\n
Note: Accept suitable labelled diagram.
\nAny two of:
surface water is warmer «so faster reaction rate»/more light/energy from the sun [✔]
more oxygen «for aerobic bacteria/oxidation of oil» [✔]
\ngreater surface area [✔]
\nAny one of:
non-hazardous/toxic to the environment/living organisms [✔]
energy requirements «during production» [✔]
\nquantity/type of waste produced «during production»
OR
atom economy [✔]
safety of process [✔]
\n\n
Note: Accept “use of solvents/toxic materials «during production»”.
\nDo not accept “more steps involved”.
\nThis question was well answered with many scoring the mark although there were quite a few incorrect responses that answered “beta-helix” rather than “beta-pleated sheet”.
\nAlmost all the candidate’s stated hydrogen bonding as the similarity between the 2 types of secondary structures but lost marks on the difference between them.
\nThis question was well answered where most candidates received one mark for identifying that the enzyme will denature with an increase in temperature. However, many candidates did not continue with the explanation that the shape of the active site changes.
\nMany candidates stated correctly that Vmax remains unchanged but only some mentioned that a higher substrate concentration was required to reach Vmax for the second mark.
\nMany candidates received two marks for this part while some candidates only suggested one reason or repeated the same reason (for example - heat and energy from the sun) even though the question clearly asked for two reasons.
\nThe candidates struggled with this part and gave journalistic or vague answers that cannot be awarded marks. Atom economy was mentioned correctly by a few candidates.
\nMonosaccharides can combine to form disaccharides and polysaccharides.
\nIdentify the functional groups which are present in only one structure of glucose.
\nSucrose is a disaccharide formed from -glucose and β-fructose.
\nDeduce the structural formula of sucrose.
\nStarch is a constituent of many plastics. Suggest one reason for including starch in plastics.
\nSuggest one of the challenges scientists face when scaling up the synthesis of a new compound.
\nOnly in straight chain form:
carbonyl
OR
aldehyde
Only in ring structure:
hemiacetal
\n
Accept functional group abbreviations (eg, CHO etc.).
\nAccept “ether”.
\n[2 marks]
\ncorrect link between the two monosaccharides
\n\n
Correct 1,4 beta link AND all bonds on the 2 carbons in the link required for mark.
\nIgnore any errors in the rest of the structure.
\nPenalize extra atoms on carbons in link.
\n[1 mark]
\nplastic «more» biodegradable/degrades into nontoxic products
OR
plastic can be produced using green technology/renewable resource
OR
reduces fossil fuel use/petrochemicals
OR
easily plasticized
OR
used to form thermoplasts
[1 mark]
\nminimize «negative» impact on environment
OR
minimize waste produced
OR
consider atom economy
OR
efficiency of synthetic process
OR
problems of side reactions/lower yields
OR
control temperature «inside large reactors»
OR
availability of starting/raw materials
OR
minimize energy costs
OR
value for money/cost effectiveness/cost of production
[1 mark]
\nAmino acids contain both acidic and basic functional groups.
\nAn aqueous solution of glutamine contains 0.600 mol dm−3 of the zwitterion and 0.300 mol dm−3 of the anionic form.
\npKa1 = 2.2 and pKa2 = 9.1
\nOutline which pKa value should be used when calculating the pH of the solution, giving your reason.
\nCalculate the pH of the glutamine solution using section 1 of the data booklet.
\nDescribe what is meant by the genetic code and how it relates to protein synthesis.
\npKa2 AND pH of solution > pH of isoelectric point «as anion present»
OR
pKa2 AND zwitterion has lost H+ to become anion «so in basic solution»
OR
pKa2 AND «only» anion «and zwitterion» present [✔]
«pH = 9.1 + log »
«= 9.1 + (–0.3)» = 8.8 [✔]
sequence of bases in DNA [✔]
\ncodon/triplet code/each set of three bases codes for an amino acid [✔]
\nSeveral candidates stated correctly that pKa2 should be used with a reason whereas others wrote pKa1, which was incorrect.
\nMajority of the candidates calculated correctly the pH of the glutamine solution while other candidates managed the ECF mark from part a (i).
\nThis part was poorly answered by many candidates and was unable to state that genetic code is a sequence of bases in DNA and each codon codes for an amino acid. Often complex answers were written which were incorrect.
\nPhosphatidylcholine is an example of a phospholipid found in lecithin.
\nPhosphatidylcholine may be formed from propane-1,2,3-triol, two lauric acid molecules, phosphoric acid and the choline cation.
\nDeduce the structural formula of phosphatidylcholine.
\nIdentify the type of reaction in (a).
\nLecithin is a major component of cell membranes. Describe the structure of a cell membrane.
\nLecithin aids the body’s absorption of vitamin E.
\nSuggest why vitamin E is fat-soluble.
\nPhospholipids are also found in lipoprotein structures.
\nDescribe one effect of increased levels of low-density lipoprotein (LDL) on health.
\nphosphodiester correctly drawn [✔]
\nboth ester groups correctly drawn [✔]
\n\n
Note: Accept protonated phosphate.
\nAccept phosphodiester in centre position.
\ncondensation [✔]
\n\n
Note: Accept “esterification”.
\nAccept “nucleophilic substitution/SN”.
\nphospholipid bilayer/double layer
OR
two layers of phospholipids [✔]
polar/hydrophilic heads facing aqueous environment AND non-polar/hydrophobic tails facing away from aqueous environment [✔]
\n\n
Note: Award [2] for a suitably labelled diagram.
\nAward [1] for a correct but unlabelled diagram.
\nAccept “polar/hydrophilic heads on outside AND non-polar/hydrophobic tails on inside” for M2.
\nlong non-polar/hydrocarbon chain «and only one hydroxyl group»
OR
forms London/dispersion/van der Waals/vdW interactions with fat [✔]
Any one of:
atherosclerosis/cholesterol deposition «in artery walls» [✔]
increases risk of heart attack/stroke/cardiovascular/heart disease/CHD [✔]
\n\n
Note: Accept “arteries become blocked/walls become thicker”, “increases blood pressure”, or “blood clots”.
\nDo not accept “high cholesterol”.
\nAlmost all the candidates struggled with this part. Although the phosphodiester was a challenging mark it could be awarded in both the protonated and deprotonated form. The two ester groups were required for the second mark. Candidates were not able to draw correctly for both marks, and many left this question blank.
\nThis part was very well answered.
\nThis question was another one where the first part was fairly well answered, but the explanation or second mark was often not correct or incomplete.
\nMany candidates missed the idea of a long or large non-polar chain when describing the structure of vitamin E. Simply stating non-polar chain was not sufficient for the mark.
\nCandidates were required to state one effect of increased LDL. Majority of the candidates scored well on this part. High cholesterol is not an acceptable answer but was frequently seen.
\nSucrose is a disaccharide.
\nState the name of the functional group forming part of the ring structure of each monosaccharide unit.
\nClassify, giving your reason, the hexose (six-membered) ring of sucrose as an α or β isomer.
\nSketch the cyclic structures of the two monosaccharides which combine to form sucrose.
\nacetal
OR
ether [✔]
\n
Note: Accept “glycosidic bond/linkage” but not “glucosidic”.
\nDo not accept “hemiacetal”.
\nα-isomer AND hydroxyl group on carbon 1 and –CH2OH are trans
OR
α-isomer AND hydroxyl group on carbon 1 is below plane of ring
OR
α-isomer AND glycosidic linkage between rings is below plane of ring [✔]
\n
Note: Accept “ether linkage” for M3.
\n [✔]
\n
[✔]
This was reasonably answered although there were some candidates who stated ester or hemiacetal, which is incorrect.
\nThis part was very poorly answered. Majority of the candidates had no idea about the reason whether the six-membered ring was an alpha or beta isomer.
\nThis question was poorly answered. Many candidates lost marks due to sloppy drawing and incorrect bond linkages. Some candidates did not separate the two monosaccharides as instructed.
\nHemoglobin contains heme groups with the porphyrin ring bound to an iron(II) ion.
\nA hemoglobin’s oxygen dissociation curve is shown.
\nOutline why the complex formed between Fe2+ and oxygen is red. Refer to the diagram above and section 17 of the data booklet.
\nExplain the shape of the curve.
\nSketch another line to show the effect of an increase in body temperature on the oxygen saturation of hemoglobin.
\nextensive conjugated system
OR
extensive delocalized bonding system
OR
extended system of alternating double and single bonds [✔]
absorbs green
OR
complementary to red light [✔]
sigmoid/S-shaped
OR
as oxygen binds to one active site, shape of other active sites change [✔]
affinity of other sites for oxygen increases/ability to bind oxygen is increased by initial binding of oxygen
OR
cooperative binding [✔]
\n
Note: Accept description of sigmoid/S-shaped curve if not stated for M1.
\nThis part was fairly well answered with the majority of the candidates managed one mark. The second mark was missed because the candidates did not write extensive conjugated system or extensive delocalised bonding system.
\nMany candidates managed one mark by describing the graph and stating that the affinity of other sites for oxygen increases/cooperative bonding. Several candidates missed that it was a sigmoid/S-shaped curve.
\nMany candidates were able to sketch another line to show the effect of an increase in body temperature on the oxygen saturation of haemoglobin.
\nWhich equation represents lattice enthalpy?
\nA. NaCl (g) → Na+ (g) + Cl− (g)
\nB. NaCl (s) → Na+ (g) + Cl− (g)
\nC. NaCl (s) → Na+ (aq) + Cl− (aq)
\nD. NaCl (s) → Na+ (s) + Cl− (s)
\nB
\n72 % of candidates chose the correct equation that represents lattice enthalpy. Many candidates chose A, where the ionic compound (NaCl) was gaseous, and others chose distractor C, where the ions produced were aqueous. The discrimination index for the question was quite high.
\nWhich change has the greatest increase in entropy?
\nA. CO2 (s) → CO2 (g)
\nB. CO2 (g) → CO2 (l)
\nC. CO2 (g) → CO2 (s)
\nD. CO2 (l) → CO2 (s)
\nA
\n94 % of the candidates chose the change with the greatest increase in entropy.
\nSuggest, in terms of its structure, why vitamin D is fat-soluble using section 35 of the data booklet.
\n«mostly» non-polar
OR
hydrocarbon backbone
OR
only 1 hydroxyl «group so mostly non-polar»
\n
Accept “alcohol/hydroxy” for “hydroxyl” but not “hydroxide”.
\n[1 mark]
\nThe same amount of two gases, X and Y, are in two identical containers at the same temperature. What is the difference between the gases?
\nA. X has the higher molar mass.
\nB. Y has the higher molar mass.
\nC. X has the higher average kinetic energy.
\nD. Y has the higher average kinetic energy.
\nA
\nThis question generated debate among teachers. It extended the concept that equal amounts of gases at the same temperature have the same distribution of kinetic energy curve, to the molecular speed distribution among particles. Candidates had to know that kinetic energy is calculated based on speed and mass of the molecule to deduce the answer.
\nSome teachers welcomed the question as a “good challenge to students’ thinking”, others thought it was difficult and a couple felt it was outside of the syllabus.
\nIt was by far the most challenging question on the paper with only 21 % of candidates obtaining the correct answer. The majority of candidates chose distractor D which did not take note of the fact that the gases were at the same temperature and hence had the same average kinetic energy.
\nSeveral reactions of calcium carbonate with dilute hydrochloric acid are carried out at the same temperature.
\nCaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)
\nWhich reaction has the greatest rate?
\nA
\n74 % of the candidates chose the correct combination to give the greatest rate of reaction. The most commonly chosen distractor was D where “smaller surface area of same mass of CaCO3(s)” was chosen. It seems these candidates confused “surface area” with “particle size”.
\nThere are many sources of energy available.
\nState one advantage and one disadvantage for each energy source in the table.
\nCalculate the specific energy of hydrogen, stating its units. Refer to sections 1, 6 and 13 of the data booklet.
\nHydrogen has a higher specific energy than petrol (gasoline) but is not used as a primary fuel source in cars. Discuss the disadvantages of using hydrogen.
\n\n
Do not award marks for converse statements for advantage and disadvantage.
\nPoints related to greenhouse gases should be counted only once for the entire question.
\nBiofuels:
\nAccept “«close to» carbon neutral”, “produce less greenhouse gases/CO2” as an advantage.
\nAccept “engines have to be modified if biodiesel used” as a disadvantage.
\nFossil Fuels:
\nAccept specific pollution examples (eg, oil spills, toxic substances released when burning crude oil, etc.) as a disadvantage.
\n[4 marks]
\n«specific energy =» 142
\nkJg–1
\n\n
Accept other correct values with the correct corresponding units.
\nM2 can be scored independently.
\n[2 marks]
\nlarge volumes of hydrogen required
OR
hydrogen has lower energy density
not easily transportable «form» as it is a gas
OR
heavy containers required to carry AND compress/regulate «hydrogen»
OR
high energy/cost required to compress hydrogen to transportable liquid form
OR
atmospheric pollution may be generated during production of hydrogen
OR
hydrogen fuel cells do not work at very low temperatures
OR
highly flammable when compressed/difficult to extinguish fires
OR
leaks not easy to detect
OR
high cost of production
OR
lack of filling stations/availability to consumer «in many countries»
\n
Accept “«hydrogen combustion contributes to» knocking in engines” OR “modified engine required” for M2.
\nAccept “explosive” but not “more dangerous” for M2.
\n[2 marks]
\nWhich statement is correct about a catalyst?
\nA. It decreases the activation energy of the forward reaction but not the reverse.
\nB. It increases the proportion of products to reactants in an equilibrium.
\nC. It decreases the enthalpy change of the reaction.
\nD. It changes the mechanism of the reaction.
\nD
\nThis was surprisingly one of the most challenging questions on the paper and discriminated very well between high scoring and low scoring candidates. It tested understanding of the role of the catalyst. 52 % of candidates chose the correct answer (a catalyst changes the mechanism of the reaction). The most commonly chosen distractor was A (a catalyst decreases the activation energy of the forward reaction but not the reverse).
\nWhat is the order with respect to each reactant?
\n2NO (g) + Cl2 (g) → 2NOCl (g)
\n\n
C
\n88 % of the candidates deduced the order of the reaction with respect to each reactant based on the experimental data of initial rate and concentrations of reactants.
\nConsider the following equilibrium reaction.
\n2N2O (g) + O2 (g) 4NO (g) ΔH = +16 kJ
\nWhich change will move the equilibrium to the right?
\nA. Decrease in pressure
\nB. Decrease in temperature
\nC. Increase in [NO]
\nD. Decrease in [O2]
\nA
\n77 % of the candidates applied Le Chatelier’s Principle correctly. The question had a high discrimination index.
\nWhat is the pH of 0.001 mol dm−3 NaOH (aq)?
\nA. 1
\nB. 3
\nC. 11
\nD. 13
\nC
\n77 % of the candidates were able to calculate the pH of the aqueous solution of NaOH. The most commonly chosen distractor was D (pH = 13) which was probably selected because NaOH is a strong base.
\nIodine and bromine gases were mixed and allowed to reach equilibrium.
\nWhat is the value of the equilibrium constant?
\nA. 0.05
\nB. 1
\nC. 4
\nD. 10
\nC
\nThis was a challenging question with the highest discrimination index on the paper. 62 % of the candidates were able to deduce the equilibrium concentration of IBr and calculate the equilibrium constant correctly. The most commonly chosen distractor was B where the stoichiometric ratio was not taken into account when calculating the equilibrium concentration of IBr.
\nThe combustion of fossil fuels produces large amounts of CO2, a greenhouse gas.
\nThe diagram below illustrates a range of wavelengths in the electromagnetic spectrum.
\nSynthesis gas, or syngas, mainly composed of CO(g) and H2(g), is an alternative form of fuel. It can be produced by coal or biomass gasification, passing steam over the source material in a low oxygen environment.
\nIdentify which region, A or B, corresponds to each type of radiation by completing the table.
\nOceans can act as a carbon sink, removing some CO2(g) from the atmosphere.
\nCO2(g) CO2(aq)
\nAqueous carbon dioxide, CO2(aq), quickly reacts with ocean water in a new equilibrium reaction. Construct the equilibrium equation for this reaction including state symbols.
\nDescribe how large amounts of CO2 could reduce the pH of the ocean using an equation to support your answer.
\nSuggest an equation for the production of syngas from coal.
\nThe Fischer-Tropsch process, an indirect coal liquefaction method, converts CO(g) and H2(g) to larger molecular weight hydrocarbons and steam.
\nDeduce the equation for the production of octane by this process.
\nSuggest a reason why syngas may be considered a viable alternative to crude oil.
\n\n
Accept “B” alone for incoming radiation from sun.
\nAll three correct answers necessary for mark.
\n[1 mark]
\nCO2(aq) + H2O(l) H2CO3(aq)
\n\n
State symbols AND equilibrium arrow required for mark.
\nAccept
\nCO2(aq) + H2O(l) H+(aq) + HCO3–(aq).
\nCO2(aq) + H2O(l) 2H+(aq) + CO32–(aq).
\n[1 mark]
\nCO2(aq) + H2O(l) 2H+(aq) + CO32–(aq)
OR
CO2(aq) + H2O(l) H+(aq) + HCO3–(aq)
OR
H2CO3(aq) + H2O(l) H3O+(aq) + HCO3–(aq)
OR
H2CO3(aq) H+(aq) + HCO3–(aq)
OR
H2CO3(aq) + 2H2O(l) 2H3O+(aq) + CO32–(aq)
OR
H2CO3(aq) 2H+(aq) + CO32–(aq)
equilibrium shifts to the right causing increase in [H3O+]/[H+ ] «thereby decreasing pH»
\n\n
Equilibrium sign needed in (b) (ii) but penalize missing equilibrium sign once only in b (i) and (ii).
\nDo not accept “CO2(aq) + H2O(l) H2CO3(aq)” unless equation was not given in b (i).
\n[2 marks]
\nC(s) + H2O(g) → CO(g) + H2(g)
OR
3C(s) + H2O(g) + O2(g) → 3CO(g) + H2(g)
OR
4C(s) + 2H2O(g) + O2(g) → 4CO(g) + 2H2(g)
OR
5C(s) + H2O(g) + 2O2(g) → 5CO(g) + H2(g)
\n
Accept other correctly balanced equations which produce both CO AND H2.
\n[1 mark]
\n8CO(g) + 17H2(g) → C8H18(l) + 8H2O(g)
\n[1 mark]
\ncoal more plentiful than crude oil
OR
syngas can be produced from biomass/renewable source
OR
syngas can undergo liquefaction to form octanes/no need to transport crude
OR
syngas can be produced by gasification underground, using carbon
OR
capture/storage «to not release CO2 to the atmosphere»
OR
coal gasification produces other usable products/slag
[1 mark]
\nWhere is the buffer region for the titration of a weak acid with a strong base?
\nB
\nThere was a mistake on this question and it had to be annulled (39 marks paper). Grade boundaries were lowered accordingly. The x-axis was incorrectly labelled as “volume of weak acid” instead of “volume of strong base”. We sincerely apologize for this mistake which will be corrected before publication.
\nThe following equation represents the dissociation of water at 25 °C.
\n2H2O (l) H3O+ (aq) + OH− (aq) ΔH = +56 kJ
\nWhich changes occur as the temperature increases?
\nA. [H3O+] increases and pH will decrease.
\nB. [H3O+] decreases and pH will increase.
\nC. [H3O+] increases and pH will increase.
\nD. [H3O+] decreases and pH will decrease.
\nA
\nThis is one of the more challenging questions on the paper. 62 % of the candidates obtained the correct answer. The most commonly chosen distractor was C, where the increase in the concentration of H3O+ was recognized by applying Le Chatelier’s principle but the effect on pH was incorrect.
\nWhat is the major reason why the pH of unpolluted rain is less than 7?
\nA. methane
\nB. carbon dioxide
\nC. nitrogen oxides
\nD. sulfur dioxide
\nB
\n69 % of the candidates identified CO2 as the gas responsible for the acidity of unpolluted rain. The majority of the candidates that answered incorrectly chose nitrogen oxides (C) or sulfur dioxide (D).
\nWhich compound contains sulfur with an oxidation state of +6?
\nA. SO2
\nB. H2S
\nC. H2SO3
\nD. H2SO4
\nD
\n94 % of candidates identified the compound that contains sulfur with an oxidation state of +6.
\nThe following reaction occurs in a voltaic (galvanic) cell.
\nMg (s) + 2Ag+ (aq) → Mg2+ (aq) + 2Ag (s)
\nWhich reaction takes place at each electrode?
\nC
\n83 % of the candidates were able to identify the electrode equations given the cell reaction. The most commonly chosen distractor B had the correct half-equations at the opposite electrodes.
\nConsider the following table of standard electrode potentials.
\nWhich is the strongest oxidizing agent?
\nA. Pb2+
\nB. Pb
\nC. Al3+
\nD. Al
\nA
\nThis was a challenging question with a high discrimination index. 57 % of the candidates identified the strongest oxidizing agent given the standard electrode potentials. The most commonly chosen distractor was Al3+ (C).
\nWhat are the products when concentrated KBr (aq) is electrolyzed?
\nD
\nAnother challenging question with a high discrimination index. 57 % of the candidates were able to identify the electrode products during the electrolysis of concentration KBr (aq). The most commonly chosen distractor was C where K was the product at the cathode (instead of H2). Some teachers commented that the data booklet was needed to solve this question and others said “concentrated” was vague. But the effect of concentration is clearly stated in the syllabus and does not need the data booklet to be determined. As for the cation, potassium is known as a reactive metal according to the periodic trends and should have been easy to recognize as more reactive than hydrogen. Similar questions have appeared in past papers.
\nWhich compound has the lowest boiling point?
\nA. CH3CH2CH2CH2CH2CH3
\nB. CH3CH2CH2CH2CH3
\nC. CH3CH(CH3)CH2CH3
\nD. CH3C(CH3)2CH3
\nD
\nSurprisingly, this was one of the challenging questions on the paper. Only 63 % of the candidates chose dimethylpropane as the compound having the lowest boiling point. The most commonly chosen distractor was pentane (B) which did not take into account the effect of branching on the strength of London dispersion forces.
\nMethane reacts with chlorine in sunlight.
\nCH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g)
\nWhich type of reaction occurs?
\nA. free-radical substitution
\nB. electrophilic substitution
\nC. nucleophilic substitution
\nD. electrophilic addition
\nA
\n85 % of the candidates chose free-radical substitution as the type of reaction occurring between methane and chlorine in sunlight.
\nWhat is the name of this compound using IUPAC rules?
\nA. 2,3-diethylbutane
\nB. 2-ethyl-3-methylpentane
\nC. 3-methyl-4-ethylpentane
\nD. 3,4-dimethylhexane
\nD
\nThis was the easiest question on the paper. 97 % of the candidates recognized the correct IUPAC name of the compound.
\nWhat must be present on a nucleophile?
\nA. Negative charge
\nB. Lone pair of electrons
\nC. Positive charge
\nD. Symmetrical distribution of electrons
\nB
\n72 % of the candidates chose B (a nucleophile must have a lone pair of electrons). The most commonly chosen distractor was A (a nucleophile must have a negative charge). Some teachers commented that both answers A and B are correct. However, the word “must” in the question means B is the only answer. Candidates are aware that some nucleophiles like water are neutral. Moreover, the definition of a nucleophile is clearly stated in the syllabus (topic 10.2).
\nWhich compound exists as two configurational isomers?
\nA. CBr2=CH2
\nB. CH2=CHBr
\nC. CHBr2CH2Br
\nD. CHBr=CHBr
\nD
\n71 % of the candidates identified 1,2-dibromoethene as having two configurational isomers. The most commonly chosen distractor was C which was the only saturated halogenoalkane, indicating that these candidates may have confused the term with “conformational” isomers.
\nWhich class of compound is formed when a ketone is reduced?
\nA. primary alcohol
\nB. secondary alcohol
\nC. ether
\nD. carboxylic acid
\nB
\n85 % of the candidates identified the secondary alcohol as the product of the reduction of a ketone. The other three distractors (primary alcohol, ether and carboxylic acid) were chosen almost equally by the remaining candidates.
\nThe following data were recorded for determining the density of three samples of silicon, Si.
\nWhich average density value, in g cm−3, has been calculated to the correct number of significant figures?
\nA. 2
\nB. 2.3
\nC. 2.27
\nD. 2.273
\nB
\nCandidates found this question relatively challenging and only 67 % chose the answer with two significant figures. The most commonly chosen distractor was C which expressed the answer to three significant figures.
\nWhat can be deduced from the infrared (IR) spectrum of a compound?
\nA. Number of hydrogens
\nB. Number of hydrogen environments
\nC. Bonds present
\nD. Molar mass
\nC
\n82 % of the candidates identified the bonds present as the information that can be deduced from an infrared spectrum. The most commonly chosen distractor was B (the number of hydrogen environments).
\nWhich technique involves breaking covalent bonds when carried out on an organic compound?
\nA. infrared spectroscopy
\nB. nuclear magnetic resonance spectroscopy
\nC. X-ray crystallography
\nD. mass spectrometry
\nD
\nThis question challenged candidates to think about analytical techniques more deeply. 67 % of the candidates recognized that mass spectrometry involves breaking covalent bonds. The most commonly chosen distractor was X-ray crystallography (option C).
\nThis question is about a mug made of a lead alloy.
\nThe rate of lead dissolving in common beverages with various pH values was analysed.
\nBromine and methanoic acid react in aqueous solution.
\nBr2 (aq) + HCOOH (aq) → 2Br− (aq) + 2H+ (aq) + CO2 (g)
\nThe reaction was monitored by measuring the volume of carbon dioxide produced as time progressed.
\nDetermine from the graph the rate of reaction at 20 s, in cm3 s−1, showing your working.
\nOutline, with a reason, another property that could be monitored to measure the rate of this reaction.
\nExamine, giving a reason, whether the rate of lead dissolving increases with acidity at 18 °C.
\nDescribe one systematic error associated with the use of the gas syringe, and how the error affects the calculated rate.
\nIdentify one error associated with the use of an accurate stopwatch.
\ntangent drawn to curve at t = 20 s [✔]
\nslope/gradient calculation [✔]
\n0.35 «cm3 s–1» [✔]
\n\n
Note: Accept values in the range 0.32–0.42 «cm3 s–1».
\nALTERNATIVE 1
colour [✔]
Br2/reactant is coloured «Br– (aq)/product is not» [✔]
\nNote: Do not accept “changes in temperature” or “number of bubbles”.
\n\n
ALTERNATIVE 2
conductivity [✔]
greater/increased concentration of ions in products [✔]
\n\n
ALTERNATIVE 3
mass/pressure [✔]
gas is evolved/produced [✔]
\nNote: Do not accept “mass of products is less than mass of reactants”.
\n
ALTERNATIVE 4
pH [✔]
methanoic acid is weak AND HBr is strong
OR
increase in [H+] [✔]
no AND experiment 7/beer has lowest rate and intermediate acidity/pH
OR
no AND experiment 6/orange juice has fastest rate but lower acidity/higher pH than experiment 5/lemonade
OR
no AND experiment 6/orange juice has highest rate and intermediate acidity/pH [✔]
\n
Note: Accept no AND any comparison, with experimental support, that concludes no pattern/increase with acidity.
eg: “rate of Pb/lead dissolving generally decreases with acidity as tap water has highest rate (after orange juice) while lemonade (lower pH) has lower rate”.
ALTERNATIVE 1
gas may leak/be lost/escape
OR
plunger may stick/friction «so pressure is greater than atmospheric pressure»
OR
syringe may be tilted «up» so plunger moves less «with gravity acting on plunger»
OR
CO2 dissolved in water [✔]
calculated rate lower [✔]
\n\n
ALTERNATIVE 2
syringe may be tilted «down» so plunger moves more «with gravity acting on plunger»
OR
syringe is held in hand so gets warmer and gas expands [✔]
calculated rate higher [✔]
\n\n
Note: Calculated rate is lower or higher must be stated for M2.
\nDo not accept “scale on syringe is inaccurate”, “errors in reading syringe”, or “bubbles in syringe”.
\nhuman reaction time/delay «starting/stopping the stopwatch» [✔]
\nNote: Do not accept “inaccurate stopwatch”.
\nThis question was challenging for many students. Quite a few candidates did draw a tangent line at 20s for 1 mark, show a slope/gradient calculation of the line for 1 mark, and had a reasonable final value for the final mark. Some candidates only found the average rate by finding the ratio of the value at that data point and received one mark (16/20=0.80 cm3 s-1). Candidates also received one mark if they had a correct answer with no work since the question clearly asked students to show their work.
\nThe reaction rate was originally monitored by measuring the volume of CO2 produced. Students needed to propose another method for this reaction, with a reason, that could be used to measure the rate. There were several possible correct answers and most students received at least one mark with many receiving both marks. The most common incorrect answer involved changes in temperature.
\nThis question required students to recognize the rate of lead dissolving did not increase with acidity and to refer to data in the table for the reason. Some students did not refer to data in the table and did not receive the mark because they did not have a reason, other students compared the rate of lead dissolving with temperature increasing which did not answer the question.
\nThis question was asking about a systematic error. There were several possible correct answers for the error, but students also needed to clearly identify a specific error and if the rate increased or decreased for the second mark. Many students confused this with the concept of a random error and identified the uncertainty of reading the syringe which is incorrect. Teachers need to reinforce the concept of systematic versus random errors.
\nThis question was well answered by most candidates although some students did not read the question clearly and commented on the stopwatch having problems or not being accurate.
\nDescribe the characteristics of the nematic liquid crystal phase and the effect that an electric field has on it.
\n
Shape of molecules:
Distribution:
\nEffect of electric field:
\nShape of molecules:
linear
OR
rod «shaped» [✔]
Distribution:
no positional order AND «some» directional order [✔]
Note: Accept “partly ordered”.
\nEffect of electric field:
«directional» order increases
OR
molecules align in same direction [✔]
Most students were able to obtain at least one mark on this question. The distribution was the most challenging part.
\nMetals are extracted from their ores by several methods, including electrolysis and reduction with carbon.
\nDetermine the mass of aluminium, in g, that could be extracted from an appropriate solution by a charge of 48250 C. Use sections 2 and 6 of the data booklet.
\nOnce extracted, the purity of the metal can be assessed using ICP-MS. Suggest two advantages of using plasma technology rather than regular mass spectrometry.
\nExplain the action of metals as heterogeneous catalysts.
\nOutline how alloys conduct electricity and why they are often harder than pure metals.
\n\n
Conduct electricity:
\nHarder than pure metals:
\nCarbon nanotubes are added to metals to increase tensile strength.
\nWrite an equation for the formation of carbon nanotubes from carbon monoxide.
\nmoles of electrons «= » = 0.5000 «mol» [✔]
\nmoles of aluminium «= » = 0.1667 «mol» [✔]
\nmass of aluminium «= 26.98 g mol–1 × 0.1667mol» = 4.50 «g» [✔]
\n\n
Note: Award [3] for correct final answer.
\nAny two of:
larger linear calibration [✔]
«accurate» detection of multiple elements/metals [✔]
\n«accurate» detection of elements in low concentration [✔]
\ntemperature around 10000 K atomises/ionises every material [✔]
\nAny two of:
reactant(s) adsorb onto active sites/surface [✔]
bonds weakened/broken/stretched «in adsorbed reactants»
OR
activation energy lowered [✔]
products desorbed [✔]
\n\n
Note: Accept “products released” for M3.
\nConduct electricity:
«delocalized/valence» electrons free to move «under potential difference» [✔]
Harder than pure metals:
atoms/ions of different sizes prevent layers «of atoms/ions» from sliding over one another [✔]
2CO(g) → C(s) + CO2 (g) [✔]
\nMany students scored at least one point typically the 1st mark with many obtaining the 2nd and 3rd marks as well. If students struggled it was typically with the second mark where the number of electrons was required.
\nThis was question was not answered well. Most candidates did not seem to understand any advantages of ICP-MS or how to describe them.
\nThis question was reasonably answered with many candidates receiving a mark for the action of a catalyst. Teachers should remind students to use the terms adsorb/desorb with this type of process, not absorb.
\nMost candidates were awarded M1 for how alloys conduct electricity. Some struggled with describing why they are harder than pure metals. Teachers should remind student to use proper terminology such as atoms or ions not nuclei for this type of answer.
\nThis was a very poorly answered question. Very few candidates knew the correct balanced equation or gave an equation that was not balanced.
\nHow many electrons form the carbon–oxygen bond in methanal, HCHO?
\nA. 2
\nB. 4
\nC. 8
\nD. 12
\nB
\nPolymers have a wide variety of uses but their disposal can be problematic.
\nDraw a section of isotactic polychloroethene (polyvinylchloride, PVC) showing all the atoms and all the bonds of four monomer units.
\nThe infrared (IR) spectrum of polyethene is given.
\nSuggest how the IR spectrum of polychloroethene would differ, using section 26 of the data booklet.
\nIdentify a hazardous product of the incineration of polychloroethene.
\nExplain how plasticizers affect the properties of plastics.
\nSuggest why the addition of plasticizers is controversial.
\ncorrect bonding [✔]
\nCl atoms all on same side and alternate [✔]
\n\n
Note: Continuation bonds must be shown.
\nAward [1 max] if less than or more than four units shown.
\nAccept a stereo formula with all atoms and bonds shown.
\n«strong additional» absorption at 600–800 «cm–1» [✔]
\nAny one of:
HCl [✔]
Cl2 [✔]
\ndioxins [✔]
\nC [✔]
\nCO [✔]
\nAny two of:
embedded/fit between chains of polymers [✔]
prevent chains from forming crystalline regions [✔]
\nkeep polymer strands/chains/molecules separated/apart [✔]
increase space/volume between chains [✔]
\nweaken intermolecular/dipole-dipole/London/dispersion/instantaneous dipoleinduced dipole/van der Waals/vdW forces «between chains» [✔]
\nincrease flexibility/durability/softness [✔]
\nmake polymers less brittle [✔]
\nleach into foodstuffs/environment
OR
«unknown» health/environmental consequences [✔]
\n
Note: Accept “plasticizers cannot be recycled”.
\nQuite a few candidates scored at least one mark although most either scored both or none for this polymer structure.
\nAlmost all students who attempted this question received the mark for identifying the correct absorption band.
\nThis was a well answered question, with most candidates identifying at least one method plasticizers affect the properties of plastic.
\nMany students received a mark for this question although some did not because their answers were too vague.
\nEthane-1,2-diol, HOCH2CH2OH, has a wide variety of uses including the removal of ice from aircraft and heat transfer in a solar cell.
\nEthane-1,2-diol can be formed according to the following reaction.
\n2CO (g) + 3H2 (g) HOCH2CH2OH (g)
\n(i) Deduce the equilibrium constant expression, Kc, for this reaction.
\n\n
(ii) State how increasing the pressure of the reaction mixture at constant temperature will affect the position of equilibrium and the value of Kc.
\nPosition of equilibrium:
\nKc:
\n\n
(iii) Calculate the enthalpy change, ΔHθ, in kJ, for this reaction using section 11 of the data booklet. The bond enthalpy of the carbon–oxygen bond in CO (g) is 1077kJmol-1.
\n\n
(iv) The enthalpy change, ΔHθ, for the following similar reaction is –233.8 kJ.
\n2CO(g) + 3H2(g) HOCH2CH2OH (l)
\nDeduce why this value differs from your answer to (a)(iii).
\nDetermine the average oxidation state of carbon in ethene and in ethane-1,2-diol.
\nEthene:
\nEthane-1,2-diol:
\nExplain why the boiling point of ethane-1,2-diol is significantly greater than that of ethene.
\nEthane-1,2-diol can be oxidized first to ethanedioic acid, (COOH)2, and then to carbon dioxide and water. Suggest the reagents to oxidize ethane-1,2-diol.
\n(i)
\n
(ii)
Position of equilibrium: moves to right OR favours product
Kc: no change OR is a constant at constant temperature
\n
(iii)
Bonds broken: 2C≡O + 3(H-H) / 2(1077kJmol-1) + 3(436kJmol-1) / 3462 «kJ»
Bonds formed: 2(C-O) + 2(O-H) + 4(C-H) + (C-C) / 2(358kJmol-1) + 2(463kJmol-1) + 4(414kJmol-1) + 346kJmol-1 / 3644 «kJ»
\n«Enthalpy change = bonds broken - bonds formed = 3462 kJ - 3644 kJ =» -182 «kJ»
\nAward [3] for correct final answer.
Award [2 max] for «+»182 «kJ».
(iv)
in (a)(iii) gas is formed and in (a)(iv) liquid is formed
OR
products are in different states
OR
conversion of gas to liquid is exothermic
OR
conversion of liquid to gas is endothermic
OR
enthalpy of vapourisation needs to be taken into account
Accept product is «now» a liquid.
Accept answers referring to bond enthalpies being means/averages.
Ethene: –2
\nEthane-1,2-diol: –1
\nDo not accept 2–, 1– respectively.
\n\n
ethane-1,2-diol can hydrogen bond to other molecules «and ethene cannot»
\nOR
\nethane-1,2-diol has «significantly» greater van der Waals forces
\nAccept converse arguments.
Award [0] if answer implies covalent bonds are broken
hydrogen bonding is «significantly» stronger than other intermolecular forces
\nacidified «potassium» dichromate«(VI)»/H+ AND K2Cr2O7/H+ AND Cr2O72-
\nOR
\n«acidified potassium» manganate(VII)/ «H+» KMnO4 /«H+» MnO4-
\nAccept Accept H2SO4 or H3PO4 for H+.
Accept “permanganate” for “manganate(VII)”.
Which species has bond angles of 90°?
\nA. AlCl4-
\nB. Cl4-
\nC. NH4+
\nD. SiCl4
\nB
\nPhosphatidylcholine is an example of a phospholipid found in lecithin.
\nPhosphatidylcholine may be formed from propane-1,2,3-triol, two lauric acid molecules, phosphoric acid and the choline cation.
\nDeduce the structural formula of phosphatidylcholine.
\nIdentify the type of reaction in (a).
\nLecithin is a major component of cell membranes. Describe the structure of a cell membrane.
\nPredict, giving a reason, the relative energy density of a carbohydrate and a lipid of similar molar mass.
\nLecithin aids the body’s absorption of vitamin E.
\nSuggest why vitamin E is fat-soluble.
\nPhospholipids are also found in lipoprotein structures.
\nDescribe two effects of increased levels of low-density lipoprotein (LDL) on health.
\nphosphodiester correctly drawn [✔]
\nboth ester groups correctly drawn [✔]
\n\n
Note: Accept protonated phosphate.
\nAccept phosphodiester in centre position.
\ncondensation [✔]
\n\n
Note: Accept “esterification”.
\nAccept “nucleophilic substitution/SN”.
\nphospholipid bilayer/double layer
OR
two layers of phospholipids [✔]
polar/hydrophilic heads facing aqueous environment AND non-polar/hydrophobic tails facing away from aqueous environment [✔]
\n\n
Note: Award [2] for a suitably labelled diagram.
\nAward [1 max] for a correct but unlabelled diagram.
\nAccept “polar/hydrophilic heads on outside AND non-polar/hydrophobic tails on inside for M2.
\ncarbohydrates less energy dense AND carbohydrates higher ratio of oxygen to carbon/more oxidized/less reduced [✔]
\nlong non-polar/hydrocarbon chain «and only one hydroxyl group»
OR
forms London/dispersion/van der Waals/vdW interactions with fat [✔]
\n
Note: Accept “alcohol/hydroxy/OH” for “hydroxyl” but not “hydroxide”.
\natherosclerosis/cholesterol deposition «in artery walls» [✔]
\nincreases risk of heart attack/stroke/cardiovascular disease/CHD [✔]
\n\n
Note: Accept “arteries become blocked/walls become thicker”, “increases blood pressure”, or “blood clots”.
\nDo not accept “high cholesterol”.
\nThis was very poorly answered. Although the phosphodiester was a challenging mark it could be awarded in both the protonated and deprotonated form. The two esters should have been much more straight forward mark, and both were required to receive the second mark. Students struggled with proper structural drawings for both marks and many students simply left this question blank. The functional groups did need to be drawn out in their full structural form to receive the mark as indicated in the question.
\nThis question was well answered.
\nThis question was another one where the first mark was fairly well answered but the explanation or second mark was often not correct or complete.
\nThis question was not well answered even though it has appeared on previous tests. In any cases the students did not give the relative energy density or the reason. It is important that candidates read question carefully and responds completely to each question as asked.
\nThis question was also not answered well even though it has appeared on previous tests. Many students missed the idea of a long or large non-polar chain when describing the structure.
\nStudents were required to state two effects of increased LDL. High cholesterol is not an accepted answer but still frequently seen. Many students also repeated similar answers that could not receive the same mark.
\nWhat is the hybridization of the numbered atoms in ethanoic acid?
\nD
\nSodium thiosulfate solution reacts with dilute hydrochloric acid to form a precipitate of sulfur at room temperature.
\nNa2S2O3 (aq) + 2HCl (aq) → S (s) + SO2 (g) + 2NaCl (aq) + X
\nIdentify the formula and state symbol of X.
\nSuggest why the experiment should be carried out in a fume hood or in a well-ventilated laboratory.
\nThe precipitate of sulfur makes the mixture cloudy, so a mark underneath the reaction mixture becomes invisible with time.
\n10.0 cm3 of 2.00 mol dm-3 hydrochloric acid was added to a 50.0 cm3 solution of sodium thiosulfate at temperature, T1. Students measured the time taken for the mark to be no longer visible to the naked eye. The experiment was repeated at different concentrations of sodium thiosulfate.
\nShow that the hydrochloric acid added to the flask in experiment 1 is in excess.
\nDraw the best fit line of against concentration of sodium thiosulfate on the axes provided.
\nA student decided to carry out another experiment using 0.075 mol dm-3 solution of sodium thiosulfate under the same conditions. Determine the time taken for the mark to be no longer visible.
\nAn additional experiment was carried out at a higher temperature, T2.
\n(i) On the same axes, sketch Maxwell–Boltzmann energy distribution curves at the two temperatures T1 and T2, where T2 > T1.
\n(ii) Explain why a higher temperature causes the rate of reaction to increase.
\nSuggest one reason why the values of rates of reactions obtained at higher temperatures may be less accurate.
\nH2O AND (l)
Do not accept H2O (aq).
SO2 (g) is an irritant/causes breathing problems
OR
SO2 (g) is poisonous/toxic
Accept SO2 (g) is acidic, but do not accept “causes acid rain”.
Accept SO2 (g) is harmful.
Accept SO2 (g) has a foul/pungent smell.
n(HCl) = «dm3 × 2.00 mol dm-3 =» 0.0200 / 2.00 × 10-2«mol»
AND
n(Na2S2O3) = «dm3 × 0.150 mol × dm-3 =» 0.00750 / 7.50 × 10-3 «mol»
0.0200 «mol» > 0.0150 «mol»
OR
2.00 × 10-2«mol» > 2 × 7.50 × 10-3 «mol»
OR
× 2.00 × 10-2 «mol» > 7.50 × 10-3 «mol»
Accept answers based on volume of solutions required for complete reaction.
Award [2] for second marking point.
Do not award M2 unless factor of 2 (or half) is used.
five points plotted correctly
best fit line drawn with ruler, going through the origin
22.5 × 10-3 «s-1»
\n«Time = =» 44.4 «s»
\nAward [2] for correct final answer.
Accept value based on candidate’s graph.
Award M2 as ECF from M1.
Award [1 max] for methods involving taking mean of appropriate pairs of values.
Award [0] for taking mean of pairs of time values.
Award [2] for answers between 42.4 and 46.4 «s».
(i)
\ncorrectly labelled axes
peak of T2 curve lower AND to the right of T1 curve
Accept “probability «density» / number of particles / N / fraction” on y-axis.
\nAccept “kinetic E/KE/EK” but not just “Energy/E” on x-axis.
\n\n
(ii)
\ngreater proportion of molecules have E ≥ Ea or E > Ea
OR
greater area under curve to the right of the Ea
greater frequency of collisions «between molecules»
OR
more collisions per unit time/second
Accept more molecules have energy greater than Ea.
Do not accept just “particles have greater kinetic energy”.
Accept “rate/chance/probability/likelihood/” instead of “frequency”.
Accept suitably shaded/annotated diagram.
Do not accept just “more collisions”.
shorter reaction time so larger «%» error in timing/seeing when mark disappears
\nAccept cooling of reaction mixture during course of reaction.
\nThis question is about fuel for engines.
\nCrude oil can be converted into fuels by fractional distillation and cracking.
\nContrast these two processes.
\nDetermine the specific energy, in kJ g−1, and energy density, in kJ cm−3, of hexane, C6H14. Give both answers to three significant figures.
\nHexane: Mr = 86.2; ΔHc = −4163 kJ mol−1; density = 0.660 g cm−3
\n\n
Specific energy:
\nEnergy density:
\nHydrocarbons need treatment to increase their octane number to prevent pre-ignition (knocking) before they can be used in internal combustion engines.
\nDescribe how this is carried out and the molecular changes that take place.
\nNote: Award [1] max for any two correct answers from one column OR one from each column.
\nAward [2] for any two correct from each column; eg: fractional distillation – any two correct award [1 max] AND cracking – any two correct award [1 max].
\nspecific energy = « =» 48.3 «kJ g–1» [✔]
\nenergy density =«48.3 kJ g–1 × 0.660 g cm–3 =» 31.9 «kJ cm–3» [✔]
\n\n
Note: Award [1 max] if either or both answers not expressed to three significant figures.
\nAny two of:
«hydrocarbons are heated with» catalyst [✔]
long chains break and reform
OR
branching/aromatization occurs
OR
isomerisation/reforming/platforming/cracking [✔]
zeolite separates branched from non-branched
OR
products are distilled
OR
«distillation» separates reformed and cracked products [✔]
\n
Note: Accept a specific catalysis name or formula for M1 such as Pt/Re/Rh/Pd/Ir.
\nThis question was not well answered. Many candidates didn’t answer the question as asked. Candidates needed two correct statements, either about fractional distillation or cracking as a process for 1 mark.
\nThis question was very well answered by most students and many answered with the correct number of significant figures as specified by the question.
\nStudents responded well to at least one mark of this question. There were several different ways to earn the 2 marks possible. The most common way students earned marks were by identifying the use of a catalyst and then the idea of the compound reforming into a smaller or branched compound. Very few students discussed the idea of purification or separation into individual compounds which is another important part of this process.
\nCompound A is in equilibrium with compound B.
\nPredict the electron domain and molecular geometries around the oxygen atom of molecule A using VSEPR
\n\n
State the type of hybridization shown by the central carbon atom in molecule B.
\nState the number of sigma () and pi () bonds around the central carbon atom in molecule B.
\nThe IR spectrum of one of the compounds is shown:
\nCOBLENTZ SOCIETY. Collection © 2018 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.
\nDeduce, giving a reason, the compound producing this spectrum.
\nCompound A and B are isomers. Draw two other structural isomers with the formula .
\nThe equilibrium constant, , for the conversion of A to B is in water at .
\nDeduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.
\nCalculate the standard Gibbs free energy change, , in , for the reaction (A to B) at . Use sections 1 and 2 of the data booklet.
\nPropanone can be synthesized in two steps from propene. Suggest the synthetic route including all the necessary reactants and steps.
Propanone can be synthesized in two steps from propene.
\nSuggest why propanal is a minor product obtained from the synthetic route in (g)(i).
\nElectron domain geometry: tetrahedral ✔
\nMolecular geometry: bent/V-shaped ✔
\n✔
\n-bonds:
AND
-bonds: ✔
B AND absorption/
OR
B AND absence of ✔
Accept any value between .
\nAccept any two isomers except for propanone and propen-2-ol:
\n✔✔
\n
Penalize missing hydrogens in displayed structural formulas once only.
\nAND is greater than /large ✔
\n
✔
/water «and » ✔
\n/propan-2-ol ✔
\n/«potassium» dichromate(VI) AND
OR
/«acidified potassium» manganate(VII) ✔
Accept .
\n\n
\n
primary carbocation «intermediate forms»
OR
minor product «of the water addition would be» propan-1-ol
OR
anti-Markovnikov addition of water ✔
primary alcohol/propan-1-ol oxidizes to an aldehyde/propanal ✔
\nThe majority of students got at least one of electron domain geometry or molecular geometry correct.
\nThe vast majority of students could identify the hybridization around a central carbon atom.
\nThe vast majority of students could identify BOTH sigma and pi bonds in a molecule.
\nMany candidates identified B having C = O and a peak at 1750.
\nA surprising number of candidates drew propanone here as an option, either failing to read the question or perhaps finding the structural formulae provided difficult to understand.
\nMost candidates identified B, the product, as being in greater concentration at equilibrium however some lost the mark because they did not include a reason.
\nMost candidates could apply the formula for Gibbs free energy change, ΔGΘ, correctly however some did not get the units correct.
\nThe mean mark was ⅔ for the required synthetic route. Some candidates failed to identify water as a reagent in the hydration reaction, or note that dichromate ion oxidation requires acidic conditions. This was also the question with most No Response.
\nThis question regarding the formation of a minor product was not well answered. Many candidates struggled to explain the formation of propan-1-ol and to then oxidize it to propanal.
\nThe concentration of a solution of a weak acid, such as ethanedioic acid, can be determined
by titration with a standard solution of sodium hydroxide, NaOH (aq).
Distinguish between a weak acid and a strong acid.
\nWeak acid:
\nStrong acid:
\nSuggest why it is more convenient to express acidity using the pH scale instead of using the concentration of hydrogen ions.
\n5.00 g of an impure sample of hydrated ethanedioic acid, (COOH)2•2H2O, was dissolved in water to make 1.00 dm3 of solution. 25.0 cm3 samples of this solution were titrated against a 0.100 mol dm-3 solution of sodium hydroxide using a suitable indicator.
\n(COOH)2 (aq) + 2NaOH (aq) → (COONa)2 (aq) + 2H2O (l)
\nThe mean value of the titre was 14.0 cm3.
\n(i) Calculate the amount, in mol, of NaOH in 14.0 cm3 of 0.100 mol dm-3 solution.
\n(ii) Calculate the amount, in mol, of ethanedioic acid in each 25.0 cm3 sample.
\n(iii) Determine the percentage purity of the hydrated ethanedioic acid sample.
\nThe Lewis (electron dot) structure of the ethanedioate ion is shown below.
\nOutline why all the C–O bond lengths in the ethanedioate ion are the same length and suggest a value for them. Use section 10 of the data booklet.
\nWeak acid: partially dissociated/ionized «in solution/water»
AND
Strong acid: «assumed to be almost» completely/100% dissociated/ionized «in solution/water»
Accept answers relating to pH, conductivity, reactivity if solutions of equal concentrations stated.
\n«log scale» reduces a wide range of numbers to a small range
OR
simple/easy to use
OR
converts exponential expressions into linear scale/simple numbers
Do not accept “easy for calculations”
\ni
\n«n(NaOH) = dm-3 x 0.100 mol dm-3 =» 1.40 x 10-3 «mol»
\n\n
ii
\n« «mol»
\n\n
iii
ALTERNATIVE 1:
«mass of pure hydrated ethanedioic acid in each titration = 7.00 × 10-4 mol × 126.08 g mol-1 =» 0.0883 / 8.83 × 10-2 «g»
mass of sample in each titration = «×5.00g=»0.125«g»
\n«% purity = × 100 =» 70.6 «%»
\nALTERNATIVE 2:
«mol of pure hydrated ethanedioic acid in 1 dm3 solution = 7.00 × 10-4 × =» 2.80×10-2 «mol»
«mass of pure hydrated ethanedioic acid in sample = 2.80 × 10-2 mol × 126.08 g mol-1 =» 3.53 «g»
«% purity = × 100 =» 70.6 «%»
ALTERNATIVE 3:
mol of hydrated ethanedioic acid (assuming sample to be pure) = = 0.03966 «mol»
actual amount of hydrated ethanedioic acid = «7.00 × 10-4 × =» 2.80 × 10-2 «mol»
«% purity = × 100 =» 70.6 «%»
\nAward suitable part marks for alternative methods.
Award [3] for correct final answer.
Award [2 max] for 50.4 % if anhydrous ethanedioic acid assumed.
electrons delocalized «across the O–C–O system»
OR
resonance occurs
Accept delocalized π-bond(s).
122 «pm» < C–O < 143 «pm»
\nAccept any answer in the range 123 «pm» to 142 «pm». Accept “bond intermediate between single and double bond” or “bond order 1.5”.
\nWhich represents the enthalpy change of hydration of the chloride ion?
\nA
\nWhich ionic compound has the largest value of lattice enthalpy?
\nA. MgS
\nB. MgO
\nC. CaBr2
\nD. NaF
\nB
\nNickel catalyses the conversion of propanone to propan-2-ol.
\nOutline how a catalyst increases the rate of reaction.
\nExplain why an increase in temperature increases the rate of reaction.
\nDiscuss, referring to intermolecular forces present, the relative volatility of propanone and propan-2-ol.
\nThe diagram shows an unlabelled voltaic cell for the reaction
\n\nLabel the diagram with the species in the equation.
\nCalculate the standard cell potential, in , for the cell at . Use section 24 of the data booklet
\nCalculate the standard free energy change, , in , for the cell using sections 1 and 2 of the data booklet.
\nSuggest a metal that could replace nickel in a new half-cell and reverse the electron flow. Use section 25 of the data booklet.
\nDescribe the bonding in metals.
\nNickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the addition of another metal to nickel.
\nprovides an alternative pathway/mechanism AND lower Ea ✔
\nAccept description of how catalyst lowers Ea (e.g. “reactants adsorb on surface «of catalyst»”, “reactant bonds weaken «when adsorbed»”).
\nmore/greater proportion of molecules with E Ea ✔
greater frequency/probability/chance of collisions «between the molecules»
OR
more collision per unit of time/second ✔
hydrogen bonding/bonds «and dipole–dipole and London/dispersion forces are present in» propan-2-ol ✔
\ndipole–dipole «and London/dispersion are present in» propanone ✔
\npropan-2-ol less volatile AND hydrogen bonding/bonds stronger «than dipole–dipole »
OR
propan-2-ol less volatile AND «sum of all» intermolecular forces stronger ✔
✔
\n✔
\n✔
\nAccept OR .
\nelectrostatic attraction ✔
\nbetween «a lattice of» metal/positive ions/cations AND «a sea of» delocalized electrons ✔
\nAccept “mobile/free electrons”.
\nAny of:
\nmalleability/hardness
OR
«tensile» strength/ductility
OR
density
OR
thermal/electrical conductivity
OR
melting point
OR
thermal expansion ✔
Do not accept corrosion/reactivity or any chemical property.
\nAccept other specific physical properties.
\nAlthough fairly well done some candidates did not mention that providing an alternate pathway to the reaction was how the activation energy was lowered and hence did not gain the mark.
\nAlmost all candidates earned at least 1 mark for the effect of temperature on rate. Some missed increase in collision frequency, others the idea that more particles reached the required activation energy.
\nThe average mark was 1.9/3. Almost all candidates could recognize hydrogen bonding in alcohol but many missed the dipole-dipole attraction in propanone. There was also some confusion on the term volatility, with some thinking stronger IMF meant higher volatility.
\nA surprising number of No Response for a question where candidates simply had to label a diagram with the species in the equation. Some candidates had the idea but did not use the species for electrolytic cell, e.g., Pb(SO4) instead of Pb2+(aq).
\n80% of candidates could correctly calculate a cell potential by using a reduction table and a balanced redox reaction.
\nThis was similar to 2f(ii) where many could apply the formula for Gibbs free energy change, ΔGө, correctly however some did not get the units correct.
\n80% could correctly pick a metal to reverse the electron flow, however some candidates thought a more reactive, rather than a less reactive metal than nickel would reverse the electron flow.
\nMost candidates were aware that metallic bonding involved a \"sea of electrons\", but were unsure about surrounding what and could not identify that it was electrostatic attraction holding the metal together.
\nAlmost all candidates could correctly identify a physical property of a metal which might be altered when alloying.
\nProteins have structural or enzyme functions.
\nOil spills are a major environmental problem.
\nSome proteins form an α-helix. State the name of another secondary protein structure.
\nCompare and contrast the bonding responsible for the two secondary structures.
\n\n
One similarity:
\nOne difference:
\nExplain why an increase in temperature reduces the rate of an enzyme-catalyzed reaction.
\nSuggest two reasons why oil decomposes faster at the surface of the ocean than at greater depth.
\nOil spills can be treated with an enzyme mixture to speed up decomposition.
\nOutline one factor to be considered when assessing the greenness of an enzyme mixture.
\nβ/beta pleated/sheet [✔]
\nOne similarity:
hydrogen bonding
OR
attractions between C=O and N–H [✔]
One difference:
α-helix has hydrogen bonds between amino acid residues that are closer than β-pleated sheet
OR
H-bonds in α-helix parallel to helix axis AND perpendicular to sheet in β-pleated sheet
OR
α-helix has one strand AND β-pleated sheet has two «or more» strands
OR
α-helix is more elastic «since H-bonds can be broken easily» AND β-pleated sheet is less elastic «since H-bonds are difficult to break» [✔]
\n
Note: Accept a diagram which shows hydrogen bonding between O of C=O and H of NH groups for M1.
\nAccept “between carbonyl/amido/amide/carboxamide” but not “between amino/amine” for M1.
\nenzyme denatured/loss of 3-D structure/conformational change
OR
«interactions responsible for» tertiary/quaternary structures altered [✔]
shape of active site changes
OR
fewer substrate molecules fit into active sites [✔]
Any two of:
surface water is warmer «so faster reaction rate»/more light/energy from the sun [✔]
more oxygen «for aerobic bacteria/oxidation of oil» [✔]
\ngreater surface area [✔]
\nAny one of:
non-hazardous/toxic to the environment/living organisms [✔]
energy requirements «during production» [✔]
\nquantity/type of waste produced «during production»
OR
atom economy [✔]
safety of process [✔]
\n\n
Note: Accept “use of solvents/toxic materials «during production»”.
\nDo not accept “more steps involved”.
\nThis question was quite well answered with many scoring the mark although there were quite a few incorrect responses that answered “beta-helix” rather than “beta-pleated sheet”.
\nThe similarity in bonding between the 2 types of secondary structures was answered well but the difference was not. Most students were not descriptive enough to receive the second mark or simply repeated the idea of proteins containing an alpha-helix and beta-pleated sheets rather than describing something different about them.
\nThis was another question where most candidates received one mark for identifying that the enzyme will denature with an increase in temperature. However, many candidates did not continue with the explanation of the active site shape changing or substrate molecules not longer fitting into the active site.
\nWhile many candidates did receive two marks for this question some candidates only suggested one reason or repeated the same reason (for example - heat and energy from the sun) even though the question clearly asked for two reasons.
\nStudents tend to struggle with these questions and end up giving journalistic or vague answers that cannot be awarded marks. It is important for teachers to instruct students to give more specific answers directly related to the topics presented.
\nMagnesium is a group 2 metal which exists as a number of isotopes and forms many compounds.
\nState the nuclear symbol notation, , for magnesium-26.
\nMass spectroscopic analysis of a sample of magnesium gave the following results:
\nCalculate the relative atomic mass, Ar, of this sample of magnesium to two decimal places.
\nMagnesium burns in air to form a white compound, magnesium oxide. Formulate an equation for the reaction of magnesium oxide with water.
\nDescribe the trend in acid-base properties of the oxides of period 3, sodium to chlorine.
\nIn addition to magnesium oxide, magnesium forms another compound when burned in air. Suggest the formula of this compound
\nDescribe the structure and bonding in solid magnesium oxide.
\nMagnesium chloride can be electrolysed.
\nDeduce the half-equations for the reactions at each electrode when molten magnesium chloride is electrolysed, showing the state symbols of the products. The melting points of magnesium and magnesium chloride are 922 K and 987 K respectively.
\nAnode (positive electrode):
\nCathode (negative electrode):
\n\n
«Ar =»
\n«= 24.3269 =» 24.33
\nAward [2] for correct final answer.
Do not accept data booklet value (24.31).
MgO(s) + H2O(l) → Mg(OH)2(s)
\nOR
\nMgO(s) + H2O(l) → Mg2+(aq) + 2OH–(aq)
\nAccept .
\nfrom basic to acidic
\nthrough amphoteric
\nAccept “alkali/alkaline” for “basic”.
Accept “oxides of Na and Mg: basic AND oxide of Al: amphoteric” for M1.
Accept “oxides of non-metals/Si to Cl acidic” for M2.
Do not accept just “become more acidic”
Mg3N2
\nAccept MgO2, Mg(OH)2, Mg(NOx)2, MgCO3.
\n«3-D/giant» regularly repeating arrangement «of ions»
OR
lattice «of ions»
Accept “giant” for M1, unless “giant covalent” stated.
electrostatic attraction between oppositely charged ions
OR
electrostatic attraction between Mg2+ and O2– ions
Do not accept “ionic” without description.
Anode (positive electrode):
2Cl– → Cl2(g) + 2e–
Cathode (negative electrode):
Mg2+ + 2e– → Mg(l)
Penalize missing/incorrect state symbols at Cl2 and Mg once only.
Award [1 max] if equations are at wrong electrodes.
Accept Mg (g).
\n
Which statement describes the characteristics of a transition state relative to the potential energy of the reactants and products?
\nA. It is an unstable species with lower potential energy.
\nB. It is an unstable species with higher potential energy.
\nC. It is a stable species with lower potential energy.
\nD. It is a stable species with higher potential energy.
\nB
\nSucrose is a disaccharide.
\nState the name of the functional group forming part of the ring structure of each monosaccharide unit.
\nSketch the cyclic structures of the two monosaccharides which combine to form sucrose.
\nacetal
OR
ether [✔]
\n
Note: Accept “glycosidic bond/linkage” but not “glucosidic”.
\nDo not accept “hemiacetal”.
\n [✔]
[✔]
This was reasonably answered although there were some students who responded with ester or hemiacetal which is incorrect.
\nThis question was very poorly answered. Many students lost marks due to sloppy drawing and incorrect bond linkages. Some students did not separate the two saccharides as instructed.
\nDecomposition of hydrogen peroxide in an aqueous solution proceeds as follows.
\n2H2O2(aq) → 2H2O(l) + O2(g)
\nThe rate expression for the reaction was found to be: rate = k [H2O2].
\nWhich graph is consistent with the given rate expression?
\nD
\nA mixture of 0.40 mol of CO (g) and 0.40 mol of H2 (g) was placed in a 1.00 dm3 vessel. The following equilibrium was established.
\nCO (g) + 2H2 (g) CH3OH (g)
At equilibrium, the mixture contained 0.25 mol of CO (g). How many moles of H2 (g) and CH3OH (g) were present at equilibrium?
\nD
\nThe regular rise and fall of sea levels, known as tides, can be used to generate energy.
\nState one advantage, other than limiting greenhouse gas emissions, and one disadvantage of tidal power.
\n\n
Advantage:
\nDisadvantage:
\nAdvantage
Any one of:
renewable [✔]
predictable supply [✔]
tidal barrage may prevent flooding [✔]
effective at low speeds [✔]
long life-span [✔]
low cost to run [✔]
Disadvantage
Any one of:
cost of construction [✔]
changes/unknown effects on marine life [✔]
changes circulation of tides in the area [✔]
power output is variable [✔]
limited locations where feasible [✔]
equipment maintenance can be challenging [✔]
difficult to store energy [✔]
\n
Note: Do not accept vague generalizations.
\nDo not accept economic issues for both advantage and disadvantage.
\nDo not accept sustainable.
\nAccept “energy” or “electricity” for “power”.
\nMany candidates performed well on this question especially when identifying an advantage of tidal power. The students who struggled tended to either give vague or journalistic answers especially for the disadvantage of tidal power.
\nWhich mixture is a buffer solution?
\nA. 25 cm3 of 0.10 mol dm-3 NH3 (aq) and 50 cm3 of 0.10 mol dm-3 HCl (aq)
\nB. 50 cm3 of 0.10 mol dm-3 NH3 (aq) and 25 cm3 of 0.10 mol dm-3 HCl (aq)
\nC. 25 cm3 of 0.10 mol dm-3 NaOH (aq) and 25 cm3 of 0.10 mol dm-3 HCl (aq)
\nD. 50 cm3 of 0.10 mol dm-3 NaOH (aq) and 25 cm3 of 0.10 mol dm-3 HCl (aq)
\nB
\nAn equation for the combustion of propane is given below.
\nC3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
Determine the standard enthalpy change, , for this reaction, using section 11 of the data booklet.
\n\n
Calculate the standard enthalpy change, , for this reaction using section 12 of the data booklet.
\nPredict, giving a reason, whether the entropy change, , for this reaction is negative or positive.
\n\n
Calculate for the reaction in , using section 12 of the data booklet.
\nThe standard molar entropy for oxygen gas is .
\nCalculate the standard Gibbs free energy change, , in , for the reaction at 5 °C, using your answers to (b) and (d). Use section 1 of the data booklet.
\n(If you did not obtain an answer to (b) or (d) use values of and respectively, although these are not the correct answers.)
\nBonds broken: 8(C–H) + 2(C–C) + 5(O=O) / 8 × 414 «kJ mol−1» + 2 × 346 «kJ mol−1» + 5 × 498 «kJ mol−1» / 6494 «kJ» ✔
\nBonds formed: 6(C=O) + 8(O–H) / 6 × 804 «kJ mol−1» + 8 × 463 «kJ mol−1» / 8528 «kJ» ✔
\n«Enthalpy changebonds brokenbonds formed ✔
\nAward [3] for correct final answer.
\n\n
AND AND ✔
✔
Award [2] for correct final answer.
\nAward [1 max] for .
\npositive AND more moles «of gas» in products ✔
\nAND AND AND ✔
\n✔
\nAward [2] for correct final answer.
\n✔
\n✔
\nAward [2] for correct final answer.
\nMany candidates had difficulty determining the number and type of bonds broken or formed and consequently this was the part of question 3 that was most poorly attempted. Those that could identify these bonds performed the calculations correctly.
\nEnthalpy calculations using enthalpy of formation data were generally well done.
\nMost knew that entropy increased however some lost the mark by not including an explanation based on increase number of mol of gaseous products.
\nCalculating ΔSө, like most other calculations, was well done.
\nΔGө calculations were also well done, with some not seeing that specific units were to be used.
\nWhich salt solution has the highest pH?
\nA. NH4Cl
\nB. Ca(NO3)2
\nC. Na2CO3
\nD. K2SO4
\nC
\nPropane and propene are members of different homologous series.
\nDraw the full structural formulas of propane and propene.
\nBoth propane and propene react with bromine.
\n(i) State an equation and the condition required for the reaction of 1 mol of propane with 1 mol of bromine.
\n(ii) State an equation for the reaction of 1 mol of propene with 1 mol of bromine.
\n(iii) State the type of each reaction with bromine.
\nPropane:
\nPropene:
\nPropane:
\nAND
Propene:
i
\nC3H8 + Br2 → C3H7Br + HBr
\n\n
«sun»light/UV/hv
OR
high temperature
Do not accept “reflux” for M2.
\n
ii
\nC3H6 + Br2 → C3H6Br2
\n\n
iii
\nPropane: «free radical» substitution / SR
AND
Propene: «electrophilic» addition / AE
Award mark even if incorrect type of substitution/ addition given.
\n
Which signs for both Eθcell and ΔGθ result in a spontaneous redox reaction occurring under standard conditions?
\nD
\nEthane-1,2-diol, HOCH2CH2OH, has a wide variety of uses including the removal of ice from aircraft and heat transfer in a solar cell.
\n(i) Calculate ΔHθ, in kJ, for this similar reaction below using data from section 12 of the data booklet. of HOCH2CH2OH(l) is –454.8kJmol-1.
\n2CO (g) + 3H2 (g) HOCH2CH2OH (l)
\n(ii) Deduce why the answers to (a)(iii) and (b)(i) differ.
\n(iii) ΔSθ for the reaction in (b)(i) is –620.1JK-1. Comment on the decrease in entropy.
\n(iv) Calculate the value of ΔGθ, in kJ, for this reaction at 298 K using your answer to (b)(i). (If you did not obtain an answer to (b)(i), use –244.0 kJ, but this is not the correct value.)
\n(v) Comment on the statement that the reaction becomes less spontaneous as temperature is increased.
\nPredict the 1HNMR data for ethanedioic acid and ethane-1,2-diol by completing the table.
\ni
«ΔH = Σ ΔHf products – ΣΔHf reactants = –454.8 kJ mol-1 – 2(–110.5 kJ mol-1) =» –233.8 «kJ»
\n
ii
in (a)(iii) gas is formed and in (b)(i) liquid is formed
OR
products are in different states
OR
conversion of gas to liquid is exothermic
OR
conversion of liquid to gas is endothermic
OR
enthalpy of vapourisation needs to be taken into account
Accept product is «now» a liquid.
Accept answers referring to bond enthalpies being means/averages.
\n
iii
«ΔS is negative because five mols of» gases becomes «one mol of» liquid
OR
increase in complexity of product «compared to reactants»
OR
product more ordered «than reactants»
Accept “fewer moles of gas” but not “fewer molecules”.
\n
iv
ΔS = «kJ K-1»
ΔG = –233.8 kJ – (298 K kJ K-1) = –49.0 «kJ»
Award [2] for correct final answer.
Award [1 max] for «+»185 × 103.
If –244.0 kJ used, answer is:
ΔG = –244.0 kJ – (298 K kJ K-1) = –59.2 «kJ»
Award [2] for correct final answer.
\n
v
increasing T makes ΔG larger/more positive/less negative
OR
–TΔS will increase
Accept “none/no splitting” for singlet.
\nAn iron rod is electroplated with silver. Which is a correct condition for this process?
\nA. The silver electrode is the positive electrode.
\nB. The iron rod is the positive electrode.
\nC. The electrolyte is iron(II) sulfate.
\nD. Oxidation occurs at the negative electrode.
\nA
\nThe concentration of a solution of a weak acid, such as ethanedioic acid, can be determined
by titration with a standard solution of sodium hydroxide, NaOH (aq).
5.00 g of an impure sample of hydrated ethanedioic acid, (COOH)2•2H2O, was dissolved in water to make 1.00 dm3 of solution. 25.0 cm3 samples of this solution were titrated against a 0.100 mol dm-3 solution of sodium hydroxide using a suitable indicator.
\n(COOH)2 (aq) + 2NaOH (aq) → (COONa)2 (aq) + 2H2O (l)
\nThe mean value of the titre was 14.0 cm3.
\n(i) Suggest a suitable indicator for this titration. Use section 22 of the data booklet.
\n(ii) Calculate the amount, in mol, of NaOH in 14.0 cm3 of 0.100 mol dm-3 solution.
\n(iii) Calculate the amount, in mol, of ethanedioic acid in each 25.0 cm3 sample.
\n(iv) Determine the percentage purity of the hydrated ethanedioic acid sample.
\nDraw the Lewis (electron dot) structure of the ethanedioate ion, –OOCCOO–.
\nOutline why all the C–O bond lengths in the ethanedioate ion are the same length and suggest a value for them. Use section 10 of the data booklet.
\nExplain how ethanedioate ions act as ligands.
\ni
phenolphthalein
OR
phenol red
\n
ii
«n(NaOH) = dm3 × 0.100 mol dm-3 =» 1.40 × 10-3 «mol»
iii
« × 1.40 × 10-3 =» 7.00 × 10-4 «mol»
\n
iv
ALTERNATIVE 1:
«mass of pure hydrated ethanedioic acid in each titration = 7.00 × 10-4 mol × 126.08 g mol-1 =» 0.0883 / 8.83 × 10-2 «g»
mass of sample in each titration = « × 5.00 g =» 0.125 «g»
\n«% purity = × 100 =» 70.6 «%»
\nALTERNATIVE 2:
«mol of pure hydrated ethanedioic acid in 1 dm3 solution = 7.00 × 10-4 × =» 2.80 × 10-2 «mol»
«mass of pure hydrated ethanedioic acid in sample = 2.80 × 10-2 mol × 126.08 g mol-1 =» 3.53 «g»
\n«% purity = × 100 =» 70.6 «%»
\nALTERNATIVE 3:
mol of hydrated ethanedioic acid (assuming sample to be pure) = = 0.03966 «mol»
actual amount of hydrated ethanedioic acid = «7.00 × 10-4 × =» 2.80 × 10-2 «mol»
\n«% purity = × 100 =» 70.6 «%»
\nAward suitable part marks for alternative methods.
Award [3] for correct final answer.
Award [2 max] for 50.4 % if anhydrous ethanedioic acid assumed.
Accept single negative charges on two O atoms singly bonded to C.
Do not accept resonance structures.
Allow any combination of dots/crosses or lines to represent electron pairs.
electrons delocalized «across the O–C–O system»
OR
resonance occurs
Accept delocalized π-bond(s).
No ECF from (d).
\n
122 «pm» < C–O < 143 «pm»
\nAccept any answer in range 123 «pm» to 142 «pm».
Accept “bond intermediate between single and double bond” or “bond order 1.5”.
coordinate/dative/covalent bond from O to «transition» metal «ion»
OR
acts as a Lewis base/nucleophile
can occupy two positions
OR
provide two electron pairs from different «O» atoms
OR
form two «coordinate/dative/covalent» bonds «with the metal ion»
OR
chelate «metal/ion»
\n
Which is correct for the conversion of propanal to propyl methanoate?
\nD
\nWhich statement is correct for a pair of enantiomers under the same conditions?
\nA. A racemic mixture of the enantiomers is optically active.
\nB. They have the same chemical properties in all their reactions.
\nC. They have the same melting and boiling points.
\nD. They rotate the plane of plane-polarized light by different angles.
\nC
\nMagnesium is a group 2 metal which exists as a number of isotopes and forms many compounds.
\nMagnesium ions produce no emission or absorption lines in the visible region of the electromagnetic spectrum. Suggest why most magnesium compounds tested in a school laboratory show traces of yellow in the flame.
\n(i) Explain the convergence of lines in a hydrogen emission spectrum.
\n(ii) State what can be determined from the frequency of the convergence limit.
\nMagnesium chloride can be electrolysed.
\n(i) Deduce the half-equations for the reactions at each electrode when molten magnesium chloride is electrolysed, showing the state symbols of the products. The melting points of magnesium and magnesium chloride are 922K and 987K respectively.
\n(ii) Identify the type of reaction occurring at the cathode (negative electrode).
\n(iii) State the products when a very dilute aqueous solution of magnesium chloride is electrolysed.
\nStandard electrode potentials are measured relative to the standard hydrogen electrode. Describe a standard hydrogen electrode.
\nA magnesium half-cell, Mg(s)/Mg2+(aq), can be connected to a copper half-cell, Cu(s)/Cu2+(aq).
\n(i) Formulate an equation for the spontaneous reaction that occurs when the circuit is completed.
\n(ii) Determine the standard cell potential, in V, for the cell. Refer to section 24 of the data booklet.
\n(iii) Predict, giving a reason, the change in cell potential when the concentration of copper ions increases.
\ncontamination with sodium/other «compounds»
\ni
energy levels are closer together at high energy / high frequency / short wavelength
\n
ii
ionisation energy
i)
\nAnode (positive electrode):
\n2Cl– → Cl2 (g) + 2e–
\nCathode (negative electrode):
\nMg2+ + 2e– → Mg (l)
\nPenalize missing/incorrect state symbols at Cl2 and Mg once only.
\nAward [1 max] if equations are at wrong electrodes.
\nAccept Mg (g).
\n\n
ii)
\nreduction
\n\n
iii)
\nAnode (positive electrode):
oxygen/O2
OR
hydogen ion/proton/H+ AND oxygen/O2
Cathode (negative electrode):
hydrogen/H2
OR
hydroxide «ion»/OH– AND hydrogen/H2
Award [1 max] if correct products given at wrong electrodes.
\n\n
Any two of:
\n«inert» Pt electrode
OR
platinum black conductor
1 mol dm–3 H+ (aq)
\nH2 (g) at 100 kPa
\nAccept 1 atm H2 (g).
Accept 1 bar H2 (g)
Accept a labelled diagram.
Ignore temperature if it is specified.
i
\nMg(s) + Cu2+ (aq) → Mg2+ (aq) + Cu(s)
\n\n
ii
\n«+0.34V – (–2.37V) = +»2.71 «V»
\n\n
iii
\ncell potential increases
\nreaction «in Q4(k)(i)» moves to the right
OR
potential of the copper half-cell increases/becomes more positive
Accept correct answers based on the Nernst equation
\nSodium thiosulfate solution reacts with dilute hydrochloric acid to form a precipitate of sulfur at room temperature.
\nNa2S2O3 (aq) + 2HCl (aq) → S (s) + SO2 (g) + 2NaCl (aq) + X
\n(i) Using the graph, explain the order of reaction with respect to sodium thiosulfate.
\n(ii) In a different experiment, this reaction was found to be first order with respect to hydrochloric acid. Deduce the overall rate expression for the reaction.
\ni
\nfirst order
«because» [Na2S2O3] is «directly» proportional to rate of reaction «»
Do not accept “linear” for M2.
\n\n
ii
\nrate = k[Na2S2O3][HCl]
\nWhich property explains why tetramethylsilane, Si(CH3)4, can be used as a reference standard in 1H NMR spectroscopy?
\nA. It has a high boiling point.
\nB. It is a reactive compound.
\nC. All its protons are in the same chemical environment.
\nD. It gives multiple signals.
\nC
\nThis question is about the weak acid methanoic acid, HCOOH.
\nCalculate the pH of 0.0100 mol dm–3 methanoic acid stating any assumption you make. Ka = 1.6 × 10–4.
\n(i) Sketch a graph of pH against volume of a strong base added to a weak acid showing how you would determine pKa for the weak acid.
\n(ii) Explain, using an equation, why the pH increases very little in the buffer region when a small amount of alkali is added.
\nCalculation:
\nALTERNATIVE 1:
[H+] = (Ka × [HA])1/2 / (1.6 × 10–4 × 0.0100)1/2 / 1.3 × 10–3 «mol dm–3»
pH = «–log10[H+] ≈» 2.9
\nALTERNATIVE 2:
pH = 0.5(pKa - log10[HA])
pH = 2.9
Award [2] for correct final answer
\nAssumption:
ionisation is << 0.0100 so 0.0100 - [A–] ≈ 0.0100
OR
[HA]eqm = [HA]initial
OR
all H+ ions in the solution come from the acid «and not from the self-ionisation of water»
OR
[H+] = [HCOO–]
Do not accept partial dissociation
\ni
\ncorrect shape of graph
pH at half neutralization/equivalence
M1: must show buffer region at pH < 7 and equivalence at pH > 7.
Accept graph starting from where two axes meet as pH scale is not specified.
\n
ii
\nALTERNATIVE 1:
\nHCOOH HCOO– + H+
H+ ions consumed in reaction with OH– are produced again as equilibrium moves to the right «so [H+] remains almost unchanged»
ALTERNATIVE 2:
HCOOH + OH– HCOO– + H2O
added OH- are neutralized by HCOOH
OR
strong base replaced by weak base
Accept HA or any other weak acid in equations.
Equilibrium sign must be included in equation for M1
The oxidation state of cobalt in the complex ion [Co(NH3)5Br]x is +3. Which of the following statements are correct?
\nI. The overall charge, x, of the complex ion is 2+.
II. The complex ion is octahedral.
III. The cobalt(III) ion has a half-filled d-subshell.
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nIn order to provide safe drinking water, a water supply is often treated with disinfectants, which aim to inactivate disease-causing bacteria in the water.
\nTo compare the effectiveness of different disinfectants, a CT value is used as a measure of the dosage of disinfectant needed to achieve a certain level of inactivation of specific bacteria.
\nCT value (mg min dm−3) = C (mg dm−3) concentration of disinfectant × T (min) contact time with water
\nThe table below compares the CT values of different disinfectants necessary to achieve 99% inactivation of two types of bacteria, listed as A and B.
\n(i) Deduce the oxidation state of chlorine in the following disinfectants.
\n(ii) From the data on CT values, justify the statement that bacterium B is generally more resistant to disinfection than bacterium A.
\n(iii) CT values can be used to determine whether a particular treatment process is adequate. Calculate the CT value, in mg min dm−3, when 1.50 × 10−5 g dm−3 of chlorine dioxide is added to a water supply with a contact time of 9.82 minutes.
\n(iv) From your answer to (a) (iii) and the data in the table, comment on whether this treatment will be sufficient to inactivate 99% of bacterium A.
\nCT values are influenced by temperature and by pH. The table below shows the CT values for chlorine needed to achieve 99% inactivation of a specific bacterium at stated values of pH and temperature.
\n(i) With reference to the temperature data in the table, suggest why it may be more difficult to treat water effectively with chlorine in cold climates.
\n(ii) Sketch a graph on the axes below to show how the CT value (at any temperature) varies with pH.
\n(iii) Comment on the relative CT values at pH 6.0 and pH 9.0 at each temperature.
\n(iv) Chlorine reacts with water as follows:
\nCl2 (g) + H2O (l) HOCl (aq) + HCl (aq)
\nHOCl (aq) OCl− (aq) + H+ (aq)
\nPredict how the concentrations of each of the species HOCl (aq) and OCl− (aq) will change if the pH of the disinfected water increases.
\nDespite widespread improvements in the provision of safe drinking water, the sale of bottled water has increased dramatically in recent years. State one problem caused by this trend.
\ni
\nHOCl: +1
AND
ClO2: +4
Accept “I” and “IV” but not “1+/1” and “4+/4” notations.
\n\n
ii
\n«most» CT values are higher for «bacterium» B
OR
«generally» higher dosage needed for «bacterium» B
Accept converse arguments. Accept “concentration” for “dosage”
\n\n
iii
\n«CT = 1.50 × 10–5 × 103 mg dm–3 × 9.82 min =» 1.47 × 10–1 «mg min dm–3»
\n\n
iv
\nlower than CT value/minimum dosage/1.8 × 10–1 «mg min dm–3»
AND
no/insufficient
Accept “concentration” for “dosage”.
\ni
\nhigher CT value at lower temperature
OR
higher dosage «of chlorine» needed at low temperature
Accept “effectiveness decreases at lower temperature”.
Accept “concentration” for “dosage”.
Accept converse arguments.
\n
ii
\nlabeled axes ( y: CT and x: pH)
AND
curve with increasing gradient
Do not accept axes the wrong way round.
Accept a linear sketch.
\n
iii
\nvalues at pH 9.0 approximately 3 times values at pH 6.0
OR
increase in CT values in same ratio
The exact ratio is 2.9 times
Do not accept just “increase in value”.
\n
iv
\n[HOCl] decreases AND [OCl−] increases
\nplastic disposal/pollution
OR
plastic bottles use up petroleum/non-renewable raw material
OR
chemicals in plastic bottle can contaminate water
OR
«prolonged» storage in plastic can cause contamination of water
OR
plastic water bottles sometimes reused without proper hygiene considerations
Accept other valid answers.
Accept economic considerations such as “greater production costs”, “greater transport costs” or “bottled water more expensive than tap water”
Aspirin is one of the most widely used drugs in the world.
\nAspirin was synthesized from 2.65 g of salicylic acid (2-hydroxybenzoic acid) (Mr = 138.13) and 2.51 g of ethanoic anhydride (Mr = 102.10).
\nCalculate the amounts, in mol, of each reactant.
\nCalculate, in g, the theoretical yield of aspirin.
\nState two techniques which could be used to confirm the identity of aspirin.
\nState how aspirin can be converted to water-soluble aspirin.
\nCompare, giving a reason, the bioavailability of soluble aspirin with aspirin.
\nn(salicylic acid) = «» 0.0192 «mol»
\nAND
\nn(ethanoic anhydride) = «» 0.0246 «mol»
\n[1 mark]
\n«mass = 0.0192 mol x 180.17 gmol–1 =» 3.46 «g»
\n\n
Award ECF mark only if limiting reagent determined in (i) has been used.
\n[1 mark]
\nAny two of:
\nmelting point
\nmass spectrometry/MS
\nhigh-performance liquid chromatography/HPLC
\nNMR/nuclear magnetic resonance
\nX-ray crystallography
\nelemental analysis «for elemental percent composition»
\n\n
Accept “spectroscopy” instead of “spectrometry” where mentioned but not “spectrum”.
\nAccept “infra-red spectroscopy/IR” OR “ultraviolet «-visible» spectroscopy/UV/UV-Vis”.
\nDo not accept “gas chromatography/GC”.
\nAccept “thin-layer chromatography/TLC” as an alternative to “HPLC”.
\n[2 marks]
\nreact with NaOH
\n\n
Accept “NaHCO3” or “Na2CO3” instead of “NaOH”.
\nAccept chemical equation OR name for reagent used.
\n[1 mark]
\n«marginally» higher AND increase rate of dispersion
OR
«marginally» higher AND increase absorption in mouth/stomach «mucosa»
OR
«approximately the» same AND ionic salt reacts with HCl/acid in stomach to produce aspirin again
\n
Do not accept “«marginally» higher AND greater solubility in blood”.
\n[1 mark]
\nPropane and propene are members of different homologous series.
\n(i) Draw diagrams to show how sigma (σ) and pi (π) bonds are formed between atoms.
\n\n
(ii) State the number of sigma (σ) and pi (π) bonds in propane and propene.
\nConstruct the mechanism of the formation of 2-bromopropane from hydrogen bromide and propene using curly arrows to denote the movement of electrons.
\ni
\n\n
ii
\nAward [1] for two or three correct answers.
Award [2] for all four correct.
curly arrow going from C=C to H of HBr and curly arrow showing Br leaving
representation of carbocation
curly arrow going from lone pair/negative charge on Br– to C+
Award [2 max] for formation of 1-bromopropane.
\nIn a class experiment, students were asked to determine the value of x in the formula of a hydrated salt, BaCl2･xH2O. They followed these instructions:
\nTheir results in three trials were as follows:
\nState and explain the further work students need to carry out in trial 2 before they can process the results alongside trial 1.
\nIn trial 3, the students noticed that after heating, the crucible had turned black on the outside. Suggest what may have caused this, and how this might affect the calculated value for x in the hydrated salt.
\nList two assumptions made in this experiment.
\nrepeat steps 3 and 4
OR
repeat step 5
OR
conduct a third heating
OR
«re»heat AND «re»weigh
water still present
OR
need two consistent readings
OR
heat to constant mass
Accept “ensure even/strong heating” for M1.
Do not accept “cleaning/washing the crucible”.
soot/carbon deposited
OR
incomplete combustion
OR
air hole of Bunsen burner closed/not fully open
Accept “using a yellow «Bunsen burner» flame” for M1.
\n\n
«value of x» lower
\nOnly award M2 if M1 correct.
\nall mass loss is due to water loss
\nall the water «of crystallization» is lost
\ncrucible does not absorb/lose water
\ncrystal/BaCl2 does not decompose/hydrolyse/oxidize/react with oxygen/air «when heated»
\nAccept “no loss of crystals/BaCl2 occurs”, “no impurities in the «weighed hydrated» salt”, “reaction goes to completion”, “heat was consistent/strong”, “crystal/BaCl2 does not absorb water during cooling”, “balance has been calibrated” or “crucible was clean at the start”.
\nDo not accept ”heat loss to surroundings” or “no carbon deposited on crucible”.
\nReference to defects in apparatus not accepted.
\nDo not penalize if BaCl2.xH2O is used for BaCl2.
\nOne structural isomer of C4H9Br is a chiral molecule.
\nDraw the three-dimensional shape of each enantiomer of this isomer showing their spatial relationship to each other.
\nWhen one enantiomer undergoes substitution by alkaline hydrolysis approximately 75 % of the product molecules show inversion of configuration. Comment on the mechanisms that occur.
\nSuggest why the rate of alkaline hydrolysis of an enantiomer of iodopropane is greater than that of an enantiomer of bromopropane.
\ncorrect isomer
mirror image shown clearly
SN2 would give inversion of configuration «almost 100%»
OR
SN1 would give «approximately» 50% of each
so mechanism is a mixture of both mechanisms
\nC–I bond «longer, so» weaker «than C–Br bond»
OR
I– is a better leaving group than Br–
The structures of morphine, diamorphine and codeine are given in section 37 of the data booklet.
\nExplain why diamorphine passes more readily than morphine through the blood-brain barrier.
\nSuggest a reagent used to prepare diamorphine from morphine.
\nSuggest one reason why codeine is available without prescription in some countries whilst morphine is administered under strict medical supervision.
\nAny two of:
\ndiamorphine has ester/ethanoate/acetate «groups» AND morphine has hydroxyl «groups»
\ndiamorphine/ester/ethanoate/acetate groups less polar
\ndiamorphine more soluble in lipids
\n\n
Accept “alcohol/hydroxy” for “hydroxyl” but not “hydroxide”.
\nAccept “diamorphine non-polar”.
\nAccept converse statements.
\n[2 marks]
\nethanoic/acetic anhydride
OR
ethanoyl/acetyl chloride
\n
Accept other possible reagents, such as ethanoic/acetic acid or acetyl bromide.
\nAccept chemical formulas.
\n[1 mark]
\nmorphine has a smaller therapeutic window
\n\n
Accept converse statements.
\nAccept “codeine has lower activity” OR “codeine has lower risk of overdose” OR “codeine is less potent”.
\nDo not accept “lower abuse potential for codeine” OR “codeine less addictive” OR “codeine has a lower bioavailability”.
\n[1 mark]
\nWhat is the major product of the reaction between 2-methylbut-2-ene and hydrogen bromide?
\nA. 3-bromo-2-methylbutane
\nB. 3-bromo-3-methylbutane
\nC. 2-bromo-3-methylbutane
\nD. 2-bromo-2-methylbutane
\nD
\nWhat is the product of the reduction of 2-methylbutanal?
\nA. 2-methylbutan-1-ol
\nB. 2-methylbutan-2-ol
\nC. 3-methylbutan-2-one
\nD. 2-methylbutanoic acid
\nA
\nThe structures of oseltamivir (Tamiflu) and zanamivir (Relenza) are given in section 37 of the data booklet.
\nCompare and contrast the structures of oseltamivir and zanamivir, stating the names of functional groups.
\nDeduce the wavenumber of one absorbance seen in the IR spectrum of only one of the compounds, using section 26 of the data booklet.
\nSuggest one ethical consideration faced by medical researchers when developing medications.
\nOne similarity:
\nboth contain amido «group»
\nOne difference:
\noseltamivir contains ester «group» AND zanamivir does not
OR
oseltamivir contains amino «group» AND zanamivir does not «but contains a guanidino group»
OR
zanamivir contains carboxyl «group» AND oseltamivir does not
OR
zanamivir contains «several» hydroxyl «groups» AND oseltamivir does not
OR
oseltamivir contains ester «group» AND zanamivir contains carboxyl «group»
OR
oseltamivir contains ester «group» AND zanamivir contains «several» hydroxyl «groups»
\n
Accept “both contain ether «group»” OR “both contain alkene/alkenyl «group»” OR “both contain carbonyl «group»” OR “both contain amino/amine «group»”. Latter cannot be given in combination with second difference alternative with respect to amino group.
\nAccept “amide/carboxamide/carbamoyl” for “amido”.
\nAccept “amine” for “amino”.
\nAccept “carboxylic acid” for “carboxyl”.
\nAccept “hydroxy/alcohol” for “hydroxyl”, but not “hydroxide”.
\n[2 marks]
\n1050–1410
OR
1620–1680
OR
1700–1750
OR
2500–3000
OR
3200–3600
OR
2850–3090
OR
3300–3500 «cm–1»
[1 mark]
\n«negative» side-effects of medication on patient/volunteers
OR
effects on environment «from all materials used and produced»
OR
potential for abuse
OR
drugs may be developed that are contrary to some religious doctrines
OR
animal testing
OR
risk to benefit ratio
OR
appropriate consent of patient volunteers
[1 mark]
\nWhich molecule is chiral?
\nA. 2-chlorobutane
\nB. 2,2-dichloropentane
\nC. Propan-2-amine
\nD. 4-hydroxybutanoic acid
\nA
\nThe molar mass of a gas, determined experimentally, is 32 g mol−1. Its literature molar mass is 40 g mol−1.
\nWhat is the percentage error?
\nA. 80%
\nB. 25%
\nC. 20%
\nD. 8%
\nC
\nA student performs a titration to determine the concentration of ethanoic acid, , in vinegar using potassium hydroxide.
\nThe pH curve for the reaction is given.
\nWrite a balanced equation for the reaction.
\nIdentify the major species, other than water and potassium ions, at these points.
\nState a suitable indicator for this titration. Use section 22 of the data booklet
\nSuggest, giving a reason, which point on the curve is considered a buffer region.
\nState the expression for ethanoic acid.
\nCalculate the of the conjugate base of ethanoic acid using sections 2 and 21 of the data booklet.
\nIn a titration, of vinegar required of potassium hydroxide to reach the end-point.
\nCalculate the concentration of ethanoic acid in the vinegar.
\nPotassium hydroxide solutions can react with carbon dioxide from the air. The solution was made one day prior to using it in the titration.
\n State the type of error that would result from the student’s approach.
Potassium hydroxide solutions can react with carbon dioxide from the air. The solution was made one day prior to using it in the titration.
\nPredict, giving a reason, the effect of this error on the calculated concentration of ethanoic acid in 5(e).
\n\n
✔
\nAccept the ionic equation.
\nB: AND ✔
\nC: ✔
\nAccept names.
\nAccept for
\nphenolphthalein ✔
\nAccept “phenol red” or “bromothymol blue”.
\nB AND the region where small additions «of the base/ » result in little or no
change in
OR
B AND the flattest region of the curve «at intermediate /before equivalence
point »
OR
B AND half the volume needed to reach equivalence point
OR
B AND similar amounts of weak acid//ethanoic acid AND conjugate base//ethanoate ✔
Accept instead of .
\n
✔
Accept answers between .
\n✔
\n
✔
Award [2] for correct final answer.
\nsystematic «error» ✔
\nwould be higher ✔
\nactual is lower «than the value in calculation»
OR
larger volume of «solution» needed to neutralize the acid ✔
Accept partially neutralised by from air.
\nMost candidates could write a balanced neutralization equation.
\nIdentifying species present at various points along a pH titration curve was one of the most poorly answered questions in the exam. Very few candidates realized there were two major species at point B even when they were able in general to realize that B was a buffer zone.
\nAlmost all candidates could identify a suitable indicator to use in a titration of a weak acid with a strong base.
\nMost students could identify a buffer zone region in a titration but very few (50%) could coherently explain why.
\nPoorly answered with only 50% correctly writing a Ka expression. The major error was in candidates trying to calculate a Ka rather than write an expression for it.
\nLike with other calculations in this exam, the majority of candidates could correctly determine a concentration from titration data.
\n80% of candidates could identify the method used as a systematic error, with some stating human or random error.
\nMost candidates identified that the systematic error would result in the concentration of the alkali being lowered but then failed to propagate this through to the effect on the concentration of the acid.
\nWhich technique is used to determine the bond lengths and bond angles of a molecule?
\nA. X-ray crystallography
\nB. Infrared (IR) spectroscopy
\nC. Mass spectroscopy
\nD. 1H NMR spectroscopy
\nA
\nA number of drugs have been developed to treat excess acidity in the stomach.
\nTwo drugs are ranitidine (Zantac) and omeprazole (Prilosec). Outline how they function to reduce stomach acidity.
\n0.500 g of solid anhydrous sodium carbonate, Na2CO3(s), is dissolved in 75.0 cm3 of 0.100 moldm−3 sodium hydrogen carbonate solution, NaHCO3(aq). Assume the volume does not change when the salt dissolves.
\nHCO3−(aq) CO32−(aq) + H+(aq) pKa = 10.35.
\nCalculate the pH of the buffer solution.
\nRanitidine:
Blocks/binds H2-histamine receptors «in cells of stomach lining»
OR
prevents histamine molecules binding to H2-histamine receptors «and triggering acid secretion»
Omeprazole:
inhibits enzyme/gastric proton pump which secretes H+ ions «into gastric juice»
\n
Accept “H2 receptor antagonist” for M1.
\n[2 marks]
\n[Na2CO3] = «=» 0.0629 «moldm−3»
\n«pH = pKa + log»
\n«pH = 10.35 – 0.201 =» 10.15
\n\n
Alternative method involving Ka may be used to deduce pH in M2.
\nAward [2] for correct final answer.
\n-2 marks]
\nThe rate of the acid-catalysed iodination of propanone can be followed by measuring how the concentration of iodine changes with time.
\nI2(aq) + CH3COCH3(aq) → CH3COCH2I(aq) + H+(aq) + I−(aq)
\nSuggest how the change of iodine concentration could be followed.
\nA student produced these results with [H+] = 0.15 moldm−3. Propanone and acid were in excess and iodine was the limiting reagent.
\nDetermine the relative rate of reaction when [H+] = 0.15 moldm−3.
\n![\"M17/4/CHEMI/SP2/ENG/TZ1/01.a.ii\"](\"../assets/uploads/tinymce_asset/asset/3718/Schermafbeelding_2017-09-22_om_15.59.41.png\"/)
The student then carried out the experiment at other acid concentrations with all other conditions remaining unchanged.
\nState and explain the relationship between the rate of reaction and the concentration of acid.
\nuse a colorimeter/monitor the change in colour
OR
take samples AND quench AND titrate «with thiosulfate»
\n
Accept change in pH.
Accept change in conductivity.
Accept other suitable methods.
Method must imply “change”.
[1 mark]
\n![\"M17/4/CHEMI/SP2/ENG/TZ1/01.a.ii/M\"](\"../assets/uploads/tinymce_asset/asset/3719/Schermafbeelding_2017-09-22_om_16.03.25.png\"/)
best fit line
\nrelative rate of reaction = « =» 0.0074/7.4 x 10−3
\n\n
Best fit line required for M1.
\nM2 is independent of M1.
Accept range from 0.0070 to 0.0080.
\n[2 marks]
\nRelationship:
rate of reaction is «directly» proportional to [H+]
OR
rate of reaction [H+]
Explanation:
more frequent collisions/more collisions per unit of time «at greater concentration»
\n
Accept \"doubling the concentration doubles the rate\".
\nDo not accept “rate increases as concentration increases”.
\nDo not accept collisions more likely.
\n[2 marks]
\nLimestone can be converted into a variety of useful commercial products through the lime cycle. Limestone contains high percentages of calcium carbonate, CaCO3.
\nThe second step of the lime cycle produces calcium hydroxide, Ca(OH)2.
\nCalcium hydroxide reacts with carbon dioxide to reform calcium carbonate.
\nCa(OH)2 (aq) + CO2 (g) → CaCO3 (s) + H2O (l)
\nCalcium carbonate is heated to produce calcium oxide, CaO.
\nCaCO3 (s) → CaO (s) + CO2 (g)
\nCalculate the volume of carbon dioxide produced at STP when 555 g of calcium carbonate decomposes. Use sections 2 and 6 of the data booklet.
\nThermodynamic data for the decomposition of calcium carbonate is given.
\nCalculate the enthalpy change of reaction, ΔH, in kJ, for the decomposition of calcium carbonate.
\nThe potential energy profile for a reaction is shown. Sketch a dotted line labelled “Catalysed” to indicate the effect of a catalyst.
\nOutline why a catalyst has such an effect.
\nWrite the equation for the reaction of Ca(OH)2 (aq) with hydrochloric acid, HCl (aq).
\nDetermine the volume, in dm3, of 0.015 mol dm−3 calcium hydroxide solution needed to neutralize 35.0 cm3 of 0.025 mol dm−3 HCl (aq).
\nSaturated calcium hydroxide solution is used to test for carbon dioxide. Calculate the pH of a 2.33 × 10−2 mol dm−3 solution of calcium hydroxide, a strong base.
\nDetermine the mass, in g, of CaCO3 (s) produced by reacting 2.41 dm3 of 2.33 × 10−2 mol dm−3 of Ca(OH)2 (aq) with 0.750 dm3 of CO2 (g) at STP.
\n2.85 g of CaCO3 was collected in the experiment in e(i). Calculate the percentage yield of CaCO3.
\n(If you did not obtain an answer to e(i), use 4.00 g, but this is not the correct value.)
\nOutline how one calcium compound in the lime cycle can reduce a problem caused by acid deposition.
\n«nCaCO3 = =» 5.55 «mol» ✓
\n«V = 5.55 mol × 22.7 dm3 mol−1 =» 126 «dm3» ✓
\n
Award [2] for correct final answer.
Accept method using pV = nRT to obtain the volume with p as either 100 kPa (126 dm3) or 101.3 kPa (125 dm3).
\nDo not penalize use of 22.4 dm3 mol–1 to obtain the volume (124 dm3).
\n«ΔH =» (−635 «kJ» – 393.5 «kJ») – (−1207 «kJ») ✓
\n«ΔH = + » 179 «kJ» ✓
\n
Award [2] for correct final answer.
Award [1 max] for −179 kJ.
\nIgnore an extra step to determine total enthalpy change in kJ: 179 kJ mol−1 x 5.55 mol = 993 kJ.
\nAward [2] for an answer in the range 990 - 993« kJ».
\nlower activation energy curve between same reactant and product levels ✓
\n
Accept curve with or without an intermediate.
Accept a horizontal straight line below current line with the activation energy with catalyst/Ecat clearly labelled.
\nprovides an alternative «reaction» pathway/mechanism ✓
\n
Do not accept “lower activation energy” only.
Ca(OH)2 (aq) + 2HCl (aq) → 2H2O (l) + CaCl2 (aq) ✓
\n«nHCl = 0.0350 dm3 × 0.025 mol dm−3 =» 0.00088 «mol»
OR
nCa(OH)2 = nHCl/0.00044 «mol» ✓
«V = =» 0.029 «dm3» ✓
\n
Award [2] for correct final answer.
Award [1 max] for 0.058 «dm3».
\nAlternative 1:
\n[OH−] = « 2 × 2.33 × 10−2 mol dm−3 =» 0.0466 «mol dm−3» ✓
\n«[H+] = = 2.15 × 10−13 mol dm−3»
pH = « −log(2.15 × 10−13) =» 12.668 ✓
\n
Alternative 2:
\n[OH−] =« 2 × 2.33 × 10−2 mol dm−3 =» 0.0466 «mol dm−3» ✓
\n«pOH = −log (0.0466) = 1.332»
\npH = «14.000 – pOH = 14.000 – 1.332 =» 12.668 ✓
\n\n
Award [2] for correct final answer.
\nAward [1 max] for pH =12.367.
\n«nCa(OH)2 = 2.41 dm3 × 2.33 × 10−2 mol dm−3 =» 0.0562 «mol» AND
«nCO2 ==» 0.0330 «mol» ✓
«CO2 is the limiting reactant»
\n«mCaCO3 = 0.0330 mol × 100.09 g mol−1 =» 3.30 «g» ✓
\n
Only award ECF for M2 if limiting reagent is used.
Accept answers in the range 3.30 - 3.35 «g».
\n« × 100 =» 86.4 «%» ✓
\n
Accept answers in the range 86.1-86.4 «%».
Accept “71.3 %” for using the incorrect given value of 4.00 g.
\n«add» Ca(OH)2/CaCO3/CaO AND to «acidic» water/river/lake/soil
OR
«use» Ca(OH)2/CaCO3/CaO in scrubbers «to prevent release of acidic pollution» ✓
Accept any correct name for any of the calcium compounds listed.
Titanium is a transition metal.
\nTiCl4 reacts with water and the resulting titanium(IV) oxide can be used as a smoke screen.
\nDescribe the bonding in metals.
\nTitanium exists as several isotopes. The mass spectrum of a sample of titanium gave the following data:
\nCalculate the relative atomic mass of titanium to two decimal places.
\nState the number of protons, neutrons and electrons in the atom.
\nState the full electron configuration of the 2+ ion.
\nExplain why an aluminium-titanium alloy is harder than pure aluminium.
\nState the type of bonding in potassium chloride which melts at 1043 K.
\nA chloride of titanium, TiCl4, melts at 248 K. Suggest why the melting point is so much lower than that of KCl.
\nFormulate an equation for this reaction.
\nSuggest one disadvantage of using this smoke in an enclosed space.
\nelectrostatic attraction
\nbetween «a lattice of» metal/positive ions/cations AND «a sea of» delocalized electrons
\n\n
Accept mobile electrons.
\nDo not accept “metal atoms/nuclei”.
\n[2 marks]
\n\n
= 47.93
\n\n
Answer must have two decimal places with a value from 47.90 to 48.00.
\nAward [2] for correct final answer.
\nAward [0] for 47.87 (data booklet value).
\n[2 marks]
\nProtons: 22 AND Neutrons: 26 AND Electrons: 22
\n[1 mark]
\n\n
1s22s22p63s23p63d2
\n[1 mark]
\ntitanium atoms/ions distort the regular arrangement of atoms/ions
OR
titanium atoms/ions are a different size to aluminium «atoms/ions»
prevent layers sliding over each other
\n\n
Accept diagram showing different sizes of atoms/ions.
\n[2 marks]
\nionic
OR
«electrostatic» attraction between oppositely charged ions
[1 mark]
\n«simple» molecular structure
OR
weak«er» intermolecular bonds
OR
weak«er» bonds between molecules
\n
Accept specific examples of weak bonds such as London/dispersion and van der Waals.
\nDo not accept “covalent”.
\n[1 mark]
\nTiCl4(l) + 2H2O(l) → TiO2(s) + 4HCl(aq)
\ncorrect products
\ncorrect balancing
\n\n
Accept ionic equation.
\nAward M2 if products are HCl and a compound of Ti and O.
\n[2 marks]
\nHCl causes breathing/respiratory problems
OR
HCl is an irritant
OR
HCl is toxic
OR
HCl has acidic vapour
OR
HCl is corrosive
\n
Accept “TiO2 causes breathing problems/is an irritant”.
\nAccept “harmful” for both HCl and TiO2.
\nAccept “smoke is asphyxiant”.
\n[1 mark]
\nVanadium, another transition metal, has a number of different oxidation states.
\nDetermine the oxidation state of vanadium in each of the following species.
\nFormulate an equation for the reaction between VO2+(aq) and V2+(aq) in acidic solution to form V3+(aq).
\nV2O5: +5
\nVO2+: +4
\n\n
Do not penalize incorrect notation twice.
\n[2 marks]
\nVO2+(aq) + V2+(aq) + 2H+(aq) → 2V3+(aq) + H2O(l)
\n\n
Accept equilibrium sign.
\n[1 mark]
\nTwo hydrides of nitrogen are ammonia and hydrazine, N2H4. One derivative of ammonia is methanamine whose molecular structure is shown below.
\n![\"M17/4/CHEMI/SP2/ENG/TZ1/04\"](\"../assets/uploads/tinymce_asset/asset/3720/Schermafbeelding_2017-09-22_om_18.03.06.png\"/)
Hydrazine is used to remove oxygen from water used to generate steam or hot water.
\nN2H4(aq) + O2(aq) → N2(g) + 2H2O(l)
\nThe concentration of dissolved oxygen in a sample of water is 8.0 × 10−3 gdm−3.
\nEstimate the H−N−H bond angle in methanamine using VSEPR theory.
\nAmmonia reacts reversibly with water.
\nNH3(g) + H2O(l) NH4+(aq) + OH−(aq)
\nExplain the effect of adding H+(aq) ions on the position of the equilibrium.
\nHydrazine reacts with water in a similar way to ammonia. Deduce an equation for the reaction of hydrazine with water.
\nOutline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride.
\nHydrazine has been used as a rocket fuel. The propulsion reaction occurs in several stages but the overall reaction is:
\nN2H4(l) → N2(g) + 2H2(g)
\nSuggest why this fuel is suitable for use at high altitudes.
\nDetermine the enthalpy change of reaction, ΔH, in kJ, when 1.00 mol of gaseous hydrazine decomposes to its elements. Use bond enthalpy values in section 11 of the data booklet.
\nN2H4(g) → N2(g) + 2H2(g)
\nThe standard enthalpy of formation of N2H4(l) is +50.6 kJmol−1. Calculate the enthalpy of vaporization, ΔHvap, of hydrazine in kJmol−1.
\nN2H4(l) → N2H4(g)
\n(If you did not get an answer to (f), use −85 kJ but this is not the correct answer.)
\nCalculate, showing your working, the mass of hydrazine needed to remove all the dissolved oxygen from 1000 dm3 of the sample.
\nCalculate the volume, in dm3, of nitrogen formed under SATP conditions. (The volume of 1 mol of gas = 24.8 dm3 at SATP.)
\n107°
\n\n
Accept 100° to < 109.5°.
\nLiterature value = 105.8°
\n[1 mark]
\nremoves/reacts with OH−
\nmoves to the right/products «to replace OH− ions»
\n\n
Accept ionic equation for M1.
\n[2 marks]
\nN2H4(aq) + H2O(l) N2H5+(aq) + OH–(aq)
\n\n
Accept N2H4(aq) + 2H2O(l) N2H62+(aq) + 2OH–(aq).
\nEquilibrium sign must be present.
\n[1 mark]
\nbubbles
OR
gas
OR
magnesium disappears
2NH4+(aq) + Mg(s) → Mg2+(aq) + 2NH3(aq) + H2(g)
\n\n
Do not accept “hydrogen” without reference to observed changes.
\nAccept \"smell of ammonia\".
\nAccept 2H+(aq) + Mg(s) → Mg2+(aq) + H2(g)
\nEquation must be ionic.
\n[2 mark]
\nno oxygen required
\n[1 mark]
\nbonds broken:
E(N–N) + 4E(N–H)
OR
158 «kJmol–1» + 4 x 391 «kJmol–1» / 1722 «kJ»
bonds formed:
E(N≡N) + 2E(H–H)
OR
945 «kJmol–1» + 2 x 436 «kJmol–1» / 1817 «kJ»
«ΔH = bonds broken – bonds formed = 1722 – 1817 =» –95 «kJ»
\n\n
Award [3] for correct final answer.
\nAward [2 max] for +95 «kJ».
\n[3 marks]
\nOR
ΔHvap= −50.6 kJmol−1 − (−95 kJmol−1)
«ΔHvap =» +44 «kJmol−1»
\n\n
Award [2] for correct final answer.
\nAward [1 max] for −44 «kJmol−1».
\nAward [2] for:
ΔHvap − = 50.6 kJmol−1 − (−85 kJmol−1) + = 34 «kJmol−1».
Award [1 max] for −34 «kJmol−1».
\n[2 marks]
\ntotal mass of oxygen «= 8.0 x 10–3 gdm–3 x 1000 dm3» = 8.0 «g»
\nn(O2) «» 0.25 «mol»
\nOR
n(N2H4) = n(O2)
«mass of hydrazine = 0.25 mol x 32.06 gmol–1 =» 8.0 «g»
\n
Award [3] for correct final answer.
\n[3 marks]
\n«n(N2H4) = n(O2) » 0.25 «mol»
\n«volume of nitrogen = 0.25 mol x 24.8 dm3mol–1» = 6.2 «dm3»
\n\n
Award [1] for correct final answer.
\n[1 mark]
\nThe properties of elements can be predicted from their position in the periodic table.
\nExplain why Si has a smaller atomic radius than Al.
\nExplain the decrease in radius from Na to Na+.
\nState the condensed electron configurations for Cr and Cr3+.
\nDescribe metallic bonding and how it contributes to electrical conductivity.
\nDeduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl2.
\nSuggest, giving reasons, the relative volatilities of SCl2 and H2O.
\nConsider the following equilibrium reaction:
\n2SO2 (g) + O2 (g) 2SO3 (g)
\nState and explain how the equilibrium would be affected by increasing the volume of the reaction container at a constant temperature.
\nnuclear charge/number of protons/Z/Zeff increases «causing a stronger pull on the outer electrons» ✓
\nsame number of shells/«outer» energy level/shielding ✓
\nNa+ has one less energy level/shell
OR
Na+ has 2 energy levels/shells AND Na has 3 ✓
less shielding «in Na+ so valence electrons attracted more strongly to nucleus»
OR
effective nuclear charge/Zeff greater «in Na+ so valence electrons attracted more strongly to nucleus» ✓
Accept “more protons than electrons «in Na+»” OR “less electron-electron repulsion «in Na+»” for M2.
Cr:
[Ar] 4s13d5 ✓
Cr3+:
[Ar] 3d3 ✓
Accept “[Ar] 3d54s1”.
Accept “[Ar] 3d34s0”.
\nAward [1 max] for two correct full electron configurations “1s22s22p63s23p64s13d5 AND 1s22s22p63s23p63d3”.
\nAward [1 max] for 4s13d5 AND 3d3.
\nelectrostatic attraction ✓
\nbetween «a lattice of» cations/positive «metal» ions AND «a sea of» delocalized electrons ✓
\n
mobile electrons responsible for conductivity
OR
electrons move when a voltage/potential difference/electric field is applied ✓
\n
Do not accept “nuclei” for “cations/positive ions” in M2.
\nAccept “mobile/free” for “delocalized” electrons in M2.
\nAccept “electrons move when connected to a cell/battery/power supply” OR “electrons move when connected in a circuit” for M3.
\nH2O forms hydrogen bonding «while SCl2 does not» ✓
\nSCl2 «much» stronger London/dispersion/«instantaneous» induced dipole-induced dipole forces ✓
\n
Alternative 1:
H2O less volatile AND hydrogen bonding stronger «than dipole–dipole and dispersion forces» ✓
Alternative 2:
SCl2 less volatile AND effect of dispersion forces «could be» greater than hydrogen bonding ✓\\
\n
Ignore reference to Van der Waals.
\nAccept “SCl2 has «much» larger molar mass/electron density” for M2.
\npressure decrease «due to larger volume» ✓
\nreactant side has more moles/molecules «of gas» ✓
\nreaction shifts left/towards reactants ✓
\n
Award M3 only if M1 OR M2 is awarded.
This question is about carbon and chlorine compounds.
\nEthane, C2H6, reacts with chlorine in sunlight. State the type of this reaction and the name of the mechanism by which it occurs.
\nFormulate equations for the two propagation steps and one termination step in the formation of chloroethane from ethane.
\nOne possible product, X, of the reaction of ethane with chlorine has the following composition by mass:
\ncarbon: 24.27%, hydrogen: 4.08%, chlorine: 71.65%
\nDetermine the empirical formula of the product.
\nThe mass and 1HNMR spectra of product X are shown below. Deduce, giving your reasons, its structural formula and hence the name of the compound.
\nChloroethene, C2H3Cl, can undergo polymerization. Draw a section of the polymer with three repeating units.
\nsubstitution AND «free-»radical
OR
substitution AND chain
\n
Award [1] for “«free-»radical substitution” or “SR” written anywhere in the answer.
\n[1 mark]
\nTwo propagation steps:
C2H6 + •Cl → C2H5• + HCl
C2H5• + Cl2 → C2H5Cl + •Cl
\nOne termination step:
C2H5• + C2H5• → C4H10
OR
C2H5• + •Cl → C2H5Cl
OR
•Cl + •Cl → Cl2
\n
Accept radical without • if consistent throughout.
\nAllow ECF from incorrect radicals produced in propagation step for M3.
\n[3 marks]
\n= 2.021 AND = 4.04 AND
\n«hence» CH2Cl
\n\n
Accept : :
\nDo not accept C2H4Cl2.
\nAward [2] for correct final answer.
\n[2 marks]
\nmolecular ion peak(s) «about» m/z 100 AND «so» C2H4Cl2 «isotopes of Cl»
\ntwo signals «in 1HNMR spectrum» AND «so» CH3CHCl2
OR
«signals in» 3:1 ratio «in 1HNMR spectrum» AND «so» CH3CHCl2
OR
one doublet and one quartet «in 1HNMR spectrum» AND «so» CH3CHCl2
1,1-dichloroethane
\n\n
Accept “peaks” for “signals”.
\nAllow ECF for a correct name for M3 if an incorrect chlorohydrocarbon is identified
\n[3 marks]
\n\n
Continuation bonds must be shown.
\nIgnore square brackets and “n”.
\nAccept ![\"M17/4/CHEMI/SP2/ENG/TZ1/05.d/M\"](\"../assets/uploads/tinymce_asset/asset/3721/Schermafbeelding_2017-09-25_om_08.18.53.png\"/)
.
Accept other versions of the polymer, such as head to head and head to tail.
\nAccept condensed structure provided all C to C bonds are shown (as single).
\n[1 mark]
\nThe production of many pharmaceutical drugs involves the use of solvents.
\nSuggest one problem associated with chlorinated organic solvents as chemical waste.
\nSuggest how the principles of green chemistry can be used to solve the environmental problems caused by organic solvents.
\nAny of:
\n«most are» toxic «to living organisms»
OR
incomplete combustion/incineration can produce toxic products/dioxins/phosgene
OR
carcinogenic
«some can be» greenhouse gases
\nozone-depleting
\ncan contribute to formation of «photochemical» smog
\naccumulate in groundwater
OR
have limited biodegradability
cost/hazards of disposal
\n\n
Do not accept “harmful to the environment”.
\nDo not accept just “pollutes water”.
\n[1 mark]
\nuse organic solvent-free synthetic methods
OR
use water as a solvent
OR
based on atom economy
OR
recover/reuse solvents
[1 mark]
\nPolymer nanocomposites often have better structural performance than conventional materials. Lithographic etching and metal coordination are two methods of assembling these nanocomposites.
\nDendrimers are highly branched nanoparticles with a wide range of usage. One such dendrimer is PAMAM, or polyamidoamine.
\nThe first step in the synthesis is to make the core by reacting ethane-1,2-diamine with methylpropenoate.
\nEstimate the atom economy of this first step.
\nSuggest, giving one reason, whether this is an addition or condensation reaction.
\nSubsequent steps proceed under differing conditions, forming the dendrimer polymer with the following repeating unit.
\nState the name of one functional group in this repeating unit.
\n100%
\n\n
Accept “almost 100%” if a catalyst is referred to.
\n[1 mark]
\naddition AND no atoms removed/all atoms accounted for/no loss of water/ammonia/inorganic by-product/small molecules
OR
addition AND there is only one «reaction» product
[1 mark]
\namido
OR
amino
\n
Accept “amide/carboxamide/carbamoyl” for “amido”.
\nAccept “amine“ for “amino”.
\nAccept “carbonyl”.
\n[1 mark]
\nOxidation and reduction reactions can have a variety of commercial uses.
\nA student decides to build a voltaic cell consisting of an aluminium electrode, Al (s), a tin electrode, Sn (s), and solutions of aluminium nitrate, Al(NO3)3 (aq) and tin(II) nitrate, Sn(NO3)2 (aq).
\nElectron flow is represented on the diagram.
\nLabel each line in the diagram using section 25 of the data booklet.
\nWrite the equation for the expected overall chemical reaction in (a).
\nAl/aluminium «electrode» AND aluminium nitrate/Al(NO3)3/Al3+ on left ✓
\nSn/tin «electrode» AND tin«(II)» nitrate/Sn(NO3)2/Sn2+ on right ✓
\nsalt bridge AND voltmeter/V/lightbulb ✓
\n
Award [1] if M1 and M2 are reversed.
Award [1] for two correctly labelled solutions OR two correctly labelled electrodes for M1 and M2.
\nAccept a specific salt for “salt bridge”.
\nAccept other circuit components such as ammeter/A, fan, buzzer, resistor/heating element/R/Ω.
\n3Sn2+ (aq) + 2Al (s) → 3Sn (s) + 2Al3+ (aq)
OR
3Sn(NO3)2 (aq) + 2Al (s) → 3Sn (s) + 2Al(NO3)3 (aq) ✓
If half cells are reversed in part-question (a) then the equation must be reversed to award the mark.
Do not penalize equilibrium arrows.
\nOrganic chemistry can be used to synthesize a variety of products.
\nCombustion analysis of an unknown organic compound indicated that it contained only carbon, hydrogen and oxygen.
\nSeveral compounds can be synthesized from but-2-ene. Draw the structure of the final product for each of the following chemical reactions.
\nDetermine the change in enthalpy, ΔH, for the combustion of but-2-ene, using section 11 of the data booklet.
\nCH3CH=CHCH3 (g) + 6O2 (g) → 4CO2 (g) + 4H2O (g)
\nWrite the equation and name the organic product when ethanol reacts with methanoic acid.
\nOxidation of ethanol with potassium dichromate, K2Cr2O7, can form two different organic products. Determine the names of the organic products and the methods used to isolate them.
\nDeduce two features of this molecule that can be obtained from the mass spectrum. Use section 28 of the data booklet.
\nNIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce
on behalf of the United States of America. All rights reserved.
Identify the bond responsible for the absorption at A in the infrared spectrum. Use section 26 of the data booklet.
\nNIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce
on behalf of the United States of America. All rights reserved.
Deduce the identity of the unknown compound using the previous information, the 1H NMR spectrum and section 27 of the data booklet.
\nSDBS, National Institute of Advanced Industrial Science and Technology (AIST).
\nPenalize missing hydrogens in displayed structural formulas once only.
\nAccept condensed structural formulas: CH3CH(OH)CH2CH3 / CH3CH2CH2CH3 or skeletal structures.
\nBonds broken:
2(C–C) + 1(C=C) + 8(C–H) + 6O=O / 2(346) + 1(614) + 8(414) + 6(498) / 7606 «kJ» ✓
Bonds formed:
8(C=O) + 8(O–H) / 8(804) + 8(463) / 10 136 «kJ» ✓
Enthalpy change:
«Bonds broken – Bonds formed = 7606 kJ – 10 136 kJ =» –2530 «kJ» ✓
Award [2 max] for «+» 2530 «kJ».
Award [3] for correct final answer.
\nEquation:
CH3CH2OH + HCOOH HCOOCH2CH3 + H2O ✓
Product name:
ethyl methanoate ✓
Accept equation without equilibrium arrows.
\nAccept equation with molecular formulas (C2H6O + CH2O2 C3H6O2 + H2O) only if product name is correct.
\nethanal AND distillation ✓
\nethanoic acid AND reflux «followed by distillation» ✓
\n\n
Award [1 max] for both products OR both methods.
\nm/z 58:
molar/«relative» molecular mass/weight/Mr «is 58 g mol−1/58» ✓
m/z 43:
«loses» methyl/CH3 «fragment»
OR
COCH3+ «fragment» ✓
Do not penalize missing charge on the fragments.
Accept molecular ion «peak»/ CH3COCH3+/C3H6O+.
\nAccept any C2H3O+ fragment/ CH3CH2CH2+/C3H7+.
\nC=O ✓
\n
Accept carbonyl/C=C.
Information deduced from 1H NMR:
\n«one signal indicates» one hydrogen environment/symmetrical structure
OR
«chemical shift of 2.2 indicates» H on C next to carbonyl ✓
Compound:
propanone/CH3COCH3 ✓
\n
Accept “one type of hydrogen”.
Accept .
The diagram shows the energy profile for a catalysed and uncatalysed reaction.
Which represents the enthalpy change, ΔH, and the activation energy, Ea, for the catalysed reaction?
A
\nExcess magnesium powder was added to a beaker containing hydrochloric acid, HCl (aq).
\nThe mass of the beaker and its contents was recorded and plotted against time (line I).
\nWhich change could give line II?
\nA. Doubling the mass of powdered Mg
\nB. Using the same mass of Mg ribbon
\nC. Increasing the temperature
\nD. Using the same volume of more concentrated HCl
\nD
\nLipids and carbohydrates contain the same elements but have different properties.
\nList the building blocks of triglycerides and carbohydrates.
\nThe drain pipe of a kitchen sink can become clogged by fatty acids, such as linoleic acid, C18H32O2, but not by the trisaccharide, raffinose, C18H32O16, containing the same number of carbon atoms.
\nExplain why raffinose is far more water soluble than linoleic acid.
\nSolid fat triglycerides can also clog kitchen sink drains.
\nExplain how sodium hydroxide unblocks the drain.
\nThe amount of proteins, fats and carbohydrates determine the energy content of foods.
\n
Explain why linoleic acid, C18H32O2, is a more efficient energy storage molecule than raffinose, C18H32O16.
Triglycerides:
organic acid/fatty acid and glycerol/propane-1,2,3-triol
AND
\nCarbohydrates:
monosaccharides
\n
Accept simple sugars.
\n[1 mark]
\n«water/aqueous solubility depends on forming many» H-bonds with water
\nraffinose has many hydroxyl/O–H/oxygen atoms/O «and forms many H-bonds» AND linoleic acid has few/one hydroxyl/O–H/oxygen atom/O/carboxyl group/ COOH/is largely non-polar «and cannot form many H-bonds»
\n\n
Accept statement which implies comparison.
\n[2 marks]
\n«base» hydrolysis/saponification
OR
«produces glycerol and» soap/salt of the «fatty» acid
«products are» water soluble «and drain away»
\n\n
Accept condensed formulas.
\nAccept non-balanced equation.
\nAccept “RCOONa”.
\n[2 marks]
\nlinoleic acid/C18H32O2 combustion/oxidation more exothermic «per mol»
\nlinoleic acid/C18H32O2 has lower proportion/number of O atoms
OR
linoleic acid/C18H32O2 is less oxidized
\n
Accept converse arguments.
\n[2 marks]
\nSugars exist in both straight chain and ring forms.
\nBiodegradable plastics produced from starch present one solution to the environmental problem created by the use of large quantities of plastics.
\nDeduce the straight chain structure of ribose from its ring structure drawn in section 34 of the data booklet.
\nUsing the partial structure given, complete the structural formula of the molecule formed from the condensation of two cyclic -glucose molecules.
\n![\"M17/4/CHEMI/SP3/ENG/TZ1/12.a.ii\"](\"../assets/uploads/tinymce_asset/asset/3724/Schermafbeelding_2017-09-25_om_11.45.47.png\"/)
Constructing models that allow visualizations of the stereochemistry of carbohydrates is essential to understand their structural roles in cells.
\nDescribe how Haworth projections help focus on the position of attached groups.
\nState one advantage of starch based polymers besides being biodegradable.
\nBiodegradable boxes made from polylactic acid, PLA, disintegrate when exposed to water.
\nState the formula of the product formed when water reacts with PLA.
\n\n
All OH groups must be on the same side.
\nAccept structures with chiral carbon atoms shown as C or C* instead of crosses.
\n[1 mark]
\n![\"M17/4/CHEMI/SP3/ENG/TZ1/12.a.ii/M\"](\"../assets/uploads/tinymce_asset/asset/3725/Schermafbeelding_2017-09-25_om_11.47.34.png\"/)
\n
Accept –O– in a straight line provided both H’s are above the plane.
\n[1 mark]
\n«allow» 3-D perspective of structures «of cyclic monosaccharide molecules»
OR
«show» cis/same side arrangement of «attached» groups
OR
«show» trans/opposite side arrangement of «attached» groups
OR
«make» carbon and hydrogen implicit
[1 mark]
\nabundant/renewable/allows use of «local» vegetation
OR
less use of fossil fuel/oil based plastics
OR
air permeable/better breathing of products
OR
«can be» mixed/blended with synthetic polymers
\n
Do not accept answers related to biodegradable examples.
\nIgnore any reference to cost.
\nAccept “carbon neutral/do not contribute to global warming”.
\nAccept “require less energy to produce”.
\nAccept “do not produce toxic products”.
\n[1 mark]
\nHO–CH(CH3)–COOH/CH3CH(OH)COOH
\n\n
Do not accept C3H6O3.
\nDo not accept OH–CH(CH3)–COOH.
\n[1 mark]
\nPeptidase enzyme in the digestive system hydrolyses peptide bonds.
\nA tripeptide Ala-Asp-Lys was hydrolysed and electrophoresis of the mixture of the amino acids was carried out at a pH of 6.0. Refer to section 33 of the data booklet.
\nIdentify the type of metabolic process that occurs in the hydrolysis of the peptide during digestion.
\nIdentify the name of the amino acid that does not move under the influence of the applied voltage.
\nDeduce, giving a reason, which amino acid will develop closest to the negative electrode.
\nThe breakdown of a dipeptide in the presence of peptidase was investigated between 18 °C and 43 °C. The results are shown below.
\nComment on the rate of reaction at temperature X in terms of the enzyme’s active site.
\nThe solubility of a vitamin depends on its structure.
\nIdentify the vitamin given in section 35 of the data booklet that is the most soluble in water.
\nPollution from heavy metal ions has become a health concern.
\nOutline how the presence of heavy metal ions decreases the action of enzymes.
\nOutline how lead ions could be removed from an individual suffering from lead poisoning.
\ncatabolism/catabolic
\n[1 mark]
\nalanine
\n\n
Do not accept ala.
\n[1 mark]
\nLys/lysine
\npH «buffer» < pI «Lys»
OR
buffer more acidic than Lys «at isoelectric point»
OR
«Lys» exists as
OR
\n«Lys» charged positively/has +1/1+ «overall» charge «and moves to negative electrode»
\n\n
Do not apply ECF from M1.
\nAccept converse argument.
\nDo not accept just “has H3N+ group” for M2 (as H3N+ is also present in zwitterion).
\nDo not penalize if COOH is given in the structure of lysine at pH 6 instead of COO–.
\n[2 marks]
\nhighest frequency of successful collisions between active site and substrate
OR
highest frequency of collisions between active site and substrate with sufficient energy/ AND correct orientation/conformation
OR
optimal shape/conformation of the active site «that matches the substrate»
OR
best ability of the active site to bind «to the substrate»
\n
Accept “number of collisions per unit time” for “frequency”.
\nDo not accept “all active sites are occupied”.
\n[1 mark]
\nascorbic acid/vitamin C
\n[1 mark]
\nreact/bind/chelate with enzyme
OR
disrupt ionic salt bridges
OR
affect shape of tertiary/quaternary structures
OR
precipitate enzymes
OR
break/disrupt disulfide bridges/bonds
\n
Do not accept “changes shape of active site” by itself.
\n[1 mark]
\n«use of» host-guest chemistry
OR
chelation «therapy»
\n
Accept specific medication/chelating agent such as EDTA, CaNa2 EDTA, succimer, D-penicillamine, dimercaprol.
\n[1 mark]
\nWhat will happen if the pressure is increased in the following reaction mixture at equilibrium?
\nCO2 (g) + H2O (l) H+ (aq) + HCO3− (aq)
\nA. The equilibrium will shift to the right and pH will decrease.
\nB. The equilibrium will shift to the right and pH will increase.
\nC. The equilibrium will shift to the left and pH will increase.
\nD. The equilibrium will shift to the left and pH will decrease.
\nA
\n10.0 cm3 of an aqueous solution of sodium hydroxide of pH = 10 is mixed with 990.0 cm3 of distilled water. What is the pH of the resulting solution?
\nA. 8
\nB. 9
\nC. 11
\nD. 12
\nA
\nCarbon is produced by fusion reactions in stars.
\nThe main fusion reaction responsible for the production of carbon is:
\nX +
\nOutline how the spectra of light from stars can be used to detect the presence of carbon.
\nDeduce the identity of X.
\nOutline why this reaction results in a release of energy.
\nNuclear fusion reactors are predicted to become an important source of electrical energy in the future. State two advantages of nuclear fusion over nuclear fission.
\npresence of dark/absorption lines corresponding to those found for carbon
OR
missing wavelengths/frequencies correspond to carbon’s spectrum
\n
Accept “presence of characteristic dark lines”.
\nDo not accept answer in terms of emission spectra.
\n[1 mark]
\n\n
[1 mark]
\nproduct «nucleus» has a greater binding energy «per nucleon than reacting nuclei»
\n\n
Accept “mass of the products is less than mass of the reactants”.
\nAccept converse arguments.
\n[1 mark]
\nfuel more abundant/cheaper
\nno «long half-life» radioisotopes/radioactive waste
\nshipment of radioactive fuels not required
\nplutonium/nuclear weapons cannot be produced from products
\nnuclear disasters less likely «as no critical mass of fuel required»
\nhigher specific energy/energy per g/kg/unit mass than fission
\n\n
Do not accept simply “fusion produces more energy than fission”.
\n[2 marks]
\nVegetable oils, such as that shown, require conversion to biodiesel for use in current internal combustion engines.
\nState two reagents required to convert vegetable oil to biodiesel.
\nDeduce the formula of the biodiesel formed when the vegetable oil shown is reacted with the reagents in (a).
\nExplain, in terms of the molecular structure, the critical difference in properties that makes biodiesel a more suitable liquid fuel than vegetable oil.
\nDetermine the specific energy, in kJg−1, and energy density, in kJcm−3, of a particular biodiesel using the following data and section 1 of the data booklet.
\nDensity = 0.850 gcm−3; Molar mass = 299 gmol−1;
\nEnthalpy of combustion = 12.0 MJmol−1.
\nmethanol
OR
ethanol
strong acid
OR
strong base
\n
Accept “alcohol”.
\nAccept any specific strong acid or strong base other than HNO3/nitric acid.
\n[3 marks]
\nCH3(CH2)16COOCH3 / CH3OCO(CH2)16CH3
OR
CH3(CH2)16COOC2H5 / C2H5OCO(CH2)16CH3
\n
Product must correspond to alcohol chosen in (a), but award mark for either structure if neither given for (a).
\n[1 mark]
\nlower viscosity
\nweaker intermolecular/dispersion/London/van der Waals’ forces
OR
smaller/shorter molecules
\n
Accept “lower molecular mass/Mr” or “lower number of electrons”.
\nAccept converse arguments.
\n[2 marks]
\nSpecific energy: «» = 40.1 «kJ g−1»
\nEnergy density: «= 40.1 kJg−1 x 0.850 gcm−3» = 34.1 «kJcm−3»
\n\n
Award [1] if both are in terms of a unit other than kJ (such as J or MJ).
\n[2 marks]
\nCoal is often converted to liquid hydrocarbon fuels through initial conversion to carbon monoxide and hydrogen.
\nState how these gases are produced, giving the appropriate equation(s).
\nOutline how the carbon monoxide is then converted to a hydrocarbon fuel.
\nheat/react with «oxygen and» water/steam
\nC + H2O → CO + H2
OR
3C + O2 + H2O → H2 + 3CO
OR
2C + O2 → 2CO AND C + H2O → H2 + CO
OR
C + O2 → CO2 AND C + CO2 → 2CO AND C + H2O → H2 + CO
\n
M1 requires concept of heat.
\n[2 marks]
\n«Fischer-Tropsch» catalytic reduction of carbon monoxide with hydrogen
OR
(2n + 1)H2 + nCO → CnH(2n + 2) + nH2O
OR
reduction of carbon monoxide to methanol AND catalytic dehydration
OR
2H2 + CO + CH3OH AND nCH3OH → CnH2n + nH2O
\n
If equation is given for a specific alkane or alkene, it must be a liquid (n > 4).
\n[1 mark]
\nA link between the combustion of fossil fuels and an increase in the temperature of the Earth’s atmosphere was proposed over a century ago.
\nSuggest why it is only in recent years that specific predictions of the future effects of fossil fuel combustion have been made.
\nCarbon dioxide has two different bond stretching modes illustrated below.
\n![\"M17/4/CHEMI/SP3/ENG/TZ1/17.b\"](\"../assets/uploads/tinymce_asset/asset/3726/Schermafbeelding_2017-09-25_om_13.09.53.png\"/)
Predict, with an explanation, whether these stretching modes will absorb infrared radiation.
\nOutline, giving the appropriate equation(s), how increasing levels of carbon dioxide will affect the pH of the oceans.
\nMany combustion processes also release particulate matter into the atmosphere. Suggest, giving your reason, how this might affect the temperature of the Earth’s surface.
\ncomputers can now carry out more complex calculations
OR
better understanding of the interactions between the various systems involved
OR
clear evidence of global warming
OR
«reliable» global temperature data now available
OR
techniques have been available to monitor carbon dioxide levels
\n
Accept “better/faster computers”.
\nAccept “better modelling”.
\nAccept “better/more reliable/consistent data”.
\nAccept “better measuring techniques”.
\nAccept other scientifically based (not politically based) reasons.
\nAccept if specific relevant data is given.
\nDo not accept “increased combustion of fossil fuels” or “increased concerns about global warming”.
\n[1 mark]
\nsymmetric stretching will not absorb IR
OR
asymmetric stretching will absorb IR
change in polarity/dipole «moment» required «to absorb IR»
\n[2 marks]
\nCO2(aq) + H2O(l) H+(aq) + HCO3–(aq) «and pH decreases»
OR
CO2(aq) + H2O(l) H2CO3(aq) AND H2CO3(aq) H+(aq) + HCO3–(aq) «and pH decreases»
\n
Accept reversible or non-reversible arrows for all.
\n[1 mark]
\nreduce it AND absorbing/reflecting sunlight
\n\n
Accept “reduce it because of global dimming”.
\nAccept “reduce it AND blocking sunlight”.
\n[1 mark]
\nWhich statement is incorrect for a 0.10 mol dm–3 HCOOH solution?
\nA. pH = 1
\nB. [H+] << 0.10 mol dm–3
\nC. [HCOO–] is approximately equal to [H+]
\nD. HCOOH is partially ionized
\nA
\nWhat are the oxidation states of chromium in (NH4)2Cr2O7 (s) and Cr2O3 (s)?
\nB
\nWhich of the following is a redox reaction?
\nA. 3Mg (s) + 2AlCl3 (aq) → 2Al (s) + 3MgCl2 (aq)
\nB. SiO2 (s) + 2NaOH (aq) → Na2SiO3 (aq) + H2O (l)
\nC. KCl (aq) + AgNO3 (aq) → AgCl (s) + KNO3 (aq)
\nD. 2NaHCO3 (aq) → Na2CO3 (aq) + CO2 (g) + H2O (l)
\nA
\nMedicines have a variety of different effects on the body and act at the molecular level.
\nMorphine and codeine are strong analgesics. Their structures are given in section 37 of the data booklet.
\nDose response curves are determined for each drug.
\n![\"M17/4/CHEMI/SP3/ENG/TZ1/XX\"](\"../assets/uploads/tinymce_asset/asset/3727/Schermafbeelding_2017-09-25_om_13.15.24.png\"/)
Outline the significance of range “a”.
\nSuggest the type of reaction used to convert morphine to codeine.
\nState and explain the action of opiates as painkillers.
\n«measures» therapeutic window/margin «of a drug»
OR
range of doses that produce a therapeutic effect without causing toxic effects
\n
Accept “difference between ED50/minimum effective/therapeutic dose «for 50% of population» AND TD50 /toxic dose «for 50% of population»”.
\nDo not accept “therapeutic index”.
\nDo not accept “lethal/fatal dose” as this is not LD50.
\n[1 mark]
\n«nucleophilic» substitution/SN
\n\n
Accept “methylation”.
\n[1 mark]
\nwork directly on opioid/pain receptors «in brain»
\nsuppress pain impulses in brain/CNS
\nresemble endorphins/enkephalins/natural chemical painkillers «produced in the brain and spinal cord»
\n\n
Do not award mark for “resemble hormones”.
\n[2 marks]
\nSolubility plays an important role in the bioavailability of drugs in the body.
\nSuggest why aspirin is slightly soluble in water. Refer to section 37 of the data booklet.
\nFormulate an equation for the conversion of aspirin to a more water soluble derivative.
\nA student prepares aspirin from salicylic acid in the laboratory, extracts it from the reaction mixture, ensures the sample is dry and determines its melting point.
\nSuggest why the melting point of the student’s sample is lower and not sharp compared to that of pure aspirin.
\nOrganic molecules can be characterized using infrared (IR) spectroscopy.
\nCompare and contrast the infrared peaks above 1500 cm−1 in pure samples of aspirin and salicylic acid using section 26 of the data booklet.
\nThe pharmaceutical industry is one of the largest producers of waste solvents.
\nState a green solution to the problem of organic solvent waste.
\npresence of «large» benzene/arene ring AND non-polar/hydrophobic
OR
presence of «large» benzene/arene ring AND cannot form H-bond with water
contain COOH/carboxyl/–OH/hydroxyl «and ester group» AND polar/hydrophilic
OR
contain COOH/carboxyl/–OH/hydroxyl «and ester group» AND can form H-bonds with water
\n
Accept “phenyl” for “benzene ring”.
\nAccept \"carboxylic acid\" for \"carboxyl\".
\nDo not accept \"alcohol\" for \"hydroxyl\".
\n[2 marks]
\nOR
C6H4(OCOCH3)COOH + NaOH → C6H4(OCOCH3)COONa + H2O
\n
Charges (O– and Na+) not necessary to score the mark.
\nAccept net ionic equation.
\nAccept any strong base in place of NaOH.
\n[1 mark]
\n«student’s» sample impure
\nlattice disrupted/not uniform «due to presence of impurities»
OR
fewer interparticle/intermolecular forces «due to presence of impurities»
\n
Accept converse arguments.
\n[2 marks]
\nOne similarity:
peak at 2500–3000 «cm–1»/peak due to O–H/hydroxyl in carboxylic acids
OR
peak at 1700–1750 «cm–1»/peak due to C=O/carbonyl
OR
peak at 2850–3090 «cm–1»/peak due to C–H of arene
One difference:
peak at 3200–3600 «cm–1» in salicylic acid/ peak due to O–H in phenol in salicylic acid
OR
«two» peaks at 1700–1750 «cm–1» in aspirin AND one peak «in the same area» in salicylic acid
\n
Accept “peak at 1600 cm–1 for arene/benzene ring” – not in the data booklet.
\nAccept “2500–3600 cm–1 «overlapping absorptions of two O–H» in salicylic acid”.
\nAccept “stronger/broader/split peak at 1700–1750 cm–1 in aspirin”.
\n[2 marks]
\n«use of» alternative solvents such as supercritical/liquid CO2
OR
use of water «as solvent»
OR
solvent-free reactions «for example, polymerization of propene»
OR
solid-state chemistry
OR
recycle «waste» solvents
OR
catalysis that leads to better/higher yield
OR
reducing number of steps
\n
Do not accept political/regulatory solutions.
\n“catalysis” not sufficient for mark.
\n[1 mark]
\nWhat is the reaction type and major product at the anode (positive electrode) when molten sodium chloride is electrolysed using platinum electrodes?
\nB
\nThe buffer formed by carbon dioxide, CO2(aq) and hydrogen carbonate ion, HCO3−(aq), plays an important role in maintaining the pH of blood.
\nCalculate the pH of the buffer from the following data and section 1 of the data booklet.
\npKa(CO2) = 6.34
\n[HCO3−(aq)] = 1.40 × 10−2 moldm−3
\n[CO2(aq)] = 1.25 × 10−3 moldm−3
\nExplain the effect of a large amount of aspirin on the pH of blood.
\n«pH = pKa + log10 » = 7.39
\n[1 mark]
\nH+ from aspirin reacts with HCO3− to form CO2 and H2O
OR
H+(aq) + HCO3−(aq) CO2(aq) + H2O(l)
OR
reverse reaction favoured «to use up some of the H+ added»
pH decreases
\n\n
No mark for “stating aspirin is a weak acid that dissociates partially to produce H+” without reference to reaction with HCO3− or to the equation.
\nReversible arrows not required for the mark.
\nDo not accept “small pH change when small amount of H+ is added”.
\n[2 marks]
\nWhat is the major product of the reaction between HCl and but-2-ene?
\nA. 1,2-dichlorobutane
\nB. 2,3-dichlorobutane
\nC. 1-chlorobutane
\nD. 2-chlorobutane
\nD
\nAntiviral drugs are designed to take different approaches to fighting viruses.
\nOutline how oseltamivir (Tamiflu®) works.
\nOseltamivir was commercially produced from shikimic acid, a precursor which is a metabolite in micro-organisms and plants.
\nOutline how green chemistry was used to develop the precursor for oseltamivir in order to overcome a shortage of the drug during the flu season.
\nSuggest why the administration of antibiotics to humans and animals can affect the environment.
\n«drug» blocks/inhibits «viral» enzyme/neuraminidase/NA «activity»
\nprevents virus from leaving/escaping host cells «thus cannot infect other cells»
\n[2 marks]
\nALTERNATIVE 1:
«using» genetically modified/GM E. Coli/bacteria/microorganisms
E. Coli/bacteria biosynthesis
OR
E. Coli/bacteria «overfed by glucose» undergo fermentation
OR
cells of the bacteria «are broken down to» form precursor/shikimic acid
ALTERNATIVE 2:
use readily available cyclic ester/lactone
forms «the correct stereoisomer of oseltamivir» in a shorter number of chemical steps
\n\n
Do not accept “planting more Chinese star anise” or “other plant sources of shikimic acid”.
\n[2 marks]
\n«can develop antibiotic» resistance in bacteria/microorganisms
OR
changes in microbial/bacterial population
\n
Accept secondary effects, such as reduced biodiversity of aquatic/soil ecosystems, denitrification of soil (due to decline in nitrogen-fixing bacteria). No mark for just stating “water contamination”.
\nNo mark for just stating “failure of aquatic/marine environment”.
\n[1 mark]
\nWhich compound can be oxidized when heated with an acidified solution of potassium dichromate(VI)?
\nA. CH3C(O)CH2CH3
\nB. CH3CH2CH(OH)CH3
\nC. (CH3)3COH
\nD. CH3(CH2)2COOH
\nB
\nWhat is the name of this compound, using IUPAC rules?
\nA. 3-methylbutan-3-ol
\nB. 2-ethylpropan-2-ol
\nC. 2-methylbutan-2-ol
\nD. 3-methylbutan-2-ol
\nC
\nAntimony oxide is widely used as a homogeneous catalyst for the reaction of benzene-1,4-dicarboxylic acid with ethane-1,2-diol in the production of polyethylene terephthalate (PETE).
\nDeduce the repeating unit of the polymer and the other product of the reaction.
\nState the class of polymer to which PETE belongs.
\nRepeating unit:
\nOther product: water/H2O
\n\n
Continuation bonds necessary for the mark.
\nAccept alternative repeating unit with O at other end.
\nDo not penalize square brackets or n.
\n[2 marks]
\ncondensation
\n\n
Accept polyester or thermoplastic.
\n[1 mark]
\nWhich pair of molecules has the same bond angles?
\nA. PCl3 and BCl3
\nB. SO2 and CO2
\nC. H2O and NH3
\nD. CCl4 and SiH4
\nD
\nThe C=N bond has a bond length of 130 pm and an average bond enthalpy of 615kJmol-1. Which values would be most likely for the C-N bond?
\nA
\nLanthanum has a hexagonal close packed (hcp) crystal structure. State the coordination number of each lanthanum atom.
\nLanthanum becomes superconducting below 5 K. Explain, in terms of Bardeen–Cooper–Schrieffer (BCS) theory, how superconductivity occurs.
\nOutline why superconductivity only occurs at low temperatures.
\ntwelve/12
\n[1 mark]
\n«moving» electron attracts «nearby» positive charges/ions/cations
\ncreates «local» regions of increased positive charge
\npositive charge/field attracts second electron «with opposite spin»
\ntwo electrons form a Cooper pair
\n«all» Cooper pairs «in sample» interact/form «electron» condensate
\n«electron» condensate/Cooper pairs move/flow «through sample» freely/without resistance
\n[3 marks]
\nreduces the band gap to zero
OR
«at high temperatures» thermal motion disrupts the formation of Cooper pairs
[1 mark]
\nAntimony and its compounds are toxic, so it is important to check that the catalyst is removed from the final product. One technique to detect antimony is Inductively Coupled Plasma Mass Spectroscopy (ICP-MS).
\nOutline the nature of the plasma state and how it is produced in ICP-MS.
