Powdered 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.
\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.
\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.
\n\n \n Identify the chiral carbon atom using an asterisk, *.\n \n
\n\n [1]\n
\n\n \n Enantiomers can be identified using a polarimeter. Outline how this instrument differentiates the enantiomers.\n \n
\n\n [2]\n
\n\n \n \n [\n \n ✔]\n \n \n
\n \n «plane-»polarized light passed through sample\n \n [✔]\n \n \n
\n\n \n
\n analyser/second polarizer determines angle of rotation of plane of plane-polarized light\n
\n \n \n OR\n \n \n
\n each enantiomer rotates plane «of plane-polarized light» in opposite directions «by the same angle»\n \n [✔]\n \n \n
\n 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.\n
\n\n Very 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.\n
\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.
\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.
\n\n \n Draw the structure of the repeating unit of starch and state the type of linkage formed between these units.\n \n
\n\n \n \n \n
\n \n Type of linkage:\n \n
\n\n [2]\n
\n\n \n Formulate the equation for the complete hydrolysis of a starch molecule, (C\n \n 6\n \n H\n \n 10\n \n O\n \n 5\n \n )\n \n n\n \n .\n \n
\n\n [1]\n
\n\n \n
\n \n continuation bonds\n \n \n AND\n \n \n −O attached to just one end\n \n \n AND\n \n \n both H atoms on end carbons must be on the same side\n \n [✔]\n \n \n
\n\n \n
\n \n Type of linkage:\n \n
\n glycosidic\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Square brackets not required.\n \n \n
\n\n \n \n Ignore “n” if given.\n \n \n
\n\n \n \n Mark may be awarded if a polymer is shown but with the repeating unit clearly identified.\n \n \n
\n\n \n \n Accept “ether”.\n \n \n
\n\n
\n\n \n (C\n \n \n 6\n \n \n H\n \n \n 10\n \n \n O\n \n \n 5\n \n \n )\n \n \n n\n \n \n (s) +\n \n n\n \n H\n \n 2\n \n O (l) →\n \n n\n \n C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n (aq)\n \n [✔]\n \n \n
\n\n
\n\n \n \n Note:\n \n Accept “(n-1)H\n \n 2\n \n O”.\n \n
\n\n \n Do\n \n not\n \n award mark if “n” not included.\n \n
\n\n Candidates 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\n \n \n \n -glucose was given, or an attempt was made to draw a polymer. Naming the type of linkage formed was answered well.\n
\n\n Also proved challenging, with many candidates unable to write an equation for the hydrolysis of a starch molecule (C\n \n 6\n \n H\n \n 10\n \n O\n \n 5\n \n )\n \n n\n \n . The n was often omitted from otherwise correct equations or the product was incorrectly given as (C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n )\n \n n\n \n .\n
\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.
\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.
\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.
\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.
\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.
\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.
\n\n \n Ascorbic acid and retinol are two important vitamins.\n \n
\n\n \n Explain why ascorbic acid is soluble in water and retinol is not. Use section 35 of the data booklet.\n \n
\n\n [2]\n
\n\n \n The melting points of cocoa butter and coconut oil are 34 °C and 25 °C respectively.\n \n
\n\n \n Explain this in terms of their saturated fatty acid composition.\n \n
\n\n [3]\n
\n\n \n \n ascorbic acid:\n \n many hydroxyl/OH groups\n \n \n AND\n \n retinol\n \n : few/one hydroxyl/OH group\n
\n \n \n OR\n \n \n
\n \n ascorbic acid\n \n : many hydroxyl/OH groups\n \n \n AND\n \n retinol\n \n : long hydrocarbon chain\n \n [✔]\n \n \n
\n \n \n ascorbic acid\n \n : «many» H-bond with water\n
\n \n \n OR\n \n \n
\n \n retinol\n \n : cannot «sufficiently» H-bond with water\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Do\n \n not\n \n accept “OH\n \n −\n \n /hydroxide”.\n \n \n
\n\n \n coconut oil has higher content of lauric/short-chain «saturated» fatty acids\n
\n \n \n OR\n \n \n
\n cocoa butter has higher content of stearic/palmitic/longer chain «saturated» fatty acids\n \n [\n \n ✔\n \n ]\n \n \n
\n \n longer chain fatty acids have greater surface area/larger electron cloud\n \n [✔]\n \n \n
\n\n \n stronger London/dispersion/instantaneous dipole-induced dipole forces «between triglycerides of longer chain saturated fatty acids»\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Do\n \n not\n \n accept arguments that relate to melting points of saturated and unsaturated fats.\n \n \n
\n\n Another 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.\n
\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.\n
\n\n \n Show that, for combustion of equal masses of fuel, ethanol (\n \n M\n \n r\n \n \n = 46 g mol\n \n −1\n \n ) has a lower carbon footprint than octane (\n \n M\n \n \n r\n \n = 114 g mol\n \n −1\n \n ).\n \n
\n\n [3]\n
\n\n \n Biodiesel containing ethanol can be made from renewable resources.\n \n
\n\n \n Suggest\n \n one\n \n environmental disadvantage of producing biodiesel from renewable resources.\n \n
\n\n [1]\n
\n\n \n \n \n Alternative 1\n \n \n
\n C\n \n 2\n \n H\n \n 5\n \n OH (l) + 3O\n \n 2\n \n (g) → 2CO\n \n 2\n \n (g) + 3H\n \n 2\n \n O (l) / 1 mol ethanol produces 2 mol CO\n \n 2\n \n
\n \n \n OR\n \n \n
\n C\n \n 8\n \n H18 (l) + 12.5O\n \n 2\n \n (g) → 8CO\n \n 2\n \n (g) + 9H\n \n 2\n \n O (l) / 1 mol octane produces 8 mol CO\n \n 2\n \n \n [✔]\n \n \n
\n \n
\n For 1 g of fuel:\n
\n «\n \n \n \n × 2 mol CO\n \n 2\n \n (g) =» 0.04 «mol CO\n \n 2\n \n (g)» from ethanol\n \n [✔]\n \n \n
\n \n «\n \n \n \n × 8 mol CO\n \n 2\n \n (g) =» 0.07 «mol CO\n \n 2\n \n (g)» from octane\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Alternative 2\n \n \n
\n ratio of C in ethanol:octane is 2:8, so ratio in carbon dioxide produced per mole will be 1:4\n \n [✔]\n \n \n
\n \n ratio amount of fuel in 1 g =\n \n \n \n = 2.5:1\n \n [✔]\n \n \n
\n\n \n 4 > 2.5 so octane produces more carbon dioxide\n
\n \n \n OR\n \n \n
\n ratio of amount of carbon dioxide = 2.5:4 = 1:1.61 so octane produces more «for combustion of same mass»\n \n [✔]\n \n \n
\n \n use of «farm» land «for production»\n
\n \n \n OR\n \n \n
\n deforestation «for crop production for fuel»\n
\n \n \n OR\n \n \n
\n can release more NO\n \n x\n \n «than normal fuel on combustion»\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Ignore any reference to cost.\n \n \n
\n\n A 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.\n
\n\n Many 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.\n
\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.
\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.
\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.
\n\n \n Draw a circle around the functional group formed between the amino acids and state its name.\n \n
\n\n \n \n \n
\n \n Name:\n \n
\n\n [2]\n
\n\n \n Calculate the energy released, in kJ g\n \n −1\n \n , 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.\n \n
\n\n [2]\n
\n\n \n
\n \n \n Name\n \n :\n
\n amide/amido/carboxamide\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “peptide bond/linkage”.\n \n \n
\n\n \n \n q\n \n = «\n \n mc\n \n ΔT = 975 g × 4.18 J g\n \n –1\n \n K\n \n –1\n \n × 15.0 K =» 61 100 «J» / 61.1 «kJ»\n \n [✔]\n \n \n
\n\n \n «heat per gram =\n \n \n \n \n =\n \n » 17.5 «kJ g\n \n –1\n \n »\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Award\n \n [2]\n \n for correct final answer.\n \n \n
\n\n Many 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.\n
\n\n The 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\n \n −1\n \n so a further calculation was required.\n
\n\n \n State\n \n one\n \n greenhouse gas, other than carbon dioxide.\n \n
\n\n [1]\n
\n\n \n Outline\n \n one\n \n approach to controlling industrial emissions of carbon dioxide.\n \n
\n\n [1]\n
\n\n \n \n Any one of:\n \n
\n methane, water, nitrous oxide/nitrogen(I) oxide, ozone, CFCs, sulfur hexafluoride\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept formulas.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “NO\n \n 2\n \n ”, “NO\n \n x\n \n ”, “oxides of sulfur”.\n \n \n
\n\n \n \n Any one of:\n \n
\n capture where produced «and stored»\n \n [✔]\n \n \n
\n \n use scrubbers to remove\n \n [✔]\n \n \n
\n\n \n use as feedstock for synthesizing other chemicals\n \n [✔]\n \n \n
\n\n \n carbon credit/tax/economic incentive/fines/country specific action\n \n [✔]\n \n
\n \n
\n \n use alternative energy\n
\n \n \n OR\n \n \n
\n stop/reduce use of fossil fuels for producing energy\n \n [✔]\n \n \n
\n \n use carbon reduced fuels «such as methane»\n \n [✔]\n \n \n
\n\n \n increase efficiency/reduce energy use\n \n [✔]\n \n \n
\n\n This question was well answered.\n
\n\n This 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.\n
\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.
\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.
\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.
\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”.
