\n
\n \n On the following Maxwell-Boltzmann distribution, which letter represents activation energy?\n
\n\n \n
\n A. A\n
\n\n B. B\n
\n\n C. C\n
\n\n D. D\n
\n\n
\n\n [1]\n
\n\n \n Outline\n \n two\n \n differences between heterogeneous and homogeneous catalysts.\n \n
\n\n [2]\n
\n\n \n \n Any two of:\n \n
\n heterogeneous catalyst is in different phase than reactants\n \n \n AND\n \n \n homogeneous catalyst in same phase\n \n [✔]\n \n \n
\n \n homogeneous catalysts chemically change/react and are reformed at end of reaction\n
\n \n \n OR\n \n \n
\n reactants adsorb onto heterogenous catalyst and products desorb\n \n [✔]\n \n \n
\n \n heterogeneous catalysts are more easily removed than homogenous catalysts\n \n [✔]\n \n \n
\n\n \n heterogeneous catalysts can function at higher temperatures\n \n [✔]\n \n \n
\n\n \n homogeneous catalysts are «generally» more selective\n \n [✔]\n \n \n
\n\n \n homogeneous catalysts offer a broader range of reactions\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n \n Note:\n \n \n Accept “state” for “phase”.\n \n \n \n \n
\n\n \n \n \n \n Accept “heterogeneous catalyst provides a surface to activate reaction”.\n \n \n \n \n
\n\n Many students were aware of the fundamental difference between homogeneous and heterogeneous catalysts, but only a few could identify another difference to obtain full marks.\n
\n\n What information can be deduced from the splitting pattern of\n \n 1\n \n H NMR signals?\n
\n\n A. total number of hydrogen atoms in a compound\n
\n\n B. number of hydrogen atoms on adjacent atom(s)\n
\n\n C. functional group on which hydrogen atoms are located\n
\n\n D. number of hydrogen atoms in a particular chemical environment\n
\n\n [1]\n
\n\n B\n
\n\n \n Write the equation for the production of the active nitrating agent from concentrated sulfuric and nitric acids.\n \n
\n\n [1]\n
\n\n \n 2H\n \n 2\n \n SO\n \n 4\n \n + HNO\n \n 3\n \n ⇌ NO\n \n 2\n \n \n +\n \n + 2HSO\n \n 4\n \n \n \n \n −\n \n + H\n \n 3\n \n O\n \n +\n \n \n \n [\n \n ✔\n \n ]\n \n
\n\n \n \n Note\n \n :\n \n Accept a single arrow instead of an equilibrium sign.\n
\n Accept “H\n \n 2\n \n SO\n \n 4\n \n + HNO\n \n 3\n \n ⇌ NO\n \n 2\n \n \n +\n \n + HSO\n \n 4\n \n \n −\n \n + H\n \n 2\n \n O”.\n
\n Accept “H\n \n 2\n \n SO\n \n 4\n \n + HNO\n \n 3\n \n ⇌ H\n \n 2\n \n NO\n \n 3\n \n \n +\n \n + HSO\n \n 4\n \n \n −\n \n ”.\n
\n 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.\n \n \n
\n The production of NO\n \n 3\n \n \n −\n \n was a common answer.\n
\n\n \n State what point\n \n Y\n \n on the graph represents.\n \n
\n\n [1]\n
\n\n \n highest recorded temperature\n
\n \n \n OR\n \n \n
\n when rate of heat production equals rate of heat loss\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “maximum temperature”.\n \n \n
\n\n \n \n Accept “completion/end point of reaction”.\n \n \n
\n\n Many 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.\n
\n\n \n Apart 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.\n \n
\n\n [2]\n
\n\n \n \n E\n \n a\n \n \n marked on graph\n \n [✔]\n \n \n
\n\n \n explanation in terms of more “particles” with\n \n E ≥ E\n \n a\n \n \n
\n \n \n OR\n \n \n
\n greater area under curve to the right of\n \n E\n \n a\n \n \n in T\n \n 2\n \n \n [✔]\n \n \n
\n Explaining 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.\n
\n\n \n The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.\n \n
\n\n \n State the formula of its conjugate base.\n \n
\n\n [1]\n
\n\n \n CO\n \n 3\n \n \n 2–\n \n \n [✔]\n \n \n
\n\n A 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 HCO\n \n 3\n \n \n -\n \n . Wrong answers included water, the hydroxide ion and carbon dioxide.\n
\n\n How many p-orbitals are occupied in a phosphorus atom?\n
\n\n
\n A. 2\n
\n B. 3\n
\n\n C. 5\n
\n\n D. 6\n
\n\n [1]\n
\n\n D\n
\n\n Suggest which of the two methods will provide a more accurate value.\n
\n\n [1]\n
\n\n line equation\n \n \n AND\n \n \n uses the values for 5 determinations ✔\n
\n\n \n What is the mechanism of the reaction between alkenes and halogens in the absence of light?\n \n
\n\n \n A. radical substitution\n \n
\n\n \n B. electrophilic substitution\n \n
\n\n \n C. electrophilic addition\n \n
\n\n \n D. nucleophilic substitution\n \n
\n\n [1]\n
\n\n C\n
\n\n Only 33 % of candidates could identify the mechanism type with a large confusion between electrophilic addition and substitution.\n
\n\n On the following Maxwell-Boltzmann distribution, which letter represents activation energy?\n
\n\n \n
\n A. A\n
\n\n B. B\n
\n\n C. C\n
\n\n D. D\n
\n\n [1]\n
\n\n B\n
\n\n Suggest what should be used as a blank for spectrophotometric reading.\n
\n\n [1]\n
\n\n water\n \n \n AND\n \n \n all samples dissolved «in water» ✔\n
\n\n Calculate the initial pH before any sodium hydroxide was added, using section 21 of the data booklet.\n
\n\n [2]\n
\n\n «\n \n K\n \n a = 10\n \n –2.87\n \n = 1.35 × 10\n \n –3\n \n »\n
\n\n «1.35 × 10\n \n –3\n \n =\n \n »\n
\n\n «x = [H\n \n +\n \n ] =\n \n =» 2.6 × 10\n \n –2\n \n «mol dm\n \n –3\n \n » ✔\n
\n\n
\n «pH = –log[H\n \n +\n \n ] = –log(2.6 × 10\n \n –2\n \n ) =» 1.59 ✔\n
\n
\n\n \n Accept final answer in range 1.58–1.60.\n \n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n Which combination has the greatest rate of reaction at room temperature?\n
\n
\n \n
\n [1]\n
\n\n A\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 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 Explain the effect of increasing temperature on the yield of SO\n \n 3\n \n .\n
\n\n [2]\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 Another 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 SO\n \n 3\n \n 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 SO\n \n 3\n \n .\n
\n\n Explain why the first ionization energy of calcium is greater than that of potassium.\n
\n\n [2]\n
\n\n increasing number of protons/nuclear charge/Z\n \n eff\n \n ✔\n
\n\n
\n «atomic» radius/size decreases\n
\n \n \n OR\n \n \n
\n same number of energy levels\n
\n \n \n OR\n \n \n
\n similar shielding «by inner electrons» ✔\n
\n It 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.\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 \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 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 At equilibrium, the concentrations of chlorine and iodine are both 0.02 mol dm\n \n –3\n \n .\n
\n\n \n Cl\n \n 2\n \n (g) +\n \n \n (g)\n \n \n Cl (g)\n \n K\n \n \n c\n \n = 454\n
\n\n What is the concentration of iodine monochloride,\n \n Cl?\n
\n\n
\n A.\n \n
\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n B\n
\n\n High scoring candidates had more success calculating the concentration of ICl in the product given equal concentrations of Cl\n \n 2\n \n and I\n \n 2\n \n in the reactants.\n
\n\n Justify why ethene has only a single signal in its\n \n 1\n \n H NMR spectrum.\n
\n\n [1]\n
\n\n hydrogen atoms/protons in same chemical environment ✔\n
\n\n \n Accept “all H atoms/protons are equivalent”.\n \n
\n \n Accept “symmetrical”\n \n
\n What is the equilibrium constant expression for the following reaction?\n
\n\n 2SO\n \n 3\n \n (g)\n \n 2SO\n \n 2\n \n (g) + O\n \n 2\n \n (g)\n
\n\n
\n A.\n \n
\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n \n Aspirin can be obtained from salicylic acid.\n \n
\n\n \n Unreacted salicylic acid may be present as an impurity in aspirin and can be detected in the infrared (IR) spectrum.\n \n
\n\n \n \n \n
\n \n \n Name the functional group and identify the absorption band that differentiates salicylic acid from aspirin. Use section 26 of the data booklet.\n \n \n
\n\n
\n\n \n \n Name:\n \n \n
\n\n \n \n Absorption band:\n \n \n
\n\n [2]\n
\n\n \n \n Name\n \n :\n
\n hydroxyl\n \n [✔]\n \n \n
\n \n \n Absorption band\n \n :\n
\n 3200–3600 «cm\n \n –1\n \n »\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “phenol”\n \n OR\n \n “alcohol” but\n \n not\n \n “hydroxide”.\n \n \n
\n\n Many students could correctly identify a difference between the given structures and state a region in the IR spectrum where this would absorb.\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 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 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
\n\n Calculate the concentration of H\n \n 3\n \n PO\n \n 4\n \n if 25.00 cm\n \n 3\n \n is completely neutralised by the addition of 28.40 cm\n \n 3\n \n of 0.5000 mol dm\n \n −3\n \n NaOH.\n
\n\n [2]\n
\n\n «\n \n »\n
\n\n «\n \n » 0.004733 «mol» ✔\n
\n\n «\n \n » 0.1893 «mol dm\n \n −3\n \n » ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Deduce the equation for the decomposition of guanidinium nitrate.\n \n
\n\n [1]\n
\n\n \n C(NH\n \n 2\n \n )\n \n 3\n \n NO\n \n 3\n \n (s) → 2N\n \n 2\n \n (g) + 3H\n \n 2\n \n O (g) + C (s) ✔\n \n
\n\n Which technique is best for determining bond lengths within a molecule?\n
\n\n
\n A.\n \n 1\n \n H NMR spectroscopy\n
\n B. infrared spectroscopy\n
\n\n C. mass spectroscopy\n
\n\n D. X-ray crystallography\n
\n\n [1]\n
\n\n D\n
\n\n \n Nitrogen monoxide reacts with oxygen gas to form nitrogen dioxide.\n \n
\n\n \n Deduce, giving a reason, whether the following mechanism is possible.\n \n
\n\n \n \n \n
\n
\n\n [1]\n
\n\n not possible\n \n \n AND\n \n \n «proposed» mechanism does not match experimental rate law\n
\n \n \n OR\n \n \n
\n not possible\n \n \n AND\n \n \n «proposed» mechanism shows zero/not first order with respect to oxygen ✔\n
\n 60% of candidates could explain why the proposed reaction mechanism was inconsistent with the empirical data given.\n
\n\n Which combination will produce an alkaline buffer in water?\n
\n\n A. 0.10 mol NH\n \n 3\n \n and 0.05 mol H\n \n 2\n \n SO\n \n 4\n \n
\n\n B. 0.50 mol NH\n \n 3\n \n and 0.10 mol H\n \n 2\n \n SO\n \n 4\n \n
\n\n C. 0.10 mol CH\n \n 3\n \n COOH and 0.05 mol NaOH\n
\n\n D. 0.10 mol CH\n \n 3\n \n COOH and 0.50 mol NaOH\n
\n\n [1]\n
\n\n B\n
\n\n \n The experiment is repeated using the same amount of dinitrogen monoxide in the same apparatus, but at a lower temperature.\n \n
\n\n \n Sketch, on the axes in question 2, the graph that you would expect.\n \n
\n\n [2]\n
\n\n \n
\n \n smaller initial gradient\n \n [✔]\n \n \n
\n\n \n initial pressure is lower\n \n \n AND\n \n \n final pressure of gas lower «by similar factor»\n \n [✔]\n \n \n
\n\n
\n\n This 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.\n
\n\n Justify why sulfur is classified as a non-metal by giving\n \n two\n \n of its chemical properties.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n\n forms acidic oxides «rather than basic oxides» ✔\n
\n\n forms covalent/bonds compounds «with other non-metals» ✔\n
\n\n forms anions «rather than cations» ✔\n
\n\n behaves as an oxidizing agent «rather than a reducing agent» ✔\n
\n\n \n
\n Award\n \n [1 max]\n \n 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.\n \n
\n \n How many moles of magnesium hydroxide are produced with 0.50 mol of ammonia?\n \n
\n\n \n Mg\n \n 3\n \n N\n \n 2\n \n (s) + 6H\n \n 2\n \n O (l) → 3Mg(OH)\n \n 2\n \n (aq) + 2NH\n \n 3\n \n (aq)\n \n
\n\n \n A. 0.25\n \n
\n\n \n B. 0.33\n \n
\n\n \n C. 0.75\n \n
\n\n D\n \n . 1.5\n \n
\n\n [1]\n
\n\n C\n
\n\n 90 % of the candidates were able to deduce the amount of a product given the amount of another product and the balanced equation.\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 «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 Compare and contrast the combustion of an s-block metal and a p-block non-metal.\n
\n\n [2]\n
\n\n both are oxidation reactions ✓\n
\n\n
\n s-block metals produce ionic oxides\n \n \n AND\n \n \n p-block non-metals produce covalent oxides\n
\n \n \n OR\n
\n \n \n s-block metals produce basic oxides\n \n \n AND\n \n \n p-block non-metals produce acid oxides ✓\n
\n \n Determine from the graph the rate of reaction at 20 s, in cm\n \n 3\n \n s\n \n −1\n \n , showing your working.\n \n
\n\n [3]\n
\n\n \n tangent drawn to curve at t = 20 s\n \n [✔]\n \n \n
\n\n \n slope/gradient calculation\n \n [✔]\n \n \n
\n\n \n 0.35 «cm\n \n 3\n \n s\n \n –1\n \n »\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept values in the range 0.32–0.42 «cm\n \n 3\n \n s\n \n –1\n \n ».\n \n \n
\n\n This question was challenging for many students. Quite a few candidates did draw a tangent line at 20s for 1 mark, show a slope/gradient calculation of the line for 1 mark, and had a reasonable final value for the final mark. Some candidates only found the average rate by finding the ratio of the value at that data point and received one mark (16/20=0.80 cm\n \n 3\n \n s\n \n -1\n \n ). Candidates also received one mark if they had a correct answer with no work since the question clearly asked students to show their work.\n
\n\n What is the correct labelling of the blocks of the periodic table?\n
\n\n
\n \n
\n [1]\n
\n\n D\n
\n\n Why is copper(II) sulfate blue?\n
\n\n
\n A. Red light is absorbed when electrons are promoted between the orbitals in the split d-sublevels.\n
\n B. Blue light is emitted when electrons fall between the orbitals in the split d-sublevels.\n
\n\n C. Red light is absorbed when electrons fall between the orbitals in the split d-sublevels.\n
\n\n D. Blue light is emitted when electrons are promoted between the orbitals in the split d-sublevels.\n
\n\n [1]\n
\n\n A\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 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 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 Which compound has hydrogen bonds between its molecules?\n \n
\n\n \n A. CH\n \n 4\n \n \n
\n\n \n B. CH\n \n 4\n \n O\n \n
\n\n \n C. CH\n \n 3\n \n Cl\n \n
\n\n \n D. CH\n \n 2\n \n O\n \n
\n\n [1]\n
\n\n B\n
\n\n 72 % of the candidates identified CH\n \n 4\n \n O as having hydrogen bonds between its molecules. The most commonly chosen distractor was CH\n \n 2\n \n O, the only other option containing oxygen.\n
\n\n \n Which compound has the lowest boiling point?\n \n
\n\n \n A. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n \n
\n\n \n B. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n \n
\n\n \n C. CH\n \n 3\n \n CH(CH\n \n 3\n \n )CH\n \n 2\n \n CH\n \n 3\n \n \n
\n\n \n D. CH\n \n 3\n \n C(CH\n \n 3\n \n )\n \n 2\n \n CH\n \n 3\n \n \n
\n\n [1]\n
\n\n D\n
\n\n Surprisingly, this was one of the challenging questions on the paper. Only 63 % of the candidates chose dimethylpropane as the compound having the lowest boiling point. The most commonly chosen distractor was pentane (B) which did not take into account the effect of branching on the strength of London dispersion forces.\n
\n\n Outline why vitamins usually need to be obtained from food sources.\n
\n\n [1]\n
\n\n cannot be synthesized «by the human body» ✓\n
\n\n State the name of the products of the reaction between benzene, C\n \n 6\n \n H\n \n 6\n \n and bromine, Br\n \n 2\n \n in presence of the catalyst, FeBr\n \n 3\n \n .\n
\n\n [1]\n
\n\n bromobenzene\n \n \n AND\n \n \n HBr ✓\n
\n\n
\n\n \n Accept structural formula of bromobenzene or correct reaction\n \n
\n\n \n 100.0 cm\n \n 3\n \n of soda water contains 3.0 × 10\n \n −2\n \n g NaHCO\n \n 3\n \n .\n \n
\n\n \n Calculate the concentration of NaHCO\n \n 3\n \n in mol dm\n \n −3\n \n .\n \n
\n\n [2]\n
\n\n \n «molar mass of NaHCO\n \n 3\n \n =» 84.01 «g mol\n \n –1\n \n »\n \n [✔]\n \n \n
\n\n \n «concentration =\n \n \n \n » 3.6 × 10\n \n –3\n \n «mol dm\n \n –3\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 Very well answered. Most candidates calculated the molar concentration correctly.\n
\n\n Outline why metals, like iron, can conduct electricity.\n
\n\n [1]\n
\n\n mobile/delocalized «sea of» electrons\n
\n\n Sketch the neutralisation curve obtained\n \n and\n \n label the equivalence point.\n
\n\n \n
\n [3]\n
\n\n \n
\n \n \n OR\n \n \n
\n\n \n
\n starts at 1.6\n \n \n AND\n \n \n finishes at 13.6 ✔\n
\n\n approximately vertical at the correct volume of alkali added ✔\n
\n\n equivalence point labelled\n \n \n AND\n \n \n above pH 7 ✔\n
\n\n
\n\n \n Accept any range from 1.1-1.9\n \n AND\n \n 13.1-13.9 for\n \n M1\n \n or ECF from 11c(i) and 11c(ii).\n \n
\n\n \n Award\n \n M2\n \n for vertical climb at 28 cm\n \n 3\n \n \n OR\n \n 15 cm\n \n 3\n \n .\n \n
\n\n \n Equivalence point must be labelled for\n \n M3\n \n .\n \n
\n\n \n Distinguish between a weak and strong acid.\n \n
\n\n \n Weak acid:\n \n
\n\n \n Strong acid:\n \n
\n\n [1]\n
\n\n \n \n Weak acid:\n \n partially dissociated/ionized «in aqueous solution/water»\n
\n \n \n AND\n \n \n
\n \n Strong acid\n \n : «assumed to be almost» completely/100 % dissociated/ionized «in aqueous solution/water»\n \n [✔]\n \n \n
\n As expected, many candidates were able to distinguish between strong and weak acids; some candidates referred to “dissolve” rather than dissociate.\n
\n\n What is the main interaction between liquid CH\n \n 4\n \n molecules?\n
\n\n A. London (dispersion) forces\n
\n\n B. Dipole–dipole forces\n
\n\n C. Hydrogen bonding\n
\n\n D. Covalent bonding\n
\n\n [1]\n
\n\n A\n
\n\n 56% of the candidates were able to identify London (dispersion) forces as the main interaction between liquid CH\n \n 4\n \n molecules. Covalent bonding was the most commonly chosen distractor. Good performance on this question correlated well with candidates who scored well overall.\n
\n\n \n Methane undergoes incomplete combustion.\n \n
\n\n \n 2CH\n \n 4\n \n (g) + 3O\n \n 2\n \n (g) → 2CO (g) + 4H\n \n 2\n \n O (g)\n \n
\n\n \n What is the enthalpy change, in kJ, using the bond enthalpy data given below?\n \n
\n\n \n \n \n
\n \n \n A. [2(1077) + 4(463)] − [2(414) + 3(498)]\n \n \n
\n\n \n \n B. [2(414) + 3(498)] − [2(1077) + 4(463)]\n \n \n
\n\n \n \n C. [8(414) + 3(498)] − [2(1077) + 8(463)]\n \n \n
\n\n \n \n D. [2(1077) + 8(463)] − [8(414) + 3(498)]\n \n \n
\n\n [1]\n
\n\n C\n
\n\n Only 70 % of the candidates chose the correct calculation of the enthalpy change using bond enthalpy data. The most commonly chosen distractor (D) reversed the signs.\n
\n\n \n What is the empirical formula of a hydrocarbon with 75 % carbon and 25 % hydrogen by mass?\n \n
\n\n \n A. C\n \n 3\n \n H\n \n
\n\n \n B. CH\n \n 2\n \n \n
\n\n \n C. C\n \n 2\n \n H\n \n 6\n \n \n
\n\n \n D. CH\n \n 4\n \n \n
\n\n [1]\n
\n\n D\n
\n\n Well answered and straight forward.\n
\n\n Which species are\n \n both\n \n Lewis and Brønsted–Lowry bases?\n
\n\n I. CN\n \n −\n \n
\n II. OH\n \n −\n \n
\n III. NH\n \n 3\n \n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n D\n
\n\n 45% of candidates chose all three species as Lewis and Brønsted-Lowry bases. The most commonly chosen distractor excluded CN\n \n -\n \n .\n
\n\n Sulfur dioxide reacts with oxygen to form sulfur trioxide.\n
\n\n 2SO\n \n 2\n \n (g) + O\n \n 2\n \n (g)\n \n 2SO\n \n 3\n \n (g) Δ\n \n H\n \n = −197 kJ\n
\n\n Which change increases the value of\n \n K\n \n \n c\n \n ?\n
\n\n A. increasing the temperature\n
\n\n B. decreasing the temperature\n
\n\n C. decreasing [SO\n \n 2\n \n (g)]\n
\n\n D. decreasing [SO\n \n 3\n \n (g)]\n
\n\n [1]\n
\n\n B\n
\n\n \n The absorption spectrum of chlorophyll a is shown below.\n \n
\n\n \n \n \n
\n \n \n Suggest how the combination of chlorophyll a and carotenoids is beneficial for photosynthesis.\n \n \n
\n\n [1]\n
\n\n \n extends energy absorption spectrum «for photosynthesis»\n \n [✔]\n \n \n
\n\n Most candidates gave good answers to suggest how a combination of chlorophyll\n \n a\n \n and carotenoids is beneficial for photosynthesis.\n
\n\n Which sample contains the fewest moles of HCl?\n
\n\n \n N\n \n \n A\n \n = 6.02 × 10\n \n 23\n \n mol\n \n –1\n \n .\n
\n\n Molar volume of an ideal gas at STP = 22.7 dm\n \n 3\n \n mol\n \n –1\n \n .\n
\n\n
\n A. 10.0 cm\n \n 3\n \n of 0.1 mol dm\n \n –3\n \n HCl (aq)\n
\n B. 6.02 × 10\n \n 24\n \n molecules of HCl (g)\n
\n\n C. 0.365 g of HCl (g)\n
\n\n D. 2.27 dm\n \n 3\n \n of HCl (g) at STP\n
\n\n [1]\n
\n\n A\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 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 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 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 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 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 State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.\n
\n\n \n
\n [4]\n
\n\n \n
\n \n
\n Award\n \n [1]\n \n for all bonding types correct.\n \n
\n \n Award\n \n [1]\n \n for\n \n each\n \n correct description.\n \n
\n\n \n Apply ECF for M2 only once.\n \n
\n\n About 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.\n
\n\n \n Deduce the hybridization of the central nitrogen atom in the molecule.\n \n
\n\n [1]\n
\n\n \n sp\n \n [✔]\n \n \n
\n\n Hybridisation of the N atom was correct in most cases.\n
\n\n What are the units of molar mass?\n
\n\n
\n A. amu\n
\n B. g\n
\n\n C. mol g\n \n −1\n \n
\n\n D. g mol\n \n −1\n \n
\n\n [1]\n
\n\n D\n
\n\n State the full electronic configuration of Fe\n \n 2+\n \n .\n
\n\n [1]\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 3d\n \n 6\n \n ✔\n
\n\n Mostly well done which was a pleasant surprise since this is not overly easy, predictably some gave [Ar] 4s\n \n 2\n \n 3d\n \n 4\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 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 \n Which molecule contains an incomplete octet of electrons?\n \n
\n\n \n A. NF\n \n 3\n \n \n
\n\n \n B. BF\n \n 3\n \n \n
\n\n \n C. BrF\n \n
\n\n \n D. SF\n \n 2\n \n \n
\n\n [1]\n
\n\n B\n
\n\n 66 % of the candidates identified BF\n \n 3\n \n as having an incomplete octet of electrons. The distractors NF\n \n 3\n \n and BrF were chosen in equal numbers, and the distractor SF\n \n 2\n \n was chosen by the least number of candidates.\n
\n\n \n Outline, with a reason, another property that could be monitored to measure the rate of this reaction.\n \n
\n\n [2]\n
\n\n \n \n \n ALTERNATIVE 1\n \n \n
\n colour\n \n [✔]\n \n \n
\n \n Br\n \n 2\n \n /reactant is coloured «Br\n \n –\n \n (aq)/product is not»\n \n [✔]\n \n \n
\n\n \n \n \n Note:\n \n Do\n \n not\n \n accept “changes in temperature” or “number of bubbles”.\n \n \n
\n\n
\n\n \n \n \n ALTERNATIVE 2\n \n \n
\n conductivity\n \n [✔]\n \n \n
\n \n greater/increased concentration of ions in products\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n ALTERNATIVE 3\n \n \n
\n mass/pressure\n \n [✔]\n \n \n
\n \n gas is evolved/produced\n \n [✔]\n \n \n
\n\n \n \n \n Note\n \n : Do\n \n not\n \n accept “mass of products is less than mass of reactants”.\n \n \n
\n\n \n
\n \n \n ALTERNATIVE 4\n \n \n
\n pH\n \n [✔]\n \n \n
\n \n methanoic acid is weak\n \n \n AND\n \n \n HBr is strong\n
\n \n \n OR\n \n \n
\n increase in [H\n \n +\n \n ]\n \n [✔]\n \n \n
\n The reaction rate was originally monitored by measuring the volume of CO\n \n 2\n \n produced. Students needed to propose another method for this reaction, with a reason, that could be used to measure the rate. There were several possible correct answers and most students received at least one mark with many receiving both marks. The most common incorrect answer involved changes in temperature.\n
\n\n Draw the structural formula of the alkene required.\n
\n\n \n
\n [1]\n
\n\n \n
\n Few candidates could correctly eliminate water to deduce the identity of the required reactant.\n
\n\n Outline the reason that sodium hydroxide is considered a Brønsted–Lowry base.\n
\n\n [1]\n
\n\n «OH\n \n −\n \n is a» proton acceptor ✔\n
\n\n Which factor does\n \n not\n \n affect the colour of a complex ion?\n
\n\n A. temperature of the solution\n
\n\n B. identity of the ligand\n
\n\n C. identity of the metal\n
\n\n D. oxidation number of the metal\n
\n\n [1]\n
\n\n A\n
\n\n Which is a renewable energy source?\n
\n\n
\n A. natural gas\n
\n B. uranium\n
\n\n C. coal\n
\n\n D. wood\n
\n\n [1]\n
\n\n D\n
\n\n \n Which diagram represents a heterogeneous mixture?\n \n
\n\n \n \n \n
\n [1]\n
\n\n A\n
\n\n A few G2 forms suggested that candidates could have chosen A or B. However, homogeneous and heterogeneous mixtures are usually represented in such a way while only option A shows a clear separation in the mixture. We will avoid schematic diagrams in the future for this type of questions.\n
\n\n \n The cell potential for the spontaneous reaction when standard magnesium and silver half-cells are connected is +3.17 V.\n \n
\n\n \n Determine the cell potential at 298 K when:\n \n
\n\n \n [Mg\n \n 2+\n \n ] = 0.0500 mol dm\n \n −3\n \n
\n [Ag\n \n +\n \n ] = 0.100 mol dm\n \n −3\n \n \n
\n \n Use sections 1 and 2 of the data booklet.\n \n
\n\n
\n\n [2]\n
\n\n \n «\n \n E = E\n \n ᶱ\n \n \n \n × ln\n \n Q\n \n »\n \n
\n\n \n ln\n \n Q\n \n = «\n \n \n \n » 1.61\n \n [✔]\n \n \n
\n\n \n E = «3.17 V\n \n \n \n »3.15 «V»\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 The cell potential was correctly calculated by several candidates with some candidates managed an ECF mark for an error in the calculation. Unfortunately, the ln\n \n Q\n \n part was frequently wrong due to candidates forgetting to square the denominator.\n
\n\n Deduce the rate expression for the reaction.\n
\n\n [1]\n
\n\n Rate=\n \n k\n \n [H\n \n 2\n \n ] [\n \n \n 2\n \n ]\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 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 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 Which technique is most likely to be used for identification of functional groups?\n
\n\n
\n A. Combustion analysis\n
\n B. Determination of melting point\n
\n\n C. Infra-red (IR) spectroscopy\n
\n\n D. Mass spectroscopy (MS)\n
\n\n [1]\n
\n\n C\n
\n\n State, with a reason, the effect of an increase in temperature on the position of this equilibrium.\n
\n\n [1]\n
\n\n «shifts» left/towards reactants\n \n \n AND\n \n \n «forward reaction is» exothermic/ΔH is negative ✔\n
\n\n \n Identify the chiral carbon atom using an asterisk, *.\n \n
\n\n [1]\n
\n\n \n \n [\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 Which molecule has the\n \n weakest\n \n nitrogen to nitrogen bond?\n
\n\n
\n A. N\n \n 2\n \n
\n B. N\n \n 2\n \n H\n \n 2\n \n
\n\n C. N\n \n 2\n \n H\n \n 4\n \n
\n\n D.\n \n
\n [1]\n
\n\n C\n
\n\n Which compound contains both ionic and covalent bonds?\n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n D\n
\n\n Calculate the change in entropy, Δ\n \n S\n \n , in J K\n \n −1\n \n , for the decomposition of calcium carbonate.\n
\n\n [1]\n
\n\n «Δ\n \n S\n \n = (40 J K\n \n −1\n \n + 214 J K\n \n −1\n \n ) − (93 J K\n \n −1\n \n ) =» 161 «J K\n \n −1\n \n » ✓\n
\n\n \n
\n Ignore an extra step to determine total entropy change in JK\n \n –1\n \n : 161 J mol\n \n –1\n \n K\n \n –1\n \n x 5.55 mol = 894 «J mol\n \n –1\n \n K\n \n –1\n \n »\n \n
\n \n Award\n \n [1]\n \n for 894 «J mol\n \n –1\n \n K\n \n –1\n \n ».\n \n
\n\n \n This question is about biofuel.\n \n
\n\n \n Evaluate the use of biodiesel in place of diesel from crude oil.\n \n
\n\n [2]\n
\n\n \n \n \n Strength\n \n \n
\n \n Any one of:\n \n
\n less flammable «than diesel»\n \n [✔]\n \n \n
\n \n recycles carbon «lower carbon footprint»\n
\n \n \n OR\n \n \n
\n lower greenhouse gas emissions\n \n [✔]\n \n \n
\n \n easily biodegradable «in case of spill»\n \n [✔]\n \n \n
\n\n \n renewable\n
\n \n \n OR\n \n \n
\n does not deplete fossil fuel reserves\n \n [✔]\n \n \n
\n \n economic security/availability in countries without crude oil\n \n [✔]\n \n \n
\n\n \n \n \n Limitation\n \n \n
\n \n Any one of\n \n :\n