\nHydrogen sulfide could be used to remove antimony(III) ions from a solution.
\nDetermine the concentration of antimony(III) ions that would be required to precipitate antimony(III) sulfide in a solution saturated with hydrogen sulfide.
\n[S2−] in water saturated with hydrogen sulfide = 1.0 × 10−14 mol dm−3
\nKsp (Sb2S3) = 1.6 × 10−93
\nIdentify a ligand that could be used to chelate antimony(III) ions in solution.
\nelectrons AND positive ions «in gaseous state»
\nhigh frequency/alternating current passed through argon
OR
«oscillating» electromagnetic/magnetic field
OR
high frequency radio waves
\n
Accept “gas” instead of “argon”.
\n[2 marks]
\nKsp = [Sb3+]2 [S2−]3
\n[Sb3+]2 (10−14)3 = 1.6 x 10−93
\n[Sb3+] «» = 4.0 x 10−26 «mol dm−3»
\n\n
Award [3] for correct final answer.
\n[3 marks]
\nEDTA/ethylenediaminetetraacetic aci
OR
H2N–CH2–CH2–HN2/ethane-1,2-diamine
\n
Accept “EDTA4–”.
\nAccept other chelating agents.
\n[1 mark]
\nWhich type of reaction occurs between an alcohol and a carboxylic acid?
\nA. Addition
\nB. Oxidation
\nC. Esterification
\nD. Polymerization
\nC
\nHow many structural isomers of C6H14 exist?
\nA. 4
\nB. 5
\nC. 6
\nD. 7
\nB
\nMaterials science involves understanding the properties of materials and applying those properties to desired structures.
\nMagnesium oxide, MgO, and silicon carbide, SiC, are examples of ceramic materials. State the name of the predominant type of bonding in each material.
\nPredict the predominant type of bonding for a binary compound AB in which the electronegativity of both atoms is low. Use section 29 of the data booklet.
\nMgO: ionic AND SiC: covalent
\nAccept “covalent network/network covalent” for “covalent” but not just “network”.
\nmetallic «bonding»
\nWhat information is provided by 1H NMR, MS and IR for an organic compound?
\nI. 1H NMR: chemical environment(s) of protons
II. MS: fragmentation pattern
III. IR: types of functional group
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nBetween which pair of molecules can hydrogen bonding occur?
\nA. CH4 and H2O
B. CH3OCH3 and CF4
C. CH4 and HF
D. CH3OH and H2O
D
\nIn which reaction do the reactants have a lower potential energy than the products?
\nA. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
B. HBr(g) → H(g) + Br(g)
C. Na+(g) + Cl-(g) → NaCl(s)
D. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
B
\n5.35g of solid ammonium chloride, NH4Cl(s), was added to water to form 25.0g of solution. The maximum decrease in temperature was 14 K. What is the enthalpy change, in kJmol-1, for this reaction? (Molar mass of NH4Cl = 53.5gmol-1; the specific heat capacity of the solution is 4.18 Jg-1K-1)
\nA.
\nB.
\nC.
\nD.
\nC
\nFor the reaction R → P, which letter represents the activation energy for the catalysed reverse reaction?
\nC
\nHydrazine reacts with oxygen.
\nN2H4(l) + O2(g) → N2(g) + 2H2O(l) ΔHθ = -623 kJ
\nWhat is the standard enthalpy of formation of N2H4(l) in kJ? The standard enthalpy of formation of H2O(l) is -286 kJ.
\nA. -623 - 286
B. -623 + 572
C. -572 + 623
D. -286 + 623
C
\nWhich experimental methods could be used to observe the progress of the following reaction?
\nCr2O72-(aq) + 6I-(aq) + 14H+(aq) → 2Cr3+(aq) + 3I2(aq) + 7H2O(l)
\nI. Change in colour
II. Change in mass
III. Change in electrical conductivity
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nWhat happens when the temperature of the following equilibrium system is increased?
\nCO(g) + 2H2(g) CH3OH(g) ΔHθ = -91kJ
A
\nA student wanted to determine the formula of indium sulfate. She applied an electrical current of 0.300A to an aqueous solution of indium sulfate for 9.00 × 103 s and found that 1.07 g of indium metal deposited on the cathode.
\nCalculate the charge, in coulombs, passed during the electrolysis.
\n\n
Calculate the amount, in mol, of electrons passed using section 2 of the data booklet.
\nCalculate the mass of indium deposited by one mole of electrons.
\nCalculate the number of moles of electrons required to deposit one mole of indium. Relative atomic mass of indium, Ar=114.82.
\nDeduce the charge on the indium ion and the formula of indium sulfate.
\n«0.300A × 9.00 × 103 s =» 2.70 × 103 «C»
\n«mol e− = =» 2.80 × 10−2 «mol»
\n«=» 38.2 «g»
\n« e− =» 3.01/3.00 «mol e−»
\nIn3+ /3+ AND In2(SO4)3
\nDo not accept “+3/3”
\nA student performs an acid-base titration using a pH meter, but forgets to calibrate it. Which type of error will occur and how will it affect the quality of the measurements?
\nA. Random error and lower precision
\nB. Systematic error and lower accuracy
\nC. Systematic error and lower precision
\nD. Random error and lower accuracy
\nB
\nWhich species behave as Brønsted–Lowry bases in the following reaction?
\nH2SO4 + HNO3 H2NO3+ + HSO4-
A. HNO3 and HSO4-
\nB. HNO3 and H2NO3+
\nC. H2SO4 and HSO4-
\nD. H2NO3+ and HSO4-
\nA
\nWhich solution neutralizes 50.0 cm3 of 0.120 mol dm–3 NaOH (aq)?
\nA. 12.5 cm3 of 0.080 mol dm–3 H3PO4
\nB. 25.0 cm3 of 0.120 mol dm–3 CH3COOH
\nC. 25.0 cm3 of 0.120 mol dm–3 H2SO4
\nD. 50.0 cm3 of 0.060 mol dm–3 HNO3
\nC
\nWhat occurs when solid sodium hydrogen carbonate reacts with aqueous sulfuric acid?
\nA. Bubbles of sulfur dioxide form.
\nB. Bubbles of both hydrogen and carbon dioxide form.
\nC. Bubbles of hydrogen form.
\nD. Bubbles of carbon dioxide form.
\nD
\nA voltaic cell is constructed from zinc and copper half-cells. Zinc is more reactive than copper. Which statement is correct when this cell produces electricity?
\nA. Electrons flow from the copper half-cell to the zinc half-cell.
\nB. The concentration of Cu2+ (aq) increases.
\nC. Electrons flow through the salt bridge.
\nD. Negative ions flow through the salt bridge from the copper half-cell to the zinc half-cell.
\nD
\nWhat is the pressure, in Pa, inside a 1.0 m3 cylinder containing 10 kg of H2 (g) at 25 ºC?
\nR = 8.31 J K–1 mol–1; pV = nRT
\nA.
\nB.
\nC.
\nD.
\nD
\nWhich is a correct statement for the reaction below?
\n2MnO4-(aq) + 6H+(aq) + 5NO2-(aq) → 2Mn2+(aq) + 5NO3-(aq) + 3H2O(l)
\nA. MnO4- is the reducing agent and the oxidation number of Mn increases.
B. MnO4- is the oxidizing agent and the oxidation number of Mn decreases.
C. NO2- is the reducing agent and the oxidation number of N decreases.
D. NO2- is the oxidizing agent and the oxidation number of N increases.
B
\nThe structure of a drug used to treat symptoms of Alzheimer’s disease is shown below. Which functional groups are present in this molecule?
\nA. Hydroxyl and ester
\nB. Hydroxide and ether
\nC. Hydroxyl and ether
\nD. Hydroxide and ester
\nC
\nThe graph represents the first ten ionisation energies (IE) of an element.
\nWhat is the element?
\nA. O
\nB. S
\nC. Ne
\nD. Cl
\nB
\nWhich alcohols are oxidized by acidified potassium dichromate(VI) solution when heated?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nA compound with Mr = 102 contains 58.8 % carbon, 9.80 % hydrogen and 31 % oxygen by mass.
What is its molecular formula?
Ar: C = 12.0; H = 1.0; O = 16.0
\nA. C2H14O4
\nB. C3H4O4
\nC. C5H10O2
\nD. C6H14O
\nC
\nWhich electron configuration is that of a transition metal atom in the ground state?
\nA. [Ne]3s23p64s1
\nB. [Ar]3d9
\nC. 1s22s22p63s23p64s23d104p2
\nD. [Ar]4s13d5
\nD
\nWhich monomer is used to form the polymer with the following repeating unit?
\nA. CH3CH=CHCH3
\nB. CH3CH2CH=CH2
\nC. CH3CH2CH2CH3
\nD. (CH3)2C=CH2
\nA
\nWhat is the hybridization state and electron domain geometry around the circled C, N and O atoms?
\nA
\nWhich type of reaction occurs when methanol and propanoic acid react together in the presence of a catalyst?
\nA. Addition
\nB. Condensation
\nC. Redox
\nD. Neutralization
\nB
\nWhat is the index of hydrogen deficiency (IHD) for this molecule?
\nA. 3
\nB. 4
\nC. 5
\nD. 6
\nD
\nHow many sigma (σ) and pi (π) bonds are present in this molecule?
\nC
\n[CoCl6]3– is orange while [Co(NH3)6]3+ is yellow. Which statement is correct?
\nA. [CoCl6]3– absorbs orange light.
\nB. The oxidation state of cobalt is different in each complex.
\nC. The different colours are due to the different charges on the complex.
\nD. The different ligands cause different splitting in the 3d orbitals.
\nD
\nWhat is always correct about the molecular ion, M+, in a mass spectrum of a compound?
\nA. The M+ ion peak has the smallest m/z ratio in the mass spectrum.
\nB. The m/z ratio of the M+ ion peak gives the relative molecular mass of the molecule.
\nC. The M+ ion is the most stable fragment formed during electron bombardment.
\nD. The M+ ion peak has the greatest intensity in the mass spectrum.
\nB
\nWhich statements are correct for ionic compounds?
\nI. Lattice energy increases as ionic radii increase.
II. Within the same group, the melting point of salts tends to decrease as the radius of the cation increases.
III. Solubility in water depends on the relative magnitude of the lattice energy compared to the hydration energy.
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nWhat is the standard enthalpy of formation, in kJ mol–1, of IF (g)?
\nIF7 (g) + I2 (s) → IF5 (g) + 2IF (g) ΔH = –89 kJ
\nΔH (IF7) = –941 kJ mol–1
\nΔH (IF5) = –840 kJ mol–1
\nA. –190
\nB. –95
\nC. +6
\nD. +95
\nB
\nA student carried out a titration to determine the concentration of an acid and found that his value had good precision but poor accuracy. Which process explains this outcome?
\nA. Consistently overshooting the volume of solution from the burette into the flask.
\nB. Collection of insufficient titration data.
\nC. Reading the meniscus in the burette at a different angle each time.
\nD. Forgetting to rinse the flask after one of the titrations.
\nA
\nThe combustion of glucose is exothermic and occurs according to the following equation:
\nC6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)
\nWhich is correct for this reaction?
\nA
\nWhich equation represents the lattice enthalpy of magnesium sulfide?
\nA. MgS (s) → Mg (g) + S (g)
\nB. MgS (s) → Mg+ (g) + S– (g)
\nC. MgS (s) → Mg2+ (g) + S2– (g)
\nD. MgS (s) → Mg (s) + S (s)
\nC
\nThe enthalpy change for the dissolution of NH4NO3 is +26 kJ mol–1 at 25 °C. Which statement about this reaction is correct?
\nA. The reaction is exothermic and the solubility decreases at higher temperature.
\nB. The reaction is exothermic and the solubility increases at higher temperature.
\nC. The reaction is endothermic and the solubility decreases at higher temperature.
\nD. The reaction is endothermic and the solubility increases at higher temperature.
\nD
\nThe rate expression for the reaction X (g) + 2Y (g) → 3Z (g) is
\nrate = k[X]0 [Y]2
\nBy which factor will the rate of reaction increase when the concentrations of X and Y are both increased by a factor of 3?
\nA. 6
\nB. 9
\nC. 18
\nD. 27
\nB
\nA student measured the change in mass on heating a sample of calcium carbonate, CaCO3(s). What is the mass loss?
\nMass before heating: 2.347 g ± 0.001
Mass after heating: 2.001 g ± 0.001
A. 0.346g ± 0.001
\nB. 0.346g ± 0.002
\nC. 0.35g ± 0.002
\nD. 0.35g ± 0.001
\nB
\nWhich substance has a giant covalent structure?
\nD
\nA period 3 element, M, forms an oxide of the type M2O. Which represents the first four successive ionization energies of M?
\n\n
A
\nWhich representation would be correct for a species, Z, which has 31 protons, 40 neutrons and 28 electrons?
\nA.
\nB.
\nC.
\nD.
\nA
\nWhich property increases down group 17, the halogens?
\nA. Electron affinity
\nB. Boiling point
\nC. First ionization energy
\nD. Reactivity
\nB
\nWhat is the correct explanation for the colour of [Cu(H2O)6]2+?
\nA. Light is absorbed when an electron moves to a d orbital of higher energy.
\nB. Light is released when an electron moves to a d orbital of higher energy.
\nC. Light is absorbed when electrons move from the ligands to the central metal ion.
\nD. Light is absorbed when electrons move between d and s orbitals.
\nA
\nWhich correctly describes the reaction between potassium and excess water?
\nA. The reaction is endothermic.
\nB. The final products of the reaction are potassium oxide and hydrogen.
\nC. The final products of the reaction are potassium hydroxide and hydrogen.
\nD. The final pH of the solution is 7.
\nC
\nWhich pair of statements explains the increase in rate of reaction when the temperature is increased or a catalyst is added?
\nC
\nAt 700 ºC, the equilibrium constant, Kc, for the reaction is 1.075 × 108.
\n2H2 (g) + S2 (g) 2H2S (g)
\nWhich relationship is always correct for the equilibrium at this temperature?
\nA. [H2S]2 < [H2]2 [S2]
\nB. [S2] = 2[H2S]
\nC. [H2S] < [S2]
\nD. [H2S]2 > [H2]2[S2]
\nD
\nSuperconductors are materials that conduct electric current with practically zero resistance.
\nDescribe the Meissner effect.
\nOutline one difference between type 1 and type 2 superconductors.
\ncreation of a mirror image magnetic field of an external field «below the critical temperature/Tc of the superconductor»
OR
expulsion of a magnetic field from a superconductor «below its critical temperature/Tc»
Accept “Type 1: «most» metals AND Type 2: alloys/metal oxide ceramics/perovskites”.
\nThe Fenton and Haber–Weiss reactions convert organic matter in waste water to carbon dioxide and water.
\nCompare and contrast the Fenton and Haber–Weiss reaction mechanisms.
\nAdsorption and chelation are two methods of removing heavy metal ion pollution from the environment.
\n(i) Describe the process of adsorption.
\n(ii) Deduce the structure of the complex ion formed by the reaction of three H2N−CH2−CH2−NH2 chelating molecules with a mercury(II) ion.
\nOne similarity:
both involve hydroxyl/•OH «radicals»
One difference:
\nAccept “hydroxy” for “hydroxyl”.
Do not penalize missing radical symbols if consistent throughout.
Accept “H2O2 → 2•OH” for the Fenton mechanism.
\n
i
molecules/ions/substances are attracted to/form «non-covalent» interactions with the surface of the adsorbent
\n
ii
\nDo not penalize missing charge or square brackets.
Bonds to Hg must be shown (in any format).
Amino acids are usually identified by their common names. Use section 33 of the data booklet.
\nAmino acids act as buffers in solution. In aspartic acid, the side chain (R group) carboxyl has pKa = 4.0. Determine the percentage of the side chain carboxyl that will be ionized (–COO–) in a solution of aspartic acid with pH = 3.0. Use section 1 of the data booklet.
\n«–COOH –COO– + H+ (–COOH = HA ; –COO– = A–)»
pH = pKa + log / 3.0 = 4.0 + log / –1.0 = log
10–1=
\n«percentage ionized/–COO– = × 100 =» 9.1 «%»
Award [3] for correct final answer.
Glucokinase and hexokinase are both enzymes that catalyse the conversion of glucose to glucose-6-phosphate. The enzymes differ, however, in their affinity for the substrate, as shown in the graph below.
\n\n
(i) Estimate the Km values of the two enzymes.
\n(ii) Suggest, with a reason, which enzyme will be more responsive to changes in the concentration of glucose in the blood.
\n(i) Outline what is meant by product inhibition as it applies to hexokinase.
\n(ii) Product inhibition of hexokinase does not affect its Km value. Using this information, deduce the type of binding site that the inhibitor attaches to.
\ni
\nKm hexokinase: approx. 1.7 «mmol dm–3»
AND
Km glucokinase: approx. 8.5 «mmol dm–3»
Accept answers in the range 1.0-2.0 for hexokinase and 7.0-9.0 for glucokinase.
\n\n
ii
\nglucokinase as it is not saturated «with substrate at normal concentration of blood glucose»
OR
glucokinase as its saturation increases with increased glucose concentration in the blood
Accept “at the normal levels of blood glucose concentration, relative velocity of glucokinase still dependent on concentration of glucose”
\ni
\nglucose-6-phosphate lowers enzyme activity/acts as enzyme inhibitor
\n\n
ii
\n«inhibitor binds at» allosteric site
\nAccept “outside/away from active site”.
\nThe structure of DNA (deoxyribonucleic acid) has been studied in many different ways.
\nState the name of the component of DNA responsible for the migration of its fragments to the positive electrode in gel electrophoresis.
\nIn 2010, scientists claimed that they had discovered a species of bacteria capable of incorporating arsenic in place of phosphorus into the bacterial DNA. This claim has since proved controversial. Suggest one technique or evidence that might help support the claim.
\nphosphato/phosphate «group»
\nDo not accept “phosphoric acid”, “phosphorus” or any formula.
\nmass spectrometry / X ray diffraction/crystallography / nuclear magnetic resonance «spectroscopy»
OR
bacteria able to grow in absence of phosphorus
OR
reproducible data
Accept abbreviations (eg, MS, NMR).
Accept “elemental analysis” or “atomic absorption spectroscopy/AA(S)”.
Anthocyanins are pigments that give colour to many flowers and fruits. The red colour of ripe strawberries is mainly due to the anthocyanin pigment whose structure is shown below.
\nOutline why this molecule absorbs visible light.
\nWith reference to its chemical structure, outline whether this pigment is found in aqueous solution in the cells or in the lipid-based membranes.
\nA student investigated the ability of anthocyanins to act as pH indicators. He extracted juice from blackberries and used a UV-vis spectrophotometer to produce absorption spectra at different pH values. His results are shown below.
\nDeduce the colour of the juice at each pH, giving your reasoning. Use section 17 of the data booklet.
\n«extensive» conjugation «of double bonds»/delocalization «of electrons»
OR
«many» alternating single/C–C AND double/multiple/C=C bonds
in aqueous solution AND hydroxyl/OH/ionic/oxonium/O+ «groups»
\nAccept “polar/hydroxy” for “hydroxyl”.
Do not accept “OH- /hydroxide/ oxygen”.
\n
pH 2: «absorption peak 520 nm» red AND pH 11: «absorption peak 620 nm» blue
\ncomplementary/opposite colour observed «to wavelength absorbed»
OR
pH 2: «absorption peak 520 nm» green absorbed AND pH 11: «absorption peak 620 nm» orange absorbed
Award [1 max] if colour absorbed and colour observed are correct for either at pH 2 or pH 11.
\nResearch has led to the discovery of new catalysts that are in high demand and used in many chemical industries.
\nExplain, with reference to their structure, the great selectivity of zeolites as catalysts.
\nNanocatalysts play an essential role in the manufacture of industrial chemicals.
\n(i) Describe the high pressure carbon monoxide (HIPCO) method for the production of carbon nanotubes.
\n(ii) Outline one benefit of using nanocatalysts compared to traditional catalysts in industry.
\npores/cavities/channels/holes/cage-like structures
\n«only» reactants with appropriate/specific size/geometry fit inside/go through/are activated/can react
\nAccept “molecules/ions” for reactants.
\ni
\niron«0»«penta»carbonyl/Fe(CO)5 catalyst decomposes
OR
Fe(CO)5 (g) → Fe(s) + 5CO(g)
OR
metal nanocatalyst/clusters/particles formed «in situ»
Accept “cobalt-molybdenum/Co-Mo/CoMo” as a catalyst
\n2CO(g) → CO2(g) +C(s)
\nAccept “conversion of CO molecules into CNTs/ SWNTs” for M2.
\n\n
ii
\nhigher efficiency per unit mass/volume of the catalyst «due to higher surface to mass/volume ratio»
OR
greater selectivity «due to metal nanoclusters/surface topology/pore size»
OR
higher stability of the catalyst «due to lower tendency to aggregation»
OR
reduced cost of the catalyst/product/chemicals «as precious metals can be replaced with nanocatalysts made of inexpensive materials»
Accept “high conversion efficiency”.
\nAccept specific examples such as use of nanocatalysts in fuel cells/catalytic converters «leading to reduced use of Pt/Rh/Pd».
\nAccept “nanocatalysts often operate under milder conditions «so less energy consumption involved/so promotes principles of green chemistry»”.
\nAccept “lower energy consumption” OR “reduced carbon footprint” OR “reduced global warming”.
\nAccept “nanocatalysts often have long lifetimes «so more economical».
\nAccept “some nanocatalysts have enzyme mimicking activities”.
\nPolymers are made up of repeating monomer units which can be manipulated in various ways to give structures with desired properties.
\n(i) Draw the structure of 2-methylpropene.
\n(ii) Deduce the repeating unit of poly(2-methylpropene).
\nDeduce the percentage atom economy for polymerization of 2-methylpropene.
\n(i) Suggest why incomplete combustion of plastic, such as polyvinyl chloride, is common in industrial and house fires.
\n(ii) Phthalate plasticizers such as DEHP, shown below, are frequently used in polyvinyl chloride.
\nWith reference to bonding, suggest a reason why many adults have measurable levels of phthalates in their bodies.
\ni
\nOR
H2C=C(CH3)2
\n
ii
\nOR
−CH2C(CH3)2−
Continuation bonds needed for mark.
No penalty if square brackets present or “n” appears after the bracket/formula.
«same mass of product as reactant, thus» 100«%»
\nAccept “less than 100%” only if a reason is given (eg, the catalyst is not converted into the product, or other reasonable answer).
\ni
\ndue to stability of plastics/strong covalent bonds
OR
low volatility preventing good mixing with oxygen «gas»
OR
lack of/insufficient oxygen
OR
plastics are often parts of devices with non-combustible components «which mechanically prevent the combustion of plastic components»
OR
PVC already partly oxidised «because some C–H bonds are replaced with C–Cl bonds», so it cannot produce enough heat for complete combustion
OR
many industrial/household materials contain additives that reduce their flammability/act as flame retardants
\n
ii
\nweakly bound to the PVC/no covalent bonds to PVC/only London/dispersion/instantaneous induced dipole-induced dipole forces between DEHP and PVC AND leach/evaporate «from PVC» to atmosphere/food chain
OR
has low polarity/contains non-polar hydrocarbon chains AND fat-soluble/deposits in the fatty tissues
OR
has unusual structural fragments/is a xenobiotic/difficult to metabolise AND stays in the body for a long time
A fuel cell is an energy conversion device that generates electricity from a spontaneous redox reaction.
\nThe Geobacter species of bacteria can be used in microbial fuel cells to oxidise aqueous ethanoate ions,
CH3COO−(aq), to carbon dioxide gas.
State the half-equations for the reactions at both electrodes.
\nA concentration cell is an example of an electrochemical cell.
\n(i) State the difference between a concentration cell and a standard voltaic cell.
\n(ii) The overall redox equation and the standard cell potential for a voltaic cell are:
\nZn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s) Eθcell = +1.10 V
\nDetermine the cell potential E at 298 K to three significant figures given the following concentrations in mol dm−3:
\n[Zn2+] = 1.00 × 10−4 [Cu2+] = 1.00 × 10−1
\nUse sections 1 and 2 of the data booklet.
\n(iii) Deduce, giving your reason, whether the reaction in (b) (ii) is more or less spontaneous than in the standard cell.
\nDye-sensitized solar cells (DSSC) convert solar energy into electrical energy.
\n(i) Describe how a DSSC converts sunlight into electrical energy.
\n(ii) Explain the role of the electrolyte solution containing iodide ions, I−, and triiodide ions, I3−, in the DSSC.
\nNegative electrode (anode): CH3COO− (aq) + 2H2O (l) → 2CO2 (g) + 7H+ (aq) + 8e−
\nPositive electrode (cathode): O2 (g) + 4H+ (aq) + 4e− → 2H2O (l)
\nAccept equilibrium signs in equations.
Award [1 max] if correct equations are given at wrong electrodes.
i
concentration cell has different concentrations of electrolyte «solutions» «but same electrodes and electrolytes»
OR
standard voltaic cell has different electrodes/electrolytes «but same concentration of electrolytes»
Accept “both half-cells in concentration cell made from same materials”.
ii
«»
(+) 1.19 «V»
3 significant figures needed for mark.
iii
more spontaneous because E > Eθcell
i
\nphoton/«sun»light absorbed by the dye/photosensitizer/«transition» metal complex
OR
dye/photosensitizer/«transition» metal complex excited by photon/«sun»light
electron«s» move«s» to conduction band
OR
electron«s» transferred to semiconductor/TiO2
\n
ii
\nI3− + 2e− → 3I− «at cathode»
OR
triiodide ions/I3− reduced into/produce iodide ions/I− «at cathode»
iodide ions/I− reduce dye/act as reducing agent AND oxidized into/produce triiodide ions/I3−
OR
dye+ + e− → dye AND 3I- → I3− + 2e−
Liquid crystals have many applications.
\nOutline how a lyotropic liquid crystal differs from a thermotropic liquid crystal.
\nExplain the effect of increasing the temperature of a nematic liquid crystal on its directional order.
\nDo not award any credit if one type only is described as the question asks how they differ.
\ndecreases AND as energy «added» overcomes interparticle forces
OR
decreases AND as energy «added» causes faster movement «of particles»
Nuclear reactions transform one nuclide into another. Fission, splitting a large nucleus into two smaller nuclei, releases vast amounts of energy.
\n(i) Uranium hexafluoride, UF6, is used in the uranium enrichment process that produces fuel for nuclear reactors.
\nState the molecular shape of uranium hexafluoride.
\n(ii) Explain why uranium dioxide, UO2, has a very high melting point whereas uranium hexafluoride vapourises easily into gas.
\ni
\noctahedral
\nAccept “square bipyramidal”
\n\n
ii
\nUO2 strong bonding throughout crystal structure
\nUF6 molecular «covalent bonds between atoms» AND London/dispersion/instantaneous induced dipole-induced dipole forces between molecules
\nAccept “UO2 has ionic lattice”
\n\n
Lipids are an important part of the human diet.
\nFatty acids react with glycerol to form fats and oils. State the name of the chemical link formed in this reaction and the name of the other product.
\nThe table below shows average figures for the percentage fatty acid composition of some common fats and oils.
\n(i) Deduce, with a reason, which fat or oil from the table above has the lowest iodine number.
\n(ii) Deduce, with a reason, which fat or oil from the table above is most likely to become rancid when exposed to the air.
\n(iii) The P/S index of a fat or oil is the ratio of polyunsaturated fat to saturated fat present. It is sometimes used to compare the relative health benefits of different lipids in the diet. Calculate the P/S index of beef fat and soybean oil.
\n(iv) Suggest why a P/S index of greater than 1 is considered beneficial to health.
\n(v) Cotton seed oil and corn oil have similar iodine numbers but the melting point of cotton seed oil is higher than that of corn oil. Suggest an explanation in terms of the structure and bonding in these two oils.
\nName of the chemical link: ester/ethoxycarbonyl
AND
Name of the other product: water
Do not accept formulas.
Do not accept “esterification”
i
coconut oil AND lowest «percentage of» unsaturated fatty acids
OR
coconut oil AND smallest number of C=C bonds
OR
coconut oil AND highest «percentage of» saturated fatty acids
Accept “fats” for “fatty acids”.
ii
soybean oil AND highest «percentage of» polyunsaturated fatty acids
OR
soybean oil AND greatest number of C=C bonds
OR
soybean oil AND lowest «percentage of» saturated fatty acids
Accept “fats” for “fatty acids”.
iii
Beef fat: «P/S = = » 0.05
AND
Soybean oil: «P/S = =» 4.1
\n
iv
«higher proportion of» polyunsaturated fatty acids decrease risk of atherosclerosis/heart disease/cardiovascular disease/CVD
OR
«higher proportion of» polyunsaturated fatty acids which are less likely to be deposited on the walls of arteries «than saturated fatty acids»
Accept converse arguments.
\nAccept correct arguments in terms of HDL and LDL but not in terms of “good” and “bad” cholesterol.
\nAccept “fats” for “fatty acids”.
\n\n
v
\nAny two of:
cotton seed oil has «a higher proportion of» longer chain/greater molar mass fatty acids
molecules of cotton seed oil have greater surface area/have higher electron density
\nAccept “molecules of cotton seed oil are packed more closely/have more regular structure” for M2.
\nstronger London/dispersion/instantaneous induced dipole-induced dipole forces between chains in cotton seed oil
\nAccept converse arguments.
\nAccept “fats” for “fatty acids”.
\n\n
Excess stomach acid leads to medical conditions that affect many people worldwide. These conditions can be treated with several types of medical drugs.
\nOmeprazole exists as a racemic mixture whereas esomeprazole is a single enantiomer. Outline how, starting from a non-chiral molecule, esomeprazole but not omeprazole, could be synthesized. Details of chemicals and conditions are not required.
\nAny two of:
\nchiral molecule/auxiliary/optically active species is used/added/connected «to the starting molecule to force reaction to follow a certain path»
\nchiral intermediate forms «only» one enantiomer
OR
auxiliary creates stereochemical condition «necessary to follow a certain pathway» / stereochemical induction
OR
existing chiral centre affects configuration of new chiral centres
«after new chiral centre created» chiral auxiliary removed «to obtain desired product»
\nTechnetium-99m is the most widely used medical radioisotope. It is usually made on-site in medical facilities from isotopes of molybdenum.
\nDeduce equations for the following nuclear reactions:
\n(i) Molybdenum-98 absorbs a neutron.
\n(ii) The isotope produced in (a) (i) decays into technetium-99m.
\nMolybdenum-99 has a half-life of 66 hours, while technetium-99m has a half-life of 6 hours. Outline why technetium-99m is made on-site.
\nOutline two reasons, other than its half-life, why technetium-99m is so useful in medical diagnosis.
\nOutline the nature of the radioactive waste that is generated by the use of technetium-99m in medical diagnosis.
\ni
\n+ →
\nAccept 98Mo + 1n/n → 99Mo.
\n
ii
\n
Accept “ ” for “ ”.
\nAccept “ ”.
\nAccept “ ” for “”.
\nDo not penalize “ ” for “ ”.
\nmolybdenum-99 can be easily transported «before it decays»/more stable
OR
«most of» technetium-99m will decay during transportation
Do not accept just “short half-life of Tc-99m”
\nemits gamma rays
OR
emissions escape from body
OR
emissions detected by gamma camera
OR
radiation dose is low
chemically reactive/versatile/transition metal bonds to a range of «biologically active» substances
\nDo not accept “short half-life of Tc-99m”.
Accept “energy of photons produced is «relatively» low” and “no high energy beta emission” for M1.
Accept “…has ability to form tracers” for “…bonds to a range of «biologically active» substances\".
low-level «radioactive» waste/LLW
OR
small amounts of ionizing radiation for short time
The use of performance-enhancing drugs presents a challenge in the world of competitive sports. New regulations have lowered the acceptable concentrations of certain drugs in athletes’ bodies.
\nSuggest what may have led to these changes in acceptable concentrations.
\nOne class of performance-enhancing drugs is the anabolic steroids. Detection of these drugs in urine samples uses a combination of gas chromatography and mass spectrometry (GC/MS).
\n(i) Describe how gas chromatography enables the components of urine to be analysed.
\n(ii) The structures of two steroids, testosterone and nandrolone, are given below.
\nWith reference to the molar masses of the two steroids, determine, with a reason, which can be identified from the mass spectrum below.
\nimprovements in technology/instrumentation/analytical techniques/precision of measurements
\nAccept “greater awareness/knowledge of the negative effects of the drugs”.
\ni
\n«components have» different affinities for/partition between 2 phases/mobile and stationary phase
\nmove at different rates through instrument
OR
have different retention times
\n
ii
\nnandrolone M = 274 «g mol–1»
OR
testosterone M = 288 «g mol–1»
nandrolone identified because «molecular ion peak of» m/z = 274
\nAccept non-integer molar masses, ie, 274.44 «g mol–1» and 288.47 «g mol–1».
Accept also “m/z = 275” for “m/z = 274” in M2.
Accept “absence of peak with m/z = 288”
\nCarbohydrates are energy-rich molecules which can be synthesized in some plant cells from inorganic compounds.
\nState the raw materials and source of energy used in the process described above.
\nThe structures of two molecules, X and Y, are shown below.
\n(i) Justify why both these molecules are carbohydrates.
\n(ii) Distinguish between these molecules in terms of their functional groups.
\nAmylose is an unbranched polysaccharide composed of repeating units of glucose.
\n(i) Draw the structure of the repeating unit of amylose. Use section 34 of the data booklet.
\n(ii) Amylose is a major component of starch. Corn starch can be used to make replacements for plastics derived from oil, especially for packaging. Discuss one potential advantage and one disadvantage of this use of starch.
\nCO2 AND H2O AND sun
\nAccept names.
Accept “sunlight/light/photons” instead of “sun”.
i
\nboth have formula Cx(H2O)y
OR
both contain several OH/hydroxyl «groups» AND a C=O/carbonyl «group»
Accept “both have the formula CnH2nOn /empirical formula CH2O” but do not accept “both have same molecular formula/have formula C3H6O3”.
\nAccept “aldehyde or ketone” for “carbonyl”.
\n\n
ii
\nAccept “alkyl” for “R”.
Accept “X: aldose/aldehyde AND Y: ketose/ketone”.
Accept “CO” for “C=O”.
i
\ncontinuation bonds AND open O on either but not both ends
\nBrackets are not necessary for the mark.
Do not accept β-isomer.
Mark may be awarded if a polymer is shown but with the repeating unit clearly identified.
3-D representation is not required.
\n
ii
\nAdvantage:
Any one of:
biodegradable / break down naturally/by bacteria
\nDo not accept just “decompose easily”.
\ncompostable
\ndoes not contribute to land-fill
\nrenewable/sustainable resource
\nstarch grains swell AND help break up plastic
\nlower greenhouse gas emissions
\nuses less fossil fuels than traditional plastics
\nless energy needed for production
\n\n
Disadvantage:
Any one of:
land use «affects biodiversity/loss of habitat»
\ngrowing corn for plastics instead of food
\n«starch» breakdown can increase acidity of soil/compost
\n«starch» breakdown can produce methane «especially when buried»
\nsensitive to moisture/bacteria/acidic foods
\n«bioplastics sometimes» degrade quickly/before end of use
\ncannot be reused
\npoor mechanical strength
\neutrophication
\nincreased use of fertilizers/pesticides/phosphorus/nitrogen «has negative environmental effects»
\nIgnore any reference to cost.
Accept “prone to site explosions/fires” or “low heat resistance” for disadvantage.
Only award [1 max] if the same example is used for the advantage and disadvantage.
\nAnalysis of amino acid and protein concentration is a key area of biological research.
\nThe titration curve of aqueous glycine zwitterions with aqueous sodium hydroxide is shown from pH 6.0 to 13.0. Refer to section 33 of the data booklet.
\nDeduce the pH range in which glycine is an effective buffer in basic solution.
\nEnzymes are biological catalysts.
\nThe data shows the effect of substrate concentration, [S], on the rate, v, of an enzyme-catalysed reaction.
\nDetermine the value of the Michaelis constant (Km) from the data. A graph is not required.
\nOutline the action of a non-competitive inhibitor on the enzyme-catalysed reaction.
\nThe sequence of nitrogenous bases in DNA determines hereditary characteristics.
\nCalculate the mole percentages of cytosine, guanine and thymine in a double helical DNA structure if it contains 17% adenine by mole.
\n«pH range» 8.6–10.6
\n\n
Accept any value between 8.2 and 11.0.
\n[1 mark]
\n«Km =» 0.67 «mmol dm–3»
\n\n
Do not penalize if a graph is drawn to determine the value.
\n[1 mark]
\ndoes not compete for active site
OR
binds to allosteric site/away from «enzyme» active site
OR
alters shape of enzyme
reduces rate/Vmax
\n[2 marks]
\n«% cytosine + % guanine = 100% – 17% – 17% = 66%»
\nCytosine: 33 «%» AND Guanine: 33 «%»
\nThymine: 17 «%»
\n[2 marks]
\nWhich compound has the greatest percentage by mass of nitrogen atoms?
\nA. N2H4
\nB. NH3
\nC. N2O4
\nD. NaNO3
\nA
\nWhich statements about mixtures are correct?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\n5.0 cm3 of 2.00 moldm–3 sodium carbonate solution, Na2CO3(aq), was added to a volumetric flask and the volume was made up to 500 cm3 with water. What is the concentration, in moldm–3, of the solution?
\nA. 0.0050
\nB. 0.0040
\nC. 0.020
\nD. 0.010
\nC
\nWhat is the expression for the volume of hydrogen gas, in dm3, produced at STP when 0.30 g of magnesium reacts with excess hydrochloric acid solution?
\nMg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
\nMolar volume of an ideal gas at STP = 22.7 dm3mol−1
\nA.
\nB.
\nC.
\nD.
\nB
\nIn which set do all the species contain more electrons than neutrons?
\nA. 14N, 16O, 11C
\nB. 14N, 16O, 11C4–
\nC. 14N3–, 16O2–, 11C
\nD. 14N3–, 16O2–, 11C4+
\nC
\nAmino acids are usually identified by their common names. Use section 33 of the data booklet.
\nState the IUPAC name for leucine.
\nA mixture of amino acids is separated by gel electrophoresis at pH 6.0. The amino acids are then stained with ninhydrin.
\n(i) On the diagram below draw the relative positions of the following amino acids at the end of the process: Val, Asp, Lys and Thr.
\n(ii) Suggest why glycine and isoleucine separate slightly at pH 6.5.
\nDetermine the number of different tripeptides that can be made from twenty different amino acids.
\nThe fibrous protein keratin has a secondary structure with a helical arrangement.
\n(i) State the type of interaction responsible for holding the protein in this arrangement.
\n(ii) Identify the functional groups responsible for these interactions.
\n2-amino-4-methylpentanoic acid
\nAccept 4-methyl-2-aminopentanoic acid.
\ni
\nLys on cathode side AND Asp on anode side
Val at origin AND Thr on anode side but closer to origin than Asp
Val and Thr need not overlap.
Accept any (reasonable) size and demarcation of position so long as position relative to origin is correct.
Accept crosses for spots.
Award [1 max] for any two correct.
Award [1 max] if net direction of spots is reversed.
Award [1 max] if the four points are in the correct order but not in a straight line.
\n
ii
\ndifferent sizes/molar masses/chain lengths «so move with different speeds»
\n«203 =» 8000
\ni
\nhydrogen bonds
\n\n
ii
\ncarboxamide/amide/amido
OR
C=O AND N–H
Accept peptide.
\nWhich electron transition in the hydrogen atom emission spectrum emits radiation with the longest wavelength?
\nA. n = 2 → n = 1
\nB. n = 1 → n = 2
\nC. n = 4 → n = 1
\nD. n = 3 → n = 2
\nD
\nThe full electron configuration of an element is:
\n1s22s22p63s23p2
\nTo which group and period does the element belong?
\nD
\nWhich oxide, when added to water, produces the solution with the highest pH?
\nA. Na2O
\nB. SO3
\nC. MgO
\nD. CO2
\nA
\nChemical energy from redox reactions can be used as a portable source of electrical energy. A hybrid car uses a lithium ion battery in addition to gasoline as fuel.
\n(i) Calculate the specific energy of the lithium ion battery, in MJ kg−1, when 80.0 kg of fuel in the battery releases 1.58 × 107 J. Use section 1 of the data booklet.
\n(ii) The specific energy of gasoline is 46.0 MJ kg−1. Suggest why gasoline may be considered a better energy source than the lithium ion battery based on your answer to part (a) (i).
\n(i) The energy density of gasoline is 34.3 MJ dm−3. Calculate the volume of gasoline, in dm3, that is equivalent to the energy in 80.0 kg of fuel in the lithium ion battery. Use section 1 of the data booklet.
\n(ii) The efficiency of energy transfer by this lithium ion battery is four times greater than that of gasoline. Determine the distance, in km, the car can travel on the lithium ion battery power alone if the gasoline-powered car uses 1.00 dm3 gasoline to travel 32.0 km.
\ni
«==» 1.98 × 10−1 «MJ kg−1»
ii
gasoline releases more energy from a given mass of fuel
OR
gasoline has higher specific energy
Do not accept volume in place of mass as question refers to specific energy, not energy density.
i
«»= 4.61 × 10−1 «dm3»
ii
«4.61 × 10−1 dm3 × 32.0 km dm−3 × 4»= 59.0/59.1 «km»
A substance has the following properties:
\nWhat is the most probable structure of this substance?
\nA. Network covalent
\nB. Polar covalent molecule
\nC. Ionic lattice
\nD. Metallic lattice
\nA
\nWhich combination describes the sulfate(IV) ion, SO32– (also known as sulfite ion)?
\nD
\nWhich correctly states the strongest intermolecular forces in the compounds below?
\nB
\nWhich two atoms form the most polar bond?
\nA. C and F
\nB. C and Cl
\nC. Si and F
\nD. Si and Cl
\nC
\nThere is concern about damage done to the ozone layer in the stratosphere by jet-propelled aircraft.
\nFormulate two equations to show how nitrogen(II) oxide, NO, catalyses the destruction of ozone.
\nSuggest why the loss of ozone is an international environmental concern.
\n\n
OR
\n
Allow representation of radicals without if consistent throughout.
\n[2 marks]
\n«loss of ozone» allows UV radiation to penetrate atmosphere/reach earth
\nUV radiation causes skin cancer
OR
UV radiation causes tissue damage
[2 marks]
\nWhich expression gives the mass, in g, of ethanol required to produce 683.5 kJ of heat upon complete combustion?
\n(Mr for ethanol = 46.0, )
\nA.
\nB.
\nC.
\nD.
\nC
\nAuto-ignition of hydrocarbon fuel in a car engine causes “knocking”. The tendency of a fuel to knock depends on its molecular structure.
\nDiscuss how the octane number changes with the molecular structure of the alkanes.
\nCatalytic reforming and cracking reactions are used to produce more efficient fuels. Deduce the equation for the conversion of heptane to methylbenzene.
\n«tends to» decrease with longer/larger/heavier alkanes
\n«tends to» increase with bulkier/more branched alkanes
\nAccept “octane number decreases with the separation between branches” OR “increases with the more central position of branches”.
\nAccept converse arguments.
\nC7H16 → C6H5CH3 + 4H2
\nAccept “C7H8” for “C6H5CH3”.
\nWhich expression gives the enthalpy change, ΔH, for the thermal decomposition of calcium carbonate?
\nA. ΔH = ΔH1 − ΔH2
\nB. ΔH = 2ΔH1 − ΔH2
\nC. ΔH = ΔH1 − 2ΔH2
\nD. ΔH = ΔH1 + ΔH2
\nA
\nCalcium carbide, CaC2, is an ionic solid.
\nDescribe the nature of ionic bonding.
\nDescribe how the relative atomic mass of a sample of calcium could be determined from its mass spectrum.
\nWhen calcium compounds are introduced into a gas flame a red colour is seen; sodium compounds give a yellow flame. Outline the source of the colours and why they are different.
\nSuggest two reasons why solid calcium has a greater density than solid potassium.
\nOutline why solid calcium is a good conductor of electricity.
\nSketch a graph of the first six ionization energies of calcium.
\nCalcium carbide reacts with water to form ethyne and calcium hydroxide.
\nCaC2(s) + H2O(l) → C2H2(g) + Ca(OH)2(aq)
\nEstimate the pH of the resultant solution.
\nDescribe how sigma (σ) and pi () bonds are formed.
\n\n
Deduce the number of σ and bonds in a molecule of ethyne.
\n\n
electrostatic attraction AND oppositely charged ions
\n[1 mark]
\nmultiply relative intensity by «m/z» value of isotope
\nOR
\nfind the frequency of each isotope
\n\n
sum of the values of products/multiplication «from each isotope»
\nOR
\nfind/calculate the weighted average
\n\n
Award [1 max] for stating “m/z values of isotopes AND relative abundance/intensity” but not stating these need to be multiplied.
\n[2 marks]
\n«promoted» electrons fall back to lower energy level
\nenergy difference between levels is different
\n\n
Accept “Na and Ca have different nuclear charge” for M2.
\n[2 marks]
\nAny two of:
\nstronger metallic bonding
\nsmaller ionic/atomic radius
\n\n
two electrons per atom are delocalized
\nOR
\ngreater ionic charge
\n\n
greater atomic mass
\n\n
Do not accept just “heavier” or “more massive” without reference to atomic mass.
\n[2 marks]
\ndelocalized/mobile electrons «free to move»
\n\n
\n
[1 mark]
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/02.e/M\"](\"../assets/uploads/tinymce_asset/asset/4479/Schermafbeelding_2018-08-07_om_13.26.19.png\"/)
general increase
\nonly one discontinuity between “IE2” and “IE3”
\n[2 marks]
\npH > 7
\n\n
Accept any specific pH value or range of values above 7 and below 14.
\n[1 mark]
\nsigma (σ):
\noverlap «of atomic orbitals» along the axial/internuclear axis
\nOR
\nhead-on/end-to-end overlap «of atomic orbitals»
\n\n
pi ():
\noverlap «of p-orbitals» above and below the internuclear axis
\nOR
\nsideways overlap «of p-orbitals»
\n\n
Award marks for suitable diagrams.
\n\n
[2 marks]
\nsigma (σ): 3
\nAND
\npi (): 2
\n[1 mark]
\nBenzene is an aromatic hydrocarbon.
\nDiscuss the physical evidence for the structure of benzene.
\nState the typical reactions that benzene and cyclohexene undergo with bromine.
\nAny two of:
\nplanar «X-ray»
\nC to C bond lengths all equal
OR
C to C bonds intermediate in length between C–C and C=C
all C–C–C bond angles equal
\n\n
Accept all C to C bonds have same bond strength/bond energy.
\n[2 marks]
\nbenzene: «electrophilic» substitution/SE
AND
cyclohexene: «electrophilic» addition/AE
\n
Accept correct equations.
\n[1 mark]
\nA class was determining the concentration of aqueous sodium hydroxide by titrating it with hydrochloric acid, whilst monitoring the pH of the solution. The sodium hydroxide solution was added into a glass beaker from a measuring cylinder and the hydrochloric acid added using a burette. One group of students accidentally used a temperature probe rather than a pH probe. Their results are given below.
\nVolume of aqueous NaOH = 25.0 ± 0.5 cm3
\nConcentration of HCl = 1.00 ± 0.01 mol dm−3
\nCalculate the percentage uncertainty of the volume of the aqueous sodium hydroxide.
\nSuggest how the precision of this measurement could be improved.
\n«» = 2 «%»
\n[1 mark]
\npipette/pipet «rather than a measuring cylinder»
\n\n
Accept “using a burette/buret”.
\nAccept “using a volumetric/measuring flask”.
\nDo not accept “use of a more precise measuring cylinder”.
\n[1 mark]
\nIn which order does the oxygen–oxygen bond enthalpy increase?
\nA. H2O2 < O2 < O3
\nB. H2O2 < O3 < O2
\nC. O2 < O3 < H2O2
\nD. O3 < H2O2 < O2
\nB
\nCarbon dioxide and water vapour are greenhouse gases produced by the combustion of fossil fuels.
\nExplain the effect of the increasing concentration of atmospheric carbon dioxide on the acidity of oceans.
\n(i) Describe the changes that occur at the molecular level when atmospheric carbon dioxide gas absorbs infrared radiation emitted from the Earth’s surface.
\n(ii) Other than changes to the acidity of oceans, suggest why the production of carbon dioxide is of greater concern than the production of water vapour.
\nAny two of:
\nCO2 (g) CO2 (aq)
\nCO2 (aq) + H2O (l) H+ (aq) + HCO3– (aq)
OR
HCO3– AND H+ are formed «by dissolved CO2»
«increasing [CO2]» shifts equilibrium to right/increases acidity/decreases pH
\nH2O (l) not required over equilibrium sign for M1.
\nState symbols required in the equation in M1.
\nAccept “H2CO3 ” at either side of the equilibrium in M2.
\nEquilibrium sign required for M1 but not for M2.
\ni
\nbond length/C=O changes
OR
«asymmetric» stretching «of bonds»
OR
bond angle/OCO changes
Accept “molecule bends” for M1.
Accept appropriate diagrams
photon re-emitted in random direction
OR
polarity/dipole «moment» changes
OR
dipole «moment» created «when molecule absorbs IR»
\n
ii
\nCO2 gas «ten times» more effective as greenhouse gas/GHG than H2O
OR
CO2 gas levels keep increasing «unlike H2O»
OR
CO2 has higher Global Warming Potential/GWP than H2O
OR
CO2 stays in the atmosphere for longer than H2O
Accept converse arguments.
\nCopper catalyses the reaction between zinc and dilute sulfuric acid.
\nZn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
\nWhy does copper affect the reaction?
\nA. Decreases the activation energy
\nB. Increases the activation energy
\nC. Increases the enthalpy change
\nD. Decreases the enthalpy change
\nA
\nA class was determining the concentration of aqueous sodium hydroxide by titrating it with hydrochloric acid, whilst monitoring the pH of the solution. The sodium hydroxide solution was added into a glass beaker from a measuring cylinder and the hydrochloric acid added using a burette. One group of students accidentally used a temperature probe rather than a pH probe. Their results are given below.
\nVolume of aqueous NaOH = 25.0 ± 0.5 cm3
\nConcentration of HCl = 1.00 ± 0.01 mol dm−3
\nDeduce why more heat was produced in mixture B than in mixture A.
\nDeduce why the temperature is higher in mixture C than in mixture D.
\nmore «moles/amount of» acid have been added/reacted
OR
more of the limiting reagent is present
OR
more «of the exothermic» reaction has occurred
[1 mark]
\nno more reaction/same energy released AND cold/colder/cooler liquid added
OR
no more reaction/same energy released AND greater total volume of liquid
\n
Accept “no more reaction/same energy released AND greater heat loss «to the surroundings in mixture D»”.
\n[1 mark]
\nA class was determining the concentration of aqueous sodium hydroxide by titrating it with hydrochloric acid, whilst monitoring the pH of the solution. The sodium hydroxide solution was added into a glass beaker from a measuring cylinder and the hydrochloric acid added using a burette. One group of students accidentally used a temperature probe rather than a pH probe. Their results are given below.
\nVolume of aqueous NaOH = 25.0 ± 0.5 cm3
\nConcentration of HCl = 1.00 ± 0.01 mol dm−3
\nSuggest how the end point of the titration might be estimated from the graph.
\nvolume «found by extrapolation of the two best fit lines» required to give the highest temperature
OR
extrapolate «two best fit» lines to the point where they meet
\n
Accept “where lines through the points meet”.
\nAccept “at maximum temperature”.
\nAccept “at 35 cm3 of HCl”.
\n[1 mark]
\n100 cm3 of 10% hydrogen peroxide solution decomposes at 298 K to form water and oxygen.
\nH2O2(aq) → H2O(l) + O2(g)
\nThe dotted line graph represents the volume of oxygen produced.
\nWhich graph represents the decomposition of an equal volume of a 20% solution under the same conditions?
\nA
\nBiofuels are renewable energy sources derived mainly from plants.
\nState the equation for the complete transesterification of the triglyceride given below with methanol.
\nOutline why the fuel produced by the reaction in (a) is more suitable for use in diesel engines than vegetable oils.
\nmethyl ester formula AND glycerol formula
\ncorrect balancing
\nAward M2 only if M1 is correct.
\n«methyl esters have» low«er» viscosity/surface tensions
OR
«methyl esters have» high«er» volatility
OR
«combustion of vegetable oils» produces carbon deposits in engine/reduces engine life
Accept converse arguments.
\nConsider the equilibrium between N2O4(g) and NO2(g).
\nN2O4(g) 2NO2(g) ΔH = +58 kJmol−1
\nWhich changes shift the position of equilibrium to the right?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nWhich is an acid-base conjugate pair?
\nA. H3O+ / OH–
\nB. H2SO4 / SO42–
\nC. CH3COOH / H3O+
\nD. CH3NH3+ / CH3NH2
\nD
\nWhich 1.0 moldm–3 solution has the highest pH?
\nA. Ammonium chloride
\nB. Sulfuric acid
\nC. Sodium chloride
\nD. Ammonia
\nD
\nThis question is about ethene, C2H4, and ethyne, C2H2.
\nEthyne, like ethene, undergoes hydrogenation to form ethane. State the conditions required.
\nOutline the formation of polyethene from ethene by drawing three repeating units of the polymer.
\nEthyne reacts with chlorine in a similar way to ethene. Formulate equations for the following reactions.
\n\n
Under certain conditions, ethyne can be converted to benzene.
\nDetermine the standard enthalpy change, ΔHΘ, for the reaction stated, using section 11 of the data booklet.
\n3C2H2(g) → C6H6(g)
\nDetermine the standard enthalpy change, ΔHΘ, for the following similar reaction, using ΔHf values in section 12 of the data booklet.
\n3C2H2(g) → C6H6(l)
\nExplain, giving two reasons, the difference in the values for (c)(i) and (ii). If you did not obtain answers, use −475 kJ for (i) and −600 kJ for (ii).
\nCalculate the standard entropy change, ΔSΘ, in J K−1, for the reaction in (ii) using section 12 of the data booklet.
\nDetermine, showing your working, the spontaneity of the reaction in (ii) at 25 °C.
\nOne possible Lewis structure for benzene is shown.
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/03.d\"](\"../assets/uploads/tinymce_asset/asset/4485/Schermafbeelding_2018-08-07_om_14.31.21.png\")
State one piece of physical evidence that this structure is incorrect.
\nnickel/Ni «catalyst»
\n\n
high pressure
\nOR
\nheat
\n\n
Accept these other catalysts: Pt, Pd, Ir, Rh, Co, Ti.
\nAccept “high temperature” or a stated temperature such as “150 °C”.
\n[2 marks]
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/03.a.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4482/Schermafbeelding_2018-08-07_om_13.51.15.png\"/)
\n
Ignore square brackets and “n”.
\nConnecting line at end of carbons must be shown.
\n[1 mark]
\nethyne: C2H2 + Cl2 → CHClCHCl
\nbenzene: C6H6 + Cl2 → C6H5Cl + HCl
\n\n
Accept “C2H2Cl2”.
\n[2 marks]
\nΔHΘ = bonds broken – bonds formed
\n«ΔHΘ = 3(C≡C) – 6(CC)benzene / 3 839 – 6 507 / 2517 – 3042 =» –525 «kJ»
\n
Award [2] for correct final answer.
\nAward [1 max] for “+525 «kJ»”.
\nAward [1 max] for:
\n«ΔHΘ = 3(C≡C) – 3(C–C) – 3(C=C) / 3 839 – 3 346 – 3 614 / 2517 – 2880 =» –363 «kJ».
\n[2 marks]
\nΔHΘ = ΣΔHf (products) – ΣΔHf (reactants)
\n«ΔHΘ = 49 kJ – 3 228 kJ =» –635 «kJ»
\n\n
Award [2] for correct final answer.
\nAward [1 max] for “+635 «kJ»”.
\n[2 marks]
\nΔHf values are specific to the compound
\nOR
\nbond enthalpy values are averages «from many different compounds»
\n\n
condensation from gas to liquid is exothermic
\n\n
Accept “benzene is in two different states «one liquid the other gas»” for M2.
\n[2 marks]
\n«ΔSΘ = 173 – 3 201 =» –430 «J K–1»
\n[1 mark]
\nT = «25 + 273 =» 298 «K»
\nΔGϴ «= –635 kJ – 298 K × (–0.430 kJ K–1)» = –507 kJ
\nΔGϴ < 0 AND spontaneous
\n\n
ΔGϴ < 0 may be inferred from the calculation.
\n[3 marks]
\nequal C–C bond «lengths/strengths»
\nOR
\nregular hexagon
\nOR
\n«all» C–C have bond order of 1.5
\nOR
\n«all» C–C intermediate between single and double bonds
\n\n
Accept “all C–C–C bond angles are equal”.
\n[1 mark]
\nWhat is the order of decreasing reactivity of the metals (most reactive first)?
\nZn(s) + Sn2+(aq) → Zn2+(aq) + Sn(s)
Cu(s) + Zn2+(aq) → No Reaction
Sn(s) + Cu2+(aq) → Sn2+(aq) + Cu(s)
Ag(s) + Cu2+(aq) → No Reaction
A. Zn > Cu > Sn > Ag
\nB. Sn > Zn > Ag > Cu
\nC. Ag > Cu > Zn > Sn
\nD. Zn > Sn > Cu > Ag
\nD
\nWhat is the oxidation half-equation in the redox reaction?
\n2S2O32–(aq) + I2(aq) → S4O62–(aq) + 2I–(aq)
\nA. I2(aq) + 2e– → 2I–(aq)
\nB. 2I–(aq) → I2(aq) + 2e–
\nC. 2S2O32–(aq) → S4O62–(aq) + 2e–
\nD. S4O62–(aq) + 2e– → 2S2O32–(aq)
\nC
\nMillerite, a nickel sulfide mineral, is an important source of nickel. The first step in extracting nickel is to roast the ore in air.
\nThe reaction for the formation of liquid tetracarbonylnickel is shown below:
\n\n
Formulate an equation for the oxidation of nickel(II) sulfide to nickel(II) oxide.
\nThe nickel obtained from another ore, nickeliferous limonite, is contaminated with iron. Both nickel and iron react with carbon monoxide gas to form gaseous complexes, tetracarbonylnickel, , and pentacarbonyliron, . Suggest why the nickel can be separated from the iron successfully using carbon monoxide.
\nCalculate the standard entropy change, , of the reaction, in , using the values given.
\nCalculate a value for in kJ.
\nUse your answers to (c)(i) and (c)(ii), to determine the temperature, in °C, at which the decomposition of liquid tetracarbonylnickel to nickel and carbon monoxide becomes favourable.
\n
(If you did not get answers to (c)(i) and (c)(ii), use and respectively but these are not the correct answers.)
Suggest why experiments involving tetracarbonylnickel are very hazardous.
\n\n
[1 mark]
\nformation of «gaseous» pentacarbonyliron is slower
OR
«gaseous» complexes form at different rates
OR
gases have different rates of diffusion «due to difference in masses»
OR
difference in thermal stability of «gaseous» complexes
OR
difference in boiling points of «gaseous» complexes
OR
difference in solubility of «gaseous» complexes
OR
difference in surface affinity «onto solid absorbent»
OR
difference in chemical properties of «gaseous» complexes
\n
Accept any other valid answer.
\n[1 mark]
\n
AND
\n
\n
Award [2] for correct final answer.
\n[2 marks]
\n\n
[1 mark]
\n«when» «forward and backward reactions are equally favourable»
\n«when , »,
\n«temperature =» 104 «°C»
\n\n
Award [3] for correct final answer. Use of –500 J K–1 and –200 kJ gives 127 °C.
\nAward [2 max] for T < 104 «°C».
\nAccept ΔG < 0 and T > 104 «°C».
\n[3 marks]
\nCO is toxic/poisonous
OR
Ni(CO)4 decomposition deposits nickel in the lungs
OR
tetracarbonylnickel is toxic/poisonous
OR
tetracarbonylnickel is highly flammable «auto-ignition temperature of 60 °C»
[1 mark]
\nNuclear reactions transform one nuclide into another. Fission, splitting a large nucleus into two smaller nuclei, releases vast amounts of energy.
\n(i) Explain why fusion, combining two smaller nuclei into a larger nucleus, releases vast amounts of energy. Use section 36 of the data booklet.
\n(ii) Outline one advantage of fusion as a source of energy.
\nRadioactive phosphorus, 33P, has a half-life of 25.3 days.
\n(i) Calculate 33P decay constant λ and state its unit. Use section 1 of the data booklet.
\n(ii) Determine the fraction of the 33P sample remaining after 101.2 days.
\ni
\nproduct has higher binding energy «per nucleon»/more stable
OR
nucleons in product more tightly bound «with one another»
lighter elements «than Fe» can fuse/combine with loss of mass/mass defect «and release vast amount of energy»
\nAccept “mass is converted to energy” for M2
\n\n
ii
\nAny one of:
deuterium/fuel is abundant/cheap
«helium» products not radioactive
fusion much less dangerous than fission
large amounts/shipments of radioactive fuel not required
far less radioactive waste «created by fast moving neutrons» has to be stored
Accept “reduces greenhouse gas emissions/global warming” OR “no radioactive waste” OR “more reliable power” OR “fewer safety issues”.
Do not accept “gives out a large amount of energy” as it is in the stem of the question.
i
«» 2.74 × 10−2 day−1
Need correct unit for mark.
\n
ii
«4 half-lives; 1 →→→→ =» / 6.25 × 10−2
OR
«» 6.25 × 10−2
Accept 6.25%.
\nPenicillin is an antibiotic which contains a beta-lactam ring. Its general structure is shown below.
\n(i) Outline what is meant by the term “ring strain”.
\n(ii) On the diagram above, label with asterisk/s (*) the carbon atom/s that experience ring strain.
\n(i) Some antibiotic-resistant bacteria produce a beta-lactamase enzyme which destroys penicillin activity. Suggest how adding clavulanic acid to penicillin enables the antibiotic to retain its activity.
\n(ii) Populations of antibiotic-resistant bacteria have increased significantly over the last 60 years. Outline why antibiotics such as penicillin should not be prescribed to people suffering from a viral infection.
\ni
\nbond angles smaller/distorted
OR
instability resulting from abnormal bond angles
OR
bond angles «approximately» 90° instead of 109.5°/120°
Accept “109/110°” for “109.5°”
\n\n
ii
\nasterisks (*) on all 3 lactam ring carbon atoms
\nMust mark all 3 carbon atoms.
Ignore asterisks on the RHS carbon atoms of the five-membered ring.
i
\nbeta-lactam/four-membered ring «in clavulanic acid» reacts with enzyme/beta lactamase
\nAccept “acts as enzyme inhibitor/suicide substrate/preferentially binds to enzyme”.
\n\n
ii
\nantibiotics not effective against viruses
OR
viruses have no cell wall/cell structure/target structures to attack
increasing exposure of bacteria «to antibiotic» increases resistance
\nAccept “antibiotics kill beneficial bacteria” for M2.
\nOseltamivir (Tamiflu) and zanamivir (Relenza) are both used as antivirals to help prevent the spread of the flu virus, but are administered by different methods.
\nZanamivir must be taken by inhalation, not orally. Deduce what this suggests about the bioavailability of zanamivir if taken orally.
\nOseltamivir does not possess the carboxyl group needed for activity until it is chemically changed in the body. Deduce the name of the functional group in oseltamivir which changes into a carboxyl group in the body. Use section 37 of the data booklet.
\nThe synthesis of oseltamivir is dependent on a supply of the precursor shikimic acid, which is available only in low yield from certain plants, notably Chinese star anise. State one alternative green chemistry source of shikimic acid.
\n«oral bioavailability is» low
OR
drug is broken down/pH too low/unable to be absorbed from gut
OR
only a small proportion of the drug «taken by mouth» reaches the target organ
ethoxycarbonyl/carbonyl attached to oxygen
\nAccept “ester”.
\nAny one of:
\nfermentation
OR
microbial production
genetically engineered bacteria/E.coli
\nsweetgum «seeds/leaves/bark»
OR
pine/fir/spruce tree «needles»
OR
Ginkgo biloba
Accept other specific examples of more plentiful plant sources.
\nThe mild analgesic aspirin can be prepared in the laboratory from salicylic acid.
\n(CH3CO)2O + HOC6H4COOH → CH3CO2C6H4COOH + CH3COOH
\nSalicylic acid Aspirin
\n\n
After the reaction is complete, the product is isolated, recrystallized, tested for purity and the experimental yield is measured. A student’s results in a single trial are as follows.
\nLiterature melting point data: aspirin = 138–140 °C
\nDetermine the percentage experimental yield of the product after recrystallization. The molar masses are as follows: M(salicylic acid) = 138.13 g mol−1, M(aspirin) = 180.17 g mol−1. (You do not need to process the uncertainties in the calculation.)
\nSuggest why isolation of the crude product involved the addition of ice-cold water.
\nJustify the conclusion that recrystallization increased the purity of the product, by reference to two differences between the melting point data of the crude and recrystallized products.
\nState why aspirin is described as a mild analgesic with reference to its site of action.
\nALTERNATIVE 1:
«theoretical yield = × 180.17 g mol−1 =» 2.024 «g»
«experimental yield = × 100 =» 55.53 «%»
\nALTERNATIVE 2:
«»= 0.01124 «mol salicylic acid/aspirin theoretical» AND
«»= 0.006239 «mol aspirin experimental»
\n«experimental yield = x 100 =» 55.51 «%»
\nAccept answers in the range 55.4 % to 55.7 %.
Award [2] for correct final answer.
low temperature gives greater difference between solubility of aspirin and impurities
OR
«product» crystallizes out from cold solution/«ice-cold water/lower temperature» speeds up crystallization process
OR
aspirin/product has low solubility «in water» at low temperatures
intercepts pain stimulus at source/acts at site of pain
OR
interferes with production of pain sensitizing substances/prostaglandins «at site of pain»
recrystallized melting point is higher
OR
recrystallized melting point is closer to pure substance/literature value
smaller range of values
\nExcess stomach acid leads to medical conditions that affect many people worldwide. These conditions can be treated with several types of medical drugs.
\nRanitidine (Zantac) is a drug that inhibits stomach acid production. Outline why the development of this drug was based on a detailed knowledge of the structure of histamine, shown below.
\nTwo other drugs, omeprazole (Prilosec) and esomeprazole (Nexium), directly prevent the release of acid into the stomach. Identify the site of action in the body.
\nA different approach to treating excess stomach acid is to neutralize it with antacids. Formulate an equation that shows the action of an antacid that can neutralize three moles of hydrogen ions, H+, per mole of antacid.
\n«ranitidine» blocks/inhibits histamine binding to «H2» receptor
OR
ranitidine binds to same «H2» receptors «as histamine»
OR
competes with histamine for binding
proton pump
OR
H+ /K+ ATPase enzyme
Accept “«secretary surface of» parietal cells”.
Do not accept “stomach/stomach wall”.
Al(OH)3(s) + 3H+ (aq) → Al3+ (aq) + 3H2O (l)
OR
Al(OH)3(s) + 3HCl (aq) → AlCl3 (aq) + 3H2O (l)
Methadone, a synthetic opioid, binds to opioid receptors in the brain.
\nCompare and contrast the functional groups present in methadone and diamorphine (heroin), giving their names. Use section 37 of the data booklet.
\nMethadone is sometimes used to help reduce withdrawal symptoms in the treatment of heroin addiction. Outline one withdrawal symptom that an addict may experience.
\nSimilarity:
\nboth contain «at least one» benzene/aromatic ring
OR
both contain amino «group»
Difference:
\ndiamorphine has one benzene/aromatic ring AND methadone has two phenyl «groups»
OR
diamorphine has one vinylene/ethenylene/1,2-ethenediyl «group» AND methadone has no vinylene/ethenylene/1,2-ethenediyl «group»
OR
diamorphine has one ether «group» AND methadone has no ether «group»
OR
diamorphine has «two» ethanoate/acetate «groups» AND methadone has no ethanoate/acetate «groups»
Accept “both contain carbonyl «groups»”.
Accept “amine” for “amino «group»”.
Accept “phenyl” for “benzene ring” in M1 and M2 although there are no phenyl groups in diamorphine, as the benzene ring in this compound is a part of a polycyclic structure.
Do not accept “arene” or “benzene” alone in M1 and M2.
Accept “alkenyl/alkene” for “vinylene/ethenylene/1,2-ethenediyl” and “ester” for “ethanoate/acetate”.
Accept “methadone has a ketone/carbonyl AND diamorphine does not/has an ester/ethanoate/acetate”.
Accept “diamorphine is a heterocycle/heterocyclic compound AND methadone is not a heterocycle/heterocyclic compound”.
feeling depressed/anxious/irritable
OR
craving for opioids/heroin
OR
experience fever/cold sweats/nausea/vomiting/insomnia/muscle pain/cramps/diarrhea/increased rate of respiration/increased heartbeat/lacrimation
Accept listed symptoms (eg, depression, anxiety, fever etc.).
Some of the most common symptoms are listed here – there may be other valid ones. Accept “headaches”.
Polymers are made up of repeating monomer units which can be manipulated in various ways to give structures with desired properties.
\nFermentation of sugars from corn starch produces propane-1,3-diol, which can be polymerized with benzene-1,4-dicarboxylic acid to produce the PTT polymer (polytrimethylene terephthalate).
\n(i) Draw the molecular structure of each monomer.
\n(ii) Deduce the name of the linkage formed on polymerization between the two monomers and the name of the inorganic product.
\ni
\nHO–CH2–CH2–CH2–OH AND HOOC–C6H4–COOH
\nAccept full or condensed structural formulas. Labelling of monomers not required but penalize incorrect labels.
\n\n
ii
\nName of linkage: ester
AND
Name of inorganic product: water
Do not accept “esterification”.
Do not accept formulas.
Chromium forms coloured compounds and is used to make stainless and hard steel. The distance between layers of chromium atoms in the metal can be obtained using X-ray crystallography.
\n(i) The diagram below shows the diffraction of two X-ray beams, y and z of wavelength λ, shining on a chromium crystal whose planes are a distance d nm apart.
\nDeduce the extra distance travelled by the second beam, z, compared to the first one, y.
\n(ii) State the Bragg’s condition for the observed diffraction to be at its strongest (constructive interference).
\n(i) The mass of one unit cell of chromium metal is 17.28 × 10−23 g. Calculate the number of unit cells in one mole of chromium. Ar(Cr) = 52.00.
\n(ii) Deduce the number of atoms of chromium per unit cell.
\ni
\n2d sin θ
OR
2|AB| / 2|BC| / |AB| + |BC| / |AB| AND |BC|
Vertical lines indicating lengths not required. Answer may be conveyed in words also.
Do not accept |AC| – reference must be made to B.
\n
ii
\nextra distance travelled/|AB| + |BC| = nλ/a whole number of wavelengths
\nAccept notations of extra distance as in (a)(i).
\ni
«» 3.009 × 1023 «unit cells mol−1»
ii
«» 2 «atoms per unit cell»
Biological pigments include a variety of chemical structures with diverse functions.
\nThe graph shows the conversion of hemoglobin to oxyhemoglobin.
\nHb(aq) + 4O2(g) Hb(O2)4(aq)
\nThe partial pressure of oxygen gas, p(O2) is proportional to its concentration.
\n![\"M17/4/CHEMI/HP3/ENG/TZ1/16\"](\"../assets/uploads/tinymce_asset/asset/3696/Schermafbeelding_2017-09-21_om_11.43.40.png\"/)
Explain the shape of the curve at low oxygen partial pressure up to about 5 kPa.
\nSketch a graph on the axes above to show the effect of decreasing pH on the binding of oxygen to hemoglobin (the Bohr Effect).
\nOutline the effect of decreasing pH on the oxygen saturation of hemoglobin.
\nbinding of O2 «to one active site» affects shape of Hb/other active sites
OR
binding of one O2 «molecule» affects binding of other O2 «molecules»
increasing affinity of Hb to O2
OR
enhanced binding of «further» O2 «molecules»
OR
cooperative binding
[2 marks]
\nsketching right shift of curve on graph
\n[1 mark]
\ndecreases «oxygen saturation»
\n\n
Accept “hemoglobin binds to O2 with less affinity\".
\n[1 mark]
\nVision is dependent on retinol (vitamin A) present in retina cells. Retinol is oxidized to the photosensitive chemical 11-cis-retinal and isomerizes to 11-trans-retinal on absorption of light.
\nOutline how the formation of 11-trans-retinal results in the generation of nerve signals to the brain.
\n11-trans retinal no longer fits into the rhodopsin/protein
OR
11-trans retinal is ejected from the rhodopsin/protein
leads to conformational change in rhodopsin/protein «to opsin generating signals»
\n[2 marks]
\nCarbon is produced by fusion reactions in stars.
\nThe main fusion reaction responsible for the production of carbon is:
\nX +
\nThe mass of X is 8.005305 amu and that of is 4.002603 amu. Determine the energy produced, in J, when one atom of is formed in this reaction. Use section 2 of the data booklet.
\nloss in mass = «8.005305 amu + 4.002603 amu – 12.000000 amu =» 0.007908 «amu»
\n= «0.007908 amu x 1.66 x 10–27 kg amu–1 =» 1.313 x 10–29 «kg»
\nE = «mc2 = 1.313 x 10–29 kg x (3.00 x 108 ms–1)2 =» 1.18 x 10–12 «J»
\n\n
Award [3] for correct final answer.
\n[3 marks]
\nDye-Sensitized Solar Cells (DSSCs) use organic dyes. Their interaction with light has some similarities to photosynthesis.
\nIdentify two ways in which the structure of the dye shown resembles the chlorophyll molecule. Use section 35 of the data booklet.
\nBoth photosynthesis and the Grätzel cell use energy from sunlight to bring about reduction. Deduce an equation for the reduction reaction in the electrolyte of a Grätzel cell.
\ndelocalized bonding/conjugated bonds
\ncontain metal atom/ion coordinated to «organic» ligand(s)
\ninvolve bonds from nitrogen to the central metal ion
\n[2 marks]
\nI3– + 2e– → 3I–
\n\n
Accept I2 + 2e– → 2I–.
\n[1 mark]
\nAs well as being burnt, methanol can also be used to provide electricity through a fuel cell. A schematic diagram of such a fuel cell, that depends on the transfer of hydrogen ions between the electrodes, is shown below.
\nEven though fuel cells, primary cells and rechargeable cells have similar fundamental characteristics, there are important differences between them.
\nDeduce half-equations for the reactions at the two electrodes and hence the equation for the overall reaction.
\nSuggest a way in which they are similar.
\nOutline the difference between primary and rechargeable cells.
\nIdentify one factor that affects the voltage of a cell and a different factor that affects the current it can deliver.
\nAnode: CH3OH(aq) + H2O(l) → CO2(aq) + 6H+(aq) + 6e–
\nCathode: O2(aq) + 4H+(aq) + 4e– → 2H2O(l)
\nOverall: 2CH3OH(aq) + 3O2(g) → 2CO2(aq) + 4H2O(l)
\n\n
Accept correctly balanced equations with multiples of the coefficients given here.
\nAccept reversible or non-reversible arrows for all.
\n[3 marks]
\n«portable» sources of electrical energy/electricity
OR
convert chemical «potential» energy to electrical energy/electricity
[1 mark]
\nprimary cells involve irreversible reactions AND rechargeable cells involve reversible reactions
\n\n
Accept “primary cells have a limited life before going ‘flat’ AND rechargeable cells can be recharged when ‘flat’”.
\n[1 mark]
\nVoltage:
chemical nature of electrodes
OR
electrode reactions
Current:
diffusion rate
OR
internal resistance/resistance of the cell
\n
Accept temperature for either but not both.
\nAccept concentration for either but not both.
\nAccept pH for either but not both.
\nAccept the current depends on the area/separation of the electrodes.
\n[2 marks]
\nSolubility plays an important role in the bioavailability of drugs in the body.
\nSuggest why aspirin is slightly soluble in water. Refer to section 37 of the data booklet.
\nA student prepares aspirin from salicylic acid in the laboratory, extracts it from the reaction mixture, ensures the sample is dry and determines its melting point.
\nSuggest why the melting point of the student’s sample is lower and not sharp compared to that of pure aspirin.
\nOrganic molecules can be characterized using infrared (IR) spectroscopy.
\nCompare and contrast the infrared peaks above 1500 cm−1 in pure samples of aspirin and salicylic acid using section 26 of the data booklet.
\nSome mild analgesics contain a solid mixture of acidic aspirin and a non-acidic organic chemical of similar polarity to asprin.
\nDiscuss how acid-base properties and the process of solvent extraction can be used to separate aspirin from the mixture.
\nThe pharmaceutical industry is one of the largest producers of waste solvents.
\nState a green solution to the problem of organic solvent waste.
\npresence of «large» benzene/arene ring AND non-polar/hydrophobic
OR
presence of «large» benzene/arene ring AND cannot form H-bond with water
contain COOH/carboxyl/–OH/hydroxyl «and ester group» AND polar/hydrophilic
OR
contain COOH/carboxyl/–OH/hydroxyl «and ester group» AND can form H-bonds with water
\n
Accept “phenyl” for “benzene ring”.
\nAccept \"carboxylic acid\" for \"carboxyl\".
\nDo not accept \"alcohol\" for \"hydroxyl\".
\n[2 marks]
\n«student’s» sample impure
\nlattice disrupted/not uniform «due to presence of impurities»
OR
fewer interparticle/intermolecular forces «due to presence of impurities»
\n
Accept converse arguments.
\n[2 marks]
\nOne similarity:
peak at 2500–3000 «cm–1»/peak due to O–H/hydroxyl in carboxylic acids
OR
peak at 1700–1750 «cm–1»/peak due to C=O/carbonyl
OR
peak at 2850–3090 «cm–1»/peak due to C–H of arene
One difference:
peak at 3200–3600 «cm–1» in salicylic acid/ peak due to O–H in phenol in salicylic acid
OR
«two» peaks at 1700–1750 «cm–1» in aspirin AND one peak «in the same area» in salicylic acid
\n
Accept “peak at 1600 cm–1 for arene/benzene ring” – not in the data booklet.
\nAccept “2500–3600 cm–1 «overlapping absorptions of two O–H» in salicylic acid”.
\nAccept “stronger/broader/split peak at 1700–1750 cm–1 in aspirin”.
\n[2 marks]
\ndissolve compounds in an organic solvent
\nadd NaOH(aq)/OH–(aq) «to the mixture» to convert aspirin to its water soluble salt
\nseparate the two «immiscible» layers
\nconvert salt «in aqueous layer» back to aspirin by reacting with acid/H+
\n«evaporate solvents and dry»
\n\n
Accept organic solvents immiscible with water such as hexane, ethyl ethanoate, butyl acetate.
\nAccept any other base.
\nNeed explanation for mark.
\n[3 marks]
\n«use of» alternative solvents such as supercritical/liquid CO2
OR
use of water «as solvent»
OR
solvent-free reactions «for example, polymerization of propene»
OR
solid-state chemistry
OR
recycle «waste» solvents
OR
catalysis that leads to better/higher yield
OR
reducing number of steps
\n
Do not accept political/regulatory solutions.
\n“catalysis” not sufficient for mark.
\n[1 mark]
\nA polarimeter can be used to determine the optical rotation of an optically active substance.
\nDescribe what happens to plane-polarized light when it passes through a solution of an optically active compound.
\nA mixture of enantiomers shows optical rotation.
\nSuggest a conclusion you can draw from this data.
\nplane of polarization is rotated
\n\n
Award zero if answer refers to plane-polarized light being bent.
\n[1 mark]
\nnot a racemic mixture
OR
two enantiomers not equimolar
OR
mixture contains optically active impurity
OR
relative proportions of enantiomers in mixture can be determined
[1 mark]
\nNuclear radiation is dangerous because of its ability to damage cells but it can also be used in nuclear medicine.
\nIodine-131 is released in nuclear explosions but is used in scanners for thyroid cancer. The half-life of iodine-131 is 8.02 days.
\nYttrium-90 is used in treating certain cancers.
\nFormulate a nuclear equation for the beta decay of yttrium-90.
\nLutetium-177 is a common isotope used for internal radiation therapy.
\nSuggest why lutetium-177 is an ideal isotope for the treatment of certain cancers based on the type of radiation emitted.
\nCalculate the rate constant, , in day−1, for the decay of iodine-131 using section 1 of the data booklet.
\nCalculate the time, in days, for 90% of the sample to decay.
\nA breathalyser measures the blood alcohol content from a breath sample. Formulate half-equations for the reactions at the anode (negative electrode) and the cathode (positive electrode) in a fuel cell breathalyser.
\n90Y → 90Zr + β–
\n\n
Accept β, e or e–.
Accept 90Y → 90Zr + β– + v
[1 mark]
\nbeta-radiation/emission AND targets tumour/cancer cells
OR
beta-radiation/emission AND limited damage to healthy cells/tissues
OR
beta-radiation/emission AND produces «small amount of» gamma-rays «for visualizing tumours/monitoring treatment»
[1 mark]
\n«day−1»
\n[1 mark]
\nALTERNATIVE 1:
«N0 = initial amount = 100%»
N «= 100 – 90» = 10% at time t
\n«»
\n«» 26.7 «days»
\n\n
Accept 26.6 or 27 «days»
Award [2] for correct final answer.
\n
ALTERNATIVE 2:
«Nt = N0(0.5)n where n = number of half-lives»
10 = 100(0.5)n
\n«»
\n«»
\n«t » 26.6 «days»
\n\n
Accept 26.7 or 27 «days»
\nAward [2] for correct final answer.
\n[2 marks]
\nAnode (negative electrode): C2H5OH + H2O → CH3COOH + 4H+ + 4e–
\nCathode (positive electrode): O2 + 4H+ + 4e– → 2H2O
\n[2 marks]
\nA class was determining the concentration of aqueous sodium hydroxide by titrating it with hydrochloric acid, whilst monitoring the pH of the solution. The sodium hydroxide solution was added into a glass beaker from a measuring cylinder and the hydrochloric acid added using a burette. One group of students accidentally used a temperature probe rather than a pH probe. Their results are given below.
\nVolume of aqueous NaOH = 25.0 ± 0.5 cm3
\nConcentration of HCl = 1.00 ± 0.01 mol dm−3
\nState and explain how the graph would differ if 1 moldm−3 sulfuric acid had been used instead of 1 moldm−3 hydrochloric acid.
\ngraph would peak/maximum at 17.5 cm3
OR
smaller volume of acid «needed to reach equivalence»
sulfuric acid is dibasic/diprotic
\nhigher temperature would be reached
\n\n
Accept “gradient/slope «of graph» is greater/steeper” for M1.
\nAccept “one mole of sulfuric acid neutralizes two moles of NaOH” for M2.
\n[2 marks]
\nWhat is the sum of the coefficients when the equation is balanced with whole numbers?
\n—C8H18(g) + —O2(g) → —CO(g) + —H2O(l)
\nA. 26.5
\nB. 30
\nC. 53
\nD. 61
\nC
\nHow many moles of oxygen atoms are there in 0.500 mol of hydrated iron(II) ammonium sulfate, (NH4)2Fe(SO4)2•6H2O(s)?
\nA. 4.00
\nB. 7.00
\nC. 8.00
\nD. 14.00
\nB
\nWhat is the maximum volume, in dm3, of CO2(g) produced when 1.00 g of CaCO3(s) reacts with 20.0 cm3 of 2.00 moldm–3 HCl(aq)?
\nCaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
\nMolar volume of gas = 22.7 dm3mol–1; Mr(CaCO3) = 100.00
\nA.
\nB.
\nC.
\nD.
\nC
\nWhich factors affect the molar volume of an ideal gas?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nWhat does represent?
\nA. An ion with 12 protons and 24 neutrons
\nB. An ion with 14 protons and 24 neutrons
\nC. An ion with 12 protons and 12 neutrons
\nD. An ion with 12 protons and 22 neutrons
\nC
\nWhich element is a lanthanide?
\nA. Hf
\nB. Tb
\nC. U
\nD. Y
\nB
\nWhat are the approximate bond angles and structure of crystalline SiO2?
\nB
\nWhich metal has the strongest metallic bond?
\nA. Li
\nB. Na
\nC. K
\nD. Rb
\nA
\nWhich electron transition emits radiation of the longest wavelength?
\nC
\nCaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
\nWhich methods can be used to monitor the progress of this reaction?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nWhich element is reduced in the following decomposition?
\n(NH4)2Cr2O7(s) → N2(g) + Cr2O3(s) + 4H2O(g)
\nA. N
\nB. H
\nC. Cr
\nD. O
\nC
\nHow many bonding electrons are there in the urea molecule?
\nA. 8
\nB. 16
\nC. 20
\nD. 24
\nB
\nA class was determining the concentration of aqueous sodium hydroxide by titrating it with hydrochloric acid, whilst monitoring the pH of the solution. The sodium hydroxide solution was added into a glass beaker from a measuring cylinder and the hydrochloric acid added using a burette. One group of students accidentally used a temperature probe rather than a pH probe. Their results are given below.
\nVolume of aqueous NaOH = 25.0 ± 0.5 cm3
\nConcentration of HCl = 1.00 ± 0.01 mol dm−3
\nThe graph of temperature against titre can be used to calculate the concentration of alkali without knowing the concentration of the hydrochloric acid, using the enthalpy of neutralization.
\nExplain how the concentration may be calculated in this way.
\nHeat losses would make this method less accurate than the pH probe method. Outline why the thermometric method would always give a lower, not a higher, concentration.
\nSuggest how heat loss could be reduced.
\nState one other assumption that is usually made in the calculation of the heat produced.
\nSuggest why scientists often make assumptions that do not correspond to reality.
\nOutline why the thermochemical method would not be appropriate for 0.001 moldm−3 hydrochloric acid and aqueous sodium hydroxide of a similar concentration.
\nheat change/evolved can be calculated from the «maximum» temperature increase and the mass of solution
OR
q = mcΔT
heat «evolved» gives the number of moles «of both acid and alkali present when neutralisation occurs»
OR
volume «of acid and the volume of alkali required to just neutralise each other» can be used to calculate the concentration«s of both»
OR
[2 marks]
\nsmaller temperature increase/ΔT
OR
heat released would «appear to» be less
amount of substance/n calculated is smaller
\n[2 marks]
\nusing «expanded» polystyrene cup
OR
insulating beaker
OR
putting a lid on beaker
\n
Do not accept calorimeter by itself.
\nAccept any other reasonable suggestion.
\n[1 mark]
\n«specific» heat capacity of the beaker/container/thermometer is ignored
OR
density of the solutions is assumed as 1.00 gcm–3/same as water
OR
specific heat capacity of the solutions is assumed as 4.18 J g–1K–1/same as water
\n
Accept “reaction goes to completion”.
\nAccept “reaction is conducted under standard conditions”.
\nAccept “no evaporation occurs”.
\nAccept any other relevant valid assumption.
\nDo not accept “heat is not released from other reactions”.
\n[1 mark]
\nallows simple theories to be applied to real life situations
OR
enables us to start to understand complex situations
OR
gives answers that are accurate to the required order of magnitude
OR
simplifies the calculations involved
\n
Do not accept “to simplify the situation” without further detail.
\nAccept “errors do not have a major impact on the results”.
\n[1 mark]
\ntemperature rise would be too small «to be accurately measured»
\n\n
Accept “heat released would be too small «to be accurately measured»”.
\n[1 mark]
\nWhich of the following is not a redox reaction?
\nA. CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
\nB. C(s) + O2(g) → CO2(g)
\nC. 2CO(g) → CO2(g) + C(s)
\nD. CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
\nD
\nWhat occurs at the anode (positive electrode) during the electrolysis of molten strontium bromide?
\nA. Formation of bromine and oxidation
\nB. Formation of bromine and reduction
\nC. Formation of strontium and oxidation
\nD. Formation of strontium and reduction
\nA
\nWhat is the equilibrium constant expression, Kc, for the following reaction?
\n2NH3(g) + 2O2(g) N2O(g) + 3H2O(g)
\nA.
\nB.
\nC.
\nD.
\nD
\nWhich property increases down Group 1, the alkali metals?
\nA. Atomic radius
\nB. Electronegativity
\nC. First ionization energy
\nD. Melting point
\nA
\nWhat can be deduced from this reaction profile?
\nA. The reactants are less stable than the products and the reaction is exothermic.
\nB. The reactants are less stable than the products and the reaction is endothermic.
\nC. The reactants are more stable than the products and the reaction is exothermic.
\nD. The reactants are more stable than the products and the reaction is endothermic.
\nA
\nWhich bonds cause the boiling point of water to be significantly greater than that of hydrogen sulfide?
\nA. London (dispersion)
\nB. Covalent
\nC. Ionic
\nD. Hydrogen
\nD
\nThere are many oxides of silver with the formula AgxOy. All of them decompose into their elements when heated strongly.
\nAfter heating 3.760 g of a silver oxide 3.275 g of silver remained. Determine the empirical formula of AgxOy.
\nSuggest why the final mass of solid obtained by heating 3.760 g of AgxOy may be greater than 3.275 g giving one design improvement for your proposed suggestion. Ignore any possible errors in the weighing procedure.
\nNaturally occurring silver is composed of two stable isotopes, 107Ag and 109Ag.
\nThe relative atomic mass of silver is 107.87. Show that isotope 107Ag is more abundant.
\nSome oxides of period 3, such as Na2O and P4O10, react with water. A spatula measure of each oxide was added to a separate 100 cm3 flask containing distilled water and a few drops of bromothymol blue indicator.
\nThe indicator is listed in section 22 of the data booklet.
\nDeduce the colour of the resulting solution and the chemical formula of the product formed after reaction with water for each oxide.
\nExplain the electrical conductivity of molten Na2O and P4O10.
\nOutline the model of electron configuration deduced from the hydrogen line emission spectrum (Bohr’s model).
\nn(Ag) = «» 0.03036 «mol»
\nAND
\nn(O) = «» 0.03031 «mol»
\n« / ratio of Ag to O approximately 1 : 1, so»
\nAgO
\n\n
Accept other valid methods for M1.
\nAward [1 max] for correct empirical formula if method not shown.
\n[2 marks]
\ntemperature too low
OR
heating time too short
OR
oxide not decomposed completely
heat sample to constant mass «for three or more trials»
\n\n
Accept “not heated strongly enough”.
\nIf M1 as per markscheme, M2 can only be awarded for constant mass technique.
\nAccept \"soot deposition\" (M1) and any suitable way to reduce it (for M2).
\nAccept \"absorbs moisture from atmosphere\" (M1) and \"cool in dessicator\" (M2).
\nAward [1 max] for reference to impurity AND design improvement.
\n[2 marks]
\nAr closer to 107/less than 108 «so more 107Ag»
OR
Ar less than the average of (107 + 109) «so more 107Ag»
\n
Accept calculations that gives greater than 50% 107Ag.
\n[1 mark]
\n\n
Do not accept name for the products.
\nAccept “Na+ + OH–” for NaOH.
\nIgnore coefficients in front of formula.
\n[3 marks]
\n«molten» Na2O has mobile ions/charged particles AND conducts electricity
\n«molten» P4O10 does not have mobile ions/charged particles AND does not conduct electricity/is poor conductor of electricity
\n\n
Do not award marks without concept of mobile charges being present.
\nAward [1 max] if type of bonding or electrical conductivity correctly identified in each compound.
\nDo not accept answers based on electrons.
\nAward [1 max] if reference made to solution.
\n[2 marks]
\nelectrons in discrete/specific/certain/different shells/energy levels
\nenergy levels converge/get closer together at higher energies
OR
energy levels converge with distance from the nucleus
\n
Accept appropriate diagram for M1, M2 or both.
\nDo not give marks for answers that refer to the lines in the spectrum.
\n[2 marks]
\nWhy is the value of the enthalpy change of this reaction calculated from bond enthalpy data less accurate than that calculated from standard enthalpies of formation?
\n2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
\nA. All the reactants and products are gases.
\nB. Bond enthalpy data are average values for many compounds.
\nC. Elements do not have standard enthalpy of formation.
\nD. Standard enthalpies of formation are per mole.
\nB
\nWhich of the following does not react with dilute HCl(aq)?
\nA. Na2CO3
\nB. Cu
\nC. Zn
\nD. CuO
\n\n
B
\nWhat can be deduced from the facts that ozone absorbs UV radiation in the region of 340 nm and molecular oxygen in the region of 242 nm?
\nA. The bond between atoms in molecular oxygen is a double bond.
\nB. The bonds in ozone are delocalized.
\nC. The bonds between atoms in ozone are stronger than those in molecular oxygen.
\nD. The bonds between atoms in molecular oxygen need more energy to break.
\nD
\nAn acidic sample of a waste solution containing Sn2+(aq) reacted completely with K2Cr2O7 solution to form Sn4+(aq).
\nState the oxidation half-equation.
\nDeduce the overall redox equation for the reaction between acidic Sn2+(aq) and Cr2O72–(aq), using section 24 of the data booklet.
\nCalculate the percentage uncertainty for the mass of K2Cr2O7(s) from the given data.
\nThe sample of K2Cr2O7(s) in (i) was dissolved in distilled water to form 0.100 dm3 solution. Calculate its molar concentration.
\n10.0 cm3 of the waste sample required 13.24 cm3 of the K2Cr2O7 solution. Calculate the molar concentration of Sn2+(aq) in the waste sample.
\nSn2+(aq) → Sn4+(aq) + 2e–
\n\n
Accept equilibrium sign.
\nAccept Sn2+(aq) – 2e– → Sn4+(aq).
\n[1 mark]
\nCr2O72–(aq) + 14H+(aq) + 3Sn2+(aq) → 2Cr3+(aq) + 7H2O(l) + 3Sn4+(aq)
\n\n
Accept equilibrium sign.
\n[1 mark]
\n«13.239 g ± 0.002 g so percentage uncertainty» 0.02 «%»
\n\n
Accept answers given to greater precision, such as 0.0151%.
\n[1 mark]
\n« [K2Cr2O7] = =» 0.450 «moldm–3»
\n[1 mark]
\nn(Sn2+) = «0.450 moldm–3 x 0.01324 dm3 x =» 0.0179 «mol»
\n«[Sn2+] = =» 1.79 «moldm–3»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\nCaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
\nWhich change does not increase the initial rate of reaction when CaCO3(s) is added to excess HCl(aq)?
\nA. Decrease in the size of the CaCO3(s) particles
\nB. Increase in the temperature of the reaction mixture
\nC. Increase in the concentration of HCl(aq), keeping the same volume
\nD. Increase in the volume of HCl(aq), keeping the same concentration
\nD
\nWhich of the following is correct?
\nA. A weak acid is a proton donor and its aqueous solution shows good conductivity.
\nB. A weak acid is a proton donor and its aqueous solution shows poor conductivity.
\nC. A weak acid is a proton acceptor and its aqueous solution shows good conductivity.
\nD. A weak acid is a proton acceptor and its aqueous solution shows poor conductivity.
\nB
\nPCl5(g) and Cl2(g) were placed in a sealed flask and allowed to reach equilibrium at 200 °C. The enthalpy change, ΔH, for the decomposition of PCl5(g) is positive.
\n![\"M17/4/CHEMI/SP2/ENG/TZ2/03\"](\"../assets/uploads/tinymce_asset/asset/3687/Schermafbeelding_2017-09-21_om_08.14.28.png\"/)
Deduce the equilibrium constant expression, Kc, for the decomposition of PCl5(g).
\nDeduce, giving a reason, the factor responsible for establishing the new equilibrium after 14 minutes.
\nDeduce the Lewis (electron dot) structure and molecular geometry of PCl3.
\n«Kc» =
\n[1 mark]
\ndecrease in temperature
\nendothermic «reaction» AND «equilibrium» shifts to the left/reactants
OR
endothermic «reaction» AND Kc decreases
OR
endothermic «reaction» AND concentration of PCl5 increased/concentration of PCl3 and Cl2 decreased
OR
«equilibrium» shifts in exothermic direction
\n
Do not accept “temperature change”.
\nAccept “ΔH positive” in place of “endothermic”.
\nAccept “products” instead of “PCl3 and Cl2”.
\n[2 marks]
\nLewis structure:
\nMolecular geometry:
\ntrigonal/triangular pyramidal
\n\n
Penalize missing lone pairs once only between this question and 4(b).
\nAccept any combination of lines, dots or crosses to represent electrons.
\nDo not apply ECF.
\n[2 marks]
\nBonds can be formed in many ways.
\nThe landing module for the Apollo mission used rocket fuel made from a mixture of hydrazine, N2H4, and dinitrogen tetraoxide, N2O4.
\nN2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g)
\nState and explain the difference in bond strength between the nitrogen atoms in a hydrazine and nitrogen molecule.
\nState why hydrazine has a higher boiling point than dinitrogen tetraoxide.
\nDetermine the oxidation state of nitrogen in the two reactants.
\nDeduce, giving a reason, which species is the reducing agent.
\nDeduce the Lewis (electron dot) structures of ozone.
\ntriple bond in nitrogen «molecule» AND single bond in hydrazine
\ntriple bond stronger than single bond
OR
more shared «pairs of» electrons make bond stronger/attract nuclei more
\n
Accept bond enthalpy values from data booklet (158 and 945 kJmol–1).
\n[2 marks]
\nhydrogen bonding «between molecules, dinitrogen tetraoxide does not»
\n[1 mark]
\nN2H4: –2 AND N2O4: +4
\n[1 mark]
\nN2H4 AND oxidized/oxidation state increases
OR
N2H4 AND loses hydrogen
OR
N2H4 AND reduces/removes oxygen from N2O4
\n
Accept “N2H4 AND gives electrons «to N2O4»”.
\n[1 mark]
\nAccept any combination of lines, dots or crosses to represent electrons.
\nDo not penalize missing lone pairs if already done in 3b.
\nDo not accept structure that represents 1.5 bonds.
\n[2 marks]
\nMagnesium reacts with sulfuric acid:
\nMg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)
\nThe graph shows the results of an experiment using excess magnesium ribbon and dilute sulfuric acid.
\n![\"M17/4/CHEMI/SP2/ENG/TZ2/05.a.i\"](\"../assets/uploads/tinymce_asset/asset/3723/Schermafbeelding_2017-09-25_om_09.45.43.png\"/)
Outline why the rate of the reaction decreases with time.
\nSketch, on the same graph, the expected results if the experiment were repeated using powdered magnesium, keeping its mass and all other variables unchanged.
\nNitrogen dioxide and carbon monoxide react according to the following equation:
\nNO2(g) + CO(g) NO(g) + CO2(g) ΔH = –226 kJ
\nCalculate the activation energy for the reverse reaction.
\nState the equation for the reaction of NO2 in the atmosphere to produce acid deposition.
\nconcentration of acid decreases
OR
surface area of magnesium decreases
\n
Accept “less frequency/chance/rate/probability/likelihood of collisions”.
\nDo not accept just “less acid” or “less magnesium”.
\nDo not accept “concentrations of reagents decrease”.
\n[1 mark]
\ncurve starting from origin with steeper gradient AND reaching same maximum volume
\n[1 mark]
\n«Ea(rev) = 226 + 132 =» 358 «kJ»
\nDo not accept –358.
\n[1 mark]
\n2NO2(g) + H2O(l) → HNO3(aq) + HNO2(aq)
OR
2NO2(g) + 2H2O(l) + O2(g) → 4HNO3(aq)
\n
Accept ionised forms of the acids.
\n[1 mark]
\nThe photochemical chlorination of methane can occur at low temperature.
\nUsing relevant equations, show the initiation and the propagation steps for this reaction.
\nBromine was added to hexane, hex-1-ene and benzene. Identify the compound(s) which will react with bromine in a well-lit laboratory.
\nPolyvinyl chloride (PVC) is a polymer with the following structure.
\nState the structural formula for the monomer of PVC.
\nInitiation:
\nCl–Cl → Cl• + Cl•
\n\n
Propagation:
\nCl• + CH4 → Cl–H + •CH3
\nCl–Cl + •CH3 → Cl–CH3 + Cl•
\n\n
Do not penalize missing electron dot on radicals if consistent throughout.
\nAccept Cl2, HCl and CH3Cl without showing bonds.
\nDo not accept hydrogen radical, H• or H, but apply ECF to other propagation steps.
\n[3 marks]
\nhexane AND hex-1-ene
\n\n
Accept “benzene AND hexane AND hex-1-ene”.
\n[1 mark]
\nOR
\n\n
Accept “CH2CHCl” or “CHClCH2”.
\nDo not accept “C2H3Cl”.
\n[1 mark]
\nSoluble acids and bases ionize in water.
\nSodium hypochlorite ionizes in water.
\nOCl–(aq) + H2O(l) OH–(aq) + HOCl(aq)
\nA solution containing 0.510 g of an unknown monoprotic acid, HA, was titrated with 0.100 mol dm–3 NaOH(aq). 25.0 cm3 was required to reach the equivalence point.
\nIdentify the amphiprotic species.
\nIdentify one conjugate acid-base pair in the reaction.
\nCalculate the amount, in mol, of NaOH(aq) used.
\nCalculate the molar mass of the acid.
\nCalculate [H+] in the NaOH solution.
\nwater/H2O
\nAccept “hydroxide ion/OH–”.
\n[1 mark]
\n[1 mark]
\n«0.100 moldm–3 x 0.0250 dm3» = 0.00250 «mol»
\n[1 mark]
\n«M = =» 204 «gmol–1»
\n[1 mark]
\n«1.00 x 10–14 = [H+] x 0.100»
\n1.00 x 10–13 «moldm–3»
\n[1 mark]
\nThe Bombardier beetle sprays a mixture of hydroquinone and hydrogen peroxide to fight off predators. The reaction equation to produce the spray can be written as:
\n| C6H4(OH)2(aq) + H2O2(aq) | \n→ | \nC6H4O2(aq) + 2H2O(l) | \n
| hydroquinone | \n\n | quinone | \n
Calculate the enthalpy change, in kJ, for the spray reaction, using the data below.
\n\n
The energy released by the reaction of one mole of hydrogen peroxide with hydroquinone is used to heat 850 cm3 of water initially at 21.8°C. Determine the highest temperature reached by the water.
\nSpecific heat capacity of water = 4.18 kJkg−1K−1.
\n(If you did not obtain an answer to part (i), use a value of 200.0 kJ for the energy released, although this is not the correct answer.)
\nIdentify the species responsible for the peak at m/z = 110 in the mass spectrum of hydroquinone.
\nIdentify the highest m/z value in the mass spectrum of quinone.
\nΔH = 177.0 – –285.5 «kJ»
\n«ΔH =» –203.1 «kJ»
\n\n
Accept other methods for correct manipulation of the three equations.
\nAward [2] for correct final answer.
\n[2 marks]
\n203.1 «kJ» = 0.850 «kg» x 4.18 «kJkg–1K–1» x ΔT «K»
OR
«ΔT =» 57.2 «K»
«Tfinal = (57.2 + 21.8) °C =» 79.0 «°C» / 352.0 «K»
If 200.0 kJ was used:
200.0 «kJ» = 0.850 «kg» x 4.18 «kJkg–1K–1» x ΔT «K»
OR
«ΔT =» 56.3 «K»
«Tfinal = (56.3 + 21.8) °C =» 78.1 «°C» / 351.1 «K»
\n
Award [2] for correct final answer.
\nUnits, if specified, must be consistent with the value stated.
\n[2 marks]
\nC6H4(OH)2+
\n\n
Accept “molecular ion”.
\nDo not accept “C6H4(OH)2” (positive charge missing).
\n[1 mark]
\n«highest m/z» 108
\n\n
Only accept exactly 108, not values close to this.
\n[1 mark]
\nWhich functional group is present in paracetamol?
\nA. Carboxyl
\nB. Amino
\nC. Nitrile
\nD. Hydroxyl
\nD
\nLanthanum, La, and antimony, Sb, form compounds with bromine that have similar formulas, LaBr3 and SbBr3.
\nDetermine the type of bond present in SbBr3, showing your method. Use sections 8 and 29 of the data booklet.
\nLanthanum has a similar electronegativity to group 2 metals. Explain, in terms of bonding and structure, why crystalline lanthanum bromide is brittle.
\npolar covalent
\naverage electronegativity «= (3.0 + 2.0)» = 2.5 AND electronegativity difference «= 3.0 – 2.0» = 1.0
\n[2 marks]
\nionic bonding
OR
electrostatic forces between ions
«slight» movement brings ions of same charge adjacent to each other «causing the crystal to break»
OR
«slight» movement results in repulsion between layers «causing the crystal to break»
[2 marks]
\nLanthanum metal may be produced by the electrolysis of molten LaBr3.
\nState why lanthanum cannot be produced by reducing its oxide with carbon.
\nCalculate the current (I), in A, required to produce 1.00 kg of lanthanum metal per hour. Use the formula and sections 2 and 6 of the data booklet.
\ntoo high/higher than carbon in the reactivity series
OR
carbon/C is a weaker reducing agent than lanthanum/La
\n
Accept “lanthanum is more reactive than carbon”.
\nAccept “lanthanum is a weaker oxidizing agent than carbon”.
\nAccept converse arguments.
\n[1 mark]
\namount of La «» = 7.20 «mol»
\nQ «= 7.20 mol x 3 x 96500 Cmol–1» = 2.08 x 106 «C»
\nI «» = 579 «A»
\n\n
Award [3] for “578 «A»” (from premature rounding) or “579 «A»”.
\n[3 marks]
\nLanthanum nanoparticles are incorporated into certain catalysts and the electrodes of some fuel cells.
\nState the major advantage that nanoparticles have in these applications.
\nSuggest why nanoparticles need to be handled with care.
\nlarge surface area
\n[1 mark]
\n«potentially» explosive
OR
small size/large surface area could give dangerously fast reactions
OR
unknown health effects
OR
potentially toxic
OR
immune system/allergy concerns
\n
Do not accept just “dangerous/poisonous/toxic”.
\nAccept other valid concerns.
\n[1 mark]
\nAntimony oxide is widely used as a homogeneous catalyst in the reaction of benzene-1,4-dicarboxylic acid with ethane-1,2-diol in the production of polyethylene terephthalate (PETE) shown below.
\n\n
Catalysts reduce the activation energy. Outline how homogeneous catalysts are involved in the reaction mechanism.
\nSuggest why it is important to know how catalysts function.
\nAntimony and its compounds are toxic, so it is important to check that the catalyst is removed from the final product. One technique to detect antimony is Inductively Coupled Plasma Mass Spectroscopy (ICP-MS).
\nOutline the nature of the plasma state and how it is produced in ICP-MS.
\ncombine with reactants to form a «temporary» activated complex/intermediate
OR
consumed in one reaction/step AND regenerated in a later reaction/step
[1 mark]
\ncan modify/improve the catalyst/reaction «by making logical predictions»
OR
science relies on models to understand physical reality
\n
Accept other reasonable, relevant answers.
\nAccept \"to predict/select
\n[1 mark]
\nelectrons AND positive ions «in gaseous state»
\nhigh frequency/alternating current passed through argon
OR
«oscillating» electromagnetic/magnetic field
OR
high frequency radio waves
\n
Accept “gas” instead of “argon”.
\n[2 marks]
\nInfrared (IR) spectroscopy is often used for the identification of polymers, such as PETE, for recycling.
\nLDPE and high density polyethene (HDPE) have very similar IR spectra even though they have rather different structures and physical properties.
\nBelow are the IR spectra of two plastics (A and B); one is PETE, the other is low density polyethene (LDPE).
\nDeduce, giving your reasons, the identity and resin identification code (RIC) of A and B using sections 26 and 30 of the data booklet.
\nDescribe the difference in their structures.
\nExplain why the difference in their structures affects their melting points.
\nA RIC: 1 AND B RIC: 4
\nALTERNATIVE 1:
«only» PETE contains carbonyl/C=O/ester/COO groups
carbonyl groups absorb at 1700–1750 «cm–1»
ALTERNATIVE 2:
LDPE contains more C–H bonds «than PETE»
C–H bonds absorb at 2850–3090 «cm–1»
\n
For either, accept specific frequencies in these ranges (eg 1735 «cm–1» or 2900 «cm–1»).
\n[3 marks]
\nHDPE less branched
OR
LDPE more branched
\n
Accept “no branching in HDPE AND branching in LDPE”.
\n[1 mark]
\nHDPE «polymer» chains/molecules can pack together more closely «than LDPE chains»
OR
HDPE «polymer» chains/molecules have a higher contact surface area «than LDPE chains»
stronger intermolecular/dispersion/London/van der Waals’ forces in HDPE AND higher melting point
\n\n
Accept converse arguments.
\n[2 marks]
\nAn acidic sample of a waste solution containing Sn2+(aq) reacted completely with K2Cr2O7 solution to form Sn4+(aq).
\nIdentify one organic functional group that can react with acidified K2Cr2O7(aq).
\nCorrosion of iron is similar to the processes that occur in a voltaic cell. The initial steps involve the following half-equations:
\nFe2+(aq) + 2e– Fe(s)
\nO2(g) + H2O(l) + 2e– 2OH–(aq)
\nCalculate E θ, in V, for the spontaneous reaction using section 24 of the data booklet.
\nCalculate the Gibbs free energy, ΔG θ, in kJ, which is released by the corrosion of 1 mole of iron. Use section 1 of the data booklet.
\nExplain why iron forms many different coloured complex ions.
\nZinc is used to galvanize iron pipes, forming a protective coating. Outline how this process prevents corrosion of the iron pipes.
\nhydroxyl/OH
OR
aldehyde/CHO
\n
Accept “hydroxy/alcohol” for “hydroxyl”.
\nAccept amino/amine/NH2.
\n[1 mark]
\n«E θ =» +0.85 «V»
\n\n
Accept 0.85 V.
\n[1 mark]
\nΔG θ «= –nFE θ» = –2 «mol e–» x 96500 «C mol–1» x 0.85 «V»
\n«ΔG θ =» –164 «kJ»
\n\n
Accept “«+»164 «kJ»” as question states energy released.
\nAward [1 max] for “+” or “–” 82 «kJ».
\nDo not accept answer in J.
\n[2 marks]
\nincompletely filled d-orbitals
\ncolour depends upon the energy difference between the split d-orbitals
\nvariable/multiple/different oxidation states
\ndifferent «nature/identity of» ligands
\ndifferent number of ligands
\n[3 marks]
\nZn/zinc is a stronger reducing agent than Fe/iron
OR
Zn/zinc is oxidized instead of Fe/iron
OR
Zn/zinc is the sacrificial anode
\n
Accept “Zn is more reactive than Fe”.
\nAccept “Zn oxide layer limits further corrosion”.
\nDo not accept “Zn layer limits further corrosion”.
\n[1 mark]
\nAnalytical chemistry uses instruments to separate, identify, and quantify matter.
\nNitric oxide reacts with chlorine.
\n2NO (g) + Cl2 (g) → 2NOCl (g)
\nThe following experimental data were obtained at 101.3 kPa and 263 K.
\nMenthol is an organic compound containing carbon, hydrogen and oxygen.
\nOutline how this spectrum is related to the energy levels in the hydrogen atom.
\nA sample of magnesium has the following isotopic composition.
\nCalculate the relative atomic mass of magnesium based on this data, giving your answer to two decimal places.
\nComplete combustion of 0.1595 g of menthol produces 0.4490 g of carbon dioxide and 0.1840 g of water. Determine the empirical formula of the compound showing your working.
\n0.150 g sample of menthol, when vaporized, had a volume of 0.0337 dm3 at 150 °C and 100.2 kPa. Calculate its molar mass showing your working.
\nDetermine the molecular formula of menthol using your answers from parts (d)(i) and (ii).
\nDeduce the order of reaction with respect to Cl2 and NO.
\nState the rate expression for the reaction.
\nCalculate the value of the rate constant at 263 K.
\nelectron transfer/transition between high«er» energy level to low«er» energy level
\nOR
\nelectron transitions into first energy level causes UV series
\nOR
\ntransition into second energy level causes visible series
\nOR
\ntransition into third energy level causes infrared series
\nAccept any of the points shown on a diagram.
\n24 x 0.786 + 25 x 0.101 + 26 x 0.113
\n24.33
\nAward [2] for correct final answer.
Award [0] for 24.31 with no working (data booklet value).
carbon: « =» 0.01020 «mol» / 0.1225 «g»
OR
hydrogen: « =» 0.02042 «mol» / 0.0206 «g»
oxygen: «0.1595 – (0.1225 + 0.0206)» = 0.0164 «g» / 0.001025 «mol»
\nempirical formula: C10H20O
\nAward [3] for correct final answer.
\nDo not award M3 for a hydrocarbon.
\n«temperature =» 423 K
OR
M
«M » 156 «g mol–1»
\nAward [1] for correct answer with no working shown.
\nAccept “pV = nRT AND n = ” for M1.
\nC10H20O
\n[1 Mark]
\nCl2: first
NO: second
rate = k [NO]2 [Cl2]
\n180 / 1.80 x 102 «dm6 mol–2 min–1»
\nPCl5(g) and Cl2(g) were placed in a sealed flask and allowed to reach equilibrium at 200 °C. The enthalpy change, ΔH, for the decomposition of PCl5(g) is positive.
\n
Deduce the Lewis (electron dot) structure and molecular geometry and the bond angles of PCl3.
\nLewis structure:
\nMolecular geometry:
\ntrigonal/triangular pyramidal
\n\n
Bond angles:
\n< 109.5°
\n\n
Penalize missing lone pairs once only between this question and 4(b)(ii).
\nAccept any combination of lines, dots or crosses to represent electrons.
\nDo not apply ECF.
\nDo not accept answer equal to or less than 90°.
Literature value is 100.1°.
\n[3 marks]
\nBonds can be formed in many ways.
\nThe equilibrium for a mixture of NO2 and N2O4 gases is represented as:
\n2NO2(g) N2O4(g)
\nAt 100°C, the equilibrium constant, Kc, is 0.21.
\nBonds can be formed in many ways.
\nDiscuss the bonding in the resonance structures of ozone.
\nDeduce one resonance structure of ozone and the corresponding formal charges on each oxygen atom.
\nThe first six ionization energies, in kJ mol–1, of an element are given below.
\n![\"M17/4/CHEMI/HP2/ENG/TZ2/04.c\"](\"../assets/uploads/tinymce_asset/asset/3691/Schermafbeelding_2017-09-21_om_08.29.16.png\"/)
Explain the large increase in ionization energy from IE3 to IE4.
\nAt a given time, the concentration of NO2(g) and N2O4(g) were 0.52 and respectively.
\nDeduce, showing your reasoning, if the forward or the reverse reaction is favoured at this time.
\nComment on the value of ΔG when the reaction quotient equals the equilibrium constant, Q = K.
\nlone pair on p orbital «of O atom» overlaps/delocalizes with pi electrons «from double bond»
\nboth O–O bonds have equal bond length
OR
both O–O bonds have same/1.5 bond order
OR
both O–O are intermediate between O–O AND O=O
both O–O bonds have equal bond energy
\n\n
Accept “p/pi/ electrons are delocalized/not localized”.
\n[3 marks]
\nALTERNATIVE 1:
\nFC: –1 AND +1 AND 0
\n\n
ALTERNATIVE 2:
\nFC: 0 AND +1 AND –1
\n\n
Accept any combination of lines, dots or crosses to represent electrons.
\nDo not accept structure that represents 1.5 bonds.
\nDo not penalize missing lone pairs if already penalized in 3(b).
\nIf resonance structure is incorrect, no ECF.
\nAny one of the structures with correct formal charges for [2 max].
\n[2 marks]
\nAny two of:
IE4: electron in lower/inner shell/energy level
OR
IE4: more stable/full electron shell
IE4: electron closer to nucleus
OR
IE4: electron more tightly held by nucleus
IE4: less shielding by complete inner shells
\n\n
Accept “increase in effective nuclear charge” for M2.
\n[2 marks]
\n«Qc = =» 0.37
reaction proceeds to the left/NO2(g) «until Q = Kc»
OR
reverse reaction «favoured»
\n
Do not award M2 without a calculation for M1 but remember to apply ECF.
\n[2 marks]
\nΔG = 0
reaction at equilibrium
OR
rate of forward and reverse reaction is the same
OR
constant macroscopic properties
[2 marks]
\nEnzyme activity depends on many factors. Explain how pH change causes loss of activity of an enzyme.
\nconformation/shape altered
\nOR
\nactive site altered
\nOR
\ntertiary structure altered
\nacidic/basic/ionizable/COOH/carboxyl/NH2/amino groups in the R groups/side chains «react»
\nexchange/lose/gain protons/H+
\nionic/H-bonds altered
\nAccept “substrate doesn't fit/fits poorly into active site” OR “enzyme denatures” for M1 but not “affects potential of
enzyme to form complex with substrate”.
One method of comparing fuels is by considering their specific energies.
\nCalculate the specific energy of octane, C8H18, in kJ kg–1 using sections 1, 6 and 13 of the data booklet.
\nA typical wood has a specific energy of 17 × 103 kJ kg–1. Comment on the usefulness of octane and wood for powering a moving vehicle, using your answer to (a).
\nIf you did not work out an answer for (a), use 45 × 103 kJ kg–1 but this is not the correct answer.
\nState the name of one renewable source of energy other than wood.
\nMr (C8H18) = 114.26 AND ΔH= –5470 «kJ mol–1»
\n«specific energy = =» 4.79 x 104/47873/47900 «kJ kg–1»
\nAward [2] for correct final answer.
\nAccept “48 x 103 «kJ kg–1»” OR “47.9 x 103 «kJ kg–1»”.
\nwood is less useful because it requires «about three times» more mass for same energy
\nAccept “octane is more useful because it has higher specific energy”.
\nAny one of:
wind
tidal/wave
hydro-electric
solar
thermal/geothermal
plant oil
Accept “biofuel/biodiesel/«bio»ethanol” but not just “water” or “fuel cells”.
\n[Max 1 Mark]
\nMuch of our energy needs are still provided by the refined products of crude oil.
\n“Knocking” in an automobile (car) engine can be prevented by increasing the octane number of the fuel. Explain, including an equation with structural formulas, how heptane, C7H16, could be chemically converted to increase its octane number.
\nMany like to refer to our “carbon footprint”. Outline one difficulty in quantifying such a concept.
\nClimate change or global warming is a consequence of increased levels of carbon dioxide in the atmosphere. Explain how the greenhouse effect warms the surface of the earth.
\nOutline how water and carbon dioxide absorb infrared radiation.
\nCH3CH2CH2CH2CH2CH2CH3 → CH3CH(CH3)CH2CH(CH3)2
\nOR
\nCH3CH2CH2CH2CH2CH2CH3 →
isomerisation/reforming/platforming/cracking
\nPt/Re/Rh/Pd/Ir
\nOR
\ncatalyst
\nA structural formula is only required for the organic product, not heptane.
\nAccept any correctly balanced equation showing increased branching or cyclization OR aromatization OR cracking.
\nSuitable supports for catalysts may be included for M3 (eg silica, alumina, zeolite) but the symbol or name of an appropriate metal must be given (typically a noble metal). Ignore temperature and other conditions.
\nAward M2 AND M3 for “catalytic isomerisation” OR “catalytic reforming” OR “catalytic cracking”.
\nwhich specific carbon-based greenhouse gases are included
\nOR
\nwhether non-carbon based greenhouse gases should be included
\nOR
\nwhether CO/incomplete combustion should be included «as can be oxidized to CO2»
\nOR
\nhow to “sum” all steps in a process creating CO2
\nOR
\ndifficult to determine both direct and indirect production of GHG/greenhouse gas emissions
\nIgnore reference to geopolitical issues (eg false recording of data by governments etc.).
\nAccept “difficult to measure all sources of CO2” but not “difficult to measure CO2 released in atmosphere”.
\nAny three of:
\nincoming solar radiation is short wavelength/high frequency/high energy/UV
\nradiated/emitted as long wavelength/low frequency/low energy/IR «radiation»
\nenergy/IR «radiation» absorbed by «bonds in» greenhouse gases
\nenergy radiated/emitted as IR «radiation» some of which returns back to Earth
\nDo not accept “reflected” OR “bounced” OR “trapped”.
\n[Max 3 Marks]
\nbond length changes
\nOR
\n«asymmetric» stretching «of bonds»
\nOR
\nbond angle changes/bends
\nOR
\npolarity/dipole «moment» changes
\nOR
\na dipole «moment» is created «when the molecule absorbs IR»
\nAccept “vibration of bonds” OR appropriate diagram
\nIn the 20th Century, both fission and fusion were considered as sources of energy but fusion was economically and technically unattainable.
\nCompare and contrast fission and fusion in terms of binding energy and the types of nuclei involved.
\nSuggest two advantages that fusion has over fission.
\nThe amount of 228Ac in a sample decreases to one eighth of its original value in about 18 hours due to β-decay. Estimate the half-life of 228Ac.
\nFission: heavy nuclei AND Fusion: light nuclei
\nboth increase in binding energy/energy yield «per nucleon»
\nAccept “large nuclei” OR “greater atomic masses of nuclei” for fission AND “small nuclei” OR “smaller atomic masses of nuclei” for fusion.
\nAward [1 max] for “Fission: heavy nuclei AND increase in binding energy «per nucleon»” OR “Fusion: light nuclei AND increase in binding energy” «per nucleon»”.
\nAny two of:
\nno/less radioactive waste produced
\nabundance/low cost of fuel
\nlarger amounts of energy released per unit mass
\ndoes not require a critical mass
\ncan be used continuously
\nfusion reactor less likely to cause large-scale technological disaster
\nDo not accept \"no/less waste produced\".
\nAccept “higher specific energy”.
\n[Max 2 Marks]
\n6 «hours»
\nNitrogen dioxide and carbon monoxide react according to the following equation:
\nNO2(g) + CO(g) NO(g) + CO2(g) ΔH = –226 kJ
\nExperimental data shows the reaction is second order with respect to NO2 and zero order with respect to CO.
\nState the rate expression for the reaction.
\nThe following mechanism is proposed for the reaction.
\n\n
Identify the rate determining step giving your reason.
\nState one method that can be used to measure the rate for this reaction.
\nSketch the relationship between the rate of reaction and the concentration of NO2.
\nThe Arrhenius equation, , gives the relationship between the rate constant and temperature.
\nState how temperature affects activation energy.
\n«rate =» k [NO2]2
\n\n
Accept rate = k [NO2]2[CO]0.
\n[1 mark]
\n«step» I AND CO does not appear in the rate law expression
OR
«step» I AND only «2 molecules of» NO2 appears in rate expression
\n
Do not allow ECF from (i).
\n[1 mark]
\n«IR or UV-vis» spectroscopy
OR
colorimetry
OR
colour change «over time»
\n
Accept GC/gas chromatography.
\n[1 mark]
\n\n
Curve must go through origin.
\n[1 mark]
\nactivation energy is independent of temperature
\n\n
Accept “no relationship”.
\n[1 mark]
\nVegetable oils can be used as a source of energy.
\nState the structural feature of chlorophyll that enables it to absorb visible light.
\nVegetable oils are too viscous for use as liquid fuels. Describe, using an equation, how a vegetable oil, such as that shown, is converted to oils with lower viscosity by reaction with methanol, CH3OH.
\n«extensive» conjugation
\nOR
\nalternating single and double bonds
\nester product
\nglycerol AND correct balancing
\nCatalyst not required for equation.
\nAward M2 only if M1 is correct.
\nWhich describes the reaction between a halogen and ethane?
\nA
\nThe photochemical chlorination of methane can occur at low temperature.
\nThe overall equation for monochlorination of methane is:
\nCH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
\nCalculate the standard enthalpy change for the reaction, ΔH θ, using section 12 of the data booklet.
\n«ΔH θ =» –82.0 «kJ» –92.3 «kJ» – (–74.0 «kJ»)
\n«ΔH θ =» –100.3 «kJ»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\nWhich conditions are used to convert ethanol to ethanal?
\nA. Excess oxidizing agent and reflux
\nB. Excess oxidizing agent and distillation
\nC. Excess ethanol and reflux
\nD. Excess ethanol and distillation
\nD
\nWhich compound contains a secondary carbon atom?
\nA. CH3CH(Cl)CH(CH3)2
\nB. (CH3)2CHCH2Cl
\nC. (CH3)3CCl
\nD. CH3CH2Cl
\nA
\nWhich information can be gained from an infrared (IR) spectrum?
\nA. Ionization energy of the most abundant element
\nB. Number of different elements in the compound
\nC. Bonds present in a molecule
\nD. Molecular formula of the compound
\nC
\nWhat can be deduced from the following 1HNMR spectrum?
\nA. There is only one hydrogen atom in the molecule.
\nB. There is only one hydrogen environment in the molecule.
\nC. The molecule is a hydrocarbon.
\nD. There is only one isotope in the element.
\nB
\nCalcium carbonate reacts with hydrochloric acid.
\nCaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
\nThe results of a series of experiments in which the concentration of HCl was varied are shown below.
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/X04.b\"](\"../assets/uploads/tinymce_asset/asset/4466/Schermafbeelding_2018-08-07_om_11.18.37.png\"/)
Outline two ways in which the progress of the reaction can be monitored. No practical details are required.
\nSuggest why point D is so far out of line assuming human error is not the cause.
\nDraw the best fit line for the reaction excluding point D.
\nSuggest the relationship that points A, B and C show between the concentration of the acid and the rate of reaction.
\nDeduce the rate expression for the reaction.
\nCalculate the rate constant of the reaction, stating its units.
\nPredict from your line of best fit the rate of reaction when the concentration of HCl is 1.00 mol dm−3.
\nDescribe how the activation energy of this reaction could be determined.
\nAny two of:
\nloss of mass «of reaction mixture/CO2»
\n«increase in» volume of gas produced
\nchange of conductivity
\nchange of pH
\nchange in temperature
\n\n
Do not accept “disappearance of calcium carbonate”.
\nDo not accept “gas bubbles”.
\nDo not accept “colour change” or “indicator”.
\n[2 marks]
\nreaction is fast at high concentration AND may be difficult to measure accurately
\nOR
\nso many bubbles of CO2 produced that inhibit contact of HCl(aq) with CaCO3(s)
\nOR
\ninsufficient change in conductivity/pH at high concentrations
\nOR
\ncalcium carbonate has been used up/is limiting reagent/ there is not enough calcium carbonate «to react with the high concentration of HCl»
\nOR
\nHCl is in excess
\nOR
\nso many bubbles of CO2 produced that inhibit contact of HCl(aq) with CaCO3(s)
\n\n
[1 mark]
\n![\"M18/4/CHEMI/HP2/ENG/TZ1/04.b.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4487/Schermafbeelding_2018-08-07_om_14.39.56.png\"/)
straight line going through the origin AND as close to A, B, C as is reasonably possible
\n[1 mark]
\n«directly» proportional
\n\n
Accept “first order” or “linear”.
\nDo not accept “rate increases as concentration increases” or “positive correlation”.
\n[1 mark]
\nrate = k [H+]
\n\n
Accept “rate = k [HCl]”.
\n[1 mark]
\n0.02
\ns–1
\n\n
[2 marks]
\n20.5 10–3 «mol dm–3 s–1»
\n\n
Accept any answer in the range 19.5–21.5.
\n[1 mark]
\nALTERNATIVE 1:
\ncarry out reaction at several temperatures
\nplot against log rate constant
\nEa = – gradient R
\n\n
ALTERNATIVE 2:
\ncarry out reaction at two temperatures
\n\n
determine two rate constants
\nOR
\ndetermine the temperature coefficient of the rate
\n\n
use the formula
\n\n
\n
Accept “gradient = ” for M3.
\nAward both M2 and M3 for the formula .
\nAccept any variation of the formula, such as .
\n[3 marks]
\nWhat is the graphical relationship between n and T in the ideal gas equation, pV = nRT, all other variables remaining constant?
\nD
\nX, Y and Z represent the successive elements, Ne, Na and Mg, but not necessarily in that order.
\nWhat is the order of increasing atomic number?
\nA. X < Y < Z
\nB. X < Z < Y
\nC. Y < Z < X
\nD. Y < X < Z
\nA
\nAmmonia is a stronger ligand than water. Which is correct when concentrated aqueous ammonia solution is added to dilute aqueous copper(II) sulfate solution?
\nA. The d-orbitals in the copper ion split.
\nB. There is a smaller splitting of the d-orbitals.
\nC. Ammonia replaces water as a ligand.
\nD. The colour of the solution fades.
\nC
\nWhich does not show resonance?
\nA. PO43–
\nB. C6H6
\nC. C6H12
\nD. O3
\nC
\nWhich is the first step in the CFC-catalysed destruction of ozone in UV light?
\nA. CCl2F2 → CClF2+ + Cl–
\nB. CCl2F2 → •CClF2 + Cl•
\nC. CCl2F2 → CCl2F+ + F–
\nD. CCl2F2 → •CCl2F + F•
\nB
\nCobalt forms the transition metal complex [Co(NH3)4 (H2O)Cl]Br.
\nTrends in physical and chemical properties are useful to chemists.
\nExplain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
\nState the shape of the complex ion.
\nDeduce the charge on the complex ion and the oxidation state of cobalt.
\nDescribe, in terms of acid-base theories, the type of reaction that takes place between the cobalt ion and water to form the complex ion.
\nAny three of:
\nGroup 1:
atomic/ionic radius increases
smaller charge density
\nOR
\nforce of attraction between metal ions and delocalised electrons decreases
\nDo not accept discussion of attraction between valence electrons and nucleus for M2.
\nAccept “weaker metallic bonds” for M2.
\nGroup 17:
number of electrons/surface area/molar mass increase
London/dispersion/van der Waals’/vdw forces increase
\nAccept “atomic mass” for “molar mass”.
\n[Max 3 Marks]
\n«distorted» octahedral
\nAccept “square bipyramid”.
\nCharge on complex ion: 1+/+
Oxidation state of cobalt: +2
Lewis «acid-base reaction»
\nH2O: electron/e– pair donor
\nOR
\nCo2+: electron/e– pair acceptor
\nLewis (electron dot) structures are useful models.
\nDraw the Lewis (electron dot) structures of PF3 and PF5 and use the VSEPR theory to deduce the molecular geometry of each species including bond angles.
\nPredict whether the molecules PF3 and PF5 are polar or non-polar.
\nState the type of hybridization shown by the phosphorus atom in PF3.
\n\n
Accept any combination of dots, crosses and lines.
\nPenalize missing lone pairs once only.
\nDo not apply ECF for molecular geometry.
\nAccept values in the range 95–109 for PF3.
\nPF3 polar AND PF5 non-polar
\nApply ECF from part (a) molecular geometry.
\nsp3
\nWhich statement is correct?
\nA. Sigma bonds are formed only by the combination of s atomic orbitals.
\nB. Pi bonds can be formed in the absence of sigma bonds.
\nC. Pi bonds are formed parallel to the axis between atoms.
\nD. Pi bonds are formed only by the combination of hybrid orbitals.
\nC
\nThe Born-Haber cycle for potassium oxide is shown below:
\nWhich expression represents the lattice enthalpy in kJ mol–1?
\nA. –361 + 428 + 838 + 612
\nB. –(–361) + 428 + 838 + 612
\nC. –361 + 428 + 838 – 612
\nD. –(–361) + 428 + 838 – 612
\nB
\nWhich ion’s hydration energy is the most exothermic?
\nA. Li+
\nB. Na+
\nC. Br–
\nD. I–
\nA
\nEthane-1,2-diol, HOCH2CH2OH, reacts with thionyl chloride, SOCl2, according to the reaction below.
\nHOCH2CH2OH (l) + 2SOCl2 (l) → ClCH2CH2Cl (l) + 2SO2 (g) + 2HCl (g)
\nCalculate the standard enthalpy change for this reaction using the following data.
\nCalculate the standard entropy change for this reaction using the following data.
\nThe standard free energy change, ΔGθ, for the above reaction is –103 kJ mol–1 at 298 K.
\nSuggest why ΔGθ has a large negative value considering the sign of ΔHθ in part (a).
\nΔHθ = [–165.2 + 2(–296.9) + 2(–92.3)] – [–454.7 + 2(–245.7)]
\n«ΔHθ = +»2.5 «kJ»
\n\n
Award [2] for correct final answer.
\nAward [1] for –2.5 «kJ».
\nDo not accept ECF for M2 if more than one error in M1.
\n«ΔSθ = [208.5 + 2(248.1) + 2(186.8)] – [166.9 + 2(278.6)]»
\n«ΔSθ = +» 354.2 «J K–1 mol–1»
\n«3 moles of» liquid to «4 moles of» gas
\nOR
\n«large» positive ΔS
\nOR
\n«large» increase in entropy
\nTΔS > ΔH «at the reaction temperature»
\nWhich is true of an Arrhenius plot of (y-axis) against ?
\nA. The graph goes through the origin.
\nB. The activation energy can be determined from the gradient.
\nC. The intercept on the x-axis is the activation energy.
\nD. The intercept on the y-axis is the frequency factor, A.
\nB
\nWhich is correct about reaction mechanisms?
\nA. A species that is zero order does not take part in the reaction.
\nB. A catalyst does not take part in the reaction.
\nC. Reactants in a fast step before the slow step are included in the rate expression.
\nD. Reactants in a fast step after the slow step are included in the rate expression.
\nC
\nRadioisotopes are used to diagnose and treat various diseases. Explain the low environmental impact of most medical nuclear waste.
\nAny two of:
\nemits weak ionising radiation
\nOR
\nlow activity/radioactivity
\ncan be stored until material becomes inactive AND then disposed with normal waste
\n«isotopes» have short lives
\nOR
\nexist for a short period of time
\nAward [1 max] for “low-level waste/LLW”.
\n[Max 2 Marks]
\nSome analgesics are derived from compounds found in plants.
\nAspirin is a mild analgesic derived from salicylic acid found in willow bark.
\nDescribe how mild analgesics function.
\nThe strong analgesics morphine and codeine are opiates. Outline how codeine can be synthesized from morphine. The structures of morphine and codeine are in section 37 of the data booklet.
\nExplain why opiates are addictive.
\nprevents/interferes with the production of prostaglandins
\nOR
\nprevents/interferes with the production of substances responsible for
\ninflammation/pain/fever
\nat the site of injury/source of pain
\nreact with CH3I/methyl iodide «in alkaline solution»
\nAccept “react with CH3Cl/methyl chloride” OR “react with methyl halide”.
\nAccept name or formula of a suitable specific methylating reagent (eg trimethylphenylammonium chloride etc.).
\nAccept “hydroxy/alcohol” but not “hydroxide” for “hydroxyl”.
\nAny two of:
\ninteract with opioid receptors in the brain
\nalter the structure of brain cells
\nOR
\nalter the way the brain works «so that it only works normally when the opiates are present»
\nOR
\nprevents transmission of pain impulses inside the brain
\nrelease dopamine «that the person craves»
\nOR
\ngive a feeling of pleasure/euphoria «that the person craves»
\nwithdrawal symptoms «prevent patient from terminating drug use»
\nAccept specific withdrawal symptoms.
\n[Max 2 Marks]
\nOseltamivir (Tamiflu) and zanamivir (Relenza) are antiviral drugs used to prevent flu.
\nState the names of two functional groups that both compounds contain, using section 37 of the data booklet.
\nExplain how oseltamivir and zanamivir can stop the spread of the flu virus in the body.
\nAny two of:
\namido
\nether
\ncarbonyl
\nAccept “amide/carboxamide”.
\nAccept “alkenyl/alkene”.
\nAccept “amino/amine”.
\n[Max 2 Marks]
\nby preventing the virus from leaving the host cell
\nby inhibiting viral enzymes/neuraminidases «needed to release virus»
\nWhich variable affects the equilibrium constant, Kc?
\nA. Atmospheric pressure
\nB. Catalyst
\nC. Concentration of reactants
\nD. Temperature
\nD
\nLimescale, CaCO3(s), can be removed from water kettles by using vinegar, a dilute solution of ethanoic acid, CH3COOH(aq).
\nPredict, giving a reason, a difference between the reactions of the same concentrations of hydrochloric acid and ethanoic acid with samples of calcium carbonate.
\nDissolved carbon dioxide causes unpolluted rain to have a pH of approximately 5, but other dissolved gases can result in a much lower pH. State one environmental effect of acid rain.
\nWrite an equation to show ammonia, NH3, acting as a Brønsted–Lowry base and a different equation to show it acting as a Lewis base.
\n\n
Determine the pH of 0.010 mol dm−3 2,2-dimethylpropanoic acid solution.
\nKa (2,2-dimethylpropanoic acid) = 9.333 × 10−6
\nExplain, using appropriate equations, how a suitably concentrated solution formed by the partial neutralization of 2,2-dimethylpropanoic acid with sodium hydroxide acts as a buffer solution.
\nslower rate with ethanoic acid
\nOR
\nsmaller temperature rise with ethanoic acid
\n\n
[H+] lower
\nOR
\nethanoic acid is weak
\nOR
\nethanoic acid is partially dissociated
\n\n
Accept experimental observations such as “slower bubbling” or “feels less warm”.
\n\n
[2 marks]
\nAny one of:
\ncorrosion of materials/metals/carbonate materials
\ndestruction of plant/aquatic life
\n«indirect» effect on human health
\n\n
Accept “lowering pH of oceans/lakes/waterways”.
\n[1 mark]
\nBrønsted–Lowry base:
\nNH3 + H+ → NH4+
\nLewis base:
\nNH3 + BF3 → H3NBF3
\n\n
Accept “AlCl3 as an example of Lewis acid”.
\nAccept other valid equations such as Cu2+ + 4NH3 → [Cu(NH3)4]2+.
\n[2 marks]
\n[H+] «» = 3.055 × 10–4 «mol dm–3»
\n«pH =» 3.51
\n\n
Accept “pH = 3.52”.
\nAward [2] for correct final answer.
\nAccept other calculation methods.
\n[2 marks]
\n(CH3)3CCOOH(aq) + OH–(aq) → (CH3)3CCOO–(aq) + H2O(l)
\nOR
\n(CH3)3CCOOH(aq) + OH–(aq) (CH3)3CCOO–(aq) + H2O(l) AND addition of alkali causes equilibrium to move to right
\n\n
(CH3)3CCOO–(aq) + H+(aq) → (CH3)3CCOOH(aq)
\nOR
\n(CH3)3CCOO–(aq) + H+(aq) (CH3)3CCOOH(aq) AND addition of acid causes equilibrium to move to right
\n\n
Accept “HA” for the acid.
\nAward [1 max] for correct explanations of buffering with addition of acid AND base without equilibrium equations.
\n[2 marks]
\nMolecules of antibiotics often contain a beta-lactam ring. Explain the importance of the betalactam ring in the action of penicillin, using section 37 of the data booklet.
\nring is «sterically» strained
\nOR
\nangles of 90° instead of 109.5/109/120° angles
\nOR
\nangles smaller than 109.5/109/120°/tetrahedral/trigonal planar/triangular planar angle
\nring breaks up/opens/reacts «easily»
\nOR
\namido/amine group «in ring» is «highly» reactive
\nbinds to/reacts with/interferes with/inactivates transpeptidase
\nOR
\nbinds to/reacts with/interferes with/inactivates enzyme responsible for bacterial cell wall formation/cross-linking
\nExcess acid in the stomach can cause discomfort and more serious health issues.
\nExplain how ranitidine (Zantac) reduces stomach acid production.
\nThe pH is maintained in different fluids in the body by the use of buffers.
\nCalculate the pH of a buffer solution of 0.0200 mol dm–3 carbonic acid, H2CO3, and 0.400 mol dm–3 sodium hydrogen carbonate, NaHCO3. The pKa of carbonic acid is 6.35.
\nblocks/binds to H2/histamine receptors «in cells of stomach lining»
\nOR
\nprevents histamine binding to H2/histamine receptors «and triggering acid secretion»
prevents parietal cells from releasing/producing acid
Accept \"H2-receptor antagonist/H2RA\" OR \"blocks/inhibits action of histamine\" for M1.
\nALTERNATIVE 1
\npH = «pKa + log =» 6.35 + log ()
\n«pH =» 7.65
\nALTERNATIVE 2
\nKa = 4.5 x 10–7
\n«Ka = 0.400 x , [H+] =» 2.3 x 10–8 «mol dm–3»
\n«pH =» 7.64
\nAward [2] for correct final answer.
\nDo not accept “pH = 8”.
\nComponents X and Y are mixed together and allowed to reach equilibrium. The concentrations of X, Y, W and Z in the equilibrium mixture are 4, 1, 4 and respectively.
\nX + 2Y 2W + Z
\nWhat is the value of the equilibrium constant, Kc?
\nA.
\nB.
\nC. 2
\nD. 8
\nD
\nNew drugs undergo thorough clinical trials before they are approved.
\nOutline the difference between the therapeutic index in animal studies and the therapeutic index in humans.
\nState the method of drug administration that gives the maximum bioavailability.
\nin animal studies AND in humans
\nOR
\nin animal studies lethal dose/LD50 AND in humans toxic dose/TD50
\nintravenous/IV «injection»
\nOR
\ninjection into the bloodstream
\nThe development of materials with unique properties is critical to advances in industry.
\nExplain why Type 2 superconductors are generally more useful than Type 1.
\nAny two of:
\nhave higher critical temperatures/Tc «than Type 1»
\nOR
\ncan act at higher temperatures
\nhave higher critical magnetic fields/Bc «than Type 1»
\nless time needed to cool to operating temperature
\nless energy required to cool down/maintain low temperature
\nDo not accept “Type 2 has a gradual transition to a superconducting state but in Type 1 it is a sharp transition”.
\n[Max 2 Marks]
\nLow density polyethene (LDPE) and high density polyethene (HDPE) are both addition polymers.
\nDescribe how the monomers of addition polymers and of condensation polymers differ.
\nIdentify the type of intermolecular bonding that is responsible for Kevlar®’s strength.
\naddition: C=C
\nAND
\ncondensation: two functional groups needed on each monomer
\nAccept \"alkene/alkenyl\" OR \"double bond\" OR \"multiple bond\".
\nhydrogen bonds
\nAccept “ stacking/interactions”.
\nMany reactions are in a state of equilibrium.
\nThe following reaction was allowed to reach equilibrium at 761 K.
\nH2 (g) + I2 (g) 2HI (g) ΔHθ < 0
\nThe pH of 0.010 mol dm–3 carbonic acid, H2CO3 (aq), is 4.17 at 25 °C.
\nH2CO3 (aq) + H2O (l) HCO3– (aq) + H3O+ (aq).
\nState the equilibrium constant expression, Kc , for this reaction.
\nThe following equilibrium concentrations in mol dm–3 were obtained at 761 K.
\nCalculate the value of the equilibrium constant at 761 K.
\nDetermine the value of ΔGθ, in kJ, for the above reaction at 761 K using section 1 of the data booklet.
\nCalculate [H3O+] in the solution and the dissociation constant, Ka , of the acid at 25 °C.
\nCalculate Kb for HCO3– acting as a base.
\nKc =
\n45.6
\nΔGθ = «– RT ln K = – (0.00831 kJ K−1 mol−1 x 761 K x ln 45.6) =» – 24.2 «kJ»
\n[H3O+] = 6.76 x 10–5 «mol dm–3»
\nKa =
\n4.6 x 10–7
\nAccept 4.57 x 10–7
\nAward [3] for correct final answer.
\n« =» 2.17 x 10–8
\nOR
\n« =» 2.19 x 10–8
\nMetals have various crystal structures. Cobalt forms a face-centred cubic (FCC) lattice. Two representations of FCC are shown.
\nCalculate the total number of cobalt atoms within its unit cell.
\nThe atomic radius, r, of cobalt is 1.18 × 10–8 cm. Determine the edge length, in cm, of the unit cell, a, using the second diagram.
\nDetermine a value for the density of cobalt, in g cm–3, using data from sections 2 and 6 of the data booklet and your answers from (a) and (b) (i).
\nIf you did not obtain an answer to (b) (i), use 3.00 × 10–8 cm but this is not the correct answer.
\n«» 4
\nface diagonal
\n«» 3.34 x 10–8 «cm»
\nmass of 4 atoms = «g»
\n«density » 10.5 «g cm–3»
\nAnswer using 3.00 x 10–8 cm:
\nmass of 4 atoms = «g»
\n«density » 14.5 «g cm–3»
\nAward [2] for correct final answer.
\nThe diagram shows an incomplete voltaic cell with a light bulb in the circuit.
\nIdentify the missing component of the cell and its function.
\nDeduce the half-equations for the reaction at each electrode when current flows.
\n\n
Annotate the diagram with the location and direction of electron movement when current flows.
\nCalculate the cell potential, in V, using section 24 of the data booklet.
\nDetermine the loss in mass of one electrode if the mass of the other electrode increases by 0.10 g.
\nsalt bridge
\n\n
movement of ions
\nOR
\nbalance charge
\n\n
Do not accept “to complete circuit” unless ion movement is mentioned for M2.
\n[2 marks]
\nPositive electrode (cathode):
\nAg+(aq) + e– → Ag(s)
\n\n
Negative electrode (anode):
\nMg(s) → Mg2+(aq) + 2e–
\n\n
Award [1 max] if correct equations given at wrong electrodes.
\n[2 marks]
\nin external wire from left to right
\n[1 mark]
\n«E = +0.80 V – (–2.37 V) = +» 3.17 «V»
\n[1 mark]
«moles of silver »
\nmoles of magnesium
\n«loss in mass of magnesium » 0.011 «g»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\nWhich type of bond is formed when a Lewis acid reacts with a Lewis base?
\nA. Covalent
\nB. Dipole-dipole
\nC. Double
\nD. Hydrogen
\nA
\nWhat is the order of increasing acidity of the following acids?
\nA. chloroethanoic < ethanoic < hydrogen fluoride < hydrogen cyanide
\nB. ethanoic < chloroethanoic < hydrogen fluoride < hydrogen cyanide
\nC. chloroethanoic < ethanoic < hydrogen cyanide < hydrogen fluoride
\nD. hydrogen cyanide < ethanoic < hydrogen fluoride < chloroethanoic
\nD
\nWhich element is reduced in the following decomposition?
\n(NH4)2Cr2O7(s) → N2(g) + Cr2O3(s) + 4H2O(g)
\nA. N
\nB. H
\nC. Cr
\nD. O
\nC
\nWhat is the standard half-cell potential of copper if the “zero potential reference electrode” is changed from the standard hydrogen electrode to a standard zinc electrode?
\n![\"M17/4/CHEMI/HPM/ENG/TZ2/30\"](\"../assets/uploads/tinymce_asset/asset/3717/Schermafbeelding_2017-09-22_om_15.19.51.png\"/)
A. –1.1
\nB. –0.34
\nC. +0.34
\nD. +1.1
\nD
\nWhat are the relative volumes of gas given off at E and F during electrolysis of the two cells in series? Assume all electrodes are inert.
\nA. 1:1
\nB. 1:2
\nC. 2:1
\nD. 5:2
\nB
\nWhich pair of isomers always shows optical activity?
\nA. Cis-trans
\nB. Enantiomers
\nC. Conformational
\nD. E/Z
\nB
\nWhich compounds can be reduced?
\nI. C2H4
II. CH3COOH
III. CH3CHO
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nIn which order should the reagents be used to convert benzene into phenylamine (aniline)?
\nC
\nWhich technique can be used to identify bond length and bond angle?