\n\n \n Draw the repeating unit of polyphenylethene.\n \n
\n\n [1]\n
\n\n \n Determine the density of calcium, in g cm\n \n −3\n \n , using section 2 of the data booklet.\n \n
\n\n \n Ar = 40.08; metallic radius (r) = 1.97 × 10\n \n −10\n \n m\n \n
\n\n [3]\n
\n\n \n Suggest\n \n two\n \n reasons why oil decomposes faster at the surface of the ocean than at greater depth.\n \n
\n\n [2]\n
\n\n \n Oil spills can be treated with an enzyme mixture to speed up decomposition.\n \n
\n\n \n Outline\n \n one\n \n factor to be considered when assessing the greenness of an enzyme mixture.\n \n
\n\n [1]\n
\n\n \n The minor product, C\n \n 6\n \n H\n \n 5\n \n –CH\n \n 2\n \n –CH\n \n 2\n \n Br, can be directly converted to an intermediate compound,\n \n X\n \n , which can then be directly converted to the acid C\n \n 6\n \n H\n \n 5\n \n –CH\n \n 2\n \n –COOH.\n \n
\n\n \n C\n \n 6\n \n H\n \n 5\n \n –CH\n \n 2\n \n –CH\n \n 2\n \n Br →\n \n X\n \n → C\n \n 6\n \n H\n \n 5\n \n –CH\n \n 2\n \n –COOH\n \n
\n\n \n Identify\n \n X\n \n .\n \n
\n\n [1]\n
\n\n \n \n [\n \n ✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n \n \n Do\n \n not\n \n penalize the use of brackets and “n”.\n \n \n
\n\n \n \n Do\n \n not\n \n award the mark if the continuation bonds are missing.\n \n \n
\n\n \n \n a\n \n = «\n \n \n \n =» 5.572 × 10\n \n –10\n \n «m»\n
\n \n \n OR\n \n \n
\n volume of unit cell = «(5.572 × 10\n \n –10\n \n m)\n \n 3\n \n × 10\n \n 6\n \n =» 1.73 × 10\n \n –22\n \n «cm\n \n 3\n \n »\n \n [✔]\n \n \n
\n \n mass of unit cell =«\n \n \n \n =» 2.66 × 10\n \n –22\n \n «g»\n \n [✔]\n \n \n
\n\n \n density = «\n \n \n \n » 1.54 «g cm\n \n –3\n \n »\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Award\n \n [3]\n \n for correct final answer.\n \n \n
\n\n \n \n Any two of:\n \n
\n surface water is warmer «so faster reaction rate»/more light/energy from the sun\n \n [✔]\n \n \n
\n \n more oxygen «for aerobic bacteria/oxidation of oil»\n \n [✔]\n \n \n
\n\n \n greater surface area\n \n [✔]\n \n \n
\n\n \n \n Any one of:\n \n
\n non-hazardous/toxic to the environment/living organisms\n \n [✔]\n \n \n
\n \n energy requirements «during production»\n \n [✔]\n \n \n
\n\n \n quantity/type of waste produced «during production»\n
\n \n \n OR\n \n \n
\n atom economy\n \n [✔]\n \n \n
\n \n safety of process\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “use of solvents/toxic materials «during production»”.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “more steps involved”.\n \n \n
\n\n \n C\n \n 6\n \n H\n \n 5\n \n –CH\n \n 2\n \n –CH\n \n 2\n \n OH\n \n [✔]\n \n \n
\n\n Most candidates were able to draw the monomer correctly. Some candidates made careless mistakes writing C\n \n 6\n \n H\n \n 6\n \n .\n
\n\n Majority of the candidates managed to get three marks in determining the density of the calcium.\n
\n\n While 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.\n
\n\n Students 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.\n
\n\n Quite 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.\n
\n\n \n The regular rise and fall of sea levels, known as tides, can be used to generate energy.\n \n
\n\n \n State\n \n one\n \n advantage, other than limiting greenhouse gas emissions, and one disadvantage of tidal power.\n \n
\n\n
\n\n \n Advantage:\n \n
\n\n \n Disadvantage:\n \n
\n\n [2]\n
\n\n \n Suggest\n \n two\n \n reasons why oil decomposes faster at the surface of the ocean than at greater depth.\n \n
\n\n [2]\n
\n\n \n Oil spills can be treated with an enzyme mixture to speed up decomposition.\n \n
\n\n \n Outline\n \n one\n \n factor to be considered when assessing the greenness of an enzyme mixture.\n \n
\n\n [1]\n
\n\n \n \n \n Advantage\n \n \n
\n \n Any one of:\n \n
\n renewable\n \n [✔]\n \n
\n predictable supply\n \n [✔]\n \n
\n tidal barrage may prevent flooding\n \n [✔]\n \n
\n effective at low speeds\n \n [✔]\n \n
\n long life-span\n \n [✔]\n \n
\n low cost to run\n \n [✔]\n \n \n
\n \n
\n \n \n Disadvantage\n \n \n
\n \n Any one of:\n \n
\n cost of construction\n \n [✔]\n \n
\n changes/unknown effects on marine life\n \n [✔]\n \n
\n changes circulation of tides in the area\n \n [✔]\n \n
\n power output is variable\n \n [✔]\n \n
\n limited locations where feasible\n \n [✔]\n \n
\n equipment maintenance can be challenging\n \n [✔]\n \n
\n difficult to store energy\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note\n \n : Do\n \n not\n \n accept vague generalizations.\n \n \n
\n\n \n \n Do\n \n not\n \n accept economic issues for both advantage and disadvantage.\n \n \n
\n\n \n \n Do\n \n not\n \n accept sustainable.\n \n \n
\n\n \n \n Accept “energy” or “electricity” for “power”.\n \n \n
\n\n \n \n Any two of:\n \n
\n surface water is warmer «so faster reaction rate»/more light/energy from the sun\n \n [✔]\n \n \n
\n \n more oxygen «for aerobic bacteria/oxidation of oil»\n \n [✔]\n \n \n
\n\n \n greater surface area\n \n [✔]\n \n \n
\n\n \n \n Any one of:\n \n
\n non-hazardous/toxic to the environment/living organisms\n \n [✔]\n \n \n
\n \n energy requirements «during production»\n \n [✔]\n \n \n
\n\n \n quantity/type of waste produced «during production»\n
\n \n \n OR\n \n \n
\n atom economy\n \n [✔]\n \n \n
\n \n safety of process\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note\n \n : Accept “use of solvents/toxic materials «during production»”.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “more steps involved”.\n \n \n
\n\n Many 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.\n
\n\n Many 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.\n
\n\n The 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.\n
\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.
\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.
\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 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.
\n\n \n Discuss the data.\n \n
\n\n [3]\n
\n\n \n Outline what is meant by the degradation of energy.\n \n
\n\n [1]\n
\n\n \n «similar specific energy and» pentane has «much» larger energy density ✔\n \n
\n\n \n \n Any two for\n \n [2 max]\n \n :\n \n
\n similar number of bonds/«C and H» atoms in 1 kg «leading to similar specific energy»\n
\n \n \n OR\n
\n \n \n only one carbon difference in structure «leading to similar specific energy» ✔\n
\n \n NOTE:\n \n \n Accept “both are alkanes” for M2.\n \n
\n \n
\n \n pentane is a liquid\n \n \n AND\n \n \n butane is a gas «at STP» ✔\n
\n \n NOTE: Accept “pentane would be easier to transport”.\n \n
\n \n
\n \n 1 m\n \n 3\n \n of pentane contains greater amount/mass than 1 m\n \n 3\n \n of butane ✔\n
\n \n NOTE: Accept “same volume” for “1 m\n \n 3\n \n ” and “more moles” for “greater amount” for M4.\n \n
\n \n
\n \n energy converted to heat\n
\n \n \n OR\n \n \n
\n energy converted to less useful/dispersed forms\n
\n \n \n OR\n \n \n
\n energy converted to forms that have lower potential to do work\n
\n \n \n OR\n \n \n
\n heat transferred to the surroundings ✔\n \n
\n \n \n NOTE: Reference to energy conversion/transfer required. Do\n \n not\n \n accept reference to loss of energy.\n \n \n
\nThe 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» ✔
\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”.
\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”.
In 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.
\n\n \n The vapour pressure of pure ethanal at\n \n \n \n is\n \n .\n \n
\n\n \n Calculate the vapour pressure of ethanal above the liquid mixture at\n \n \n \n .\n \n
\n\n [1]\n
\n\n \n Describe how this mixture is separated by fractional distillation.\n \n
\n\n [2]\n
\n\n \n ✔\n
\n\n \n Any two of:\n \n
\n continuous evaporation and condensation\n
\n \n \n OR\n \n \n
\n increased surface area in column helps condensation ✔\n
\n \n Accept “glass «beads» aid condensation «in fractionating column»”.\n \n
\n temperature decreases up the fractionating column ✔\n
\n\n liquids condense at different heights\n
\n \n \n OR\n \n \n
\n liquid of lowest boiling point collected first\n
\n \n \n OR\n \n \n
\n liquid with weakest intermolecular forces collected first\n
\n \n \n OR\n \n \n
\n most volatile component collected first\n
\n \n \n OR\n \n \n
\n fractions/liquids collected in order of boiling point/volatility ✔\n
\n \n Accept “liquids collected in order of molar mass”.\n \n
\n 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.\n
\n\n In 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.\n
\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”.
Nitrogen 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.
\n\n \n Calculate the energy released, in\n \n , from the complete combustion of\n \n \n \n \n \n of ethanol.\n \n
\n\n [1]\n
\n\n \n Outline the advantages and disadvantages of using biodiesel instead of gasoline as fuel for a car. Exclude any discussion of cost.\n \n
\n\n \n \n \n
\n [4]\n
\n\n \n A 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
\n\n
\n\n [1]\n
\n\n \n Methane is another greenhouse gas. Contrast the reasons why methane and carbon dioxide are considered significant greenhouse gases.\n \n
\n\n [2]\n
\n\n \n ✔\n
\n\n \n Advantages:\n \n [2 max]\n \n \n
\n\n renewable ✔\n
\n\n uses up waste «such as used cooking oil» ✔\n
\n\n lower carbon footprint/carbon neutral ✔\n
\n\n higher flashpoint ✔\n
\n\n produces less\n \n /\n \n
\n \n \n OR\n \n \n
\n less polluting emissions ✔\n
\n has lubricating properties\n
\n \n \n OR\n \n \n
\n preserves/increases lifespan of engine ✔\n
\n increases the life of the catalytic converter ✔\n
\n\n eliminates dependence on foreign suppliers ✔\n
\n\n does not require pipelines/infrastructure «to produce» ✔\n
\n\n relatively less destruction of habitat compared to obtaining petrochemicals ✔\n
\n\n
\n\n \n Accept “higher energy density” OR “biodegradable” for advantage.\n \n
\n\n
\n \n Disadvantages:\n \n [2 max]\n \n \n
\n needs conversion/transesterification ✔\n
\n\n takes time to produce/grow plants ✔\n
\n\n takes up land\n
\n \n \n OR\n \n \n
\n deforestation ✔\n
\n fertilizers/pesticides/phosphates/nitrates «used in production of crops» have negative environmental effects ✔\n
\n\n biodiversity affected\n
\n \n \n OR\n \n \n
\n loss of habitats «due to energy crop plantations» ✔\n
\n cannot be used at low temperatures ✔\n
\n\n variable quality «in production» ✔\n
\n\n high viscosity/can clog/damage engines ✔\n
\n\n
\n \n Accept “lower specific energy” as disadvantage.\n \n
\n \n Do\n \n not\n \n accept “lower octane number” as disadvantage”.\n \n
\n\n \n Any one:\n \n
\n\n uses up fossil fuels more slowly ✔\n
\n\n lower carbon footprint/CO2 emissions ✔\n
\n\n undergoes more complete combustion ✔\n
\n\n produces fewer particulates ✔\n
\n\n higher octane number/rating\n
\n \n \n OR\n \n \n
\n less knocking ✔\n
\n prevents fuel injection system build up\n
\n \n \n OR\n \n \n
\n helps keep engine clean ✔\n
\n
\n \n Accept an example of a suitable advantage even if repeated from 11c.\n \n
\n carbon dioxide is highly/more abundant «in the atmosphere» ✔\n
\n\n methane is more effective/potent «as a greenhouse gas»\n
\n \n \n OR\n \n \n
\n methane/better/more effective at absorbing\n \n «radiation»\n
\n \n \n OR\n \n \n
\n methane has greater greenhouse factor\n
\n \n \n OR\n \n \n
\n methane has greater global warming potential/GWP✔\n
\n
\n \n Accept “carbon dioxide contributes more to global warming” for M1.\n \n
\n Almost all were able to calculate the energy released from the complete combustion of ethanol.\n
\n\n Most 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.\n
\n\n Most scored the one mark for this question, with \"less knocking or higher octane number/rating\" the most common correct answer seen.\n
\n\n This 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.\n
\n\n \n Calculate the energy released, in\n \n , from the complete combustion of\n \n \n \n \n \n of ethanol.\n \n
\n\n [1]\n
\n\n \n Outline the advantages and disadvantages of using biodiesel instead of gasoline as fuel for a car. Exclude any discussion of cost.\n \n
\n\n \n \n \n
\n [4]\n
\n\n \n A 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
\n\n
\n\n [1]\n
\n\n \n Methane is another greenhouse gas. Contrast the reasons why methane and carbon dioxide are considered significant greenhouse gases.\n \n
\n\n [2]\n
\n\n \n ✔\n
\n\n \n Advantages:\n \n [2 max]\n \n \n
\n\n renewable ✔\n
\n\n uses up waste «such as used cooking oil» ✔\n
\n\n lower carbon footprint/carbon neutral ✔\n
\n\n higher flashpoint ✔\n
\n\n produces less\n \n /\n \n
\n \n \n OR\n \n \n
\n less polluting emissions ✔\n
\n has lubricating properties\n
\n \n \n OR\n \n \n
\n preserves/increases lifespan of engine ✔\n
\n increases the life of the catalytic converter ✔\n
\n\n eliminates dependence on foreign suppliers ✔\n
\n\n does not require pipelines/infrastructure «to produce» ✔\n
\n\n relatively less destruction of habitat compared to obtaining petrochemicals ✔\n
\n\n
\n\n \n Accept “higher energy density” OR “biodegradable” for advantage.\n \n
\n\n
\n \n Disadvantages:\n \n [2 max]\n \n \n
\n needs conversion/transesterification ✔\n
\n\n takes time to produce/grow plants ✔\n
\n\n takes up land\n
\n \n \n OR\n \n \n
\n deforestation ✔\n
\n fertilizers/pesticides/phosphates/nitrates «used in production of crops» have negative environmental effects ✔\n
\n\n biodiversity affected\n
\n \n \n OR\n \n \n
\n loss of habitats «due to energy crop plantations» ✔\n
\n cannot be used at low temperatures ✔\n
\n\n variable quality «in production» ✔\n
\n\n high viscosity/can clog/damage engines ✔\n
\n\n
\n \n Accept “lower specific energy” as disadvantage.\n \n
\n \n Do\n \n not\n \n accept “lower octane number” as disadvantage”.\n \n
\n\n \n Any one:\n \n
\n\n uses up fossil fuels more slowly ✔\n
\n\n lower carbon footprint/CO2 emissions ✔\n
\n\n undergoes more complete combustion ✔\n
\n\n produces fewer particulates ✔\n
\n\n higher octane number/rating\n
\n \n \n OR\n \n \n
\n less knocking ✔\n
\n prevents fuel injection system build up\n
\n \n \n OR\n \n \n
\n helps keep engine clean ✔\n
\n
\n \n Accept an example of a suitable advantage even if repeated from 9c.\n \n
\n carbon dioxide is highly/more abundant «in the atmosphere» ✔\n
\n\n methane is more effective/potent «as a greenhouse gas»\n
\n \n \n OR\n \n \n
\n methane/better/more effective at absorbing\n \n «radiation»\n
\n \n \n OR\n \n \n
\n methane has greater greenhouse factor\n
\n \n \n OR\n \n \n
\n methane has greater global warming potential/GWP✔\n
\n
\n \n Accept “carbon dioxide contributes more to global warming” for M1.\n \n
\n Even rather weak candidates answered this one correctly.\n
\n\n There 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.\n
\n\n Another 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.\n
\n\n We 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.\n
\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 ✔
\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.