\n more difficult to ignite inside the engine «than diesel»\n \n [✔]\n \n \n
\n \n more viscous «than diesel»\n \n [✔]\n \n \n
\n\n \n lower energy content/specific energy/energy density\n \n [✔]\n \n \n
\n\n \n uses food sources\n
\n \n \n OR\n \n \n
\n uses land that could be used for food\n \n [✔]\n \n \n
\n \n «production is» more expensive\n \n [✔]\n \n \n
\n\n \n less suitable in low temperatures\n \n [✔]\n \n \n
\n\n \n increased NO\n \n x\n \n emissions for biodiesel\n \n [✔]\n \n \n
\n\n \n greenhouse gases still produced\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “«close to» carbon neutral”, “produce less greenhouse gases/CO2”.\n \n \n
\n\n \n \n Accept “engines have to be modified if biodiesel used” as limitation.\n \n \n
\n\n \n \n Do\n \n not\n \n award marks for strength and limitation that are the same topic/concept.\n \n \n
\n\n This question was well answered, and many candidates received either one or both marks.\n
\n\n At which temperature could Δ\n \n H\n \n , Δ\n \n S\n \n , and Δ\n \n G\n \n all be positive?\n
\n
\n
\n A. High temperatures\n
\n\n B. Low temperatures\n
\n\n C. Any temperature\n
\n\n D. No temperature\n
\n\n [1]\n
\n\n B\n
\n\n Nearly 70% of the candidates managed to deduce that Δ\n \n H\n \n , Δ\n \n S\n \n , and Δ\n \n G\n \n can only all be positive at low temperatures, when the contribution of entropy to free energy is minimal. The distractors were all about equally favoured.\n
\n\n A student was investigating rates of reaction. In which of the following cases would a colorimeter show a change in absorbance?\n
\n\n A. KBr (aq) + Cl\n \n 2\n \n (aq)\n
\n\n B. Cu (s) + Na\n \n 2\n \n SO\n \n 4\n \n (aq)\n
\n\n C. HCl (aq) + NaOH (aq)\n
\n\n D. (CH\n \n 3\n \n )\n \n 3\n \n COH (aq) + K\n \n 2\n \n Cr\n \n 2\n \n O\n \n 7\n \n (aq)\n
\n\n [1]\n
\n\n A\n
\n\n Question 17 was the question about the colorimeter), which many teachers felt was unfair, and only 22% of students gave the correct answer. The complaints about the question were often based on the lack of familiarity with the colorimeter. However, Chemistry is a practical course and student should be exposed to a piece of equipment such as this. Also, it was felt that there was a lot to process in this question. Although answer C could fairly easily be eliminated as there is no colour change. Students then had to recognize that in D, alcohol oxidation would result in a colour change from orange to green, except that it is a tertiary alcohol. This left A and B. A is of course the right answer as chlorine replaces bromine, but if B actually happened, there would also be a colour change from clear to blue. So, students needed to remember that copper is below sodium on the activity series.\n
\n\n \n Identify the type of bonding in sodium hydrogencarbonate.\n \n
\n\n \n \n Between sodium and hydrogencarbonate:\n \n \n
\n\n \n \n \n Between hydrogen and oxygen in hydrogencarbonate:\n \n \n \n
\n\n [2]\n
\n\n \n \n Between sodium and hydrogencarbonate:\n \n
\n ionic\n \n [✔]\n \n
\n \n
\n \n \n Between hydrogen and oxygen in hydrogencarbonate:\n \n
\n «polar» covalent\n \n [✔]\n \n \n
\n This 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.\n
\n\n \n \n Justify this hypothesis.\n \n \n
\n\n [1]\n
\n\n \n sulfuric acid is diprotic/contains two H\n \n +\n \n «while nitric acid contains one H\n \n +\n \n »/releases more H\n \n +\n \n «so reacts with more limestone»\n
\n \n \n OR\n \n
\n \n higher concentration of protons/H\n \n +\n \n ✔\n \n
\n \n \n NOTE: Ignore any reference to the relative strengths of sulfuric acid and nitric acid.\n
\n Accept “sulfuric acid has two hydrogens «whereas nitric has one»”.\n
\n Accept \"dibasic\" for \"diprotic\".\n \n \n
\n Which instrument would best monitor the rate of this reaction?\n
\n\n 2KI (aq) + Cl\n \n 2\n \n (aq) → 2KCl (aq) + I\n \n 2\n \n (aq)\n
\n\n
\n A. Balance\n
\n B. Colorimeter\n
\n\n C. Volumetric flask\n
\n\n D. Gas syringe\n
\n\n [1]\n
\n\n B\n
\n\n Outline why metals, like iron, can conduct electricity.\n
\n\n [1]\n
\n\n mobile/delocalized «sea of» electrons\n
\n\n Which compound has the largest value of lattice enthalpy?\n
\n\n A. Na\n \n 2\n \n O\n
\n\n B. K\n \n 2\n \n O\n
\n\n C. Na\n \n 2\n \n S\n
\n\n D. K\n \n 2\n \n S\n
\n\n [1]\n
\n\n A\n
\n\n A majority of students recognized the ionic compound with the highest lattice enthalpy.\n
\n\n Which properties depend on the movement of the delocalized electrons in a metal?\n
\n\n
\n I. Electrical conductivity\n
\n II. Thermal conductivity\n
\n III. Density\n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\n
\n\n Which statements are correct about the action of a catalyst in a chemical reaction?\n
\n\n I. It increases the energy of each collision.\n
\n II. It alters the mechanism of the reaction.\n
\n III. It remains unchanged at the end of the reaction.\n
\n A. I and II only\n
\n\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n C\n
\n\n \n 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.\n \n
\n\n [2]\n
\n\n \n E\n \n a\n \n marked on graph\n \n [✔]\n \n \n
\n\n \n explanation in terms of more “particles” with E ≥ E\n \n a\n \n \n
\n\n \n \n \n OR\n \n \n \n
\n\n \n greater area under curve to the right of E\n \n a\n \n in T\n \n 2\n \n \n [✔]\n \n \n
\n\n The majority of candidates earned at least one mark, many both marks. Errors included not annotating the graph with\n \n E\n \n \n a\n \n and referring to increase of kinetic energy as reason for higher rate at T2.\n
\n\n \n MnO\n \n 2\n \n is another possible catalyst for the reaction. State the IUPAC name for MnO\n \n 2\n \n .\n \n
\n\n [1]\n
\n\n \n manganese(IV) oxide\n
\n \n \n OR\n \n \n
\n manganese dioxide\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “manganese(IV) dioxide”.\n \n \n
\n\n Many could correctly name manganese(IV)oxide, but there were answers of magnesium(IV) oxide or manganese(II) oxide.\n
\n\n Identify the type of reaction.\n
\n\n [1]\n
\n\n «nucleophilic» substitution\n
\n \n \n OR\n \n \n
\n SN2 ✔\n
\n \n
\n Accept “hydrolysis”.\n \n
\n \n Accept SN1\n \n
\n\n About 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\n
\n\n \n Predict, giving a reason, how the final enthalpy of reaction calculated from this experiment would compare with the theoretical value.\n \n
\n\n [1]\n
\n\n \n lower/less exothermic/less negative\n \n \n AND\n \n \n heat loss/some heat not accounted for\n
\n \n \n OR\n \n \n
\n lower/less exothermic/less negative\n \n \n AND\n \n \n mass of reaction mixture greater than 25.00 g\n
\n \n \n OR\n \n \n
\n greater/more exothermic/more negative\n \n \n AND\n \n \n specific heat of solution less than water\n \n [✔]\n \n \n
\n Most scored a mark for predicting how the calculated enthalpy value would compare with the theoretical value.\n
\n\n What is the relative atomic mass of an element with the following mass spectrum?\n
\n\n \n
\n A. 23\n
\n\n B. 24\n
\n\n C. 25\n
\n\n D. 28\n
\n\n [1]\n
\n\n B\n
\n\n 69% of the candidates determined the relative atomic mass of the element from its mass spectrum.\n
\n\n Which technique is used to purify a solid obtained from a chemical reaction?\n
\n\n
\n A. distillation\n
\n B. evaporation\n
\n\n C. recrystallization\n
\n\n D. filtration\n
\n\n [1]\n
\n\n C\n
\n\n State the formula of its conjugate base.\n
\n\n [1]\n
\n\n HS\n \n −\n \n ✔\n
\n\n Which E\n \n ⦵\n \n value, in V, for the reaction Mn (s) + Zn\n \n 2+\n \n (aq) → Mn\n \n 2+\n \n (aq) + Zn (s) can be deduced from the following equations?\n
\n\n Mn (s) + 2Ag\n \n +\n \n (aq) → Mn\n \n 2+\n \n (aq) + 2Ag (s) E\n \n ⦵\n \n = 1.98 V\n
\n\n Zn (s) + Cu\n \n 2+\n \n (aq) → Zn\n \n 2+\n \n (aq) + Cu (s) E\n \n ⦵\n \n = 1.10 V\n
\n\n Cu (s) + 2Ag\n \n +\n \n (aq) → Cu\n \n 2+\n \n (aq) + 2Ag (s) E\n \n ⦵\n \n = 0.46 V\n
\n\n A. 0.42\n
\n\n B. 1.34\n
\n\n C. 2.62\n
\n\n D. 3.54\n
\n\n [1]\n
\n\n A\n
\n\n 56% of candidates were able to deduce the E\n \n Ɵ\n \n of the reaction.\n
\n\n Which molecule has a tetrahedral molecular geometry?\n
\n\n
\n A. HNO\n \n 3\n \n
\n B. SF\n \n 4\n \n
\n\n C. XeF\n \n 4\n \n
\n\n D. XeO\n \n 4\n \n
\n\n [1]\n
\n\n D\n
\n\n Less than half of the candidates were able to identify XeO\n \n 4\n \n as having a tetrahedral shape and there was a strong link with overall performance. Nitric acid was the most commonly chosen distractor and the fact that XeF\n \n 4\n \n was the least favoured seems to indicate that students are well aware of its unusual geometry.\n
\n\n \n Predict, giving a reason, how the final enthalpy of reaction calculated from this experiment would compare with the theoretical value.\n \n
\n\n [1]\n
\n\n \n lower/less exothermic/less negative\n \n \n AND\n \n \n heat loss/some heat not accounted for\n
\n \n \n OR\n \n \n
\n lower/less exothermic/less negative\n \n \n AND\n \n \n mass of reaction mixture greater than 25.00 g\n
\n \n \n OR\n \n \n
\n greater/more exothermic/more negative\n \n \n AND\n \n \n specific heat of solution less than water\n \n [✔]\n \n \n
\n Most scored a mark for predicting how the calculated enthalpy value would compare with the theoretical value.\n
\n\n Determine the molar enthalpy of combustion of an alkane if 8.75 × 10\n \n −4\n \n moles are burned, raising the temperature of 20.0 g of water by 57.3 °C.\n
\n\n [2]\n
\n\n «\n \n q\n \n =\n \n mc\n \n Δ\n \n T\n \n = 20.0 g × 4.18 J g\n \n −1\n \n °C\n \n −1\n \n × 57.3 °C =» 4790 «J» ✔\n
\n\n «\n \n » –5470 «kJ mol\n \n –1\n \n » ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Accept answers in the range –5470 to –5480 «kJ mol\n \n −1\n \n ».\n \n
\n\n \n Accept correct answer in any units, e.g. –5.47 «MJ mol\n \n −1\n \n » or 5.47 x 10\n \n 6\n \n «J mol\n \n −1\n \n ».\n \n
\n\n \n Which is correct for\n \n \n \n ?\n \n
\n\n \n \n \n
\n [1]\n
\n\n D\n
\n\n A well answered question. 88 % of the candidates deduced the correct numbers of protons, neutrons and electrons in the sulfide ion.\n
\n\n Explain\n \n two\n \n different ways antiviral medications work.\n
\n\n [2]\n
\n\n Any two of:\n
\n\n alters «viral» enzyme\n \n \n AND\n \n \n prevent virus from entering the cell ✓.\n
\n\n alter the cell DNA\n \n \n AND\n \n \n virus cannot multiply ✓\n
\n\n block «cell» enzyme activity\n \n \n AND\n \n \n prevent virus multiplication ✓\n
\n\n alters «viral» enzyme\n \n \n AND\n \n \n prevents release of «new» viral particles «from the cell» ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “just interferes with viral reproductive cycle”.\n \n
\n\n \n Award\n \n [1 max]\n \n for two partial answers.\n \n
\n\n Deduce the ratio of Fe\n \n 2+\n \n :Fe\n \n 3+\n \n in Fe\n \n 3\n \n O\n \n 4\n \n .\n
\n\n [1]\n
\n\n 1:2 ✔\n
\n\n \n Accept 2 Fe\n \n 3+\n \n : 1 Fe\n \n 2+\n \n \n
\n \n Do\n \n not\n \n accept 2:1 only\n \n
\n \n Describe\n \n one\n \n systematic error associated with the use of the gas syringe, and how the error affects the calculated rate.\n \n
\n\n [2]\n
\n\n \n \n \n ALTERNATIVE 1\n \n \n
\n gas may leak/be lost/escape\n
\n \n \n OR\n \n \n
\n plunger may stick/friction «so pressure is greater than atmospheric pressure»\n
\n \n \n OR\n \n \n
\n syringe may be tilted «up» so plunger moves less «with gravity acting on plunger»\n
\n \n \n OR\n \n \n
\n CO\n \n 2\n \n dissolved in water\n \n [✔]\n \n
\n calculated rate lower\n \n [✔]\n \n
\n \n
\n \n \n \n ALTERNATIVE 2\n \n \n
\n syringe may be tilted «down» so plunger moves more «with gravity acting on plunger»\n
\n \n \n OR\n \n \n
\n syringe is held in hand so gets warmer and gas expands\n \n [✔]\n \n
\n calculated rate higher\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Calculated rate is lower or higher must be stated for M2.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “scale on syringe is inaccurate”, “errors in reading syringe”, or “bubbles in syringe”.\n \n \n
\n\n This part was about the systematic error and was answered correctly, but many candidates failed to state how the error affected the calculated rate. Many candidates confused this with the concept of a random error and identified the uncertainty of reading the syringe, which is incorrect.\n
\n\n How should the difference between 27.0 ± 0.3 and 9.0 ± 0.2 be shown?\n
\n\n A. 18.0 ± 0.1\n
\n\n B. 18.0 ± 0.3\n
\n\n C. 18.0 ± 0.5\n
\n\n D. 18.0 ± 0.6\n
\n\n [1]\n
\n\n C\n
\n\n \n What is the sum of the coefficients when the equation is balanced with the smallest whole numbers?\n \n
\n\n \n __BaCl\n \n 2\n \n (aq) + __Fe\n \n 2\n \n (SO\n \n 4\n \n )\n \n 3\n \n (aq) → __FeCl\n \n 3\n \n (aq) + __BaSO\n \n 4\n \n (s)\n \n
\n\n \n A. 4\n \n
\n\n \n B. 6\n \n
\n\n \n C. 8\n \n
\n\n \n D. 9\n \n
\n\n [1]\n
\n\n D\n
\n\n This question was well answered while one teacher commented that using 1 as a coefficient in a “sum the coefficients” question seemed tricky.\n
\n\n \n What is the enthalpy change of reaction for the following equation?\n \n
\n\n \n C\n \n 2\n \n H\n \n 4\n \n (g) + H\n \n 2\n \n (g) → C\n \n 2\n \n H\n \n 6\n \n (g)\n \n
\n\n \n \n C\n \n 2\n \n H\n \n 4\n \n (g) + 3O\n \n 2\n \n (g) → 2CO\n \n 2\n \n (g) + 2H\n \n 2\n \n O (l)\n \n ΔH = x\n \n \n \n
\n\n \n \n \n C\n \n 2\n \n H\n \n 6\n \n (g) +\n \n \n \n O\n \n 2\n \n (g) → 2CO\n \n 2\n \n (g) + 3H\n \n 2\n \n O (l)\n \n \n \n ΔH = y\n \n \n \n \n
\n\n \n \n \n \n H\n \n 2\n \n (g) +\n \n \n \n \n \n O\n \n 2\n \n (g) → H\n \n 2\n \n O (l)\n \n \n \n \n \n ΔH = z\n \n \n \n \n \n
\n\n \n \n \n \n \n \n \n \n
\n\n \n \n \n \n \n A.\n \n x + y + z\n \n \n \n \n \n \n
\n\n \n \n \n \n \n B.\n \n −x − y + z\n \n \n \n \n \n \n
\n\n \n \n \n \n \n C.\n \n x − y − z\n \n \n \n \n \n \n
\n\n \n \n \n \n \n D.\n \n x − y + z\n \n \n \n \n \n \n
\n\n [1]\n
\n\n D\n
\n\n Nearly 89 % of candidates answered this Hess Law question correctly. This topic was obviously well covered.\n
\n\n State the full electron configuration of the sulfide ion.\n
\n\n [1]\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 ✔\n
\n\n \n
\n Do\n \n not\n \n accept “[Ne] 3s\n \n 2\n \n 3p\n \n 6\n \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 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 What is the pH of 0.01 mol dm\n \n −3\n \n KOH (aq)?\n
\n\n
\n A. 1.0\n
\n B. 2.0\n
\n\n C. 12.0\n
\n\n D. 13.0\n
\n\n [1]\n
\n\n C\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 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 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
\n\n What is the enthalpy of combustion of propan-1-ol, in\n \n kJ mol\n \n −1\n \n \n , according to the following calorimetry data?\n
\n| \n \n Mass of water in calorimeter\n \n | \n\n 75 g\n | \n
| \n \n Amount of propan-1-ol burned\n \n | \n\n 0.015 mol\n | \n
| \n \n Temperature rise\n \n | \n\n 24 °C\n | \n
| \n \n Specific heat capacity of water\n \n | \n\n 4.2 J g\n \n −1\n \n K\n \n −1\n \n | \n
\n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n D\n
\n\n \n What is the ground state electron configuration of an atom of chromium, Cr (Z = 24)?\n \n
\n\n \n A. [Ar]3d\n \n 6\n \n \n
\n\n \n B. [Ar]4s\n \n 2\n \n 3d\n \n 4\n \n \n
\n\n \n C. [Ar]4s\n \n 1\n \n 3d\n \n 5\n \n \n
\n\n \n D. [Ar]4s\n \n 2\n \n 4p\n \n 4\n \n \n
\n\n [1]\n
\n\n C\n
\n\n A large number of the candidates had ground state configuration of Cr as 4s\n \n 2\n \n 3d\n \n 4\n \n rather than 4s\n \n 1\n \n 3d\n \n 5\n \n
\n\n \n Which formula is correct?\n \n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n C\n
\n\n 64% of candidates picked the correct formula for ammonium phosphate with (NH\n \n 4\n \n )\n \n 2\n \n PO\n \n 4\n \n being the incorrect answer chosen most frequently. NH\n \n 4\n \n PO\n \n 4\n \n or (NH\n \n 4\n \n )\n \n 3\n \n (PO\n \n 4\n \n )\n \n 2\n \n was rarely selected.\n
\n\n Deduce the half-equations for the reaction at each electrode.\n
\n\n \n
\n [2]\n
\n\n Cathode (negative electrode):\n
\n\n Zn\n \n 2+\n \n + 2e\n \n −\n \n → Zn (l) ✔\n
\n\n
\n\n Anode (positive electrode):\n
\n\n 2Cl\n \n −\n \n → Cl\n \n 2\n \n (g) + 2e\n \n −\n \n
\n\n \n \n OR\n \n \n
\n\n Cl\n \n −\n \n → ½ Cl\n \n 2\n \n (g) + e\n \n −\n \n ✔\n
\n\n The 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\n \n +\n \n instead of Zn\n \n 2+\n \n . Some candidates reversed the anode and cathode equations earning only 1 of the 2 marks.\n
\n\n What would be the electrode potential,\n \n E\n \n \n ⦵\n \n , of the Mn\n \n 2+\n \n (aq)|Mn (s) half-cell if Fe\n \n 3+\n \n (aq)|Fe\n \n 2+\n \n (aq) is used as the reference standard?\n
\n\n Mn\n \n 2+\n \n (aq) + 2e\n \n −\n \n \n Mn (s)\n \n E\n \n \n ⦵\n \n = −1.18 V\n
\n
\n Fe\n \n 3+\n \n (aq) + e\n \n −\n \n \n Fe\n \n 2+\n \n (aq)\n \n E\n \n \n ⦵\n \n = +0.77 V\n
\n A. −1.95 V\n
\n\n B. −0.41 V\n
\n\n C. +0.41 V\n
\n\n D. +1.95 V\n
\n\n [1]\n
\n\n A\n
\n\n \n Phenylethene is manufactured from benzene and ethene in a two-stage process. The overall reaction can be represented as follows with ΔG\n \n θ\n \n = +10.0 kJ mol\n \n −1\n \n at 298 K.\n \n
\n\n \n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5991/6b.PNG\")
\n \n
\n \n \n Calculate the equilibrium constant for the overall conversion at 298 K, using section 1 of the data booklet.\n \n \n
\n\n [2]\n
\n\n \n ln\n \n k\n \n «=\n \n \n \n » = –4.04\n \n [✔]\n \n \n
\n\n \n \n k\n \n = 0.0176\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 Another calculation which most candidates were able to work out, though some failed to convert Δ\n \n G\n \n given value in kJ mol\n \n -1\n \n to J mol\n \n -1\n \n or forgot the negative sign. Some used an inappropriate expression of\n \n R\n \n .\n
\n\n Deduce, giving reasons, whether the reaction of magnesium nitride with water is an acid–base reaction, a redox reaction, neither or both.\n
\n\n \n
\n [2]\n
\n\n \n Acid–base:\n \n
\n yes\n \n AND\n \n N\n \n 3-\n \n accepts H\n \n +\n \n /donates electron pair«s»\n
\n \n \n OR\n \n \n
\n yes\n \n AND\n \n H\n \n 2\n \n O loses H\n \n +\n \n «to form OH\n \n -\n \n »/accepts electron pair«s» ✔\n
\n \n Redox:\n \n
\n no\n \n AND\n \n no oxidation states change ✔\n
\n
\n\n \n Accept “yes\n \n AND\n \n proton transfer takes place”\n \n
\n\n \n Accept reference to the oxidation state of specific elements not changing.\n \n
\n\n \n Accept “not redox as no electrons gained/lost”.\n \n
\n\n \n Award\n \n [1 max]\n \n for Acid–base: yes\n \n AND\n \n Redox: no without correct reasons, if no other mark has been awarded\n \n
\n\n Probably 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.\n
\n\n Deduce the hybridization of the central carbon atom in Compound A.\n
\n\n [1]\n
\n\n sp\n \n 2\n \n ✔\n
\n\n sp\n \n 2\n \n hybridization of the central carbon atom in the ketone was very done well.\n
\n\n \n Suggest why the existence of salts containing an ion with this formula could be predicted. Refer to section 7 of the data booklet.\n \n
\n\n [1]\n
\n\n \n same group as Mn «which forms M\n \n n\n \n O\n \n 4\n \n \n -\n \n »\n
\n \n \n OR\n \n \n
\n in group 7/has 7 valence electrons, so its «highest» oxidation state is +7\n \n [✔]\n \n \n
\n This 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.\n
\n\n \n Which combination correctly describes the geometry of the carbonate ion,\n \n \n \n ?\n \n
\n\n \n
\n [1]\n
\n\n C\n
\n\n Identifying correct electron domain and molecular geometries around a central atom was somewhat challenging and answered slighter better by higher scoring candidates.\n
\n\n \n State what the presence of alternative Lewis structures shows about the nature of the bonding in the molecule.\n \n
\n\n [1]\n
\n\n \n delocalization\n \n
\n\n \n OR\n \n
\n\n \n delocalized\n \n π\n \n -electrons\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “resonance”.\n \n \n
\n\n Most candidates identified resonance for this given Lewis representation.\n
\n\n Which change causes the greatest increase in the initial rate of reaction between nitric acid and magnesium?\n
\n\n 2HNO\n \n 3\n \n (aq) + Mg (s) → Mg(NO\n \n 3\n \n )\n \n 2\n \n (aq) + H\n \n 2\n \n (g)\n
\n\n \n
\n [1]\n
\n\n A\n
\n\n Which species contains a coordination bond?\n
\n\n
\n A. CO\n \n 2\n \n
\n B. HCN\n
\n\n C. NO\n \n 2\n \n \n +\n \n
\n\n D. NO\n \n 3\n \n \n −\n \n
\n\n [1]\n
\n\n D\n
\n\n Which molecule has the same empirical formula as molecular formula?\n
\n\n A. CH\n \n 3\n \n COOH\n
\n\n B. C\n \n 2\n \n H\n \n 5\n \n OH\n
\n\n C. C\n \n 2\n \n H\n \n 4\n \n
\n\n D. C\n \n 4\n \n H\n \n 10\n \n
\n\n [1]\n
\n\n B\n
\n\n Predict the shape of the hydrogen sulfide molecule.\n
\n\n [1]\n
\n\n bent/non-linear/angular/v-shaped✔\n
\n\n \n Suggest the basis of these predictions.\n \n
\n\n [2]\n
\n\n \n gap in the periodic table\n
\n \n \n OR\n \n \n
\n element with atomic number «75» unknown\n
\n \n \n OR\n \n \n
\n break/irregularity in periodic trends\n \n [✔]\n \n \n
\n \n «periodic table shows» regular/periodic trends «in properties»\n \n [✔]\n \n \n
\n\n This 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.\n
\n\n The rate equation for a reaction is:\n
\n\n rate = k[A][B]\n
\n\n Which mechanism is consistent with this rate equation?\n
\n\n
\n A. 2A\n \n I Fast\n
\n I + B → P Slow\n
\n B. A + B\n \n I Fast\n
\n I + A → P Slow\n
\n C. A → I Slow\n
\n I + B → P Fast\n
\n D. B\n \n I Fast\n
\n I + A → P Slow\n
\n [1]\n
\n\n D\n
\n\n All models have limitations. Suggest\n \n two\n \n limitations to this model of the electron energy levels.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n\n does not represent sub-levels/orbitals ✔\n
\n\n only applies to atoms with one electron/hydrogen ✔\n
\n\n does not explain why only certain energy levels are allowed ✔\n
\n\n the atom is considered to be isolated ✔\n
\n\n does not take into account the interactions between atoms/molecules/external fields ✔\n
\n\n does not consider the number of electrons the energy level can fit ✔\n
\n\n does not consider probability of finding electron at different positions/\n \n OWTTE\n \n ✔\n
\n\n
\n\n \n Do\n \n not\n \n accept “does not represent distance «from nucleus»”.\n \n
\n\n \n What can be deduced from the infrared (IR) spectrum of a compound?\n \n
\n\n \n A. Number of hydrogens\n \n
\n\n \n B. Number of hydrogen environments\n \n
\n\n \n C. Bonds present\n \n
\n\n \n D. Molar mass\n \n
\n\n [1]\n
\n\n C\n
\n\n 71 % of the candidates identified the bonds present as the information that can be deduced from an infrared spectrum. The most commonly chosen distractor was B (the number of hydrogen environments).\n
\n\n Justify whether a 1.0 dm\n \n 3\n \n solution made from 0.10 mol NH\n \n 3\n \n and 0.20 mol HCl will form a buffer solution.\n
\n\n [1]\n
\n\n no\n \n \n AND\n \n \n is not a weak acid conjugate base system\n
\n\n \n \n OR\n \n \n
\n\n no\n \n \n AND\n \n \n weak base «totally» neutralized/ weak base is not in excess\n
\n\n \n \n OR\n \n \n
\n\n no\n \n \n AND\n \n \n will not neutralize small amount of acid ✔\n
\n\n
\n\n \n Accept “no\n \n AND\n \n contains 0.10 mol NH\n \n 4\n \n Cl + 0.10 mol HCl”.\n \n
\n\n Poorly answered revealing little understanding of buffering mechanisms, which is admittedly a difficult topic.\n
\n\n What is the coefficient of\n \n (aq) when the equation is balanced using the smallest possible whole numbers?\n
\n\n \n
\n\n A. 1\n
\n\n B. 2\n
\n\n C. 3\n
\n\n D. 4\n
\n\n
\n\n [1]\n
\n\n B\n
\n\n The potential energy profile for a “coffee cup” calorimetry experiment is shown.\n
\n\n \n
\n
\n What is the correct interpretation of this reaction?\n
\n
\n \n
\n [1]\n
\n\n A\n
\n\n \n Calculate the percentage, by mass, of rhenium in ReCl\n \n 3\n \n .\n \n
\n\n [2]\n
\n\n \n «M\n \n r\n \n ReCl\n \n 3\n \n = 186.21 + (3 × 35.45) =» 292.56\n \n [✔]\n \n
\n «100 ×\n \n \n \n =» 63.648 «%»\n \n \n \n [✔]\n \n \n
\n Most candidates were able to answer this stoichiometric question correctly.\n
\n\n \n Another airbag reactant produces nitrogen gas and sodium.\n \n
\n\n \n Suggest, including an equation, why the products of this reactant present a safety hazard.\n \n
\n\n [2]\n
\n\n \n 2Na (s) + 2H\n \n 2\n \n O (l) → 2NaOH (aq) + H\n \n 2\n \n (g) ✔\n \n
\n\n \n hydrogen explosive\n
\n \n \n OR\n \n \n
\n highly exothermic reaction\n
\n \n \n OR\n \n \n
\n sodium reacts violently with water\n
\n \n \n OR\n \n \n
\n forms strong alkali ✔\n \n
\n \n \n NOTE: Accept the equation of combustion of hydrogen.\n
\n Do\n \n not\n \n accept just “sodium is reactive/dangerous”.\n \n \n
\n Identify\n \n three\n \n structural differences between DNA and RNA. Use section 34 of the data booklet.\n
\n| \n \n DNA\n \n | \n\n \n RNA\n \n | \n
| \n ................................................\n | \n\n ................................................\n | \n
| \n ................................................\n | \n\n ................................................\n | \n
| \n ................................................\n | \n\n ................................................\n | \n
\n
\n\n
\n\n [3]\n
\n| \n \n DNA\n \n | \n\n | \n\n \n RNA\n \n | \n
| \n thymine\n | \n\n \n \n AND\n \n \n | \n\n uracil ✓\n | \n
| \n deoxyribose\n | \n\n \n \n AND\n \n \n | \n\n ribose ✓\n | \n
| \n double stranded\n | \n\n \n \n AND\n \n \n | \n\n single stranded ✓\n | \n
\n
\n\n \n Accept “additional methyl/CH3 group” for DNA\n \n OR\n \n “one less methyl group” for RNA for M1.\n
\n Accept “one less hydroxyl/OH group” for DNA\n \n OR\n \n “additional hydroxyl/OH group” for RNA for M2.\n \n
\n Some nitride ions are\n \n 15\n \n N\n \n 3–\n \n . State the term that describes the relationship between\n \n 14\n \n N\n \n 3–\n \n and\n \n 15\n \n N\n \n 3–\n \n .\n
\n\n [1]\n
\n\n \n isotope\n \n «s» ✔\n
\n\n Identification of isotopes was answered correctly by most students.\n
\n\n Which monomer would produce the polymer shown?\n
\n\n \n
\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n C\n
\n\n Outline the meaning of homologous series.\n
\n\n [1]\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 \n State the name of this compound, applying IUPAC rules.\n \n
\n\n [1]\n
\n\n \n rhenium(III) chloride\n
\n \n \n OR\n \n \n
\n rhenium trichloride\n \n [✔]\n \n \n
\n More than half of the candidates named ReCl\n \n 3\n \n correctly. Common mistakes included “rhenium chloride” and “trichlororhenium”.\n
\n\n Which is most likely to hydrolyse via a S\n \n N\n \n 1 mechanism?\n
\n\n A. CH\n \n 3\n \n CHBrCH\n \n 2\n \n CH\n \n 3\n \n
\n\n B. (CH\n \n 3\n \n )\n \n 2\n \n CHBr\n
\n\n C. (CH\n \n 3\n \n )\n \n 3\n \n CBr\n
\n\n D. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n Br\n
\n\n [1]\n
\n\n C\n
\n\n Identify the number of sigma and pi bonds in HCN.\n
\n\n \n
\n [1]\n
\n\n Sigma (\n \n ): 2\n \n \n AND\n \n \n Pi (\n \n ): 2 ✔\n
\n\n \n Outline why ethanoic acid is classified as a weak acid.\n \n
\n\n [1]\n
\n\n \n partial dissociation «in aqueous solution»\n \n [✔]\n \n \n
\n\n The definition of a weak acid was generally correct.\n
\n\n \n The graph below shows the Maxwell–Boltzmann distribution of molecular energies at a particular temperature.\n \n
\n\n \n \n \n
\n \n \n The rate at which dinitrogen monoxide decomposes is significantly increased by a metal oxide catalyst.\n \n \n
\n\n \n \n Annotate and use the graph to outline why a catalyst has this effect.\n \n \n
\n\n [2]\n
\n\n \n
\n \n catalysed and uncatalysed Ea marked on graph AND with the catalysed being at lower energy\n \n [✔]\n \n \n
\n\n \n «for catalysed reaction» greater proportion of/more molecules have E ≥ E\n \n a\n \n / E > E\n \n a\n \n
\n \n \n OR\n \n \n
\n «for catalysed reaction» greater area under curve to the right of the E\n \n a\n \n \n [✔]\n \n \n