\nA. 1H NMR spectroscopy
\nB. IR spectroscopy
\nC. Mass spectroscopy
\nD. X-ray crystallography
\nD
\nHeavy metal ions are an important environmental concern.
\nState the name of one method, other than precipitation, of removing heavy metal ions from solution in water.
\nThe solubility product, Ksp , of cadmium sulfide, CdS, is 8.0 × 10–27. Determine the concentration of cadmium ions in 1.0 dm3 of a saturated solution of cadmium sulfide to which 0.10 mol of solid sodium sulfide has been added, stating any assumption you make.
\nadsorption
\nOR
\nchelation
\nOR
\nion exchange
\nAccept other valid methods such as “phytoremediation” OR “Fenton reaction” OR “electrolysis”.
\nCalculation:
\nKsp = [Cd2+] x [S2–] ✔
\n[Cd2+] = 8.0 x 10–26 «mol dm–3» ✔
\nAssumption:
\nvolume of solution remains 1.0 dm3
\nOR
\nconcentration of sulfide ions in original solution is negligible
\nOR
\nhydrolysis of sulfide ions is negligible
\nAward [2] for correct numerical answer of [Cd2+] for M1 and M2.
\nAccept “0.10 + x ∼ 0.10 «mol dm–3»”.
\nAn enzyme catalyses the conversion of succinate to fumarate ions in a cell, as part of the process of respiration.
\nThe rate of the reaction was monitored and the following graph was plotted.
\nDetermine the value of the Michaelis constant, Km, by annotating the graph.
\nThe malonate ion acts as an inhibitor for the enzyme.
\nSuggest, on the molecular level, how the malonate ion is able to inhibit the enzyme.
\nDraw a curve on the graph above showing the effect of the presence of the malonate ion inhibitor on the rate of reaction.
\nKm labelled on x-axis as the [succinate ion] at ½Vmax
\nOR
\nhorizontal line at ½Vmax AND vertical line down to x-axis
\n«Km ＝» 6.5 x 10–3 «mol dm–3»
\nAnnotation of graph required for M1.
\nAccept any specific value in the range 6.0 x 10–3 to 7.5 x 10–3 «mol dm–3».
\nsimilar shape/size/structure «as succinate ion/substrate»
\ncompetes for the active site «with the succinate ion/substrate»
\nAccept “competitive inhibitor” for M2.
\nAward [1 max] if non-competitive inhibition is correctly described.
\nsame Vmax reached at higher [substrate]
\nThis question is about the reactions of halogenoalkanes.
\nCompare and contrast the mechanisms by which 1-chlorobutane, CH3CH2CH2CH2Cl, and 2-chloro-2-methylpropane, (CH3)3CCl, react with aqueous sodium hydroxide, giving two similarities and one difference.
\n\n
Outline why the rate of reaction of the similar bromo-compounds is faster.
\nState the organic product of the reaction between 1-chlorobutane, CH3CH2CH2CH2Cl, and aqueous sodium hydroxide.
\nSuggest how this product could be synthesized in one step from butanoic acid.
\nDeduce the name of the class of compound formed when the product of (c)(i) reacts with butanoic acid.
\nAny two similarities:
\nheterolytic bond breaking
\nOR
\nchloride ions leave
\n\n
nucleophilic/OH– substitution
\nboth first order with regard to [halogenoalkane]
\n\n
One difference:
\nCH3CH2CH2CH2Cl is second order/bimolecular/SN2 AND (CH3)3CCl is first order/unimolecular/SN1
\nOR
\nCH3CH2CH2CH2Cl rate depends on [OH–] AND (CH3)3CCl does not
\nOR
\nCH3CH2CH2CH2Cl is one step AND (CH3)3CCl is two steps
\nOR
\n(CH3)3CCl involves an intermediate AND CH3CH2CH2CH2Cl does not
\nOR
\nCH3CH2CH2CH2Cl has inversion of configuration AND (CH3)3CCl has c. 50 : 50 retention and inversion
\n\n
Do not accept “produces alcohol” or “produces NaCl”.
\nAccept “substitution in 1-chlorobutane and «some» elimination in 2-chloro-2-methylpropane”.
\n[3 marks]
\nC–Br bond weaker than C–Cl bond
\n\n
Accept “Br– is a better leaving group”.
\nDo not accept \"bromine is more reactive\".
\nDo not accept “C–Br bond is longer than C–Cl” alone.
\n[1 mark]
\nbutan-1-ol/CH3CH2CH2CH2OH
\n\n
Do not accept “butanol” for “butan-1-ol”.
\nAccept “1-butanol”.
\nDo not penalize for name if correct formula is drawn.
\n[1 mark]
\n«reduction with» lithium aluminium hydride/LiAlH4
\n\n
Do not accept “sodium borohydride/NaBH4”.
\n[1 mark]
\nester
\n[1 mark]
\nBoth HDPE (high density polyethene) and LDPE (low density polyethene) are produced by the polymerization of ethene.
\nAn alternative method of polymerizing molecules is condensation polymerization. One of the earliest condensation polymers was nylon-6. A short section of the polymer chain of nylon-6 is shown below.
\n![\"M18/4/CHEMI/HP3/ENG/TZ1/04.c\"](\"../assets/uploads/tinymce_asset/asset/4494/Schermafbeelding_2018-08-08_om_08.31.45.png\"/)
Draw the structure of the monomer from which nylon-6 is produced by a condensation reaction.
\n\n
Deduce, giving a reason, whether the atom economy of a condensation polymerization, such as this, would be greater or less than an addition polymerization, such as the formation of HDPE.
\n![\"M18/4/CHEMI/HP3/ENG/TZ1/04.c.i/M\"](\"../assets/uploads/tinymce_asset/asset/4497/Schermafbeelding_2018-08-08_om_09.02.17.png\"/)
–NH2 AND –COOH
\nsix C-atoms
\n\n
Accept –COCl instead of –COOH.
\n[2 marks]
\nless AND a second molecule/product formed
\n\n
Accept “not all the reactant molecules «in the equation» are converted «to product molecules»”.
\n[1 mark]
\nOrganomagnesium compounds can react with carbonyl compounds. One overall equation is:
\nCompound B can also be prepared by reacting an alkene with water.
\nIodomethane is used to prepare CH3Mg. It can also be converted into methanol:
\nCH3 + HO– → CH3OH + –
\nState the name of Compound B, applying International Union of Pure and Applied Chemistry (IUPAC) rules.
\nCompound A and Compound B are both liquids at room temperature and pressure. Identify the strongest intermolecular force between molecules of Compound A.
\nState the number of (sigma) and (pi) bonds in Compound A.
\nDeduce the hybridization of the central carbon atom in Compound A.
\nIdentify the isomer of Compound B that exists as optical isomers (enantiomers).
\nDraw the structural formula of the alkene required.
\nExplain why the reaction produces more (CH3)3COH than (CH3)2CHCH2OH.
\nDeduce the structural formula of the repeating unit of the polymer formed from this alkene.
\nDeduce what would be observed when Compound B is warmed with acidified aqueous potassium dichromate (VI).
\nIdentify the type of reaction.
\nOutline the requirements for a collision between reactants to yield products.
\nExplain the mechanism of the reaction using curly arrows to represent the movement of electron pairs.
\nThe polarity of the carbon–halogen bond, C–X, facilitates attack by HO–.
\nOutline, giving a reason, how the bond polarity changes going down group 17.
\n2-methylpropan-2-ol /2-methyl-2-propanol ✔
\n\n
Accept methylpropan-2-ol/ methyl-2-propanol.
\nDo not accept 2-methylpropanol.
\ndipole-dipole ✔
\n\n
Do not accept van der Waals’ forces.
\n: 9
AND
: 1 ✔
sp2 ✔
\nbutan-2-ol/CH3CH(OH)C2H5 ✔
\ncarbocation formed from (CH3)3COH is more stable / (CH3)3C+ is more stable than (CH3)2CHCH2+ ✔
\n
«because carbocation has» greater number of alkyl groups/lower charge on the atom/higher e- density
OR
«greater number of alkyl groups» are more electron releasing
OR
«greater number of alkyl groups creates» greater inductive/+I effect ✔
\n
Do not award any marks for simply quoting Markovnikov’s rule.
\nDo not penalize missing brackets or n.
\nDo not award mark if continuation bonds are not shown.
\nno change «in colour/appearance/solution» ✔
\n«nucleophilic» substitution
OR
SN2 ✔
Accept “hydrolysis”.
Accept SN1
\nenergy/E ≥ activation energy/Ea ✔
\ncorrect orientation «of reacting particles»
OR
correct geometry «of reacting particles» ✔
curly arrow going from lone pair/negative charge on O in -OH to C ✔
\ncurly arrow showing I leaving ✔
\nrepresentation of transition state showing negative charge, square brackets and partial bonds ✔
\n\n
Accept OH- with or without the lone pair.
\nDo not allow curly arrows originating on H, rather than the -, in OH-.
\nAccept curly arrows in the transition state.
\nDo not penalize if HO and I are not at 180°.
\nDo not award M3 if OH–C bond is represented.
\nAward [2 max] if SN1 mechanism shown.
\ndecreases/less polar AND electronegativity «of the halogen» decreases ✔
\n\n
Accept “decreases” AND a correct comparison of the electronegativity of two halogens.
\nAccept “decreases” AND “attraction for valence electrons decreases”.
\nNaming the organic compound using IUPAC rules was generally done well.
\nMediocre performance in stating the number of σ (sigma) and π (pi) bonds in propanone; the common answer was 3 σ and 1 π instead of 9 σ and 1 π, suggesting the three C-H σ bonds in each of the two methyl groups were ignored.
\nsp2 hybridization of the central carbon atom in the ketone was very done well.
\nMediocre performance; some identified 2-methylpropan-1-ol or -2-ol, instead butan-2-ol/CH3CH(OH)C2H5 as the isomer that exists as an optical isomer.
\nGood performance; some had a H and CH3 group on each C atom across double bond instead of having two H atoms on one C and two CH3 groups on the other.
\nPoor performance, particularly in light of past feedback provided in similar questions since there was repeated reference simply to Markovnikov's rule, without any explanation.
\nMediocre performance; deducing structural formula of repeating unit of the polymer was challenging in which continuation bonds were sometimes missing, or structure included a double bond or one of the CH3 group was missing.
\nMediocre performance; deducing whether the tertiary alcohol could be oxidized solicited mixed responses ranging from the correct one, namely no change (in colour, appearance or solution), to tertiary alcohol will be reduced, or oxidized, or colour will change will occur, and such.
\nExcellent performance on the type of reaction but with some incorrect answers such as alkane substitution, free radical substitution or electrophilic substitution.
\nGood performance. For the requirements for a collision between reactants to yield products, some suggested necessary, sufficient or enough energy or even enough activation energy instead of energy/E ≥ activation energy/Ea.
\nMechanism for SN2 not done well. Often the negative charge on OH was missing, the curly arrow was not going from lone pair/negative charge on O in -OH to C, or the curly arrow showing I leaving placed incorrectly and specially the negative charge was missing in the transition state. Formation of a carbocation intermediate indicating SN1 mechanism could score a maximum of 2 marks.
\nGood performance on how the polarity of C-X bond changes going down group 17.
\nNitric acid is usually produced by the oxidation of ammonia.
\nA mixture of nitric acid and sulfuric acid can be used to convert benzene to nitrobenzene, C6H5NO2.
\nDraw arrows in the boxes to represent the electron configuration of a nitrogen atom.
\nDeduce a Lewis (electron dot) structure of the nitric acid molecule, HNO3, that obeys the octet rule, showing any non-zero formal charges on the atoms.
\nExplain the relative lengths of the three bonds between N and O in nitric acid.
\nState a technique used to determine the length of the bonds between N and O in solid HNO3.
\nWrite an equation for the reaction between the acids to produce the electrophile, NO2+.
\nDraw the structural formula of the carbocation intermediate produced when this electrophile attacks benzene.
\nDeduce the number of signals that you would expect in the 1H NMR spectrum of nitrobenzene and the relative areas of these.
\n
Accept all 2p electrons pointing downwards.
Accept half arrows instead of full arrows.
\nbonds and non-bonding pairs correct ✔
\nformal charges correct ✔
\n\n
Accept dots, crosses or lines to represent electron pairs.
\nDo not accept resonance structures with delocalised bonds/electrons.
\nAccept + and – sign respectively.
\nDo not accept a bond between nitrogen and hydrogen.
\nFor an incorrect Lewis structure, allow ECF for non-zero formal charges.
\nAny three of:
\ntwo N-O same length/order ✔
delocalization/resonance ✔
N-OH longer «than N-O»
OR
N-OH bond order 1 AND N-O bond order 1½ ✔
\n
Award [2 max] if bond strength, rather than bond length discussed.
\nAccept N-O between single and double bond AND N-OH single bond.
\nX-ray crystallography ✔
\nHNO3 + 2H2SO4 NO2+ + H3O+ + 2HSO4- ✔
\n\n
Accept “HNO3 + H2SO4 NO2+ + H2O + HSO4-”.
\nAccept “HNO3 + H2SO4 H2NO3+ + HSO4-” AND “H2NO3+ NO2+ + H2O”.
\nAccept single arrows instead of equilibrium signs.
\n\n
Accept any of the five structures.
\nDo not accept structures missing the positive charge.
\nNumber of signals: three/3 ✔
\nRelative areas: 2 : 2 : 1 ✔
\nDrawing arrows in the boxes to represent the electron configuration of a nitrogen atom was done extremely well.
\nDrawing the Lewis structure of HNO3 was performed extremely poorly with structures that included H bonded to N, no double bond or a combination of single, double and even a triple bond or incorrect structures with dotted lines to reflect resonance. Many did not calculate non-zero formal charges.
\nPoorly done; some explained relative bond strengths between N and O in HNO3, not relative lengths; others included generic answers such as triple bond is shortest, double bond is longer, single longest.
\nA majority could not state the technique to determine length of bonds; answers included NMR, IR, and such instead of X-ray crystallography.
\nMany had difficulties writing the balanced equation(s) for the formation of the nitronium ion.
\nAgain, many had difficulty drawing the structural formula of the carbocation intermediate produced in the reaction.
\nDeducing the number of signals in the 1H NMR spectrum of nitrobenzene, which depend on the number of different hydrogen environments, was done poorly. Also, instead of relative areas, the common answer included chemical shift (ppm) values.
\nLewis (electron dot) structures are useful models.
\nDraw the Lewis (electron dot) structures of PF3 and PF4+ and use the VSEPR theory to deduce the molecular geometry of each species.
\nPredict with a reason, whether the molecule PF3 is polar or non-polar.
\nAccept any combination of dots, crosses and lines.
\nIgnore missing brackets and positive charge.
\nPenalize missing lone pairs once only.
\nDo not apply ECF for molecular geometry.
\npolar AND bond polarities/dipoles do not cancel out
\nOR
\npolar AND unsymmetrical distribution of charge
\nApply ECF from part (a) molecular geometry.
\nAluminium is produced by the electrolysis of a molten electrolyte containing bauxite.
\nThe graph of the resistance of aluminium with temperature is shown below.
\n![\"M18/4/CHEMI/HP3/ENG/TZ1/05.d\"](\"../assets/uploads/tinymce_asset/asset/4496/Schermafbeelding_2018-08-08_om_08.38.12.png\"/)
The diagram illustrates the crystal structure of aluminium metal with the unit cell indicated. Outline the significance of the unit cell.
\n![\"M18/4/CHEMI/HP3/ENG/TZ1/05.b\"](\"../assets/uploads/tinymce_asset/asset/4498/Schermafbeelding_2018-08-08_om_09.07.28.png\"/)
When X-rays of wavelength 0.154 nm are directed at a crystal of aluminium, the first order diffraction pattern is observed at 18°. Determine the separation of layers of aluminium atoms in the crystal, in m, using section 1 of the data booklet.
\nDeduce what the shape of the graph indicates about aluminium.
\nOutline why the resistance of aluminium increases above 1.2 K.
\nThe concentration of aluminium in drinking water can be reduced by precipitating aluminium hydroxide. Calculate the maximum concentration of aluminium ions in water of pH 7 at 298 K. Solubility product of aluminium hydroxide = 3.3 × 10−34 at 298 K.
\nthe smallest repeating unit «from which the crystal structure can be derived»
\n\n
Accept “building block that the structure is made from”.
\n[1 mark]
\n«nλ = 2d sin θ»
\n1 × 1.54 × 10–10 = 2 × d × sin 18
\nd «» = 2.49 × 10–10 «m»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\ntype 1
\nsuperconductor
\n[2 marks]
\ncollisions between electrons and «lattice of metal» ions become more frequent
\nOR
\nthermal oscillations/vibrations disrupt the Cooper electron pairs
\n[1 mark]
\nKsp = [Al3+][OH–]3 «= 3.3 × 10–34»
\n[Al3] = «» 3.3 × 10–13 «mol dm–3»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\nPolymers of glucose include starch and cellulose.
\nOutline why cellulose fibres are strong.
\nAny two of:
\nlong straight/unbranched chains
\nmultiple hydrogen bonds «between chains»
\n\n
microfibrils
\nOR
\nrigid/cable structure
\n\n
[2 marks]
\nAn inhibitor reduces the rate, V, of an enzyme-catalysed reaction.
\nExplain with reference to the binding site on the enzyme how a non-competitive inhibitor lowers the value of Vmax.
\nOutline the significance of the value of the Michaelis constant, Km.
\nbinds at allosteric site
\nOR
\nbinds away from active site
\n\n
changes shape of active site
\nOR
\nrenders active sites ineffective
\n\n
[2 marks]
\nKm is inverse measure of affinity of enzyme for a substrate
\nOR
\nKm is inversely proportional to enzyme activity
\nOR
\nhigh value of Km indicates higher substrate concentration needed for enzyme saturation
\nOR
\nlow value of Km means reaction is fast at low substrate concentration
\n\n
Idea of inverse relationship must be conveyed.
\nAccept “high value of Km indicates low affinity of enzyme for substrate/less stable ES complex/lower enzyme activity”.
\nAccept “low value of Km indicates high affinity of enzyme for substrate/stable ES complex/greater enzyme activity”.
\n[1 mark]
\nInsulin was the first protein to be sequenced. It was determined that the end of one chain had the primary structure Phe–Val–Asn–Gln.
\nDescribe how DNA determines the primary structure of a protein such as insulin.
\ntriplet/genetic code
\nOR
\nsequence of three bases/nucleotides
\n\n
instruction for «particular» amino acid
\n\n
[2 marks]
\nMenthol is an organic compound containing carbon, hydrogen and oxygen.
\nComplete combustion of 0.1595 g of menthol produces 0.4490 g of carbon dioxide and 0.1840 g of water. Determine the empirical formula of the compound showing your working.
\n0.150 g sample of menthol, when vaporized, had a volume of 0.0337 dm3 at 150 °C and 100.2 kPa. Calculate its molar mass showing your working.
\ncarbon: «» = 0.01020 «mol» / 0.1225 «g»
\nOR
\nhydrogen: «» = 0.02042 «mol» / 0.0206 «g»
\noxygen: «0.1595 – (0.1225 + 0.0206)» = 0.0164 «g» / 0.001025 «mol»
\nempirical formula: C10H20O
\nAward [3] for correct final answer.
\ntemperature = 423 K
\nOR
\n«M
\n«M » 156 «g mol–1»
\nAward [1] for correct answer with no working shown.
\nAccept “pV = nRT AND n = ” for M1.
\nAnthocyanins are naturally occurring plant pigments. Depending on the solution pH, they can exist as quinoidal bases or flavylium cations as shown in section 35 of the data booklet.
\nOutline why anthocyanins are coloured.
\nExplain why the blue colour of a quinoidal base changes to the red colour of a flavylium cation as pH decreases.
\nhighly conjugated systems
\nOR
\nalternating single and double bonds
\nOR
\nmany delocalized electrons
\n\n
electron transitions occur when visible light is absorbed
\n\n
[2 marks]
\ngaining protons
\ndecreases electron density/extent of conjugation «in aromatic backbone»
\nincreases energy of electron transitions
\n\n
[3 marks]
\nModern electric cars store their energy in lithium ion batteries.
\nThe diagram represents a cell in such a battery delivering a current.
\nThe carbon footprint of electric cars depends on how the electricity is produced. Nuclear fission of 235U is one source of electrical energy that has a minimal carbon footprint.
\nComplete the half-equations on the diagram and identify the species moving between the electrodes.
\n![\"M18/4/CHEMI/HP3/ENG/TZ1/14.a.i\"](\"../assets/uploads/tinymce_asset/asset/4501/Schermafbeelding_2018-08-08_om_09.45.58.png\"/)
State the factor that limits the maximum current that can be drawn from this cell and how electrodes are designed to maximize the current.
\nExplain how the proportion of 235U in natural uranium is increased.
\n\n
Accept any balanced equation which shows Li oxidized to Li+ for M3, such as
\nLiC6 → Li+ + C6 + e– or
\nLixC6 → xLi+ + 6C + xe–
\n[3 marks]
\nLimiting factor:
\ninternal resistance «of the cell»
\nElectrodes design:
\nlarge surface area
\n\n
Accept “time it takes ions to diffuse between electrodes”.
\nAccept specific ways of increasing surface area, such as “porous electrodes”.
\nAccept “close together/small separation”.
\n[2 marks]
\nuranium converted to uranium hexafluoride/UF6 gas
\n\n
ALTERNATIVE 1:
\ngas «allowed to» diffuse
\nlower mass isotope/235U passes through more rapidly
\n\n
ALTERNATIVE 2:
\nuse of centrifuge
\n\n
higher mass isotope/238U moves/closer to outside of centrifuge
\nOR
\nlower mass isotope/235U stays in/removed from middle of centrifuge
\n\n
[3 marks]
\nMany reactions are in a state of equilibrium.
\nThe equations for two acid-base reactions are given below.
\nHCO3– (aq) + H2O (l) H2CO3 (aq) + OH– (aq)
HCO3– (aq) + H2O (l) CO32– (aq) + H3O+ (aq)
The following reaction was allowed to reach equilibrium at 761 K.
\nH2 (g) + I2 (g) 2HI (g) ΔHθ < 0
\nOutline the effect, if any, of each of the following changes on the position of equilibrium, giving a reason in each case.
\nIdentify two different amphiprotic species in the above reactions.
\nState what is meant by the term conjugate base.
\nState the conjugate base of the hydroxide ion, OH–.
\nA student working in the laboratory classified HNO3, H2SO4, H3PO4 and HClO4 as acids based on their pH. He hypothesized that “all acids contain oxygen and hydrogen”.
\nEvaluate his hypothesis.
\nAward [1 max] if both effects are correct.
\nReason for increasing volume:
\nAccept “concentration of all reagents reduced by an equal amount so cancels out in Kc expression”.
\nAccept “affects both forward and backward rates equally”.
\nHCO3– AND H2O
\nspecies that has one less proton/H+ ion «than its conjugate acid»
\nOR
\nspecies that forms its conjugate acid by accepting a proton
\nOR
\nspecies that is formed when an acid donates a proton
\nDo not accept “differs by one proton/H+ from conjugate acid”.
\noxide ion/O2–
\ninsufficient data to make generalization
\nOR
\nneed to consider a «much» larger number of acids
\nOR
\nhypothesis will continue to be tested with new acids to see if it can stand the test of time
\n«hypothesis is false as» other acids/HCl/HBr/HCN/transition metal ion/BF3 do not contain oxygen
\nOR
\nother acids/HCl/HBr/HCN/transition metal ion/BF3 falsify hypothesis
\ncorrect inductive reasoning «based on limited sample»
\n«hypothesis not valid as» it contradicts current/accepted theories/Brønsted-Lowry/Lewis theory
\n[Max 2 Marks]
\nOne suggestion for the reduction of carbon footprints is the use of biofuels, such as vegetable oils, as a substitute for petroleum based fuels.
\nOutline the major technical problem affecting the direct use of vegetable oils as fuels in internal combustion engines and the chemical conversion that has overcome this.
\nState the formula of a fuel that might be produced from the vegetable oil whose formula is shown.
\n ![\"M18/4/CHEMI/HP3/ENG/TZ1/13,b\"](\"../assets/uploads/tinymce_asset/asset/4499/Schermafbeelding_2018-08-08_om_09.42.54.png\"/)
viscosity «of vegetable oils is too high»
\n\n
transesterification
\nOR
\n«conversion into» alkyl/methyl/ethyl esters
\n\n
[2 marks]
\nR–CO–O–CH3 / RCOOMe
\nOR
\nR–CO–O–C2H5 / RCOOEt
\n[1 mark]
\nPhotovoltaic cells are much less hazardous than nuclear fission.
\nEarly photovoltaic cells were based on silicon containing traces of other elements. State the type of semiconductor produced by doping silicon with indium, In, giving a reason that refers to its electronic structure.
\nDye-sensitized solar cells, DSSCs, use a dye to absorb the sunlight. State two advantages that DSSCs have over traditional silicon based photovoltaic cells.
\nThe structure of two dyes used in DSSCs are shown.
\n![\"M18/4/CHEMI/HP3/ENG/TZ1/15.c\"](\"../assets/uploads/tinymce_asset/asset/4505/Schermafbeelding_2018-08-08_om_10.05.44.png\"/)
Predict, giving a reason, which dye will absorb light of longer wavelength.
\np-type AND has 3 «valence» electrons
\nOR
\np-type AND fewer electrons «than silicon»
\n\n
\n
Do not accept “it is in group 3/13” as reason.
\n[1 mark]
\nAny two of:
\ncheaper
\nOR
\nease of fabrication
\n\n
use light of lower energy/lower frequency/longer wavelength
\nabsorb wider range of wavelengths
\ndye converts most/all absorbed photons into electrons
\nplentiful /renewable resources «to construct DSSC cells»
\noperate at lower «internal» temperatures/better at radiating heat away «since constructed with thin front layer of conductive plastic compared to glass box in photovoltaic cell»
\nuse of nanoparticles provides large surface area exposure to sunlight/sun/light
\ncan absorb better under cloudy/low light conditions
\nbetter conductivity
\nmore flexible
\n\n
[2 marks]
\nB AND has greater/more «extensive» conjugation
\n\n
Accept “more alternating single and double bonds”.
\n[1 mark]
\nMany drugs, including aspirin, penicillin, codeine and taxol, have been modified from compounds that occur naturally.
\nMany drugs are chiral. Explain how a polarimeter can be used to determine the relative proportion of two enantiomers.
\n«pure» enantiomers rotate the plane «of plane-»polarized light «by equal angles» in opposite directions
\n\n
Any two of:
\nfind angle of rotation of pure enantiomers
\nmeasure angle of rotation of mixture
\nmixture has angle between that of two enantiomers
\nratio of angles gives purity
\n\n
[3 marks]
\nRadiotherapy is one type of treatment for cancer.
\nDescribe how ionizing radiation destroys cancer cells.
\nOutline how Targeted Alpha Therapy (TAT) is used for treating cancers that have spread throughout the body.
\nAny two of:
\nradiation causes breaks in DNA chains
\nOR
\nradiation causes errors in DNA sequences
\n\n
«damage accumulates and» cells cannot multiply
\nrapidly dividing/cancer cells more susceptible
\n\n
Accept “alters DNA”.
\n[2 marks]
\nAny two of:
\nradiation source delivered directly to «targeted» cancer cells
\nby a carrier drug/protein/antibody
\nseveral sites in body can be targeted «at same time»
\n\n
[2 marks]
\nOrganic solvents are commonly used in the pharmaceutical industry.
\nHexane and propanone have vapour pressures of 17 kPa and 24 kPa respectively at 20 °C.
\nCalculate the vapour pressure, in kPa, at 20 °C of a mixture containing 60% hexane and 40% propanone by mole fraction, using Raoult’s law and assuming the mixture is ideal.
\nExplain how hexane and propanone may be separated by fractional distillation.
\n«vapour pressure = 0.6 × 17 + 0.4 × 24 =»
\n19.8 «kPa»
\n[1 mark]
\nAny three of:
\ndifferent molar masses
\nOR
\ndifferent strength of intermolecular forces
\n\n
different boiling points
\ntemperature in «fractionating» column decreases upwards
\n\n
«components» condense at different temperatures/heights
\nOR
\n«component with» lower boiling point leaves column first
\n\n
[3 marks]
\nThe reactivity of organic compounds depends on the nature and positions of their functional groups.
\nThe structural formulas of two organic compounds are shown below.
\nDeduce the type of chemical reaction and the reagents used to distinguish between these compounds.
\nState the observation expected for each reaction giving your reasons.
\nDeduce the number of signals and the ratio of areas under the signals in the 1H NMR spectra of the two compounds.
\nExplain, with the help of equations, the mechanism of the free-radical substitution reaction of ethane with bromine in presence of sunlight.
\noxidation/redox AND acidified «potassium» dichromate(VI)
\nOR
\noxidation/redox AND «acidified potassium» manganate(VII)
\nAccept “acidified «potassium» dichromate” OR “«acidified potassium» permanganate”.
\nAccept name or formula of the reagent(s).
\nALTERNATIVE 1 using K2Cr2O7:
\nCompound A: orange to green AND secondary hydroxyl
\nOR
\nCompound A: orange to green AND hydroxyl oxidized «by chromium(VI) ions»
\nCompound B: no change AND tertiary hydroxyl «not oxidized by chromium(VI) ions»
\nAward [1] for “A: orange to green AND B: no change”.
\nAward [1] for “A: secondary hydroxyl AND B: tertiary hydroxyl”.
\nALTERNATIVE 2 using KMnO4:
\nCompound A: purple to colourless AND secondary hydroxyl
\nOR
\nCompound A: purple to colourless AND hydroxyl oxidized «by manganese(VII) ions»
\nCompound B: no change AND tertiary hydroxyl «not oxidized by manganese(VII) ions»
\nAccept “alcohol” for “hydroxyl”.
\nAward [1] for “A: purple to colourless AND B: no change”
\nAward [1] for “A: secondary hydroxyl AND B: tertiary hydroxyl”.
\nAccept “purple to brown” for A.
\nAccept ratio of areas in any order.
\nDo not apply ECF for ratios.
\nInitiation:
Br2 2Br•
Propagation:
Br• + C2H6 → C2H5• + HBr
C2H5• + Br2 → C2H5Br + Br•
\nTermination:
Br• + Br• → Br2
OR
\nC2H5• + Br• → C2H5Br
\nOR
\nC2H5• + C2H5• → C4H10
\nReference to UV/hν/heat not required.
\nAccept representation of radical without • (eg, Br, C2H5) if consistent throughout mechanism.
\nAccept further bromination.
\nAward [3 max] if initiation, propagation and termination are not stated or are incorrectly labelled for equations.
\nAward [3 max] if methane is used instead of ethane, and/or chlorine is used instead of bromine.
\nA student titrated two acids, hydrochloric acid, HCl (aq) and ethanoic acid, CH3COOH (aq), against 50.0 cm3 of 0.995 mol dm–3 sodium hydroxide, NaOH (aq), to determine their concentration. The temperature of the reaction mixture was measured after each acid addition and plotted against the volume of each acid.
\nUsing the graph, estimate the initial temperature of the solutions.
\nDetermine the maximum temperature reached in each experiment by analysing the graph.
\nSuggest why the enthalpy change of neutralization of CH3COOH is less negative than that of HCl.
\n21.4 °C
\nAccept values in the range of 21.2 to 21.6 °C.
Accept two different values for the two solutions from within range.
HCl: 30.4 «°C»
\nAccept range 30.2 to 30.6 °C.
\n
CH3COOH: 29.0 «°C»
Accept range 28.8 to 29.2 °C.
\nCH3COOH is weak acid/partially ionised
\nenergy used to ionize weak acid «before reaction with NaOH can occur»
\nThe graph shows the first ionization energies of some consecutive elements.
\n![\"M18/4/CHEMI/HPM/ENG/TZ2/05\"](\"../assets/uploads/tinymce_asset/asset/4506/Schermafbeelding_2018-08-08_om_11.01.14.png\")
Which statement is correct?
\nA. Y is in group 3
\nB. Y is in group 10
\nC. X is in group 5
\nD. X is in group 18
\nD
\nPart of the spectrochemical series is shown for transition metal complexes.
\nI−< Cl− < H2O < NH3
\nWhich statement can be correctly deduced from the series?
\nA. H2O increases the p–d separation more than Cl−.
\nB. H2O increases the d–d separation more than Cl−.
\nC. A complex with Cl− is more likely to be blue than that with NH3.
\nD. Complexes with water are always blue.
\nB
\nWhich molecule has an expanded octet?
\nA. CO
\nB. CO2
\nC. SF2
\nD. SF4
\nD
\nThe stability of DNA is due to interactions of its hydrophilic and hydrophobic components.
\nOutline the interactions of the phosphate groups in DNA with water and with surrounding proteins (histones).
\nWater:
hydrogen/H-bonds
OR
\nion–dipole interactions
\nProteins:
ionic bonds/interactions
OR
\nhydrogen/H-bonds
\nOR
\nion–dipole interactions
\nIgnore “London/dispersion/vdW/dipole–dipole interactions” stated for water and/or proteins.
\nWhich overlap of atomic orbitals leads to the formation of only a sigma (σ) bond?
\nI. s − p
\nII. p − p
\nIII. s − s
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nThe heme groups in cytochromes contain iron ions that are involved in the reduction of molecular oxygen.
\nState the half-equation for the reduction of molecular oxygen to water in acidic conditions.
\nOutline the change in oxidation state of the iron ions in heme groups that occurs when molecular oxygen is converted to water.
\nO2 + 4H+ (aq) + 4e– → 2H2O (l)
\nAccept any balanced equation with any integer or fractional coefficients.
\n+2 to +3
\nOR
\n+1
\nOR
\nincreases «by 1»
\nWhich value represents the lattice enthalpy, in kJ mol−1, of strontium chloride, SrCl2?
\n![\"M18/4/CHEMI/HPM/ENG/TZ2/16\"](\"../assets/uploads/tinymce_asset/asset/4507/Schermafbeelding_2018-08-08_om_11.13.49.png\"/)
A. – (–829) + 164 + 243 + 550 + 1064 – (–698)
\nB. –829 + 164 + 243 + 550 + 1064 – 698
\nC. – (–829) + 164 + 243 + 550 + 1064 – 698
\nD. –829 + 164 + 243 + 550 + 1064 – (–698)
\nC
\nVitamins can be water-soluble or fat-soluble.
\nRetinal is the key molecule involved in vision. Explain the roles of cis and trans-retinal in vision and how the isomers are formed in the visual cycle.
\nAny three of:
\ncis-retinal binds to «the protein» opsin
\nOR
\ncis-retinal «binds to opsin and» forms rhodopsin
\nrhodopsin extends conjugation in retinal
\nOR
\nrhodopsin allows absorption of visible/blue/green light
\nwhen visible light is absorbed cis-retinal changes to trans-retinal
\nchange «to trans-retinal» triggers an electrical/nerve signal
\ntrans-retinal detaches from opsin AND is converted back to cis-retinal
\nOR
\ntrans-retinal is converted back to cis-retinal through enzyme activity
\nDo not accept “cis-retinal to trans-retinal” alone without reference to absorption of visible light.
\n[Max 3 Marks]
\nWhich system has the most negative entropy change, ΔS, for the forward reaction?
\nA. N2(g) + 3H2(g) 2NH3(g)
\nB. CaCO3(s) → CaO(s) + CO2(g)
\nC. 2S2O32−(aq) + I2(aq) → S4O62−(aq) + 2I–(aq)
\nD. H2O(l) → H2O(g)
\nA
\nIn the 20th Century, both fission and fusion were considered as sources of energy but fusion was economically and technically unattainable.
\nCalculate the loss in mass, in kg, and the energy released, in J, when 0.00100 mol of 228Ac decays, each atom losing an electron. Use section 2 of the data booklet and E = mc2.
\n228Ac → + 228Th
\nDetermine the energy released, in J, by 0.00100 mol of 228Ac over the course of 18 hours.
\nOutline how nuclear ionising radiation can damage DNA and enzymes in living cells.
\nLoss in mass:
\n«(3.78532 x 10–25 kg – 9.109383 x 10–31 kg – 3.78528 x 10–25 kg) x 0.00100 x 6.02 x 1023 =»1.86 x 10–9 «kg»
\nEnergy released:
\n«E = mc2 = 1.86 x 10–9 kg x (3.00 x 108 m s–1)2 =» 1.67 x 108 «J»
\n«1.67 x 108 J x =» 1.46 x 108 «J»
\nproduction of radicals/•O2–/•OH
\nOR
\ndirect effect such as breaking bonds/atom migration
\nIgnore missing dots on radical species.
\nAccept named radical eg “superoxide radical” OR “hydroxyl radical”.
\nAn example must be given for second alternative.
\nWhen X reacts with Y to give Z, the following graph is plotted. What can be deduced from the graph?
\n![\"M18/4/CHEMI/HPM/ENG/TZ2/20\"](\"../assets/uploads/tinymce_asset/asset/4508/Schermafbeelding_2018-08-08_om_11.19.32.png\")
A. The concentration of X is directly proportional to time.
\nB. The reaction is first order overall.
\nC. The reaction is zero order with respect to X.
\nD. The reaction is first order with respect to X.
\nC
\nVegetable oils can be used as a source of energy.
\nThe natural absorption of light by chlorophyll has been copied by those developing dye-sensitized solar cells (DSSCs). Outline how a DSSC works.
\nAny three of:
\ndye has conjugated system
\ndye absorbs a photon «and injects an electron into TiO2»
\nelectrons transferred to semiconductor «and dye ionized»
\ndye oxidizes/takes electron from electrolyte
\nelectron flows through external circuit «to reduce electrolyte»
\nM4 may also be scored from more detailed answers involving iodide species (eg “iodide/I– oxidized to I3–/triiodide” OR “I–/iodide reduces dye” OR “I–/iodide releases electron to dye” OR “I3–/triiodide reduced to I–/iodide”).
\n[Max 3 Marks]
\nWhich statement is correct?
\nA. The value of the rate constant, k, is independent of temperature and is deduced from the equilibrium constant, Kc.
\nB. The value of the rate constant, k, is independent of temperature and the overall reaction order determines its units.
\nC. The value of the rate constant, k, is temperature dependent and is deduced from the equilibrium constant, Kc.
\nD. The value of the rate constant, k, is temperature dependent and the overall reaction order determines its units.
\nD
\nA fuel cell converts chemical energy directly to electrical energy.
\nDeduce the half-equations and the overall equation for the reactions taking place in a direct methanol fuel cell (DMFC) under acidic conditions.
\nOutline one advantage and one disadvantage of the methanol cell (DMFC) compared with a hydrogen-oxygen fuel cell.
\nNegative electrode (anode):
\nCH3OH (aq) + H2O (l) → CO2 (g) + 6H+ (aq) + 6e–
\nPositive electrode (cathode):
O2 (g) + 4H+ (aq) + 4e– → 2H2O (l)
Overall equation:
\n2CH3OH (aq) + 3O2 (g) → 2CO2 (g) + 4H2O (l)
\nAccept any whole or fractional coefficients in balanced equations.
\nAward [1 max] for correct half-equations at wrong electrodes for M1 and M2.
\nAdvantage:
\nAny one of:
\nliquid methanol is easier to transport/store than gaseous hydrogen
\nOR
\nhydrogen is explosive
\nlonger membrane life «as it operates in aqueous environment»
\nmethanol has greater energy density than hydrogen
\nDisadvantage:
\nAny one of:
\nlower voltage
\nlower power per unit mass «of the cell»
\nlower efficiency
\ntoxic/can be mistaken for ethanol
\nlower specific energy
\nIgnore any cost references throughout.
\nAccept “CO2/greenhouse gas produced” OR “requires a more highly efficient catalyst”.
\nDo not award marks for converse statements for the advantage and disadvantage.
\nWhat occurs when the pressure on the given equilibrium is increased at constant temperature?
\nN2(g) + O2(g) 2NO(g) ΔH = +180 kJ
\nA. Kc increases and the position of equilibrium moves to the right.
\nB. Kc stays the same and the position of equilibrium is unchanged.
\nC. Kc stays the same and the position of equilibrium moves to the left.
\nD. Kc decreases and the position of equilibrium moves to the left.
\nB
\nWhich is an example of a Lewis base?
\nA. an electrophile
\nB. BF3
\nC. CH4
\nD. a nucleophile
\nD
\nTwo cells undergoing electrolysis are connected in series.
\n![\"M18/4/CHEMI/HPM/ENG/TZ2/30\"](\"../assets/uploads/tinymce_asset/asset/4510/Schermafbeelding_2018-08-08_om_11.27.51.png\"/)
If g of silver are deposited in cell 1, what volume of oxygen, in dm3 at STP, is given off in cell 2?
\nAr(Ag) = 108; Molar volume of an ideal gas at STP = 22.7 dm3 mol−1
\nA.
\nB.
\nC.
\nD.
\nA
\nWhat is the order of increasing acidity?
\n![\"M18/4/CHEMI/HPM/ENG/TZ2/27\"](\"../assets/uploads/tinymce_asset/asset/4509/Schermafbeelding_2018-08-08_om_11.25.10.png\"/)
A. HClO < CH3CH2COOH < HF < HIO3
\nB. HClO < HF < CH3CH2COOH < HIO3
\nC. HIO3 < HF < CH3CH2COOH < HClO
\nD. HIO3 < CH3CH2COOH < HF < HClO
\nA
\nWhat are the major products of electrolysing concentrated aqueous potassium iodide, KI(aq)?
\n![\"M18/4/CHEMI/HPM/ENG/TZ2/31\"](\"../assets/uploads/tinymce_asset/asset/4511/Schermafbeelding_2018-08-08_om_11.32.44.png\"/)
B
\nWhich is the correct combination of substitution reaction mechanisms?
\n![\"M18/4/CHEMI/HPM/ENG/TZ2/35\"](\"../assets/uploads/tinymce_asset/asset/4512/Schermafbeelding_2018-08-08_om_11.33.47.png\"/)
A
\nThe emission spectrum of an element can be used to identify it.
\nHydrogen spectral data give the frequency of 3.28 × 1015 s−1 for its convergence limit.
\nCalculate the ionization energy, in J, for a single atom of hydrogen using sections 1 and 2 of the data booklet.
\nCalculate the wavelength, in m, for the electron transition corresponding to the frequency in (a)(iii) using section 1 of the data booklet.
\nDeduce any change in the colour of the electrolyte during electrolysis.
\nDeduce the gas formed at the anode (positive electrode) when graphite is used in place of copper.
\nExplain why transition metals exhibit variable oxidation states in contrast to alkali metals.
\n\n
IE «= ΔE = hν = 6.63 × 10–34 J s × 3.28 × 1015 s–1» = 2.17 × 10–18 «J»
\n[1 mark]
\n«» 9.15 × 10–8 «m»
\n[1 mark]
\nno change «in colour»
\n\n
Do not accept “solution around cathode will become paler and solution around the anode will become darker”.
\n[1 mark]
\noxygen/O2
\n\n
Accept “carbon dioxide/CO2”.
\n[1 mark]
\nTransition metals:
\n«contain» d and s orbitals «which are close in energy»
\nOR
\n«successive» ionization energies increase gradually
\n\n
Alkali metals:
\nsecond electron removed from «much» lower energy level
\nOR
\nremoval of second electron requires large increase in ionization energy
\n[2 marks]
\nIn acidic solution, bromate ions, BrO3−(aq), oxidize iodide ions, I−(aq).
\nBrO3−(aq) + 6H+(aq) + 6e− Br−(aq) + 3H2O(l)
\n2I−(aq) I2(s) + 2e−
\nFormulate the equation for the redox reaction.
\nThe change in the free energy for the reaction under standard conditions, ΔGΘ, is −514 kJ at 298 K.
\nDetermine the value of EΘ, in V, for the reaction using sections 1 and 2 of the data booklet.
\nCalculate the standard electrode potential, in V, for the BrO3−/Br− reduction half‑equation using section 24 of the data booklet.
\nBrO3–(aq) + 6H+(aq) + 6I–(aq) Br–(aq) + 3I2(s) + 3H2O(l)
\n\n
Accept → for .
\n[1 mark]
\nn = 6
\n«ΔGΘ = –nFEΘ»
\n«» 0.888 «V»
\n[2 marks]
\n«EΘ = EΘ(BrO3–/Br–) – EΘ(I2/I–)»
\n«EΘ(BrO3–/Br–) = EΘ + EΘ(I2/I–) = 0.888 + 0.54 =» «+» 1.43 «V»
\n[1 mark]
\nPropene is reacted first with hydrogen chloride to produce X which is then reacted with aqueous sodium hydroxide to give Y. Finally, Y is reacted with excess acidified potassium dichromate solution.
\n\n
What is the major product, Z?
\nA. CH3CH(OH)CH3
\nB. CH3COCH3
\nC. CH3CH2CHO
\nD. CH3(CH2)2COOH
\nB
\nWhich isomers exist as non-superimposable mirror images?
\nA. cis-trans isomers
\nB. diastereomers
\nC. enantiomers
\nD. structural isomers
\nC
\nA student determined the percentage of the active ingredient magnesium hydroxide, Mg(OH)2, in a 1.24 g antacid tablet.
\nThe antacid tablet was added to 50.00 cm3 of 0.100 mol dm−3 sulfuric acid, which was in excess.
\nOutline why repeating quantitative measurements is important.
\nto reduce random errors
\nOR
\nto increase precision
\n\n
Accept “to ensure reliability”.
\n[1 mark]
\nGraphing is an important tool in the study of rates of chemical reactions.
\nThe graph represents the titration of 25.00 cm3 of 0.100 mol dm−3 aqueous ethanoic acid with 0.100 mol dm−3 aqueous sodium hydroxide.
\n![\"M18/4/CHEMI/HP2/ENG/TZ2/02.d.i_01\"](\"../assets/uploads/tinymce_asset/asset/4515/Schermafbeelding_2018-08-08_om_13.41.48.png\"/)
Deduce the major species, other than water and sodium ions, present at points A and B during the titration.
\n\n
Calculate the pH of 0.100 mol dm−3 aqueous ethanoic acid.
\nKa = 1.74 × 10−5
\nOutline, using an equation, why sodium ethanoate is basic.
\nPredict whether the pH of an aqueous solution of ammonium chloride will be greater than, equal to or less than 7 at 298 K.
\nFormulate the equation for the reaction of nitrogen dioxide, NO2, with water to form two acids.
\nFormulate the equation for the reaction of one of the acids produced in (e)(i) with calcium carbonate.
\nA: CH3COOH/ethanoic/acetic acid AND CH3COO–/ethanoate/acetate ions
\nB: CH3COO–/ethanoate/acetate ions
\n\n
Penalize “sodium ethanoate/acetate” instead of “ethanoate/acetate ions” only once.
\n[2 marks]
\n\n
OR
\n[H+] = 1.32 × 10–3 «mol dm–3»
\n«pH =» 2.88
\n\n
Accept [2] for correct final answer.
\n[2 marks]
\n«forms weak acid and strong base, thus basic»
\nCH3COO–(aq) + H2O(l) CH3COOH(aq) + OH–(aq)
\n\n
Accept → for .
\n[1 mark]
\nless than 7
\n[1 mark]
\n2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq)
\n[1 mark]
\n2HNO2(aq) + CaCO3(s) → Ca(NO2)2(aq) + CO2(g) + H2O(l)
\nOR
\n2HNO3(aq) + CaCO3(s) → Ca(NO3)2(aq) + CO2(g) + H2O(l)
\n[1 mark]
\nOrganic molecules can be visualized using three-dimensional models built from kits such as that pictured below.
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/02\"](\"../assets/uploads/tinymce_asset/asset/4637/Schermafbeelding_2018-08-10_om_11.20.15.png\"/)
Describe two differences, other than the number of atoms, between the models of ethane and ethene constructed from the kit shown.
\nThe above ball and stick model is a substituted pyridine molecule (made of carbon, hydrogen, nitrogen, bromine and chlorine atoms). All atoms are shown and represented according to their relative atomic size.
\nLabel each ball in the diagram, excluding hydrogens, as a carbon, C, nitrogen, N, bromine, Br, or chlorine, Cl.
\nSuggest one advantage of using a computer generated molecular model compared to a ball and stick 3-D model.
\nPyridine, like benzene, is an aromatic compound.
\nOutline what is meant by an aromatic compound.
\nAny two of:
\nEthene: «carbon–carbon» double bond AND Ethane: «carbon–carbon» single bond
\nethene has a shorter carbon–carbon bond «than ethane»
\n\n
Ethene: planar/two-dimensional/2-D AND Ethane: tetrahedral «carbons»/three-dimensional/3-D
\nOR
\nEthene: each carbon surrounded by three electron domains AND Ethane: each carbon surrounded by four electron domains
\nOR
\ndifferent molecular geometries/shapes
\n\n
rotation about carbon–carbon inhibited/blocked in ethene AND not in ethane
\n\n
«H–C–C/H–C–H» bond angles different
\nOR
\nEthene: «bond angles approximately» 120° AND Ethane: 109.5/109°
\n\n
Do not accept “different number of atoms/hydrogens/bonds” etc.
\nAccept “Ethene: unsaturated AND Ethane: saturated” OR “Ethene: has a double bond AND Ethane: does not” OR “Ethene: two flexible bonds between carbon atoms AND Ethane: one”.
\nAccept any reasonable physical description of the two different molecular models based on a variety of kits for M1.
\nFor ethene, accept any bond angle in the range 117–122°.
\nAward [2] if any two of the concepts listed are shown in a correctly labelled or annotated diagram.
\nAward [1 max] for two correct statements for either molecule but with no comparison given to the other.
\nAward [1 max] for suitable unlabeled diagrams of both compounds.
\n[2 marks]
\n6 carbon atoms labelled in correct positions
\nboth nitrogen atoms labelled in correct positions
\nbromine AND chlorine atoms labelled in correct positions
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/02.b.i/M\"](\"../assets/uploads/tinymce_asset/asset/4644/Schermafbeelding_2018-08-10_om_12.53.07.png\"/)
[3 marks]
\naccurate bond angles/lengths can be measured
\nOR
\n«using mathematical functions» can calculate expected shapes based on energy minimizations
\nOR
\nbetter visualization of possible bond rotations/conformation/modes of vibration
\nOR
\ncan visualize macromolecules/proteins/DNA
\nOR
\nhydrogen bonding «networks» can be generated/allows intermolecular forces «of attraction» to be simulated
\nOR
\nmore variety of visualization representations/can observe space filling
\nOR
\ncan produce an electron density map/electrostatic potential map
\nOR
\nonce model is generated file can be saved for future use/computer models can be shared globally by scientists
\nOR
\nhelps design molecules of biological significance/assists in drug design «using libraries»
\nOR
\ncan predict molecular interactions with solvents/can predict physical properties/can predict spectral data/can examine crystal structures
\nOR
\n«often» easier to construct/modify «model»
\n\n
Accept “precise” for “accurate”.
\nAccept “computer generated structural representation is normally what is expected in order to be published «in a scientific journal»”.
\nAccept “easier to see different sizes of atoms/atomic radii”.
\n[1 mark]
\nbonds within ring have resonance
\nOR
\ncontains delocalized «conjugated pi» electrons in ring
\n\n
There must be reference to a ring or cyclic structure.
\nAccept “alternating single and double bonds in a ring”.
\nAccept “ring which shows resonance/delocalization”.
\nAccept “follows Hückel/4n +2 rule”.
\nDo not accept “contains one or more benzene rings”.
\n[1 mark]
\nWhat is the percentage yield when 7 g of ethene produces 6 g of ethanol?
\nMr(ethene) = 28 and Mr(ethanol) = 46
\nC2H4(g) + H2O(g) → C2H5OH(g)
\nA.
\nB.
\nC.
\nD.
\nD
\nWhich shows the number of subatomic particles in 31P3−?
\n![\"M18/4/CHEMI/SPM/ENG/TZ1/05\"](\"../assets/uploads/tinymce_asset/asset/4552/Schermafbeelding_2018-08-09_om_11.44.09.png\"/)
A
\nWhich are correct statements about the emission spectrum of hydrogen in the visible region?
\nI. The red line has a lower energy than the blue line.
\nII. The lines converge at longer wavelength.
\nIII. The frequency of the blue line is greater than the frequency of the red line.
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nWhich describes the oxide of sodium, Na2O?
\n![\"M18/4/CHEMI/SPM/ENG/TZ1/07\"](\"../assets/uploads/tinymce_asset/asset/4553/Schermafbeelding_2018-08-09_om_11.47.12.png\"/)
D
\nWhich statement is correct?
\nA. Atomic radius decreases down group 17.
\nB. First ionization energy decreases down group 1.
\nC. Atomic radius increases across period 3 from Na to Cl.
\nD. First ionization energy decreases across period 3 from Na to Cl.
\nB
\nWhich form of carbon is the poorest electrical conductor?
\nA. Graphite
\nB. Graphene
\nC. Diamond
\nD. Carbon nanotube
\nC
\nWhat is the formula of ammonium phosphate?
\nA. (NH3)3PO4
\nB. (NH4)3PO4
\nC. (NH4)2PO4
\nD. (NH3)2PO3
\nB
\nWhat is the molecular geometry and bond angle in the molecular ion NO3−?
\n![\"M18/4/CHEMI/SPM/ENG/TZ1/11\"](\"../assets/uploads/tinymce_asset/asset/4554/Schermafbeelding_2018-08-09_om_11.54.44.png\"/)
B
\nWhat are the strongest intermolecular forces between molecules of propanone, CH3COCH3, in the liquid phase?
\nA. London (dispersion) forces
\nB. Covalent bonding
\nC. Hydrogen bonding
\nD. Dipole–dipole forces
\nD
\nThe enthalpy of combustion of ethanol is determined by heating a known mass of tap water in a glass beaker with a flame of burning ethanol.
\nWhich will lead to the greatest error in the final result?
\nA. Assuming the density of tap water is 1.0 g cm−3
\nB. Assuming all the energy from the combustion will heat the water
\nC. Assuming the specific heat capacity of the tap water is 4.18 J g−1 K−1
\nD. Assuming the specific heat capacity of the beaker is negligible
\nB
\nWhat is the enthalpy of combustion of butane in kJ mol−1?
\n2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
\n\n
A. 4x + 5y − z
\nB. 4x + 5y + z
\nC. 8x + 10y − 2z
\nD. 8x + 5y + 2z
\nA
\nWhich statement is correct?
\nA. In an exothermic reaction, the products have more energy than the reactants.
\nB. In an exothermic reversible reaction, the activation energy of the forward reaction is greater than that of the reverse reaction.
\nC. In an endothermic reaction, the products are more stable than the reactants.
\nD. In an endothermic reversible reaction, the activation energy of the forward reaction is greater than that of the reverse reaction.
\nD
\nWhich change increases the rate of formation of hydrogen when zinc reacts with excess hydrochloric acid, assuming all other conditions remain the same?
\nZn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
\nA. Adding water to the hydrochloric acid
\nB. Decreasing the temperature
\nC. Increasing the volume of hydrochloric acid
\nD. Decreasing the size of the zinc particles while keeping the total mass of zinc the same
\nD
\nWhich statements are correct?
\nI. The activation energy of a reaction is not affected by temperature.
\nII. A catalyst reduces the enthalpy change of a reaction.
\nIII. Catalysts provide alternative reaction pathways.
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nThe equilibrium constant for N2(g) + 3H2(g) 2NH3(g) is K.
\nWhat is the equilibrium constant for this equation?
\n2N2(g) + 6H2(g) 4NH3(g)
\nA. K
\nB. 2K
\nC. K2
\nD. 2K2
\nC
\nWhich classification is correct for the reaction?
\nH2PO4−(aq) + H2O(l) → HPO42−(aq) + H3O+(aq)
\n![\"M18/4/CHEMI/SPM/ENG/TZ1/19\"](\"../assets/uploads/tinymce_asset/asset/4555/Schermafbeelding_2018-08-09_om_12.48.11.png\"/)
D
\nA mixture of 1.00 mol SO2(g), 2.00 mol O2(g) and 1.00 mol SO3(g) is placed in a 1.00 dm3 container and allowed to reach equilibrium.
\n2SO2(g) + O2(g) 2SO3(g)
\nNitrogen oxide is in equilibrium with dinitrogen dioxide.
\n2NO(g) N2O2(g) ΔHΘ < 0
\nDeduce, giving a reason, the effect of increasing the temperature on the concentration of N2O2.
\nA two-step mechanism is proposed for the formation of NO2(g) from NO(g) that involves an exothermic equilibrium process.
\nFirst step: 2NO(g) N2O2(g) fast
\nSecond step: N2O2(g) + O2 (g) → 2NO2(g) slow
\nDeduce the rate expression for the mechanism.
\nThe rate constant for a reaction doubles when the temperature is increased from 25.0 °C to 35 °C.
\nCalculate the activation energy, Ea, in kJ mol−1 for the reaction using section 1 and 2 of the data booklet.
\n[N2O2] decreases AND exothermic «thus reverse reaction favoured»
\n\n
Accept “product” for [N2O2].
\nDo not accept just “reverse reaction favoured/shift to left” for “[N2O2] decreases”.
\n[1 mark]
\nALTERNATIVE 1:
\n«from equilibrium, step 1»
\n\n
OR
\n[N2O2] = Kc[NO]2
\n«from step 2, rate «= k1[N2O2][O2] = k2K[NO]2[O2]»
\nrate = k[NO]2[O2]
\n\n
ALTERNATIVE 2:
\n«from step 2» rate = k2[N2O2][O2]
\n«from step 1, rate(1) = k1[NO]2 = k–1[N2O2], [N2O2] = [NO]2»
\n«rate = k2[NO]2[O2]»
\nrate = k[NO]2[O2]
\n\n
Award [2] for correct rate expression.
\n[2 marks]
\n«»
\nT2 = «273 + 35 =» 308 K AND T1 = «273 + 25 =» 298 K
\nEa = 52.9 «kJ mol–1»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\nWhat are the products of electrolysis when molten calcium bromide is electrolysed using graphite electrodes?
\n![\"M18/4/CHEMI/SPM/ENG/TZ1/22\"](\"../assets/uploads/tinymce_asset/asset/4556/Schermafbeelding_2018-08-09_om_12.55.34.png\"/)
A
\nWhat are the products of the reaction between sulfuric acid and sodium hydrogen carbonate?
\nA. NaSO4 + H2O + CO2
\nB. Na2SO4 + CO2
\nC. Na2SO4 + H2O + CO2
\nD. NaSO4 + H2CO3
\nC
\nWhich coefficients correctly balance this redox equation?
\naFe2+(aq) + MnO4−(aq) + bH+(aq) → cFe3+(aq) + Mn2+(aq) + dH2O(l)
\n![\"M18/4/CHEMI/SPM/ENG/TZ1/23\"](\"../assets/uploads/tinymce_asset/asset/4557/Schermafbeelding_2018-08-09_om_12.56.56.png\"/)
D
\nConsider the following half-cell reactions and their standard electrode potentials.
\nDeduce a balanced equation for the overall reaction when the standard nickel and iodine half-cells are connected.
\nPredict, giving a reason, the direction of movement of electrons when the standard nickel and manganese half-cells are connected.
\nCalculate the cell potential, in V, when the standard iodine and manganese half-cells are connected.
\nIdentify the best reducing agent in the table above.
\nState and explain the products of electrolysis of a concentrated aqueous solution of sodium chloride using inert electrodes. Your answer should include half-equations for the reaction at each electrode.
\nNi (s) + I2 (aq) → 2I– (aq) + Ni2+ (aq)
\nelectron movement «in the wire» from Mn(s) to Ni(s)
\nEθ «for reduction» of Ni2+ is greater/less negative than Eθ «for reduction» of Mn2+
\nOR
\nNi2+ is stronger oxidizing agent than Mn2+
\nOR
\nMn is stronger reducing agent than Ni
\n«0.54 V – (–1.18 V) = +»1.72 «V»
\nDo not accept –1.72 V.
\nMn «(s)»
\nPositive electrode (anode):
2Cl– (aq) → Cl2 (g) + 2e–
Cl– oxidized because higher concentration
\nOR
\nelectrode potential/E depends on concentration
\nOR
\nelectrode potential values «of H2O and Cl–» are close
\nNegative electrode (cathode):
2H2O (l) + 2e– → H2 (g) + 2OH– (aq)
OR
\n2H+ (aq) + 2e– → H2 (g)
\nH2O/H+ reduced because Na+ is a weaker oxidizing agent
\nOR
\nNa+ not reduced to Na in water
\nOR
\nH+ easier to reduce than Na+
OR
H lower in activity series «than Na»
\nAccept .
\nWhat are possible names of a molecule with molecular formula C4H10O?
\nI. 1-Methoxypropane
\nII. 2-Methylpropan-2-ol
\nIII. Butanal
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nSome physical properties of molecular substances result from the different types of forces between their molecules.
\nResonance structures exist when a molecule can be represented by more than one Lewis structure.
\nCarbon dioxide can be represented by at least two resonance structures, I and II.
\n![\"M18/4/CHEMI/HP2/ENG/TZ2/07.c.i_01\"](\"../assets/uploads/tinymce_asset/asset/4524/Schermafbeelding_2018-08-08_om_16.23.50.png\"/)
Calculate the formal charge on each oxygen atom in the two structures.
\n![\"M18/4/CHEMI/HP2/ENG/TZ2/07.c.i_02\"](\"../assets/uploads/tinymce_asset/asset/4525/Schermafbeelding_2018-08-08_om_16.25.45.png\"/)
Deduce, giving a reason, the more likely structure.
\nAbsorption of UV light in the ozone layer causes the dissociation of oxygen and ozone.
\nIdentify, in terms of bonding, the molecule that requires a longer wavelength to dissociate.
\nCarbon and silicon are elements in group 14.
\nExplain why CO2 is a gas but SiO2 is a solid at room temperature.
\n![\"M18/4/CHEMI/HP2/ENG/TZ2/07.c.i/M\"](\"../assets/uploads/tinymce_asset/asset/4536/Schermafbeelding_2018-08-08_om_16.59.05.png\"/)
\n
Award [1] for any two correctly filled cells.
\n[2 marks]
\nstructure I AND no formal charges
\nOR
\nstructure I AND no charge transfer «between atoms»
\n[1 mark]
\nO3 has bond between single and double bond AND O2 has double bond
\nOR
\nO3 has bond order of 1.5 AND O2 has bond order of 2
\nOR
\nbond in O3 is weaker/longer than in O2
\n\n
O3 requires longer wavelength
\n\n
M1: Do not accept “ozone has one single and one double bond”.
\n[2 marks]
\nCO2 «non-polar» «weak» London/dispersion forces/instantaneous induced dipole-induced dipole forces between molecules
\nSiO2 network/lattice/3D/giant «covalent» structure
\n\n
M1: The concept of “between” is essential.
\n[2 marks]
\nWhich equation shows oxygen undergoing reduction?
\nA. 2F2 + O2 → 2F2O
\nB. Na2O + H2O → 2NaOH
\nC. H2O2 + 2HI → 2H2O + I2
\nD. 2CrO42− + 2H+ Cr2O72− + H2O
\nC
\nWhat is the product of the reaction between hex-3-ene and steam?
\nA. Hexan-1-ol
\nB. Hexan-2-ol
\nC. Hexan-3-ol
\nD. Hexan-4-ol
\nC
\nWhich of these reactions proceeds by a free radical mechanism in the presence of UV light?
\nA. C6H6 + Cl2 → C6H5Cl + HCl
\nB. C6H6 + 3H2 → C6H12
\nC. CH2CH2 + HBr → CH3CH2Br
\nD. CH3CH3 + Cl2 → CH3CH2Cl + HCl
\nD
\nWhich compound could be formed when CH3CH2CH2OH is heated with acidified potassium dichromate(VI)?
\nI. CH3CH2CHO
\nII. CH3CH2COOH
\nIII. CH3COCH3
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nWhich value of q, in J, has the correct number of significant figures?
\nq = mcΔT
\nwhere m = 2.500 g, c = 4.18 J g−1 K−1 and ΔT = 0.60 K.
\nA. 6
\nB. 6.3
\nC. 6.27
\nD. 6.270
\nB
\nWhat is the index of hydrogen deficiency, IHD, of 3-methylcyclohexene?
\n![\"M18/4/CHEMI/SPM/ENG/TZ1/29\"](\"../assets/uploads/tinymce_asset/asset/4558/Schermafbeelding_2018-08-09_om_13.09.54.png\"/)
A. 0
\nB. 1
\nC. 2
\nD. 3
\nC
\nThe reactivity of organic compounds depends on the nature and positions of their functional groups.
\nThe structural formulas of two organic compounds are shown below.
\nDeduce, giving a reason, which of the two compounds can show optical activity.
\nDraw three-dimensional representations of the two enantiomers.
\nState the reagents used in the nitration of benzene.
\nState an equation for the formation of NO2+.
\nExplain the mechanism of the reaction between 2-bromo-2-methylpropane, (CH3)3CBr, and aqueous sodium hydroxide, NaOH (aq), using curly arrows to represent the movement of electron pairs.
\nA AND it has a chiral centre/asymmetric carbon atom/carbon with 4 different substituents
\nAccept structures without tapered bonds.
\nconcentrated HNO3 AND concentrated H2SO4
\n“concentrated” must occur at least once (with either acid).
\nHNO3 + 2H2SO4 H3O+ + NO2+ + 2HSO4–
Accept: HNO3 + 2H2SO4 NO2+ + HSO4– + H2O.
\nAccept: HNO3 + 2H2SO4 H2NO3+ + HSO4–.
\nAccept single arrow instead of equilibrium sign.
\nAccept equivalent two step reactions in which sulfuric acid first behaves as strong acid and protonates nitric acid, before behaving as dehydrating agent removing water from it.
\ncurly arrow showing Br– leaving
\nrepresentation of tertiary carbocation
\ncurly arrow going from lone pair/negative charge on O in –OH to C+
\nformation of (CH3)3COH AND Br–
\nDo not accept curly arrow originating from C of C–Br bond.
\nDo not accept arrow originating on H in –OH.
\nAccept Br– anywhere on product side in the reaction scheme.
\nAward [2 max] for an SN2 type mechanism.
\nWhat is the ratio of the areas of the signals in the 1H NMR spectrum of pentan-3-ol?
\nA. 6:4:1:1
\nB. 6:2:2:2
\nC. 5:5:1:1
\nD. 3:3:2:2:1:1
\nA
\nUrea, (H2N)2CO, is excreted by mammals and can be used as a fertilizer.
\nCalculate the percentage by mass of nitrogen in urea to two decimal places using section 6 of the data booklet.
\nSuggest how the percentage of nitrogen affects the cost of transport of fertilizers giving a reason.
\nThe structural formula of urea is shown.
\n![\"M18/4/CHEMI/SP2/ENG/TZ1/01.b_01\"](\"../assets/uploads/tinymce_asset/asset/4562/Schermafbeelding_2018-08-09_om_13.51.02.png\"/)
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
\n![\"M18/4/CHEMI/SP2/ENG/TZ1/01.b_02\"](\"../assets/uploads/tinymce_asset/asset/4563/Schermafbeelding_2018-08-09_om_13.52.37.png\"/)
Urea can be made by reacting potassium cyanate, KNCO, with ammonium chloride, NH4Cl.
\nKNCO(aq) + NH4Cl(aq) → (H2N)2CO(aq) + KCl(aq)
\nDetermine the maximum mass of urea that could be formed from 50.0 cm3 of 0.100 mol dm−3 potassium cyanate solution.
\nUrea can also be made by the direct combination of ammonia and carbon dioxide gases.
\n2NH3(g) + CO2(g) (H2N)2CO(g) + H2O(g) ΔH < 0
\nPredict, with a reason, the effect on the equilibrium constant, Kc, when the temperature is increased.
\nSuggest one reason why urea is a solid and ammonia a gas at room temperature.
\nSketch two different hydrogen bonding interactions between ammonia and water.
\nThe combustion of urea produces water, carbon dioxide and nitrogen.
\nFormulate a balanced equation for the reaction.
\nThe mass spectrum of urea is shown below.
\n![\"M18/4/CHEMI/SP2/ENG/TZ1/01.g_01\"](\"../assets/uploads/tinymce_asset/asset/4567/Schermafbeelding_2018-08-09_om_14.15.23.png\"/)
Identify the species responsible for the peaks at m/z = 60 and 44.
\n\n
The IR spectrum of urea is shown below.
\n![\"M18/4/CHEMI/SP2/ENG/TZ1/01.h_01\"](\"../assets/uploads/tinymce_asset/asset/4570/Schermafbeelding_2018-08-09_om_14.21.30.png\"/)
Identify the bonds causing the absorptions at 3450 cm−1 and 1700 cm−1 using section 26 of the data booklet.
\n\n
Predict the number of signals in the 1H NMR spectrum of urea.
\nmolar mass of urea «= 4 × 1.01 + 2 × 14.01 + 12.01 + 16.00» = 60.07 «g mol–1»
\n«% nitrogen = × 100 =» 46.65 «%»
\n\n
\n
Award [2] for correct final answer.
\nAward [1 max] for final answer not to two decimal places.
\n[2 marks]
\n«cost» increases AND lower N% «means higher cost of transportation per unit of nitrogen»
\nOR
\n«cost» increases AND inefficient/too much/about half mass not nitrogen
\n\n
Accept other reasonable explanations.
\nDo not accept answers referring to safety/explosions.
\n[1 mark]
\n![\"M18/4/CHEMI/SP2/ENG/TZ1/01.b/M\"](\"../assets/uploads/tinymce_asset/asset/4564/Schermafbeelding_2018-08-09_om_13.53.16.png\"/)
\n
Note: Urea’s structure is more complex than that predicted from VSEPR theory.
\n[3 marks]
\nn(KNCO) «= 0.0500 dm3 × 0.100 mol dm–3» = 5.00 × 10–3 «mol»
\n«mass of urea = 5.00 × 10–3 mol × 60.07 g mol–1» = 0.300 «g»
\n\n
\n
Award [2] for correct final answer.
\n[2 marks]
\n«Kc» decreases AND reaction is exothermic
\nOR
\n«Kc» decreases AND ΔH is negative
\nOR
\n«Kc» decreases AND reverse/endothermic reaction is favoured
\n[1 mark]
\nAny one of:
\nurea has greater molar mass
\nurea has greater electron density/greater London/dispersion
\nurea has more hydrogen bonding
\nurea is more polar/has greater dipole moment
\n\n
Accept “urea has larger size/greater van der Waals forces”.
\nDo not accept “urea has greater intermolecular forces/IMF”.
\n[1 mark]
\n![\"M18/4/CHEMI/SP2/ENG/TZ1/01.e.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4565/Schermafbeelding_2018-08-09_om_14.09.21.png\"/)
\n
Award [1] for each correct interaction.
\nIf lone pairs are shown on N or O, then the lone pair on N or one of the lone pairs on O MUST be involved in the H-bond.
\nPenalize solid line to represent H-bonding only once.
\n[2 marks]
\n2(H2N)2CO(s) + 3O2(g) → 4H2O(l) + 2CO2(g) + 2N2(g)
\ncorrect coefficients on LHS
\ncorrect coefficients on RHS
\n\n
Accept (H2N)2CO(s) + O2(g) → 2H2O(l) + CO2(g) + N2(g).
\nAccept any correct ratio.
\n[2 marks]
\n60: CON2H4+
\n44: CONH2+
\n\n
Accept “molecular ion”.
\n[2 marks]
\n3450 cm–1: N–H
\n1700 cm–1: C=O
\n\n
Do not accept “O–H” for 3450 cm–1.
\n[2 marks]
\n1
\n[1 mark]
\nCalcium carbide, CaC2, is an ionic solid.
\nDescribe the nature of ionic bonding.
\nState the electron configuration of the Ca2+ ion.
\nWhen calcium compounds are introduced into a gas flame a red colour is seen; sodium compounds give a yellow flame. Outline the source of the colours and why they are different.
\nSuggest two reasons why solid calcium has a greater density than solid potassium.
\nOutline why solid calcium is a good conductor of electricity.
\nCalcium carbide reacts with water to form ethyne and calcium hydroxide.
\nCaC2(s) + H2O(l) → C2H2(g) + Ca(OH)2(aq)
\nEstimate the pH of the resultant solution.
\nelectrostatic attraction AND oppositely charged ions
\n[1 mark]
\n1s22s22p63s23p6
\nOR
\n[Ar]
\n[1 mark]
\n«promoted» electrons fall back to lower energy level
\nenergy difference between levels is different
\n\n
Accept “Na and Ca have different nuclear charge” for M2.
\n[2 marks]
\nAny two of:
\nstronger metallic bonding
\nsmaller ionic/atomic radius
\n\n
two electrons per atom are delocalized
\nOR
\ngreater ionic charge
\n\n
greater atomic mass
\n\n
Do not accept just “heavier” or “more massive” without reference to atomic mass.
\n[2 marks]
\ndelocalized/mobile electrons «free to move»
\n[1 mark]
\npH > 7
\n\n
Accept any specific pH value or range of values above 7 and below 14.
\n[1 mark]
\nThis question is about ethene, C2H4, and ethyne, C2H2.
\nEthyne, like ethene, undergoes hydrogenation to form ethane. State the conditions required.
\nOutline the formation of polyethene from ethene by drawing three repeating units of the polymer.
\nUnder certain conditions, ethyne can be converted to benzene.
\nDetermine the standard enthalpy change, ΔHϴ, for the reaction stated, using section 11 of the data booklet.
\n3C2H2(g) → C6H6(g)
\nDetermine the standard enthalpy change, ΔHΘ, for the following similar reaction, using ΔHf values in section 12 of the data booklet.
\n3C2H2(g) → C6H6(l)
\nExplain, giving two reasons, the difference in the values for (b)(i) and (ii). If you did not obtain answers, use −475 kJ for (i) and −600 kJ for (ii).
\nOne possible Lewis structure for benzene is shown.
\n ![\"M18/4/CHEMI/SP2/ENG/TZ1/03.c\"](\"../assets/uploads/tinymce_asset/asset/4574/Schermafbeelding_2018-08-09_om_15.01.32.png\"/)
State one piece of physical evidence that this structure is incorrect.
\nState the characteristic reaction mechanism of benzene.
\nnickel/Ni «catalyst»
\n\n
high pressure
\nOR
\nheat
\n\n
Accept these other catalysts: Pt, Pd, Ir, Rh, Co, Ti.
\nAccept “high temperature” or a stated temperature such as “150 °C”.
\n[2 marks]
\n![\"M18/4/CHEMI/SP2/ENG/TZ1/03.a.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4573/Schermafbeelding_2018-08-09_om_14.47.44.png\"/)
\n
Ignore square brackets and “n”.
\nConnecting line at end of carbons must be shown.
\n[1 mark]
\nΔHϴ = bonds broken – bonds formed
\n«ΔHϴ = 3(C≡C) – 6(C=C)benzene/3 × 839 – 6 × 507 / 2517 – 3042 =»
\n–525 «kJ»
\n\n
\n
Award [2] for correct final answer.
\nAward [1 max] for +525 «kJ»
\nAward [1 max] for:
\n«ΔHϴ = 3(C≡C) – 3(C–C) – 3(C=C) / 3 × 839 – 3 × 346 – 3 × 614 / 2517 – 2880 =» –363 «kJ».
\n[2 marks]
\nΔHΘ = ΣΔHf(products) – ΣΔHf(reactants)
\n«ΔHΘ = 49 kJ – 3 × 228 kJ =» –635 «kJ»
\n\n
Award [2] for correct final answer.
\nAward [1 max] for “+635 «kJ»”.
\n[2 marks]
\nΔHf values are specific to the compound
\nOR
\nbond enthalpy values are averages «from many different compounds»
\n\n
condensation from gas to liquid is exothermic
\n\n
Accept “benzene is in two different states «one liquid the other gas»“ for M2.
\n[2 marks]
\nequal C–C bond «lengths/strengths»
\nOR
\nregular hexagon
\nOR
\n«all» C–C have» bond order of 1.5
\nOR
\n«all» C–C intermediate between single and double bonds
\n\n
Accept “all C–C–C bond angles are equal”.
\n[1 mark]
\nelectrophilic substitution
\nOR
\nSE
\n[1 mark]
\nCalcium carbonate reacts with hydrochloric acid.
\nCaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
\nThe results of a series of experiments in which the concentration of HCl was varied are shown below.
\n![\"M18/4/CHEMI/SP2/ENG/TZ1/04.b\"](\"../assets/uploads/tinymce_asset/asset/4559/Schermafbeelding_2018-08-09_om_13.37.39.png\"/)
Outline two ways in which the progress of the reaction can be monitored. No practical details are required.
\nSuggest why point D is so far out of line assuming human error is not the cause.
\nSuggest the relationship that points A, B and C show between the concentration of the acid and the rate of reaction.
\nAny two of:
\nloss of mass «of reaction mixture/CO2»
\n«increase in» volume of gas produced
\nchange of conductivity
\nchange of pH
\nchange in temperature
\n\n
Do not accept “disappearance of calcium carbonate”.
\nDo not accept “gas bubbles”.
\nDo not accept “colour change” or “indicator”.
\n[2 marks]
\nreaction is fast at high concentration AND may be difficult to measure accurately
\nOR
\nso many bubbles of CO2 produced that inhibit contact of HCl(aq) with CaCO3(s)
\nOR
\ninsufficient change in conductivity/pH at high concentrations
\nOR
\ncalcium carbonate has been used up/is limiting reagent/there is not enough calcium carbonate «to react with the high concentration of HCl»
\nOR
\nHCl is in excess
\nOR
\nso many bubbles of CO2 produced that inhibit contact of HCl(aq) with CaCO3(s)
\n[1 mark]
\n«directly» proportional
\n\n
Accept “first order” or “linear”.
\nDo not accept “rate increases as concentration increases” or “positive correlation”
\n[1 mark]
\nOrganic compounds often have isomers.
\nA straight chain molecule of formula C5H10O contains a carbonyl group. The compound cannot be oxidized by acidified potassium dichromate(VI) solution.
\nA tertiary halogenoalkane with three different alkyl groups, (R1R2R3)C−X, undergoes a SN1 reaction and forms two isomers.
\nDeduce the structural formulas of the two possible isomers.
\nMass spectra A and B of the two isomers are given.
\n![\"M18/4/CHEMI/HP2/ENG/TZ2/09.a.ii_01\"](\"../assets/uploads/tinymce_asset/asset/4529/Schermafbeelding_2018-08-08_om_16.37.09.png\"/)
Explain which spectrum is produced by each compound using section 28 of the data booklet.
\n\n
State the type of bond fission that takes place in a SN1 reaction.
\nState the type of solvent most suitable for the reaction.
\nDraw the structure of the intermediate formed stating its shape.
\nSuggest, giving a reason, the percentage of each isomer from the SN1 reaction.
\nNitrobenzene, C6H5NO2, can be converted to phenylamine via a two-stage reaction.
\nIn the first stage, nitrobenzene is reduced with tin in an acidic solution to form an intermediate ion and tin(II) ions. In the second stage, the intermediate ion is converted to phenylamine in the presence of hydroxide ions.
\nFormulate the equation for each stage of the reaction.
\n\n
![\"M18/4/CHEMI/HP2/ENG/TZ2/09.a.i/M\"](\"../assets/uploads/tinymce_asset/asset/4534/Schermafbeelding_2018-08-08_om_16.55.28.png\"/)
\n
Accept condensed formulas.
\n[2 marks]
\nA:
\nCH3CH2COCH2CH3 AND «peak at» 29 due to
\n(CH3CH2)+/(C2H5)+/(M – CH3CH2CO)+
\nOR
\nCH3CH2COCH2CH3 AND «peak at» 57 due to
\n(CH3CH2CO)+/(M – CH3CH2)+/(M – C2H5)+
\n\n
B:
\nCH3COCH2CH2CH3 AND «peak at» 43 due to
\n(CH3CH2CH2)+/(CH3CO)+/(C2H3O)+/(M – CH3CO)+
\n\n
Penalize missing “+” sign once only.
\nAccept “CH3COCH2CH2CH3 by elimination since fragment CH3CO is not listed” for M2.
\n[2 marks]
\nheterolytic/heterolysis
\n\n
[1 mark]
\npolar protic
\n\n
[1 mark]
\n![\"M18/4/CHEMI/HP2/ENG/TZ2/09.b.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4533/Schermafbeelding_2018-08-08_om_16.52.34.png\"/)
Shape: triangular/trigonal planar
\n[2 marks]
\n«around» 50% «each»
\nOR
\nsimilar/equal percentages
\n\n
nucleophile can attack from either side «of the planar carbocation»
\n\n
Accept “racemic mixture/racemate” for M1.
\n[2 marks]
\nStage one:
\nC6H5NO2(l) + 3Sn(s) + 7H+(aq) → C6H5NH3+(aq) + 3Sn2+(aq) + 2H2O(l)
\n\n
Stage two:
\nC6H5NH3+(aq) + OH–(aq) → C6H5NH2(l) + H2O(l)
\n\n
[2 marks]
\nLutetium-177 is used in radiotherapy. It emits beta radiation when it decays.
\nState a nuclear equation to show the decay of lutetium-177.
\nThe half-life of lutetium-177 is 6.73 days. Determine the percentage of a sample of lutetium-177 remaining after 14.0 days.
\nExplain the low environmental impact of most medical nuclear waste.
\n«+ »
\nHf
\ncorrect A and Z AND beta product
\nAccept “β/ β–/e/e–” for “”.
\nAccept “177Lu → 177Hf + e– «+ »”.
\nnumber of half-lives = = 2.08
\nOR
\n\n
OR
\n«» 0.103 «day–1»
\nOR
\n\n
23.6 «%»
\nAward [2] for correct final answer.
\nAny two of:
\nemits weak ionising radiation
\nOR
\nlow activity/radioactivity
\ncan be stored until material becomes inactive AND then disposed with normal waste
\n«isotopes» have short lives
\nOR
\nexist for a short period of time
\nAward [1 max] for “low-level waste/LLW”.
\n[Max 2 Marks]
\nVanadium forms a body centred cubic (BCC) crystal structure with an edge length of 303 pm, (303 × 10−12 m).
\nDeduce the number of atoms per unit cell in vanadium.
\nCalculate the expected first order diffraction pattern angle, in degrees, if x-rays of wavelength 150 pm are directed at a crystal of vanadium. Assume the edge length of the crystal to be the same as separation of layers of vanadium atoms found by x-ray diffraction. Use section 1 of the data booklet.
\nCalculate the average mass, in g, of a vanadium atom by using sections 2 and 6 of the data booklet.
\nDetermine the volume, in cm3, of a vanadium unit cell.
\nDetermine the density, in g cm−3, of vanadium by using your answers to (a)(i), (a)(iii) and (a)(iv).
\nVanadium and other transition metals can interfere with cell metabolism.
\nState and explain one process, other than by creating free radicals, by which transition metals interfere with cell metabolism.
\nVanadium(IV) ions can create free radicals by a Fenton reaction.
\nDeduce the equation for the reaction of V4+ with hydrogen peroxide.
\n2
\n[1 mark]
\nnλ = 2dsinθ
\nOR
\n\n
\n
θ = «» 14.3 «°»
\n\n
Award [2] for correct final answer.
\n[2 marks]
\nm = « » 8.46 × 10–23 «g»
\n[1 mark]
\n«303 pm = 303 × 10–10 cm»
\nV = «(303 × 10–10)3 =» 2.78 × 10–23 «cm3 »
\n[1 mark]
\n«8.46 × 10–23 g × 2 =» 1.69 × 10–22 «g»
\nd = «» 6.08 «g cm–3»
\n\n
Accept any value in the range 6.07–6.09 «g cm–3».
\nAward [2] for correct final answer.
\n[2 marks]
\nAny one of these alternatives:
\nALTERNATIVE 1
\ndisrupt enzyme binding sites
\nwhich can inhibit/over-stimulate enzymes
\n\n
ALTERNATIVE 2
\ndisrupt endocrine system
\nbecause they compete for active sites of enzymes/cellular receptors
\n\n
ALTERNATIVE 3
\nform complexes/coordination compounds
\nwhich can bind to enzymes
\n\n
ALTERNATIVE 4
\nact as oxidizing/reducing agents
\nOR
\nact as catalysts
\n\n
which can initiate unwanted reactions
\n\n
Accept “can undergo oxidation–reduction reactions” for M1 in Alternative 4.
\n[2 marks]
\nV4+(aq) + H2O2(aq) → V5+(aq) + OH–(aq) + •OH(aq)
\n\n
Do not accept • on H.
\nAccept answer without •
\n[1 mark]
\nLimescale, CaCO3(s), can be removed from water kettles by using vinegar, a dilute solution of ethanoic acid, CH3COOH(aq).
\nPredict, giving a reason, a difference between the reactions of the same concentrations of hydrochloric acid and ethanoic acid with samples of calcium carbonate.
\nDissolved carbon dioxide causes unpolluted rain to have a pH of approximately 5, but other dissolved gases can result in a much lower pH. State one environmental effect of acid rain.
\nslower rate with ethanoic acid
\nOR
\nsmaller temperature rise with ethanoic acid
\n\n
[H+] lower
\nOR
\nethanoic acid is partially dissociated
\nOR
\nethanoic acid is weak
\n\n
Accept experimental observations such as “slower bubbling” or “feels less warm”.
\n[2 marks]
\nAny one of:
\ncorrosion of materials/metals/carbonate materials
\ndestruction of plant/aquatic life
\n«indirect» effect on human health
\n\n
Accept “lowering pH of oceans/lakes/waterways”.
\n[1 mark]
\nThe diagram shows an incomplete voltaic cell with a light bulb in the circuit.
\nIdentify the missing component of the cell and its function.
\nDeduce the half-equations for the reaction at each electrode when current flows.
\nAnnotate the diagram with the location and direction of electron movement when current flows.
\nsalt bridge
\n\n
movement of ions
\nOR
\nbalance charge
\n\n
Do not accept “to complete circuit” unless ion movement is mentioned for M2.
\n[2 marks]
\nPositive electrode (cathode):
\nAg+(aq) + e– → Ag(s)
\nNegative electrode (anode):
\nMg(s) → Mg2+(aq) + 2e–
\n\n
Award [1 max] if correct equations given at wrong electrodes.
\n[2 marks]
\nin external wire from left to right
\n[1 mark]
\nPropene can polymerize to form polypropene.
\nPropene monomer: ![\"M18/4/CHEMI/HP3/ENG/TZ2/05\"](\"../assets/uploads/tinymce_asset/asset/4539/Schermafbeelding_2018-08-08_om_17.53.44.png\"/)
Distinguish between the manufacture of polyester and polyethene.
\nAny one of these alternatives:
\nALTERNATIVE 1
\nPolyester: produced by condensation/esterification polymerization
\nPolyethene: produced by addition polymerization
\n\n
ALTERNATIVE 2
\nPolyester: reaction between monomers/molecules containing two functional groups per molecule
\nPolyethene: reaction between monomers/molecules containing a carbon–carbon double bond/C=C
\n\n
ALTERNATIVE 3
\npolyester polymerization forms a by-product/H2O
\npolyethene has no by-products/100% atom economy
\n\n
Accept the names of different catalysts used for each polymerization as an alternative answer.
\n[2 marks]
\nChemical vapour deposition (CVD) produces multi-walled carbon nanotubes (MWCNT) of a more appropriate size for use in liquid crystals than production by arc discharge.
\nMWCNT are very small in size and can greatly increase switching speeds in a liquid crystal allowing the liquid crystal to change orientation quickly.
\nDiscuss two other properties a substance should have to be suitable for use in liquid crystal displays.
\nAny two from:
\nchemically stable AND does not «chemically» degrade over time
\nstable over range of temperatures AND to avoid «voltage/random shift» fluctuations
\npolar AND influenced by an electric field
\nstrong intermolecular forces AND allow molecule to align in specific orientations
\n\n
Award [1 max] for identifying two correct properties without any discussion given or incorrect interpretation of suitability.
\nAccept “voltage” for “electric field”.
\n[2 marks]
\nExplain how the structure of vitamin A is important to vision using section 35 of the data booklet.
\nvitamin A oxidized to «11-cis-»retinal
\n\n
extended conjugation
\nOR
\nextensive delocalization
\n\n
cis-retinal converts to trans-retinal through absorption of light
\n\n
Accept “vitamin A/hydroxyl/hydroxy/alcohol/CH2OH group oxidized to aldehyde/CHO «group in retinal»”.
\n[3 marks]
\nHemoglobin contains an iron ion that can bind to oxygen as part of the process of respiration.
\nHemoglobin’s oxygen dissociation curve is shown at a given temperature. Sketch the curve on the graph at a higher temperature.
\n![\"M18/4/CHEMI/HP3/ENG/TZ2/11.a\"](\"../assets/uploads/tinymce_asset/asset/4543/Schermafbeelding_2018-08-08_om_18.58.13.png\"/)
Outline two differences between normal hemoglobin and foetal hemoglobin.
\ncurve below original curve «showing lower affinity for oxygen» beginning at 0
\n\n
Award mark if end of student curve does not finish at same location as original curve.
\n[1 mark]
\nAny two of:
\nfoetal hemoglobin has higher affinity for oxygen «than normal hemoglobin»
\nfoetal hemoglobin is less sensitive to inhibitors/2,3-bisphosphoglycerate/2,3-BPG/DPG «than normal hemoglobin»
\nfoetal hemoglobin contains two gamma units instead of the two beta units found in adult hemoglobin
\n[2 marks]
\nDNA is a biopolymer made up of nucleotides. List two components of a nucleotide.
\nAny two of:
\npentose «sugar»
\nOR
\ndeoxyribose
\n\n
phosphate «group»
\n\n
«organic» nitrogenous base
\nOR
\nnucleobase
\nOR
\nnucleic base
\nOR
\npurine
\nOR
\npyrimidine
\n\n
Accept names or formulas.
\nAccept “ribose” for M1.
\nDo not accept “phosphoric acid”.
\nAccept the four bases together: “adenine, cytosine, thymine, guanine”.
\n[2 marks]
\nCrude oil is a useful energy resource.
\nFuel cells have a higher thermodynamic efficiency than octane. The following table gives some information on a direct methanol fuel cell.
\n![\"M18/4/CHEMI/HP3/ENG/TZ2/13.c\"](\"../assets/uploads/tinymce_asset/asset/4547/Schermafbeelding_2018-08-08_om_19.08.02.png\"/)
Determine the thermodynamic efficiency of a methanol fuel cell operating at 0.576 V.
\nUse sections 1 and 2 of the data booklet.
\nn = 6
\n«ΔGΘ = –nFEΘ = 6 mol × 96 500 C mol–1 × 0.576 V =» –333 504 J/–334 kJ
\n«Efficiency = =» 0.459/45.9%
\n\n
Award [3] for correct final answer.
\n[3 marks]
\nPalmitic acid has a molar mass of 256.5 g mol−1.
\nThe apparatus in the diagram measures the surface pressure created by palmitic acid molecules on the surface of water. This pressure is caused by palmitic acid molecules colliding with the fixed barrier. The pressure increases as the area, A, available to the palmitic acid is reduced by the movable barrier.
\n![\"M18/4/CHEMI/SP3/ENG/TZ1/01.b_01\"](\"../assets/uploads/tinymce_asset/asset/4577/Schermafbeelding_2018-08-09_om_15.37.49.png\"/)
When a drop of a solution of palmitic acid in a volatile solvent is placed between the barriers, the solvent evaporates leaving a surface layer. The graph of pressure against area was obtained as the area A was reduced.
\n![\"M18/4/CHEMI/SP3/ENG/TZ1/01.b_02\"](\"../assets/uploads/tinymce_asset/asset/4578/Schermafbeelding_2018-08-09_om_15.39.43.png\"/)
Part of this molecule is hydrophilic (bonds readily to water) and part hydrophobic (does not bond readily to water). Draw a circle around all of the hydrophilic part of the molecule.
\nWhen a small amount of palmitic acid is placed in water it disperses to form a layer on the surface that is only one molecule thick. Explain, in terms of intermolecular forces, why this occurs.
\nSuggest why there is a small increase in the surface pressure as the area is reduced to about 240 cm2, but a much faster increase when it is further reduced.
\nThe solution of palmitic acid had a concentration of 0.0034 mol dm−3. Calculate the number of molecules of palmitic acid present in the 0.050 cm3 drop, using section 2 of the data booklet.
\nAssuming the sudden change in gradient occurs at 240 cm2, calculate the area, in cm2, that a single molecule of palmitic acid occupies on surface of the water.
\nIf you did not obtain an answer for (b)(ii) use a value of 8.2 × 1016, but this is not the correct answer.
\n![\"M18/4/CHEMI/SP3/ENG/TZ1/01.a.i/M\"](\"../assets/uploads/tinymce_asset/asset/4581/Schermafbeelding_2018-08-09_om_16.46.42.png\"/)
\n
Must cut CH2–CO bond AND enclose all of the –COOH group.
\n[1 mark]
\nAny two of:
\n–COOH/CO/OH/carboxylate/carboxyl/hydroxyl/hydroxy group forms hydrogen bonds/H-bonds to water
\nLondon/dispersion/instantaneous induced dipole-induced dipole forces occur between hydrocarbon chains
\nhydrocarbon chain cannot form hydrogen bonds/H-bonds to water
\nstrong hydrogen bonds/H-bonds between water molecules exclude hydrocarbon chains «from the body of the water»
\n\n
\n
Accept “hydrophilic part/group forms hydrogen bonds/H-bonds to water”.
\nAccept “hydrophobic section” instead of “hydrocarbon chain”.
\nAward [1 max] for answers based on “the –COOH group being polar AND the hydrocarbon chain being non-polar”.
\n[2 marks]
\nAbove about 240 cm2:
\ngreater collision frequency/collisions per second between «palmitic acid» molecules and the barrier «as area reduced»
\n\n
At less than about 240 cm2:
\nmolecules completely cover the surface
\nOR
\nthere is no space between molecules
\nOR
\nforce from movable barrier transmitted directly through the molecules to the fixed barrier
\nOR
\n«palmitic acid» molecules are pushed up/down/out of layer
\n\n
For both M1 and M2 accept “particles” for “molecules”.
\nFor M1 accept “space/area between molecules reduced” OR “molecules moving closer together”.
\n[2 marks]
\namount of acid = «5.0 × 10–5 dm3 × 0.0034 mol dm–3» = 1.7 × 10–7 «mol»
\nnumber of molecules = «1.7 × 10–7 mol × 6.02 × 1023 mol–1 =» 1.0 × 1017
\n\n
Award [2] for correct final answer.
\nAward [1] for “1.0 × 1020”.
\n[2 marks]
\n«area = » 2.4 × 10–15 «cm2»
\n[1 mark]
\nNuclear power is another source of energy.
\n235U atoms can be used in nuclear reactors whereas 238U cannot. A centrifuge is used to separate isotopes.
\nCalculate the relative rate of effusion of 235UF6(g) to 238UF6(g) using sections 1 and 6 of the data booklet.
\nExplain, based on molecular structure and bonding, why diffusion or centrifuging can be used for enrichment of UF6 but not UO2.
\nMr(235UF6) = 235 + (19.00 × 6)/349
\nOR
\nMr(238UF6) = 238 + (19.00 × 6)/352
\n\n
«» 1.004
\n\n
Award [2] for correct final answer.
\nDo not accept “1.00” OR “0.996”.
\n[2 marks]
\nUF6: Structure: octahedral «solid»/square bipyramidal «solid»/«simple» molecular solid/simple molecule AND Bonding: covalent
\nUO2: Structure: crystal/lattice/network «solid»/«resembles» fluorite AND Bonding: «partly» covalent
\nUF6 sublimes/evaporates/boils at low temperature
\n\n
Accept “UF6: Structure: octahedral «solid»/square bipyramidal «solid»/«simple» molecular solid/simple molecule AND weak intermolecular/London/dispersion/van der Waals’/vdW forces”.
\nAccept “non-polar molecule” for “«simple» molecular solid”.
\nAccept “giant molecular” OR “macromolecular” for “network”.
\nAccept “ionic/electrostatic attractions «between ions»” for bonding in UO2.
\nAward M2 for “UO2: network covalent/covalent network/giant covalent” OR “UO2: network ionic/giant ionic”.
\nFor M1 and M2 award [1 max] for two correct structures OR two bonding types.
\nAccept any specified low temperature in the range 56–65 °C.
\nThe conductivity of a germanium semiconductor can be increased by doping.
\nA dye-sensitized solar cell uses a ruthenium(II)–polypyridine complex as the dye. Two ruthenium(II) complexes, A and B, absorb light of wavelengths 665 nm and 675 nm respectively.
\nDraw the Lewis (electron dot) structure for an appropriate doping element in the box in the centre identifying the type of semiconductor formed.
\n![\"M18/4/CHEMI/HP3/ENG/TZ2/18.a\"](\"../assets/uploads/tinymce_asset/asset/4549/Schermafbeelding_2018-08-08_om_19.31.34.png\"/)
State the feature of the molecules responsible for the absorption of light.
\nOutline why complex B absorbs light of longer wavelength than complex A.
\nALTERNATIVE 1
\nB/Ga in circle AND Type of semiconductor: p-type
\nshowing 3 electron pairs AND one lone electron «and hole»
\n\n
ALTERNATIVE 2
\nP/As in circle AND Type of semiconductor: n-type
\nshowing 4 electron pairs AND one non-bonded electron
\n\n
Accept any group 13 element labelled as p-type.
\nAccept showing 7 electrons.
\nAccept any group 15 element labelled as n-type.
\nAccept showing 9 electrons.
\nAccept dots or crosses for electrons.
\n[2 marks]
\nconjugated C=C/carbon–carbon double bonds
\nOR
\n«multiple» alternating C=C/carbon–carbon double bonds
\nOR
\n«extensive electron» conjugation/delocalization
\nOR
\n«many» fused/conjugated aromatic/benzene rings
\n[1 mark]
\ncomplex B has greater conjugation/delocalization
\n[1 mark]
\nInductively Coupled Plasma (ICP) used with Mass Spectrometry (MS) or Optical Emission Spectrometry (OES) can be used to identify and quantify elements in a sample.
\nThe following graphs represent data collected by ICP-OES on trace amounts of vanadium in oil.
\nGraph 1: Calibration graph and signal for 10 μg kg−1 of vanadium in oil
\nGraph 2: Calibration of vanadium in μg kg−1
\n[Source: © Agilent Technologies, Inc.1998. Reproduced with Permission, Courtesy of Agilent Technologies, Inc.]
\nICP-OES/MS can be used to analyse alloys and composites. Distinguish between alloys and composites.
\nICP-MS is a reference mode for analysis. The following correlation graphs between ICP-OES and ICP-MS were produced for yttrium and nickel.
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/03.b\"](\"../assets/uploads/tinymce_asset/asset/4645/Schermafbeelding_2018-08-10_om_13.02.42.png\"/)
Each y-axis shows concentrations calculated by ICP-OES; each x-axis shows concentrations for the same sample as found by ICP-MS.
\nThe line in each graph is y = x.
\nDiscuss the effectiveness of ICP-OES for yttrium and nickel.
\nIdentify the purpose of each graph.
\n\n
Calculate, to four significant figures, the concentration, in μg kg−1, of vanadium in oil giving a signal intensity of 14 950.
\nVanadium(V) oxide is used as the catalyst in the conversion of sulfur dioxide to sulfur trioxide.
\nSO2(g) + V2O5(s) → SO3(g) + 2VO2(s)
\nO2(g) + 2VO2(s) → V2O5(s)
\nOutline how vanadium(V) oxide acts as a catalyst.
\nAlloy:
\nmixture of metal with other metals/non-metals
\nOR
\nmixture of elements that retains the properties of a metal
\n\n
Composite:
\nreinforcing phase embedded in matrix phase
\n\n
Award [1 max] for implying “composites only have heterogeneous/nonhomogeneous compositions”.
\n[2 marks]
\neffective for yttrium «but less/not for nickel»
\n\n
points on nickel graph do not lie on «y = x» line
\nOR
\ncannot be used for low concentrations of nickel
\nOR
\nconcentration of nickel is lower than recorded value
\n\n
Accept “ICP-OES is more accurate for lower yttrium concentrations than higher concentrations” for M1.
\nAccept [Ni] and [Y] for concentrations of nickel and yttrium.
\nAccept “detection limit for yttrium is lower than for nickel” for M2.
\nAward [1 max] for “more accurate for yttrium at lower concentrations AND nickel at higher concentrations”.
\n[2 marks]
\nGraph 1: determines wavelength of maximum absorption/maximum intensity «for vanadium»
\n\n
Graph 2: determines absorption of known concentrations «at that wavelength»
\nOR
\nestimates [V]/concentration in a sample using «the signal» intensity
\n\n
Do not accept just “determines maximum wavelength/λmax” for M1.
\nDo not accept “calibration curve” for M2.
\n[2 marks]
\n«14 950 = 392.19x + 147.62»
\nx = 37.74 «μg kg–1»
\n\n
Answer must be given to four significant figures.
\nDo not accept values obtained directly from the graph.
\n[1 mark]
\nvanadium reduced in first reaction AND oxidized in second reaction
\nOR
\nV2O5 oxidizes SO2 in first reaction AND VO2 reduces O2 in second reaction
\nOR
\nvanadium returns to original oxidation state «after reaction»
\n\n
provides an alternative reaction pathway/mechanism «with a lower activation energy» ✔
\n\n
Do not accept “reactants adsorb onto surface AND products desorb”.
\nAccept “oxidation number” for “oxidation state”.
\n[2 marks]
\nTaxol was originally obtained from the bark of the Pacific yew tree.
\nOutline how Green Chemistry has improved the process of obtaining Taxol.
\nAny two of:
\nstripping the bark kills Pacific yew tree
\n\n
plant cell fermentation «and extraction»/PCF technology/use of plant cell cultures/Taxol «precursors» produced by biosynthesis/fungi/yeast/e-coli/use of natural enzymes «more sustainable process»
\nOR
\nTaxol produced semi-synthetically/Taxol from 10-DAB/10-deacetylbaccatin
\n\n
uses renewable resources
\nOR
\nuse «needles/leaves/twigs of» European/common yew/yew from Himalayas
\n\n
«sustainable» process has eliminated «high proportion of» hazardous chemicals/waste
\nOR
\n«sustainable» process has eliminated several solvents/«sustainable» process uses greener solvents/«sustainable» process recycles/reuses solvents
\nOR
\n«sustainable» process has eliminated several «drying» steps/«sustainable» process has eliminated lots of the work-up after the synthesis
\nOR
\n«sustainable» process has increased energy efficiency
\nOR
\n«sustainable» process has no intermediates
\nOR
\n«sustainable» process uses more efficient catalysts
\n\n
Accept “Pacific yew rare/slowgrowing/takes 100/200 years to mature” for M1.
\nAccept “synthesis of Taxol using chiral auxiliaries increases efficiency of process as single enantiomer formed” for M4.
\n[2 marks]
\nWhich represents a reduction?
\n\n
A. SO3 to SO42−
\nB. Mn2O3 to MnO2
\nC. H2O2 to OH−
\nD. CrO42− to Cr2O72−
\nC
\nWhich compounds cause the colour of acidified potassium manganate(VII) solution to change from purple to colourless?
\nI. CH3CH2CH2CH2OH
\nII. (CH3)3CCH2OH
\nIII. CH3CH2CH(OH)CH3
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nWhich is correct for benzene?
\n\n
A. It readily undergoes addition reactions and decolourises bromine water.
\nB. It contains alternate single and double carbon–carbon bonds and is planar.
\nC. Its 1H NMR spectrum shows six signals and it readily undergoes substitution reactions.
\nD. Its 1H NMR spectrum shows one signal and it forms a single C6H5Br isomer.
\nD
\nWhat is the order of increasing boiling point for the isomers of C5H12?
\n\n
A. CH3CH2CH2CH2CH3 < CH3CH(CH3)CH2CH3 < CH3C(CH3)3
\nB. CH3C(CH3)3 < CH3CH(CH3)CH2CH3 < CH3CH2CH2CH2CH3
\nC. CH3C(CH3)3 < CH3CH2CH2CH2CH3 < CH3CH(CH3)CH2CH3
\nD. CH3CH(CH3)CH2CH3 < CH3C(CH3)3 < CH3CH2CH2CH2CH3
\nB
\nWhich compounds react to form CH3CH2CH2COOCH(CH3)2?
\n\n
A. propanoic acid and propan-2-ol
\nB. propanoic acid and butan-2-ol
\nC. butanoic acid and propan-1-ol
\nD. butanoic acid and propan-2-ol
\nD
\nWhich is correct for the spectra of organic compounds?
\n\n
A. Mass spectroscopy provides information about bond vibrations.
\nB. 1H NMR spectroscopy provides the values of carbon–hydrogen bond lengths.
\nC. Infrared spectroscopy provides the number of hydrogen atoms.
\nD. Mass spectroscopy provides information about the structure.
\nD
\nStudents were asked to investigate how a change in concentration of hydrochloric acid, HCl, affects the initial rate of its reaction with marble chips, CaCO3.
\nThey decided to measure how long the reaction took to complete when similar chips were added to 50.0 cm3 of 1.00 mol dm−3 acid and 50.0 cm3 of 2.00 mol dm−3 acid.
\nTwo methods were proposed:
\n(1) using small chips, keeping the acid in excess, and recording the time taken for the solid to disappear
\n(2) using large chips, keeping the marble in excess, and recording the time taken for bubbles to stop forming.
\nA group recorded the following results with 1.00 mol dm−3 hydrochloric acid:
\n![\"M18/4/CHEMI/SP3/ENG/TZ1/02.d\"](\"../assets/uploads/tinymce_asset/asset/4579/Schermafbeelding_2018-08-09_om_15.47.04.png\"/)
Annotate the balanced equation below with state symbols.
\nCaCO3(__) + 2HCl(__) → CaCl2(__) + CO2(__) + H2O(__)
\nNeither method actually gives the initial rate. Outline a method that would allow the initial rate to be determined.
\nDeduce, giving a reason, which of the two methods would be least affected by the chips not having exactly the same mass when used with the different concentrations of acid.
\nState a factor, that has a significant effect on reaction rate, which could vary between marble chips of exactly the same mass.
\nJustify why it is inappropriate to record the uncertainty of the mean as ±0.01 s.
\nIf doubling the concentration doubles the reaction rate, suggest the mean time you would expect for the reaction with 2.00 mol dm−3 hydrochloric acid.
\nAnother student, working alone, always dropped the marble chips into the acid and then picked up the stopwatch to start it. State, giving a reason, whether this introduced a random or systematic error.
\nCaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
\n\n
Accept “CO2(aq)”.
\n[1 mark]
\nmeasure the volume of gas at different times «plot a graph and extrapolate»
\nOR
\nmeasure the mass of the reaction mixture at different times «plot a graph and extrapolate»
\n\n
Accept other techniques that yield data which can be plotted and extrapolated.
\n[1 mark]
\nmethod 2 AND marble is in excess «so a little extra has little effect»
\nOR
\nlarge chips AND marble is in excess «so a little extra has little effect»
\nOR
\nmethod 2 AND HCl is limiting reagent «so a little extra marble has little effect»
\nOR
\nlarge chips AND HCl is limiting reagent «so a little extra marble has little effect»
\n\n
Accept, as a reason, that “as the mass is greater the percentage variation will be lower”.
\n[1 mark]
\nsurface area
\nOR
\npurity «of the marble»
\n\n
Accept “shape of the chip”.
\n[1 mark]
\nvariation of individual values is much greater «than this uncertainty»
\nOR
\n«uncertainty» does not take into account «student» reaction time
\n[1 mark]
\n« = 60.98 s» = 61 «s»
\n[1 mark]
\nsystematic AND always makes the time shorter «than the actual value»
\nOR
\nsystematic AND it is an error in the method used «not an individual measurement»
\nOR
\nsystematic AND more repetitions would not reduce the error
\n\n
Accept, as reason, “it always affects the value in the same direction” OR “the error is consistent”.
\n[1 mark]
\nAluminium and high density polyethene (HDPE) are both materials readily found in the kitchen, for example as saucepans and mixing bowls respectively. In these applications it is important that they are impermeable to water.
\nBoth materials are also used in other applications that are more demanding of their physical properties. Carbon nanotubes are often incorporated into their structures to improve certain properties.
\nDiscuss, in terms of its structure, why an aluminium saucepan is impermeable to water.
\nState the name given to a material composed of two distinct solid phases.
\nState one physical property of HDPE that will be affected by the incorporation of carbon nanotubes.
\nDescribe how carbon nanotubes are produced by chemical vapour deposition (CVD).
\nState the property of carbon nanotubes that enables them to form a nematic liquid crystal phase.
\n«close packed» lattice of metal atoms/ions
\nno spaces for water molecules to pass though the structure
\n[2 marks]
\ncomposite
\n[1 mark]
\nmelting point
\nOR
\npermeability
\nOR
\ndensity
\nOR
\nconductivity
\nOR
\nelasticity/stiffness
\nOR
\nbrittleness/flexibility
\nOR
\n«tensile» strength
\n\n
Accept “colour/transparency”.
\n[1 mark]
\nAny three of:
\nhydrocarbon/carbon-containing gas/compound
\nmixed with inert gas
\nheat/high temperature
\n«transition» metal catalyst
\nhydrocarbon/carbon compound decomposes to form carbon «nanotubes»
\nnanotubes form on catalyst surface
\n\n
Accept “ethanol” or specific hydrocarbons.
\nAccept “N2”, “H2”, “NH3” or specific inert gases.
\nAccept temperature or range within 600–800 °C.
\nAccept specific metals such as Ni, Co or Fe.
\n[3 marks]
\nrod shaped molecules
\n[1 mark]
\nBoth HDPE (high density polyethene) and LDPE (low density polyethene) are produced by the polymerization of ethene.
\nBoth of these are thermoplastic polymers. Outline what this term means.
\nCompare and contrast the structures of HDPE and LDPE.
\nState one way in which a physical property of HDPE, other than density, differs from that of LDPE as a result of this structural difference.
\nThe production of HDPE involves the use of homogeneous catalysts. Outline how homogeneous catalysts reduce the activation energy of reactions.
\nTrace amounts of metal from the catalysts used in the production of HDPE sometimes remain in the product. State a technique that could be used to measure the concentration of the metal.
\nSuggest two of the major obstacles, other than collection and economic factors, which have to be overcome in plastic recycling.
\nSuggest why there are so many different ways in which plastics can be classified. HDPE can, for example, be categorized thermoplastic, an addition polymer, having Resin Identification Code (RIC) 2, etc.
\nsoften/melt when heated
\nOR
\ncan be melted and moulded
\n\n
Accept “low melting point” OR “can be moulded when heated”.
\n[1 mark]
\nboth have «long» hydrocarbon chains
\nOR
\nboth have chains comprising CH2 units
\n\n
HDPE has little/no branching AND LDPE has «more» branching
\n\n
Accept “CH2–CH2 units”.
\nAccept “HDPE more crystalline”.
\n[2 marks]
\nHDPE is more rigid/less flexible
\nOR
\nHDPE has a higher melting point
\nOR
\nHDPE has greater «tensile» strength
\n\n
Accept “HDPE has lower ductility”.
\n[1 mark]
\nform «temporary» activated complexes/reaction intermediates
\n\n
Accept “consumed in one reaction/step AND regenerated in a later reaction/step”.
\nAccept “provides alternative mechanism”.
\n[1 mark]
\ninductively coupled plasma/ICP spectroscopy using mass spectroscopy/mass spectrometry/MS/ICP-MS
\nOR
\ninductively coupled plasma/ICP spectroscopy using optical emission spectroscopy/OES/ICP-OES
\n\n
Accept “atomic absorption/aa spectroscopy” or “MS/massspectroscopy/mass spectrometry”.
\n[1 mark]
\nAny two of:
\nmany types «of plastics» exist
\nOR
\n«plastics» require sorting «by type»
\n\n
«plastics» need to be separated from non-plastic materials
\nOR
\n«often» composites/moulded on/bound to non-plastic/other components
\n\n
Accept other valid factors such as thermal decomposition of some plastics, production of toxic fumes, etc.
\n[2 marks]
\n«different classifications are appropriate for» different properties/applications/purposes
\n[1 mark]
\nAluminium is produced by the electrolysis of a molten electrolyte containing bauxite.
\nDetermine the mass, in g, of aluminium produced by the passage of a charge of 1.296 × 1013 C. Use sections 2 and 6 of the data booklet.
\nratio of electrons : aluminium ions = 3 : 1
\namount Al « » = 4.48 × 107 «mol»
\nmass Al «= 4.48 × 107 mol × 26.98 g mol–1» = 1.21 × 109 «g»
\n\n
Award [3] for correct final answer.
\n[3 marks]
\nInsulin was the first protein to be sequenced. It was determined that the end of one chain had the primary structure Phe–Val–Asn–Gln.
\nPaper chromatography can be used to identify the amino acids in insulin.
\nDraw the structural formula of a dipeptide containing the residues of valine, Val, and asparagine, Asn, using section 33 of the data booklet.
\nDeduce the strongest intermolecular forces that would occur between the following amino acid residues in a protein chain.
\n\n
State the name of the process used to break down the insulin protein into its constituent amino acids.
\nOutline how the amino acids may be identified from a paper chromatogram.
\n![\"M18/4/CHEMI/SP3/ENG/TZ1/06.a/M\"](\"../assets/uploads/tinymce_asset/asset/4584/Schermafbeelding_2018-08-09_om_18.08.31.png\"/)
correct structures of Val AND Asn
\ncorrect amide link
\n[2 marks]
\nPhenylalanine and valine:
\nLondon/dispersion/instantaneous induced dipole-induced dipole forces
\nOR
\npermanent dipole-induced dipole «interactions»
\n\n
Glutamine and asparagine:
\nhydrogen bonds
\n\n
Do not accept dipole-dipole interactions.
\n[2 marks]
\nhydrolysis
\n[1 mark]
\ncompare Rf with known amino acids
\nOR
\ncompare distance moved with known amino acids
\n\n
Accept “from Rf”.
\n[1 mark]
\nSaturated lipids found in butter and unsaturated lipids found in fish oil readily become rancid.
\nIdentify the type of rancidity occurring in saturated lipids and the structural feature that causes it.
\nState one factor that increases the rate at which saturated lipids become rancid.
\nButter contains varying proportions of oleic, myristic, palmitic and stearic acids. Explain in terms of their structures why stearic acid has a higher melting point than oleic acid, using section 34 of the data booklet.
\nFish oil is an excellent dietary source of omega-3 fatty acids. Outline one impact on health of consuming omega-3 fatty acids.
\nPredict the solubility of retinol (vitamin A) in body fat, giving a reason. Use section 35 of the data booklet.
\nExplain why sharks and swordfish sometimes contain high concentrations of mercury and polychlorinated biphenyls (PCBs).
\nPlastics are another source of marine pollution. Outline one way in which plastics can be made more biodegradable.
\nhydrolytic «rancidity»
\nester group
\n\n
Accept a formula for ester group.
\n[2 marks]
\n«presence of» moisture/water
\nOR
\n«increase in» temperature
\nOR
\n«presence of» enzymes/bacteria/fungi/mould
\nOR
\nlow pH/«presence of» acid
\n\n
Accept “heat”.
\n[1 mark]
\n«stearic acid» straight chain/chain has no kinks/more regular structure
\nOR
\n«stearic acid» saturated/no «carbon–carbon» double bonds
\n\n
«stearic acid» chains pack more closely together
\nstronger London/dispersion/instantaneous induced dipole-induced dipole forces «between molecules»
\n\n
Accept “«stearic acid» greater surface area/electron density”.
\n[3 marks]
\nlowers risk of heart disease/atherosclerosis
\nOR
\nlowers LDL cholesterol
\nOR
\nincreases HDL cholesterol
\nOR
\naids brain/neurological development «in children»
\nOR
\nrelieves rheumatoid arthritis
\n[1 mark]
\nsoluble AND non-polar hydrocarbon chain
\n\n
Accept as reasons “«predominantly» non-polar” OR “long hydrocarbon chain”.
\n[1 mark]
\nnot biodegradable
\nOR
\nstored/accumulate in fat
\n\n
biomagnification occurs
\nOR
\nconcentration increases along food chain
\n\n
Accept “stored/accumulate in bodies of prey/animals eaten”.
\nAccept “not excreted”.
\n[2 marks]
\nadd starch/cellulose/carbohydrates/additives/catalysts «to plastic during manufacture to allow digestion by micro-organisms»
\nOR
\nreplace traditional plastics with polylactic acid/PLA-based ones
\nOR
\nblend traditional and polylactic acid/PLA-based plastics
\n\n
Accept reference to biodegradable plastics other than PLA; for example polyhydroxyalkanoates (PHA), poly(butylene succinate) (PBS), polybutylene adipate terephthalate (PBAT) and polycaprolactone (PCL).
\n[1 mark]
\nWhat is the ratio of areas under each signal in the 1H NMR spectrum of 2-methylbutane?
\n\n
A. 6 : 1 : 2 : 3
\nB. 3 : 3 : 1 : 5
\nC. 6 : 1 : 5
\nD. 3 : 3 : 1 : 2 : 3
\nA
\nWhat are the absolute and percentage uncertainties for the change in mass?
\nInitial mass: 22.35 ±0.05 g
\nFinal mass: 42.35 ±0.05 g
\n\n
B
\nThe values for the first three successive ionization energies for two elements X and Z are given.
\nWhich pair of elements represents X and Z?
\n\n
B
\nWhich is correct for the complex ion in [Fe(H2O)5Cl]SO4?
\n\n
D
\nWhat is the number of sigma (σ) and pi (π) bonds in the molecule (NC)2C=C(CN)2?
\n\n
A
\nWhat is the hybridization of the circled carbon, oxygen and nitrogen atoms?
\n\n
D
\nWhat are the signs of ΔHΘ and ΔSΘ for the reaction, which is spontaneous at low temperature and non-spontaneous at very high temperature?
\nΔGΘ = ΔHΘ − TΔSΘ
\nSO3 (g) + CaO (s) → CaSO4 (s)
\n\n
C
\nWhich change is exothermic?
\n\n
A. Cl2 (g) → Cl (g)
\nB. K (g) → K+ (g) + e−
\nC. KCl (s) → K+ (g) + Cl− (g)
\nD. Cl (g) + e− → Cl− (g)
\nD
\nCompounds X and Y were mixed and the time taken for a colour to appear was recorded at various reactant concentrations.
\nWhat are the orders of reaction with respect to X and Y?
\n\n
A
\n3.26 g of iron powder are added to 80.0 cm3 of 0.200 mol dm−3 copper(II) sulfate solution. The following reaction occurs:
\nFe (s) + CuSO4 (aq) → FeSO4 (aq) + Cu (s)
\nDetermine the limiting reactant showing your working.
\nThe mass of copper obtained experimentally was 0.872 g. Calculate the percentage yield of copper.
\nThe reaction was carried out in a calorimeter. The maximum temperature rise of the solution was 7.5 °C.
\nCalculate the enthalpy change, ΔH, of the reaction, in kJ, assuming that all the heat released was absorbed by the solution. Use sections 1 and 2 of the data booklet.
\nState another assumption you made in (b)(i).
\nThe only significant uncertainty is in the temperature measurement.
\nDetermine the absolute uncertainty in the calculated value of ΔH if the uncertainty in the temperature rise was ±0.2 °C.
\nSketch a graph of the concentration of iron(II) sulfate, FeSO4, against time as the reaction proceeds.
\nOutline how the initial rate of reaction can be determined from the graph in part (c)(i).
\nExplain, using the collision theory, why replacing the iron powder with a piece of iron of the same mass slows down the rate of the reaction.
\nnCuSO4 «= 0.0800 dm3 × 0.200 mol dm–3» = 0.0160 mol AND
\nnFe «» = 0.0584 mol ✔
\nCuSO4 is the limiting reactant ✔
\n\n
Do not award M2 if mole calculation is not shown.
\nALTERNATIVE 1:
«0.0160 mol × 63.55 g mol–1 =» 1.02 «g» ✔
«» 85.5 «%» ✔
\n\n
ALTERNATIVE 2:
«» 0.0137 «mol» ✔
«» 85.6 «%» ✔
\n\n
Accept answers in the range 85–86 %.
\nAward [2] for correct final answer.
\nALTERNATIVE 1:
\nq = «80.0 g × 4.18 J g–1 K–1 × 7.5 K =» 2.5 × 103 «J»/2.5 «kJ» ✔
\n«per mol of CuSO4 = kJ mol–1»
\n«for the reaction» ΔH = –1.6 × 102 «kJ» ✔
\n\n
ALTERNATIVE 2:
\nq = «80.0 g × 4.18 J g–1 K–1 × 7.5 K =» 2.5 × 103 «J»/2.5 «kJ» ✔
\n«nCu = = 0.0137 mol»
\n«per mol of CuSO4 = kJ mol–1»
\n«for the reaction» ΔH = –1.8 × 102 «kJ» ✔
\n\n
Award [2] for correct final answer.
\ndensity «of solution» is 1.00 g cm−3
\nOR
\nspecific heat capacity «of solution» is 4.18 J g−1 K−1/that of «pure» water
\nOR
\nreaction goes to completion
\nOR
\niron/CuSO4 does not react with other substances ✔
\n\n
The mark for “reaction goes to completion” can only be awarded if 0.0160 mol was used in part (b)(i).
\nDo not accept “heat loss”.
\nALTERNATIVE 1:
\n«» 3 %/0.03 ✔
\n«0.03 × 160 kJ» = «±» 5 «kJ» ✔
\n\n
ALTERNATIVE 2:
\n«» 3 %/0.03 ✔
\n«0.03 × 180 kJ» = «±» 5 «kJ» ✔
\n\n
Accept values in the range 4.1–5.5 «kJ».
\nAward [2] for correct final answer.
\n\n
\n
initial concentration is zero AND concentration increases with time ✔
\ndecreasing gradient as reaction proceeds ✔
\n«draw a» tangent to the curve at time = 0 ✔
\n«rate equals» gradient/slope «of the tangent» ✔
\n\n
Accept suitable diagram.
\npiece has smaller surface area ✔
\n\n
lower frequency of collisions
\nOR
\nfewer collisions per second/unit time ✔
\n\n
Accept “chance/probability” instead of “frequency”.
\nDo not accept just “fewer collisions”.
\nThe rate expression for the reaction is: rate = k [NO]2[O2].
\n2NO (g) + O2 (g) → 2NO2 (g)
\nWhich mechanism is not consistent with this rate expression?
\n\n
C
\nPropene can polymerize to form polypropene.
\nPropene monomer:
Sketch four repeating units of the polymer to show atactic and isotactic polypropene.
\nState the chemical reason why plastics do not degrade easily.
\nCompare two ways in which recycling differs from reusing plastics.
\nCivilizations are often characterized by the materials they use.
\nSuggest an advantage polymers have over materials from the iron age.
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/04.a_01/M\"](\"../assets/uploads/tinymce_asset/asset/4648/Schermafbeelding_2018-08-10_om_13.19.54.png\"/)
Do not accept syndiotactic (alternating orientation of the CH3 groups), eg,
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/04.a_02/M\"](\"../assets/uploads/tinymce_asset/asset/4649/Schermafbeelding_2018-08-10_om_13.21.18.png\"/)
for M1 or M2.
\nAccept any correct atactic ordering of CH3 groups.
\nPenalize missing hydrogens or incorrect bond connectivities once only.
\nAccept skeletal structures.
\nIgnore continuation bonds, brackets and “n” indices in structures.
\n[2 marks]
\nstrong covalent bonds
\n\n
Accept “moisture cannot get inside the plastic matrix, and bacteria cannot live without moisture, so they cannot attack the polymer chains”.
\nAccept “bacteria lack the enzymes required to break down the hydrocarbon chains”.
\n[1 mark]
\nAny two of:
\nRecycling: shredded/melted/reformed AND Reuse: used in its current form
\nrecycling is more energy intensive «than reusing»
\nrecycling degrades the quality of plastic but reusing «typically» does not
\nrecycling breaks down original product to form a new product whereas reuse extends product life
\n[2 marks]
\nmore pliable/flexible materials
\nOR
\nmore durable/non-corrosive/longer-lasting materials
\nOR
\ngreater variety of materials
\nOR
\nlower density
\nOR
\ncan be clear/translucent
\n\n
Accept “more adaptable”.
\nDo not accept just “more useful”.
\n[1 mark]
\n3.26 g of iron powder are added to 80.0 cm3 of 0.200 mol dm−3 copper(II) sulfate solution. The following reaction occurs:
\nFe (s) + CuSO4 (aq) → FeSO4 (aq) + Cu (s)
\nDetermine the limiting reactant showing your working.
\nThe mass of copper obtained experimentally was 0.872 g. Calculate the percentage yield of copper.
\nThe reaction was carried out in a calorimeter. The maximum temperature rise of the solution was 7.5 °C.
\nCalculate the enthalpy change, ΔH, of the reaction, in kJ, assuming that all the heat released was absorbed by the solution. Use sections 1 and 2 of the data booklet.
\nState another assumption you made in (b)(i).
\nThe only significant uncertainty is in the temperature measurement.
\nDetermine the absolute uncertainty in the calculated value of ΔH if the uncertainty in the temperature rise was ±0.2 °C.
\nSketch a graph of the concentration of iron(II) sulfate, FeSO4, against time as the reaction proceeds.
\nOutline how the initial rate of reaction can be determined from the graph in part (c)(i).
\nExplain, using the collision theory, why replacing the iron powder with a piece of iron of the same mass slows down the rate of the reaction.
\nA student electrolyzed aqueous iron(II) sulfate, FeSO4 (aq), using platinum electrodes. State half-equations for the reactions at the electrodes, using section 24 of the data booklet.
\nnCuSO4 «= 0.0800 dm3 × 0.200 mol dm–3» = 0.0160 mol AND
\nnFe «» = 0.0584 mol ✔
\nCuSO4 is the limiting reactant ✔
\n\n
Do not award M2 if mole calculation is not shown.
\nALTERNATIVE 1:
«0.0160 mol × 63.55 g mol–1 =» 1.02 «g» ✔
«» 85.5 «%» ✔
\n\n
ALTERNATIVE 2:
«» 0.0137 «mol» ✔
«» 85.6 «%» ✔
\n\n
Accept answers in the range 85–86 %.
\nAward [2] for correct final answer.
\nALTERNATIVE 1:
\nq = «80.0 g × 4.18 J g–1 K–1 × 7.5 K =» 2.5 × 103 «J»/2.5 «kJ» ✔
\n«per mol of CuSO4 = kJ mol–1»
\n«for the reaction» ΔH = –1.6 × 102 «kJ» ✔
\n\n
ALTERNATIVE 2:
\nq = «80.0 g × 4.18 J g–1 K–1 × 7.5 K =» 2.5 × 103 «J»/2.5 «kJ» ✔
\n«nCu = = 0.0137 mol»
\n«per mol of CuSO4 = kJ mol–1»
\n«for the reaction» ΔH = –1.8 × 102 «kJ» ✔
\n\n
Award [2] for correct final answer.
\ndensity «of solution» is 1.00 g cm−3
\nOR
\nspecific heat capacity «of solution» is 4.18 J g−1 K−1/that of «pure» water
\nOR
\nreaction goes to completion
\nOR
\niron/CuSO4 does not react with other substances ✔
\n\n
The mark for “reaction goes to completion” can only be awarded if 0.0160 mol was used in part (b)(i).
\nDo not accept “heat loss”.
\nALTERNATIVE 1:
\n«» 3 %/0.03 ✔
\n«0.03 × 160 kJ» = «±» 5 «kJ» ✔
\n\n
ALTERNATIVE 2:
\n«» 3 %/0.03 ✔
\n«0.03 × 180 kJ» = «±» 5 «kJ» ✔
\n\n
Accept values in the range 4.1–5.5 «kJ».
\nAward [2] for correct final answer.
\n\n
\n
initial concentration is zero AND concentration increases with time ✔
\ndecreasing gradient as reaction proceeds ✔
\n«draw a» tangent to the curve at time = 0 ✔
\n«rate equals» gradient/slope «of the tangent» ✔
\n\n
Accept suitable diagram.
\npiece has smaller surface area ✔
\n\n
lower frequency of collisions
\nOR
\nfewer collisions per second/unit time ✔
\n\n
Accept “chance/probability” instead of “frequency”.
\nDo not accept just “fewer collisions”.
\nAnode (positive electrode):
\n2H2O (l) → O2 (g) + 4H+ (aq) + 4e– ✔
\n\n
Cathode (negative electrode):
\n2H2O (l) + 2e– → H2 (g) + 2OH– (aq)
OR
2H+ (aq) + 2e– → H2 (g) ✔
\n
Accept “4OH– (aq) → O2 (g) + 2H2O (l) + 4e–” OR “Fe2+ (aq) → Fe3+ (aq) + e–” for M1.
\nAccept “Fe2+ (aq) + 2e– → Fe (s)” OR “SO42- (aq) 4H+ (aq) + 2e– → 2H2SO3(aq) + H2O (l)”
for M2.
Which combination describes the system at equilibrium?
\n\n
B
\nWhat is the order of increasing pH for the following solutions of the same concentration?
\n\n
A. NaCl < NH4Cl < Na2CO3 < CH3COONa
\nB. CH3COONa < NH4Cl < NaCl < Na2CO3
\nC. NH4Cl < NaCl < CH3COONa < Na2CO3
\nD. Na2CO3 < CH3COONa < NaCl < NH4Cl
\nC
\nWhich species is not a Lewis base?
\n\n
A. OH−
\nB. NH4+
\nC. H2O
\nD. PH3
\nB
\nAn indicator, HIn, has a pKa of 5.1.
\nHIn (aq) H+ (aq) + In− (aq)
\ncolour A colour B
\nWhich statement is correct?
\n\n
A. At pH = 7, colour B would be observed
B. At pH = 3, colour B would be observed
C. At pH = 7, [HIn] = [In−]
D. At pH = 3, [HIn] < [In−]
A
\nWhich is correct for a redox reaction where the standard electrode potential is negative?
\nΔGΘ = −nFEΘ and ΔGΘ = −RT ln K
\n\n
A. ΔGΘ is negative and K is less than 1.
\nB. ΔGΘ is negative and K is greater than 1.
\nC. ΔGΘ is positive and K is less than 1.
\nD. ΔGΘ is positive and K is greater than 1.
\nC
\nPropan-2-ol is a useful organic solvent.
\nDraw the structural formula of propan-2-ol.
\nCalculate the number of hydrogen atoms in 1.00 g of propan-2-ol.
\nClassify propan-2-ol as a primary, secondary or tertiary alcohol, giving a reason.
\nState a suitable oxidizing agent for the oxidation of propan-2-ol in an acidified aqueous solution.
\nDeduce the average oxidation state of carbon in propan-2-ol.
\nDeduce the product of the oxidation of propan-2-ol with the oxidizing agent in (d)(i).
\nCH3CH(OH)CH3
\n\n
Accept the full or condensed structural formula.
\n«» 0.0166 «mol CH3CH(OH)CH3» ✔
\n«0.0166 mol × 6.02 × 1023 molecules mol−1 × 8 atoms molecule−1 =» 8.01 × 1022 «atoms of hydrogen» ✔
\n\n
Accept answers in the range 7.99 × 1022 to 8.19 × 1022.
\nAward [2] for correct final answer.
\nsecondary AND OH/hydroxyl is attached to a carbon bonded to one hydrogen
\nOR
\nsecondary AND OH/hydroxyl is attached to a carbon bonded to two C/R/alkyl/CH3 «groups» ✔
\n\n
Accept “secondary AND OH is attached to the second carbon in the chain”.
\n«potassium/sodium» manganate(VII)/permanganate/KMnO4/NaMnO4/MnO4−
\nOR
\n«potassium/sodium» dichromate(VI)/K2Cr2O7/Na2Cr2O7/Cr2O72− ✔
\n−2 ✔
\npropanone/propan-2-one/CH3COCH3 ✔
\nConsider the standard electrode potentials:
\nCr3+ (aq) + 3e− Cr (s) EΘ = −0.74 V
\nHg2+ (aq) + 2e− Hg (l) EΘ = +0.85 V
\nWhat is the cell potential, in V, for the voltaic cell?
\n2Cr (s) + 3Hg2+ (aq) → 3Hg (l) + 2Cr3+ (aq)
\n\n
A. −1.59
\nB. +0.11
\nC. +1.07
\nD. +1.59
\nD
\nChemical vapour deposition (CVD) produces multi-walled carbon nanotubes (MWCNT) of a more appropriate size for use in liquid crystals than production by arc discharge.
\nState the source of carbon for MWCNT produced by arc discharge and by CVD.
\nDiscuss three properties a substance should have to be suitable for use in liquid crystal displays.
\nArc discharge:
\ngraphite electrode
\nOR
\nhydrocarbon solvent
\n\n
CVD:
\ngaseous hydrocarbons
\n\n
Accept “carbon electrode”.
\nAccept specific examples of suitable hydrocarbon solvents (eg, methyl benzene/toluene OR cyclohexane).
\nAccept specific examples of suitable gaseous hydrocarbons (eg, methane, ethane, ethyne/acetylene) OR carbon monoxide OR carbon dioxide.
\n[2 marks]
\nAny three from:
\nchemically stable AND does not «chemically» degrade over time
\nstable over range of temperatures AND to avoid «voltage/random shift» fluctuations ✔
\npolar AND influenced by an electric field
\nstrong intermolecular forces AND allow molecule to align in specific orientations ✔
\nrapid switching speed/low viscosity AND change orientation «quickly» when electric field is applied/reversed
\n\n
Award [1 max] for identifying three correct properties without any discussion or incorrect interpretation of suitability.
\nAccept “voltage” for “electric field”.
\n[3 marks]
\nLipids provide energy and are an important part of a balanced diet.
\nIdentify the type of chemical reaction that occurs between fatty acids and glycerol to form lipids and the by-product of the reaction.
\n\n
Arachidonic acid is a polyunsaturated omega-6 fatty acid found in peanut oil.
\nDetermine the number of carbon–carbon double bonds present if the iodine number for the compound is 334. (Arachidonic acid Mr = 304.5)
\nDeduce the structure of the lipid formed by the reaction between lauric acid and glycerol (propane-1,2,3-triol) using section 34 of the data booklet.
\nOutline one impact food labelling has had on the consumption of foods containing different types of lipids.
\nDetermine, to the correct number of significant figures, the energy produced by the respiration of 29.9 g of C5H10O5.
\nΔHc (C5H10O5) = 205.9 kJ mol−1
\nExplain why lipids provide more energy than carbohydrates and proteins.
\nType of reaction:
\ncondensation
\nOR
\nesterification/triesterification
\nOR
\nnucleophilic substitution/nucleophilic displacement/SN2
\n\n
By-product:
\nwater/H2O
\n\n
Do not accept just “substitution/displacement”.
\n[2 marks]
\nALTERNATIVE 1
\n« =» 1.32 AND « =» 0.328
\n« ≈» 4
\n\n
ALTERNATIVE 2
\n«334 × ≈» 1017
\n« ≈» 4
\n\n
Award [2] for correct final answer.
\n[2 marks]
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/06.c/M\"](\"../assets/uploads/tinymce_asset/asset/4654/Schermafbeelding_2018-08-10_om_13.46.59.png\"/)
glycerol backbone
\nester formula AND linkage
\n\n
Accept a skeletal structure.
\nPenalize missing hydrogens or incorrect bond connectivities once only in Option B.
\nAccept condensed formula for ester.
\n[2 marks]
\nhas affected consumption of trans-fats/cis-fats/saturated fats/unsaturated fats/hydrogenated/artificially altered fats
\nOR
\nreduce/eliminate trans-fats/increase in cis-fats
\nOR
\nreduce/eliminate saturated fats
\nOR
\nincrease unsaturated fats
\n\n
Do not accept “decrease in fat” alone.
\nAccept “lipid” for “fats”.
\n[1 mark]
\n« =» 0.199 «mol»
\n«0.199 mol × 205.9 kJ mol–1 =» 41.0 «kJ»
\n\n
Ignore significant figures in M1.
\nAward [2] for correct final answer.
\nAward [1 max] for incorrect significant figures in final answer.
\n[2 marks]
\nratio of oxygen to carbon in lipids lower
\nOR
\nlipids less oxidized
\nOR
\nlipids more reduced
\n\n
more energy per mass/g released when lipids are oxidized
\n\n
Accept “«average» oxidation number of carbon in linoleic acid is lower” for M1.
\n[2 marks]
\nWhich statement about the reaction of a hydroxide ion with the organic reagent is correct?
\n\n
A. 1-bromopentane predominantly follows an SN1 mechanism.
\nB. 2-bromo-2-methylbutane predominantly follows an SN2 mechanism.
\nC. Reaction with 1-bromopentane occurs at a slower rate than with 1-chloropentane.
\nD. Reaction with 1-bromopentane occurs at a slower rate than with 2-bromo-2-methylbutane.
\nD
\nWhat is the major product of the reaction of HBr with but-1-ene?
\n\n
A. 1-bromobutane
\nB. 2-bromobutane
\nC. 1,2-dibromobutane
\nD. 2,2-dibromobutane
\nB
\nHow many chiral carbon atoms are present in one molecule of (CH3)2CHCHClCHBrCH3?
\n\n
A. 0
\nB. 1
\nC. 2
\nD. 3
\nC
\nBromine can form the bromate(V) ion, BrO3−.
\nState the electron configuration of a bromine atom.
\nSketch the orbital diagram of the valence shell of a bromine atom (ground state) on the energy axis provided. Use boxes to represent orbitals and arrows to represent electrons.
\nDraw the Lewis (electron dot) structure for BrO3− that obeys the octet rule.
\nPredict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
\nBromate(V) ions act as oxidizing agents in acidic conditions to form bromide ions.
\nDeduce the half-equation for this reduction reaction.
\nBromate(V) ions oxidize iron(II) ions, Fe2+, to iron(III) ions, Fe3+.
\nDeduce the equation for this redox reaction.
\n1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
\nOR
\n[Ar] 4s2 3d10 4p5 ✔
\n\n
Accept 3d before 4s.
\nAccept double-headed arrows.
\nAccept dots, crosses or lines to represent electron pairs.
\nGeometry:
trigonal/pyramidal ✔
Reason:
three bonds AND one lone pair
OR
four electron domains ✔
O−Br−O angle:
107° ✔
\n
Accept “charge centres” for “electron domains”.
\nAccept answers in the range 104–109°.
\nBrO3− (aq) + 6e− + 6H+ (aq) → Br− (aq) + 3H2O (l)
\ncorrect reactants and products ✔
\nbalanced equation ✔
\n\n
Accept reversible arrows.
\nBrO3− (aq) + 6Fe2+ (aq) + 6H+ (aq) → Br− (aq) + 3H2O (l) + 6Fe3+ (aq) ✔
\n\n
Which technique may be used to find the bond lengths and bond angles within a molecule?
\n\n
A. X-ray crystallography
\nB. 1H NMR spectroscopy
\nC. Infrared spectroscopy
\nD. Mass spectroscopy
\nA
\nButanoic acid, CH3CH2CH2COOH, is a weak acid and ethylamine, CH3CH2NH2, is a weak base.
\nState the equation for the reaction of each substance with water.
\nDraw a diagram showing the delocalization of electrons in the conjugate base of butanoic acid.
\nDeduce the average oxidation state of carbon in butanoic acid.
\nA 0.250 mol dm−3 aqueous solution of butanoic acid has a concentration of hydrogen ions, [H+], of 0.00192 mol dm−3. Calculate the concentration of hydroxide ions, [OH−], in the solution at 298 K.
\nDetermine the pH of a 0.250 mol dm−3 aqueous solution of ethylamine at 298 K, using section 21 of the data booklet.
\nSketch the pH curve for the titration of 25.0 cm3 of ethylamine aqueous solution with 50.0 cm3 of butanoic acid aqueous solution of equal concentration. No calculations are required.
\nExplain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room temperature.
\nState a suitable reagent for the reduction of butanoic acid.
\nDeduce the product of the complete reduction reaction in (e)(i).
\nButanoic acid:
CH3CH2CH2COOH (aq) + H2O (l) CH3CH2CH2COO− (aq) + H3O+ (aq) ✔
\n
Ethylamine:
CH3CH2NH2 (aq) + H2O (l) CH3CH2NH3+ (aq) + OH− (aq) ✔
Diagram showing:
dotted line along O–C–O AND negative charge
\n
Accept correct diagrams with pi clouds.
\n–1 ✔
\n«» = 5.21 × 10–12 «mol dm–3» ✔
\n«pKb = 3.35, Kb = 10–3.35 = 4.5 × 10–4»
\n«C2H5NH2 + H2O C2H5NH3+ + OH–»
\n\n
Kb =
\nOR
\n«Kb =» 4.5 × 10–4 =
\nOR
\n«Kb =» 4.5 × 10–4 = ✔
\n
« x = [OH–] =» 0.011 «mol dm–3» ✔
\n
«pH = –log» 12.04
\nOR
\n«pH = 14.00 – (–log 0.011)=» 12.04 ✔
\n\n
Award [3] for correct final answer.
\ndecreasing pH curve ✔
\npH close to 7 (6–8) at volume of 25 cm3 butanoic acid ✔
\nweak acid/base shape with no flat «strong acid/base» parts on the curve ✔
\nAny two of:
butanoic acid forms more/stronger hydrogen bonds ✔
butanoic acid forms stronger London/dispersion forces ✔
butanoic acid forms stronger dipole–dipole interaction/force ✔
\n
Accept “butanoic acid forms dimers”
\nAccept “butanoic acid has larger Mr/hydrocarbon chain/number of electrons” for M2.
\nAccept “butanoic acid has larger «permanent» dipole/more polar” for M3.
\nlithium aluminium hydride/LiAlH4 ✔
\nbutan-1-ol/1-butanol/CH3CH2CH2CH2OH ✔
\nProperties of elements and their compounds can be related to the position of the elements in the periodic table.
\nExplain the decrease in atomic radius from Na to Cl.
\nExplain why the radius of the sodium ion, Na+, is smaller than the radius of the oxide ion, O2−.
\nSketch a graph to show the relative values of the successive ionization energies of boron.
\nPredict, giving your reasons, whether Mn2+ or Fe2+ is likely to have a more exothermic enthalpy of hydration.
\nnuclear charge/number of protons/Zeff increases «causing a stronger pull on the outer electrons» ✔
\nsame number of shells/«outer» energy level/shielding ✔
\n\n
Accept “atomic number” for “number of protons”.
\nisoelectronic/same electronic configuration/«both» have 2.8 ✔
\nmore protons in Na+ ✔
\nSketch showing:
\nlargest increase between third and fourth ionization energies ✔
\nIE1 < IE2 < IE3 < IE4 < IE5 ✔
\nFe2+ AND smaller size/radius
\nOR
\nFe2+ AND higher charge density ✔
\n\n
stronger interaction with «polar» water molecules ✔
\n\n
M1 not needed for M2.
\nRadioisotopes can be used to treat a wide variety of diseases.
\nPhosphorous-32 undergoes beta decay. Formulate a balanced nuclear equation for this process.
\nThe half-life of phosphorus-32 is 14.3 days. Calculate the mass, in g, of 32P remaining after 57.2 days if the initial sample contains 2.63 × 10−8 mol. Use table 1 of the data booklet and Mr = 31.97 g mol−1.
\nExplain the targeted alpha therapy (TAT) technique and why it is useful.
\n32P → 32S + β
\n\n
Accept “e–/e/β” instead of “β”.
\n[1 mark]
\nALTERNATIVE 1
\n«λ = =» 0.04847 «day–1»
\n«m(32P) = 2.63 × 10–8 mol × 31.97 g mol–1 × e–0.04847 × 57.2 =» 5.26 × 10–8 «g»
\n\n
ALTERNATIVE 2
\n« =» 4 «half-lives passed»
\nOR
\n«n(32P) = =» 1.64 × 10–9 «mol»
\n\n
«m(32P) = 1.64 × 10–9 mol × 31.97 g mol–1 =» 5.26 × 10–8 «g»
\n\n
Award [2] for correct final answer.
\nAccept any value in the range “5.24–5.26 × 10–8 «g»”.
\n[2 marks]
\nalpha-emitting isotopes/212Pb/225Ac attached to drugs/antibodies/chelating ligands/carriers
\n\n
Award [2 max] for any two of:
\nabsorbed by «cancer/growing» cells
\nOR
\nbind to «cancer/growing» cell receptors
\n\n
alpha particles have high ionizing density/power
\n\n
short-range of emission «of alpha-particles»
\nOR
\nhealthy tissues less affected «as slower cell growth»
\nOR
\nlocal effect «on dispersed/spread/metastasised cancers»
\n\n
Accept “radionuclide” for “isotope”.
\nAccept “alpha particles are highly ionizing”.
\nAccept “alpha particles have low penetrating power”.
\nAccept “used to treat dispersed/spread/metastasised cancers” OR “can be used to map the distribution of cancer cells in the body”.
\n[3 marks]
\nAn organic compound containing carbon, hydrogen and oxygen has 62.02 % carbon and 10.43 % hydrogen by mass.
\nDetermine the empirical formula of the compound, showing your working.
\nThe infrared spectrum of the compound is shown. Deduce the functional group of the compound.
\nThe mass spectrum of the compound is shown. Deduce the relative molecular mass of the compound.
\nThe compound could not be oxidized using acidifi ed potassium dichromate(VI).
\nDeduce the structural formula of the compound.
\n«in 100 g sample» AND
\nOR
\n«in 100 g sample» 5.164 mol C AND 10.33 mol H ✔
\n\n
27.55 %
\nOR
\n1.722 mol O ✔
\n\n
«empirical formula» C3H6O ✔
\n«absorption at wavenumber 1700−1750 cm–1» C=O/carbonyl ✔
\n\n
Do not accept “ketone” or “aldehyde”.
\n«m/z =» 58 ✔
\nThis reaction is used in the manufacture of sulfuric acid.
\n2SO2 (g) + O2 (g) 2SO3 (g) Kc = 280 at 1000 K
\nState why this equilibrium reaction is considered homogeneous.
\n0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm3 flask at 1000 K.
\nPredict the direction of the reaction showing your working.
\nall «species» are in same phase ✔
\n\n
Accept “all species are in same state”.
\nAccept “all species are gases”.
\n«reaction quotient/Q =» ✔
\n\n
reaction quotient/Q/20.8/answer < Kc/280
\nOR
\nmixture needs more product for the number to equal Kc ✔
\n\n
reaction proceeds to the right/products ✔
\n\n
Do not award M3 without valid reasoning.
\nWhich graph shows the relationship between the volume and pressure of a fixed mass of an ideal gas?
\n![\"M18/4/CHEMI/SPM/ENG/TZ1/03\"](\"../assets/uploads/tinymce_asset/asset/4550/Schermafbeelding_2018-08-09_om_11.36.39.png\")
A
\nBromine can form the bromate(V) ion, BrO3−.
\nState the electron configuration of a bromine atom.
\nSketch the orbital diagram of the valence shell of a bromine atom (ground state) on the energy axis provided. Use boxes to represent orbitals and arrows to represent electrons.
\nDraw two Lewis (electron dot) structures for BrO3−.
\nDetermine the preferred Lewis structure based on the formal charge on the bromine atom, giving your reasons.