\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
Limestone 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.
\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 ✓
\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
\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 ✔
\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”.
Magnetite, 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) ✔
\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 ✔
\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”.
\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.
The 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.
\n State the equilibrium constant expression,\n \n K\n \n \n c\n \n , for the reaction above.\n
\n\n [1]\n
\n\n State and explain how the equilibrium would be affected by increasing the volume of the reaction container at a constant temperature.\n
\n\n [3]\n
\n\n «\n \n K\n \n \n c\n \n =\n \n » ✓\n
\n\n \n
\n Square brackets required for the mark.\n \n
\n pressure decrease «due to larger volume» ✓\n
\n\n reaction shifts to side with more moles/molecules «of gas» ✓\n
\n\n reaction shifts left/towards reactants ✓\n
\n\n \n
\n Award M3 only if M1\n \n OR\n \n M2 awarded.\n \n
50.00 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.
\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» ✔
\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 ✔
\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 ✔
Explain 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» ✔
\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 ✔
\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 ✔
\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·.
Fast 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.
\n\n Suggest what can be concluded about the gold atom from this experiment.\n
\n\n \n
\n [2]\n
\n\n Subsequent experiments showed electrons existing in energy levels occupying various orbital shapes.\n
\n\n Sketch diagrams of 1s, 2s and 2p.\n
\n\n \n
\n [2]\n
\n\n State the electron configuration of copper.\n
\n\n [1]\n
\n\n \n Most\n \n 4\n \n He\n \n 2+\n \n passing straight through:\n \n
\n\n most of the atom is empty space\n
\n \n \n OR\n \n \n
\n the space between nuclei is much larger than\n \n 4\n \n He\n \n 2+\n \n particles\n
\n \n \n OR\n \n \n
\n nucleus/centre is «very» small «compared to the size of the atom» ✔\n
\n \n
\n Very few\n \n 4\n \n He\n \n 2+\n \n deviating largely from their path:\n \n
\n nucleus/centre is positive «and repels\n \n 4\n \n He\n \n 2+\n \n particles»\n
\n \n \n OR\n \n \n
\n nucleus/centre is «more» dense/heavy «than\n \n 4\n \n He\n \n 2+\n \n particles and deflects them»\n
\n \n \n OR\n \n \n
\n nucleus/centre is «very» small «compared to the size of the atom» ✔\n
\n
\n\n \n Do\n \n not\n \n accept the same reason for both\n \n M1\n \n and\n \n M2\n \n .\n \n
\n\n \n Accept “most of the atom is an electron cloud” for\n \n M1\n \n .\n \n
\n\n \n Do not accept only “nucleus repels\n \n 4\n \n He\n \n 2+\n \n particles” for\n \n M2\n \n .\n \n
\n\n \n
\n 1s\n \n \n AND\n \n \n 2s as spheres ✔\n
\n\n one or more 2p orbital(s) as figure(s) of 8 shape(s) of any orientation (p\n \n x\n \n , p\n \n y\n \n p\n \n z\n \n ) ✔\n
\n\n 1s\n \n 2\n \n 2s\n \n 2\n \n 2p\n \n 6\n \n 3s\n \n 2\n \n 3p\n \n 6\n \n 4s\n \n 1\n \n 3d\n \n 10\n \n
\n \n \n OR\n \n \n
\n [Ar] 4s\n \n 1\n \n 3d\n \n 10\n \n ✔\n
\n \n
\n Accept configuration with 3d before 4s.\n \n
Ammonia 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.
\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.
\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.
\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.
\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.
\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 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
\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.
\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.
\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.
\n\n State the name of Compound B, applying International Union of Pure and Applied Chemistry (IUPAC) rules.\n
\n\n [1]\n
\n\n Compound A and Compound B are both liquids at room temperature and pressure. Identify the strongest intermolecular force between molecules of Compound A.\n
\n\n [1]\n
\n\n Draw the structural formula of the alkene required.\n
\n\n \n
\n [1]\n
\n\n Deduce the structural formula of the repeating unit of the polymer formed from this alkene.\n
\n\n [1]\n
\n\n Outline the requirements for a collision between reactants to yield products.\n
\n\n [2]\n
\n\n The polarity of the carbon–halogen bond, C–X, facilitates attack by HO\n \n –\n \n .\n
\n\n Outline, giving a reason, how the bond polarity changes going down group 17.\n
\n\n [1]\n
\n\n 2-methylpropan-2-ol /2-methyl-2-propanol ✔\n
\n\n
\n\n \n Accept methylpropan-2-ol/ methyl-2-propanol.\n \n
\n\n \n Do\n \n not\n \n accept 2-methylpropanol.\n \n
\n\n dipole-dipole ✔\n
\n\n
\n\n \n Do not accept van der Waals’ forces.\n \n
\n\n \n
\n \n
\n \n Do\n \n not\n \n penalize missing brackets or n.\n \n
\n\n \n Do\n \n not\n \n award mark if continuation bonds are not shown.\n \n
\n\n energy/E ≥ activation energy/E\n \n a\n \n ✔\n
\n\n correct orientation «of reacting particles»\n
\n \n \n OR\n \n \n
\n correct geometry «of reacting particles» ✔\n
\n decreases/less polar\n \n \n AND\n \n \n electronegativity «of the halogen» decreases ✔\n
\n\n
\n\n \n Accept “decreases”\n \n AND\n \n a correct comparison of the electronegativity of two halogens.\n \n
\n\n \n Accept “decreases”\n \n AND\n \n “attraction for valence electrons decreases”.\n \n
\n\n Naming the organic compound using IUPAC rules was generally done well.\n
\n\n Good performance; some had a H and CH\n \n 3\n \n group on each C atom across double bond instead of having two H atoms on one C and two CH\n \n 3\n \n groups on the other.\n
\n\n Mediocre 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 CH\n \n 3\n \n group was missing.\n
\n\n Good 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/\n \n E ≥\n \n activation energy/\n \n E\n \n \n a\n \n .\n
\n\n Good performance on how the polarity of C-X bond changes going down group 17.\n
\n\n Draw arrows in the boxes to represent the electron configuration of a nitrogen atom.\n
\n\n \n
\n [1]\n
\n\n Explain the relative lengths of the three bonds between N and O in nitric acid.\n
\n\n [3]\n
\n\n \n
\n \n
\n Accept\n \n all\n \n 2p electrons pointing downwards.\n \n
\n \n Accept half arrows instead of full arrows.\n \n
\n\n \n Any three of:\n \n
\n\n two N-O same length/order ✔\n
\n delocalization/resonance ✔\n
\n N-OH longer «than N-O»\n
\n \n \n OR\n \n \n
\n N-OH bond order 1\n \n \n AND\n \n \n N-O bond order 1½ ✔\n
\n
\n\n \n Award\n \n [2 max]\n \n if bond strength, rather than bond length discussed.\n \n
\n\n \n Accept N-O between single and double bond\n \n AND\n \n N-OH single bond.\n \n
\n\n Drawing arrows in the boxes to represent the electron configuration of a nitrogen atom was done extremely well.\n
\n\n Poorly done; some explained relative bond strengths between N and O in HNO\n \n 3\n \n , not relative lengths; others included generic answers such as triple bond is shortest, double bond is longer, single longest.\n
\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.
\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.
\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.
\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.
\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.