\n \n \n \n Note:\n \n Accept “more molecules have the activation energy”.\n \n \n
\n\n This 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\n \n E\n \n \n a\n \n 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.\n
\n\n Suggest\n \n two\n \n possible products of the incomplete combustion of ethene that would not be formed by complete combustion.\n
\n\n [1]\n
\n\n carbon monoxide/CO\n \n \n AND\n \n \n carbon/C/soot ✔\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 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 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 20 cm\n \n 3\n \n of 1 mol dm\n \n −3\n \n sulfuric acid was added dropwise to 20 cm\n \n 3\n \n of 1 mol dm\n \n −3\n \n barium hydroxide producing a precipitate of barium sulfate.\n
\n\n H\n \n 2\n \n SO\n \n 4\n \n (aq) + Ba(OH)\n \n 2\n \n (aq) → 2H\n \n 2\n \n O (l) + BaSO\n \n 4\n \n (s)\n
\n\n Which graph represents a plot of conductivity against volume of acid added?\n
\n\n \n
\n [1]\n
\n\n B\n
\n\n The question involved the formation of a precipitate – not a familiar concept to the candidates. It required candidates (to work out how the number of ions and hence conductivity were changing. 37% of the candidates got the correct answer (conductivity decreased as volume of H\n \n 2\n \n SO\n \n 4\n \n increased). The majority increased conductivity as H\n \n 2\n \n SO\n \n 4\n \n was added.\n
\n\n Calculate the Gibbs free energy change (Δ\n \n G\n \n ), in kJ mol\n \n −1\n \n , for this reaction at 25 °C. Use section 1 of the data booklet.\n
\n\n If you did not obtain an answer in c(i) or c(ii) use −87.6 kJ mol\n \n −1\n \n and −150.5 J mol\n \n −1\n \n K\n \n −1\n \n respectively, but these are not the correct answers.\n
\n\n [2]\n
\n\n «ΔS =» –0.1702 «kJ mol\n \n –1\n \n K\n \n –1\n \n »\n
\n \n \n OR\n \n \n
\n 298 «K» ✔\n
\n «ΔG = –92.5 kJ mol\n \n –1\n \n – (298 K × –0.1702 kJ mol\n \n –1\n \n K\n \n –1\n \n ) =» –41.8 «kJ mol\n \n –1\n \n » ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n If –87.6 and -150.5 are used then –42.8.\n \n
\n\n Which products are formed from the neutralization of nitric acid by calcium hydroxide?\n
\n
\n
\n A. Calcium oxide and ammonia\n
\n\n B. Calcium nitrate and water\n
\n\n C. Calcium nitrate and ammonia\n
\n\n D. Calcium nitrate and hydrogen\n
\n\n [1]\n
\n\n B\n
\n\n Which diagram shows the enthalpy changes for dissolving a solid,\n \n MX\n \n , in water, if the process increases the temperature of the solution?\n
\n\n
\n A.\n \n
\n B.\n \n
\n C.\n \n
\n D.\n \n
\n [1]\n
\n\n D\n
\n\n What are the products when concentrated aqueous copper (II) chloride is electrolysed using platinum electrodes?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n \n Comment on the magnitudes of random and systematic errors in this experiment using the answers in (e).\n \n
\n\n [2]\n
\n\n \n Any two:\n \n
\n\n large percentage error means large systematic error «in procedure» ✔\n
\n\n small percentage uncertainty means small random errors ✔\n
\n\n random errors smaller than systematic error ✔\n
\n\n \n
\n \n \n Award\n \n [2]\n \n for “both random and systematic errors are significant.”\n \n
\n Of the many students that obtained the mark most did through the first alternative and a lesser percentage through the third. Many students were unable to relate their calculations from 2e (percentage error and percentage uncertainty) to systematic error and random error. They either compared the calculations to incorrect errors or in some cases did not discuss the errors at all. Once again this points to a general lack on experimental understanding.\n
\n\n Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl\n \n 2\n \n .\n
\n\n \n
\n [2]\n
\n\n \n
\n Which is correct?\n
\n\n A. Electrophiles are Brønsted–Lowry acids.\n
\n\n B. Nucleophiles are Brønsted–Lowry acids.\n
\n\n C. Electrophiles are Lewis acids.\n
\n\n D. Nucleophiles are Lewis acids.\n
\n\n [1]\n
\n\n C\n
\n\n \n Which statements about bond strength and activation energy are correct for this reaction?\n \n
\n\n \n \n
\n\n \n
\n [1]\n
\n\n B\n
\n\n More than 60% could select relative bond strength of products and reactants in an exothermic reaction, and correctly apply the magnitude of activation energy between forward and reverse reactions for this.\n
\n\n \n Which molecule can be oxidized to a carboxylic acid by acidified potassium dichromate(VI)?\n \n
\n\n \n A. Propan-1-ol\n
\n \n
\n \n B. Propan-2-ol\n
\n \n
\n \n C. 2-methylpropan-2-ol\n
\n \n
\n \n D. Propanone\n \n
\n\n [1]\n
\n\n A\n
\n\n Another well answered question with the vast majority understanding that only propan-1-ol can be oxidized to a carboxylic acid.\n
\n\n Which of the 0.001 mol dm\n \n −3\n \n solutions is most likely to have a pH of 11.3?\n
\n\n A. Ca(OH)\n \n 2\n \n (aq)\n
\n\n B. H\n \n 3\n \n PO\n \n 4\n \n (aq)\n
\n\n C. NaOH (aq)\n
\n\n D. NH\n \n 4\n \n OH (aq)\n
\n\n [1]\n
\n\n A\n
\n\n This required calculation of pH of a base – not an easy task for SL candidates without a calculator. 28% of the candidates answered the question correctly.\n
\n\n Which of the following is the correct skeletal formula of butanoic acid?\n
\n\n \n
\n [1]\n
\n\n A\n
\n\n In which of the following compounds does ionic bonding predominate?\n
\n\n A. HCl\n
\n\n B. NaF\n
\n\n C. NH\n \n 4\n \n Br\n
\n\n D. NaOH\n
\n\n [1]\n
\n\n B\n
\n\n A well answered question. 70% of the candidates selected NaF as the compound in which ionic bonding predominates. Several teachers expressed concern regarding the unusual wording of this question, but thankfully the candidates understood it in the right way.\n
\n\n Which is the preferred Lewis formula of nitrous oxide, N\n \n 2\n \n O, as deduced by formal charges?\n
\n\n
\n \n
\n [1]\n
\n\n B\n
\n\n \n Which is correct?\n \n
\n\n \n \n \n
\n [1]\n
\n\n B\n
\n\n \n Predict, giving\n \n two\n \n reasons, how the first ionization energy of\n \n 15\n \n N compares with that of\n \n 14\n \n N.\n \n
\n\n [2]\n
\n\n \n \n Any two\n \n :\n
\n same\n \n \n AND\n \n \n have same nuclear charge/number of protons/Z\n \n eff\n \n \n [✔]\n \n \n
\n \n same\n \n \n AND\n \n \n neutrons do not affect attraction/ionization energy/Z\n \n eff\n \n
\n \n \n OR\n \n \n
\n same\n \n \n AND\n \n \n neutrons have no charge\n \n [✔]\n \n \n
\n \n same\n \n \n AND\n \n \n same attraction for «outer» electrons\n \n [✔]\n \n \n
\n\n \n same\n \n \n AND\n \n \n have same electronic configuration/shielding\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n \n Note:\n \n Accept “almost the same”.\n
\n “same” only needs to be stated once.\n \n \n \n
\n This 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.\n
\n\n \n Nitrogen monoxide reacts with oxygen gas to form nitrogen dioxide.\n \n
\n\n \n The following experimental data was obtained.\n
\n \n
\n \n \n \n
\n \n Deduce the partial order of reaction with respect to nitrogen monoxide and oxygen.\n
\n \n
\n \n \n \n
\n
\n\n
\n\n [2]\n
\n\n \n : second ✔\n
\n \n : first ✔\n
\n Most candidates could correctly deduce the order of each reactant from rate experimental rate data.\n
\n\n \n Determine the specific energy, in kJ g\n \n −1\n \n , and energy density, in kJ cm\n \n −3\n \n , of hexane, C\n \n 6\n \n H\n \n 14\n \n . Give both answers to three significant figures.\n \n
\n\n \n Hexane:\n \n M\n \n r\n \n \n = 86.2; ΔH\n \n c\n \n = −4163 kJ mol\n \n −1\n \n ; density = 0.660 g cm\n \n −3\n \n \n
\n\n
\n\n Specific energy:\n
\n\n Energy density:\n
\n\n [2]\n
\n\n \n specific energy = «\n \n \n \n =» 48.3 «kJ g\n \n –1\n \n »\n \n [✔]\n \n \n
\n\n \n energy density =«48.3 kJ g\n \n –1\n \n × 0.660 g cm\n \n –3\n \n =» 31.9 «kJ cm\n \n –3\n \n »\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Award\n \n [1 max]\n \n if either or both answers not expressed to three significant figures.\n \n \n
\n\n This question was very well answered by most students and many answered with the correct number of significant figures as specified by the question.\n
\n\n \n Which statement is correct for a spontaneous reaction?\n \n
\n\n \n
\n [1]\n
\n\n A\n
\n\n Nearly all candidates knew that\n \n is negative for a spontaneous reaction, however some were confused about the effect on the equilibrium constant.\n
\n Saturated calcium hydroxide solution is used to test for carbon dioxide. Calculate the pH of a 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n solution of calcium hydroxide, a strong base.\n
\n\n [2]\n
\n\n \n \n Alternative 1:\n \n \n
\n\n [OH\n \n −\n \n ] = « 2 × 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n =» 0.0466 «mol dm\n \n −3\n \n » ✓\n
\n\n «[H\n \n +\n \n ] =\n \n = 2.15 × 10\n \n −13\n \n mol dm\n \n −3\n \n »\n
\n\n pH = « −log (2.15 × 10−13) =» 12.668 ✓\n
\n\n
\n\n \n \n Alternative 2:\n \n \n
\n\n [OH\n \n −\n \n ] =« 2 × 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n =» 0.0466 «mol dm\n \n −3\n \n » ✓\n
\n\n «pOH = −log (0.0466) = 1.332»\n
\n\n pH = «14.000 – pOH = 14.000 – 1.332 =» 12.668 ✓\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Award\n \n [1 max]\n \n for pH =12.367.\n \n
\n\n \n Discuss the data.\n \n
\n\n [3]\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 Non-polar solvents can be toxic. Suggest a modification to the experiment which allows the evaporated solvent to be collected.\n \n
\n\n [1]\n
\n\n distillation «instead of evaporation» ✔\n
\n\n \n Accept “pass vapour through a condenser and collect liquid”.\n \n
\n\n \n Do\n \n not\n \n accept “condensation” without experimental details.\n \n
\n\n A 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.\n
\n\n Outline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.\n
\n\n [1]\n
\n\n «valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔\n
\n\n \n
\n Accept 2,8 (for O\n \n 2–\n \n ) and 2,8,8 (for S\n \n 2–\n \n )\n \n
\n Which of these acids has the weakest conjugate base?\n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n More than 59% of the candidates understood that strong acids have weak conjugate bases.\n
\n\n Sketch the Lewis (electron dot) structure of the P\n \n 4\n \n molecule, containing only single bonds.\n
\n\n
\n\n [1]\n
\n\n \n
\n \n Accept any diagram with each P joined to the other three.\n
\n Accept any combination of dots, crosses and lines.\n \n
\n Explain the decrease in radius from Na to Na\n \n +\n \n .\n
\n\n [2]\n
\n\n Na\n \n +\n \n has one less energy level/shell\n
\n \n \n OR\n \n \n
\n Na\n \n +\n \n has 2 energy levels/shells\n \n \n AND\n \n \n Na has 3 ✓\n
\n less shielding «in Na\n \n +\n \n so valence electrons attracted more strongly to nucleus»\n
\n \n \n OR\n \n \n
\n effective nuclear charge/Z\n \n eff\n \n greater «in Na\n \n +\n \n so valence electrons attracted more strongly to nucleus» ✓\n
\n \n
\n Accept “more protons than electrons «in Na\n \n +\n \n »”\n \n OR\n \n “less electron-electron repulsion «in Na\n \n +\n \n »” for M2.\n \n
\n Write an equation that shows how sulfur dioxide can produce acid rain.\n
\n\n [1]\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 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 Write an equation for the reaction of white phosphorus (P\n \n 4\n \n ) with chlorine gas to form phosphorus trichloride (PCl\n \n 3\n \n ).\n
\n\n [1]\n
\n\n P\n \n 4\n \n (s) + 6Cl\n \n 2\n \n (g) → 4PCl\n \n 3\n \n (l) ✔\n
\n\n Describe metallic bonding and how it contributes to electrical conductivity.\n
\n\n [3]\n
\n\n electrostatic attraction ✓\n
\n\n between «a lattice of» cations/positive «metal» ions\n \n \n AND\n \n \n «a sea of» delocalized electrons ✓\n
\n\n
\n mobile electrons responsible for conductivity\n
\n \n \n OR\n \n \n
\n electrons move when a voltage/potential difference/electric field is applied ✓\n
\n
\n\n \n Do\n \n not\n \n accept “nuclei” for “cations/positive ions” in M2.\n \n
\n\n \n Accept “mobile/free” for “delocalized” electrons in M2.\n \n
\n\n \n Accept “electrons move when connected to a cell/battery/power supply”\n \n OR\n \n “electrons move when connected in a circuit” for M3.\n \n
\n\n What is the hydroxide ion concentration in a solution of pH = 4 at 298 K?\n
\n\n
\n A. 4\n
\n B. 10\n
\n\n C. 10\n \n −4\n \n
\n\n D. 10\n \n −10\n \n
\n\n [1]\n
\n\n D\n
\n\n \n When equal masses of X and Y absorb the same amount of energy, their temperatures rise by 5 °C and 10 °C respectively. Which is correct?\n \n
\n\n \n A. The specific heat capacity of X is twice that of Y.\n \n
\n\n \n B. The specific heat capacity of X is half that of Y.\n \n
\n\n \n C. The specific heat capacity of X is one fifth that of Y.\n \n
\n\n \n D. The specific heat capacity of X is the same as Y.\n \n
\n\n [1]\n
\n\n A\n
\n\n Answered correctly by 53 %, this question relating specific heat capacity to temperature rise was handled better by higher scoring candidates.\n
\n\n \n Which is correct for\n \n \n \n ?\n \n
\n\n \n \n \n
\n [1]\n
\n\n D\n
\n\n 93 % of the candidates deduced the correct numbers of protons, neutrons and electrons in the sulfide ion.\n
\n\n \n Draw a Lewis (electron dot) structure of chloramine.\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 Accept any combination of dots/crosses or lines to represent electron pairs.\n \n \n
\n\n The Lewis structure of chloramine was correct for strong candidates, but many made the mistake of omitting electron pairs on N and Cl.\n
\n\n What is the formula of the compound formed from Ca\n \n 2+\n \n and PO\n \n 4\n \n \n 3−\n \n ?\n
\n\n A. CaPO\n \n 4\n \n
\n\n B. Ca\n \n 3\n \n (PO\n \n 4\n \n )\n \n 2\n \n
\n\n C. Ca\n \n 2\n \n (PO\n \n 4\n \n )\n \n 3\n \n
\n\n D. Ca(PO\n \n 4\n \n )\n \n 2\n \n
\n\n [1]\n
\n\n B\n
\n\n \n Which properties can be monitored to determine the rate of the reaction?\n \n
\n\n \n Fe (s) + CuSO\n \n 4\n \n (aq) → Cu (s) + FeSO\n \n 4\n \n (aq)\n \n
\n\n \n I. change in volume\n
\n II. change in temperature\n
\n III. change in colour\n
\n \n
\n \n A. I and II only\n \n
\n\n \n B. I and III only\n \n
\n\n \n C. II and III only\n \n
\n\n \n D. I, II and III\n \n
\n\n [1]\n
\n\n C\n
\n\n One G2 form queried if visible spectroscopy is on the core syllabus and therefore should candidates be aware of monitoring a reaction via colour change. However, 6.1 clearly states that following change in colour is one method of following reactions.\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 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 Which atom has an expanded octet?\n
\n\n A. C in CO\n \n 2\n \n
\n\n B. S in SCl\n \n 4\n \n
\n\n C. O in H\n \n 2\n \n O\n \n 2\n \n
\n\n D. P in PCl\n \n 3\n \n
\n\n [1]\n
\n\n B\n
\n\n Suggest why this process might raise environmental concerns.\n
\n\n [1]\n
\n\n sulfur dioxide/SO\n \n 2\n \n causes acid rain ✔\n
\n\n \n Accept sulfur dioxide/SO\n \n 2\n \n /dust causes respiratory problems\n \n
\n \n Do\n \n not\n \n accept just “causes respiratory problems” or “causes acid rain”.\n \n
\n Which contains the most atoms of oxygen?\n
\n\n A. 64 g of O\n \n 2\n \n
\n\n B. 1.2 × 10\n \n 24\n \n molecules of O\n \n 2\n \n
\n\n C. 64 g of C\n \n 3\n \n H\n \n 5\n \n O\n \n 3\n \n
\n\n D. 1.2 × 10\n \n 24\n \n molecules of C\n \n 3\n \n H\n \n 5\n \n O\n \n 3\n \n
\n\n [1]\n
\n\n D\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 \n State\n \n one\n \n analytical technique that could be used to determine the ratio of\n \n 14\n \n N :\n \n 15\n \n N.\n \n
\n\n [1]\n
\n\n \n mass spectrometry/MS\n \n \n [✔]\n \n
\n\n All candidates, with very few exceptions, answered this correctly.\n
\n\n What happens to the average kinetic energy, KE, of the particles in a gas when the absolute temperature is doubled?\n
\n\n \n
\n\n
\n A. Increases by a factor of 2\n
\n B. Decreases by a factor of 2\n
\n\n C. Increases by a factor of 4\n
\n\n D. Decreases by a factor of 4\n
\n\n [1]\n
\n\n A\n
\n\n What is the strongest acid in the equation below?\n
\n\n H\n \n 3\n \n AsO\n \n 4\n \n + H\n \n 2\n \n O\n \n H\n \n 2\n \n AsO\n \n 4\n \n \n −\n \n + H\n \n 3\n \n O\n \n +\n \n \n K\n \n \n c\n \n = 4.5 × 10\n \n −4\n \n
\n\n A. H\n \n 3\n \n AsO\n \n 4\n \n
\n\n B. H\n \n 2\n \n O\n
\n\n C. H\n \n 2\n \n AsO\n \n 4\n \n \n −\n \n
\n\n D. H\n \n 3\n \n O\n \n +\n \n
\n\n [1]\n
\n\n D\n
\n\n A challenging question that combined interpreting the implications of the value of the equilibrium constant with the Brønsted-Lowry concept. Most candidates focused on BL theory and the forward reaction only. 36% of the candidates answered the question correctly by selecting the acid involved in the backward reaction.\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 q\n \n = «\n \n mcΔT\n \n = 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 =» 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 Nearly half the students correctly calculated the enthalpy change and some of these went on to find the value in kJ g\n \n -1\n \n . The most common mistakes were to use the mass of starch rather than the mass of water and adding 273 to the temperature change.\n
\n\n \n Draw the Lewis structures of oxygen, O\n \n 2\n \n , and ozone, O\n \n 3\n \n .\n \n
\n\n [2]\n
\n\n \n
\n \n NOTES: Coordinate bond may be represented by an arrow.\n \n
\n\n \n Do\n \n not\n \n accept delocalized structure for ozone.\n \n
\n\n \n What is the order of increasing boiling point?\n \n
\n\n \n A.\n \n CH\n \n \n 3\n \n \n CH\n \n \n 2\n \n \n CH\n \n \n 2\n \n \n CH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n CH(OH)CH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n COCH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n CO\n \n \n 2\n \n \n H\n \n \n
\n\n \n B.\n \n CH\n \n \n 3\n \n \n CH\n \n \n 2\n \n \n CH\n \n \n 2\n \n \n CH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n COCH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n CH(OH)CH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n CO\n \n \n 2\n \n \n H\n \n \n
\n\n \n C.\n \n CH\n \n \n 3\n \n \n CO\n \n \n 2\n \n \n H\n \n <\n \n CH\n \n \n 3\n \n \n COCH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n CH(OH)CH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n CH\n \n \n 2\n \n \n CH\n \n \n 2\n \n \n CH\n \n \n 3\n \n \n
\n\n \n D.\n \n CH\n \n \n 3\n \n \n CH\n \n \n 2\n \n \n CH\n \n \n 2\n \n \n CH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n COCH\n \n \n 3\n \n <\n \n CH\n \n \n 3\n \n \n CO\n \n \n 2\n \n \n H\n \n <\n \n CH\n \n \n 3\n \n \n CH(OH)CH\n \n \n 3\n \n \n
\n\n [1]\n
\n\n B\n
\n\n Ranking compounds in order of increasing boiling point proved challenging. There existed a number of misconceptions based on the incorrect answers chosen being close to evenly distributed.\n
\n\n State the equilibrium constant expression,\n \n K\n \n \n c\n \n , for this reaction.\n
\n\n [1]\n
\n\n \n ✔\n
\n\n One of the best answered questions.\n
\n\n Outline how host–guest chemistry mimics enzymes in the removal of xenobiotics.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n\n host molecule/supramolecule forms complex/bond with guest/xenobiotic ✓\n
\n\n binding between host and guest specific ✓\n
\n\n bonding «usually» non-covalent «in both cases» ✓\n
\n\n
\n\n \n Accept “supermolecule” for “supramolecule”.\n \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 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 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 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 «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 Deduce what would be observed when Compound B is warmed with acidified aqueous potassium dichromate (VI).\n
\n\n [1]\n
\n\n no change «in colour/appearance/solution» ✔\n
\n\n Some 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\n \n observation\n \n - \"it would remain orange\".\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 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 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
\n\n Formulate equations for the two propagation steps and one termination step in the formation of chloroethane from ethane.\n
\n\n [3]\n
\n\n Cl\n \n ·\n \n + C\n \n 2\n \n H\n \n 6\n \n →\n \n ·\n \n C\n \n 2\n \n H\n \n 5\n \n + HCl ✔\n
\n\n \n ·\n \n C\n \n 2\n \n H\n \n 5\n \n + Cl\n \n 2\n \n → Cl\n \n ·\n \n + C\n \n 2\n \n H\n \n 5\n \n Cl ✔\n
\n\n
\n \n ·\n \n C\n \n 2\n \n H\n \n 5\n \n + Cl\n \n ·\n \n → C\n \n 2\n \n H\n \n 5\n \n Cl\n
\n \n \n OR\n \n \n
\n Cl\n \n ·\n \n + Cl\n \n ·\n \n → Cl\n \n 2\n \n
\n \n \n OR\n \n \n
\n \n ·\n \n C\n \n 2\n \n H\n \n 5\n \n +\n \n ·\n \n C\n \n 2\n \n H\n \n 5\n \n → C\n \n 4\n \n H\n \n 10\n \n ✔\n
\n \n
\n Do not penalize incorrectly placed radical sign, eg\n \n C\n \n 2\n \n H\n \n 5\n \n \n \n ·\n \n .\n \n
\n Calculate the change in entropy, Δ\n \n S\n \n , in J K\n \n −1\n \n , for the decomposition of calcium carbonate.\n
\n\n [1]\n
\n\n «Δ\n \n S\n \n = (40 J K\n \n −1\n \n + 214 J K\n \n −1\n \n ) − (93 J K\n \n −1\n \n ) =» 161 «J K\n \n −1\n \n » ✓\n
\n\n \n
\n Ignore an extra step to determine total entropy change in JK\n \n –1\n \n : 161 J mol\n \n –1\n \n K\n \n –1\n \n x 5.55 mol = 894 «J mol\n \n –1\n \n K\n \n –1\n \n »\n \n
\n \n Award\n \n [1]\n \n for 894 «J mol\n \n –1\n \n K\n \n –1\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 \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 \n The following data were recorded for determining the density of three samples of silicon, Si.\n \n
\n\n \n \n \n
\n \n \n Which average density value, in g cm\n \n −3\n \n , has been calculated to the correct number of significant figures?\n \n \n
\n\n \n \n A. 2\n \n \n
\n\n \n \n B. 2.3\n \n \n
\n\n \n \n C. 2.27\n \n \n
\n\n \n \n D. 2.273\n \n \n
\n\n [1]\n
\n\n B\n
\n\n This question about choosing the answer to a calculation with the appropriate number of significant figures discriminated well between high-scoring and low-scoring candidates. Candidates found this question relatively challenging and only 62 % chose the answer with two significant figures. The most commonly chosen distractor was C which expressed the answer to three significant figures.\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 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 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
\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 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 The Lewis structure of methylamine is shown.\n
\n\n \n
\n What is the molecular geometry around N?\n
\n\n A. Square planar\n
\n\n B. Tetrahedral\n
\n\n C. Trigonal planar\n
\n\n D. Trigonal pyramidal\n
\n\n [1]\n
\n\n D\n
\n\n Explain the electron domain geometry of NO\n \n 3\n \n \n −\n \n .\n
\n\n [2]\n
\n\n three electron domains repel\n
\n\n \n \n OR\n \n \n
\n\n three electron domains as far away as possible ✔\n
\n\n
\n\n trigonal planar\n
\n\n \n \n OR\n \n \n
\n\n «all» angles are 120° ✔\n
\n\n The 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.\n
\n\n \n Discuss the data.\n \n
\n\n [3]\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 A gas storage tank of fixed volume V contains N molecules of an ideal gas at 300 K with a pressure of 40 kPa.\n \n
\n molecules are removed, and the temperature is changed to 450 K.\n
\n
\n What is the new pressure of the gas in kPa?\n
\n
\n A. 15\n
\n B. 30\n
\n\n C. 45\n
\n\n D. 60\n
\n\n [1]\n
\n\n C\n
\n\n Which graph shows the relationship between the pressure and volume of a sample of gas at constant temperature?\n
\n\n \n
\n [1]\n
\n\n B\n
\n\n \n Nitrogen monoxide reacts with oxygen gas to form nitrogen dioxide.\n \n
\n\n \n Deduce, giving a reason, whether the following mechanism is possible.\n \n
\n\n \n \n \n
\n
\n\n [1]\n
\n\n not possible\n \n \n AND\n \n \n «proposed» mechanism does not match experimental rate law\n
\n \n \n OR\n \n \n
\n not possible\n \n \n AND\n \n \n «proposed» mechanism shows zero/not first order with respect to oxygen ✔\n
\n 60% of candidates could explain why the proposed reaction mechanism was inconsistent with the empirical data given.\n
\n\n \n The experiment gave an error in the rate because the pressure gauge was inaccurate.\n \n
\n\n \n Outline whether repeating the experiment, using the same apparatus, and averaging the results would reduce the error.\n \n
\n\n [1]\n
\n\n \n no\n \n \n AND\n \n \n it is a systematic error/not a random error\n \n
\n\n \n \n \n OR\n \n \n \n
\n\n \n no\n \n \n AND\n \n \n «a similar magnitude» error would occur every time\n \n [✔]\n \n \n
\n\n Almost all candidates identified the inaccurate pressure gauge as a systematic error, thus relating accuracy to this type of error.\n
\n\n The efficiency of a nuclear power plant is approximately 33 %.\n
\n\n
\n\n 33 % of energy input/released «in fission» is converted to electricity ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept generalizations, such as “Only produces 33 % useful energy.”\n \n
\n\n 0.50 mol of\n \n (g) and 0.50 mol of Br\n \n 2\n \n (g) are placed in a closed flask. The following equilibrium is established.\n
\n\n \n \n \n (g) + Br\n \n 2\n \n (g)\n \n 2\n \n Br (g)\n
\n\n The equilibrium mixture contains 0.80 mol of\n \n Br (g). What is the value of\n \n K\n \n \n c\n \n ?\n
\n\n A. 0.64\n
\n\n B. 1.3\n
\n\n C. 2.6\n
\n\n D. 64\n
\n\n [1]\n
\n\n D\n
\n\n 41% of the candidates were able to calculate\n \n K\n \n \n c\n \n given the initial amounts of reactants and the amount of product at equilibrium.\n
\n\n \n Which species has delocalized electrons?\n \n
\n\n \n A. OH\n \n −\n \n \n
\n\n \n B. H\n \n 2\n \n CO\n \n
\n\n \n C. CO\n \n 2\n \n \n
\n\n \n D. CO\n \n 3\n \n \n 2−\n \n \n
\n\n [1]\n
\n\n D\n
\n\n This question on delocalized electrons was also well answered.\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 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 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 systematic error\n \n \n AND\n \n \n all values «equally» inaccurate ✔\n
\n\n Deduce the change in the oxidation state of sulfur.\n
\n\n [1]\n
\n\n +6\n
\n \n \n OR\n \n \n
\n −2 to +4 ✔\n
\n \n Accept “6/VI”.\n \n
\n \n Accept “−II, 4//IV”.\n \n
\n Do\n \n not\n \n accept 2- to 4+.\n
\n What can be deduced from the period number of an element?\n
\n\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\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 \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 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 \n Mass spectrometry analysis of a sample of iron gave the following results:\n \n
\n\n \n \n \n
\n \n \n Calculate the relative atomic mass, A\n \n r\n \n , of this sample of iron to two decimal places.\n \n \n
\n\n [2]\n
\n\n \n «\n \n A\n \n \n r\n \n =» 54 × 0.0584 + 56\n \n ×\n \n 0.9168 + 57\n \n ×\n \n 0.0217 + 58\n \n ×\n \n 0.0031\n
\n \n \n OR\n \n \n
\n «\n \n A\n \n \n r\n \n =» 55.9111\n \n [✔]\n \n \n
\n \n «\n \n A\n \n \n r\n \n =» 55.91\n \n [✔]\n \n \n
\n\n \n \n Notes:\n \n \n
\n\n \n \n Award [2] for correct final answer.\n
\n Do not accept data booklet value (55.85).\n \n \n
\n Calculation 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.\n
\n\n Explain how ranitidine (Zantac®) can also relieve excess stomach acid.\n
\n\n [2]\n
\n\n blocks/binds to H2/histamine receptors «in cells of stomach lining»\n
\n \n \n OR\n \n \n
\n prevents histamine binding to H2/histamine receptors «and triggering acid secretion» ✓\n
\n prevents «parietal cells from» releasing/producing acid ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “antihistamine” by itself.\n \n
\n\n \n Accept “H2-receptor antagonist/H2RA”\n
\n \n OR\n \n
\n “blocks/inhibits action of histamine” for M1.\n \n