\nPredict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
\nBromate(V) ions act as oxidizing agents in acidic conditions to form bromide ions.
\nDeduce the half-equation for this reduction reaction.
\nBromate(V) ions oxidize iron(II) ions, Fe2+, to iron(III) ions, Fe3+.
\nDeduce the equation for this redox reaction.
\nCalculate the standard Gibbs free energy change, ΔGΘ, in J, of the redox reaction in (ii), using sections 1 and 24 of the data booklet.
\nEΘ (BrO3− / Br−) = +1.44 V
\n\n
State and explain the magnetic property of iron(II) and iron(III) ions.
\n\n
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
\nOR
\n[Ar] 4s2 3d10 4p5 ✔
\n\n
Accept 3d before 4s.
\nAccept double-headed arrows.
\nStructure I - follows octet rule:
\nStructure II - does not follow octet rule:
\n\n
Accept dots, crosses or lines to represent electron pairs.
\n«structure I» formal charge on Br = +2
\nOR
\n«structure II» formal charge on Br = 0/+1 ✔
\n\n
structure II is preferred AND it produces formal charge closer to 0 ✔
\n\n
Ignore any reference to formal charge on oxygen.
\nGeometry:
trigonal/pyramidal ✔
Reason:
three bonds AND one lone pair
OR
four electron domains ✔
O−Br−O angle:
107° ✔
\n
Accept “charge centres” for “electron domains”.
\nAccept answers in the range 104–109°.
\nBrO3− (aq) + 6e− + 6H+ (aq) → Br− (aq) + 3H2O (l)
\ncorrect reactants and products ✔
\nbalanced equation ✔
\n\n
Accept reversible arrows.
\nBrO3− (aq) + 6Fe2+ (aq) + 6H+ (aq) → Br− (aq) + 3H2O (l) + 6Fe3+ (aq) ✔
\n\n
EΘreaction = «+1.44 V – 0.77 V =» 0.67 «V» ✔
\nΔGΘ = «–nFEΘreaction = – 6 × 96500 C mol–1 × 0.67 V =» –3.9 × 105 «J» ✔
\n\n
both are paramagnetic ✔
\n«both» contain unpaired electrons ✔
\n\n
Accept orbital diagrams for both ions showing unpaired electrons.
\n\n
Ethanol can be detected by a variety of instruments.
\nFuel cells use an electrochemical process to determine the concentration of ethanol.
\nFormulate the overall equation for this process.
\nPredict the chemical shifts and integration for each signal in the 1H NMR spectrum for ethanol using section 27 of the data booklet.
\nC2H5OH(g) + O2(g) → CH3COOH(aq) + H2O(l)
\n\n
Accept any correct formula for reactants and products.
\n[1 mark]
\nR–OH:
\n1.0–6.0 «ppm» AND 1 H
\nR–O–CH2–:
\n3.3–3.7 «ppm» AND 2 H
\n–CH3:
\n0.9–1.0 «ppm» AND 3 H
\n\n
Award [1] for the ratio of 1:2:3 (in any order).
\nAward [2] for three correct chemical shifts without integration.
\nAward [1] for two correct chemical shifts without integration.
\nFor each chemical shift accept a specific value within the range.
\nAssignment of proton to fragment (eg, R–OH) is not required in each case.
\n[3 marks]
\nConsider the following Hess’s law cycle:
\nIdentify the type of reaction in step 1.
\nCalculate the standard enthalpy change, ΔHΘ, of step 2 using section 13 of the data booklet.
\nDetermine the standard enthalpy change, ΔHΘ, of step 1.
\nSuggest one reason why the calculated value of ΔHΘ using Hess’s Law in part (c) can be considered accurate and one reason why it can be considered approximate.
\n«electrophilic» addition/AE
OR
reduction ✔
Accept “hydrogenation”.
\n«(−286 kJ) + (−1411 kJ) =» −1697 «kJ» ✔
\n«−1697 kJ + 1561 kJ =» −136 «kJ»
\nOR
\n«ΔHΘ = Δ (products) − Δ (reactants) = −84 kJ − 52 kJ =» −136 «kJ» ✔
\nAccurate:
no approximations were made in the cycle
OR
values are specific to the compounds
OR
Hess’s law is a statement of conservation of energy
OR
method is based on a law
OR
data in table has small uncertainties ✔
\n
Approximate:
values were experimentally determined/had uncertainties
OR
each value has been determined to only three/four significant figures
OR
different sources have «slightly» different values for enthalpy of combustion
OR
law is valid until disproved
OR
law of conservation of energy is now conservation of mass–energy
OR
small difference between two quite large terms «leads to high percentage uncertainty» ✔
Which statement is correct?
\nA. A strong acid is a good proton donor and has a strong conjugate base.
\nB. A weak acid is a poor proton acceptor and has a strong conjugate base.
\nC. A strong acid is a good proton donor and has a weak conjugate base.
\nD. A strong base is a good proton donor and has a weak conjugate acid.
\nC
\nWhich can describe oxidation?
\nA. Loss of hydrogen
\nB. Decrease in oxidation number
\nC. Gain of electrons
\nD. Loss of oxygen
\nA
\nWhat are the products of the electrolysis of molten zinc bromide?
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/23\"](\"../assets/uploads/tinymce_asset/asset/4608/Schermafbeelding_2018-08-10_om_07.36.58.png\"/)
A
\nWhich is a homogeneous mixture?
\nA. Oil and water
\nB. Sand and water
\nC. Ethanol and water
\nD. Chalk and sand
\nC
\nWhich element has the same oxidation number in both species?
\nA. C in C2H4 and CO2
\nB. H in H2O and NaH
\nC. S in SO42−and SO3
\nD. O in H2O2 and H2O
\nC
\nWhat is the molecular formula of a hydrocarbon containing 84.6% carbon by mass with a molar mass of 142.3 g mol−1?
\nA. C20H44
\nB. C11H10
\nC. C10H22
\nD. C5H11
\nC
\nWhich compounds belong to the same homologous series?
\nA. CHCCH2CH3, CHCCH2CH2CH3
\nB. CH3CH2CH2CH2OH, CH3CH2OCH2CH3
\nC. CH2CHCH3, CH3CH2CH2CH3
\nD. CH3COCH3, CH3CH2OCH3
\nA
\nPolymers of α-glucose include the disaccharide maltose and the polysaccharide amylose, a type of starch. The cyclic structure of α-glucose is shown in section 34 of the data booklet.
\nState the specific type of linkage formed between α-glucose fragments in both maltose and amylose.
\nA person with diabetes suffering very low blood sugar (hypoglycaemia) may be advised to consume glucose immediately and then eat a small amount of starchy food such as a sandwich. Explain this advice in terms of the properties of glucose and starch.
\n«α-1,4-»glycosidic
\n\n
Accept «α-1,4-»glycoside.
\nAccept “ether”.
\n[1 mark]
\nGlucose:
\nreadily passes through intestine wall/dissolves in blood
\nOR
\nis immediately available for energy/respiration
\nOR
\ntransported rapidly around body
\n\n
Starch:
\nmust be hydrolysed/broken down «into smaller molecules» first
\n[2 marks]
\nProperties of elements and their compounds can be related to the position of the elements in the periodic table.
\nExplain the decrease in atomic radius from Na to Cl.
\nExplain why the radius of the sodium ion, Na+, is smaller than the radius of the oxide ion, O2−.
\nState a physical property of sodium oxide.
\nnuclear charge/number of protons/Zeff increases «causing a stronger pull on the outer electrons» ✔
\nsame number of shells/«outer» energy level/shielding ✔
\n\n
Accept “atomic number” for “number of protons”.
\nisoelectronic/same electronic configuration/«both» have 2.8 ✔
\nmore protons in Na+ ✔
\nAny one of:
brittle ✔
high melting point/crystalline/solid «at room temperature» ✔
low volatility ✔
conducts electricity when molten ✔
does not conduct electricity at room temperature ✔
\n
Do not accept soluble in water.
\nIgnore any chemical properties.
\nGreenhouse gases absorb infrared radiation.
\nIdentify one naturally occurring greenhouse gas, other than carbon dioxide or water vapour, and its natural source.
\nFormulate an equation that shows how aqueous carbon dioxide produces hydrogen ions, H+(aq).
\nThe concentrations of oxygen and nitrogen in the atmosphere are much greater than those of greenhouse gases. Outline why these gases do not absorb infrared radiation.
\n![\"M18/4/CHEMI/SP3/ENG/TZ1/09.a/M\"](\"../assets/uploads/tinymce_asset/asset/4594/Schermafbeelding_2018-08-09_om_18.35.29.png\"/)
\n
Accept “nitrous oxide”.
\nAccept “electrical discharges/lightning”.
\n[2 marks]
\nCO2(aq) + H2O(l) H+(aq) + HCO3–(aq)
\nOR
\nCO2(aq) + H2O(l) H2CO3(aq) AND H2CO3(aq) H+(aq) + HCO3–(aq)
\n\n
Accept CO2(aq) + H2O(l) 2H+(aq) + CO32–(aq).
\nAccept equations with single arrow.
\n[1 mark]
\nno change in polarity/dipole «moment when molecule vibrates»
\n\n
Do not accept “non-polar” or “no dipole moment” – idea of change must be there.
\n[1 mark]
\nThis reaction is used in the manufacture of sulfuric acid.
\n2SO2 (g) + O2 (g) 2SO3 (g) Kc = 280 at 1000 K
\nState why this equilibrium reaction is considered homogeneous.
\nPredict, giving your reason, the sign of the standard entropy change of the forward reaction.
\nCalculate the standard Gibbs free energy change, ΔGΘ, in kJ, for this reaction at 1000 K. Use sections 1 and 2 of the data booklet.
\nPredict, giving your reasons, whether the forward reaction is endothermic or exothermic. Use your answers to (b) and (c).
\n0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm3 flask at 1000 K.
\nPredict the direction of the reaction showing your working.
\nall «species» are in same phase ✔
\n\n
Accept “all species are in same state”.
\nAccept “all species are gases”.
\nnegative AND fewer moles/molecules «of gas» in the products ✔
\nΔGΘ =«–RT ln Kc =» –8.31 J K–1 mol–1 × 1000 K × ln 280
\nOR
\nΔGΘ = – 4.7 × 104 «J» ✔
\n\n
«ΔGΘ =» – 47 «kJ» ✔
\n\n
Award [2] for correct final answer.
\nΔGΘ < 0/spontaneous AND ΔSΘ < 0/unfavourable ✔
\nexothermic AND ΔHΘ «must be» negative/favourable ✔
\n«reaction quotient/Q =» ✔
\n\n
reaction quotient/Q/20.8/answer < Kc/280
\nOR
\nmixture needs more product for the number to equal Kc ✔
\n\n
reaction proceeds to the right/products ✔
\n\n
Do not award M3 without valid reasoning.
\nButanoic acid, CH3CH2CH2COOH, is a weak acid and ethylamine, CH3CH2NH2, is a weak base.
\nState the equation for the reaction of each substance with water.
\nExplain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room temperature.
\nState the formula of the salt formed when butanoic acid reacts with ethylamine.
\nButanoic acid:
CH3CH2CH2COOH (aq) + H2O (l) CH3CH2CH2COO− (aq) + H3O+ (aq) ✔
\n
Ethylamine:
CH3CH2NH2 (aq) + H2O (l) CH3CH2NH3+ (aq) + OH− (aq) ✔
Any two of:
butanoic acid forms more/stronger hydrogen bonds ✔
butanoic acid forms stronger London/dispersion forces ✔
butanoic acid forms stronger dipole–dipole interaction/force ✔
\n
Accept “butanoic acid forms dimers”
\nAccept “butanoic acid has larger Mr/hydrocarbon chain/number of electrons” for M2.
\nAccept “butanoic acid has larger «permanent» dipole/more polar” for M3.
\nCH3CH2NH3+ CH3CH2CH2COO−
OR
CH3CH2CH2COO− CH3CH2NH3+
OR
CH3CH2CH2COO− H3N+CH2CH3 ✔
\n
The charges are not necessary for the mark.
\n\n
\n
Carbon dioxide contributes significantly to global warming. It can be used as a raw material with methyloxirane to form polymers.
\nSuggest why the three-membered ring in methyloxirane is unstable.
\nDraw two structural isomers of methyloxirane.
\nState, giving a reason, whether methyloxirane can form cis-trans isomers.
\nPredict the chemical shift and splitting pattern of the signal produced by the hydrogen atoms labelled X in the 1H NMR spectrum of the polymer. Use section 27 of the data booklet.
\nangle between bonds is 60°/strained/smaller than 109.5° ✔
\nAny two of:
CH3COCH3 ✔
CH3CH2CHO ✔
\nCH2=CHCH2OH ✔
\nCH3OCH=CH2 ✔
\n\n
Accept displayed or condensed structural formulas or skeletal formulas.
\nAccept CH(OH)=CHCH3 and CH2=C(OH)CH3.
\nno AND only one «axial/methyl/CH3» substituent «at the ring»
\nOR
\nno AND two «axial» substituents required «for cis/trans-isomers» ✔
\n\n
Accept “no AND «O in the ring and» one carbon has two H atoms”.
\nChemical shift:
3.7–4.8 «ppm» ✔
Splitting pattern:
doublet ✔
The increased concentration of carbon dioxide in the atmosphere is thought to result from the increased combustion of fossil fuels such as petroleum.
\nIdentify an element, other than carbon and hydrogen, found at significant concentrations in fossil fuels.
\nPetroleum contains many hydrocarbons. Explain how these are separated by fractional distillation.
\nDetermine the specific energy and energy density of petrol (gasoline), using data from sections 1 and 13 of the data booklet. Assume petrol is pure octane, C8H18. Octane: molar mass = 114.26 g mol−1, density = 0.703 g cm−3.
\nOutline why the energy available from an engine will be less than these theoretical values.
\nnitrogen/N
\nOR
\noxygen/O
\nOR
\nsulfur/S
\n\n
Accept “phosphorus/P”.
\n[1 mark]
\nAny three of:
\ndifferent molar masses
\nOR
\ndifferent strengths of intermolecular forces
\n\n
different boiling points
\ntemperature in «fractionating» column decreases upwards
\n\n
«components» condense at different temperatures/heights
\nOR
\n«component with» lower boiling point leaves column first
\n\n
[3 marks]
\nspecific energy «= = » = 47.9 «kJ g–1»
\nenergy density « = specific energy × density = 47.9 kJ g–1 × 0.703 g cm–3» = 33.7 «kJ cm–3 »
\n\n
Do not accept “–47.9 «kJ g–1»”.
\nDo not accept “–33.7 «kJ cm–3»” unless “–47.9 «kJ g–1»” already penalized.
\n[2 marks]
\nenergy is lost «to the surroundings» as heat/sound/friction
\nOR
\nenergy is lost to the surroundings «as heat/sound/friction»
\nOR
\nincomplete combustion
\n\n
Do not accept just “energy is lost”.
\n[1 mark]
\nOne suggestion for the reduction of carbon footprints is the use of biofuels, such as vegetable oils, as a substitute for petroleum based fuels.
\nOutline the major technical problem affecting the direct use of vegetable oils as fuels in internal combustion engines and the chemical conversion that has overcome this.
\nState the formula of a fuel that might be produced from the vegetable oil whose formula is shown.
\n
Outline why biofuels are considered more environmentally friendly, even though they produce more carbon dioxide per kJ of energy than petroleum based fuels.
\nviscosity «of vegetable oils is too high»
\n\n
transesterification
\nOR
\n«conversion into» alkyl/methyl/ethyl esters
\n\n
[2 marks]
\nR–CO–O–CH3/RCOOMe
\nOR
\nR–CO–O–C2H5/RCOOEt
\n[1 mark]
\n«growing oil producing» plants absorbs carbon dioxide from the atmosphere
\nOR
\n«combustion of» petroleum based fuels releases carbon stored «for millions of years»
\n\n
Accept “biofuels renewable” OR “petroleum based fuels non-renewable”.
\nAccept “waste vegetable oils can be converted to biofuels/biodiesel”.
\nAccept “biofuels do not contain sulfur”.
\n[1 mark]
\nNuclear fission of 235U is one source of electrical energy that has a minimal carbon footprint.
\nNatural uranium needs to be enriched to increase the proportion of 235U. Suggest a technique that would determine the relative abundances of 235U and 238U.
\nExplain how 235U fission results in a chain reaction, including the concept of critical mass.
\nSuggest one reason why there is opposition to the increased use of nuclear fission reactors.
\nmass spectrometry/mass spectroscopy/MS
\n\n
Accept “analysis of radiation emitted”.
\n[1 mark]
\ncritical mass: mass required so that «on average» each fission/reaction results in a further fission/reaction
\nAny two for [2 max]:
\nneutron captured by «235U» nucleus
\nfission/reaction produces many neutrons/more than one neutron
\nif these cause further fission/reaction a chain reaction occurs
\n\n
Accept “minimum mass of fuel needed for the reaction to be self-sustaining”.
\nAccept answers in the form of suitable diagrams/equations.
\n[3 marks]
\nproduce long lived/long half-life radioisotopes/radioactivity
\nOR
\ncould be used to produce nuclear weapons
\nOR
\n«nuclear» accidents/meltdowns can occur
\n\n
Accept “long lived/long half-life radioactive waste”.
\n[1 mark]
\nThe following mechanism is proposed for a reaction:
\nA + B → C + D slow step
D + B → A + E fast step
Classify substances B and D as reactant, product, catalyst, or intermediate, based on the proposed mechanism.
\nDeduce the rate expression.
\nCalculate the initial rate of reaction for experiment 2, if measured under the same conditions.
\nB: reactant ✔
\nD: intermediate ✔
\nrate = k[A][B] ✔
\n1.80 «mol dm–3 s–1» ✔
\nAlloys containing at least 60 % copper reduce the presence of bacteria on their surface.The percentage of copper in brass, an alloy of copper and zinc, can be determined by UV-vis spectrometry.
\nA sample of brass is dissolved in concentrated nitric acid and then made up to 250.0 cm3 with water before analysis.
\nCu (s) + 4HNO3 (aq) → Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O (l)
\n3Zn (s) + 8HNO3 (aq) → 3Zn(NO3)2 (aq) + 2NO (g) + 4H2O (l)
\nThe concentration of copper(II) ions in the resulting solution is then determined from a calibration curve, which is plotted by measuring the light absorbance of standard solutions.
\nTitration is another method for analysing the solution obtained from adding brass to nitric acid.
\nOutline why the initial reaction should be carried out under a fume hood.
\nDeduce the equation for the relationship between absorbance and concentration.
\nOutline how a solution of 0.0100 mol dm−3 is obtained from a standard 1.000 mol dm−3 copper(II) sulfate solution, including two essential pieces of glassware you would need.
\nThe original piece of brass weighed 0.200 g. The absorbance was 0.32.
\nCalculate, showing your working, the percentage of copper by mass in the brass.
\nDeduce the appropriate number of significant figures for your answer in (d)(i).
\nComment on the suitability of using brass of this composition for door handles in hospitals.
\nIf you did not obtain an answer to (d)(i), use 70 % but this is not the correct answer.
\nSuggest another property of brass that makes it suitable for door handles.
\nCopper(II) ions are reduced to copper(I) iodide by the addition of potassium iodide solution, releasing iodine that can be titrated with sodium thiosulfate solution, Na2S2O3 (aq). Copper(I) iodide is a white solid.
\n4I− (aq) + 2Cu2+ (aq) → 2CuI (s) + I2 (aq)
\nI2 (aq) + 2S2O32− (aq) → 2I− (aq) + S4O62− (aq)
\nDeduce the overall equation for the two reactions by combining the two equations.
\nSuggest why the end point of the titration is difficult to determine, even with the addition of starch to turn the remaining free iodine black.
\nNO2/NO/NOx/HNO3/gas is poisonous/toxic/irritant ✔
\n\n
Accept formula or name.
\nAccept “HNO3 is corrosive” OR “poisonous/toxic gases produced”.
\nAccept “reaction is harmful/hazardous”.
\nSlope (gradient):
\n40 ✔
\n\n
Equation:
\nabsorbance = 40 × concentration
\nOR
\ny = 40x ✔
\n\n
Accept any correct relationship for slope such as .
\nAward [2] if equation in M2 is correct.
\ndilute 1.00 cm3 «of the standard solution with water» to 100 cm3
\nOR
\ndilute sample of standard solution «with water» 100 times ✔
\n\n
«graduated/volumetric» pipette/pipet ✔
\nvolumetric flask ✔
\n\n
Accept any 1 : 100 ratio for M1.
\nAccept “mix 1 cm3 of the standard solution with 99 cm3 of water” for M1.
\nDo not accept “add 100 cm3 of water to 1.00 cm3 of standard solution” for M1.
\nAccept “burette/buret” for M2.
\nAccept “graduated/measuring flask” for M3 but not “graduated/measuring cylinder” or “conical/Erlenmeyer flask”.
\nconcentration of copper = 0.0080 «mol dm–3» ✔
\n\n
mass of copper in 250.0 cm3 = «0.0080 mol dm–3 × 0.2500 dm3 × 63.55 g mol–1 =» 0.127 «g»
\nOR
\nmass of brass in 1 dm3 = «4 × 0.200 g =» 0.800 g AND [Cu2+] = «0.0080 mol dm–3 × 63.55 g mol–1 =» 0.5084 g dm–3 ✔
\n\n
«% copper in this sample of brass » 64 «%»
\nOR
\n«% copper in this sample of brass » 64 «%» ✔
\n\n
Accept any value in range 0.0075–0.0085 «mol dm–3» for M1.
\nAccept annotation on graph for M1.
\nAward [3] for correct final answer.
\nAccept “65 «%»”.
\ntwo ✔
\n\n
Do not apply ECF from 1(d)(i).
\n«since it is greater than 60%» it will reduce the presence of bacteria «on door handles» ✔
\nresistant to corrosion/oxidation/rusting
\nOR
\nlow friction surface «so ideal for connected moving components» ✔
\n\n
Accept “hard/durable”, “«high tensile» strength”, “unreactive”, “malleable” or any reference to the appearance/colour of brass (eg “gold-like”, “looks nice” etc.).
\nDo not accept irrelevant properties, such as “high melting/boiling point”, “non-magnetic”, “good heat/electrical conductor”, “low volatility”, etc.
\nDo not accept “ductile”.
\n2I− (aq) + 2Cu2+ (aq) + 2S2O32− (aq) → 2CuI (s) + S4O62− (aq)
\ncorrect reactants and products ✔
\nbalanced equation ✔
\n\n
M2 can only be awarded if M1 is correct.
\nprecipitate/copper(I) iodide/CuI makes colour change difficult to see
\nOR
\nrelease of I2/iodine from starch-I2 complex is slow so titration must be done slowly ✔
\nMany drugs, including aspirin, penicillin, codeine and taxol, have been modified from compounds that occur naturally.
\nAspirin is often taken to reduce pain, swelling or fever. State one other use of aspirin.
\nState what is meant by the bioavailability of a drug.
\nOutline how the bioavailability of aspirin may be increased.
\nCompare and contrast the IR spectrum of aspirin with that of salicylic acid, using section 26 of the data booklet.
\n\n
Describe how penicillin combats bacterial infections.
\nOutline two consequences of prescribing antibiotics such as penicillin unnecessarily.
\nState how penicillins may be modified to increase their effectiveness.
\nMorphine and codeine are strong analgesics. Outline how strong analgesics function.
\nSuggest one reason why codeine is more widely used than morphine as an analgesic.
\nAny one of:
\nanticoagulant
\nlower risk of heart attack/strokes
\nprevent recurrence of heart attack/stroke
\nprevents cancer of colon/oesophagus/stomach
\n\n
Accept “prevents/reduces blood clots” OR “blood thinner”.
\n[1 mark]
\nfraction/proportion/percentage «of administered dosage» that reaches target «part of human body»
\nOR
\nfraction/ proportion/percentage «of administered dosage» that reaches blood «plasma»/systemic circulation
\n\n
Accept “the ability of the drug to be absorbed by the body” OR “the extent to which the drug is absorbed by the body”.
\nDo not accept “the amount/quantity of the drug absorbed”.
\n[1 mark]
\n«intravenous» injection/IV
\n\n
Accept “parenterally”.
\nAccept “react with alkali/NaOH” OR “convert to ionic form/salt”.
\n[1 mark]
\nOne absorption found in both spectra:
\nAny one of:
\n1050–1410 cm–1 «C–O in alcohols, esters, ethers»
\n1700–1750 cm–1 «C=O in carboxylic acids, esters»
\n2500–3000 cm–1 «O–H in carboxylic acids»
\n2850–3090 cm–1 «C–H in alkanes, alkenes, arenes»
\n\n
One absorption found in only one of the spectra:
\n3200–3600 cm–1 «O–H in alcohols, phenols»
\n\n
Award [1 max] if candidate states bonds (C=O in both, O–H in salicylic acid only) but doesn’t quote wavelength ranges.
\nAccept a second/additional absorption at 1700–1750 cm–1 from the C=O in ester.
\n[2 marks]
\nAny two of:
\nring is «sterically» strained
\nOR
\nring breaks up/opens/reacts «easily»
\nOR
\namide/amido group «in ring» is «highly» reactive
\n\n
«irreversibly» binds/bonds to enzyme/transpeptidase
\nOR
\ninhibits enzyme/transpeptidase «in bacteria» that produces cell walls
\nOR
\nprevents cross-linking of bacterial cell walls
\n\n
cells absorb water AND burst
\nOR
\ncells cannot reproduce
\n\n
Award [1 max] for “interferes with cell wall production”.
\nDo not accept “cell membrane” instead of “cell wall”.
\n[2 marks]
\nAny two of:
\nleads to «bacterial» resistance/proportion of resistant bacteria increases
\nOR
\nleads to penicillinase-producing bacteria
\ndamage to/contamination of bodies of water/ecosystems
\ndestroys useful/beneficial bacteria
\ndestroyed bacteria replaced by more harmful bacteria
\n\n
Accept “endocrine disruptor”.
\nDo not accept “increased cost of developing antibiotics”.
\n[2 marks]
\nmodify side chain
\n[1 mark]
\ntemporarily bind to/block/interfere with receptor sites in brain
\nOR
\nprevent transmission of pain impulses within CNS/central nervous system
\n[1 mark]
\ncodeine has a wider therapeutic window
\n\n
Accept “codeine has lower activity” OR “codeine has lower risk of overdose” OR “codeine is less potent” OR “codeine has less side-effects”.
\nDo not accept “lower abuse potential for codeine” OR “less addictive «than morphine»” OR “codeine has a lower bioavailability” OR “available without prescription” OR “cheaper”.
\n[1 mark]
\nHalogenoalkanes undergo nucleophilic substitution reactions with sodium hydroxide.
\nState a reason why most halogenoalkanes are more reactive than alkanes.
\nClassify 1-bromopropane as a primary, secondary or tertiary halogenoalkane, giving a reason.
\nExplain the mechanism of the reaction between 1-bromopropane with aqueous sodium hydroxide using curly arrows to represent the movement of electron pairs.
\nState, giving your reason, whether the hydroxide ion acts as a Lewis acid, a Lewis base, or neither in the nucleophilic substitution.
\nSuggest two advantages of understanding organic reaction mechanisms.
\npolarity/polar «molecule/bond»
OR
carbon–halogen bond is weaker than C–H bond ✔
primary AND Br/bromine is attached to a carbon bonded to two hydrogens
OR
primary AND Br/bromine is attached to a carbon bonded to one C/R/alkyl «group» ✔
\n
Accept “primary AND Br/bromine is attached to the first carbon in the chain”.
\ncurly arrow going from lone pair/negative charge on O in HO– to C ✔
\ncurly arrow showing Br leaving ✔
\nrepresentation of transition state showing negative charge, square brackets and partial bonds ✔
\nformation of organic product CH3CH2CH2OH AND Br– ✔
\n\n
Do not allow curly arrow originating on H in HO–.
\nAccept curly arrow either going from bond between C and Br to Br in 1-bromopropane or in the transition state.
\nDo not penalize if HO and Br are not at 180° to each other.
\nDo not award M3 if OH–C bond is represented.
\n«Lewis» base AND donates a pair of electrons ✔
\nAny two of:
choose «most» appropriate reaction «for preparing the target compound» ✔
design/discover new reactions/reagents ✔
apply this knowledge to other areas of chemistry/science ✔
«retro-»synthesis «more effective» ✔
control/predict «desired» products ✔
control rate of reaction «more effectively» ✔
satisfy intellectual curiosity ✔
predicting how changing reagents/conditions might affect reaction ✔
suggesting intermediates/transition states ✔
\n
Accept other reasonable answers.
\nExcess stomach acid can be counteracted by a range of medications.
\nAn antacid tablet contains 680 mg of calcium carbonate, CaCO3, and 80 mg of magnesium carbonate, MgCO3.
\nState the equation for the reaction of magnesium carbonate with hydrochloric acid.
\nDetermine the amount, in mol, of hydrochloric acid neutralized by one antacid tablet.
\nExplain how omeprazole (Prilosec) reduces stomach acidity.
\nMgCO3(s) + 2HCl(aq) → CO2(g) + H2O(l) + MgCl2(aq)
\n\n
Do not accept “H2CO3”.
\n[1 mark]
\nn(HCl) = 2 n(CaCO3) + 2 n(MgCO3)
\nOR
\nn(HCl) =
\n\n
«n(HCl) = 0.0136 mol + 0.0019 mol =» 0.016 «mol»
\n\n
Award [2] for correct final answer.
\nAward [1 max] for correctly calculating amount of acid neutralized by just CaCO3 (0.014 «mol») or MgCO3 (0.002 «mol»).
\n[2 marks]
\ninhibits the secretion of stomach acid/H+
\n«active metabolites» bind «irreversibly» to «receptors of the» proton pump
\n\n
Accept “PPI/proton pump inhibitor”.
\nDo not award mark for “binds to H2/histamine receptors”. (Ranitidine mode of action.)
\nAccept “H+/K+ ATPase” for “proton pump”.
\n[2 marks]
\nAntiviral drugs are a major research focus.
\nOseltamivir (Tamiflu) and zanamivir (Relenza) are used against flu viruses. Explain how these drugs function.
\nShikimic acid, the precursor for oseltamivir (Tamiflu), was originally extracted from star anise, and is now produced using genetically modified E. coli bacteria.
\nSuggest one difficulty associated with synthesizing oseltamivir (Tamiflu) from star anise.
\nblocks/inhibits neuraminidase/NA/«viral» enzyme which allows viruses to pass through cell membrane
\nprevent virus from leaving/escaping host cell «thus it cannot infect other cells»
\n[2 marks]
\nAny one of:
\nlimited supply of star anise/plant
\n«star anise» takes time to grow
\ntime-consuming/multi-step extraction
\nlow concentration in plan
\n\n
Accept “low yield for extraction/conversion” OR “requires environmentally damaging solvents”.
\n[1 mark]
\nWhat is the sum of the coefficients when the equation is balanced with the lowest whole number ratio?
\n__Na2S2O3(aq) + __HCl(aq) → __S(s) + __SO2(g) + __NaCl(aq) + __H2O(l)
\nA. 6
\nB. 7
\nC. 8
\nD. 9
\nC
\nWhat is the number of atoms of oxygen in 2.0 mol of hydrated sodium carbonate, Na2CO3•10H2O? Avogadro’s constant, L or NA: 6.02 × 1023 mol–1
\nA. 6
\nB. 26
\nC. 3.6 × 1024
\nD. 1.6 × 1025
\nD
\nWhat is the volume, in cm3, of the final solution if 100 cm3 of a solution containing 1.42 g of sodium sulfate, Na2SO4, is diluted to the concentration of 0.020 mol dm–3?
\nMr(Na2SO4) = 142
\nA. 50
\nB. 400
\nC. 500
\nD. 600
\nC
\nOne way of classifying materials is based on the type of bonding present.
\nOne reaction to convert cyclohexanone to caprolactam using concentrated sulfuric acid as a catalyst is shown.
\nOutline why this type of classification is not entirely satisfactory by using magnesium diboride, MgB2, as an example. Refer to sections 8 and 29 of the data booklet.
\nStructures of poly(methyl acrylate), PMA, and Bakelite® are shown.
\nSuggest, giving reasons, which is the thermoplastic polymer and which is the thermosetting polymer.
\nIn an incomplete combustion of the polyvinyl chloride, PVC, it was found that hydrogen chloride, carbon monoxide, carbon dioxide, and water vapour were released.
\nFormulate an equation for this reaction using the formula of the PVC repeating unit.
\nA zeolite is an alternative catalyst for this reaction.
\nExplain how zeolites act as selective catalysts.
\nIdentify another advantage of using a zeolite instead of concentrated sulfuric acid.
\nRepeating units of several polymers are listed.
\nThe infrared (IR) spectrum of one of these polymers is shown.
\nDeduce, giving a reason, the name of this polymer and its Resin Identification Code (RIC), using sections 26 and 30 in the data booklet.
\n= 0.7 AND average = 1.7 ✔
\n\n
bonding between metallic and ionic
\nOR
\nmore than one type of bonding present
\nOR
\nbond type difficult to determine as close to several regions/several types/named bonding types «eg ionic and covalent etc.»
\nOR
\nbond is mostly covalent «based on % covalent scale on diagram»
\nOR
\nbond has « » 22% ionic character ✔
\n\n
Accept “EN” for “\".
\nAccept “bond is ionic but close to several regions/several types/other named bonding type(s) (eg covalent, metallic and covalent etc.)”.
\nDo not accept just “bond is ionic”.
\nAccept any value for % ionic character in range 15–24% or % covalent character in range 76–85%.
\nThermoplastic polymer:
\nPMA AND «weak» intermolecular/IMFs/London/dispersion/van der Walls/vdW/dipole-dipole forces «between layers/chains»
\nOR
\nPMA AND no/few cross-links «between layers/chains» ✔
\n\n
Thermosetting polymer:
\nBakelite® AND «strong» covalent bonds «between layers/chains»
\nOR
\nBakelite® AND extensive cross-links «between layers/chains» ✔
\n\n
Do not accept “hydrogen bonding” for M1.
\nAward [1 max] for correct reasons for both polymer classes even if named polymers are incorrectly classified.
\nCH2CHCl (s) + 2O2 (g) → HCl (g) + CO (g) + CO2 (g) + H2O (g)
\nOR
\nCH2CHCl (s) + O2 (g) → HCl (g) + 2CO (g) + H2O (g) AND 2CO (g) + O2 (g) → CO2 (g) ✔
\n\n
Accept any correctly balanced equation that includes the products specified.
\npores/cavities/channels/holes/cage-like structures «in zeolites» have specific shape/size ✔
\nonly reactants «with appropriate size/geometry» fit inside/go through/are activated/can react ✔
\ndoes not require corrosive acid/«concentrated» sulfuric acid/H2SO4
\nOR
\nzeolite can be recycled «more easily»
\nOR
\nproduct can be «more» easily separated from a zeolite «than from sulfuric acid»
\nOR
\nminimal/less impact on environment
\nOR
\nsynthesis of specific isomers as products ✔
\nName and reason:
\nPET/PETE AND peak for C=O «at 1700–1750 cm–1» ✔
\n\n
RIC:
\n1 ✔
\n\n
Accept “PET/PETE AND peak for C–O «at 1050–1410 cm–1»” for M1.
\nAccept “PET/PETE AND peak(s) for COO” for M1.
\nAccept name or abbreviation for polymer.
\nNo ECF for M2.
\nWhat is the percentage yield when 2.0 g of ethene, C2H4, is formed from 5.0 g of ethanol, C2H5OH?
Mr(ethene) = 28; Mr(ethanol) = 46
A.
\nB.
\nC.
\nD.
\nB
\nWhat is the composition of the nucleus of 26Mg?
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/05\"](\"../assets/uploads/tinymce_asset/asset/4601/Schermafbeelding_2018-08-10_om_06.15.06.png\"/)
D
\nWhich electron transition emits energy of the longest wavelength?
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/06\"](\"../assets/uploads/tinymce_asset/asset/4602/Schermafbeelding_2018-08-10_om_06.16.15.png\"/)
A
\nWhich increase across a period from left to right?
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/07\"](\"../assets/uploads/tinymce_asset/asset/4603/Schermafbeelding_2018-08-10_om_06.17.52.png\"/)
D
\nPowdered zinc was reacted with 25.00 cm3 of 1.000 mol dm−3 copper(II) sulfate solution in an insulated beaker. Temperature was plotted against time.
\nEstimate the time at which the powdered zinc was placed in the beaker.
\nState what point Y on the graph represents.
\nThe maximum temperature used to calculate the enthalpy of reaction was chosen at a point on the extrapolated (dotted) line.
\nState the maximum temperature which should be used and outline one assumption made in choosing this temperature on the extrapolated line.
\n\n
Maximum temperature:
\nAssumption:
\nTo determine the enthalpy of reaction the experiment was carried out five times. The same volume and concentration of copper(II) sulfate was used but the mass of zinc was different each time. Suggest, with a reason, if zinc or copper(II) sulfate should be in excess for each trial.
\nThe formula q = mcΔT was used to calculate the energy released. The values used in the calculation were m = 25.00 g, c = 4.18 J g−1 K−1.
\nState an assumption made when using these values for m and c.
\nPredict, giving a reason, how the final enthalpy of reaction calculated from this experiment would compare with the theoretical value.
\n100 «s» [✔]
\n\n
Note: Accept 90 to 100 s.
\nhighest recorded temperature
OR
when rate of heat production equals rate of heat loss [✔]
Maximum temperature:
73 «°C» [✔]
Assumption:
«temperature reached if» reaction instantaneous
OR
«temperature reached if reaction occurred» without heat loss [✔]
\n
Note: Accept “rate of heat loss is constant” OR “rate of temperature decrease is constant”.
\nAny one of:
copper(II) sulfate AND mass/amount of zinc is independent variable/being changed
OR
copper(II) sulfate AND with zinc in excess there is no independent variable «as amount of copper(II) sulfate is fixed» [✔]
copper(II) sulfate AND having excess zinc will not yield different results in each trial [✔]
\nzinc AND results can be used to see if amount of zinc affects temperature rise «so this can be allowed for» [✔]
\nzinc AND reduces variables/keeps the amount reacting constant [✔]
\n\n
Note: Accept “copper(II) sulfate/zinc sulfate” for “solution”.
\nlower/less exothermic/less negative AND heat loss/some heat not accounted for
OR
lower/less exothermic/less negative AND mass of reaction mixture greater than 25.00 g
OR
greater/more exothermic/more negative AND specific heat of solution less than water [✔]
Nearly everyone correctly estimated 100s as the time when powdered zinc was added to the beaker.
\nMost scored the mark in for stating that point Y either indicated the maximum temperature or the end of the reaction.
\nStating the maximum temperature that should be used in calculations was less well answered, with answers between 63 and 65, or 78 commonly given instead of the correct answer of 73°C. Most candidates managed to score the second mark for stating “no heat loss”.
\nMany candidates struggled to explain their choice of which reagent should be in excess. This question proved quite difficult with many candidates seeming to confuse independent and dependent variables.
\nExplaining the assumptions made when using values for m and c was challenging in. Many referred to the accuracy of the data when using m = 25.00g or said that no mass was lost during the reaction. Most knew that the value of c used was for water and suggested that the water was pure, but did not say that the specific heat of solution was assumed to be the same as that of water.
\nMost scored a mark for predicting how the calculated enthalpy value would compare with the theoretical value.
\nWhich element is in the p-block?
\nA. Pb
\nB. Pm
\nC. Pt
\nD. Pu
\nA
\nWhat is the formula of magnesium nitride?
\nA. MgN
\nB. Mg2N3
\nC. Mg3N
\nD. Mg3N2
\nD
\nWhich species has the longest carbon to oxygen bond length?
\nA. CO
\nB. CH3OH
\nC. CH3CO2−
\nD. H2CO
\nB
\nWhat are the predicted electron domain geometries around the carbon and both nitrogen atoms in urea, (NH2)2CO, applying VSEPR theory?
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/11\"](\"../assets/uploads/tinymce_asset/asset/4604/Schermafbeelding_2018-08-10_om_07.00.48.png\"/)
B
\nSolar energy, which is freely available, is indispensable to life on earth.
\nSuggest another advantage and one disadvantage of solar energy.
\nLight can be absorbed by chlorophyll and other pigments.
\nConsider molecules A and B represented below.
\nIdentify, with a reason, the molecule that absorbs visible light.
\nState a physical property of vegetable oils that makes them very difficult to use as fuel in internal combustion engines.
\nDescribe how vegetable oils can be converted to a more suitable fuel.
\nContrast the importance of carbon dioxide and methane as greenhouse gases.
\nExplain, using an equation, the effect of increased carbon dioxide in the atmosphere on the pH of lake water.
\nAdvantage:
\nrenewable «energy source»
\nOR
\ndoes not produce greenhouse gases
\nOR
\ncan be installed «almost» anywhere
\nOR
\nlow maintenance costs ✔
\n\n
Disadvantage:
\nwidely dispersed/not concentrated «form of energy»
\nOR
\ngeography/weather/seasonal dependent
\nOR
\nnot available at night
\nOR
\nenergy storage is difficult/expensive
\nOR
\ntoxic/hazardous materials used in production
\nOR
\nconcerns about space/aesthetics/local environment where installed
\nOR
\nneed to be «constantly» cleaned ✔
\n\n
Accept “can be used for passive/active heating”, “can be converted to electric energy”.
\nAccept any specific greenhouse gas name or formula for “greenhouse gases”.
\nAccept “solar cells require large areas”, “solar cell manufacture produces pollution/greenhouse gases”, “higher cost of solar cells «compared with traditional sources such as fossil fuels or hydroelectric»”.
\nB AND larger/more extensive «electronic» conjugation
\nOR
\nB AND «contains» more alternate single and double bonds ✔
\n\n
Accept more specific statements, such as “sp3 carbon in A prevents conjugation between aromatic rings”.
\nhigh viscosity ✔
\n\n
Accept “low volatility”, just “viscous/viscosity” OR “does not flow easily”.
\nconvert to esters of monoatomic alcohols
\nOR
\nreact with short-chain alcohols «in the presence of acid or base» ✔
\n\n
Accept “convert to shorter «carbon chain» esters” OR “transesterification”.
\nAccept specific alcohols, such as methanol or ethanol.
\ncarbon dioxide/CO2 more/most abundant «GHG than methane/CH4»
\nOR
\ncarbon dioxide/CO2 has «much» longer atmospheric life «than methane/CH4» ✔
\n\n
methane/CH4 «much» better/more effective at absorbing IR radiation «than carbon dioxide/CO2»
\nOR
\nmethane/CH4 has a greater greenhouse factor «than carbon dioxide/CO2»
\nOR
\nmethane/CH4 has a greater global warming potential/GWP «than carbon dioxide/CO2» ✔
\n\n
Accept “carbon dioxide/CO2 contributes more to global warming «than methane/CH4»”.
\nCO2 (g) + H2O (l) H+ (aq) + HCO3– (aq)
\nOR
\nCO2 (g) CO2 (aq) AND CO2 (aq) + H2O (l) H+ (aq) + HCO3– (aq) ✔
\n«increasing [CO2 (g)]» shifts equilibrium/reaction to right AND pH decreases ✔
\n\n
Accept “H2CO3 (aq)” for “CO2 (aq) + H2O (l)”.
\nEquilibrium arrows required for M1.
\nState symbols required for CO2 (g) CO2 (aq) equation only for M1.
\nAccept “concentration of H+/[H+] increases AND pH decreases” for M2.
\nThe compounds shown below have similar relative molecular masses. What is the correct order of increasing boiling point?
\nA. CH3COOH < (CH3)2CO < (CH3)2CHOH
\nB. CH3COOH < (CH3)2CHOH < (CH3)2CO
\nC. (CH3)2CO < CH3COOH < (CH3)2CHOH
\nD. (CH3)2CO < (CH3)2CHOH < CH3COOH
\nD
\nWhich describes the reaction shown in the potential energy profile?
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/13\"](\"../assets/uploads/tinymce_asset/asset/4605/Schermafbeelding_2018-08-10_om_07.10.46.png\")
A. The reaction is endothermic and the products have greater enthalpy than the reactants.
\nB. The reaction is endothermic and the reactants have greater enthalpy than the products.
\nC. The reaction is exothermic and the products have greater enthalpy than the reactants.
\nD. The reaction is exothermic and the reactants have greater enthalpy than the products.
\nA
\nWhat is the enthalpy change of combustion of urea, (NH2)2CO, in kJ mol−1?
\n2(NH2)2CO(s) + 3O2(g) → 2CO2(g) + 2N2(g) + 4H2O(l)
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/14\"](\"../assets/uploads/tinymce_asset/asset/4606/Schermafbeelding_2018-08-10_om_07.13.07.png\")
A. 2 × (−333) −2 × (−394) −4 × (−286)
\nB. [2 × (−394) + 4 × (−286) −2 × (−333)]
\nC. 2 × (−394) + 4 × (−286) −2 × (−333)
\nD. [2 × (−333) −2 × (−394) −4 × (−286)]
\nB
\nTwo 100 cm3 aqueous solutions, one containing 0.010 mol NaOH and the other 0.010 mol HCl, are at the same temperature.
\nWhen the two solutions are mixed the temperature rises by y °C.
\nAssume the density of the final solution is 1.00 g cm−3.
\nSpecific heat capacity of water = 4.18 J g−1 K−1
\nWhat is the enthalpy change of neutralization in kJ mol−1?
\nA.
\nB.
\nC.
\nD.
\nB
\nThe potential energy profile for the reversible reaction, X + Y Z is shown.
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/16\"](\"../assets/uploads/tinymce_asset/asset/4607/Schermafbeelding_2018-08-10_om_07.23.02.png\")
Which arrow represents the activation energy for the reverse reaction, Z → X + Y, with a catalyst?
\nC
\nWhich factors can affect the rate of reaction?
\nI. Particle size of solid reactant
\nII. Concentration of reacting solution
\nIII. Pressure of reacting gas
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nWhich factor does not affect the position of equilibrium in this reaction?
\n2NO2(g) N2O4(g) ΔH = −58 kJ mol−1
\nA. Change in volume of the container
\nB. Change in temperature
\nC. Addition of a catalyst
\nD. Change in pressure
\nC
\nThe structure of penicillin is shown in section 37 of the data booklet.
\nState the internal bond angles in the β-lactam ring and the expected bond angles for the same atoms in an open structure.
\nExplain how the open β-lactam ring kills bacteria.
\nOutline one effect of over-prescription of penicillin.
\nState how the structure of penicillin can be changed to combat this effect.
\nSuggest why human cells are not affected by penicillin.
\nAccept “109º”.
\n«irreversibly» binds/bonds to enzyme/transpeptidase
\nOR
\ninhibits enzyme/transpeptidase «in bacteria» that produces cell walls
\nOR
\nprevents cross-linking of bacterial cell walls ✔
\n\n
cells absorb water AND burst
\nOR
\ncells cannot reproduce ✔
\n\n
Accept “reacts with” for “bonds to” for M1.
\nDo not accept “cell membrane” for “cell wall” for M1.
\nAccept “cells burst due to osmotic pressure” for M2.
\nAccept “bacteria” for “cells” for M2.
\nAny one of:
\nleads to «bacterial» resistance «to antibiotics»
\nOR
\nmakes antibiotics less effective
\nOR
\nincreased side effects due to larger dosages/over time ✔
\n\n
increases proportion of resistant bacteria ✔
\n\n
destroys useful/beneficial bacteria
\nOR
\ndestroyed bacteria replaced by more harmful bacteria ✔
\n\n
resistant bacteria pass on their mutation to next generation ✔
\ndamage to ecosystems ✔
\n\n
Accept “superbugs such as MRSA develop” but superbug must be identified.
\nAccept “immune” for resistant but do not accept “tolerance”
\n«modify» side-chain ✔
\n\n
Accept “«modify» R”.
\nno cell walls
\nOR
\nhumans do not have transpeptidase ✔
\nLithium has many uses.
\nThe emission spectra obtained by ICP-OES for a mixture containing the isotope 6Li (Li-6) and naturally occurring lithium (Li (N)) is shown.
\n\n
Identify the type of bonding in lithium hydride, using sections 8 and 29 of the data booklet.
\nIdentify the colour of the emission spectrum of lithium using section 17 of the data booklet.
\nSuggest why ICP-OES does not give good quantitative results for distinguishing 6Li from naturally occurring lithium.
\nSuggest a better method.
\nLithium is obtained by electrolysis of molten lithium chloride. Calculate the time, in seconds, taken to deposit 0.694 g Li using a current of 2.00 A.
\nQ (charge) = I (current) × t (time)
\nLithium has shown some superconductive properties when doped into graphene or when under high pressure. Under high pressure, however, the Meissner effect is absent.
\nDescribe the Meissner effect.
\nAt very low temperatures, lithium atoms enhance the phonon binding of electrons in graphene suggesting the formation of Cooper pairs.
\nExplain how Cooper pairs are formed.
\nLithium forms a crystalline lattice with the unit cell structure shown below.
\nX-ray diffraction shows that the length of the edge of the unit cell is 3.51 × 10−8 cm.
\nDetermine the density of lithium, in g cm−3, using sections 2 and 6 of the data booklet.
\nionic [✔]
\nred [✔]
\nemission spectra of both «6Li and natural Li» give same colour/produce same «range of» wavelengths
OR
they have same electron transitions/same nuclear charge [✔]
\n
Note: Accept “the spectra are almost identical”.
\nICP-MS [✔]
\n\n
Note: Accept “MS/mass spectrometry”.
\nn «= » =0.100«mol»
\n « »
4830 «s» [✔]
\ncreation of mirror image/opposing magnetic field of external field «below critical temperature/T of superconductor»
OR
expulsion of magnetic field from superconductor «below critical temperature/T» [✔]
Any three of:
positive ions/cations in lattice are attracted to passing electron [✔]
lattice is distorted «by this passing electron» [✔]
\ncreates «local» regions of increased positive charge [✔]
\nsecond electron is attracted to deformation AND a coupling occurs [✔]
\nmass of Li in unit cell = « » 2.31 × 10–23 g [✔]
\nvolume of unit cell = «(3.51 × 10–8 cm)3 =» 4.32 × 10–23 cm3 [✔]
\n«density = =» 0.535 «g cm–3» [✔]
\n\n
Note: Award [3] for correct final answer.
\nMost candidates correctly identified the type of bonding and the colour of the emission spectrum of lithium.
\nMost candidates correctly identified the type of bonding and the colour of the emission spectrum of lithium.
\nMost candidates correctly identified the type of bonding and the colour of the emission spectrum of lithium, but frequently referred to the ICP-OES spectra of 6Li and naturally occurring lithium as being the same, rather than similar and thus failed to score the mark in (b)(ii).
\nA better method was selected by most candidates.
\nThe calculation in was done well.
\nCandidates had some difficulty describing the Meissner effect, with several responses using the terms repelling or repulsion instead of opposing and expulsion. Correct terminology is required.
\nPoor expression was also evident in responses explaining the formation of Cooper pairs, with very few candidates scoring full marks.
\nMost candidates had difficulty determining the number of atoms in lithium in a unit cell, even with a diagram provided. However, ECF marks were frequently scored.
\nThe presence of very small amounts of lead in calcium-based antacids can be determined using inductively coupled plasma-mass spectroscopy (ICP-MS).
\nState the type of particle present in the plasma formed.
\nAn unknown antacid sample has a lead ion concentration of 0.50 μg dm‒3.
\nCalculate the concentration of lead ions in the sample in mol dm‒3.
\nElectrolysis is used to obtain lead from Pb2+ (aq) solution.
\nDetermine the time, in hours, required to produce 0.0500 mol lead using a current (I) of 1.34 A. Use section 2 of the data booklet and the equation, charge (Q) = current (I) × time (t, in seconds).
\npositive ions/cations/Pb2+
\nOR
\nfree electrons ✔
\n\n
Accept “ions” OR “charged species/particle”.
\n[Pb2+] = 0.50 × 10‒6/5.0 × 10–7 «g dm–3» ✔
\n[Pb2+] «» = 2.4 × 10‒9 «mol dm‒3» ✔
\n\n
Award [2] for correct final answer.
\n«Faraday’s constant, F = 9.65 × 104 C mol‒1 and 1 A = 1 C s–1»
Q «= 0.0500 mol × 2 × 96500 C mol‒1» = 9650 «C» ✔
t « so » = 2.00 «hours» ✔
\n\n
Award [2] for correct final answer.
\nDietary recommendations are made by scientists.
\nThe formation of proteins from amino acids is an example of an anabolic reaction in the human body.
\nState the source of energy for such a synthetic reaction.
\nSuggest why it is advisable for those living in northerly or southerly latitudes (that is away from the equator) to take vitamin D supplements during the winter.
\nExplain how a xenobiotic is biomagnified.
\ncatabolism «of food/nutrients»
\nOR
\n«cellular» respiration ✔
\n\n
Accept “ATP” but not “burning of food/nutrients”.
\nnot enough sunlight/UV light «for synthesis of vitamin D in the skin» ✔
\ncannot be metabolized/broken down
\nOR
\nnot biodegradable
\nOR
\naccumulates in lipid/fat tissues ✔
\n\n
increased concentration as one species feeds on another «in the food chain» ✔
\nOpiates are strong analgesics.
\nExplain why diamorphine (heroin) crosses the blood–brain barrier more easily than morphine.
\nDescribe the analgesic action of an opiate.
\nOutline the meaning of the bioavailability of a drug.
\nblood-brain barrier is hydrophobic/non-polar/made of lipids ✔
\nmorphine has hydroxyl/OH «groups»/is more polar AND diamorphine has ester/ethanoate/OCOCH3/acetate «groups»/is less polar/is lipid soluble ✔
\n\n
Accept “fats” for “lipid”.
\nAccept “alcohol/hydroxy” for “hydroxyl” but not “hydroxide”.
\nAccept “non-polar” for “less polar” in M2.
\n«temporarily» binds to «opioid» receptor sites in the brain/CNS
\nOR
\n«temporarily» suppresses pain impulses in/to the brain/CNS ✔
\nfraction/proportion/percentage of «administered dosage» enters blood/plasma/circulation ✔
\n\n
Accept “fraction/proportion/percentage of «administered dosage» that reaches target «part of human body»”.
\nWhile heating solid cholesteryl benzoate, Reinitzer discovered the liquid crystal phase.
\nOutline two observations that he could have made.
\nThe structure of biphenyl nitrile is shown.
\nDescribe, giving a reason, a feature of the molecular structure, other than its polarity, that allows biphenyl nitrile to show liquid crystal behaviour.
\nArc discharge, consisting of two inert metal electrodes in a liquid solvent, is one method of producing carbon nanotubes (CNTs).
\nPredict, giving a reason, the electrode at which the solvent cyclohexane, C6H12, will decompose to form CNTs.
\nAny two of:
\ncloudy/foggy/hazy phase «at first melting point» ✔
\nclear liquid phase «at second melting point/higher temperature» ✔
\n\n
two «different» melting points
\nOR
\nnew phase observed over a wide temperature range ✔
\n\n
Accept “exhibit both liquid and solid properties at the same time” for M3.
\nALTERNATIVE 1:
\n«bulky/long» C5H11/R/alkyl «group/chain» AND prevents molecules from packing closer together «to form solid state» ✔
\n\n
ALTERNATIVE 2:
\nbiphenyl «fragment»/two benzene rings/two aromatic rings AND «makes molecule» rigid/rod-shaped ✔
\n\n
Accept “rigid/rod-shaped molecule, so aligns with other molecules” for ALTERNATIVE 2.
\n«average» oxidation state of C in C6H12/cyclohexane = –2 AND in CNTs = 0
\nOR
\noxidation state of C in CNTs is higher than in C6H12/cyclohexane
\nOR
\nloss of H’s/hydrogens ✔
\n\n
«oxidation at» anode/positive/+ «electrode» ✔
\n\n
Accept “oxidation number” for “oxidation state”.
\nEnzymes are mainly globular proteins.
\nDescribe the interaction responsible for the secondary structure of a protein.
\nExplain the action of an enzyme and state one of its limitations.
\nEnzymes are widely used in washing detergents. Outline how they improve the efficiency of the process.
\nhydrogen bonding ✔
\nbetween C=O and H–N «groups» ✔
\n\n
Accept a diagram which shows hydrogen bonding for M1 and which shows the interaction between O of C=O and H of NH groups for M2.
\nAccept “between amido/amide/carboxamide” but not “between amino/amine” for M2.
\nEnzyme action:
\nAny two of:
\nsubstrate binds to active site ✔
\nweakens bonds in substrate ✔
\n\n
lowers activation energy
\nOR
\nprovides alternate pathway ✔
\n\n
increases rate of reaction
\nOR
\nacts as catalyst ✔
\n\n
substrate specific ✔
\n\n
Limitation:
\nAny one of:
\ntemperature dependent ✔
\npH dependent ✔
\ncan be sensitive to heavy metal ions ✔
\nsensitive to denaturation ✔
\ncan be inhibited ✔
\nsubstrate specific ✔
\n\n
Accept “favourable orientation/conformation of the substrate «enforced by enzyme»” for M1.
\nDo not accept “substrate specific” as both an enzyme action and a limitation.
\nAny one of:
\n«increase rate of» hydrolyse/break down lipids/oils/fats/proteins ✔
\n«wash at» lower temperature/consume less energy ✔
\nViruses and bacteria both cause diseases and are frequently confused.
\nState one way in which viruses differ from bacteria.
\nOutline two different ways in which antiviral medications work.
\nbacteria perform living functions «on their own and viruses do not without host cell»
\nOR
\nbacteria have cell walls «and viruses do not»
\nOR
\nbacteria do not have a capsid «and viruses do»
\nOR
\nbacteria larger than viruses
\nOR
\nbacteria reproduce by fission/budding «and viruses reproduce within a living host cell»
\nOR
\nbacteria affected by antibiotics «while viruses are not» ✔
\n\n
Accept “bacteria have flagella/ cytoplasm/ribosome «and virus can have head/protein tail/double stranded RNA/single stranded DNA»”, “asexual reproduction for bacteria”, other specific structural differences between bacteria and viruses, and examples of living functions that bacteria perform (such as excretion, reproduction etc.) that viruses do not.
\nAny two of:
\nprevents virus attaching to host cell ✔
\nalters cell’s genetic material/DNA «so that virus cannot use it to multiply» ✔
\nblocks enzyme activity in the host cell «so that virus cannot use it to multiply» ✔
\nprevents removal of protein coat/capsid ✔
\nprevents injection of viral DNA/RNA into cell ✔
\nprevents release of «replicated» viruses from host cell ✔
\n\n
Accept “prevents synthesis of virus by host cell”.
\nAccept “alters RNA/DNA/genetic material of virus”.
\nDo not accept just “mimics nucleotides”.
\nBuffer systems control pH in the body.
\nDetermine the pH of a buffer solution that is 0.0100 mol dm−3 sodium hydrogen carbonate and 0.0200 mol dm−3 sodium carbonate, using section 1 of the data booklet.
\nKa (hydrogen carbonate ion) = 4.8 × 10−11
\nState the equation for the reaction of calcium carbonate, the active ingredient in some antacids, with stomach acid.
\nSuggest a technique for measuring the percentage mass of calcium carbonate in this type of antacid tablet.
\nALTERNATIVE 1:
\nUsing: pH = pKa + log
\npKa = 10.32 ✔
\npH = «10.32 + log =» 10.62 ✔
\n\n
ALTERNATIVE 2:
\n[H+] «= Ka × » = 2.4 × 10−11 ✔
\npH = 10.62 ✔
\n\n
Award [2] for correct final answer.
\nAccept answers for M2 between 10.6 and 10.7.
\nAward [1 max] for pH = 10.02.
\nCaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)
\nOR
\nCaCO3 (s) + 2H+ (aq) → Ca2+ (aq) + H2O (l) + CO2 (g) ✔
\n«back» titration
\nOR
\nthermal decomposition
\nOR
\natomic absorption/AA ✔
\n\n
Accept “gravimetric analysis”.
\nDo not accept description of a technique without proper term given for the technique.
\nPolybutadiene, used in truck tyres, is a polymer of buta-1,3-diene. The spatial arrangement of atoms in the polymer depends on the type of catalyst used.
\nOutline two differences between heterogeneous and homogeneous catalysts.
\nSuggest, giving a reason, how elastomers used for the tyre tread can increase the traction between the tyre and the road.
\nTyre fires emit trace quantities of polychlorinated dibenzofurans and polychlorinated dibenzo-p-dioxin.
\nOutline, using section 31 of the data booklet, why polychlorinated dibenzofuran is not classed chemically as a dioxin but considered “dioxin-like”.
\nClassify polybutadiene as either an addition or condensation polymer, giving a reason.
\nState one factor considered when making green chemistry polymers.
\nAny two of:
heterogeneous catalyst is in different phase than reactants AND homogeneous catalyst in same phase [✔]
homogeneous catalysts chemically change/react and reformed at end of reaction
OR
reactants adsorb onto heterogenous catalyst and products desorb [✔]
heterogeneous catalysts are more easily removed than homogenous catalysts [✔]
\nheterogeneous catalysts can function at higher temperatures [✔]
\nhomogeneous catalysts are «generally» more selective [✔]
\nhomogeneous catalysts offer a broader range of reactions [✔]
\n\n
Note: Accept “state” for “phase”.
\nAccept “heterogeneous catalyst provides a surface to activate reaction”.
\n\n
elastomers bend under force «and return to original form when force is released»
OR
elastomers make tyre more flexible [✔]
allows greater contact with road [✔]
\ndoes not contain heterocyclic ring with 2 oxygen atoms
OR
middle ring has only 1 oxygen atom [✔]
produces similar toxic effects to dioxins [✔]
\naddition AND not two different functional groups reacting
OR
addition AND formed by breaking one bond of the carbon–carbon double bonds
OR
addition AND empirical formula of monomer equals empirical formula of polymer
OR
addition AND no atoms removed/all atoms accounted for/no loss of water/ammonia/inorganic by-product/small molecules
OR
addition AND atom economy/efficiency is 100 %
OR
addition AND there is only one «reaction» product [✔]
Any one of:
high content of raw materials in product/high atom economy [✔]
use of low toxic chemicals/catalysts/materials/solvents [✔]
\nrenewable feedstock/raw materials [✔]
\nuse of renewable/clean/low carbon energy source [✔]
\nhigh safety standards [✔]
\nincrease energy efficiency [✔]
\nwaste recycling [✔]
\n\n
Note: Accept other reasonable answers.
\nMost candidates correctly stated one difference between heterogeneous and homogeneous catalysts. Few gave a second difference even though the question is worth 2 marks.
\nMost explained well how elastomers increase tyre traction.
\nBut had difficulty applying their knowledge to outline why polychlorinated dibenzofuran is considered dioxin-like but is not classified as a dioxin.
\nSome candidates failed to score the mark as they did not give a reason for classifying polybutadiene as an addition polymer.
\nMost candidates were able to state a factor considered when making green chemistry polymers.
\nSuggest two reasons why chlorinated solvents should neither be released into the atmosphere nor incinerated (burnt).
\nAny two of:
\n«weak» C–Cl bonds break/produce radicals ✔
\ncontribute to ozone depletion ✔
\ncontribute to «photochemical» smog ✔
\ncause cancers ✔
\ndamage respiratory system ✔
\ncause organ failure ✔
\nproduce toxic chemicals/phosgene/dioxins ✔
\n\n
Accept “chlorinated solvents are toxic”.
\nAlloys containing at least 60 % copper reduce the presence of bacteria on their surface.The percentage of copper in brass, an alloy of copper and zinc, can be determined by UV-vis spectrometry.
\nA sample of brass is dissolved in concentrated nitric acid and then made up to 250.0 cm3 with water before analysis.
\nCu (s) + 4HNO3 (aq) → Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O (l)
\n3Zn (s) + 8HNO3 (aq) → 3Zn(NO3)2 (aq) + 2NO (g) + 4H2O (l)
\nThe concentration of copper(II) ions in the resulting solution is then determined from a calibration curve, which is plotted by measuring the light absorbance of standard solutions.
\nYou may find the following chart and diagram helpful.
\n\n
Outline why the initial reaction should be carried out under a fume hood.
\nDeduce the equation for the relationship between absorbance and concentration.
\nCopper(II) ion solutions are blue. Suggest, giving your reason, a suitable wavelength of light for the analysis.
\nOutline how a solution of 0.0100 mol dm−3 is obtained from a standard 1.000 mol dm−3 copper(II) sulfate solution, including two essential pieces of glassware you would need.
\nThe original piece of brass weighed 0.200 g. The absorbance was 0.32.
\nCalculate, showing your working, the percentage of copper by mass in the brass.
\nDeduce the appropriate number of significant figures for your answer in (e)(i).
\nComment on the suitability of using brass of this composition for door handles in hospitals.
\nIf you did not obtain an answer to (e)(i), use 70 % but this is not the correct answer.
\nSuggest another property of brass that makes it suitable for door handles.
\nTitration is another method for analysing the solution obtained from adding brass to nitric acid.
\nCopper(II) ions are reduced to copper(I) iodide by the addition of potassium iodide solution, releasing iodine that can be titrated with sodium thiosulfate solution, Na2S2O3 (aq). Copper(I) iodide is a white solid.
\n4I− (aq) + 2Cu2+ (aq) → 2CuI (s) + I2 (aq)
\nI2 (aq) + 2S2O32− (aq) → 2I− (aq) + S4O62− (aq)
\nSuggest why the end point of the titration is difficult to determine, even with the addition of starch to turn the remaining free iodine black.
\nNO2/NO/NOx/HNO3/gas is poisonous/toxic/irritant ✔
\n\n
Accept formula or name.
\nAccept “HNO3 is corrosive” OR “poisonous/toxic gases produced”.
\nAccept “reaction is harmful/hazardous”.
\nSlope (gradient):
\n40 ✔
\n\n
Equation:
\nabsorbance = 40 × concentration
\nOR
\ny = 40x ✔
\n\n
Accept any correct relationship for slope such as .
\nAward [2] if equation in M2 is correct.
\norange is opposite blue «in the colour wheel»
\nOR
\nthe complementary colour «blue» is seen/transmitted ✔
\n\n
585–647 «nm would be absorbed» ✔
\n\n
Accept any value or range within 550–680 «nm» for M2.
\ndilute 1.00 cm3 «of the standard solution with water» to 100 cm3
\nOR
\ndilute sample of standard solution «with water» 100 times ✔
\n\n
«graduated/volumetric» pipette/pipet ✔
\nvolumetric flask ✔
\n\n
Accept any 1 : 100 ratio for M1.
\nAccept “mix 1 cm3 of the standard solution with 99 cm3 of water” for M1.
\nDo not accept “add 100 cm3 of water to 1.00 cm3 of standard solution” for M1.
\nAccept “burette/buret” for M2.
\nAccept “graduated/measuring flask” for M3 but not “graduated/measuring cylinder” or “conical/Erlenmeyer flask”.
\nconcentration of copper = 0.0080 «mol dm–3» ✔
\n\n
mass of copper in 250.0 cm3 = «0.0080 mol dm–3 × 0.2500 dm3 × 63.55 g mol–1 =» 0.127 «g»
\nOR
\nmass of brass in 1 dm3 = «4 × 0.200 g =» 0.800 g AND [Cu2+] = «0.0080 mol dm–3 × 63.55 g mol–1 =» 0.5084 g dm–3 ✔
\n\n
«% copper in this sample of brass » 64 «%»
\nOR
\n«% copper in this sample of brass » 64 «%» ✔
\n\n
Accept any value in range 0.0075–0.0085 «mol dm–3» for M1.
\nAccept annotation on graph for M1.
\nAward [3] for correct final answer.
\nAccept “65 «%»”.
\ntwo ✔
\n\n
Do not apply ECF from 1(e)(i).
\n«since it is greater than 60%» it will reduce the presence of bacteria «on door handles» ✔
\nresistant to corrosion/oxidation/rusting
\nOR
\nlow friction surface «so ideal for connected moving components» ✔
\n\n
Accept “hard/durable”, “«high tensile» strength”, “unreactive”, “malleable” or any reference to the appearance/colour of brass (eg “gold-like”, “looks nice” etc.).
\nDo not accept irrelevant properties, such as “high melting/boiling point”, “non-magnetic”, “good heat/electrical conductor”, “low volatility”, etc.
\nDo not accept “ductile”.
\nprecipitate/copper(I) iodide/CuI makes colour change difficult to see
\nOR
\nrelease of I2/iodine from starch-I2 complex is slow so titration must be done slowly ✔
\nLiquid-crystal displays (LCDs) have many uses.
\nA molecule which acts as a chiral nematic thermotropic liquid-crystal is given.
\nLabel with an asterisk, *, the chiral carbon atom.
\nExplain the effects of very low and high temperatures on the liquid-crystal behaviour of this molecule.
\n\n
Low temperature:
\nHigh temperature:
\n [✔]
Low temperature:
intermolecular forces prevent molecules moving AND solid/«normal» crystal formation [✔]
High temperature:
«above a critical temperature» disrupts alignment of molecules AND behaves as fluid/liquid [✔]
\n
Note: Accept “weak intermolecular forces break AND behaves as fluid/liquid”.
\nIdentifying a chiral carbon atom was answered reasonably well.
\nExplaining effects of very low and very high temperatures on liquid-crystal behaviour proved difficult for most candidates. Responses lacked the required detail about intermolecular forces between molecules.
\nStarch is a natural polymer of glucose.
\nDraw the structure of the repeating unit of starch and state the type of linkage formed between these units.
\nType of linkage:
\nFormulate the equation for the complete hydrolysis of a starch molecule, (C6H10O5)n.
\nCalculate the energy released, in kJ g−1, when 3.49 g of starch are completely combusted in a calorimeter, increasing the temperature of 975 g of water from 21.0 °C to 36.0 °C. Use section 1 of the data booklet.
\nExplain how the inclusion of starch in plastics makes them biodegradable.
\ncontinuation bonds AND −O attached to just one end AND both H atoms on end carbons must be on the same side [✔]
\n
Type of linkage:
glycosidic [✔]
\n
Note: Square brackets not required.
\nIgnore “n” if given.
\nMark may be awarded if a polymer is shown but with the repeating unit clearly identified.
\nAccept “ether”.
\n\n
(C6H10O5)n (s) + nH2O (l) → nC6H12O6 (aq) [✔]
\n\n
Note: Accept “(n-1)H2O”.
\nDo not award mark if “n” not included.
\nq = «mcΔT = 975 g × 4.18 J g–1 K–1 × 15.0 K =» 61 100 «J» / 61.1 «kJ» [✔]
\n«heat per gram = =» 17.5 «kJ g–1» [✔]
\n\n
Note: Award [2] for correct final answer.
\nAny two of:
carbohydrate grains swell/break plastic into smaller pieces [✔]
inclusion of carbohydrate makes the plastic more hydrophilic/water soluble [✔]
\ncarbohydrates are broken down/hydrolysed/digested by bacteria/micro-organisms [✔]
\nplastic becomes more accessible to bacteria as holes/channels are created in it [✔]
\n«presence of» carbohydrate weakens intermolecular/London/dispersion forces between polymer chains in the plastic [✔]
\n\n
Note: Accept “starch” for “carbohydrate” throughout.
\nDo not accept “carbohydrates are broken down/hydrolyzed”.
\nCandidates were required to draw the structure of the repeating unit of starch given the ring structure as a starting point. This proved extremely difficult with very few candidates scoring a mark. Commonly, the structure of -glucose was given, or an attempt was made to draw a polymer. Naming the type of linkage formed was answered well.
\nAlso proved challenging, with many candidates unable to write an equation for the hydrolysis of a starch molecule (C6H10O5)n. The n was often omitted from otherwise correct equations or the product was incorrectly given as (C6H12O6)n.
\nThe incorrect mass was frequently used when calculating energy released from combustion of starch in a calorimeter. Those who used the mass of water correctly frequently stopped when energy in kJ or J was calculated, and did not seem to notice that the question asked for the energy to be calculated in kJg−1 so a further calculation was required.
\nResponses to explain how including starch in plastics makes them biodegradable were sometimes lacking in detail. Some candidates simply said that “starch is soluble in water”. Some said that “starch can be broken down/hydrolyzed” but omitted the key words “by bacteria or microorganisms”.
\nNanotechnology has allowed the manipulation of materials on the atomic level.
\nDescribe the structure and bonding of a carbon nanotube.
\n\n
Structure:
\nBonding:
\nSuggest one application for carbon nanotubes.
\nStructure:
giant covalent/network covalent [✔]
Bonding:
each carbon covalently bonded to 3 other carbons
OR
each bond has order of 1.5 [✔]
\n
Note: Accept “cylindrical/tube shaped”.
\nAccept “has delocalized electrons” OR “has sp2 hybridization”.
\nAny one of:
3D electrodes [✔]
catalysts [✔]
\nbiosensors [✔]
\nmolecular stents [✔]
\nbody armour [✔]
\nsynthetic muscles [✔]
\nmicro transistors/circuitry/capacitors/electrodes [✔]
\nreinforcing phase in a matrix/composite material «such as concrete» [✔]
\nmicro antenna [✔]
\nstealth technology [✔]
\nwater/air filtration [✔]
\nsolar cells [✔]
\ntennis racquets [✔]
\nmicroelectronic circuits [✔]
\n\n
Note: Do not accept just general answers such as “medicine” or “defence”.
\nDescribing the structure and bonding of a carbon nanotube was generally answered satisfactorily, although some candidates simply said the bonding was covalent with no further detail.
\nThere were some vague responses for applications of carbon nanotubes when specific details were needed to score the mark in (b).
\nAspartame is a derivative of a dipeptide formed between two amino acids, phenylalanine (Phe) and aspartic acid (Asp).
\nDraw a circle around the functional group formed between the amino acids and state its name.
\nName:
\nA mixture of phenylalanine and aspartic acid is separated by gel electrophoresis with a buffer of pH = 5.5.
\nDeduce their relative positions after electrophoresis, annotating them on the diagram. Use section 33 of the data booklet.
\nAspartic acid is obtained synthetically as a racemic mixture. Draw the three‑dimensional shape of each isomer showing their spatial relationship to each other. Use section 33 of the data booklet.
\nName:
amide/amido/carboxamide [✔]
\n
Note: Accept “peptide bond/linkage”.
\nPhe: must be on the origin [✔]
\nAsp: any position on the left/anode/+ side [✔]
\n [✔]
Candidates were challenged to draw a circle around the functional group formed between the two amino acids in a dipeptide. Commonly the central C=O or the N-H groups were circled instead of the amide. Some candidates then named the functional group as a ketone or amine, however most candidates stated “peptide” here and scored a mark.
\nMost candidates correctly deduced the positions of the two amino acids after electrophoresis.
\nWas very poorly answered. Many candidates had difficulty drawing a 3D structure and showing the spatial relationship of two enantiomers. An incorrect amino acid residue was often used, or incorrect bond connections drawn.
\nEDTA chelates with Ni2+ (aq).
\n[Ni(H2O)6]2+ (aq) + EDTA4− (aq) [Ni(EDTA)]2− (aq) + 6H2O (l)
\nExplain how entropy affects this equilibrium.
\nState the number of coordinate covalent bonds EDTA forms with Ni2+.
\nentropy increases «and the reaction proceeds to the right» [✔]
\nmore species / free molecules are formed
OR
more ways of distributing energy [✔]
six [✔]
\nThere were several incorrect responses that products were more ordered than the reactants.
\nProved very challenging with very few candidates knowing the number of coordinate covalent bonds EDTA forms with a nickel ion.
\nEnzymes are mainly globular proteins.
\nDescribe the interaction responsible for the secondary structure of a protein.
\nExplain the action of an enzyme and state one of its limitations.
\nContrast the actions of non-competitive and competitive inhibitors of an enzyme and state their effects on the maximum rate of reaction, Vmax, and the Michaelis–Menten constant, Km.
\nhydrogen bonding ✔
\nbetween C=O and H–N «groups» ✔
\n\n
Accept a diagram which shows hydrogen bonding for M1 and which shows the interaction between O of C=O and H of NH groups for M2.
\nAccept “between amido/amide/carboxamide” but not “between amino/amine” for M2.
\nEnzyme action:
\nAny two of:
\nsubstrate binds to active site ✔
\nweakens bonds in substrate ✔
\n\n
lowers activation energy
\nOR
\nprovides alternate pathway ✔
\n\n
increases rate of reaction
\nOR
\nacts as catalyst ✔
\n\n
substrate specific ✔
\n\n
Limitation:
\nAny one of:
\ntemperature dependent ✔
\npH dependent ✔
\ncan be sensitive to heavy metal ions ✔
\nsensitive to denaturation ✔
\ncan be inhibited ✔
\nsubstrate specific ✔
\n\n
Accept “favourable orientation/conformation of the substrate «enforced by enzyme»” for M1.
\nDo not accept “substrate specific” as both an enzyme action and a limitation.
\n\n
Award [1] for each action.
\nAward [1] for any two effects stated correctly.
\nAward [2 max] if both actions and effects are switched to incorrect inhibitor types.
\nThe main fatty acid composition of cocoa butter and coconut oil is detailed below.
\nThe melting points of cocoa butter and coconut oil are 34 °C and 25 °C respectively.
\nExplain this in terms of their saturated fatty acid composition.
\nFats contain triglycerides that are esters of glycerol and fatty acids. Deduce an equation for the acid hydrolysis of the following triglyceride.
\nThe addition of partially hydrogenated cocoa butter to chocolate increases its melting point and the content of trans-fatty acids (trans-fats).
\nOutline one effect of trans-fatty acids on health.
\ncoconut oil has higher content of lauric/short-chain «saturated» fatty acids
OR
cocoa butter has higher content of stearic/palmitic/longer chain «saturated» fatty acids [✔]
longer chain fatty acids have greater surface area/larger electron cloud [✔]
\nstronger London/dispersion/instantaneous dipole-induced dipole forces «between triglycerides of longer chain saturated fatty acids» [✔]
\n\n
Note: Do not accept arguments that relate to melting points of saturated and unsaturated fats.
\ncorrect products [✔]
\ncorrectly balanced [✔]
\nAny one of:
«increased risk of» coronary/heart disease [✔]
«increased risk of» stroke [✔]
\n«increased risk of» atherosclerosis [✔]
\n«increased risk of type-2» diabetes [✔]
\nincrease in LDL cholesterol [✔]
\ndecrease in HDL cholesterol [✔]
\n«increased risk of» obesity [✔]
\n\n
Candidates had difficulty explaining the melting points of fats in terms of length of carbon chain, and referred instead to an explanation of saturated and unsaturated fat structures.
\nQuite a few candidates had no idea how to write an equation for the acid hydrolysis of a given triglyceride. Many struggled with the structure or molecular formula of glycerol. Some created products of (R-O-CO-OH)3 and alkanes, even though the question told them that triglycerides are esters of glycerol and fatty acids. Some got the products almost correct but wrote O-H-CO-R for the fatty acid so failed to score the first mark. Some forgot to balance the equation.
\nOutlining an effect of trans-fatty acids on health was generally answered well, with only a few vague answers.
\nLipids play several roles in our bodies.
\nA phospholipid generally consists of two hydrophobic fatty acids and a hydrophilic group.
\nFatty acids are products of the acidic hydrolysis of phospholipids. Deduce the names of the other two products.
\nThe iodine number is the maximum mass of iodine that reacts with 100 g of an unsaturated compound.
\nDetermine the iodine number of stearidonic acid, C17H27COOH.
\nState two functions of lipids in the body.
\nOutline one effect of increased levels of low-density lipoproteins in the blood.
\nphosphoric acid ✔
\nglycerol/propane-1,2,3-triol ✔
\n\n
Do not accept formulas.
\nALTERNATIVE 1:
\n4 C=C bonds/4 carbon to carbon double bonds ✔
\nmass of iodine per mole of acid = «4 × 253.80 g mol–1 =» 1015.2 «g mol–1» ✔
\niodine number «= » = 367 ✔
\n\n
ALTERNATIVE 2:
\n4 C=C bonds/4 carbon to carbon double bonds ✔
\n« =» 1.447 mol of I2 «reacts with 100 g» ✔
\niodine number «= 1.447 mol × 253.80 g mol–1» = 367 ✔
Award [3] for correct final answer.
\nAny two of:
\n«structural» components of cell membranes ✔
\nenergy storage/utilization ✔
\n«thermal/electrical» insulation ✔
\ntransport «of lipid-soluble molecules» ✔
\nhormones/chemical messengers ✔
\n\n
Accept other specific functions, such as “prostaglandin/cytokine/bile acid synthesis”, “cell differentiation/growth”, “myelination”, “storage of vitamins/biomolecules”, “signal transmission”, “protection/padding of organs”, “precursors/starting materials for the biosynthesis of other lipid”.
\nAny one of:
\natherosclerosis/cholesterol deposition «in artery walls» ✔
\nheart/cardiovascular disease ✔
\nstroke ✔
\n\n
Accept “arteries become blocked/walls become thicker”.
\nLipids play several roles in our bodies.
\nThe iodine number is the maximum mass of iodine that reacts with 100 g of an unsaturated compound.
\nDetermine the iodine number of stearidonic acid, C17H27COOH.
\nState two functions of lipids in the body.
\nOutline one effect of increased levels of low-density lipoproteins in the blood.
\nALTERNATIVE 1:
\n4 C=C bonds/4 carbon to carbon double bonds ✔
\nmass of iodine per mole of acid = «4 × 253.80 g mol–1 =» 1015.2 «g mol–1» ✔
\niodine number «» = 367 ✔
\n\n
ALTERNATIVE 2:
\n4 C=C bonds/4 carbon to carbon double bonds ✔
\n« =» 1.447 mol of I2 «reacts with 100 g» ✔
\niodine number «= 1.447 mol × 253.80 g mol–1» = 367 ✔
\n\n
Award [3] for correct final answer.
\nAny two of:
\n«structural» components of cell membranes ✔
\nenergy storage/utilization ✔
\n«thermal/electrical» insulation ✔
\ntransport «of lipid-soluble molecules» ✔
\nhormones/chemical messengers ✔
\n\n
Accept other specific functions, such as “prostaglandin/cytokine/bile acid synthesis”, “cell differentiation/growth”, “myelination”, “storage of vitamins/biomolecules”, “signal transmission”, “protection/padding of organs”, “precursors/starting materials for the biosynthesis of other lipid”.
\nAny one of:
\natherosclerosis/cholesterol deposition «in artery walls» ✔
\nheart/cardiovascular disease ✔
\nstroke ✔
\n\n
Accept “arteries become blocked/walls become thicker”.
\nActivity series of selected elements:
\n\n
Which react with dilute sulfuric acid?
\nI. Cu
\nII. CuO
\nIII. CuCO3
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nLactose, found in milk and dairy products, is a disaccharide formed from two different monosaccharides. The structure of lactose is shown with numbered carbon atoms.
\nName the type of link between the two monosaccharide residues.
\nOutline how the two monomer structures, galactose and glucose, differ.
\n«1,4-»glycosidic ✔
\n\n
Do not accept “glucosidic”.
\nH and OH are reversed/in different positions on C-4 ✔
\n\n
C-4 must be specified.
\nDo not penalize if reference is made to H and OH above and below ring/in alpha and beta positions on C-4 incorrectly.
\nThe Sun’s energy is produced by the fusion of hydrogen nuclei.
\nUranium-238 produces plutonium-239, which is used as fuel in breeder reactors.
\nExplain fusion reactions with reference to binding energy.
\nOutline why the term breeder is used for the reactors.
\nDeduce the fission reaction when 239Pu is bombarded with a neutron to produce 133Xe and 103Zr.
\nNuclear disasters release radioactive caesium into the atmosphere, which presents serious health risks.
\nCs-137 has a half-life of 30 years.
\nCalculate the percentage of Cs-137 remaining in the atmosphere after 240 years.
\nsmall/lighter nuclei combine to form larger/heavier nuclei ✔
\nproduct has higher binding energy «per nucleon» ✔
\n\n
Accept binding energy curve with explanation.
\nconverts non-fissile «238U» material into fissile «239Pu» material
\nOR
\nproduces more fissile material than it consumes ✔
\n239Pu + 1n → 133Xe + 103Zr + 41n ✔
\n\n
Accept equation with correct atomic numbers included.
\nAccept notation for neutrons of “n”.
\nAccept a correctly described equation in words.
\nALTERNATIVE 1:
\n«» 8 /8 half-lives «required» ✔
\n% remaining = «0.508 × 100 =» 0.39 «%» ✔
\n\n
ALTERNATIVE 2:
\nλ = «» 0.023 ✔
\n% remaining = «100 × e–0.023 × 240 =» 0.39 «%» ✔
\n\n
Award [2] for correct final answer.
\nCoal can be converted to clean-burning synthetic natural gas.
\nAutomobile companies use hydrogen as an alternative to fossil fuels. Some properties of fuels are shown.
\nFormulate equation(s) for the conversion of coal and steam to methane.
\nCalculate the specific energy, in kJ g−1, of methane.
\nComment on the specific energies of hydrogen and methane.
\nCalculate the mass, in kg, of carbon dioxide produced by the complete combustion of 72.0 dm3 octane, C8H18.
\nDensity of C8H18 = 703 g dm−3
\nC8H18 (l) + 12.5O2 (g) → 8CO2 (g) + 9H2O (g)
\nALTERNATIVE 1:
\n2C (s) + 2H2O (g) → CH4 (g) + CO2 (g) ✔
\n\n
ALTERNATIVE 2:
\nC (s) + 2H2O (g) → CO (g) + H2 (g) AND 3H2 (g) + CO (g) → CH4 (g) + H2O (g) ✔
\n\n
Accept “3C (s) + 2H2O (g) → CH4 (g) + 2CO (g)”.
\n« =» 55.5 «kJ g–1» ✔
\n\n
Do not penalize negative sign.
\nDo not accept energy density at STP/density at STP = = 55.06 «kJ g–1».
\n«» hydrogen/H2 produces 2.6 times/more than twice the energy of methane/CH4 «per mass/g»
\nOR
\nless mass of hydrogen/H2 required «to produce same amount of energy»
\nOR
\nhydrogen/H2 more energy efficient ✔
\n\n
Accept “hydrogen/H2 produces «nearly» three times more energy than methane/CH4 «per mass/g»”.
\nmoctane «= 72.0 dm3 × 703 g dm–3» = 50600 «g»/50.6 «kg» ✔
\nmcarbon dioxide «= » = 156 «kg» ✔
\n\n
Award [2] for correct final answer.
\nThe thermal decomposition of dinitrogen monoxide occurs according to the equation:
\n2N2O (g) → 2N2 (g) + O2 (g)
\nThe reaction can be followed by measuring the change in total pressure, at constant temperature, with time.
\nThe x-axis and y-axis are shown with arbitrary units.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5982/2.PNG\")
This decomposition obeys the rate expression:
\n= k[N2O]
\nExplain why, as the reaction proceeds, the pressure increases by the amount shown.
\nOutline, in terms of collision theory, how a decrease in pressure would affect the rate of reaction.
\nDeduce how the rate of reaction at t = 2 would compare to the initial rate.
\nIt has been suggested that the reaction occurs as a two-step process:
\nStep 1: N2O (g) → N2 (g) + O (g)
\nStep 2: N2O (g) + O (g) → N2 (g) + O2 (g)
\nExplain how this could support the observed rate expression.
\nThe experiment is repeated using the same amount of dinitrogen monoxide in the same apparatus, but at a lower temperature.
\nSketch, on the axes in question 2, the graph that you would expect.
\nThe experiment gave an error in the rate because the pressure gauge was inaccurate.
\nOutline whether repeating the experiment, using the same apparatus, and averaging the results would reduce the error.
\nThe graph below shows the Maxwell–Boltzmann distribution of molecular energies at a particular temperature.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5983/2f.PNG\")
The rate at which dinitrogen monoxide decomposes is significantly increased by a metal oxide catalyst.
\nAnnotate and use the graph to outline why a catalyst has this effect.
\nDetermine the standard entropy change, in J K−1, for the decomposition of dinitrogen monoxide.
\n2N2O (g) → 2N2 (g) + O2 (g)
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5984/2gi.PNG\")
Dinitrogen monoxide has a positive standard enthalpy of formation, ΔHfθ.
\nDeduce, giving reasons, whether altering the temperature would change the spontaneity of the decomposition reaction.
\nincrease in the amount/number of moles/molecules «of gas» [✔]
\nfrom 2 to 3/by 50 % [✔]
\n«rate of reaction decreases»
concentration/number of molecules in a given volume decreases
OR
more space between molecules [✔]
collision rate/frequency decreases
OR
fewer collisions per unit time [✔]
\n
Note: Do not accept just “larger space/volume” for M1.
\nhalf «of the initial rate» [✔]
\n\n
Note: Accept “lower/slower «than initial rate»”.
\n1 slower than 2
OR
1 rate determinant step/RDS [✔]
\n
1 is unimolecular/involves just one molecule so it must be first order
OR
if 1 faster/2 RDS, second order in N2O
OR
if 1 faster/2 RDS, first order in O [✔]
smaller initial gradient [✔]
\ninitial pressure is lower AND final pressure of gas lower «by similar factor» [✔]
\nno AND it is a systematic error/not a random error
\nOR
\nno AND «a similar magnitude» error would occur every time [✔]
\ncatalysed and uncatalysed Ea marked on graph AND with the catalysed being at lower energy [✔]
\n
«for catalysed reaction» greater proportion of/more molecules have E ≥ Ea / E > Ea
OR
«for catalysed reaction» greater area under curve to the right of the Ea [✔]
\n
Note: Accept “more molecules have the activation energy”.
\nΔSθ = 2(Sθ(N2)) + Sθ(O2) – 2(Sθ(N2O))
OR
ΔSθ = 2 × 193 «J mol-1 K-1» + 205 «J mol-1 K-1» – 2 × 220 «J mol-1 K-1» [✔]
«ΔSθ = +»151 «J K-1» [✔]
\n\n
Note: Award [2] for correct final answer.
\nexothermic decomposition
OR
ΔH(decomposition) < 0 [✔]
TΔSθ > ΔHθ
OR
ΔGθ «= ΔHθ – TΔSθ» < 0 «at all temperatures» [✔]
reaction spontaneous at all temperatures [✔]
\nStudents were able in general to relate more moles of gas to increase in pressure.
\nFew students were able to relate the effect of reduced pressure at constant volume with a decrease in concentration of gas molecules and mostly did not even refer to this, but rather concentrated on lower rate of reaction and frequency of collisions. Many candidates lost a mark by failing to explain rate as collisions per unit time, frequency, etc.
\nThough the differential equation was considered to be misleading by teachers, most candidates attempted to answer this question, and more than half did so correctly, considering they had the graph to visualize the gradient.
\nMost students were able to identity step 1 as the RDS/slow but few mentioned unimolecularity or referred vaguely to NO2 as the only reagent (which was obvious) and got only 1 mark.
\nMany students drew a lower initial gradient, but most did not reflect the effect of lower temperature on pressure at constant volume and started and finished the curve at the same pressure as the original one.
\n\n
Almost all candidates identified the inaccurate pressure gauge as a systematic error, thus relating accuracy to this type of error.
\nThe graph was generally well done, but in quite a few cases, candidates did not mention that increase of rate in the catalyzed reaction was due to E (particles) > Ea or did so too vaguely.
\nCandidates were able to calculate the ΔS of the reaction, though in some cases they failed to multiply by the number of moles.
\nThough the question asked for decomposition (in bold), most candidates ignored this and worked on the basis of a the ΔH of formation. However, many did write a sound explanation for that situation. On the other hand, in quite a number of cases, they did not state the sign of the ΔH (probably taking it for granted) nor explicitly relate ΔG and spontaneity, which left the examiner with no possibility of evaluating their reasoning.
\nWhat is the name of this compound, using IUPAC rules?
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/25\"](\"../assets/uploads/tinymce_asset/asset/4609/Schermafbeelding_2018-08-10_om_07.41.06.png\")
A. 1,1-dimethylpropanoic acid
\nB. 3,3-dimethylpropanoic acid
\nC. 2-methylbutanoic acid
\nD. 3-methylbutanoic acid
\nD
\nCarotenoids are pigments found in leaves.
\nThe absorption spectrum of β-carotene is shown below.
\nExplain its colour in terms of its absorption bands. Use section 17 of the data booklet.
\nThe absorption spectrum of chlorophyll a is shown below.
\nSuggest how the combination of chlorophyll a and carotenoids is beneficial for photosynthesis.
\n400–424 «nm» absorption band/violet AND 424–490 «nm» absorption band/blue [✔]
\ncomplementary/opposite colour observed
OR
yellow/orange observed [✔]
\n
Note: Accept “400–500 «nm» absorption band” for M1.
\nextends energy absorption spectrum «for photosynthesis» [✔]
\nVery few candidates scored the first mark. Two absorption bands were required but most candidates only mentioned one. However, most candidates scored the second mark for stating the colour of β-carotene.
\nMost candidates gave good answers to suggest how a combination of chlorophyll a and carotenoids is beneficial for photosynthesis.
\nAlcohol dehydrogenase (ADH) catalyses the oxidation of methanol. The products of oxidation, methanal and methanoic acid, are toxic.
\nA Michaelis–Menten plot for an enzyme-catalysed reaction is shown.
\nSketch a curve to show the effect of a competitive inhibitor.
\nSuggest, based on the Michaelis–Menten plot, how a competitive inhibitor such as ethanol reduces the toxicity of methanol.
\nEnzymatic activity is studied in buffered aqueous solutions.
\nCalculate the ratio in which 0.1 mol dm−3 NaH2PO4 (aq) and 0.1 mol dm−3 Na2HPO4 (aq) should be mixed to obtain a buffer with pH = 6.10. Use section 1 of the data booklet.
\npKa (NaH2PO4) = 7.20
\n [✔]
\n
Note: Line must start at origin, to the right of original line and bending toward the same Vmax.
\nKm is higher /same Vmax reached at higher [substrate] [✔]
\nslower reaction rate
OR
gives time to excrete/eliminate methanol [✔]
«pH = pKa + log / 6.10 = 7.20 + log »
\nlog = «6.10 – 7.20 =» –1.10
OR
= «10–1.10 =» 0.079 [✔]
NaH2PO4 : Na2HPO4 = 12.6 : 1 [✔]
\n\n
Note: Award [2] for correct final answer.
\nMost candidates correctly sketched a curve to show the effect of a competitive inhibitor on a Michaelis-Menten plot.
\nMany scored 1 out of 2 for stating that the inhibitor causes a slower reaction rate.
\nThe calculation of the ratio of a conjugate acid-base pair to create a buffer with a specific pH was poorly done. Some candidates wrote a ratio without indicating which compound each value referred to and thus could not score. Many candidates used the concentrations of 0.1 mol dm−3 for both compounds and could not proceed.
\nAscorbic acid and retinol are two important vitamins.
\nExplain why ascorbic acid is soluble in water and retinol is not. Use section 35 of the data booklet.
\nascorbic acid: many hydroxyl/OH groups AND retinol: few/one hydroxyl/OH group
OR
ascorbic acid: many hydroxyl/OH groups AND retinol: long hydrocarbon chain [✔]
ascorbic acid: «many» H-bond with water
OR
retinol: cannot «sufficiently» H-bond with water [✔]
\n
Note: Do not accept “OH−/hydroxide”.
\nExplanations of why ascorbic acid is soluble in water and retinol is not were poor. Very few referred to the ability to form hydrogen bonds with water. Some said “hydroxide” instead of “hydroxyl” and thus failed to score a mark. Commonly, candidates scored 1 mark for saying that ascorbic acid has many hydroxyl groups and retinol has a long hydrocarbon chain.
\nWhat is the mechanism for the reaction of propene with iodine in the dark?
\nA. electrophilic addition
\nB. electrophilic substitution
\nC. free radical substitution
\nD. nucleophilic substitution
\nA
\nWhich are structural isomers?
\nI. CH3CH2OH and CH3OCH3
\nII. HOCH2CH3 and CH3CH2OH
\nIII. CH3COOH and HCOOCH3
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nWhich feature of a molecule does infrared spectrometry detect?
\nA. molecular mass
\nB. bonds present
\nC. total number of protons
\nD. total number of proton environments
\nB
\nDinitrogen monoxide, N2O, causes depletion of ozone in the stratosphere.
\nDifferent sources of N2O have different ratios of 14N : 15N.
\nThe Lewis (electron dot) structure of the dinitrogen monoxide molecule can be represented as:
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5985/3d.PNG\")
Outline why ozone in the stratosphere is important.
\nDinitrogen monoxide in the stratosphere is converted to nitrogen monoxide, NO (g).
\nWrite two equations to show how NO (g) catalyses the decomposition of ozone.
\nState one analytical technique that could be used to determine the ratio of 14N : 15N.
\nA sample of gas was enriched to contain 2 % by mass of 15N with the remainder being 14N.
\nCalculate the relative molecular mass of the resulting N2O.
\nPredict, giving two reasons, how the first ionization energy of 15N compares with that of 14N.
\nExplain why the first ionization energy of nitrogen is greater than both carbon and oxygen.
\nNitrogen and carbon:
\nNitrogen and oxygen:
\nState what the presence of alternative Lewis structures shows about the nature of the bonding in the molecule.
\nState, giving a reason, the shape of the dinitrogen monoxide molecule.
\nDeduce the hybridization of the central nitrogen atom in the molecule.
\nabsorbs UV/ultraviolet light «of longer wavelength than absorbed by O2» [✔]
\nNO (g) + O3 (g) → NO2 (g) + O2 (g) [✔]
NO2 (g) + O3 (g) → NO (g) + 2O2 (g) [✔]
\n
Note: Ignore radical signs.
\nAccept equilibrium arrows.
\nAward [1 max] for NO2 (g) + O (g) → NO (g) + O2 (g).
\nmass spectrometry/MS [✔]
\n« =» 14.02 [✔]
\n«Mr = (14.02 × 2) + 16.00 =» 44.04 [✔]
\nAny two:
\nsame AND have same nuclear charge /number of protons/Zeff [✔]
\nsame AND neutrons do not affect attraction/ionization energy/Zeff
OR
same AND neutrons have no charge [✔]
same AND same attraction for «outer» electrons [✔]
\nsame AND have same electronic configuration/shielding [✔]
\n\n
Note: Accept “almost the same”.
\n“Same” only needs to be stated once.
\nNitrogen and carbon:
\nN has greater nuclear charge/«one» more proton «and electrons both lost from singly filled p-orbitals» [✔]
\n
Nitrogen and oxygen:
\nO has a doubly filled «p-»orbital
OR
N has only singly occupied «p-»orbitals [✔]
\n
Note: Accept “greater e– - e- repulsion in O” or “lower e– - e- repulsion in N”.
\nAccept box annotation of electrons for M2.
\ndelocalization
\nOR
\ndelocalized π-electrons [✔]
\n\n
Note: Accept “resonance”.
\nlinear AND 2 electron domains
\nOR
\nlinear AND 2 regions of electron density [✔]
\n\n
Note: Accept “two bonds AND no lone pairs” for reason.
\nsp [✔]
\nCandidates sometimes failed to identify how ozone works in chemical terms, referring to protects/deflects, i.e., the consequence rather than the mechanism.
\nMany candidates recalled the first equation for NO catalyzed decomposition of ozone only. Some considered other radical species.
\nAll candidates, with very few exceptions, answered this correctly.
\nMost candidates were able to calculate the accurate mass of N2O, though quite a few candidates just calculated the mass of N and didn’t apply it to N2O, losing an accessible mark.
\nMany students realized that neutrons had no charge and could not affect IE significantly, but many others struggled a lot with this question since they considered that 15N would have a higher IE because they considered the greater mass of the nucleus would result in an increase of attraction of the electrons.
\nMixed responses here; the explanation of higher IE for N with respect to C was less well explained, though it should have been the easiest. It was good to see that most candidates could explain the difference in IE of N and O, either mentioning paired/unpaired electrons or drawing box diagrams.
\nMost candidates identified resonance for this given Lewis representation.
\nThough quite a number of candidates suggested a linear shape correctly, they often failed to give a complete correct explanation, just mentioning the absence of lone pairs but not two bonds, instead of referring to electron domains.
\nHybridisation of the N atom was correct in most cases.
\nHow are the uncertainties of two quantities combined when the quantities are multiplied together?
\nA. Uncertainties are added.
\nB. % uncertainties are multiplied.
\nC. Uncertainties are multiplied.
\nD. % uncertainties are added.
\nD
\nThe rate of a reaction is studied at different temperatures.
\nWhich is the best way to plot the data?
\n![\"M18/4/CHEMI/SPM/ENG/TZ2/30\"](\"../assets/uploads/tinymce_asset/asset/4610/Schermafbeelding_2018-08-10_om_07.47.54.png\"/)
C
\nA student determined the percentage of the active ingredient magnesium hydroxide, Mg(OH)2, in a 1.24 g antacid tablet.
\nThe antacid tablet was added to 50.00 cm3 of 0.100 mol dm−3 sulfuric acid, which was in excess.
\nCalculate the amount, in mol, of H2SO4.
\nFormulate the equation for the reaction of H2SO4 with Mg(OH)2.
\nThe excess sulfuric acid required 20.80 cm3 of 0.1133 mol dm−3 NaOH for neutralization.
\nCalculate the amount of excess acid present.
\nCalculate the amount of H2SO4 that reacted with Mg(OH)2.
\nDetermine the mass of Mg(OH)2 in the antacid tablet.
\nCalculate the percentage by mass of magnesium hydroxide in the 1.24 g antacid tablet to three significant figures.
\nn(H2SO4) «= 0.0500 dm3 × 0.100 mol dm–3» = 0.00500/5.00 × 10–3«mol»
\n[1 mark]
\nH2SO4(aq) + Mg(OH)2(s) → MgSO4(aq) + 2H2O(l)
\n\n
Accept an ionic equation.
\n[1 mark]
\n«n(H2SO4) = × n(NaOH) = (0.02080 dm3 × 0.1133 mol dm–3)»
\n0.001178/1.178 × 10–3 «mol»
\n[1 mark]
\nn(H2SO4) reacted «= 0.00500 – 0.001178» = 0.00382/3.82 × 10–3 «mol»
\n[1 mark]
\nn(Mg(OH)2) «= n(H2SO4) =» = 0.00382/3.82 × 10–3 «mol»
\nm(Mg(OH)2) «= 0.00382 mol × 58.33 g mol–1» = 0.223 «g»
\n\n
\n
Award [2] for correct final answer.
\n[2 marks]
\n% Mg(OH)2 «= × 100» = 18.0 «%»
\n\n
Answer must show three significant figures.
\n[1 mark]
\nGraphing is an important tool in the study of rates of chemical reactions.
\nExcess hydrochloric acid is added to lumps of calcium carbonate. The graph shows the volume of carbon dioxide gas produced over time.
\nSketch a Maxwell–Boltzmann distribution curve for a chemical reaction showing the activation energies with and without a catalyst.
\nSketch a curve on the graph to show the volume of gas produced over time if the same mass of crushed calcium carbonate is used instead of lumps. All other conditions remain constant.
\nState and explain the effect on the rate of reaction if ethanoic acid of the same concentration is used in place of hydrochloric acid.
\nOutline why pH is more widely used than [H+] for measuring relative acidity.
\nOutline why H3PO4/HPO42− is not a conjugate acid-base pair.
\n![\"M18/4/CHEMI/SP2/ENG/TZ2/02.a/M\"](\"../assets/uploads/tinymce_asset/asset/4613/Schermafbeelding_2018-08-10_om_08.59.39.png\"/)
both axes correctly labelled
\ncorrect shape of curve starting at origin
\nEa(catalyst) < Ea(without catalyst) on x-axis
\n\n
M1:
\nAccept “speed” for x-axis label.
\nAccept “number of particles”, “N”, “frequency” or “probability «density»” for y-axis label.
\nDo not accept “potential energy” for x-axis label.
\n\n
M2:
\nDo not accept a curve that touches the x-axis at high energy.
\nDo not award M2 if two curves are drawn.
\n\n
M3:
\nIgnore any shading under the curve.
\n[3 marks]
\n![\"M18/4/CHEMI/SP2/ENG/TZ2/02.b.i/M\"](\"../assets/uploads/tinymce_asset/asset/4614/Schermafbeelding_2018-08-10_om_09.03.30.png\"/)
curve starting from origin with steeper gradient AND reaching same maximum volume
\n[1 mark]
\nrate decreases
\nOR
\nslower reaction
\n\n
«ethanoic acid» partially dissociated/ionized «in solution/water»
\nOR
\nlower [H+]
\n\n
Accept “weak acid” or “higher pH”.
\n[2 marks]
\n«pH» converts «wide range of [H+]» into simple «log» scale/numbers
\nOR
\n«pH» avoids need for exponential/scientific notation
\nOR
\n«pH» converts small numbers into values «typically» between 0/1 and 14
\nOR
\n«pH» allows easy comparison of values of [H+]
\n\n
Accept “uses values between 0/1 and 14”.
\nDo not accept “easier to use”.
\nDo not accept “easier for calculations”.
\n[1 mark]
\n«species» do not differ by a «single» proton/H+
\nOR
\nconjugate base of H3PO4 is H2PO4– «not HPO42–»
\nOR
\nconjugate acid of HPO42– is H2PO4– «not H3PO4»
\n\n
Do not accept “hydrogen/H” for “H+/proton”.
\n[1 mark]
\nRhenium, Re, was the last element with a stable isotope to be isolated.
\nBefore its isolation, scientists predicted the existence of rhenium and some of its properties.
\nOne chloride of rhenium has the empirical formula ReCl3.
\nRhenium forms salts containing the perrhenate(VII) ion, ReO4−.
\nThe stable isotope of rhenium contains 110 neutrons.
\nState the nuclear symbol notation for this isotope.
\nSuggest the basis of these predictions.
\nA scientist wants to investigate the catalytic properties of a thin layer of rhenium metal on a graphite surface.
Describe an electrochemical process to produce a layer of rhenium on graphite.
\nPredict two other chemical properties you would expect rhenium to have, given its position in the periodic table.
\nDescribe how the relative reactivity of rhenium, compared to silver, zinc, and copper, can be established using pieces of rhenium and solutions of these metal sulfates.
\nState the name of this compound, applying IUPAC rules.
\nCalculate the percentage, by mass, of rhenium in ReCl3.
\nSuggest why the existence of salts containing an ion with this formula could be predicted. Refer to section 6 of the data booklet.
\nDeduce the coefficients required to complete the half-equation.
\nReO4− (aq) + ____H+ (aq) + ____e− ⇌ [Re(OH)2]2+ (aq) + ____H2O (l) Eθ = +0.36 V
\nPredict, giving a reason, whether the reduction of ReO4− to [Re(OH)2]2+ would oxidize Fe2+ to Fe3+ in aqueous solution. Use section 24 of the data booklet.
\n[✔]
\ngap in the periodic table
OR
element with atomic number «75» unknown
OR
break/irregularity in periodic trends [✔]
«periodic table shows» regular/periodic trends «in properties» [✔]
\nelectrolyze «a solution of /molten» rhenium salt/Ren+ [✔]
\ngraphite as cathode/negative electrode
OR
rhenium forms at cathode/negative electrode [✔]
\n
Note: Accept “using rhenium anode” for M1.
\nAny two of:
variable oxidation states [✔]
forms complex ions/compounds [✔]
\ncoloured compounds/ions [✔]
\n«para»magnetic compounds/ions [✔]
\n\n
Note: Accept other valid responses related to its chemical metallic properties.
\nDo not accept “catalytic properties”.
\nplace «pieces of» Re into each solution [✔]
\nif Re reacts/is coated with metal, that metal is less reactive «than Re» [✔]
\n\n
Note: Accept other valid observations such as “colour of solution fades” or “solid/metal appears” for “reacts”.
\nrhenium(III) chloride
OR
rhenium trichloride [✔]
«Mr ReCl3 = 186.21 + (3 × 35.45) =» 292.56 [✔]
«100 × =» 63.648 «%» [✔]
same group as Mn «which forms MnO4-»
OR
in group 7/has 7 valence electrons, so its «highest» oxidation state is +7 [✔]
ReO4− (aq) + 6H+ (aq) + 3e− ⇌ [Re(OH)2]2+ (aq) + 2H2O (l) [✔]
\nno AND ReO4− is a weaker oxidizing agent than Fe3+
OR
no AND Fe3+ is a stronger oxidizing agent than ReO4−
OR
no AND Fe2+ is a weaker reducing agent than [Re(OH)2]2+
OR
no AND [Re(OH)2]2+ is a stronger reducing agent than Fe2+
OR
no AND cell emf would be negative/–0.41 V [✔]
It was expected that this question would be answered correctly by all HL candidates. However, many confused the A-Z positions or calculated very unusual numbers for A, sometimes even with decimals.
\nThis is a NOS question which required some reflection on the full meaning of the periodic table and the wealth of information contained in it. But very few candidates understood that they were being asked to explain periodicity and the concept behind the periodic table, which they actually apply all the time. Some were able to explain the “gap” idea and other based predictions on properties of nearby elements instead of thinking of periodic trends. A fair number of students listed properties of transition metals in general.
\nGenerally well done; most described the cell identifying the two electrodes correctly and a few did mention the need for Re salt/ion electrolyte.
\nGenerally well answered though some students suggested physical properties rather than chemical ones.
\nMany candidates chose to set up voltaic cells and in other cases failed to explain the actual experimental set up of Re being placed in solutions of other metal salts or the reaction they could expect to see.
\nAlmost all candidates were able to name the compound according to IUPAC.
\nMost candidates were able to answer this stoichiometric question correctly.
\nThis should have been a relatively easy question but many candidates sometimes failed to see the connection with Mn or the amount of electrons in its outer shell.
\nSurprisingly, a great number of students were unable to balance this simple half-equation that was given to them to avoid difficulties in recall of reactants/products.
\nMany students understood that the oxidation of Fe2+ was not viable but were unable to explain why in terms of oxidizing and reducing power; many students simply gave numerical values for EΘ often failing to realise that the oxidation of Fe2+ would have the inverse sign to the reduction reaction.
\nAmino acids are the building blocks of proteins.
\nDraw the dipeptide represented by the formula Ala-Gly using section 33 of the data booklet.
\nDeduce the number of 1H NMR signals produced by the zwitterion form of alanine.
\nOutline why amino acids have high melting points.
\n![\"M18/4/CHEMI/SP3/ENG/TZ2/07.a/M\"](\"../assets/uploads/tinymce_asset/asset/4655/Schermafbeelding_2018-08-10_om_13.57.39.png\"/)
peptide bond
\norder of amino acids
\n\n
Accept zwitterion form of dipeptide.
\nAccept a condensed structural formula or a skeletal structure.
\nPenalize missing hydrogens or incorrect bond connectivities once only in Option B.
\n[2 marks]
\n3
\n[1 mark]
\nform zwitterions
\n\n
«strong» ionic bonding
\nOR
\n«strong» ionic lattice
\nOR
\n«strong» electrostatic attraction/forces
\n\n
Do not accept hydrogen bonding or IMFs for M2.
\n[2 mark]
\nCrude oil is a useful energy resource.
\nOutline two reasons why oil is one of the world’s significant energy sources.
\nFormulate an equation for the cracking of C16H34 into two products with eight carbon atoms each.
\nIdentify, giving a reason, which product in (b)(i) could be used in petrol (gasoline).
\nOutline how higher octane fuels help eliminate “knocking” in engines.
\nThe performance of hydrocarbons as fuels can be improved by catalytic reforming.
\nOutline how catalytic reforming increases a fuel’s octane rating.
\nAny two of:
\nhigh energy content/high energy density/high specific energy
\nOR
\nhigh enthalpy of combustion/very exothermic enthalpy of combustion
\n\n
shortage of alternatives
\nOR
\nalternatives are expensive
\nOR
\noil is relatively cheap
\nOR
\noil is «still» abundant/common
\n\n
well-established technology
\nOR
\neasy for consumers to obtain
\nOR
\ncommonly used
\n\n
easy to store
\nOR
\neasy to transport
\nOR
\neasy to extract
\n\n
produces energy at a reasonable rate
\n\n
Accept “high potential energy” for M1.
\n[2 marks]
\nC16H34(g) → C8H16(g) + C8H18(g)
\nOR
\nC16H34(g) + H2(g) → 2 C8H18(g)
\n[1 mark]
\nC8H18 AND is an alkane
\nOR
\nC8H18 AND petrol does not contain alkenes
\n\n
[1 mark]
\nfuels can be compressed more without undergoing «unwanted» auto-ignition
\n\n
Accept “burns smoother without undergoing «unwanted» auto-ignition” OR “fuel does not auto-ignite”.
\n[1 mark]
\nproduces more branched chain hydrocarbons «with higher octane rating»
\nOR
\nproduces aromatics «which have higher octane rating»
\nOR
\nproduces cyclohexanes «which have higher octane rating»
\n\n
\n
Accept “increase branches”.
\nDo not accept “produces benzene”.
\nDo not penalize for “benzene” if penalty applied in 2.b.iii.
\nAccept “produces cyclic structures”.
\n[1 mark]
\nNew genetically modified organisms (GMO), especially plants, are continually being developed in research laboratories.
\nOutline what is meant by genetically modified organisms.
\nOutline one benefit of the use of these products.
\norganism whose genetic material/DNA has been altered by genetic engineering techniques «involving transferring DNA between species» [✔]
\n\n
Note: Accept “any living organism that possesses a novel combination of genetic material obtained through the use of modern biotechnology”.
\nAny one of:
increased resistance to pests/micro-organisms [✔]
increased shelf-life of food [✔]
\nincreased nutritional value [✔]
\ngreater crop yield [✔]
\ngreater tolerance of crops to adverse climatic/soil/growing condition [✔]
\nCandidates had difficulty outline the meaning of genetically modified organisms with many simply repeating the question by stating modified DNA without including technology or engineering techniques.
\nOutlining one benefit of using GMOs was well answered.
\nGreen Chemistry reduces the production of hazardous materials and chemical waste.
\nOutline two specific examples or technological processes of how Green Chemistry has accomplished this environmental impact.
\nAny two of:
\nreplaces plastics with biodegradable/starch/cellulose based plastics
\n\n
use enzymes instead of polluting detergents/phosphates
\nOR
\nuse of enzymes means lower temperatures can be used
\nOR
\nuse enzymes instead of emulsifiers to treat oil spills
\nOR
\nuse enzymes to produce esters at lower temperatures/without sulfuric acid
\n\n
replace organic/toxic solvents with carbon dioxide
\nreplace polymers from fossil fuel with bamboo/renewable resources
\ndevelop paint resins reducing production of volatile compounds «when paint is applied»
\nindustrial synthesis of ethanoic/acetic acid from methanol and carbon monoxide has 100% atom economy
\nenergy recovery
\n\n
Accept formulas for names.
\nAward mark for any other reasonable specific green chemistry example that prevents the release of pollutants/toxic chemicals into the environment by changing the method or the materials used.
\nDo not award mark for methods that involve clean-up of pollutants from the environment such as host-guest chemistry or alternative energy sources.
\n[2 marks]
\nCarbon dioxide is a product of the combustion of petrol.
\nExplain the molecular mechanism by which carbon dioxide acts as a greenhouse gas.
\nDiscuss the significance of two greenhouse gases, other than carbon dioxide, in causing global warming or climate change.
\nAny three of:
\nIR/long wavelength/low frequency radiation radiated/emitted by the Earth’s «surface absorbed in the bonds»
\n\n
bond length/C=O changes
\nOR
\n«asymmetric» stretching of bonds
\nOR
\nbond angle/OCO changes
\n\n
polarity/dipole «moment» changes
\nOR
\ndipole «moment» created «when molecule absorbs IR»
\n\n
«some of» energy is then re-radiated towards «the surface of the» Earth
\n\n
Do not accept terms such as “reflect” OR “bounced” OR “trapped”.
\n[3 marks]
\nAny two of:
\nH2O AND «relatively» greater abundance/stable concentration/less effective at absorbing radiation/lower GWP so not much overall effect on global warming/climate change
\nCH4/N2O/CFCs/SF6/O3/HCFCs AND more effective «than CO2» at absorbing radiation/higher GWP so could contribute to global warming/climate change
\nPFCs/SF6/NF3/Some CFCs AND have very long life in atmosphere so could contribute «in the future» to global warming/climate change
\n\n
Accept names or formulas.
\nAccept two different gases with the same effect for [2].
\nAward [1 max] for identifying the names/formulas of two greenhouse gases.
\nAccept “greenhouse factor” for “GWP” but not just “greenhouse effect”.
\nFor M3, do not allow “CFC” alone as only some have long lifetimes (eg, CFC-115, CFC-113).
\n[2 marks]
\nNatural gas is an energy source composed mainly of methane.
\nNatural gas is burned to produce steam which turns turbines in an electricity generating power plant.
\nThe efficiency of several sources for power plants is given below.
\nCalculate the specific energy of methane, in MJ kg−1, using sections 1, 6 and 13 of the data booklet.
\nCalculate the maximum electric energy output, in MJ, which can be obtained from burning 1.00 kg of methane by using your answer from (a).
\nHydroelectric power plants produced 16% of the world’s energy in 2015, down from 21% in 1971.
\nSuggest why hydroelectric power production has a higher efficiency than the other sources given in (b) and why its relative use has decreased despite the high efficiency.
\nReason for higher efficiency:
\nReason for decreased use:
\nMethane can also be obtained by fractional distillation of crude oil.
\n[Source: Image used with kind permission of science-resources.co.uk]
\nDraw a circle on the diagram to show where the methane fraction is withdrawn.
\nList the following products, which are also obtained by fractional distillation, according to decreasing volatility: asphalt, diesel, gasoline, lubricating motor oil.
\nExplain how methane absorbs infrared (IR) radiation by referring to its molecular geometry and dipole moment.
\nCompare methane’s atmospheric abundance and greenhouse effect to that of carbon dioxide.
\n« =» 55.5 kJ g–1 =» 55.5 «MJ kg–1» ✔
\n«55.5 MJ × 58 % =» 32.2 «MJ» [✔]
\nReason for higher efficiency:
no heat/energy loss in producing steam
OR
no need to convert chemical energy of the fuel into heat and then heat into mechanical energy
OR
direct conversion of «gravitational» potential energy to mechanical energy [✔]
Note: Accept “less energy lost as heat” but do not accept “no energy lost”.
\n\n
Reason for decreased use:
limited supply of available hydroelectric sites
OR
rapid growth of electrical supply in countries with little hydroelectric potential
OR
not building «new hydroelectric» dams because of environmental concerns [✔]
Note: Accept “new/alternative/solar/wind power sources «have taken over some of the demand»”.
Accept “lower output from existing stations due to limited water supplies”.
\n [✔]
gasoline > diesel > lubricating motor oil > asphalt [✔]
\n\n
Note: Accept products written in this order whether separated by >, comma, or nothing.
\nmethane is tetrahedral
OR
methane has zero dipole moment/is non-polar/bond polarities cancel [✔]
Any two of:
IR absorption can result in increased vibrations/bending/stretching [✔]
only modes that cause change in dipole absorb IR [✔]
\nfor methane this is asymmetric bending/stretching [✔]
\nmethane is less abundant AND has a greater effect «per mol» [✔]
\nCalculations of specific energy of methane and the maximum electric energy output in parts (a) and (b)(i) were done well.
\nCalculations of specific energy of methane and the maximum electric energy output in parts (a) and (b)(i) were done well.
\nSuggesting reasons for hydroelectric power having higher efficiency but lower relative use than other energy sources in was not answered well by most candidates. Often the reasons for higher efficiency were given in vague terms that did not meet the detail required.
\nRequired candidates to circle a fractionating tower to show where the methane fraction could be withdrawn. Despite the expectation that candidates know methane is a gas at room temperature, there were many varied answers to this question.
\nRequired products of fractional distillation of crude oil to be ranked according to decreasing volatility. This should have been able to be worked out from first principles and did not have to be memorized as one G2 respondent suggested.
\nMany candidates scored the first mark for stating that methane is tetrahedral. Further details to explain how methane absorbs IR radiation were generally insufficient. Many candidates referred to “dipole movements” despite dipole moment being in the stem of the question.
\nMost candidates correctly answered comparing methane’s atmospheric abundance and greenhouse effect to that of carbon dioxide.
\nExplain the solubility of vitamins A and C using section 35 of the data booklet.
\nVitamin A:
\nfat soluble/soluble in non-polar solvents AND non-polar/long hydrocarbon backbone/chain
\n\n
Vitamin C:
\nwater soluble AND contains 4 hydroxyl groups/contains many hydroxyl groups/forms «many» H-bonds with water
\n\n
Accept “Vitamin A: fat soluble/soluble in non-polar solvents as it contains only one hydroxyl group whose H-bonds with water are not strong enough to overcome London/dispersion/vdW forces between Vitamin A molecules”.
\nAccept “lipid” for “fats”.
\nAccept “alcohol” OR “hydroxy” OR “OH groups” for “hydroxyl” but not “hydroxide”.
\nAward [1 max] for “Vitamin A: fat soluble AND Vitamin C: water soluble” with no or incomplete explanation.
\n[2 marks]
\nLactose, found in milk and dairy products, is a disaccharide formed from two different monosaccharides. The structure of lactose is shown with numbered carbon atoms.
\nStarch and cellulose are both formed from glucose and have similar structures.
\nName the type of link between the two monosaccharide residues.
\nOutline how the two monomer structures, galactose and glucose, differ.
\nOutline the difference between their structures.
\nOutline why cellulose is an essential part of human diet.
\n«1,4-»glycosidic ✔
\n\n
Do not accept “glucosidic”.
\nH and OH are reversed/in different positions on C-4 ✔
\n\n
C-4 must be specified.
\nDo not penalize if reference is made to H and OH above and below ring/in alpha and beta positions on C-4 incorrectly.
\nStarch: α«-glucose/links»
\nAND
\nCellulose: β«-glucose/links» ✔
\n\n
Accept “Starch: coiled/spiral structure OR cross-links AND Cellulose: uncoiled OR straight chains/linear polymer OR no/few cross-links”.
\nAny two of:
\nhelps food pass through intestine
\nOR
\nadds bulk/dietary fibre ✔
\n\n
reduces appetite
\nOR
\nhelps prevent obesity ✔
\n\n
prevents constipation
\nOR
\nreduces risk of hemorrhoids/diverticulosis/Crohn’s disease/irritable bowel syndrome/bowel cancer ✔
\nOne way of classifying materials is based on the type of bonding present.
\nCaprolactam reacts with water to form compound X, a monomer.
\nOne way of classifying materials is based on the type of bonding present.
\nOne reaction to convert cyclohexanone to caprolactam using concentrated sulfuric acid as a catalyst is shown.
\nOutline why this type of classification is not entirely satisfactory by using magnesium diboride, MgB2, as an example. Refer to sections 8 and 29 of the data booklet.
\nStructures of poly(methyl acrylate), PMA, and Bakelite® are shown.
\nSuggest, giving reasons, which is the thermoplastic polymer and which is the thermosetting polymer.
\nA zeolite is an alternative catalyst for this reaction.
\nExplain how zeolites act as selective catalysts.
\nState the names of the two terminal functional groups in X.
\nDeduce the repeating unit of the polymer of X.
\nRepeating units of several polymers are listed.
\nThe infrared (IR) spectrum of one of these polymers is shown.
\nDeduce, giving a reason, the name of this polymer and its Resin Identification Code (RIC), using sections 26 and 30 in the data booklet.
\n= 0.7 AND average = 1.7 ✔
\n\n
bonding between metallic and ionic
\nOR
\nmore than one type of bonding present
\nOR
\nbond type difficult to determine as close to several regions/several types/named bonding types «eg ionic and covalent etc.»
\nOR
\nbond is mostly covalent «based on % covalent scale on diagram»
\nOR
\nbond has « » 22% ionic character ✔
\n\n
Accept “EN” for “\".
\nAccept “bond is ionic but close to several regions/several types/other named bonding type(s) (eg covalent, metallic and covalent etc.)”.
\nDo not accept just “bond is ionic”.
\nAccept any value for % ionic character in range 15–24% or % covalent character in range 76–85%.
\nThermoplastic polymer:
\nPMA AND «weak» intermolecular/IMFs/London/dispersion/van der Walls/vdW/dipole-dipole forces «between layers/chains»
\nOR
\nPMA AND no/few cross-links «between layers/chains» ✔
\n\n
Thermosetting polymer:
\nBakelite® AND «strong» covalent bonds «between layers/chains»
\nOR
\nBakelite® AND extensive cross-links «between layers/chains» ✔
\n\n
Do not accept “hydrogen bonding” for M1.
\nAward [1 max] for correct reasons for both polymer classes even if named polymers are incorrectly classified.
\npores/cavities/channels/holes/cage-like structures «in zeolites» have specific shape/size ✔
\nonly reactants «with appropriate size/geometry» fit inside/go through/are activated/can react ✔
\namino AND carboxyl ✔
\n\n
Do not accept “carbonyl”, “hydroxyl”.
\n\n
Continuation bonds at NH and CO are required for mark.
\nIgnore any brackets and n.
\nName and reason:
\nPET/PETE AND peak for C=O «at 1700–1750 cm–1» ✔
\n\n
RIC:
\n1 ✔
\n\n
Accept “PET/PETE AND peak for C–O «at 1050–1410 cm–1»” for M1.
\nAccept “PET/PETE AND peak(s) for COO” for M1.
\nAccept name or abbreviation for polymer.
\nNo ECF for M2.
\nGenetic information is stored in DNA.
\nState the feature of DNA that determines the primary structure of proteins synthesised by a cell.
\nSuggest one concern about the use of genetically modified, GM, food.
\n«triplet» sequence/«specific» order of «nitrogenous» bases
\nOR
\ncodon ✔
\nAny one of:
\nlong-term «health» effects unknown ✔
\ncan cause allergic reactions ✔
\npossible transfer of genetic material to other/wild species ✔
\n\n
concern that power over farming is concentrated in hands of multinationals
\nOR
\ndependent on multinationals ✔
\n\n
labelling differences between countries «means informed choice not possible» ✔
\n\n
Accept “outcrossing”.
\nCarbonated water is produced when carbon dioxide is dissolved in water under pressure. The following equilibria are established.
\nEquilibrium (1) CO2 (g) ![\"\"](\"../assets/uploads/tinymce_asset/asset/5986/5.PNG\")
CO2 (aq)
Equilibrium (2) CO2 (aq) + H2O (l) H+ (aq) + HCO3− (aq)
Carbon dioxide acts as a weak acid.
\nSoda water has sodium hydrogencarbonate, NaHCO3, dissolved in the carbonated water.
\nDistinguish between a weak and strong acid.
\nWeak acid:
\nStrong acid:
\nThe hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
\nState the formula of its conjugate base.
\nWhen a bottle of carbonated water is opened, these equilibria are disturbed.
\nState, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).
\nAt 298 K the concentration of aqueous carbon dioxide in carbonated water is 0.200 mol dm−3 and the pKa for Equilibrium (2) is 6.36.
\nCalculate the pH of carbonated water.
\nIdentify the type of bonding in sodium hydrogencarbonate.
\nBetween sodium and hydrogencarbonate:
\nBetween hydrogen and oxygen in hydrogencarbonate:
\nPredict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)
\n100.0cm3 of soda water contains 3.0 × 10−2g NaHCO3.
\nCalculate the concentration of NaHCO3 in mol dm−3.
\nThe uncertainty of the 100.0cm3 volumetric flask used to make the solution was ±0.6cm3.
\nCalculate the maximum percentage uncertainty in the mass of NaHCO3 so that the concentration of the solution is correct to ±1.0 %.
\nThe reaction of the hydroxide ion with carbon dioxide and with the hydrogencarbonate ion can be represented by Equations 3 and 4.
\nEquation (3) OH− (aq) + CO2 (g) → HCO3− (aq)
Equation (4) OH− (aq) + HCO3− (aq) → H2O (l) + CO32− (aq)
Discuss how these equations show the difference between a Lewis base and a Brønsted–Lowry base.
\n\n
Equation (3):
\nEquation (4):
\nAqueous sodium hydrogencarbonate has a pH of approximately 7 at 298 K.
\nSketch a graph of pH against volume when 25.0cm3 of 0.100 mol dm−3 NaOH (aq) is gradually added to 10.0cm3 of 0.0500 mol dm−3 NaHCO3 (aq).
\nWeak acid: partially dissociated/ionized «in aqueous solution/water»
AND
Strong acid: «assumed to be almost» completely/100 % dissociated/ionized «in aqueous solution/water» [✔]
CO32- [✔]
\nshifts to left/reactants AND to increase amount/number of moles/molecules of gas/CO2 (g) [✔]
\n\n
Note: Accept “shifts to left/reactants AND to increase pressure”.
\n«Ka =» 10–6.36/4.37 × 10–7 =
OR
«Ka =» 10–6.36/4.37 × 10–7 = [✔]
\n
[H+] « » = 2.95 × 10–4 «mol dm–3» [✔]
«pH =» 3.53 [✔]
\n
Note: Award [3] for correct final answer.
\nBetween sodium and hydrogencarbonate:
ionic [✔]
Between hydrogen and oxygen in hydrogencarbonate:
«polar» covalent [✔]
«additional HCO3-» shifts position of equilibrium to left [✔]
\npH increases [✔]
\n\n
Note: Do not award M2 without any justification in terms of equilibrium shift in M1.
\n«molar mass of NaHCO3 =» 84.01 «g mol-1» [✔]
\n«concentration = =» 3.6 × 10–3 «mol dm-3» [✔]
\n\n
Note: Award [2] for correct final answer.
\n«1.0 – 0.6 = ± » 0.4 «%» [✔]
\nEquation (3):
OH- donates an electron pair AND acts as a Lewis base [✔]
Equation (4):
OH- accepts a proton/H+/hydrogen ion AND acts as a Brønsted–Lowry base [✔]
S-shaped curve from ~7 to between 12 and 14 [✔]
\nequivalence point at 5 cm3 [✔]
\n\n
Note: Accept starting point >6~7.
\nAs expected, many candidates were able to distinguish between strong and weak acids; some candidates referred to “dissolve” rather than dissociate.
\nMore than half the candidates were able to deduce that carbonate was the conjugate base but a significant proportion of those that did, wrote the carbonate ion with an incorrect charge.
\nMany students gave generic responses referring to a correct shift without conveying the idea of compensation or restoration of pressure or moles of gas. This generic reply reflects the difficulty in applying a theoretical concept to the practical situation described in the question.
\nMost candidates calculated the pH of the aqueous CO2. Some candidates attempted to use the Henderson-Hasselback equation and others used the quadratic expression to calculate [H+] (these two options were very common in the Spanish scripts) getting incorrect solutions. These answers usually ended in pH of approx. 1 which candidates should realize cannot be correct for soda water.
\nThis was an easy question, especially the identification of the type of bond between H and O, yet some candidates interpreted that the question referred to intermolecular bonding.
\nA significant number of candidates omitted the “equilibrium” involved in the dissolution of a weak base.
\nThis is another stoichiometry question that most candidates were able to solve well, with occasional errors when calculating Mr of hydrogen carbonate.
\nMixed responses, more attention should be given to this simple calculation which is straightforward and should be easy as required for IA reports.
\nThis was a good way to test this topic because answers showed that, while candidates usually knew the topic in theory, they could not apply this to identify the Lewis and Bronsted-Lowry bases in the context of a reaction that was given to them. In some cases, they failed to specify the base, OH- or also lost marks referring just to electrons, an electron or H instead of hydrogen ions or H+ for example.
\nMost students that got 1mark for this titration curve was for the general shape, because few realized they had the data to calculate the equivalence point. There were also some difficulties in establishing the starting point even if it was specified in the stem.
\nUranium-235, 235U, is bombarded with a neutron causing a fission reaction.
\nTwo products of the fission of 235U are 144Ba and 89Kr.
\nWrite the nuclear equation for this fission reaction.
\nOutline why the reaction releases energy.
\nThe masses of the particles involved in this fission reaction are shown below.
\nMass of neutron = 1.00867 amu
Mass of U-235 nucleus = 234.99346 amu
Mass of Ba-144 nucleus = 143.89223 amu
Mass of Kr-89 nucleus = 88.89788 amu
Determine the energy released, in J, when one uranium-235 nucleus undergoes fission. Use this data and information from sections 1 and 2 of the data booklet.
\nThe critical mass for weapons-grade uranium can be as small as 15 kg. Outline what is meant by critical mass by referring to the equation in (a)(i).
\nThe daughter product, 89Kr, has a half-life of 3.15 min.
\nCalculate the time required, in minutes, for its radioactivity to fall to 10% of its initial value, using section 1 of the data booklet.
\n235U + 1n → 144Ba + 89Kr + 31n [✔]
\ngreater binding energy per nucleon in products than reactants [✔]
\n\n
Note: Accept “mass of products less than mass of reactants” OR “mass converted to energy/E = mc2”.
\n«Δm mass of reactants-mass of products»
\nΔm = «234.99346 – 143.89223 – 88.89788 – (2 × 1.00867) =» 0.18601 «amu» [✔]
\nΔm = «0.18601 amu × 1.66 × 10−27 kg amu–1 =» 3.09 × 10–28 «kg» [✔]
\nE = «mc2 = 3.09 × 10–28 kg × (3.00 × 108 m s–1)2 =» 2.78 × 10–11 «J» [✔]
\n\n
Note: Award [3] for correct final answer.
\nmass/amount/quantity required so that «on average» each fission/reaction results in a further fission/reaction [✔]
\nat least one of the «3» neutrons produced must cause another reaction [✔]
\n\n
Note: Accept “minimum mass of nuclear fuel needed for the reaction to be selfsustaining”.
\n«min–1» [✔]
\n«min» [✔]
\n\n
Note: Award [2] for correct final answer.
\nRequired candidates to write a nuclear equation for a fission reaction with the question indicating that 235U was bombarded with a neutron to produce 144Ba and 89Kr. Despite this, some candidates did not include the initial neutron or omitted neutrons completely.
\nOutlining why the fission reaction releases energy was challenging. Some candidates simply said the reaction was exothermic. Some said that the products were smaller than the reactant. Very few candidates referred to binding energy per nucleon.
\nThe calculation of the energy released in was done reasonably well. Some candidates answered this very well and scored full marks. However, many scored 1 or 2 marks out of 3 through ECF marks. Common errors were incorrect calculation of mass in amu, or omission of converting amu to kg. Here, it was apparent that good setting out of calculations was effective in scoring for partially correct responses.
\nThe meaning of critical mass was answered reasonably well with most candidates scoring at least 1 out of 2.
\nThe calculation of time taken for radioactivity to fall to 10% of its initial value was answered very well.
\nThe transport of oxygen around the body is essential for life.
\nA graph showing saturation of oxygen against partial pressure of oxygen is shown.
\nExplain the shape of the graph from 0 to 50 % saturation.
\nExplain why carbon monoxide is very toxic and how it may be possible to treat carbon monoxide poisoning.
\nbinding of oxygen/O2 «to one active site» affects shape of Hb/other active sites
\nOR
\nbinding of one oxygen/O2 «molecule» affects binding of other oxygen/O2 «molecules» ✔
\n\n
increasing affinity of Hb to oxygen/O2
\nOR
\nenhanced binding of «further» oxygen/O2 «molecules»
\nOR
\ncooperative binding ✔
\nToxicity:
\ncarboxyhemoglobin/Hb–CO does not readily dissociate
\nOR
\nCO + Hb Hb–CO AND forward reaction favoured
\nOR
\naffinity of carbon monoxide/CO for hemoglobin is «200 times/much» higher than that of oxygen/O2
\nOR
\ncompetitive inhibitor of oxygen/O2 binding ✔
\n\n
Treatment:
\nmoving patient to fresh air
\nOR
\n«in severe cases» inhaling pure oxygen/O2
\nOR
\nhigh pressure oxygen/O2 chamber ✔
\n\n
Accept “move away from carbon monoxide/CO source” OR “remove carbon monoxide/CO source”.
\nPhenylethene can be polymerized to form polyphenylethene (polystyrene, PS).
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5990/6.PNG\")
The major product of the reaction with hydrogen bromide is C6H5–CHBr–CH3 and the minor product is C6H5–CH2–CH2Br.
\nDraw the repeating unit of polyphenylethene.
\nPhenylethene is manufactured from benzene and ethene in a two-stage process. The overall reaction can be represented as follows with ΔGθ = +10.0 kJ mol−1 at 298 K.
\n![\"\"](\"../assets/uploads/tinymce_asset/asset/5991/6b.PNG\")
Calculate the equilibrium constant for the overall conversion at 298 K, using section 1 of the data booklet.
\nThe benzene ring of phenylethene reacts with the nitronium ion, NO2+, and the C=C double bond reacts with hydrogen bromide, HBr.
\nCompare and contrast these two reactions in terms of their reaction mechanisms.
\n\n
Similarity:
\nDifference:
\nOutline why the major product, C6H5–CHBr–CH3, can exist in two forms and state the relationship between these forms.
\n\n
Two forms:
\nRelationship:
\nThe minor product, C6H5–CH2–CH2Br, can exist in different conformational forms (isomers).
\nOutline what this means.
\nThe minor product, C6H5–CH2–CH2Br, can be directly converted to an intermediate compound, X, which can then be directly converted to the acid C6H5–CH2–COOH.
\nC6H5–CH2–CH2Br → X → C6H5–CH2–COOH
\nIdentify X.
\n [✔]
\n
Note: Do not penalize the use of brackets and “n”.
\nDo not award the mark if the continuation bonds are missing.
\nln k «= » = –4.04 [✔]
\nk = 0.0176 [✔]
\n\n
Note: Award [2] for correct final answer.
\nSimilarity:
«both» involve an electrophile
OR
«both» electrophilic [✔]
\n
Difference:
first/reaction of ring/with NO2+ is substitution/S«E» AND second/reaction of C=C/with HBr is addition/A«E» [✔]
\n
Note: Answer must state which is substitution and which is addition for M2.
\nTwo forms:
chiral/asymmetric carbon
OR
carbon atom attached to 4 different groups [✔]
\n
Relationship:
mirror images
OR
enantiomers/optical isomers [✔]
\n
Note: Accept appropriate diagrams for either or both marking points.
\nbenzene ring «of the C6H5–CH2» and the bromine «on the CH2–Br» can take up different relative positions by rotating about the «C–C, σ–»bond [✔]
\n\n
Note: Accept “different parts of the molecule can rotate relative to each other”.
Accept “rotation around σ–bond”.
\nC6H5–CH2–CH2OH [✔]
\nMost candidates were able to draw the monomer correctly. Some candidates made careless mistakes writing C6H6.
\nAnother calculation which most candidates were able to work out, though some failed to convert ΔG given value in kJ mol-1 to J mol-1 or forgot the negative sign. Some used an inappropriate expression of R.
\nThe strong candidates were generally able to see the similarity between the two reactions but unexpectedly some could not identify “electrophilic” as a similarity even if they referred to the differences as electrophilic substitution/addition, so probably were unable to understand what was being asked.
\nCandidates were given the products of the addition reaction and asked about the major product. Perhaps they were put off by the term “forms” and thus failed to “see” the chiral C that allowed the existence of enantiomers. There was some confusion with the type of isomerism and some even suggested cis/trans isomers.
\nIf candidates seemed rather confused in the previous question, they seemed more so in this one. Most simply referred to isomers in general, not seeming to be slightly aware of what conformational isomerism is, even if it is in the curriculum.
\nQuite well answered though some candidates suggested an aldehyde rather than the alcohol, or forgot that C has two hydrogens apart from the -OH. In other cases, they left a Br there.
\nThe Sun’s energy is produced by the fusion of hydrogen nuclei.
\nUranium-238 produces plutonium-239, which is used as fuel in breeder reactors.
\nNuclear energy produces ionizing radiation which leads to the formation of free radicals.
\nExplain fusion reactions with reference to binding energy.
\nOutline why the term breeder is used for the reactors.
\nDeduce the fission reaction when 239Pu is bombarded with a neutron to produce 133Xe and 103Zr.
\nNuclear disasters release radioactive caesium into the atmosphere, which presents serious health risks.
\nCs-137 has a half-life of 30 years.
\nCalculate the percentage of Cs-137 remaining in the atmosphere after 240 years.
\nDeduce a Lewis (electron dot) structure of the superoxide, O2–, free radical.
\nExplain why free radicals are harmful to living cells.
\nsmall/lighter nuclei combine to form larger/heavier nuclei ✔
\nproduct has higher binding energy «per nucleon» ✔
\n\n
Accept binding energy curve with explanation.
\nconverts non-fissile «238U» material into fissile «239Pu» material
\nOR
\nproduces more fissile material than it consumes ✔
\n239Pu + 1n → 133Xe + 103Zr + 41n ✔
\n\n
Accept equation with correct atomic numbers included.
\nAccept notation for neutrons of “n”.
\nAccept a correctly described equation in words.
\nALTERNATIVE 1:
\n«» 8 /8 half-lives «required» ✔
\n% remaining = «0.508 × 100 =» 0.39 «%» ✔
\n\n
ALTERNATIVE 2:
\nλ = «» 0.023 ✔
\n% remaining = «100 × e–0.023 × 240 =» 0.39 «%» ✔
\n\n
Award [2] for correct final answer.
\nOR
\n\n
Accept any combination of dots, crosses and lines to represent electrons.
\nDo not penalize missing brackets.
\nPenalize missing negative charge.
\nhighly reactive
\nOR
\nstart redox reactions ✔
\n\n
damage/mutate DNA
\nOR
\ncause cancer
\nOR
\ndamage enzymes ✔
\nThe presence of very small amounts of lead in calcium-based antacids can be determined using inductively coupled plasma-mass spectroscopy (ICP-MS).
\nAn unknown antacid sample has a lead ion concentration of 0.50 μg dm‒3.
\nChelating agents can be used to treat heavy metal poisoning.
\nState the type of particle present in the plasma formed.
\nCalculate the concentration of lead ions in the sample in mol dm‒3.
\nLead ions are toxic and can be precipitated using hydroxide ions.
\nPb2+ (aq) + 2OH‒ (aq) Pb(OH)2 (s)
\nSufficient sodium hydroxide solid is added to the antacid sample to produce a 1.0 × 10‒2 mol dm‒3 hydroxide ion solution at 298 K.
\nDeduce if a precipitate will be formed, using section 32 of the data booklet.
\nIf you did not calculate the concentration of lead ions in (b)(i), use the value of 2.4 × 10−4 mol dm‒3, but this is not the correct value.
\nElectrolysis is used to obtain lead from Pb2+ (aq) solution.
\nDetermine the time, in hours, required to produce 0.0500 mol lead using a current (I) of 1.34 A. Use section 2 of the data booklet and the equation, charge (Q) = current (I) × time (t, in seconds).
\nState one feature of a chelating agent.
\nAn aqueous lead(II) ion reacts with three ethane-1,2-diamine molecules to form an octahedral chelate ion.
\nOutline why the chelate ion is more stable than the reactants.
\npositive ions/cations/Pb2+
\nOR
\nfree electrons ✔
\n\n
Accept “ions” OR “charged species/particle”.
\n[Pb2+] = 0.50 × 10‒6/5.0 × 10–7 «g dm–3» ✔
\n[Pb2+] «» = 2.4 × 10‒9 «mol dm‒3» ✔
\n\n
Award [2] for correct final answer.
\n«Ksp = 1.43 × 10–20»
\n\n
ALTERNATIVE 1:
\n«Q = [Pb2+] [OH–]2 = 2.4 × 10–9 × (1.0 × 10–2)2» = 2.4 × 10–13 ✔
\n\n
Q > Ksp AND precipitate will form
\nOR
\n2.4 × 10–13 > 1.43 × 10–20 AND precipitate will form ✔
\n\n
ALTERNATIVE 2:
\ncritical [Pb2+] for hydroxide solution «» = 1.4 × 10–16 ✔
\n\n
initial concentration > critical concentration AND precipitate will form
\nOR
\n2.4 × 10–9 > 1.4 × 10–16 AND precipitate will form ✔
\n\n
If value given is used:
\nALTERNATIVE 3:
\n«Q = [Pb2+] [OH–]2 = 2.4 × 10–4 × (1.0 × 10–2)2» = 2.4 × 10–8 ✔
\n\n
Q > Ksp AND precipitate will form
\nOR
\n2.4 × 10–8 > 1.43 × 10–20 AND precipitate will form ✔
\n«Faraday’s constant, F = 9.65 × 104 C mol‒1 and 1 A = 1 C s–1»
Q «= 0.0500 mol × 2 × 96500 C mol‒1» = 9650 «C» ✔
t « so » = 2.00 «hours» ✔
\n\n
Award [2] for correct final answer.
\nAny one of:
\ntwo «or more» lone/non-bonding pairs on different atoms
\nOR
\ntwo «or more» atoms/centres that act as Lewis bases ✔
\n\n
form «at least» two coordination/coordinate bonds
\nOR
\n«at least» two atoms can form coordination/coordinate bonds ✔
\n\n
Reference to “on DIFFERENT atoms” required.
\nAccept “dative «covalent» bond” for “coordination/coordinate bond”.
\nincrease in entropy
\nOR
\nΔS > 0/ΔS positive ✔
\n\n
Accept “ΔG < 0” but not “ΔH < 0”.
\nThe regular rise and fall of sea levels, known as tides, can be used to generate energy.
\nState one advantage, other than limiting greenhouse gas emissions, and one disadvantage of tidal power.
\nAdvantage:
\nDisadvantage:
\nAdvantage
Any one of:
renewable [✔]
predictable supply [✔]
tidal barrage may prevent flooding [✔]
effective at low speeds [✔]
long life-span [✔]
low cost to run [✔]
Disadvantage
Any one of:
cost of construction [✔]
changes/unknown effects on marine life [✔]
changes circulation of tides in the area [✔]
power output is variable [✔]
limited locations where feasible [✔]
equipment maintenance can be challenging [✔]
difficult to store energy [✔]
\n
Note: Do not accept vague generalisations.
\nDo not accept economic issues for both advantage and disadvantage.
\nDo not accept sustainable.
\nAccept “energy” or “electricity” for “power”.
\nMany candidates performed well on this question, especially when identifying an advantage of tidal power. The candidates who struggled tended to either give vague or journalistic answers especially for the disadvantage of tidal power.
\nThis question is about fuel for engines.
\nCrude oil can be converted into fuels by fractional distillation and cracking.
\nContrast these two processes.
\nDetermine the specific energy, in kJ g−1, and energy density, in kJ cm−3, of hexane, C6H14. Give both answers to three significant figures.
\nHexane: Mr = 86.2; ΔHc = −4163 kJ mol−1; density = 0.660 g cm−3
\nSpecific energy:
\nEnergy density:
\nHydrocarbons need treatment to increase their octane number to prevent pre-ignition (knocking) before they can be used in internal combustion engines.
\nDescribe how this is carried out and the molecular changes that take place.
\nNote: Award [1] for any two correct answers from one column OR one from each column.
\nAward [2] for any two correct from each column; eg: fractional distillation – any two correct award [1 max] AND
cracking – any two correct, award [1 max].
specific energy = « =» 48.3 «kJ g–1» [✔]
\nenergy density = «48.3 kJ g–1 × 0.660 g cm–3 =» 31.9 «kJ cm–3» [✔]
\n\n
Note: Award [1 max] if either or both answers not expressed to three significant figures.