\n\n Outline, giving a reason, the effect of a catalyst on a reaction.\n
\n\n [2]\n
\n\n On the axes, sketch Maxwell–Boltzmann energy distribution curves for the reacting species at two temperatures T\n \n 1\n \n \n and\n \n T\n \n 2\n \n , where T\n \n 2\n \n > T\n \n 1\n \n .\n
\n\n \n
\n [3]\n
\n\n Explain the effect of increasing temperature on the yield of SO\n \n 3\n \n .\n
\n\n [2]\n
\n\n State the product formed from the reaction of SO\n \n 3\n \n with water.\n
\n\n [1]\n
\n\n State the meaning of a strong Brønsted–Lowry acid.\n
\n\n [2]\n
\n\n increases rate\n \n \n AND\n \n \n lower\n \n E\n \n \n a\n \n ✔\n
\n\n provides alternative pathway «with lower\n \n E\n \n \n a\n \n »\n
\n \n \n OR\n \n \n
\n more/larger fraction of molecules have the «lower»\n \n E\n \n \n a\n \n ✔\n
\n
\n\n \n Accept description of how catalyst lowers E\n \n a\n \n for M2 (e.g. “reactants adsorb on surface «of catalyst»”, “reactant bonds weaken «when adsorbed»”, “helps favorable orientation of molecules”).\n \n
\n\n \n
\n both axes correctly labelled ✔\n
\n\n peak of T\n \n 2\n \n curve lower\n \n \n AND\n \n \n to the right of T\n \n 1\n \n curve ✔\n
\n\n lines begin at origin\n \n \n AND\n \n \n correct shape of curves\n \n \n AND\n \n \n T\n \n 2\n \n must finish above T\n \n 1\n \n ✔\n
\n\n
\n\n \n Accept “probability «density» / number of particles / N / fraction” on y-axis.\n \n
\n\n \n Accept “kinetic E/KE/E\n \n k\n \n ” but not just “Energy/E” on x-axis.\n \n
\n\n decrease\n \n \n AND\n \n \n equilibrium shifts left / favours reverse reaction ✔\n
\n\n «forward reaction is» exothermic / ΔH is negative ✔\n
\n\n sulfuric acid/H\n \n 2\n \n SO\n \n 4\n \n ✔\n
\n\n \n
\n Accept “disulfuric acid/H\n \n 2\n \n S\n \n 2\n \n O\n \n 7\n \n ”.\n \n
\n fully ionizes/dissociates ✔\n
\n\n proton/H\n \n +\n \n «donor »✔\n
\n\n Overall 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.\n
\n\n Few 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.\n
\n\n Many candidates got one mark at least, sometimes failing to state the effect on the production of SO\n \n 3\n \n 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.\n
\n\n 6(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.\n
\n\n Outline\n \n two\n \n differences between the bonding of carbon atoms in C\n \n 60\n \n and diamond.\n
\n\n [2]\n
\n\n Explain why C\n \n 60\n \n and diamond sublime at different temperatures and pressures.\n
\n\n [2]\n
\n\n State two features showing that propane and butane are members of the same homologous series.\n
\n\n [2]\n
\n\n Describe a test and the expected result to indicate the presence of carbon–carbon double bonds.\n
\n\n \n
\n [2]\n
\n\n Write the equation for the reaction between but-2-ene and hydrogen bromide.\n
\n\n [1]\n
\n\n State the type of reaction.\n
\n\n [1]\n
\n\n Calculate the enthalpy change of the reaction, Δ\n \n H\n \n , using section 11 of the data booklet.\n
\n\n [3]\n
\n\n Draw and label an enthalpy level diagram for this reaction.\n
\n\n \n
\n [2]\n
\n\n Any\n \n two\n \n of:\n
\n\n C\n \n 60\n \n fullerene: bonded to 3 C\n \n \n AND\n \n \n diamond: bonded to 4 C ✔\n
\n\n C\n \n 60\n \n fullerene: delocalized/resonance\n \n \n AND\n \n \n diamond: not delocalized / no resonance ✔\n
\n\n C\n \n 60\n \n fullerene:\n \n sp\n \n 2\n \n \n AND\n \n \n diamond:\n \n sp\n \n 3\n \n \n ✔\n
\n\n C\n \n 60\n \n fullerene: bond angles between 109–120°\n \n \n AND\n \n \n diamond: 109° ✔\n
\n\n
\n\n \n Accept \"bonds in fullerene are shorter/stronger/have higher bond order\n \n OR\n \n bonds in diamond longer/weaker/have lower bond order\".\n \n
\n\n diamond giant/network covalent\n \n \n AND\n \n \n sublimes at higher temperature ✔\n
\n\n C\n \n 60\n \n molecular/London/dispersion/intermolecular «forces» ✔\n
\n\n
\n\n \n Accept “diamond has strong covalent bonds\n \n AND\n \n require more energy to break «than intermolecular forces»” for M1.\n \n
\n\n same general formula / C\n \n n\n \n H\n \n 2n+2\n \n ✔\n
\n\n differ by CH\n \n 2\n \n /common structural unit ✔\n
\n\n
\n\n \n Accept \"similar chemical properties\".\n \n
\n\n \n Accept “gradation/gradual change in physical properties”.\n \n
\n\n \n ALTERNATIVE 1:\n \n
\n\n \n Test:\n \n
\n\n add bromine «water»/Br\n \n 2\n \n (aq) ✔\n
\n\n \n Result:\n \n
\n\n «orange/brown/yellow» to colourless/decolourised ✔\n
\n\n \n
\n Do not accept “clear” for M2.\n \n
\n \n
\n ALTERNATIVE 2:\n \n
\n \n Test:\n \n
\n\n add «acidified» KMnO\n \n 4\n \n ✔\n
\n\n \n Result:\n \n
\n\n «purple» to colourless/decolourised/brown ✔\n
\n\n \n
\n Accept “colour change” for M2.\n \n
\n \n
\n ALTERNATIVE 3:\n \n
\n \n Test:\n \n
\n\n add iodine /\n \n ✔\n
\n\n \n Result:\n \n
\n\n «brown» to colourless/decolourised ✔\n
\n
\n
\n CH\n \n 3\n \n CH=CHCH\n \n 3\n \n + HBr (g) → CH\n \n 3\n \n CH\n \n 2\n \n CHBrCH\n \n 3\n \n
\n\n Correct reactants ✔\n
\n\n Correct products ✔\n
\n\n
\n\n \n Accept molecular formulas for both reactants and product\n \n
\n\n «electrophilic» addition/EA ✔\n
\n\n
\n\n \n Do\n \n not\n \n accept nucleophilic or free radical addition.\n \n
\n\n bond breaking: C–H + Cl–Cl / 414 «kJ mol\n \n –1\n \n » + 242 «kJ mol\n \n –1\n \n »/656 «kJ»\n
\n \n \n OR\n \n \n
\n bond breaking: 4C–H + Cl–Cl / 4 × 414 «kJ mol\n \n –1\n \n » + 242 «kJ mol\n \n –1\n \n » / 1898 «kJ» ✔\n
\n
\n\n bond forming: «C–Cl + H–Cl / 324 kJ mol\n \n –1\n \n + 431 kJ mol\n \n –1\n \n » / 755 «kJ»\n
\n \n \n OR\n \n \n
\n bond forming: «3C–H + C–Cl + H–Cl / 3 × 414 «kJ mol\n \n –1\n \n » + 324 «kJ mol\n \n –1\n \n » + 431 kJ mol\n \n –1\n \n » / 1997 «kJ» ✔\n
\n
\n\n «ΔH = bond breaking – bond forming = 656 kJ – 755 kJ» = –99 «kJ» ✔\n
\n\n
\n\n \n Award\n \n [3]\n \n for correct final answer.\n \n
\n\n \n Award\n \n [2 max]\n \n for 99 «kJ».\n \n
\n\n \n
\n reactants at higher enthalpy than products ✔\n
\n\n
\n ΔH/-99 «kJ» labelled on arrow from reactants to products\n
\n \n \n OR\n \n \n
\n activation energy/\n \n E\n \n \n a\n \n labelled on arrow from reactant to top of energy profile ✔\n
\n
\n\n \n Accept a double headed arrow between reactants and products labelled as ΔH for M2.\n \n
\n\n A 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.\n
\n\n Again, 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.\n
\n\n This easy question was quite well answered; same/similar physical properties and empirical formula were common errors.\n
\n\n Candidates misinterpreted the question and mentioned CH3\n \n +\n \n , 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.\n
\n\n Well 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.\n
\n\n Well done in general and most candidates wrote correct reagents, eventually losing a mark when considering H\n \n 2\n \n to be a product alongside 2-bromobutane.\n
\n\n Very well answered, some mentioned halogenation which is a different reaction.\n
\n\n Only the very weak candidates were unable to calculate the enthalpy change correctly, eventually missing 1 mark for inverted calculations.\n
\n\n Most 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.\n
\n\n Deduce the structural and empirical formulas of\n \n B\n \n .\n
\n\n \n
\n [3]\n
\n\n Explain, with reference to Le Châtelier’s principle, the effect of using dilute rather than concentrated sulfuric acid as the catalyst on the yield of the reaction.\n
\n\n [2]\n
\n\n Explain, with reference to intermolecular forces, why\n \n B\n \n is more volatile than\n \n A\n \n .\n
\n\n [2]\n
\n\n Compound\n \n A\n \n can also react with bromine. Describe the change observed if\n \n A\n \n is reacted with bromine.\n
\n\n [1]\n
\n\n \n Structure:\n \n
\n\n \n
\n ester functional group ✔\n
\n\n rest of structure ✔\n
\n\n
\n \n Empirical Formula:\n \n
\n C\n \n 3\n \n H\n \n 5\n \n O ✔\n
\n\n
\n\n \n Accept condensed/skeletal formula.\n \n
\n\n \n Structure:\n \n
\n\n \n
\n ester functional group ✔\n
\n\n rest of structure ✔\n
\n\n
\n \n Empirical Formula:\n \n
\n C\n \n 3\n \n H\n \n 5\n \n O ✔\n
\n\n
\n\n \n Accept condensed/skeletal formula.\n \n
\n\n dilute adds «excess» water\n
\n\n \n \n OR\n \n \n
\n\n water is a product ✔\n
\n\n
\n\n shift left\n \n \n AND\n \n \n decreases yield ✔\n
\n\n \n A\n \n has hydrogen bonding/bonds «and dipole-dipole and London/dispersion forces»\n \n \n AND\n \n B\n \n has dipole-dipole «and London/dispersion forces»\n
\n\n \n \n OR\n \n \n
\n\n \n A\n \n has hydrogen bonding/bonds\n \n \n AND\n \n \n \n B\n \n does not ✔\n
\n\n
\n\n intermolecular forces are weaker in\n \n B\n \n
\n\n \n \n OR\n \n \n
\n\n hydrogen bonding/bonds stronger «than dipole-dipole» ✔\n
\n\n brown/orange/red/yellow to colourless ✔\n
\n\n
\n \n Do\n \n not\n \n accept clear for colourless.\n \n
\n A question that discriminated well between high-scoring and low-scoring candidates. The average mark on this three-mark question was 1.3. The majority of candidates did not recognize it as an esterification reaction and the ester functional group was only seen in a small proportion of the scripts. Some candidates earned a mark for the remainder of the structure. Only about half of the candidates earned error carried forward for the mark allocated for the empirical formula. Some candidates had the molecular formula instead, and some candidates miscounted the numbers of atoms in the structure they drew.\n
\n\n A question that discriminated well between high-scoring and low-scoring candidates. The average mark on this three-mark question was 1.3. The majority of candidates did not recognize it as an esterification reaction and the ester functional group was only seen in a small proportion of the scripts. Some candidates earned a mark for the remainder of the structure. Only about half of the candidates earned error carried forward for the mark allocated for the empirical formula. Some candidates had the molecular formula instead, and some candidates miscounted the numbers of atoms in the structure they drew.\n