\n \n Do\n \n not\n \n accept “blocks receptors” alone for M1.\n \n
\n\n \n Do\n \n not\n \n accept “proton pump/ATPase inhibitor”.\n \n
\n\n Which statement explains why the\n \n second\n \n ionization energy of aluminium is higher than the\n \n first\n \n ionization energy of magnesium?\n
\n\n
\n A. Ionization energy increases along period 3.\n
\n B. 3p electrons are at a higher energy level than 3s electrons.\n
\n\n C. 3p electrons are further away from the nucleus than 2p electrons.\n
\n\n D. Both have the same number of electrons and aluminium has one more proton.\n
\n\n [1]\n
\n\n D\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 \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 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 \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 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 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 \n Which is correct for the reaction H\n \n 2\n \n O (g) → H\n \n 2\n \n O (l) ?\n \n
\n\n
\n\n \n A. Enthalpy increases and entropy increases.\n \n
\n\n \n B. Enthalpy decreases and entropy increases.\n \n
\n\n \n C. Enthalpy increases and entropy decreases.\n \n
\n\n \n D. Enthalpy decreases and entropy decreases.\n \n
\n\n [1]\n
\n\n D\n
\n\n Higher scoring candidates had more success understanding enthalpy and entropy decreases.\n
\n\n Which substance is the reducing agent in the given reaction?\n
\n\n H\n \n +\n \n (aq) + 2H\n \n 2\n \n O (l) + 2MnO\n \n 4\n \n \n −\n \n (aq) + 5SO\n \n 2\n \n (g) → 2Mn\n \n 2+\n \n (aq) + 5HSO\n \n 4\n \n \n −\n \n (aq)\n
\n\n
\n A. H\n \n +\n \n
\n B. H\n \n 2\n \n O\n
\n\n C. MnO\n \n 4\n \n \n −\n \n
\n\n D. SO\n \n 2\n \n
\n\n [1]\n
\n\n D\n
\n\n 50% of the candidates identified SO\n \n 2\n \n as the reducing agent. The most commonly chosen distractor was MnO\n \n 4\n \n \n −\n \n . The question had good discrimination between high-scoring and low-scoring candidates.\n
\n\n \n What must be present on a nucleophile?\n \n
\n\n \n A. Negative charge\n \n
\n\n \n B. Lone pair of electrons\n \n
\n\n \n C. Positive charge\n \n
\n\n \n D. Symmetrical distribution of electrons\n \n
\n\n [1]\n
\n\n B\n
\n\n 72 % of the candidates chose B (a nucleophile must have a lone pair of electrons). The most commonly chosen distractor was A (a nucleophile must have a negative charge). Some teachers commented that both answers A and B are correct. However, the word “must” in the question means B is the only answer. Candidates are aware that some nucleophiles like water are neutral. Moreover, the definition of a nucleophile is clearly stated in the syllabus (topic 10.2).\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 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 Which graph represents a second order reaction with respect to X?\n
\n\n X → Y\n
\n\n \n
\n [1]\n
\n\n D\n
\n\n \n The following reaction occurs in a voltaic (galvanic) cell.\n \n
\n\n \n Mg (s) + 2Ag\n \n +\n \n (aq) → Mg\n \n 2+\n \n (aq) + 2Ag (s)\n \n
\n\n \n Which reaction takes place at each electrode?\n \n
\n\n \n \n \n
\n [1]\n
\n\n C\n
\n\n The question about electrode reactions discriminated well between high scoring and low scoring candidates. 70 % of the candidates selected the correct electrode reactions given the cell reaction. The most commonly chosen distractor had the same reactions at the opposite electrodes.\n
\n\n \n Describe the effect of increasing the voltage on the chemical yield of:\n \n
\n\n \n Ethanal using Pt/C:\n \n
\n\n \n Carbon dioxide using PtRu/C:\n \n
\n\n [2]\n
\n\n \n \n Ethanal using Pt/C:\n \n
\n decreases ✔\n \n
\n \n \n Carbon dioxide using PtRu/C\n \n :\n
\n «generally» increases\n \n \n AND\n \n \n then decreases ✔\n \n
\n \n \n NOTE: Accept “no clear trend/pattern”\n \n OR\n \n “increases and decreases”\n \n OR\n \n “increases, reaches a plateau and «then» decreases” for M2.\n \n \n
\n\n The orange colour disappears when bromine water is added to compound X in the dark.\n
\n\n Which compound is X?\n
\n\n
\n A. Ethene\n
\n B. Ethane\n
\n\n C. Ethanol\n
\n\n D. Ethanoic acid\n
\n\n [1]\n
\n\n A\n
\n\n \n The reaction of the hydroxide ion with carbon dioxide and with the hydrogencarbonate ion can be represented by Equations 3 and 4.\n \n
\n\n \n Equation (3) OH\n \n −\n \n (aq) + CO\n \n 2\n \n (g) → HCO\n \n 3\n \n \n −\n \n (aq)\n
\n Equation (4) OH\n \n −\n \n (aq) + HCO\n \n \n 3\n \n \n −\n \n \n (aq) → H\n \n 2\n \n O (l) + CO\n \n 3\n \n \n 2−\n \n (aq)\n
\n \n
\n \n Discuss how these equations show the difference between a Lewis base and a Brønsted–Lowry base.\n \n
\n\n
\n\n \n Equation (3):\n \n
\n\n \n Equation (4):\n \n
\n\n [2]\n
\n\n \n \n Equation (3):\n \n
\n OH\n \n -\n \n donates an electron pair\n \n \n AND\n \n \n acts as a Lewis base\n \n [✔]\n \n \n
\n \n \n Equation (4):\n \n
\n OH\n \n -\n \n accepts a proton/H\n \n +\n \n /hydrogen ion\n \n \n AND\n \n \n acts as a Brønsted–Lowry base\n \n \n \n [✔]\n \n \n
\n This 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\n \n -\n \n or also lost marks referring just to electrons, an electron or H instead of hydrogen ions or H\n \n +\n \n for example.\n
\n\n The nitride ion and the magnesium ion are isoelectronic (they have the same electron configuration). Determine, giving a reason, which has the greater ionic radius.\n
\n\n [1]\n
\n\n nitride\n \n \n AND\n \n \n smaller nuclear charge/number of protons/atomic number ✔\n
\n\n In 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.\n
\n\n \n Calculate the pressure, in kPa, of this gas in a 10.0 dm\n \n 3\n \n air bag at 127°C, assuming no gas escapes.\n \n
\n\n [1]\n
\n\n \n «\n \n \n \n » = 136 «kPa» ✔\n \n
\n\n Which substance has the following\n \n 1\n \n H NMR spectrum?\n
\n\n \n
\n \n SDBS, National Institute of Advanced Science and Technology.\n \n
\n\n
\n A. Propane\n
\n B. Propanal\n
\n\n C. Butanoic acid\n
\n\n D. Ethyl ethanoate\n
\n\n [1]\n
\n\n D\n
\n\n \n Which solution is basic at 25 °C?\n \n
\n\n \n K\n \n w\n \n = 1.0 × 10\n \n −14\n \n \n
\n\n \n A. [H\n \n +\n \n ] = 1.0 × 10\n \n −3\n \n mol dm\n \n −3\n \n \n
\n\n \n B. [OH\n \n −\n \n ] = 1.0 × 10\n \n −13\n \n mol dm\n \n −3\n \n \n
\n\n \n C. solution of pH = 4.00\n \n
\n\n \n D. [H\n \n 3\n \n O\n \n +\n \n ] = 1.0 × 10\n \n −13\n \n mol dm\n \n −3\n \n \n
\n\n [1]\n
\n\n D\n
\n\n This had the highest discriminatory index on the exam. Most of the incorrect answers indicated that a basic solution would have\n \n [OH\n \n -\n \n ] = 1.0 x 10\n \n -13\n \n mol dm\n \n -3\n \n rather than [H\n \n 3\n \n O\n \n +\n \n ] = 1.0 x 10\n \n -13\n \n mol dm\n \n -3\n \n \n .\n
\n\n Write the equation for the reaction between but-2-ene and hydrogen bromide.\n
\n\n [1]\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 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 Formulate an equation for the reaction of one mole of phosphoric acid with one mole of sodium hydroxide.\n
\n\n [1]\n
\n\n H\n \n 3\n \n PO\n \n 4\n \n (aq) + NaOH (aq) → NaH\n \n 2\n \n PO\n \n 4\n \n (aq) + H\n \n 2\n \n O (l) ✔\n
\n\n \n
\n Accept net ionic equation.\n \n
\n Which changes produce the greatest increase in the percentage conversion of methane?\n
\n\n CH\n \n 4\n \n (g) + H\n \n 2\n \n O (g)\n \n CO (g) + 3H\n \n 2\n \n (g)\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n \n Suggest a concern about the disposal of solvents from drug manufacturing.\n \n
\n\n [1]\n
\n\n \n \n Any one of:\n \n
\n «most are» toxic «to living organisms»\n
\n \n \n OR\n \n \n
\n incomplete combustion/incineration can produce toxic products/dioxins/phosgene\n
\n \n \n OR\n \n \n
\n carcinogenic/can cause cancer ✔\n
\n \n
\n \n accumulate in groundwater\n
\n \n \n OR\n \n \n
\n have limited biodegradability ✔\n
\n \n
\n \n cost of disposal ✔\n \n
\n\n \n \n NOTE: Do\n \n not\n \n accept “harmful to the environment”.\n
\n Do\n \n not\n \n accept just “pollutes water”.\n
\n Do\n \n not\n \n accept “hazard of disposal”.\n
\n Accept “ozone depletion” only if there is some reference to chlorinated solvents.\n \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 +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 \n The stable isotope of rhenium contains 110 neutrons.\n \n
\n\n \n State the nuclear symbol notation\n \n \n \n for this isotope.\n \n
\n\n [1]\n
\n\n \n \n \n \n \n [✔]\n \n \n
\n\n 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.\n
\n\n What is represented by\n \n “2−”\n \n in\n \n ?\n
\n\n A. loss of electron\n
\n\n B. gain of electron\n
\n\n C. loss of proton\n
\n\n D. gain of proton\n
\n\n [1]\n
\n\n B\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 «\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 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 Which quantities are different between two species represented by the notation\n \n and\n \n \n −\n \n ?\n
\n
\n
\n A. The number of protons only\n
\n\n B. The number of protons and electrons only\n
\n\n C. The number of protons and neutrons only\n
\n\n D. The number of protons, neutrons and electrons\n
\n\n [1]\n
\n\n D\n
\n\n A well answered question. 62% of the candidates compared the numbers of protons, neutrons and electrons in the two species correctly.\n
\n\n The diagram shows the energy profile of a reaction.\n
\n\n \n
\n
\n Which combination is correct?\n
\n
\n \n
\n [1]\n
\n\n B\n
\n\n What effect does a catalyst have on the position of equilibrium and the value of the equilibrium constant,\n \n K\n \n \n c\n \n , for an exothermic reaction?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n \n Explain why, as the reaction proceeds, the pressure increases by the amount shown.\n \n
\n\n [2]\n
\n\n \n increase in the amount/number of moles/molecules «of gas»\n \n [✔]\n \n \n
\n\n \n from 2 to 3/by 50 %\n \n [✔]\n \n \n
\n\n About 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 (\n \n i.e.\n \n how 200 & 300 connect to coefficients) or rates (\n \n i.e.\n \n 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.\n
\n\n Which allotrope, oxygen or ozone, has the stronger bond between O atoms, and which absorbs higher frequency UV radiation in the atmosphere?\n
\n\n
\n| \n | \n\n \n Stronger bond between O atoms\n \n | \n\n \n Absorbs higher frequency UV\n \n | \n
| \n A.\n | \n\n ozone\n | \n\n ozone\n | \n
| \n B.\n | \n\n ozone\n | \n\n oxygen\n | \n
| \n C.\n | \n\n oxygen\n | \n\n oxygen\n | \n
| \n D.\n | \n\n oxygen\n | \n\n ozone\n | \n
\n [1]\n
\n\n C\n
\n\n Which formula represents an ether?\n
\n\n A. C\n \n 6\n \n H\n \n 5\n \n OH\n
\n\n B. CH\n \n 3\n \n CHO\n
\n\n C. CH\n \n 3\n \n COCH\n \n 3\n \n
\n\n D. CH\n \n 3\n \n OCH\n \n 3\n \n
\n\n [1]\n
\n\n D\n
\n\n \n Which solvent is aprotic?\n \n
\n\n \n A. H\n \n 2\n \n O\n \n
\n\n \n B. C\n \n 6\n \n H\n \n 5\n \n CH\n \n 3\n \n \n
\n\n \n C. CH\n \n 3\n \n OH\n \n
\n\n \n D. CH\n \n 3\n \n NH\n \n 2\n \n \n
\n\n [1]\n
\n\n B\n
\n\n Identifying protic from aprotic solvents was poorly done my most candidates.\n
\n\n Which illustrates the correct intermediate formed in the nitration of benzene by NO\n \n 2\n \n \n +\n \n ?\n
\n\n
\n \n
\n [1]\n
\n\n A\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 \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 Correct 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.\n
\n\n Which is the correct structural formula of this compound?\n
\n\n \n
\n
\n A. OCOCH\n \n 2\n \n CH\n \n 3\n \n
\n B. HCOOC\n \n 2\n \n H\n \n 5\n \n
\n\n C. HCOOCH\n \n 2\n \n CH\n \n 3\n \n
\n\n D. OCOHC\n \n 2\n \n CH\n \n 3\n \n
\n\n [1]\n
\n\n C\n
\n\n \n Draw the Lewis structures of oxygen, O\n \n 2\n \n , and ozone, O\n \n 3\n \n .\n \n
\n\n [2]\n
\n\n \n
\n \n NOTES: Coordinate bond may be represented by an arrow.\n \n
\n\n \n Do\n \n not\n \n accept delocalized structure for ozone.\n \n
\n\n Which graph represents the relationship between the amount of gas, n, and the absolute temperature, T, with all other variables in the ideal gas equation,\n \n PV = nRT\n \n , held constant?\n
\n\n \n
\n [1]\n
\n\n B\n
\n\n 56% of the candidates selected the correct graph representing the relationship between the amount of gas and its absolute temperature. The most commonly chosen distractor gave a directly proportional relationship.\n
\n\n \n The benzene ring of phenylethene reacts with the nitronium ion, NO\n \n 2\n \n \n +\n \n , and the C=C double bond reacts with hydrogen bromide, HBr.\n \n
\n\n \n Compare and contrast these two reactions in terms of their reaction mechanisms.\n \n
\n\n
\n\n \n Similarity:\n \n
\n\n \n Difference:\n \n
\n\n [2]\n
\n\n \n \n Similarity:\n \n
\n «both» involve an electrophile\n
\n \n \n OR\n \n \n
\n «both» electrophilic\n \n [✔]\n \n
\n \n
\n
\n\n \n \n Difference\n \n :\n
\n first/reaction of ring/with NO\n \n 2\n \n \n +\n \n is substitution/S\n \n «E»\n \n \n \n AND\n \n \n second/reaction of C=C/with HBr is addition/A\n \n «E»\n \n \n \n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Answer must state which is substitution and which is addition for M2.\n \n \n
\n\n The 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.\n
\n\n Iron 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.\n
\n\n Determine the specific heat capacity of iron, in J g\n \n −1\n \n K\n \n −1\n \n . Use section 1 of the data booklet.\n
\n\n [1]\n
\n\n specific heat capacity « =\n \n » = 0.45 «J g\n \n −1\n \n K\n \n −1\n \n » ✔\n
\n\n Write an equation for the reaction of white phosphorus (P\n \n 4\n \n ) with chlorine gas to form phosphorus trichloride (PCl\n \n 3\n \n ).\n
\n\n [1]\n
\n\n P\n \n 4\n \n (s) + 6Cl\n \n 2\n \n (g) → 4PCl\n \n 3\n \n (l) ✔\n
\n\n Which solution has a pH of 9?\n
\n\n A. 1.0 × 10\n \n −9\n \n mol dm\n \n −3\n \n \n (aq)\n
\n\n B. 1.0 × 10\n \n −5\n \n mol dm\n \n −3\n \n \n (aq)\n
\n\n C. 1.0 × 10\n \n −9\n \n mol dm\n \n −3\n \n \n (aq)\n
\n\n D. 1.0 × 10\n \n −5\n \n mol dm\n \n −3\n \n \n (aq)\n
\n\n [1]\n
\n\n B\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 Incorrect\n \n \n AND\n \n \n two readings, uncertainty is ±0.1 ✔\n
\n\n \n Outline 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.\n \n
\n\n \n Reason:\n \n
\n\n \n Length:\n \n
\n\n [2]\n
\n\n \n resonance «structures»\n
\n \n \n OR\n \n \n
\n delocalization of «the double/pi bond» electrons ✔\n
\n 121 «pm» < length < 148 «pm» ✔\n \n
\n \n \n NOTE: Accept any length between these two values.\n \n \n
\n\n \n Identify\n \n one\n \n error associated with the use of an accurate stopwatch.\n \n
\n\n [1]\n
\n\n \n human reaction time/delay «starting/stopping the stopwatch»\n \n [✔]\n \n \n
\n\n \n \n Note\n \n :\n \n Do\n \n not\n \n accept “inaccurate stopwatch”.\n \n \n
\n\n This question was well answered by most candidates although some students did not read the question clearly and commented on the stopwatch having problems or not being accurate.\n
\n\n \n Calculate the standard Gibbs free energy change,\n \n , in kJ mol\n \n −1\n \n , for the first dissociation of citric acid at 298 K, using section 1 of the data booklet.\n \n
\n\n [1]\n
\n\n \n \n «\n \n \n \n \n = −\n \n RT\n \n ln\n \n K\n \n = −8.31 J K\n \n –1\n \n mol\n \n –1\n \n × 298 K × ln(5.01 × 10\n \n –4\n \n ) ÷ 1000 =» 18.8 «kJ mol\n \n –1\n \n » ✔\n \n
\n\n Justify why ethene has only a single signal in its\n \n 1\n \n H NMR spectrum.\n
\n\n [1]\n
\n\n hydrogen atoms/protons in same chemical environment ✔\n
\n\n \n Accept “all H atoms/protons are equivalent”.\n \n
\n \n Accept “symmetrical”\n \n
\n Which molecule is polar?\n
\n\n
\n A. BeH\n \n 2\n \n
\n B. AlH\n \n 3\n \n
\n\n C. PH\n \n 3\n \n
\n\n D. SiH\n \n 4\n \n
\n\n [1]\n
\n\n C\n
\n\n A well-answered question - 61% of the candidates selected PH\n \n 3\n \n as the polar molecule as the dipoles in the other symmetrical molecules would cancel out due to the molecular geometry. The question also discriminated well between high-scoring and low-scoring candidates. Some teachers were unhappy about the question as the electronegativities of P and H in the data booklet (which the students did not have access to in this paper) are identical. Apologies for this choice - a better example should have been used.\n
\n\n Write an equation for the complete combustion of C\n \n 17\n \n H\n \n 36\n \n .\n
\n\n [1]\n
\n\n C\n \n 17\n \n H\n \n 36\n \n (l) + 26O\n \n 2\n \n (g) → 17CO\n \n 2\n \n (g) + 18H\n \n 2\n \n O(l)\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 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 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 pressure, in kPa, of this gas in a 10.0 dm\n \n 3\n \n air bag at 127°C, assuming no gas escapes.\n \n
\n\n [1]\n
\n\n \n «\n \n \n \n » = 136 «kPa» ✔\n \n
\n\n \n State, giving a reason, the shape of the dinitrogen monoxide molecule.\n \n
\n\n [1]\n
\n\n \n linear\n \n \n AND\n \n \n 2 electron domains\n \n
\n\n \n \n \n OR\n \n \n \n
\n\n \n linear\n \n \n AND\n \n \n 2 regions of electron density\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “two bonds\n \n \n \n AND\n \n \n \n no lone pairs” for reason.\n \n \n
\n\n Though 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.\n
\n\n State\n \n one\n \n problem associated with chlorinated organic solvents as chemical waste.\n
\n\n [1]\n
\n\n \n Any one of:\n \n
\n\n «most are» toxic «to living organisms»\n
\n \n \n OR\n \n \n
\n incomplete combustion/incineration can produce toxic products/dioxins/phosgene\n
\n \n \n OR\n \n \n
\n carcinogenic ✓\n
\n «some can be» greenhouse gases ✓\n
\n\n ozone-depleting ✓\n
\n\n can contribute to formation of «photochemical» smog ✓\n
\n\n accumulate in groundwater\n
\n \n \n OR\n \n \n
\n have limited biodegradability ✓\n
\n cost/hazards of disposal ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “harmful to the environment”.\n \n
\n\n \n Do\n \n not\n \n accept just “pollutes water”.\n \n
\n\n \n Do\n \n not\n \n accept “increases acid rain/acidity/acid deposition”.\n \n
\n\n Calculate the entropy change of reaction, Δ\n \n S\n \n \n ⦵\n \n , in J K\n \n −1\n \n mol\n \n −1\n \n .\n
\n\n \n
\n [1]\n
\n\n «Δ\n \n = 2 × 206.6 – (130.6 + 116.1) =» 166.5 «J K\n \n –1\n \n mol\n \n –1\n \n » ✔\n
\n\n ΔS was well calculated in general except for some inverted calculations or failure to consider the ratios of the reactants.\n
\n
\n
\n \n What is the correct trend going down groups 1\n \n \n \n and\n \n \n \n 17?\n \n
\n\n \n A. Melting points increase\n
\n \n
\n \n B. Boiling points decrease\n
\n \n
\n \n C. Electronegativities increase\n
\n \n
\n \n D. Ionization energies decrease\n \n
\n\n [1]\n
\n\n D\n
\n\n Most candidates selected ionization energy as the property that decreases down both group 1 and 17.\n
\n\n \n Additional experiments were carried out at an elevated temperature. On the axes below, sketch Maxwell–Boltzmann energy distribution curves at two temperatures T\n \n 1\n \n and T\n \n 2\n \n , where T\n \n 2\n \n > T\n \n 1\n \n .\n \n
\n\n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5963/4biii.PNG\")
\n
\n [2]\n
\n\n \n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5964/4biii_m.PNG\")
\n \n
\n \n peak of T\n \n 2\n \n to right of\n \n \n AND\n \n \n lower than T\n \n 1\n \n \n [✔]\n \n \n
\n\n \n lines begin at origin\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 Fair 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.\n
\n\n \n Describe how this mixture is separated by fractional distillation.\n \n
\n\n [2]\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 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
\n\n \n Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled, while nitrogen does not.\n \n
\n\n [2]\n
\n\n \n \n Any\n \n two\n \n of:\n \n
\n nitrogen non-polar/London/dispersion forces\n \n \n AND\n \n \n water polar/H-bonding ✔\n
\n water has «much» stronger intermolecular forces ✔\n
\n water molecules attract/condense/occupy smaller volume «and therefore deviate from ideal behaviour» ✔\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 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 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 Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)\n \n
\n\n [2]\n
\n\n \n «additional HCO\n \n 3\n \n \n -\n \n » shifts position of equilibrium to left\n \n [✔]\n \n \n
\n\n \n pH increases\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Do\n \n not\n \n award M2 without any justification in terms of equilibrium shift in M1.\n \n \n
\n\n A significant number of candidates omitted the “equilibrium” involved in the dissolution of a weak base.\n
\n\n \n Identify the type of bonding in sodium hydrogencarbonate.\n \n
\n\n
\n\n \n \n Between sodium and hydrogencarbonate:\n \n \n
\n\n \n \n Between hydrogen and oxygen in hydrogencarbonate:\n \n \n
\n\n [2]\n
\n\n \n \n Between sodium and hydrogencarbonate:\n \n
\n ionic\n \n [✔]\n \n \n
\n \n \n Between hydrogen and oxygen in hydrogencarbonate:\n \n
\n «polar» covalent\n \n [✔]\n \n \n
\n Many 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 Suggest why this process might raise environmental concerns.\n
\n\n [1]\n
\n\n sulfur dioxide/SO\n \n 2\n \n causes acid rain ✔\n
\n\n \n Accept sulfur dioxide/SO\n \n 2\n \n /dust causes respiratory problems\n \n
\n \n Do\n \n not\n \n accept just “causes respiratory problems” or “causes acid rain”.\n \n
\n Which species can act as an electrophile?\n
\n\n
\n A. CH\n \n 4\n \n
\n B. Cl\n \n 2\n \n
\n\n C. Cl\n \n −\n \n
\n\n D. OH\n \n −\n \n
\n\n [1]\n
\n\n B\n
\n\n Which change involves oxidation of N?\n
\n
\n
\n A. NH\n \n 3\n \n to N\n \n 2\n \n
\n\n B. NO\n \n 2\n \n to NO\n
\n\n C. N\n \n 2\n \n to AlN\n
\n\n D. NO\n \n 2\n \n to N\n \n 2\n \n O\n \n 4\n \n
\n\n
\n\n A\n
\n\n Draw the skeletal formula of 5-chloro-2-methylpentan-2-ol.\n
\n\n [1]\n
\n\n \n ✓\n
\n Which reaction mechanisms are typical for alcohols and halogenoalkanes?\n
\n\n \n
\n [1]\n
\n\n D\n
\n\n Only 25% of the candidates deduced the reaction mechanisms that are typical for alcohols and halogenoalkanes. Candidates were expected to deduce that the reactions would be substitutions since both compounds are saturated.\n
\n\n The table shows data for the hydrolysis of a halogenoalkane, RCl.\n
\n\n \n
\n Which statements are correct?\n
\n\n I. The reaction is first order with respect to RCl.\n
\n II. The reaction is second order overall.\n
\n III. The reaction proceeds by an S\n \n N\n \n 2 mechanism.\n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n D\n
\n\n Average performance on a high discriminatory question on rate of reaction and mechanism question, with no clear misconception based on the other choices.\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 0.050−1.000 ✔\n
\n\n State the number of subatomic particles in this ion.\n
\n\n \n
\n [1]\n
\n\n \n Protons\n \n : 7\n \n \n AND\n \n Neutrons\n \n : 7\n \n \n AND\n \n Electrons\n \n : 10 ✔\n
\n\n Most candidates could answer the question about subatomic particles correctly.\n
\n\n \n Which reaction becomes more spontaneous as temperature increases?\n
\n \n
\n \n A.\n \n \n
\n\n \n B.\n \n \n
\n\n \n C.\n \n \n
\n\n \n D.\n \n \n
\n\n [1]\n
\n\n A\n
\n\n The majority of candidates could see that reactions which form more mol of gas become spontaneous at higher temperature.\n
\n\n State the electron configuration of copper.\n
\n\n [1]\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
\n Which is the correct order based on\n \n increasing\n \n strength?\n
\n\n A. covalent bonds < hydrogen bonds < dipole–dipole forces < dispersion forces\n
\n\n B. dipole–dipole forces < dispersion forces < hydrogen bonds < covalent bonds\n
\n\n C. dispersion forces < dipole–dipole forces < hydrogen bonds < covalent bonds\n
\n\n D. dispersion forces < dipole–dipole forces < covalent bonds < hydrogen bonds\n
\n\n [1]\n
\n\n C\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 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 \n Which change has the greatest increase in entropy?\n \n
\n\n \n A. CO\n \n 2\n \n (s) →\n \n CO\n \n \n 2\n \n (g)\n \n
\n\n \n B.\n \n CO\n \n \n 2\n \n (g) →\n \n CO\n \n \n 2\n \n (l)\n \n
\n\n \n C.\n \n CO\n \n \n 2\n \n (g) →\n \n CO\n \n \n 2\n \n (s)\n \n
\n\n \n D.\n \n CO\n \n \n 2\n \n (l) →\n \n CO\n \n \n 2\n \n (s)\n \n
\n\n [1]\n
\n\n A\n
\n\n 94 % of the candidates chose the change with the greatest increase in entropy.\n
\n\n Draw and label an enthalpy level diagram for this reaction.\n
\n\n \n
\n [2]\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 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 what would be observed when Compound B is warmed with acidified aqueous potassium dichromate (VI).\n
\n\n [1]\n
\n\n no change «in colour/appearance/solution» ✔\n
\n\n Some 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\n \n observation\n \n - \"it would remain orange\".\n
\n\n Explain why the addition of small amounts of carbon to iron makes the metal harder.\n
\n\n [2]\n
\n\n disrupts the regular arrangement «of iron atoms/ions»\n
\n \n \n OR\n \n \n
\n carbon different size «to iron atoms/ions» ✔\n
\n prevents layers/atoms sliding over each other ✔\n
\n\n Identify the strongest force between the molecules of Compound B.\n
\n\n [1]\n
\n\n hydrogen bonds ✔\n
\n\n Probably 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.\n
\n\n Which is a propagation step in the free-radical substitution mechanism of ethane with chlorine?\n
\n\n A. C\n \n \n 2\n \n → 2 •C\n \n
\n\n B. •C\n \n 2\n \n H\n \n 5\n \n + C\n \n \n 2\n \n → C\n \n 2\n \n H\n \n 5\n \n C\n \n + •C\n \n
\n\n C. •C\n \n 2\n \n H\n \n 5\n \n + •C\n \n → C\n \n 2\n \n H\n \n 5\n \n C\n \n
\n\n D. C\n \n 2\n \n H\n \n 6\n \n + •C\n \n → C\n \n 2\n \n H\n \n 5\n \n C\n \n + •H\n
\n\n [1]\n
\n\n B\n
\n\n \n The same amount of two gases, X and Y, are in two identical containers at the same temperature. What is the difference between the gases?\n \n
\n\n \n \n \n
\n \n \n A.\n \n X\n \n has the higher molar mass.\n \n \n
\n\n \n \n B.\n \n Y\n \n has the higher molar mass.\n \n \n
\n\n \n \n C.\n \n X\n \n has the higher average kinetic energy.\n \n \n
\n\n \n \n D.\n \n Y\n \n has the higher average kinetic energy.\n \n \n
\n\n [1]\n
\n\n A\n
\n\n This was by far the most challenging question on the paper, answered correctly by only 19 % of the candidates. Some teachers thought this was beyond the scope of the syllabus while others thought it was a question requiring thought. To be able to answer, candidates needed to connect kinetic energy to the speed and mass of the molecule.\n
\n\n \n A sample of ethanoic acid was titrated with sodium hydroxide solution, and the following pH curve obtained.\n \n
\n\n \n \n \n
\n \n \n Annotate the graph to show the buffer region and the volume of sodium hydroxide at the equivalence point.\n \n \n
\n\n [2]\n
\n\n \n
\n \n buffer region on graph ✔\n
\n equivalence point/V\n \n eq\n \n on graph ✔\n \n
\n \n \n NOTE: Construction lines not required.\n \n \n
\n\n Which species can act both as a Lewis acid and a Lewis base?\n
\n\n
\n A. H\n \n 2\n \n O\n
\n B. NH\n \n 4\n \n \n +\n \n
\n\n C. Cu\n \n 2+\n \n
\n\n D. CH\n \n 4\n \n
\n\n [1]\n
\n\n A\n
\n\n Nearly 65% of candidates correctly identified water as being able to act as both a Lewis acid and a Lewis base. Cu\n \n 2+\n \n (C) was the most favoured of the distractors and gaining this mark seemed to be only weakly related to overall performance.\n
\n\n Which is a d-block element?\n
\n\n A. Ca\n
\n\n B. Cf\n
\n\n C. C\n \n
\n\n D. Co\n
\n\n [1]\n
\n\n D\n