\nAny two of:
«hydrocarbons are heated with» catalyst [✔]
long chains break and reform
OR
branching/aromatisation occurs
OR
isomerisation/reforming/platforming/cracking [✔]
zeolite separates branched from non-branched
OR
products are distilled
OR
«distillation» separates reformed and cracked products [✔]
\n
Note: Accept a specific catalyst name or formula for M1 such as Pt/Re/Rh/Pd/Ir.
\nThis part was not well answered. Many candidates didn’t answer the question as instructed. Candidates required two correct statements, either about fractional distillation or cracking as a process for one mark.
\nThis part was very well answered by most candidates with the correct number of significant digits as specified in the question.
\nCandidates responded well to at least one mark of this question. There were several different ways to earn the two marks possible. The most common way candidates earned marks were by identifying the use of a catalyst and then the idea of the compound reforming into a smaller or branched compound. Very few candidates discussed the idea of purification or separation into individual compounds, which is another important part of this process.
\nE10 is composed of 10% ethanol and 90% normal unleaded fuel.
\nEthanol has a Research Octane Number (RON) of 108.6.
\nOutline how higher octane fuels affect engine performance.
\nEthanol can be used in a direct-ethanol fuel cell (DEFC) as illustrated by the flow chart.
\nDeduce the half-equations occurring at electrodes A and B.
\nElectrode A:
\nElectrode B:
\nState the name and function of X in the diagram in (b)(i).
\n\n
Name:
\nFunction:
\nOutline why aqueous ethanol, rather than pure ethanol, is used in a DEFC.
\nBiodiesel containing ethanol can be made from renewable resources.
\nSuggest one environmental disadvantage of producing biodiesel from renewable resources.
\nincreased AND fuels can be compressed more «before ignition» [✔]
\n\n
Note: Accept “engines can be designed with higher compression ratio” OR “less chance of pre-ignition/auto-ignition/knocking occurring”.
\nElectrode A: C2H6O (aq) + 3H2O (l) → 12H+ (aq) + 12e– + 2CO2 (g) [✔]
\nElectrode B: 3O2 (g) + 12H+ (aq) + 12e– → 6H2O (l) [✔]
\n\n
Note: Accept balanced equations with integer or fractional coefficients.
\nPenalize equilibrium arrows once only.
\nName:
PEM/proton-exchange membrane/polymer exchange membrane/polymer electrolyte membrane [✔]
Function:
allows the passage of protons/H+ ions «from anode to cathode but not electrons or molecules» [✔]
Any one of:
water is a reactant/allows the cell to operate at a higher concentration of protons/H+ ions
OR
water is a stronger electrolyte and thus produces higher electric current [✔]
less dangerous/flammable [✔]
\nuse of «farm» land «for production»
OR
deforestation «for crop production for fuel»
OR
can release more NOx «than normal fuel on combustion» [✔]
\n
Note: Ignore any reference to cost.
\nOne G2 respondent was concerned that “Research Octane Number” was used instead of “Octane Rating”, but most candidates correctly outlined how higher octane fuels affect engine performance.
\nSome candidates did very well and deduced the half-equations occurring in a DEFC as illustrated in a flow chart. A few used equilibrium arrows and lost a mark. Some candidates failed to balance the half-equations.
\nWas answered well.
\nFew candidates could outline why aqueous ethanol is used in a DEFC.
\nMost could suggest one environmental disadvantage of producing biodiesel from renewable resources.
\nThis question is about nuclear reactions.
\nFission of a nucleus can be initiated by bombarding it with a neutron.
\nDetermine the other product of the fission reaction of plutonium-239.
\n\n
Outline the concept of critical mass with respect to fission reactions.
\nOutline one advantage of allowing all countries access to the technology to generate electricity by nuclear fission.
\nState one advantage of using fusion reactions rather than fission to generate electrical power.
\nOutline how the energy of a fission reaction can be calculated.
\nCalculate the half-life of an isotope whose mass falls from 5.0 × 10−5 g to 4.0 × 10−5 g in 31.4 s, using section 1 of the data booklet.
\n[✔]
\nminimum mass to «self-»sustain chain reaction
OR
if mass of fissile material is too small, too many neutrons produced pass out of the nuclear fuel
OR
at least one neutron produced causes further reaction [✔]
Any one of:
reduction in emission of greenhouse gases «from burning fossil fuels» [✔]
economic independence/self-sufficiency «from crude oil/producing states» [✔]
uranium is more abundant on Earth «in terms of total energy that can be produced from this fuel» than fossil fuels [✔]
\n\n
Note: Accept specific greenhouse gases (such as carbon dioxide/CO2) but not pollutants or other general statements.
\nAny one of:
fuel is inexpensive/readily available [✔]
no/less radioactive waste is formed [✔]
lower risk of accidents/large-scale disasters [✔]
impossible/harder to use for making materials for nuclear weapons [✔]
larger amounts of energy released per unit mass [✔]
does not require a critical mass [✔]
can be used continuously [✔]
\n
Note: Accept “higher specific energy for fusion”.
\nDo not accept “no/less waste produced for fusion”.
\nAccept specific example for a disaster.
\nmass difference between reactants and products AND E = mc2 [✔]
\n«N = N0e–λt»
\nλ«= »
\n= 7.106 × 10–3 s–1 [✔]
\n« =» 98/97.5 «s» [✔]
\n\n
Note: Award [2] for correct final answer.
\nThis part was well answered.
\nThis part was also fairly well answered although some candidates missed the concept of minimum mass to sustain a chain reaction.
\nThis part saw some reasonable answers, but some other candidates wrote very vague or general answers.
\nThis was a well-answered question with most candidates referring to fusion having less or no radioactive waste.
\nMost of the candidates were able to state correctly the mass difference between reactants and products and E = mc2.
\nMany candidates were able to calculate the half-life of an isotope correctly.
\nCoal can be converted to clean-burning synthetic natural gas.
\nAutomobile companies use hydrogen as an alternative to fossil fuels. Some properties of fuels are shown.
\nFormulate equation(s) for the conversion of coal and steam to methane.
\nComment on the specific energies of hydrogen and methane.
\nCalculate the mass, in kg, of carbon dioxide produced by the complete combustion of 72.0 dm3 octane, C8H18.
\nDensity of C8H18 = 703 g dm−3
\nC8H18 (l) + 12.5O2 (g) → 8CO2 (g) + 9H2O (g)
\nALTERNATIVE 1:
\n2C (s) + 2H2O (g) → CH4 (g) + CO2 (g) ✔
\n\n
ALTERNATIVE 2:
\nC (s) + 2H2O (g) → CO (g) + H2 (g) AND 3H2 (g) + CO (g) → CH4 (g) + H2O (g) ✔
\n\n
Accept “3C (s) + 2H2O (g) → CH4 (g) + 2CO (g)”.
\n«» hydrogen/H2 produces 2.6 times/more than twice the energy of methane/CH4 «per mass/g»
\nOR
\nless mass of hydrogen/H2 required «to produce same amount of energy»
\nOR
\nhydrogen/H2 more energy efficient ✔
\n\n
Accept “hydrogen/H2 produces «nearly» three times more energy than methane/CH4 «per mass/g»”.
\nmoctane «= 72.0 dm3 × 703 g dm–3» = 5.06 × 104 «g»/50.6 «kg» ✔
\nmcarbon dioxide «= » = 156 «kg» ✔
\n\n
Award [2] for correct final answer.
\nA representation of the unit cell of gold is shown.
\nState the name of the crystal structure of gold.
\nCalculate the number of atoms per unit cell of gold, showing your working.
\nThe edge length of the gold unit cell is 4.08 × 10‒8 cm.
\nDetermine the density of gold in g cm‒3, using sections 2 and 6 of the data booklet.
\nface-centred cube/fcc
\nOR
\ncubic close packed/ccp ✔
\n«atom per face» × 6 «faces per cube» × 3 «atoms» AND «atom per corner» × 8 «corners per cube» = 1 «atom» ✔
\n«atoms per unit cell = 3 + 1 =» 4 ✔
\n\n
Award [1 max] for “4” without working shown.
\n«4 atoms per unit cell»
\nmass of 4 atoms «» 1.31 × 10–21 «g»
\nvolume of unit cell «= (4.08 × 10‒8)3 cm3» = 6.79 × 10–23 «cm3»
\ndensity = «» = 1.93 × 101/19.3 «g cm‒3»
\n\n
Award [3] for correct final answer.
\n\n
Electricity can be generated in a variety of ways.
\nOutline how a microbial fuel cell produces an electric current from glucose.
\nC6H12O6 (aq) + 6O2 (g) → 6CO2 (g) + 6H2O (l)
\nThe cell potential for the spontaneous reaction when standard magnesium and silver half-cells are connected is +3.17 V.
\nDetermine the cell potential at 298 K when:
\n [Mg2+] = 0.0500 mol dm−3
[Ag+] = 0.100 mol dm−3
Use sections 1 and 2 of the data booklet.
\n\n
Outline one difference between a primary and a secondary cell.
\nAny three of:
C6H12O6 (aq) + 6H2O (l) → 6CO2 (g) + 24H+ (aq) + 24e–
OR
electrons released by oxidation of glucose [✔]
enzymes «in bacteria» oxidize glucose
OR
«bacteria» transfer «released» electrons directly to anode [✔]
24H+ (aq) + 6O2 (g) + 24e– → 12H2O (l)
OR
electrons consumed by reduction of oxygen [✔]
PEM/membrane separates two half reactions
OR
PEM/membrane allows proton/H+ transfer from anode to cathode [✔]
electrons flow though external circuit [✔]
\n\n
Note: Accept 4H+ (aq) + O2 (g) + 4e– → 2H2O (l).
\n«E = Eᶱ × lnQ»
\nlnQ = « » 1.61 [✔]
\nE = «3.17 V »3.15 «V» [✔]
\n\n
Note: Award [2] for correct final answer.
\nAny one of:
\nNote: Accept “primary cannot be recharged AND “secondary can be recharged”.
\nThe question on microbial cell invited varied responses. Half equations for the oxidation of glucose or reduction of oxygen were rarely written. PEM/membrane separates two half- reactions and allows proton transfer from anode to cathode was missed by most of the candidates.
\nThe cell potential was correctly calculated by several candidates with some candidates managed an ECF mark for an error in the calculation. Unfortunately, the ln Q part was frequently wrong due to candidates forgetting to square the denominator.
\nMost candidates were able to state one difference between a primary and a secondary cell.
\nLithium reacts with water to form an alkaline solution.
\nA 0.200 g piece of lithium was placed in 500.0 cm3 of water.
\nDetermine the coefficients that balance the equation for the reaction of lithium with water.
\nCalculate the molar concentration of the resulting solution of lithium hydroxide.
\nCalculate the volume of hydrogen gas produced, in cm3, if the temperature was 22.5 °C and the pressure was 103 kPa. Use sections 1 and 2 of the data booklet.
\nSuggest a reason why the volume of hydrogen gas collected was smaller than predicted.
\nThe reaction of lithium with water is a redox reaction. Identify the oxidizing agent in the reaction giving a reason.
\nDescribe two observations that indicate the reaction of lithium with water is exothermic.
\n2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 (g) ✔
\n✔
\n«nLiOH = nLi»
\n✔
\n\n
Award [2] for correct final answer.
\n✔
\n✔
\n\n
Award [2] for correct final answer.
\nAccept answers in the range 334 – 344 cm3.
\nAward [1 max] for 0.343 «cm3/dm3/m3».
\nAward [1 max] for 26.1 cm3 obtained by using 22.5 K.
\nAward [1 max] for 687 cm3 obtained by using 0.0288 mol.
\nlithium was impure/«partially» oxidized
\nOR
\ngas leaked/ignited ✔
\n\n
Accept “gas dissolved”.
\nH2O AND hydrogen gains electrons «to form H2»
\nOR
\nH2O AND H oxidation state changed from +1 to 0 ✔
\n\n
Accept “H2O AND H/H2O is reduced”.
\nAny two:
\ntemperature of the water increases ✔
\nlithium melts ✔
\npop sound is heard ✔
\n\n
Accept “lithium/hydrogen catches fire”.
\nDo not accept “smoke is observed”.
\nThis part-question was better answered than part (ii). 50% of the candidates drew a correct arrow between n=2 and n=3. Both absorption and emission transitions were accepted since the question did not specify which type of spectrum was required. Some teachers commented on this in their feedback. Mistakes often included transitions between higher energy levels.
\nThis question is about solar cells.
\nSome solar cells use photovoltaic semi-conductors. Compare, giving reasons, the electrical conductivity of metals and semi-conductors as temperature increases.
\nSuggest one advantage of a dye-sensitized solar cell (DSSC) over a silicon based photovoltaic cell.
\nmetal conductivity decreases AND semi-conductor conductivity increases [✔]
\nmetal: collisions between «already free moving» electrons/vibrating lattice ions and electrons increase [✔]
\nsemi-conductor:
provides sufficient energy for electrons to move to conduction band
OR
allows semiconductors to ionize forming freely moving electrons [✔]
Any one of:
\ncheaper [✔]
\nuses light of lower energy [✔]
\nplentiful resources [✔]
\nrenewable resources [✔]
\nuse of nanoparticles provides large surface area exposure to sunlight [✔]
\ncan absorb better under cloudy conditions [✔]
\nbetter conductivity [✔]
\nmore flexible [✔]
\nCandidates struggled to express themselves adequately in comparing the change in electrical conductivity of metals and semi-conductors as temperature increases. Many scored 1 mark for stating the effect of increasing temperature, but very few scored any marks for the explanation.
\nGenerally answered well with most candidates able to suggest one advantage of a DSSC over a silicon based photovoltaic cell.
\nWhich conditions are required for the reaction between two molecules?
\n I. a collision
II. E ≥ Ea
III. proper orientation
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nThis question on collision theory was one of the best answered in the exam.
\nPhysical properties of elements vary according to atomic number. Sections 6 to 9 of the data
booklet list some of these properties.
Melting points and boiling points of elements 1 to 95
\nDeduce, giving a reason, the group of elements in the periodic table most likely to undergo sublimation.
\nDescribe the density trend across periods 4 and 5 of the periodic table.
\nSuggest, with a reason, whether the lanthanoids or actinoids of the f-block would have the higher density.
\nCompare the ease of oxidation of s-block and d-block metals to their melting points and densities. Use section 25 of the data booklet.
\nSketch how the first ionization energies of elements vary with their atomic radius.
\ngroup 18/noble gases [✔]
\n\n
smallest difference between melting and boiling points
OR
weakest intermolecular forces «in that period» [✔]
\n
Note: Accept “group 17/halogens”.
\ndensity increases «to a maximum in the transition elements» AND then decreases [✔]
\nactinoids AND density increases down all groups «due to large increase in atomic mass for small increase in atomic volume»
OR
actinoids AND «much» greater atomic mass with similar type of bonding
OR
actinoids AND density «of actinoids» atomic number 90 to 95 is greater than corresponding lanthanoids [✔]
\n
Note: Accept “actinoids AND on graph actinoids have «much» greater density than lanthanoids”.
\nAlternative 1:
«metals with» low densities oxidize easier [✔]
«metals with» low melting points oxidize easier [✔]
\nAlternative 2:
in s-block «metals with» high densities oxidize easier
OR
in s-block «metals with» low melting points oxidize easier [✔]
in d-block «metals with» low densities oxidize easier
OR
in d-block «metals with» low melting points oxidize easier [✔]
\n
Note: Award [1 max] for “s-block metals more easily oxidized” OR “s-block metals have lower melting points” OR “s-block metals have lower densities”.
\nAccept “have greater activity” for “oxidize easier”.
\n [✔]
\n
Note: Accept any negative sloping line.
\nDo not award mark if line touches either axis.
\nMost candidates correctly identified the group of elements most likely to undergo sublimation but did not score for the reason as they referred to low melting and boiling points, rather than the smallest difference between these temperatures. There were several G2 comments that “the group of elements” was a confusing requirement as elements could be grouped in many ways, including for instance, from B to Ne. The Chemistry Guide clearly states that a group on the periodic table refers to a vertical column of elements. A few complaints were received about the inclusion of a question on sublimation, but the question was designed to make candidates think, and did not require knowledge of phase diagrams.
\nRequired candidates to consider density trends. Most candidates correctly described trends across periods 4 and 5 but had difficulty predicting and explaining whether lanthanoids or actinoids would have the higher density. Many said that actinoids would have higher density because they have more protons and neutrons / greater atomic number / greater mass with no further detail about having similar bonding and hence similar volume. Some G2 comments complained about the inclusion of lanthanoids and actinoids in this question. However, the Chemistry Guide clearly states that these terms should be known.
\nMost candidates scored at least 1 mark for comparing the s-block and d-block metals and most drew a line with a negative slope.
\nAlthough many here failed to score because the line crossed or touched one of the axes. A few sketched a graph reminiscent of first ionization energy against atomic number.
\nAspirin can be obtained from salicylic acid.
\nAdditional information can be obtained from the 1H NMR spectrum of aspirin.
\nUnreacted salicylic acid may be present as an impurity in aspirin and can be detected in the infrared (IR) spectrum.
\nName the functional group and identify the absorption band that diff erentiates salicylic acid from aspirin. Use section 26 of the data booklet.
\nName:
\nAbsorption band:
\nDeduce the protons responsible for signals X and Y by marking them on the structure of aspirin in (a). Use section 27 of the data booklet.
\nIdentify the splitting pattern of signals X and Y.
\n\n
X:
\nY:
\nName:
hydroxyl [✔]
Absorption band:
3200–3600 «cm–1» [✔]
\n
Note: Accept “phenol” OR “alcohol” but not “hydroxide”.
\ncorrect X [✔]
correct Y [✔]
\n
Note: X and Y must be near the Hs.
\nX: singlet AND Y: singlet [✔]
\nMany candidates scored a mark for naming a functional group that differentiates salicylic acid from aspirin. Some incorrectly said ether or carboxylic acid. Many candidates also scored for identifying the absorption band although 1700-1750 was a popular incorrect answer.
\nThere was considerable confusion with indicating protons responsible for 1H NMR signals. Often entire functional groups were circled or carbon atoms and not hydrogen atoms were circled.
\nThere was also great difficulty in identifying the splitting pattern of the signals. It was rare to see both signals identified as singlets.
\nThis question is about a mug made of a lead alloy.
\nThe rate of lead dissolving in common beverages with various pH values was analysed.
\nIdentify the experiment with the highest rate of lead dissolving.
\nSuggest why the relationship between time and lead concentration for Cola at 16 °C is not linear.
\nExamine, giving a reason, whether the rate of lead dissolving increases with acidity at 18 °C.
\nLead(II) chloride, PbCl2, has very low solubility in water.
\nPbCl2 (s) Pb2+ (aq) + 2Cl− (aq)
\nExplain why the presence of chloride ions in beverages affects lead concentrations.
\nA mean daily lead intake of greater than 5.0 × 10−6 g per kg of body weight results in increased lead levels in the body.
\nCalculate the volume, in dm3, of tap water from experiment 8 which would exceed this daily lead intake for an 80.0 kg man.
\n6 [✔]
\n\n
Note: Accept “orange juice”.
\nequilibrium is being established «between lead in solution and in mug»
OR
solution becoming saturated
OR
concentration of lead ions/[Pb2+] in the solution has increased «over time»
OR
acid concentration has decreased «as reacted with lead»
OR
surface lead has decreased/formed a compound/forms insoluble layer on surface
OR
acid reacts with other metals «because it is an alloy» [✔]
\n
Note: Do not accept “concentration of cola, orange juice, etc… has decreased”
\nDo not accept a response that only discusses mathematical or proportional relationships.
\nno AND experiment 7/beer has lowest rate and intermediate acidity/pH
OR
no AND experiment 6/orange juice has fastest rate but lower acidity/higher pH than lemonade
OR
no AND experiment 6/orange juice has highest rate and intermediate acidity/pH [✔]
\n
Note: Accept no AND any comparison, with experimental support, that concludes no pattern/increase with acidity
\neg: “rate of Pb/lead dissolving generally decreases with acidity as tap water has highest rate (after orange juice) while lemonade (lower pH) has lower rate”.
\nequilibrium shifts to the left/towards reactants [✔]
\nlead «compounds/ions» precipitate
OR
concentration of lead «ions»/[Pb2+] decreases [✔]
\n
Note: Award [2] for “equilibrium shifts to the left/towards reactants due to common ion effect”.
\nAccept “lead ions/[Pb2+] removed from solution” for M2.
\n«daily limit = 5.0 × 10–6 g kg–1 × 80.0 kg =» 4.0 × 10–4 «g of lead» [✔]
\n«volume » 2.7 × 10–2/0.027 «dm3» [✔]
\n\n
Note: Award [2] for correct final answer
\nThis part was correctly answered by the majority of the candidates.
\nA surprising number of candidates gave evidence for the non-linearity but then did not go on to explain why, giving no reasons or causes rooted in chemical theory. The command term \"suggest\" involves proposing a solution or hypothesis. Here the instruction \"suggest why\" indicates that the reason has to be explained.
\nThe candidates who examined the data and quoted it in their answer generally scored the mark but several candidates did not refer to the data table.
\nSeveral candidates missed that this question was based on the equilibrium and it will shift to the left in presence of chloride ions.
\nMajority of the candidates scored two marks but some struggled with the conversion of grams to milligrams.
\nWhich graph is obtained from a first order reaction?
\nB
\nMore than 78 % of candidates correctly identified a first order reaction graph.
\nThis question is about biofuel.
\nEvaluate the use of biodiesel in place of diesel from crude oil.
\nStrength
Any one of:
less flammable «than diesel» [✔]
recycles carbon «lower carbon footprint»
OR
lower greenhouse gas emissions [✔]
easily biodegradable «in case of spill» [✔]
\nrenewable
OR
does not deplete fossil fuel reserves [✔]
economic security/availability in countries without crude oil [✔]
\nLimitation
Any one of:
more difficult to ignite inside the engine «than diesel» [✔]
more viscous «than diesel» [✔]
\nlower energy content/specific energy/energy density [✔]
\nuses food sources
OR
uses land that could be used for food [✔]
«production is» more expensive [✔]
\nless suitable in low temperatures [✔]
\nincreased NOx emissions for biodiesel [✔]
\ngreenhouse gases still produced [✔]
\n\n
Note: Accept “«close to» carbon neutral”, “produce less greenhouse gases/CO2”.
\nAccept “engines have to be modified if biodiesel used” as limitation.
\nDo not award marks for strength and limitation that are the same topic/concept.
\nThis question was well answered, and many candidates received either one or both marks.
\nSolar energy, which is freely available, is indispensable to life on earth.
\nSuggest another advantage and one disadvantage of solar energy.
\nState a physical property of vegetable oils that makes them very difficult to use as fuel in internal combustion engines.
\nDescribe how vegetable oils can be converted to a more suitable fuel.
\nContrast the importance of carbon dioxide and methane as greenhouse gases.
\nExplain, using an equation, the effect of increased carbon dioxide in the atmosphere on the pH of lake water.
\nAdvantage:
\nrenewable «energy source»
\nOR
\ndoes not produce greenhouse gases
\nOR
\ncan be installed «almost» anywhere
\nOR
\nlow maintenance costs ✔
\n\n
Disadvantage:
\nwidely dispersed/not concentrated «form of energy»
\nOR
\ngeography/weather/seasonal dependent
\nOR
\nnot available at night
\nOR
\nenergy storage is difficult/expensive
\nOR
\ntoxic/hazardous materials used in production
\nOR
\nconcerns about space/aesthetics/local environment where installed
\nOR
\nneed to be «constantly» cleaned ✔
\n\n
Accept “can be used for passive/active heating”, “can be converted to electric energy”.
\nAccept any specific greenhouse gas name or formula for “greenhouse gases”.
\nAccept “solar cells require large areas”, “solar cell manufacture produces pollution/greenhouse gases”, “higher cost of solar cells «compared with traditional sources such as fossil fuels or hydroelectric»”.
\nhigh viscosity ✔
\n\n
Accept “low volatility”, just “viscous/viscosity” OR “does not flow easily”.
\nconvert to esters of monoatomic alcohols
\nOR
\nreact with short-chain alcohols «in the presence of acid or base» ✔
\n\n
Accept “convert to shorter «carbon chain» esters” OR “transesterification”.
\nAccept specific alcohols, such as methanol or ethanol.
\ncarbon dioxide/CO2 more/most abundant «GHG than methane/CH4»
\nOR
\ncarbon dioxide/CO2 has «much» longer atmospheric life «than methane/CH4» ✔
\n\n
methane/CH4 «much» better/more effective at absorbing IR radiation «than carbon dioxide/CO2»
\nOR
\nmethane/CH4 has a greater greenhouse factor «than carbon dioxide/CO2»
\nOR
\nmethane/CH4 has a greater global warming potential/GWP «than carbon dioxide/CO2» ✔
\n\n
Accept “carbon dioxide/CO2 contributes more to global warming «than methane/CH4»”.
\nCO2 (g) + H2O (l) H+ (aq) + HCO3– (aq)
\nOR
\nCO2 (g) CO2 (aq) AND CO2 (aq) + H2O (l) H+ (aq) + HCO3– (aq) ✔
\n«increasing [CO2 (g)]» shifts equilibrium/reaction to right AND pH decreases ✔
\n\n
Accept “H2CO3 (aq)” for “CO2 (aq) + H2O (l)”.
\nEquilibrium arrows required for M1.
\nState symbols required for CO2 (g) CO2 (aq) equation only for M1.
\nAccept “concentration of H+/[H+] increases AND pH decreases” for M2.
\nElectrons are arranged in energy levels around the nucleus of an atom.
\nThe diagram represents possible electron energy levels in a hydrogen atom.
\nExplain why the first ionization energy of calcium is greater than that of potassium.
\nAll models have limitations. Suggest two limitations to this model of the electron energy levels.
\nDraw an arrow, labelled X, to represent the electron transition for the ionization of a hydrogen atom in the ground state.
\nDraw an arrow, labelled Z, to represent the lowest energy electron transition in the visible spectrum.
\nincreasing number of protons/nuclear charge/Zeff ✔
\n
«atomic» radius/size decreases
OR
same number of energy levels
OR
similar shielding «by inner electrons» ✔
Any two of:
\ndoes not represent sub-levels/orbitals ✔
\nonly applies to atoms with one electron/hydrogen ✔
\ndoes not explain why only certain energy levels are allowed ✔
\nthe atom is considered to be isolated ✔
\ndoes not take into account the interactions between atoms/molecules/external fields ✔
\ndoes not consider the number of electrons the energy level can fit ✔
\ndoes not consider probability of finding electron at different positions/OWTTE ✔
\n\n
Do not accept “does not represent distance «from nucleus»”.
\nupward arrow X AND starting at n = 1 extending to n = ∞ ✔
\ndownward or upward arrow between n = 3 and n = 2 ✔
\nIt was surprising that this question that appears regularly in IB chemistry papers was not better answered. Many candidates only obtained one of the two marks for identifying one factor (often the larger nuclear charge of calcium or that the number of shells was the same for Ca and K). However, a few candidates did write thorough answers reflecting a good understanding of the factors affecting ionization energy. This question had a strong correlation between candidates who scored well and those who had a high score overall. Some candidates did not score any marks by focusing on trends in the Periodic Table without offering an explanation, or by discussing the number of electrons in Ca and K instead of the number of protons.
\nOnly 30% of the candidates drew the correct arrow on the diagram representing the ionization of hydrogen. A few candidates missed the mark by having the arrow pointing downwards. The most common incorrect answer was a transition between n=1 and n=2.
\nThis question is about global warming.
\nDescribe the effect of infrared (IR) radiation on carbon dioxide molecules.
\nOutline one approach to controlling industrial emissions of carbon dioxide.
\nbond length/C=O distance changes
OR
«asymmetric» stretching «of bonds»
OR
bond angle/OCO changes [✔]
polarity/dipole «moment» changes
OR
dipole «moment» created «when molecule absorbs IR» [✔]
\n
Note: Accept appropriate diagrams.
\nAny one of:
capture where produced «and store» [✔]
use scrubbers to remove [✔]
\nuse as feedstock for synthesising other chemicals [✔]
\ncarbon credit/tax/economic incentive/fines/country specific action [✔]
\nuse alternative energy
OR
stop/reduce use of fossil fuels for producing energy [✔]
use carbon reduced fuels «such as methane» [✔]
\nincrease efficiency and reduce energy use [✔]
\n\n
Note: Do not accept “planting more trees”.
\nAccept specific correct examples.
\nThis part was fairly well answered with most candidates receiving one of the two marks. There were many candidates who stated asymmetric stretching and bonds vibrate but missed writing polarity and dipole changes, which deprived them of the second mark.
\nThis part was reasonably answered although there were many candidates who gave vague answers that did not receive marks.
\nSemiconductors and light-sensitive dyes are used in photovoltaic cells.
\nSketch graphs to show the general effect of increasing temperature on the electrical conductivity of semiconductors and metals on the axes below.
\nExplain the function of dyes in a dye-sensitized solar cell (DSSC).
\nSemiconductors:
increases [✔]
Metals:
decreases [✔]
\n
Note: Accept any graph showing general increase for semiconductor.
\nAccept any graph showing general decrease for metal.
\nAccept a graph showing vertical section below transition temperature for a superconducting metal.
\ndye absorbs light [✔]
\nelectrons from «excited» dye pass to TiO2/semiconductor/electrolyte/cell
OR
dye undergoes photo-oxidation [✔]
Several candidates managed one mark to show conductivity of semiconductors on increasing the temperature but were unable to show that generally conductivity decreases for metals when the temperature is increased.
\nThe question on dye-sensitized solar cell invited mixed responses. While most candidates correctly stated that dyes absorb light but several failed to mention that electrons from the excited dye pass to TiO2/semiconductor.
\nStaphylococcus aureus (S. aureus) infections have been successfully treated with penicillin and penicillin derivatives.
\nIdentify the feature in penicillin responsible for its antibiotic activity.
\nThe widespread use of penicillin and its derivatives has led to the appearance of resistant S. aureus strains.
\nOutline how these bacteria inactivate the antibiotics.
\nOutline how the structure of penicillin has been modified to overcome this resistance.
\n«four-membered» beta-lactam ring [✔]
\n\n
Note: Accept a diagram showing a structural representation of the beta-lactam ring.
\nproduce penicillinase/enzyme that deactivates penicillin [✔]
\nside-chain changed «preserving beta-lactam ring» [✔]
\n\n
Note: Accept “R group changed”.
\nMost candidates correctly identified the feature in penicillin responsible for its antibiotic activity.
\nMost candidates could outline how bacteria inactivate the antibiotics.
\nOutlining how the structure of penicillin has been modified was less well answered, with many candidates referring to functional groups rather than the side chain or R group.
\nSulfur trioxide is produced from sulfur dioxide.
\n2SO2 (g) + O2 (g) 2SO3 (g) ΔH = −196 kJ mol−1
\nThe reaction between sulfur dioxide and oxygen can be carried out at different temperatures.
\nNitric acid, HNO3, is another strong Brønsted–Lowry acid. Its conjugate base is the nitrate ion, NO3−
\nOutline, giving a reason, the effect of a catalyst on a reaction.
\nOn the axes, sketch Maxwell–Boltzmann energy distribution curves for the reacting species at two temperatures T1 and T2, where T2 > T1.
\nExplain the effect of increasing temperature on the yield of SO3.
\nState the product formed from the reaction of SO3 with water.
\nState the meaning of a strong Brønsted–Lowry acid.
\nDraw the Lewis structure of NO3−.
\nExplain the electron domain geometry of NO3−.
\nincreases rate AND lower Ea ✔
\nprovides alternative pathway «with lower Ea»
OR
more/larger fraction of molecules have the «lower» Ea ✔
\n
Accept description of how catalyst lowers Ea for M2 (e.g. “reactants adsorb on surface «of catalyst»”, “reactant bonds weaken «when adsorbed»”, “helps favorable orientation of molecules”).
\nboth axes correctly labelled ✔
\npeak of T2 curve lower AND to the right of T1 curve ✔
\nlines begin at origin AND correct shape of curves AND T2 must finish above T1 ✔
\n\n
Accept “probability «density» / number of particles / N / fraction” on y-axis.
\nAccept “kinetic E/KE/Ek” but not just “Energy/E” on x-axis.
\ndecrease AND equilibrium shifts left / favours reverse reaction ✔
\n«forward reaction is» exothermic / ΔH is negative ✔
\nsulfuric acid/H2SO4 ✔
\n\n
Accept “disulfuric acid/H2S2O7”.
\nfully ionizes/dissociates ✔
\nproton/H+ «donor » ✔
\n\n
Do not accept the delocalised structure.
\nAccept any combination of dots, crosses and lines.
\nCoordinate/dative bond may be represented by an arrow.
\nthree electron domains repel
\nOR
\nthree electron domains as far away as possible ✔
\n\n
trigonal planar
\nOR
\n«all» angles are 120° ✔
\nA generally well-answered question. Most candidates explained the effect of a catalyst on a reaction correctly. A small proportion of candidates thought the catalyst increased the frequency of collisions. Some candidates focussed on the effect of the catalyst on an equilibrium since the equation above the question was that of a reversible reaction. These candidates usually still managed to gain at least the first marking point by stating that both forward and reverse reaction rates were increased due to the lower activation energy. Most candidates mentioned the alternative pathway for the second mark, and some gave a good discussion about the increase in the number of molecules or collisions with E≥Ea. A few candidates lost one of the marks for not explicitly stating the effect of a catalyst (that it increases the rate of the reaction).
\nThe average mark scored for the Maxwell-Boltzmann distribution curves sketch was 1.5 out of 3 marks and the question had a strong correlation with the candidates who did well overall. The majority of candidates were familiar with the shapes of the curves. The most commonly lost mark was missing or incorrect labels on the axes. Sometimes candidates added the labels but did not specify “kinetic” energy for the x-axis. As for the curves, some candidates reversed the labels T1 and T2, some made the two curves meet at high energy or even cross, and some did not have the correct relationship between the peaks of T1 and T2.
\nAnother question that showed a strong correlation with the candidates who did well overall. The average mark was 1 out of 2 marks. Many candidates explained the effect of an increase in temperature on the yield of SO3 correctly and thoroughly. One of the common mistakes was to miss the fact that it was an equilibrium and reason that yield would not change due to an increase in the rate of reaction. Unfortunately, a number of candidates also deduced that yield would increase due to the increase in rate. Other candidates recognized that it was an exothermic reaction but deduced the equilibrium would shift to the right giving a higher yield of SO3.
\nA very well answered question. 70% of the candidates stated H2SO4 as the product from the reaction of SO3 with water.
\nWhile a straightforward question, many candidates only answered part of the question - either focussing on the “strong” or on the “Brønsted-Lowry acid”. The average mark on this question was 1.2 out of 2 marks.
\nOnly 20% of the candidates scored the mark for the Lewis structure of NO3-. Mistakes included: missing charge, missing lone pairs, 3 single bonds, 2 double bonds.
\nThe majority of candidates deduced the correct electron domain geometry scoring the first mark including cases of ECF. Only a small number of candidates satisfied the requirements of the markscheme for the explanation.
\nMedicines and drugs are tested for effectiveness and safety.
\nDistinguish between therapeutic window and therapeutic index in humans.
\nTherapeutic window:
\nTherapeutic index:
\nExplain why diamorphine (heroin) is more potent than morphine using section 37 of the data booklet.
\nTherapeutic window:
range of dosage «over which a drug» provides the therapeutic/desired effect without causing adverse/toxic effects [✔]
Therapeutic index:
toxic dose of drug for 50 % of population divided by minimum effective dose for 50 % of population
OR
[✔]
\n
Note: M1 may be scored from a correctly labelled diagram.
\nDo not accept reference to lethal dose used in therapeutic index in animal studies.
\nmorphine has «two» hydroxyl groups AND diamorphine has «two» ester/ethanoate/acetate groups
OR
molecule of diamorphine is less polar than morphine
OR
groups in morphine are replaced with less polar/non-polar groups in diamorphine [✔]
«less polar molecules» cross the blood–brain barrier faster/more easily
OR
diamorphine is more soluble in non-polar environment of CNS/central nervous system than morphine [✔]
\n
Note: Accept “alcohol/hydroxy” for “hydroxyl” but not “hydroxide”.
\nAccept “fats” for “lipid”.
\nAccept “heroin” for “diamorphine”.
\nMost candidates receive one mark for this question, mainly for the therapeutic window. Some candidates inverted the ratio as ED50/TD50 for therapeutic index.
\nThis part was reasonably well answered with some very good answers.
\nChemical energy from redox reactions can be used as a source of electrical energy.
\nThe chemical structure of a photosensitive dye found in blueberries and a schematic diagram of a solar cell are shown.
\nOutline how a rechargeable battery differs from a primary cell.
\nFormulate half-equations for the reactions at the anode (negative electrode) and cathode (positive electrode) during discharge of a lithium-ion battery.
\nA voltaic cell consists of a nickel electrode in 1.0 mol dm−3 Ni2+ (aq) solution and a cadmium electrode in a Cd2+ (aq) solution of unknown concentration.
\nCd (s) + Ni2+ (aq) → Cd2+ (aq) + Ni (s) EΘcell = 0.14 V
\nDetermine the concentration of the Cd2+ (aq) solution if the cell voltage, E, is 0.19 V at 298 K. Use section 1 of the data booklet.
\nIdentify the structural feature of the dye that allows the conversion of solar energy into electrical energy.
\nOutline the effect of sunlight on the dye in the solar cell.
\nState the purpose of TiO2.
\nDeduce the reduction half-equation at the cathode.
\n«redox» reaction in rechargeable battery is reversible «but not in a primary cell»
\nOR
\nsecondary cells need to be charged before use
\nOR
\nsecondary cells have greater rate of self-discharge ✔
\n\n
Accept “primary cells cannot be recharged/reused”, “primary cells can be used only once” OR “lithium batteries may explode”.
\nAnode (negative electrode):
\nLi (graphite) → Li+ (electrolyte) + e–
\nOR
\nLiC6 (s) → 6C (s) + Li+ (electrolyte) + e– ✔
\n\n
Cathode (positive electrode):
\nLi+ (electrolyte) + e– + MnO2 (s) → LiMnO2 (s)
\nOR
\nLi+ (electrolyte) + e– + NiO2 (s) → LiNiO2 (s)
\nOR
\nLi+ (electrolyte) + e– + CoO2 (s) → LiCoO2 (s)
\nOR
\n2Li+ (electrolyte) + 2e– + 2CoO2 (s) → Co2O3 (s) + Li2O (s) ✔
\n\n
Accept “polymer” for “electrolyte”.
\nAward [1 max] if electrodes are reversed.
\nDo not accept “CO” for “Co”.
\n«E = EΘ − lnQ»
\n\n
OR
\n0.05 = –0.01283 ln [Cd2+]
\nOR
\nln[Cd2+] = – 3.897 ✔
\n[Cd2+] = 0.020 «mol dm–3» ✔
\n\n
Award [2] for correct final answer.
\n«extensive» conjugation
\nOR
\n«extensive» alternate single and double bonds ✔
\n\n
Accept “delocalization”.
\nelectrons excited/released «from dye» ✔
\n\n
Accept “photooxidation/oxidizes dye”.
\ntransfers e– to external circuit ✔
\n\n
Accept “provides large surface area”.
\nI3– (aq) + 2e– → 3I– (aq) ✔
\n\n
Accept “3I2 (aq) + 2e– → 2I3– (aq)”.
\nA student synthesized aspirin, acetylsalicylic acid, in a school laboratory.
\n0.300 g of crude aspirin was dissolved in ethanol and titrated with sodium hydroxide solution, NaOH (aq).
\nNaOH (aq) + C9H8O4 (in ethanol) → NaC9H7O4 (aq) + H2O (l)
\nPredict one absorption band present in an infrared (IR) spectrum of aspirin, using section 26 of the data booklet.
\nDetermine the mass of aspirin which reacted with 16.25 cm3 of 0.100 mol dm−3 NaOH solution.
\nDetermine the percentage purity of the synthesized aspirin.
\nOutline how aspirin can be chemically modified to increase its solubility in water.
\nState why aspirin should not be taken with alcohol.
\nAny one of:
1050–1410 «cm–1 due to C–O» [✔]
1700–1750 «cm–1 due to C=O in acids and esters» [✔]
2500–3000 «cm–1 due to O–H in acids» [✔]
2850–3090 «cm–1 due to C–H in alkanes and arenes» [✔]
n(aspirin) «= n(NaOH) = × 0.100 mol dm–3 »= 1.625 × 10–3 «mol» [✔]
\nm(aspirin) «= 1.625 × 10–3 mol × 180.17 g mol–1 »= 0.293 «g» [✔]
\n\n
Note: Award [2] for correct final answer.
\n« × 100 % »= 97.7 «%» [✔]
\nconvert to a salt
OR
react with sodium hydroxide [✔]
\n
Note: Accept other reactions forming soluble salts.
\nAccept “to ionize” but not “more polar”.
\nsynergistic effect/increased toxicity
OR
increased risk of stomach/intestines bleeding/ulcers/heartburn
OR
increased risk of liver toxicity/damage
OR
increased risk of nausea/vomiting [✔]
This was a very well answered question. Even weak candidates were able to identify one correct absorption band present in an infrared spectrum of aspirin.
\nA significant number of candidates were able to calculate the mass of aspirin correctly.
\nA significant number of candidates were able to calculate the percentage purity of aspirin correctly although some managed an ECF mark.
\nThis part was reasonably answered by most candidates.
\nThis part was well answered by the majority of the candidates.
\nMild heartburn is treated with antacids such as calcium carbonate.
\nFormulate an equation for the neutralization of stomach acid with calcium carbonate, CaCO3 (s).
\nAcid secretion can be regulated by other types of drugs such as omeprazole and ranitidine. Outline how each of these drugs acts to reduce excess stomach acid.
\nOmeprazole:
\nRanitidine:
\nCaCO3 (s) + 2HCl (aq) → CO2 (g) + CaCl2 (aq) + H2O (l) [✔]
\n\n
Note: Accept balanced ionic equations involving “H+” or “H3O+”.
\nDo not accept “H2CO3”.
\nOmeprazole:
inhibits enzyme/«gastric» proton pump «which secretes H+ ions into gastric juice»
OR
inhibits the H+/K+-ATPase system [✔]
Ranitidine:
inhibits/blocks H2/histamine receptors «in cells of stomach lining»
OR
prevents histamine binding to H2/histamine receptors «and triggering acid secretion» [✔]
\n
Note: Accept “H2-receptor antagonist” for M2.
\nResponses were mixed with some candidates easily writing an equation for the neutralization of stomach acid with CaCO3. Others failed to score for having incorrect products such as H2CO3 or CaCl, or for using sulfuric acid instead of hydrochloric for stomach acid.
\nMany candidates correctly outlined how drugs such as omeprazole and ranitidine regulate acid secretion.
\nThis question is about antiviral drugs.
\nOseltamivir, used for the treatment of severe flu, is inactive until converted in the liver to its active carboxylate form.
\nDraw a circle around the functional group that can be converted to the carboxylate by hydrolysis.
\nThe resulting active metabolite of oseltamivir can be detected by mass spectrometry (MS) analysis.
\nDeduce the mass of the expected carboxylate ion.
\nMr oseltamivir = 312
\nSuggest a reason for using a phosphate salt of oseltamivir in oral tablets.
\nAnti-HIV drugs, such as zidovudine, often become less effective over time.
\nExplain the development of resistant virus strains in the presence of antiviral drugs.
\n [✔]
\n
Note: Accept circles that include the alkyl side chain.
\n283 [✔]
\nmore soluble «in water» [✔]
\nviruses undergo «rapid» mutation [✔]
\nmutation causes a change in viral protein
OR
drug no longer binds to virus [✔]
\n
Note: Accept “rapid reproduction «allows resistant viruses to multiply»”.
\nRequired candidates to identify the functional group in a diagram of the structure of oseltamivir that can be converted to a carboxylate by hydrolysis. This was very challenging with many varied parts of the structure circled. Many circled the amide group. Candidates who selected the ester had to be careful to not include the ring structure as well.
\nThe challenge continued where the expected mass of the carboxylate ion was required. Some candidates chose values that exceeded the molar mass of oseltamivir itself, and some chose 45. There were very few correct answers.
\nMany candidates referred correctly to increased solubility of the salt in water while some mentioned bioavailability but did not realize that a salt will also form in the stomach.
\nMost candidates scored the first marking point when explaining the development of resistant virus strains but almost no-one scored the second mark. Many candidates were confused between bacteria and viruses and gave explanations about bacterial resistance and natural selection.
\nCarbon forms many compounds.
\nC60 and diamond are allotropes of carbon.
\nBut-2-ene reacts with hydrogen bromide.
\nChlorine reacts with methane.
\nCH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g)
\nOutline one difference between the bonding of carbon atoms in C60 and diamond.
\nState two features showing that propane and butane are members of the same homologous series.
\nDescribe a test and the expected result to indicate the presence of carbon–carbon double bonds.
\nDraw the full structural formula of but-2-ene.
\nWrite the equation for the reaction between but-2-ene and hydrogen bromide.
\nState the type of reaction.
\nSuggest two differences in the 1H NMR of but-2-ene and the organic product from (d)(ii).
\nCalculate the enthalpy change of the reaction, ΔH, using section 11 of the data booklet.
\nDraw and label an enthalpy level diagram for this reaction.
\nC60 fullerene: «each carbon is» bonded to 3 C AND diamond: bonded to 4 C
OR
C60 fullerene: delocalized/resonance AND diamond: not delocalized/no resonance
OR
C60 fullerene: single and double bonds AND diamond: single bonds ✔
\n
Accept “C60 fullerene: sp2 AND diamond: sp3”.
\nAccept “C60 fullerene: trigonal planar geometry / bond angles between 109.5°/109°/108°–120° AND diamond: tetrahedral geometry / bond angle 109.5°/109°”.
\nAccept \"bonds in fullerene are shorter/stronger/have higher bond order\".
\nsame general formula / CnH2n+2 ✔
\ndiffer by CH2/common structural unit ✔
\n\n
Accept \"similar chemical properties\".
\nAccept “gradation/gradual change in physical properties”.
\nALTERNATIVE 1:
\nTest:
\nadd bromine «water»/Br2 (aq) ✔
\nResult:
\n«orange/brown/yellow» to colourless/decolourised ✔
\n\n
Do not accept “clear” for M2.
\n\n
ALTERNATIVE 2:
\nTest:
\nadd «acidified» KMnO4 ✔
\nResult:
\n«purple» to colourless/decolourised/brown ✔
\n\n
Accept “colour change” for M2.
\n\n
ALTERNATIVE 3:
\nTest:
\nadd iodine / ✔
\nResult:
\n«brown» to colourless/decolourised ✔
\n\n
Accept
\nCH3CH=CHCH3 (g) + HBr (g) → CH3CH2CHBrCH3 (l)
\nOR
\nC4H8 (g) + HBr (g) → C4H9Br (l) ✔
\n«electrophilic» addition/EA ✔
\n
Do not accept nucleophilic or free radical addition.
ALTERNATIVE 1: Any two of:
\nbut-2-ene: 2 signals AND product: 4 signals ✔
\nbut-2-ene: «area ratio» 3:1/6:2 AND product: «area ratio» 3:3:2:1 ✔
\nproduct: «has signal at» 3.5-4.4 ppm «and but-2-ene: does not» ✔
\nbut-2-ene: «has signal at» 4.5-6.0 ppm «and product: does not» ✔
\n\n
ALTERNATIVE 2:
\nbut-2-ene: doublet AND quartet/multiplet/4 ✔
\nproduct: doublet AND triplet AND quintet/5/multiplet AND sextet/6/multiplet ✔
\n\n
Accept “product «has signal at» 1.3–1.4 ppm «and but-2-ene: does not»”.
\nbond breaking: C–H + Cl–Cl / 414 «kJ mol–1» + 242 «kJ mol–1»/656 «kJ»
OR
bond breaking: 4C–H + Cl–Cl / 4 × 414 «kJ mol–1» + 242 «kJ mol–1» / 1898 «kJ» ✔
bond forming: «C–Cl + H–Cl / 324 kJ mol–1 + 431 kJ mol–1» / 755 «kJ»
OR
bond forming: «3C–H + C–Cl + H–Cl / 3 × 414 «kJ mol–1» + 324 «kJ mol–1» + 431 kJ mol–1» / 1997 «kJ» ✔
«ΔH = bond breaking – bond forming = 656 kJ – 755 kJ» = –99 «kJ» ✔
Award [3] for correct final answer.
Award [2 max] for 99 «kJ».
\nreactants at higher enthalpy than products ✔
\n
ΔH/-99 «kJ» labelled on arrow from reactants to products
OR
activation energy/Ea labelled on arrow from reactant to top of energy profile ✔
\n
Accept a double headed arrow between reactants and products labelled as ΔH for M2.
\nThis was a challenging question that asked about the difference between the bonding of carbon atoms in C60 and diamond. 20% of the candidates gained the mark. The majority of the candidates did not have a specific enough answer for C60 and mentioned the pentagons and hexagons but not the number of bonds or the geometry or the bond order or the electron delocalisation. Diamond was better known to candidates as expected.
\nAbout two-thirds of the candidates scored one of the two marks and stronger candidates scored both. The most common answers were the same general formula/CnH2n+2, the difference between the compounds was CH2 and similar chemical properties. The same functional group was not accepted since alkanes do not have a functional group. Some candidates only stated that they are saturated hydrocarbons not gaining any marks.
\nAbout half of the candidates gave the bromine water test with the correct results. Iodine and KMnO4 were rarely seen in the scripts. There were candidates who used the term “clear” to mean “colourless” which was not accepted. Some candidates referred to the presence of UV light in a correct way and others in an incorrect way which was not penalized in this case. 10% of the candidates left the question blank. The most common incorrect answer was in terms of the IR absorptions. Other candidates referred to enthalpies of combustion and formation.
\nA well answered question. 70% of the candidates gave the correct structural formula for but-2-ene. Mistakes included too many hydrogens in the structure and an incorrect position of the C=C. Candidates should be reminded that the full structural formula requires all covalent bonds to be shown.
\nHalf of the candidates wrote the correct equation for the reaction of but-2-ene with hydrogen bromide. Incorrect answers included hydrogen as a product. As expected, the question correlated well with highly achieving candidates.
\nWell answered. 60% of candidates identified the type of reaction between but-2-ene and HBr, some of them including the term “electrophilic”. ECF was generously awarded when substitution was stated based on the equation where H2 was produced in part (ii). Candidates lost the mark if they only stated “hydrobromination” without mentioning addition. Some candidates lost the mark for stating “nucleophilic” or “free radical” addition.
\nThe comparison of the 1H NMR spectra of the two organic compounds was more challenging and 10% of the candidates left this question blank. The average mark was 0.7 out of 2 marks. Mistakes included non-specific answers that just stated “more signals” or “higher chemical shift”, and stating 3 signals in 2-bromobutane instead of 4 signals. Standard level candidates were expected to use the number of signals and the ratio of the areas under the signals to answer the question since they do not cover chemical shift, however, many of them did use the 1H NMR section in the data booklet to obtain correct answers in terms of chemical shift.
\nThis was the best answered question on the paper. Candidates identified the bonds and used bond enthalpies to calculate the enthalpy of reaction accurately. The most common mistakes were reversing the signs of bonds broken and bonds formed, assuming two Cl-Cl bonds were broken and using an incorrect value of bond enthalpy for one of the bonds.
\nThe majority of candidates drew the enthalpy level diagram and labelled it correctly based on their answer to part (i). Some candidates reversed the products and reactants. A few candidates did not add any labels which prevented the awarding of the second mark. With 2 marks allocated to the question the second mark was awarded for correct labeling of either ΔH or Ea.
\nWhich is correct for the reaction mechanism shown?
\nD
\nAnother well answered question on rate mechanisms.
\nExcess acid in the stomach can cause breakdown of the stomach lining.
\nOutline how ranitidine (Zantac) inhibits stomach acid production.
\nOutline two advantages of taking ranitidine instead of an antacid which neutralizes excess acid.
\nSome antacids contain carbonates.
\nDetermine the pH of a buffer solution which contains 0.160 mol dm−3 CO32− and 0.200 mol dm−3 HCO3−, using section 1 of the data booklet.
\npKa (HCO3−) = 10.32
\nblocks/binds H2/histamine receptors «in cells of stomach lining»
OR
prevents histamine molecules binding to H2/histamine receptors «and triggering acid secretion» [✔]
Any two of:
ranitidine can be effective in treating ulcers «but antacid is not» [✔]
ranitidine can prevent long-term damage «from overproduction of acid and antacid does not» [✔]
ranitidine has a long-term effect «and antacid has short-term effect only» [✔]
ranitidine does not affect ionic balance in body «and antacid does» [✔]
ranitidine does not produce bloating/flatulence [✔]
\n
Note: Accept “ranitidine stops the over production of acid in the stomach while antacids neutralise the excess acid
giving temporary relief” for M2.
«pH = pKa »
«pH =»10.22 [✔]
Some candidates were not confident enough in their answers to receive a mark while others confused the action of ranitidine which blocks H2 receptors with omeprazole which is a proton pump inhibitor.
\nWhile most candidates were awarded at least one of the two marks possible for this question some of the descriptions were too vague or incomplete to receive a mark.
\nThis was generally a well-answered question. Most candidates who did not receive the mark inverted the concentration of the conjugate base/concentration of the acid in the calculation.
\nOpium and its derivatives have been used for thousands of years as strong analgesics.
\nExplain how opiates act to provide pain relief.
\nDiscuss how the difference in structure of two opiates, codeine and morphine, affect their ability to cross the blood–brain barrier. Use section 37 of the data booklet.
\n«temporarily» bond/bind to «opioid» receptors in the brain/CNS [✔]
\nblock the transmission of pain impulses [✔]
\n«codeine crosses blood–brain barrier more easily» morphine has more hydroxyl/OH «groups than codeine» [✔]
\ncodeine/ether group is less polar
OR
hydroxyl/OH «groups in morphine» H-bond to water [✔]
\n
Note: Award [1 max] if no statement or an incorrect statement about the blood–brain barrier.
\nThe question asked candidates to explain how opiates provide pain relief. This was difficult and was poorly answered by many. As this has been asked many times over the years, it would be an advantage to candidates to practise answering past examination questions.
\nThe explanation that required a discussion of the difference in structure of codeine and morphine, and how this affects their ability to cross the blood-brain barrier was challenging. Many scored for saying “codeine is less polar”. Some also scored for saying that “morphine has more hydroxyl groups” but others provided less detail and could not be awarded any marks.
\nAntiviral medications have recently been developed for some viral infections.
\nOutline one way in which antiviral drugs work.
\nDiscuss two difficulties associated with solving the AIDS problem.
\nAny one of:
alter cell’s genetic material so that virus cannot use it to multiply [✔]
prevent viruses from multiplying by blocking enzyme activity within host cell
OR
inhibit the synthesis of viral components by blocking enzymes inside the cell [✔]
prevent viruses from entering «host» cell
OR
bind to cellular receptors targeted by viruses
OR
bind to virus-associated proteins/VAPs which target cellular receptors
OR
prevents removal of protein coat/capsid
OR
prevents injection of viral DNA/RNA into cell [✔]
prevent/hinder the release of viruses from the cell [✔]
\n\n
Note: Accept “prevents synthesis of virus by host cell”.
\nAccept “alters RNA/DNA/genetic material of virus”.
\nDo not accept just “mimics nucleotides”.
\nAny two of:
viruses lack cell structure so difficult to target with drugs [✔]
HIV is a retrovirus
OR
HIV genetic material is in the form of RNA instead of DNA [✔]
HIV affects/destroys helper/T-cells which are necessary to fight infection [✔]
\nHIV has great genetic diversity so difficult to produce «a» vaccine [✔]
\nanti-retroviral agents are expensive so not everyone/country can afford them [✔]
\nsocio-cultural issues deter people from seeking treatment/prevention/diagnosis
OR
lack of education/conversation/stigma associated with being HIV-positive [✔]
mutation of virus/HIV [✔]
\nvirus/HIV metabolism linked to that of host cell [✔]
\ndrugs harm host cell as well as virus/HIV [✔]
\nHIV difficult to detect/remains dormant [✔]
\nCandidates responded fairly well to this question. Candidates who did not receive a mark were either too vague or discussed anti-bacterial methods.
\nMost candidates were awarded at least one of the two marks possible for this question. Some student responses were too vague or discussed the social and political issues surrounding the AIDS crisis. There were also some responses, which only talked about AIDS extensively with no mention of the virus.
\nKc for 2N2O (g) 2N2 (g) + O2 (g) is 7.3 × 1034.
\nWhat is Kc for the following reaction, at the same temperature?
\nN2 (g) + O2 (g) N2O (g)
\n\n
A. 7.3 × 1034
\nB.
\nC.
\nD.
\nB
\nCandidates scored well while one teacher thought it was a trick question and particularly tough. Relationship between K values on reversed equations should be understood.
\nThe structure of penicillin is shown in section 37 of the data booklet.
\nState the internal bond angles in the b-lactam ring and the expected bond angles in sp2 and sp3 hybridised atoms.
\nExplain how the open β-lactam ring kills bacteria.
\nState how the structure of penicillin can be modified to combat the effect of resistance caused by over prescription.
\nSuggest why human cells are not affected by penicillin.
\nAccept “109º”.
\n«irreversibly» binds/bonds to enzyme/transpeptidase
\nOR
\ninhibits enzyme/transpeptidase «in bacteria» that produces cell walls
\nOR
\nprevents cross-linking of bacterial cell walls ✔
\n\n
cells absorb water AND burst
\nOR
\ncells cannot reproduce ✔
\n\n
Accept “reacts with” for “bonds to” for M1.
\nDo not accept “cell membrane” for “cell wall” for M1.
\nAccept “cells burst due to osmotic pressure” for M2.
\nAccept “bacteria” for “cells” for M2.
\n«modify» side-chain ✔
\n\n
Accept “«modify» R”.
\nno cell walls
\nOR
\nhumans do not have transpeptidase ✔
\nThis question is about peroxides.
\nHydrogen peroxide decomposes to water and oxygen when a catalyst such as potassium iodide, KI, is added.
\n2H2O2 (aq) O2 (g) + 2H2O (l)
\nSuggest why many chemicals, including hydrogen peroxide, are kept in brown bottles instead of clear colourless bottles.
\nIn a laboratory experiment solutions of potassium iodide and hydrogen peroxide were mixed and the volume of oxygen generated was recorded. The volume was adjusted to 0 at t = 0.
\nThe data for the first trial is given below.
\nPlot a graph on the axes below and from it determine the average rate of formation of oxygen gas in cm3 O2 (g) s−1.
\nAverage rate of reaction:
\nAdditional experiments were carried out at an elevated temperature. On the axes below, sketch Maxwell–Boltzmann energy distribution curves at two temperatures T1 and T2, where T2 > T1.
\nApart from a greater frequency of collisions, explain, by annotating your graphs in (b)(ii), why an increased temperature causes the rate of reaction to increase.
\nMnO2 is another possible catalyst for the reaction. State the IUPAC name for MnO2.
\nComment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.
\nH2O2 (aq) + CH3COOH (aq) CH3COOOH (aq) + H2O (l)
\nSodium percarbonate, 2Na2CO3•3H2O2, is an adduct of sodium carbonate and hydrogen peroxide and is used as a cleaning agent.
\nMr (2Na2CO3•3H2O2) = 314.04
\nCalculate the percentage by mass of hydrogen peroxide in sodium percarbonate, giving your answer to two decimal places.
\ndecomposes in light [✔]
\n\n
Note: Accept “sensitive to light”.
\npoints correctly plotted [✔]
\nbest fit line AND extended through (to) the origin [✔]
\nAverage rate of reaction:
«slope (gradient) of line =» 0.022 «cm3 O2 (g) s−1» [✔]
\n
Note: Accept range 0.020–0.024cm3 O2 (g) s−1.
\npeak of T2 to right of AND lower than T1 [✔]
\nlines begin at origin AND T2 must finish above T1 [✔]
\nEa marked on graph [✔]
\nexplanation in terms of more “particles” with E ≥ Ea
OR
greater area under curve to the right of Ea in T2 [✔]
manganese(IV) oxide
OR
manganese dioxide [✔]
\n
Note: Accept “manganese(IV) dioxide”.
\nmove «position of» equilibrium to right/products [✔]
\n\n
Note: Accept “reactants are always present as the reaction is in equilibrium”.
\nM (H2O2) «= 2 × 1.01 + 2 × 16.00» = 34.02 «g» [✔]
\n«% H2O2 = 3 × × 100 =» 32.50 «%» [✔]
\n\n
Note: Award [2] for correct final answer.
\nThe explanation that the brown bottle prevented light causing a decomposition of the chemical was well answered but some incorrectly suggested it helped to stop mixing up of chemicals e.g. acid/water/peroxide.
\nThe graphing was disappointing with a surprising number of students missing at least one mark for failing to draw a straight line or for failing to draw the line passing through the origin. Also some were unable to calculate the gradient.
\nThe drawing of the two curves at T1 and T2 was generally poorly done.
\nExplaining why temperature increase caused an increase in reaction rate was generally incorrectly answered with most students failing to mention “activation energy” in their answer or failing to annotate the graph.
\nMany could correctly name manganese(IV)oxide, but there were answers of magnesium(IV) oxide or manganese(II) oxide.
\nSuggesting why peractic acid was sold in solution was very poorly answered and only a few students mentioned equilibrium and, if they did, they thought it would move to the left to restore equilibrium.
\nCalculating the % by mass was generally well answered although some candidates started by using rounded values of atomic masses which made their final answer unprecise.
\nTaxol is a drug that was once obtained from yew trees and is now produced using chiral auxiliaries.
\nExamine the synthesis of taxol in terms of green chemistry criteria.
\nOutline the operation of a polarimeter used to distinguish between enantiomers.
\nAny two of:
produced by genetically engineered/modified bacteria/E. coli
OR
sustainable because synthesized and not obtained from yew trees [✔]
chiral auxiliaries «isolated and» reused [✔]
\none enantiomer produced [✔]
\ntoxicity/recycling of solvents/materials used [✔]
\noverall yield/atom economy/waste generated [✔]
\nAny two of:
«plane-» polarized light
OR
light passes through polarizer/polarizing filter [✔]
enantiomers rotate plane of «plane-» polarized light «by equal angles» in opposite directions [✔]
\nmeasure angle/direction of rotation [✔]
\nThe synthesis of taxol in terms of green chemistry criteria invited varied responses. While some candidates were precise in their answers but others lost focus and wrote something about green chemistry.
\nThe operation of a polarimeter to distinguish between enantiomers was generally well handled by the candidates while some missed stating to measure the angle/direction of rotation.
\nWhich is correct for a reaction with a positive change in Gibbs free energy, ΔGθ?
\nA. The formation of reactants is favoured.
\nB. The formation of products is favoured.
\nC. The reaction is at equilibrium.
\nD. The reaction is spontaneous.
\nA
\nThe higher scoring candidates did better on identifying the direction of spontaneity given a positive ΔG
\nWhich solution is basic at 25 °C?
\nKw = 1.0 × 10−14
\nA. [H+] = 1.0 × 10−3 mol dm−3
\nB. [OH−] = 1.0 × 10−13 mol dm−3
\nC. solution of pH = 4.00
\nD. [H3O+] = 1.0 × 10−13 mol dm−3
\nD
\nThis had the highest discriminatory index on the exam. Most of the incorrect answers indicated that a basic solution would have [OH- ] = 1.0 x 10-13 mol dm-3 rather than [H3O+ ] = 1.0 x 10-13 mol dm-3 .
\nTechnetium-99m, a widely used radionuclide, has a half-life of 6.0 hours and undergoes gamma decay to technetium-99.
\nMost of the nuclear waste generated in a hospital is low-level waste (LLW).
\nDetermine the percentage of technetium-99m remaining after 24.0 hours.
\nTechnetium-99 decays further, emitting beta radiation. Formulate the equation for the decay of technetium-99.
\nOutline what is meant by low-level waste.
\nOutline the disposal of LLW.
\nMagnetic resonance imaging (MRI) is an application of NMR technology using radiowaves.
\nSuggest why MRI is much less dangerous than imaging techniques such as X-rays and radiotracers. Use section 3 of the data booklet.
\nAlternative 1
half-lives = « =» 4.0 [✔]
«N(t) (%) = 100(0.5)4 =» 6.3 «%» [✔]
\nNote: Accept “6.25 «%»”.
\n
Alternative 2
λ =«» 0.116 hour–1 OR =e–0.116 × 24 [✔]
6.3 «%» [✔]
\nNote: Award [2] for correct final answer.
\n99Tc → 99Ru + β–
\n
Ru [✔]
mass number of Ru AND beta product [✔]
\n\n
Note: Accept “e/e–/ e” for “β–”.
\nsmall/low amounts of radiation AND for a short time [✔]
\n\n
Note: Accept “weakly ionizing radiation” instead of “small amounts of radiation”.
\nAccept “short half-lives” instead of “for a short time”.
\nstored in shielded containers until radiation level drops «to a safe level» [✔]
\nlower frequency/longer wavelength/lower energy
OR
does not use ionizing radiation/radionuclides [✔]
\n
Note: Do not accept “does not cause cancer”.
\nMost candidates correctly determined the % of technetium-99m remaining after 24.0 hours. Some candidates did not read the question properly and forgot to convert to a percentage.
\nA common incorrect answer was to give Tc as a product or to give an incorrect symbol for beta radiation. Most candidates scored a mark for the correct mass number of the product and beta radiation.
\nLow level nuclear waste was poorly outlined with many superficial responses. Many candidates only gave half the required answer.
\nOutlining the disposal of LLW was also challenging with many candidates saying that it should be put in a container without saying the container should be shielded.
\nSuggesting why MRI is less dangerous than using X-rays and radiotracers was mostly answered well, but some candidates were confused and linked longer wavelengths with higher energy.
\nBoth vinegar (a dilute aqueous solution of ethanoic acid) and bleach are used as cleaning agents.
\nBleach reacts with ammonia, also used as a cleaning agent, to produce the poisonous compound chloramine, NH2Cl.
\nOutline why ethanoic acid is classified as a weak acid.
\nA solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.
\nCl2 (g) + 2NaOH (aq) NaOCl (aq) + NaCl (aq) + H2O (l)
\nSuggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.
\nDraw a Lewis (electron dot) structure of chloramine.
\nDeduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
\n\n
Molecular geometry:
\nH–N–H bond angle:
\npartial dissociation «in aqueous solution» [✔]
\nethanoic acid/vinegar reacts with NaOH [✔]
\nmoves equilibrium to left/reactant side [✔]
\nreleases Cl2 (g)/chlorine gas
OR
Cl2 (g)/chlorine gas is toxic [✔]
\n
Note: Accept “ethanoic acid produces H+ ions”.
\nAccept “ethanoic acid/vinegar reacts with NaOCl”.
\nDo not accept “2CH3COOH + NaOCl + NaCl → 2CH3COONa + Cl2 + H2O” as it does not refer to equilibrium.
Accept suitable molecular or ionic equations for M1 and M3.
\n [✔]
\n
Note: Accept any combination of dots/crosses or lines to represent electron pairs.
\nMolecular geometry:
«trigonal» pyramidal [✔]
H–N–H bond angle:
107° [✔]
\n
Note: Accept angles in the range of 100–109.
\nThe definition of a weak acid was generally correct.
\nExplaining why it was dangerous to mix chlorine with vinegar was not well answered but most students gained at least one mark for stating that “chlorine gas will be produced”, but couldn’t link it to equilibrium ideas.
\nThe Lewis structure of chloramine was correct for strong candidates, but many made the mistake of omitting electron pairs on N and Cl.
\nThe molecular geometry and bond angles often did not correspond to each other with quite a few candidates stating trigonal planar and then 107 for the angle.
\nTechnetium-99m, Tc-99m, is a gamma-ray emitter commonly used as a medical tracer.
\nIts half-life is 6.0 hours.
\nEvaluate the suitability of technetium-99m for this use.
\nCalculate the percentage of technetium-99m remaining after 10.0 hours. Use section 1 of the data booklet.
\nAny two of:
can be readily “tagged” to variety of biologically active carriers «which will deliver it to specific locations for imaging uses» [✔]
frequency of radiation is compatible with existing X-ray detection apparatus [✔]
\nproduct of decay has low radioactivity/relatively short half-life/low total exposure to patient [✔]
\n«but small» increased risk of cancer to patient [✔]
\nmust be made on site [✔]
\n\n
Note: Accept other valid answers outlining advantages or limitations of Tc-99m, such as “produces only LLW”, “Tc is a transition element forming compounds in a variety of oxidation states”, “gamma-radiation «can escape the body and» be detected by external sensors”, “activity decreases quickly, so dose must be calculated prior to each injection”.
\nALTERNATIVE 1
\n [✔]
31 «% remaining» [✔]
\n
ALTERNATIVE 2
\nλ «= »= 0.1155 hours–1 [✔]
\n« × 100 = e–λt ×100 = 0.31498 × 100»
\n31 «% remaining» [✔]
\n\n
Note: M1 is for correct substitution of values.
\nAward [2] for correct final answer.
\n\n
The most frequent response was the short half-life, followed by the emission and detection of gamma radiation.
\nMost candidates were able to calculate the percentage of technetium-99m correctly.
\nWith which do most acids react?
\n I. sodium hydrogen carbonate
II. magnesium
III. calcium sulfate
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nThis question on reactivity of acids with carbonates and metals (first bullet point in Topic 8.2) confused higher and lower scoring candidates alike and was the worst answered question in paper 1. One teacher commented “The question does not seem clear. They would all react with acids, but what does “most acids” mean?”.
\nOpiates are strong analgesics.
\nExplain why diamorphine (heroin) crosses the blood–brain barrier more easily than morphine.
\nOutline the meaning of the bioavailability of a drug.
\nblood-brain barrier is hydrophobic/non-polar/made of lipids ✔
\nmorphine has hydroxyl/OH «groups»/is more polar AND diamorphine has ester/ethanoate/OCOCH3/acetate «groups»/is less polar/is lipid soluble ✔
\n\n
Accept “fats” for “lipid”.
\nAccept “alcohol/hydroxy” for “hydroxyl” but not “hydroxide”.
\nAccept “non-polar” for “less polar” in M2.
\nfraction/proportion/percentage of «administered dosage» enters blood/plasma/circulation ✔
\n\n
Accept “fraction/proportion/percentage of «administered dosage» that reaches target «part of human body»”.
\nWhich is a Lewis acid but not a Brønsted−Lowry acid?
\nA. AlCl3
\nB. CH3CO2H
\nC. HF
\nD. CCl4
\nA
\nIdentifying a Lewis acid seemed relatively easy for the majority of candidates.
\nXylene is a derivative of benzene. One isomer is 1,4-dimethylbenzene.
\nBromine reacts with alkanes.
\nState the number of 1H NMR signals for this isomer of xylene and the ratio in which they appear.
\nNumber of signals:
Ratio:
\nDraw the structure of one other isomer of xylene which retains the benzene ring.
\nIdentify the initiation step of the reaction and its conditions.
\n1,4-dimethylbenzene reacts as a substituted alkane. Draw the structures of the two products of the overall reaction when one molecule of bromine reacts with one molecule of 1,4-dimethylbenzene.
\nNumber of signals:
2 [✔]
Ratio:
3 : 2
OR
6 : 4 [✔]
\n
Note: Accept any correct integer or fractional ratio.
\nAccept ratios in reverse order.
\n [✔]
Br2 → 2Br• [✔]
\n«sun»light/UV/hv
OR
high temperature [✔]
\n
Note: Do not penalize missing radical symbol on Br.
\nAccept “homolytic fission of bromine” for M1.
\n [✔]
HBr [✔]
\n\n
Note: Accept condensed formulae, such as CH3C6H4CH2Br.
\nAccept skeletal structures.
\nMost students gained M1 but very few gained M2, suggesting that the correct answer of 2 signals may have been a guess.
\nAnother isomer of xylene was generally correctly drawn, but some candidates drew the original compound.
\nDrawing or describing the homolytic fission of bromine was generally done well.
\nVery few students gained 2 marks finding hard to apply their knowledge of free radical substitution to a benzene containing compound. Many thought that the bromine will attach to the benzene ring or would substitute the alkyl group twice and not produce HBr.
\nWhich has the strongest conjugate base?
\nA. HCOOH (Ka = 1.8 × 10−4)
\nB. HNO2 (Ka = 7.2 × 10−4)
\nC. HCN (Ka = 6.2 × 10−10)
\nD. HIO3 (Ka = 1.7 × 10−1)
\nC
\nStrength of conjugate bases appeared to be another good discriminator, with higher scoring candidates performing better.
\nViruses and bacteria both cause diseases and are frequently confused.
\nOutline two different ways in which antiviral medications work.
\nAny two of:
\nprevents virus attaching to host cell ✔
\nalters cell’s genetic material/DNA «so that virus cannot use it to multiply» ✔
\nblocks enzyme activity in the host cell «so that virus cannot use it to multiply» ✔
\nprevents removal of protein coat/capsid ✔
\nprevents injection of viral DNA/RNA into cell ✔
\nprevents release of «replicated» viruses from host cell ✔
\n\n
Accept “prevents synthesis of virus by host cell”.
\nAccept “alters RNA/DNA/genetic material of virus”.
\nDo not accept just “mimics nucleotides”.
\nTaxol is a chiral cancer drug which is synthesized using a chiral auxiliary.
\nThe diagram shows part of a Taxol molecule in skeletal form.
\nDraw a circle around each chiral carbon.
\nOutline how chiral auxiliaries are used to synthesize the desired enantiomer.
\nExplain the process of solvent extraction by which Taxol is isolated.
\n\n
Do not penalize any other notation (eg *) used for a circle.
\nchiral auxiliary creates stereochemical condition necessary to follow a certain pathway
\nOR
\nstereochemical induction
\nOR
\nexisting chiral centre affects configuration of new chiral centres ✔
\n\n
chiral molecule/auxiliary/optically active species is used/added/connected to the starting molecule «to force reaction to follow a certain path»
\nOR
\n«after new chiral centre created» chiral auxiliary removed «to obtain desired product» ✔
\nAny two of:
\nimmiscible solvents ✔
\npartitioning of Taxol between the two solvents
\nTaxol more soluble in one solvent ✔
\nextraction carried out multiple times «to improve extraction» ✔
\nshaking/stirring the mixture ✔
\nseparating the two layers ✔
\nevaporation of the solvent from the final solution «to obtain pure Taxol» ✔
\nWhich product will be obtained at the anode (positive electrode) when molten NaCl is electrolysed?
\nA. Na (l)
\nB. Cl (g)
\nC. Cl2 (g)
\nD. Na (s)
\nC
\n57 % of candidates correctly identified products of electrolysis at the anode with the incorrect answers being split by the remaining 43 %
\nThe presence of alcohol in the breath can be detected using a breathalyzer.
\nDescribe how a fuel cell breathalyser works.
\nAlcohol levels in the breath can also be determined using IR spectroscopy.
\nSuggest, giving a reason, which bond’s absorbance is most useful for detecting ethanol in breath.
\nBond:
\nReason:
\nAny three of:
ethanol «in breath» is oxidized «to ethanoic acid» [✔]
electrons pass through external circuit/meter [✔]
«to cathode where» O2 is reduced [✔]
current is proportional to alcohol concentration [✔]
\n
Note: Accept equations for oxidation of ethanol or reduction of oxygen.
\nBond:
C–O
OR
C–H [✔]
Reason:
cannot use O–H bonds as in water «found in breath» [✔]
\n
Note: Accept “C–O/C–H «bonds in molecules in breath» most likely to be in ethanol”.
\nDo not apply ECF here.
\nWhile many scored the first marking point, full marks were rarely seen. Many candidates mixed up this and a dichromate breathalyser.
\nMost candidates incorrectly identified O-H, failing to realise it is unsuitable due to its abundant presence in the breath.
\nWhere does oxidation occur in a voltaic cell?
\nA. positive electrode and anode
\nB. negative electrode and anode
\nC. positive electrode and cathode
\nD. negative electrode and cathode
\nB
\nOne G2 form thought the question was awkwardly worded, implying oxidation is occurring in two places rather than one. This was one of the better answered question with a relatively low discriminatory index.
\nMethadone is a synthetic opiate administered as a racemic mixture to treat strong pain and morphine or heroin dependence.
\nIdentify the chiral carbon atom using an asterisk, *.
\nEnantiomers can be identified using a polarimeter. Outline how this instrument differentiates the enantiomers.
\n [✔]
«plane-»polarized light passed through sample [✔]
\n
analyser/second polarizer determines angle of rotation of plane of plane-polarized light
OR
each enantiomer rotates plane «of plane-polarized light» in opposite directions «by the same angle» [✔]
Some candidates had difficulty identifying the chiral carbon in a methadone structure, with quite a few varied answers. However, many managed to mark the correct carbon.
\nVery poorly answered. Few scored any marks at all when outlining how a polarimeter can be used to differentiate between enantiomers. Many referred to the light or the enantiomers themselves being rotated.
\nThis question is about iron.
\nState the nuclear symbol notation, , for iron-54.
\nMass spectrometry analysis of a sample of iron gave the following results:
\nCalculate the relative atomic mass, Ar, of this sample of iron to two decimal places.
\nAn iron nail and a copper nail are inserted into a lemon.
\nExplain why a potential is detected when the nails are connected through a voltmeter.
\n[✔]
\n«Ar =» 54 × 0.0584 + 56 × 0.9168 + 57 × 0.0217 + 58 × 0.0031
OR
«Ar =» 55.9111 [✔]
«Ar =» 55.91 [✔]
\nNotes:
\nAward [2] for correct final answer.
Do not accept data booklet value (55.85).
lemon juice is the electrolyte
OR
lemon juice allows flow of ions
OR
each nail/metal forms a half-cell with the lemon juice [✔]
Note: Accept “lemon juice acts as a salt bridge”.
\nAny one of:
iron is higher than copper in the activity series
OR
each half-cell/metal has a different redox/electrode potential [✔]
Note: Accept “iron is more reactive than copper”.
\niron is oxidized
OR
Fe → Fe2+ + 2e–
OR
Fe → Fe3+ + 3e−
OR
iron is anode/negative electrode of cell [✔]
copper is cathode/positive electrode of cell
OR
reduction occurs at the cathode
OR
2H+ + 2e− → H2 [✔]
electrons flow from iron to copper [✔]
\n\n
Notes:
Accept “lemon juice acts as a salt bridge”.
Accept “iron is more reactive than copper”.
The nuclear symbol notation was generally correct. However, some students swapped atomic and mass numbers and hence lost the mark.
\nCalculation of RAM was generally correctly calculated, but some candidates did not give their answer to two decimal places while they should use the provided periodic table.
\nVery few students gained the 2 marks available for explaining the potential generated in the lemon as they didn’t realise it was the lemon that acted as the electrolyte and allowed ions to flow. Some were able to gain a mark for explaining that electrons moved from iron to copper as iron is more reactive.
\nNuclear isotopes are used in the treatment of cancer.
\nGamma radiation is also used in radiotherapy.
\nAlpha particles are more damaging to human cells than any other nuclear radiation and yet they are used in targeted alpha therapy (TAT).
\nExplain how TAT is relatively safe to use in the treatment of dispersed cancers.
\nTechnetium-99m () has a half-life of 6.0 hours. Calculate the percentage of remaining in a sample of the radioisotope after two days.
\nSuggest why the percentage of technetium-99m remaining in the human body two days after injection will be lower than that calculated in (b)(i).
\n«alpha emitter» carried to/selectively absorbed by cancer cells «by antibody, carrier drug, protein» ✔
\n\n
low penetrating power
\nOR
\nshort range ✔
\n\n
Do not accept just “targets cancer cells and does not affect healthy cells”.
\nALTERNATIVE 1:
\n« =» 8 /8 half-lives «required» ✔
\n% remaining = «(0.5)8 × 100 =» 0.39 «%» ✔
\n\n
ALTERNATIVE 2:
\nλ = « =» 0.1155 ✔
\n% remaining = «100 × e–0.1155 × 48 =» 0.39 «%» ✔
\n\n
Award [2] for correct final answer.
\nAccept “0.32 «%»” in ALTERNATIVE 2.
\nremoved by excretion ✔
\n\n
Accept any method of excretion.
\nThe thermal decomposition of dinitrogen monoxide occurs according to the equation:
\n2N2O (g) → 2N2 (g) + O2 (g)
\nThe reaction can be followed by measuring the change in total pressure, at constant temperature, with time.
\nThe x-axis and y-axis are shown with arbitrary units.
\nExplain why, as the reaction proceeds, the pressure increases by the amount shown.
\nOutline, in terms of collision theory, how a decrease in pressure would affect the rate of reaction.
\nThe experiment is repeated using the same amount of dinitrogen monoxide in the same apparatus, but at a lower temperature.
\nSketch, on the axes in question 2, the graph that you would expect.
\nThe experiment gave an error in the rate because the pressure gauge was inaccurate. Outline whether repeating the experiment, using the same apparatus, and averaging the results would reduce the error.
\nThe graph below shows the Maxwell–Boltzmann distribution of molecular energies at a particular temperature.
\nThe rate at which dinitrogen monoxide decomposes is significantly increased by a metal oxide catalyst.
\nAnnotate and use the graph to outline why a catalyst has this effect.
\nincrease in the amount/number of moles/molecules «of gas» [✔]
\nfrom 2 to 3/by 50 % [✔]
\n«rate of reaction decreases»
concentration/number of molecules in a given volume decreases
OR
more space between molecules [✔]
collision rate/frequency decreases
OR
fewer collisions per second/unit time [✔]
\n
Note: Do not accept just “larger space/volume” for M1.
\nsmaller initial gradient [✔]
\ninitial pressure is lower AND final pressure of gas lower «by similar factor» [✔]
\n\n
no AND it is a systematic error/not a random error
OR
no AND «a similar magnitude» error would occur every time [✔]
catalysed and uncatalysed Ea marked on graph AND with the catalysed being at lower energy [✔]
\n«for catalysed reaction» greater proportion of/more molecules have E ≥ Ea / E > Ea
OR
«for catalysed reaction» greater area under curve to the right of the Ea [✔]
Note: Accept “more molecules have the activation energy”.
\nAbout a quarter of the candidates gave the full answer. Some only gained the first marking point (M1) by recognizing the increase in the number of moles of gas. Some candidates wrote vague answers that did not receive credit such as “pressure increases as more gaseous products form” without explicitly recognizing that the reactants have fewer moles of gas than the products. Some candidates mistook it for a system at equilibrium when the pressure stops changing (although a straight arrow is shown in the equation). A teacher commented that the wording of the question was rather vague “not clear if question is asking about stoichiometry (i.e. how 200 & 300 connect to coefficients) or rates (i.e. explain graph shape)”. We did not see a discussion of the slope of the graph with time and most candidates understood the question as it was intended.
\nMore than half of the candidates obtained the mark allocated for “less frequent collisions” at lower pressure, but only strong candidates explained that this was due to the lower concentration or increased spacing between molecules. Some candidates talked about a decrease in kinetic energy and they did not show a good understanding of collision theory. Some candidates lost M1 for stating “fewer collisions” without reference to time or probability.
\nThis was a challenging question. Candidates usually obtained only one of the two marks allocated for the answer. Most of them scored the mark for a lower initial slope at low temperature, while others scored a mark for sketching their curve below the original curve as all pressures (initial and final) will be lower at the lower temperature. A teacher commented that the wording was unclear “sketch on the axes in question 2”, and it would have been better to label the graph instead.
\nThis question was well answered by nearly 70 % of the candidates reflecting a good understanding of the impact of systematic errors. Some students did not gain the mark because of an incomplete answer. The question raised much debate among teachers. They worried if the error was clearly a systematic one. However, a high proportion of candidates had very clear and definite answers. In Spanish and French, the wording was a bit ambiguous which caused the markscheme in these languages to be more opened.
\nThis question discriminated very well between high-scoring and low-scoring candidates. About half of the candidates annotated the Maxwell-Boltzmann distribution to show the effect of the catalyst. Some left it blank and some sketched a new distribution that would be obtained at a higher temperature instead. The majority of candidates knew that the catalyst provided an alternative route with lower Ea but only stronger candidates related it to the annotation of the graph and used the accurate language needed to score M2. A common mistake was stating that molecules have higher kinetic energy when a catalyst is added.
\nWhich diagram represents a heterogeneous mixture?
\nA
\nA few G2 forms suggested that candidates could have chosen A or B. However, homogeneous and heterogeneous mixtures are usually represented in such a way while only option A shows a clear separation in the mixture. We will avoid schematic diagrams in the future for this type of questions.
\nWhat volume of carbon dioxide, CO2 (g), can be obtained by reacting 1 dm3 of methane, CH4 (g), with 1 dm3 of oxygen, O2 (g)?
\nCH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
\nA. 0.5 dm3
\nB. 1 dm3
\nC. 2 dm3
\nD. 6 dm3
\nA
\nThe concept of using volume ratios of gases as mole ratios in balanced equations was well answered.
\nEthyne, C2H2, reacts with oxygen in welding torches.
\nEthyne reacts with steam.
\nC2H2 (g) + H2O (g) → C2H4O (g)
\nTwo possible products are:
\nProduct B, CH3CHO, can also be synthesized from ethanol.
\nWrite an equation for the complete combustion of ethyne.
\nDeduce the Lewis (electron dot) structure of ethyne.
\nCompare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
\nIdentify the type of interaction that must be overcome when liquid ethyne vaporizes.
\nProduct A contains a carbon–carbon double bond. State the type of reactions that compounds containing this bond are likely to undergo.
\nState the name of product B, applying IUPAC rules.
\nDetermine the enthalpy change for the reaction, in kJ, to produce A using section 11 of the data booklet.
\nThe enthalpy change for the reaction to produce B is −213 kJ. Predict, giving a reason, which product is the most stable.
\nThe IR spectrum and low resolution 1H NMR spectrum of the actual product formed are shown.
\nDeduce whether the product is A or B, using evidence from these spectra together with sections 26 and 27 of the data booklet.
\n\n
Identity of product:
\nOne piece of evidence from IR:
\nOne piece of evidence from 1H NMR:
\nSuggest the reagents and conditions required to ensure a good yield of product B.
\n\n
Reagents:
\nConditions:
\nDeduce the average oxidation state of carbon in product B.
\nExplain why product B is water soluble.
\nC2H2 (g) + 2.5O2 (g) → 2CO2 (g) + H2O (l)
OR
2C2H2 (g) + 5O2 (g) → 4CO2 (g) + 2H2O (l) [✔]
[✔]
\n
Note: Accept any valid combination of lines, dots and crosses.
\n«ethyne» shorter AND a greater number of shared/bonding electrons
OR
«ethyne» shorter AND stronger bond [✔]
London/dispersion/instantaneous dipole-induced dipole forces [✔]
\n\n
Note: Do not accept just “intermolecular forces” or “van der Waals’ forces”.
\n«electrophilic» addition/A«E» [✔]
\n\n
Note: Accept “polymerization”.
\nethanal [✔]
\n«sum of bond enthalpies of reactants =» 2(C–H) + C≡C + 2(O–H)
OR
2 × 414 «kJ mol–1» + 839 «kJ mol–1» + 2 × 463 «kJ mol–1»
OR
2593 «kJ» [✔]
«sum of bond enthalpies of A =» 3(C–H) + C=C + C–O + O–H
OR
3 × 414 «kJ mol–1» + 614 «kJ mol–1» + 358 «kJ mol–1» + 463 «kJ mol–1»
OR
2677 «kJ» [✔]
«enthalpy of reaction = 2593 kJ – 2677 kJ» = –84 «kJ» [✔]
\n\n
Note: Award [3] for correct final answer.
\nB AND it has a more negative/lower enthalpy/«potential» energy
OR
B AND more exothermic «enthalpy of reaction from same starting point» [✔]
Identity of product: «B»
\nIR spectrum:
1700–1750 «cm–1 band» AND carbonyl/CO group present
OR
no «band at» 1620–1680 «cm–1» AND absence of double bond/C=C
OR
no «broad band at» 3200–3600 «cm–1» AND absence of hydroxyl/OH group [✔]
Note: Accept a specific value or range of wavenumbers and chemical shifts.
\n1H NMR spectrum:
«only» two signals AND A would have three
OR
«signal at» 9.4–10.0 «ppm» AND «H atom/proton of» aldehyde/–CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of alkyl/CH next to» aldehyde/CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of» RCOCH2- present
OR
no «signal at» 4.5–6.0 «ppm» AND absence of «H-atom/proton next to» double bond/C=C [✔]
Note: Accept “two signals with areas 1:3”.
\nReagents:
acidified/H+ AND «potassium» dichromate«(VI)»/K2Cr2O7/Cr2O72– [✔]
Conditions:
distil «the product before further oxidation» [✔]
\n
Note: Accept “«acidified potassium» manganate(VII)/KMnO4/MnO4–/permanganate”.
\nAccept “H2SO4” or “H3PO4” for “H+”.
\nAccept “more dilute dichromate(VI)/manganate(VII)” or “excess ethanol”.
\nAward M1 if correct reagents given under “Conditions”.
\n–1 [✔]
\nAny three of:
has an oxygen/O atom with a lone pair [✔]
that can form hydrogen bonds/H-bonds «with water molecules» [✔]
\nhydrocarbon chain is short «so does not disrupt many H-bonds with water molecules» [✔]
\n«large permanent» dipole-dipole interactions with water [✔]
\nAlmost all candidates recognized the products of the complete combustion of ethyne, and over two thirds managed to balance the equation. It was good to see candidates using integers for the balancing.
\nThe majority of candidates drew the Lewis structure of ethyne. A few teachers commented that they did not cover alkynes assuming they are not included in the syllabus. Please check the current syllabus carefully when preparing students.
\nA very well answered question. The vast majority of candidates understood that triple bonds are stronger than single bonds and result in a shorter bond length. It was disappointing, however, to see a considerable number of candidates stating that ethane has a double bond.
\nSome candidates could not relate evaporation of a liquid to the breaking of its intermolecular forces and gave irrelevant answers such as “evaporation”. Other candidates gave general answers such as “the intermolecular forces” or used the term “van der Waals’ forces” which did not gain credit as too vague. The current guide is clear that “London/dispersion forces” is the appropriate term to use for instantaneous dipole-induced dipole forces. Less than 40 % of the candidates scored the mark. It was disappointing to see some candidates state “covalent bonding” as the type of interaction that must be overcome when liquid ethyne vaporizes. Some teachers thought the wording of the question may have been vague and candidates may have been confused about what was meant by the “type of interaction”.
\nAbout 60 % of the candidates stated “addition” as the type of reactions that compounds containing carbon-carbon double bonds underwent. It was disappointing to see a variety of answers including substitution, condensation and combustion showing a total lack of understanding. Some candidates gave specific types such as \"bromination\" or “hydration” which did not receive the mark.
\n60 % of the candidates were able to name compound B as ethanal. Some candidates did not recognize it as an aldehyde and gave names related to carboxylic acids or other homologous series. Other candidates called it methanal.
\nCandidates were confident in using average bond enthalpies for calculating the enthalpy change for the reaction. Mistakes included forgetting to include the breaking of the O-H bonds in water and reversing the signs.
\nReasonably well answered. About half of the candidates showed understanding of the relation between stability and the enthalpy change from the same starting materials. ECF was applied in this question based on the answer in part (iii).
\nThe majority of candidates handled this question competently and nearly half of the candidates obtained both marks. They obtained the value of the absorption from the spectra provided and compared it to the values in the data booklet to deduce the identity of the product. Common mistakes included not identifying the peaks and signals precisely (for example C=O instead of CHO for 1H NMR signal at 9.4-10.0 ppm). Some teachers commented that the TMS signal should not have been included as the SL do not know about it. Other teachers commented that using the 'actual' rather than an ‘idealized’ IR spectrum may have caused confusion due to the peak at around 3400 cm-1 which could be confused for O-H in alcohols. Thankfully both of these answers were hardly seen in the scripts. The peak at 3400 cm-1 was not at all broad and did not confuse the majority of students. Please note that real spectra are usually used in examination papers, and it is worth encouraging students to check more than one peak to confirm their deductions.
Surprisingly, this question was not answered well by the majority of the candidates. However, it did discriminate well between high-scoring and low-scoring candidates. Common mistakes included incorrect formulas (such as K2CrO7), missing the acidic conditions and stating “reflux” instead of “distillation”. Many candidates gave completely irrelevant reagents and conditions such as “oxygen, pressure and a nickel catalyst”. It is possible that some candidates did not think of “distillation” as a “condition”.
\nAbout 60 % of the candidates determined the average oxidation state of carbon in ethanal. A couple of teachers commented that asking SL students to determine an “average oxidation state” seems a little difficult. Please note that this term has been used in recent papers whenever there are two or more atoms of the element in different parts of the compound. There was no evidence of confusion on the part of the candidates and most answered the question well.
\nThis was a challenging question with a demanding markscheme. Most students missed the fact that ethanal can form hydrogen bonds with water. And students who did state this often achieved only 1 out of the 3 marks because they did not offer a full explanation. Some candidates stating \"hydrogen bonding\" showed confusion by mentioning the hydrogen of the aldehyde group. Few identified the lone pairs on oxygen as the reason for the ability to hydrogen bond. Most candidates just stated that ethanal is polar and dissolves in polar water achieving no marks. However, one mark was awarded for “dipole-dipole interactions with water”.
\nWhat is the empirical formula of a hydrocarbon with 75 % carbon and 25 % hydrogen by mass?
\nA. C3H
\nB. CH2
\nC. C2H6
\nD. CH4
\nD
\nWell answered and straight forward.
\nScientists have developed various analytical techniques.
\nState an analytical technique used to separate anabolic steroids from other compounds in an athlete’s urine or blood.
\nEthanol in breath can be detected by a redox reaction. Outline this method of detection. An equation is not required.
\ngas chromatography/GC
\nOR
\nhigh performance liquid chromatography/HPLC ✔
\n\n
Accept “chromatography”, “TLC/thin-layer chromatography”, “paper chromatography” OR “extraction”.
\nDo not accept just “mass spectrometry/MS” but do not penalize any reference to MS with HPLC or GC (eg GC-MS).
\nALTERNATIVE 1:
\nAny two of:
\n«blow through tube of» acidified «orange» potassium dichromate(VI)/K2Cr2O7/dichromate/Cr2O72– ✔
\nCr(VI)/Cr6+/Cr2O72– reduced to Cr(III)/Cr3+✔
\n\n
colour changes «from orange» to green
\nOR
\ncolour change is monitored ✔
\n\n
ALTERNATIVE 2:
\noxygen reduced to water
\nOR
\nethanol oxidized to ethanoic/acetic acid ✔
\n\n
current measured ✔
\n\n
Accept “ethanol oxidized to ethanal/acetaldehyde”.
\nBenzoic acid, C6H5COOH, is another derivative of benzene.
\nDraw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
\nThe pH of an aqueous solution of benzoic acid at 298 K is 2.95. Determine the concentration of hydroxide ions in the solution, using section 2 of the data booklet.
\nFormulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.
\nSuggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
\n [✔]
\n
Note: Accept Kekulé structures.
\nNegative sign must be shown in correct position- on the O or delocalised over the carboxylate.
\nALTERNATIVE 1:
[H+] «= 10−2.95» = 1.122 × 10−3 «mol dm−3» [✔]
«[OH−] = =» 8.91 × 10−12 «mol dm−3» [✔]
\n\n
ALTERNATIVE 2:
pOH = «14 − 2.95 =» 11.05 [✔]
«[OH−] = 10−11.05 =» 8.91 × 10−12 «moldm−3» [✔]
\n\n
Note: Award [2] for correct final answer.
\nAccept other methods.
\n2C6H5COOH(s) + 15O2 (g) → 14CO2 (g) + 6H2O(l)
\ncorrect products [✔]
\ncorrect balancing [✔]
\n«intermolecular» hydrogen bonding [✔]
\n\n
Note: Accept diagram showing hydrogen bonding.
\nMost failed to score a mark for the conjugate base of benzoic acid as either they didn’t show all bonds and atoms in the ring and/or they did not put the minus sign in the correct place. Some didn't read the question carefully so gave the structure of the acid form.
\nMany students could correctly calculate the hydroxide concentration, but some weaker students calculated hydrogen ion concentration only.
\nMost students earned at least one mark for writing the correct products of the combustion of benzoic acid but the balancing appeared to be difficult for some.
\nVery few students answered this question correctly, thinking benzoic would bond with the hexane even though it was a non-polar solvent. It was very rare for a student to realize there was intermolecular hydrogen bonding.
\nWhat is the ground state electron configuration of an atom of chromium, Cr (Z = 24)?
\nA. [Ar]3d6
\nB. [Ar]4s23d4
\nC. [Ar]4s13d5
\nD. [Ar]4s24p4
\nC
\nSome candidates had ground state configuration of Cr as 4s2 3d4 rather than 4s1 3d5
\nWhich element is represented by the first eight successive ionization energies on the graph?
\nA. Mg
\nB. S
\nC. Cl
\nD. Ar
\nB
\n52 % of candidates could identify the element for successive ionization energies with no clear misconception based on the other choices.
\nThis question is about compounds of sodium.
\nSodium peroxide is used in diving apparatus to produce oxygen from carbon dioxide.
\n2Na2O2 (s) + 2CO2 (g) → 2Na2CO3 (s) + O2 (g)
\nDescribe the structure and bonding in solid sodium oxide.
\nWrite equations for the separate reactions of solid sodium oxide and solid phosphorus(V) oxide with excess water and differentiate between the solutions formed.
\nSodium oxide, Na2O:
Phosphorus(V) oxide, P4O10:
Differentiation:
Sodium peroxide, Na2O2, is formed by the reaction of sodium oxide with oxygen.
\n2Na2O (s) + O2 (g) → 2Na2O2 (s)
\nCalculate the percentage yield of sodium peroxide if 5.00 g of sodium oxide produces 5.50 g of sodium peroxide.
\nDetermine the enthalpy change, ΔH, in kJ, for this reaction using data from the table and section 12 of the data booklet.
\nOutline why bond enthalpy values are not valid in calculations such as that in (c)(i).
\nThe reaction of sodium peroxide with excess water produces hydrogen peroxide and one other sodium compound. Suggest the formula of this compound.
\nState the oxidation number of carbon in sodium carbonate, Na2CO3.
\n«3-D/giant» regularly repeating arrangement «of ions»
OR
lattice «of ions» [✔]
electrostatic attraction between oppositely charged ions
OR
electrostatic attraction between Na+ and O2− ions [✔]
\n
Note: Do not accept “ionic” without description.
\nSodium oxide:
Na2O(s) + H2O(l) → 2NaOH (aq) [✔]
Phosphorus(V) oxide:
P4O10 (s) + 6H2O(l) → 4H3PO4 (aq) [✔]
Differentiation:
NaOH / product of Na2O is alkaline/basic/pH > 7 AND H3PO4 / product of P4O10 is acidic/pH < 7 [✔]
n(Na2O2) theoretical yield «= » = 0.0807/8.07 × 10−2 «mol»
OR
mass Na2O2 theoretical yield «= × 77.98 gmol−1» = 6.291 «g» [✔]
% yield «= × 100» OR « x 100» = 87.4 «%» [✔]
\n\n
Note: Award [2] for correct final answer.
\nΣΔHf products = 2 × (−1130.7) / −2261.4 «kJ» [✔]
\nΣΔHf reactants = 2 × (−510.9) + 2 × (−393.5) / −1808.8 «kJ» [✔]
\nΔH = «ΣΔHf products − ΣΔHf reactants = −2261.4 −(−1808.8) =» −452.6 «kJ» [✔]
\n\n
Note: Award [3] for correct final answer.
\nAward [2 max] for “+452.6 «kJ»”.
\nonly valid for covalent bonds
OR
only valid in gaseous state [✔]
NaOH [✔]
\n\n
Note: Accept correct equation showing NaOH as a product.
\nIV [✔]
\nDisappointingly many students did not realise that sodium oxide was held by ionic bonds, many said it was covalent or metallic bonding. The ones that knew it was ionic failed to describe it adequately to earn the 2 marks.
\nVery few students could correctly write out the two equations and so often were unable to realise it was acid/base behaviour that would differentiate the oxides.
\nMany candidates were able to correctly calculate the % yield but some weaker candidates just used 5.0/5.5 to find %.
\nThe calculation of the enthalpy change using enthalpies of formation was generally answered well but common mistakes were students forgetting to multiply by 2 or adding extra terms for oxygen.
\nMost students didn’t gain a mark and “values are average” was the most common incorrect answer. The fact this was an ionic compound did not register with them. Some students did gain a mark for stating that the substances were not in a gaseous state.
\nSome students correctly identified sodium hydroxide as the correct product, but hydrogen, oxygen and sodium oxide were common answers.
\nOxidation number of +4 was often correctly identified.
\nPowdered zinc was reacted with 25.00 cm3 of 1.000 mol dm−3 copper(II) sulfate solution in an insulated beaker. Temperature was plotted against time.
\nEstimate the time at which the powdered zinc was placed in the beaker.
\nState what point Y on the graph represents.
\nThe maximum temperature used to calculate the enthalpy of reaction was chosen at a point on the extrapolated (dotted) line.
\nState the maximum temperature which should be used and outline one assumption made in choosing this temperature on the extrapolated line.
\n\n
Maximum temperature:
\nAssumption:
\nTo determine the enthalpy of reaction the experiment was carried out five times. The same volume and concentration of copper(II) sulfate was used but the mass of zinc was different each time. Suggest, with a reason, if zinc or copper(II) sulfate should be in excess for each trial.
\nThe formula q = mcΔT was used to calculate the energy released. The values used in the calculation were m = 25.00 g, c = 4.18 J g−1 K−1.
\nState an assumption made when using these values for m and c.
\nPredict, giving a reason, how the final enthalpy of reaction calculated from this experiment would compare with the theoretical value.
\n100 «s» [✔]
\n\n
Note: Accept 90 to 100 s.
\nhighest recorded temperature
OR
when rate of heat production equals rate of heat loss [✔]
\n
Note: Accept “maximum temperature”.
\nAccept “completion/end point of reaction”.
\nMaximum temperature:
73 «°C» [✔]
Assumption:
«temperature reached if» reaction instantaneous
OR
«temperature reached if reaction occurred» without heat loss [✔]
\n
Note: Accept “rate of heat loss is constant” OR “rate of temperature decrease is constant”.
\nAny one of:
copper(II) sulfate AND mass/amount of zinc is independent variable/being changed.
OR
copper(II) sulfate AND with zinc in excess there is no independent variable «as amount of copper(II) sulfate is fixed» [✔]
copper(II) sulfate AND having excess zinc will not yield different results in each trial [✔]
\nzinc AND results can be used to see if amount of zinc affects temperature rise «so this can be allowed for» [✔]
\nzinc AND reduces variables/keeps the amount reacting constant [✔]
\n\n
Note: Accept “copper(II) sulfate/zinc sulfate” for “solution”.
\nlower/less exothermic/less negative AND heat loss/some heat not accounted for
OR
lower/less exothermic/less negative AND mass of reaction mixture greater than 25.00 g
OR
greater/more exothermic /more negative AND specific heat of solution less than water [✔]
\n
Note: Accept “temperature is lower” instead of “heat loss”.
\nAccept “similar to theoretical value AND heat losses have been compensated for”.
\nAccept “greater/more exothermic/more negative AND linear extrapolation overestimates heat loss”.
\nAlmost all candidates identified 100 s as the time at which the reaction was initiated.
\nMany students gained this mark through stating this was the highest temperature recorded, though even more took advantage of the acceptance of the completion of the reaction, expressed in many different ways. Very few answered that it was when heat loss equalled heat production.
\nEven though almost all students recognised 100 seconds as the start time of the reaction less than 50% chose the extrapolated temperature at this time. Predictably the most common answer was the maximum of the graph, followed closely by the intercept with the y-axis. With regard to reasons, again relatively few gained the mark, though most who did wrote “no loss of heat”, even though it was rare to find this preceded by “the temperature that would have been attained if …”.
\nThe correct answer depended on whether students considered the object of the additional trials was to investigate the effect of a new independent variable (excess copper(II) sulphate) or to obtain additional values of the same enthalpy change so they could be averaged (excess zinc). Answers that gave adequate reasons were rare.
\nAgain relatively few gained these marks for stating that it was assumed the density and specific heat of the solution were the same as water.
\nOnly about a third of the students correctly deduced that loss of heat to the environment means that the experimental value is lower than the theoretical one, though other answers, such as “higher because linear extrapolation over-compensates for the heat losses” were also accepted.
\nLithium has many uses.
\nThe emission spectra obtained by ICP-OES for a mixture containing the isotope 6Li (Li-6) and naturally occurring lithium (Li (N)) is shown.
\nIdentify the type of bonding in lithium hydride, using sections 8 and 29 of the data booklet.
\nExplain why lithium is paramagnetic while lithium hydride is diamagnetic by referring to electron configurations.
\nSuggest why ICP-OES does not give good quantitative results for distinguishing 6Li from naturally occurring lithium.
\nSuggest a better method.
\nLithium is obtained by electrolysis of molten lithium chloride. Calculate the time, in seconds, taken to deposit 0.694 g Li using a current of 2.00 A.
\nQ (charge) = I (current) × t (time)
\nionic [✔]
\nlithium has an unpaired electron [✔]
\nall electrons in lithium hydride are paired [✔]
\n\n
Note: Award [1 max] for correct electron configurations of Li AND Li+ AND H- without discussion of pairing.
\nemission spectra of both «6Li and natural Li» give same colour/produce same «range of» wavelengths
OR
they have same electron transitions/same nuclear charge [✔]
\n
Note: Accept “the spectra are almost identical”.
\nICP-MS [✔]
\n\n
Note: Accept “MS/mass spectrometry”.
\nn«= » = 0.100 «mol» [✔]
\n«t = =»
4830 «s» [✔]
\n
Note: Accept “4820” OR “4825 «s»”.
\nAward [2] for correct final answer.
\nMany students correctly used the data tables and chart to deduce the bonding is ionic.
\nOnly a few students were aware of the connection between electron pairing and magnetic properties.
\nSurprisingly few candidates referred to the fact that the two spectra shown are almost identical.
\nAgain relatively few students were able to suggest ICP-MS as a more appropriate alternative.
\nAbout half the candidates who attempted this were able to calculate the amount of lithium being formed and hence the time required.
\nThis question is about a mug made of a lead alloy.
\nThe rate of lead dissolving in common beverages with various pH values was analysed.
\nIdentify the experiment with the highest rate of lead dissolving.
\nSuggest why the relationship between time and lead concentration for Cola at 16 °C is not linear.
\nLead(II) chloride, PbCl2, has very low solubility in water.
\nPbCl2 (s) Pb2+ (aq) + 2Cl− (aq)
\nExplain why the presence of chloride ions in beverages affects lead concentrations.
\nA mean daily lead intake of greater than 5.0 × 10−6 g per kg of body weight results in increased lead levels in the body.
\nCalculate the volume, in dm3, of tap water from experiment 8 which would exceed this daily lead intake for an 80.0 kg man.
\n6 [✔]
\n\n
Note: Accept “orange juice”.
\nequilibrium is being established «between lead in solution and in mug»
OR
solution becoming saturated
OR
concentration of lead ions/[Pb2+] has increased «over time»
OR
acid concentration has decreased «as reacted with lead»
OR
surface lead has decrease/formed a compound/forms insoluble layer on surface
OR
acid reacts with other metals «because it is an alloy» [✔]
\n
Note: Do not accept “concentration of cola, orange juice, etc… has decreased”.
\nDo not accept responses that only discusses mathematical or proportional relationships.
\nequilibrium shifts to the left/towards reactants [✔]
\nlead «compounds/ions» precipitate
OR
concentration of lead «ions»/[Pb2+] decreases [✔]
\n
Note: Award [2] for “equilibrium shifts to the left/towards reactants due to common ion effect”.
\nAccept “lead ions/[Pb2+] removed from solution” for M2.
\n«daily limit = 5.0 × 10–6 g kg–1 × 80.0kg =» 4.0 × 10–4 «g of lead» [✔]
\n«volume =» 2.7 × 10–2/0.027 «dm3» [✔]
\n\n
Note: Award [2] for correct final answer.
\nMost candidates did well on this question, identifying the correct experiment by number or beverage.
\nMany candidates struggled with this question, answering it from a mathematical perspective rather than explaining why the rate would decrease over time from a chemical perspective. There were several possible correct answers (reaching equilibrium, acid concentration decreasing, solution becoming saturated with lead ions, etc.…)
\nThis question was an equilibrium question. Many students received 1 mark for either concentration of lead decreased, or lead chloride was produced and quite a few recognized that the explanation was the reaction shifted to the reactant or left side for the second mark.
\nMost students receive one mark for this question, and many receive both marks. The most common mistakes involved incorrect conversions from gram to milligrams or milligrams to grams.
\nPolybutadiene, used in truck tyres, is a polymer of buta-1,3-diene. The spatial arrangement of atoms in the polymer depends on the type of catalyst used.
\nOutline two differences between heterogeneous and homogeneous catalysts.
\nSuggest, giving a reason, how elastomers used for the tyre tread can increase the traction between the tyre and the road.
\nTyre fires emit trace quantities of polychlorinated dibenzofurans and polychlorinated dibenzo-p-dioxin.
\nOutline, using section 31 of the data booklet, why polychlorinated dibenzofuran is not classed chemically as a dioxin but considered “dioxin-like”.
\nThe trace quantities of dioxins from tyre fires are rarely inhaled and instead settle on the ground.
\nDescribe why this is a health concern.
\nAny two of:
heterogeneous catalyst is in different phase than reactants AND homogeneous catalyst in same phase [✔]
homogeneous catalysts chemically change/react and are reformed at end of reaction
OR
reactants adsorb onto heterogenous catalyst and products desorb [✔]
heterogeneous catalysts are more easily removed than homogenous catalysts [✔]
\nheterogeneous catalysts can function at higher temperatures [✔]
\nhomogeneous catalysts are «generally» more selective [✔]
\nhomogeneous catalysts offer a broader range of reactions [✔]
\n\n
Note: Accept “state” for “phase”.
\nAccept “heterogeneous catalyst provides a surface to activate reaction”.
\nelastomers bend under force «and return to original form when force is released»
OR
elastomers make tyre more flexible [✔]
allows greater contact with road [✔]
\ndoes not contain heterocyclic ring with 2 oxygen atoms
OR
middle ring has only 1 oxygen atom [✔]
produces similar toxic effects to dioxins [✔]
\n\n
Note: Accept “does not contain dioxin ring” for M1.
\ntaken up by plants, which are eaten by animals «and then further dispersed»
OR
passed on in food chain [✔]
\n
Note: Accept “do not break down and can be remobilised as dust”.
\nMany students were aware of the fundamental difference between homogeneous and heterogeneous catalysts, but only a few could identify another difference to obtain full marks.
\nWhilst some students realised that elastomers would increase the area of road in contact, hardly any linked this to their fundamental property of deforming under stress.
\nThis was a question that highlighted a perennial problem; students appearing to know the answer but not having sufficient command of the English language to express it accurately. A number did identify similar toxic effects as the reason for being “dioxin like”.
\nOnly a few students linked this to the toxin entering the food chain.
\nLiquid-crystal displays (LCDs) have many uses.
\nA molecule which acts as a thermotropic liquid crystal is shown.
\nState the name of the functional group which allows the molecule to be responsive to applied electric fields.
\nExplain the effects of very low and high temperatures on the liquid-crystal behaviour of this molecule.
\n\n
Low temperature:
\nHigh temperature:
\nnitrile [✔]
\nNote: Accept “cyano\".
\nLow temperature:
intermolecular forces prevent molecules moving AND solid/«normal» crystal formation [✔]
High temperature:
«above a critical temperature» disrupts alignment of molecules AND behaves as fluid/liquid [✔]
\n
Note: Accept “weak intermolecular forces break AND behaves as fluid/liquid”.
\nA few students correctly named the nitrile group as being the one responsible for the polarity of the liquid crystal molecule.
\nVery poorly answered with candidates failing to link the change in liquid crystal properties to changes in molecular motion.
\nStarch is a natural polymer of glucose.
\nDraw the structure of the repeating unit of starch and state the type of linkage formed between these units.
\nType of linkage:
\nFormulate the equation for the complete hydrolysis of a starch molecule, (C6H10O5)n.
\nCalculate the energy released, in kJ g−1, when 3.49 g of starch are completely combusted in a calorimeter, increasing the temperature of 975 g of water from 21.0 °C to 36.0 °C. Use section 1 of the data booklet.
\nExplain how the inclusion of starch in plastics makes them biodegradable.
\ncontinuation bonds AND -O- attached to just one end AND both H-atoms on end carbons must be on the same side [✔]
\nNote: Square brackets not required.
Ignore “n” if given.
Mark may be awarded if a polymer is shown but with the repeating unit clearly identified.
\n
Type of linkage:
glycosidic [✔]
Note: Accept “ether”.
\n(C6H10O5)n(s) + nH2O (l) → nC6H12O6 (aq) [✔]
\n\n
Note: Accept “(n-1)H2O”.
\nDo not award mark if “n” not included.
\nq = «mcΔT = 975 g × 4.18 J g–1 K–1 × 15.0 K =» 61 100 «J» / 61.1 «kJ» [✔]
\n«heat per gram= =» 17.5 «kJ g–1» [✔]
\n\n
Note: Award [2] for correct final answer.
\nAny two of:
carbohydrate grains swell/break plastic into smaller pieces [✔]
inclusion of carbohydrate makes the plastic more hydrophilic/water soluble [✔]
\ncarbohydrates are broken down/hydrolysed/digested by bacteria/micro-organisms [✔]
\nplastic becomes more accessible to bacteria as holes/channels are created in it [✔]
\n«presence of» carbohydrate weakens intermolecular/London/dispersion forces between polymer chains in the plastic [✔]
\n\n
Note: Accept “starch” for “carbohydrate” throughout. Do not accept carbohydrates are broken down/hydrolyzed.
\nHardly any students were able to draw the required repeating unit, but in contrast almost all knew that the monomers were joined by a glycosidic linkage.
\nIt was very unusual to find a candidate who could give a correct equation for starch hydrolysis.
\nNearly half the students correctly calculated the enthalpy change and some of these went on to find the value in kJ g-1. The most common mistakes were to use the mass of starch rather than the mass of water and adding 273 to the temperature change.
\nWhilst many could quote from 7b, that starch undergoes hydrolysis, very few linked this to a biochemical mechanism. Other factors, relating to the reduction of intermolecular forces between the polymer chains were also rarely encountered.
\nAspartame is a derivative of a dipeptide formed between two amino acids, phenylalanine (Phe) and aspartic acid (Asp).
\nDraw a circle around the functional group formed between the amino acids and state its name.
\nName:
\nA mixture of phenylalanine and aspartic acid is separated by gel electrophoresis with a buffer of pH = 5.5.
\nDeduce their relative positions after electrophoresis, annotating them on the diagram. Use section 33 of the data booklet.
\nName:
amide/amido/carboxamide [✔]
\n
Note: Accept “peptide bond/linkage”.
\nPhe: must be on the origin [✔]
\nAsp: any position on the left/anode/+ side [✔]
\nMany candidates correctly circled the bond between the amino acid residues, though in some cases their circle missed out key atoms. Many correctly identified it as a peptide or amide linkage.
\nMost candidates seemed to realise that phenylalanine would be neutral and hence unaffected by the field, but many failed to realise that the negative charge of the aspartic acid anion would cause it to move to the left, not the right.
\nNanotechnology has allowed the manipulation of materials on the atomic level.
\nDescribe the structure and bonding of a carbon nanotube.
\n\n
Structure:
\nBonding:
\nSuggest one application for carbon nanotubes.
\nStructure:
giant covalent/network covalent [✔]
Note: Accept “cylindrical/tube shaped”.
Bonding:
each carbon covalently bonded to 3 other carbons
OR
each bond has order of 1.5
Note: Accept “has delocalized electrons” OR “has sp2 hybridization”.
\nAny one of:
3D electrodes [✔]
catalysts [✔]
biosensors [✔]
molecular stents [✔]
body armour [✔]
synthetic muscles [✔]
micro transistors/circuitry/capacitors/electrodes [✔]
reinforcing phase in a matrix/composite material «such as concrete» [✔]
micro antenna [✔]
stealth technology [✔]
water/air filtration [✔]
solar cells [✔]
tennis racquets [✔]
microelectronic circuits [✔]
\n
Note: Do not accept just general answerssuch as “medicine” or “defence”.
\nMost students were aware that nanotubes have a tubular structure, but answers to the bonding were rarely detailed enough to gain the second mark.
\nOnly a few students gained this mark and they usually gave the use of nanotubes for a reinforcing.
\nThe main fatty acid composition of cocoa butter and coconut oil is detailed below.
\n\n
The melting points of cocoa butter and coconut oil are 34 °C and 25 °C respectively.
\nExplain this in terms of their saturated fatty acid composition.
\nFats contain triglycerides that are esters of glycerol and fatty acids. Deduce an equation for the acid hydrolysis of the following triglyceride.
\nThe addition of partially hydrogenated cocoa butter to chocolate increases its melting point and the content of trans-fatty acids (trans-fats).
\nOutline two effects of trans-fatty acids on health.
\ncoconut oil has higher content of lauric/short-chain «saturated» fatty acids
OR
cocoa butter has higher content of stearic/palmitic/longer chain «saturated» fatty acids [✔]
longer chain fatty acids have greater surface area/larger electron cloud [✔]
\nstronger London/dispersion/instantaneous dipole-induced dipole forces «between triglycerides of longer chain saturated fatty acids» [✔]
\n\n
Note: Do not accept arguments that relate to the melting points of saturated and unsaturated fats.
\n\n
correct products [✔]
\ncorrectly balanced [✔]
\nAny two of:
«increased risk of» coronary/heart disease [✔]
«increased risk of» stroke [✔]
\n«increased risk of» atherosclerosis [✔]
\n«increased risk of type-2» diabetes [✔]
\nincrease in LDL cholesterol [✔]
\ndecrease in HDL cholesterol [✔]
\n«increased risk of» obesity [✔]
\nA classic instance of candidates answering the question they thought (or hoped?) they had been asked rather than the one that was asked. Almost all answers referred to the differing amounts of saturated and unsaturated fatty acids present, totally ignoring the fact that the question clearly stated “their saturated fatty acid composition”, where the relative lengths of the chains was the key point. Nevertheless some who went on to discuss the nature of the intermolecular forces between the chains gained some credit.
\nA disappointingly small number of candidates gained any marks for deducing the equation for the hydrolysis of the given lipid.
\nAlmost all students were aware of negative health effects of trans-fats, though quite a few lost marks by just stating “cholesterol” without specifying HDL or LDL.
\nAscorbic acid and retinol are two important vitamins.
\nExplain why ascorbic acid is soluble in water and retinol is not. Use section 35 of the data booklet.
\nascorbic acid: many hydroxyl/OH groups AND retinol: few/one hydroxyl/OH group
OR
ascorbic acid: many hydroxyl/OH groups AND retinol: long hydrocarbon chain [✔]
ascorbic acid: «many» H-bond with water
OR
retinol: cannot «sufficiently» H-bond with water [✔]
\n
Note: Do not accept “OH−/hydroxide”.
\nAnother instance where candidates insist on discussing water solubility in terms of polarity or hydrophilicity rather than its fundamental dependence on the presence of sufficient groups that can form hydrogen bonds to water. A few however gained a mark through pointing out the significance of the –OH groups in ascorbic acid and the long hydrocarbon chain in retinol.
\nNatural gas is an energy source composed mainly of methane.
\nNatural gas is burned to produce steam which turns turbines in an electricity generating power plant.
\nThe efficiency of several sources for power plants is given below.
\nCalculate the specific energy of methane, in MJ kg−1, using sections 1, 6 and 13 of the data booklet.
\nCalculate the maximum electric energy output, in MJ, which can be obtained from burning 1.00 kg of methane by using your answer from (a).
\nHydroelectric power plants produced 16 % of the world’s energy in 2015, down from 21 % in 1971.
\nSuggest why hydroelectric power production has a higher efficiency than the other sources given in (b) and why its relative use has decreased despite the high efficiency.
\n\n
Reason for higher efficiency:
\nReason for decreased use:
\nMethane can also be obtained by fractional distillation of crude oil.
\nDraw a circle on the diagram to show where the methane fraction is withdrawn.
\nList the following products, which are also obtained by fractional distillation, according to decreasing volatility: asphalt, diesel, gasoline, lubricating motor oil.
\nExplain how methane absorbs infrared (IR) radiation by referring to its molecular geometry and dipole moment.
\nCompare methane’s atmospheric abundance and greenhouse effect to that of carbon dioxide.
\n« = 55.5 kJ g–1 =» 55.5 «MJ kg–1» [✔]
\n«55.5 MJ × 58 % =» 32.2 «MJ» [✔]
\nReason for higher efficiency:
no heat/energy loss in producing steam
OR
no need to convert chemical energy of the fuel into heat and then heat into mechanical energy
OR
direct conversion of «gravitational» potential energy to mechanical energy [✔]
Note: Accept “less energy lost as heat” but do not accept \"no energy lost”.
\n\n
Reason for decreased use:
limited supply of available hydroelectric sites
OR
rapid growth of electrical supply in countries with little hydroelectric potential
OR
not building «new hydroelectric» dams because of environmental concerns [✔]
Note: Accept “new/alternative/solar/wind power sources «have taken over some of the demand»”.
Accept “lower output from existing stations due to limited water supplies”.
[✔]
gasoline > diesel > lubricating motor oil > asphalt [✔]
\n\n
Note: Accept products written in this order whether separated by >, comma, or nothing.
\nmethane is tetrahedral
OR
methane has zero dipole moment/is non-polar/bond polarities cancel [✔]
Any two of:
IR absorption can result in increased vibrations/bending/stretching [✔]
only modes that cause change in dipole absorb IR [✔]
\nfor methane this is asymmetric bending/stretching [✔]
\nmethane is less abundant AND has a greater effect «per mol» [✔]
\nAbout half the candidates were able to locate the appropriate data and use it to calculate the specific energy of methane.
\nMany students were aware that methane is the major component of natural gas and could use the efficiency data to calculate the electrical energy available from methane.
\nThis seemed to cause quite a lot of difficulties, especially as some students appeared totally unaware of what hydroelectric power was, with a number discussing it as if it were some kind of fuel. The most usual mark gained was from discussing environmental concerns as a reason for its decreased use.
\nHaving been given it is a gas, it is difficult to know why probably only about a third of the candidates could identify where methane would appear on a fractionating column.
\nAgain surprisingly poorly done. Firstly there appeared to be some confusion about the term “volatility” with listing in the reverse order being quite common. Secondly many seemed unaware of the nature of “asphalt” as it was the one most frequently misplaced.
\nComprehensive answers were rare. Many students gained a mark for correct statements about methane’s molecular geometry or polarity, though quite a few totally disregarded the instruction to refer to these. Some seemed aware of the link to vibrational motion and the better ones also identified the need for a change in dipole moment.
\nQuite a few candidates were aware of the relative atmospheric abundances of carbon dioxide and methane as well as their relative potency for enhancing the greenhouse effect.
\nUranium-235, 235U, is bombarded with a neutron causing a fission reaction.
\nTwo products of the fission of 235U are 144Ba and 89Kr.
\nWrite the nuclear equation for this fission reaction.
\nOutline why the reaction releases energy.
\nThe critical mass for weapons-grade uranium can be as small as 15 kg. Outline what is meant by critical mass by referring to the equation in (a)(i).
\nThe daughter product, 89Kr, has a half-life of 3.15 min.
\nCalculate the time required, in minutes, for the mass of 89Kr to fall to 6.25 % of its initial value.
\n235U + 1n → 144Ba + 89Kr + 3 1n [✔]
\ngreater binding energy per nucleon in products than reactant [✔]
\n\n
Note: Accept “mass of products less than reactants” OR “mass converted to energy/E = mc2”.
\nmass/amount/quantity required so that «on average» each fission/reaction results in a further fission/reaction [✔]
\nat least one of the «3» neutrons produced must cause another reaction [✔]
\n\n
Note: Accept “minimum mass of fuel needed for the reaction to be self-sustaining”.
\n«6.25 % = 4 half-lives, so 4 × 3.15 =» 12.6 «min» [✔]
\nOnly about a third of the candidates could write the nuclear equation for the requested fission reaction.
\nOnly about a quarter of the candidates could explain the release of energy, usually in terms of the change in nuclear binding energy.
\nLess than half the candidates could correctly explain “critical mass” with even fewer combining the definition as the minimum mass for a sustainable fission reaction with the required clarification in terms of neutrons creating a chain reaction.
\nAnother question that created far more problems than anticipated with only about half gaining the mark. Many students appeared not to realise that 6.25 % is 1/16, hence the amount remaining after 4 half lives.
\nE10 is composed of 10 % ethanol and 90 % normal unleaded fuel.
\nEthanol has a Research Octane Number (RON) of 108.6.
\nOutline how higher octane fuels affect engine performance.
\nShow that, for combustion of equal masses of fuel, ethanol (Mr = 46 g mol−1) has a lower carbon footprint than octane (Mr = 114 g mol−1).
\nBiodiesel containing ethanol can be made from renewable resources.
\nSuggest one environmental disadvantage of producing biodiesel from renewable resources.
\nincreased AND fuels can be compressed more «before ignition» [✔]
\n\n
Note: Accept “engines can be designed with higher compression ratio” OR “less chance of pre-ignition/autoignition/
knocking occurring”.
Alternative 1
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l) / 1 mol ethanol produces 2 mol CO2
OR
C8H18 (l) + 12.5O2 (g) → 8CO2 (g) + 9H2O (l) / 1 mol octane produces 8 mol CO2 [✔]
For 1 g of fuel:
« × 2 mol CO2 (g) =» 0.04 «mol CO2 (g)» from ethanol [✔]
« × 8 mol CO2 (g) =» 0.07 «mol CO2 (g)» from octane [✔]
\n\n
Alternative 2
ratio of C in ethanol:octane is 2:8, so ratio in carbon dioxide produced per mole will be 1:4 [✔]
ratio amount of fuel in 1 g = = 2.5:1 [✔]
\n4 > 2.5 so octane produces more carbon dioxide
OR
ratio of amount of carbon dioxide = 2.5:4 = 1:1.61 so octane produces more «for combustion of same mass» [✔]
use of «farm» land «for production»
OR
deforestation «for crop production for fuel»
OR
can release more NOx «than normal fuel on combustion» [✔]
\n
Note: Ignore any reference to cost.
\nOver half the students could relate a high octane number to a reduced tendency for auto-ignition, though this was usually referred to as “knocking”.
\nA question that gave the opportunity for a variety of different approaches. This challenge was beyond all but the best students, though there were a number of well argued responses.
\nMany students did not take into account “production from renewable resources” and answered in terms of the combustion of biodiesel, though about a third correctly identified the area of land biofuel crops require.
\nAspirin can be obtained from salicylic acid.
\nUnreacted salicylic acid may be present as an impurity in aspirin and can be detected in the infrared (IR) spectrum.
\nName the functional group and identify the absorption band that differentiates salicylic acid from aspirin. Use section 26 of the data booklet.
\n\n
Name:
\nAbsorption band:
\nName:
hydroxyl [✔]
Absorption band:
3200–3600 «cm–1» [✔]
\n
Note: Accept “phenol” OR “alcohol” but not “hydroxide”.
\nMany students could correctly identify a difference between the given structures and state a region in the IR spectrum where this would absorb.
\nThis question is about nuclear reactions.
\nFission of a nucleus can be initiated by bombarding it with a neutron.
\nDetermine the other product of the fission reaction of plutonium-239.
\nOutline the concept of critical mass with respect to fission reactions.
\nOutline one advantage of allowing all countries access to the technology to generate electricity by nuclear fission.
\nState one advantage of using fusion reactions rather than fission to generate electrical power.
\n90Sr, a common product of fission, has a half-life of 28.8 years.
\nDetermine the number of years for the activity of a sample of 90Sr to fall to one eighth () of its initial value.
\n[✔]
\nminimum mass to «self-»sustain chain reaction
OR
if mass of fissile material is too small, too many neutrons produced pass out of the nuclear fuel
OR
at least one neutron produced causes further reaction [✔]
Any one of:
reduction in emission of greenhouse gases «from burning fossil fuels» [✔]
economic independence/self-sufficiency «from crude oil/producing states» [✔]
\nuranium is more abundant on Earth «in terms of total energy that can be produced from this fuel» than fossil fuels [✔]
\nAny one of:
fuel is inexpensive/readily available [✔]
no/less radioactive waste is formed [✔]
lower risk of accidents/large-scale disasters [✔]
impossible/harder to use for making materials for nuclear weapons [✔]
larger amounts of energy released per unit mass [✔]
does not require a critical mass [✔]
can be used continuously [✔]
\n
Note: Accept “higher specific energy for fusion”.
\nDo not accept “no/less waste produced for fusion”.
\nAccept specific example for disasters.
\n86.4 «years» [✔]
\nThis question was well answered.
\nThis question was also fairly well answered although some students missed the concept of maintaining a chain reaction.
\nThis question was reasonable answered by many students, but some gave very vague or general answers.
\nThis was a well answered question with most student referring to fusion having less or no nuclear waste. There were many different possible correct answers.
\nThis was a well answered question with most students solving for the number of years correctly.
\nThis question is about biofuel.
\nThe structure of chlorophyll is given in section 35 of the data booklet.
\nState the feature of the chlorophyll molecule that enables it to absorb light in the visible spectrum.
\nEvaluate the use of biodiesel in place of diesel from crude oil.
\n\n
Strength:
\nLimitation:
\nlarge/extensive «electronic» conjugation
OR
«contains» many alternate single and double bonds
OR
extended system of alternating double and single bonds [✔]
\n
Note: Student response must indicate a large or extended system to award mark.
\nStrength
Any one of:
less flammable «than diesel» [✔]
recycles carbon «lower carbon footprint»
OR
lower greenhouse gas emissions [✔]
easily biodegradable «in case of spill» [✔]
\nrenewable
OR
does not deplete fossil fuel reserves [✔]
economic security/availability in countries without crude oil [✔]
\nLimitation
Any one of:
more difficult to ignite inside the engine «than diesel» [✔]
more viscous «than diesel» [✔]
\nlower energy content/specific energy/energy density [✔]
\nuses food sources
OR
uses land that could be used for food [✔]
«production is» more expensive [✔]
\nless suitable in low temperatures [✔]
\nincreased NOx emissions for biodiesel [✔]
\ngreenhouse gases still produced [✔]
\n\n
Note: Accept “«close to» carbon neutral”, “produce less greenhouse gases/CO2”.
\nAccept “engines have to be modified if biodiesel used” as limitation.
\nDo not award marks for strength and limitation that are the same topic/concept.
\nThis question was not well answered as many students missed the idea of the system being a large/extensive/extended or containing many alternating single and double bonds. Although fewer students than expected received a mark, examiners did note there were more student who wrote about conjugation or alternating single and double bonds (but were missing the idea of the system being large) which is an improvement from previous sessions.
\nThis question was well answered, and many candidates received one or both marks. Some candidate who did not receive marks were too vague, especially with the limitation.
\nThis question is about global warming.
\nState one greenhouse gas, other than carbon dioxide.
\nDescribe the effect of infrared (IR) radiation on carbon dioxide molecules.
\nOutline one approach to controlling industrial emissions of carbon dioxide.
\nAny one of:
methane, water, nitrous oxide/nitrogen(I) oxide, ozone, CFCs, sulfur hexafluoride [✔]
\n
Note: Accept formulas.
\nDo not accept “NO2”, “NOx”, “oxides of sulfur”.
\nbond length/C=O distance changes
OR
«asymmetric» stretching «of bonds»
OR
bond angle/OCO changes [✔]
\n
polarity/dipole «moment» changes
OR
dipole «moment» created «when molecule absorbs IR» [✔]
\n
Note: Accept appropriate diagrams.
\nAny one of:
capture where produced «and stored» [✔]
use scrubbers to remove [✔]
\nuse as feedstock for synthesizing other chemicals [✔]
\ncarbon credit/tax/economic incentive/fines/country specific action [✔]
use alternative energy
OR
stop/reduce use of fossil fuels for producing energy [✔]
use carbon reduced fuels «such as methane» [✔]
\nincrease efficiency/reduce energy use [✔]
\nThis question was well answered.
\nThis question was fairly well answered with most students receiving one of the two marks. There were many students who mentioned the information in M1 (asymmetric stretching and bonds vibrate) or M2 (polarity and dipole changes) more than one time but could only receive one mark. Teachers need to remind students each mark is a different topic or concept.
\nThis question was reasonably answered although there were many students who gave vague answers that did not receive marks. Carbon cannot be “filtered out” and the process of “carbon capture or scrubbing” is different from filtering.
\nA student synthesized aspirin, acetylsalicylic acid, in a school laboratory.
\n0.300 g of crude aspirin was dissolved in ethanol and titrated with sodium hydroxide solution, NaOH (aq).
\nNaOH (aq) + C9H8O4 (in ethanol) → NaC9H7O4 (aq) + H2O (l)
\nPredict one absorption band present in an infrared (IR) spectrum of aspirin, using section 26 of the data booklet.
\nDetermine the mass of aspirin which reacted with 16.25 cm3 of 0.100 mol dm−3 NaOH solution.
\nDetermine the percentage purity of the synthesized aspirin.
\nOutline how aspirin can be chemically modified to increase its solubility in water.
\nState why aspirin should not be taken with alcohol.
\nOutline two factors which must be considered to assess the greenness of any chemical process.
\nAny one of:
1050–1410 «cm–1 due to C–O» [✔]
1700–1750 «cm–1 due to C=O in acids and esters» [✔]
\n2500–3000 «cm–1 due to O–H in acids» [✔]
\n2850–3090 «cm–1 due to C–H in alkanes and arenes» [✔]
\nn(aspirin) «= n(NaOH) = × 0.100 mol dm–3 » = 1.625 × 10–3 «mol» [✔]
\nm(aspirin) «= 1.625 × 10–3 mol × 180.17 g mol–1 » = 0.293 «g» [✔]
\n\n
Note: Award [2] for correct final answer.
\n« × 100% » = 97.7 «%» [✔]
\nconvert to a salt
OR
react with sodium hydroxide/NaOH [✔]
\n
Note: Accept other reactions forming soluble salts.
\nAccept “to ionize” but not “more polar”.
\nsynergistic effect/increased toxicity
OR
increased risk of stomach/intestines bleeding/ulcers/heartburn
OR
increased risk of liver toxicity/damage
OR
increased risk of nausea/vomiting [✔]
Any two of:
energy requirements «during production» [✔]
use of toxic materials «during production» [✔]
\nuse of solvents «that are not recycled» [✔]
\nemission of toxic by-products [✔]
\nquantity of waste produced
OR
atom economy [✔]
\n
Note: Accept “E-factor/carbon efficiency/% of carbon in reactants vs products” for M1.
\nAccept references to materials being/not being recycled for M3.
\nThis was a very well answered question. Even weak candidates were able to identify one correct wavenumber.
\nThis was a fairly well answered question with most students receiving both marks and many receiving one mark.
\nThis was a reasonably well answered question but quite a few students still struggled with what should have been a basic calculation, even when using ECF from the previous question.
\nThis question was reasonably answered by most students although some candidates confused the modification of aspirin into an ionic salt with the modification of the side chain of penicillin.
\nThis question was well answered.
\nThis question was not very well answered by some students. Some students gave vague answers or responses not related to the concept of Green chemistry.
\nMedicines and drugs are tested for effectiveness and safety.
\nDistinguish between therapeutic window and therapeutic index in humans.
\n\n
Therapeutic window:
\nTherapeutic index:
\nState one advantage of using morphine as an analgesic.
\nExplain why diamorphine (heroin) is more potent than morphine using section 37 of the data booklet.
\nTherapeutic window:
range of dosage «over which a drug» provides the therapeutic/desired effect without causing adverse/toxic effects [✔]
\n
Therapeutic index:
toxic dose of drug for 50 % of population divided by minimum effective dose for 50 % of population
OR
[✔]
\n
Note: M1 may be scored from a correctly labelled diagram.
\nAccept “difference between ED50/minimum effective/therapeutic dose «for 50% of population» AND TD50/toxic dose «for 50% of population»” for M1.
\nDo not accept reference to lethal dose used in therapeutic index in animal studies.
\nblocks pain impulses/binds with «opioid» receptors in brain/CNS
OR
effective against strong pain
OR
sedate patients to reduce trauma [✔]
\n
Note: Accept “effective against pain after surgery/cancer/following serious injury”.
\nAccept “relieves anxiety/stress associated with severe/terminal illness”.
\nmorphine has «two» hydroxyl groups AND diamorphine has «two» ester/ethanoate/acetate groups
OR
molecule of diamorphine is less polar than morphine
OR
groups in morphine are replaced with less polar/non-polar groups in diamorphine [✔]
«less polar molecules» cross the blood–brain barrier faster/more easily
OR
diamorphine is more soluble in non-polar environment of CNS/central nervous system than morphine [✔]
\n
Note: Accept “alcohol/hydroxy” for “hydroxyl” but not ”hydroxide”.
\nAccept “fats” for “lipid”.
\nAccept “heroin” for “diamorphine”.
\nMost students receive one mark for this question and therapeutic window was probably the more successfully mark. Teachers need to remind students not to refer to lethal dose for therapeutic index. Some students forgot to mention the 50/50 % when explaining therapeutic index or inverted the ration as ED50/TD50.
\nThis was not a particularly well answered question even though it has been seen in previous exams. Many students confused it with the idea of solubility and passing through the blood brain barrier which was 1(b) (ii).
\nThis question was reasonably well answered with many students receiving at least one of the two marks.
\nExcess acid in the stomach can cause breakdown of the stomach lining.
\nOutline how ranitidine (Zantac) inhibits stomach acid production.
\nOutline two advantages of taking ranitidine instead of an antacid which neutralizes excess acid.
\nSome antacids contain carbonates.
\nDetermine the pH of a buffer solution which contains 0.160 mol dm−3 CO32− and 0.200 mol dm−3 HCO3−, using section 1 of the data booklet.
\npKa (HCO3−) = 10.32
\nblocks/binds H2/histamine receptors «in cells of stomach lining»
OR
prevents histamine molecules binding to H2/histamine receptors «and triggering acid secretion» [✔]
Any two of:
ranitidine can be effective in treating ulcers «but antacid is not» [✔]
ranitidine can prevent long-term damage «from overproduction of acid and antacid does not» [✔]
\nranitidine has a long-term effect «and antacid has short-term effect only» [✔]
\nranitidine does not affect ionic balance in body «and antacid does» [✔]
\nranitidine does not produce bloating/flatulence [✔]
\n\n
Note: Accept “ranitidine stops the over production of acid in the stomach while antacids neutralize the excess acid giving temporary relief”.
\n«pH = pKa + log = 10.32 + log = 10.32 – 0.097»
\n«pH =»10.22 [✔]
\nThis question was not very well answered even though it has been seen in previous scripts. Some students were not through enough in their answers to receive a mark while others confused the action of ranitidine which blocks H2 receptors with omeprazole which is a proton pump inhibitor.
\nWhile most students were awarded at least one of the two marks possible for this question some of the descriptions were too vague or incomplete to receive a mark.
\nThis was in general a well answered question. Most candidates who did not receive the mark inverted the concentration of the conjugate base/concentration of the acid in the calculation.
\nWhat is the order of increasing boiling point?
\nA. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
\nB. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
\nC. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
\nD. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
\nB
\nRanking compounds in order of increasing boiling points proved challenging. There existed a number of misconceptions based on the incorrect answers chosen. Wrong answers were evenly distributed.
\nWhich describes an atom of bismuth, Bi (Z = 83)?
\nC
\nA well answered question regarding finding principal energy level and valence electrons from position on the periodic table.
\nAntiviral medications have recently been developed for some viral infections.
\nOutline one way in which antiviral drugs work.
\nDiscuss two difficulties associated with solving the AIDS problem.
\nAny one of:
alter cell’s genetic material «so that virus cannot use it to multiply» [✔]
prevent viruses from multiplying by blocking enzyme activity within host cell
OR
inhibit the synthesis of viral components by blocking enzymes inside the cell [✔]
prevent viruses from entering «host» cell
OR
bind to cellular receptors targeted by viruses
OR
bind to virus-associated proteins/VAPs which target cellular receptors
OR
prevents removal of protein coat/capsid OR prevents injection of viral DNA/RNA into cell [✔]
prevent/hinder the release of viruses from the cell [✔]
\n\n
Note: Accept “prevents synthesis of virus by host cell”.
\nAccept “alters RNA/DNA/genetic material of virus”.
\nDo not accept just “mimics nucleotides”.
\nAny two of:
viruses lack cell structure «so difficult to target with drugs» [✔]
HIV is a retrovirus
OR
HIV genetic material is in the form of RNA instead of DNA [✔]
HIV affects/destroys helper/T-cells which are necessary to fight infection [✔]
\nHIV has great genetic diversity so difficult to produce «a» vaccine [✔]
\nanti-retroviral agents are expensive so not everyone/country can afford them [✔]
\nsocio-cultural issues deter people from seeking treatment/prevention/diagnosis
OR
lack of education/conversation/stigma associated with being HIV-positive [✔]
mutation of virus/HIV [✔]
\nvirus/HIV metabolism linked to that of host cell [✔]
\ndrugs harm host cell as well as virus/HIV [✔]
\nHIV difficult to detect/remains dormant [✔]
\nCandidates responded fairly well to this question. Students who did not receive a mark were either too vague or discussed anti-bacterial methods.
\nMost candidates were awarded at least one of the two marks possible for this question. Some student responses were too vague or discussed the social and political issues surrounding the AIDS crisis. There were also some responses that did not refer to a virus but AIDS.
\nWhat is the sum of the coefficients when the equation is balanced with the smallest whole numbers?
\n__BaCl2 (aq) + __Fe2(SO4)3 (aq) → __FeCl3 (aq) + __BaSO4 (s)
\nA. 4
\nB. 6
\nC. 8
\nD. 9
\nD
\nThis question was well answered while one teacher commented that using 1 as a coefficient in a “sum the coefficients” question seemed tricky.
\nWhat is the IUPAC name of NiCO3?
\nA. nickel(II) carbonate
\nB. nickel carbonate
\nC. nickel(I) carbonate
\nD. nitrogen(I) carbonate
\nA
\nMore than 81 % of candidates could correctly apply IUPAC nomenclature for an inorganic substance.
\nWhich series represents atoms in order of decreasing atomic radius?
\nA. N > C > Be > Mg
\nB. Mg > N > C > Be
\nC. Be > C > N > Mg
\nD. Mg > Be > C > N
\nD
\nAnother well answered question with 75 % correctly ranking elements according to atomic radii. One G2 form stated that this question was repeated in paper 2.
\nWhich diagram represents a heterogeneous mixture?
\nA
\nA few G2 forms suggested that candidates could have chosen A or B. However, homogeneous and heterogeneous mixtures are usually represented in such a way while only option A shows a clear separation in the mixture. We will avoid schematic diagrams in the future for this type of questions.
\nWhat is the empirical formula of a hydrocarbon with 75 % carbon and 25 % hydrogen by mass?
\nA. C3H
\nB. CH2
\nC. C2H6
\nD. CH4
\nD
\nWell answered and straight forward.
\nWhich graph would not show a linear relationship for a fixed mass of an ideal gas with all other variables constant?
\nA. P against V
\nB. P against
\nC. P against T
\nD. V against T
\nA
\nThe idea of P vs 1/V as being non-linear was the most common mistake, with just over 50 % of candidates earning this mark.
\nWhich electrons are removed from iron (Z = 26) to form iron(II)?
\nA. two 3d electrons
\nB. two 4s electrons
\nC. one 4s electron and one 3d electron
\nD. two 4p electrons
\nB
\nStating the 4s electrons are lost first in forming the Fe(II) ion was done correctly by 58 % but with a low discrimination index.
\nWhich combination corresponds to a strong metallic bond?
\nB
\nThe concept of charge and ionic radii size on the strength of the bond was answered well but had one of the lowest discriminatory indices on the paper.
\nWhich species has delocalized electrons?
\nA. OH−
\nB. H2CO
\nC. CO2
\nD. CO32−
\nD
\nThis question on delocalized electrons was also well answered.
\nHow many carbon atoms are sp3, sp2 and sp hybridized in the molecule?
\nA
\nIdentify carbons with sp, sp2 , and sp3 hybridization was well answered.
\nWhen equal masses of X and Y absorb the same amount of energy, their temperatures rise by 5 °C and 10 °C respectively. Which is correct?
\nA. The specific heat capacity of X is twice that of Y.
\nB. The specific heat capacity of X is half that of Y.
\nC. The specific heat capacity of X is one fifth that of Y.
\nD. The specific heat capacity of X is the same as Y.
\nA
\nAnswered correctly by 66 %, this question relating specific heat capacity to temperature rise was handled better by higher scoring candidates.
\nWhich is correct for the reaction H2O (g) → H2O (l) ?
\n\n
A. Enthalpy increases and entropy increases.
\nB. Enthalpy decreases and entropy increases.
\nC. Enthalpy increases and entropy decreases.
\nD. Enthalpy decreases and entropy decreases.
\nD
\nHigher scoring candidates had more success understanding enthalpy and entropy decreases.
\nWhich equation represents the standard enthalpy of atomization of bromine, Br2?
\nA. Br2 (l) → Br (g)
\nB. Br2 (l) → 2Br (g)
\nC. Br2 (l) → 2Br (l)
\nD. Br2 (l) → Br (l)
\nA
\nOne G2 comment asked if enthalpy of atomization was on the syllabus. This Topic 15.1 question was answered correctly by 54 % of candidates but did not differentiate well between higher and lower scoring candidates.
\nWhat is the enthalpy change of reaction for the following equation?
\nC2H4 (g) + H2 (g) → C2H6 (g)
\nC2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l) ΔH = x
\nC2H6 (g) + O2 (g) → 2CO2 (g) + 3H2O (l) ΔH = y
\nH2 (g) + O2 (g) → H2O (l) ΔH = z
\n\n
A. x + y + z
\nB. −x − y + z
\nC. x − y − z
\nD. x − y + z
\nD
\nNearly 89 % of candidates answered this Hess Law question correctly. This topic was obviously well covered.
\nWhich properties can be monitored to determine the rate of the reaction?
\nFe (s) + CuSO4 (aq) → Cu (s) + FeSO4 (aq)
\n I. change in volume
II. change in temperature
III. change in colour
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nOne G2 form queried if visible spectroscopy is on the core syllabus and therefore should candidates be aware of monitoring a reaction via colour change. However, 6.1 clearly states that following change in colour is one method of following reactions.
\nWhich factors affect the amount of product formed at the cathode during electrolysis of molten salts?
\n I. current
II. time
III. charge on the cation
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nMajority of candidates answered this correctly with the most common mistake being omission of time as a factor affecting electrolysis quantities.
\nWhich is not a requirement of the standard hydrogen electrode (SHE)?