\n\n This was the most challenging question on the paper according to the difficulty index. Many candidates stated that catalysts do not affect the position of an equilibrium and hence the yield is not changed. Some candidates stated that the rate of reaction would be slower and the yield per unit time would be lower. Only a few candidates recognized that the dilute sulfuric acid catalyst would introduce more water, and since water is a product it would shift the equilibrium to the left and lower the yield of the ester. 23% of the candidates did not answer the question. Some teachers commented in their feedback that it was not fair to expect the students to know about the dehydrating property of H\n \n 2\n \n SO\n \n 4\n \n , but this was not intended. The students were expected to deduce the effect.\n
\n\n This question about intermolecular forces discriminated well between high-achieving and low-achieving candidates. Stronger candidates showed excellent understanding of the types of intermolecular forces found between the molecules of each compound and how they compared in strength. They gave more detail than the markscheme required. The average mark on the question was 0.8 out of 2. 21% of the candidates left the question blank. Error carried forward was applied whenever it was possible based on the answer in (a)(i).\n
\n\n Although it was a straightforward organic question, 22% of the candidates left it blank, indicating less confidence in answering the organic chemistry questions. 40% of the candidates gained the mark for the decolourization of bromine. One of the common mistakes was reversing the colour change and another was using the term \"clear\" instead of \"colourless\".\n
\n\n Calculate the amount, in mol, of sulfur dioxide produced when 500.0 g of lignite undergoes combustion.\n
\n\n S (s) + O\n \n 2\n \n (g) → SO\n \n 2\n \n (g)\n
\n\n [2]\n
\n\n Write an equation that shows how sulfur dioxide can produce acid rain.\n
\n\n [1]\n
\n\n Sodium thiosulfate reacts with hydrochloric acid as shown:\n
\n\n Na\n \n 2\n \n S\n \n 2\n \n O\n \n 3\n \n (aq) + 2HCl (aq) → S (s) + SO\n \n 2\n \n (aq) + 2NaCl (aq) + H\n \n 2\n \n O (l)\n
\n\n The precipitate of sulfur makes the mixture cloudy, so a mark underneath the reaction mixture becomes invisible with time.\n
\n\n \n
\n Suggest\n \n two\n \n variables, other than concentration, that should be controlled when comparing relative rates at different temperatures.\n
\n\n [2]\n
\n\n Discuss\n \n two\n \n different ways to reduce the environmental impact of energy production from coal.\n
\n\n [2]\n
\n\n «\n \n » 2.0 «g» ✔\n
\n\n «\n \n » = 0.062 «mol of SO\n \n 2\n \n » ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Accept 0.063 «mol».\n \n
\n\n «\n \n » 2.0 «g» ✔\n
\n\n «\n \n » = 0.062 «mol of SO\n \n 2\n \n » ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Accept 0.063 «mol».\n \n
\n\n SO\n \n 2\n \n (g) + H\n \n 2\n \n O (l) → H\n \n 2\n \n SO\n \n 3\n \n (aq)\n
\n\n \n \n OR\n \n \n
\n\n SO\n \n 2\n \n (g) + ½O\n \n 2\n \n (g) → SO\n \n 3\n \n (g)\n \n \n AND\n \n \n SO\n \n 3\n \n (g) + H\n \n 2\n \n O (l) → H\n \n 2\n \n SO\n \n 4\n \n (aq)\n
\n\n \n \n OR\n \n \n
\n\n SO\n \n 2\n \n (g) + ½O\n \n 2\n \n (g) + H\n \n 2\n \n O (l) → H\n \n 2\n \n SO\n \n 4\n \n (aq) ✔\n
\n\n
\n\n \n Accept ionized forms of acids.\n \n
\n\n \n Any two of:\n \n
\n\n depth/volume «of solution» ✔\n
\n\n colour/darkness/thickness/size/background of mark ✔\n
\n\n intensity of lighting in the lab ✔\n
\n\n
\n\n \n Accept same size flask.\n \n
\n\n \n Accept position of observation/person observing.\n \n
\n\n \n Accept same equipment/apparatus.\n \n
\n\n \n Do\n \n not\n \n accept catalyst/particle size/pressure/time.\n \n
\n\n \n Any\n \n two\n \n of:\n \n
\n\n remove sulfur from coal ✔\n
\n\n add lime during combustion ✔\n
\n\n not allow sulfur oxides to be released into the environment ✔\n
\n\n reduce proportion/percentage of energy/power produced by «the combustion of» coal ✔\n
\n\n
\n\n \n Accept any valid method to wash coal and remove sulfur content for M1.\n \n
\n\n \n Accept any valid combustion/post-combustion method to remove sulfur oxides.\n \n
\n\n \n Accept any suggestion that would reduce the amount of coal that is burnt or would reduce the damage caused.\n \n
\n\n \n Do\n \n not\n \n accept answers that only reduce production of SO\n \n 2\n \n /CO\n \n 2\n \n from other fuels.\n \n
\n\n \n Accept “improve efficiency of energy production from coal”.\n \n
\n\n \n Accept “use coal of lower sulfur content”\n \n OR\n \n “cleaner coal”.\n \n
\n\n A question that discriminated well between high-achieving and low-achieving candidates. The majority of the candidates were able to achieve one mark for determining the number of moles using 500g, while stronger candidates determined 0.40% of 500g to determine the correct number of moles. A number of candidates had a power of ten error in the first step.\n
\n\n A question that discriminated well between high-achieving and low-achieving candidates. The majority of the candidates were able to achieve one mark for determining the number of moles using 500g, while stronger candidates determined 0.40% of 500g to determine the correct number of moles. A number of candidates had a power of ten error in the first step.\n
\n\n This question was poorly answered and only 30% of the candidates wrote a correct equation for the formation of acid rain from SO\n \n 2\n \n . Mistakes included unbalanced equations and hydrogen added as a product.\n
\n\n Most candidates mentioned \"volume\" as a variable that should be controlled gaining one of the two marks, while only a small proportion of candidates seemed to understand how the experiment worked and discussed the lighting in the room and the thickness of the mark. The most common incorrect answer was \"pressure\" which was irrelevant to this experiment.\n
\n\n Some candidates referred to the pre-combustion and post combustion methods of minimizing the release of SO\n \n 2\n \n . Some candidates focused on using cleaner coal, other fuels or renewable energy sources. Some mentioned increasing the efficiency of power stations to reduce the amount of coal burned. Some focused on removing the CO\n \n 2\n \n released by planting trees. All these options were valid with sufficient detail. But answers that were not relevant to coal, such as fitting catalytic converters on cars, were not accepted. 15% of the candidates did not answer the question, and the average mark was 0.8 out of 2 marks.\n
\n\n Deduce the ionic equation, including state symbols, for the reaction of hydrogen chloride gas with water.\n
\n\n [2]\n
\n\n H\n \n 2\n \n O\n \n (l)\n \n + HCl\n \n (g)\n \n → Cl\n \n −\n \n \n (aq)\n \n + H\n \n 3\n \n O\n \n +\n \n \n (aq) ✓✓\n \n
\n\n
\n\n \n One for the equation and one for the state symbols.\n
\n Do not accept\n \n H\n \n 2\n \n O\n \n (l)\n \n + H\n \n +\n \n \n (g)\n \n →\n \n \n H\n \n 3\n \n O\n \n +\n \n \n (aq)\n \n \n
\n Do not accept equilibrium sign.\n \n
\n Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.\n
\n\n \n
\n [1]\n
\n\n The results are given where ✓ = reaction occurred and\n \n = no reaction.\n
\n| \n \n Metal\n \n | \n\n \n ASO\n \n 4\n \n (aq)\n \n | \n\n \n BSO\n \n 4\n \n (aq)\n \n | \n\n \n CSO\n \n 4\n \n (aq)\n \n | \n\n \n DSO\n \n 4\n \n (aq)\n \n | \n\n \n ESO\n \n 4\n \n (aq)\n \n | \n
| \n \n A\n \n | \n\n —\n | \n\n ✓\n | \n\n ✗\n | \n\n ✓\n | \n\n ✓\n | \n
| \n \n B\n \n | \n\n ✗\n | \n\n —\n | \n\n ✗\n | \n\n ✓\n | \n\n ✓\n | \n
| \n \n C\n \n | \n\n ✓\n | \n\n ✓\n | \n\n —\n | \n\n ✓\n | \n\n ✓\n | \n
| \n \n D\n \n | \n\n ✗\n | \n\n ✗\n | \n\n ✗\n | \n\n —\n | \n\n ✓\n | \n
| \n \n E\n \n | \n\n ✗\n | \n\n ✗\n | \n\n ✗\n | \n\n ✗\n | \n\n —\n | \n
\n
\n\n \n
\n arrows\n \n \n AND\n \n \n identifies 2s\n \n \n AND\n \n \n 2p sub orbitals ✓\n
\n\n
\n\n \n Accept “hooks” to represent the electrons.\n \n
\n\n +2/II ✓\n
\n\n
\n \n Do\n \n not\n \n accept A\n \n 2+\n \n , A\n \n +2\n \n , 2\n \n OR\n \n 2+.\n \n
\n Outline the meaning of homologous series.\n
\n\n [1]\n
\n\n High-pressure carbon monoxide disproportionation (HiPco) produces carbon atoms that react with nano catalysts to produce carbon nanotubes.\n
\n\n
\n\n compounds of the same family\n \n \n AND\n \n \n general formula\n
\n \n \n OR\n \n \n
\n compounds of the same family\n \n \n AND\n \n \n differ by a common structural unit/\n \n CH\n \n \n 2\n \n ✓\n
\n
\n\n \n Accept contains the same functional group for same family.\n \n
\n\n 2CO(g) → C(s) + CO\n \n 2\n \n (g) ✓\n
\n\n
\n \n Accept reversible arrows.\n \n
\n State the oxidation state of sulfur in copper (II) sulfate.\n
\n\n [1]\n
\n\n Explain why metals alloyed with another metal are usually harder and stronger but poorer conductors than the pure metal.\n
\n\n [3]\n
\n\n +6/VI ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept 6/6\n \n +\n \n .\n \n
\n\n metal ions/atoms have different sizes ✓\n
\n cations/atoms/layers do not slide over each other as easily ✓\n
\n «irregularities» obstruct free movement of electrons ✓\n
\n
\n \n Accept electrons move less easily/less delocalized for M3.\n \n
\n Draw\n \n one\n \n Lewis (electron dot) structure of the sulfate ion.\n
\n\n [1]\n
\n\n The\n \n K\n \n \n sp\n \n of copper (II) hydroxide is 2.2 × 10\n \n −20\n \n . Calculate the molar solubility of Cu\n \n 2+\n \n (aq) ions in a solution of pH 9.\n
\n\n [2]\n
\n\n \n ✓\n
\n
\n\n \n Accept any combination of dots, crosses and lines.\n
\n Double bonds do not have to be opposite each other.\n