\n\n The following compounds have similar relative molecular masses. What is the order of increasing boiling point?\n
\n\n
\n A. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n OH < CH\n \n 3\n \n CH\n \n 2\n \n CHO < CH\n \n 3\n \n COOH\n
\n B. CH\n \n 3\n \n CH\n \n 2\n \n CHO < CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n OH < CH\n \n 3\n \n COOH\n
\n\n C. CH\n \n 3\n \n CH\n \n 2\n \n CHO < CH\n \n 3\n \n COOH < CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n OH\n
\n\n D. CH\n \n 3\n \n COOH < CH\n \n 3\n \n CH\n \n 2\n \n CHO < CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n OH\n
\n\n [1]\n
\n\n B\n
\n\n Burette readings for a titration are shown.\n
\n\n
\n\n \n
\n What is the mean titre?\n
\n\n A. 11.1 cm\n \n 3\n \n ± 0.1 cm\n \n 3\n \n
\n\n B. 11.15 cm\n \n 3\n \n ± 0.05 cm\n \n 3\n \n
\n\n C. 11.2 cm\n \n 3\n \n ± 0.05 cm\n \n 3\n \n
\n\n D. 11.2 cm\n \n 3\n \n ± 0.1 cm\n \n 3\n \n
\n\n [1]\n
\n\n D\n
\n\n Outline why many elements have atomic volumes greater than 10 000 cm\n \n 3\n \n mol\n \n −1\n \n .\n
\n\n [1]\n
\n\n gases «and others are solids» ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “lower density” alone.\n \n
\n\n A white solid was formed when ethene was subjected to high pressure.\n
\n\n Deduce the type of reaction that occurred.\n
\n\n [1]\n
\n\n «addition» polymerization ✔\n
\n\n \n When a bottle of carbonated water is opened, these equilibria are disturbed.\n \n
\n\n \n State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).\n \n
\n\n [1]\n
\n\n \n shifts to left/reactants\n \n \n AND\n \n \n to increase amount/number of moles/molecules of gas/CO\n \n 2\n \n (g)\n \n [✔]\n \n \n
\n\n This 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.\n
\n\n What is the number of hydrogen atoms in 2.00 moles of Ca(HCO\n \n 3\n \n )\n \n 2\n \n ?\n
\n\n Avogadro’s constant,\n \n L\n \n or\n \n N\n \n A\n \n \n : 6.02 × 10\n \n 23\n \n mol\n \n −1\n \n
\n\n A. 2.00\n
\n\n B. 4.00\n
\n\n C. 1.20 × 10\n \n 24\n \n
\n\n D. 2.41 × 10\n \n 24\n \n
\n\n [1]\n
\n\n D\n
\n\n Bromobenzene, C\n \n 6\n \n H\n \n 5\n \n Br, can be formed by the reaction of benzene, C\n \n 6\n \n H\n \n 6\n \n and bromine, Br\n \n 2\n \n in presence of the catalyst, FeBr\n \n 3\n \n .\n
\n\n State the name of the mechanism of the reaction.\n
\n\n [1]\n
\n\n electrophilic substitution ✓\n
\n\n A reaction proceeds by the following mechanism:\n
\n\n step 1:\n \n
\n step 2:\n \n
\n Which rate equation is consistent with this mechanism?\n
\n\n A. Rate =\n \n k\n \n [B]\n \n 2\n \n [C]\n
\n\n B. Rate =\n \n k\n \n [A]\n \n 2\n \n [B][C]\n
\n\n C. Rate =\n \n k\n \n [A]\n \n 2\n \n
\n\n D. Rate =\n \n k\n \n [A][C]\n
\n\n [1]\n
\n\n C\n
\n\n \n State the number of 1H NMR signals for this isomer of xylene and the ratio in which they appear.\n \n
\n\n \n Number of signals:\n \n \n
\n \n
\n \n Ratio:\n \n
\n\n [2]\n
\n\n \n Number of signals:\n
\n 2\n \n [✔]\n \n \n
\n \n Ratio:\n
\n 3 : 2\n
\n \n \n OR\n \n \n
\n 6 : 4\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept any correct integer or fractional ratio.\n \n \n
\n\n \n \n Accept ratios in reverse order.\n \n \n
\n\n Most students gained M1 but very few gained M2, suggesting that the correct answer of 2 signals may have been a guess.\n
\n\n \n What is correct in an electrolytic cell?\n \n
\n\n \n
\n [1]\n
\n\n D\n
\n\n A good majority of candidates understood that electrons are gained at the negative cathode causing reduction during electrolysis.\n
\n\n Which equation represents the standard enthalpy of formation of lithium oxide?\n
\n\n
\n A. 4Li (s) + O\n \n 2\n \n (g) → 2Li\n \n 2\n \n O (s)\n
\n B. 2Li (s) +\n \n O\n \n 2\n \n (g) → Li\n \n 2\n \n O (s)\n
\n\n C. Li (s) +\n \n O\n \n 2\n \n (g) →\n \n Li\n \n 2\n \n O (s)\n
\n\n D. Li (g) +\n \n O\n \n 2\n \n (g) →\n \n Li\n \n 2\n \n O (g)\n
\n\n [1]\n
\n\n B\n
\n\n In which reaction does entropy decrease?\n
\n\n A. NaCl (s) → NaCl (aq)\n
\n\n B. Zn (s) + H\n \n 2\n \n SO\n \n 4\n \n (aq) → ZnSO\n \n 4\n \n (aq) + H\n \n 2\n \n (g)\n
\n\n C. NH\n \n 3\n \n (g) + HCl (g) → NH\n \n 4\n \n Cl (s)\n
\n\n D. CuCO\n \n 3\n \n (s) → CuO (s) + CO\n \n 2\n \n (g)\n
\n\n [1]\n
\n\n C\n
\n\n Second best answered question in the exam with 85% of candidates identifying the correct answering for reaction in which entropy decreases.\n
\n\n \n The 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.\n \n
\n\n [1]\n
\n\n \n no\n \n \n AND\n \n \n it is a systematic error/not a random error\n
\n \n \n OR\n \n \n
\n no\n \n \n AND\n \n \n «a similar magnitude» error would occur every time\n \n [✔]\n \n \n
\n This 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.\n
\n\n \n What is the difference between a conjugate Brønsted–Lowry acid–base pair?\n \n
\n\n \n A. Electron pair\n
\n \n
\n \n B. Positive charge\n
\n \n
\n \n C. Proton\n
\n \n
\n \n D. Hydrogen atom\n \n
\n\n [1]\n
\n\n C\n
\n\n Which of the following elements yields a basic oxide when combusted?\n
\n\n
\n A. Au\n
\n B. P\n
\n\n C. Ca\n
\n\n D. N\n
\n\n [1]\n
\n\n C\n
\n\n \n What is the enthalpy of combustion, Δ\n \n H\n \n \n c\n \n , of ethanol in kJ mol\n \n −1\n \n ?\n
\n Maximum temperature of water: 30.0°C\n
\n Initial temperature of water: 20.0°C\n
\n Mass of water in beaker: 100.0 g\n
\n Loss in mass of ethanol: 0.230 g\n
\n \n M\n \n \n r\n \n (ethanol): 46.08\n
\n Specific heat capacity of water: 4.18 J g\n \n −1\n \n K\n \n −1\n \n
\n \n q = mc\n \n Δ\n \n T\n \n \n
\n
\n\n \n A.\n \n \n
\n\n \n B.\n \n \n
\n\n \n C.\n \n \n
\n\n \n D.\n \n \n
\n\n [1]\n
\n\n C\n
\n\n \n Which graph represents the relationship between the rate constant,\n \n \n \n \n \n , and temperature,\n \n \n \n \n \n , in kelvin?\n \n
\n\n \n
\n [1]\n
\n\n C\n
\n\n 60% of candidates selected the appropriate graph showing how rate constant, k, varies with temperature. Incorrect answers were equally split between a linear and inverse relationship.\n
\n\n Draw the structural formula of the carbocation intermediate produced when this electrophile attacks benzene.\n
\n\n [1]\n
\n\n \n
\n
\n\n \n Accept any of the five structures.\n \n
\n\n \n Do\n \n not\n \n accept structures missing the positive charge.\n \n
\n\n Again, many had difficulty drawing the structural formula of the carbocation intermediate produced in the reaction.\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 \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 Correct 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.\n
\n\n Write the equation for the fermentation of glucose, C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n , to produce ethanol.\n
\n\n [1]\n
\n\n C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n → 2C\n \n 2\n \n H\n \n 5\n \n OH + 2CO\n \n 2\n \n ✓\n
\n\n Which compounds are members of the same homologous series?\n
\n\n A. propanal, propanone, propanoic acid\n
\n\n B. propane, propene, propyne\n
\n\n C. hexan-1-ol, hexan-2-ol, hexan-3-ol\n
\n\n D. ethanol, propan-1-ol, butan-1-ol\n
\n\n [1]\n
\n\n D\n
\n\n \n What is the activation energy of the reverse reaction?\n \n
\n\n \n \n \n
\n
\n\n [1]\n
\n\n C\n
\n\n 81 % of the candidates identified the activation energy of the reverse reaction. The most commonly chosen distractor was B, the activation energy of the forward reaction.\n
\n\n \n What is the structure and bonding in SiO\n \n 2\n \n (s)?\n \n
\n\n \n
\n [1]\n
\n\n A\n
\n\n What can be deduced from the period number of an element?\n
\n\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\n
\n\n \n Which compound exists as two configurational isomers?\n \n
\n\n \n A. CBr\n \n 2\n \n =CH\n \n 2\n \n \n
\n\n \n B. CH\n \n 2\n \n =CHBr\n \n
\n\n \n C. CHBr\n \n 2\n \n CH\n \n 2\n \n Br\n \n
\n\n \n D. CHBr=CHBr\n \n
\n\n [1]\n
\n\n D\n
\n\n 71 % of the candidates identified 1,2-dibromoethene as having two configurational isomers. The most commonly chosen distractor was C which was the only saturated halogenoalkane, indicating that these candidates may have confused the term with “conformational” isomers.\n
\n\n \n Which compound is\n \n not\n \n in the same homologous series as the others?\n \n
\n\n A. C\n \n 5\n \n H\n \n 12\n \n
\n\n \n B. C\n \n 6\n \n H\n \n 12\n \n
\n \n
\n \n C. C\n \n 7\n \n H\n \n 16\n \n
\n \n
\n \n D. C\n \n 8\n \n H\n \n 18\n \n \n
\n\n [1]\n
\n\n B\n
\n\n \n A student obtained the following data to calculate\n \n \n \n ,\n \n \n using\n \n \n .\n
\n\n \n
\n\n \n
\n\n \n
\n\n \n What is the percentage uncertainty in the calculated value of\n \n \n \n \n \n ?\n
\n \n
\n \n A.\n \n
\n \n
\n \n B.\n \n
\n \n
\n \n C.\n \n
\n \n
\n \n D.\n \n \n
\n\n [1]\n
\n\n C\n
\n\n Just over ½ of the candidates could propagate uncertainties from absolute to relative.\n
\n\n \n What is the pH of 0.001 mol dm\n \n −3\n \n NaOH (aq)?\n \n
\n\n \n A. 1\n \n
\n\n \n B. 3\n \n
\n\n \n C. 11\n \n
\n\n \n D. 13\n \n
\n\n [1]\n
\n\n C\n
\n\n The question discriminated well between high scoring and low scoring candidates. 58 % of the candidates were able to calculate the pH of the aqueous solution of NaOH. The most commonly chosen distractor was B (pH = 3) where the students determined pOH but did not complete the calculation. It is interesting that these candidates did not seem to notice that NaOH is a base and should have a higher pH.\n
\n\n Which change involves oxidation of N?\n
\n\n A. NH\n \n 3\n \n to N\n \n 2\n \n
\n\n B. NO\n \n 2\n \n to NO\n
\n\n C. N\n \n 2\n \n to AlN\n
\n\n D. NO\n \n 2\n \n to N\n \n 2\n \n O\n \n 4\n \n
\n\n [1]\n
\n\n A\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 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 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 Deduce a balanced equation for the oxidation of Fe2+ by acidified hydrogen peroxide.\n
\n\n [1]\n
\n\n H\n \n 2\n \n O\n \n 2\n \n (aq) + 2H\n \n +\n \n (aq) + 2Fe\n \n 2+\n \n (aq) → 2H\n \n 2\n \n O(l) + 2Fe\n \n 3+\n \n (aq) ✔\n
\n\n Formulate an equation for the reaction of one mole of phosphoric acid with one mole of sodium hydroxide.\n
\n\n [1]\n
\n\n H\n \n 3\n \n PO\n \n 4\n \n (aq) + NaOH (aq) → NaH\n \n 2\n \n PO\n \n 4\n \n (aq) + H\n \n 2\n \n O (l) ✔\n
\n\n \n
\n Accept net ionic equation.\n \n
\n \n Which product will be obtained at the anode (positive electrode) when molten NaCl is electrolysed?\n \n
\n\n \n A. Na (l)\n \n
\n\n \n B. Cl (g)\n \n
\n\n \n C. Cl\n \n 2\n \n (g)\n \n
\n\n \n D. Na (s)\n \n
\n\n [1]\n
\n\n C\n
\n\n 57 % of candidates correctly identified products of electrolysis at the anode with the incorrect answers being split by the remaining 43 %\n
\n\n Explain how ranitidine (Zantac®) can also relieve excess stomach acid.\n
\n\n [2]\n
\n\n blocks/binds to H2/histamine receptors «in cells of stomach lining»\n
\n \n \n OR\n \n \n
\n prevents histamine binding to H2/histamine receptors «and triggering acid secretion» ✓\n
\n prevents «parietal cells from» releasing/producing acid ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “antihistamine” by itself.\n \n
\n\n \n Accept “H2-receptor antagonist/H2RA”\n
\n \n OR\n \n
\n “blocks/inhibits action of histamine” for M1.\n \n
\n \n Do\n \n not\n \n accept “blocks receptors” alone for M1.\n \n
\n\n \n Do\n \n not\n \n accept “proton pump/ATPase inhibitor”.\n \n
\n\n \n Which functional groups are present in this molecule?\n
\n \n
\n \n \n \n
\n \n A. carbonyl, ether, nitrile\n
\n \n
\n \n B. carbonyl, ester, nitrile\n
\n \n
\n \n C. carboxyl, ether, amine\n
\n \n
\n \n D. carboxyl, ester, amine\n \n
\n\n [1]\n
\n\n A\n
\n\n This question on identifying functional groups was the most discriminating question on the paper. 58% correctly identified all three groups but it was answered much better by higher scoring candidates.\n
\n\n Which alkane has the lowest standard entropy,\n \n S\n \n \n ⦵\n \n ?\n
\n\n
\n A. CH\n \n 4\n \n (g)\n
\n B. C\n \n 2\n \n H\n \n 6\n \n (g)\n
\n\n C. C\n \n 3\n \n H\n \n 8\n \n (g)\n
\n\n D. C\n \n 4\n \n H\n \n 10\n \n (g)\n
\n
\n
\n [1]\n
\n\n A\n
\n\n Only just over half of the candidates were aware that entropy increases with molecular complexity, with the most favoured distractor (D) indicating that many consider the reverse to be the case. This question had a negative discrimination index, showing that top students actually performed worse than their peers.\n
\n\n Outline, giving a reason, the effect of a catalyst on a reaction.\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 A 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≥E\n \n a\n \n . 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).\n
\n\n \n Which of these molecular formulae are also empirical formulae?\n \n
\n\n A. I and II only\n
\n\n \n B. I and III only\n
\n \n
\n \n C. II and III only\n
\n \n
\n \n D. I, II and III\n \n
\n\n [1]\n
\n\n B\n
\n\n One of the best answered questions on paper 1. Very few of the candidates struggled with distinguishing molecular and empirical formula.\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 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 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 \n How should a measurement of 5.00 g from a balance be recorded?\n \n
\n\n \n A. 5.00 ± 0.1 g\n \n
\n\n \n B. 5.00 ± 0.01 g\n \n
\n\n \n C. 5.00 ± 1 g\n \n
\n\n \n D. 5.00 ± 0.001 g\n \n
\n\n [1]\n
\n\n B\n
\n\n More than 85 % of candidates were able to identify a correct uncertainty, however not a good discriminating question.\n
\n\n State why NH\n \n 3\n \n is a Lewis base.\n
\n\n [1]\n
\n\n donates «lone/non-bonding» pair of electrons ✔\n
\n\n The 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.\n
\n\n Suggest, giving reasons, the relative volatilities of SCl\n \n 2\n \n and H\n \n 2\n \n O.\n
\n\n [3]\n
\n\n H\n \n 2\n \n O forms hydrogen bonding «while SCl\n \n 2\n \n does not» ✓\n
\n\n SCl\n \n 2\n \n «much» stronger London/dispersion/«instantaneous» induced dipole-induced dipole forces ✓\n
\n\n \n \n
\n Alternative 1:\n \n \n
\n H\n \n 2\n \n O less volatile\n \n \n AND\n \n \n hydrogen bonding stronger «than dipole–dipole and dispersion forces» ✓\n
\n \n \n
\n Alternative 2:\n \n \n
\n SCl\n \n 2\n \n less volatile\n \n \n AND\n \n \n effect of dispersion forces «could be» greater than hydrogen bonding ✓\\\n
\n
\n\n \n Ignore reference to Van der Waals.\n \n
\n\n \n Accept “SCl\n \n 2\n \n has «much» larger molar mass/electron density” for M2.\n \n
\n\n \n An aqueous solution of silver nitrate, AgNO\n \n 3\n \n (aq), can be electrolysed using platinum electrodes.\n \n
\n\n \n Formulate the half-equations for the reaction at each electrode during electrolysis.\n \n
\n\n \n \n Cathode (negative electrode):\n \n \n
\n\n \n \n \n Anode (positive electrode):\n \n \n \n
\n\n [2]\n
\n\n \n \n Cathode (negative electrode):\n \n
\n Ag\n \n +\n \n (aq) + e\n \n −\n \n → Ag (s)\n \n [✔]\n \n \n
\n \n
\n \n Anode (positive electrode):\n \n
\n 2H\n \n 2\n \n O(l) → O\n \n 2\n \n (g) + 4H\n \n +\n \n (aq) + 4e\n \n −\n \n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept 4OH\n \n −\n \n (aq) → O\n \n 2\n \n (g) + 2H\n \n 2\n \n O(l) + 4e\n \n −\n \n \n \n
\n\n \n \n Accept multiple or fractional coefficients in both half-equations.\n \n \n
\n\n Very 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 H\n \n 2\n \n O. Some candidates lost one of the marks for using equilibrium arrows in an electrolysis equation.\n
\n\n \n Draw the structure of one other isomer of xylene which retains the benzene ring.\n \n
\n\n [1]\n
\n\n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5943/1a_m.PNG\")
\n \n [\n \n \n ✔\n \n \n ]\n \n
\n Generally well done, although some candidates repeated the formula of the 1,4-isomer structure or drew the wrong bond,\n \n e.g.\n \n benzene ring to H rather than C on CH\n \n 3\n \n .\n
\n\n Given equimolar concentrations, which substance would produce the strongest signal in a\n \n 1\n \n H NMR spectrum?\n
\n\n
\n A. (CH\n \n 3\n \n )\n \n 3\n \n CH\n
\n B. C\n \n 6\n \n H\n \n 6\n \n
\n\n C. C\n \n 8\n \n H\n \n 18\n \n
\n\n D. Si(CH\n \n 3\n \n )\n \n 4\n \n
\n\n [1]\n
\n\n D\n
\n\n 47% of the candidates identified TMS as the substance producing the strongest signal in a\n \n 1\n \n H NMR spectrum. The wording of this question was not ideal and several teachers commented on this. We do not usually refer to the strength but the area under the signal.\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 \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 Which statement best explains the first ionization energy of sulfur being lower than that of phosphorus?\n
\n
\n
\n A. Sulfur has more protons than phosphorus.\n
\n\n B. Phosphorus does not have paired electrons in the outer p sub-level.\n
\n\n C. Sulfur has an unpaired electron in the outer p sub-level.\n
\n\n D. Phosphorus is more reactive than sulfur.\n
\n\n [1]\n
\n\n B\n
\n\n \n Which of the following can be both formed from bromoethane and converted directly into ethanal?\n \n
\n\n \n CH\n \n 3\n \n CH\n \n 2\n \n Br →\n \n X\n \n
\n \n X\n \n → CH\n \n 3\n \n CHO\n \n
\n \n A. CH\n \n 3\n \n CH\n \n 2\n \n OH\n \n
\n\n \n B. CH\n \n 3\n \n OCH\n \n 3\n \n \n
\n\n \n C. CH\n \n 3\n \n COOH\n \n
\n\n \n D. H\n \n 2\n \n C=CHBr\n \n
\n\n [1]\n
\n\n A\n
\n\n The question involving a sequence of organic reactions discriminated well between high scoring and low scoring candidates. 68 % chose the correct compound.\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 \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 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 Draw the skeletal structure of ethyl propanoate.\n
\n\n [1]\n
\n\n \n ✓\n
\n For which species can resonance structures be drawn?\n
\n\n A. HCOOH\n
\n\n B. HCOO\n \n –\n \n
\n\n C. CH\n \n 3\n \n OH\n
\n\n D. H\n \n 2\n \n CO\n \n 3\n \n
\n\n [1]\n
\n\n B\n
\n\n \n What is the effect of a stronger ligand?\n \n
\n\n \n
\n [1]\n
\n\n A\n
\n\n Explain how strong analgesics like morphine work.\n
\n\n [2]\n
\n\n «temporarily» binding to receptors in the nervous system/spinal cord/brain ✓\n
\n\n preventing transmission of pain impulses ✓\n
\n\n
\n\n \n Accept “bonding” for “binding” in M1.\n \n
\n\n \n Accept “without depressing the central nervous system” for M2.\n \n
\n\n Outline the reason why PCl\n \n 5\n \n is a non-polar molecule, while PCl\n \n 4\n \n F is polar.\n
\n\n \n
\n [3]\n
\n\n \n PCl\n \n 5\n \n is non-polar\n \n :\n
\n\n symmetrical\n
\n \n \n OR\n \n \n
\n dipoles cancel ✔\n
\n
\n\n \n PCl\n \n 4\n \n F is polar:\n \n
\n\n P–Cl has a different bond polarity than P–F ✔\n
\n\n non-symmetrical «dipoles»\n
\n \n \n OR\n \n \n
\n dipoles do not cancel ✔\n
\n \n
\n \n \n Accept F more electronegative than/different electronegativity to Cl for\n \n M2\n \n .\n \n
\n \n Outline\n \n two\n \n laboratory methods of distinguishing between solutions of citric acid and hydrochloric acid of equal concentration, stating the expected observations.\n \n
\n\n [2]\n
\n\n \n \n Any two of:\n \n
\n \n
\n \n «electrical» conductivity\n \n \n AND\n \n \n HCl greater ✔\n
\n \n
\n \n pH\n \n \n AND\n \n \n citric acid higher ✔\n
\n \n
\n \n titrate with strong base\n \n \n AND\n \n \n pH at equivalence higher for citric acid ✔\n
\n \n
\n \n add reactive metal/carbonate/hydrogen carbonate\n \n \n AND\n \n \n stronger effervescence/faster reaction with HCl ✔\n
\n \n
\n \n titration\n \n \n AND\n \n \n volume of alkali for complete neutralisation greater for citric acid ✔\n
\n \n
\n \n titrate with strong base\n \n \n AND\n \n \n more than one equivalence point for complete neutralisation of citric acid ✔\n
\n \n
\n \n titrate with strong base\n \n \n AND\n \n \n buffer zone with citric acid ✔\n
\n \n
\n
\n\n \n \n NOTE: Accept “add universal indicator\n \n \n \n \n AND\n \n \n \n \n HCl more red/pink” for M2.\n \n \n
\n\n \n \n Accept any acid reaction\n \n \n \n \n AND\n \n \n \n \n HCl greater rise in temperature.\n \n \n
\n\n \n \n Accept specific examples throughout.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “smell” or “taste”.\n \n \n
\n\n \n Determine the specific energy, in kJ g\n \n −1\n \n , and energy density, in kJ cm\n \n −3\n \n , of hexane, C\n \n 6\n \n H\n \n 14\n \n . Give both answers to three significant figures.\n \n
\n\n \n Hexane:\n \n M\n \n \n r\n \n = 86.2; Δ\n \n H\n \n c\n \n \n = −4163 kJ mol\n \n −1\n \n ; density = 0.660 g cm\n \n −3\n \n \n
\n\n Specific energy:\n
\n\n Energy density:\n
\n\n [2]\n
\n\n \n specific energy = «\n \n \n \n =» 48.3 «kJ g\n \n –1\n \n »\n \n [✔]\n \n \n
\n\n \n energy density = «48.3 kJ g\n \n –1\n \n × 0.660 g cm\n \n –3\n \n =» 31.9 «kJ cm\n \n –3\n \n »\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Award\n \n [1 max]\n \n if either or both answers not expressed to three significant figures.\n \n \n
\n\n This part was very well answered by most candidates with the correct number of significant digits as specified in the question.\n
\n\n Calculate the hydroxide ion concentration in saturated aqueous hydrogen sulfide.\n
\n\n [1]\n
\n\n 10\n \n −10\n \n «mol dm\n \n −3\n \n » ✔\n
\n\n Outline why the value obtained in (b)(i) might differ from a value calculated using Δ\n \n H\n \n \n f\n \n data.\n
\n\n [1]\n
\n\n «N-H» bond enthalpy is an average «and may not be the precise value in NH\n \n 3\n \n » ✔\n
\n\n
\n\n \n Accept ΔH\n \n f\n \n data are more accurate / are not average values.\n \n
\n\n Outlining why Δ\n \n H\n \n \n rxn\n \n based on BE values differ due to being average compared to using Δ\n \n H\n \n \n f\n \n values was generally done well.\n
\n\n State the electron configuration of copper.\n
\n\n [1]\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 \n OR\n \n \n
\n\n [Ar] 4s\n \n 1\n \n 3d\n \n 10\n \n ✔\n
\n\n
\n\n \n Accept configuration with 3d before 4s.\n \n
\n\n 2.85 g of CaCO\n \n 3\n \n was collected in the experiment in d(i). Calculate the percentage yield of CaCO\n \n 3\n \n .\n
\n\n (If you did not obtain an answer to d(i), use 4.00 g, but this is not the correct value.)\n
\n\n [1]\n
\n\n «\n \n × 100 =» 86.4 «%» ✓\n
\n\n
\n\n \n Accept answers in the range 86.1-86.4 «%».\n \n
\n\n \n Accept “71.3 %” for using the incorrect given value of 4.00 g.\n \n
\n\n A series of experiments were carried out at different temperatures and the rate of reaction, in mol dm\n \n −3\n \n s\n \n −1\n \n , was determined for each. The rate constant for the reaction of propanone (CH\n \n 3\n \n COCH\n \n 3\n \n ) with iodine (I\n \n 2\n \n ) was calculated and the processed data is represented in the following graph.\n
\n\n \n
\n Determine the activation energy for this reaction, stating the units. Use sections 1 and 2 of the data booklet.\n
\n\n [3]\n
\n\n \n
\n «two construction lines shown on the graph, and slope calculated:\n
\n «(−1.0 − (−3.0))/(0.0032 − 0.0035) =» −6700 ✓\n
\n
\n
\n «gradient x R = -E\n \n a\n \n »\n
\n «6700 × 8.31 J K\n \n −1\n \n mol\n \n −1\n \n /1000=» 56\n
\n \n \n OR\n \n \n
\n «6700 × 8.31 J K\n \n −1\n \n mol\n \n −1\n \n =» 56000 ✓\n
\n kJ «mol\n \n −1\n \n » ✓\n
\n \n \n OR\n \n \n
\n J «mol\n \n −1\n \n » ✓\n
\n
\n \n Accept range 6400–7000 for M1.\n
\n Accept range 53–59 or 53000–59000 for M2.\n
\n Accept the unit as kJ or J without reference to per mol.\n
\n Award\n \n [2]\n \n for final answer without units.\n
\n Accept use of\n \n ln k1/k2 = −EA/R (1/T2−1/T1)\n \n .\n \n
\n Which compound will have only one\n \n 1\n \n H NMR signal and show a carbonyl group in the IR spectrum?\n
\n\n
\n A. CH\n \n 3\n \n CHO\n
\n B. CH\n \n 3\n \n COOH\n
\n\n C. CH\n \n 3\n \n OCH\n \n 3\n \n
\n\n D. CH\n \n 3\n \n COCH\n \n 3\n \n
\n\n [1]\n
\n\n D\n
\n\n SO\n \n 2\n \n (g), O\n \n 2\n \n (g) and SO\n \n 3\n \n (g) are mixed and allowed to reach equilibrium at 600 °C.\n
\n\n \n
\n Determine the value of\n \n K\n \n \n c\n \n at 600 °C.\n
\n\n [2]\n
\n\n [O\n \n 2\n \n ] = 1.25 «mol dm\n \n −3\n \n »\n \n \n AND\n \n \n [SO\n \n 3\n \n ] = 3.50 «mol dm\n \n −3\n \n » ✓\n
\n\n «\n \n K\n \n \n c\n \n =\n \n =» 4.36 ✓\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer\n \n
\n\n Which d block element has the highest number of different oxidation states?\n
\n\n
\n A. Ti\n
\n B. Mn\n
\n\n C. Cu\n
\n\n D. Zn\n
\n\n [1]\n
\n\n B\n
\n\n What is correct for this redox reaction?\n
\n\n MnO\n \n 2\n \n (s) + 2\n \n \n −\n \n (aq) + 4H\n \n +\n \n (aq) → Mn\n \n 2+\n \n (aq) +\n \n \n 2\n \n (aq) + 2H\n \n 2\n \n O (l)\n
\n\n \n
\n [1]\n
\n\n A\n
\n\n \n Which factors affect the amount of product formed at the cathode during electrolysis of molten salts?\n \n
\n\n \n I. current\n
\n II. time\n
\n III. charge on the cation\n \n
\n \n A. I and II only\n \n
\n\n \n B. I and III only\n \n
\n\n \n C. II and III only\n \n
\n\n \n D. I, II and III\n \n
\n\n [1]\n
\n\n D\n
\n\n Majority of candidates answered this correctly with the most common mistake being omission of time as a factor affecting electrolysis quantities.\n
\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 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 Naming the organic compound using IUPAC rules was generally done well.\n
\n\n What are the products of the electrolysis of molten potassium chloride,\n \n ?\n
\n\n \n
\n [1]\n
\n\n B\n
\n\n Distinguish between the hazards of high-level and low-level nuclear waste.\n
\n\n [2]\n
\n\n level has large amounts of «ionizing» radiation ✓\n
\n\n high-level has long half-lives\n
\n \n \n OR\n \n \n
\n high-level last longer/persists ✓\n
\n
\n\n \n Accept converse statements for low-level.\n \n
\n\n \n Accept “high radioactivity for high-level\" for M1.\n \n
\n\n \n Do\n \n not\n \n accept “high-level has ionizing radiation” alone for M1.\n \n
\n\n \n Do\n \n not\n \n accept answers based on storage or disposal differences alone.\n \n
\n\n \n Accept “high-level releases heat” for M2.\n \n
\n\n \n Do\n \n not\n \n accept “high-level has more penetrating radiation”\n
\n \n OR\n \n
\n “high-level has higher frequency radiation” for M1.\n \n
\n \n Which electrons are removed from iron (Z = 26) to form iron(II)?\n \n
\n\n \n A. two 3d electrons\n \n
\n\n \n B. two 4s electrons\n \n
\n\n \n C. one 4s electron and one 3d electron\n \n
\n\n \n D. two 4p electrons\n \n
\n\n [1]\n
\n\n B\n
\n\n Stating the 4s electrons are lost first in forming the Fe(II) ion was done correctly by 58 % but with a low discrimination index.\n
\n\n \n The maximum temperature used to calculate the enthalpy of reaction was chosen at a point on the extrapolated (dotted) line.\n \n
\n\n \n State the maximum temperature which should be used and outline\n \n one\n \n assumption made in choosing this temperature on the extrapolated line.\n \n
\n\n
\n\n \n Maximum temperature:\n \n
\n\n \n Assumption:\n \n
\n\n [2]\n
\n\n \n \n Maximum temperature:\n \n
\n 73 «°C»\n \n [✔]\n \n \n
\n \n \n Assumption\n \n :\n
\n «temperature reached if» reaction instantaneous\n
\n \n \n OR\n \n \n
\n «temperature reached if reaction occurred» without heat loss\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “rate of heat loss is constant”\n \n OR\n \n “rate of temperature decrease is constant”.\n \n \n
\n\n Even 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 …”.\n
\n\n \n Deduce the equation for the decomposition of guanidinium nitrate.\n \n
\n\n [1]\n
\n\n \n C(NH\n \n 2\n \n )\n \n 3\n \n NO\n \n 3\n \n (s) → 2N\n \n 2\n \n (g) + 3H\n \n 2\n \n O (g) + C (s) ✔\n \n
\n\n \n Lecithin aids the body’s absorption of vitamin E.\n \n
\n\n \n \n \n
\n \n \n Suggest why vitamin E is fat-soluble.\n \n \n
\n\n [1]\n
\n\n \n long non-polar/hydrocarbon chain «and only one hydroxyl group»\n
\n \n \n OR\n \n \n