\nA. V = 1 dm3
\nB. p(H2) = 100 kPa
\nC. use of platinum as the electrode material
\nD. [H3O+] = 1 mol dm−3
\nA
\nThis was a poorly answered question on the standard hydrogen electrode. Many candidates believed what was not required was a Pt electrode material
\nWhat is the IUPAC name of the following molecule?
\nA. 2-bromo-3-ethylbutane
\nB. 3-methyl-4-bromopentane
\nC. 2-ethyl-3-bromobutane
\nD. 2-bromo-3-methylpentane
\nD
\nAnother relatively high scoring question on IUPAC nomenclature.
\nBromine consists of two stable isotopes that exist in approximately a 1 : 1 ratio. The relative atomic mass, Ar, of bromine is 79.90. Which are the stable isotopes of bromine?
\nA. 79Br and 81Br
\nB. 80Br and 81Br
\nC. 78Br and 80Br
\nD. 79Br and 80Br
\nA
\nAlmost 56 % of candidates could find a 1:1 ratio as an average of 2 items, however many had 79Br and 80Br with an average mass of 79.90
\nWhat is the order of increasing boiling point?
\nA. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
\nB. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
\nC. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
\nD. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
\nB
\nRanking compounds in order of increasing boiling point proved challenging. There existed a number of misconceptions based on the incorrect answers chosen being close to evenly distributed.
\nSuperconductors have no resistance below a critical temperature.
\nOutline how resistance to electric currents occurs in metals.
\nSuggest why the resistance of metals increases with temperature.
\nState two differences between Type I and Type II superconductors.
\nelectrons collide with cations/positive ions ✔
\nincreased vibrations of «lattice» ions ✔
increased «probability of» collisions «between electrons and cations» ✔
NOTE: Accept “increases lattice vibrations” for M1.
\nAny two of:
Type I have sharper transitions to superconductivity «than Type II» ✔
Type I have lower critical/operating temperatures «than Type II» ✔
Type I have lower critical magnetic field «strength than Type II» ✔
Type I carry lower currents «than Type II» ✔
Type I are «pure» metals/metalloids AND Type II are alloys/metal oxide ceramics/perovskites/metallic compounds ✔
Type II exist in a mixed state/are partly permeable to the magnetic field AND Type I do not/are not ✔
Which species does not have resonance structures?
\nA. C6H6
\nB. NH4+
\nC. CO32−
\nD. O3
\nB
\nIdentifying resonance structure had one of the highest discriminatory indices. Only the higher achieving candidates had much success with this.
\nWhich describes an ionic compound?
\nB
\nFairly well answered with the most widely held misconception being that ionic compounds conduct electricity in the solid state.
\nWhich is a major product of the electrophilic addition of hydrogen chloride to propene?
\nA. ClCH2CH=CH2
\nB. CH3CH(Cl)CH3
\nC. CH3CH2CH2Cl
\nD. CH3CH=CHCl
\nB
\nMarkovnikov addition was handled much better by higher scoring candidates.
\nWhich solvent is aprotic?
\nA. H2O
\nB. C6H5CH3
\nC. CH3OH
\nD. CH3NH2
\nB
\nIdentifying protic from aprotic solvents was poorly done my most candidates.
\nWhat is the ground state electron configuration of an atom of chromium, Cr (Z = 24)?
\nA. [Ar]3d6
\nB. [Ar]4s23d4
\nC. [Ar]4s13d5
\nD. [Ar]4s24p4
\nC
\nA large number of the candidates had ground state configuration of Cr as 4s2 3d4 rather than 4s1 3d5
\nWhich alcohol would produce a carboxylic acid when heated with acidified potassium dichromate(VI)?
\nA. propan-2-ol
\nB. butan-1-ol
\nC. 2-methylpropan-2-ol
\nD. pentan-3-ol
\nB
\n75 % of candidates could correctly identify the oxidation of a primary alcohol to form carboxylic acid.
\nWhich statement is not correct regarding benzene?
\nA. It is planar.
\nB. The ring contains delocalized electrons.
\nC. It always reacts in the same way as alkenes.
\nD. The carbon–carbon bond has a bond order of 1.5.
\nC
\n73 % of candidates knew that benzene did not react the same way as alkenes.
\nWhich describes an atom of bismuth, Bi (Z = 83)?
\nC
\nA well answered question regarding finding principal energy level and valence electrons from position on the periodic table.
\nWhen equal masses of X and Y absorb the same amount of energy, their temperatures rise by 5 °C and 10 °C respectively. Which is correct?
\nA. The specific heat capacity of X is twice that of Y.
\nB. The specific heat capacity of X is half that of Y.
\nC. The specific heat capacity of X is one fifth that of Y.
\nD. The specific heat capacity of X is the same as Y.
\nA
\nAnswered correctly by 53 %, this question relating specific heat capacity to temperature rise was handled better by higher scoring candidates.
\nWhich combination corresponds to a strong metallic bond?
\n\n
B
\nThe concept of charge and ionic radii size on the strength of the bond was answered well, but had the lowest discriminatory index on the paper.
\nKevlar® is used to make racing tires.
\nDraw the structure of the monomers of Kevlar® if the by-product of the condensation polymerization is hydrogen chloride.
\nState and explain why plasticizers are added to polymers.
\nDiscuss why the recycling of plastics is an energy intensive process.
\nOR
H2NC6H4NH2 ✔
OR
Cl(O)CC6H4C(O)Cl ✔
increases flexibility/softness/plasticity ✔
break/weaken intermolecular forces/IMF/H-bonds «between chains» ✔
Any two of:
collection/transportation of plastic waste ✔
separation/sorting of different types «of plastic»
OR
separation/sorting of plastic from other materials ✔
melting plastic ✔
\nprocessing/washing/cleaning/drying/manufacture of recycled plastic ✔
\nThe dotted line represents the formation of oxygen, O2 (g), from the uncatalysed complete decomposition of hydrogen peroxide, H2O2 (aq).
\nWhich curve represents a catalysed reaction under the same conditions?
\n\n
B
\nGraphical representation of catalysis was also well answered.
\nWhich can be identified using infrared (IR) spectroscopy?
\nA. functional groups
\nB. molar mass
\nC. 3-D configuration
\nD. bond angle
\nA
\nThe vast majority of candidates knew that IR spectroscopy can be used to identify functional groups.
\nWhat is the enthalpy change of reaction for the following equation?
\nA. x + y + z
\nB. −x − y + z
\nC. x − y − z
\nD. x − y + z
\n\n
D
\n83 % of candidates answered this Hess Law question correctly. This topic was obviously well covered.
\nHow many moles of magnesium hydroxide are produced with 0.50 mol of ammonia?
\nMg3N2 (s) + 6H2O (l) → 3Mg(OH)2 (aq) + 2NH3 (aq)
\nA. 0.25
\nB. 0.33
\nC. 0.75
\nD. 1.5
\nC
\n90 % of the candidates were able to deduce the amount of a product given the amount of another product and the balanced equation.
\nWhich is correct for the reaction?
\n2Al (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H2 (g) ΔH = −1049 kJ
\nA. Reactants are less stable than products and the reaction is endothermic.
\nB. Reactants are more stable than products and the reaction is endothermic.
\nC. Reactants are more stable than products and the reaction is exothermic.
\nD. Reactants are less stable than products and the reaction is exothermic.
\nD
\nThis question on enthalpy change was also well answered.
\nWhich compound can exist as cis- and trans-isomers?
\nC
\nCis and trans isomerism on cyclic alkanes was poorly answered. This had the lowest discriminatory index on the test and all incorrect answers were fairly evenly distributed.
\nWhich properties can be monitored to determine the rate of the reaction?
\nFe (s) + CuSO4 (aq) → Cu (s) + FeSO4 (aq)
\n I. change in volume
II. change in temperature
III. change in colour
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nOne G2 form queried if visible spectroscopy is on the core syllabus and therefore should candidates be aware of monitoring a reaction via colour change. However, 6.1 clearly states that following change in colour is one method of following reactions.
\nMethane undergoes incomplete combustion.
\n2CH4 (g) + 3O2 (g) → 2CO (g) + 4H2O (g)
\nWhat is the enthalpy change, in kJ, using the bond enthalpy data given below?
\nA. [2(1077) + 4(463)] − [2(414) + 3(498)]
\nB. [2(414) + 3(498)] − [2(1077) + 4(463)]
\nC. [8(414) + 3(498)] − [2(1077) + 8(463)]
\nD. [2(1077) + 8(463)] − [8(414) + 3(498)]
\nC
\nOnly 70 % of the candidates chose the correct calculation of the enthalpy change using bond enthalpy data. The most commonly chosen distractor (D) reversed the signs.
\nHow should a measurement of 5.00 g from a balance be recorded?
\nA. 5.00 ± 0.1 g
\nB. 5.00 ± 0.01 g
\nC. 5.00 ± 1 g
\nD. 5.00 ± 0.001 g
\nB
\nMore than 85 % of candidates were able to identify a correct uncertainty, however not a good discriminating question.
\nWhat is the sum of the integer coefficients when propene undergoes complete combustion?
\n__C3H6 (g) + __O2 (g) → __CO2 (g) + __H2O (l)
\nA. 11
\nB. 17
\nC. 21
\nD. 23
\nD
\n93 % of the candidates balanced the equation for the combustion of propene and obtained the sum of the integer coefficients.
\nWhat are typical characteristics of metals?
\nA
\nVery disappointing results with a large percentage of the candidates thinking metals have a high electron affinity.
\nWhat is the activation energy of the reverse reaction?
\n\n
C
\n81 % of the candidates identified the activation energy of the reverse reaction. The most commonly chosen distractor was B, the activation energy of the forward reaction.
\nWhat is the volume of gas when the pressure on 100 cm3 of gas is changed from 400 kPa to 200 kPa at constant temperature?
\nA. 50.0 cm3
\nB. 100 cm3
\nC. 200 cm3
\nD. 800 cm3
\nC
\n85 % of the candidates were able to deduce the new volume of a sample of gas after the pressure was halved. The most commonly chosen distractor was the value that assumed a direct proportionality between volume and pressure.
\nWhich is correct for ?
\nD
\n93 % of the candidates deduced the correct numbers of protons, neutrons and electrons in the sulfide ion.
\nWhich of the following transitions in the hydrogen atom emits the least energy?
\nA. n = 2 to n = 1
\nB. n = 3 to n = 1
\nC. n = 4 to n = 2
\nD. n = 4 to n = 3
\nD
\n81 % of the candidates identified the transition the emits the least energy in the given four transitions in the hydrogen atom (which was n=4 to n=3). The most commonly chosen distractor was A (n=2 to n=1) which means these candidates did not realize that the energy levels in the hydrogen atom converge at higher energy.
\nDinitrogen monoxide, N2O, causes depletion of ozone in the stratosphere.
\nDifferent sources of N2O have different ratios of 14N:15N.
\nOutline why ozone in the stratosphere is important.
\nState one analytical technique that could be used to determine the ratio of 14N:15N.
\nA sample of gas was enriched to contain 2 % by mass of 15N with the remainder being 14N.
\nCalculate the relative molecular mass of the resulting N2O.
\nPredict, giving two reasons, how the first ionization energy of 15N compares with that of 14N.
\nSuggest why it is surprising that dinitrogen monoxide dissolves in water to give a neutral solution.
\nabsorbs UV/ultraviolet light «of longer wavelength than absorbed by O2» [✔]
\nmass spectrometry/MS [✔]
\n« » 14.02 [✔]
\n«Mr = (14.02 × 2) + 16.00 =» 44.04 [✔]
\nAny two:
same AND have same nuclear charge/number of protons/Zeff [✔]
same AND neutrons do not affect attraction/ionization energy/Zeff
OR
same AND neutrons have no charge [✔]
same AND same attraction for «outer» electrons [✔]
\nsame AND have same electronic configuration/shielding [✔]
\n\n
Note: Accept “almost the same”.
“same” only needs to be stated once.
oxides of nitrogen/non-metals are «usually» acidic [✔]
\n60 % of the candidates were aware that ozone in the atmosphere absorbs UV light. Some candidates did not gain the mark for not specifying the type of radiation absorbed.
\nWell answered. More than half of the candidates stated mass spectrometry is used to determine the ratio of the isotopes.
\nMany candidates successfully calculated the relative atomic mass of nitrogen in the sample. M2 was awarded independently of M1, so candidates who calculated the relative molecular mass using the Ar of nitrogen in the data booklet (14.01) were awarded M2. Many candidates scored both marks.
\nThis was a challenging question for many candidates, while stronger candidates often showed clarity of thinking and were able to conclude that the ionization energies of the two isotopes must be the same and to provide two different reasons for this. Some candidates did realize that the ionization energies are similar but did not give the best reasons to support their answer. Many candidates thought the ionization energies would be different because the size of the nucleus was different. Some teachers commented that the question was difficult while others liked it because it made students apply their knowledge in an unfamiliar situation. The question had a good discrimination index.
\nOnly a quarter of the candidates answered correctly. Some simply stated that N2O forms HNO3 with water which did not gain the mark.
\nHow is colour produced in transition metal complexes?
\nA. Light is absorbed when electrons are promoted between split d-orbitals.
\nB. Light is emitted when electrons fall between split d-orbitals.
\nC. Light is absorbed when electrons escape from the complex.
\nD. Light is emitted when the complex returns to ground state.
\nA
\n69 % of the candidates understood how colour is produced in transition metal complexes. The most commonly chosen distractor was B, which also recognized the involvement of the split d-orbitals, however stated that colour is produced when light is emitted when electrons fall between split d-orbitals.
\nHow do the following properties change down Group 17 of the periodic table?
\nC
\n85 % of the candidates know the trends in ionization energy and ionic radius down Group 17. The other distractors were equally chosen by the remainder of the candidates.
\nWhat is the oxidation state of the metal ion and charge of the complex ion in [Co(NH3)4Cl2]Cl?
\nC
\nThis was one of the most challenging questions on the paper and it discriminated well between high scoring and low scoring candidates. 57 % of the candidates were able to use the formula of the compound to deduce the oxidation state of the metal ion and the charge of the complex ion. The most commonly chosen distractor was B where the charge of the complex ion was correct but the charge of the metal ion was not. Some teachers commented that the question was challenging but reasonable.
\nHow does a lithium atom form the most stable ion?
\nA. The atom gains a proton to form a positive ion.
\nB. The atom loses a proton to form a negative ion.
\nC. The atom loses an electron to form a positive ion.
\nD. The atom gains an electron to form a negative ion.
\nC
\nOne of the most straight-forward questions on the paper about how lithium forms its most stable ion.
\nRhenium, Re, was the last element with a stable isotope to be isolated.
\nOne chloride of rhenium has the empirical formula ReCl3.
\nBefore its isolation, scientists predicted the existence of rhenium and some of its properties.
\nSuggest the basis of these predictions.
\nDescribe how the relative reactivity of rhenium, compared to silver, zinc, and copper, can be established using pieces of rhenium and solutions of these metal sulfates.
\nState the name of this compound, applying IUPAC rules.
\nCalculate the percentage, by mass, of rhenium in ReCl3.
\ngap in the periodic table
OR
element with atomic number «75» unknown
OR
break/irregularity in periodic trends [✔]
«periodic table shows» regular/periodic trends «in properties» [✔]
\nplace «pieces of» Re into each solution [✔]
\nif Re reacts/is coated with metal, that metal is less reactive «than Re» [✔]
\n\n
Note: Accept other valid observations such as “colour of solution fades” or “solid/metal appears” for “reacts”.
\nrhenium(III) chloride
OR
rhenium trichloride [✔]
«Mr ReCl3 = 186.21 + (3 × 35.45) =» 292.56 [✔]
\n«100 × =» 63.648 «%» [✔]
\nThis nature of science question generated a lot of discussion among teachers. Some in support of such questions and others concerned that it takes a lot of time for candidates to know how to answer. Some teachers thought it was unclear what the question was asking. It is pleasing that about a quarter of the candidates answered both parts successfully and many candidates gained one mark usually for “periodic trends”. However, some candidates only focused on one part of the question. Quite a few candidates discussed isotopes, probably thrown off by the stem. A teacher was concerned that since transition metals are not part of the SL syllabus that Re was a bad choice, however, the question did not really require any transition metal chemistry to be answered.
\nThis question was a good discriminator between high-scoring and low-scoring candidates. It was well answered by more than half of the candidates who had obviously carried out such displacement reactions and interpreted the outcomes during the course. Some candidates did not state the obvious of dipping the metal into the sulfates.
\nMore than half of the candidates named ReCl3 correctly. Common mistakes included “rhenium chloride” and “trichlororhenium”.
\nThe majority of candidates calculated the percentage, by mass, of rhenium in ReCl3 correctly. Some rounding errors were seen that students should be more careful with.
\nWhich combination causes the strength of metallic bonding to increase?
\nD
\nA straight-forward question about factors affecting the strength of metallic bonding answered correctly by 86 % of candidates. The distractors were almost equally chosen by the rest of the candidates.
\nCarbonated water is produced when carbon dioxide is dissolved in water under pressure.
\nThe following equilibria are established.
\nCarbon dioxide acts as a weak acid.
\nSoda water has sodium hydrogencarbonate, NaHCO3, dissolved in the carbonated water.
\nDistinguish between a weak and strong acid.
\n\n
Weak acid:
\nStrong acid:
\nThe hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
\nState the formula of its conjugate base.
\nWhen a bottle of carbonated water is opened, these equilibria are disturbed.
\nState, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).
\nPredict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)
\n100.0 cm3 of soda water contains 3.0 × 10−2 g NaHCO3.
\nCalculate the concentration of NaHCO3 in mol dm−3.
\nIdentify the type of bonding in sodium hydrogencarbonate.
\n\n
Between sodium and hydrogencarbonate:
\nBetween hydrogen and oxygen in hydrogencarbonate:
\nWeak acid: partially dissociated/ionized «in solution/water»
AND
Strong acid: «assumed to be almost» completely/100 % dissociated/ionized «in solution/water» [✔]
CO32– [✔]
\nshifts to left/reactants AND to increase amount/number of moles/molecules of gas/CO2 (g) [✔]
\n«additional HCO3–» shifts position of equilibrium to left [✔]
\npH increases [✔]
\n\n
Note: Do not award M2 without any justification in terms of equilibrium shift in M1.
\n«molar mass of NaHCO3 =» 84.01 «g mol–1» [✔]
\n«concentration = » 3.6 × 10–3 «mol dm–3» [✔]
\n\n
Note: Award [2] for correct final answer.
\nBetween sodium and hydrogencarbonate:
ionic [✔]
Between hydrogen and oxygen in hydrogencarbonate:
«polar» covalent [✔]
It was rather disappointing that less than 70 % of the candidates could distinguish between weak and strong acids. Many candidates referred to pH differences.
\nA poorly answered question, though it discriminated very well between high-scoring and low-scoring candidates. Less than 40 % of the candidates were able to deduce the formula of the conjugate base of HCO3-. Wrong answers included water, the hydroxide ion and carbon dioxide.
\nThis was a relatively challenging question. Only about a quarter of the candidates explained how a decrease in pressure affected the equilibrium. Some candidates stated there was no shift in the equilibrium as the number of moles is the same on both sides of the equation, not acknowledging that only gaseous substances need to be considered when deciding the direction of shift in equilibrium due to a change in pressure. Some candidates wrote that the equilibrium shifts right because the gas escapes.
\nThis was one of the most challenging questions on the paper that required application of Le Chatelier’s Principle in an unfamiliar situation. Most candidates did not refer to equilibrium (2), as directed by the question, and hence could not gain any marks. Some candidates stated that NaHCO3 was an acid and decreased pH. Some answers had contradictions that showed poor understanding of the pH concept.
\nVery well answered. Most candidates calculated the molar concentration correctly.
\nMany candidates identified the bonding between sodium and hydrogencarbonate as ionic. A much smaller proportion of candidates identified the bonding between hydrogen and oxygen in hydrogencarbonate as covalent. The most common mistake was “hydrogen bonding”.
\nWhich species has a square planar molecular geometry?
\nA. SF4
\nB. XeF4
\nC. CF4
\nD. PF4+
\nB
\nThis was a highly discriminating question. 67 % of the candidates identified the species that has a square planar molecular geometry. One teacher commented that this was a difficult recall question. However, the intention here is not to memorize shapes but rather to use VSEPR theory to deduce the shape.
\nHow many sigma (σ) and pi (π) bonds are present in hydrogen cyanide, HCN?
\nB
\n89 % of the candidates deduced the numbers of sigma and pi bonds in HCN correctly. The most commonly chosen distractor (C) had the correct number of sigma bonds but only one pi bond.
\nWhat is the hybridization of carbon and oxygen in methanol?
\nA
\n77 % of the candidates deduced the hydridization of carbon and oxygen in methanal. A teacher commented that it was unusual to ask for the hydridization of oxygen, however, this is a reasonable application of the concept.
\nPhysical properties of elements vary according to atomic number. Sections 6 to 9 of the data booklet list some of these properties.
\nDeduce, giving a reason, the group of elements in the periodic table most likely to undergo sublimation.
\nDescribe the density trend across periods 4 and 5 of the periodic table.
\nSuggest, with a reason, whether the lanthanoids or actinoids of the f-block would have the higher density.
\nCompare the ease of oxidation of s-block and d-block metals to their melting points and densities. Use section 25 of the data booklet.
\nSketch how the first ionization energies of elements vary with their atomic radius.
\n\n
group 18/noble gases [✔]
\nsmallest difference between melting and boiling points
OR
weakest intermolecular forces «in that period» [✔]
\n
Note: Accept “group 17/halogens”.
\ndensity increases «to a maximum in the transition elements» AND then decreases [✔]
\nactinoids AND density increases down all groups «due to large increase in atomic mass for small increase in atomic volume»
OR
actinoids AND «much» greater atomic mass with similar type of bonding
OR
actinoids AND density «of actinoids» atomic number 90 to 95 is greater than corresponding lanthanoids [✔]
\n
Note: Accept “actinoids AND on graph actinoids have «much» greater density than lanthanoids”.
\nAlternative 1:
«metals with» low densities oxidize easier [✔]
«metals with» low melting points oxidize easier [✔]
\nAlternative 2:
in s-block «metals with» high densities oxidize easier
OR
in s-block «metals with» low melting points oxidize easier [✔]
in d-block «metals with» low densities oxidize easier
OR
in d-block «metals with» low melting points oxidize easier [✔]
\n
Note: Award [1 max] for “s-block metals more easily oxidized” OR “s-block metals have lower melting points” OR “s-block metals have lower densities”.
\nAccept “have greater activity” for “oxidize easier”.
\n [✔]
\n
Note: Accept any negative sloping line.
\nDo not award mark if line touches either axis.
\nSome candidates appeared to be unfamiliar with the term “sublimation”. Whilst most students correctly identified the noble gases as the group most likely to sublime there were a variety of other answers. Many students referred to “low melting and boiling points” rather than identifying the difference between these as the key factor.
\nMost students realised that density peaks around the middle of each of these periods, though a significant minority seemed unaware of the significance of “periods” and just reported the overall general increase in density.
\nPoorly answered. Some students seemed unaware of the terms “lanthanoids” and “actinoids”. Many others just stated the actinoids because they had greater atomic masses, without adding that the bonding, and hence the volume occupied by each atom, would be similar to the lanthanoids. Others responded in terms of the given data, but this required rather more justification than just stating “as can be seen from the graph”.
\nAlmost all candidates gained some credit on this question and many obtained full marks. Students were generally aware that s-block elements are more reactive than d-block ones and hence are more easily oxidised. Many correctly linked this to lower melting points and densities. Often a causal relationship was implied (more reactive because of their low density/mp) but this was not penalised. A significant minority of students referred to only one of the physical properties; not reading the question fully?
\nMost students sketched a negative correlation between IE and radius, but then many lost the mark by drawing a line that met an axis; it is not possible for either to have a value of zero.
\nWhich will increase the rate of reaction between calcium carbonate and hydrochloric acid?
\n I. an increase in temperature
II. an increase in concentration of hydrochloric acid
III. an increase in particle size of calcium carbonate
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nOne teacher pointed out that the question should have stated for a fixed mass of calcium carbonate. However, this question was one of the highest scoring so it did not cause confusion.
\nKc for 2N2O (g) 2N2 (g) + O2 (g) is 7.3 × 1034.
\nWhat is Kc for the following reaction, at the same temperature?
\nN2 (g) + O2 (g) N2O (g)
\nA. 7.3 × 1034
\nB.
\nC.
\nD.
\nB
\nNot scored well and a couple of teachers thought it was a trick question and particularly tough. Relationship between K values on reversed equations should be understood.
\nWhich solution is basic at 25 °C?
\nKw = 1.0 × 10−14
\nA. [H+] = 1.0 × 10−3 mol dm−3
\nB. [OH−] = 1.0 × 10−13 mol dm−3
\nC. solution of pH = 4.00
\nD. [H3O+] = 1.0 × 10−13 mol dm−3
\nD
\nThis was very poorly answered and had one of the highest discriminatory indices. More candidates thought that a basic solution would have [OH- ] = 1.0 x 10-13 mol dm-3 , rather than [H3O+ ] = 1.0 x 10-13 mol dm-3
\nWhich is not a source of oxides of sulfur and nitrogen?
\nA. burning coal
\nB. internal combustion engines
\nC. burning methane
\nD. volcanic eruptions
\nC
\nOne teacher thought this was a tricky question if candidates did not think of impurities in the combustion. However, 64 % of candidates answered correctly.
\nWhere does oxidation occur in a voltaic cell?
\nA. positive electrode and anode
\nB. negative electrode and anode
\nC. positive electrode and cathode
\nD. negative electrode and cathode
\nB
\nOne G2 form thought the question was awkwardly worded, implying oxidation is occurring in two places rather than one. This question also had a relatively high discriminatory index.
\nWhich is the species oxidized and the oxidizing agent in the reaction?
\nMnO2 (s) + 4HCl (aq) → MnCl2 (aq) + Cl2 (g) + 2H2O (l)
\nD
\nThis question on identifying oxidizing agents was well answered by high scoring candidates, but not handled well by low scoring candidates.
\nMetals are extracted from their ores by various means.
\nAluminium is produced by the electrolysis of alumina (aluminium oxide) dissolved in cryolite.
\nDiscuss why different methods of reduction are needed to extract metals.
\nDetermine the percentage of ionic bonding in alumina using sections 8 and 29 of the data booklet.
\nWrite half-equations for the electrolysis of molten alumina using graphite electrodes, deducing the state symbols of the products.
\nAnode (positive electrode):
Cathode (negative electrode):
\nions of more reactive metals are harder to reduce
OR
more reactive metals have more negative electrode potentials ✔
electrolysis is needed/used for most reactive metals
OR
carbon is used to reduce metal oxides of intermediate reactivity/less reactive than carbon
OR
heating ore is sufficient for less reactive metals ✔
NOTE: Award [1 max] for “«ease of reduction/extraction» depends on reactivity”.
\nelectronegativity difference = 1.8 «and average electronegativity = 2.5» ✔
57 «%» ✔
NOTE: Accept any value in the range 52−65 %.
Award [2] for correct final answer.
Anode (positive electrode):
2O2− → 4e− + O2(g)
OR
2O2− + C → 4e− + CO2 (g) ✔
NOTE: Award [1 max] for M1 and M2 if correct half-equations are given at the wrong electrodes OR if incorrect reversed half-equations are given at the correct electrodes.
\n\n
Cathode (negative electrode):
Al3+ + 3e− → Al (l) ✔
O2 gas AND Al liquid ✔
NOTE: Only state symbols of products required, which might be written as (g) and (l) in half-equations. Ignore any incorrect or missing state symbols for reactants.
\nWhich product will be obtained at the anode (positive electrode) when molten NaCl is electrolysed?
\nA. Na (l)
\nB. Cl (g)
\nC. Cl2 (g)
\nD. Na (s)
\nC
\n46 % of candidates correctly identified products of electrolysis at the anode with the incorrect answers being split by the remaining 54 %
\nWhich compound has the lowest boiling point?
\nA. CH3CH2CH2CH2CH2CH3
\nB. CH3CH2CH2CH2CH3
\nC. CH3CH(CH3)CH2CH3
\nD. CH3C(CH3)2CH3
\nD
\nSurprisingly, this was one of the challenging questions on the paper. Only 57 % of the candidates chose dimethylpropane as the compound having the lowest boiling point. The most commonly chosen distractor was pentane (B) which did not take into account the effect of branching on the strength of London dispersion forces.
\nWhich functional group is surrounded in the molecule?
\nA. hydroxyl
\nB. carboxyl
\nC. carbonyl
\nD. ether
\nC
\nOne G2 comment stated “the functional group by definition is an aldehyde, however, students must assume that the H should be considered as an 'R group in order to call it a carbonyl”. However, aldehyde was not an alternative answer, with the main misconception considering the C=O as a carboxyl rather than carbonyl.
\nWhich of the following can be both formed from bromoethane and converted directly into ethanal?
\nCH3CH2Br → X
X → CH3CHO
A. CH3CH2OH
\nB. CH3OCH3
\nC. CH3COOH
\nD. H2C=CHBr
\nA
\nThe question involving a sequence of organic reactions discriminated well between high scoring and low scoring candidates. 68 % chose the correct compound.
\nWhat is the IUPAC name of the following molecule?
\nA. 2-bromo-3-ethylbutane
\nB. 3-methyl-4-bromopentane
\nC. 2-ethyl-3-bromobutane
\nD. 2-bromo-3-methylpentane
\nD
\n57 % of candidates could correctly apply IUPAC nomenclature.
\nPropene is an important starting material for many products. The following shows some compounds which can be made from propene, C3H6.
\n
Propene (C3H6) → C3H7Cl → C3H8O → C3H6O
Consider the conversion of propene to C3H7Cl.
\nAn experiment was carried out to determine the order of reaction between one of the isomers of C3H7Cl and aqueous sodium hydroxide. The following results were obtained.
\nState the type of reaction.
\nState the IUPAC name of the major product.
\nOutline why it is the major product.
\nWrite an equation for the reaction of the major product with aqueous sodium hydroxide to produce a C3H8O compound, showing structural formulas.
\nDetermine the rate expression from the results, explaining your method.
\nDeduce the type of mechanism for the reaction of this isomer of C3H7Cl with aqueous sodium hydroxide.
\nSketch the mechanism using curly arrows to represent the movement of electrons.
\nWrite an equation for the complete combustion of the compound C3H8O formed in (a)(iv).
\nDetermine the enthalpy of combustion of this compound, in kJ mol−1, using data from section 11 of the data booklet.
\nState the reagents for the conversion of the compound C3H8O formed in (a)(iv) into C3H6O.
\nExplain why the compound C3H8O, produced in (a)(iv), has a higher boiling point than compound C3H6O, produced in d(i).
\nExplain why the 1H NMR spectrum of C3H6O, produced in (d)(i), shows only one signal.
\nPropene is often polymerized. Draw a section of the resulting polymer, showing two repeating units.
\n«electrophilic» addition ✔
\nNOTE: Do not accept “nucleophilic addition” or “free radical addition”.
Do not accept “halogenation”.
2-chloropropane ✔
\nsecondary carbocation/carbonium «ion» is more stable
OR
carbocation/carbonium «ion» stabilized by two/more alkyl groups ✔
CH3CHClCH3 (l) + OH− (aq) → CH3CH(OH)CH3 (aq) + Cl− (aq)
OR
CH3CHClCH3 (l) + NaOH (aq) → CH3CH(OH)CH3 (aq) + NaCl (aq) ✔
Rate = k [C3H7Cl] [OH−] ✔
«[OH−] held constant and» [C3H7Cl] triples AND rate triples «so first order wrt C3H7Cl» ✔
[C3H7Cl] doubles AND [OH−] doubles AND rate quadruples «so first order wrt OH−» ✔
\nSN2 ✔
\nNOTE: Accept ‘bimolecular nucleophilic substitution.’
\ncurly arrow going from lone pair on O/negative charge on OH– to C ✔
curly arrow showing C–Cl bond breaking ✔
representation of transition state showing negative charge, square brackets and partial bonds ✔
formation of CH3CH(OH)CH3 AND Cl– ✔
\n\n
NOTE: Do not allow arrow originating on H in OH–.
\nAllow curly arrow going from bond between C and Cl to Cl in either reactant or transition state.
\nDo not award M3 if OH–C bond is represented.
\nAccept formation of NaCl instead of Cl–.
\n2C3H8O (l) + 9O2 (g) → 6CO2 (g) + 8H2O (g)
OR
C3H8O (l) + 4.5O2 (g) → 3CO2 (g) + 4H2O (g) ✔
bonds broken:
7(C–H) + C–O + O–H + 2(C–C) + 4.5(O=O)
OR
7(414 «kJ mol−1») + 358 «kJ mol−1» + 463 «kJ mol−1» + 2(346 «kJ mol−1») + 4.5(498 «kJ mol−1») / 6652 «kJ» ✔
bonds formed:
6(C=O) + 8(O–H)
OR
6(804 «kJ mol−1») + 8(463 «kJ mol−1») / 8528 «kJ» ✔
«ΔH = bonds broken − bonds formed = 6652 – 8528 =» −1876 «kJ mol−1» ✔
\n
NOTE: Award [3] for correct final answer.
\nK2Cr2O7/Cr2O72–/«potassium» dichromate «(VI)» AND acidified/H+
OR
«acidified potassium» manganate(VII) / «H+ and» KMnO4 / «H+ and» MnO4– ✔
\n
NOTE: Accept “H2SO4” or “H3PO4” for “H+”.
Do not accept HCl.
Accept “permanganate” for “manganate(VII)”.
C3H8O/propan-2-ol: hydrogen-bonding AND C3H6O/propanone: no hydrogen bonding/«only» dipole–dipole/dispersion forces ✔
hydrogen bonding stronger «than dipole–dipole» ✔
\nonly one hydrogen environment
OR
methyl groups symmetrical «around carbonyl group» ✔
NOTE: Accept “all hydrogens belong to methyl groups «which are in identical positions»”.
\n✔
\n
NOTE: Continuation bonds must be shown.
\nMethyl groups may be drawn on opposite sides of the chain or head to tail.
\nIgnore square brackets and “n”.
\nMethane reacts with chlorine in sunlight.
\nCH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g)
\nWhich type of reaction occurs?
\nA. free-radical substitution
\nB. electrophilic substitution
\nC. nucleophilic substitution
\nD. electrophilic addition
\nA
\n74 % of the candidates chose free-radical substitution as the type of reaction occurring between methane and chlorine in sunlight.
\nA molecule of citric acid, C6H8O7, is shown.
\nThe equation for the first dissociation of citric acid in water is
\nC6H8O7 (aq) + H2O (l) C6H7O7− (aq) + H3O+ (aq)
\nIdentify a conjugate acid–base pair in the equation.
\nThe value of Ka at 298 K for the first dissociation is 5.01 × 10−4.
\nState, giving a reason, the strength of citric acid.
\nThe dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H+], and on Ka, of increasing the temperature.
\nCalculate the standard Gibbs free energy change, , in kJ mol−1, for the first dissociation of citric acid at 298 K, using section 1 of the data booklet.
\nComment on the spontaneity of the reaction at 298 K.
\nOutline two laboratory methods of distinguishing between solutions of citric acid and hydrochloric acid of equal concentration, stating the expected observations.
\nC6H8O7 AND C6H7O7−
OR
H2O AND H3O+ ✔
weak acid AND partially dissociated
OR
weak acid AND equilibrium lies to left
OR
weak acid AND Ka < 1 ✔
« = −RT ln K = −8.31 J K–1 mol–1 × 298 K × ln(5.01 × 10–4) ÷ 1000 =» 18.8 «kJ mol–1» ✔
\nnon-spontaneous AND positive ✔
\nAny two of:
«electrical» conductivity AND HCl greater ✔
pH AND citric acid higher ✔
titrate with strong base AND pH at equivalence higher for citric acid ✔
add reactive metal/carbonate/hydrogen carbonate AND stronger effervescence/faster reaction with HCl ✔
titration AND volume of alkali for complete neutralisation greater for citric acid ✔
titrate with strong base AND more than one equivalence point for complete neutralisation of citric acid ✔
titrate with strong base AND buffer zone with citric acid ✔
\n
NOTE: Accept “add universal indicator AND HCl more red/pink” for M2.
\nAccept any acid reaction AND HCl greater rise in temperature.
\nAccept specific examples throughout.
\nDo not accept “smell” or “taste”.
\nAnother common acid found in food is ethanoic acid.
\nA sample of ethanoic acid was titrated with sodium hydroxide solution, and the following pH curve obtained.
\nAnnotate the graph to show the buffer region and the volume of sodium hydroxide at the equivalence point.
\nIdentify the most suitable indicator for the titration using section 22 of the data booklet.
\nDescribe, using a suitable equation, how the buffer solution formed during the titration resists pH changes when a small amount of acid is added.
\nbuffer region on graph ✔
equivalence point/Veq on graph ✔
NOTE: Construction lines not required.
\nphenolphthalein ✔
\nNOTE: Accept phenol red.
\nALTERNATIVE 1:
H+ (aq) + CH3COO– (aq) → CH3COOH (aq) ✔
added acid neutralised by ethanoate ions
OR
«weak» CH3COOH (aq)/ethanoic acid replaces H+ (aq)
OR
CH3COOH/CH3COO– ratio virtually/mostly unchanged ✔
ALTERNATIVE 2:
CH3COOH (aq) H+ (aq) + CH3COO– (aq) ✔
equilibrium shifts to the ethanoic acid side
OR
CH3COOH/CH3COO− ratio virtually/mostly unchanged ✔
The following data were recorded for determining the density of three samples of silicon, Si.
\nWhich average density value, in g cm−3, has been calculated to the correct number of significant figures?
\nA. 2
\nB. 2.3
\nC. 2.27
\nD. 2.273
\nB
\nThis question about choosing the answer to a calculation with the appropriate number of significant figures discriminated well between high-scoring and low-scoring candidates. Candidates found this question relatively challenging and only 62 % chose the answer with two significant figures. The most commonly chosen distractor was C which expressed the answer to three significant figures.
\nCopper forms two chlorides, copper(I) chloride and copper(II) chloride.
\nTwo electrolysis cells were assembled using graphite electrodes and connected in series as shown.
\n\n
Copper(I) chloride undergoes a disproportionation reaction, producing copper(II) chloride and copper.
\n2Cu+ (aq) → Cu (s) + Cu2+ (aq)
\nDilute copper(II) chloride solution is light blue, while copper(I) chloride solution is colourless.
\nState the electron configuration of the Cu+ ion.
\nCopper(II) chloride is used as a catalyst in the production of chlorine from hydrogen chloride.
\n4HCl (g) + O2 (g) → 2Cl2 (g) + 2H2O (g)
\nCalculate the standard enthalpy change, ΔHθ, in kJ, for this reaction, using section 12 of the data booklet.
\nThe diagram shows the Maxwell–Boltzmann distribution and potential energy profile for the reaction without a catalyst.
\nAnnotate both charts to show the activation energy for the catalysed reaction, using the label Ea (cat).
\nExplain how the catalyst increases the rate of the reaction.
\nSolid copper(II) chloride absorbs moisture from the atmosphere to form a hydrate of formula CuCl2•H2O.
\nA student heated a sample of hydrated copper(II) chloride, in order to determine the value of . The following results were obtained:
\nMass of crucible = 16.221 g
Initial mass of crucible and hydrated copper(II) chloride = 18.360 g
Final mass of crucible and anhydrous copper(II) chloride = 17.917 g
Determine the value of .
\nState how current is conducted through the wires and through the electrolyte.
\nWires:
\nElectrolyte:
\nWrite the half-equation for the formation of gas bubbles at electrode 1.
\nBubbles of gas were also observed at another electrode. Identify the electrode and the gas.
\nElectrode number (on diagram):
\nName of gas:
\nDeduce the half-equation for the formation of the gas identified in (c)(iii).
\nDetermine the enthalpy of solution of copper(II) chloride, using data from sections 18 and 20 of the data booklet.
\nThe enthalpy of hydration of the copper(II) ion is −2161 kJ mol−1.
\nCalculate the cell potential at 298 K for the disproportionation reaction, in V, using section 24 of the data booklet.
\nComment on the spontaneity of the disproportionation reaction at 298 K.
\nCalculate the standard Gibbs free energy change, ΔGθ, to two significant figures, for the disproportionation at 298 K. Use your answer from (e)(i) and sections 1 and 2 of the data booklet.
\nSuggest, giving a reason, whether the entropy of the system increases or decreases during the disproportionation.
\nDeduce, giving a reason, the sign of the standard enthalpy change, ΔHθ, for the disproportionation reaction at 298 K.
\nPredict, giving a reason, the effect of increasing temperature on the stability of copper(I) chloride solution.
\nDescribe how the blue colour is produced in the Cu(II) solution. Refer to section 17 of the data booklet.
\nDeduce why the Cu(I) solution is colourless.
\nWhen excess ammonia is added to copper(II) chloride solution, the dark blue complex ion, [Cu(NH3)4(H2O)2]2+, forms.
\nState the molecular geometry of this complex ion, and the bond angles within it.
\n\n
Molecular geometry:
\nBond angles:
\nExamine the relationship between the Brønsted–Lowry and Lewis definitions of a base, referring to the ligands in the complex ion [CuCl4]2−.
\n[Ar] 3d10
OR
1s2 2s2 2p6 3s2 3p6 3d10 ✔
ΔHθ = ΣΔHθf (products) − ΣΔHθf (reactants) ✔
ΔHθ = 2(−241.8 «kJ mol−1») − 4(−92.3 «kJ mol−1») = −114.4 «kJ» ✔
\n
NOTE: Award [2] for correct final answer.
\nEa (cat) to the left of Ea ✔
\n\n
peak lower AND Ea (cat) smaller ✔
\n«catalyst provides an» alternative pathway ✔
\n«with» lower Ea
OR
higher proportion of/more particles with «kinetic» E ≥ Ea(cat) «than Ea» ✔
mass of H2O = «18.360 g – 17.917 g =» 0.443 «g» AND mass of CuCl2 = «17.917 g – 16.221 g =» 1.696 «g» ✔
\n\n
moles of H2O = «=» 0.0246 «mol»
OR
moles of CuCl2 =«= » 0.0126 «mol» ✔
\n
«water : copper(II) chloride = 1.95 : 1»
« =» 2 ✔
\nNOTE: Accept « =» 1.95.
\nNOTE: Award [3] for correct final answer.
\nWires:
«delocalized» electrons «flow» ✔
Electrolyte:
«mobile» ions «flow» ✔
2Cl− → Cl2 (g) + 2e−
OR
Cl− → Cl2 (g) + e− ✔
NOTE: Accept e for e−.
\n«electrode» 3 AND oxygen/O2 ✔
\nNOTE: Accept chlorine/Cl2.
\n2H2O (l) → 4H+ (aq) + O2 (g) + 4e– ✔
\nNOTE: Accept 2Cl– (aq) → Cl2 (g) + 2e–.
Accept 4OH− → 2H2O + O2 + 4e−
enthalpy of solution = lattice enthalpy + enthalpies of hydration «of Cu2+ and Cl−» ✔
\n«+2824 kJ mol–1 − 2161 kJ mol–1 − 2(359 kJ mol–1) =» −55 «kJ mol–1» ✔
\nNOTE: Accept enthalpy cycle.
Award [2] for correct final answer.
Eθ = «+0.52 – 0.15 = +» 0.37 «V» ✔
\nspontaneous AND Eθ positive ✔
\nΔGθ = «−nFE = −1 mol × 96 500 C Mol–1 × 0.37 V=» −36 000 J/−36 kJ ✔
\n\n
NOTE: Accept “−18 kJ mol–1 «per mole of Cu+»”.
Do not accept values of n other than 1.
Apply SF in this question.
Accept J/kJ or J mol−1/kJ mol−1 for units.
\n2 mol (aq) → 1 mol (aq) AND decreases ✔
\nNOTE: Accept “solid formed from aqueous solution AND decreases”.
Do not accept 2 mol → 1 mol without (aq).
ΔGθ < 0 AND ΔSθ < 0 AND ΔHθ < 0
OR
ΔGθ + TΔSθ < 0 AND ΔHθ < 0 ✔
TΔS more negative «reducing spontaneity» AND stability increases ✔
\nNOTE: Accept calculation showing non-spontaneity at 433 K.
\n«ligands cause» d-orbitals «to» split ✔
\nlight absorbed as electrons transit to higher energy level «in d–d transitions»
OR
light absorbed as electrons promoted ✔
energy gap corresponds to «orange» light in visible region of spectrum ✔
\ncolour observed is complementary ✔
\nfull «3»d sub-level/orbitals
OR
no d–d transition possible «and therefore no colour» ✔
octahedral AND 90° «180° for axial» ✔
\nNOTE: Accept square-based bi-pyramid.
\nAny two of:
ligand/chloride ion Lewis base AND donates e-pair ✔
not Brønsted–Lowry base AND does not accept proton/H+ ✔
Lewis definition extends/broader than Brønsted–Lowry definition ✔
What is the name of this compound using IUPAC rules?
\nA. 2,3-diethylbutane
\nB. 2-ethyl-3-methylpentane
\nC. 3-methyl-4-ethylpentane
\nD. 3,4-dimethylhexane
\nD
\nThis was the easiest question on the paper. 94% of the candidates recognized the correct IUPAC name of the compound.
\nData collected from a larger number of silicon samples could also be plotted to determine the density using the following axes.
\nWhich statements are correct?
\n I. The density is the slope of the graph.
II. The data will show that mass is proportional to volume.
III. The best-fit line should pass through the origin.
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nThis question was answered in two different ways. 33 % of the candidates chose the correct answer (D) considering what would be true about a mass-volume graph for silicon samples. However, 39 % chose distractor (B) as they considered the silicon samples in the previous question which did not show a directly proportional relationship. Both answers were accepted as we agree that the wording was ambiguous (it will be amended before publication).
\nWhat can be deduced from the infrared (IR) spectrum of a compound?
\nA. Number of hydrogens
\nB. Number of hydrogen environments
\nC. Bonds present
\nD. Molar mass
\nC
\n71 % of the candidates identified the bonds present as the information that can be deduced from an infrared spectrum. The most commonly chosen distractor was B (the number of hydrogen environments).
\nWhat is the mechanism of the reaction between alkenes and halogens in the absence of light?
\nA. radical substitution
\nB. electrophilic substitution
\nC. electrophilic addition
\nD. nucleophilic substitution
\nC
\nOnly 33 % of candidates could identify the mechanism type with a large confusion between electrophilic addition and substitution.
\nWhich alcohol would produce a carboxylic acid when heated with acidified potassium dichromate(VI)?
\nA. propan-2-ol
\nB. butan-1-ol
\nC. 2-methylpropan-2-ol
\nD. pentan-3-ol
\nB
\nThe largest gap between high scoring and low scoring candidates was on this question about oxidation of alcohols.
\nX-ray crystallography of a metal crystal produces a diffraction pattern of bright spots.
\nUsing X-rays of wavelength 1.54 × 10−10 m, the first bright spots were produced at an angle θ of 22.3° from the centre.
\nCalculate the separation between planes of atoms in the lattice, in meters, using section 1 of the data booklet.
\n«d = »
\nd = «» 2.03 × 10−10 «m» ✔
\nAspartame is formed from the two amino acids aspartic acid (Asp) and phenylalanine (Phe).
\nDraw the structure of the dipeptide Asp–Phe using section 33 of the data booklet.
\nThe isoelectric point of amino acids is the intermediate pH at which an amino acid is electrically neutral.
\nSuggest why Asp and Phe have different isoelectric points.
\namide link (eg, CONH) ✔
\ncorrect order and structures of amino acids ✔
\n NOTE: Accept a skeletal formula or a full or condensed structural formula.
Accept zwitterion form of dipeptide.
Accept CO–NH but not CO–HN for amide link.
«Asp isoelectric point lower than Phe and » Phe has a neutral/hydrocarbon side chain ✔
Asp side chain contains −COOH/carboxyl ✔
NOTE: Award [1 max] for “Asp has two carboxyl/−COOH groups and Phe has one carboxyl/−COOH group”.
Accept “Asp has an acidic side chain” for M2
How should a measurement of 5.00 g from a balance be recorded?
\nA. 5.00 ± 0.1 g
\nB. 5.00 ± 0.01 g
\nC. 5.00 ± 1 g
\nD. 5.00 ± 0.001 g
\nB
\n80 % of candidates could correctly identify uncertainty.
\nThe equations show steps in the formation and decomposition of ozone in the stratosphere, some of which absorb ultraviolet light.
\n
Step 1 O2 → 2O•
Step 2 O• + O2 → O3
\nStep 3 O3 → O• + O2
\nStep 4 O• + O3 → 2O2
\nDraw the Lewis structures of oxygen, O2, and ozone, O3.
\nOutline why both bonds in the ozone molecule are the same length and predict the bond length in the ozone molecule. Refer to section 10 of the data booklet.
\nReason:
\nLength:
\nDistinguish ultraviolet light from visible light in terms of wavelength and energy.
\nDiscuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer affect radiation reaching the Earth’s surface.
\nNOTES: Coordinate bond may be represented by an arrow.
\nDo not accept delocalized structure for ozone.
\nresonance «structures»
OR
delocalization of «the double/pi bond» electrons ✔
121 «pm» < length < 148 «pm» ✔
NOTE: Accept any length between these two values.
\n«UV» shorter wavelength AND higher energy «than visible» ✔
\n«bond» in O2 stronger than in O3 ✔
\n
ozone absorbs lower frequency/energy «radiation than oxygen»
OR
ozone absorbs longer wavelength «radiation than oxygen» ✔
\n
NOTE: Accept ozone «layer» absorbs a range of frequencies.
\nThe dotted line represents the formation of oxygen, O2(g), from the uncatalysed complete decomposition of hydrogen peroxide, H2O2 (aq).
\nWhich curve represents a catalysed reaction under the same conditions?
\nB
\nGraphical representation of catalysis was also well answered.
\nAspartame is formed from the two amino acids aspartic acid (Asp) and phenylalanine (Phe).
\nChromatography is used in the analysis of proteins in the food and pharmaceutical industry.
\nDraw the structure of the dipeptide Asp–Phe using section 33 of the data booklet.
\nDescribe, using another method, how a mixture of four amino acids, alanine, arginine, glutamic acid and glycine, could be separated when placed in a buffer solution of pH 6.0.
\nSuggest why alanine and glycine separate slightly at pH 6.5.
\nCalculate the ratio of [A−] : [HA] in a buffer of pH 6.0 given that pKa for the acid is 4.83, using section 1 of the data booklet.
\namide link (eg, CONH) ✔
\ncorrect order and structures of amino acids ✔
\n NOTE: Accept a skeletal formula or a full or condensed structural formula.
Accept zwitterion form of dipeptide.
Accept CO–NH but not CO–HN for amide link.
Any three of:
«gel» electrophoresis «technique»
OR
mixture «in buffer solution» placed on gel/paper ✔
voltage/potential «difference» applied ✔
\namino acids move differently «depending on pH/isoelectric point» ✔
\ncompare/measure distances travelled/Rf values ✔
\nNOTE: Accept “mixture placed on plate covered with polyacrylamide «gel» OR “mixture put in a gel «medium»”.
\ndifferent sizes/molar masses/chain lengths «so move with different speeds» ✔
\nNOTE: Do not accept “different side-chains/R-groups/number of carbons”.
\n«6.0 = 4.83 + log »
\n«log= 1.17»
\n«[A−] : [HA] =» 14.8 : 1 ✔
\n NOTE: Accept “15:1”.
Do not accept 1:14.8.
1.40 × 10−3 g of NaOH (s) are dissolved in 250.0 cm3 of 1.00 × 10−11 mol dm−3 Pb(OH)2 (aq) solution.
\nDetermine the change in lead ion concentration in the solution, using section 32 of the data booklet.
\n«[OH−] = =» 1.40 × 10−4 «mol dm−3» ✔
\n«[OH−] from dissolved Pb(OH)2 is negligible»
\nNOTE: Accept «ratio =» 13.7 OR «ratio =» 0.0730 for M4.
\n\n
Ksp = [Pb2+][OH−]2
OR
1.43 × 10−20 = [Pb2+] × (1.40 × 10−4)2 ✔
[Pb2+]final = 7.30 × 10−13 «mol dm−3» ✔
\nNOTE: Award [4] for correct final answer.
\n\n
«change in [Pb2+] = 1.00 × 10−11 − 7.30 × 10−13 =» 9.27 × 10−12 «mol dm−3» ✔
\nNOTE: Award [3] for correct [Pb2+]final.
\nWhich is correct?
\n
A. Mixtures are either homogeneous or heterogeneous and their chemical properties are an average of the individual component properties.
B. Mixtures are never heterogeneous and their chemical properties are an average of the individual component properties.
\nC. Mixtures are either homogeneous or heterogeneous and the components retain their individual chemical properties.
\nD. Mixtures are never homogeneous and the components retain their individual chemical properties.
\nC
\nThe equations show steps in the formation and decomposition of ozone in the stratosphere, some of which absorb ultraviolet light.
\n
Step 1 O2 → 2O•
Step 2 O• + O2 → O3
\nStep 3 O3 → O• + O2
\nStep 4 O• + O3 → 2O2
\nDraw the Lewis structures of oxygen, O2, and ozone, O3.
\nOutline why both bonds in the ozone molecule are the same length and predict the bond length in the ozone molecule. Refer to section 10 of the data booklet.
\nReason:
\nLength:
\nPredict the bond angle in the ozone molecule.
\nDiscuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer affect radiation reaching the Earth’s surface.
\nIdentify the steps which absorb ultraviolet light.
\nDetermine, showing your working, the wavelength, in m, of ultraviolet light absorbed by a single molecule in one of these steps. Use sections 1, 2 and 11 of the data booklet.
\nOzone depletion is catalysed by nitrogen monoxide, NO, which is produced in aircraft and motor vehicle engines, and has the following Lewis structure.
\nShow how nitrogen monoxide catalyses the decomposition of ozone, including equations in your answer.
\nNOTES: Coordinate bond may be represented by an arrow.
\nDo not accept delocalized structure for ozone.
\nresonance «structures»
OR
delocalization of «the double/pi bond» electrons ✔
121 «pm» < length < 148 «pm» ✔
NOTE: Accept any length between these two values.
\nany value from 110°–119° ✔
\n«bond» in O2 stronger than in O3 ✔
\n
ozone absorbs lower frequency/energy «radiation than oxygen»
OR
ozone absorbs longer wavelength «radiation than oxygen» ✔
\n
NOTE: Accept ozone «layer» absorbs a range of frequencies.
\nsteps 1 AND 3 ✔
\nALTERNATIVE 1:
for oxygen:
✔
\n✔
\n\n
ALTERNATIVE 2:
for ozone:
similar calculation using 200 < bond enthalpy < 400 for ozone, such as
\n✔
\n✔
\n\n
NOTE: Award [2] for correct final answer.
\n•NO + O3 → •NO2 + O2 ✔
\n•NO2 + O3 → •NO + 2O2 ✔
\nNOTE: Accept •NO2 → •NO + •O AND •O + O3 → 2O2 for M2.
\nWhat is the degree of unsaturation (index of hydrogen deficiency) for the molecule?
\nA. 1
\nB. 2
\nC. 4
\nD. 5
\n\n
D
\nIHD was the worst answered question on Paper 1, with a low discriminatory index so handled poorly by all candidates.
\nWhat is represented by A in ?
\nA. Number of electrons
B. Number of neutrons
C. Number of nucleons
D. Number of protons
\nC
\nA soap solution can form a liquid-crystal state.
\nDescribe the arrangement of soap molecules in the nematic liquid crystal phase.
\nState how liquid crystals are affected by an electric field.
\nmolecules point/align in same direction/orientation
OR
molecules have directional order ✔
molecules randomly distributed
OR
molecules not in a layered arrangement
OR
molecules do not have positional order ✔
NOTE: Accept suitable diagram for M1 and M2.
\nmolecules align with field ✔
\nWhich is correct for all solid ionic compounds?
\nA. High volatility
B. Poor electrical conductivity
C. Low melting point
D. Good solubility in water
\nB
\nThe Michaelis–Menten equation describes the kinetics of enzyme-catalysed reactions.
\nOutline the significance of the Michaelis constant Km.
\nCompare the effects of competitive and non-competitive inhibitors.
\nKm is inverse measure of affinity of enzyme for a substrate
OR
Km is inversely proportional to enzyme activity
OR
high value of Km indicates higher substrate concentration needed for enzyme saturation
OR
low value of Km means reaction is fast at low substrate concentration ✔
NOTE: Idea of inverse relationship must be conveyed.
Accept “high value of Km indicates low affinity of enzyme for substrate/less stable ES complex/lower enzyme activity”.
Accept “low value of Km indicates high affinity of enzyme for substrate/stable ES complex/greater enzyme activity”.
NOTE: Accept “outside/away from active site” for “allosteric site”.
Award [1] for any two correct effects from any of the six listed.
Questions 13 and 14 are about an experiment to measure the enthalpy of combustion, ΔHc, of ethanol, using the apparatus and setup shown.
\nWhat is the enthalpy of combustion, ΔHc, of ethanol in kJ mol−1?
Maximum temperature of water: 30.0°C
Initial temperature of water: 20.0°C
Mass of water in beaker: 100.0 g
Loss in mass of ethanol: 0.230 g
Mr (ethanol): 46.08
Specific heat capacity of water: 4.18 J g−1 K−1
q = mcΔT
\n
A.
\nB.
\nC.
\nD.
\nC
\nAutomobile air bags inflate by a rapid decomposition reaction. One typical compound used is guanidinium nitrate, C(NH2)3NO3, which decomposes very rapidly to form nitrogen, water vapour and carbon.
\nDeduce the equation for the decomposition of guanidinium nitrate.
\nCalculate the total number of moles of gas produced from the decomposition of 10.0 g of guanidinium nitrate.
\nCalculate the pressure, in kPa, of this gas in a 10.0 dm3 air bag at 127°C, assuming no gas escapes.
\nSuggest why water vapour deviates significantly from ideal behaviour when the gases are cooled, while nitrogen does not.
\nAnother airbag reactant produces nitrogen gas and sodium.
\nSuggest, including an equation, why the products of this reactant present a safety hazard.
\nC(NH2)3NO3 (s) → 2N2 (g) + 3H2O (g) + C (s) ✔
\nmoles of gas = « » 0.409 «mol» ✔
\n«» = 136 «kPa» ✔
\nAny two of:
nitrogen non-polar/London/dispersion forces AND water polar/H-bonding ✔
water has «much» stronger intermolecular forces ✔
water molecules attract/condense/occupy smaller volume «and therefore deviate from ideal behaviour» ✔
2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g) ✔
\nhydrogen explosive
OR
highly exothermic reaction
OR
sodium reacts violently with water
OR
forms strong alkali ✔
NOTE: Accept the equation of combustion of hydrogen.
Do not accept just “sodium is reactive/dangerous”.
Questions 13 and 14 are about an experiment to measure the enthalpy of combustion, ΔHc, of ethanol, using the apparatus and setup shown.
\nWhich quantity is likely to be the most inaccurate due to the sources of error in this experiment?
\nA. Mass of ethanol burnt
B. Molecular mass of ethanol
C. Mass of water
D. Temperature change
\nD
\nStearic acid (Mr = 284.47) and oleic acid (Mr = 282.46) have the same number of carbon atoms. The structures of both lipids are shown in section 34 of the data booklet.
\nThe iodine number is the number of grams of iodine which reacts with 100 g of fat. Calculate the iodine number of oleic acid.
\nState one impact on health of the increase in LDL cholesterol concentration in blood.
\nExplain why stearic acid has a higher melting point than oleic acid.
\nState one similarity and one difference in composition between phospholipids and triglycerides.
\nSimilarity:
\nDifference:
\nIdentify a reagent that hydrolyses triglycerides.
\n«one C=C bond»
«1 mole iodine : 1 mole oleic acid»
« =» 89.85 «g of I2» ✔
\nNOTE: Accept “90 «g of I2»”.
\natherosclerosis/cholesterol deposition «in artery walls»/increases risk of heart attack/stroke/cardiovascular disease/CHD ✔
\nNOTE: Accept “arteries become blocked/walls become thicker”, “increases blood pressure”, OR “blood clots”.
Do not accept “high cholesterol” OR \"obesity\"
no kinks in chain/more regular structure
OR
straight chain
OR
no C=C/carbon to carbon double bonds
OR
saturated
OR
chains pack more closely together ✔
NOTE: Accept “greater surface area/electron density” for M1.
\n
stronger London/dispersion/instantaneous induced dipole-induced dipole forces «between molecules» ✔
NOTE: Accept “stronger intermolecular/van der Waals’/vdW forces” for M2.
Similarity:
«derived from» propane-1,2,3-triol/glycerol/glycerin/glycerine
OR
«derived from» at least two fatty acids
OR
contains ester linkages
OR
long carbon chains ✔
NOTE: Do not accept “two fatty acids as both a similarity and a difference”.
Do not accept just “hydrocarbon/carbon chains”.
\n
Difference:
phospholipids contain two fatty acids «condensed onto glycerol» AND triglycerides three
OR
phospholipids contain phosphate/phosphato «group»/residue of phosphoric acid AND triglycerides do not ✔
NOTE: Accept “phospholipids contain phosphorus AND triglycerides do not\".
Accept “phospholipids are amphiphilic AND triglycerides are not” OR “phospholipids have hydrophobic tails and hydrophilic heads AND triglycerides do not”.
«concentrated» NaOH (aq)/sodium hydroxide
OR
«concentrated» HCl (aq)/hydrochloric acid
OR
enzymes/lipases ✔
NOTE: Accept other strong acids or bases.
\nA student investigated how the type of acid in acid deposition affects limestone, a building material mainly composed of calcium carbonate.
\nThe student monitored the mass of six similarly sized pieces of limestone. Three were placed in beakers containing 200.0 cm3 of 0.100 mol dm−3 nitric acid, HNO3 (aq), and the other three in 200.0 cm3 of 0.100 mol dm−3 sulfuric acid, H2SO4 (aq).
\nThe limestone was removed from the acid, washed, dried with a paper towel and weighed every day at the same time and then replaced in the beakers.
\nThe student plotted the mass of one of the pieces of limestone placed in nitric acid against time.
\n[Source: © International Baccalaureate Organization 2019]
\nThe student hypothesized that sulfuric acid would cause a larger mass loss than nitric acid.
\nDraw a best-fit line on the graph.
\nDetermine the initial rate of reaction of limestone with nitric acid from the graph.
\nShow your working on the graph and include the units of the initial rate.
\nExplain why the rate of reaction of limestone with nitric acid decreases and reaches zero over the period of five days.
\nSuggest a source of error in the procedure, assuming no human errors occurred and the balance was accurate.
\nJustify this hypothesis.
\nThe student obtained the following total mass losses.
\nShe concluded that nitric acid caused more mass loss than sulfuric acid, which did not support her hypothesis.
\nSuggest an explanation for the data, assuming that no errors were made by the student.
\nbest-fit smooth curve ✔
\nNOTE: Do not accept a series of connected lines that pass through all points OR any straight line representation.
\ntangent drawn at time zero ✔
g day−1 ✔
0.16 ✔
\n
NOTE: Accept other reasonable units for initial rate eg, mol dm−3 s−1, mol dm−3 min−1, g s−1 OR g min−1.
\nM3 can only be awarded if the value corresponds to the correct unit given in M2.
Accept values for the initial rate for M3 in the range: 0.13 − 0.20 g day−1 OR 1.5 × 10−6 g s−1 − 2.3 × 10−6 g s−1 OR 7.5 × 10−8 − 1.2 × 10−7 mol dm−3 s−1 OR 4.5 × 10−6 − 6.9 × 10−6 mol dm−3 min−1 OR 9.0 × 10−5 − 1.4 × 10−4 g min−1 OR a range based on any other reasonable unit for rate.
Ignore any negative rate value.
Award [2 max] for answers such as 0.12/0.11 g day−1, incorrectly obtained by using the first two points on the graph (the average rate between t = 0 and 1 day).
Award [1 max] for correctly calculating any other average rate.
acid used up
OR
acid is the limiting reactant ✔
concentration of acid decreases
OR
less frequent collisions ✔
NOTE: Award [1 max] for \"surface area decreases\" if the idea that CaCO3 is used up/acts as the limiting reactant” is conveyed for M1.
\nDo not accept “reaction reaches equilibrium” for M2.
\nsurface area not uniform
NOTE: Accept “acids impure.
OR
limestone pieces do not have same composition/source
NOTE: Accept “«limestone» contains impurities”.
OR
limestone absorbed water «which increased mass»
OR
acid removed from solution when limestone removed
NOTE: Accept “loss of limestone when dried\" OR \"loss of limestone due to crumbling when removed from beaker”.
OR
«some» calcium sulfate deposited on limestone lost
OR
pieces of paper towel may have stuck to limestone
OR
beakers not covered/evaporation
OR
temperature was not controlled ✔
sulfuric acid is diprotic/contains two H+ «while nitric acid contains one H+»/releases more H+ «so reacts with more limestone»
OR
higher concentration of protons/H+ ✔
NOTE: Ignore any reference to the relative strengths of sulfuric acid and nitric acid.
Accept “sulfuric acid has two hydrogens «whereas nitric has one»”.
Accept \"dibasic\" for \"diprotic\".
calcium sulfate remained/deposited on limestone «in sulfuric acid»
OR
reaction prevented/stopped by slightly soluble/deposited/layer of calcium sulfate ✔
NOTE: Answer must refer to calcium sulfate.
\nWhich equation represents the N–H bond enthalpy in NH3?
\nA. NH3 (g) → N (g) + 3H (g)
\nB. NH3 (g) → N (g) + H (g)
C. NH3 (g) → N2 (g) + H2 (g)
D. NH3 (g) → •NH2 (g) + •H (g)
\nB
\nStearic acid (Mr = 284.47) and oleic acid (Mr = 282.46) have the same number of carbon atoms. The structures of both lipids are shown in section 34 of the data booklet.
\nThe iodine number is the number of grams of iodine which reacts with 100 g of fat. Calculate the iodine number of oleic acid.
\nThe chemical change in stored fats causes rancidity characterized by an unpleasant smell or taste.
\nCompare hydrolytic and oxidative rancidity.
\nState one similarity and one difference in composition between phospholipids and triglycerides.
\nSimilarity:
\nDifference:
\n«one C=C bond»
«1 mole iodine : 1 mole oleic acid»
« =» 89.85 «g of I2» ✔
\nNOTE: Accept “90 «g of I2»”.
\nNOTE: Award [1] for any two sites or conditions from any of the four listed.
Accept “high temperature” for “heat”. Accept \"lipase\" for \"enzyme\".
Do not accept just “double bond”.
Accept “air” for “oxygen” and “UV/sun” for “light”.
Ignore any reference to heat/high temperature as a condition for oxidative.
Similarity:
«derived from» propane-1,2,3-triol/glycerol/glycerin/glycerine
OR
«derived from» at least two fatty acids
OR
contains ester linkages
OR
long carbon chains ✔
NOTE: Do not accept “two fatty acids as both a similarity and a difference”.
Do not accept just “hydrocarbon/carbon chains”.
\n
Difference:
phospholipids contain two fatty acids «condensed onto glycerol» AND triglycerides three
OR
phospholipids contain phosphate/phosphato «group»/residue of phosphoric acid AND triglycerides do not ✔
NOTE: Accept “phospholipids contain phosphorus AND triglycerides do not\".
Accept “phospholipids are amphiphilic AND triglycerides are not” OR “phospholipids have hydrophobic tails and hydrophilic heads AND triglycerides do not”.
The biochemical oxygen demand of a water sample can be determined by the following series of reactions. The final step is titration of the sample with sodium thiosulfate solution, Na2S2O3 (aq).
\n2Mn2+ (aq) + O2 (aq) + 4OH− (aq) → 2MnO2 (s) + 2H2O (l)
\nMnO2 (s) + 2I− (aq) + 4H+ (aq) → Mn2+ (aq) + I2 (aq) + 2H2O (l)
2S2O32− (aq) + I2 (aq) → 2I− (aq) + S4O62− (aq)
\nA student analysed two 300.0 cm3 samples of water taken from the school pond: one immediately (day 0), and the other after leaving it sealed in a dark cupboard for five days (day 5). The following results were obtained for the titration of the samples with 0.0100 mol dm−3 Na2S2O3 (aq).
\nDetermine the mole ratio of S2O32− to O2, using the balanced equations.
\nCalculate the number of moles of oxygen in the day 0 sample.
\nThe day 5 sample contained 5.03 × 10−5 moles of oxygen.
Determine the 5-day biochemical oxygen demand of the pond, in mg dm−3 (“parts per million”, ppm).
\nCalculate the percentage uncertainty of the day 5 titre.
\nSuggest a modification to the procedure that would make the results more reliable.
\n4 : 1 ✔
\n✔
\n« ✔
\nNOTE: Award [2] for correct final answer.
\n«difference in moles per dm3 = (6.45 × 10−5 − 5.03 × 10−5) × =»
\n4.73 × 10−5 «mol dm−3» ✔
\n«convert to mg per dm3: 4.73 × 10−5 mol dm−3 × 32.00 g mol−1 × 1000 mg g–1 = » 1.51 «ppm/mg dm−3» ✔
\nNOTE: Award [2] for correct final answer.
\n« «%»✔
\nrepetition / take several samples «and average» ✔
\nWhat is formed at the electrodes during the electrolysis of molten sodium bromide?
\n\n
D
\nWhat type of reaction occurs when C6H13Br becomes C6H13OH?
A. Nucleophilic substitution
B. Electrophilic substitution
C. Radical substitution
D. Addition
\nA
\nWhich technique is used to detect the isotopes of an element?
\nA. Mass spectrometry
B. Infrared spectroscopy
C. Titration
D. Recrystallization
\nA
\nWhat is the molar mass, in , of a compound if of the compound has a mass of ?
\nA.
B.
C.
D.
\nA
\n70% of candidates correctly found the molar mass of a substance given the mass of 0.200 mol. Good to see that very few were fooled by the same answer but incorrect significant figures.
\nWhat is the number of carbon atoms in of ethanoic acid , ?
\nA.
\nB.
\nC.
\nD.
\nD
\nApproximately 57% of candidates could correctly apply Avogadro's number in calculating number of carbon atoms in 12 g of ethanoic acid. Many candidates did not identify that there were 2 carbon atoms per molecule.
\nWhich of these molecular formulae are also empirical formulae?
\nA. I and II only
\nB. I and III only
C. II and III only
D. I, II and III
\nB
\nOne of the best answered questions on paper 1. Very few of the candidates struggled with distinguishing molecular and empirical formula.
\nWhich volume of ethane gas, in , will produce of carbon dioxide gas when mixed with of oxygen gas, assuming the reaction goes to completion?
\n\n
A.
B.
C.
D.
\nB
\nMore than 65% of candidates were able to apply coefficient ratios in a stoichiometry question involving only gaseous reactants and products.
\nEthanol was electrolysed at different voltages. The products at the anode, ethanoic acid, ethanal and carbon dioxide, were collected and analysed.
\nThe percentages of products obtained using three different catalysts mounted on a carbon anode, platinum (Pt/C), platinum and ruthenium alloy (PtRu/C) and platinum and tin alloy (PtSn/C) are shown.
\nChemical yields of ethanoic acid, ethanal and carbon dioxide as a function of voltage for
oxidation of 0.100 mol dm−3 ethanol at Pt/C, PtRu/C and PtSn/C anodes at 80°C.
[Source: Product Distributions and Efficiencies for Ethanol Oxidation in a Proton Exchange Membrane Electrolysis Cell, Rakan M. Altarawneh and Peter G. Pickup, Journal of the Electrochemical Society, 2017, volume 164, issue 7, http://jes.ecsdl.org/. Distributed under the terms of the Creative Commons Attribution 4.0 License (CC BY, http://creativecommons.org/licenses/by/4.0/)]
Describe the effect of increasing the voltage on the chemical yield of:
\nEthanal using Pt/C:
\nCarbon dioxide using PtRu/C:
\nDetermine the change in the average oxidation state of carbon.
\nFrom ethanol to ethanal:
\nFrom ethanol to carbon dioxide:
\nList the three products at the anode from the least to the most oxidized.
\nDeduce, giving your reason, which catalyst is most effective at fully oxidizing ethanol.
\nEthanal using Pt/C:
decreases ✔
Carbon dioxide using PtRu/C:
«generally» increases AND then decreases ✔
NOTE: Accept “no clear trend/pattern” OR “increases and decreases” OR “increases, reaches a plateau and «then» decreases” for M2.
\nFrom ethanol to ethanal:
−2 to −1
OR
+1/increases by 1 ✔
NOTE: Do not accept “2− to 1−”.
\nFrom ethanol to carbon dioxide:
−2 to +4
OR
+6/increases by 6 ✔
NOTE: Do not accept “2− to 4+”.
Do not penalize incorrect notation twice.
\nPenalize incorrect oxidation state value of carbon in ethanol once only.
\nethanal < ethanoic acid < carbon dioxide ✔
\nNOTE: Accept formulas.
No ECF from 2aii calculations.
Pt/platinum/PtC AND highest yield of CO2 «at all voltages» ✔
\nNOTE: ECF from 2aiii.
\nWhat is the relative atomic mass, , of an element with this mass spectrum?
\nA.
B.
C.
D.
\nB
\nWell answered question with more than 70% of candidates able to find the RAM of an element from isotope relative abundances.
\nWhat is the maximum number of electrons that can occupy the 4th main energy level in an atom?
\nA.
B.
C.
D.
\nD
\nPoorly answered questions from candidates in general. Only 38% of candidates could state the maximum number of electrons in the 4th principal energy level is 32, with the most common incorrect answer being 18.
\nWhich of the following shows a general increase across period 3 from to ?
\nA. Ionic radius
B. Atomic radius
C. Ionization energy
D. Melting point
\nC
\nAlthough 75% of candidates had the correct answer, this question about trends across period 3 was not a very good discriminator.
\nWhich oxide will dissolve in water to give the solution with the lowest pH?
\nA.
\nB.
\nC.
\nD.
\nA
\nFairly well answered. Almost 70% of candidates could identify that non-metallic oxides would dissolve in water to yield a substance of low pH.
\nWhich formula is correct?
\nA.
\nB.
\nC.
\nD.
\nC
\n64% of candidates picked the correct formula for ammonium phosphate with (NH4)2PO4 being the incorrect answer chosen most frequently. NH4PO4 or (NH4)3(PO4)2 was rarely selected.
\nCatalysts are commonly used in industry.
\nDescribe how a heterogeneous catalyst provides an alternative pathway for a reaction.
\nDistinguish between heterogeneous and homogeneous catalysts, giving one difference.
\nNanotubes are used to support the active material in nanocatalysts.
\nExplain why oxygen cannot be used for the chemical vapour deposition (CVD) preparation of carbon nanotubes.
\nreactant(s) adsorb onto active sites/surface ✔
\n NOTE: Do not accept “absorb” for “adsorb” for M1.
Accept “bonds to” for “adsorb” for M1.
«reactant» bonds weakens «and products are desorbed» ✔
NOTE: Accept “bonds break/stretch «and products are desorbed»” for M2.
Award [1 max] for “lowers activation energy”.
Any one of the following:
\nNOTE: Accept “heterogeneous adsorb reactants and homogeneous” but do not accept “absorb” for “adsorb”.
Accept “heterogeneous have active sites and homogeneous do not”.
high temperature used ✔
\noxygen/O2 reacts with carbon/C
OR
carbon dioxide/CO2 can form ✔
Which molecule is most polar?
\nA.
\nB.
\nC.
\nD.
\nA
\n63% of the candidates could identify CHF3 as more polar than CClF3. A surprising number selected either CCl4 or CF4 as the most polar.
\nWhich combination correctly describes the geometry of the carbonate ion, ?
\nC
\nIdentifying correct electron domain and molecular geometries around a central atom was somewhat challenging and answered slighter better by higher scoring candidates.
\nWhich series shows the correct order of metallic bond strength from strongest to weakest?
\nA.
\nB.
\nC.
\nD.
\nD
\nA small majority of candidates could identify the relative strength of metallic bonding amongst group 1 and 2 metals.
\nWhich equation shows the enthalpy of formation, , of ethanol?
\nA.
\nB.
\nC.
\nD.
\nC
\nLess than half of the candidates selected the correct equation for the enthalpy of formation of ethanol. The most common wrong answer involved ethanol formed as a gas.
\nWhich combination will give you the enthalpy change for the hydrogenation of ethene to ethane, ?
\nA.
\nB.
\nC.
\nD.
\nA
\nA good majority of candidates could select the Hess' Law cycle for hydrogenation of ethene to ethane.
\nWhat is the bond enthalpy, in , in the molecule?
\n\n\nA.
\nB.
\nC.
\nD.
\nB
\nThere was no correct answer to this question as the value was given instead of . The majority of candidates answered as if it was
\nWhich apparatus can be used to monitor the rate of this reaction?
\n\n
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
\nB
\nIt was pleasing to see that a vast majority of candidates could select equipment necessary to monitor rates of reaction. This was one of the best answered questions in the exam.
\nWhich arrow shows the activation energy of the uncatalysed forward reaction for this equilibrium?
\n\n
C
\nIdentifying the activation energy of an uncatalyzed reaction from an energy profile diagram was the best answered question in the exam.
\nWhat is correct when temperature increases in this reaction at equilibrium?
\n\n
D
\nA good majority of candidates understood how position of equilibrium and rate constant of an endothermic reaction are affected by temperature increase.
\nWhich substance will not produce copper(II) chloride when added to dilute hydrochloric acid?
\nA.
B.
C.
D.
\nA
\nPoorly answered with less than ½ of candidates able to identify that copper metal would not produce
copper II chloride when added to dilute HCl.
Which of these acids has the weakest conjugate base?
\nA.
\nB.
\nC.
\nD.
\nA
\nMore than 59% of the candidates understood that strong acids have weak conjugate bases.
\nIron is a stronger reducing agent than silver.
\nWhat is correct when this voltaic cell is in operation?
\nD
\n53% of candidates identified the anode, cathode, AND direction of electron flow in a voltaic cell, with misconceptions being equally distributed. This was answered much better by higher scoring candidates.
\nWhat is correct in an electrolytic cell?
\nD
\nA good majority of candidates understood that electrons are gained at the negative cathode causing reduction during electrolysis.
\nA proton-exchange membrane (PEM) fuel cell uses pure hydrogen gas as the fuel and a proton exchange membrane as the electrolyte.
\n\n
A dye-sensitized solar cell (DSSC) uses light energy to produce electricity.
\nDeduce the half-equations for the reactions occurring at the electrodes.
\n
Anode (negative electrode):
Cathode (positive electrode):
\nCalculate the cell potential, Eθ, in V, using section 24 of the data booklet.
\nSuggest how PEM fuel cells can be used to produce a larger voltage than that calculated in (b)(i).
\nSuggest an advantage of the PEM fuel cell over the lead-acid battery for use in cars.
\nOutline the functions of the dye, TiO2 and the electrolyte in the operation of the DSSC.
\nDye:
\nTiO2:
\nElectrolyte:
\nSuggest an advantage of the DSSC over silicon-based photovoltaic cells.
\nAnode (negative electrode):
H2 (g) → 2H+ (aq) + 2e− ✔
Cathode (positive electrode):
O2 (g) + 4H+ (aq) + 4e− → 2H2O (l) ✔
NOTE: Accept any correct integer or fractional coefficients. Award [1 max] for M1 and M2 if correct half-equations are given at the wrong electrodes OR if incorrect reversed half-equations are given at the correct electrodes.
\n(+)1.23 «V» ✔
\nNOTE: Do not accept “-1.23 «V»”.
\nconnect several fuel cells in series
OR
increase pressure/concentration of reactant/hydrogen/oxygen ✔
NOTE: Do not accept changes in [H+]/pH as they do not affect cell potential in this case.
Do not accept reference to quantity for “concentration”.
liquid in cell is less/not corrosive
OR
does not contain lead/toxic chemicals
OR
larger energy density/charge capacity/current per unit mass
OR
does not have to be charged prior to use / is always ready for use «as long as fuel is available» ✔
Dye:
absorbs photons/light
OR
releases electrons ✔
TiO2:
conducts current/electricity
OR
semiconductor ✔
Electrolyte:
reduces/regenerates «the oxidized» dye ✔
Any one of:
cheaper/ease of manufacture
OR
plentiful and renewable resources «to construct DSSC cells» ✔
use light of lower energy/lower frequency/longer wavelength
OR
use of nanoparticles provides large surface area for exposure to sunlight/sun/light
OR
can absorb better under cloudy conditions ✔
operate at lower «internal» temperatures
OR
better at radiating heat away «since constructed with thin front layer of conductive plastic compared to glass box in photovoltaic cells» ✔
better conductivity ✔
\nmore flexible/durable ✔
\nNOTE: Accept “lower mass/lighter «so greater flexibility to integrate into windows etc.»” OR “greater power-conversion efficiency «with latest DSSC models»”.
\nBody fluids have different pH values.
\nIdentify the compound responsible for the acidity of gastric juice, and state whether it is a strong or weak acid.
\nAn antacid contains calcium carbonate and magnesium carbonate.
\nWrite the equation for the reaction of magnesium carbonate with excess stomach acid.
\nOutline how ranitidine reduces stomach acidity.
\nCalculate the pH of a buffer solution which contains 0.20 mol dm−3 ethanoic acid and 0.50 mol dm−3 sodium ethanoate. Use section 1 of the data booklet.
\npKa (ethanoic acid) = 4.76
\nhydrochloric acid/HCl «(aq)» AND strong «acid» ✔
\nMgCO3 (s) + 2HCl (aq) → MgCl2 (aq) + CO2 (g) + H2O (l) ✔
\nNOTE: Accept ionic equation.
\nblocks/binds to H2-histamine receptors «in cells of stomach lining»
OR
prevents histamine molecules binding to H2-histamine receptors «and triggering acid secretion»
OR
prevents parietal cells from releasing/producing acid ✔
NOTE: Do not accept “antihistamine” by itself.
Accept “H2-receptor antagonist/H2RA” OR “blocks/inhibits action of histamine”.
Accept “blocks receptors in parietal cells «from releasing/producing acid»”.
Do not accept “proton pump/ATPase inhibitor”.
«pKa = 4.76»
«pH = pKa + log »
«pH = 4.76 + 0.40 =» 5.16 ✔
The structure of aspirin is shown in section 37 of the data booklet.
\nSuggest one reactant used to prepare aspirin from salicylic acid.
\nAspirin, C6H4(OCOCH3)COOH, is only slightly soluble in water.
\nOutline, including an equation, how aspirin can be made more water-soluble. Use section 37 in the data booklet.
\nNOTE: Accept condensed structural formulas.
Accept “ethanoic acid/acetic acid/CH3COOH”.
Accept “C4H6O3” OR “C2H3OCl”.
react with sodium hydroxide/NaOH/«strong» base
OR
convert to «ionic» salt ✔
NOTE: Accept other suitable bases (eg, KOH/NaHCO3/Na2CO3) with corresponding equation for chosen base for M2.
Accept “CaCO3”, although calcium salicylate is not water soluble.
Accept ionic equation.
C6H4(OCOCH3)COOH (s) + NaOH (aq) → C6H4(OCOCH3)COONa (aq) + H2O (l) ✔
NOTE: Award [2] for M2.
\n
Codeine, morphine and diamorphine (heroin) are derived from opium.
\nState the names of two functional groups present in all three molecules, using section 37 of the data booklet.
\nExplain why diamorphine has greater potency than morphine.
\nAny two of:
benzene/aromatic ring ✔
«tertiary» amino «group» ✔
ethenylene/1,2-ethenediyl «group» ✔
ether «group» ✔
NOTE: Accept “phenyl” for “benzene ring” although there are no phenyl groups as the benzene ring in this compound is a part of a polycyclic structure.
Do not accept “arene” or “benzene” alone.
Accept “amine” for “amino «group»”.
Accept “alkenyl/alkene/vinylene” for ethenylene/1,2-ethenediyl «group».
Any three of:
morphine has «two» hydroxyl «groups» AND diamorphine has «two» ester/ethanoate/acetate «groups» ✔
NOTE: Accept “heroin” for “diamorphine”.
Accept formulas.
Accept “hydroxy” for “hydroxyl” but not “hydroxide”.
Accept “acyl” for “ester «groups»”.
\n
morphine is more polar than diamorphine
OR
groups in morphine are replaced with less polar/non-polar groups in diamorphine ✔
NOTE: Do not accept just “diamorphine is non-polar” for M2.
\n
morphine is «more» soluble in blood «plasma»
NOTE: Accept “water” for “blood”.
OR
diamorphine is «more» soluble in lipids
NOTE: Accept “fats” for “lipid”.
OR
diamorphine is more soluble in non-polar environment of CNS/central nervous system than morphine ✔
\n
diamorphine crosses the blood–brain barrier/BBB «easily» ✔
\n\n
Steroids are lipids with a steroidal backbone. The structure of cholesterol is shown in section 34 of the data booklet.
\nInfrared (IR) spectroscopy is used to identify functional groups in organic compounds.
\nDeduce the wavenumber, in cm−1, of an absorption peak found in the IR spectrum of testosterone but not in that of cholesterol.
\nDescribe a technique for the detection of steroids in blood and urine.
\nExplain how redox chemistry is used to measure the ethanol concentration in a breathalyser.
\n1700−1750 «cm−1» ✔
NOTE: Accept a specific wavenumber value within range.
Any three of:
sample/liquids vaporized «in oven/at high temperature»
OR
sample injected into mobile phase/inert gas
OR
nitrogen/helium/inert gas acts as mobile phase
OR
sample carried by inert gas «through column» ✔
NOTE: Award [1 max] for identifying suitable technique (eg GC-MS etc.).
Do not accept just “gas”.
Accept description of HPLC using liquid mobile phase.
stationary phase consists of a packed column
OR
packing/solid support acts as stationary phase ✔
NOTE: Accept named stationary phase, such as «long-chain» hydrocarbon/polysiloxane/silica.
components separated by partition «between mobile phase and stationary phase»
OR
gases/liquids/components have different retention times/Rf
OR
gases/liquids/components move through tube/column at different speeds/rates ✔
detector/mass spectrometer/MS «at end of column»
OR
databases/library of known fragmentation patterns can be used ✔
NOTE: Accept “area under peak proportional to quantity/amount/concentration of component present «in mixture»”.
ALTERNATIVE 1
oxidizing agent/«acidified» potassium dichromate(VI) converts ethanol to ethanoic acid ✔
colour change «from orange to green» is measured/analysed «using photocell» ✔
ALTERNATIVE 2
ethanol is oxidized to ethanoic acid «at anode and oxygen is reduced to water at cathode» ✔
current/voltage/potential is measured «by computer»
OR
current/voltage/potential is proportional to ethanol concentration ✔
NOTE: Accept names or formulas for reagents.
Accept “«acidified» dichromate/Cr2O72−” for “K2Cr2O7”.
Award [1 max] for \"Cr(VI) going to Cr(III) AND colour changing/colour changing from orange to green\".
Do not penalize incorrect oxidation state notation here.
Accept \"EMF\" for \"voltage\".
Codeine, morphine and diamorphine (heroin) are derived from opium.
\nExplain why diamorphine has greater potency than morphine.
\nExperimental research on both animals and humans contributes to the development of pharmaceuticals.
\nState the meaning of the term therapeutic index in human studies.
\nAny three of:
morphine has «two» hydroxyl «groups» AND diamorphine has «two» ester/ethanoate/acetate «groups» ✔
NOTE: Accept “heroin” for “diamorphine”.
Accept formulas.
Accept “hydroxy” for “hydroxyl” but not “hydroxide”.
Accept “acyl” for “ester «groups»”.
\n
morphine is more polar than diamorphine
OR
groups in morphine are replaced with less polar/non-polar groups in diamorphine ✔
NOTE: Do not accept just “diamorphine is non-polar” for M2.
\n
morphine is «more» soluble in blood «plasma»
NOTE: Accept “water” for “blood”.
OR
diamorphine is «more» soluble in lipids
NOTE: Accept “fats” for “lipid”.
OR
diamorphine is more soluble in non-polar environment of CNS/central nervous system than morphine ✔
\n
diamorphine crosses the blood–brain barrier/BBB «easily» ✔
\n\n
toxic dose for 50% of population divided by «minimum» effective dose for 50 % of population ✔
\nNOTE: Accept “TD50/ED50”.
Reference to 50% required.
The discovery of penicillins contributed to the development of antibiotics.
\nExplain how the beta-lactam ring is responsible for the antibiotic properties of penicillin. Refer to section 37 of the data booklet.
\nOutline the impact of antibiotic waste on the environment.
\nSuggest a concern about the disposal of solvents from drug manufacturing.
\nDiscuss two difficulties, apart from socio-economic factors, associated with finding a cure for AIDS.
\nring is «sterically» strained
OR
angles of 90° instead of 109.5/109/120° angles
OR
angles smaller than 109.5/109/120°/tetrahedral/trigonal planar/triangular planar angle ✔
ring breaks up/opens/reacts «easily»
OR
amido/amide group «in ring» is «highly» reactive ✔
«irreversibly» binds/bonds to enzyme/transpeptidase
OR
inhibits enzyme/transpeptidase «in bacteria» that produces cell walls
OR
prevents cross-linking of «bacterial» cell walls ✔
NOTE: Accept arguments using correct descriptions of hybridization for M1.
Do not accept \"breaks/binds to cell walls\" – a reference to the enzyme is needed for alternatives 1 and 2 for M3.
Do not accept \"cell membrane\" for \"cell wall\".
«leads to bacterial» resistance «to antibiotics»
OR
destroys useful/beneficial bacteria
OR
useful/beneficial/less harmful bacteria replaced with «more» harmful bacteria ✔
NOTE: Accept “affects/disturbs micro-ecosystems”.
\nAny one of:
«most are» toxic «to living organisms»
OR
incomplete combustion/incineration can produce toxic products/dioxins/phosgene
OR
carcinogenic/can cause cancer ✔
NOTE: Do not accept “harmful to the environment”.
accumulate in groundwater
OR
have limited biodegradability ✔
NOTE: Do not accept just “pollutes water”.
cost of disposal ✔
NOTE: Do not accept “hazard of disposal”.
NOTE: Accept “ozone depletion” only if there is some reference to chlorinated solvents.
\nAny two of:
HIV difficult to detect/remains dormant ✔
HIV mutates rapidly/quickly ✔
HIV replicates rapidly/quickly ✔
HIV destroys «T-»helper cells/white blood cells/lymphocytes
OR
HIV attacks immune system ✔
HIV has several «significantly different» strains/subtypes ✔
\n NOTE: Accept “virus” for “HIV”.
Do not accept “AIDS mutates” without mention of the HIV/virus.
Penalize the use of “AIDS” for “HIV” once only.
Accept “HIV metabolism linked to that of host cell” OR “drugs harm host cell as well as HIV”.
Taxol is an anticancer drug.
\nState the feature of Taxol that is a major challenge in its synthesis. Use section 37 of the data booklet.
\nDescribe how the challenge in (a) was resolved by pharmaceutical companies.
\nnumerous stereoisomers/chiral carbons/chiral centres/stereocentres/optical isomers ✔
\nNOTE: Accept exact number of chiral carbons ie 11, but do not accept just “chiral”.
\nchiral auxiliaries/molecule binds to reactant blocking one reaction site «by steric hindrance»
OR
asymmetric synthesis / enantioselective catalysis «producing a specific enantiomer»
OR
biosynthesis / genetically modified bacteria/microorganisms ✔
NOTE: Accept “use of a chiral auxiliary leading to «the synthesis of» the desired enantiomer”.
\nNuclear medicine uses small amounts of radioisotopes to diagnose and treat some diseases.
\nState two common side effects of radiotherapy.
\nExplain why technetium-99m is the most common radioisotope used in nuclear medicine.
\n25.0 μg of iodine-131, with a half-life of 8.00 days, was left to decay.
\nCalculate the mass of iodine-131, in μg, remaining after 32.0 days. Use section 1 of the data booklet.
\nAny two of:
hair loss
fatigue
nausea
sterility
skin reaction
diarrhoea
vomiting
damage to lymph system
urinary/bladder changes
anxiety/emotional problems
joint/muscular stiffness
loss of appetite
sore/dry mouth
loss of weight
secondary cancer ✔
Any two of:
half-life is 6 hours/long enough for a scan to occur
OR
half-life short enough not to remain in body ✔
NOTE: Accept “short half-life so patient is not exposed to lots of ionizing radiation”.
decay releases «low energy» gamma rays
OR
gamma rays less likely to be absorbed by cells ✔
can form several «coordination» complexes ✔
NOTE: Accept \"can exist in many oxidation states «so can form multiple complexes»\" OR \"chemically versatile «so can act as a tracer by bonding to several bioactive compounds»”.
«low-energy» radiation/gamma-rays can be detected by common X-ray equipment ✔
\nALTERNATIVE 1
4 half-lives ✔
1.56 «μg of iodine-131 remain» ✔
\n
ALTERNATIVE 2
m = 25.0 ✔
1.56 «μg of iodine-131 remain» ✔
\n
ALTERNATIVE 3
λ = « » = 8.66 × 10−2 «day−1» ✔
m = « » 1.56 «μg of iodine-131 remain» ✔
\n
Award [2] for correct final answer.
\n0.10 mol of hydrochloric acid is mixed with 0.10 mol of calcium carbonate.
\n2HCl (aq) + CaCO3 (s) → CaCl2 (aq) + H2O (l) + CO2 (g)
\nWhich is correct?
\nC
\nDNA, deoxyribonucleic acid, is made up of nucleotides.
\nList two components of nucleotides.
\nExplain how the double-helical structure of DNA is stabilized once formed.
\nAny two correct for [1]:
pentose «sugar»
OR
deoxyribose ✔
phosphate/phosphato «group»/residue of phosphoric acid ✔
\nNOTE: Accept “−OPO32−/−OPO3H−/−OPO3H2” but not “PO43−”.
\n«organic» nitrogenous base
OR
nucleobase
OR
nucleic base
OR
purine
OR
pyrimidine ✔
NOTE: Accept the four bases together: “adenine/A, guanine/G, cytosine/C, thymine/T”.
Accept names or formulas.
Any two of:
H-bonding between bases in each pair ✔
hydrophobic interactions/π-stacking between bases ✔
polar/charged/hydrophilic groups in sugar-phosphate backbone interactions with aqueous solution/water
OR
H-bonding AND ion-dipole interactions between phosphato «groups» andwater/histones ✔
Accept \"phosphate groups are hydrophilic and form H-bonds with water\".
Accept “H-bonding with histones”.
What are the oxidation states of oxygen?
\nC
\nLess than 50% of candidates could correctly identify the oxidation of oxygen in various compounds, with the greatest misconception being oxygen still had a −2 charge in OF2.
\nWhich functional groups are present in this molecule?
A. carbonyl, ether, nitrile
B. carbonyl, ester, nitrile
C. carboxyl, ether, amine
D. carboxyl, ester, amine
\nA
\nThis question on identifying functional groups was the most discriminating question on the paper. 58% correctly identified all three groups but it was answered much better by higher scoring candidates.
\nWhat is the sum of the coefficients when the equation is balanced with whole numbers?
\n__MnO2 (s) + __HCl (aq) → __MnCl2 (aq) + __H2O (l) + __Cl2 (g)
A. 6
B. 7
C. 8
D. 9
\nD
\nWhich molecule will decolorize bromine water in the dark?
\nA. cyclohexane
B. hexane
C. hex-1-ene
D. hexan-1-ol
\nC
\nTwo-thirds of candidates knew that hex-1-ene would decolourize bromine water in the dark.
\nWhich contains the greatest number of moles of oxygen atoms?
\nA. 0.05 mol Mg(NO3)2
B. 0.05 mol C6H4(NO2)2
C. 0.1 mol H2O
D. 0.1 mol NO2
\nA
\nStaphylococcus aureus (S. aureus) infections have been successfully treated with penicillin and penicillin derivatives.
\nIdentify the feature in penicillin responsible for its antibiotic activity.
\nThe widespread use of penicillin and its derivatives has led to the appearance of resistant S. aureus strains.
\nOutline how these bacteria inactivate the antibiotics.
\nOutline how the structure of penicillin has been modified to overcome this resistance.
\n«four-membered» beta-lactam ring [✔]
\n\n
Note: Accept a diagram showing a structural representation of the beta-lactam ring.
\nproduce penicillinase/enzyme that deactivates penicillin [✔]
\nside-chain changed «preserving beta-lactam ring» [✔]
\n\n
Note: Accept “R group changed”.
\nMost students could identify the beta-lactam ring as the structural feature responsible for penicillin’s activity.
\nThe action of penicillinase/beta-lactamase was less well known with only about a third of students gaining this mark.
\nMost students were also aware that changing to the side chain of penicillin is often used to counter resistance, though many benefitted from the decision to accept “R-” without any mention of the data booklet structure.
\nChlorine undergoes many reactions.
\nof manganese(IV) oxide was added to of .
\n\nChlorine gas reacts with water to produce hypochlorous acid and hydrochloric acid.
\n\nis a common chlorofluorocarbon, .
\nState the full electron configuration of the chlorine atom.
\nState, giving a reason, whether the chlorine atom or the chloride ion has a larger radius.
\nOutline why the chlorine atom has a smaller atomic radius than the sulfur atom.
\nThe mass spectrum of chlorine is shown.
\n NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.
Outline the reason for the two peaks at and .
Explain the presence and relative abundance of the peak at .
\nCalculate the amount, in , of manganese(IV) oxide added.
\nDetermine the limiting reactant, showing your calculations.
\nDetermine the excess amount, in , of the other reactant.
\nCalculate the volume of chlorine, in , produced if the reaction is conducted at standard temperature and pressure (STP). Use section 2 of the data booklet.
\nState the oxidation state of manganese in and .
\nDeduce, referring to oxidation states, whether is an oxidizing or reducing agent.
\nHypochlorous acid is considered a weak acid. Outline what is meant by the term weak acid.
\n\n
State the formula of the conjugate base of hypochlorous acid.
\nCalculate the concentration of in a solution with a .
\nState the type of reaction occurring when ethane reacts with chlorine to produce chloroethane.
\nPredict, giving a reason, whether ethane or chloroethane is more reactive.
\nWrite the equation for the reaction of chloroethane with a dilute aqueous solution of sodium hydroxide.
\nDeduce the nucleophile for the reaction in d(iii).
\nEthoxyethane (diethyl ether) can be used as a solvent for this conversion. Draw the structural formula of ethoxyethane
\nDeduce the number of signals and their chemical shifts in the spectrum of ethoxyethane. Use section 27 of the data booklet.
\nCalculate the percentage by mass of chlorine in .
\nComment on how international cooperation has contributed to the lowering of emissions responsible for ozone depletion.
\n✔
\nDo not accept condensed electron configuration.
\nAND more «electron–electron» repulsion ✔
\n
Accept AND has an extra electron.
has a greater nuclear charge/number of protons/ «causing a stronger pull on the outer electrons» ✔
\nsame number of shells
OR
same «outer» energy level
OR
similar shielding ✔
«two major» isotopes «of atomic mass and » ✔
\n«diatomic» molecule composed of «two» chlorine-37 atoms ✔
\nchlorine-37 is the least abundant «isotope»
OR
low probability of two «isotopes» occurring in a molecule ✔
✔
\n✔
\nAND is the limiting reactant ✔
\nAccept other valid methods of determining the limiting reactant in M2.
\n✔
\n✔
\n
Accept methods employing .
✔
\n✔
\noxidizing agent AND oxidation state of changes from to /decreases ✔
\npartially dissociates/ionizes «in water» ✔
\n✔
\n✔
\n«free radical» substitution/ ✔
\n
Do not accept electrophilic or nucleophilic substitution.
chloroethane AND bond is weaker/ than bond/
OR
chloroethane AND contains a polar bond ✔
Accept “chloroethane AND polar”.
OR
✔
Accept use of and in the equation.
\nhydroxide «ion»/ ✔
\n
Do not accept .
/ ✔
Accept .
\n«signals» ✔
\nAND ✔
\n
Accept any values in the ranges.
Award [1 max] for two incorrect chemical shifts.
\n✔
\n✔
\n
Award [2] for correct final answer.
Any of:
\nresearch «collaboration» for alternative technologies «to replace s»
OR
technologies «developed»/data could be shared
OR
political pressure/Montreal Protocol/governments passing legislations ✔
\n
Do not accept just “collaboration”.
\nDo not accept any reference to as greenhouse gas or product of fossil fuel combustion.
\nAccept reference to specific measures, such as agreement on banning use/manufacture of s.
\nMost candidates wrote the electron configuration of chlorine correctly.
\nOnly half of the candidates deduced that the chloride ion has a larger radius than the chlorine atom with a valid reason. Many candidates struggled with this question and decided that the extra electron in the chloride ion caused a greater attraction between the nucleus and the outer electrons.
\nOnly about a third of the candidates identified the extra proton in the chlorine nucleus as the cause of the smaller atomic radius when compared to the sulfur atom, and only the stronger candidates also compared the shielding or the number of shells in the two atoms. Many candidates had a poor understanding of factors affecting atomic radius and could not explain the difference.
\nAbout 60% of the candidates recognized that the peaks at m/z 35 and 37 in the mass spectrum of chlorine are due to its isotopes. A few students wrote 'isomers' instead of 'isotopes'.
\nThis was the lowest scoring question on the paper, that was also left blank by 10% of the candidates. About 20% of the candidates identified the peak at m/z = 74 to be due to a molecule made up of two 37Cl atoms. And only very few candidates commented that the low abundance of the peak was due to the low abundance of the 37Cl isotope. A common incorrect answer was that chlorine has an isotope of mass number 74.
\nMost candidates were able to determine the number of moles of MnO2 using the mass.
\nIt was pleasing that the majority of the candidates were able to determine the limiting reactant by using the stoichiometric ratio.
\nHalf of the candidates were able to determine the amount of excess reactant. Some candidates who determined the limiting reactant in the previous part correctly, forgot to use the stoichiometric ratio in this part, and ended up with incorrect answers.
\n60% of the candidates determined the volume of chlorine produced correctly. Some candidates made mistakes in the units when using PV = nRT and had a power of 10 error.
\nThe majority of candidates were able to determine the oxidation states of Mn in the two compounds correctly.
\nLess than half of the candidates were awarded the mark. Some did identify MnO2 as the oxidizing agent but did not give the explanation in terms of oxidation state as required in the question. Other candidates did not have an understanding of oxidizing and reducing agents.
\nA very well answered question - 80% of candidates understood what is meant by the term weak acid. Incorrect answers included 'acids that have high pH'.
\nHalf of the candidates deduced the formula of the conjugate base of hypochlorous acid. Incorrect answers included H2O and HCl.
\nA well answered question. It was pleasing to see that 70% of the candidates were able to calculate [H+] from the given pH.
\nMore than half of the candidates identified the type of reaction between ethane and chlorine as a substitution reaction. A few candidates lost the marks for writing 'electrophilic substitution' or 'nucleophilic substitutions'.
\nThis was a challenging question that was answered correctly by only 30% of the candidates. A variety of incorrect answers were seen such as 'chlorine is a halogen and hence it is reactive', and 'ethane is more reactive because it is an alkane'. For students who answered correctly, the polarity was the most frequently given reason.
\nHalf of the candidates wrote the correct equation for the hydrolysis of chloroethane. Incorrect answers often included carbon dioxide and water as the products.
\nThis was a highly discriminating question. Only 30% of the candidates were able to identify the hydroxide ion as the nucleophile in the hydrolysis of chloroethane. Incorrect answers included NaOH where the ion was not specified. 14% of the candidates left this question blank.
\nHalf of the candidates were able to give the structural formula of ethoxyethane. Incorrect answers included methoxymethane, ketones and esters.
\nNearly half of the candidates were able to identify the number of signals obtained in the 1H NMR spectrum of ethoxyethane, obtaining the first mark of this question. Many candidates were awarded the mark as 'error carried forward' from an incorrect structure of ethoxyethane. The second mark for this question required candidates to look up values of chemical shift from the data booklet. Nearly a third of the candidates were able to match the chemical environments of the hydrogen atoms in ethoxyethane to those listed in the data booklet successfully.
\nThis was the highest scoring question in the paper. The majority of candidates were able to calculate the percentage by mass of chlorine in CCl2F2. Mistakes included incorrect rounding and arithmetic errors.
\nThis nature of science question was well answered by half of the candidates. Some teachers commented that the wording was rather vague. Incorrect answers were mainly assuming that CFCs were related to the combustion of fuels and greenhouse gas emissions.
\nWhich represents the shape of an s atomic orbital?
\nA
\nWhat is the IUPAC name of this molecule?
\nA. 1,1,2,4-tetramethylpent-1-ene
B. 2,4,5-trimethylhex-4-ene
C. 2,4,5,5-tetramethylpent-4-ene
D. 2,3,5-trimethylhex-2-ene
\nD
\nIt was good to see that 2/3 of candidates could pick a preferred IUPAC name from a skeletal structure.
\nWhat is the enthalpy change of the reaction?
\nC6H14 (l) → C2H4 (g) + C4H10 (g)
\nA. + 1411 + 2878 + 4163
B. + 1411 − 2878 − 4163
C. + 1411 + 2878 − 4163
D. − 1411 − 2878 + 4163
\nC
\nWhich reaction has the greatest increase in entropy of the system?
\nA. HCl (g) + NH3 (g) → NH4Cl (s)
B. (NH4)2Cr2O7 (s) → Cr2O3 (s) + N2 (g) + 4H2O (g)
C. CaCO3 (s) → CaO (s) + CO2 (g)
D. I2 (g) → I2 (s)
\nB
\nWhich shows the first ionization energies of successive elements across period 2, from left to right?
\n\n
C
\nWhat is the order of increasing (more exothermic) enthalpy of hydration?
\nXn+ (g) → Xn+ (aq)
\nA. Ca2+, Mg2+, K+, Na+
B. Na+, K+, Mg2+, Ca2+
C. K+, Na+, Ca2+, Mg2+
D. Mg2+, Ca2+, Na+, K+
\nC
\nChanges in physiology can impact living creatures.
\nThe graph shows the change in oxygen partial pressure in blood, measured at different pH values.
\nExplain the effect of changing pH on the percentage saturation of hemoglobin at a given partial pressure of oxygen.
\nExplain the biomagnification of the pesticide DDT.
\nVitamins are organic compounds essential in small amounts.
\nState the name of one functional group common to all three vitamins shown in section 35 of the data booklet.
\nas pH decreases, protons/CO2 bind to allosteric sites
OR
as pH decreases, protons/CO2 act as non-competitive inhibitor
OR
active/binding site changes shape ✔
saturation decreases
OR
more oxygen released
OR
affinity to oxygen decreases ✔
accumulates in fat/tissues/living organisms
OR
cannot be metabolized/does not break down «in living organisms»
OR
not excreted / excreted «very» slowly ✔
passes «unchanged» up the food chain
OR
increased concentration as one species feeds on another «up the food chain» ✔
NOTE: Accept “lipids” for “fat”.
\nhydroxyl ✔
\nNOTE: Accept “hydroxy” but not “hydroxide”.
Accept “alkenyl”.
Do not accept formula.
\nThe dotted line represents the volume of carbon dioxide evolved when excess calcium carbonate is added to hydrochloric acid.
\nWhich graph represents the production of carbon dioxide when excess calcium carbonate is added to the same volume of hydrochloric acid of double concentration?
\nD
\nWhich property shows a general increase from left to right across period 2, Li to F?
\nA. Melting point
B. Electronegativity
C. Ionic radius
D. Electrical conductivity
\nB
\nHow many moles of magnesium hydroxide are produced with 0.50 mol of ammonia?
\nMg3N2 (s) + 6H2O (l) → 3Mg(OH)2 (aq) + 2NH3 (aq)
\nA. 0.25
\nB. 0.33
\nC. 0.75
\nD. 1.5
\nC
\nThis was one of the easier questions on the paper. 79 % of the candidates were able to deduce the amount of a product given the amount of another product and the balanced equation.
\nThe graph shows the Maxwell–Boltzmann energy distribution curve for a given gas at a certain temperature.
\nHow will the curve change if the temperature of the gas is increased, while other conditions remain constant?
\nA. The maximum is higher and to the left of A.
B. The maximum is higher and to the right of A.
C. The maximum is lower and to the right of A.
D. The maximum is lower and to the left of A.
\nC
\nMild heartburn is treated with antacids such as calcium carbonate.
\nFormulate an equation for the neutralization of stomach acid with calcium carbonate, CaCO3 (s).
\nDetermine the volume of CO2 (g), in dm3, produced at STP, when 1.00 g of CaCO3 (s) reacts completely with stomach acid.
\nMr CaCO3 = 100.09
\nAcid secretion can be regulated by other types of drugs such as omeprazole and ranitidine. Outline how each of these drugs acts to reduce excess stomach acid.
\n\n
Omeprazole:
\nRanitidine:
\nCaCO3 (s) + 2HCl (aq) → CO2 (g) + CaCl2 (aq) + H2O (l) [✔]
\n\n
Note: Accept balanced ionic equations involving “H+” or “H3O+”.
\nDo not accept “H2CO3”.
\nn CaCO3 = « =» 0.00999 «mol» [✔]
\nvolume CO2 = «0.00999 mol × 22.7 dm3 mol–1 = » 0.227 «dm3» [✔]
\n\n
Note: Accept 0.224 «dm3» if 22.4 dm3 mol–1 is used as molar volume.
\nAward [2] for correct answer.
\nOmeprazole:
inhibits enzyme/«gastric» proton pump «which secretes H+ ions into gastric juice»
OR
inhibits the H+/K+-ATPase system [✔]
Ranitidine:
inhibits/blocks H2/histamine receptors «in cells of stomach lining»
OR
prevents histamine binding to H2/histamine receptors «and triggering acid secretion» [✔]
Note: Accept “H2-receptor antagonist” for M2.
\nAbout two-thirds of the candidates could write an equation for the reaction of calcium carbonate with an acid. It was refreshing to find that “H2CO3” was only rarely encountered as a product.
\nIt was encouraging that over half the students were able to calculate the volume of gas produced from a given mass of calcium carbonate.
\nMany students correctly remembered the modes of action of Omeprazole and Ranitidine, though needless to say a few confused these and weaker students thought they acted like conventional antacids.
\nWhat is the sum of the integer coefficients when propene undergoes complete combustion?
\n__C3H6 (g) + __O2 (g) → __CO2 (g) + __H2O (l)
\nA. 11
\nB. 17
\nC. 21
\nD. 23
\nD
\nVery well answered. 88 % of the candidates balanced the equation and added up the integer coefficients correctly.
\nWhich is correct?
\nB
\nWhat is the volume of gas when the pressure on 100 cm3 of gas is changed from 400 kPa to 200 kPa at constant temperature?
\nA. 50.0 cm3
\nB. 100 cm3
\nC. 200 cm3
\nD. 800 cm3
\nC
\n80 % of the candidates were able to deduce the new volume of a sample of gas after the pressure was halved. The most commonly chosen distractor (A) was the value that assumed a direct proportionality between volume and pressure.
\nWhat is the intercept on the y-axis when a graph of lnk is plotted against on the x-axis?
\n\n
A. lnA
\nB.
\nC.
\nD.
\nA
\nWhat effect does increasing both pressure and temperature have on the equilibrium constant, Kc?
\nN2 (g) + 3H2 (g) 2NH3 (g) ΔH = −45.9 kJ
\nA. Decreases
B. Increases
C. Remains constant
D. Cannot be predicted as effects are opposite
\nA
\nWhich is an f-block element?
\nA. Sc
B. Sm
C. Sn
D. Sr
\nB
\nWhat is the concentration, in mol dm−3, of 20.0 g of NaOH (Mr = 40.0) in 500.0 cm3?
\nA. 0.250
\nB. 0.500
\nC. 1.00
\nD. 4.00
\nC
\n75 % of the candidates were able to calculate the molar concentration of the NaOH solution. The question had a good discrimination index.
\nWhich is correct for ?
\nD
\nA well answered question. 88 % of the candidates deduced the correct numbers of protons, neutrons and electrons in the sulfide ion.
\nWhich transition in the hydrogen atom emits visible light?
\nA. n = 1 to n = 2
\nB. n = 2 to n = 3
\nC. n = 2 to n = 1
\nD. n = 3 to n = 2
\nD
\nThis question discriminated well between high scoring and low scoring candidates. 68% of the candidates chose the correct transition that emits visible light (n = 3 to n = 2). The most commonly chosen distractor C was the only other option that involved emission.
\nWhat is the effect of a stronger ligand?
\nA
\nWhich compound has the shortest C to O bond?
\nA. CH3CHO
B. CO
C. CO2
D. C2H5OC2H5
\nB
\nWhich of the following would have the same numerical value for all elements in the same period?
\nA. Highest energy levels occupied
\nB. Energy sub-levels occupied
\nC. Orbitals occupied
\nD. Valence electrons
\nA
\nA challenging question with a high discrimination index. 46 % of the candidates chose the correct answer A (the highest energy levels occupied have the same numerical value for all elements in the same period). A high proportion of candidates chose C (orbitals occupied have the same value ….) and a significant proportion chose B (energy sub-levels occupied have the same value …).
\nHow do the following properties change down Group 17 of the periodic table?
\n\n
C
\n74 % of the candidates chose the correct trends for ionization energy and ionic radius down Group 17.
\nWhich describes a resonance structure?
\nA. Double bond can be drawn in alternative positions.
B. Bonds vibrate by absorbing IR radiation.
C. A double and a single bond in the molecule
D. A Lewis structure
\nA
\nHow does a lithium atom form the most stable ion?
\nA. The atom gains a proton to form a positive ion.
\nB. The atom loses a proton to form a negative ion.
\nC. The atom loses an electron to form a positive ion.
\nD. The atom gains an electron to form a negative ion.
\nC
\nOne of the most straight-forward questions on the paper about how lithium forms its most stable ion. It was answered correctly by 87 % of the candidates.
\nWhich combination causes the strength of metallic bonding to increase?
\nD
\nA question about factors affecting the strength of metallic bonding answered correctly by 68 % of candidates. The distractors were almost equally chosen by the rest of the candidates.
\nWhich corresponds to a system at equilibrium?
\nB
\nWhich molecule contains an incomplete octet of electrons?
\nA. NF3
\nB. BF3
\nC. BrF
\nD. SF2
\nB
\n66 % of the candidates identified BF3 as having an incomplete octet of electrons. The distractors NF3 and BrF were chosen in equal numbers, and the distractor SF2 was chosen by the least number of candidates.
\nWhat is the structure and bonding in SiO2 (s)?
\nA
\nWhat is the difference between a conjugate Brønsted–Lowry acid–base pair?
\nA. Electron pair
B. Positive charge
C. Proton
D. Hydrogen atom
\nC
\nThis question is about antiviral drugs.
\nOseltamivir, used for the treatment of severe flu, is inactive until converted in the liver to its active carboxylate form.
\nDraw a circle around the functional group that can be converted to the carboxylate by hydrolysis.
\nSuggest a reason for using a phosphate salt of oseltamivir in oral tablets.
\nAnti-HIV drugs, such as zidovudine, often become less effective over time.
\nExplain the development of resistant virus strains in the presence of antiviral drugs.
\n [✔]
\n
Note: Accept circles that include the alkyl side chain.
\nmore soluble «in water» [✔]
\nviruses undergo «rapid» mutation [✔]
\n mutation causes a change in viral protein
OR
drug to no longer binds to virus [✔]
\n
Note: Accept “rapid reproduction «allows resistant viruses to multiply»”.
\nMany students erroneously identified the amide as the required group, failing to realise that its hydrolysis would give the carboxylate ion of the side chain lost to the drug.
\nAbout a third of the candidates realised that producing a salt would increase the drug’s aqueous solubility, though many just stated “increased bioavailability” without explaining how this came about.
\nAnother question where well argued responses were rare, though many students gained credit for mentioning the ease of mutation and the speed of reproduction of viruses.
\nWhich is an example of an amphiprotic species?
\nA. Al2O3
B. CO32−
C. P4O10
D. HPO42−
\nD
\nIn which species does sulfur have the same oxidation state as in SO32–?
A. S2O32–
B. SO42–
C. H2S
D. SOCl2
\nD
\nThe following occurs when metal X is added to Y sulfate solution and Z sulfate solution. (X, Y and Z represent metal elements but not their symbols.)
\nX (s) + YSO4 (aq) → XSO4 (aq) + Y (s)
X (s) + ZSO4 (aq): no reaction
What is the order of increasing reactivity?
\nA. X < Y < Z
B. Y < X < Z
C. Z < Y < X
D. Z < X < Y
\nB
\nWhich atom is sp2 hybridized?
A. C in H2CO
B. C in CO2
C. N in CH3NH2
D. O in H2O
\nA
\nWhich can act as a Lewis acid but not a Brønsted–Lowry acid?
\nA. BF3
B. H2O
C. NF3
D. NH3
\nA
\nWhat are the products of electrolysis of concentrated aqueous sodium bromide?
\nD
\nWhat is the order, in increasing pH, of the following solutions of equal concentration?
\nA. H3BO3 < CH3CH2CH2COOH < CH3CH2COOH < CHCl2COOH
B. H3BO3 < CH3CH2CH2COOH < CHCl2COOH < CH3CH2COOH
C. CH3CH2CH2COOH < CH3CH2COOH < CHCl2COOH < H3BO3
D. CHCl2COOH < CH3CH2CH2COOH < CH3CH2COOH < H3BO3
\nD
\nWhich atom does not obey the octet rule?
\nA. C in CO2
B. F in BF3
C. O in H2O
D. S in SF6
\nD
\nWhich compound has hydrogen bonds between its molecules?
\nA. CH4
\nB. CH4O
\nC. CH3Cl
\nD. CH2O
\nB
\n72 % of the candidates identified CH4O as having hydrogen bonds between its molecules. The most commonly chosen distractor was CH2O, the only other option containing oxygen.
\nThree cells with platinum electrodes are connected in series to a DC power supply.
\nWhat is the ratio of moles formed at each cathode (negative electrode)?
\nB
\nThe following shows some compounds which can be made from ethene, C2H4.
\nethene (C2H4) → C2H5Cl → C2H6O → C2H4O
\nState the type of reaction which converts ethene into C2H5Cl.
\nWrite an equation for the reaction of C2H5Cl with aqueous sodium hydroxide to produce a C2H6O compound, showing structural formulas.
\nWrite an equation for the complete combustion of the organic product in (b).
\nDetermine the enthalpy of combustion of the organic product in (b), in kJ mol−1, using data from section 11 of the data booklet.
\nState the reagents and conditions for the conversion of the compound C2H6O, produced in (b), into C2H4O.
\nExplain why the compound C2H6O, produced in (b), has a higher boiling point than compound C2H4O, produced in d(i).
\nEthene is often polymerized. Draw a section of the resulting polymer, showing two repeating units.
\n«electrophilic» addition ✔
\n NOTE: Do not accept “nucleophilic addition” or “free radical addition”.
Do not accept “halogenation”.
CH3CH2Cl (g) + OH− (aq) → CH3CH2OH (aq) + Cl− (aq)
OR
CH3CH2Cl (g) + NaOH (aq) → CH3CH2OH (aq) + NaCl (aq) ✔
C2H6O (g) + 3O2 (g) → 2CO2 (g) + 3H2O (g)
OR
CH3CH2OH (g) + 3O2 (g) → 2CO2 (g) + 3H2O (g) ✔
bonds broken:
5(C–H) + C–C + C–O + O–H + 3(O=O)
OR
5(414«kJ mol−1») + 346«kJ mol−1» + 358«kJ mol−1» + 463«kJ mol−1» + 3(498«kJ mol−1») / 4731 «kJ» ✔
bonds formed:
4(C=O) + 6(O–H)
OR
4(804«kJ mol−1») + 6(463«kJ mol−1») / 5994 «kJ» ✔
«ΔH = bonds broken − bonds formed = 4731 − 5994 =» −1263 «kJ mol−1» ✔
NOTE: Award [3] for correct final answer.
\nK2Cr2O7/Cr2O72−/«potassium» dichromate «(VI)» AND acidified/H+
OR
«acidified potassium» manganate(VII) / «H+» KMnO4 / «H+» MnO4− ✔
NOTE: Accept “H2SO4” or “H3PO4” for “H+”.
Do not accept “HCl”.
Accept “permanganate” for “manganate(VII)”.
\n
distil ✔
\n\n
C2H6O/ethanol: hydrogen-bonding AND C2H4O/ethanal: no hydrogen-bonding/«only» dipole–dipole forces ✔
hydrogen bonding stronger «than dipole–dipole» ✔
\nNOTE: Continuation bonds must be shown.
Ignore square brackets and “n”.
Opium and its derivatives have been used for thousands of years as strong analgesics.
\nExplain how opiates act to provide pain relief.
\nDiscuss how the difference in structure of two opiates, codeine and morphine, affect their ability to cross the blood–brain barrier. Use section 37 of the data booklet.
\n«temporarily» bond/bind to «opioid» receptors in the brain/CNS [✔]
\nblock the transmission of pain impulses [✔]
\n«codeine crosses blood–brain barrier more easily»
morphine has more hydroxyl/OH «groups than codeine» [✔]
codeine/ether group is less polar
OR
hydroxyl/OH «groups in morphine» H-bond to water [✔]
\n
Note: Award [1 max] if no statement or an incorrect statement about the blood–brain barrier.
\nMost candidates correctly remembered the mode of action strong analgesics, though some had difficulty expressing it clearly.
\nMany candidates identified the relevant structural difference from the data booklet structures and could use these to explain the difference in the ability of the drugs to reach the brain, though some were confused as to whether this was favoured by aqueous solubility. This again was more frequently discussed in terms of polarity than the ability of –OH groups to form hydrogen bonds.
\nCompound A is in equilibrium with compound B.
\nPredict the electron domain and molecular geometries around the oxygen atom of molecule A using VSEPR.
\nThe IR spectrum of one of the compounds is shown:
\nCOBLENTZ SOCIETY. Collection © 2018 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.
\n
Deduce, giving a reason, the compound producing this spectrum.
Compound A and B are isomers. Draw two other structural isomers with the formula .
\nThe equilibrium constant, , for the conversion of A to B is in water at .
\nDeduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.
\nElectron domain geometry: tetrahedral ✔
\nMolecular geometry: bent/V-shaped ✔
\nB AND absorption/
OR
B AND absence of ✔
Accept any value between .
Accept any two isomers except for propanone and propen-2-ol:
\n✔✔
\n
Penalize missing hydrogens in displayed structural formulas once only.
\nAND is greater than /large ✔
\nHalf of the candidates answered correctly. The rest of the candidates often answered the question in terms of the carbon atom indicating that they did not read the question carefully.
\nAbout 50% of the candidates answered correctly. Quite a few, however, gave compounds other than A or B, indicating not reading the question properly or being confused by the skeletal formulas given in the question.
\nNearly half of the candidates gave two correct isomers. Propanal was often given as one of the isomers. Some candidates repeated the compounds given in the question and a few gave structures with 5 bonds on a carbon atom.
\nHalf of the candidates answered correctly. A common mistake was K > 0 instead of K > 1.
\nWhich compound is not in the same homologous series as the others?
\nA. C5H12
\nB. C6H12
C. C7H16
D. C8H18
\nB
\nConsider the following equations.
\n2Al (s) + O2 (g) → Al2O3 (s) ΔHƟ = −1670 kJ
Mn (s) + O2 (g) → MnO2 (s) ΔHƟ = −520 kJ
What is the standard enthalpy change, in kJ, of the reaction below?
\n4Al (s) + 3MnO2 (s) → 2Al2O3 (s) + 3Mn (s)
\nA. −1670 + 520
\nB. (−1670) + 3(520)
\nC. 2(−1670) + 3(−520)
\nD. 2(−1670) + 3(520)
\nD
\nIt is pleasing that 83 % of the candidates were confident in applying Hess’s law to obtain an expression for the standard enthalpy change of reaction. The most commonly chosen distractor (C) had the correct coefficients but one of the signs was incorrect.
\nVitamins are organic compounds essential in small amounts.
\nState the name of one functional group common to all three vitamins shown in section 35 of the data booklet.
\nExplain the biomagnification of the pesticide DDT.
\nExplain why maltose, C12H22O11, is soluble in water.
\nhydroxyl ✔
\nNOTE: Accept “hydroxy” but not “hydroxide”.
Accept “alkenyl”.
Do not accept formula.
accumulates in fat/tissues/living organisms
OR
cannot be metabolized/does not break down «in living organisms»
OR
not excreted / excreted «very» slowly ✔
passes «unchanged» up the food chain
OR
increased concentration as one species feeds on another «up the food chain» ✔
NOTE: Accept “lipids” for “fat”.
\n«solubility depends on forming many» H-bonds with water ✔
maltose has many hydroxyl/OH/oxygen atom/O «and forms many H-bonds» ✔
NOTE: Reference to “with water” required.
Accept “hydroxy” for “hydroxyl” but not “hydroxide/OH–”.
Reference to many/several OH groups/O atoms required for M2.
Consider the following data for butane and pentane at STP.
\nDiscuss the data.
\nIn a natural gas power station, 1.00 tonne of natural gas produces 2.41 × 104 MJ of electricity.
\nCalculate the percentage efficiency of the power station.
\n1 tonne = 1000 kg
Specific energy of natural gas used = 55.4 MJ kg−1
«similar specific energy and» pentane has «much» larger energy density ✔
\nAny two for [2 max]:
similar number of bonds/«C and H» atoms in 1 kg «leading to similar specific energy»
OR
only one carbon difference in structure «leading to similar specific energy» ✔
NOTE: Accept “both are alkanes” for M2.
pentane is a liquid AND butane is a gas «at STP» ✔
NOTE: Accept “pentane would be easier to transport”.
1 m3 of pentane contains greater amount/mass than 1 m3 of butane ✔
NOTE: Accept “same volume” for “1 m3” and “more moles” for “greater amount” for M4.
«energy input =» 5.54 ×104 «MJ» ✔
«efficiency = » 43.5 «%» ✔
NOTE: Award [2] for correct final answer.
\nWhich will react with a halogen by an electrophilic substitution mechanism?
\nA
\nA molecule of citric acid, C6H8O7, is shown.
\nThe equation for the first dissociation of citric acid in water is
\nC6H8O7 (aq) + H2O (l) C6H7O7− (aq) + H3O+ (aq)
\nIdentify a conjugate acid–base pair in the equation.
\nThe value of the equilibrium constant for the first dissociation at 298 K is 5.01 × 10−4.
\nState, giving a reason, the strength of citric acid.
\nThe dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H+], and on the equilibrium constant, of increasing the temperature.
\nOutline one laboratory methods of distinguishing between solutions of citric acid and hydrochloric acid of equal concentration, stating the expected observations.
\nC6H8O7 AND C6H7O7−
OR
H2O AND H3O+ ✔
weak acid AND partially dissociated
OR
weak acid AND equilibrium lies to left
OR
weak acid AND Kc/Ka<1 ✔
Any one of:
«electrical» conductivity AND HCl greater ✔
pH AND citric acid higher ✔
titrate with strong base AND pH at equivalence higher for citric acid ✔
add reactive metal/carbonate/hydrogen carbonate AND stronger effervescence/faster reaction with HCl ✔
titration AND volume of alkali for complete neutralisation greater for citric acid ✔
titrate with strong base AND more than one equivalence point for complete neutralisation of citric acid ✔
titrate with strong base AND buffer zone with citric acid ✔
\n
NOTE: Accept “add universal indicator AND HCl more red/pink” for M2.
\nAccept any acid reaction AND HCl greater rise in temperature.
\nAccept specific examples throughout.
\nDo not accept “smell” or “taste”.
\nEthanol is a biofuel that can be mixed with gasoline.
\nWrite the equation for the complete combustion of ethanol.
\nOutline the evidence that relates global warming to increasing concentrations of greenhouse gases in the atmosphere.
\nExplain, including a suitable equation, why biofuels are considered to be carbon neutral.
\nState the type of reaction that occurs when ethanol reacts with vegetable oil to form biodiesel.
\nC2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l) ✔
\nAny two of:
«showing strong» correlation between «atmospheric» CO2 concentration/greenhouse gas concentration and average «global/surface/ocean» temperature ✔
lab evidence that greenhouse gases/CO2 absorb(s) infrared radiation ✔
\n«advanced» computer modelling ✔
\nice core data ✔
\ntree ring data ✔
\nocean sediments / coral reefs / sedimentary rocks data ✔
\nNOTE: Do not accept “global warming” for “average temperature”.
Do not accept “traps/reflects heat” OR “thermal energy”.
Evidence must be outlined and connected to data.
Accept references to other valid greenhouse gases other than carbon dioxide/CO2, such as methane/CH4 or nitrous oxide/N2O.
biofuel raw material/sugar/glucose formed by photosynthesis
OR
biofuel raw material/sugar/glucose uses up carbon dioxide during its formation
OR
biofuel from capturing gases due to decaying organic matter formed from photosynthesis ✔
6CO2 (g) + 6H2O (l) → C6H12O6 (aq) + 6O2 (g) ✔
\nNOTE: Accept arguments based on material coming from plant sources consuming carbon dioxide/carbon for M1.
\ntransesterification
OR
«nucleophilic» substitution/SN ✔
Which compound cannot undergo addition polymerization?
\nC
\nIn which compound is the halogen substituted the most rapidly by aqueous hydroxide ions?
\nA. (CH3)3CCl
B. (CH3)3CI
C. CH3CH2CH2CH2Cl
D. CH3CH2CH2CH2I
\nB
\nMethane undergoes incomplete combustion.
\n2CH4 (g) + 3O2 (g) → 2CO (g) + 4H2O (g)
\nWhat is the enthalpy change, in kJ, using the bond enthalpy data given below?
\n\n
A. [2(1077) + 4(463)] − [2(414) + 3(498)]
\nB. [2(414) + 3(498)] − [2(1077) + 4(463)]
\nC. [8(414) + 3(498)] − [2(1077) + 8(463)]
\nD. [2(1077) + 8(463)] − [8(414) + 3(498)]
\nC
\n65 % of the candidates chose the correct expression for determining the enthalpy change using bond enthalpies. The most commonly chosen distractor (C) had the signs reversed. The question had a good discrimination index.
\nMost of the nuclear waste generated in a hospital is low-level waste (LLW).
\nOutline what is meant by low-level waste.
\nOutline the disposal of LLW.
\nsmall/low amounts of radiation AND for a short time [✔]
\n\n
Note: Accept “weakly ionizing radiation” instead of “small amounts of radiation”.
\nAccept “short half-lives” instead of “for a short time”.
\nstored in shielded containers until radiation drops «to a safe level» [✔]
\nMost candidates just listed items that low-level nuclear waste was likely to comprise, rather than giving two of its radiation characteristics (low activity, non-penetrating radiation, short half-life).
\nThere appeared to be very limited awareness of the need to store low-level waste in shielded containers until the radiation falls to a negligible level.
\nInfectious diseases can be caused by bacteria or viruses.
\nState one difference between bacteria and viruses.
\nDiscuss two difficulties, apart from socio-economic factors, associated with finding a cure for AIDS.
\nThe discovery of penicillins contributed to the development of antibiotics.
\nExplain how the beta-lactam ring is responsible for the antibiotic properties of penicillin. Refer to section 37 of the data booklet.
\nAny one of:
bacteria perform living functions «on their own and viruses do not without host cell»
OR
bacteria have cell walls «and viruses do not»
OR
bacteria do not have a capsid «and viruses do»
OR
bacteria larger than viruses
OR
bacteria reproduce by fission/budding «and viruses reproduce within a living host cell»
OR
bacteria affected by antibiotics «while viruses are not» ✔
NOTE: Accept “bacteria have flagella/cytoplasm/ribosome «and virus can have head/protein tail/double stranded RNA/single stranded DNA»”
Accept “asexual reproduction” for bacteria.
Accept other specific structural differences between bacteria and viruses, and examples of living functions that bacteria perform (such as excretion, reproduction etc.) that viruses do not.
Accept “bacteria are living and viruses are not”
\nAny two of:
HIV difficult to detect/remains dormant ✔
HIV mutates rapidly/quickly ✔
NOTE: Do not accept “AIDS mutates” without mention of the HIV/virus.
HIV replicates rapidly/quickly ✔
\nHIV destroys «T-» helper cells/white blood cells/lymphocytes
OR
HIV attacks immune system ✔
HIV has several «significantly different» strains/subtypes ✔
\nNOTE: Accept “virus” for “HIV”.
Penalize the use of “AIDS” for “HIV” once only.
Accept “HIV metabolism linked to that of host cell” OR “drugs harm host cell as well as HIV”.
ring is «sterically» strained
OR
angles of 90° instead of 109.5/109/120° angles
OR
angles smaller than 109.5/109/120°/tetrahedral/trigonal planar/triangular planar angle ✔
NOTE: Accept arguments using correct descriptions of hybridization for M1.
ring breaks up/opens/reacts «easily»
OR
amido/amide group «in ring» is «highly» reactive ✔
«irreversibly» binds/bonds to enzyme/transpeptidase
OR
inhibits enzyme/transpeptidase «in bacteria» that produces cell walls
OR
prevents cross-linking of «bacterial» cell walls ✔
NOTE: Do not accept \"breaks/binds to cell walls\" – a reference to the enzyme is needed for alternatives 1 and 2 for M3.
Do not accept “cell membrane” for “cell wall” for M3.
Copper forms two chlorides, copper(I) chloride and copper(II) chloride.
\nAn electrolysis cell was assembled using graphite electrodes and connected as shown.
\nState the electron configuration of the Cu+ ion.
\nCopper(II) chloride is used as a catalyst in the production of chlorine from hydrogen chloride.
\n4HCl (g) + O2 (g) → 2Cl2 (g) + 2H2O (g)
\nCalculate the standard enthalpy change, ΔHθ, in kJ, for this reaction, using section 12 of the data booklet.
\nThe diagram shows the Maxwell–Boltzmann distribution and potential energy profile for the reaction without a catalyst.
\nAnnotate both charts to show the activation energy for the catalysed reaction, using the label Ea (cat).
\nExplain how the catalyst increases the rate of the reaction.
\nSolid copper(II) chloride absorbs moisture from the atmosphere to form a hydrate of formula CuCl2•H2O.
\nA student heated a sample of hydrated copper(II) chloride, in order to determine the value of . The following results were obtained:
\nMass of crucible = 16.221 g
Initial mass of crucible and hydrated copper(II) chloride = 18.360 g
Final mass of crucible and anhydrous copper(II) chloride = 17.917 g
Determine the value of .
\nState how current is conducted through the wires and through the electrolyte.
\nWires:
\nElectrolyte:
\nWrite the half-equation for the formation of gas bubbles at electrode 1.
\n[Ar] 3d10
OR
1s2 2s2 2p6 3s2 3p6 3d10 ✔
ΔHθ = ΣΔHθf (products) − ΣΔHθf (reactants) ✔
ΔHθ = 2(−241.8 «kJ mol−1») − 4(−92.3 «kJ mol−1») = −114.4 «kJ» ✔
\n
NOTE: Award [2] for correct final answer.
\nEa (cat) to the left of Ea ✔
\n\n
peak lower AND Ea (cat) smaller ✔
\n«catalyst provides an» alternative pathway ✔
\n«with» lower Ea
OR
higher proportion of/more particles with «kinetic» E ≥ Ea(cat) «than Ea» ✔
mass of H2O = «18.360 g – 17.917 g =» 0.443 «g» AND mass of CuCl2 = «17.917 g – 16.221 g =» 1.696 «g» ✔
\nmoles of H2O = «=» 0.0246 «mol»
OR
moles of CuCl2 =«= » 0.0126 «mol» ✔
«water : copper(II) chloride = 1.95 : 1»
« =» 2 ✔
NOTE: Accept « =» 1.95.
\nNOTE: Award [3] for correct final answer.
\nWires:
«delocalized» electrons «flow» ✔
Electrolyte:
«mobile» ions «flow» ✔
2Cl− → Cl2 (g) + 2e−
OR
Cl− → Cl2 (g) + e− ✔
NOTE: Accept e for e−.
\nWhich is the activation energy of the forward reaction?
\nB
\nWell answered. 88 % of the candidates chose the correct arrow representing the activation energy of the forward reaction.
\nWhich can be reduced to an aldehyde?
\nA. Butanone
B. Butan-1-ol
C. Butanoic acid
D. Butan-2-ol
\nC
\nThe same amount of two gases, X and Y, are in two identical containers at the same temperature. What is the difference between the gases?
\nA. X has the higher molar mass.
\nB. Y has the higher molar mass.
\nC. X has the higher average kinetic energy.
\nD. Y has the higher average kinetic energy.
\nA
\nThis was by far the most challenging question on the paper, answered correctly by only 19 % of the candidates. Some teachers thought this was beyond the scope of the syllabus while others thought it was a question requiring thought. To be able to answer, candidates needed to connect kinetic energy to the speed and mass of the molecule.
\nWhich can show optical activity?
\nA. CHBrCHCl
B. CH3CH2CHBrCH2CH3
C. (CH3)2CBrCl
D. CH3CH2CH(CH3)Br
\nD
\nSeveral reactions of calcium carbonate with dilute hydrochloric acid are carried out at the same temperature.
\nCaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)
\nWhich reaction has the greatest rate?
\n\n
A
\n65 % of the candidates chose the correct combination to give the greatest rate of reaction. The most commonly chosen distractor was D where “smaller surface area of same mass of CaCO3(s)” was chosen. It seems these candidates confused “surface area” with “particle size”.
\nWhat is the value of the temperature change?
\nInitial temperature: 2.0 ± 0.1 °C
Final temperature: 15.0 ± 1.0 °C
\nA. 13.0 ± 0.1 °C
B. 13.0 ± 0.9 °C
C. 13.0 ± 1.0 °C
D. 13.0 ± 1.1 °C
\nD
\nDisposal of chemical waste is a growing problem in industry.
\nOutline the impact of antibiotic waste on the environment.
\nSuggest a concern about the disposal of solvents from drug manufacturing.
\n«leads to bacterial» resistance «to antibiotics»
OR
destroys useful/beneficial bacteria
OR
useful/beneficial/less harmful bacteria replaced with «more» harmful bacteria ✔
NOTE: Accept \"affects/disturbs micro-ecosystems\".
\nAny one of:
«most are» toxic «to living organisms»
OR
incomplete combustion/incineration can produce toxic products/dioxins/phosgene
OR
carcinogenic/can cause cancer ✔
accumulate in groundwater
OR
have limited biodegradability ✔
cost of disposal ✔
\nNOTE: Do not accept “harmful to the environment”.
Do not accept just “pollutes water”.
Do not accept “hazard of disposal”.
Accept “ozone depletion” only if there is some reference to chlorinated solvents.
Which is the 1H NMR spectrum of tetramethylsilane, TMS, (CH3)4Si?
\nC
\nWhat is the equilibrium constant expression for the following equation?
\n2NO2 (g) + F2 (g) 2NO2F (g)
\nA.
\nB.
\nC.
\nD.
\nD
\nOne of the easiest questions on the paper. 89 % of the candidates chose the correct Kc expression.
\nWhat is the pH of 0.001 mol dm−3 NaOH (aq)?
\nA. 1
\nB. 3
\nC. 11
\nD. 13
\nC
\nThe question discriminated well between high scoring and low scoring candidates. 58 % of the candidates were able to calculate the pH of the aqueous solution of NaOH. The most commonly chosen distractor was B (pH = 3) where the students determined pOH but did not complete the calculation. It is interesting that these candidates did not seem to notice that NaOH is a base and should have a higher pH.
\nWhat is the major reason why the pH of unpolluted rain is less than 7?
\nA. methane
\nB. carbon dioxide
\nC. nitrogen oxides
\nD. sulfur dioxide
\nB
\nThis was one of the challenging questions on the paper. 58 % of the candidates identified carbon dioxide as the cause of acidity in unpolluted rain. The rest mainly chose either nitrogen oxides (option C) or sulfur dioxide (option D), the causes of acidity in polluted rain.
\nPolypropene is used to make many objects including carpets, stationery and laboratory equipment.
\nDraw a section of an isotactic polypropene polymer chain containing four repeating units.
\nPredict, with a reason, whether isotactic or atactic polypropene has the higher melting point.
\nPolypropene is a thermoplastic. Outline what is meant by thermoplastic.
\nDiscuss why the recycling of plastics is an energy intensive process.
\n\n
NOTE: Continuation bonds must be shown.
Ignore square brackets and “n”.
Do not accept one repeating unit in square brackets with a subscript of 4.
Accept condensed structure provided all C to C bonds are shown and CH3 groups on same side.
Accept
Do not accept syndiotactic (alternating orientation of the CH3 groups).
\nisotactic «has higher melting point» AND ordered chains pack more closely
OR
isotactic «has higher melting point» AND stronger intermolecular/London/dispersion forces ✔
NOTE: Accept “van der Waals’ forces”/”vdW”.
\nsoftens when heated «and hardens when cooled» ✔
\nAny two of:
collection/transportation of plastic waste ✔
separation of different types «of plastic»
OR
separation of plastic from other materials ✔
melting plastic ✔
processing/washing/cleaning/drying/manufacture of recycled plastic ✔
An equation for the combustion of propane is given below.
\nC3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
Determine the standard enthalpy change, , for this reaction, using section 11 of the data booklet.
\n\n
Calculate the standard enthalpy change, , for this reaction using section 12 of the data booklet.
\nBonds broken: ✔
\nBonds formed: ✔
\n«Enthalpy changebonds brokenbonds formed ✔
\n\n
Award [3] for correct final answer.
\nAND AND ✔
\n✔
\n\n
Award [2] for correct final answer.
\nAward [1 max] for .
\nThis question was generally well answered. The mean mark on the question was 1.8 out of 3 marks. Mistakes included three C–C (instead of two), missing C–C bonds completely, subtracting bonds formed bonds broken, 3 and 4 instead of 6 and 8 for bonds formed coefficients, and bond enthalpies of C=O and O=O double bonds treated as single bonds.
\nAbout half of the candidates obtained both marks. In the incorrect answers coefficients were sometimes ignored, some candidates used the wrong state for water, and some candidates did not realize the heat of formation of O2 was zero and inserted the bond enthalpy for O2 in the calculation.
\nDescribe the function of chlorophyll in photosynthesis.
\nCompare and contrast the structures of starch and cellulose.
\nOne similarity:
\nOne difference:
\nExplain why maltose, C12H22O11, is soluble in water.
\nabsorbs/traps light «energy» ✔
\ninitiates redox reactions
OR
transfers electrons ✔
One similarity:
1−4/glycosidic linkage
OR
glucose monomers/residues ✔
NOTE: Accept “both are polysaccharides”.
\nOne difference:
starch has α-glucose AND cellulose has β-glucose «monomers»
OR
starch can form coiled/spiral/helical chains «and straight chains» AND cellulose cannot/can only form straight chains/can only form a linear structure
OR
starch «in amylopectin» also has 1−6 glycosidic links AND cellulose does not ✔
NOTE: Accept \"cellulose has alternate glucose monomers upside down with respect to each other AND starch does not\".
\n«solubility depends on forming many» H-bonds with water ✔
maltose has many hydroxyl/OH/oxygen atom/O «and forms many H-bonds» ✔
NOTE: Reference to “with water” required.
Accept “hydroxy” for “hydroxyl” but not “hydroxide/OH–”.
Reference to many/several OH groups/O atoms required for M2.
Which species contains nitrogen with the highest oxidation state?
\nA. NO3−
\nB. NO2−
\nC. NO2
\nD. N2O
\nA
\nIt is pleasing that 85 % of the candidates identified the species containing nitrogen with the highest oxidation state.
\nConsider the following electrochemical cell.
\nWhat happens to the ions in the salt bridge when a current flows?
\nA. Na+ ions flow to the zinc half-cell and SO42− ions flow to the copper half-cell.
\nB. Na+ ions flow to the copper half-cell and SO42− ions flow to the zinc half-cell.
\nC. Na+ and SO42− ions flow to the copper half-cell.
\nD. Na+ and SO42− ions flow to the zinc half-cell.
\nB
\nA challenging question about direction of ion flow in a salt bridge. 60 % answered correctly and the option with the opposite directions of flow of ions (A) was the most commonly chosen distractor.
\nNickel catalyses the conversion of propanone to propan-2-ol.
\nOutline how a catalyst increases the rate of reaction.
\nExplain why an increase in temperature increases the rate of reaction.
\nDiscuss, referring to intermolecular forces present, the relative volatility of propanone and propan-2-ol.
\nThe diagram shows an unlabelled voltaic cell for the reaction
\n\nLabel the diagram with the species in the equation.
\n\n
Suggest a metal that could replace nickel in a new half-cell and reverse the electron flow. Use section 25 of the data booklet.
\nDescribe the bonding in metals.
\nNickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the addition of another metal to nickel.
\nprovides an alternative pathway/mechanism AND lower ✔
\nAccept description of how catalyst lowers (e.g. “reactants adsorb on surface «of catalyst»”, “reactant bonds weaken «when adsorbed»”).
\nmore/greater proportion of molecules with ✔
\ngreater frequency/probability/chance of collisions «between the molecules»
OR
more collision per unit of time/second ✔
hydrogen bonding/bonds «and dipole–dipole and London/dispersion forces are present in» propan-2-ol ✔
\ndipole–dipole «and London/dispersion are present in» propanone ✔
\npropan-2-ol less volatile AND hydrogen bonding/bonds stronger «than dipole–dipole »
OR
propan-2-ol less volatile AND «sum of all» intermolecular forces stronger ✔
✔
\nAccept OR .
\nelectrostatic attraction ✔
\nbetween «a lattice of» metal/positive ions/cations AND «a sea of» delocalized electrons ✔
\n
Accept “mobile/free electrons”.
Any of:
\nmalleability/hardness
OR
«tensile» strength/ductility
OR
density
OR
thermal/electrical conductivity
OR
melting point
OR
thermal expansion ✔
Do not accept corrosion/reactivity or any chemical property.
Accept other specific physical properties.
\nA straight-forward question, however, half of the candidates only mentioned the lower activation energy and did not mention that this is through an alternative mechanism, so did not score the mark.
\nHalf of the candidates gained the mark about the increased frequency of collision. Fewer candidates also clarified that a larger proportion of molecules have the activation energy.
\nMost candidates had the correct structure in their answers identifying the type of intermolecular forces in each compound and then comparing the strength of the two and reaching a conclusion. Some candidates did not know what was meant by volatile. Some candidates stated London dispersion forces in propanone instead of dipole-dipole.
\n60% of the candidates obtained the mark. Some candidates labelled the electrodes as ions indicating they do not understand the structure of a voltaic cell.
\n70% of the candidates answered correctly. The common mistake was to select a more reactive metal instead.
\nThe mean mark on the question was 1.0 out of 2 marks. Mistakes included not mentioning the 'electrostatic attraction' and talking about 'nuclei attracting the delocalised electrons'. The weakest candidates discussed aspects of ionic and/or covalent bonding.
\n80% obtained the mark. Many candidates wrote more than one property, which should be discouraged. Incorrect answers included chemical properties such as reactivity.