\n Do\n \n not\n \n penalise missing square brackets.\n \n
\n \n K\n \n \n sp\n \n =[Cu\n \n 2+\n \n ][OH\n \n −\n \n ]\n \n 2\n \n
\n \n \n OR\n \n \n
\n 2.2 × 10\n \n −20\n \n =[Cu\n \n 2+\n \n ] × (10\n \n −5\n \n )\n \n 2\n \n ✓\n
\n [Cu2+] «= Ksp/[OH−]2 =2.2 x 10−20/(10−5)2»\n
\n = 2.2 ×10\n \n −10\n \n «mol dm\n \n −3\n \n » ✓\n
\n
\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n Determine the standard enthalpy of reaction (\n \n ), in kJ mol\n \n −1\n \n , for the oxidation of SO\n \n 2\n \n to SO\n \n 3\n \n .\n
\n| \n \n Substance\n \n | \n\n \n Enthalpy of formation, (\n \n ), in kJ mol\n \n −1\n \n \n | \n
| \n SO\n \n 2\n \n | \n\n −296.8\n | \n
| \n SO\n \n 3\n \n | \n\n −395.8\n | \n
\n [1]\n
\n\n Outline why vitamins usually need to be obtained from food sources.\n
\n\n [1]\n
\n\n «Δ\n \n H\n \n °\n \n rxn\n \n = ΣΔ\n \n H\n \n °f (Products) − ΣΔ\n \n H\n \n °f (Reactants) =»\n
\n −395.8 − (−296.8)» = −99.0«kJ mol\n \n −1\n \n » ✓\n
\n cannot be synthesized «by the human body» ✓\n
\n\n Deduce the number of signals you would expect to find in the\n \n 1\n \n H NMR spectrum of each compound.\n
\n\n \n
\n [1]\n
\n\n Glucose, an isomer of fructose, exists as two isomeric ring forms. Annotate the diagram below to complete the structure of β-glucose. Use section 34 of the data booklet.\n
\n\n \n
\n [1]\n
\n| \n Name\n | \n\n Number of signals\n | \n
| \n Ethyl methanoate\n | \n\n 3\n | \n
| \n Methyl ethanoate\n | \n\n \n \n AND\n \n \n 2\n | \n
\n ✓\n
\n\n \n
\n
\n\n \n Entire structure must be correct to score the mark.\n \n
\n\n \n Ignore incorrect connectivity.\n \n
\n\n Explain why a colorimeter set at a wavelength of 500 nm is not suitable to investigate reactions of Zn\n \n 2+\n \n compounds. Use section 3 of the data booklet.\n
\n\n [2]\n
\n\n State the\n \n three\n \n components of a monomer of DNA (a nucleotide).\n
\n\n [1]\n
\n\n Zn\n \n 2+\n \n does not form coloured compounds/ has a complete d subshell/orbital ✓\n
\n
\n 500 nm/«the setting on the colorimeter» in visible region\n \n \n AND\n \n \n no absorbance will be seen ✓\n
\n phosphate\n \n \n AND\n \n \n deoxyribose\n \n \n AND\n \n \n nitrogenous base ✓\n
\n\n
\n \n Accept named base.\n
\n Do\n \n not\n \n accept ‘sugar’ or ‘pentose sugar’ in place of deoxyribose.\n \n
\n An unknown organic compound,\n \n X\n \n , comprising of only carbon, hydrogen and oxygen was found to contain 48.6 % of carbon and 43.2 % of oxygen.\n
\n\n
\n\n «n(C) =» 4.05 «mol»\n
\n \n \n AND\n \n \n
\n «n(O) =» 2.70 «mol» ✓\n
\n «% H =» 8.2 %\n
\n \n \n OR\n \n \n
\n «n(H) =» 8.12 «mol» ✓\n
\n «empirical formula =» C\n \n 3\n \n H\n \n 6\n \n O\n \n 2\n \n ✓\n
\n\n \n Award\n \n [2]\n \n for the simplest ratio ″1.5 C: 3 H: 1 O″.\n \n
\n\n Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N\n \n 2\n \n O\n \n 4\n \n (g), which is colourless.\n
\n\n 2NO\n \n 2\n \n (g) ⇌ N\n \n 2\n \n O\n \n 4\n \n (g)\n
\n\n
\n\n All species are almost colourless except for MnO\n \n 4\n \n \n −\n \n , which has an intense purple colour, though the kale extract is coloured by the chlorophyll present.\n
\n\n
\n\n reaction hardly proceeds\n
\n \n \n OR\n \n \n
\n reverse reaction/formation of NO\n \n 2\n \n is favoured\n
\n \n \n OR\n \n \n
\n «concentration of» reactants greater than «concentration of» products «at equilibrium» ✓\n
\n \n Accept equilibrium lies to the left.\n \n
\n\n green to purple\n
\n \n \n OR\n \n \n
\n green to brown\n
\n \n \n OR\n \n \n
\n green to purple-green ✓\n
\n
\n\n \n Accept “colourless to purple”.\n
\n Accept “green to grey/blueish”.\n
\n Do\n \n not\n \n accept “clear” for “colourless”.\n
\n Do\n \n not\n \n accept “purple to “brown”.\n
\n Do\n \n not\n \n accept blue as final colour.\n \n
\n An electrolytic cell was set up using inert electrodes and a dilute aqueous solution of magnesium chloride, MgCl\n \n 2\n \n (aq).\n
\n\n
\n\n \n
\n electron flow from anode to battery\n \n \n OR\n \n \n from battery to cathode ✓\n
\n\n Mg\n \n 2+\n \n /H\n \n +\n \n ions to − electrode\n
\n \n \n AND\n \n \n
\n Cl\n \n −\n \n /OH\n \n −\n \n ions to + electrode ✓\n
\n \n Do not award M1 if electrons are shown in electrolyte.\n \n
\n\n Bismuth has atomic number 83. Deduce\n \n two\n \n pieces of information about the electron configuration of bismuth from its position on the periodic table.\n
\n\n [2]\n
\n\n Explain how a substance in the same phase as the reactants can reduce the activation energy and act as a catalyst.\n
\n\n [2]\n
\n\n \n Any two of the following:\n \n
\n «group 15 so Bi has» 5 valence electrons ✓\n
\n «period 6 so Bi has» 6 «occupied» electron shells/energy levels ✓\n
\n «in p-block so» p orbitals are highest occupied ✓\n
\n occupied d/f orbitals ✓\n
\n has unpaired electrons ✓\n
\n has incomplete shell(s)/subshell(s) ✓\n
\n \n Award\n \n [1]\n \n for full or condensed electron configuration, [Xe] 4f\n \n 14\n \n 5d\n \n 10\n \n 6s\n \n 2\n \n 6p\n \n 3\n \n .\n
\n \n
\n \n Accept other valid statements about the electron configuration.\n \n
\n\n forms an intermediate/activated complex ✓\n
\n «intermediate/activated complex» dissociates to form product «\n \n \n AND\n \n \n catalyst» ✓\n
\n
\n\n \n Accept correct annotated energy profile for either mark.\n \n
\n\n The concentration of methanoic acid was found by titration with a 0.200 mol dm\n \n −\n \n 3\n \n \n standard solution of sodium hydroxide, NaOH (aq), using an indicator to determine the end point.\n
\n\n
\n\n The unit cell of lead (II) sulfide is shown:\n
\n\n \n
\n
\n\n «[OH\n \n −\n \n ] = 0.200 mol dm\n \n −3\n \n »\n
\n\n \n \n ALTERNATIVE 1:\n \n \n
\n «pOH = −log\n \n 10\n \n (0.200) =» 0.699 ✓\n
\n «pH = 14.000 − 0.699 =» 13.301 ✓\n
\n \n \n ALTERNATIVE 2:\n \n \n
\n «[H\n \n +\n \n ] =\n \n = » 5.00 × 10\n \n −14\n \n «mol dm\n \n −3\n \n » ✓\n
\n «pH = −log\n \n 10\n \n (5.00 × 10\n \n −14\n \n )» = 13.301 ✓\n
\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n 6 ✓\n
\n\n Identify the type of reaction.\n
\n\n [1]\n
\n\n Describe the interactions between amino acids occurring at the primary, secondary and tertiary levels within a protein.\n
\n| \n \n Structure Level\n \n | \n\n \n Interactions between amino acids\n \n | \n
| \n Primary\n | \n\n ...........................................................\n | \n
| \n Secondary\n | \n\n ...........................................................\n | \n
| \n Tertiary\n | \n\n ...........................................................\n | \n
\n [3]\n
\n\n «electrophilic» addition/A\n \n E\n \n ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept nucleophilic addition.\n \n
\n| \n \n Structure Level\n \n | \n\n \n Interactions between amino acids\n \n | \n
| \n Primary\n | \n\n covalent bonding\n \n \n \n OR\n \n \n \n peptide bond\n \n \n \n OR\n \n \n \n amide bond ✓\n | \n
| \n Secondary\n | \n\n hydrogen bonding ✓\n | \n
| \n Tertiary\n | \n\n interactions between R groups/side chains\n \n \n \n OR\n \n \n \n ionic/electrostatic «attraction»\n \n \n \n OR\n \n \n \n hydrogen bonding\n \n \n \n OR\n \n \n \n hydrophobic interactions\n \n \n \n OR\n \n \n \n disulfide bridges\n \n \n \n OR\n \n \n \n London/dispersion/van der Waals/«instantaneous» induced dipole-induced dipole ✓\n | \n
\n
\n\n \n Do\n \n not\n \n accept “amino acid sequence” for M1.\n
\n \n
\n \n Do\n \n not\n \n accept “alpha helix”\n \n OR\n \n “beta sheets” for M2.\n
\n \n
\n \n Accept “covalent bonding” for M3.\n \n
\n\n Enthalpy of solution, enthalpy of hydration and lattice enthalpy are related in an energy cycle.\n
\n\n
\n\n Compare the hydrolytic and oxidative rancidity and contrast the site where the chemical changes occur.\n
\n\n
\n\n Compare rancidity: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n Contrast reaction site: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n [2]\n
\n\n \n
\n correct boxes ✓\n
\n A: enthalpy of solution / Δ\n \n H\n \n solution / Δ\n \n H\n \n sol\n
\n \n \n AND\n \n \n
\n B: lattice enthalpy / Δ\n \n H\n \n lattice\n
\n \n \n AND\n \n \n
\n C: enthalpy of hydration / Δ\n \n H\n \n hydration ✓\n
\n \n Compare rancidity:\n \n
\n «both produce» disagreeable smell/taste/texture/appearance ✓\n
\n \n Contrast reaction site:\n \n
\n hydrolytic reaction occurs at ester link/COOC link\n \n \n AND\n \n \n oxidative reaction occurs at carbon-carbon double bond/C=C ✓\n
\n
\n\n \n Do\n \n not\n \n accept “double bond” alone for oxidative reaction site.\n \n
\n\n Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.\n
\n\n \n
\n [1]\n
\n\n The results are given where ✓ = reaction occurred and\n \n = no reaction.\n
\n| \n \n Metal\n \n | \n\n \n ASO\n \n 4\n \n (aq)\n \n | \n\n \n BSO\n \n 4\n \n (aq)\n \n | \n\n \n CSO\n \n 4\n \n (aq)\n \n | \n\n \n DSO\n \n 4\n \n (aq)\n \n | \n\n \n ESO\n \n 4\n \n (aq)\n \n | \n
| \n \n A\n \n | \n\n —\n | \n\n ✓\n | \n\n ✗\n | \n\n ✓\n | \n\n ✓\n | \n
| \n \n B\n \n | \n\n ✗\n | \n\n —\n | \n\n ✗\n | \n\n ✓\n | \n\n ✓\n | \n
| \n \n C\n \n | \n\n ✓\n | \n\n ✓\n | \n\n —\n | \n\n ✓\n | \n\n ✓\n | \n
| \n \n D\n \n | \n\n ✗\n | \n\n ✗\n | \n\n ✗\n | \n\n —\n | \n\n ✓\n | \n
| \n \n E\n \n | \n\n ✗\n | \n\n ✗\n | \n\n ✗\n | \n\n ✗\n | \n\n —\n | \n
\n
\n\n \n
\n arrows\n \n \n AND\n \n \n identifies 2s\n \n \n AND\n \n \n 2p sub orbitals ✓\n
\n\n
\n\n \n Accept “hooks” to represent the electrons.\n \n
\n\n +2/II ✓\n
\n\n
\n \n Do\n \n not\n \n accept A\n \n 2+\n \n , A\n \n +2\n \n , 2\n \n OR\n \n 2+.\n \n
\n Outline the meaning of homologous series.\n
\n\n [1]\n
\n\n High-pressure carbon monoxide disproportionation (HiPco) produces carbon atoms that react with nano catalysts to produce carbon nanotubes.\n
\n\n
\n\n compounds of the same family\n \n \n AND\n \n \n general formula\n