\n forms London/dispersion/van der Waals/vdW interactions with fat\n \n [✔]\n \n \n
\n Many candidates missed the idea of a long or large non-polar chain when describing the structure of vitamin E. Simply stating non-polar chain was not sufficient for the mark.\n
\n\n What is the product of the reaction of benzene with a mixture of concentrated nitric and sulfuric acids?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n \n Outline how alloys conduct electricity and why they are often harder than pure metals.\n \n
\n\n
\n\n \n Conduct electricity:\n \n
\n\n \n Harder than pure metals:\n \n
\n\n [2]\n
\n\n \n \n Conduct electricity:\n \n
\n «delocalized/valence» electrons free to move «under potential difference»\n \n [✔]\n \n \n
\n \n \n Harder than pure metals\n \n :\n
\n atoms/ions of different sizes prevent layers «of atoms/ions» from sliding over one another\n \n [✔]\n \n \n
\n Most candidates were awarded M1 for how alloys conduct electricity. Some struggled with describing why they are harder than pure metals. Teachers should remind student to use proper terminology such as atoms or ions not nuclei for this type of answer.\n
\n\n Which compound is acidic in aqueous solution?\n
\n\n A. KBr\n
\n\n B. CH\n \n 3\n \n COONa\n
\n\n C. NH\n \n 4\n \n Cl\n
\n\n D. Na\n \n 2\n \n CO\n \n 3\n \n
\n\n [1]\n
\n\n C\n
\n\n \n The value of the equilibrium constant for the first dissociation at 298 K is 5.01 × 10\n \n −4\n \n .\n \n
\n\n \n State, giving a reason, the strength of citric acid.\n \n
\n\n [1]\n
\n\n \n weak acid\n \n \n AND\n \n \n partially dissociated\n
\n \n \n OR\n \n \n
\n weak acid\n \n \n AND\n \n \n equilibrium lies to left\n
\n \n \n OR\n \n \n
\n weak acid\n \n \n AND\n \n K\n \n \n c\n \n /\n \n K\n \n \n a\n \n <1 ✔\n \n
\n What is the electron domain geometry of Si in SiO\n \n 2\n \n ?\n
\n\n A. bent\n
\n\n B. linear\n
\n\n C. square planar\n
\n\n D. tetrahedral\n
\n\n [1]\n
\n\n D\n
\n\n Curve 1 shows the mass change when marble chips are added to excess hydrochloric acid in an open beaker.\n
\n\n \n
\n Which changes would produce curve 2?\n
\n\n A. Powdering the marble chips and heating\n
\n\n B. Powdering the marble chips and doubling their mass\n
\n\n C. Doubling the volume of acid and heating\n
\n\n D. Doubling the acid concentration and powdering the marble chips\n
\n\n [1]\n
\n\n B\n
\n\n Calculate the concentration of H\n \n 3\n \n PO\n \n 4\n \n if 25.00 cm\n \n 3\n \n is completely neutralised by the addition of 28.40 cm\n \n 3\n \n of 0.5000 mol dm\n \n −3\n \n NaOH.\n
\n\n [2]\n
\n\n «NaOH\n \n »\n
\n\n «\n \n » 0.004733 «mol» ✔\n
\n\n «\n \n » 0.1893 «mol dm\n \n −3\n \n » ✔\n
\n\n \n
\n Award\n \n [2]\n \n for correct final answer.\n \n
\n Which is the correct structural formula of this compound?\n
\n\n \n
\n
\n A. OCOCH\n \n 2\n \n CH\n \n 3\n \n
\n B. HCOOC\n \n 2\n \n H\n \n 5\n \n
\n\n C. HCOOCH\n \n 2\n \n CH\n \n 3\n \n
\n\n D. OCOHC\n \n 2\n \n CH\n \n 3\n \n
\n\n [1]\n
\n\n C\n
\n\n \n Calculate the total number of moles of gas produced from the decomposition of 10.0 g of guanidinium nitrate.\n \n
\n\n [1]\n
\n\n \n moles of gas = «\n \n \n \n » 0.409 «mol» ✔\n \n
\n\n Which of the following is the correct skeletal formula of butanoic acid?\n
\n\n \n
\n [1]\n
\n\n A\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 \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 Which can act as a Lewis acid but not a Brønsted–Lowry acid?\n \n
\n\n \n A. BF\n \n 3\n \n
\n \n
\n \n B. H\n \n 2\n \n O\n
\n \n
\n \n C. NF\n \n 3\n \n
\n \n
\n \n D. NH\n \n 3\n \n \n
\n\n [1]\n
\n\n A\n
\n\n \n What is the sum of the integer coefficients when propene undergoes complete combustion?\n \n
\n\n \n __C\n \n 3\n \n H\n \n 6\n \n (g) + __O\n \n 2\n \n (g) → __CO\n \n 2\n \n (g) + __H\n \n 2\n \n O (l)\n \n
\n\n \n A. 11\n \n
\n\n \n B. 17\n \n
\n\n \n C. 21\n \n
\n\n \n D. 23\n \n
\n\n [1]\n
\n\n D\n
\n\n 93 % of the candidates balanced the equation for the combustion of propene and obtained the sum of the integer coefficients.\n
\n\n \n Identify a conjugate acid–base pair in the equation.\n \n
\n\n [1]\n
\n\n \n C\n \n 6\n \n H\n \n 8\n \n O\n \n 7\n \n \n \n AND\n \n \n C\n \n 6\n \n H\n \n 7\n \n O\n \n 7\n \n \n −\n \n
\n \n \n OR\n \n \n
\n H\n \n 2\n \n O\n \n \n AND\n \n \n H\n \n 3\n \n O\n \n +\n \n ✔\n \n
\n \n Which describes an ionic compound?\n \n
\n\n \n \n \n
\n [1]\n
\n\n B\n
\n\n Fairly well answered with the most widely held misconception being that ionic compounds conduct electricity in the solid state.\n
\n\n \n State what point\n \n Y\n \n on the graph represents.\n \n
\n\n [1]\n
\n\n \n highest recorded temperature\n
\n \n \n OR\n \n \n
\n when rate of heat production equals rate of heat loss\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “maximum temperature”.\n \n \n
\n\n \n \n Accept “completion/end point of reaction”.\n \n \n
\n\n Many 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.\n
\n\n Write the equation for this reaction.\n
\n\n [1]\n
\n\n 4FeS(s) + 7O\n \n 2\n \n (g) → 2Fe\n \n 2\n \n O\n \n 3\n \n (s) + 4SO\n \n 2\n \n (g) ✔\n
\n\n \n
\n Accept any correct ratio.\n \n
\n Determine the molar enthalpy of combustion of an alkane if 8.75 × 10\n \n −4\n \n moles are burned, raising the temperature of 20.0 g of water by 57.3 °C.\n
\n\n [2]\n
\n\n «\n \n q\n \n =\n \n mc\n \n Δ\n \n T\n \n = 20.0 g × 4.18 J g\n \n −1\n \n °C\n \n −1\n \n × 57.3 °C =» 4790 «J» ✔\n
\n\n «\n \n » –5470 «kJ mol\n \n –1\n \n » ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Accept answers in the range –5470 to –5480 «kJ mol\n \n −1\n \n ».\n \n
\n\n \n Accept correct answer in any units, e.g. –5.47 «MJ mol\n \n −1\n \n » or 5.47 x 10\n \n 6\n \n «J mol\n \n −1\n \n ».\n \n
\n\n State the meaning of a strong Brønsted–Lowry acid.\n
\n\n [2]\n
\n\n fully ionizes/dissociates ✔\n
\n\n proton/H\n \n +\n \n «donor »✔\n
\n\n \n The stable isotope of rhenium contains 110 neutrons.\n \n
\n\n \n State the nuclear symbol notation\n \n \n \n for this isotope.\n \n
\n\n [1]\n
\n\n \n \n \n \n \n [✔]\n \n \n
\n\n 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.\n
\n\n Suggest an experimental method that could be used to determine the rate of reaction.\n
\n\n [2]\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 How many sigma (σ) and pi (π) bonds are present in hydrogen cyanide, HCN?\n \n
\n\n \n \n \n
\n [1]\n
\n\n B\n
\n\n 89 % of the candidates deduced the numbers of sigma and pi bonds in HCN correctly. The most commonly chosen distractor (C) had the correct number of sigma bonds but only one pi bond.\n
\n\n Explain the general increase in trend in the first ionization energies of the period 3 elements, Na to Ar.\n
\n\n [2]\n
\n\n increasing number of protons\n
\n \n \n OR\n \n \n
\n increasing nuclear charge ✔\n
\n «atomic» radius/size decreases\n
\n \n \n OR\n \n \n
\n same number of shells/electrons occupy same shell\n
\n \n \n OR\n \n \n
\n similar shielding «by inner electrons» ✔\n
\n \n Outline the\n \n \n \n two\n \n \n \n distinct phases of this composite.\n \n
\n\n
\n\n [2]\n
\n\n carbon fibre reinforcing phase ✔\n
\n\n «in a»\n \n matrix\n \n phase of epoxy ✔\n
\n\n
\n \n Award\n \n [1 max]\n \n for “reinforcing phase «embedded» in a\n \n matrix\n \n ”.\n \n
\n Approximately 1% of the candidates this session attempted this option so feedback on this section by question is not possible due to the minimal number of scripts available for marking. The students that did attempt this option did not perform very well with several of them leaving some questions unanswered.\n
\n\n The equilibrium 2H\n \n 2\n \n (g) + N\n \n 2\n \n (g)\n \n N\n \n 2\n \n H\n \n 4\n \n (g) has an equilibrium constant,\n \n K\n \n , at 150 °C.\n
\n\n What is the equilibrium constant at 150 °C, for the reverse reaction?\n
\n\n N\n \n 2\n \n H\n \n 4\n \n (g)\n \n 2H\n \n 2\n \n (g) + N\n \n 2\n \n (g)\n
\n\n
\n A.\n \n K\n \n
\n B.\n \n K\n \n \n −1\n \n
\n\n C. −\n \n K\n \n
\n\n D. 2\n \n K\n \n
\n\n [1]\n
\n\n B\n
\n\n Which pairs of reactants could produce the following intermediate?\n
\n\n \n
\n \n
\n
\n\n A. I and II only\n
\n\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\n
\n\n State why NH\n \n 3\n \n is a Lewis base.\n
\n\n [1]\n
\n\n donates «lone/non-bonding» pair of electrons ✔\n
\n\n The 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.\n
\n\n \n What is the pH of an ammonia solution that has\n \n \n \n ?\n \n
\n\n \n A.\n \n
\n \n
\n \n B.\n \n
\n \n
\n \n C.\n \n
\n \n
\n \n D.\n \n \n
\n\n [1]\n
\n\n C\n
\n\n More than 86% of candidates could find a pH from a given hydroxide concentration.\n
\n\n Which is the enthalpy change of reaction, Δ\n \n H\n \n ?\n
\n\n \n
\n [1]\n
\n\n A\n
\n\n \n Which is correct for a reaction with a positive change in Gibbs free energy, ΔG\n \n θ\n \n ?\n \n
\n\n \n A. The formation of reactants is favoured.\n \n
\n\n \n B. The formation of products is favoured.\n \n
\n\n \n C. The reaction is at equilibrium.\n \n
\n\n \n D. The reaction is spontaneous.\n \n
\n\n [1]\n
\n\n A\n
\n\n The higher scoring candidates did better on identifying the direction of spontaneity given a positive Δ\n \n G\n \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 \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 Suggest what can be concluded about the gold atom from this experiment.\n
\n\n \n
\n [2]\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 Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.\n
\n\n \n
\n [1]\n
\n\n HCN has stronger dipole–dipole attraction ✔\n
\n\n
\n\n \n Do\n \n not\n \n accept reference to H-bonds.\n \n
\n\n This 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.\n
\n\n Explain\n \n two\n \n ways antiviral medications prevent the replication of viruses.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n «bind to receptors to» prevent virus entering cell ✓\n
\n alter genetic material of cell/cytoplasm «to prevent replication» ✓\n
\n block enzyme activity in cell «to prevent replication» ✓\n
\n prevent virus leaving cell «by inhibiting viral enzyme/neuraminidase» ✓\n
\n \n The formula\n \n q = mcΔT\n \n was used to calculate the energy released. The values used in the calculation were\n \n m\n \n = 25.00 g,\n \n c\n \n = 4.18 J g\n \n −1\n \n K\n \n −1\n \n .\n \n
\n\n \n State an assumption made when using these values for\n \n m\n \n and\n \n c\n \n .\n \n
\n\n \n \n \n
\n [2]\n
\n\n \n
\n
\n\n \n \n Note:\n \n \n \n \n Accept “copper(II) sulfate/zinc sulfate” for “solution”.\n \n \n
\n\n Again relatively few gained these marks for stating that it was assumed the density and specific heat of the solution were the same as water.\n
\n\n Saturated aqueous hydrogen sulfide has a concentration of 0.10 mol dm\n \n −3\n \n and a pH of 4.0. Demonstrate whether it is a strong or weak acid.\n
\n\n [1]\n
\n\n weak\n \n \n AND\n \n \n strong acid of this concentration/[H\n \n +\n \n ] = 0.1 mol dm\n \n −3\n \n would have pH = 1\n
\n \n \n OR\n \n \n
\n weak\n \n \n AND\n \n \n [H\n \n +\n \n ] = 10\n \n −4\n \n < 0.1 «therefore only fraction of acid dissociated» ✔\n
\n Which sequence has the oxides arranged in order of increasing acidity?\n
\n
\n
\n A. Na\n \n 2\n \n O < Al\n \n 2\n \n O\n \n 3\n \n < SO\n \n 3\n \n
\n\n B. Al\n \n 2\n \n O\n \n 3\n \n < SO\n \n 3\n \n < Na\n \n 2\n \n O\n
\n\n C. SO\n \n 3\n \n < Na\n \n 2\n \n O <Al\n \n 2\n \n O\n \n 3\n \n
\n\n D. SO\n \n 3\n \n < Al\n \n 2\n \n O\n \n 3\n \n < Na\n \n 2\n \n O\n
\n\n [1]\n
\n\n A\n
\n\n The observed specific optical rotation, [α], of a compound is +7.00 °. What is the specific optical rotation of a racemate of this compound?\n
\n\n
\n A. −7.00 °\n
\n B. 0.00 °\n
\n\n C. +7.00 °\n
\n\n D. +14.00 °\n
\n\n [1]\n
\n\n B\n
\n\n \n The uncertainty of the 100.0cm\n \n 3\n \n volumetric flask used to make the solution was ±0.6cm\n \n 3\n \n .\n \n
\n\n \n Calculate the maximum percentage uncertainty in the mass of NaHCO\n \n 3\n \n so that the concentration of the solution is correct to ±1.0 %.\n \n
\n\n [1]\n
\n\n \n «1.0 – 0.6 = ± » 0.4 «%»\n \n [✔]\n \n \n
\n\n Mixed responses, more attention should be given to this simple calculation which is straightforward and should be easy as required for IA reports.\n
\n\n State, with a reason, the effect of an increase in temperature on the position of this equilibrium.\n
\n\n [1]\n
\n\n «shifts» left/towards reactants\n \n \n AND\n \n \n «forward reaction is» exothermic/ΔH is negative ✔\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 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 Which alcohol is\n \n least\n \n soluble in water?\n
\n\n
\n A. CH\n \n 3\n \n OH\n
\n B. CH\n \n 3\n \n CH\n \n 2\n \n OH\n
\n\n C. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n OH\n
\n\n D. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n OH\n
\n\n [1]\n
\n\n D\n
\n\n Combustion of coal emits particulates into the atmosphere.\n
\n\n
\n\n reflects «sun» light ✓\n
\n\n
\n\n \n Accept “results in global dimming”\n
\n \n OR\n \n
\n “reduces the amount of energy reaching the Earth”\n
\n \n OR\n \n
\n “acts as nucleation points for cloud formation”.\n \n
\n \n Do\n \n not\n \n accept answers that only indicate increases in global temperatures.\n \n
\n\n Which functional groups are present in this molecule?\n
\n\n \n
\n
\n A. amino, alkoxy, ester\n
\n B. ether, carboxyl, amino\n
\n\n C. carboxyl, alkoxy, ester\n
\n\n D. ester, amino, carboxyl\n
\n\n [1]\n
\n\n A\n
\n\n What is the enthalpy change of the following reaction?\n
\n\n CH\n \n 2\n \n CHCH\n \n 2\n \n CH\n \n 3\n \n + HBr → CH\n \n 3\n \n CHBrCH\n \n 2\n \n CH\n \n 3\n \n
\n\n \n
\n A. –119.6 kJ\n
\n\n B. +119.6 kJ\n
\n\n C. –119.8 kJ\n
\n\n D. +119.8 kJ\n
\n\n [1]\n
\n\n C\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 «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 Write the half-equation for the reduction of hydrogen peroxide to water in acidic solution.\n
\n\n [1]\n
\n\n H\n \n 2\n \n O\n \n 2\n \n (aq) + 2H\n \n +\n \n (aq) + 2e\n \n −\n \n → 2H\n \n 2\n \n O(l) ✔\n
\n\n Which is the correct equation for the electrolysis of molten sodium chloride?\n
\n
\n
\n A. 2NaCl (l) → 2Na (l) + Cl\n \n 2\n \n (g)\n
\n\n B. 2NaCl (s) → 2Na (s) + Cl\n \n 2\n \n (g)\n
\n\n C. 2NaCl (l) → 2Na (s) + Cl\n \n 2\n \n (g)\n
\n\n D. 2NaCl (aq) → 2Na (s) + Cl\n \n 2\n \n (g)\n
\n\n [1]\n
\n\n A\n
\n\n The first eight successive ionization energies for an element are shown. In which group is the element?\n
\n\n \n
\n Copyright Prof Mark Winter.\n
\n\n A. 6\n
\n\n B. 7\n
\n\n C. 8\n
\n\n D. 17\n
\n\n [1]\n
\n\n A\n
\n\n State the type of spectroscopy that could be used to determine their relative abundances.\n
\n\n [1]\n
\n\n mass «spectroscopy»/MS ✔\n
\n\n Predict, giving a reason, the major product of reaction between but-1-ene and steam.\n
\n\n [2]\n
\n\n CH\n \n 3\n \n CH\n \n 2\n \n CH(OH)CH\n \n 3\n \n ✔\n
\n\n «secondary» carbocation/CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n +\n \n CH\n \n 3\n \n more stable ✔\n
\n\n
\n\n \n Do\n \n not\n \n accept “Markovnikov’s rule” without reference to carbocation stability.\n \n
\n\n Product 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.\n
\n\n Deduce the order of reaction with respect to hydrogen.\n
\n\n [1]\n
\n\n first order ✔\n
\n\n 4(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.\n
\n\n \n Dinitrogen monoxide in the stratosphere is converted to nitrogen monoxide, NO (g).\n \n
\n\n \n Write\n \n two\n \n equations to show how NO (g) catalyses the decomposition of ozone.\n \n
\n\n [2]\n
\n\n \n NO (g) + O\n \n 3\n \n (g) → NO\n \n 2\n \n (g) + O\n \n 2\n \n (g)\n \n [✔]\n \n
\n NO\n \n 2\n \n (g) + O\n \n 3\n \n (g)\n \n →\n \n NO (g) + 2O\n \n 2\n \n (g)\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note\n \n : Ignore radical signs.\n \n \n
\n\n \n \n Accept equilibrium arrows.\n \n \n
\n\n \n \n Award\n \n [1 max]\n \n for NO\n \n 2\n \n (g) + O (g)\n \n →\n \n NO (g) + O\n \n 2\n \n (g).\n \n \n
\n\n Many candidates recalled the first equation for NO catalyzed decomposition of ozone only. Some considered other radical species.\n
\n\n \n Which series represents atoms in order of decreasing atomic radius?\n \n
\n\n \n A. N > C > Be > Mg\n \n
\n\n \n B. Mg > N > C > Be\n \n
\n\n \n C. Be > C > N > Mg\n \n
\n\n \n D. Mg > Be > C > N\n \n
\n\n [1]\n
\n\n D\n
\n\n Another well answered question with 75 % correctly ranking elements according to atomic radii. One G2 form stated that this question was repeated in paper 2.\n
\n\n Which is the product when but-1-yne reacts with excess hydrogen gas?\n
\n\n
\n A. But-1-ene\n
\n B. Butane\n
\n\n C. But-2-ene\n
\n\n D. No reaction\n
\n\n [1]\n
\n\n B\n
\n\n \n Which molecule is most polar?\n \n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n 63% of the candidates could identify CHF\n \n 3\n \n as more polar than CClF\n \n 3\n \n . A surprising number selected either CCl\n \n 4\n \n or CF\n \n 4\n \n as the most polar.\n
\n\n \n What is the oxidation state of the metal ion and charge of the complex ion in [Co(NH\n \n 3\n \n )\n \n 4\n \n Cl\n \n 2\n \n ]Cl?\n \n
\n\n \n \n \n
\n [1]\n
\n\n C\n
\n\n This was one of the most challenging questions on the paper and it discriminated well between high scoring and low scoring candidates. 57 % of the candidates were able to use the formula of the compound to deduce the oxidation state of the metal ion and the charge of the complex ion. The most commonly chosen distractor was B where the charge of the complex ion was correct but the charge of the metal ion was not. Some teachers commented that the question was challenging but reasonable.\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 \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 \n Determine the mole ratio of S\n \n 2\n \n O\n \n 3\n \n \n 2−\n \n to O\n \n 2\n \n , using the balanced equations.\n \n
\n\n [1]\n
\n\n \n 4 : 1 ✔\n \n
\n\n \n Which class of compound is formed when a ketone is reduced?\n \n
\n\n \n A. primary alcohol\n \n
\n\n \n B. secondary alcohol\n \n
\n\n \n C. ether\n \n
\n\n \n D. carboxylic acid\n \n
\n\n [1]\n
\n\n B\n
\n\n 85 % of the candidates identified the secondary alcohol as the product of the reduction of a ketone. The other three distractors (primary alcohol, ether and carboxylic acid) were chosen almost equally by the remaining candidates.\n
\n\n Photosynthesis enables green plants to store energy from sunlight as glucose.\n
\n\n
\n\n 6CO\n \n 2\n \n (g) + 6H\n \n 2\n \n O (l) → C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n (aq) + 6O\n \n 2\n \n (g) ✓\n
\n\n A reaction has an activation energy of 40 kJ mol\n \n −1\n \n and an enthalpy change of −60 kJ mol\n \n −1\n \n .\n
\n\n Which potential energy diagram illustrates this reaction?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n Very well answered. 82% of the candidates selected the potential energy diagram with the correct activation energy and enthalpy change to represent the reaction.\n
\n\n \n Which is\n \n not\n \n a source of oxides of sulfur and nitrogen?\n \n
\n\n \n A. burning coal\n \n
\n\n \n B. internal combustion engines\n \n
\n\n \n C. burning methane\n \n
\n\n \n D. volcanic eruptions\n \n
\n\n [1]\n
\n\n C\n
\n\n One teacher thought this was a tricky question if candidates did not think of impurities in the combustion. However, 64 % of candidates answered correctly.\n
\n\n \n What are the units of the rate constant,\n \n \n \n \n \n , if the rate equation is\n \n \n \n ?\n \n
\n\n \n A.\n \n \n \n
\n \n
\n \n B.\n \n \n \n
\n \n
\n \n C.\n \n \n \n
\n \n
\n \n D.\n \n \n
\n\n [1]\n
\n\n C\n
\n\n Nearly three quarters of candidates could identify the units of a rate constant for a 3rd order reaction.\n
\n\n Evaluate whether this, rather than the loss of product, could explain the yield found in (b)(iii).\n
\n\n [1]\n
\n\n yes\n
\n \n \n AND\n \n \n
\n «each Mg combines with\n \n N, so» mass increase would be 14x\n \n which is less than expected increase of 16x\n
\n \n \n OR\n \n \n
\n 3 mol Mg would form 101g of Mg\n \n 3\n \n N\n \n 2\n \n but would form 3 x MgO = 121 g of MgO\n
\n \n \n OR\n \n \n
\n 0.2614 mol forms 10.536 g of MgO, but would form 8.796 g of Mg\n \n 3\n \n N\n \n 2\n \n ✔\n
\n
\n\n \n Accept Yes\n \n AND\n \n “the mass of N/N\n \n 2\n \n that combines with each g/mole of Mg is lower than that of O/O\n \n 2\n \n ”\n \n
\n\n \n Accept YES\n \n AND\n \n “molar mass of nitrogen less than of oxygen”.\n \n
\n\n This proved to be a very difficult question to answer in the quantitative manner required, with hardly any correct responses.\n
\n\n \n Which species is a Lewis acid but\n \n \n \n not\n \n \n \n a Brønsted–Lowry acid?\n \n
\n\n \n A.\n \n \n \n
\n \n
\n \n B.\n \n \n \n
\n \n
\n \n C.\n \n
\n \n
\n \n D.\n \n \n
\n\n [1]\n
\n\n A\n
\n\n Vast majority of candidates understood the difference between Lewis and Brønsted Lowry acids.\n
\n\n \n Mass spectrometry analysis of a sample of iron gave the following results:\n \n
\n\n \n \n \n
\n \n \n Calculate the relative atomic mass, A\n \n r\n \n , of this sample of iron to two decimal places.\n \n \n
\n\n [2]\n
\n\n \n «\n \n A\n \n \n r\n \n =» 54 × 0.0584 + 56\n \n ×\n \n 0.9168 + 57\n \n ×\n \n 0.0217 + 58\n \n ×\n \n 0.0031\n
\n \n \n OR\n \n \n
\n «\n \n A\n \n \n r\n \n =» 55.9111\n \n [✔]\n \n \n
\n \n «\n \n A\n \n \n r\n \n =» 55.91\n \n [✔]\n \n \n
\n\n \n \n Notes:\n \n \n
\n\n \n \n Award [2] for correct final answer.\n
\n Do not accept data booklet value (55.85).\n \n \n
\n Calculation 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.\n
\n\n Outline, giving a reason, the effect of a catalyst on a reaction.\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 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 Which compound has a chiral carbon?\n
\n
\n
\n A. Bromoethane\n
\n\n B. 2-bromopropane\n
\n\n C. 2-bromobutane\n
\n\n D. 3-bromopentane\n
\n\n [1]\n
\n\n C\n
\n\n A well answered question, with over 83% deducing the name of the compound that contains a chiral carbon. The numbers choosing the distractors were all quite similar.\n
\n\n \n The uncertainty of the 100.0cm\n \n 3\n \n volumetric flask used to make the solution was ±0.6cm\n \n 3\n \n .\n \n
\n\n \n Calculate the maximum percentage uncertainty in the mass of NaHCO\n \n 3\n \n so that the concentration of the solution is correct to ±1.0 %.\n \n
\n\n [1]\n
\n\n \n «1.0 – 0.6 = ± » 0.4 «%»\n \n [✔]\n \n \n
\n\n Mixed responses, more attention should be given to this simple calculation which is straightforward and should be easy as required for IA reports.\n
\n\n Explain in terms of their metallic bonding why aluminium, Al and not calcium, Ca, can be used to make tent frames.\n
\n\n [3]\n
\n\n metallic bonding in Al is stronger ✓\n
\n\n
\n charge on cation /Al\n \n 3+\n \n is higher\n
\n \n \n OR\n \n \n
\n more valence electrons «than in Ca» ✓\n
\n
\n ionic radius of Al\n \n 3+\n \n is smaller ✓\n
\n Which statement is correct about configurational isomers?\n
\n\n
\n A. Configurational isomers can only be interconverted by breaking and reforming bonds.\n
\n B. Configurational isomers have different molecular formulas but the same structural formulas.\n
\n\n C. Configurational isomers are not distinct compounds.\n
\n\n D. Configurational isomers always have identical physical properties.\n
\n\n [1]\n
\n\n A\n
\n\n \n State the nuclear symbol notation,\n \n \n \n , for iron-54.\n \n
\n\n [1]\n
\n\n \n \n \n \n \n \n [\n \n ✔\n \n ]\n \n
\n\n The nuclear symbol notation was generally correct. However, some students swapped atomic and mass numbers and hence lost the mark.\n
\n\n \n Which molecule has an enantiomer?\n \n
\n\n \n A.\n \n \n \n
\n \n
\n \n B.\n \n \n \n
\n \n
\n \n C.\n \n \n \n
\n \n
\n \n D.\n \n \n
\n\n [1]\n
\n\n D\n
\n\n The majority of candidates could correctly identify the molecule which has an enantiomer.\n
\n\n \n What is the ground state electron configuration of an atom of chromium, Cr (Z = 24)?\n \n
\n\n \n A. [Ar]3d\n \n 6\n \n \n
\n\n \n B. [Ar]4s\n \n 2\n \n 3d\n \n 4\n \n \n
\n\n \n C. [Ar]4s\n \n 1\n \n 3d\n \n 5\n \n \n
\n\n \n D. [Ar]4s\n \n 2\n \n 4p\n \n 4\n \n \n
\n\n [1]\n
\n\n C\n
\n\n Some candidates had ground state configuration of Cr as 4s\n \n 2\n \n 3d\n \n 4\n \n rather than 4s\n \n 1\n \n 3d\n \n 5\n \n
\n\n Magnesium can be produced by the electrolysis of molten magnesium chloride.\n
\n\n Write the half-equation for the formation of magnesium.\n
\n\n [1]\n
\n\n Mg\n \n 2+\n \n + 2 e\n \n -\n \n → Mg ✔\n
\n\n
\n\n \n Do\n \n not\n \n penalize missing charge on electron.\n \n
\n\n \n Accept equation with equilibrium arrows.\n \n
\n\n Unfortunately, only 40% of the students could write this quite straightforward half equation.\n
\n\n \n The following equation represents the dissociation of water at 25 °C.\n \n
\n\n \n 2H\n \n 2\n \n O (l)\n \n \n \n H\n \n 3\n \n O\n \n +\n \n (aq) + OH\n \n −\n \n (aq)\n \n ΔH\n \n = +56 kJ\n \n
\n\n \n Which changes occur as the temperature increases?\n \n
\n\n \n A. [\n \n H\n \n \n 3\n \n \n O\n \n \n +\n \n ] increases and pH will decrease.\n \n
\n\n \n B. [\n \n H\n \n \n 3\n \n \n O\n \n \n +\n \n ] decreases and pH will increase.\n \n
\n\n \n C. [\n \n H\n \n \n 3\n \n \n O\n \n \n +\n \n ] increases and pH will increase.\n \n
\n\n \n D. [\n \n H\n \n \n 3\n \n \n O\n \n \n +\n \n ] decreases and pH will decrease.\n \n
\n\n [1]\n
\n\n A\n
\n\n This is one of the more challenging questions on the paper. 62 % of the candidates obtained the correct answer. The most commonly chosen distractor was C, where the increase in the concentration of H\n \n 3\n \n O\n \n +\n \n was recognized by applying Le Chatelier’s principle but the effect on pH was incorrect.\n
\n\n What is the overall charge,\n \n , of the chromium (III) complex?\n
\n\n \n
\n\n A. 0\n
\n\n B. 1+\n
\n\n C. 2−\n
\n\n D. 3+\n
\n\n [1]\n
\n\n B\n
\n\n \n Identify the experiment with the highest rate of lead dissolving.\n \n
\n\n [1]\n
\n\n \n 6\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note\n \n : Accept “orange juice”.\n \n \n
\n\n Most candidates did well on this question, identifying the correct experiment by number or beverage.\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 \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 Write an equation that shows how sulfur dioxide can produce acid rain.\n
\n\n [1]\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 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 State the formula of its conjugate base.\n
\n\n [1]\n
\n\n HS\n \n −\n \n ✔\n
\n\n \n To 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.\n \n
\n\n [1]\n
\n\n \n \n Any one of:\n \n
\n copper(II) sulfate\n \n \n AND\n \n \n mass/amount of zinc is independent variable/being changed\n
\n \n \n OR\n \n \n
\n copper(II) sulfate\n \n \n AND\n \n \n with zinc in excess there is no independent variable «as amount of copper(II) sulfate is fixed»\n \n [✔]\n \n \n
\n \n copper(II) sulfate\n \n \n AND\n \n \n having excess zinc will not yield different results in each trial\n \n [✔]\n \n \n
\n\n \n zinc\n \n \n AND\n \n \n results can be used to see if amount of zinc affects temperature rise «so this can be allowed for»\n \n \n \n [✔]\n \n \n
\n\n \n zinc\n \n \n AND\n \n \n reduces variables/keeps the amount reacting constant\n \n \n \n [✔]\n \n \n
\n\n Many candidates struggled to explain their choice of which reagent should be in excess. This question proved quite difficult with many candidates seeming to confuse independent and dependent variables.\n