\nThe following reaction occurs in a voltaic (galvanic) cell.
\nMg (s) + 2Ag+ (aq) → Mg2+ (aq) + 2Ag (s)
\nWhich reaction takes place at each electrode?
\nC
\nThe question about electrode reactions discriminated well between high scoring and low scoring candidates. 70 % of the candidates selected the correct electrode reactions given the cell reaction. The most commonly chosen distractor had the same reactions at the opposite electrodes.
\nOctane number is a measure of the performance of engine fuel.
\nSuggest why a high-octane number fuel is preferable.
\nReforming reactions are used to increase the octane number of a hydrocarbon fuel.
\nSuggest the structural formulas of two possible products of the reforming reaction of heptane, C7H16.
\nThe 1H NMR spectrum of one of the products has four signals. The integration trace shows a ratio of the areas under the signals of 9 : 3 : 2 : 2.
\nDeduce the structural formula of the product.
\nlow knocking/auto-ignition
NOTE: Do not accept “pre-ignition”.
OR
more efficient fuel
NOTE: Accept “less CO2 emissions since knocking engine uses more fuel «to produce the same power»”.
OR
high compression
OR
more power extracted
OR
more air going into engine / turbocharging
OR
less engine damage ✔
Any two of:
\nNOTE: Accept skeletal formulas or full or condensed structural formulas.
Accept any other branched cycloalkane that contains 7 carbons.
Do not accept any alkenes.
Penalise missing hydrogens or bond connectivities once only in Option C.
Accept hydrogen as the second product if the first product is toluene or a cycloalkane.
NOTE: Accept a skeletal formula or a full or condensed structural formula.
Penalise missing hydrogens or bond connectivities once only in Option C.
Consider the following data for butane and pentane at STP.
\nDiscuss the data.
\nOutline what is meant by the degradation of energy.
\n«similar specific energy and» pentane has «much» larger energy density ✔
\nAny two for [2 max]:
similar number of bonds/«C and H» atoms in 1 kg «leading to similar specific energy»
OR
only one carbon difference in structure «leading to similar specific energy» ✔
NOTE: Accept “both are alkanes” for M2.
pentane is a liquid AND butane is a gas «at STP» ✔
NOTE: Accept “pentane would be easier to transport”.
1 m3 of pentane contains greater amount/mass than 1 m3 of butane ✔
NOTE: Accept “same volume” for “1 m3” and “more moles” for “greater amount” for M4.
energy converted to heat
OR
energy converted to less useful/dispersed forms
OR
energy converted to forms that have lower potential to do work
OR
heat transferred to the surroundings ✔
NOTE: Reference to energy conversion/transfer required. Do not accept reference to loss of energy.
\nOxidation and reduction reactions can have a variety of commercial uses.
\nA student decides to build a voltaic cell consisting of an aluminium electrode, Al (s), a tin electrode, Sn (s), and solutions of aluminium nitrate, Al(NO3)3 (aq) and tin(II) nitrate, Sn(NO3)2 (aq).
\nElectron flow is represented on the diagram.
\nLabel each line in the diagram using section 25 of the data booklet.
\nWrite the equation for the expected overall chemical reaction in (a).
\nCalculate the cell potential using section 24 of the data booklet.
\nCalculate the Gibbs free energy change, ΔG⦵, in kJ, for the cell, using section 1 of the data booklet.
\nAl/aluminium «electrode» AND aluminium nitrate/Al(NO3)3/Al3+ on left ✓
\nSn/tin «electrode» AND tin«(II)» nitrate/Sn(NO3)2/Sn2+ on right ✓
\nsalt bridge AND voltmeter/V/lightbulb ✓
\n
Award [1] if M1 and M2 are reversed.
Award [1] for two correctly labelled solutions OR two correctly labelled electrodes for M1 and M2.
\nAccept a specific salt for “salt bridge”.
\nAccept other circuit components such as ammeter/A, fan, buzzer, resistor/heating element/R/Ω.
\n3Sn2+ (aq) + 2Al (s) → 3Sn (s) + 2Al3+ (aq)
OR
3Sn(NO3)2 (aq) + 2Al (s) → 3Sn (s) + 2Al(NO3)3 (aq) ✓
If half cells are reversed in part-question (a) then the equation must be reversed to award the mark.
\nDo not penalize equilibrium arrows.
\n«1.66 + (−0.14) = +»1.52 «V» ✓
\n\n
Calculation must be consistent with equation given in 3 b.
\n«ΔG⦵ = −nFE⦵ = −6 × 9.65 × 104 × 1.52 =» −880080 «J mol−1»
OR
6 «electrons» ✓
«=» −880 «kJ» ✓
\n\n
Award [1] for “«+»880”.
\nAward [2] for correct final answer.
\nEnzymes are biological catalysts.
\nThe graph shows the relationship between the temperature and the rate of an enzyme-catalysed reaction.
\nState one reason for the decrease in rate above the optimum temperature.
\nExplain why a change in pH affects the tertiary structure of an enzyme in solution.
\nState one use of enzymes in reducing environmental problems.
\nenzyme denatures
OR
change of conformation/shape of active site
OR
substrate cannot bind to active site/binds less efficiently ✔
NOTE: Accept “change in structure” or “substrate doesn't fit/fits poorly into active site”
\nAny two of:
acidic/basic/ionizable/COOH/carboxyl/NH2/amino groups in the R groups/side chains «react» ✔
exchange/lose/gain protons/H+ ✔
change in H-bonds/ionic interactions/intermolecular forces/London dispersion forces ✔
NOTE: Do not accept “enzyme denatures” OR “change of conformation/tertiary structure” OR “substrate cannot bind to active site/binds less efficiently” as this was the answer to 8(a).
breakdown of oil spills/industrial/sewage waste/plastics
OR
production of alternate sources of energy «such as bio diesel»
OR
involve less toxic chemical pathway «in industry» ✔
NOTE: Accept “«enzymes in» biological detergents can improve energy efficiency”.
\nRed supergiant stars contain carbon-12 formed by the fusion of helium-4 nuclei with beryllium-8 nuclei.
\nMass of a helium-4 nucleus = 4.002602 amu
Mass of a beryllium-8 nucleus = 8.005305 amu
Mass of a carbon-12 nucleus = 12.000000 amu
State the nuclear equation for the fusion reaction.
\nExplain why fusion is an exothermic process.
\nCalculate the heat energy released, in J, by the fusion reaction producing one atom of carbon-12. Use section 2 of the data booklet and E = mc2.
\nBeryllium-8 is a radioactive isotope with a half-life of 6.70 × 10−17 s.
\nCalculate the mass of beryllium-8 remaining after 2.01 × 10−16 s from a sample initially containing 4.00 g of beryllium-8.
\n✔
\nNOTE: Do not penalize missing atomic numbers.
\nALTERNATIVE 1
binding energy per nucleon is larger in carbon-12/product «than beryllium-8 and helium-4/reactants» ✔
difference in «total» binding energy is released «during fusion» ✔
\n\n
ALTERNATIVE 2
mass of carbon-12/product «nucleus» is less than «the sum of» the masses of helium-4 and beryllium-8 «nuclei»/reactants
OR
two smaller nuclei form a lager nucleus ✔
mass lost/difference is converted to energy «and released»
OR
E = mc2 ✔
Δm = «12.000000 amu − (4.002602 amu + 8.005305 amu) =» −0.007907 «amu» ✔
«0.007907 amu × 1.66 × 10−27 kg amu−1 =» 1.31 × 10−29 «kg» ✔
«E = mc2 = 1.31 × 10−29 kg × (3.00 × 108 m s−1)2 =» 1.18 × 10−12 «J» ✔
NOTE: Accept “0.007907 «amu»”.
Award [2 max] for “7.12 x 1014 «J»”.
Award [3] for correct final answer.
ALTERNATIVE 1
3 half-lives ✔
0.500 g «of beryllium-8 remain» ✔
\n
ALTERNATIVE 2
✔
0.500 g «of beryllium-8 remain» ✔
\n
ALTERNATIVE 3
λ = « »= 1.03 × 1016 «s−1» ✔
m = « » 0.500 «g» ✔
\n
NOTE: Award [2] for correct final answer.
\nRed supergiant stars contain carbon-12 formed by the fusion of helium-4 nuclei with beryllium-8 nuclei.
\nMass of a helium-4 nucleus = 4.002602 amu
Mass of a beryllium-8 nucleus = 8.005305 amu
Mass of a carbon-12 nucleus = 12.000000 amu
State the nuclear equation for the fusion reaction.
\nExplain why fusion is an exothermic process.
\nBeryllium-8 is a radioactive isotope with a half-life of 6.70 × 10−17 s.
\nCalculate the mass of beryllium-8 remaining after 2.01 × 10−16 s from a sample initially containing 4.00 g of beryllium-8.
\n✔
\nNOTE: Do not penalize missing atomic numbers.
\nALTERNATIVE 1
binding energy per nucleon is larger in carbon-12/product «than beryllium-8 and helium-4/reactants» ✔
difference in «total» binding energy is released «during fusion» ✔
\n\n
ALTERNATIVE 2
mass of carbon-12/product «nucleus» is less than «the sum of» the masses of helium-4 and beryllium-8 «nuclei»/reactants
OR
two smaller nuclei form a lager nucleus ✔
mass lost/difference is converted to energy «and released»
OR
E = mc2 ✔
ALTERNATIVE 1
3 half-lives ✔
0.500 g «of beryllium-8 remain» ✔
\n
ALTERNATIVE 2
✔
0.500 g «of beryllium-8 remain» ✔
\n
ALTERNATIVE 3
λ = « »= 1.03 × 1016 «s−1» ✔
m = « » 0.500 «g» ✔
\n
NOTE: Award [2] for correct final answer.
\nWhich mechanism does benzene most readily undergo?
\nA. Nucleophilic substitution
B. Electrophilic substitution
C. Electrophilic addition
D. Free radical substitution
\nB
\nVast majority of candidates understood benzene undergoes substitution rather than addition, with some thinking of it as nucleophilic rather than electrophilic substitution.
\nA student obtained the following data to calculate , using .
\n\n\n\nWhat is the percentage uncertainty in the calculated value of ?
A.
B.
C.
D.
\nC
\nJust over ½ of the candidates could propagate uncertainties from absolute to relative.
\nWhich region of the electromagnetic spectrum is used to identify hydrogen environments in a molecule?
\nA. X-ray
B. UV
C. IR
D. radio waves
\nD
\nNot even a quarter of candidates knew the region of the electromagnetic spectrum associated with identifying hydrogen environments. This was the lowest scoring question on the exam with the most popular answer
being IR region.
What is the index of hydrogen deficiency (IHD) in cyclohexanol?
\nA.
\nB.
\nC.
\nD.
\nB
\nSlightly more than half of candidates could deduce an index of hydrogen deficiency. It was a poor discriminating question with distractors being chosen equally.
\nWhich element is in group 13?
\nB
\nThe majority of candidates could identify the group an element belongs in based on its successive ionization energies.
\nChlorine undergoes many reactions.
\nof manganese(IV) oxide was added to of .
\n\nChlorine gas reacts with water to produce hypochlorous acid and hydrochloric acid.
\n\nis a common chlorofluorocarbon, .
\nState the full electron configuration of the chlorine atom.
\nState, giving a reason, whether the chlorine atom or the chloride ion has a larger radius.
\nOutline why the chlorine atom has a smaller atomic radius than the sulfur atom.
\nThe mass spectrum of chlorine is shown.
\n NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.
Outline the reason for the two peaks at and .
Explain the presence and relative abundance of the peak at .
\nCalculate the amount, in , of manganese(IV) oxide added.
\nDetermine the limiting reactant, showing your calculations.
\nDetermine the excess amount, in , of the other reactant.
\nCalculate the volume of chlorine, in , produced if the reaction is conducted at standard temperature and pressure (STP). Use section 2 of the data booklet.
\nState the oxidation state of manganese in and .
\nDeduce, referring to oxidation states, whether is an oxidizing or reducing agent.
\nHypochlorous acid is considered a weak acid. Outline what is meant by the term weak acid.
\n\n
State the formula of the conjugate base of hypochlorous acid.
\nCalculate the concentration of in a solution with a .
\nState the type of reaction occurring when ethane reacts with chlorine to produce chloroethane.
\nPredict, giving a reason, whether ethane or chloroethane is more reactive.
\nExplain the mechanism of the reaction between chloroethane and aqueous sodium hydroxide, , using curly arrows to represent the movement of electron pairs.
\nEthoxyethane (diethyl ether) can be used as a solvent for this conversion.
Draw the structural formula of ethoxyethane
Deduce the number of signals and chemical shifts with splitting patterns in the 1H NMR spectrum of ethoxyethane. Use section 27 of the data booklet.
\nCalculate the percentage by mass of chlorine in .
\nComment on how international cooperation has contributed to the lowering of emissions responsible for ozone depletion.
\ns produce chlorine radicals. Write two successive propagation steps to show how chlorine radicals catalyse the depletion of ozone.
\n✔
\n
Do not accept condensed electron configuration.
AND more «electron–electron» repulsion ✔
\n
Accept AND has an extra electron.
has a greater nuclear charge/number of protons/ «causing a stronger pull on the outer electrons» ✔
\nsame number of shells
OR
same «outer» energy level
OR
similar shielding ✔
«two major» isotopes «of atomic mass and » ✔
\n«diatomic» molecule composed of «two» chlorine-37 atoms ✔
\nchlorine-37 is the least abundant «isotope»
OR
low probability of two «isotopes» occurring in a molecule ✔
✔
\n✔
\nAND is the limiting reactant ✔
\n
Accept other valid methods of determining the limiting reactant in M2.
✔
\n✔
\n
Accept methods employing .
✔
\n✔
\noxidizing agent AND oxidation state of changes from to /decreases ✔
\npartially dissociates/ionizes «in water» ✔
\n✔
\n✔
\n«free radical» substitution/ ✔
\n
Do not accept electrophilic or nucleophilic substitution.
chloroethane AND C–Cl bond is weaker/ than C–H bond/
OR
chloroethane AND contains a polar bond ✔
Accept “chloroethane AND polar”.
curly arrow going from lone pair/negative charge on in −OH to ✔
\ncurly arrow showing leaving ✔
\nrepresentation of transition state showing negative charge, square brackets and partial bonds ✔
\n\n
Accept with or without the lone pair.
\nDo not accept curly arrows originating on in .
\nAccept curly arrows in the transition state.
\nDo not penalize if and are not at 180°.
\nDo not award M3 if bond is represented.
\n / ✔
Accept .
2 «signals» ✔
\n0.9−1.0 AND triplet ✔
\n3.3−3.7 AND quartet ✔
\nAccept any values in the ranges.
\nAward [1] for two correct chemical shifts or two correct splitting patterns.
\n✔
\n✔
\n
Award [2] for correct final answer.
Any of:
\nresearch «collaboration» for alternative technologies «to replace s»
OR
technologies «developed»/data could be shared
OR
political pressure/Montreal Protocol/governments passing legislations ✔
\n
Do not accept just “collaboration”.
\nDo not accept any reference to as greenhouse gas or product of fossil fuel combustion.
\nAccept reference to specific measures, such as agreement on banning use/manufacture of s.
\n✔
\n
OR
✔
Penalize missing/incorrect radical dot (∙) once only.
\nWell answered question with 90% of candidates correctly identifying the complete electron configuration for chlorine.
\nMost candidates could correctly explain the relative sizes of chlorine atom and chloride ion.
\nFairly well answered though some candidates missed M2 for not recognizing the same number of shells affected.
\nMore than 80% could identify that the two peaks in the MS of chlorine are due to different isotopes.
\nNot well answered. Some candidates were able to identify m/z 74 being due to the m/z of two Cl-37 atoms, however fewer candidates were able to explain the relative abundance of the isotope.
\nStoichiometric calculations were generally well done and over 90% could calculate mol from a given mass.
\n90% of candidates earned full marks on this 2-mark question involving finding a limiting reactant.
\nSurprisingly, quite a number of candidates struggled with the quantity of excess reactant despite correctly identifying limiting reactant previously.
\nMost candidates could find the volume of gas produced in a reaction under standard conditions.
\nMore than 90% could identify the oxidation number of manganese in both MnO2 and MnCl2.
\nMost candidates stated that MnO2 is an oxidizing agent in the reaction but many did not get the mark because there was no reference to oxidation states.
\nAnother well answered 1-mark question where candidates correctly identified a weak acid as an acid which partially dissociates in water.
\nRoughly ⅓ of the candidates failed to identify the conjugate base, perhaps distracted by the fact it was not contained in the equation given.
\nVast majority of candidates could calculate the concentration of H+ (aq) in a HClO (aq) solution with a pH =3.61.
\nMany identified the reaction of chlorine with ethane as free-radical substitution, or just substitution, with some erroneously stating nucleophilic or electrophilic substitution.
\nThe underlying reasons for the relative reactivity of ethane and chloroethane were not very well known with a few giving erroneous reasons and some stating ethane more reactive.
\nFew earned full marks for the curly arrow mechanism of the reaction between sodium hydroxide and chloroethane. Mistakes being careless curly arrow drawing, inappropriate –OH notation, curly arrows from the hydrogen or from the carbon to the C–Cl bond, or a method that missed the transition state.
\nApproximately 60% could draw ethoxyethane however many demonstrated little knowledge of structure of an ether molecule.
\nA poorly answered question with some getting full marks on this 1HNMR spectrum of ethoxyethane question. Very few could identify all 3 of number of signals, chemical shift, and splitting pattern.
\nAnother good example of candidates being well rehearsed in calculations with 90% earning 2/2 on this question of calculation percentage by mass composition.
\nSomewhat disappointing answers on this question about how international cooperation has contributed to the lowering of CFC emissions. Many gave vague answers and some referred to carbon emissions and global warming.
\nFew could construct the propagation equations showing how CFCs affect ozone, and many lost marks by failing to identify ClO· as a radical.
\nProteins are polymers of amino acids. A paper chromatogram of two amino acids, A1 and A2, is obtained using a non-polar solvent.
\n© International Baccalaureate Organization 2020.
\nDetermine the value of A1.
\nProteins are polymers of amino acids.
\nThe mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
\nDeduce, referring to relative affinities and , the identity of A1.
\nProteins are polymers of amino acids.
\nGlycine is one of the amino acids in the primary structure of hemoglobin.
\nState the type of bonding responsible for the α-helix in the secondary structure.
\nProteins are polymers of amino acids.
\nDescribe how the tertiary structure differs from the quaternary structure in hemoglobin.
\n✔
\nAccept any value within the range “”.
\nIle AND larger Rf ✔
\nmore soluble in non-polar solvent «mobile phase»
OR
not as attracted to polar «stationary» phase ✔
Only award M2 if Ile is identified in M1.
hydrogen/ bonding «between amido hydrogen and carboxyl oxygen atoms» ✔
\ntertiary: folding/shape of a single «polypeptide/protein» chain ✔
\nquaternary: arrangement/folding of four/several chains/proteins/polypeptides «held together by » ✔
\n
Accept “two or more polypeptides” for M2.
Many students scored this mark. The students who missed this mark that were close either measured from the top or bottom of the spot rather than the middle. A few students had answers that were greater than 1 which indicated a clear lack of understanding of this concept.
\nNot well answered. Many candidates referred to glycine even when they had obtained the correct Rf value in 5a. Answers referring to Molar mass and isoelectric point were quite common. Some candidates that identified Ile correctly lost the first mark as didn't make any reference to the Rf. There was a clear lack of understanding that the Rf value was related to polarity not molar mass.
\nA well answered question.
\nMany candidates didn't understand the question and provided long answers referring to the interactions but failing to identify those took place within the same/one chain. More candidates were able to score the second mark referring to multiple chains.
\nOrganic chemistry can be used to synthesize a variety of products.
\nCombustion analysis of an unknown organic compound indicated that it contained only carbon, hydrogen and oxygen.
\nSeveral compounds can be synthesized from but-2-ene. Draw the structure of the final product for each of the following chemical reactions.
\nDetermine the change in enthalpy, ΔH, for the combustion of but-2-ene, using section 11 of the data booklet.
\nCH3CH=CHCH3 (g) + 6O2 (g) → 4CO2 (g) + 4H2O (g)
\nState the hybridization of the carbon I and II atoms in but-2-ene.
\nDraw diagrams to show how sigma (σ) and pi (π) bonds are formed between atoms.
\nSketch the mechanism for the reaction of 2-methylbut-2-ene with hydrogen bromide using curly arrows.
\nExplain why the major organic product is 2-bromo-2-methylbutane and not 2-bromo-3-methylbutane.
\nDeduce two features of this molecule that can be obtained from the mass spectrum. Use section 28 of the data booklet.
\nNIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce
on behalf of the United States of America. All rights reserved.
Identify the bond responsible for the absorption at A in the infrared spectrum. Use section 26 of the data booklet.
\nNIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce
on behalf of the United States of America. All rights reserved.
\n
Deduce the identity of the unknown compound using the previous information, the 1H NMR spectrum and section 27 of the data booklet.
\nSDBS, National Institute of Advanced Industrial Science and Technology (AIST).
Draw the stereoisomers of butan-2-ol using wedge-dash type representations.
\nOutline how two enantiomers can be distinguished using a polarimeter.
\nPenalize missing hydrogens in displayed structural formulas once only.
\nAccept condensed structural formulas: CH3CH(OH)CH2CH3 / CH3CH2CH2CH3 or skeletal structures.
\nBonds broken:
2(C–C) + 1(C=C) + 8(C–H) + 6O=O / 2(346) + 1(614) + 8(414) + 6(498) / 7606 «kJ» ✓
Bonds formed:
8(C=O) + 8(O–H) / 8(804) + 8(463) / 10 136 «kJ» ✓
Enthalpy change:
«Bonds broken – Bonds formed = 7606 kJ – 10 136 kJ =» –2530 «kJ» ✓
\n
Award [3] for correct final answer.
\nAward [2 max] for «+» 2530 «kJ».
\nSigma (σ):
\nAccept any diagram showing end to end/direct overlap of atomic/hybridized orbitals and electron density concentrated between nuclei.
\n\n
Pi (π):
\nAccept any diagram showing sideways overlap of unhybridized p/atomic orbitals and electron density above and below plane of bond axis.
\n\n
Alternative 1
\n
Penalize incorrect bond e.g., -CH-H3C or –CH3C only once in the paper.
Alternative 2
\n
curly arrow going from C=C to H of HBr AND curly arrow showing Br leaving ✓
\nrepresentation of carbocation ✓
\ncurly arrow going from lone pair/negative charge on Br− to C+ ✓
\n«2-bromo-2-methylbutane involves» formation of more stable «tertiary» carbocation/intermediate
OR
«2-bromo-3-methylbutane involves» formation of less stable «secondary» carbocation/intermediate ✓
«intermediate» more stable due to «increased positive» inductive/electron-releasing effect of extra –R/alkyl group/–CH3/methyl ✓
\n
Do not award marks for quoting Markovnikov’s rule without any explanation.
m/z 58:
molar/«relative» molecular mass/weight/Mr «is 58 g mol−1/58» ✓
m/z 43:
«loses» methyl/CH3 «fragment»
OR
COCH3+ «fragment» ✓
Do not penalize missing charge on the fragments.
Accept molecular ion «peak»/ CH3COCH3+/C3H6O+.
\nAccept any C2H3O+ fragment/ CH3CH2CH2+/C3H7+.
\nC=O ✓
\n
Accept carbonyl/C=C.
Information deduced from 1H NMR:
\n«one signal indicates» one hydrogen environment/symmetrical structure
OR
«chemical shift of 2.2 indicates» H on C next to carbonyl ✓
Compound:
propanone/CH3COCH3 ✓
\n\n
Accept “one type of hydrogen”.
\nAccept .
enantiomers rotate «plane of» plane-polarized light ✓
\nequal degrees/angles/amounts AND opposite directions/rotation ✓
\n
Accept “optical isomers” for “enantiomers”.
| Phospholipids are a main component of cell membranes. | \n
\n
Deduce the products of the hydrolysis of a non-substituted phospholipid, where and represent long alkyl chains.
\nA representation of a phospholipid bilayer cell membrane is shown:
\n© International Baccalaureate Organization 2020.
\nIdentify the components of the phospholipid labelled A and B.
\nState the most significant intermolecular forces in the phospholipid in b(i).
\nPhospholipids help maintain cellular environments while fatty acid lipids have important roles in energy storage and electrical insulation. Discuss the structural properties of saturated fats needed for these roles.
\nglycerol ✔
\nboth fatty acids AND phosphoric acid ✔
\nAccept either names OR structures.
Accept “long chain carboxylic acid” for “fatty acid”.
Penalize once only if an incorrect name is given for a correct structure or vice-versa.
\nA: phosphate/ionic group
AND
B: alkyl/hydrocarbon «chain» ✔
Accept \"glycerol «fragment»\" OR \"glycerophosphate\" OR “ester” for A.
Accept “fatty acid «tail»” for B.
\nDo not accept terms such as “polar head”, “non-polar tail”, “hydrophilic” OR “hydrophobic” for components alone.
\nForces occurring between components labelled A:
hydrogen/ bonding
OR
ion–dipole
OR
ionic/electrostatic «repulsion and/or attraction» ✔
Accept “dipole-dipole” for M1.
Do not accept “van der Waals/vdW” for M1.
Forces occurring between components labelled B:
dispersion/London/instantaneous dipoles/temporary dipoles ✔
Accept “van der Waals/vdW” for M2.
\nEnergy storage:
not water-soluble/no hydrogen/ bonding
OR
less oxidized/more reduced
OR
high energy stored in bonds
OR
high «negative» enthalpy of combustion/oxidation ✔
Accept “potential energy” for “stored energy”.
\nElectrical insulator:
no delocalized electrons/conjugation ✔
Not as well answered as expected. However, many candidates managed to score at least one point. Quite a few lost the only mark due to wrong linkage. Some students drew fatty acid structures with aldehydes or phosphoric acid with incorrect bond linkages in the structure (OH-).
\nMostly well answered. Weaker candidates used the terms hydrophobic (non-polar) tail and/or hydrophilic (polar) head and therefore lost the mark. Students are expected to know the names of these structures.
\nMany good answers. Those that failed to score often provided the inverted answer. Dipole-dipole was fairly common for M1 but VdW forces less so for M2.
\nNot well answered. In particular the first mark seemed particularly challenging for students. Even when at times wording wasn't enough to allow BOD for M2 it was evident the candidate had some idea but none for the first one. Non-polar allowed many students to score the second mark.
\nWhat is the correct trend going down groups 1 and 17?
\nA. Melting points increase
B. Boiling points decrease
C. Electronegativities increase
D. Ionization energies decrease
\nD
\nMost candidates selected ionization energy as the property that decreases down both group 1 and 17.
\nWhich series is in order of increasing boiling point?
\nA. CH2CH2CH3OH CH3COCH3 CH3CH2CH3
\nB. CH3CH2CH3 CH3COCH3 CH2CH2CH3OH
\nC. CH3COCH3 CH2CH2CH3OH CH3CH2CH3
\nD. CH3CH2CH3 CH2CH2CH3OH CH3COCH3
\nB
\nThere has been significant growth in the use of carbon nanotubes, CNT.
\nExplain these properties of carbon nanotubes.
\nCNT can act as Type 2 superconductors. Outline why Type 2 superconductors are generally more useful than Type 1.
\nExplain the role of electrons in superconducting materials in terms of the Bardeen–Cooper–Schrieffer (BCS) theory.
\nAlloying metals changes their properties. Suggest one property of magnesium that could be improved by making a magnesium–CNT alloy.
\n Pure magnesium needed for making alloys can be obtained by electrolysis of molten magnesium chloride.
© International Baccalaureate Organization 2020
\nCalculate the theoretical mass of magnesium obtained if a current of 3.00 A is used for hours. Use charge :(Q) = current (I) × time (t) and section 2 of the data booklet.
\nSuggest a gas which should be continuously passed over the molten magnesium in the electrolytic cell.
\nZeolites can be used as catalysts in the manufacture of CNT. Explain, with reference to their structure, the high selectivity of zeolites.
\nExperiments have been done to explore the nematic liquid crystal behaviour of CNT. Justify how CNT molecules could be classified as nematic.
\nExcellent strength: defect-free AND rigid/regular 2D/3D ✔
\nExcellent conductivity: delocalized electrons ✔
\nAccept “carbons/atoms are all covalently bonded to each other” for M1.
\nAny two of:
have higher critical temperatures/Tc «than Type 1»
OR
can act at higher temperatures ✔
have higher critical magnetic fields/Bc «than Type 1» ✔
\nless time needed to cool to operating temperature ✔
\nless energy required to cool down/maintain low temperature ✔
\nAny three of:
\npassing electrons «slightly» deform lattice/displace positive ions/cations ✔
\nelectrons couple/form Cooper pairs/condense with other electrons ✔
\nenergy propagates along the lattice in wave-like manner/as phonons ✔
\nCooper pair/electron condensate/pair of electrons moves through lattice freely
OR
phonons are «perfectly» elastic/cause no energy loss ✔
Any of:
ductility ✔
strength/resistance to deformation ✔
malleability ✔
hardness ✔
resistance to corrosion/chemical resistance ✔
range of working temperatures ✔
density ✔
Do not accept “conductivity”.
\n✔
\n✔
\n
✔
Award [3] for correct final answer.
argon//helium/ ✔
\nAccept any identified noble/inert gas.
Accept name OR formula.
Do not accept “nitrogen/“.
\npores/cavities/channels/holes/cage-like structures ✔
\n«only» reactants with appropriate/specific size/geometry/structure fit inside/go through/are activated/can react ✔
\nAccept “molecules/ions” for “reactants” in M2.
\nrod-shaped molecules
OR
«randomly distributed but» generally align
OR
no positional order AND have «some» directional order/pattern ✔
Accept “linear” for “rod-shaped”.
\nThe stronger candidates knew that the excellent conductivity associated with CNTs is associated with delocalised electrons but few scored the mark for citing the property associated with excellent strength, which can be attributed to being defect-free and having a rigid/regular 2D/3D structure.
\nMost gained at least one mark here for stating that Type 2 superconductors have higher critical temperatures than Type 1.
\nThe role of electrons in superconducting materials in terms of the Bardeen-Cooper-Schrieffer (BCS) theory was very well understood and many scored all three marks.
\nThis question proved to be difficult and few could suggest a suitable property (such as ductility) of magnesium that could be improved by making a magnesium-CNT alloy.
\nThe better candidates scored all three marks for the electrolysis calculation. Even the weaker candidates managed to score at least one mark for calculating Q = 108,000 C.
\nThe most common error here was \"nitrogen\" as the gas that should be continuously passed over the molten magnesium in the electrolytic cell. Magnesium can react with nitrogen forming magnesium nitride, which makes this choice of gas unsuitable (unlike argon for example).
\nThe explanation of the high selectivity of zeolites, in terms of their structure, was very well answered and many scored both marks. A thorough understanding of zeolites was much better conveyed in N20 compared to previous sessions.
Most gained the one mark here, justifying how CNT molecules can be classified as nematic, by stating that they are \"rod-shaped molecules\".
\nWhich of these statements are correct?
I. Zinc is not a transition element.
II. Ligands are Lewis bases.
III. Manganese(II) chloride is paramagnetic.
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
\nD
\nThis was one of only two questions where less than 50% of candidates got the correct answer. Many either didn't understand manganese chloride as paramagnetic or classified zinc as a transition element.
\nWhat is the name of this compound, applying IUPAC rules?
\nA. 4-methylhex-2-ene
\nB. 4-ethylpent-2-ene
\nC. 2-ethylpent-3-ene
\nD. 3-methylhex-4-ene
\nA
\nWhich of these species contains the shortest carbon to oxygen bond length?
\nA.
\nB.
\nC.
\nD.
\nD
\nMore than 60% of the candidates identified the compound with the shortest C to O bond as ethanoic acid, with the most common incorrect answer being the ethanoate ion.
\nWhat is formed in a propagation step of the substitution reaction between bromine and ethane?
\nA. CH3CH2•
\nB. CH3CH2CH2CH3
\nC. H•
\nD. Br−
\nA
\nWhich monomer would produce the polymer shown?
\nA.
\nB.
\nC.
\nD.
\nC
\nWhich combination correctly describes the geometry of ?
\nC
\nIdentifying correct electron domain and molecular geometries around a central atom was answered well by the better candidates. This had the highest discrimination index of all the questions.
\nA reaction proceeds by the following mechanism:
\nstep 1:
step 2:
Which rate equation is consistent with this mechanism?
\nA. Rate = k[B]2[C]
\nB. Rate = k[A]2[B][C]
\nC. Rate = k[A]2
\nD. Rate = k[A][C]
\nC
\nThe diverse functions of biological molecules depend on their structure and shape.
\nClassify vitamins A, C and D as either mainly fat- or water-soluble, using section 35 of the data booklet.
\nThe diverse functions of biological molecules depend on their structure and shape.
\nDeduce the straight chain structure of deoxyribose from its ring structure drawn in section 34 of the data booklet.
\nThe diverse functions of biological molecules depend on their structure and shape.
\n Sucrose is a disaccharide formed in the reaction of glucose with fructose.
Identify the reaction type and the newly formed functional group that joins the monosaccharide units in the product.
all three correct ✔
\nmust be placed next to AND s on central carbons must be on same side (LHS or RHS) ✔
\nAccept crosses in place of on three middle carbons.
\nReaction type:
condensation ✔
Accept “nucleophilic substitution/” for M1.
\nFunctional group:
acetal/ether/glycosidic «linkage» ✔
Accept “glycoside” for M2.
\nOnly a few very weak candidates answered this one wrongly.
\nVery disappointing. I had only a few correct answers and the wrong ones very often showed no understanding at all. Some candidates drew ring structures even though a straight chain was requested.
\nMany candidates scored both marks here. Ether was the most common answer for M2 with some responding glycosidic.
\nWhich graphs show a first order reaction?
\nA. V and X
\nB. V and Y
\nC. W and X
\nD. W and Y
\nB
\nThe enthalpy of combustion of a fuel was determined using the calorimeter shown. The final result was lower than the literature value.
\nWhich factors could have contributed to this error?
\nI. Not all heat from the combustion was transferred to the calorimeter.
II. Incomplete combustion occurred.
III. The temperature probe touched the bottom of the calorimeter.
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nWhich statement is correct?
\nA. bond dissociation occurs at a longer wavelength of light than bond dissociation.
B. bond dissociation occurs at a higher energy than bond dissociation.
C. bond lengths are shorter than bond lengths.
D. bond dissociation occurs at a higher frequency of light than bond dissociation.
\nA
\nAnother well answered question where a good majority of candidates understood how ozone and oxygen bond dissociation energies would be affected by various wavelengths or frequencies of light.
\nThe electron configuration of copper makes it a useful metal.
\nDetermine the frequency of a photon that will cause the first ionization of copper. Use sections 1, 2 and 8 of the data booklet.
\n\n
The electron configuration of copper makes it a useful metal.
\nExplain why a copper(II) solution is blue, using section 17 of the data booklet.
\nThe electron configuration of copper makes it a useful metal.
\nCopper plating can be used to improve the conductivity of an object.
\nState, giving your reason, at which electrode the object being electroplated should be placed.
✔
\n
✔
Award [2] for correct final answer.
Award [1] for 1.12 × 1039 «Hz».
orange light is absorbed «and the complementary colour is observed» ✔
\nAny TWO from:
partially filled d-orbitals ✔
«ligands/water cause» d-orbitals «to» split ✔
light is absorbed as electrons move to a higher energy orbital «in d–d transitions»
OR
light is absorbed as electrons are promoted ✔
energy gap corresponds to «orange» light in the visible region of the spectrum ✔
cathode/negative «electrode» AND reduced «at that electrode» ✔
\nAccept cathode/negative «electrode» AND copper forms «at that electrode».
\nDetermining the frequency of a photon that will cause the first ionization of copper was the most challenging question on the exam. Many could not do it all, although some came up with the answer that came from using the result that would arise from the ionization energy in J/mole (and frequently kJ/mole) rather than J/atom.
\nMany students were able to fully explain why solutions containing Cu2+ appear blue, however the misconception between absorption and emission spectra is still quite evident.
\nSurprisingly not that well answered. Most students identified the cathode as the electrode where electroplating occurs but few could adequately justify why.
\nPrecipitation is one method used to treat waste water.
\nPhosphates, , in waste water can be removed by precipitation with magnesium ions. of magnesium phosphate is .
\n
Calculate the maximum solubility of phosphate ions in a solution containing magnesium ions.
Precipitation is one method used to treat waste water.
\nZinc, cadmium, nickel, and lead are metal ions which can be removed by precipitation. Explain why waste water is adjusted to a pH of 9−10 to remove these ions by referring to section 32 of the data booklet.
\n✔
\n✔
\n
Accept “” for M1.
Award [2] for correct final answer.
Any two of:
precipitation occurs with a base/carbonate//hydroxide/ ✔
is high enough to cause metal hydroxide precipitation at that ✔
\nthese ions are slightly acidic/more soluble in acidic conditions ✔
\nonly small amounts of carbonate/hydroxides/anion needed at that ✔
\nsolubility products of the hydroxides are very small ✔
\nDo not accept “hydroxyl” for “hydroxide”.
\nThe Ksp problem was very well answered by the stronger candidates, who had no difficulty working out the maximum solubility of phosphate ions in solution as 1.02 x 10-9 mol dm-3.
\nThis question although somewhat challenging was well answered by the stronger candidates.
\n1.0 mol each of sulfur dioxide, oxygen, and sulfur trioxide are in equilibrium.
\n\nWhich change in the molar ratio of reactants will cause the greatest increase in the amount of sulfur trioxide?
\nAssume volume and temperature of the reaction mixture remain constant.
\nD
\nProteins are polymers of amino acids.
\nA paper chromatogram of two amino acids, A1 and A2, is obtained using a non-polar solvent.
\n© International Baccalaureate Organization 2020.
\nDetermine the value of A1.
\nProteins are polymers of amino acids.
\nThe mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
\nDeduce, referring to relative affinities and , the identity of A1.
\nProteins are polymers of amino acids.
\nGlycine is one of the amino acids in the primary structure of hemoglobin.
\nState the type of bonding responsible for the α-helix in the secondary structure.
\nProteins are polymers of amino acids.
\nSketch and label two oxygen dissociation curves, one for adult hemoglobin and one for foetal hemoglobin.
\nProteins are polymers of amino acids.
\nExplain why the affinity for oxygen of foetal hemoglobin differs from that of adult hemoglobin.
\n\n
✔
\nAccept any value within the range “”.
\nIle AND larger Rf ✔
\nmore soluble in non-polar solvent «mobile phase»
OR
not as attracted to polar «stationary» phase ✔
Only award M2 if Ile is identified in M1.
\nhydrogen/ bonding «between amido hydrogen and carboxyl oxygen atoms» ✔
\nboth curves sigmoidal shape AND starting at zero ✔
\nfoetal hemoglobin showing greater affinity/steeper/higher gradient ✔
\n
Do not penalize if convergence is not approached for M1.
Both curves must be labelled to score M2.
\nAny two of:
\ncontains two gamma/ units «instead of two beta/ units found in adults»
OR
differs in amino acid sequence «from the two beta// units found in adults» ✔
less sensitive to inhibitors/2,3-BPG/DPG ✔
\nreceives from «partly deoxygenated» blood so can work at low ✔
\nlow in foetal blood increases affinity for ✔
\nhemoglobin concentration in foetal blood greater than in the mother ✔
\nIt was surprising that more did not manage to determine the Rf value of A1 within the acceptable range of \"0.67 to 0.73\". An out of range value of 0.75 was frequently seen.
\nFew candidates recognised that A1 was Ile so no marks were scored. Even of those candidates that did identify A1 as IIe, many did not mention larger Rf and therefore M1 was lost. There appeared to be in general, poor understanding of the basic principles of paper chromatography.
\nA majority stated that hydrogen bonding is the bonding responsible for the alpha-helix in the secondary structure.
\nThe better candidates scored both marks in this question on oxygen dissociation curves. Many scored M2 for showing foetal hemoglobin having a higher gradient. Frequently sigmoidal curves were not drawn which lost M1 and often both curves were not explicitly labelled which was necessary to score M2.
\nThis question required an explanation as to why the affinity for oxygen of foetal hemoglobin differs from that of adult hemoglobin. This question was poorly answered and only the stronger candidates managed to score both marks, usually by stating that foetal hemoglobin contains two gamma units instead of the two beta units found in adult hemoglobin, required for M1 and often by stating that there is less sensitivity to inhibitors for M2.
\nWhich statements about bond strength and activation energy are correct for this reaction?
\n\n
B
\nMore than 60% could select relative bond strength of products and reactants in an exothermic reaction, and correctly apply the magnitude of activation energy between forward and reverse reactions for this.
\nBiomagnification factor, BMF, can be defined as the concentration of a chemical, , in a predator, relative to the concentration found in its prey.
\nBMF, where ( per body weight)
\n| [Franklin, J., 2015. How reliable are field-derived biomagnification factors and trophic magnification factors as indicators of bioaccumulation potential? Conclusions from a case study on per- and polyfluoroalkyl substances. Available at: https://setac.onlinelibrary.wiley.com/doi/full/10.1002/ieam.1642.] | \n
\n
Calculate the BMF if a shark consumes mackerel in one year. Each mackerel weighs on average. The per body weight. Assume chemical remains in the shark’s body for two years.
\nSuggest, with a reason, if fat-soluble or water-soluble xenobiotics would have a larger BMF.
\n✔
\n✔
\n
Award [2] for correct final answer.
M2 may also be correctly expressed to SF.
\nfat-soluble AND pass through lipid membranes/accumulate in cells/fatty tissues
OR
fat-soluble AND less easily excreted/metabolized ✔
Accept “water-soluble” only if an organometallic–protein interaction is mentioned.
\nMany candidates produced the correct answer for M1 but not as many fully scored. Students did not appear to understand the concept, and many missed the idea of 2 years in the calculation. Students should always clearly show their calculations so examiners can award marks throughout the question and potentially award ECF if possible. It is very difficult to do this when students do not show work clearly.
\nA well answered question although some did not give a reason.
\nBurette readings for a titration are shown.
\n\n
What is the mean titre?
\nA. 11.1 cm3 ± 0.1 cm3
\nB. 11.15 cm3 ± 0.05 cm3
\nC. 11.2 cm3 ± 0.05 cm3
\nD. 11.2 cm3 ± 0.1 cm3
\nD
\nWhich is a Lewis acid, but not a Brønsted-Lowry acid?
\nA.
\nB.
\nC.
\nD.
\nA
\nWhich combination gives the standard hydration enthalpy of ?
\nA.
B.
C.
D.
\nB
\nAbout 60% of candidates could select values to put in an energy cycle for hydration enthalpy. The higher scoring candidates performed better on this question than lower scoring ones.
\nWhich reaction becomes more spontaneous as temperature increases?
A.
\nB.
\nC.
\nD.
\nA
\nThe majority of candidates could see that reactions which form more mol of gas become spontaneous at higher temperature.
\nDetermine the index of hydrogen deficiency (IHD) of paracetamol.
\nA. 3
\nB. 4
\nC. 5
\nD. 6
\nC
\nGasoline (petrol), biodiesel and ethanol are fuels.
\n[U.S. Department of Energy. https://afdc.energy.gov/]
Calculate the energy released, in , from the complete combustion of of ethanol.
\nState a class of organic compounds found in gasoline.
\nOutline the advantages and disadvantages of using biodiesel instead of gasoline as fuel for a car. Exclude any discussion of cost.
\nA mixture of gasoline and ethanol is often used as a fuel. Suggest an advantage of such a mixture over the use of pure gasoline. Exclude any discussion of cost.
\n\n
Contrast the molecular structures of biodiesel and the vegetable oil from which it is formed.
\nWhen combusted, all three fuels can release carbon dioxide, a greenhouse gas, as well as particulates. Contrast how carbon dioxide and particulates interact with sunlight.
\nMethane is another greenhouse gas. Contrast the reasons why methane and carbon dioxide are considered significant greenhouse gases.
\nSuggest a wavenumber absorbed by methane gas.
\n✔
\nalkane
OR
cycloalkane
OR
arene ✔
Accept “alkene”.
Do not accept just “hydrocarbon”, since given in stem.
Do not accept “benzene/aromatic” for “arene”.
Advantages: [2 max]
\nrenewable ✔
\nuses up waste «such as used cooking oil» ✔
\nlower carbon footprint/carbon neutral ✔
\nhigher flashpoint ✔
\nproduces less /
OR
less polluting emissions ✔
has lubricating properties
OR
preserves/increases lifespan of engine ✔
increases the life of the catalytic converter ✔
\neliminates dependence on foreign suppliers ✔
\ndoes not require pipelines/infrastructure «to produce» ✔
\nrelatively less destruction of habitat compared to obtaining petrochemicals ✔
\n\n
Accept “higher energy density” OR “biodegradable” for advantage.
\n
Disadvantages: [2 max]
needs conversion/transesterification ✔
\ntakes time to produce/grow plants ✔
\ntakes up land
OR
deforestation ✔
fertilizers/pesticides/phosphates/nitrates «used in production of crops» have negative environmental effects ✔
\nbiodiversity affected
OR
loss of habitats «due to energy crop plantations» ✔
cannot be used at low temperatures ✔
\nvariable quality «in production» ✔
\nhigh viscosity/can clog/damage engines ✔
\n
Accept “lower specific energy” as disadvantage.
Do not accept “lower octane number” as disadvantage”.
\nAny one:
\nuses up fossil fuels more slowly ✔
\nlower carbon footprint/CO2 emissions ✔
\nundergoes more complete combustion ✔
\nproduces fewer particulates ✔
\nhigher octane number/rating
OR
less knocking ✔
prevents fuel injection system build up
OR
helps keep engine clean ✔
Accept an example of a suitable advantage even if repeated from 9c.
Any two:
biodiesel has smaller molecules/single «hydrocarbon» chain AND oil has larger molecules/multiple «hydrocarbon» chains ✔
biodiesel is methyl/ethyl ester AND oil has «backbone of» glycerol joined to fatty acids ✔
\nbiodiesel contains one ester group AND oil contains three ester groups ✔
\n
Do not accept properties such as “less viscous” or “lower ignition point”.
carbon dioxide allows sunlight/short wavelength radiation to pass through AND particulates reflect/scatter/absorb sunlight ✔
\nAccept “particulates reflect/scatter/absorb sunlight AND carbon dioxide does not”.
Accept “ absorbs «radiation» AND particulates reflect/scatter/absorb sunlight”.
Do not accept “traps” for “absorbs”.
\ncarbon dioxide is highly/more abundant «in the atmosphere» ✔
\nmethane is more effective/potent «as a greenhouse gas»
OR
methane/better/more effective at absorbing «radiation»
OR
methane has greater greenhouse factor
OR
methane has greater global warming potential/GWP✔
Accept “carbon dioxide contributes more to global warming” for M1.
any value or range within ✔
\nEven rather weak candidates answered this one correctly.
\nMost candidates answered alkanes with a lower number stating hydrocarbons or benzene and therefore lost the mark.
\nThere were many good answers, but few candidates fully scored. Higher energy density and lower specific energy were quite common, and so references to damaging engines. Many students spent more time explaining each advantage rather than simply outlining. There were fewer journalistic and generic answers for this type of question than in the past.
\nAnother question where many candidates obtained the mark. In quite a few cases students repeated the argument for (c) and this allowed them to get two points for the same answer.
\nQuite disappointing with few candidates producing answers that showed deep understanding. Answers such as less viscous or lower ignition point were common. This question specifically asks for contrasts in the structures not the properties of the compounds. Students need to be reminded that a contrast statement requires something about each substance.
\nShowed a wide variety of answers but is was worrying that many students limited to explain the greenhouse effect. There were many responses that did not answer the question or only gave a response for one of the 2 substances.
\nWe received many good answers, but it was worrying the number of students that still provided general and shallow comments. Of the 3 contrast question this had the best response.
\nMany good answers with some students losing the mark as didn't read or understand the question correctly and provided answers in terms of wavelengths.
\nWhich combination will produce an alkaline buffer in water?
\nA. 0.10 mol NH3 and 0.05 mol H2SO4
\nB. 0.50 mol NH3 and 0.10 mol H2SO4
\nC. 0.10 mol CH3COOH and 0.05 mol NaOH
\nD. 0.10 mol CH3COOH and 0.50 mol NaOH
\nB
\nA sample of a compound contains approximately 24.0 g C, 3.0 g H, and 1.6 g O. What is the empirical formula of the compound?
\nA. C20H30O
\nB. C84H10O6
\nC. C2H3O
\nD. C24H30O2
\nA
\nWhich gives the equation and cell potential of the spontaneous reaction?
\nD
\nThe diverse functions of biological molecules depend on their structure and shape.
\nClassify vitamins A, C and D as either mainly fat- or water-soluble, using section 35 of the data booklet.
\nThe diverse functions of biological molecules depend on their structure and shape.
\nDeduce the straight chain structure of deoxyribose from its ring structure drawn in section 34 of the data booklet.
\nThe diverse functions of biological molecules depend on their structure and shape.
\nDraw the nitrogenous base that is paired with guanine in DNA, showing the hydrogen bonds between the bases. Use section 34 of the data booklet.
\nThe diverse functions of biological molecules depend on their structure and shape.
\nRetinal is the key molecule involved in vision. Explain the roles of cis- and trans-retinal in vision and how the isomers are formed in the visual cycle.
\nall three correct ✔
\n–CH2– must be placed next to AND 2OHs on central carbons must be on same side (LHS or RHS) ✔
\nAccept crosses in place of on three middle carbons.
\ncytosine drawn ✔
\nappropriate representation of three hydrogen bonds AND between correct atoms ✔
\n
Structure of cytosine must be given for M1.
Ignore missing hydrogens on carbon atoms in cytosine.
\nDashed lines (horizontal or vertical) OR dots can be used to represent hydrogen
bonds.
Only award M2 if M1 correct.
\nAny three of:
cis-retinal binds to «the protein» opsin
OR
cis-retinal «binds to opsin and» forms rhodopsin ✔
opsin extends conjugation in retinal
OR
conjugation in rhodopsin is larger/more extended/involves more atoms than that in retinal
OR
rhodopsin allows absorption of visible/blue/green light ✔
when visible light is absorbed cis-retinal changes to trans-retinal ✔
\nchange «to trans-retinal» triggers an electrical/nerve signal ✔
\ntrans-retinal detaches from opsin AND is converted back to cis-retinal
OR
trans-retinal is converted back to cis-retinal through enzyme activity ✔
Classification of the three vitamins A, C and D as fat-soluble, water-soluble and fat-soluble was very well answered and nearly all scored the single mark here.
\nThe straight chain structure of deoxyribose was poorly represented. All sorts of errors were seen, such as not having –CHO at one end, not having the methylene group adjacent to –CHO etc.
\nThis question proved to be very challenging and although many were able to identify and draw cytosine correctly, few were able to score M2 which required the representation of the multiple hydrogen bonds between guanine and cytosine. It was very disappointing that such a high percentage of candidates did not understand the nature of a hydrogen bond and equally tried to represent a hydrogen bond using a full line notation (typical of a covalent bond), instead of the more customary dashed lines (either horizontal or vertical) or dots type representation, used to represent hydrogen bonds.
\nThis question which centred on retinal as the key molecule in vision was well answered and many scored all three marks. The most common incomplete answer was \"cis-retinal changing to trans-retinal\" without stating that this occurs when visible light is absorbed. Some also were unclear of the difference between opsin and rhodopsin.
\nWhich change does not increase the rate of this reaction?
\n\nA. Increasing the particle size of the
B. Increasing the temperature
C. Increasing the concentration of
D. Stirring the reaction mixture
\nA
\nPerhaps asking which factor does not affect a rate threw some candidates. 80% did get this correct but it was not a very discriminating question with higher and lower scorers performing equally well/poor.
\nThe diagram shows the first ionisation energies of consecutive elements in the same period of the periodic table.
\nWhich factor explains why element X has a higher first ionisation energy than element Y?
\nA. Element Y loses an electron from a different sub-level.
\nB. Element Y has a smaller atomic radius.
\nC. Element X has a full octet.
\nD. Element Y has a greater nuclear charge.
\nA
\nWhat is the overall charge, , of the chromium (III) complex?
\n\nA. 0
\nB. 1+
\nC. 2−
\nD. 3+
\nB
\nWhat are the products when concentrated aqueous copper (II) chloride is electrolysed using platinum electrodes?
\nC
\nWhat are the units of the rate constant, , if the rate equation is ?
\nA.
B.
C.
D.
\nC
\nNearly three quarters of candidates could identify the units of a rate constant for a 3rd order reaction.
\nWhich is most likely to hydrolyse via a SN1 mechanism?
\nA. CH3CHBrCH2CH3
\nB. (CH3)2CHBr
\nC. (CH3)3CBr
\nD. CH3CH2CH2CH2Br
\nC
\nWhich graph represents the relationship between the rate constant, , and temperature, , in kelvin?
\nC
\n60% of candidates selected the appropriate graph showing how rate constant, k, varies with temperature. Incorrect answers were equally split between a linear and inverse relationship.
\nWhich contain delocalised electrons?
\nI. C6H5OH
II. CH3COO−
III. CO32−
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nEthanol is obtained by the hydration of ethene, C2H4.
\nState the class of compound to which ethene belongs.
\nState the molecular formula of the next member of the homologous series to which ethene belongs.
\nJustify why ethene has only a single signal in its 1H NMR spectrum.
\nSuggest two possible products of the incomplete combustion of ethene that would not be formed by complete combustion.
\nA white solid was formed when ethene was subjected to high pressure.
\nDeduce the type of reaction that occurred.
\nalkene ✔
\nC3H6 ✔
\nAccept structural formula.
\nhydrogen atoms/protons in same chemical environment ✔
\nAccept “all H atoms/protons are equivalent”.
Accept “symmetrical”
carbon monoxide/CO AND carbon/C/soot ✔
\n«addition» polymerization ✔
\nThe kinetics of an enzyme-catalysed reaction are studied in the absence and presence of an inhibitor. The graph represents the initial rate as a function of substrate concentration.
\n
Identify the type of inhibition shown in the graph.
\nDetermine the value of and in the absence and presence of the inhibitor.
\nOutline the significance of the value of the Michaelis constant, .
\nnon-competitive «inhibition» ✔
\n
✔✔✔
Award [3] for four values correct.
Award [2] for three values correct.
Award [1] for two values correct.
Ignore units.
Accept for and . No ECF applied.
is an inverse measure of affinity of substrate for enzyme
OR
higher indicates higher substrate concentration is needed for enzyme saturation
OR
low value of means reaction is fast at low substrate concentration ✔
Idea of “inverse relationship” must be conveyed.
Most scored the one mark here for identifying the correct type of inhibition, namely, non-competitive inhibition.
\nThe stronger candidates scored all three marks here for the four correct values. Most scored at least one mark for the Vmax values in the absence and presence of the inhibitor. The Km values sometimes were outside the permitted ±0.1 range.
\nThe stronger candidates scored all three marks here for the four correct values. Most scored at least one mark for the Vmax values in the absence and presence of the inhibitor. The Km values sometimes were outside the permitted ±0.1 range.
\nWhich statement is correct for a spontaneous reaction?
\nA
\nNearly all candidates knew that is negative for a spontaneous reaction, however some were confused about the effect on the equilibrium constant.
Which is the enthalpy change of reaction, ΔH?
\nA
\nWhich of these oxides contribute to acid deposition?
\nI.
II.
III.
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
\nA
\nNot well answered. Less than half knew which oxides contribute to acid deposition, with the majority including CO2 in the list. This misconception was held by higher and lower scoring candidates equally.
\nNitrogen monoxide reacts with oxygen gas to form nitrogen dioxide.
\n The following experimental data was obtained.
Deduce the partial order of reaction with respect to nitrogen monoxide and oxygen.
\n
\n
Nitrogen monoxide reacts with oxygen gas to form nitrogen dioxide.
\nDeduce, giving a reason, whether the following mechanism is possible.
\n\n
: second ✔
: first ✔
not possible AND «proposed» mechanism does not match experimental rate law
OR
not possible AND «proposed» mechanism shows zero/not first order with respect to oxygen ✔
Most candidates could correctly deduce the order of each reactant from rate experimental rate data.
\n60% of candidates could explain why the proposed reaction mechanism was inconsistent with the empirical data given.
\nWhat is the product of the reaction of benzene with a mixture of concentrated nitric and sulfuric acids?
\nC
\nWhich species is a Lewis acid but not a Brønsted–Lowry acid?
\nA.
B.
C.
D.
\nA
\nVast majority of candidates understood the difference between Lewis and Brønsted Lowry acids.
\nIn which series are all carbon atoms sp2 hybridized?
\nA. C2H2 H2CO HCOOH
\nB. C2H4 H2CO HCOOH
\nC. C2H2 CO HCN
\nD. C2H6 CH3OH CH3OCH3
\nB
\nof the mass of a rock weighing is uranium(IV) oxide, . of the uranium atoms in the rock are uranium-238, .
\nShow that the mass of the isotope in the rock is .
\nThe half-life of 238U is years. Calculate the mass of 238U that remains after has decayed for years.
\nOutline a health risk produced by exposure to radioactive decay.
\nDeduce the nuclear equation for the decay of uranium-238 to thorium-234.
\nThorium-234 has a higher binding energy per nucleon than uranium-238. Outline what is meant by the binding energy of a nucleus.
\n// ✔
\n✔
\nAward [1 max] for omitting mass composition (giving ).
\nM2 is for numerical setup, not for final value of .
\nAlternative 1
✔
✔
\n
Alternative 2
✔
✔
\n
Award [2] for correct final answer.
Any one:
\n«genetic» mutations ✔
\n«could cause» cancer ✔
Accept specific named types of cancer.
cells «in body» altered ✔
\ncells «in body» cannot function ✔
\ndamaged DNA/proteins/enzymes/organs/tissue ✔
\n«radiation» burns ✔
\nhair loss ✔
\ndamage in foetuses ✔
\ndamages/weakens immune system ✔
\n\n
✔
\nDo not penalize missing atomic numbers in the equation.
\nAccept “” for \"”.
\nenergy required to separate a nucleus into protons and neutrons/nucleons
OR
energy released when nucleus was formed from «individual/free/isolated» protons and neutrons/nucleons ✔
Do not accept “energy released when atom was formed”.
This was a very different question as student were given the answer and it was the work that was being marked. Students should always clearly show their calculations so examiners can award marks throughout the question and potentially award ECF if possible. It is very difficult to do this when students do not show work clearly.
\nThis continues to be a topic that students understand well, and probably more used alternative 1 to calculate the half-life. Students should always clearly show their calculations so examiners can award marks throughout the question and potentially award ECF if possible. It is very difficult to do this when students do not show work clearly.
\nA well answered question.
\nOnly very weak candidates lost this mark.
\nMany students appeared to have portion of the answer but not the entire concept. They seemed to have studied previous MS by heart and entered an answer that was mostly correct but didn't address the question. It is important that student understand the material and not try to memorize their way through the topics.
\nHow many chiral centres are there in the following molecule?
\nA. 2
\nB. 3
\nC. 4
\nD. 6
\nC
\nWhich change causes the greatest increase in the initial rate of reaction between nitric acid and magnesium?
\n2HNO3 (aq) + Mg (s) → Mg(NO3)2 (aq) + H2 (g)
\nA
\nThe graph shows the Maxwell–Boltzmann energy distribution curve for a given gas at a certain temperature.
\nHow would the curve change if the temperature of the gas decreases while the other conditions remain constant?
\nA. The maximum would be lower and to the left of M.
\nB. The maximum would be lower and to the right of M.
\nC. The maximum would be higher and to the left of M.
\nD. The maximum would be higher and to the right of M.
\nC
\nIn order to determine the oil content of different types of potato crisps (chips), a student weighed of crushed crisps and mixed them with of non-polar solvent.
\nShe assumed all the oil in the crisps dissolved in the solvent.
\nThe student then filtered the mixture to remove any solids, and gently heated the solution on a hot plate to evaporate the solvent.
\nShe measured the mass of the oil that remained from each type of crisps
\nSuggest why a non-polar solvent was needed.
\nState one reason why the mixture was not heated strongly.
\nNon-polar solvents can be toxic. Suggest a modification to the experiment which allows the evaporated solvent to be collected.
\nSuggest one source of error in the experiment, excluding faulty apparatus and human error, that would lead to the following:
\noil is non-polar «and dissolves best in non-polar solvents»
OR
oil does not dissolve in polar solvents ✔
Do not accept “like dissolves like” only.
\nsolvent/oil is flammable
OR
solvent/oil must be kept below its flash point
OR
oxidation/decomposition of oil
OR
mixture has a low boiling point ✔
Accept “to prevent evaporation of oil”.
\ndistillation «instead of evaporation» ✔
\nAccept “pass vapour through a condenser and collect liquid”.
\nDo not accept “condensation” without experimental details.
\nExperimental mass greater than actual mass of oil in crisps:
other substances «in the crisps» are soluble in the solvent
OR
not all the solvent evaporates ✔
Experimental mass less than actual mass of oil in crisps:
not all oil dissolved/extracted ✔
Accept “oil evaporated” OR “oil burned/decomposed” OR “oil absorbed by the filter” OR “assumption «all oil dissolved» was wrong” for M2.
\nDo not accept examples of human errors OR faulty apparatus.
\nA well answered question where replies used all the alternatives provided. Very few candidates limited their answer to \"like dissolves like\" and while this expression was used most student elaborated with higher quality answer. Some common incorrect responses included students talking about dissolving the crisps (chips) or indicating the oil was a polar compound.
\nAnother correctly answered question. As accepted by notes many candidates scored by stating \"to prevent evaporation of oil\". This resulted in the same argument scoring twice as often used for 1d as well. Some students incorrectly indicated the problem was to prevent the evaporation of the solvent which was the point of this step in the experiment. This could indicate a general lack of understanding of experimental methods.
\nA bit disappointing as the number of correct answers were substantially lower than expected. Many students responded using a fume hood or other method to remove the solvent. Once again this indicates a general misunderstanding about experimental methods.
\nEven weak candidates scored at least one point and often both. One common pitfall was to invert the arguments or provide answers excluded by the stem. A frequent incorrect answer was identification of faulty apparatus and human error which was specifically excluded in the question.
\nWhich compound produces the following 1H NMR spectrum?
\n
SDBS, National Institute of Advanced Industrial Science and Technology (AIST).
A. Propane
B. Propanone
\nC. Propanal
\nD. 2,2-dimethylpropane
\nA
\nWhat is the pH of an ammonia solution that has ?
\nA.
B.
C.
D.
\nC
\nMore than 86% of candidates could find a pH from a given hydroxide concentration.
\nWhat effect does a catalyst have on the position of equilibrium and the value of the equilibrium constant, Kc, for an exothermic reaction?
\nC
\n0.20 mol of magnesium is mixed with 0.10 mol of hydrochloric acid.
\n\nWhich is correct?
\nC
\nThe table shows the variation of standard Gibbs energy with temperature for a reversible reaction.
\n\n\nWhat can be concluded about the reaction?
\nA. Equilibrium shifts left as temperature increases.
\nB. The forward reaction is more spontaneous below 300 K.
\nC. Entropy is higher in the products than in the reactants.
\nD. Kc decreases as temperature increases.
\nC
\nWhich conditions deposit the greatest mass of copper when solutions containing copper ions are electrolysed for 10 minutes?
\nD
\nNearly all candidates understood that larger current would produce more mass in electrolysis, however quite a few also thought that more copper would be deposited if the ion had a +2 rather than +1 charge.
\nWhich amount, in mol, of sodium chloride is needed to make 250 cm3 of 0.10 mol dm−3 solution?
\nA. 4.0 × 10−4
\nB. 0.025
\nC. 0.40
\nD. 25
\nB
\nWhich substance has the highest lattice enthalpy?
\nA.
\nB.
\nC.
\nD.
\nD
\nGasoline (petrol), biodiesel and ethanol are fuels.
\n[U.S. Department of Energy. https://afdc.energy.gov/]
Calculate the energy released, in , from the complete combustion of of ethanol.
\nState a class of organic compounds found in gasoline.
\nOutline the advantages and disadvantages of using biodiesel instead of gasoline as fuel for a car. Exclude any discussion of cost.
\nA mixture of gasoline and ethanol is often used as a fuel. Suggest an advantage of such a mixture over the use of pure gasoline. Exclude any discussion of cost.
\n\n
When combusted, all three fuels can release carbon dioxide, a greenhouse gas, as well as particulates. Contrast how carbon dioxide and particulates interact with sunlight.
\nMethane is another greenhouse gas. Contrast the reasons why methane and carbon dioxide are considered significant greenhouse gases.
\nSuggest a wavenumber absorbed by methane gas.
\nDetermine the relative rate of effusion of methane () to carbon dioxide (), under the same conditions of temperature and pressure. Use section 1 of the data booklet.
\n✔
\nalkane
OR
cycloalkane
OR
arene ✔
Accept “alkene”.
Do not accept just “hydrocarbon”, since given in stem.
Do not accept “benzene/aromatic” for “arene”.
Advantages: [2 max]
\nrenewable ✔
\nuses up waste «such as used cooking oil» ✔
\nlower carbon footprint/carbon neutral ✔
\nhigher flashpoint ✔
\nproduces less /
OR
less polluting emissions ✔
has lubricating properties
OR
preserves/increases lifespan of engine ✔
increases the life of the catalytic converter ✔
\neliminates dependence on foreign suppliers ✔
\ndoes not require pipelines/infrastructure «to produce» ✔
\nrelatively less destruction of habitat compared to obtaining petrochemicals ✔
\n\n
Accept “higher energy density” OR “biodegradable” for advantage.
\n
Disadvantages: [2 max]
needs conversion/transesterification ✔
\ntakes time to produce/grow plants ✔
\ntakes up land
OR
deforestation ✔
fertilizers/pesticides/phosphates/nitrates «used in production of crops» have negative environmental effects ✔
\nbiodiversity affected
OR
loss of habitats «due to energy crop plantations» ✔
cannot be used at low temperatures ✔
\nvariable quality «in production» ✔
\nhigh viscosity/can clog/damage engines ✔
\n
Accept “lower specific energy” as disadvantage.
Do not accept “lower octane number” as disadvantage”.
\nAny one:
\nuses up fossil fuels more slowly ✔
\nlower carbon footprint/CO2 emissions ✔
\nundergoes more complete combustion ✔
\nproduces fewer particulates ✔
\nhigher octane number/rating
OR
less knocking ✔
prevents fuel injection system build up
OR
helps keep engine clean ✔
Accept an example of a suitable advantage even if repeated from 11c.
carbon dioxide allows sunlight/short wavelength radiation to pass through AND particulates reflect/scatter/absorb sunlight ✔
\nAccept “particulates reflect/scatter/absorb sunlight AND carbon dioxide does not”.
Accept “ absorbs «radiation» AND particulates reflect/scatter/absorb sunlight”.
Do not accept “traps” for “absorbs”.
\ncarbon dioxide is highly/more abundant «in the atmosphere» ✔
\nmethane is more effective/potent «as a greenhouse gas»
OR
methane/better/more effective at absorbing «radiation»
OR
methane has greater greenhouse factor
OR
methane has greater global warming potential/GWP✔
Accept “carbon dioxide contributes more to global warming” for M1.
any value or range within ✔
\n«rate of effusion of ✔
\nAlmost all were able to calculate the energy released from the complete combustion of ethanol.
\nThe majority cited correctly that alkanes are a class of organic compounds found in gasoline.
\nMost gained at least one mark for an advantage of using biodiesel instead of gasoline as fuel for a car and most scored one mark at least for a disadvantage of biodiesel. Many conveyed solid understanding, though the disadvantages were not as well articulated as the advantages. Some incorrectly based their responses on cost factors which were excluded as outlined in the stem of the question.
\nMost scored the one mark for this question, with \"less knocking or higher octane number/rating\" the most common correct answer seen.
\nThe wording of this question was critical which involved contrasting how carbon dioxide and particulates interact with sunlight. Some missed the \"Contrast\" command term as the action verb. Loose, non-scientific syntax was often seen such as stating \"traps\" instead of \"absorbs\".
\nThis was another \"Contrast-type\" question, which was better answered compared to (e)(i). Many scored both marks by stating that carbon dioxide is more abundant in the atmosphere whereas methane is more effective at absorbing IR radiation.
\nThe main issue with this question was that a high percentage of candidates did not realise that wavenumber is the reciprocal of wavelength and hence wavenumber has typical units of cm-1. Many incorrectly gave wavelength values, in nm, which did not answer the question posed.
\nThe determination of the relative rate of effusion of methane to carbon dioxide was almost universally correctly computed as 1.656.
\nWhich statement is correct when a zinc spoon is electroplated with silver?
\nA. The cathode (negative electrode) is made of silver.
B. The anode (positive electrode) is the zinc spoon.
C. The anode (positive electrode) is made of silver.
D. The electrolyte is zinc sulfate solution.
\nC
\nVery pleasing to see that nearly 80% could identify the material best suited as anode in an electrolytic cell. Many candidates, however, thought that the material being electroplated should be the anode.
\nAn investigation was carried out to determine the effect of chain length of the alcohol on the equilibrium constant, , for the reversible reaction:
\n\n
The reactants, products and the catalyst form a homogeneous mixture.
\nFixed volumes of each alcohol, the ethanoic acid and the sulfuric acid catalyst were placed in sealed conical flasks.
\nAt equilibrium, the flasks were placed in an ice bath, and samples of each flask titrated with to determine the ethanoic acid concentration present in the equilibrium mixture.
\nThe following processed results were obtained.
\n © International Baccalaureate Organization 2020
Identify the independent and dependent variables in this experiment.
\nThe ice bath is used at equilibrium to slow down the forward and reverse reactions. Explain why adding a large amount of water to the reaction mixture would also slow down both reactions.
\nSuggest why the titration must be conducted quickly even though a low temperature is maintained.
\nAn additional experiment was conducted in which only the sulfuric acid catalyst was titrated with . Outline why this experiment was necessary.
\nCalculate the percentage uncertainty and percentage error in the experimentally determined value of for methanol.
\nComment on the magnitudes of random and systematic errors in this experiment using the answers in (e).
\nSuggest a risk of using sulfuric acid as the catalyst.
\nIndependent variable:
chain length OR number of carbon «atoms in alcohol»
AND
Dependent variable:
volume of OR /equilibrium constant OR equilibrium concentration/moles of ✔
dilution/lower concentrations ✔
\nless frequent collisions «per unit volume» ✔
\nAccept “lowers concentration of acid catalyst” for M1. M2 must refer to increase in activation energy or different pathway.
\nDo not accept responses referring to equilibrium.
\nequilibrium shifts to left
OR
more ethanoic acid is produced «as ethanoic acid is neutralized»
OR
prevents/slows down ester hydrolysis ✔
Accept “prevents equilibrium shift” if described correctly without direction.
\nto determine volume/moles of used up by the catalyst/sulfuric acid «in the titration»
OR
to eliminate/reduce «systematic» error caused by acid catalyst ✔
Do not accept “control” OR “standard” alone.
Percentage uncertainty:
✔
Percentage error:
✔
Award [1 max] if calculations are reversed OR if incorrect alcohol is used.
\nAny two:
\nlarge percentage error means large systematic error «in procedure» ✔
\nsmall percentage uncertainty means small random errors ✔
\nrandom errors smaller than systematic error ✔
\n
Award [2] for “both random and systematic errors are significant.”
corrosive/burns/irritant/strong oxidizing agent/carcinogenic
OR
disposal is an environmental issue
OR
causes other side reactions/dehydration/decomposition ✔
Do not accept just “risk of accidents” OR “health risks” OR “hazardous”.
Aspirin is formed by reacting salicylic acid with ethanoic anhydride. The structure of aspirin is given in section 37 of the data booklet.
\nDeduce the structural formula of the by-product of this reaction.
\nAspirin crystals are rinsed with water after recrystallization to remove impurities.
Suggest why cold water is used.
The solubility of aspirin is increased by converting it to an ionic form. Draw the structure of the ionic form of aspirin.
\nComment on the risk of overdose when taking aspirin as an analgesic, referring to the following values, for a person weighing :
\nMinimum therapeutic dose
\nEstimated minimum lethal dose
\nOR
\n✔
\n
Accept full OR condensed structural formula.
to avoid dissolving the crystals/aspirin ✔
\nAccept “to avoid loss of product” OR “aspirin is less soluble in cold water”.
\n✔
Accept a positive metal ion next to the such as “”.
Accept “” without showing the charges.
\nAccept notations such as “” OR “” OR “” but not “” OR “” OR “”.
\nlow/medium risk «of overdosing» AND «estimated» lethal dose is times/much larger than therapeutic dose
\nOR
\ntimes the dose results in chance of dying ✔
\nAccept “ and low/medium risk due to large therapeutic index”.
Do not accept “low/medium risk AND large therapeutic window”.
Do not accept “ times the dose” alone for the mark.
A well answered question. Most candidates chose to enter the full structure. Some incorrect answers gave the aspirin product or the salicylic acid rather than the acetic acid.
\nWhile there were many good answers it was worrying to correct as many where the student clearly didn't establish a connection between solubility and purification. Many incorrect responses indicated the cold water was to \"stay below the melting point of the aspirin\" rather than relate it to the solubility of the final product.
\nNot well answered. Students evidenced familiarity with the content but failed to provide correct structures. Showing lines used to represent covalent bonds to show an ionic interaction, using convention for complexes, showing only one ion were some of the many mistakes. Many students also had an incorrect product having the aspirin lose the -OH group from the carboxylic acid or the entire carboxylic acid functional group rather than just the H+.
\nThis is a topic that continues to challenge students and \"low/medium risk AND large therapeutic window\" was worryingly common. There were also many students who did not calculate the ratio correctly.
\nWhich molecule has the same empirical formula as molecular formula?
\nA. CH3COOH
\nB. C2H5OH
\nC. C2H4
\nD. C4H10
\nB
\nWhich molecule can be oxidized to a carboxylic acid by acidified potassium dichromate(VI)?
\nA. Propan-1-ol
B. Propan-2-ol
C. 2-methylpropan-2-ol
D. Propanone
\nA
\nAnother well answered question with the vast majority understanding that only propan-1-ol can be oxidized to a carboxylic acid.
\nof the mass of a rock weighing is uranium(IV) oxide, . of the uranium atoms in the rock are uranium-238, .
\nShow that the mass of the 238U isotope in the rock is .
\nThe half-life of 238U is years. Calculate the mass of 238U that remains after has decayed for years.
\nOutline a health risk produced by exposure to radioactive decay.
\nDeduce the nuclear equation for the decay of uranium-238 to thorium-234.
\nThorium-234 has a higher binding energy per nucleon than uranium-238. Outline what is meant by the binding energy of a nucleus.
\nDetermine the nuclear binding energy, in , of using sections 2 and 4 of the data booklet.
\nThe mass of the nucleus is .
\n// ✔
\n✔
\nAward [1 max] for omitting mass composition (giving ).
\nM2 is for numerical setup, not for final value of .
\nAlternative 1
✔
✔
\n
Alternative 2
✔
✔
\n
Award [2] for correct final answer.
Any one:
\n«genetic» mutations ✔
\n«could cause» cancer ✔
Accept specific named types of cancer.
cells «in body» altered ✔
\ncells «in body» cannot function ✔
\ndamaged DNA/proteins/enzymes/organs/tissue ✔
\n«radiation» burns ✔
\nhair loss ✔
\ndamage in foetuses ✔
\ndamages/weakens immune system ✔
\n\n
✔
\nDo not penalize missing atomic numbers in the equation.
\nAccept “” for \"”.
\nenergy required to separate a nucleus into protons and neutrons/nucleons
OR
energy released when nucleus was formed from «individual/free/isolated» protons and neutrons/nucleons ✔
Do not accept “energy released when atom was formed”.
OR
✔
OR
✔
✔
\n
Accept answers in the range “”.
Award [3] for correct final answer.
\nThis question was generally well answered. Many candidates approached this question using amount, in mol, and converting to mass at the end. Some candidates omitted the mass composition however, resulting in a mass of 0.725 kg, which yielded [1 max].
\nThe question involving half-life was very well answered and most scored both marks giving a final answer of 0.0200 kg, via different methods of calculation.
\nThis question on outlining a health risk produced by exposure to radioactive decay posed no difficulty and the most common, correct answers included \"could cause cancer\" and \"can damage DNA\".
Most scored the one mark for the nuclear equation for the decay of uranium-238 to thorium-234. The most common error was including a neutron on the product side
\nAlthough many scored the one mark for outlining what is meant by the binding energy of a nucleus, the question was often answered as energy involved in the formation of an atom. Some candidates did not understand the difference between nucleus and nucleons.
\nThis demanding question on the determination of the nuclear binding energy was well executed and many scored all three marks. Even the weaker candidates still managed to gain an ECF mark for M3 by using the E = mc2 equation. The general performance on this type of calculation was much better than in previous sessions.
\nWhat is the sum of the coefficients when the equation is balanced with whole numbers?
\n__Sn(OH)4 (aq) + __NaOH (aq) → __Na2SnO3 (aq) + __H2O (l)
\nA. 4
\nB. 5
\nC. 6
\nD. 7
\nD
\nWhich is the electrophile in the nitration of benzene?
\nA.
\nB.
\nC.
\nD.
\nB
\nThe vast majority selected the nitronium ion as the electrophile in the nitration of benzene.
\nWhat is represented by “2−” in ?
\nA. loss of electron
\nB. gain of electron
\nC. loss of proton
\nD. gain of proton
\nB
\nAspirin is formed by reacting salicylic acid with ethanoic anhydride. The structure of aspirin is given in section 37 of the data booklet.
\nDeduce the structural formula of the by-product of this reaction.
\nAspirin crystals are rinsed with water after recrystallization to remove impurities.
Suggest why cold water is used.
The solubility of aspirin is increased by converting it to an ionic form. Draw the structure of the ionic form of aspirin.
\nComment on the risk of overdose when taking aspirin as an analgesic, referring to the following values, for a person weighing :
\nMinimum therapeutic dose
\nEstimated minimum lethal dose
\nExplain how IR spectroscopy can be used to distinguish aspirin from salicylic acid.
\nOR
\n✔
\n
Accept full OR condensed structural formula.
to avoid dissolving the crystals/aspirin ✔
\nAccept “to avoid loss of product” OR “aspirin is less soluble in cold water”.
\n✔
Accept a positive metal ion next to the such as “”.
Accept “” without showing the charges.
\nAccept notations such as “” OR “” OR “” but not “” OR “” OR “”.
\nlow/medium risk «of overdosing» AND «estimated» lethal dose is times/much larger than therapeutic dose OR
times the dose results in chance of dying ✔
Accept “ and low/medium risk due to large therapeutic index”.
Do not accept “low/medium risk AND large therapeutic window”.
Do not accept “ times the dose” alone for the mark.
salicylic acid contains absorption in the range ✔
\ndue to phenol/hydroxyl/ group not present in aspirin ✔
\n
Award [2] for “additional «stretch» in IR for salicylic acid at higher wavenumber than corresponding «stretch» in aspirin” OR “aspirin has two absorption bands/one stronger absorption band in while salicylic acid has one/weaker absorption band in that region”.
Award [1 max] for “fingerprint regions will be different for both”.
\nMost candidates were able to deduce a correct structural formula (either full or condensed) for ethanoic acid. A minority did not read the question fully and gave the structure of aspirin instead of the by-product of the reaction. Another incorrect answer cited as the by-product was water.
\nMany were unable to explain why aspirin should be washed with cold water, namely, to avoid dissolving crystals. Surprisingly, the incorrect term \"melt\" was frequently used instead of \"dissolve\".
\nA drawing of the structure of the ionic form of aspirin was required for this question. This question was poorly answered by a significant number of candidates, and lots of basic chemical errors were seen, such as incorrect valencies, writing RCO- instead of RCOO-, showing a cationic structure instead of an anionic structure etc. A couple of candidates also lost the mark by drawing square brackets with a negative charge both inside and outside the bracket.
\nFew scored this mark. Most knew the overdose risk was low but referred to a large therapeutic window instead of a large therapeutic index. Many also did not quantify the therapeutic index by working out that the estimated lethal dose is actually 30 times the therapeutic dose.
\nThis question which asked for an explanation of how IR spectroscopy can be used to distinguish aspirin from salicyclic acid was generally very well answered. The majority stated that salicyclic acid contains an absorption in the IR spectrum in the 3200-3600 cm-1 range due to the phenolic OH group, which is not present in aspirin. A few stated that aspirin has a methyl group and hence the CH stretch will appear in the 2850-3090 cm-1 region of the IR spectrum in aspirin (using Section 26 of the Data Booklet) which will not appear in the corresponding IR spectrum for salicyclic acid. This is somewhat incorrect as in salicyclic acid the benzene ring will also have CH bonds and the CH stretch for the benzene ring will occur in a similar region of the IR spectrum (as indicated in Section 26 of the Data Booklet) and hence cannot be used to distinguish fully between the two structures per se if using the Data Booklet range. Of course, in practice the alkyl CH stretch would be at a slightly lower wavenumber (e.g. 2850-2950 cm-1) in the IR spectrum compared to the aromatic CH stretch (3030 cm-1), but virtually no candidate gave this type of precise detail.
\nHow are emission spectra formed?
\nA. Photons are absorbed when promoted electrons return to a lower energy level.
\nB. Photons are absorbed when electrons are promoted to a higher energy level.
\nC. Photons are emitted when electrons are promoted to a higher energy level.
\nD. Photons are emitted when promoted electrons return to a lower energy level.
\nD
\nWhat will be the major product in the reaction between but-1-ene and ?
\nA. 2-bromobut-1-ene
B. 1-bromobut-1-ene
C. 2-bromobutane
D. 1-bromobutane
\nC
\nThe vast majority selected the correct product for the reaction of but-1-ene and HBr although quite a few selected the minor product of 1-bromobutane.
\nWhich molecule has an enantiomer?
\nA.
B.
C.
D.
\nD
\nThe majority of candidates could correctly identify the molecule which has an enantiomer.
\nWhich compound with the molecular formula has this high resolution ?
\nFrom: libretexts.org. Courtesy of Chris Schaller, Professor (Chemistry)
at College of Saint Benedict/Saint John’s University.
\n
A. but-3-en-2-ol,
B. butanal,
C. butanone,
D. but-3-en-1-ol,
\nB
\nJust over 50% of candidates could identify the correct compound from high resolution 1H NMR data. The question had a relatively high discriminatory index.
\nWhich property increases down group 1?
\nA. atomic radius
\nB. electronegativity
\nC. first ionization energy
\nD. melting point
\nA
\nTechnetium-99m is the most commonly used isotope for diagnostic medicine.
\nState the type of radiation technetium-99m emits.
\nDiscuss the properties that make a radioisotope suitable for diagnosis.
\nDescribe the proper disposal of low-level radioactive waste in hospitals.
\nTechnetium-99m has a half-life of hours. Calculate the amount of of technetium-99m remaining after hours.
\ngamma/ ✔
\nAny three of:
\n«easily» detected/traced
OR
«gamma-radiation of approximately» same frequency as X-rays «so can be detected using existing X-ray equipment» ✔
short/intermediate half-life «hence does not remain in body for long time» ✔
\nweak ionizing radiation «less harmful»
OR
low amount of radiation produced «so less harmful»
OR
energy of photons is low ✔
form «variety of» compounds that are absorbed by «different» organs
OR
«chemically» binds to many biologically active compounds ✔
excreted quickly «from body» ✔
\nstore until material becomes inactive/radiation levels drop ✔
\ndispose with other waste
OR
dispose in landfills ✔
Only award M2 if M1 correct.
Accept “dispose by incineration” for M2.
\nAlternative 1:
\n✔
\n✔
\n
Alternative 2:
✔
✔
\n
Award [2] for correct final answer.
This question was well answered. Most were able to state that gamma radiation is emitted from technetium-99m. The most common incorrect answer was beta radiation.
\nSome excellent answers were seen for this question; often candidates were hitting four of the assigned marking points, though a few candidates confused diagnosis and radiotherapy. Nearly everyone got a mark for \"short half-life\", however. This question was much better answered than in previous sessions.
\nSurprisingly, this question caught out several candidates and the marks varied from zero to one to two. To score full marks, candidates first had to state that the proper disposal of low-level radioactive waste (LLW) in hospitals involves storing the material until such time as radiation levels drop. Then the material can be disposed of in landfills for example. A number failed to outline the first point and some also mixed up LLW with HLW.
\nThe question on half-life was very well answered and nearly all scored full marks, often via different methods of calculation of the amount of technetium-99m remaining after a period of 48 HRS.
\nConsider the following antacids:
\nShow that antacid X is more effective, per tablet, than antacid Y.
\nsame reactant mole ratio «in both equations»
OR
AND
✔
AND
✔
«tablet of» X neutralizes AND «tablet of» Y neutralizes ✔
\nAward [3] for correct final statement AND values in M3.
\nVery poorly answered. Many students scored at least 1 point. This was usually M2 by determining the mols
of each antacid present. However, many students totally ignored the corresponding balanced chemical equations or entered wrong ones. A worrying number of candidates didn't therefore use the correct molar ratio to calculate the values in M3. Showing the calculations and how X is more effective than Y was the point of the question.
Which is a d-block element?
\nA. Ca
\nB. Cf
\nC. C
\nD. Co
\nD
\nWhich compound has the greatest volatility under the same conditions?
\nA. SO2
\nB. SiO2
\nC. SnO2
\nD. SrO
\nA
\nDescribe the proper disposal of low-level radioactive waste in hospitals.
\nOutline a green chemistry solution for problems generated by the use of organic solvents.
\nstore until material becomes inactive/radiation levels drop ✔
\ndispose with other waste
OR
dispose in landfills ✔
Only award M2 if M1 correct.
Accept “dispose by incineration” for M2.
«use of» alternative solvents such as supercritical/liquid
Do not accept political or regulatory solutions.
OR
use of water «as solvent»
OR
solvent-free reactions «for example, polymerization of propene»
OR
solid-state chemistry
OR
recycle «waste» solvents
OR
catalysis that leads to better/higher yield
“catalysis” alone not sufficient for mark.
OR
reducing number of steps ✔
Mostly well answered. Some students lost a mark due to not giving a complete answer. In Spanish quite a few students referred to isotopes 'decomposing' and therefore lost their mark.
\nMany correct answers. The most common correct answer was using water as a solvent. Some students lost the mark with responses like \" less dangerous organic solvents\", \"alternative solvents (but not giving a specific example)\" or \"proper disposal of organic solvents\".
\nWhich compound has the shortest C to N bond?
\nA. HCN
\nB. CH3CH2NH2
\nC. CH3CHNH
\nD. (CH3)2NH
\nA
\nWhen dinitrogen pentoxide, N2O5, is heated the colourless gas undergoes thermal decomposition to produce brown nitrogen dioxide:
\nN2O5 (g) → 2NO2 (g) + O2 (g)
\nData for the decomposition at constant temperature is given.
\nSuggest how the extent of decomposition could be measured.
\nPlot the missing point on the graph and draw the best-fit line.
\nDeduce the relationship between the concentration of N2O5 and the rate of reaction.
\nOutline why increasing the concentration of N2O5 increases the rate of reaction.
\nuse colorimeter
OR
change in colour
OR
change in volume
OR
change in pressure ✔
Accept suitable instruments, e.g. pressure probe/oxygen sensor.
\npoint correct ✔
\nstraight line passing close to all points AND through origin ✔
\n
Accept free hand drawn line as long as attempt to be linear and meets criteria for M2.
« rate of reaction is directly» proportional to/∝[N2O5]
OR
doubling concentration doubles rate ✔
Do not accept “rate increases as concentration increases”/ positive correlation
Accept linear
\ngreater frequency of collisions «as concentration increases»
OR
more collisions per unit time «as concentration increases» ✔
Accept “rate/chance/probability/likelihood” instead of “frequency”.
Do not accept just “more collisions”.
\n| Carbon fibre reinforced plastic (CFRP) is a useful composite. Epoxy is a thermoset polymer that is used as a binding polymer when making CFRP. | \n
\n
Outline the two distinct phases of this composite.
\n\n
Thermoplastic composites are increasingly replacing thermosets.
\nSuggest one advantage of thermoplastic polymers over thermosets.
\nExplain how thermoplastics, such as polyvinylchloride, PVC, can be made more flexible by the addition of phthalate ester plasticizers.
\nExplain why phthalates are replaced by other plasticizers in the production of plastics.
\n\n
carbon fibre reinforcing phase ✔
\n«in a» matrix phase of epoxy ✔
\n
Award [1 max] for “reinforcing phase «embedded» in a matrix”.
can be recycled
OR
can be reformed when hot
OR
high impact/chemical/abrasion resistance ✔
Any three of:
\nplasticizers embed/fit between «polymer» chains ✔
\nkeep polymer strands/chains/molecules separated/apart ✔
\nweaken intermolecular/London/dispersion/attractive/forces/instantaneous induced dipole-induced dipole/forces «between chains» ✔
\nprevent chains from packing closely/forming regular packing/structure ✔
\n
Accept “van der Waals/vdW” for “London”.
Any two of:
\nreadily released into environment
OR
have weak intermolecular forces «rather than covalent bonds between chains» ✔
get into biological systems by ingestion/inhalation ✔
\ninterrupt endocrine systems
OR
affect release of hormones
OR
effect development of male reproductive system ✔
considered carcinogenic
OR
can cause cellular damage ✔
can cause early puberty in females ✔
\ncan cause thyroid effects ✔
\ncan cause asthma ✔
\n
Do not accept just “are a health concern”.
Approximately 1% of the candidates this session attempted this option so feedback on this section by question is not possible due to the minimal number of scripts available for marking. The students that did attempt this option did not perform very well with several of them leaving some questions unanswered.
\nApproximately 1% of the candidates this session attempted this option so feedback on this section by question is not possible due to the minimal number of scripts available for marking. The students that did attempt this option did not perform very well with several of them leaving some questions unanswered.
\nApproximately 1% of the candidates this session attempted this option so feedback on this section by question is not possible due to the minimal number of scripts available for marking. The students that did attempt this option did not perform very well with several of them leaving some questions unanswered.
\nApproximately 1% of the candidates this session attempted this option so feedback on this section by question is not possible due to the minimal number of scripts available for marking. The students that did attempt this option did not perform very well with several of them leaving some questions unanswered.
\nConsider the structures of medicinal molecules in section 37 of the data booklet.
\nName two functional groups that both zanamivir and oseltamivir contain.
\nExplain how zanamivir works as a preventative agent against flu viruses.
\nCircle the side-chain in penicillin on the structure below.
\nExplain, with reference to the action of penicillin, why new penicillins with different side-chains need to be produced.
\nState and explain the relative solubility of codeine in water compared to morphine and diamorphine.
\nState the natural source from which codeine, morphine and diamorphine are obtained.
\nAny two:
«secondary» carboxamide/amido ✔
ether ✔
carbonyl ✔
Accept amide
Accept amino/amine.
Accept alkenyl/alkene.
Do not accept formula.
\n«drug» blocks/inhibits «viral» enzyme/neuraminidase/ «activity» ✔
prevents virus from leaving/escaping host cells «thus cannot infect other cells» ✔
Do not accept other anti-viral methods (as question is specific to Zanamivir).
✔
Accept a circle that does not surround the amido group.
\nDo not accept a circle that only surrounds the phenol group.
\nbacterial resistance «to older penicillins/antibiotics» ✔
\nprevent penicillinase/beta-lactamase/enzyme in bacterium to deactivate/open penicillin/beta-lactam ring ✔
\n
Accept “antibiotic resistant bacteria” but not “antibiotic resistance” for M1.
Accept “reduce allergic reactions from from penicillin” for M2.
Award [1 max] for “increased efficiency” OR “increased stability in GIT”.
Do not accept “bacteria develop tolerance”.
codeine less soluble «in water» than morphine AND more soluble than diamorphine
OR
morphine > codeine > diamorphine «in terms of solubility in water» ✔
more/stronger/greater hydrogen/H bonding «due to more hydroxyl groups leads to greater solubility» ✔
\nopium poppy/plants/seeds ✔
\nAccept “poppy” OR “opioid”.
\nMost students scored both marks here. Some students with incorrect answers gave carboxylic acid/carboxyl, benzene/arene, and ester.
\nEven when a relevant number of students obtained at least one mark, the quality of the answers was in general low and many evidenced very shallow understanding just repeating information
\nMany students scored here but a relevant percentage didn't include the amide group. Quite a few students did not know where the side chain is and circled other parts of the molecule.
\nMany good answers for M1 but only stronger students managed to score the second one. The answers evidenced again shallow understanding that may result from independent learning. Many students gave answers related to how penicillin causes bacteria to burst which is not related to this question. Once again it shows students perhaps memorizing answers without a true understanding of concepts.
\nM1 was correctly presented but most students focused on the polarity of the molecules and presented long arguments where intermolecular forces were ignored. The concept of water solubility and hydrogen bonding was largely ignored.
\nMostly well answered. Student who missed this question were clearly guessing with other natural products presented with the Option D material.
\nWhat is the formula of the compound formed from Ca2+ and PO43−?
\nA. CaPO4
\nB. Ca3(PO4)2
\nC. Ca2(PO4)3
\nD. Ca(PO4)2
\nB
\nWhich is the correct order based on increasing strength?
\nA. covalent bonds < hydrogen bonds < dipole–dipole forces < dispersion forces
\nB. dipole–dipole forces < dispersion forces < hydrogen bonds < covalent bonds
\nC. dispersion forces < dipole–dipole forces < hydrogen bonds < covalent bonds
\nD. dispersion forces < dipole–dipole forces < covalent bonds < hydrogen bonds
\nC
\nWhich describes an exothermic reaction?
\nD
\nA mixture of ethanal, ethanol and ethanoic acid is fractionally distilled.
\nCalculate the mole fraction of ethanal in the mixture.
\nThe vapour pressure of pure ethanal at is .
\nCalculate the vapour pressure of ethanal above the liquid mixture at .
\nDescribe how this mixture is separated by fractional distillation.
\n✔
\n
Accept “”.
✔
\nAny two of:
continuous evaporation and condensation
OR
increased surface area in column helps condensation ✔
Accept “glass «beads» aid condensation «in fractionating column»”.
temperature decreases up the fractionating column ✔
\nliquids condense at different heights
OR
liquid of lowest boiling point collected first
OR
liquid with weakest intermolecular forces collected first
OR
most volatile component collected first
OR
fractions/liquids collected in order of boiling point/volatility ✔
Accept “liquids collected in order of molar mass”.
This question involving Raoult's Law was very well answered and most were able to calculate the mole fraction of ethanal in the mixture (0.250) and the corresponding vapour pressure of ethanal above the liquid mixture at 20 °C (25.3 kPa). There was one G2 comment on this question. One teacher stated that the diagram shows four fractions but the stem of the question specifically states only three components and hence the fourth test tube is not required. The teacher commented that some students may have been distracted by this.
\nThis question involving Raoult's Law was very well answered and most were able to calculate the mole fraction of ethanal in the mixture (0.250) and the corresponding vapour pressure of ethanal above the liquid mixture at 20 °C (25.3 kPa). There was one G2 comment on this question. One teacher stated that the diagram shows four fractions but the stem of the question specifically states only three components and hence the fourth test tube is not required. The teacher commented that some students may have been distracted by this.
\nIn this question candidates were required to describe how the mixture can be separated by fractional distillation. Only the better candidates scored both marks, though most gained at least one mark, usually for stating that the most volatile component is collected first. Many did not convey the idea that there is continuous evaporation and condensation in the process or the fact that the temperature decreases up the fractionating column.
\nWhat is the heat change, in kJ, when 100.0 g of aluminium is heated from 19.0 °C to 32.0 °C?
\nSpecific heat capacity of aluminium: 0.90 J g−1 K−1
\nA.
\nB.
\nC.
\nD.
\nC
\nIron may be extracted from iron (II) sulfide, FeS.
\nIron (II) sulfide, FeS, is ionically bonded.
\nThe first step in the extraction of iron from iron (II) sulfide is to roast it in air to form iron (III) oxide and sulfur dioxide.
\nOutline why metals, like iron, can conduct electricity.
\nJustify why sulfur is classified as a non-metal by giving two of its chemical properties.
\nSketch the first eight successive ionisation energies of sulfur.
\nDescribe the bonding in this type of solid.
\nState a technique that could be used to determine the crystal structure of the solid compound.
\nState the full electron configuration of the sulfide ion.
\nOutline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.
\nSuggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
\nWrite the equation for this reaction.
\nDeduce the change in the oxidation state of sulfur.
\nSuggest why this process might raise environmental concerns.
\nExplain why the addition of small amounts of carbon to iron makes the metal harder.
\nmobile/delocalized «sea of» electrons
\nAny two of:
\nforms acidic oxides «rather than basic oxides» ✔
\nforms covalent/bonds compounds «with other non-metals» ✔
\nforms anions «rather than cations» ✔
\nbehaves as an oxidizing agent «rather than a reducing agent» ✔
\n
Award [1 max] for 2 correct non-chemical properties such as non-conductor, high ionisation energy, high electronegativity, low electron affinity if no marks for chemical properties are awarded.
two regions of small increases AND a large increase between them✔
\nlarge increase from 6th to 7th ✔
\n
Accept line/curve showing these trends.
electrostatic attraction ✔
\nbetween oppositely charged ions/between Fe2+ and S2− ✔
\nX-ray crystallography ✔
\n1s2 2s2 2p6 3s2 3p6 ✔
\n
Do not accept “[Ne] 3s2 3p6”.
«valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔
\n
Accept 2,8 (for O2–) and 2,8,8 (for S2–)
allows them to explain the properties of different compounds/substances
OR
enables them to generalise about substances
OR
enables them to make predictions ✔
Accept other valid answers.
4FeS(s) + 7O2(g) → 2Fe2O3(s) + 4SO2(g) ✔
\n
Accept any correct ratio.
+6
OR
−2 to +4 ✔
Accept “6/VI”.
Accept “−II, 4//IV”.
Do not accept 2- to 4+.
sulfur dioxide/SO2 causes acid rain ✔
\nAccept sulfur dioxide/SO2/dust causes respiratory problems
Do not accept just “causes respiratory problems” or “causes acid rain”.
disrupts the regular arrangement «of iron atoms/ions»
OR
carbon different size «to iron atoms/ions» ✔
prevents layers/atoms sliding over each other ✔
\nThere has been significant growth in the use of carbon nanotubes, CNT.
\nExplain these properties of carbon nanotubes.
\nAlloying metals changes their properties. Suggest one property of magnesium that could be improved by making a magnesium–CNT alloy.
\nPure magnesium needed for making alloys can be obtained by electrolysis of molten magnesium chloride.
\n © International Baccalaureate Organization 2020.
Write the half-equations for the reactions occurring in this electrolysis.
\nCalculate the theoretical mass of magnesium obtained if a current of is used for hours. Use charge and section 2 of the data booklet
\nSuggest a gas which should be continuously passed over the molten magnesium in the electrolytic cell.
\nZeolites can be used as catalysts in the manufacture of CNT. Explain, with reference to their structure, the high selectivity of zeolites.
\nExperiments have been done to explore the nematic liquid crystal behaviour of CNT. Justify how CNT molecules could be classified as nematic.
\nExcellent strength: defect-free AND rigid/regular 2D/3D ✔
\nExcellent conductivity: delocalized electrons ✔
\n
Accept “carbons/atoms are all covalently bonded to each other” for M1.
Any of:
ductility ✔
strength/resistance to deformation ✔
malleability ✔
hardness ✔
resistance to corrosion/chemical resistance ✔
range of working temperatures ✔
density ✔
Do not accept “conductivity”.
Anode: ✔
\nCathode: ✔
\n
Accept .
Award [1 max] for correct equations at incorrect electrodes.
\n✔
\n✔
\n
✔
Award [3] for correct final answer.
argon//helium/ ✔
\n
Accept any identified noble/inert gas.
Accept name OR formula.
Do not accept “nitrogen/“.
\npores/cavities/channels/holes/cage-like structures ✔
\n«only» reactants with appropriate/specific size/geometry/structure fit inside/go through/are activated/can react ✔
\n
Accept “molecules/ions” for “reactants” in M2.
rod-shaped molecules
OR
«randomly distributed but» generally align
OR
no positional order AND have «some» directional order/pattern ✔
Accept “linear” for “rod-shaped”.
This was a very popular option with approximately 34% of candidates attempting Option B. Many students appeared well prepared for this option. Some candidates continue to provide answers with a heavy Biology bias that often make them lose valuable points.
\nThis was a very popular option with approximately 34% of candidates attempting Option B. Many students appeared well prepared for this option. Some candidates continue to provide answers with a heavy Biology bias that often make them lose valuable points.
\nThis was a very popular option with approximately 34% of candidates attempting Option B. Many students appeared well prepared for this option. Some candidates continue to provide answers with a heavy Biology bias that often make them lose valuable points.
\nThis was a very popular option with approximately 34% of candidates attempting Option B. Many students appeared well prepared for this option. Some candidates continue to provide answers with a heavy Biology bias that often make them lose valuable points.
\nThis was a very popular option with approximately 34% of candidates attempting Option B. Many students appeared well prepared for this option. Some candidates continue to provide answers with a heavy Biology bias that often make them lose valuable points.
\nThis was a very popular option with approximately 34% of candidates attempting Option B. Many students appeared well prepared for this option. Some candidates continue to provide answers with a heavy Biology bias that often make them lose valuable points.
\nThis was a very popular option with approximately 34% of candidates attempting Option B. Many students appeared well prepared for this option. Some candidates continue to provide answers with a heavy Biology bias that often make them lose valuable points.
\nConsider the structures of medicinal molecules in section 37 of the data booklet.
\nExplain how zanamivir works as a preventative agent against flu viruses.
\nCircle the side-chain in penicillin on the structure below.
\nExplain, with reference to the action of penicillin, why new penicillins with different side-chains need to be produced.
\n\n
State and explain the relative solubility of codeine in water compared to morphine and diamorphine.
\nState the natural source from which codeine, morphine and diamorphine are obtained.
\nCircle two chiral carbons in the section of the Taxol structure below.
\n«drug» blocks/inhibits «viral» enzyme/neuraminidase/ «activity» ✔
\nprevents virus from leaving/escaping host cells «thus cannot infect other cells» ✔
\n
Do not accept other anti-viral methods (as question is specific to Zanamivir).
✔
Accept a circle that does not surround the amido group.
\nDo not accept a circle that only surrounds the phenol group.
\nbacterial resistance «to older penicillins/antibiotics» ✔
\nprevent penicillinase/beta-lactamase/enzyme in bacterium to deactivate/open penicillin/beta-lactam ring ✔
\n
Accept “antibiotic resistant bacteria” but not “antibiotic resistance” for M1.
Accept “reduce allergic reactions from penicillin” for M2.
\nAward [1 max] for “increased efficiency” OR “increased stability in GIT”.
\nDo not accept “bacteria develop tolerance”.
\ncodeine less soluble «in water» than morphine AND more soluble than diamorphine
OR
morphine > codeine > diamorphine «in terms of solubility in water» ✔
more/stronger/greater hydrogen/H bonding «due to more hydroxyl groups leads to greater solubility» ✔
\nopium poppy/plants/seeds ✔
\nAccept “poppy” OR “opioid”.
\nany two chiral carbons identified ✔
\nIn this question candidates were required to describe how the mixture can be separated by fractional distillation. Only the better candidates scored both marks, though most gained at least one mark, usually for stating that the most volatile component is collected first. Many did not convey the idea that there is continuous evaporation and condensation in the process or the fact that the temperature decreases up the fractionating column.
\nMost were able to circle the side-chain in penicillin. Common errors included circling only the phenolic group, the four-membered ring or the five-membered ring.
\nThis question was well answered though many lost M1 for stating \"antibiotic resistance\" instead of mentioning \"bacterial resistance\".
\nOnly the better candidates were able to state and explain the relative solubility of codeine in water compared to morphine and diamorphine. The majority scored M1 for the correct order of relative solubility. Few gave a comprehensive explanation outlining that there is greater hydrogen bonding due to more hydroxyl groups, which results in greater solubility.
\nThe natural source from which codeine, morphine and diamorphine are obtained had to be stated. Most scored the single mark here for \"poppy\". Common errors included \"willow tree\" or \"opium\" alone, which was not deemed sufficient to score the mark.
\nThe idea of a chiral carbon was very well understood and nearly all scored the mark for identifying any two chiral carbons in Taxol.
\n| Carbon fibre reinforced plastic (CFRP) is a useful composite. Epoxy is a thermoset polymer that is used as a binding polymer when making CFRP. | \n
\n
Outline the two distinct phases of this composite.
\n\n
Thermoplastic composites are increasingly replacing thermosets.
\nSuggest one advantage of thermoplastic polymers over thermosets.
\nExplain how thermoplastics, such as polyvinylchloride, PVC, can be made more flexible by the addition of phthalate ester plasticizers.
\nExplain why phthalates are replaced by other plasticizers in the production of plastics.
\nClassify PVC and polyethene terephthalate, PET, as addition or condensation polymers and deduce the structural formulas.
\ncarbon fibre reinforcing phase ✔
\n«in a» matrix phase of epoxy ✔
\n
Award [1 max] for “reinforcing phase «embedded» in a matrix”.
can be recycled
OR
can be reformed when hot
OR
high impact/chemical/abrasion resistance ✔
Any three of:
\nplasticizers embed/fit between «polymer» chains ✔
\nkeep polymer strands/chains/molecules separated/apart ✔
\nweaken intermolecular/London/dispersion/attractive/forces/instantaneous induced dipole-induced dipole/forces «between chains» ✔
\nprevent chains from packing closely/forming regular packing/structure ✔
\n
Accept “van der Waals/vdW” for “London”.
Any two of:
readily released into environment
OR
have weak intermolecular forces «rather than covalent bonds between chains» ✔
get into biological systems by ingestion/inhalation ✔
\ninterrupt endocrine systems
OR
affect release of hormones
OR
effect development of male reproductive system ✔
considered carcinogenic
OR
can cause cellular damage ✔
can cause early puberty in females ✔
\ncan cause thyroid effects ✔
\ncan cause asthma ✔
\n
Do not accept just “are a health concern”.
PVC: addition AND PET: condensation ✔
\nstructure of PVC monomer ✔
\nstructure of PET monomers ✔
\n
Accept full OR condensed structural formulas.
Most had some idea of a composite and tended to gain at least one mark for stating \"reinforcing phase embedded in a matrix\".
\nThis was well answered and most stated that one advantage of thermoplastic polymers over thermosets is that they can be recycled.
\nThis also was well done and some managed to score all three marks.
\nThis question required an explanation of why phthalates are replaced by other plasticizers in the production of plastics. This question was found to be very challenging and even the better candidates failed to score both marks. The weaker candidates typically stated that they are just a health concern, which was deemed insufficient to warrant even a salvage mark.
\nThis question on PVC and PET was very well answered. All three marks for their correct classification and the corresponding structures of their monomers was frequently scored.
\nMagnetite, Fe3O4, is another ore of iron that contains both Fe2+ and Fe3+.
\nIron exists as several isotopes.
\nDeduce the ratio of Fe2+:Fe3+ in Fe3O4.
\nState the type of spectroscopy that could be used to determine their relative abundances.
\nState the number of protons, neutrons and electrons in each species.
\nIron has a relatively small specific heat capacity; the temperature of a 50 g sample rises by 44.4°C when it absorbs 1 kJ of heat energy.
\nDetermine the specific heat capacity of iron, in J g−1 K−1. Use section 1 of the data booklet.
\nA voltaic cell is set up between the Fe2+ (aq) | Fe (s) and Fe3+ (aq) | Fe2+ (aq) half-cells.
\nDeduce the equation and the cell potential of the spontaneous reaction. Use section 24 of the data booklet.
\nThe figure shows an apparatus that could be used to electroplate iron with zinc. Label the figure with the required substances.
\nOutline why, unlike typical transition metals, zinc compounds are not coloured.
\nTransition metals like iron can form complex ions. Discuss the bonding between transition metals and their ligands in terms of acid-base theory.
\n1:2 ✔
\nAccept 2 Fe3+: 1 Fe2+
Do not accept 2:1 only
mass «spectroscopy»/MS ✔
\n
Award [1 max] for 4 correct values.
specific heat capacity « = » = 0.45 «J g−1 K−1» ✔
\nEquation:
2Fe3+(aq) + Fe(s) → 3Fe2+(aq) ✔
Cell potential:
«+0.77 V − (−0.45 V) = +»1.22 «V» ✔
Do not accept reverse reaction or equilibrium arrow.
Do not accept negative value for M2.
\nleft electrode/anode labelled zinc/Zn AND right electrode/cathode labelled iron/Fe ✔
\nelectrolyte labelled as «aqueous» zinc salt/Zn2+ ✔
\n
Accept an inert conductor for the anode.
Accept specific zinc salts such as ZnSO4.
\n« Zn2+» has a full d-shell
OR
does not form « ions with» an incomplete d-shell ✔
Do not accept “Zn is not a transition metal”.
Do not accept zinc atoms for zinc ions.
\nligands donate pairs of electrons to metal ions
OR
forms coordinate covalent/dative bond✔
ligands are Lewis bases
AND
metal «ions» are Lewis acids ✔
Which species has the same electron configuration as argon?
\nA. Br−
\nB. Ca2+
\nC. Al3+
\nD. Si4+
\nB
\nThe Lewis structure of methylamine is shown.
\nWhat is the molecular geometry around N?
\nA. Square planar
\nB. Tetrahedral
\nC. Trigonal planar
\nD. Trigonal pyramidal
\nD
\nWhich trend is correct, going down group 1?
\nA. Melting point increases
\nB. Reactivity decreases
\nC. First ionisation energy increases
\nD. Electronegativity decreases
\nD
\nWhich substance is most likely to be ionic?
\nD
\nWhich compound contains both ionic and covalent bonds?
\nA.
\nB.
\nC.
\nD.
\nD
\nAlong which series is the bond angle increasing?
\nA. NH3 H2O CH4
\nB. CH4 NH3 H2O
\nC. H2O NH3 CH4
\nD. H2O CH4 NH3
\n\n
C
\nWhen sodium carbonate powder is added to ethanoic acid, the beaker becomes cooler.
\nPossible enthalpy diagrams are shown.
\nWhich correctly describes the reaction?
\nC
\nWhich contains the most atoms of oxygen?
\nA. 64 g of O2
\nB. 1.2 × 1024 molecules of O2
\nC. 64 g of C3H5O3
\nD. 1.2 × 1024 molecules of C3H5O3
\nD
\nWhat is the resulting concentration, in mol dm−3, when 1.0 cm3 of 0.500 mol dm−3 nitric acid solution is diluted to 50.0 cm3 with water?
\nA. 0.002
\nB. 0.01
\nC. 0.04
\nD. 0.1
\nB
\nWhat is the enthalpy change, in J, when 5 g of water is heated from 10°C to 18°C?
\nSpecific heat capacity of water: 4.18 kJ kg−1 K−1
\nA. 5 × 4.18 × 8
\nB. 5 × 10−3 × 4.18 × 8
\nC. 5 × 4.18 × (273 + 8)
\nD. 5 × 10−3 × 4.18 × (273 + 8)
\nA
\nWhat volume of oxygen, in dm3 at STP, is needed when 5.8 g of butane undergoes complete combustion?
\n\nA.
\nB.
\nC.
\nD.
\nB
\nWhat is the enthalpy change of the reaction, in kJ?
\n2C (graphite) + O2 (g) → 2CO (g)
\nA. −394 − 283
\nB. 2(−394) + 2(−283)
\nC. −394 + 283
\nD. 2(−394) + 2(283)
\n\n
D
\nCurve 1 shows the mass change when marble chips are added to excess hydrochloric acid in an open beaker.
\nWhich changes would produce curve 2?
\nA. Powdering the marble chips and heating
\nB. Powdering the marble chips and doubling their mass
\nC. Doubling the volume of acid and heating
\nD. Doubling the acid concentration and powdering the marble chips
\nB
\nOn the following Maxwell-Boltzmann distribution, which letter represents activation energy?
\nA. A
\nB. B
\nC. C
\nD. D
\nB
\nWhat is the coefficient of (aq) when the equation is balanced using the smallest possible whole numbers?
\n\nA. 1
\nB. 2
\nC. 3
\nD. 4
\n\n
B
\nWhich changes produce the greatest increase in the percentage conversion of methane?
\nCH4 (g) + H2O (g) CO (g) + 3H2 (g)
\nC
\nWhich statements are correct for electrolysis?
\nI. An exothermic reaction occurs.
II. Oxidation occurs at the anode (positive electrode).
III. The reaction is non-spontaneous.
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nA voltaic cell is made up of nickel and magnesium half-cells.
\n\nWrite the balanced equation for the reaction in this voltaic cell.
\nCalculate the cell potential for and at . Use sections 1, 2 and 24 of the data booklet.
\nPredict, giving a reason, how an increase in temperature affects the potential of this cell.
\n✔
\nAccept a balanced molecular equation such as “”.
\n✔
\nAND ✔
\n✔
\n
Award [3] for correct final answer.
cell potential/ increases AND increasing temperature favours forward reaction
OR
cell potential/ increases AND becomes more negative
OR
cell potential/ increases AND becomes more negative ✔
Accept any correct mathematical explanation using the Nernst equation.
Nearly all were able to write the balanced equation for the reaction occurring in the voltaic cell.
\nThe cell potential numerical problem was well executed and many scored all three marks, by calculating the final answer as +2.17 V. The most common errors related to either incorrectly calculating Q or identifying n = 2 in the intermediate, numerical step.
\nThis was often gained as an easy subsequent mark for those that scored full marks in (b). The majority approached this problem by applying a mathematical explanation using the Nernst equation.
\nWhich is amphiprotic?
\nA. NH4+
\nB. PO43−
\nC. H2O
\nD. H3O+
\nC
\nWhat is the relative molecular mass of bromine, according to the following mass spectrum?
\nNIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce
on behalf of the United States of America. All rights reserved.
A.
B.
\nC.
\nD.
\nA
\nWhich cannot act as a Brønsted–Lowry base?
\nA. HPO42−
\nB. H2O
\nC. CH4
\nD. NH3
\nC
\nWhich represents a p orbital?
\nC
\nWhich solution has a pH of 9?
\nA. 1.0 × 10−9 mol dm−3 (aq)
\nB. 1.0 × 10−5 mol dm−3 (aq)
\nC. 1.0 × 10−9 mol dm−3 (aq)
\nD. 1.0 × 10−5 mol dm−3 (aq)
\nB
\nA student performed displacement reactions using metals W and X and solutions of salts of metals W, X, Y and Z. The results are summarized in the table.
\nWhich of the four metals is most reactive?
\nA. W
\nB. X
\nC. Y
\nD. Z
\nD
\nWhich statements are correct about the action of a catalyst in a chemical reaction?
\nI. It increases the energy of each collision.
II. It alters the mechanism of the reaction.
III. It remains unchanged at the end of the reaction.
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nDoping of silicon increases the conductivity in semiconductors.
\n Describe the doping in p-type and n-type semiconductors.
Doping of silicon increases the conductivity in semiconductors.
\nExplain how doping improves the conductivity of silicon.
\np-type:
«small amount of» Group 13 element produces holes ✔
n-type:
«small amount of» Group 15 element adds extra electrons ✔
Award [1 max] for correct element type for p AND n OR p-type: “produces holes” AND n-type: adds extra electrons”.
electrons and holes flow in opposite directions
OR
electrons can flow into holes
OR
gap between valence and conduction bands becomes smaller ✔
The description of doping in p- and n-type semiconductors was well expressed and the majority scored both marks.
\nThe better candidates were able to explain how doping improves the conductivity of silicon and typically stated that electrons can flow into holes. The weaker candidates had some idea of what was involved but often were not able to articulate precisely in a scientific manner the process and hence usually failed to score the mark.
\nSulfur dioxide reacts with oxygen to form sulfur trioxide.
\n2SO2 (g) + O2 (g) 2SO3 (g) ΔH = −197 kJ
\nWhich change increases the value of Kc?
\nA. increasing the temperature
\nB. decreasing the temperature
\nC. decreasing [SO2 (g)]
\nD. decreasing [SO3 (g)]
\nB
\nWhich causes acid deposition?
\nA. SO2
\nB. SiO2
\nC. SrO
\nD. CO2
\nA
\nWhat is correct for this redox reaction?
\nMnO2 (s) + 2− (aq) + 4H+ (aq) → Mn2+ (aq) + 2 (aq) + 2H2O (l)
\nA
\nEthanol is obtained by the hydration of ethene, C2H4.
\nAlternative synthetic routes exist to produce alcohols.
\nEthanol is obtained by the hydration of ethene, C2H4.
\nState the class of compound to which ethene belongs.
\nState the molecular formula of the next member of the homologous series to which ethene belongs.
\nJustify why ethene has only a single signal in its 1H NMR spectrum.
\nDeduce the chemical shift of this signal. Use section 27 of the data booklet.
\nSuggest two possible products of the incomplete combustion of ethene that would not be formed by complete combustion.
\nA white solid was formed when ethene was subjected to high pressure.
\nDeduce the type of reaction that occurred.
\nSketch the mechanism for the reaction of propene with hydrogen bromide using curly arrows.
\nExplain why the major organic product is 2-bromopropane and not 1-bromopropane.
\nExplain why the major organic product is 2-bromopropane and not 1-bromopropane.
\n2-bromopropane can be converted directly to propan-2-ol. Identify the reagent required.
\nPropan-2-ol can also be formed in one step from a compound containing a carbonyl group.
\nState the name of this compound and the type of reaction that occurs.
\nalkene ✔
\nC3H6 ✔
\nAccept structural formula.
\nhydrogen atoms/protons in same chemical environment ✔
\nAccept “all H atoms/protons are equivalent”.
Accept “symmetrical”
4.5 to 6.0 «ppm» ✔
\nAccept a single value within this range.
\ncarbon monoxide/CO AND carbon/C/soot ✔
\n«addition» polymerization ✔
\ncurly arrow going from C=C to H of HBr AND curly arrow showing Br leaving ✔
\nrepresentation of carbocation ✔
\ncurly arrow going from lone pair/negative charge on Br− to C+ ✔
\n
Award [2 max] for mechanism producing 1-brompropane.
«2-bromopropane involves» formation of more stable «secondary» carbocation/carbonium ion/intermediate
OR
1-bromopropane involves formation of less stable «primary» carbocation/carbonium ion/intermediate ✔
«increased» positive inductive/electron-releasing effect of extra–R group/–CH3/methyl «increases stability of secondary carbocation» ✔
\nAward [1] for “more stable due to positive inductive effect”.
\nDo not award marks for quoting Markovnikov’s rule without any explanation.
\n«2-bromopropane involves» formation of more stable «secondary» carbocation/carbonium ion/intermediate
OR
1-bromopropane involves formation of less stable «primary» carbocation/carbonium ion/intermediate ✔
«increased» positive inductive/electron-releasing effect of extra–R group/–CH3/methyl «increases stability of secondary carbocation» ✔
\n
Award [1] for “more stable due to positive inductive effect”.
Do not award marks for quoting Markovnikov’s rule without any explanation.
sodium hydroxide/NaOH/potassium hydroxide/KOH ✔
\nAccept «aqueous» hydroxide ions/OH−
\nName of carbonyl compound:
propanone ✔
Type of reaction:
reduction ✔
Accept other valid alternatives, such as “2-propyl ethanoate” for M1 and “hydrolysis” for M2.
What is the oxidation state of oxygen in H2O2?
\nA. −2
\nB. −1
\nC. +1
\nD. +2
\nB
\nWhich is correct?
\nA. Electrophiles are Brønsted–Lowry acids.
\nB. Nucleophiles are Brønsted–Lowry acids.
\nC. Electrophiles are Lewis acids.
\nD. Nucleophiles are Lewis acids.
\nC
\nWhat are the products of the electrolysis of molten potassium chloride, ?
\nB
\nWhich compound is acidic in aqueous solution?
\nA. KBr
\nB. CH3COONa
\nC. NH4Cl
\nD. Na2CO3
\nC
\nHydrogen peroxide can react with methane and oxygen to form methanol. This reaction can occur below 50°C if a gold nanoparticle catalyst is used.
\nNow consider the second stage of the reaction.
\nCO (g) + 2H2 (g) CH3OH (l) ΔH⦵ = –129 kJ
\nHydrogen peroxide can react with methane and oxygen to form methanol. This reaction can occur below 50°C if a gold nanoparticle catalyst is used.
\nMethanol is usually manufactured from methane in a two-stage process.
\nCH4 (g) + H2O (g) CO (g) + 3H2 (g)
CO (g) + 2H2 (g) CH3OH (l)
Consider the first stage of the reaction.
\nCH4 (g) + H2O (g) CO (g) + 3H2 (g)
\nThe diagram shows the Maxwell-Boltzmann curve for the uncatalyzed reaction.
\nDraw a distribution curve at a lower temperature (T2) and show on the diagram how the addition of a catalyst enables the reaction to take place more rapidly than at T1.
\nThe hydrogen peroxide could cause further oxidation of the methanol. Suggest a possible oxidation product.
\nDetermine the overall equation for the production of methanol.
\n8.00 g of methane is completely converted to methanol. Calculate, to three significant figures, the final volume of hydrogen at STP, in dm3. Use sections 2 and 6 of the data booklet.
\nDetermine the enthalpy change, ΔH, in kJ. Use section 11 of the data booklet.
\nBond enthalpy of CO = 1077 kJ mol−1.
\nState one reason why you would expect the value of ΔH calculated from the values, given in section 12 of data booklet, to differ from your answer to (d)(i).
\nState the expression for Kc for this stage of the reaction.
\nState and explain the effect of increasing temperature on the value of Kc.
\nThe equilibrium constant, Kc, has a value of 1.01 at 298 K.
\nCalculate ΔG⦵, in kJ mol–1, for this reaction. Use sections 1 and 2 of the data booklet.
\nCalculate a value for the entropy change, ΔS⦵, in J K–1 mol–1 at 298 K. Use your answers to (e)(i) and section 1 of the data booklet.
\nIf you did not get answers to (e)(i) use –1 kJ, but this is not the correct answer.
\nJustify the sign of ΔS with reference to the equation.
\nPredict, giving a reason, how a change in temperature from 298 K to 273 K would affect the spontaneity of the reaction.
\ncurve higher AND to left of T1 ✔
\nnew/catalysed Ea marked AND to the left of Ea of curve T1 ✔
\n
Do not penalize curve missing a label, not passing exactly through the origin, or crossing x-axis after Ea.
Do not award M1 if curve drawn shows significantly more/less molecules/greater/smaller area under curve than curve 1.
Accept Ea drawn to T1 instead of curve drawn as long as to left of marked Ea.
methanoic acid/HCOOH/CHOOH
OR
methanal/HCHO ✔
Accept “carbon dioxide/CO2”.
\nCH4(g) + H2O(g) CH3OH(l) + H2(g) ✔
\n
Accept arrow instead of equilibrium sign.
amount of methane = « = » 0.498 «mol» ✔
\namount of hydrogen = amount of methane / 0.498 «mol» ✔
\nvolume of hydrogen = «0.498 mol × 22.7 dm3 mol−1 = » 11.3 «dm3» ✔
\n
Award [3] for final correct answer.
Award [2 max] for 11.4 «dm3 due to rounding of mass to 16/moles to 0.5. »
Σbonds broken = 4 × 414 «kJ» + 2 × 463 «kJ» / 2582 «kJ» ✔
\nΣbonds formed = 1077 «kJ» + 3 × 436 «kJ» / 2385 «kJ» ✔
\nΔH «= Σbonds broken − Σbonds formed =( 2582 kJ − 2385 kJ)» = «+»197«kJ» ✔
\n
Award [3] for final correct answer.
Award [2 Max] for final answer of −197 «kJ»
bond energies are average values «not specific to the compound» ✔
\n✔
\nKc increases AND «forward» reaction endothermic ✔
\n«ΔG⦵ = − RT lnKc»
ΔG⦵ = − 8.31 «J K−1 mol−1» × 298 «K» × ln (1.01) / −24.6 «J mol−1» ✔
= −0.0246 «kJ mol–1» ✔
\n
Award [2] for correct final answer.
Award [1 max] for +0.0246 «kJ mol–1».
\n«ΔG⦵ = ΔH⦵ − TΔS⦵»
\nΔG⦵ = −129 «kJ mol–1» − (298 «K» × ΔS) = −0.0246 «kJ mol–1» ✔
\nΔS⦵ = « = » −433 «J K–1 mol–1» ✔
\n
Award [2] for correct final answer.
Award [1 max] for “−0.433 «kJ K–1 mol–1»”.
\nAward [1 max] for “433” or “+433” «J K–1 mol–1».
\nAward [2] for −430 «J K–1 mol–1» (result from given values).
\n«negative as» product is liquid and reactants gases
OR
fewer moles of gas in product ✔
reaction «more» spontaneous/ΔG negative/less positive AND effect of negative entropy decreases/TΔS increases/is less negative/more positive
OR
reaction «more» spontaneous/ΔG negative/less positive AND reaction exothermic «so Kc increases » ✔
Award mark if correct calculation shown.
\nWhat would be the electrode potential, E⦵, of the Mn2+ (aq)|Mn (s) half-cell if Fe3+ (aq)|Fe2+ (aq) is used as the reference standard?
\nMn2+ (aq) + 2e− Mn (s) E⦵ = −1.18 V
Fe3+ (aq) + e− Fe2+ (aq) E⦵ = +0.77 V
A. −1.95 V
\nB. −0.41 V
\nC. +0.41 V
\nD. +1.95 V
\nA
\nWhat happens to the mass of each copper electrode when aqueous copper(II) sulfate solution is electrolysed?
\nC
\nWhen dinitrogen pentoxide, N2O5, is heated the colourless gas undergoes thermal decomposition to produce brown nitrogen dioxide:
\nN2O5 (g) → 2NO2 (g) + O2 (g)
\nData for the decomposition at constant temperature is given.
\nSuggest how the extent of decomposition could be measured.
\nPlot the missing point on the graph and draw the best-fit line.
\nOutline why increasing the concentration of N2O5 increases the rate of reaction.
\nWrite the rate expression for this reaction.
\nCalculate the value of the rate constant, k, giving its units.
\nuse colorimeter
OR
change in colour
OR
change in volume
OR
change in pressure ✔
Accept suitable instruments, e.g. pressure probe/oxygen sensor.
\npoint correct ✔
\nstraight line passing close to all points AND through origin ✔
\n
Accept free hand drawn line as long as attempt to be linear and meets criteria for M2.
greater frequency of collisions «as concentration increases»
OR
more collisions per unit time «as concentration increases» ✔
Accept “rate/chance/probability/likelihood” instead of “frequency”.
Do not accept just “more collisions”.
\nrate = k[N2O5] ✔
\nk = ✔
\n«k = = » 0.030 «min–1» ✔
\nmin–1 ✔
\n
M1 can be awarded from correct M2 if not explicitly stated.
Accept k = gradient.
\nAccept values in the range 0.028–0.032.
\nAward [3] for correct final answer.
\nWhich compound shows cis-trans isomerism?
\nA. CH3CH=CCl2
\nB. CCl2=CH2
\nC.
D.
D
\nWhich compound rotates the plane of plane-polarized light?
\nA. CH3C(CH3)ClCH3
\nB. CH3CH2CHClCH3
\nC. CH3C(Cl)2CH3
\nD. CH3CClBrCH3
\nB
\nWhich can be reduced to a secondary alcohol?
\nA. C2H5COOH
\nB. CH3CH2OCH3
\nC. (CH3)2CHCHO
\nD. CH3COC2H5
\nD
\nWhat information can be deduced from the splitting pattern of 1H NMR signals?
\nA. total number of hydrogen atoms in a compound
\nB. number of hydrogen atoms on adjacent atom(s)
\nC. functional group on which hydrogen atoms are located
\nD. number of hydrogen atoms in a particular chemical environment
\nB
\nIron may be extracted from iron (II) sulfide, FeS.
\nIron (II) sulfide, FeS, is ionically bonded.
\nThe first step in the extraction of iron from iron (II) sulfide is to roast it in air to form iron (III) oxide and sulfur dioxide.
\nOutline why metals, like iron, can conduct electricity.
\nJustify why sulfur is classified as a non-metal by giving two of its chemical properties.
\nDescribe the bonding in this type of solid.
\nState the full electron configuration of the sulfide ion.
\nOutline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.
\nSuggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
\nWrite the equation for this reaction.
\nDeduce the change in the oxidation state of sulfur.
\nSuggest why this process might raise environmental concerns.
\nExplain why the addition of small amounts of carbon to iron makes the metal harder.
\nmobile/delocalized «sea of» electrons
\nAny two of:
\nforms acidic oxides «rather than basic oxides» ✔
\nforms covalent/bonds compounds «with other non-metals» ✔
\nforms anions «rather than cations» ✔
\nbehaves as an oxidizing agent «rather than a reducing agent» ✔
\n
Award [1 max] for 2 correct non-chemical properties such as non-conductor, high ionisation energy, high electronegativity, low electron affinity if no marks for chemical properties are awarded.
electrostatic attraction ✔
\nbetween oppositely charged ions/between Fe2+ and S2− ✔
\n1s2 2s2 2p6 3s2 3p6 ✔
\n
Do not accept “[Ne] 3s2 3p6”.
«valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔
\n
Accept 2,8 (for O2–) and 2,8,8 (for S2–)
allows them to explain the properties of different compounds/substances
OR
enables them to generalise about substances
OR
enables them to make predictions ✔
Accept other valid answers.
4FeS(s) + 7O2(g) → 2Fe2O3(s) + 4SO2(g) ✔
\n
Accept any correct ratio.
+6
OR
−2 to +4 ✔
Accept “6/VI”.
Accept “−II, 4//IV”.
Do not accept 2− to 4+.
sulfur dioxide/SO2 causes acid rain ✔
\nAccept sulfur dioxide/SO2/dust causes respiratory problems
Do not accept just “causes respiratory problems” or “causes acid rain”.
disrupts the regular arrangement «of iron atoms/ions»
OR
carbon different size «to iron atoms/ions» ✔
prevents layers/atoms sliding over each other ✔
\nWhat occurs at an anode?
\nD
\nWhich is in the same homologous series as CH3OCH3?
\nA. CH3COCH3
\nB. CH3COOCH3
\nC. CH3CH2CH2OH
\nD. CH3CH2CH2OCH3
\nD
\nWhat is the IUPAC name of the molecule shown?
\nA. 2,4-dimethylhexane
\nB. 3,5-dimethylhexane
\nC. 2-methyl-4-ethylpentane
\nD. 2-ethyl-4-methylpentane
\nA
\nWhich monomer forms the polymer shown?
\nA.
\nB.
\nC.
\nD.
\nA
\nWhich is a propagation step in the free-radical substitution mechanism of ethane with chlorine?
\nA. C2 → 2 •C
\nB. •C2H5 + C2 → C2H5C + •C
\nC. •C2H5 + •C → C2H5C
\nD. C2H6 + •C → C2H5C + •H
\nB
\nOxygen exists as two allotropes, diatomic oxygen, O2, and ozone, O3.
\nDraw a Lewis (electron dot) structure for ozone.
\nDiscuss the relative length of the two O−O bonds in ozone.
\nExplain why there are frequencies of UV light that will dissociate O3 but not O2.
\nExplain, using equations, how the presence of results in a chain reaction that decreases the concentration of ozone in the stratosphere.
\n✔
Accept any combination of lines, dots or crosses to represent electrons.
\nDo not accept structures that represent 1.5 bonds.
\nboth equal ✔
\ndelocalization/resonance ✔
\n
Accept bond length between 121 and 148 pm/ that of single O−O bond and double O=O bond for M1.
bond in O3 is weaker
OR
O3 bond order 1.5/< 2 ✔
Do not accept bond in O3 is longer for M1.
lower frequency/longer wavelength «UV light» has enough energy to break the O–O bond in O3 «but not that in O2» ✔
Accept “lower frequency/longer wavelength «UV light» has lower energy”.
✔
\n
AND
✔
Do not penalize missing radical.
Accept:for M2:
AND
Which spectra would show the difference between propan-2-ol, CH3CH(OH)CH3, and propanal, CH3CH2CHO?
\nI. mass
II. infrared
III. 1H NMR
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nHow should the difference between 27.0 ± 0.3 and 9.0 ± 0.2 be shown?
\nA. 18.0 ± 0.1
\nB. 18.0 ± 0.3
\nC. 18.0 ± 0.5
\nD. 18.0 ± 0.6
\nC
\nIron (II) sulfide reacts with hydrochloric acid to form hydrogen sulfide, H2S.
\nIn aqueous solution, hydrogen sulfide acts as an acid.
\nDraw the Lewis (electron dot) structure of hydrogen sulfide.
\nPredict the shape of the hydrogen sulfide molecule.
\nState the formula of its conjugate base.
\nSaturated aqueous hydrogen sulfide has a concentration of 0.10 mol dm−3 and a pH of 4.0. Demonstrate whether it is a strong or weak acid.
\nCalculate the hydroxide ion concentration in saturated aqueous hydrogen sulfide.
\nA gaseous sample of nitrogen, contaminated only with hydrogen sulfide, was reacted with excess sodium hydroxide solution at constant temperature. The volume of the gas changed from 550 cm3 to 525 cm3.
\nDetermine the mole percentage of hydrogen sulfide in the sample, stating one assumption you made.
\n OR
✔
Accept any combination of lines, dots or crosses to represent electrons.
\nbent/non-linear/angular/v-shaped✔
\nHS− ✔
\nweak AND strong acid of this concentration/[H+] = 0.1 mol dm−3 would have pH = 1
OR
weak AND [H+] = 10−4 < 0.1 «therefore only fraction of acid dissociated» ✔
10−10 «mol dm−3» ✔
\nMole percentage H2S:
volume of H2S = «550 − 525 = » 25 «cm3» ✔
mol % H2S = « = » 4.5 «%» ✔
Award [2] for correct final answer of 4.5 «%»
\n\n
Assumption:
«both» gases behave as ideal gases ✔
Accept “volume of gas mol of gas”.
Accept “reaction goes to completion”.
Accept “nitrogen is insoluble/does not react with NaOH/only H2S reacts with NaOH”.
A liquid was added to a graduated cylinder. What can be deduced from the graph?
\nB
\nPropanoic acid, CH3CH2COOH, is a weak organic acid.
\nCalculate the pH of 0.00100 mol dm–3 propanoic acid solution. Use section 21 of the data booklet.
\nSketch the general shape of the variation of pH when 50 cm3 of 0.001 mol dm–3 NaOH (aq) is gradually added to 25 cm3 of 0.001 mol dm–3 CH3CH2COOH (aq).
\nKa = 10−4.87 / 1.35 × 10−5 ✔
\n[H+] = «» 1.16 × 10−4 «mol dm−3» ✔
\npH = 3.94 ✔
\n
Accept alternative methods of calculation.
Award [3] for correct final answer.
\nAward [3] for 3.96 {answer if solved by quadratic}.
\nAny three of:
\ncorrect “S” shape ✔
\nequivalence point at 25 cm3 ✔
\nfinal pH tends to 11 ✔
\npH at equivalence point >7 ✔
\nstarting pH between 3.8 - 4 ✔
\npH at half equivalence approx. 5 ✔
\n
Do not penalize for incorrect points.
Award any 3 correct.
The first eight successive ionization energies for an element are shown. In which group is the element?
\nCopyright Prof Mark Winter.
\nA. 6
\nB. 7
\nC. 8
\nD. 17
\nA
\nMagnetite, Fe3O4, is another ore of iron that contains both Fe2+ and Fe3+.
\nIron exists as several isotopes.
\nIn acidic solution, hydrogen peroxide, H2O2, will oxidize Fe2+.
\nFe2+ (aq) → Fe3+ (aq) + e−
\nDeduce the ratio of Fe2+:Fe3+ in Fe3O4.
\nState the type of spectroscopy that could be used to determine their relative abundances.
\nState the number of protons, neutrons and electrons in each species.
\nIron has a relatively small specific heat capacity; the temperature of a 50 g sample rises by 44.4°C when it absorbs 1 kJ of heat energy.
\nDetermine the specific heat capacity of iron, in J g−1 K−1. Use section 1 of the data booklet.
\nWrite the half-equation for the reduction of hydrogen peroxide to water in acidic solution.
\nDeduce a balanced equation for the oxidation of Fe2+ by acidified hydrogen peroxide.
\n1:2 ✔
\nAccept 2 Fe3+: 1 Fe2+
Do not accept 2:1 only
mass «spectroscopy»/MS ✔
\nAward [1 max] for 4 correct values.
\nspecific heat capacity « = » = 0.45 «J g−1 K−1» ✔
\nH2O2(aq) + 2H+(aq) + 2e−→ 2H2O(l) ✔
\nH2O2(aq) + 2H+(aq) + 2Fe2+(aq) → 2H2O(l) + 2Fe3+(aq) ✔
\nWhich factor does not affect the colour of a complex ion?
\nA. temperature of the solution
\nB. identity of the ligand
\nC. identity of the metal
\nD. oxidation number of the metal
\nA
\nHydrogen peroxide can react with methane and oxygen to form methanol. This reaction can occur below 50°C if a gold nanoparticle catalyst is used.
\nMethanol is usually manufactured from methane in a two-stage process.
\nCH4 (g) + H2O (g) CO (g) + 3H2 (g)
CO (g) + 2H2 (g) CH3OH (l)
Consider the first stage of the reaction.
\nCH4 (g) + H2O (g) CO (g) + 3H2 (g)
\nThe diagram shows the Maxwell-Boltzmann curve for the uncatalyzed reaction.
\nDraw a distribution curve at a lower temperature (T2) and show on the diagram how the addition of a catalyst enables the reaction to take place more rapidly than at T1.
\nThe hydrogen peroxide could cause further oxidation of the methanol. Suggest a possible oxidation product.
\nDetermine the overall equation for the production of methanol.
\n8.00 g of methane is completely converted to methanol. Calculate, to three significant figures, the final volume of hydrogen at STP, in dm3. Use sections 2 and 6 of the data booklet.
\nDetermine the enthalpy change, ΔH, in kJ. Use section 11 of the data booklet.
\nBond enthalpy of CO = 1077 kJ mol−1.
\nState the expression for Kc for this stage of the reaction.
\nState and explain the effect of increasing temperature on the value of Kc.
\ncurve higher AND to left of T1 ✔
\nnew/catalysed Ea marked AND to the left of Ea of curve T1 ✔
\nDo not penalize curve missing a label, not passing exactly through the origin, or crossing x-axis after Ea.
Do not award M1 if curve drawn shows significantly more/less molecules/greater/smaller area under curve than curve 1.
Accept Ea drawn to T1 instead of curve drawn as long as to left of marked Ea.
methanoic acid/HCOOH/CHOOH
OR
methanal/HCHO ✔
Accept “carbon dioxide/CO2”.
\nCH4(g) + H2O(g) CH3OH(l) + H2(g) ✔
\nAccept arrow instead of equilibrium sign.
\namount of methane = « = » 0.498 «mol» ✔
\namount of hydrogen = amount of methane / 0.498 «mol» ✔
\nvolume of hydrogen = «0.498 mol × 22.7 dm3 mol−1 = » 11.3 «dm3» ✔
\n
Award [3] for final correct answer.
Award [2 max] for 11.4 «dm3 due to rounding of mass to 16/moles to 0.5. »
Σbonds broken = 4 × 414 «kJ» + 2 × 463 «kJ» / 2582 «kJ» ✔
\nΣbonds formed = 1077 «kJ» + 3 × 436 «kJ» / 2385 «kJ» ✔
\nΔH «= Σbonds broken − Σbonds formed =( 2582 kJ − 2385 kJ)» = «+»197«kJ» ✔
\n
Award [3] for final correct answer.
Award [2 Max] for final answer of −197 «kJ»
✔
\nKc increases AND «forward» reaction endothermic ✔
\nWhich atom has an expanded octet?
\nA. C in CO2
\nB. S in SCl4
\nC. O in H2O2
\nD. P in PCl3
\nB
\nWhat is the electron domain geometry of Si in SiO2?
\nA. bent
\nB. linear
\nC. square planar
\nD. tetrahedral
\nD
\nWhich instrument would best monitor the rate of this reaction?
\n2KI (aq) + Cl2 (aq) → 2KCl (aq) + I2 (aq)
\n
A. Balance
B. Colorimeter
\nC. Volumetric flask
\nD. Gas syringe
\nB
\nLimestone can be converted into a variety of useful commercial products through the lime cycle. Limestone contains high percentages of calcium carbonate, CaCO3.
\nThermodynamic data for the decomposition of calcium carbonate is given.
\nThe second step of the lime cycle produces calcium hydroxide, Ca(OH)2.
\nCalcium hydroxide reacts with carbon dioxide to reform calcium carbonate.
\nCa(OH)2 (aq) + CO2 (g) → CaCO3 (s) + H2O (l)
\nCalcium carbonate is heated to produce calcium oxide, CaO.
\nCaCO3 (s) → CaO (s) + CO2 (g)
\nCalculate the volume of carbon dioxide produced at STP when 555 g of calcium carbonate decomposes. Use sections 2 and 6 of the data booklet.
\nCalculate the enthalpy change of reaction, ΔH, in kJ, for the decomposition of calcium carbonate.
\nCalculate the change in entropy, ΔS, in J K−1, for the decomposition of calcium carbonate.
\nDetermine the temperature, in K, at which the decomposition of calcium carbonate becomes spontaneous, using b(i), b(ii) and section 1 of the data booklet.
\n(If you do not have answers for b(i) and b(ii), use ΔH = 190 kJ and ΔS = 180 J K−1, but these are not the correct answers.)
\nSketch an energy profile for the decomposition of calcium carbonate based on your answer to b(i), labelling the axes and activation energy, Ea.
\nState how adding a catalyst to the reaction would impact the enthalpy change of reaction, ΔH, and the activation energy, Ea.
\nWrite the equation for the reaction of Ca(OH)2 (aq) with hydrochloric acid, HCl (aq).
\nDetermine the volume, in dm3, of 0.015 mol dm−3 calcium hydroxide solution needed to neutralize 35.0 cm3 of 0.025 mol dm−3 HCl (aq).
\nSaturated calcium hydroxide solution is used to test for carbon dioxide. Calculate the pH of a 2.33 × 10−2 mol dm−3 solution of calcium hydroxide, a strong base.
\nDetermine the mass, in g, of CaCO3 (s) produced by reacting 2.41 dm3 of 2.33 × 10−2 mol dm−3 of Ca(OH)2 (aq) with 0.750 dm3 of CO2 (g) at STP.
\n2.85 g of CaCO3 was collected in the experiment in d(i). Calculate the percentage yield of CaCO3.
\n(If you did not obtain an answer to d(i), use 4.00 g, but this is not the correct value.)
\nOutline how one calcium compound in the lime cycle can reduce a problem caused by acid deposition.
\n«nCaCO3 = =» 5.55 «mol» ✓
\n«V = 5.55 mol × 22.7 dm3 mol−1 =» 126 «dm3» ✓
\n
Award [2] for correct final answer.
Accept method using pV = nRT to obtain the volume with p as either 100 kPa (126 dm3) or 101.3 kPa (125 dm3).
\nDo not penalize use of 22.4 dm3 mol–1 to obtain the volume (124 dm3).
\n«ΔH =» (−635 «kJ» – 393.5 «kJ») – (−1207 «kJ») ✓
\n«ΔH = + » 179 «kJ» ✓
\n
Award [2] for correct final answer.
Award [1 max] for −179 kJ.
\nIgnore an extra step to determine total enthalpy change in kJ: 179 kJ mol-1 x 5.55 mol = 993 kJ.
\nAward [2] for an answer in the range 990 - 993« kJ».
\n«ΔS = (40 J K−1 + 214 J K−1) − (93 J K−1) =» 161 «J K−1» ✓
\n
Ignore an extra step to determine total entropy change in JK–1: 161 J mol–1K–1 x 5.55 mol = 894 «J mol–1K–1»
Award [1] for 894 «J mol–1K–1».
\n«spontaneous» if ΔG = ΔH − TΔS < 0
OR
ΔH < TΔS ✓
«T >=» 1112 «K» ✓
\n
Award [2] for correct final answer.
Accept “1056 K” if both of the incorrect values are used to solve the problem.
\nDo not award M2 for any negative T value.
\nendothermic sketch ✓
\nx-axis labelled “extent of reaction/progress of reaction/reaction coordinate/reaction pathway” AND y-axis labelled “potential energy/energy/enthalpy✓
\nactivation energy/Ea ✓
\n
Do not accept “time” for x-axis.
ΔH same AND lower Ea ✓
\nCa(OH)2 (aq) + 2HCl (aq) → 2H2O (l) + CaCl2 (aq) ✓
\n«nHCl = 0.0350 dm3 × 0.025 mol dm−3 =» 0.00088 «mol»
\nOR
nCa(OH)2 = nHCl/0.00044 «mol» ✓
«V = =» 0.029 «dm3» ✓
Award [2] for correct final answer.
Award [1 max] for 0.058 «dm3».
\nAlternative 1:
\n[OH−] = « 2 × 2.33 × 10−2 mol dm−3 =» 0.0466 «mol dm−3» ✓
\n«[H+] = = 2.15 × 10−13 mol dm−3»
\npH = « −log (2.15 × 10−13) =» 12.668 ✓
\n\n
Alternative 2:
\n[OH−] =« 2 × 2.33 × 10−2 mol dm−3 =» 0.0466 «mol dm−3» ✓
\n«pOH = −log (0.0466) = 1.332»
\npH = «14.000 – pOH = 14.000 – 1.332 =» 12.668 ✓
\n\n
Award [2] for correct final answer.
\nAward [1 max] for pH =12.367.
\n«nCa(OH)2 = 2.41 dm3 × 2.33 × 10−2 mol dm−3 =» 0.0562 «mol» AND
\n«nCO2 ==» 0.0330 «mol» ✓
\n«CO2 is the limiting reactant»
\n«mCaCO3 = 0.0330 mol × 100.09 g mol−1 =» 3.30 «g» ✓
\n\n
Only award ECF for M2 if limiting reagent is used.
\nAccept answers in the range 3.30 - 3.35 «g».
\n« × 100 =» 86.4 «%» ✓
\n\n
Accept answers in the range 86.1-86.4 «%».
\nAccept “71.3 %” for using the incorrect given value of 4.00 g.
\n«add» Ca(OH)2/CaCO3/CaO AND to «acidic» water/river/lake/soil
OR
«use» Ca(OH)2/CaCO3/CaO in scrubbers «to prevent release of acidic pollution» ✓
\n
Accept any correct name for any of the calcium compounds listed.
\nThe potential energy profile of a reaction is shown.
\nWhat can be determined about stability and energy change from the potential energy profile shown?
\nC
\nWhich combination has the greatest rate of reaction at room temperature?
A
\nThe properties of elements can be predicted from their position in the periodic table.
\nExplain why Si has a smaller atomic radius than Al.
\nExplain why the first ionization energy of sulfur is lower than that of phosphorus.