\n \n \n OR\n \n \n
\n compounds of the same family\n \n \n AND\n \n \n differ by a common structural unit/\n \n CH\n \n \n 2\n \n ✓\n
\n
\n\n \n Accept contains the same functional group for same family.\n \n
\n\n 2CO(g) → C(s) + CO\n \n 2\n \n (g) ✓\n
\n\n
\n \n Accept reversible arrows.\n \n
\n State the oxidation state of sulfur in copper (II) sulfate.\n
\n\n [1]\n
\n\n Explain why metals alloyed with another metal are usually harder and stronger but poorer conductors than the pure metal.\n
\n\n [3]\n
\n\n +6/VI ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept 6/6\n \n +\n \n .\n \n
\n\n metal ions/atoms have different sizes ✓\n
\n cations/atoms/layers do not slide over each other as easily ✓\n
\n «irregularities» obstruct free movement of electrons ✓\n
\n
\n\n \n Accept electrons move less easily/less delocalized for M3.\n \n
\n\n Calculate the percentage of oxygen present in the double salt.\n
\n\n [1]\n
\n\n Outline why vitamins usually need to be obtained from food sources.\n
\n\n [1]\n
\n\n «100 − (7.09 + 5.11 + 16.22 + 14.91) =» 56.67 «%» ✓\n
\n\n cannot be synthesized «by the human body» ✓\n
\n\n An unknown organic compound,\n \n X\n \n , comprising of only carbon, hydrogen and oxygen was found to contain 48.6 % of carbon and 43.2 % of oxygen.\n
\n\n Determine the empirical formula.\n
\n\n [3]\n
\n\n «n(C) =» 4.05 «mol»\n
\n \n \n AND\n \n \n
\n «n(O) =» 2.70 «mol» ✓\n
\n «% H =» 8.2 %\n
\n \n \n OR\n \n \n
\n «n(H) =» 8.12 «mol» ✓\n
\n «empirical formula =» C\n \n 3\n \n H\n \n 6\n \n O\n \n 2\n \n ✓\n
\n\n
\n\n \n Award\n \n [2]\n \n for the simplest ratio ″1.5 C: 3 H: 1 O″.\n \n
\n\n Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N\n \n 2\n \n O\n \n 4\n \n (g), which is colourless.\n
\n\n 2NO\n \n 2\n \n (g) ⇌ N\n \n 2\n \n O\n \n 4\n \n (g)\n
\n\n
\n\n All species are almost colourless except for MnO\n \n 4\n \n \n −\n \n , which has an intense purple colour, though the kale extract is coloured by the chlorophyll present.\n
\n\n
\n\n reaction hardly proceeds\n
\n \n \n OR\n \n \n
\n reverse reaction/formation of NO\n \n 2\n \n is favoured\n
\n \n \n OR\n \n \n
\n «concentration of» reactants greater than «concentration of» products «at equilibrium» ✓\n
\n \n Accept equilibrium lies to the left.\n \n
\n\n green to purple\n
\n \n \n OR\n \n \n
\n green to brown\n
\n \n \n OR\n \n \n
\n green to purple-green ✓\n
\n
\n\n \n Accept “colourless to purple”.\n
\n Accept “green to grey/blueish”.\n
\n Do\n \n not\n \n accept “clear” for “colourless”.\n
\n Do\n \n not\n \n accept “purple to “brown”.\n
\n Do\n \n not\n \n accept blue as final colour.\n \n
\n An electrolytic cell was set up using inert electrodes and molten magnesium chloride, MgCl\n \n 2\n \n (l).\n
\n\n \n
\n
\n\n
\n\n magnesium/Mg «metal» ✓\n
\n\n
\n\n \n Do not accept magnesium ions/Mg\n \n 2+\n \n .\n \n
\n\n Bismuth has atomic number 83. Deduce\n \n two\n \n pieces of information about the electron configuration of bismuth from its position on the periodic table.\n
\n\n [2]\n
\n\n Explain how a substance in the same phase as the reactants can reduce the activation energy and act as a catalyst.\n
\n\n [2]\n
\n\n \n Any two of the following:\n \n
\n «group 15 so Bi has» 5 valence electrons ✓\n
\n «period 6 so Bi has» 6 «occupied» electron shells/energy levels ✓\n
\n «in p-block so» p orbitals are highest occupied ✓\n
\n occupied d/f orbitals ✓\n
\n has unpaired electrons ✓\n
\n has incomplete shell(s)/subshell(s) ✓\n
\n \n Award\n \n [1]\n \n for full or condensed electron configuration, [Xe] 4f\n \n 14\n \n 5d\n \n 10\n \n 6s\n \n 2\n \n 6p\n \n 3\n \n .\n
\n \n
\n \n Accept other valid statements about the electron configuration.\n \n
\n\n forms an intermediate/activated complex ✓\n
\n «intermediate/activated complex» dissociates to form product «\n \n \n AND\n \n \n catalyst» ✓\n
\n
\n\n \n Accept correct annotated energy profile for either mark.\n \n
\n\n Suggest an experimental method that could be used to determine the rate of reaction.\n
\n\n [2]\n
\n\n Describe the interactions between amino acids occurring at the primary, secondary and tertiary levels within a protein.\n
\n| \n \n Structure Level\n \n | \n\n \n Interactions between amino acids\n \n | \n
| \n Primary\n | \n\n ...........................................................\n | \n
| \n Secondary\n | \n\n ...........................................................\n | \n
| \n Tertiary\n | \n\n ...........................................................\n | \n
\n [3]\n
\n\n «measure change in»\n
\n mass\n
\n \n \n OR\n \n \n
\n pressure\n
\n \n \n OR\n \n \n
\n volume of gas/CO\n \n 2\n \n produced\n
\n \n \n OR\n \n \n
\n «intensity of» colour\n
\n \n \n OR\n \n \n
\n «electrical» conductivity\n
\n \n \n OR\n \n \n
\n pH ✓\n
\n with time ✓\n
\n\n
\n\n \n Accept any of the following for M1:\n
\n perform experiment on balance\n
\n \n OR\n \n
\n use pressure probe\n
\n \n OR\n \n
\n collect gas/gas syringe\n
\n \n OR\n \n
\n use colorimeter\n
\n \n OR\n \n
\n use conductivity meter\n
\n \n OR\n \n
\n use pH meter\n \n
\n \n Do\n \n not\n \n accept “measure rate of change” for M2.\n \n
\n| \n \n Structure Level\n \n | \n\n \n Interactions between amino acids\n \n | \n
| \n Primary\n | \n\n covalent bonding\n \n \n \n OR\n \n \n \n peptide bond\n \n \n \n OR\n \n \n \n amide bond ✓\n | \n
| \n Secondary\n | \n\n hydrogen bonding ✓\n | \n
| \n Tertiary\n | \n\n interactions between R groups/side chains\n \n \n \n OR\n \n \n \n ionic/electrostatic «attraction»\n \n \n \n OR\n \n \n \n hydrogen bonding\n \n \n \n OR\n \n \n \n hydrophobic interactions\n \n \n \n OR\n \n \n \n disulfide bridges\n \n \n \n OR\n \n \n \n London/dispersion/van der Waals/«instantaneous» induced dipole-induced dipole ✓\n | \n
\n
\n\n \n Do\n \n not\n \n accept “amino acid sequence” for M1.\n
\n \n
\n \n Do\n \n not\n \n accept “alpha helix”\n \n OR\n \n “beta sheets” for M2.\n
\n \n
\n \n Accept “covalent bonding” for M3.\n \n
\n\n The student reported the volumes of titrant used per trial for samples collected each day in the following table:\n
\n\n \n
\n
\n\n The candidate used a 25 cm\n \n 3\n \n burette with an uncertainty of ±0.05 cm\n \n 3\n \n . Comment on the uncertainty recorded for the titrant volumes.\n
\n\n [1]\n
\n\n Calculate the initial % content of Fe\n \n 2+\n \n in the raw spinach, showing your working.\n
\n\n [3]\n
\n\n The results calculated for the subsequent days are shown.\n
\n\n \n
\n Comment on the significance of the difference in Fe\n \n 2+\n \n content measured for day 4 and 5.\n
\n\n [1]\n
\n\n Suggest\n \n two\n \n flaws in the design that could have contributed to the random error in the investigation.\n
\n\n [2]\n
\n\n The student did not standardise the KMnO\n \n 4\n \n solution used for titration. Suggest what type of error this may have caused, giving your reasons.\n
\n\n [1]\n
\n\n The titration of Fe(II) with MnO\n \n 4\n \n in acid medium is a redox reaction. State the oxidised and reduced species, including their change in oxidation states.\n
\n\n
\n Oxidised: ........................................................................................................................................\n
\n
\n Reduced: ........................................................................................................................................\n
\n [2]\n
\n\n Spinach contains a large amount of antioxidant compounds, including ascorbic acid and oxalic acid. Predict how this will affect the accuracy of the results, mentioning the type and direction of the error.\n
\n\n [2]\n
\n\n Incorrect\n \n \n AND\n \n \n two readings, uncertainty is ±0.1 ✔\n
\n\n Incorrect\n \n \n AND\n \n \n two readings, uncertainty is ±0.1 ✔\n
\n\n «mol» MnO\n \n 4−\n \n «=0.00339 × 0.01» = 3.38 × 10\n \n −5\n \n «mol» ✓\n
\n\n «mol MnO\n \n 4−\n \n : mol Fe\n \n +2\n \n = 1:5»\n
\n\n «mol» Fe\n \n 2+\n \n = 1.95 × 10\n \n −4\n \n ✓\n
\n\n % Fe\n \n 2+\n \n «= 1.95 × 10\n \n −4\n \n × 55.8 × 100/2.0»= 0.47«%» ✓\n
\n\n
\n\n \n Must show working for the marks.\n \n
\n\n no difference\n \n \n AND\n \n \n uncertainty larger than difference ✔\n
\n\n
\n\n \n Accept other explanations referred to overlapping.\n \n
\n\n dilute the titrant to use larger volumes «of titrant» ✔\n
\n\n ensure spinach leaf fragments are the same size ✔\n
\n\n systematic error\n \n \n AND\n \n \n all values «equally» inaccurate ✔\n
\n\n Oxidised: Fe\n \n +2\n \n → Fe\n \n +3\n \n ✔\n
\n\n Reduced: Mn\n \n 7+\n \n → Mn\n \n 2+\n \n ✔\n
\n\n systematic error\n \n \n AND\n \n \n lower accuracy ✔\n
\n\n overestimation of [Fe(II)] ✔\n
\n\n A calibration curve with pure ascorbic acid and appropriate amounts of the reagent, R, was prepared by dilutions with water of an initial aqueous solution of\n \n 100\n \n \n μg/cm\n \n −3\n \n \n ascorbic acid in water, AA (aq)\n
\n\n
\n\n Calculate the volumes of pure ascorbic acid solution required for each point of the calibration curve; point 4 of the curve is shown as an example.\n
\n\n \n
\n [2]\n
\n\n The resulting calibration curve is shown:\n
\n\n \n
\n Suggest a range of absorbance values for which this curve can be used to calculate ascorbic acid of broccoli accurately.\n
\n\n [1]\n
\n\n Suggest what should be used as a blank for spectrophotometric reading.\n
\n\n [1]\n
\n\n Discuss why it is important to obtain a value of R\n \n 2\n \n close to 1 for a calibration curve.\n
\n\n [2]\n
\n\n The sample stored at 5 °C showed an absorbance of 0.600. Determine the concentration of ascorbic acid in the sample solution by interpolation and using the line equation.\n
\n\n
\n interpolation in graph: ..........................................................................................................................\n