\n\n Which statement is correct about the electrolysis of molten lead(II) bromide, PbBr\n \n 2\n \n ?\n
\n\n
\n A. Br\n \n −\n \n ions accept electrons at the cathode (negative electrode).\n
\n B. Pb\n \n 2+\n \n ions accept electrons at the anode (positive electrode).\n
\n\n C. Br\n \n −\n \n ions lose electrons at the anode (positive electrode).\n
\n\n D. Pb\n \n 2+\n \n ions lose electrons at the cathode (negative electrode).\n
\n\n [1]\n
\n\n C\n
\n\n \n Distinguish between a weak and strong acid.\n \n
\n\n \n Weak acid:\n \n
\n\n \n Strong acid:\n \n
\n\n [1]\n
\n\n \n \n Weak acid:\n \n partially dissociated/ionized «in aqueous solution/water»\n
\n \n \n AND\n \n \n
\n \n Strong acid\n \n : «assumed to be almost» completely/100 % dissociated/ionized «in aqueous solution/water»\n \n [✔]\n \n \n
\n As expected, many candidates were able to distinguish between strong and weak acids; some candidates referred to “dissolve” rather than dissociate.\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 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 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
\n\n \n Draw the repeating unit of polyphenylethene.\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 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 \n 100.0cm\n \n 3\n \n of soda water contains 3.0 × 10\n \n −2\n \n g NaHCO\n \n 3\n \n .\n \n
\n\n \n Calculate the concentration of NaHCO\n \n 3\n \n in mol dm\n \n −3\n \n .\n \n
\n\n [2]\n
\n\n \n «molar mass of NaHCO\n \n 3\n \n =» 84.01 «g mol\n \n -1\n \n »\n \n [✔]\n \n \n
\n\n \n «concentration =\n \n \n \n =» 3.6 × 10\n \n –3\n \n «mol dm\n \n -3\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 This is another stoichiometry question that most candidates were able to solve well, with occasional errors when calculating\n \n M\n \n \n r\n \n of hydrogen carbonate.\n
\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
\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 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 Deduce the number of signals that you would expect in the\n \n 1\n \n H NMR spectrum of nitrobenzene and the relative areas of these.\n
\n\n \n
\n [2]\n
\n\n \n Number of signals\n \n : three/3 ✔\n
\n\n \n Relative areas\n \n : 2 : 2 : 1 ✔\n
\n\n Deducing 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.\n
\n\n Which energy profile diagram represents an exothermic S\n \n N\n \n 1 reaction?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n \n Describe 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.\n \n
\n\n [2]\n
\n\n \n place «pieces of» Re into each solution\n \n [✔]\n \n \n
\n\n \n if Re reacts/is coated with metal, that metal is less reactive «than Re»\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept other valid observations such as “colour of solution fades” or “solid/metal appears” for “reacts”.\n \n \n
\n\n Many 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.\n
\n\n \n In 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 \n
\n\n \n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5958/4bi_1.PNG\")
\n \n
\n \n \n The data for the first trial is given below.\n \n \n
\n\n \n \n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5959/4bi_2.PNG\")
\n \n \n
\n \n \n \n Plot a graph on the axes below and from it determine the average rate of\n
\n formation of oxygen gas in cm\n \n 3\n \n O\n \n 2\n \n (g) s\n \n −1\n \n .\n \n \n \n
\n \n \n \n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5960/4bi_3.PNG\")
\n \n \n \n
\n \n \n \n \n Average rate of reaction:\n \n \n \n \n
\n\n [3]\n
\n\n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5961/4bi_m.PNG\")
\n
\n \n points correctly plotted\n \n [✔]\n \n \n
\n\n \n best fit line\n \n \n AND\n \n \n extended through (to) the origin\n \n [✔]\n \n \n
\n\n \n \n Average rate of reaction:\n \n
\n «slope (gradient) of line =» 0.022 «cm\n \n 3\n \n O\n \n 2\n \n (g) s\n \n −1\n \n »\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note\n \n : Accept range 0.020–0.024cm\n \n 3\n \n O\n \n 2\n \n (g) s\n \n −1\n \n .\n \n \n
\n\n Most candidates could plot a best fit line and find the slope to calculate an average rate of reaction.\n
\n\n Write the rate expression for this reaction.\n
\n\n [1]\n
\n\n rate =\n \n k\n \n [N\n \n 2\n \n O\n \n 5\n \n ] ✔\n
\n\n \n Which is correct?\n \n
\n\n \n
\n A. Mixtures are either homogeneous or heterogeneous and their chemical properties are an average of the individual component properties.\n \n
\n \n B. Mixtures are never heterogeneous and their chemical properties are an average of the individual component properties.\n \n
\n\n \n C. Mixtures are either homogeneous or heterogeneous and the components retain their individual chemical properties.\n \n
\n\n \n D. Mixtures are never homogeneous and the components retain their individual chemical properties.\n \n
\n\n [1]\n
\n\n C\n
\n\n Which are isomers of C\n \n 5\n \n H\n \n 12\n \n ?\n
\n\n \n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\n
\n\n It was pleasing to see this question well answered despite the use of skeletal formulas. 74% of the candidates identified the isomers of C\n \n 5\n \n H\n \n 12\n \n correctly.\n
\n\n Which substance is most likely to be ionic?\n
\n\n \n
\n [1]\n
\n\n D\n
\n\n Which species has the same electron configuration as argon?\n
\n\n A. Br\n \n −\n \n
\n\n B. Ca\n \n 2+\n \n
\n\n C. Al\n \n 3+\n \n
\n\n D. Si\n \n 4+\n \n
\n\n [1]\n
\n\n B\n
\n\n Draw an arrow, labelled\n \n X\n \n , to represent the electron transition for the ionization of a hydrogen atom in the ground state.\n
\n\n [1]\n
\n\n \n
\n upward arrow X\n \n \n AND\n \n \n starting at n = 1 extending to n = ∞ ✔\n
\n\n Only 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.\n
\n\n Which of the following is most likely to be a transition metal?\n
\n\n
\n \n
\n [1]\n
\n\n C\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 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 Along which series is the bond angle increasing?\n
\n\n A. NH\n \n 3\n \n H\n \n 2\n \n O CH\n \n 4\n \n
\n\n B. CH\n \n 4\n \n NH\n \n 3\n \n H\n \n 2\n \n O\n
\n\n C. H\n \n 2\n \n O NH\n \n 3\n \n CH\n \n 4\n \n
\n\n D. H\n \n 2\n \n O CH\n \n 4\n \n NH\n \n 3\n \n
\n\n
\n\n [1]\n
\n\n C\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 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 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
\n\n Suggest what can be concluded about the gold atom from this experiment.\n
\n\n \n
\n [2]\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 Suggest why many chemicals, including hydrogen peroxide, are kept in brown bottles instead of clear colourless bottles.\n \n
\n\n [1]\n
\n\n \n decomposes in light\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “sensitive to light”.\n \n \n
\n\n There 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.\n
\n\n \n Methane reacts with chlorine in sunlight.\n \n
\n\n \n CH\n \n 4\n \n (g) + Cl\n \n 2\n \n (g) → CH\n \n 3\n \n Cl (g) + HCl (g)\n \n
\n\n \n Which type of reaction occurs?\n \n
\n\n \n A. free-radical substitution\n \n
\n\n \n B. electrophilic substitution\n \n
\n\n \n C. nucleophilic substitution\n \n
\n\n \n D. electrophilic addition\n \n
\n\n [1]\n
\n\n A\n
\n\n 85 % of the candidates chose free-radical substitution as the type of reaction occurring between methane and chlorine in sunlight.\n
\n\n Calculate the entropy change, Δ\n \n S\n \n , in J K\n \n −1\n \n mol\n \n −1\n \n , for this reaction.\n
\n\n \n
\n
\n\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/.\n \n
\n\n [1]\n
\n\n «ΔS = 364.5 J K\n \n –1\n \n mol\n \n –1\n \n – (311.7 J K\n \n –1\n \n mol\n \n –1\n \n + 223.0 J K\n \n –1\n \n mol\n \n –1\n \n )=» –170.2 «J K\n \n –1\n \n mol\n \n –1\n \n » ✔\n
\n\n 2.67 g of lead (II) carbonate is decomposed by heating until constant mass.\n
\n\n PbCO\n \n 3\n \n (s) → PbO (s) + CO\n \n 2\n \n (g)\n
\n\n What is the final mass of solid?\n
\n\n A. 0.44 g\n
\n\n B. 2.23 g\n
\n\n C. 2.67 g\n
\n\n D. 3.11 g\n
\n\n [1]\n
\n\n B\n
\n\n 76% of candidates correctly identified final mass of solid in the decomposition reaction with little difficulty in using mental math to arrive at the answer.\n
\n\n \n The dotted line represents the volume of carbon dioxide evolved when excess calcium carbonate is added to hydrochloric acid.\n \n
\n\n \n Which graph represents the production of carbon dioxide when excess calcium carbonate is added to the same volume of hydrochloric acid of double concentration?\n \n
\n\n \n \n \n
\n [1]\n
\n\n D\n
\n\n Sketch the shape of one sigma (\n \n ) and one pi (\n \n ) bond.\n
\n\n \n
\n [2]\n
\n\n \n Sigma (\n \n ):\n \n
\n\n \n
\n
\n\n \n Pi (\n \n ):\n \n
\n\n \n
\n
\n\n \n Accept overlapping p-orbital(s) with both lobes of equal size/shape.\n \n
\n\n \n Shaded areas are not required in either diagram.\n \n
\n\n State the hybridization of the carbon atom in HCN.\n
\n\n [1]\n
\n\n sp ✔\n
\n\n Explain the mechanism of the reaction between 1-bromopropane, CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n Br, and aqueous sodium hydroxide, NaOH (aq), using curly arrows to represent the movement of electron pairs.\n
\n\n [4]\n
\n\n \n
\n curly arrow going from lone pair/negative charge on O in HO\n \n –\n \n to C ✔\n
\n\n curly arrow showing Br breaking ✔\n
\n\n representation of transition state showing negative charge, square brackets and partial bonds ✔\n
\n\n formation of organic product CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n OH\n \n \n AND\n \n \n Br– ✔\n
\n\n
\n\n \n Do not allow curly arrow originating on H in HO\n \n –\n \n .\n \n
\n\n \n Accept curly arrow either going from bond between C and Br to Br in 1-bromopropane or in the transition\n \n
\n \n state.\n \n
\n \n Do\n \n not\n \n penalize if HO and Br are not at 180° to each other.\n \n
\n\n \n Award\n \n [3 max]\n \n for S\n \n N\n \n 1 mechanism.\n \n
\n\n As 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.\n
\n\n What is the main reason for an increase in rate of reaction when the temperature is raised?\n
\n\n
\n A. A greater proportion of collisions are successful.\n
\n B. Particles collide more frequently.\n
\n\n C. The bonds in the reactants are weakened.\n
\n\n D. The activation energy of the reaction decreases.\n
\n\n [1]\n
\n\n A\n
\n\n Saturated aqueous hydrogen sulfide has a concentration of 0.10 mol dm\n \n −3\n \n and a pH of 4.0. Demonstrate whether it is a strong or weak acid.\n
\n\n [1]\n
\n\n weak\n \n \n AND\n \n \n strong acid of this concentration/[H\n \n +\n \n ] = 0.1 mol dm\n \n −3\n \n would have pH = 1\n
\n \n \n OR\n \n \n
\n weak\n \n \n AND\n \n \n [H\n \n +\n \n ] = 10\n \n −4\n \n < 0.1 «therefore only fraction of acid dissociated» ✔\n
\n Draw the Lewis (electron dot) structure of hydrogen sulfide.\n
\n\n [1]\n
\n\n \n \n \n OR\n
\n ✔\n \n \n
\n \n Accept any combination of lines, dots or crosses to represent electrons.\n \n
\n\n Which products could be obtained by heating isomers of C\n \n 3\n \n H\n \n 8\n \n O under reflux with acidified potassium dichromate (VI)?\n
\n\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n C\n
\n\n In which set are the salts arranged in order of increasing pH?\n
\n\n A. HCOONH\n \n 4\n \n < KBr < NH\n \n 4\n \n Br < HCOOK\n
\n\n B. KBr < NH\n \n 4\n \n Br < HCOOK < HCOONH\n \n 4\n \n
\n\n C. NH\n \n 4\n \n Br < HCOONH\n \n 4\n \n < KBr < HCOOK\n
\n\n D. HCOOK < KBr < HCOONH\n \n 4\n \n < NH\n \n 4\n \n Br\n
\n\n [1]\n
\n\n C\n
\n\n Not very well done in identifying the set of ionic compounds that were arranged in correct order of increasing pH, with no clear misconception based on the other choices.\n
\n\n \n Which compound can exist as cis- and trans-isomers?\n \n
\n\n \n \n \n
\n [1]\n
\n\n C\n
\n\n \n Cis\n \n and\n \n trans\n \n isomerism on cyclic alkanes was poorly answered. This had the lowest discriminatory index on the test and all incorrect answers were fairly evenly distributed.\n
\n\n A liquid was added to a graduated cylinder. What can be deduced from the graph?\n
\n\n \n
\n \n
\n [1]\n
\n\n B\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 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 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 Suggest\n \n two\n \n reasons why the penicillin side-chain is modified.\n
\n\n [2]\n
\n\n bacterial resistance «to older penicillin’s/antibiotics» ✓\n
\n\n prevent penicillinase/beta-lactamase/enzyme in bacterium to deactivate/open penicillin/beta-lactam ring ✓\n
\n\n
\n\n \n Accept “antibiotic resistant bacteria” but\n \n not\n \n “antibiotic resistance” for M1.\n \n
\n\n \n Accept “reduce allergic reactions from penicillin” for M2.\n \n
\n\n \n Award\n \n [1 max]\n \n for \"increased efficiency/bioavailability\"\n
\n \n OR\n \n
\n \"increased stability in GIT\".\n \n
\n \n Do\n \n not\n \n accept \"bacterial tolerance\".\n \n
\n\n \n Calculate the percentage, by mass, of rhenium in ReCl\n \n 3\n \n .\n \n
\n\n [2]\n
\n\n \n «\n \n M\n \n r\n \n \n ReCl\n \n 3\n \n = 186.21 + (3 × 35.45) =» 292.56\n \n [✔]\n \n \n
\n\n \n «100 ×\n \n \n \n =» 63.648 «%»\n \n [✔]\n \n \n
\n\n The majority of candidates calculated the percentage, by mass, of rhenium in ReCl\n \n 3\n \n correctly. Some rounding errors were seen that students should be more careful with.\n
\n\n \n Outline why ozone in the stratosphere is important.\n \n
\n\n [1]\n
\n\n \n absorbs\n \n UV/ultraviolet\n \n light «of longer wavelength than absorbed by O\n \n 2\n \n »\n \n [✔]\n \n \n
\n\n Candidates sometimes failed to identify how ozone works in chemical terms, referring to protects/deflects,\n \n i.e.\n \n , the consequence rather than the mechanism.\n
\n\n Calculate the Gibbs free energy change, Δ\n \n G\n \n \n ⦵\n \n , in kJ mol\n \n −1\n \n , for the reaction at 298 K. Use section 1 of the data booklet.\n
\n\n [1]\n
\n\n «Δ\n \n G\n \n \n ⦵\n \n = 53.0 kJ mol\n \n –1\n \n – (298K × 0.1665 kJ K\n \n –1\n \n mol\n \n –1\n \n ) =» 3.4 «kJ mol\n \n –1\n \n » ✔\n
\n\n \n Which species contains nitrogen with the highest oxidation state?\n \n
\n\n \n A. NO\n \n 3\n \n \n −\n \n \n
\n\n \n B. NO\n \n 2\n \n \n −\n \n \n
\n\n \n C. NO\n \n 2\n \n \n
\n\n \n D. N\n \n 2\n \n O\n \n
\n\n [1]\n
\n\n A\n
\n\n It is pleasing that 85 % of the candidates identified the species containing nitrogen with the highest oxidation state.\n
\n\n Which changes would increase the rate of an exothermic reaction?\n
\n\n
\n \n
\n [1]\n
\n\n A\n
\n\n This question had a poor discrimination index as it appears to have confused the high-scoring candidates. Half of the candidates selected an increase in temperature and a decrease in particle size to increase the rate of an exothermic reaction. The most commonly chosen distractor was a decrease in temperature and particle size.\n
\n\n \n Outline the\n \n \n \n two\n \n \n \n distinct phases of this composite.\n \n
\n\n
\n\n [2]\n
\n\n carbon fibre reinforcing phase ✔\n
\n\n «in a»\n \n matrix\n \n phase of epoxy ✔\n
\n\n
\n \n Award\n \n [1 max]\n \n for “reinforcing phase «embedded» in a\n \n matrix\n \n ”.\n \n
\n Approximately 1% of the candidates this session attempted this option so feedback on this section by question is not possible due to the minimal number of scripts available for marking. The students that did attempt this option did not perform very well with several of them leaving some questions unanswered.\n
\n\n What is the IUPAC name of the molecule shown?\n
\n\n \n
\n A. 2,4-dimethylhexane\n
\n\n B. 3,5-dimethylhexane\n
\n\n C. 2-methyl-4-ethylpentane\n
\n\n D. 2-ethyl-4-methylpentane\n
\n\n [1]\n
\n\n A\n
\n\n \n Deduce how the rate of reaction at\n \n t\n \n = 2 would compare to the initial rate.\n \n
\n\n [1]\n
\n\n \n half «of the initial rate»\n \n [✔]\n \n \n
\n\n
\n\n \n \n Note:\n \n \n Accept “lower/slower «than initial rate»”.\n \n \n
\n\n Though 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.\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 «\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 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 \n Where is the buffer region for the titration of a weak acid with a strong base?\n \n
\n\n \n \n \n
\n [1]\n
\n\n B\n
\n\n There was a mistake on this question and it had to be annulled (39 marks paper). Grade boundaries were lowered accordingly. The x-axis was incorrectly labelled as “volume of weak acid” instead of “volume of strong base”. We sincerely apologize for this mistake which will be corrected before publication.\n
\n\n \n Outline 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.\n \n
\n\n \n Reason:\n \n
\n\n \n Length:\n \n
\n\n [2]\n
\n\n \n resonance «structures»\n
\n \n \n OR\n \n \n
\n delocalization of «the double/pi bond» electrons ✔\n
\n 121 «pm» < length < 148 «pm» ✔\n \n
\n \n \n NOTE: Accept any length between these two values.\n \n \n
\n\n Write the equation for the reaction of Ca(OH)\n \n 2\n \n (aq) with hydrochloric acid, HCl (aq).\n
\n\n [1]\n
\n\n Ca(OH)\n \n 2\n \n (aq) + 2HCl (aq) → 2H\n \n 2\n \n O (l) + CaCl\n \n 2\n \n (aq) ✓\n
\n\n Saturated calcium hydroxide solution is used to test for carbon dioxide. Calculate the pH of a 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n solution of calcium hydroxide, a strong base.\n
\n\n [2]\n
\n\n \n \n Alternative 1:\n \n \n
\n\n [OH\n \n −\n \n ] = « 2 × 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n =» 0.0466 «mol dm\n \n −3\n \n » ✓\n
\n\n «[H\n \n +\n \n ] =\n \n = 2.15 × 10\n \n −13\n \n mol dm\n \n −3\n \n »\n
\n\n pH = « −log (2.15 × 10−13) =» 12.668 ✓\n
\n\n
\n\n \n \n Alternative 2:\n \n \n
\n\n [OH\n \n −\n \n ] =« 2 × 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n =» 0.0466 «mol dm\n \n −3\n \n » ✓\n
\n\n «pOH = −log (0.0466) = 1.332»\n
\n\n pH = «14.000 – pOH = 14.000 – 1.332 =» 12.668 ✓\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Award\n \n [1 max]\n \n for pH =12.367.\n \n
\n\n \n In which species does sulfur have the same oxidation state as in SO\n \n 3\n \n \n 2–\n \n ?\n
\n \n
\n \n A. S\n \n 2\n \n O\n \n 3\n \n \n 2–\n \n
\n \n
\n \n B. SO\n \n 4\n \n \n 2–\n \n
\n \n
\n \n C. H\n \n 2\n \n S\n
\n \n
\n \n D. SOCl\n \n 2\n \n \n
\n\n [1]\n
\n\n D\n
\n\n \n Identify the initiation step of the reaction and its conditions.\n \n
\n\n [2]\n
\n\n \n Br2 → 2Br•\n \n [✔]\n \n \n
\n\n \n «sun»light/UV/hv\n
\n \n \n OR\n \n \n
\n high temperature\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Do\n \n not\n \n penalize missing radical symbol on Br.\n \n \n
\n\n \n \n Accept “homolytic fission of bromine” for M1.\n \n \n
\n\n Drawing or describing the homolytic fission of bromine was generally done well.\n
\n\n \n Before its isolation, scientists predicted the existence of rhenium and some of its properties.\n \n
\n\n \n Suggest the basis of these predictions.\n \n
\n\n [2]\n
\n\n \n gap in the periodic table\n
\n \n \n OR\n \n \n
\n element with atomic number «75» unknown\n
\n \n \n OR\n \n \n
\n break/irregularity in periodic trends\n \n [✔]\n \n \n
\n \n «periodic table shows» regular/periodic trends «in properties»\n \n [✔]\n \n \n
\n\n This 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.\n
\n\n Outline, with reference to the reaction equation, why this sign for the entropy change is expected.\n
\n\n [1]\n
\n\n «forward reaction involves» decrease in number of moles «of gas» ✔\n
\n\n Average performance for sign of the entropy change expected for the reaction. Some answers were based on Δ\n \n G\n \n value rather than in terms of decrease in number of moles of gas or had no idea how to address the question.\n
\n\n \n Classify polybutadiene as either an addition or condensation polymer, giving a reason.\n \n
\n\n [1]\n
\n\n \n addition\n \n \n AND\n \n \n not two different functional groups reacting\n
\n \n \n OR\n \n \n
\n addition\n \n \n AND\n \n \n formed by breaking one bond of the carbon–carbon double bonds\n
\n \n OR\n \n
\n addition\n \n \n AND\n \n \n empirical formula of monomer equals empirical formula of polymer\n
\n \n OR\n \n
\n addition\n \n \n AND\n \n \n no atoms removed/all atoms accounted for/no loss of water/ammonia/inorganic by-product/small molecules\n
\n \n OR\n \n
\n addition\n \n \n AND\n \n \n atom economy/efficiency is 100 %\n
\n \n OR\n \n
\n addition\n \n \n AND\n \n \n there is only one «reaction» product\n \n [✔]\n \n \n
\n Some candidates failed to score the mark as they did not give a reason for classifying polybutadiene as an addition polymer.\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 \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 The 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 T\n \n 1\n \n and T\n \n 2\n \n , some made the two curves meet at high energy or even cross, and some did not have the correct relationship between the peaks of T\n \n 1\n \n and T\n \n 2\n \n .\n
\n\n \n Non-polar solvents can be toxic. Suggest a modification to the experiment which allows the evaporated solvent to be collected.\n \n
\n\n [1]\n
\n\n distillation «instead of evaporation» ✔\n
\n\n \n Accept “pass vapour through a condenser and collect liquid”.\n \n
\n\n \n Do\n \n not\n \n accept “condensation” without experimental details.\n \n
\n\n A 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.\n
\n\n \n Suggest a source of error in the procedure, assuming no human errors occurred and the balance was accurate.\n \n
\n\n [1]\n
\n\n \n surface area not uniform\n
\n \n \n \n NOTE: Accept “acids impure.\n \n
\n \n
\n \n \n \n OR\n \n \n
\n limestone pieces do not have same composition/source\n
\n \n NOTE: Accept “«limestone» contains impurities”.\n \n \n \n
\n \n
\n \n \n \n OR\n \n \n
\n limestone absorbed water «which increased mass»\n \n
\n \n \n \n OR\n \n \n
\n acid removed from solution when limestone removed\n
\n \n NOTE: Accept “loss of limestone when dried\"\n \n OR\n \n \"loss of limestone due to crumbling when removed from beaker”.\n \n
\n \n
\n \n \n \n OR\n \n \n
\n «some» calcium sulfate deposited on limestone lost\n \n
\n \n \n \n OR\n \n \n
\n pieces of paper towel may have stuck to limestone\n \n
\n \n \n \n OR\n \n \n
\n beakers not covered/evaporation\n \n
\n \n \n \n OR\n \n \n
\n temperature was not controlled ✔\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 \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 Majority of the candidates managed to get three marks in determining the density of the calcium.\n
\n\n \n Which has the strongest conjugate base?\n \n
\n\n \n A. HCOOH (K\n \n a\n \n = 1.8 × 10\n \n −4\n \n )\n \n
\n\n \n B. HNO\n \n 2\n \n (\n \n K\n \n \n a\n \n = 7.2 × 10\n \n −4\n \n )\n \n
\n\n \n C. HCN (\n \n K\n \n \n a\n \n = 6.2 × 10\n \n −10\n \n )\n \n
\n\n \n D. HIO\n \n 3\n \n (\n \n K\n \n \n a\n \n = 1.7 × 10\n \n −1\n \n )\n \n
\n\n [1]\n
\n\n C\n
\n\n Strength of conjugate bases appeared to be another good discriminator, with higher scoring candidates performing better.\n
\n\n Which gives the equation and cell potential of the spontaneous reaction?\n
\n\n \n
\n [1]\n
\n\n D\n
\n\n Which reagents and conditions are best for converting propan-1-ol into propanoic acid?\n
\n\n A. Reflux with acidified potassium dichromate (VI)\n
\n\n B. Reflux with LiAlH\n \n 4\n \n
\n\n C. Distil with acidified potassium dichromate (VI)\n
\n\n D. Distil with LiAlH\n \n 4\n \n
\n\n [1]\n
\n\n A\n
\n\n The vast majority of students identified the best reagents and conditions for converting propan-1-ol into propanoic acid.\n
\n\n Deduce the expression for the equilibrium constant,\n \n K\n \n \n c\n \n , for this equation.\n
\n\n [1]\n
\n\n \n ✔\n
\n\n Deducing the equilibrium constant expression for the given equation was done very well.\n
\n\n \n What can be deduced from the infrared (IR) spectrum of a compound?\n \n
\n\n \n A. Number of hydrogens\n \n
\n\n \n B. Number of hydrogen environments\n \n
\n\n \n C. Bonds present\n \n
\n\n \n D. Molar mass\n \n
\n\n [1]\n
\n\n C\n
\n\n 82 % of the candidates identified the bonds present as the information that can be deduced from an infrared spectrum. The most commonly chosen distractor was B (the number of hydrogen environments).\n
\n\n \n Deduce the products of the hydrolysis of a non-substituted phospholipid, where\n \n \n \n and\n \n \n \n represent long alkyl chains.\n \n
\n\n \n
\n [2]\n
\n\n \n
\n glycerol ✔\n
\n\n both fatty acids\n \n \n AND\n \n \n phosphoric acid ✔\n
\n\n \n Accept either names\n \n OR\n \n structures.\n \n
\n \n Accept “long chain carboxylic acid” for “fatty acid”.\n \n
\n \n Penalize once only if an incorrect name is given for a correct structure or vice-versa.\n \n
\n\n Not as well answered as expected. However, many candidates managed to score at least one point. Quite a few lost the only mark due to wrong linkage. Some students drew fatty acid structures with aldehydes or phosphoric acid with incorrect bond linkages in the structure (OH\n \n -\n \n ).\n
\n\n Outline why we need vitamins/micronutrients in our diets.\n
\n\n [1]\n
\n\n «mostly» not synthesized by body «and needed for proper growth/metabolism» ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “needed for proper growth/metabolism” alone.\n \n
\n\n What is produced when chlorobutane is treated with aqueous sodium hydroxide solution?\n
\n\n A. butane\n
\n\n B. butanoic acid\n
\n\n C. butanal\n
\n\n D. butan-1-ol\n
\n\n [1]\n
\n\n D\n
\n\n State\n \n two\n \n conditions necessary for a successful collision between reactants.\n
\n\n [1]\n
\n\n \n E\n \n ≥\n \n E\n \n \n a\n \n \n \n AND\n \n \n appropriate «collision» geometry/correct orientation ✔\n
\n\n Generally 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\n \n E\n \n a\n \n \n .\n
\n\n \n Which quantity is likely to be the most inaccurate due to the sources of error in this experiment?\n \n
\n\n \n A. Mass of ethanol burnt\n
\n \n
\n \n B. Molecular mass of ethanol\n
\n \n
\n \n C. Mass of water\n
\n \n
\n \n D. Temperature change\n \n
\n\n [1]\n
\n\n D\n
\n\n A student performed an experiment to find the melting point of sulfur, obtaining 118.0 °C. The literature value is 115.2 °C. What was the percentage error?\n
\n\n
\n A.\n \n
\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n State the hybridization of the carbon atom in HCN.\n
\n\n [1]\n
\n\n sp ✔\n
\n\n Draw an arrow, labelled\n \n Z\n \n , to represent the lowest energy electron transition in the visible spectrum.\n
\n\n [1]\n
\n\n \n
\n downward or upward arrow between n = 3 and n = 2 ✔\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 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 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 Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.\n
\n\n \n
\n [1]\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 \n Which substance will\n \n \n \n not\n \n \n \n produce copper(II) chloride when added to dilute hydrochloric acid?\n \n
\n\n \n A.\n \n
\n \n
\n \n B.\n \n \n \n
\n \n
\n \n C.\n \n \n \n
\n \n
\n \n D.\n \n \n
\n\n [1]\n
\n\n A\n
\n\n Poorly answered with less than ½ of candidates able to identify that copper metal would not produce\n
\n copper II chloride when added to dilute HCl.\n
\n Which species could be reduced to form NO\n \n 2\n \n ?\n
\n\n A. N\n \n 2\n \n
\n\n B. NO\n \n 3\n \n \n −\n \n
\n\n C. HNO\n \n 2\n \n
\n\n D. NO\n
\n\n [1]\n
\n\n B\n
\n\n 58% of the candidates were able to identify the species that could be reduced to form NO\n \n 2\n \n . The most commonly chosen distractor was HNO\n \n 2\n \n .\n
\n\n Which compound produces the following\n \n 1\n \n H NMR spectrum?\n