\nState the condensed electron configurations for Cr and Cr3+.
\nDescribe metallic bonding and how it contributes to electrical conductivity.
\nDeduce, giving a reason, which complex ion [Cr(CN)6]3− or [Cr(OH)6]3− absorbs higher energy light. Use section 15 of the data booklet.
\n[Cr(OH)6]3− forms a green solution. Estimate a wavelength of light absorbed by this complex, using section 17 of the data booklet.
\nDeduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF4, and sulfur dichloride, SCl2.
\nSuggest, giving reasons, the relative volatilities of SCl2 and H2O.
\nnuclear charge/number of protons/Z/Zeff increases «causing a stronger pull on the outer electrons» ✓
\nsame number of shells/«outer» energy level/shielding ✓
\nP has «three» unpaired electrons in 3p sub-level AND S has one full 3p orbital «and two 3p orbitals with unpaired electrons»
OR
P: [Ne]3s23px13py13pz1 AND S: [Ne]3s23px23py13pz1 ✓
Accept orbital diagrams for 3p sub-level for M1. Ignore other orbitals or sub-levels.
\n
repulsion between paired electrons in sulfur «and therefore easier to remove» ✓
\n
Accept “removing electron from S gives more stable half-filled sub-level\" for M2.
Cr:
[Ar] 4s13d5 ✓
Cr3+:
[Ar] 3d3 ✓
\n
Accept “[Ar] 3d54s1”.
\nAccept “[Ar] 3d34s0”.
\nAward [1 max] for two correct full electron configurations “1s22s22p63s23p64s13d5 AND 1s22s22p63s23p63d3”.
\nAward [1 max] for 4s13d5 AND 3d3.
\nelectrostatic attraction ✓
\nbetween «a lattice of» cations/positive «metal» ions AND «a sea of» delocalized electrons ✓
\nmobile electrons responsible for conductivity
OR
electrons move when a voltage/potential difference/electric field is applied ✓
\n
Do not accept “nuclei” for “cations/positive ions” in M2.
\nAccept “mobile/free” for “delocalized” electrons in M2.
\nAccept “electrons move when connected to a cell/battery/power supply” OR “electrons move when connected in a circuit” for M3.
\n[Cr(CN)6]3− AND CN−/ligand causes larger splitting «in d-orbitals compared to OH−»
OR
[Cr(CN)6]3− AND CN−/ligand associated with a higher Δ/«crystal field» splitting energy/energy difference «in the spectrochemical series compared to OH− » ✓
\n
Accept “[Cr(CN)6]3− AND «CN−» strong field ligand”.
\nany value or range between 647 and 700 nm ✓
\n\n
SF4/SCl2 structure does not have to be 3-D for mark.
Penalize missing lone pairs of electrons on halogens once only.
Accept any combination of dots, lines or crosses for bonds/lone pairs.
Accept “non-linear” for SCl2 molecular geometry.
Award [1] for two correct electron domain geometries, e.g. trigonal bipyramidal for SF4 and tetrahedral for SCl2.
H2O forms hydrogen bonding «while SCl2 does not» ✓
\nSCl2 «much» stronger London/dispersion/«instantaneous» induced dipole-induced dipole forces ✓
\n
Alternative 1:
H2O less volatile AND hydrogen bonding stronger «than dipole–dipole and dispersion forces» ✓
Alternative 2:
SCl2 less volatile AND effect of dispersion forces «could be» greater than hydrogen bonding ✓
Ignore reference to Van der Waals.
Accept “SCl2 has «much» larger molar mass/electron density” for M2.
\nWhich represents electron affinity?
\nA. Al2+ (g) → Al3+ (g) + e−
\nB. C (g) + e− → C− (g)
\nC. Cl2 (g) → 2Cl (g)
\nD. S (s) → S+ (g) + e−
\nB
\nThe equilibrium 2H2 (g) + N2 (g) N2H4 (g) has an equilibrium constant, K, at 150 °C.
\nWhat is the equilibrium constant at 150 °C, for the reverse reaction?
\nN2H4 (g) 2H2 (g) + N2 (g)
\n
A. K
B. K−1
\nC. −K
\nD. 2K
\nB
\nWhich change results in the largest negative value of ΔS?
\nA. C2H5OH (l) + SOCl2 (l) → C2H5Cl (l) + SO2 (g) + HCl (g)
\nB. CaCO3 (s) → CaO (s) + CO2 (g)
\nC. H2O (l) → H2O (s)
\nD. NH3 (g) + HCl (g) → NH4Cl (s)
\nD
\nWhich ions are present in an aqueous solution of Na2CO3?
\nI. HCO3−
II. OH−
III. CO32−
A. I and II only
B. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nWhich explains increasing rate of reaction with increasing temperature?
\nB
\nWhat is the conjugate acid of HS−?
\n
A. H2S
B. S2−
\nC. H2SO3
\nD. H2SO4
\nA
\nWhich graph represents a second order reaction with respect to X?
\nX → Y
\nD
\nWhat is the change in the oxidation state of oxygen?
\n2Fe2+ (aq) + H2O2 (aq) + 2H+ (aq) → 2H2O (l) + 2Fe3+ (aq)
\n
A. +1
B. 0
\nC. −1
\nD. −2
\nC
\nMolten zinc chloride undergoes electrolysis in an electrolytic cell at 450 °C.
\nDeduce the half-equations for the reaction at each electrode.
\nDeduce the overall cell reaction including state symbols. Use section 7 of the data booklet.
\nCathode (negative electrode):
\nZn2+ + 2e− → Zn (l) ✔
\n\n
Anode (positive electrode):
\n2Cl− → Cl2 (g) + 2e−
\nOR
\nCl− → ½ Cl2 (g) + e− ✔
\nZnCl2 (l) → Zn (l) + Cl2 (g)
\n
balanced equation ✔
correct state symbols ✔
\n
Accept ionic equation.
\nThe half-equations were often incorrect. The average mark was 0.8 out of 2, and the correlation to high scoring candidates was strong as expected. Many candidates started the half-equations with the elements and gave the ions as products. We also saw some scripts with Cl instead of Cl2 as the product. Some of the candidates thought the zinc ion was Zn+ instead of Zn2+. Some candidates reversed the anode and cathode equations earning only 1 of the 2 marks.
\nThe performance was weak on this part-question as well. The overall equations did not balance atoms or charges on many of the incorrect answers. For the state symbols, many candidates used the aqueous state symbol, some gave the chloride ion a gaseous state symbol, and some candidates still had a solid zinc product even though they were directed to use the melting point of zinc in the data booklet. 12% of the candidates did not answer the question and the average mark was 0.5 out of 2 marks.
\nWhich statement is correct about the electrolysis of molten lead(II) bromide, PbBr2?
\n
A. Br− ions accept electrons at the cathode (negative electrode).
B. Pb2+ ions accept electrons at the anode (positive electrode).
\nC. Br− ions lose electrons at the anode (positive electrode).
\nD. Pb2+ ions lose electrons at the cathode (negative electrode).
\nC
\nEthanol and methanoic acid are important industrial products.
\nEthanol is used as a fuel.
\nWrite the chemical equation for the complete combustion of ethanol.
\nDeduce the change in enthalpy, ΔH, in kJ, when 56.00 g of ethanol is burned. Use section 13 in the data booklet.
\nOxidation of ethanol with potassium dichromate, K2Cr2O7, can form two different organic products. Determine the names of the organic products and the methods used to isolate them.
\nWrite the equation and name the organic product when ethanol reacts with methanoic acid.
\nSketch the titration curve of methanoic acid with sodium hydroxide, showing how you would determine methanoic acid pKa.
\nIdentify an indicator that could be used for the titration in 5(d)(i), using section 22 of the data booklet.
\nDetermine the concentration of methanoic acid in a solution of pH = 4.12. Use section 21 of the data booklet.
\nIdentify if aqueous solutions of the following salts are acidic, basic, or neutral.
\nCH3CH2OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (g) ✓
\n«n = =» 1.215 «mol» ✓
\n«1.215mol × (−1367 kJ mol−1) =» −1661 «kJ» ✓
\n\n
Award [2] for correct final answer.
\nAward [1 max] for “«+»1661 «kJ»”.
\nethanal AND distillation ✓
\nethanoic acid AND reflux «followed by distillation» ✓
\n
Award [1] for both products OR both methods.
Equation:
CH3CH2OH + HCOOH HCOOCH2CH3 + H2O ✓
Product name:
ethyl methanoate ✓
Accept equation without equilibrium arrows.
Accept equation with molecular formulas (C2H6O + CH2O2 C3H6O2 + H2O) only if product name is correct.
\nincreasing S-shape pH curve ✓
\npKa: pH at half neutralization/equivalence ✓
\n
M1: Titration curve must show buffer region at pH <7 and equivalence at pH >7.
Ignore other parts of the curve, i.e., before buffer region, etc.
\nAccept curve starting from where two axes meet as pH scale is not specified.
\nphenolphthalein
OR
phenol red ✓
Alternative 1:
Ka =
OR
[HCOOH] = ✓
«[HCOOH] =» 3.24 × 10−5 «mol dm−3» ✓
\n\n
Alternative 2:
«pH = pKa + log »
4.12 = 3.75 + log ✓
«[HCOOH] =» 3.24 × 10−5 «mol dm−3» ✓
\n\n
Award [2] for correct final answer.
\nSodium methanoate: basic
\nAmmonium chloride: acidic
\nSodium nitrate: neutral ✓ ✓
\n
Award [2] for three correct.
Award [1] for two correct.
\nStandard electrode potential values, E⦵, can be used to predict spontaneity.
\nIron(II) is oxidized by bromine.
\n2Fe2+ (aq) + Br2 (l) 2Fe3+ (aq) + 2Br− (aq)
\nCalculate the E⦵cell, in V, for the reaction using section 24 of the data booklet.
\nDetermine, giving a reason, if iodine will also oxidize iron(II).
\nMolten zinc chloride undergoes electrolysis in an electrolytic cell at 450 °C.
\nDeduce the half-equations for the reaction at each electrode.
\n\n
Deduce the overall cell reaction including state symbols. Use section 7 of the data booklet.
\n« = 1.09 – 0.77 =» 0.32 «V» ✔
\n«2Fe2+ (aq) + 2 (s) → 2Fe3+ (aq) + 2– (aq) »
\nno/non-spontaneous AND «= 0.54 – 0.77 »= –0.23 «V»/
OR
no AND reduction potential of 2 lower «than Fe3+ »/ 0.54 <0.77 ✔
\n
Accept “standard electrode potential of 2 lower /less positive than iron”.
\nCathode (negative electrode):
\nZn2+ + 2e− → Zn (l) ✔
\n\n
Anode (positive electrode):
\n2Cl− → Cl2 (g) + 2e−
OR
Cl− → ½ Cl2 (g) + e− ✔
ZnCl2 (l) → Zn (l) + Cl2 (g)
\nbalanced equation ✔
\ncorrect state symbols ✔
\n\n
Accept ionic equation.
\nOnly 50% got this straightforward calculation right, the most common error being to multiply both E0 values by 2, reflecting a lack of practice with this type of exercises.
\nOnly 10% were able to correctly justify the feasibility of the reaction with I2; the MS showed the best answer using the E(v) values but also allowed simpler explanations referring to E0 of iron; even then many candidates wrote Fe+2 instead of Fe+3, understandably perhaps as this was mentioned in the question. However, it also revealed some difficulty in using and understanding data from the E0 table in the data booklet.
\n3(bi)/(bii) Answers to both these questions revealed that many candidates struggle to conceptualize the equations that describe electrolysis. The question asked for products of the easiest case of electrolysis, a molten salt. However, many candidates proposed oxidation or reduction equations at both electrodes, or Zn and Cl2 (with no charge) as the initial species rather than the product; the average mark was 1.2/2 as only 55% answered correctly.
\n3(bi)/(bii) Answers to both these questions revealed that many candidates struggle to conceptualize the equations that describe electrolysis. The question asked for products of the easiest case of electrolysis, a molten salt. However, many candidates proposed oxidation or reduction equations at both electrodes, or Zn and Cl2 (with no charge) as the initial species rather than the product; the average mark was 1.2/2 as only 55% answered correctly.
\nThe determination of the states proved to be even more difficult, with many stating the ions were aqueous in spite of the fact that the question is clearly about molten zinc chloride. Allowing ECF for the overall equation allowed marks for many candidates, but very few realised that both ionic species in ZnCl2 were actually liquid (being a molten salt). As a result, correct answers were below 45% and the average mark was 0.9/2.
\nConsider this voltaic cell, where Cu is a more reactive metal than Ag:
\nWhich combination describes the movement of charge in this cell?
\nD
\nWhat is the name of this substance using IUPAC rules?
\nA. 2-ethyl-1-methylbutan-1-ol
\nB. 1-methyl-2-ethylbutan-1-ol
\nC. 3-ethylpentan-2-ol
\nD. 3-ethylpentan-4-ol
\nC
\nWhich pair of compounds are structural isomers?
\n
A. Propane and propene
B. Propanal and propanone
\nC. Propan-1-ol and propanal
\nD. Propyl propanoate and propanoic acid
\nB
\nWhat is the slope of the graph?
\n
A. −0.0025 mol dm−3 s−1
B. −0.0025 mol dm−3 s
\nC. −0.0033 mol dm−3 s−1
\nD. −0.0033 mol dm−3 s
\nA
\nWhat is the general formula of alkynes?
\n
A. CnH2n+2
B. CnH2n
\nC. CnH2n−2
\nD. CnHn
\nC
\nThe rate equation for a reaction is:
\nrate = k[A][B]
\nWhich mechanism is consistent with this rate equation?
\n
A. 2A I Fast
I + B → P Slow
B. A + B I Fast
I + A → P Slow
C. A → I Slow
I + B → P Fast
D. B I Fast
I + A → P Slow
D
\nA reversible reaction has a reaction quotient, Q, of 4.5 and equilibrium constant, Kc, of 6.2.
\n2A (g) A2 (g)
\nWhich statement describes the reaction at this time?
\n
A. The system has reached equilibrium.
B. The rate of the forward reaction is greater than the rate of the reverse reaction.
\nC. The concentration of reactant is greater than the concentration of product.
\nD. At equilibrium, the concentration of reactant is greater than the concentration of product.
\nB
\nWhich graph shows the relationship between the pressure and volume of a sample of gas at constant temperature?
\nB
\nThe graph shows Gibbs free energy of a mixture of N2O4 (g) and NO2 (g) in different proportions.
\nN2O4 (g) 2NO2 (g)
\nWhich point shows the system at equilibrium?
\nB
\nWhat is the pH of 0.01 mol dm−3 KOH (aq)?
\n
A. 1.0
B. 2.0
\nC. 12.0
\nD. 13.0
\nC
\nHydrogen and iodine react to form hydrogen iodide.
\nH2 (g) + 2 (g) 2H (g)
\nThe following experimental data was obtained.
\nConsider the reaction of hydrogen with solid iodine.
\nH2 (g) + 2 (s) 2H (g) ΔH⦵ = +53.0 kJ mol−1
\nDeduce the order of reaction with respect to hydrogen.
\nDeduce the rate expression for the reaction.
\nCalculate the value of the rate constant stating its units.
\nState two conditions necessary for a successful collision between reactants.
\nState the equilibrium constant expression, Kc, for this reaction.
\nCalculate the entropy change of reaction, ΔS⦵, in J K−1 mol−1.
\nPredict, giving a reason, how the value of the ΔS⦵reaction would be affected if (g) were used as a reactant.
\nCalculate the Gibbs free energy change, ΔG⦵, in kJ mol−1, for the reaction at 298 K. Use section 1 of the data booklet.
\nCalculate the equilibrium constant, Kc, for this reaction at 298 K. Use your answer to (d)(iii) and sections 1 and 2 of the data booklet.
\n(If you did not obtain an answer to (d)(iii) use a value of 2.0 kJ mol−1, although this is not the correct answer).
\nfirst order ✔
\nRate=k [H2] [2]
\n✔
\nmol–1 dm3 s–1 ✔
\nE ≥ Ea AND appropriate «collision» geometry/correct orientation ✔
\n✔
\n«Δ = 2 × 206.6 – (130.6 + 116.1) =» 166.5 «J K–1 mol–1» ✔
\nΔ lower/less positive AND same number of moles of gas
\nOR
\nΔ lower/less positive AND a solid has less entropy than a gas ✔
\n«ΔG⦵ = 53.0 kJ mol–1 – (298K × 0.1665 kJ K–1 mol–1) =» 3.4 «kJ mol–1» ✔
\n«ln Kc= – (3.4 × 103 J mol–1 /8.31 J K–1 mol–1 × 298 K)» = –1.37 ✔
\n«Kc =» 0.25 ✔
\nAward [2] for “0.45” for the use of 2.0 kJ mol–1 for ΔG⦵.
\n4(a)(i)-(iii): Deduction of rate orders and rate expression were very well done overall, with occasional errors in the units of the rate constant, but clearly among the best answered questions.
\nGenerally well answered by all but very weak candidates. Some teachers thought this should be a 2-mark question but actually the marks were generally missed when students mentioned both required conditions but failed to refer the necessary energy to Ea.
\nOne of the best answered questions.
\nΔS was well calculated in general except for some inverted calculations or failure to consider the ratios of the reactants.
Some candidates confused the entropy change in this situation with absolute entropy of a solid and gas, or having realised that entropy would decrease lacked clarity in their explanations and lost the mark.
\n4(d)(ii)-(d)(iv): marks were lost due to inconsistency of units throughout, i.e., not because answers were given in different units to those required, but because candidates failed to convert all data to the same unit for calculations.
\nWhat can be deduced from the mass spectrum of CH3COCH2CH2CH3?
\nNIST Mass Spectrometry Data Center Collection (C) 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved. 2-Pentanone Mass Spectrum, MS Number 291264. [graph] Available at: https://webbook.nist.gov/cgi/cbook.cgi?ID=C107879&Units=SI&Mask=200#Mass-Spec2-pentanone [Accessed 4 May 2020]. source adapted.
\n
A. The molar mass is 43 g mol−1.
B. The atoms have many isotopes.
\nC. The most likely bond to break is C–C between carbons 2 and 3.
\nD. The signal with the largest mass is due to the oxidation of the ketone in the spectrometer.
\nC
\n1-chloropentane reacts with aqueous sodium hydroxide.
\nThe reaction was repeated at a lower temperature.
\nIdentify the type of reaction.
\nOutline the role of the hydroxide ion in this reaction.
\nSuggest, with a reason, why 1-iodopentane reacts faster than 1-chloropentane under the same conditions. Use section 11 of the data booklet for consistency.
\nSketch labelled Maxwell–Boltzmann energy distribution curves at the original temperature (T1) and the new lower temperature (T2).
\nExplain the effect of lowering the temperature on the rate of the reaction.
\n«nucleophilic» substitution/SN2 ✔
\n
Do not accept if “electrophilic” or “free radical” substitution is stated.
«acts as a» nucleophile/Lewis base
OR
donates/provides lone pair «of electrons»
OR
attacks the «partially» positive carbon ✔
bond enthalpy C–I lower than C–Cl
OR
C–I bond weaker than C–Cl ✔
«weaker bond» broken more easily/with less energy
OR
lower Ea «for weaker bonds» ✔
Accept the bond enthalpy values for C–I and C–Cl for M1.
peak at T1 to right of AND lower than T2 ✔
\nlines begin at origin AND T1 must finish above T2 ✔
\n«rate is» lower AND «average» kinetic energy of molecules is lower
OR
«rate is» lower AND less frequent collisions
OR
«rate is» lower AND fewer collisions per unit time ✔
«rate is» lower AND fewer/smaller fraction of molecules/collisions have the E ≥ Ea ✔
\n
Lower «rate» needs to be mentioned once only.
Do not accept “fewer collisions” without reference to time/frequency/probability for M1.
\nHow much ethanol contains 1.20 × 1024 atoms of carbon?
\nAvogadro’s constant, L or NA: 6.02 × 1023 mol−1
\n
A. 0.333 mol
B. 0.500 mol
\nC. 1.00 mol
\nD. 2.00 mol
\nC
\nWhat is a possible value of pH at the equivalence point in the titration of a strong acid with a weak base?
\n
A. 5
B. 7
\nC. 9
\nD. 11
\nA
\nPhosphoric acid, H3PO4, can undergo stepwise neutralization, forming amphiprotic species.
\nFormulate an equation for the reaction of one mole of phosphoric acid with one mole of sodium hydroxide.
\nFormulate two equations to show the amphiprotic nature of H2PO4−.
\nCalculate the concentration of H3PO4 if 25.00 cm3 is completely neutralised by the addition of 28.40 cm3 of 0.5000 mol dm−3 NaOH.
\nOutline the reason that sodium hydroxide is considered a Brønsted–Lowry base.
\nH3PO4 (aq) + NaOH (aq) → NaH2PO4 (aq) + H2O (l) ✔
\n
Accept net ionic equation.
H2PO4− (aq) + H+ (aq) → H3PO4 (aq) ✔
\nH2PO4− (aq) + OH− (aq) → HPO42− (aq) + H2O (l) ✔
\n
Accept reactions of H2PO4− with any acidic, basic or amphiprotic species, such as H3O+, NH3 or H2O.
Accept H2PO4− (aq) → HPO42− (aq) + H+ (aq) for M2.
\n«NaOH »
\n«» 0.004733 «mol» ✔
\n«» 0.1893 «mol dm−3» ✔
\n
Award [2] for correct final answer.
«OH− is a» proton acceptor ✔
\n3.00 mol of C3H8 is mixed with 20.00 mol of O2. Which quantity is present at the end of the reaction?
\nC3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)
\n
A. 1.00 mol of C3H8
B. 5.00 mol of O2
\nC. 12.00 mol of CO2
\nD. 16.00 mol of H2O
\nB
\nBromate and bromide ions react in acidic aqueous solution.
\nBrO3− (aq) + 5Br− (aq) + 6H+ (aq) → 3Br2 (l) + 3H2O (l)
\nThe following rate data was collected.
\nDetermine the rate expression for the reaction.
\nDetermine the value and unit of the rate constant using the rate expression in (a).
\nBrO3–: 1/first AND Br–: 1/first AND H+: 2/second ✓
\n«Rate =» k[BrO3−][Br−][H+]2 ✓
\n
M2: Square brackets required for the mark.
«k ==» 8.0 ✓
\nmol−3 dm9 s−1 ✓
\nWhat is correct for pure hot water?
B
\nConsider the following standard electrode potentials:
\nWhich species will react with each other spontaneously under standard conditions?
\n
A. Zn2+ (aq) + Pb (s)
B. Pb2+ (aq) + Br2 (l)
\nC. Zn (s) + Br− (aq)
\nD. Pb (s) + Br2 (l)
\nD
\nWhich species has two more neutrons than electrons?
\n\n
A.
B.
\nC.
\nD.
\nC
\nWhich statement explains why the second ionization energy of aluminium is higher than the first ionization energy of magnesium?
\n
A. Ionization energy increases along period 3.
B. 3p electrons are at a higher energy level than 3s electrons.
\nC. 3p electrons are further away from the nucleus than 2p electrons.
\nD. Both have the same number of electrons and aluminium has one more proton.
\nD
\nConsider the following equilibrium reaction:
\n2SO2 (g) + O2 (g) 2SO3 (g)
\nState the equilibrium constant expression, Kc, for the reaction above.
\nState and explain how the equilibrium would be affected by increasing the volume of the reaction container at a constant temperature.
\nSO2 (g), O2 (g) and SO3 (g) are mixed and allowed to reach equilibrium at 600 °C.
\nDetermine the value of Kc at 600 °C.
\n«Kc = » ✓
\n
Square brackets required for the mark.
pressure decrease «due to larger volume» ✓
\nreaction shifts to side with more moles/molecules «of gas» ✓
\nreaction shifts left/towards reactants ✓
\n
Award M3 only if M1 OR M2 awarded.
[O2] = 1.25 «mol dm−3» AND [SO3] = 3.50 «mol dm−3» ✓
\n«Kc ==» 4.36 ✓
\n\n
Award [2] for correct final answer
\nConsider the Born–Haber cycle for the formation of sodium oxide:
\nWhat is the lattice enthalpy, in kJ mol−1, of sodium oxide?
\n
A. 414 + 2(108) + 249 + 2(496) − 141 + 790
B. 414 + 2(108) + 249 + 2(496) + 141 + 790
\nC. −414 + 2(108) + 249 + 2(496) − 141 + 790
\nD. −414 − 2(108) − 249 − 2(496) + 141 − 790
\nA
\nWhich ion has the largest radius?
\n
A. Na+
B. Mg2+
\nC. P3−
\nD. S2−
\nC
\nWhich graph shows a first order reaction?
\nA
\nWhite phosphorus is an allotrope of phosphorus and exists as P4.
\nAn equilibrium exists between PCl3 and PCl5.
\nPCl3 (g) + Cl2 (g) PCl5 (g)
\nSketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
\n\n
Write an equation for the reaction of white phosphorus (P4) with chlorine gas to form phosphorus trichloride (PCl3).
\nDeduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl bond angle in PCl3.
\nOutline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
\nCalculate the standard enthalpy change (ΔH⦵) for the forward reaction in kJ mol−1.
\nΔH⦵f PCl3 (g) = −306.4 kJ mol−1
\nΔH⦵f PCl5 (g) = −398.9 kJ mol−1
\nCalculate the entropy change, ΔS, in J K−1 mol−1, for this reaction.
\n\n
Chemistry 2e, Chpt. 21 Nuclear Chemistry, Appendix G: Standard Thermodynamic Properties for Selected Substances https://openstax.org/books/chemistry-2e/pages/g-standard-thermodynamic-properties-for- selectedsubstances# page_667adccf-f900-4d86-a13d-409c014086ea © 1999-2021, Rice University. Except where otherwise noted, textbooks on this site are licensed under a Creative Commons Attribution 4.0 International License. (CC BY 4.0) https://creativecommons.org/licenses/by/4.0/.
\nCalculate the Gibbs free energy change (ΔG), in kJ mol−1, for this reaction at 25 °C. Use section 1 of the data booklet.
\nIf you did not obtain an answer in c(i) or c(ii) use −87.6 kJ mol−1 and −150.5 J mol−1 K−1 respectively, but these are not the correct answers.
\nDetermine the equilibrium constant, K, for this reaction at 25 °C, referring to section 1 of the data booklet.
\nIf you did not obtain an answer in (c)(iii), use ΔG = –43.5 kJ mol−1, but this is not the correct answer.
\nState the equilibrium constant expression, Kc, for this reaction.
\nState, with a reason, the effect of an increase in temperature on the position of this equilibrium.
\nAccept any diagram with each P joined to the other three.
Accept any combination of dots, crosses and lines.
P4 (s) + 6Cl2 (g) → 4PCl3 (l) ✔
\nElectron domain geometry: tetrahedral ✔
\nMolecular geometry: trigonal pyramidal ✔
\nBond angle: 100«°» ✔
\n
Accept any value or range within the range 91−108«°» for M3.
PCl5 is non-polar:
\nsymmetrical
OR
dipoles cancel ✔
\n
PCl4F is polar:
\nP–Cl has a different bond polarity than P–F ✔
\nnon-symmetrical «dipoles»
OR
dipoles do not cancel ✔
Accept F more electronegative than/different electronegativity to Cl for M2.
«−398.9 kJ mol−1 − (−306.4 kJ mol−1) =» −92.5 «kJ mol−1» ✔
\n«ΔS = 364.5 J K–1 mol–1 – (311.7 J K–1 mol–1 + 223.0 J K–1 mol–1)=» –170.2 «J K–1 mol–1» ✔
\n«ΔS =» –0.1702 «kJ mol–1 K–1»
OR
298 «K» ✔
«ΔG = –92.5 kJ mol–1 – (298 K × –0.1702 kJ mol–1 K–1) =» –41.8 «kJ mol–1» ✔
\n\n
Award [2] for correct final answer.
\nIf –87.6 and -150.5 are used then –42.8.
\n«ΔG = –41.8 kJ mol–1 = × 298 K × lnK»
OR
«ΔG = –41800 J mol–1 = –8.31 J mol–1 K–1 × 298 K × lnK»
«lnK = =» 16.9 ✔
«K = e16.9 =» 2.19 × 107 ✔
\n\n
Award [2] for correct final answer.
\nAccept range of 1.80 × 106–2.60 × 107.
\nIf –43.5 is used then 4.25 × 107.
\n«Kc =» ✔
\n«shifts» left/towards reactants AND «forward reaction is» exothermic/ΔH is negative ✔
\nWhat is the molecular formula of a compound with an empirical formula of CHO2 and a relative molecular mass of 90?
\nA. CHO2
\nB. C2H2O4
\nC. C3H6O3
\nD. C4H10O2
\nB
\nWhich complex ion contains a central ion with an oxidation state of +3?
\n
A. [PtCl6]2−
B. [Cu(H2O)4(OH)2]
\nC. [Ni(NH3)4(H2O)2]2+
\nD. [Co(NH3)4Cl2]+
\nD
\nIn which of the following situations is the forward reaction spontaneous?
\n
A. The equilibrium constant is greater than one under standard conditions.
B. The cell potential is negative.
\nC. The Gibbs free energy change of the reverse reaction is negative.
\nD. The entropy change of the universe for the forward reaction is negative.
\nA
\nWhich aqueous solutions produce oxygen gas during electrolysis?
\nI. Dilute CuCl2 (aq) with inert electrodes
II. Dilute FeSO4 (aq) with inert electrodes
III. Dilute CuCl2 (aq) with copper electrodes
The standard electrode potentials are provided in the table:
\n
A. I and II only
B. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nWhich molecules contain two pi () bonds?
\nI. HCN
II. H2CO3
III. H2C2O4
A. I and II only
B. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\n8.8 g of an oxide of nitrogen contains 3.2 g of oxygen. What is the empirical formula of the compound?
\nA. N2O5
\nB. N2O
\nC. NO2
\nD. NO
\nB
\nWhat is the number of hydrogen atoms in 2.00 moles of Ca(HCO3)2?
\nAvogadro’s constant, L or NA: 6.02 × 1023 mol−1
\nA. 2.00
\nB. 4.00
\nC. 1.20 × 1024
\nD. 2.41 × 1024
\nD
\nNaturally occurring gallium consists of the isotopes 71Ga and 69Ga. What is the approximate percentage abundance of 69Ga?
\nMr (Ga) = 69.72.
\n
A. 40 %
B. 50 %
\nC. 60 %
\nD. 75 %
\nC
\nWhat is the hybridization of nitrogen and chlorine in NCl3?
\nD
\nWhich statement is correct about configurational isomers?
\n
A. Configurational isomers can only be interconverted by breaking and reforming bonds.
B. Configurational isomers have different molecular formulas but the same structural formulas.
\nC. Configurational isomers are not distinct compounds.
\nD. Configurational isomers always have identical physical properties.
\nA
\nWhich product is formed when CH3COCH2CH3 is reduced with sodium borohydride?
\n
A. CH3CH2CH2CHO
B. CH3CH2CH2CH2OH
\nC. CH3CH(OH)CH2CH3
\nD. CH3CH2CH2COOH
\nC
\nA student was investigating rates of reaction. In which of the following cases would a colorimeter show a change in absorbance?
\nA. KBr (aq) + Cl2 (aq)
\nB. Cu (s) + Na2SO4 (aq)
\nC. HCl (aq) + NaOH (aq)
\nD. (CH3)3COH (aq) + K2Cr2O7 (aq)
\nA
\nQuestion 17 was the question about the colorimeter), which many teachers felt was unfair, and only 22% of students gave the correct answer. The complaints about the question were often based on the lack of familiarity with the colorimeter. However, Chemistry is a practical course and student should be exposed to a piece of equipment such as this. Also, it was felt that there was a lot to process in this question. Although answer C could fairly easily be eliminated as there is no colour change. Students then had to recognize that in D, alcohol oxidation would result in a colour change from orange to green, except that it is a tertiary alcohol. This left A and B. A is of course the right answer as chlorine replaces bromine, but if B actually happened, there would also be a colour change from clear to blue. So, students needed to remember that copper is below sodium on the activity series.
\nWhich statement describes all homogeneous mixtures?
\n
A. Any sample has the same ratio of the components.
B. The components are covalently bonded together.
\nC. The components cannot be easily separated.
\nD. The mixture needs a specific ratio of components to form.
\nA
\nWhat is the maximum number of electrons that can occupy a p-orbital?
\nA. 2
\nB. 3
\nC. 6
\nD. 8
\nA
\nQuestion 6 was poorly answered as it asked for the number of electrons in a p orbital. Very few students gave the correct answer of 2. The majority chose answer C (6) for the maximum number of electrons that can occupy a p-orbital, rather than A (2). It appears candidates reflexively conflated p-orbitals with the entire p subshell in any given period.
\nBiochemical oxygen demand (BOD) can be determined by the Winkler Method.
\nA 25.00 cm3 sample of water was treated according to the Winkler Method.
\nStep I: 2Mn2+ (aq) + O2 (g) + 4OH− (aq) → 2MnO2 (s) + 2H2O (l)
\nStep II: MnO2 (s) + 2I− (aq) + 4H+ (aq) → Mn2+ (aq) + I2 (aq) + 2H2O (l)
\nStep III: 2S2O32− (aq) + I2 (aq) → 2I− (aq) + S4O62− (aq)
\nThe iodine produced was titrated with 37.50 cm3 of 5.000 × 10−4 mol dm−3 Na2S2O3.
\nOutline what is measured by BOD.
\nA student dissolved 0.1240 ± 0.0001 g of Na2S2O3 to make 1000.0 ± 0.4 cm3 of solution to use in the Winkler Method.
\nDetermine the percentage uncertainty in the molar concentration.
\nCalculate the amount, in moles of Na2S2O3 used in the titration.
\nDeduce the mole ratio of O2 consumed in step I to S2O32− used in step III.
\nCalculate the concentration of dissolved oxygen, in mol dm−3, in the sample.
\nThe three steps of the Winkler Method are redox reactions.
\nDeduce the reduction half-equation for step II.
\n«amount of» oxygen used to decompose the organic matter in water ✔
\n«» 0.08 «%»
OR
«» 0.04 «%» ✔
«0.08 % + 0.04 % =» 0.12/0.1 «%» ✔
Award [2] for correct final answer.
Accept fractional uncertainties for M1, i.e., 0.0008 OR 0.0004.
\n«× 5.000 × 10−4 mol dm−3 =» 1.875 × 10−5 «mol» ✔
\n1:4 ✔
\nAccept “4 mol S2O32– :1 mol O2“, but not just 4:1.
\n«» 4.688 × 10−6 «mol» ✔
\n«» 1.875 × 10−4 «mol dm−3» ✔
\n
Award [2] for correct final answer.
MnO2 (s) + 2e− + 4H+ (aq) → Mn2+ (aq) + 2H2O (l) ✔
\nWhich attacking species is matched with its mechanism of reaction?
\nD
\nWhich gases are acidic?
\nI. nitrogen dioxide
II. carbon dioxide
III. sulfur dioxide
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nWhich combination is correct?
\nB
\nPhosphoric acid, H3PO4 can form three different salts depending on the extent of neutralisation by sodium hydroxide.
\nFormulate an equation for the reaction of one mole of phosphoric acid with one mole of sodium hydroxide.
\nFormulate two equations to show the amphiprotic nature of H2PO4−.
\nCalculate the concentration of H3PO4 if 25.00 cm3 is completely neutralised by the addition of 28.40 cm3 of 0.5000 mol dm−3 NaOH.
\nOutline the reasons that sodium hydroxide is considered a Brønsted–Lowry and Lewis base.
\nH3PO4 (aq) + NaOH (aq) → NaH2PO4 (aq) + H2O (l) ✔
\n
Accept net ionic equation.
H2PO4− (aq) + H+ (aq) → H3PO4 (aq) ✔
\nH2PO4− (aq) + OH− (aq) → HPO42− (aq) + H2O (l) ✔
\n
Accept reactions of H2PO4− with any acidic, basic or amphiprotic species, such as H3O+, NH3 or H2O.
Accept H2PO4− (aq) → HPO42− (aq) + H+ (aq) for M2.
\n«»
\n«» 0.004733 «mol» ✔
\n«» 0.1893 «mol dm−3» ✔
\n\n
Award [2] for correct final answer.
\nBrønsted–Lowry base:
proton acceptor
AND
\nLewis Base:
e– pair donor/nucleophile ✔
Which of the following is the electron configuration of a metallic element?
\nA. [Ne] 3s2 3p2
\nB. [Ne] 3s2 3p4
\nC. [Ne] 3s2 3p6 3d3 4s2
\nD. [Ne] 3s2 3p6 3d10 4s2 4p5
\nC
\nThe two containers shown are connected by a valve. What is the total pressure after the valve is opened and the two gas samples are allowed to mix at constant temperature?
\n
A. 1.5 × 105 Pa
B. 2.3 × 105 Pa
\nC. 2.5 × 105 Pa
\nD. 5.0 × 105 Pa
\nB
\nWhich substance has the following 1H NMR spectrum?
\nSDBS, National Institute of Advanced Science and Technology.
\n
A. Propane
B. Propanal
\nC. Butanoic acid
\nD. Ethyl ethanoate
\nD
\nConsider the mass spectrum of an element:
\nWhat is the relative atomic mass of this element?
\n
A. 10.2
B. 10.5
\nC. 10.8
\nD. 10.9
\nC
\nA compound consists of the ions Ca2+ and PO43–. What are the name and formula of the compound?
\n
D
\nBiochemical oxygen demand (BOD) can be determined by the Winkler Method.
\nA 25.00 cm3 sample of water was treated according to the Winkler Method.
\nStep I: 2Mn2+ (aq) + O2 (g) + 4OH− (aq) → 2MnO2 (s) + 2H2O (l)
\nStep II: MnO2 (s) + 2I− (aq) + 4H+ (aq) → Mn2+ (aq) + I2 (aq) + 2H2O (l)
\nStep III: 2S2O32− (aq) + I2 (aq) → 2I− (aq) + S4O62− (aq)
\nThe iodine produced was titrated with 37.50 cm3 of 5.000 × 10−4 mol dm−3 Na2S2O3.
\nOutline what is measured by BOD.
\nA student dissolved 0.1240 ± 0.0001 g of Na2S2O3 to make 1000.0 ± 0.4 cm3 of solution to use in the Winkler Method.
\nDetermine the percentage uncertainty in the molar concentration.
\nCalculate the amount, in moles of Na2S2O3 used in the titration.
\nDeduce the mole ratio of O2 consumed in step I to S2O32− used in step III.
\nCalculate the concentration of dissolved oxygen, in mol dm−3, in the sample.
\nThe three steps of the Winkler Method are redox reactions.
\nDeduce the reduction half-equation for step II.
\nSuggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.
\n«amount of» oxygen used to decompose the organic matter in water ✔
\n«» 0.08 «%»
OR
«» 0.04 «%» ✔
«0.08 % + 0.04 % =» 0.12/0.1 «%» ✔
\n
Award [2] for correct final answer.
\nAccept fractional uncertainties for M1, i.e., 0.0008 OR 0.0004.
\n«× 5.000 × 10−4 mol dm−3 =» 1.875 × 10−5 «mol» ✔
\n1:4 ✔
\n
Accept “4 mol S2O32– :1 mol O2“, but not just 4:1.
«» 4.688 × 10−6 «mol» ✔
\n«» 1.875 × 10−4 «mol dm−3» ✔
\n
Award [2] for correct final answer.
MnO2 (s) + 2e− + 4H+ (aq) → Mn2+ (aq) + 2H2O (l) ✔
\nrate of reaction of oxygen with impurities depends on temperature
OR
rate at which bacteria/organisms grow/respire depends on temperature ✔
A 4.406 g sample of a compound containing only C, H and O was burnt in excess oxygen. 8.802 g of CO2 and 3.604 g of H2O were produced.
\nThe following spectrums show the Infrared spectra of propan-1-ol, propanal and propanoic acid.
\nNIST Mass Spectrometry Data Center Collection © 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved. Available at: https://webbook.nist.gov/cgi/cbook.cgi?ID=C71238&Units=SI&Type=IRSPEC&Index=3#IR-SPEC [Accessed 6 May 2020]. Source adapted.
\nNIST Mass Spectrometry Data Center Collection © 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. Available at: https://webbook.nist.gov/cgi/cbook.cgi?ID=C79094&Units=SI&Mask=80#IR-Spec [Accessed 6 May 2020]. Source adapted.
\nNIST Mass Spectrometry Data Center Collection © 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. Available at: https://webbook.nist.gov/cgi/cbook.cgi?Name=propanal&Units=SI&cIR=on&cTZ=on#IRSpec [Accessed 6 May 2020]. Source adapted.
\nDetermine the empirical formula of the compound using section 6 of the data booklet.
\nDetermine the molecular formula of this compound if its molar mass is 88.12 g mol−1. If you did not obtain an answer in (a) use CS, but this is not the correct answer.
\nIdentify each compound from the spectra given, use absorptions from the range of 1700 cm−1 to 3500 cm−1. Explain the reason for your choice, referring to section 26 of the data booklet.
\n«» 0.2000 «mol of C/CO2»
\nAND «» 0.2000 «mol of H2O»/0.4000 «mol of H»
\nOR
\n«» 2.402 «g of C»
\nOR
\n«» 0.404 «g of H» ✔
\n\n
«4.406 g − 2.806 g» = 1.600 «g of O» ✔
\n«»
\nC2H4O ✔
\n\n
Award [3] for correct final answer.
\n«» C4H8O2 ✔
\n\n
C2S2 if CS used.
\n\n
Award [1 max] for correctly identifying all 3 compounds without valid reasons given.
\nAccept specific values of wavenumbers within each range.
\nCl2 (g) + (g) Cl (g) Kc = 454
\nWhat is the Kc value for the reaction below?
\n2 Cl (g) Cl2 (g) + (g)
\nA.
\nB.
\nC.
\nD.
\nD
\nQuestion 18 was identified as being mathematically challenging. However, as all answers were shown as calculations rather than as final answers, this does not seem a reasonable complaint.
\nAlkanes undergo combustion and substitution.
\nDetermine the molar enthalpy of combustion of an alkane if 8.75 × 10−4 moles are burned, raising the temperature of 20.0 g of water by 57.3 °C.
\nFormulate equations for the two propagation steps and one termination step in the formation of chloroethane from ethane.
\n«q = mcΔT = 20.0 g × 4.18 J g−1 °C−1 × 57.3 °C =» 4790 «J» ✔
\n«» –5470 «kJ mol–1» ✔
\n\n
Award [2] for correct final answer.
\nAccept answers in the range –5470 to –5480 «kJ mol−1».
\nAccept correct answer in any units, e.g. –5.47 «MJ mol−1» or 5.47 x 106 «J mol−1».
\nCl· + C2H6 → ·C2H5 + HCl ✔
\n·C2H5 + Cl2 → Cl· + C2H5Cl ✔
\n
·C2H5 + Cl· → C2H5Cl
OR
Cl· + Cl· → Cl2
OR
·C2H5 + ·C2H5 → C4H10 ✔
Do not penalize incorrectly placed radical sign, eg C2H5·.
What is the explanation for the high melting point of sodium chloride?
\nA. The covalent bond between sodium and chlorine atoms is strong.
\nB. Electrostatic attraction between sodium and chloride ions is strong.
\nC. Intermolecular forces in sodium chloride are strong.
\nD. Delocalized electrons cause strong bonding in sodium chloride.
\nB
\nHow many p-orbitals are occupied in a phosphorus atom?
\n
A. 2
B. 3
\nC. 5
\nD. 6
\nD
\nWhich species are acids in the equilibrium below?
\nCH3NH2 + H2O CH3NH3+ + OH–
\nA. CH3NH2 and H2O
\nB. H2O and CH3NH3+
\nC. H2O and OH–
\nD. CH3NH2 and CH3NH3+
\nB
\nWhich compound produces this mass spectrum?
\n[Spectral Database for Organic Compounds, SDBS. SDBS Compounds and Spectral Search. [graph] Available at:
https://sdbs.db.aist.go.jp [Accessed 3 January 2019].]
\n
C
\nQuestion 29 contained a mass spectrum and 44% of students chose the correct answer of C - ethanoic acid. There were comments that this was unfair without access to the data booklet. However, as the other molecules are clearly wrong this did not seem so difficult to do without the data booklet as they did have the periodic table.
\nFast moving helium nuclei (4He2+) were fired at a thin piece of gold foil with most passing undeflected but a few deviating largely from their path. The diagram illustrates this historic experiment.
\nFigure from PPLATO / FLAP (Flexible Learning Approach To Physics), http://www.met.reading.ac.uk/pplato2/h-flap/
phys8_1.html#top 1996 The Open University and The University of Reading.
Suggest what can be concluded about the gold atom from this experiment.
\nSubsequent experiments showed electrons existing in energy levels occupying various orbital shapes.
\nSketch diagrams of 1s, 2s and 2p.
\nState the electron configuration of copper.
\nMost 4He2+ passing straight through:
\nmost of the atom is empty space
OR
the space between nuclei is much larger than 4He2+ particles
OR
nucleus/centre is «very» small «compared to the size of the atom» ✔
\n
Very few 4He2+ deviating largely from their path:
\nnucleus/centre is positive «and repels 4He2+ particles»
OR
nucleus/centre is «more» dense/heavy «than 4He2+ particles and deflects them»
OR
nucleus/centre is «very» small «compared to the size of the atom» ✔
\n
Do not accept the same reason for both M1 and M2.
\nAccept “most of the atom is an electron cloud” for M1.
\nDo not accept only “nucleus repels 4He2+ particles” for M2.
\n1s AND 2s as spheres ✔
\none or more 2p orbital(s) as figure(s) of 8 shape(s) of any orientation (px, py pz) ✔
\n1s22s22p63s23p64s13d10
\nOR
\n[Ar] 4s13d10 ✔
\n\n
Accept configuration with 3d before 4s.
\nWhich 0.01 mol dm–3 aqueous solution has the highest pH?
\nA. HCl
\nB. H2SO4
\nC. NaOH
\nD. NH3
\nC
\nWhich element has the highest metallic character in Group 14?
\n
A. C
B. Si
\nC. Ge
\nD. Sn
\nD
\nWhich molecule is most polar?
\nA. CF4
\nB. CCl4
\nC. CHF3
\nD. CClF3
\nC
\nQuestion 11 was identified in the G2s as a tricky question, because students could have difficulty resolving the conflict between the greater polarity of the C-Cl bond compared with the C-H bond, and their contrast with the polarity of the C-F bond.) However, 56% of student did identify the correct answer of C.
\nIn which of the following species would sulfur be reduced if converted to SCl2?
\nA. S2O32–
\nB. H2S
\nC. S
\nD. SO2
\nD
\nWhich combination describes the acid–base nature of aluminium and phosphorus oxides?
\nA
\nWhich statement is correct for both voltaic and electrolytic cells?
\nA. The oxidation reaction releases electrons.
\nB. The oxidation reaction occurs at the positive electrode.
\nC. The cathode is negative.
\nD. Electrons flow through the electrolyte.
\nA
\nFor which species can resonance structures be drawn?
\nA. HCOOH
\nB. HCOO–
\nC. CH3OH
\nD. H2CO3
\nB
\nThe energy from burning 0.250 g of ethanol causes the temperature of 150 cm3 of water to rise by 10.5 °C. What is the enthalpy of combustion of ethanol, in kJ mol–1?
\nSpecific heat capacity of water: 4.18 J g–1 K–1.
A.
\nB.
\nC.
\nD.
\n\n
B
\nThe standard electrode potential of zinc can be measured using a standard hydrogen electrode (SHE).
\nDraw and annotate the diagram to show the complete apparatus required to measure the standard electrode potential of zinc.
\nH2 (g) entering at «298 K and» 100 kPa ✔
\nplatinum electrode on left ✔
\nvoltmeter connecting electrodes AND salt bridge connecting electrolytes ✔
\n1 mol dm–3 H+ on the left AND 1 mol dm–3 Zn2+ on the right ✔
\n\n
Voltmeter and salt bridge need to be drawn but not necessarily annotated for M3.
\nConcentrations, but not state symbols, required for M4.
\nWhich molecule has the weakest nitrogen to nitrogen bond?
\n
A. N2
B. N2H2
\nC. N2H4
\nD.
C
\nA 4.406 g sample of a compound containing only C, H and O was burnt in excess oxygen. 8.802 g of CO2 and 3.604 g of H2O were produced.
\nThe following spectrums show the Infrared spectra of propan-1-ol, propanal and propanoic acid.
\nNIST Mass Spectrometry Data Center Collection © 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved. Available at: https://webbook.nist.gov/cgi/cbook.cgi?ID=C71238&Units=SI&Type=IRSPEC&Index=3#IR-SPEC [Accessed 6 May 2020]. Source adapted.
\nNIST Mass Spectrometry Data Center Collection © 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. Available at: https://webbook.nist.gov/cgi/cbook.cgi?ID=C79094&Units=SI&Mask=80#IR-Spec [Accessed 6 May 2020]. Source adapted.
\nNIST Mass Spectrometry Data Center Collection © 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. Available at: https://webbook.nist.gov/cgi/cbook.cgi?Name=propanal&Units=SI&cIR=on&cTZ=on#IRSpec [Accessed 6 May 2020]. Source adapted.
\nDetermine the empirical formula of the compound using section 6 of the data booklet.
\nDetermine the molecular formula of this compound if its molar mass is 88.12 g mol−1. If you did not obtain an answer in (a) use CS, but this is not the correct answer.
\nIdentify each compound from the spectra given, use absorptions from the range of 1700 cm−1 to 3500 cm−1. Explain the reason for your choice, referring to section 26 of the data booklet.
\nPredict the number of 1H NMR signals, and splitting pattern of the –CH3 seen for propanone (CH3COCH3) and propanal (CH3CH2CHO).
\nPredict the fragment that is responsible for a m/z of 31 in the mass spectrum of propan‑1‑ol. Use section 28 of the data booklet.
\n«» 0.2000 «mol of C/CO2»
\nAND «» 0.2000 «mol of H2O»/0.4000 «mol of H»
\nOR
\n«» 2.402 «g of C»
\nOR
\n«» 0.404 «g of H» ✔
\n\n
«4.406 g − 2.806 g» = 1.600 «g of O» ✔
\n
«»
C2H4O ✔
\n\n
Award [3] for correct final answer.
\n«» C4H8O2 ✔
\n
C2S2 if CS used.
Award [1 max] for correctly identifying all 3 compounds without valid reasons given.
Accept specific values of wavenumbers within each range.
\nCH3O+ ✔
\n
Accept any structure i.e. “CH2OH+”.
How many electrons are needed when the following half-equation is balanced using the lowest possible whole numbers?
\n__ NO3– (aq) + __ H+ (aq) + __ e– → __ NO (g) + __ H2O (l)
\n
A. 1
B. 2
\nC. 3
\nD. 5
\nC
\nWhich combination would create the strongest ionic bond?
\nC
\nWhat is the index of hydrogen deficiency (IHD) of this molecule?
\nParacetamol (acetaminophen)
\nA. 3
\nB. 4
\nC. 5
\nD. 6
\nC
\nWhat is the enthalpy change of the following reaction?
\nCH2CHCH2CH3 + HBr → CH3CHBrCH2CH3
\nA. –119.6 kJ
\nB. +119.6 kJ
\nC. –119.8 kJ
\nD. +119.8 kJ
\nC
\nWhich functional groups are present in serine?
\nA. nitro, carbonyl and carboxyl
\nB. amino, hydroxyl and carbonyl
\nC. nitro, carboxyl and hydroxyl
\nD. amino, carboxyl and hydroxyl
\nD
\nWhich compound contains both ionic and covalent bonds?
\n
A. CH3COONa
B. CH3COOH
\nC. K2O
\nD. CaCl2
\nA
\nThe following compounds have similar relative molecular masses. What is the order of increasing boiling point?
\n
A. CH3CH2CH2OH < CH3CH2CHO < CH3COOH
B. CH3CH2CHO < CH3CH2CH2OH < CH3COOH
\nC. CH3CH2CHO < CH3COOH < CH3CH2CH2OH
\nD. CH3COOH < CH3CH2CHO < CH3CH2CH2OH
\nB
\nWhat is the correct interpretation of the following potential energy profile?
\n
A. Endothermic reaction; products more stable than reactants.
B. Exothermic reaction; products more stable than reactants.
\nC. Endothermic reaction; products less stable than reactants.
\nD. Exothermic reaction; products less stable than reactants.
\n\n
C
\nWhich compounds are members of the same homologous series?
\nA. propanal, propanone, propanoic acid
\nB. propane, propene, propyne
\nC. hexan-1-ol, hexan-2-ol, hexan-3-ol
\nD. ethanol, propan-1-ol, butan-1-ol
\nD
\nWhich alcohol is least soluble in water?
\n
A. CH3OH
B. CH3CH2OH
\nC. CH3CH2CH2OH
\nD. CH3CH2CH2CH2OH
\nD
\nA sample of calcium carbonate reacts with excess hydrochloric acid in a beaker. The solid line shows how the mass of the beaker changes with time.
\nWhich dashed line represents the results obtained when the acid concentration is doubled?
\nB
\nWhich reagents and conditions are best for converting propan-1-ol into propanoic acid?
\nA. Reflux with acidified potassium dichromate (VI)
\nB. Reflux with aqueous sodium hydroxide
\nC. Distil with acidified potassium dichromate (VI)
\nD. Distil with aqueous sodium hydroxide
\nA
\nExplain the general increase in trend in the first ionization energies of the period 3 elements, Na to Ar.
\nincreasing number of protons
\nOR
\nincreasing nuclear charge ✔
\n\n
«atomic» radius/size decreases
\nOR
\nsame number of shells/electrons occupy same shell
\nOR
\nsimilar shielding «by inner electrons» ✔
\nWhich combustion reaction releases the least energy per mole of C3H8?
\nApproximate bond enthalpy / kJ mol−1
O=O 500
C=O 800
C≡O 1000
A. C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)
B. C3H8 (g) + O2 (g) → 2CO2 (g) + CO (g) + 4H2O (g)
\nC. C3H8 (g) + 4O2 (g) → CO2 (g) + 2CO (g) + 4H2O (g)
\nD. C3H8 (g) + O2 (g) → 3CO (g) + 4H2O (g)
\n\n
Chemistry: Atoms First 2e, https://openstax.org/books/chemistry-atoms-first-2e/pages/9-4-strengths-of-ionic-andcovalent-bonds © 1999–2021, Rice University. Except where otherwise noted, textbooks on this site are licensed under a Creative Commons Attribution 4.0 International License.
(CC BY 4.0) https://creativecommons.org/licenses/ by/4.0/.
D
\n2.67 g of lead (II) carbonate is decomposed by heating until constant mass.
\nPbCO3 (s) → PbO (s) + CO2 (g)
\nWhat is the final mass of solid?
\nA. 0.44 g
\nB. 2.23 g
\nC. 2.67 g
\nD. 3.11 g
\nB
\n76% of candidates correctly identified final mass of solid in the decomposition reaction with little difficulty in using mental math to arrive at the answer.
\nWhat is produced when chlorobutane is treated with aqueous sodium hydroxide solution?
\nA. butane
\nB. butanoic acid
\nC. butanal
\nD. butan-1-ol
\nD
\nWhite phosphorus is an allotrope of phosphorus and exists as P4.
\nAn equilibrium exists between PCl3 and PCl5.
\nPCl3 (g) + Cl2 (g) PCl5 (g)
\nSketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
\n\n
Write an equation for the reaction of white phosphorus (P4) with chlorine gas to form phosphorus trichloride (PCl3).
\nDeduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl bond angle in PCl3.
\nExplain the polarity of PCl3.
\nCalculate the standard enthalpy change (ΔH⦵) for the forward reaction in kJ mol−1.
\nΔH⦵f PCl3 (g) = −306.4 kJ mol−1
\nΔH⦵f PCl5 (g) = −398.9 kJ mol−1
\nState the equilibrium constant expression, Kc, for this reaction.
\nState, with a reason, the effect of an increase in temperature on the position of this equilibrium.
\n
Accept any diagram with each P joined to the other three.
Accept any combination of dots, crosses and lines.
\nP4 (s) + 6Cl2 (g) → 4PCl3 (l) ✔
\nElectron domain geometry: tetrahedral ✔
\nMolecular geometry: trigonal pyramidal ✔
\nBond angle: 100«°» ✔
\n\n
Accept any value or range within the range 91−108«°» for M3.
\npolar AND unsymmetrical distribution of charge
OR
polar AND dipoles do not cancel
OR
«polar as» dipoles «add to» give a «partial» positive «charge» at P and a «partial» negative «charge» at the opposite/Cl side of the molecule ✔
Accept “polar AND unsymmetrical molecule”.
\n«−398.9 kJ mol−1 − (−306.4 kJ mol−1) =» −92.5 «kJ mol−1» ✔
\n«Kc =» ✔
\n«shifts» left/towards reactants AND «forward reaction is» exothermic/ΔH is negative ✔
\n0.02 mol of zinc is added to 10.0 cm3 of 1.0 mol dm–3 hydrochloric acid.
\nZn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g)
\nHow many moles of hydrogen are produced?
\nA. 0.005
\nB. 0.01
\nC. 0.02
\nD. 0.04
\nA
\nHigher scoring candidates managed to estimate moles of gas produced in the reaction of zinc with hydrochloric acid solution.
\nWhich equation represents the standard enthalpy of formation of lithium oxide?
\n
A. 4Li (s) + O2 (g) → 2Li2O (s)
B. 2Li (s) + O2 (g) → Li2O (s)
\nC. Li (s) + O2 (g) → Li2O (s)
\nD. Li (g) + O2 (g) → Li2O (g)
\nB
\nWhich statement describes an endothermic reaction?
\n
A. The bonds broken are stronger than the bonds formed.
B. The enthalpy of the reactants is higher than the enthalpy of the products.
\nC. The temperature of the surroundings increases.
\nD. The products are more stable than the reactants.
\nA
\nA student performed an experiment to find the melting point of sulfur, obtaining 118.0 °C. The literature value is 115.2 °C. What was the percentage error?
\n
A.
B.
\nC.
\nD.
\nA
\nThe graph shows the first six ionization energies of an element.
\n[Ionization energies of the elements (data page) Available at: https://en.wikipedia.org/wiki/Ionization_energies_of_the_
elements_(data_page) Text is available under the Creative Commons Attribution-ShareAlike License 3.0 (CC BY-SA
3.0) https://creativecommons.org/licenses/by-sa/3.0/deed.en.]
In which group is the element?
A. 13
\nB. 14
\nC. 15
\nD. 16
\nA
\nA good number of candidates recognized the element based on the graph showing the first six ionization of the element.
\nFast moving helium nuclei (4He2+) were fired at a thin piece of gold foil with most passing undeflected but a few deviating largely from their path. The diagram illustrates this historic experiment.
\nFigure from PPLATO / FLAP (Flexible Learning Approach To Physics), http://www.met.reading.ac.uk/pplato2/h-flap/
phys8_1.html#top 1996 The Open University and The University of Reading.
Suggest what can be concluded about the gold atom from this experiment.
\nSubsequent experiments showed electrons existing in energy levels occupying various orbital shapes.
\nSketch diagrams of 1s, 2s and 2p.
\nState the electron configuration of copper.
\nCopper is a transition metal that forms different coloured complexes. A complex [Cu(H2O)6]2+ (aq) changes colour when excess Cl− (aq) is added.
\nExplain the cause of this colour change, using sections 3 and 15 from the data booklet.
\nMost 4He2+ passing straight through:
\nmost of the atom is empty space
OR
the space between nuclei is much larger than 4He2+ particles
OR
nucleus/centre is «very» small «compared to the size of the atom» ✔
Very few 4He2+ deviating largely from their path:
nucleus/centre is positive «and repels 4He2+ particles»
OR
nucleus/centre is «more» dense/heavy «than 4He2+ particles and deflects them»
OR
nucleus/centre is «very» small «compared to the size of the atom» ✔
\n
Do not accept the same reason for both M1 and M2.
\nAccept “most of the atom is an electron cloud” for M1.
\nDo not accept only “nucleus repels 4He2+ particles” for M2.
\n1s AND 2s as spheres ✔
\none or more 2p orbital(s) as figure(s) of 8 shape(s) of any orientation (px, py pz) ✔
\n1s22s22p63s23p64s13d10
OR
[Ar] 4s13d10 ✔
Accept configuration with 3d before 4s.
chloride is lower in the spectrochemical series ✔
\n«ligand cause» decreased/lesser splitting «in d-orbitals compared to H2O» ✔
\n
frequency/energy of light absorbed is decreased
OR
wavelength of light absorbed is increased ✔
Accept ·chloride a weaker ligand than water/produces a smaller energy difference than water for M1.
Award [2 max] for mentioning splitting of orbitals is changed AND frequency/ wavelength/energy of light absorbed
are different/changed without mentioning correct decrease or increase.
Why is hydrated copper (II) sulfate blue?
\nA. Blue light is emitted when electrons return to lower d-orbitals.
\nB. Light complimentary to blue is absorbed when electrons return to lower d-orbitals.
\nC. Blue light is emitted when electrons are promoted between d-orbitals.
\nD. Light complimentary to blue is absorbed when electrons are promoted between d-orbitals.
\nD
\nHigher scoring candidates managed to identify why hydrated copper(II) sulfate is blue in colour.
\nWhich reaction involves homolytic fission?
\nA. CH4 + Cl2
\nB. CH3Br + NaOH
\nC. (CH3)3CBr + NaOH
\nD. C6H6 + HNO3 + H2SO4
\nA
\nA well answered question. 66% of the candidates identified the chlorination of methane as the reaction involving homolytic bond fission.
\nIn which compound are all carbon atoms sp3 hybridized?
\nA. C2H2
\nB. C2H2Cl2
\nC. C2Cl4
\nD. C2Cl6
\nD
\n76% of candidates correctly identified the organic compound in which all carbon atoms were sp2 hybridized.
\nElectron transitions are related to trends in the periodic table.
\nExplain the general increase in trend in the first ionization energies of the period 3 elements, Na to Ar.
\nSodium emits yellow light with a frequency of 5.09 × 1014 Hz when electrons transition from 3p to 3s orbitals.
\nCalculate the energy difference, in J, between these two orbitals using sections 1 and 2 of the data booklet.
\n\n
Darling, D, n.d. D lines (of sodium). [online] Available at <https://www.daviddarling.info/encyclopedia/D/D_lines.html> [Accessed 6 May 2020].
\nincreasing number of protons
OR
increasing nuclear charge ✔
«atomic» radius/size decreases
OR
same number of shells/electrons occupy same shell
OR
similar shielding «by inner electrons» ✔
«ΔE = hν = 6.63 × 10–34 J s × 5.09 × 1014 s–1 =» 3.37 × 10–19 «J» ✔
\nWhat is the product of the reaction of propanal with lithium aluminium hydride, LiAlH4?
\nA. Propanoic acid
\nB. Propanone
\nC. Propan-1-ol
\nD. Propan-2-ol
\nC
\n69% of the candidates identified the product of the reaction of propanal with LiAlH4. The question had strong correlation with high-scoring candidates.
\nWhat are the electron domain and molecular geometries of the XeF4 molecule?
\nC
\nHigher scoring candidates had greater success in correctly identifying electron domain and molecular geometry of the XeF4 molecule.
\nGiven equimolar concentrations, which substance would produce the strongest signal in a 1H NMR spectrum?
\n
A. (CH3)3CH
B. C6H6
\nC. C8H18
\nD. Si(CH3)4
\nD
\n47% of the candidates identified TMS as the substance producing the strongest signal in a 1H NMR spectrum. The wording of this question was not ideal and several teachers commented on this. We do not usually refer to the strength but the area under the signal.
\nIron(II) disulfide, FeS2, has been mistaken for gold.
\nState the full electronic configuration of Fe2+.
\nExplain why there is a large increase from the 8th to the 9th ionization energy of iron.
\nCalculate the oxidation state of sulfur in iron(II) disulfide, FeS2.
\nDescribe the bonding in iron, Fe (s).
\n1s2 2s2 2p6 3s2 3p6 3d6 ✔
\nAny two of:
\nIE9: electron in lower energy level
OR
IE9: more stable/full electron level ✔
IE9: electron closer to nucleus
OR
IE9: electron more tightly held by nucleus ✔
IE9: less shielding by «complete» inner levels ✔
–1 ✔
\n\n
Accept “– I”.
\nelectrostatic attraction/hold between «lattice of» positive ions/cations AND delocalized «valence» electrons ✔
\nMostly well done which was a pleasant surprise since this is not overly easy, predictably some gave [Ar] 4s2 3d4.
\nDespite some confusion regarding which sub-level the electrons were being removed from, many candidates were able to make at least one valid point, commonly in terms of lower energy/ full sub level/closer to nucleus.
\nThis was an easy question, yet 30% of the candidates were unable to work it out; some wrote the oxidation state in the conventionally incorrect format, 1- and lost the mark.
\nMost candidates knew the bonding in Fe is metallic but some did not “describe” it or missed the type of attraction, a minor mistake; others referred to nuclei or protons instead of cations/positive ions. In some cases, candidates referred too ionic bonding, probably still thinking of FeS2 (not reading the question well). Overall, only 30% answered satisfactorily.
\nWhen heated in air, magnesium ribbon reacts with oxygen to form magnesium oxide.
\nThe reaction in (a)(i) was carried out in a crucible with a lid and the following data was recorded:
\nMass of crucible and lid = 47.372 ±0.001 g
\nMass of crucible, lid and magnesium ribbon before heating = 53.726 ±0.001 g
\nMass of crucible, lid and product after heating = 56.941 ±0.001 g
\n\n
When magnesium is burnt in air, some of it reacts with nitrogen to form magnesium nitride according to the equation:
\n3 Mg (s) + N2 (g) → Mg3N2 (s)
\nThe presence of magnesium nitride can be demonstrated by adding water to the product. It is hydrolysed to form magnesium hydroxide and ammonia.
\nMost nitride ions are 14N3–.
\nWrite a balanced equation for the reaction that occurs.
\nState the block of the periodic table in which magnesium is located.
\nIdentify a metal, in the same period as magnesium, that does not form a basic oxide.
\nCalculate the amount of magnesium, in mol, that was used.
\nDetermine the percentage uncertainty of the mass of product after heating.
\nAssume the reaction in (a)(i) is the only one occurring and it goes to completion, but some product has been lost from the crucible. Deduce the percentage yield of magnesium oxide in the crucible.
\n\n
Evaluate whether this, rather than the loss of product, could explain the yield found in (b)(iii).
\nSuggest an explanation, other than product being lost from the crucible or reacting with nitrogen, that could explain the yield found in (b)(iii).
\nCalculate coefficients that balance the equation for the following reaction.
\n__ Mg3N2 (s) + __ H2O (l) → __ Mg(OH)2 (s) + __ NH3 (aq)
\nDetermine the oxidation state of nitrogen in Mg3N2 and in NH3.
\nDeduce, giving reasons, whether the reaction of magnesium nitride with water is an acid–base reaction, a redox reaction, neither or both.
\nState the number of subatomic particles in this ion.
\nSome nitride ions are 15N3–. State the term that describes the relationship between 14N3– and 15N3–.
\nThe nitride ion and the magnesium ion are isoelectronic (they have the same electron configuration). Determine, giving a reason, which has the greater ionic radius.
\nSuggest two reasons why atoms are no longer regarded as the indivisible units of matter.
\nState the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
\n2 Mg(s) + O2(g) → 2 MgO(s) ✔
\n
Do not accept equilibrium arrows. Ignore state symbols
s ✔
\n
Do not allow group 2
aluminium/Al ✔
\n«mol» ✔
\nmass of product ✔
\n✔
\n\n
Award [2] for correct final answer
\nAccept 0.021%
\n✔
\n✔
\n\n
Award «0.2614 mol x 40.31 g mol–1»
\nAccept alternative methods to arrive at the correct answer.
\nAccept final answers in the range 91-92%
\n[2] for correct final answer.
\nyes
AND
«each Mg combines with N, so» mass increase would be 14x which is less than expected increase of 16x
OR
3 mol Mg would form 101g of Mg3N2 but would form 3 x MgO = 121 g of MgO
OR
0.2614 mol forms 10.536 g of MgO, but would form 8.796 g of Mg3N2 ✔
\n
Accept Yes AND “the mass of N/N2 that combines with each g/mole of Mg is lower than that of O/O2”
\nAccept YES AND “molar mass of nitrogen less than of oxygen”.
\nincomplete reaction
OR
Mg was partially oxidised already
OR
impurity present that evaporated/did not react ✔
\n
Accept “crucible weighed before fully cooled”.
\nAccept answers relating to a higher atomic mass impurity consuming less O/O2.
\nAccept “non-stoichiometric compounds formed”.
\nDo not accept \"human error\", \"wrongly calibrated balance\" or other non-chemical reasons.
\nIf answer to (b)(iii) is >100%, accept appropriate reasons, such as product absorbed moisture before being weighed.
\n«1» Mg3N2 (s) + 6 H2O (l) → 3 Mg(OH)2 (s) + 2 NH3 (aq)
\nMg3N2: -3
AND
NH3: -3 ✔
Do not accept 3 or 3-
Acid–base:
yes AND N3- accepts H+/donates electron pair«s»
OR
yes AND H2O loses H+ «to form OH-»/accepts electron pair«s» ✔
Redox:
no AND no oxidation states change ✔
\n
Accept “yes AND proton transfer takes place”
\nAccept reference to the oxidation state of specific elements not changing.
\nAccept “not redox as no electrons gained/lost”.
\nAward [1 max] for Acid–base: yes AND Redox: no without correct reasons, if no other mark has been awarded
\nProtons: 7 AND Neutrons: 7 AND Electrons: 10 ✔
\nisotope«s» ✔
\nnitride AND smaller nuclear charge/number of protons/atomic number ✔
\nAny two of:
subatomic particles «discovered»
OR
particles smaller/with masses less than atoms «discovered»
OR
«existence of» isotopes «same number of protons, different number of neutrons» ✔
charged particles obtained from «neutral» atoms
OR
atoms can gain or lose electrons «and become charged» ✔
atom «discovered» to have structure ✔
fission
OR
atoms can be split ✔
\n
Accept atoms can undergo fusion «to produce heavier atoms»
\nAccept specific examples of particles.
\nAward [2] for “atom shown to have a nucleus with electrons around it” as both M1 and M3.
\n
Award [1] for all bonding types correct.
Award [1] for each correct description.
\nApply ECF for M2 only once.
\nThis was not as well done as one might have expected with the most common errors being O instead of O2 oxygen and MgO rather than MgO2.
\nMany students did not know what \"block\" meant, and often guessed group 2 etc.
\nMany students confused \"period\" and \"group\" and also many did not read metal, so aluminium was not chosen by the majority.
\nA number of students were not able to interpret the results and hence find the gain in mass and calculate the moles correctly.
\nOnly a handful could work out the correct answer. Most had no real idea and quite a lot of blank responses. There also seems to be significant confusion between \"percent uncertainty\" and \"percent error\".
\nThis was not well answered, but definitely better than the previous question with quite a few gaining some credit for correctly determining the theoretical yield.
\nThis proved to be a very difficult question to answer in the quantitative manner required, with hardly any correct responses.
\nQuite a few students realised that incomplete reaction would lead to this, but only 30% of students gave a correct answer rather than a non-specific guess, such as \"misread balance\" or \"impurities\".
\nThis was generally very well done with almost all candidates being able to determine the correct coefficients.
\nAbout 40% of students managed to correctly determine both the oxidation states, as -3, with errors being about equally divided between the two compounds.
\nProbably only about 10% could explain why this was an acid-base reaction. Rather more made valid deductions about redox, based on their answer to the previous question.
\nMost candidates could answer the question about subatomic particles correctly.
\nIdentification of isotopes was answered correctly by most students.
\nIn spite of being given the meaning of \"isoelectronic\", many candidates talked about the differing number of electrons and only about 30% could correctly analyse the situation in terms of nuclear charge.
\nThe question was marked quite leniently so that the majority of candidates gained at least one of the marks by mentioning a subatomic particle. A significant number read \"indivisible\" as \"invisible\" however.
\nAbout a quarter of the students gained full marks and probably a similar number gained no marks. Metallic bonding was the type that seemed least easily recognised and least easily described. Another common error was to explain ionic bonding in terms of attraction of ions rather than describing electron transfer.
\nWhich compound has the largest value of lattice enthalpy?
\nA. Na2O
\nB. K2O
\nC. Na2S
\nD. K2S
\nA
\nA majority of students recognized the ionic compound with the highest lattice enthalpy.
\nIn which reaction does entropy decrease?
\nA. NaCl (s) → NaCl (aq)
\nB. Zn (s) + H2SO4 (aq) → ZnSO4 (aq) + H2 (g)
\nC. NH3 (g) + HCl (g) → NH4Cl (s)
\nD. CuCO3 (s) → CuO (s) + CO2 (g)
\nC
\nSecond best answered question in the exam with 85% of candidates identifying the correct answering for reaction in which entropy decreases.
\nSulfur trioxide is produced from sulfur dioxide.
\n2SO2 (g) + O2 (g) 2SO3 (g) ΔH = −196 kJ mol−1
\nThe reaction between sulfur dioxide and oxygen can be carried out at different temperatures.
\nOutline, giving a reason, the effect of a catalyst on a reaction.
\nOn the axes, sketch Maxwell–Boltzmann energy distribution curves for the reacting species at two temperatures T1 and T2, where T2 > T1.
\nExplain the effect of increasing temperature on the yield of SO3.
\nDraw the Lewis structure of SO3.
\nExplain the electron domain geometry of SO3.
\nState the product formed from the reaction of SO3 with water.
\nState the meaning of a strong Brønsted–Lowry acid.
\nincreases rate AND lower Ea ✔
\nprovides alternative pathway «with lower Ea»
OR
more/larger fraction of molecules have the «lower» Ea ✔
\n
Accept description of how catalyst lowers Ea for M2 (e.g. “reactants adsorb on surface «of catalyst»”, “reactant bonds weaken «when adsorbed»”, “helps favorable orientation of molecules”).
\nboth axes correctly labelled ✔
\npeak of T2 curve lower AND to the right of T1 curve ✔
\nlines begin at origin AND correct shape of curves AND T2 must finish above T1 ✔
\n\n
Accept “probability «density» / number of particles / N / fraction” on y-axis.
\nAccept “kinetic E/KE/Ek” but not just “Energy/E” on x-axis.
\ndecrease AND equilibrium shifts left / favours reverse reaction ✔
\n«forward reaction is» exothermic / ΔH is negative ✔
\n✔
\n
Note:
\nAccept any of the above structures as formal charge is not being assessed.
\nthree electron domains «attached to the central atom» ✔
\nrepel/as far away as possible /120° «apart» ✔
\nsulfuric acid/H2SO4 ✔
\n
Accept “disulfuric acid/H2S2O7”.
fully ionizes/dissociates ✔
\nproton/H+ «donor »✔
\nOverall well answered though some answers were directed to explain the specific example rather than the simple and standard definition of the effect of a catalyst.
\nFew got the 3 marks for this standard question (average mark 1.7), the most common error being incomplete/incorrect labelling of axes, curves beginning above 0 on y-axis or inverted curves.
\nMany candidates got one mark at least, sometimes failing to state the effect on the production of SO3 though they knew this quite obviously. This failure to read the question properly also resulted in an incorrect prediction based exclusively on kinetics instead of using the information provided to guide their answers.
\nDrawing the Lewis structure of SO3 proved to be challenging, with lots of incorrect shapes, lone pair on S, etc.; accepting all resonant structures allowed many candidates to get the mark which was fair considering no formal charge estimation was required.
\nMost were focussed on the shape itself instead of explaining what led them to suggest that shape; number of electron domains allowed most candidates to get one mark and eventually a mention of bond angles resulted in only 35% getting both marks. In general, students were not able to provide clear explanations for the shape (not a language issue) but rather were happy to state the molecular geometry which they knew, but wasn't what was actually required for the mark.
\n6(d)(i)-(ii): These simple questions could be expected to be answered by all HL candidates. However 20% of the candidates suggested hydroxides or hydrogen as products of an aqueous dissolution of sulphur oxide. In the case of the definition of a strong Brønsted-Lowry acid, only 50% got both marks, often failing to define \"strong\" but in other cases defining them as bases even.
\nThe table shows data for the hydrolysis of a halogenoalkane, RCl.
\nWhich statements are correct?
\nI. The reaction is first order with respect to RCl.
II. The reaction is second order overall.
III. The reaction proceeds by an SN2 mechanism.
A. I and II only
B. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nAverage performance on a high discriminatory question on rate of reaction and mechanism question, with no clear misconception based on the other choices.
\nWhat are the products when dilute aqueous copper (II) nitrate is electrolysed using platinum electrodes?
\nE⦵ (Cu | Cu2+) = –0.34 V.
\nA
\nAverage performance with a lower discrimination index and no clear misconception based on the incorrect choices.
\nIn the electrolysis apparatus shown, 0.59 g of Ni is deposited on the cathode of the first cell.
\nWhat is the mass of Ag deposited on the cathode of the second cell?
\n
A. 0.54 g
B. 0.59 g
\nC. 1.08 g
\nD. 2.16 g
\nD
\nA poorly answered question with low discrimination factor. Several teachers asked whether this question of \"cells in series\" was in the Chemistry Guide (see Topic 19.1, Electrochemical cells, on page 96) and it is likely that this topic may not have been covered in class by some teachers.
\nWhat is the activation energy according to the following plot of the linear form of the Arrhenius equation?
\nArrhenius equation: .
\nA.
\nB.
\nC.
\nD.
\nA
\nA discriminating question in which high scoring candidates had much greater success in transforming the Arrhenius equation into the units used in the graph by taking its natural log and using the dimensions of the slope to arrive at the correct answer.
\nAt equilibrium, the concentrations of chlorine and iodine are both 0.02 mol dm–3.
\nCl2 (g) + (g) Cl (g) Kc = 454
\nWhat is the concentration of iodine monochloride, Cl?
\n
A.
B.
\nC.
\nD.
\nB
\nHigh scoring candidates had more success calculating the concentration of ICl in the product given equal concentrations of Cl2 and I2 in the reactants.
\nWhich reagents and conditions are best for converting propan-1-ol into propanoic acid?
\nA. Reflux with acidified potassium dichromate (VI)
\nB. Reflux with LiAlH4
\nC. Distil with acidified potassium dichromate (VI)
\nD. Distil with LiAlH4
\nA
\nThe vast majority of students identified the best reagents and conditions for converting propan-1-ol into propanoic acid.
\nWhich statement explains the Lewis acid–base nature of the chloride ion in this reaction?
\nC2H5+ + Cl– → C2H5Cl
\nA. Lewis base because it donates a pair of electrons
\nB. Lewis base because it accepts a pair of electrons
\nC. Lewis acid because it donates a pair of electrons
\nD. Lewis acid because it accepts a pair of electrons
\nA
\nA discriminating question in which high scoring candidates had much greater success in explaining the Lewis acid-base nature of the chloride ion in the given reaction. The next two common incorrect answers were B and D showing a lack of understanding of the Lewis acid-base concept.
\nMagnesium is a reactive metal often found in alloys.
\nOrganomagnesium compounds can react with carbonyl compounds. One overall equation is:
\nCompound B can also be prepared by reacting an alkene with water.
\nIodomethane is used to prepare CH3Mg. It can also be converted into methanol:
\nCH3 + HO– → CH3OH + –
\nMagnesium can be produced by the electrolysis of molten magnesium chloride.
\nWrite the half-equation for the formation of magnesium.
\nSuggest an experiment that shows that magnesium is more reactive than zinc, giving the observation that would confirm this.
\nState the name of Compound A, applying International Union of Pure and Applied Chemistry (IUPAC) rules.
\nIdentify the strongest force between the molecules of Compound B.
\nDraw the structural formula of the alkene required.
\nDeduce the structural formula of the repeating unit of the polymer formed from this alkene.
\nDeduce what would be observed when Compound B is warmed with acidified aqueous potassium dichromate (VI).
\nIdentify the type of reaction.
\nOutline the requirements for a collision between reactants to yield products.
\nThe polarity of the carbon–halogen bond, C–X, facilitates attack by HO–.
\nOutline, giving a reason, how the bond polarity changes going down group 17.
\nMg2+ + 2 e- → Mg ✔
\n\n
Do not penalize missing charge on electron.
\nAccept equation with equilibrium arrows.
\nAlternative 1
\nput Mg in Zn2+(aq) ✔
\nZn/«black» layer forms «on surface of Mg» ✔
\n
Award [1 max] for “no reaction when Zn placed in Mg2+(aq)”.
\n
Alternative 2
\nplace both metals in acid ✔
\nbubbles evolve more rapidly from Mg
OR
Mg dissolves faster ✔
\n
Alternative 3
\nconstruct a cell with Mg and Zn electrodes ✔
\nbulb lights up
OR
shows (+) voltage
OR
size/mass of Mg(s) decreases «over time»
OR
size/mass of Zn increases «over time»
Accept “electrons flow from Mg to Zn”.
Accept Mg is negative electrode/anode
OR
Zn is positive electrode/cathode
Accept other correct methods.
propanone ✔
\n
Accept 2-propanone and propan-2-one.
hydrogen bonds ✔
\n\n
Do not penalize missing brackets or n.
\nDo not award mark if continuation bonds are not shown.
\nno change «in colour/appearance/solution» ✔
\n«nucleophilic» substitution
OR
SN2 ✔
Accept “hydrolysis”.
Accept SN1
\nenergy/E ≥ activation energy/Ea ✔
\ncorrect orientation «of reacting particles»
OR
correct geometry «of reacting particles» ✔
decreases/less polar AND electronegativity «of the halogen» decreases ✔
\n\n
Accept “decreases” AND a correct comparison of the electronegativity of two halogens.
\nAccept “decreases” AND “attraction for valence electrons decreases”.
\nUnfortunately, only 40% of the students could write this quite straightforward half equation.
\nMany candidates gained some credit by suggesting voltaic cell or a displacement reaction, but most could not gain the second mark and the reason was often a failure to be able to differentiate between \"what occurs\" and \"what is observed\".
\nEven though superfluous numbers (2-propanone, propan-2-one) were overlooked, only about half of the students could correctly name this simple molecule.
\nProbably just over half the students correctly identified hydrogen bonding, with dipole-dipole being the most common wrong answer, though a significant number identified an intramolecular bond.
\nFew candidates could correctly eliminate water to deduce the identity of the required reactant.
\nCorrect answers to this were very scarce and even when candidates had an incorrect alkene for the previous part, they were unable to score an ECF mark, by deducing the formula of the polymer it would produce.
\nSome students deduced that, as it was a tertiary alcohol, there would be no reaction, but almost all were lucky that this was accepted as well as the correct observation - \"it would remain orange\".
\nAbout a quarter of the students identified this as a substitution reaction, though quite a number then lost the mark by incorrectly stating it was either \"free radical\" or \"electrophilic\". A very common wrong answer was \"displacement\" or \"single displacement\" and this makes one wonder whether this terminology is being taught instead of substitution
\nGenerally well done with the vast majority of students correctly citing \"correct orientation\" and many only failed to gain the second mark through failing to equate the energy required to the activation energy.
\nAnother question that was not well answered with probably only a quarter of candidates stating that the polarity would decrease because of decreasing electronegativity down the group.
\nWhat are the type of reaction and role of the nitronium ion, NO2+, in the following reaction?
\nC6H6 + NO2+ → C6H5NO2 + H+
\nA
\nGood performance on a discriminating question in the type of reaction and role of the nitronium ion, NO2+, in the nitration of benzene.
\nIn which set are the salts arranged in order of increasing pH?
\nA. HCOONH4 < KBr < NH4Br < HCOOK
\nB. KBr < NH4Br < HCOOK < HCOONH4
\nC. NH4Br < HCOONH4 < KBr < HCOOK
\nD. HCOOK < KBr < HCOONH4 < NH4Br
\nC
\nNot very well done in identifying the set of ionic compounds that were arranged in correct order of increasing pH, with no clear misconception based on the other choices.
\nThe equilibrium constant, Kc, for the reaction 2A + 4B 2C + 4D has a value of 4.0. What is the value of Kc for the reaction below at the same temperature?
\nC + 2D A + 2B
\nA. 0.25
\nB. 0.50
\nC. 1.0
\nD. 16
\nB
\n60% of the candidates deduced the value of Kc for related equations. The question had a strong discrimination between high-scoring and low-scoring candidates.
\nWhat is molecule Z that is formed in step 1 of this synthetic route?
\nB
\nHigher scoring candidates did better at identifying the molecule that is formed in the synthesis of aminobenzene from nitrobenzene.
\nWhich of the 0.001 mol dm−3 solutions is most likely to have a pH of 11.3?
\nA. Ca(OH)2 (aq)
\nB. H3PO4 (aq)
\nC. NaOH (aq)
\nD. NH4OH (aq)
\nA
\nThis required calculation of pH of a base – not an easy task for SL candidates without a calculator. 28% of the candidates answered the question correctly.
\nWhat are the E/Z designations of these stereoisomers?
\nA
\nMediocre performance with a good number of students who showed a lack of understanding of the E/Z designation for the alkene isomers.
\nWhich compound produces the following 1H NMR spectrum?
\n[Spectral Database for Organic Compounds, SDBS. SDBS Compounds and Spectral Search. [graph] Available at:
https://sdbs.db.aist.go.jp [Accessed 3 January 2019].]
A. propanal
B. propanone
\nC. propane
\nD. methlypropane
\nA
\nA majority of students identified the correct compound for the 1H NMR spectrum given.
\nWhat is the strongest acid in the equation below?
\nH3AsO4 + H2O H2AsO4− + H3O+ Kc = 4.5 × 10−4
\nA. H3AsO4
\nB. H2O
\nC. H2AsO4−
\nD. H3O+
\nD
\nA challenging question that combined interpreting the implications of the value of the equilibrium constant with the Brønsted-Lowry concept. Most candidates focused on BL theory and the forward reaction only. 36% of the candidates answered the question correctly by selecting the acid involved in the backward reaction.
\nWhich species could be reduced to form NO2?
\nA. N2
\nB. NO3−
\nC. HNO2
\nD. NO
\nB
\n58% of the candidates were able to identify the species that could be reduced to form NO2. The most commonly chosen distractor was HNO2.
\nWhat is the concentration of chloride ions, in mol dm−3, in a solution formed by mixing 200 cm3 of 1 mol dm−3 HCl with 200 cm3 of 5 mol dm−3 NaCl?
\nA. 1
\nB. 2
\nC. 3
\nD. 6
\nC
\n52% of the candidates calculated the concentration of chloride ions in the titration. The distractors were chosen almost equally.
\nWhich combination best describes what is happening to chloromethane, CH3Cl, in the equation below?
\nCH3Cl (g) + H2 (g) CH4 (g) + HCl (g)
\nA. Oxidation and addition
\nB. Oxidation and substitution
\nC. Reduction and addition
\nD. Reduction and substitution
\nD
\n57% of the candidates were able to identify the type of reaction (oxidation or reduction & addition or substitution) correctly.
\nWhen heated in air, magnesium ribbon reacts with oxygen to form magnesium oxide.
\nThe reaction in (a)(i) was carried out in a crucible with a lid and the following data was recorded:
\nMass of crucible and lid = 47.372 ±0.001 g
\nMass of crucible, lid and magnesium ribbon before heating = 53.726 ±0.001 g
\nMass of crucible, lid and product after heating = 56.941 ±0.001 g
\n\n
When magnesium is burnt in air, some of it reacts with nitrogen to form magnesium nitride according to the equation:
\n3 Mg (s) + N2 (g) → Mg3N2 (s)
\nThe presence of magnesium nitride can be demonstrated by adding water to the product. It is hydrolysed to form magnesium hydroxide and ammonia.
\nMost nitride ions are 14N3–.
\nWrite a balanced equation for the reaction that occurs.
\nIdentify a metal, in the same period as magnesium, that does not form a basic oxide.
\nCalculate the amount of magnesium, in mol, that was used.
\nDetermine the percentage uncertainty of the mass of product after heating.
\nAssume the reaction in (a)(i) is the only one occurring and it goes to completion, but some product has been lost from the crucible. Deduce the percentage yield of magnesium oxide in the crucible.
\n\n
Evaluate whether this, rather than the loss of product, could explain the yield found in (b)(iii).
\nSuggest an explanation, other than product being lost from the crucible or reacting with nitrogen, that could explain the yield found in (b)(iii).
\nCalculate coefficients that balance the equation for the following reaction.
\nAmmonia is added to water that contains a few drops of an indicator. Identify an indicator that would change colour. Use sections 21 and 22 of the data booklet.
\nDetermine the oxidation state of nitrogen in Mg3N2 and in NH3.
\nDeduce, giving reasons, whether the reaction of magnesium nitride with water is an acid–base reaction, a redox reaction, neither or both.
\nState the number of subatomic particles in this ion.
\nSome nitride ions are 15N3–. State the term that describes the relationship between 14N3– and 15N3–.
\nThe nitride ion and the magnesium ion are isoelectronic (they have the same electron configuration). Determine, giving a reason, which has the greater ionic radius.
\nSuggest, giving a reason, whether magnesium or nitrogen would have the greater sixth ionization energy.
\nSuggest two reasons why atoms are no longer regarded as the indivisible units of matter.
\nState the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
\n2 Mg(s) + O2(g) → 2 MgO(s) ✔
\n\n
Do not accept equilibrium arrows. Ignore state symbols
\naluminium/Al ✔
\nmass of product ✔
\n✔
\n
Award [2] for correct final answer
Accept 0.021%
\n✔
\n✔
\n\n
Award «0.2614 mol x 40.31 g mol–1»
\nAccept alternative methods to arrive at the correct answer.
\nAccept final answers in the range 90.5-91.5%
\n[2] for correct final answer.
\nyes
AND
«each Mg combines with N, so» mass increase would be 14x which is less than expected increase of 16x
OR
3 mol Mg would form 101g of Mg3N2 but would form 3 x MgO = 121 g of MgO
OR
0.2614 mol forms 10.536 g of MgO, but would form 8.796 g of Mg3N2 ✔
\n
Accept Yes AND “the mass of N/N2 that combines with each g/mole of Mg is lower than that of O/O2”
\nAccept YES AND “molar mass of nitrogen less than of oxygen”.
\nincomplete reaction
OR
Mg was partially oxidised already
OR
impurity present that evaporated/did not react ✔
\n
Accept “crucible weighed before fully cooled”.
\nAccept answers relating to a higher atomic mass impurity consuming less O/O2.
\nAccept “non-stoichiometric compounds formed”.
\nDo not accept \"human error\", \"wrongly calibrated balance\" or other non-chemical reasons.
\nIf answer to (b)(iii) is >100%, accept appropriate reasons, such as product absorbed moisture before being weighed.
\n«1» Mg3N2 (s) + 6 H2O (l) → 3 Mg(OH)2 (s) + 2 NH3 (aq) ✔
\nphenol red ✔
\n
Accept bromothymol blue or phenolphthalein.
Mg3N2: -3
AND
NH3: -3 ✔
Do not accept 3 or 3-
Acid–base:
yes AND N3- accepts H+/donates electron pair«s»
OR
yes AND H2O loses H+ «to form OH-»/accepts electron pair«s» ✔
Redox:
no AND no oxidation states change ✔
\n
Accept “yes AND proton transfer takes place”
\nAccept reference to the oxidation state of specific elements not changing.
\nAccept “not redox as no electrons gained/lost”.
\nAward [1 max] for Acid–base: yes AND Redox: no without correct reasons, if no other mark has been awarded
\nProtons: 7 AND Neutrons: 7 AND Electrons: 10 ✔
\nisotope«s» ✔
\nnitride AND smaller nuclear charge/number of protons/atomic number ✔
\nnitrogen AND electron lost from first «energy» level/s sub-level/s-orbital AND magnesium from p sub-level/p-orbital/second «energy» level
OR
nitrogen AND electron lost from lower level «than magnesium» ✔
\n
Accept “nitrogen AND electron lost closer to the nucleus «than magnesium»”.
\nAny two of:
\nsubatomic particles «discovered»
OR
particles smaller/with masses less than atoms «discovered»
OR
«existence of» isotopes «same number of protons, different number of neutrons» ✔
charged particles obtained from «neutral» atoms
OR
atoms can gain or lose electrons «and become charged» ✔
atom «discovered» to have structure ✔
fission
OR
atoms can be split ✔
\n
Accept atoms can undergo fusion «to produce heavier atoms»
\nAccept specific examples of particles.
\nAward [2] for “atom shown to have a nucleus with electrons around it” as both M1 and M3.
\n\n
Award [1] for all bonding types correct.
\nAward [1] for each correct description.
\nApply ECF for M2 only once.
\nDone very well. However, it was disappointing to see the formula of oxygen molecule as O and the oxide as Mg2O and MgO2 at HL level.
\nAverage performance; the question asked to identify a metal; however, answers included S, Si, P and even noble gases besides Be and Na. The only choice of aluminium; however, since its oxide is amphoteric, it could not be the answer in the minds of some.
\nVery good performance; some calculated the mass of oxygen instead of magnesium for the calculation of the amount, in mol, of magnesium. Others calculated the mass, but not the amount in mol as required.
\nMediocre performance; instead of calculating percentage uncertainty, some calculated percentage difference.
\nSatisfactory performance; however, a good number could not answer the question correctly on determining the percentage yield.
\nPoorly done. The question asked to evaluate and explain but instead many answers simply agreed with the information provided instead of assessing its strength and limitation.
\nMediocre performance; explaining the yield found was often a challenge by not recognizing that incomplete reaction or Mg partially oxidized or impurities present that evaporated or did not react would explain the yield.
\nCalculating coefficients that balance the given equation was done very well.
\nWell done; some chose bromocresol green or methyl red as the indicator that would change colour, instead of phenol red, bromothymol blue or phenolphthalein.
\nGood performance; however, surprising number of candidates could not determine one or both oxidation states correctly or wrote it as 3 or 3−, instead of −3.
\nAverage performance; choosing the given reaction as an acid-base or redox reaction was not done well. Often answers were contradictory and the reasoning incorrect.
\nStating the number of subatomic particles in a 14N3- was done very well. However, some answers showed a lack of understanding of how to calculate the number of relevant subatomic particles given formula of an ion with charge and mass number.
\nExceptionally well done; A few candidates referred to isomers, rather than isotopes.
\nThere was reference to nitrogen and magnesium, rather than nitride and magnesium ions. Also, instead identifying smaller nuclear charge in nitride ion, some referred to core electrons, Zeff, increased electron-electron repulsion or shielding.
\nCommon error in suggesting nitrogen would have the greater sixth ionization energy was that for nitrogen, electron is lost from first energy level without making reference to magnesium losing it from second energy level.
\nGood performance; some teachers were concerned about the expected answers. However, generally, students were able to suggest two reasons why matter is divisible.
\nOne teacher commented that not asking to describe bonding in terms of electrostatic attractions as in earlier papers would have been confusing and some did answer in terms of electrostatic forces of attractions involved. However, the question was clear in its expectation that the answer had to be in terms of how the valence electrons produce the three types of bonds and the overall performance was good. Some had difficulty identifying the bond type for Mg, O2 and MgO.
\n30 g of an organic compound produces 44 g CO2 and 18 g H2O as the only combustion products. Which of the following is the empirical formula for this compound?
\nMr CO2 = 44 Mr H2O = 18
\nA. CH2
\nB. CH3
\nC. CHO
\nD. CH2O
\nD
\n63% of the candidates determined the empirical formula of the organic compound correctly using the masses of the sample and the combustion products.
\nThe arrows represent electron flow in the diagram. What does terminal X on the battery represent?
\nA. Anode and positive terminal
\nB. Anode and negative terminal
\nC. Cathode and positive terminal
\nD. Cathode and negative terminal
\nC
\n44% of the candidates identified the electrode and its charge based on the direction of the electron flow in the circuit. Some teachers commented that this question is more of a physics than chemistry question.
\nWhich graph represents the relationship between the amount of gas, n, and the absolute temperature, T, with all other variables in the ideal gas equation, PV = nRT, held constant?
\nB
\n56% of the candidates selected the correct graph representing the relationship between the amount of gas and its absolute temperature. The most commonly chosen distractor gave a directly proportional relationship.
\nWhich equation represents the deposition of iodine?
\nA. I2 (g) → I2 (l)
\nB. I2 (g) → I2 (s)
\nC. I2 (l) → I2 (g)
\nD. I2 (s) → I2 (g)
\nB
\n58% of the candidates identified the equation that represented the deposition of iodine. The most commonly chosen distractor was the condensation of gaseous iodine. Sublimation was also quite commonly chosen.
\nWhich experimental results support the theory that electrons exist in discrete energy levels?
\nA. 1H NMR
\nB. X-ray diffraction pattern
\nC. Emission spectra
\nD. IR spectra
\nC
\nA well answered question. 84% of the candidates identified emission spectra as the experimental results that support the theory that electrons exist in discrete energy levels.
\nAmmonia, NH3, is industrially important for the manufacture of fertilizers, explosives and plastics.
\nAmmonia is produced by the Haber–Bosch process which involves the equilibrium:
\nN2 (g) + 3 H2 (g) 2 NH3 (g)
\nThe effect of temperature on the position of equilibrium depends on the enthalpy change of the reaction.
\nAmmonia is soluble in water and forms an alkaline solution:
\nNH3 (g) + H2O (l) NH4+ (aq) + HO– (aq)
\nDraw arrows in the boxes to represent the electron configuration of a nitrogen atom.
\nDraw the Lewis (electron dot) structure of the ammonia molecule.
\nDeduce the expression for the equilibrium constant, Kc, for this equation.
\nExplain why an increase in pressure shifts the position of equilibrium towards the products and how this affects the value of the equilibrium constant, Kc.
\nState how the use of a catalyst affects the position of the equilibrium.
\nDetermine the enthalpy change, ΔH, for the Haber–Bosch process, in kJ. Use Section 11 of the data booklet.
\nCalculate the enthalpy change, ΔH⦵, for the Haber–Bosch process, in kJ, using the following data.
\n.
\nSuggest why the values obtained in (d)(i) and (d)(ii) differ.
\nState the relationship between NH4+ and NH3 in terms of the Brønsted–Lowry theory.
\nDetermine the concentration, in mol dm–3, of the solution formed when 900.0 dm3 of NH3 (g) at 300.0 K and 100.0 kPa, is dissolved in water to form 2.00 dm3 of solution. Use sections 1 and 2 of the data booklet.
\nCalculate the concentration of hydroxide ions in an ammonia solution with pH = 9.3. Use sections 1 and 2 of the data booklet.
\n\n
Accept all 2p electrons pointing downwards.
\nAccept half arrows instead of full arrows.
\n\n
Accept lines or dots or crosses for electrons, or a mixture of these
\n\n
✔
\nshifts to the side with fewer moles «of gas»
OR
shifts to right as there is a reduction in volume✔
«value of » Kc unchanged ✔
\n\n
Accept “Kc only affected by changes in temperature”.
\nsame/unaffected/unchanged ✔
\nbonds broken: N≡N + 3(H–H) / «1 mol×»945 «kJ mol–1» + 3«mol»×436 «kJ mol–1» / 945 «kJ» + 1308 «kJ» / 2253 «kJ» ✔
\nbonds formed: 6(N–H) / 6«mol»×391 «kJ mol–1» / 2346 «kJ» ✔
\nΔH = «2253 kJ – 2346 kJ = » –93 «kJ» ✔
\n\n
Award [2 max] for (+)93 «kJ»
\n–92.4 «kJ» ✔
\n«N-H» bond enthalpy is an average «and may not be the precise value in NH3» ✔
\n
Accept it relies on average values not specific to NH3
conjugate «acid and base» ✔
\namount of ammonia ✔
\nconcentration ✔
\n\n
Award [2] for correct final answer.
\n[OH−] ✔
\nMost students realised that the three p-orbitals were all singly filled.
\nEven more candidates could draw the correct Lewis structure of ammonia, with omission of the lone pair being the most common error.
\nMost students could deduce the equilibrium constant expression from the equilibrium equation.
\nMany students realised that increasing pressure shifts an equilibrium to the side with the most moles of gas (though the \"of gas\" was frequently omitted!) but probably less than half realised that, even though the equilibrium position changes, the value of the equilibrium constant remains constant.
\nIt was pleasing to see that about a third of students gaining full marks and an equal number only lost a single mark because they failed to locate the correct bond enthalpy for molecular nitrogen.
\nVery few students could determine the enthalpy change from enthalpy of formation data, with many being baffled by the absence of values for the elemental reactants and more than half who overcame this obstacle failed to note that 2 moles of ammonia are produced.
\nAbout half the candidates recognised the species as a conjugate acid-base pair, though some lost the mark by confusing the acid and base, even though this information was not asked for.
\nAbout 40% of candidates gained full marks for the calculation and a significant number of others gained the second mark to calculate the concentration as an ECF.
\nThis question was very poorly answered with many candidates calculating the [H+] instead of [OH-].
\nWhat is the relative atomic mass of an element with the following mass spectrum?
\nA. 23
\nB. 24
\nC. 25
\nD. 28
\nB
\n69% of the candidates determined the relative atomic mass of the element from its mass spectrum.
\nThree elements, X, Y, and Z are in the same period of the periodic table. The relative sizes of their atoms are represented by the diagram.
\nWhich general trends are correct?
\nC
\nThe question involved the identification of three trends. 43% of the candidates selected the correct trends in ionization energy, effective nuclear charge and acidity of the oxides. The most commonly chosen distractor reversed the trend in the acidity of the oxides.
\nWhich element is found in the 4th group, 6th period of the periodic table?
\nA. Selenium
\nB. Lead
\nC. Chromium
\nD. Hafnium
\nD
\nA very well answered question. 76% of the candidates were able to identify the element in the fourth group and sixth period of the periodic table as hafnium. The most commonly chosen distractor was lead which is in group 14 not 4.
\nWhich statement best describes the intramolecular bonding in HCN (l)?
\nA. Electrostatic attractions between H+ and CN− ions
\nB. Hydrogen bonding
\nC. Van der Waals forces and hydrogen bonding
\nD. Electrostatic attractions between pairs of electrons and positively charged nuclei
\nD
\nThe majority of the candidates thought the bonding in HCN is ionic (electrostatic attraction between H+ and CN- ions) rather than covalent. The description of covalent bonding was the second most commonly chosen response.
\nWhat is the type of bonding in a compound that has high boiling and melting points, poor electrical conductivity, and low solubility in water?
\nA. Ionic
\nB. Molecular covalent
\nC. Metallic
\nD. Giant covalent
\nD
\n58% of the candidates identified giant covalent bonding as the bonding in compounds with high melting and boiling points, poor electrical conductivity and low solubility in water. The most commonly selected distractor was ionic bonding. The question discriminated well between high-scoring and low-scoring candidates.
\nMagnesium is a reactive metal often found in alloys.
\nMagnesium is sometimes used as a sacrificial anode to protect steel from corrosion.
\nA graph of the volume of gas produced by reacting magnesium with a large excess of 1 mol dm–3 hydrochloric acid is shown.
\nSuggest an experiment that shows that magnesium is more reactive than zinc, giving the observation that would confirm this.
\nCalculate the standard potential, in V, of a cell formed by magnesium and steel half-cells. Use section 24 of the data booklet and assume steel has the standard electrode potential of iron.
\nCalculate the free energy change, ΔG⦵, in kJ, of the cell reaction. Use sections 1 and 2 of the data booklet.
\nThis cell causes the electrolytic reduction of water on the steel. State the half-equation for this reduction.
\nUse the graph to deduce the dependence of the reaction rate on the amount of Mg.
\nThe reaction is first order with respect to HCl. Calculate the time taken, in seconds (s), for half of the Mg to dissolve when [HCl] = 0.5 mol dm–3.
\nCarbonates also react with HCl and the rate can be determined by graphing the mass loss. Suggest why this method is less suitable for the reaction of Mg with HCl.
\nAlternative 1
\nput Mg in Zn2+(aq) ✔
\nZn/«black» layer forms «on surface of Mg» ✔
\n
Award [1 max] for “no reaction when Zn placed in Mg2+(aq)”.
\n
Alternative 2
\nplace both metals in acid ✔
\nbubbles evolve more rapidly from Mg
OR
Mg dissolves faster ✔
\n
Alternative 3
\nconstruct a cell with Mg and Zn electrodes ✔
\n
Accept “electrons flow from Mg to Zn”.
Accept Mg is negative electrode/anode
OR
Zn is positive electrode/cathode
bulb lights up
OR
shows (+) voltage
OR
size/mass of Mg(s) decreases <<over time>>
OR
size/mass of Zn increases <<over time>>
Accept other correct methods.
Cell potential: «(–0.45 V – (–2.37 V)» = «+»1.92 «V» ✔
\n«ΔGº = -nFEº»
n = 2
OR
ΔGº = «-»2×96500×1.92 / «-»370,560 «J» ✔
-371 «kJ» ✔
\n\n
For n = 1, award [1] for –185 «kJ».
\nAward [1 max] for (+)371 «kJ»
\n2 H2O + 2 e- → H2 + 2 OH- ✔
\n
Accept equation with equilibrium arrows.
independent / not dependent ✔
\n\n
Accept “zero order in Mg”.
\n«2×170 s» = 340 «s» ✔
\n\n
Accept 320 – 360 «s».
\nAccept 400 – 450 «s» based on no more gas being produced after 400 to 450s.
\n«relative/percentage» decrease in mass is «too» small/«much» less ✔
\n
Accept “«relative/percentage» uncertainty in mass loss «too» great”. OR “density/molar mass of H2 is «much» less than CO2”.
Mediocre performance; some experiments would not have worked such as adding magnesium to zinc salt without reference to aqueous environment, adding Zn to magnesium ions, or Mg combustion reaction being more exothermic. In the last one, an inference wad made instead of identifying an observation or measuring temperature using a thermometer or a temperature probe.
\nGood performance; instead of E° = 1.92 V, answer such as −1.92 V + or −2.82 V showed a lack of understanding of how to calculate E° cell.
\nSatisfactory performance; two major challenges in applying the equation ΔG° = −nFE° from the data booklet included:
\nUsing n = 1, not 2, the number of electrons transferred in the redox reaction.
\nΔG° unit from the equation is in J; some did not convert J to kJ as asked for.
\nMediocre performance; some candidates had difficulty writing the reduction half-equation for water, the typical error included O2(g) gas in the reactant or product, rather than H2(g) in the product or including an equation with Fe(s) and H2O(l) as reactants.
\nCandidates found this to be a tough question (see comments for parts (ii) and (iii)).
\nMediocre performance in calculating time from the graph for the data provided. Some wrote the rate expression, which only contains [HCl] and not mass or amount in mol Mg (as a solid, [Mg] is constant). This presented a challenge in arriving at a reasonable answer.
\nPoorly done; many candidates did not grasp the question and answer it appropriately. Candidates generally did not realize that decrease in mass (due to H2(g) as a product for the reaction of Mg with HCl) is «too» small/«much» less compared to that of CO2(g) from the reaction of carbonates with HCl.
\nCarbon forms many compounds.
\nC60 and diamond are allotropes of carbon.
\nChlorine reacts with methane.
\nCH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g)
\nOutline two differences between the bonding of carbon atoms in C60 and diamond.
\nExplain why C60 and diamond sublime at different temperatures and pressures.
\nState two features showing that propane and butane are members of the same homologous series.
\nDescribe a test and the expected result to indicate the presence of carbon–carbon double bonds.
\nDraw the full structural formula of (Z)-but-2-ene.
\nWrite the equation for the reaction between but-2-ene and hydrogen bromide.
\nState the type of reaction.
\nSuggest two differences in the 1H NMR of but-2-ene and the organic product from (d)(ii).
\nPredict, giving a reason, the major product of reaction between but-1-ene and steam.
\nExplain the mechanism of the reaction between 1-bromopropane, CH3CH2CH2Br, and aqueous sodium hydroxide, NaOH (aq), using curly arrows to represent the movement of electron pairs.
\nDeduce the splitting pattern in the 1H NMR spectrum for 1-bromopropane.
\nCalculate the enthalpy change of the reaction, ΔH, using section 11 of the data booklet.
\nDraw and label an enthalpy level diagram for this reaction.
\nAny two of:
\nC60 fullerene: bonded to 3 C AND diamond: bonded to 4 C ✔
\nC60 fullerene: delocalized/resonance AND diamond: not delocalized / no resonance ✔
\nC60 fullerene: sp2 AND diamond: sp3 ✔
\nC60 fullerene: bond angles between 109–120° AND diamond: 109° ✔
\n\n
Accept \"bonds in fullerene are shorter/stronger/have higher bond order OR bonds in diamond longer/weaker/have lower bond order\".
\ndiamond giant/network covalent AND sublimes at higher temperature ✔
\nC60 molecular/London/dispersion/intermolecular «forces» ✔
\n\n
Accept “diamond has strong covalent bonds AND require more energy to break «than intermolecular forces»” for M1.
\nsame general formula / CnH2n+2 ✔
\ndiffer by CH2/common structural unit ✔
\n\n
Accept \"similar chemical properties\".
\nAccept “gradation/gradual change in physical properties”.
\nALTERNATIVE 1:
\nTest:
\nadd bromine «water»/Br2 (aq) ✔
\nResult:
\n«orange/brown/yellow» to colourless/decolourised ✔
\n
Do not accept “clear” for M2.
ALTERNATIVE 2:
Test:
\nadd «acidified» KMnO4 ✔
\nResult:
\n«purple» to colourless/decolourised/brown ✔
\n
Accept “colour change” for M2.
ALTERNATIVE 3:
Test:
\nadd iodine / ✔
\nResult:
\n«brown» to colourless/decolourised ✔
\n
Accept
\nCH3CH=CHCH3 + HBr (g) → CH3CH2CHBrCH3
\nCorrect reactants ✔
\nCorrect products ✔
\n\n
Accept molecular formulas for both reactants and product
\n«electrophilic» addition/EA ✔
\n\n
Do not accept nucleophilic or free radical addition.
\nALTERNATIVE 1: Any two of:
\nbut-2-ene: 2 signals AND product: 4 signals ✔
\nbut-2-ene: «area ratio» 3:1/6:2 AND product: «area ratio» 3:3:2:1 ✔
\nproduct: «has signal at» 3.5-4.4 ppm «and but-2-ene: does not» ✔
\nbut-2-ene: «has signal at» 4.5-6.0 ppm «and product: does not» ✔
\n\n
ALTERNATIVE 2:
\nbut-2-ene: doublet AND quartet/multiplet/4 ✔
\nproduct: doublet AND triplet AND quintet/5/multiplet AND sextet/6/multiplet ✔
\n\n
Accept “product «has signal at» 1.3–1.4 ppm «and but-2-ene: does not»”.
\nCH3CH2CH(OH)CH3 ✔
\n«secondary» carbocation/CH3CH2CH+CH3 more stable ✔
\n\n
Do not accept “Markovnikov’s rule” without reference to carbocation stability.
\ncurly arrow going from lone pair/negative charge on O in HO– to C ✔
\ncurly arrow showing Br breaking ✔
\nrepresentation of transition state showing negative charge, square brackets and partial bonds ✔
\nformation of organic product CH3CH2CH2OH AND Br– ✔
\n\n
Do not allow curly arrow originating on H in HO–.
\nAccept curly arrow either going from bond between C and Br to Br in 1-bromopropane or in the transition
state.
Do not penalize if HO and Br are not at 180° to each other.
\nAward [3 max] for SN1 mechanism.
\ntriplet/3 AND multiplet/6 AND triplet/3 ✔
\nbond breaking: C–H + Cl–Cl / 414 «kJ mol–1» + 242 «kJ mol–1»/656 «kJ»
OR
bond breaking: 4C–H + Cl–Cl / 4 × 414 «kJ mol–1» + 242 «kJ mol–1» / 1898 «kJ» ✔
\n
bond forming: «C–Cl + H–Cl / 324 kJ mol–1 + 431 kJ mol–1» / 755 «kJ»
OR
bond forming: «3C–H + C–Cl + H–Cl / 3 × 414 «kJ mol–1» + 324 «kJ mol–1» + 431 kJ mol–1» / 1997 «kJ» ✔
\n
«ΔH = bond breaking – bond forming = 656 kJ – 755 kJ» = –99 «kJ» ✔
\n\n
Award [3] for correct final answer.
\nAward [2 max] for 99 «kJ».
\nreactants at higher enthalpy than products ✔
\n
ΔH/-99 «kJ» labelled on arrow from reactants to products
OR
activation energy/Ea labelled on arrow from reactant to top of energy profile ✔
\n
Accept a double headed arrow between reactants and products labelled as ΔH for M2.
\nA challenging question, requiring accurate knowledge of the bonding in these allotropes (some referred to graphite, clearly the most familiar allotrope). The most frequent (correct) answer was the difference in number of bonded C atoms and hybridisation in second place. However, only 30% got a mark.
\nAgain, this was a struggle between intermolecular forces and covalent bonds and this proved to be even harder than (a)(i) with only 25% of candidates getting full marks. The distinction between giant covalent/covalent network in diamond and molecular in C60 and hence resultant sublimation points, was rarely explained. There were many general and vague answers given, as well as commonly (incorrectly) stating that intermolecular forces are present in diamond. As another example of insufficient attention to the question itself, many candidates failed to say which would sublime at a higher temperature and so missed even one mark.
\nThis easy question was quite well answered; same/similar physical properties and empirical formula were common errors.
\nCandidates misinterpreted the question and mentioned CH3+, i.e., the lost fragment; the other very common error was -COOH which shows a complete lack of understanding of MS considering the question is about butane so O should never appear.
\nWell answered by most, but some basic chemistry was missing when reporting results, perhaps as a result of little practical work due to COVID. A significant number suggested IR spectrometry, very likely because the question followed one on H NMR spectroscopy, thus revealing a failure to read the question properly (which asks for a test). Some teachers felt that adding \"chemical\" would have avoided some confusion.
\nMost were able to draw this isomer correctly, though a noticeable number of students included the Z as an atom in the structural formula, showing they were completely unfamiliar with E/Z notation.
\nWell done in general and most candidates wrote correct reagents, eventually losing a mark when considering H2 to be a product alongside 2-bromobutane.
\nVery well answered, some mentioned halogenation which is a different reaction.
\nA considerable number of students (40%) got at least 1 mark here, but marks were low (average mark 0.9/2). Common errors were predicting 3 peaks, rather than 4 for 2 -bromobutane and vague / unspecific answers, such as ‘different shifts’ or ‘different intensities’. It is surprising that more did not use H NMR data from the booklet; they were not directed to the section as is generally done in this type of question to allow for more general answers regarding all information that can be obtained from an H NMR spectrum.
\nProduct was correctly predicted by many, but most used Markovnikov's Rule to justify this, failing to mention the stability of the secondary carbocation, i.e., the chemistry behind the rule.
\nAs usual, good to excellent candidates (47.5%) were able to get 3/4 marks for this mechanism, while most lost marks for carelessness in drawing arrows and bond connectivity, issues with the lone pair or negative charge on the nucleophile, no negative charge on transition state, or incorrect haloalkane. The average mark was thus 1.9/4.
\nAnother of the very poorly answered questions where most candidates (90%) failed to predict 3 peaks and when they did, considered there would be a quartet instead of multiplet/sextet; other candidates seemed to have no idea at all. This is strange because the compound is relatively simple and while some teachers considered that predicting a sextet may be beyond the current curriculum or just too difficult, they could refer to a multiplet; a quartet is clearly incorrect.
\nOnly the very weak candidates were unable to calculate the enthalpy change correctly, eventually missing 1 mark for inverted calculations.
\nMost candidates drew correct energy profiles, consistent with the sign of the energy change calculated in the previous question. And again, only very weak candidate failed to get at least 1 mark for correct profiles.
\nWhat is the name of the compound with formula Ti3(PO4)2?
\nA. Titanium phosphate
\nB. Titanium(II) phosphate
\nC. Titanium(III) phosphate
\nD. Titanium(IV) phosphate
\nB
\n71% of candidates were able to name Ti3(PO4)2 correctly. The most commonly chosen distractor was titanium(III) phosphate.
\nThe overall equation for the production of hydrogen cyanide, HCN, is shown below.
\nCH4 (g) + NH3 (g) +O2 (g) → HCN (g) + 3H2O (g)
\nState why NH3 is a Lewis base.
\nCalculate the pH of a 1.00 × 10−2 mol dm−3 aqueous solution of ammonia.
\npKb = 4.75 at 298 K.
\nJustify whether a 1.0 dm3 solution made from 0.10 mol NH3 and 0.20 mol HCl will form a buffer solution.
\nSketch the shape of one sigma () and one pi () bond.
\nIdentify the number of sigma and pi bonds in HCN.
\nState the hybridization of the carbon atom in HCN.
\nSuggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
\nExplain why transition metal cyanide complexes are coloured.
\ndonates «lone/non-bonding» pair of electrons ✔
\nKb = 10-4.75 /1.78 x 10-5
OR
Kb = ✔
\n
[OH–] = « =» 4.22 × 10–4 «(mol dm–3)» ✔
\n\n
pOH« = –log10 (4.22 × 10–4)» = 3.37
AND
pH = «14 – 3.37» = 10.6
OR
[H+]« =» = 2.37 × 10–11
AND
pH« = –log10 2.37 × 10–11» = 10.6 ✔
\n
Award [3] for correct final answer.
\nno AND is not a weak acid conjugate base system
\nOR
\nno AND weak base «totally» neutralized/ weak base is not in excess
\nOR
\nno AND will not neutralize small amount of acid ✔
\n\n
Accept “no AND contains 0.10 mol NH4Cl + 0.10 mol HCl”.
\nSigma ():
\n
\n
Pi ():
\n\n
Accept overlapping p-orbital(s) with both lobes of equal size/shape.
\nShaded areas are not required in either diagram.
\nSigma (): 2 AND Pi (): 2 ✔
\nsp ✔
\nHCN has stronger dipole–dipole attraction ✔
\n\n
Do not accept reference to H-bonds.
\nAny three from:
\npartially filled d-orbitals ✔
\n«CN- causes» d-orbitals «to» split ✔
\nlight is absorbed as electrons transit to a higher energy level «in d–d transitions»
OR
light is absorbed as electrons are promoted ✔
energy gap corresponds to light in the visible region of the spectrum ✔
\n\n
Do not accept “colour observed is the complementary colour” for M4.
\nThe main error was the omission of lone electron \"pair\", though there was also a worrying amount of very confused answers for a very basic chemistry concept where 40% provided very incorrect answers.
\nRather surprisingly, many students got full marks for this multi-step calculation; others went straight to the pH/pKa acid/base equation so lost at least one of the marks: students often seem less prepared for base calculations, as opposed to acid calculations.
\nPoorly answered revealing little understanding of buffering mechanisms, which is admittedly a difficult topic.
\nThis proved to be the most challenging question (10%). It was a good question, where candidates had to explain a huge difference in boiling point of two covalent compounds, requiring solid understanding of change of state where breaking bonds cannot be involved). Yet most considered the triple bonds in HCN as the cause, suggesting covalent bonds break when substance boil, which is very worrying. Others considered H-bonds which at least is an intermolecular force, but shows they are not too familiar with the conditions necessary for H-bonding.
\nThis question appears frequently in exams but with slightly different approaches. In general candidates ignore the specific question and give the same answers, failing in this case to describe why complexes are coloured rather than what colour is seen. These answers appear to reveal that many candidates don't really understand this phenomenon, but learn the answer by heart and make mistakes when repeating it, for example, stating that the ‘d-orbitals of the ligands were split’- an obvious misconception. The average mark was 1.6/3, with a MS providing 4 ideas that would merit a mark
\nAmmonia is soluble in water and forms an alkaline solution:
\nNH3 (g) + H2O (l) NH4+ (aq) + HO– (aq)
\nState the relationship between NH4+ and NH3 in terms of the Brønsted–Lowry theory.
\nDetermine the concentration, in mol dm–3, of the solution formed when 900.0 dm3 of NH3 (g) at 300.0 K and 100.0 kPa, is dissolved in water to form 2.00 dm3 of solution. Use sections 1 and 2 of the data booklet.
\nCalculate the concentration of hydroxide ions in an ammonia solution with pH = 9.3. Use sections 1 and 2 of the data booklet.
\nCalculate the concentration, in mol dm–3, of ammonia molecules in the solution with pH = 9.3. Use section 21 of the data booklet.
\nAn aqueous solution containing high concentrations of both NH3 and NH4+ acts as an acid-base buffer solution as a result of the equilibrium:
\nNH3 (aq) + H+ (aq) NH4+ (aq)
\nReferring to this equilibrium, outline why adding a small volume of strong acid would leave the pH of the buffer solution almost unchanged.
\nMagnesium salts form slightly acidic solutions owing to equilibria such as:
\nMg2+ (aq) + H2O (l) Mg(OH)+ (aq) + H+ (aq)
\nComment on the role of Mg2+ in forming the Mg(OH)+ ion, in acid-base terms.
\nMg(OH)+ is a complex ion, but Mg is not regarded as a transition metal. Contrast Mg with manganese, Mn, in terms of one characteristic chemical property of transition metals, other than complex ion formation.
\nconjugate «acid and base» ✔
\namount of ammonia ✔
\nconcentration ✔
\n\n
Award [2] for correct final answer.
\n✔
\n✔
\n✔
\n\n
Accept other methods of carrying out the calculation.
\nAward [2] for correct answer.
\nequilibrium shifts to right/H+ reacts with NH3 ✔
\n«as large excess» ratio [NH3]:[NH4+] «and hence pH» almost unchanged ✔
\n\n
Accept “strong acid/H+ converted to a weak acid/NH4+ «and hence pH almost unchanged».
\nLewis acid ✔
\naccepts «a lone» electron pair «from the hydroxide ion» ✔
\n\n
Do not accept electron acceptor without mention of electron pair.
\nALTERNATIVE 1
\nProperty: variable oxidation state ✔
\nComparison: Mn compounds can exist in different valencies/oxidation states AND Mg has a valency/oxidation state of +2 in all its compounds ✔
\n
Accept valency.
Accept for second statement “Mg «always» has the same oxidation state”.
\n
ALTERNATIVE 2
\nProperty: coloured ions/compounds/complexes ✔
\nComparison: Mn ions/compounds/complexes coloured AND Mg ions/compounds white/«as solids»/colourless «in aqueous solution» ✔
\n
Accept Mn forms coloured ions/compounds/complexes and Mg does not.
\n
ALTERNATIVE 3
\nProperty: catalytic activity ✔
\nComparison: «many» Mn compounds act as catalysts AND Mg compounds do not «generally» catalyse reactions ✔
For any property accept a correct specific example, for example manganate(VII) is purple.
Do not accept differences in atomic structure, such as partially filled d sub-levels, but award ECF for a correct discussion.
Well done; However, instead of identifying the conjugate acid-base relationship, some simply identified these as Brønsted–Lowry base and acid.
\nGood performance. Some teachers suggested the question had an error in units, but this was not the case. The question had to be solved, first by using the data provided for application of gas law to determine the number of moles of gas. Next, given volume of solution, V = 2.00 dm3, determine its concentration.
\nConcentration of [OH˗] was asked for but some calculated [H3O+] instead. On the whole, question was done well.
\nMediocre performance. Since a mark was given for the Kb expression, that mark could also be scored for the Henderson Hasselbalch (HH) equation, provided it is specific to the equilibrium reaction. Unfortunately, there was poor understanding of the application of the equation in most cases. Students should be strongly encouraged to use the HH equation only when a buffer is involved. Appropriate Ka or Kb expressions should be used when buffer solutions are not involved.
\nMediocre performance. One mark was scored for suggesting equilibrium shifts to right or H+ reacts with NH3. However, some made reference to ammonia being a strong base or no reference to the strong acid, H+ being converted to a weak acid, NH4+.
\nMediocre performance; although some Mg2+ was identified as a Lewis acid, the reasoning given was that it accepts an electron, rather than an electron pair or references were made to Bronsted-Lowry theory.
\nWhat is the main interaction between liquid CH4 molecules?
\nA. London (dispersion) forces
\nB. Dipole–dipole forces
\nC. Hydrogen bonding
\nD. Covalent bonding
\nA
\n56% of the candidates were able to identify London (dispersion) forces as the main interaction between liquid CH4 molecules. Covalent bonding was the most commonly chosen distractor. Good performance on this question correlated well with candidates who scored well overall.
\nWhat is correct about energy changes during bond breaking and bond formation?
\nC
\n55% of the candidates selected the correct statements about bond breaking and bond formation. The most commonly chosen distractor reversed them. The question discriminated well between high-scoring and low-scoring candidates.
\nWhich combination of ΔH1, ΔH2, and ΔH3 would give the enthalpy of the reaction?
\nCS2 (l) + 3O2 (g) → CO2 (g) + 2SO2 (g)
\nΔH1 C (s) + O2 (g) → CO2 (g)
ΔH2 S (s) + O2 (g) → SO2 (g)
ΔH3 C (s) + 2S (s) → CS2 (l)
A. ΔH = ΔH1 + ΔH2 + ΔH3
\nB. ΔH = ΔH1 + ΔH2 − ΔH3
\nC. ΔH = ΔH1 + 2(ΔH2) + ΔH3
\nD. ΔH = ΔH1 + 2(ΔH2) − ΔH3
\nD
\nA very well answered question using Hess’s law. 85% of the candidates used the enthalpies given to calculate the unknown enthalpy of reaction.
\nWhich statement is correct about identical pieces of magnesium added to two solutions, X and Y, containing hydrochloric acid at the same temperature?
\nA. Solution X will reach a higher maximum temperature.
\nB. Solution Y will reach a higher maximum temperature.
\nC. Solutions X and Y will have the same temperature rise.
\nD. It is not possible to predict whether X or Y will have the higher maximum temperature because we cannot identify the limiting reactant.
\nA
\nA nice question that required thinking. There were two factors to consider: amount of reactants and temperature rise. Most candidates focused on the same amount of Mg and HCl hence concluded that the temperature rise would be the same. In fact, it is the same amount of heat that is released, however with a larger volume of solution in Y, the temperature rise is higher for X. 40% of the candidates answered correctly.
\nWhy does a reaction for a sample of gases, at constant temperature, occur faster at higher pressure?
\nA. Collisions are more frequent.
\nB. Collisions are more energetic.
\nC. High pressure lowers activation energy.
\nD. The reaction is more exothermic at high pressure.
\nA
\nA very well-answered question in which 90% of the candidates related the effect of higher pressure on the rate of reaction of gaseous reactants to the collision theory correctly.
\nA reaction has an activation energy of 40 kJ mol−1 and an enthalpy change of −60 kJ mol−1.
\nWhich potential energy diagram illustrates this reaction?
\nC
\nVery well answered. 82% of the candidates selected the potential energy diagram with the correct activation energy and enthalpy change to represent the reaction.
\nHybridization of hydrocarbons affects their reactivity.
\nExperiments were carried out to investigate the mechanism of reaction between 2-chloropentane and aqueous sodium hydroxide.
\nDistinguish between a sigma and pi bond.
\nIdentify the hybridization of carbon in ethane, ethene and ethyne.
\nState, giving a reason, if but-1-ene exhibits cis-trans isomerism.
\nState the type of reaction which occurs between but-1-ene and hydrogen iodide at room temperature.
\nExplain the mechanism of the reaction between but-1-ene with hydrogen iodide, using curly arrows to represent the movement of electron pairs.
\nState, giving a reason, if the product of this reaction exhibits stereoisomerism.
\nDeduce the rate expression for this reaction.
\nDeduce the units of the rate constant.
\nDetermine the initial rate of reaction in experiment 4.
\nDeduce, with a reason, the mechanism of the reaction between 2-chloropentane and sodium hydroxide.
\nDiscuss the reason benzene is more reactive with an electrophile than a nucleophile.
\nSigma (σ) bond:
\noverlap «of atomic orbitals» along the axial / intermolecular axis / electron density is between nuclei
OR
head-on/end-to-end overlap «of atomic orbitals» ✔
\n
Pi (π) bond:
\noverlap «of p-orbitals» above and below the internuclear axis/electron density above and below internuclear axis
OR
sideways overlap «of p-orbitals» ✔
\n
Accept a suitable diagram.
\n
All 3 required for mark.
no AND 2 groups on a carbon «in the double bond» are the same/hydrogen «atoms»
\nOR
\nno AND molecule produced by rearranging atoms bonded on a carbon «in the double bond» is the same as the original ✔
\n«electrophilic» addition ✔
\n\n
Do not allow nucleophilic addition.
\ncurly arrow going from C=C to H of HI AND curly arrow showing I leaving ✔
\nrepresentation of carbocation ✔
\ncurly arrow going from lone pair/negative charge on I– to C+ ✔
\n2-iodobutane formed ✔
\n\n
Penalize incorrect bond, e.g. –CH–H3C or –CH3C once only.
\nyes AND has a carbon attached to four different groups
OR
yes AND it contains a chiral carbon ✔
Accept yes AND mirror image of molecule different to original/non-superimposable on original.
«rate =» k[NaOH][C5H11Cl] ✔
\nmol–1 dm3 s–1 ✔
\nALTERNATIVE 1:
\n«k = » 1.25 «mol–1 dm3 s–1» ✔
\n\n
«rate = 1.25 mol–1 dm3 s–1 × 0.60 mol dm–3 × 0.25 mol dm–3»
\n1.9 x 10–1 «mol dm–3 s–1» ✔
\n\n
ALTERNATIVE 2:
\n«[NaOH] exp. 4 is 3 × exp. 1»
\n«[C5H11Cl] exp. 4 is 2.5 × exp. 1»
\n«exp. 4 will be » 7.5× faster ✔
\n1.9 x 10–1 «mol dm–3 s–1» ✔
\n\n
Award [2] for correct final answer.
\nSN2 AND rate depends on both OH– and 2-chloropentane ✔
\n
Accept E2 AND rate depends on both OH– and 2-chloropentane.
delocalized electrons/pi bonds «around the ring»
OR
molecule has a region of high electron density/negative charge ✔
electrophiles are attracted/positively charged AND nucleophiles repelled/negatively charged ✔
\n\n
Do not accept just “nucleophiles less attracted” for M2.
\nAccept “benzene AND nucleophiles are both electron rich” for “repels nucleophiles”.
\nHow many dichlorinated butane isomers can be formed by the halogenation of CH3CH2CH2CH3 with excess Cl2 in the presence of UV light?
\nA. 4
\nB. 6
\nC. 8
\nD. 10
\nB
\nCandidates had to work out the possible number of isomers of dichlorobutane and the majority only thought of 4 of the 6 possible isomers. 32% of the candidates obtained the correct answer.
\nAmmonia is produced by the Haber–Bosch process which involves the equilibrium:
\nN2 (g) + 3 H2 (g) 2 NH3 (g)
\nThe percentage of ammonia at equilibrium under various conditions is shown:
\n[The Haber Bosch Process [graph] Available at: https://commons.wikimedia.org/wiki/File:Ammonia_yield.png
[Accessed: 16/07/2022].]
One factor affecting the position of equilibrium is the enthalpy change of the reaction.
\nThe standard free energy change, ΔG⦵, for the Haber–Bosch process is –33.0 kJ at 298 K.
\nDeduce the expression for the equilibrium constant, Kc, for this equation.
\nState how the use of a catalyst affects the position of the equilibrium.
\nWith reference to the reaction quotient, Q, explain why the percentage yield increases as the pressure is increased at constant temperature.
\nDetermine the enthalpy change, ΔH, for the Haber–Bosch process, in kJ. Use Section 11 of the data booklet.
\nOutline why the value obtained in (b)(i) might differ from a value calculated using ΔHf data.
\nDemonstrate that your answer to (b)(i) is consistent with the effect of an increase in temperature on the percentage yield, as shown in the graph.
\nState, giving a reason, whether the reaction is spontaneous or not at 298 K.
\nCalculate the value of the equilibrium constant, K, at 298 K. Use sections 1 and 2 of the data booklet.
\nCalculate the entropy change for the Haber–Bosch process, in J mol–1 K–1 at 298 K. Use your answer to (b)(i) and section 1 of the data booklet.
\nOutline, with reference to the reaction equation, why this sign for the entropy change is expected.
\n✔
\nsame/unaffected/unchanged ✔
\nincreasing pressure increases «all» concentrations
OR
increasing pressure decreases volume ✔
Q becomes less than Kc
OR
affects the lower line/denominator of Q expression more than upper line/numerator ✔
«for Q to once again equal Kc,» ratio of products to reactants increases
OR
«for Q to once again equal Kc,» equilibrium shifts to right/products ✔
\n
Award [2 max] for answers that do not refer to Q.
\nbonds broken: N≡N + 3(H-H) /«1 mol×»945 «kJ mol–1» + 3«mol»×436 «kJ mol–1» / 945 «kJ» + 1308 «kJ» / 2253 «kJ» ✔
\nbonds formed: 6(N-H) / 6«mol»×391 «kJ mol–1» / 2346 «kJ» ✔
\nΔH = «2253 kJ - 2346 kJ = » -93 «kJ» ✔
\n\n
Award [2 max] for (+)93 «kJ».
\n«N-H» bond enthalpy is an average «and may not be the precise value in NH3» ✔
\n\n
Accept ΔHf data are more accurate / are not average values.
\nincreased temperature decreases yield «as shown on graph» ✔
\nshifts equilibrium in endothermic/reverse direction ✔
\nspontaneous AND ΔG < 0 ✔
\n✔
\n✔
\n\n
Award [2] for correct final answer.
\nAccept answers in the range 4.4×105 to 6.2×105 (arises from rounding of ln K).
\nΔG = «ΔH – TΔS =» –93000 «J» – 298«K» × ΔS = –33000 ✔
\nΔS = 〈〈〉〉 = –201 «J mol–1 K–1» ✔
\n\n
Do not penalize failure to convert kJ to J in both (c)(ii) and (c)(iii).
\nAward [2] for correct final answer
\nAward [1 max] for (+) 201 «J mol–1 K–1».
\nAward [2] for –101 or –100.5 «J mol–1 K–1».
\n«forward reaction involves» decrease in number of moles «of gas» ✔
\nDeducing the equilibrium constant expression for the given equation was done very well.
\nGood performance; however, some misread the question as asking for the effect of a catalyst on equilibrium, rather than on the position of equilibrium.
\nMediocre performance; very few identified the effect of increasing pressure on all concentrations. Consequently, Q becomes less than Kc (it affects the denominator of Q expression more than the numerator) was not addressed. Question was often answered with respect to kinetics, namely greater frequency of collisions and speed of reaction rather than from equilibrium perspective based on effect of increase in pressure on concentrations.
\nGood performance; often the bond energy for single N–N bond instead of using it for the triple bond and not taking into consideration the coefficient of three in calculation of bond enthalpies of ammonia. Also, instead of using BE of bonds broken minus those that were formed, the operation was often reversed. Students should be encouraged to draw the Lewis structures in the equations first to determine the bonds being broken and formed.
\nOutlining why ΔHrxn based on BE values differ due to being average compared to using ΔHf values was generally done well.
\nGood performance; some did not relate that increased temperature decreases yield «as shown on graph» and others arrived at incorrect shift in equilibrium for the reaction.
\nReason for the reaction being spontaneous was generally very done well indeed.
\nGood performance; for lnK calculation in the equation ΔG = RTlnK, ΔG unit had to be converted from kJ to J. This led to an error of 1000 in the value of lnK for some.
\nVery good performance; since the unit for S is J mol˗1 K˗1, ΔG and ΔH needed to be converted from kJ to J, but that was not done in some cases.
\nAverage performance for sign of the entropy change expected for the reaction. Some answers were based on ΔG value rather than in terms of decrease in number of moles of gas or had no idea how to address the question.
\nWhich structure represents a repeating unit of a polymer formed from propene?
\nA. –CH2–CH(CH3)–
\nB. –CH2–CH2–CH2–
\nC. –CH(CH3)–CH(CH3)–
\nD. –CH2–CH2–
\nA
\n44% of the candidates identified the repeating unit of the polymer formed from propene. The most commonly chosen distractor was a straight chain.
\nWhich is a homologous series?
\nA. C2H4, C3H5, C4H6
\nB. C2H2, C3H4, C4H6
\nC. C2H2, C2H4, C2H6
\nD. C2H2, C4H4, C6H6
\nB
\n45% of the candidates identified the members of the same homologous series (ethyne, propyne and butyne). The most commonly chosen distractor included the compounds C2H2, C4H4 and C6H6. The question had good discrimination between high-scoring and low-scoring candidates.
\nWhich reaction mechanisms are typical for alcohols and halogenoalkanes?
\nD
\nOnly 25% of the candidates deduced the reaction mechanisms that are typical for alcohols and halogenoalkanes. Candidates were expected to deduce that the reactions would be substitutions since both compounds are saturated.
\nHow many signals are observed in the 1H NMR spectrum of this compound?
A. 1
B. 2
\nC. 3
\nD. 4
\nA
\nThe compound used for this 1H NMR question was unfamiliar to the candidates and it was a skeletal formula. They were expected to notice its symmetry and that all H’s had the same chemical environment. The majority of candidates saw only part of this symmetry and decided on two chemical environments. 29% of the candidates obtained the correct answer. Some teachers commented on the difficulty of this question.
\nWhat is the uncertainty, in cm3, of this measurement?
A. ±0.01
B. ±0.1
\nC. ±0.15
\nD. ±1
\nC
\nThis question generated a lot of debate among teachers and examiners. The answer B is the best answer but it is a rather low value. It was selected by 34% of the candidates. Many teachers argued that the correct answer should be 0.5 cm3.
\n20 cm3 of 1 mol dm−3 sulfuric acid was added dropwise to 20 cm3 of 1 mol dm−3 barium hydroxide producing a precipitate of barium sulfate.
\nH2SO4 (aq) + Ba(OH)2 (aq) → 2H2O (l) + BaSO4 (s)
\nWhich graph represents a plot of conductivity against volume of acid added?
\nB
\nThe question involved the formation of a precipitate – not a familiar concept to the candidates. It required candidates (to work out how the number of ions and hence conductivity were changing. 37% of the candidates got the correct answer (conductivity decreased as volume of H2SO4 increased). The majority increased conductivity as H2SO4 was added.
\nWhich equation represents the bond enthalpy for H–Br in hydrogen bromide?
\nA. HBr (g) → H+ (g) + Br− (g)
\nB. HBr (g) → H (g) + Br (g)
\nC. HBr (g) → H2 (g) + Br2 (l)
\nD. HBr (g) → H2 (g) + Br2 (g)
\nB
\nOnly a quarter of the candidates identified the equation representing the bond enthalpy of HBr. The most commonly chosen distractor had the elements H2 and Br2 as the products.
\nWhat is the correct order for increasing first ionization energy?
\nA. Na < Mg < Al
\nB. Na < Al < Mg
\nC. Al < Mg < Na
\nD. Al < Na < Mg
\nB
\nThe majority of candidates focussed on the general trend of ionization energy increasing across the period and missed out the finer details due to sub-levels. The correct order for increasing first ionization energy was Na < Al < Mg rather than Na < Mg < Al.
\nWhich term in the expression ΔG⦵ = ΔH⦵ − TΔS⦵ is an indirect measure of the entropy change of the surroundings when divided by T?
\nA. ΔG⦵
\nB. ΔH⦵
\nC. ΔS⦵
\nD. −TΔS⦵
\nB
\nThis challenging question raised some debate and several teachers did not think it was suitable for this level. Most candidates chose -TΔS as “an indirect measure of the entropy change of the surroundings when divided by T”. While candidates are not expected to be familiar with the equation ΔS surroundings = -ΔHsystem/T, they could solve the question by recognizing that ΔH is the value that affects the surroundings while ΔS relates to the system. The wording of the question could have been simplified.
\nWhich are the most reactive elements of the alkali metals and halogens?
\nA. Lithium and fluorine
\nB. Lithium and iodine
\nC. Caesium and fluorine
\nD. Caesium and iodine
\nC
\n70% of candidates selected the most reactive alkali metal and halogen. The most commonly chosen distractor was lithium and fluorine.
\n50.00 cm3 of 0.75 mol dm−3 sodium hydroxide was added in 1.00 cm3 portions to 22.50 cm3 of 0.50 mol dm−3 chloroethanoic acid.
\nCalculate the initial pH before any sodium hydroxide was added, using section 21 of the data booklet.
\nThe concentration of excess sodium hydroxide was 0.362 mol dm−3. Calculate the pH of the solution at the end of the experiment.
\nSketch the neutralisation curve obtained and label the equivalence point.
\n«Ka = 10–2.87 = 1.35 × 10–3 »
\n«1.35 × 10–3 = »
\n«x = [H+] ==» 2.6 × 10–2 «mol dm–3» ✔
\n
«pH = –log[H+] = –log(2.6 × 10–2) =» 1.59 ✔
\n
Accept final answer in range 1.58–1.60.
\nAward [2] for correct final answer.
\n«pOH = –log(0.362) = 0.441»
\n«pH = 14.000 – 0.441 =» 13.559 ✔
\nOR
\nstarts at 1.6 AND finishes at 13.6 ✔
\napproximately vertical at the correct volume of alkali added ✔
\nequivalence point labelled AND above pH 7 ✔
\n\n
Accept any range from 1.1-1.9 AND 13.1-13.9 for M1 or ECF from 11c(i) and 11c(ii).
\nAward M2 for vertical climb at 28 cm3 OR 15 cm3.
\nEquivalence point must be labelled for M3.
\nWhich of these ions are likely to be paramagnetic?
\nI. Ti3+
II. Cr3+
III. Fe3+
A. I and II only
B. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nHalf of the candidates selected the correct combination of paramagnetic ions. The most commonly chosen distractor excluded Ti3+.
\nWhich energy profile diagram represents an exothermic SN1 reaction?
\nC
\n0.50 mol of (g) and 0.50 mol of Br2 (g) are placed in a closed flask. The following equilibrium is established.
\n(g) + Br2 (g) 2Br (g)
\nThe equilibrium mixture contains 0.80 mol of Br (g). What is the value of Kc?
\nA. 0.64
\nB. 1.3
\nC. 2.6
\nD. 64
\nD
\n41% of the candidates were able to calculate Kc given the initial amounts of reactants and the amount of product at equilibrium.
\n[Cr(OH2)6]3+ is violet and [Cr(NH3)6]3+ is yellow. What is correct?
\nThe Colour Wheel
\nB
\n61% of the candidates were able to determine the relative d-level splitting and the wavelength of light absorbed by complex ions with different ligands, given the colours of the complex ions and a colour wheel labelled with the wavelengths of light.
\n0.2 mol of sodium hydrogencarbonate is decomposed by heating until constant mass.
\n2 NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g)
\nHow many moles of gas are produced?
\nA. 0.1
\nB. 0.2
\nC. 0.3
\nD. 0.4
\nB
\nWhat happens to the amount of hydroxide ions and hydroxide ion concentration when water is added to a solution of NH3 (aq)?
\nC
\n35% of the candidates deduced the change in amount of OH- and [OH-] when water is added to NH3(aq). The majority of candidates recognized the increase in the amount of OH- due to the shift in the equilibrium position but assumed that the concentration of OH- increased as well ignoring the effect of the dilution. Some teachers commented that this question was too complicated.
\nIn which of the following compounds does ionic bonding predominate?
\nA. HCl
\nB. NaF
\nC. NH4Br
\nD. NaOH
\nB
\nA well answered question. 70% of the candidates selected NaF as the compound in which ionic bonding predominates. Several teachers expressed concern regarding the unusual wording of this question, but thankfully the candidates understood it in the right way.
\nWhich sample contains the fewest moles of HCl?
\nNA = 6.02 × 1023 mol–1.
\nMolar volume of an ideal gas at STP = 22.7 dm3 mol–1.
\n
A. 10.0 cm3 of 0.1 mol dm–3 HCl (aq)
B. 6.02 × 1024 molecules of HCl (g)
\nC. 0.365 g of HCl (g)
\nD. 2.27 dm3 of HCl (g) at STP
\nA
\nA weak base is titrated with a strong acid. Which value of pKb can be estimated from this titration curve?
\nA. 11.3
\nB. 9.2
\nC. 4.8
\nD. 1.8
\nC
\nOnly 29% of candidates determined pKb from the titration curve. The majority of the candidates made the mistake of using the pH without calculating the corresponding pOH.
\nWhat is the formal charge of the oxygen atom in H3O+?
\nA. −2
\nB. −1
\nC. 0
\nD. +1
\nD
\n44% of candidates were able to determine the formal charge of the oxygen atom in H3O+. The most commonly chosen distractor was -2 (the oxidation state of the oxygen).
\nWhat is the molecular geometry of SF4?
\nA. Tetrahedral
\nB. Trigonal bipyramidal
\nC. See-saw
\nD. Square planar
\nC
\n54% of the candidates deduced the molecular geometry of SF4. The distractors were chosen with equal frequency by the rest of the candidates. This question discriminated well between high-scoring and low-scoring candidates.
\nWhich species are both Lewis and Brønsted–Lowry bases?
\nI. CN−
II. OH−
III. NH3
A. I and II only
B. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\n45% of candidates chose all three species as Lewis and Brønsted-Lowry bases. The most commonly chosen distractor excluded CN-.
\nWhich equation represents hydration enthalpy?
\nA. Na+ (g) → Na+ (aq)
\nB. Na+ (aq) → Na+ (g)
\nC. NaCl (s) → NaCl (aq)
\nD. NaCl (aq) → NaCl (s)
\nA
\n65% of the candidates recognized the equation that represents hydration enthalpy. The most commonly chosen distractor was an enthalpy of solution.
\nWhich E⦵ value, in V, for the reaction Mn (s) + Zn2+ (aq) → Mn2+ (aq) + Zn (s) can be deduced from the following equations?
\nMn (s) + 2Ag+ (aq) → Mn2+ (aq) + 2Ag (s) E⦵ = 1.98 V
\nZn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s) E⦵ = 1.10 V
\nCu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s) E⦵ = 0.46 V
\nA. 0.42
\nB. 1.34
\nC. 2.62
\nD. 3.54
\nA
\n56% of candidates were able to deduce the EƟ of the reaction.
\nWhat is the order of increasing mass deposited by this electrolytic cell?
\nAr Ag = 108, Cu = 64, Sb = 122
\nA. Ag < Cu < Sb
\nB. Sb < Ag < Cu
\nC. Cu < Ag < Sb
\nD. Cu < Sb < Ag
\nD
\nA rather challenging question. Only 30% of the candidates were able to deduce the relative masses of metals deposited in three electrolytic cells in series. Candidates had to take into account the charge of the metal ions and the molar masses of the metals therefore it was a time-consuming question. The most commonly chosen distractor, C, only took the molar masses into account.
\nWhich sequence of reagents converts propene to propanone?
\nA
\n66% of the candidates were able to identify the reagents needed to convert propene to propanone in three steps. This question discriminated well between high-scoring and low-scoring candidates.
\nWhat are the signs of ΔH and ΔS for a reaction that is non-spontaneous at low temperatures but spontaneous at high temperatures?
\nD
\nTwo-thirds of the candidates related information about the spontaneity of a reaction at different temperatures to the signs of ΔH and ΔS correctly. This question discriminated well between high-scoring and low-scoring candidates.
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