\n
\n using line equation: .............................................................................................................................\n
\n [2]\n
\n\n Suggest which of the two methods will provide a more accurate value.\n
\n\n [1]\n
\n\n Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to its structure.\n
\n\n [2]\n
\n\n Estimate the % change in ascorbic acid concentration when stored for 3 days storage at 5 °C and 20 °C.\n
\n\n [1]\n
\n\n Estimate how much ascorbic acid will remain after 6 days storage at 20 °C in the same experimental conditions, stating any assumption made for the calculation.\n
\n\n [3]\n
\n\n \n 1-\n \n 1.0 cm\n \n 3\n \n
\n\n \n 2-\n \n 4.0 cm\n \n 3\n \n \n \n ✔\n
\n\n \n 3-\n \n 8.0 cm\n \n 3\n \n
\n\n \n 5-\n \n 4.0 cm\n \n 3\n \n \n \n \n \n ✔\n
\n\n
\n\n \n Award [1] for 2 correct answers\n \n
\n\n \n 1-\n \n 1.0 cm\n \n 3\n \n
\n\n \n 2-\n \n 4.0 cm\n \n 3\n \n \n \n ✔\n
\n\n \n 3-\n \n 8.0 cm\n \n 3\n \n
\n\n \n 5-\n \n 4.0 cm\n \n 3\n \n \n \n \n \n ✔\n
\n\n
\n\n \n Award [1] for 2 correct answers\n \n
\n\n 0.050−1.000 ✔\n
\n\n water\n \n \n AND\n \n \n all samples dissolved «in water» ✔\n
\n\n ensures the line is best-fit ✔\n
\n\n line/equation of the line will be used for quantitation ✔\n
\n\n
\n\n \n Accept any other explanations referring to accuracy.\n \n
\n\n \n Interpolation:\n \n 9.8 μg cm\n \n −3\n \n ✔\n
\n\n \n using equation:\n \n 0.600/0.06283 = 9.55 «μg cm\n \n −3\n \n »\n \n \n ✔\n
\n\n line equation\n \n \n AND\n \n \n uses the values for 5 determinations ✔\n
\n\n «ascorbic acid» has multiple −OH/hydroxyl groups ✔\n
\n\n can H-bond with water ✔\n
\n\n
\n\n \n Do not accept OH−/hydroxide for M1\n \n
\n\n Δ % = «95.5 - 50.0/95.5 × 100»= 47.6 % ✔\n
\n\n
\n\n \n Accept calculations using line equation\n \n
\n\n assumption: linear decrease ✔\n
\n\n rate «mg 100 g\n \n −1\n \n day\n \n −1\n \n = 95.5 − 50.0 / 3 » = 15 «mg 100 g\n \n −1\n \n day\n \n −1\n \n » ✔\n
\n\n «95.5 − 15 × 6 = 5.5 «mg 100 g\n \n −1\n \n day\n \n −1\n \n » ✔\n
\n\n The student reported the volumes of titrant used per trial for samples collected each day in the following table:\n
\n\n \n
\n
\n\n The candidate used a 25 cm\n \n 3\n \n burette with an uncertainty of ±0.05 cm\n \n 3\n \n . Comment on the uncertainty recorded for the titrant volumes.\n
\n\n [1]\n
\n\n Calculate the initial % content of Fe\n \n 2+\n \n in the raw spinach, showing your working.\n
\n\n [3]\n
\n\n The results calculated for the subsequent days are shown.\n
\n\n \n
\n Comment on the significance of the difference in Fe\n \n 2+\n \n content measured for day 4 and 5.\n
\n\n [1]\n
\n\n Suggest\n \n two\n \n flaws in the design that could have contributed to the random error in the investigation.\n
\n\n [2]\n
\n\n The student did not standardise the KMnO\n \n 4\n \n solution used for titration. Suggest what type of error this may have caused, giving your reasons.\n
\n\n [1]\n
\n\n The titration of Fe(II) with MnO\n \n 4\n \n in acid medium is a redox reaction. State the oxidised and reduced species, including their change in oxidation states.\n
\n\n
\n Oxidised: ........................................................................................................................................\n
\n
\n Reduced: ........................................................................................................................................\n
\n [2]\n
\n\n Spinach contains a large amount of antioxidant compounds, including ascorbic acid and oxalic acid. Predict how this will affect the accuracy of the results, mentioning the type and direction of the error.\n
\n\n [2]\n
\n\n Incorrect\n \n \n AND\n \n \n two readings, uncertainty is ±0.1 ✔\n
\n\n Incorrect\n \n \n AND\n \n \n two readings, uncertainty is ±0.1 ✔\n
\n\n «mol» MnO\n \n 4−\n \n «=0.00339 × 0.01» = 3.38 × 10\n \n −5\n \n «mol» ✓\n
\n\n «mol MnO\n \n 4−\n \n : mol Fe\n \n +2\n \n = 1:5»\n
\n\n «mol» Fe\n \n 2+\n \n = 1.95 × 10\n \n −4\n \n ✓\n
\n\n % Fe\n \n 2+\n \n «= 1.95 × 10\n \n −4\n \n × 55.8 × 100/2.0»= 0.47«%» ✓\n
\n\n
\n\n \n Must show working for the marks.\n \n
\n\n no difference\n \n \n AND\n \n \n uncertainty larger than difference ✔\n
\n\n
\n\n \n Accept other explanations referred to overlapping.\n \n
\n\n dilute the titrant to use larger volumes «of titrant» ✔\n
\n\n ensure spinach leaf fragments are the same size ✔\n
\n\n systematic error\n \n \n AND\n \n \n all values «equally» inaccurate ✔\n
\n\n Oxidised: Fe\n \n +2\n \n → Fe\n \n +3\n \n ✔\n
\n\n Reduced: Mn\n \n 7+\n \n → Mn\n \n 2+\n \n ✔\n
\n\n systematic error\n \n \n AND\n \n \n lower accuracy ✔\n
\n\n overestimation of [Fe(II)] ✔\n
\n\n A calibration curve with pure ascorbic acid and appropriate amounts of the reagent, R, was prepared by dilutions with water of an initial aqueous solution of\n \n 100\n \n \n μg/cm\n \n −3\n \n \n ascorbic acid in water, AA (aq)\n
\n\n
\n\n Calculate the volumes of pure ascorbic acid solution required for each point of the calibration curve; point 4 of the curve is shown as an example.\n
\n\n \n
\n [2]\n
\n\n The resulting calibration curve is shown:\n
\n\n \n
\n Suggest a range of absorbance values for which this curve can be used to calculate ascorbic acid of broccoli accurately.\n
\n\n [1]\n
\n\n Suggest what should be used as a blank for spectrophotometric reading.\n
\n\n [1]\n
\n\n Discuss why it is important to obtain a value of R\n \n 2\n \n close to 1 for a calibration curve.\n
\n\n [2]\n
\n\n After 3 days, the broccoli samples were removed from storage and 1.0 g of each sample was blended with 100.0 cm\n \n 3\n \n of water. The filtered solution was mixed with the reactant in the same proportions as that used for the calibration curve in a cuvette and measured.\n
\n\n The sample stored at 5 °C showed an absorbance of 0.600. Determine the concentration of ascorbic acid in the sample solution by interpolation and using the line equation.\n
\n\n
\n interpolation in graph: ..........................................................................................................................\n
\n
\n using line equation: .............................................................................................................................\n
\n
\n\n
\n\n [2]\n
\n\n Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to its structure.\n
\n\n [2]\n
\n\n Estimate the % change in ascorbic acid concentration when stored for 3 days storage at 5 °C and 20 °C.\n
\n\n [1]\n
\n\n Estimate how much ascorbic acid will remain after 6 days storage at 20 °C in the same experimental conditions, stating any assumption made for the calculation.\n
\n\n [3]\n
\n\n \n 1-\n \n 1.0 cm\n \n 3\n \n
\n\n \n 2-\n \n 4.0 cm\n \n 3\n \n \n \n ✔\n
\n\n \n 3-\n \n 8.0 cm\n \n 3\n \n
\n\n \n 5-\n \n 4.0 cm\n \n 3\n \n \n \n \n \n ✔\n
\n\n
\n\n \n Award [1] for 2 correct answers\n \n
\n\n \n 1-\n \n 1.0 cm\n \n 3\n \n
\n\n \n 2-\n \n 4.0 cm\n \n 3\n \n \n \n ✔\n
\n\n \n 3-\n \n 8.0 cm\n \n 3\n \n
\n\n \n 5-\n \n 4.0 cm\n \n 3\n \n \n \n \n \n ✔\n
\n\n
\n\n \n Award [1] for 2 correct answers\n \n
\n\n 0.050−1.000 ✔\n
\n\n water\n \n \n AND\n \n \n all samples dissolved «in water» ✔\n
\n\n ensures the line is best-fit ✔\n
\n\n line/equation of the line will be used for quantitation ✔\n
\n\n
\n\n \n Accept any other explanations referring to accuracy.\n \n
\n\n \n Interpolation:\n \n 9.8\n \n μg\n \n cm\n \n −3\n \n ✔\n
\n\n \n using equation:\n \n 0.600/0.06283 = 9.55 «\n \n μg\n \n cm\n \n −3\n \n »\n \n \n ✔\n
\n\n «ascorbic acid» has multiple −OH/hydroxyl groups ✔\n
\n\n can H-bond with water ✔\n
\n\n
\n\n \n Do not accept OH−/hydroxide for M1\n \n
\n\n Δ % = «95.5 - 50.0/95.5 × 100»= 47.6 % ✔\n
\n\n
\n\n \n Accept calculations using line equation\n \n
\n\n assumption: linear decrease ✔\n
\n\n rate «mg 100 g\n \n −1\n \n day\n \n −1\n \n = 95.5 − 50.0 / 3 » = 15 «mg 100 g\n \n −1\n \n day\n \n −1\n \n » ✔\n
\n\n «95.5 − 15 × 6 = 5.5 «mg 100 g\n \n −1\n \n day\n \n −1\n \n » ✔\n
\n\n Determine the ratio in which 0.1 mol dm\n \n –3\n \n NaH\n \n 2\n \n PO\n \n 4\n \n and 0.1 mol dm\n \n –3\n \n Na\n \n 2\n \n HPO\n \n 4\n \n should be mixed to obtain a buffer with pH= 7.8.\n
\n\n p\n \n K\n \n \n a\n \n NaH\n \n 2\n \n PO\n \n 4\n \n = 7.20\n
\n\n
\n\n [3]\n
\n\n Suggest, giving your reasons, the effect of diluting the buffer 1/100 with water on its pH and reaction to the addition of acids or bases.\n
\n\n
\n\n change in pH:\n
\n
\n .............................................................................................................................................................\n
\n .............................................................................................................................................................\n
\n\n
\n\n reaction to addition of bases/acids:\n
\n\n .............................................................................................................................................................\n
\n\n .............................................................................................................................................................\n
\n\n [2]\n
\n\n \n
\n\n \n «\n \n » 0.6 ✓\n
\n\n \n «\n \n » 3.98 ✓\n
\n\n Na\n \n 2\n \n PO\n \n 4\n \n to Na\n \n 2\n \n HPO\n \n 4\n \n = 3.98:1 ✓\n
\n\n \n
\n\n \n «\n \n » 0.6 ✓\n
\n\n \n «\n \n » 3.98 ✓\n
\n\n Na\n \n 2\n \n PO\n \n 4\n \n to Na\n \n 2\n \n HPO\n \n 4\n \n = 3.98:1 ✓\n
\n\n \n change in pH:\n \n
\n\n no «significant» effect as ratio of salt/acid are unchanged ✓\n
\n\n
\n\n \n reaction to addition of bases/acids:\n \n
\n\n resistance to change/buffering capacity decreases ✓\n
\n