\n\n \n
\n \n [Spectral Database for Organic Compounds, SDBS. SDBS Compounds and Spectral Search. [graph] Available at:\n \n
\n \n https://sdbs.db.aist.go.jp [Accessed 3 January 2019].]\n \n
\n
\n A. propanal\n
\n B. propanone\n
\n\n C. propane\n
\n\n D. methlypropane\n
\n\n [1]\n
\n\n A\n
\n\n A majority of students identified the correct compound for the\n \n 1\n \n H NMR spectrum given.\n
\n\n \n What is the relative atomic mass,\n \n \n \n , of an element with this mass spectrum?\n \n
\n\n \n
\n \n A.\n \n
\n \n
\n \n B.\n \n
\n \n
\n \n C.\n \n
\n \n
\n \n D.\n \n \n
\n\n [1]\n
\n\n B\n
\n\n Well answered question with more than 70% of candidates able to find the RAM of an element from isotope relative abundances.\n
\n\n Plot the missing point on the graph and draw the best-fit line.\n
\n\n \n
\n [2]\n
\n\n \n
\n point correct ✔\n
\n\n straight line passing close to all points\n \n \n AND\n \n \n through origin ✔\n
\n\n \n
\n Accept free hand drawn line as long as attempt to be linear and meets criteria for M2.\n \n
\n What happens to the average kinetic energy, KE, of the particles in a gas when the absolute temperature is doubled?\n
\n\n \n
\n\n
\n A. Increases by a factor of 2\n
\n B. Decreases by a factor of 2\n
\n\n C. Increases by a factor of 4\n
\n\n D. Decreases by a factor of 4\n
\n\n [1]\n
\n\n A\n
\n\n State\n \n two\n \n principles of green (sustainable) chemistry in drug manufacturing, other than using less hazardous or toxic reactants.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n energy efficiency ✓\n
\n prevention of waste/recycling ✓\n
\n atom economy/more efficient processes ✓\n
\n safer/reduced use of solvents ✓\n
\n design for degradation ✓\n
\n more sustainable sourcing of reactants/feedstock ✓\n
\n
\n\n \n Accept specific examples.\n \n
\n\n Which compound contains both ionic and covalent bonds?\n
\n\n
\n A. CH\n \n 3\n \n COONa\n
\n B. CH\n \n 3\n \n COOH\n
\n\n C. K\n \n 2\n \n O\n
\n\n D. CaCl\n \n 2\n \n
\n\n [1]\n
\n\n A\n
\n\n \n Which is the electrophile in the nitration of benzene?\n \n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n B\n
\n\n The vast majority selected the nitronium ion as the electrophile in the nitration of benzene.\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 \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 Which value increases when the temperature of a reaction increases?\n
\n\n
\n A. Activation energy\n
\n B. Rate constant\n
\n\n C. Enthalpy of reaction\n
\n\n D. Equilibrium constant for exothermic reaction\n
\n\n [1]\n
\n\n B\n
\n\n Which is a correct alternative representation of this molecule?\n
\n
\n
\n \n
\n
\n A. C\n \n 3\n \n H\n \n 6\n \n Cl\n \n 2\n \n
\n B.\n \n
\n C. 2,2-dichlorobut-1-ene\n
\n\n D. CH\n \n 3\n \n CHClC(Cl)=CH\n \n 2\n \n
\n\n [1]\n
\n\n D\n
\n\n Which combination would create the strongest ionic bond?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n \n Dinitrogen monoxide in the stratosphere is converted to nitrogen monoxide, NO (g).\n \n
\n\n \n Write\n \n two\n \n equations to show how NO (g) catalyses the decomposition of ozone.\n \n
\n\n [2]\n
\n\n \n NO (g) + O\n \n 3\n \n (g) → NO\n \n 2\n \n (g) + O\n \n 2\n \n (g)\n \n [✔]\n \n
\n NO\n \n 2\n \n (g) + O\n \n 3\n \n (g)\n \n →\n \n NO (g) + 2O\n \n 2\n \n (g)\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note\n \n : Ignore radical signs.\n \n \n
\n\n \n \n Accept equilibrium arrows.\n \n \n
\n\n \n \n Award\n \n [1 max]\n \n for NO\n \n 2\n \n (g) + O (g)\n \n →\n \n NO (g) + O\n \n 2\n \n (g).\n \n \n
\n\n Many candidates recalled the first equation for NO catalyzed decomposition of ozone only. Some considered other radical species.\n
\n\n The exothermic reaction\n \n \n 2\n \n (g) + 3Cl\n \n 2\n \n (g)\n \n 2\n \n Cl\n \n 3\n \n (g) is at equilibrium in a fixed volume. What is correct about the reaction quotient,\n \n Q\n \n , and shift in position of equilibrium the instant temperature is raised?\n
\n\n
\n A.\n \n Q\n \n >\n \n K\n \n , equilibrium shifts right towards products.\n
\n B.\n \n Q\n \n >\n \n K\n \n , equilibrium shifts left towards reactants.\n
\n\n C.\n \n Q\n \n <\n \n K\n \n , equilibrium shifts right towards products.\n
\n\n D.\n \n Q\n \n <\n \n K\n \n , equilibrium shifts left towards reactants.\n
\n\n [1]\n
\n\n B\n
\n\n This question required thinking and reflection. The majority of candidates deduced correctly that the exothermic reaction would shift to the left when the temperature was increased, however, the relationship between Q and K was confusing. Only 37% of the candidates deduced that Q must be larger than the new K at the higher temperature in order for the reaction to shift to the left. The question did not discriminate well between high-scoring and low-scoring candidates.\n
\n\n \n Which combination causes the strength of metallic bonding to increase?\n \n
\n\n \n \n \n
\n [1]\n
\n\n D\n
\n\n A straight-forward question about factors affecting the strength of metallic bonding answered correctly by 86 % of candidates. The distractors were almost equally chosen by the rest of the candidates.\n
\n\n Iron 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.\n
\n\n Determine the specific heat capacity of iron, in J g\n \n −1\n \n K\n \n −1\n \n . Use section 1 of the data booklet.\n
\n\n [1]\n
\n\n specific heat capacity « =\n \n » = 0.45 «J g\n \n −1\n \n K\n \n −1\n \n » ✔\n
\n\n \n Determine the change in the average oxidation state of carbon.\n \n
\n\n \n From ethanol to ethanal:\n \n
\n\n \n From ethanol to carbon dioxide:\n \n
\n\n [2]\n
\n\n \n \n From ethanol to ethanal:\n \n
\n −2 to −1\n
\n \n \n OR\n \n \n
\n +1/increases by 1 ✔\n \n
\n \n \n NOTE: Do\n \n not\n \n accept “2− to 1−”.\n \n \n
\n\n \n \n From ethanol to carbon dioxide:\n \n
\n −2 to +4\n
\n \n \n OR\n \n \n
\n +6/increases by 6 ✔\n \n
\n \n \n NOTE: Do\n \n not\n \n accept “2− to 4+”.\n
\n \n \n
\n \n \n Do\n \n not\n \n penalize incorrect notation twice.\n \n \n
\n\n \n \n Penalize incorrect oxidation state value of carbon in ethanol once only.\n \n \n
\n\n \n State\n \n one\n \n factor considered when making green chemistry polymers.\n \n
\n\n [1]\n
\n\n \n \n Any one of:\n \n
\n high content of raw materials in product/high atom economy\n \n [✔]\n \n \n
\n \n use of low toxic chemicals/catalysts/materials/solvents\n \n \n \n [✔]\n \n \n
\n\n \n renewable feedstock/raw materials\n \n \n \n [✔]\n \n \n
\n\n \n use of renewable/clean/low carbon energy source\n \n \n \n [✔]\n \n \n
\n\n \n high safety standards\n \n \n \n [✔]\n \n \n
\n\n \n increase energy efficiency\n \n \n \n [✔]\n \n \n
\n\n \n waste recycling\n \n \n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept other reasonable answers.\n \n \n
\n\n Most candidates were able to state a factor considered when making green chemistry polymers.\n
\n\n Explain the general increase in trend in the first ionization energies of the period 3 elements, Na to Ar.\n
\n\n [2]\n
\n\n increasing number of protons\n
\n \n \n OR\n \n \n
\n increasing nuclear charge ✔\n
\n «atomic» radius/size decreases\n
\n \n \n OR\n \n \n
\n same number of shells/electrons occupy same shell\n
\n \n \n OR\n \n \n
\n similar shielding «by inner electrons» ✔\n
\n State the condensed electron configurations for Cr and Cr3\n \n +\n \n .\n
\n\n \n
\n [2]\n
\n\n \n Cr:\n \n
\n [Ar] 4s\n \n 1\n \n 3d\n \n 5\n \n ✓\n
\n \n
\n Cr\n \n 3+\n \n :\n \n
\n [Ar] 3d\n \n 3\n \n ✓\n
\n \n
\n Accept “[Ar] 3d\n \n 5\n \n 4s\n \n 1\n \n ”.\n \n
\n \n Accept “[Ar] 3d\n \n 3\n \n 4s\n \n 0\n \n ”.\n \n
\n\n \n Award\n \n [1 max]\n \n for two correct full electron configurations “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 5\n \n \n AND\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 3d\n \n 3\n \n ”.\n \n
\n\n \n Award\n \n [1 max]\n \n for 4s\n \n 1\n \n 3d\n \n 5\n \n \n AND\n \n 3d\n \n 3\n \n .\n \n
\n\n The graph shows the first six ionization energies of an element.\n
\n\n \n
\n \n [Ionization energies of the elements (data page) Available at:\n \n https://en.wikipedia.org/wiki/Ionization_energies_of_the_\n \n \n
\n \n \n elements_(data_page)\n \n Text is available under the Creative Commons Attribution-ShareAlike License 3.0 (CC BY-SA\n \n
\n \n 3.0)\n \n https://creativecommons.org/licenses/by-sa/3.0/deed.en.\n \n ]\n \n
\n
\n In which group is the element?\n
\n A. 13\n
\n\n B. 14\n
\n\n C. 15\n
\n\n D. 16\n
\n\n [1]\n
\n\n A\n
\n\n A good number of candidates recognized the element based on the graph showing the first six ionization of the element.\n
\n\n \n Which element is represented by the first eight successive ionization energies on the graph?\n \n
\n\n \n \n \n
\n \n \n A. Mg\n \n \n
\n\n \n \n B. S\n \n \n
\n\n \n \n C. Cl\n \n \n
\n\n \n \n D. Ar\n \n \n
\n\n [1]\n
\n\n B\n
\n\n 52 % of candidates could identify the element for successive ionization energies with no clear misconception based on the other choices.\n
\n\n In the electrolysis apparatus shown, 0.59 g of Ni is deposited on the cathode of the first cell.\n
\n\n \n
\n What is the mass of Ag deposited on the cathode of the second cell?\n
\n\n
\n A. 0.54 g\n
\n B. 0.59 g\n
\n\n C. 1.08 g\n
\n\n D. 2.16 g\n
\n\n [1]\n
\n\n D\n
\n\n A poorly answered question with low discrimination factor. Several teachers asked whether this question of \"cells in series\" was in the Chemistry Guide (see Topic 19.1, Electrochemical cells, on page 96) and it is likely that this topic may not have been covered in class by some teachers.\n
\n\n When 100 cm\n \n −3\n \n of 1.0 mol dm\n \n −3\n \n HCl is mixed with 100 cm\n \n −3\n \n of 1.0 mol dm\n \n −3\n \n NaOH, the temperature of the resulting solution increases by 5.0 °C. What will be the temperature change, in °C, when 50 cm\n \n −3\n \n of 2.0 mol dm\n \n −3\n \n HCl is mixed with 50 cm\n \n −3\n \n of 2.0 mol dm\n \n −3\n \n NaOH?\n
\n\n
\n A. 2.5\n
\n B. 5.0\n
\n\n C. 10\n
\n\n D. 20\n
\n\n [1]\n
\n\n C\n
\n\n Enthalpy of solution, enthalpy of hydration and lattice enthalpy are related in an energy cycle.\n
\n\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 Which is an f-block element?\n \n
\n\n \n A. Sc\n
\n \n
\n \n B. Sm\n
\n \n
\n \n C. Sn\n
\n \n
\n \n D. Sr\n \n
\n\n [1]\n
\n\n B\n
\n\n \n Which is\n \n not\n \n a requirement of the standard hydrogen electrode (SHE)?\n \n
\n\n \n A. V = 1 dm\n \n 3\n \n \n
\n\n \n B. p(H\n \n 2\n \n ) = 100 kPa\n \n
\n\n \n C. use of platinum as the electrode material\n \n
\n\n \n D. [H\n \n 3\n \n O\n \n +\n \n ] = 1 mol dm\n \n −3\n \n \n
\n\n [1]\n
\n\n A\n
\n\n This was a poorly answered question on the standard hydrogen electrode. Many candidates believed what was\n \n not\n \n required was a Pt electrode material\n
\n\n Which molecule is optically active?\n
\n\n A. 2,2-dichloropropane\n
\n\n B. 1,2-dichloropropane\n
\n\n C. 1,3-dichloropropane\n
\n\n D. 1,2,3-trichloropropane\n
\n\n [1]\n
\n\n B\n
\n\n Which pair of compounds are structural isomers?\n
\n\n
\n A. Propane and propene\n
\n B. Propanal and propanone\n
\n\n C. Propan-1-ol and propanal\n
\n\n D. Propyl propanoate and propanoic acid\n
\n\n [1]\n
\n\n B\n
\n\n Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.\n
\n\n [1]\n
\n\n allows them to explain the properties of different compounds/substances\n
\n \n \n OR\n \n \n
\n enables them to generalise about substances\n
\n \n \n OR\n \n \n
\n enables them to make predictions ✔\n
\n \n
\n Accept other valid answers.\n \n
\n \n Predict\n \n two\n \n other chemical properties you would expect rhenium to have, given its position in the periodic table.\n \n
\n\n [2]\n
\n\n \n \n Any two of:\n \n
\n variable oxidation states\n \n \n \n [✔]\n \n \n
\n \n forms complex ions/compounds\n \n \n \n [✔]\n \n \n
\n\n \n coloured compounds/ions\n \n \n \n [✔]\n \n \n
\n\n \n «para»magnetic compounds/ions\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept other valid responses related to its\n \n chemical\n \n metallic properties.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “catalytic properties”.\n \n \n
\n\n Generally well answered though some students suggested physical properties rather than chemical ones.\n
\n\n What can be deduced from the mass spectrum of CH\n \n 3\n \n COCH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n ?\n
\n\n \n
\n \n NIST Mass Spectrometry Data Center Collection (C) 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved. 2-Pentanone Mass Spectrum, MS Number 291264. [graph] Available at: https://webbook.nist.gov/cgi/cbook.cgi?ID=C107879&Units=SI&Mask=200#Mass-Spec2-pentanone [Accessed 4 May 2020]. source adapted.\n \n
\n\n
\n A. The molar mass is 43 g mol\n \n −1\n \n .\n
\n B. The atoms have many isotopes.\n
\n\n C. The most likely bond to break is C–C between carbons 2 and 3.\n
\n\n D. The signal with the largest mass is due to the oxidation of the ketone in the spectrometer.\n
\n\n [1]\n
\n\n C\n
\n\n Determine the mass, in g, of CaCO\n \n 3\n \n (s) produced by reacting 2.41 dm\n \n 3\n \n of 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n of Ca(OH)\n \n 2\n \n (aq) with 0.750 dm\n \n 3\n \n of CO\n \n 2\n \n (g) at STP.\n
\n\n [2]\n
\n\n «\n \n n\n \n \n Ca(OH)2\n \n = 2.41 dm\n \n 3\n \n × 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n =» 0.0562 «mol»\n \n \n AND\n \n \n
\n\n «\n \n n\n \n \n CO2\n \n =\n \n =» 0.0330 «mol» ✓\n
\n\n «CO\n \n 2\n \n is the limiting reactant»\n
\n\n «\n \n m\n \n \n CaCO3\n \n = 0.0330 mol × 100.09 g mol\n \n −1\n \n =» 3.30 «g» ✓\n
\n\n
\n\n \n Only award ECF for M2 if limiting reagent is used.\n \n
\n\n \n Accept answers in the range 3.30 - 3.35 «g».\n \n
\n\n What is the explanation for the high melting point of sodium chloride?\n
\n\n A. The covalent bond between sodium and chlorine atoms is strong.\n
\n\n B. Electrostatic attraction between sodium and chloride ions is strong.\n
\n\n C. Intermolecular forces in sodium chloride are strong.\n
\n\n D. Delocalized electrons cause strong bonding in sodium chloride.\n
\n\n [1]\n
\n\n B\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 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 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 Which of the following can act as a nucleophile?\n
\n\n I. Benzene\n
\n\n II. Water\n
\n\n III. Bromine\n
\n\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n C\n
\n\n Describe the bonding in this type of solid.\n
\n\n [2]\n
\n\n electrostatic attraction ✔\n
\n\n between oppositely charged ions/between Fe\n \n 2+\n \n and S\n \n 2−\n \n ✔\n
\n\n Which combination of ΔH\n \n 1\n \n , ΔH\n \n 2\n \n , and ΔH\n \n 3\n \n would give the enthalpy of the reaction?\n
\n\n CS\n \n 2\n \n (l) + 3O\n \n 2\n \n (g) → CO\n \n 2\n \n (g) + 2SO\n \n 2\n \n (g)\n
\n\n ΔH\n \n 1\n \n C (s) + O\n \n 2\n \n (g) → CO\n \n 2\n \n (g)\n
\n ΔH\n \n 2\n \n S (s) + O\n \n 2\n \n (g) → SO\n \n 2\n \n (g)\n
\n ΔH\n \n 3\n \n C (s) + 2S (s) → CS\n \n 2\n \n (l)\n
\n A. ΔH = ΔH\n \n 1\n \n + ΔH\n \n 2\n \n + ΔH\n \n 3\n \n
\n\n B. ΔH = ΔH\n \n 1\n \n + ΔH\n \n 2\n \n − ΔH\n \n 3\n \n
\n\n C. ΔH = ΔH\n \n 1\n \n + 2(ΔH\n \n 2\n \n ) + ΔH\n \n 3\n \n
\n\n D. ΔH = ΔH\n \n 1\n \n + 2(ΔH\n \n 2\n \n ) − ΔH\n \n 3\n \n
\n\n [1]\n
\n\n D\n
\n\n A very well answered question using Hess’s law. 85% of the candidates used the enthalpies given to calculate the unknown enthalpy of reaction.\n
\n\n Why does a reaction for a sample of gases, at constant temperature, occur faster at higher pressure?\n
\n\n A. Collisions are more frequent.\n
\n\n B. Collisions are more energetic.\n
\n\n C. High pressure lowers activation energy.\n
\n\n D. The reaction is more exothermic at high pressure.\n
\n\n [1]\n
\n\n A\n
\n\n A very well-answered question in which 90% of the candidates related the effect of higher pressure on the rate of reaction of gaseous reactants to the collision theory correctly.\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\n \n one\n \n disadvantage of tidal power.\n \n
\n\n \n Advantage:\n \n
\n\n \n Disadvantage:\n \n
\n\n [2]\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 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 generalisations.\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 Many candidates performed well on this question, especially when identifying an advantage of tidal power. The candidates who struggled tended to either give vague or journalistic answers especially for the disadvantage of tidal power.\n
\n\n \n Which compound\n \n cannot\n \n undergo addition polymerization?\n \n
\n\n \n \n \n
\n [1]\n
\n\n C\n
\n\n 30 g of an organic compound produces 44 g CO\n \n 2\n \n and 18 g H\n \n 2\n \n O as the only combustion products. Which of the following is the empirical formula for this compound?\n
\n\n \n M\n \n \n r\n \n CO\n \n 2\n \n = 44\n \n M\n \n \n r\n \n H\n \n 2\n \n O = 18\n
\n\n A. CH\n \n 2\n \n
\n\n B. CH\n \n 3\n \n
\n\n C. CHO\n
\n\n D. CH\n \n 2\n \n O\n
\n\n [1]\n
\n\n D\n
\n\n 63% of the candidates determined the empirical formula of the organic compound correctly using the masses of the sample and the combustion products.\n
\n\n \n Suggest one source of error in the experiment, excluding faulty apparatus and human error, that would lead to the following:\n \n
\n\n \n
\n [2]\n
\n\n \n Experimental mass greater than actual mass of oil in crisps:\n \n
\n other substances «in the crisps» are soluble in the solvent\n
\n \n \n OR\n \n \n
\n not all the solvent evaporates ✔\n
\n \n Experimental mass less than actual mass of oil in crisps:\n \n
\n not all oil dissolved/extracted ✔\n
\n \n Accept “oil evaporated”\n \n OR\n \n “oil burned/decomposed”\n \n OR\n \n “oil absorbed by the filter”\n \n OR\n \n “assumption «all oil dissolved» was wrong” for M2.\n \n
\n\n \n Do\n \n not\n \n accept examples of human errors\n \n OR\n \n faulty apparatus.\n \n
\n\n Even 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 Calculate the pH of a 1.00 × 10\n \n −2\n \n mol dm\n \n −3\n \n aqueous solution of ammonia.\n
\n\n p\n \n K\n \n \n b\n \n = 4.75 at 298 K.\n
\n\n [3]\n
\n\n \n Kb\n \n = 10\n \n -4.75\n \n /1.78 x 10\n \n -5\n \n
\n \n \n OR\n \n \n
\n \n Kb\n \n =\n \n ✔\n
\n
\n\n [OH\n \n –\n \n ] = «\n \n =» 4.22 × 10\n \n –4\n \n «(mol dm\n \n –3\n \n )» ✔\n
\n\n
\n\n pOH« = –log\n \n 10\n \n (4.22 × 10\n \n –4\n \n )» = 3.37\n
\n \n \n AND\n \n \n
\n pH = «14 – 3.37» = 10.6\n
\n \n \n
\n OR\n \n \n
\n
\n [H\n \n +\n \n ]« =\n \n » = 2.37 × 10\n \n –11\n \n
\n \n \n AND\n \n \n
\n pH« = –log\n \n 10\n \n 2.37 × 10\n \n –11\n \n » = 10.6 ✔\n
\n
\n\n \n Award\n \n [3]\n \n for correct final answer.\n \n
\n\n Rather 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.\n
\n\n Predict the structures of\n \n three\n \n possible fragments you would expect from the mass spectrum of ethanol.\n
\n\n [3]\n
\n\n \n Any three of:\n \n
\n CH\n \n 3\n \n CH\n \n 2\n \n OH\n \n +\n \n ✓\n
\n CH\n \n 3\n \n CH\n \n 2\n \n \n +\n \n ✓\n
\n CH\n \n 3\n \n \n +\n \n ✓\n
\n CH\n \n 2\n \n OH\n \n +\n \n /CH\n \n 3\n \n O\n \n +\n \n ✓\n
\n OH\n \n +\n \n ✓\n
\n
\n\n \n Penalize missing charge once only.\n \n
\n\n \n Accept any other reasonable fragments or m/z values.\n \n
\n\n Draw the structural formula of the alkene required.\n
\n\n \n
\n [1]\n
\n\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 \n Calculate the total number of moles of gas produced from the decomposition of 10.0 g of guanidinium nitrate.\n \n
\n\n [1]\n
\n\n \n moles of gas = «\n \n \n \n » 0.409 «mol» ✔\n \n
\n\n \n The dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H\n \n +\n \n ], and on the equilibrium constant, of increasing the temperature.\n \n
\n\n \n
\n [2]\n
\n\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 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
\n Which compound shows\n \n cis-trans\n \n isomerism?\n
\n\n A. CH\n \n 3\n \n CH=CCl\n \n 2\n \n
\n\n B. CCl\n \n 2\n \n =CH\n \n 2\n \n
\n\n C.\n \n
\n D.\n \n
\n [1]\n
\n\n D\n
\n\n \n Formulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.\n \n
\n\n [2]\n
\n\n \n 2C\n \n 6\n \n H\n \n 5\n \n COOH(s) + 15O\n \n 2\n \n (g) → 14CO\n \n 2\n \n (g) + 6H\n \n 2\n \n O(l)\n \n
\n\n \n correct products\n \n [✔]\n \n \n
\n\n \n correct balancing\n \n [✔]\n \n \n
\n\n Most students earned at least one mark for writing the correct products of the combustion of benzoic acid but the balancing appeared to be difficult for some.\n
\n\n Explain why there is a large increase from the 8th to the 9th ionization energy of iron.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n\n IE\n \n 9\n \n : electron in lower energy level\n
\n \n \n OR\n \n \n
\n IE\n \n 9\n \n : more stable/full electron level ✔\n
\n
\n IE\n \n 9\n \n : electron closer to nucleus\n
\n \n \n OR\n \n \n
\n IE\n \n 9\n \n : electron more tightly held by nucleus ✔\n
\n
\n IE\n \n 9\n \n : less shielding by «complete» inner levels ✔\n
\n Despite 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.\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 \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 Drawing arrows in the boxes to represent the electron configuration of a nitrogen atom was done extremely well.\n
\n\n Which reactions involve the transfer of a proton?\n
\n\n
\n I. 2HCl (aq) + Mg (s) → MgCl\n \n 2\n \n (aq) + H\n \n 2\n \n (g)\n
\n II. 2HCl (aq) + MgO (s) → MgCl\n \n 2\n \n (aq) + H\n \n 2\n \n O (l)\n
\n III. 2HCl (aq) + MgCO\n \n 3\n \n (s) → MgCl\n \n 2\n \n (aq) + H\n \n 2\n \n O (l) + CO\n \n 2\n \n (g)\n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n C\n
\n\n Deduce the relationship between the concentration of N\n \n 2\n \n O\n \n 5\n \n and the rate of reaction.\n
\n\n [1]\n
\n\n « rate of reaction is directly» proportional to/∝[N\n \n 2\n \n O\n \n 5\n \n ]\n
\n \n \n OR\n \n \n
\n doubling concentration doubles rate ✔\n
\n \n
\n Do\n \n not\n \n accept “rate increases as concentration increases”/ positive correlation\n \n
\n \n Accept linear\n \n
\n\n \n Consider the following equations.\n \n
\n\n \n 2Al (s) +\n \n \n \n O\n \n 2\n \n (g) → Al\n \n 2\n \n O\n \n 3\n \n (s) Δ\n \n H\n \n \n Ɵ\n \n = −1670 kJ\n
\n Mn (s) + O\n \n 2\n \n (g) → MnO\n \n 2\n \n (s) Δ\n \n H\n \n \n Ɵ\n \n = −520 kJ\n \n
\n \n What is the standard enthalpy change, in kJ, of the reaction below?\n \n
\n\n \n 4Al (s) + 3MnO\n \n 2\n \n (s) → 2Al\n \n 2\n \n O\n \n 3\n \n (s) + 3Mn (s)\n \n
\n\n \n A. −1670 + 520\n \n
\n\n \n B.\n \n \n \n \n \n (−1670) + 3(520)\n \n
\n\n \n C. 2(−1670) + 3(−520)\n \n
\n\n \n D. 2(−1670) + 3(520)\n \n
\n\n [1]\n
\n\n D\n
\n\n It is pleasing that 83 % of the candidates were confident in applying Hess’s law to obtain an expression for the standard enthalpy change of reaction. The most commonly chosen distractor (C) had the correct coefficients but one of the signs was incorrect.\n
\n\n Which explains increasing rate of reaction with increasing temperature?\n
\n\n \n
\n [1]\n
\n\n B\n
\n\n \n What is the volume of gas when the pressure on 100 cm\n \n 3\n \n of gas is changed from 400 kPa to 200 kPa at constant temperature?\n \n
\n\n \n A. 50.0\n \n cm\n \n \n 3\n \n \n
\n\n \n B. 100\n \n cm\n \n \n 3\n \n \n
\n\n \n C. 200\n \n cm\n \n \n 3\n \n \n
\n\n \n D. 800\n \n cm\n \n \n 3\n \n \n
\n\n [1]\n
\n\n C\n
\n\n 80 % of the candidates were able to deduce the new volume of a sample of gas after the pressure was halved. The most commonly chosen distractor (A) was the value that assumed a direct proportionality between volume and pressure.\n
\n\n Outline why the value obtained in (b)(i) might differ from a value calculated using Δ\n \n H\n \n \n f\n \n data.\n
\n\n [1]\n
\n\n «N-H» bond enthalpy is an average «and may not be the precise value in NH\n \n 3\n \n » ✔\n
\n\n
\n\n \n Accept ΔH\n \n f\n \n data are more accurate / are not average values.\n \n
\n\n Outlining why Δ\n \n H\n \n \n rxn\n \n based on BE values differ due to being average compared to using Δ\n \n H\n \n \n f\n \n values was generally done well.\n
\n\n Which statement is correct about identical pieces of magnesium added to two solutions, X and Y, containing hydrochloric acid at the same temperature?\n
\n\n \n
\n
\n
\n A. Solution X will reach a higher maximum temperature.\n
\n\n B. Solution Y will reach a higher maximum temperature.\n
\n\n C. Solutions X and Y will have the same temperature rise.\n
\n\n D. It is not possible to predict whether X or Y will have the higher maximum temperature because we cannot identify the limiting reactant.\n
\n\n [1]\n
\n\n A\n
\n\n A nice question that required thinking. There were two factors to consider: amount of reactants and temperature rise. Most candidates focused on the same amount of Mg and HCl hence concluded that the temperature rise would be the same. In fact, it is the same amount of heat that is released, however with a larger volume of solution in Y, the temperature rise is higher for X. 40% of the candidates answered correctly.\n
\n\n \n What is the name of this compound using IUPAC rules?\n \n
\n\n \n \n \n
\n \n \n A. 2,3-diethylbutane\n \n \n
\n\n \n \n B. 2-ethyl-3-methylpentane\n \n \n
\n\n \n \n C. 3-methyl-4-ethylpentane\n \n \n
\n\n \n \n D. 3,4-dimethylhexane\n \n \n
\n\n [1]\n
\n\n D\n
\n\n This was the easiest question on the paper. 97 % of the candidates recognized the correct IUPAC name of the compound.\n
\n\n \n Explain how entropy affects this equilibrium.\n \n
\n\n [2]\n
\n\n \n entropy increases «and the reaction proceeds to the right»\n \n [✔]\n \n \n
\n\n \n more species / free molecules are formed\n
\n \n \n OR\n \n \n
\n more ways of distributing energy\n \n [✔]\n \n \n
\n There were several incorrect responses that products were more ordered than the reactants.\n
\n\n State, giving a reason, whether the reaction is spontaneous or not at 298 K.\n
\n\n [1]\n
\n\n spontaneous\n \n \n AND\n \n \n Δ\n \n G\n \n < 0 ✔\n
\n\n Reason for the reaction being spontaneous was generally very done well indeed.\n
\n\n State the type of reaction.\n
\n\n [1]\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 Very well answered, some mentioned halogenation which is a different reaction.\n
\n\n What is the sum of the coefficients when the equation is balanced with whole numbers?\n
\n\n __Sn(OH)\n \n 4\n \n (aq) + __NaOH (aq) → __Na\n \n 2\n \n SnO\n \n 3\n \n (aq) + __H\n \n 2\n \n O (l)\n
\n\n A. 4\n
\n\n B. 5\n
\n\n C. 6\n
\n\n D. 7\n
\n\n [1]\n
\n\n D\n
\n\n Deduce the nuclear equation for the beta decay of cobalt-60.\n
\n\n [1]\n
\n\n \n ✓\n
\n\n
\n\n \n Accept “e/e-/β for beta particle.\n \n
\n\n \n Penalize incorrect Z even if elemental symbol is correct.\n
\n Penalize incorrectly placed A and Z on nuclear symbol.\n \n
\n \n Do\n \n not\n \n penalize missing Z on nuclear symbol.\n \n
\n\n Which is correct when Δ\n \n H–\n \n \n T\n \n Δ\n \n S\n \n = 0?\n
\n\n
\n A. Forward reaction is favoured.\n
\n B. Reverse reaction is favoured.\n
\n\n C. Reaction is in a state of equilibrium.\n
\n\n D. No chemical changes can occur.\n
\n\n [1]\n
\n\n C\n
\n\n \n Explain how IR spectroscopy can be used to distinguish aspirin from salicylic acid.\n \n
\n\n \n \n \n
\n [2]\n
\n\n salicylic acid contains absorption in the range\n \n ✔\n
\n\n due to phenol/hydroxyl/\n \n group not present in aspirin ✔\n
\n\n
\n \n Award\n \n [2]\n \n for “additional\n \n «stretch» in IR for salicylic acid at higher wavenumber than corresponding\n \n «stretch» in aspirin”\n \n OR\n \n “aspirin has two absorption bands/one stronger absorption band in\n \n while salicylic acid has one/weaker absorption band in that region”.\n \n
\n \n Award\n \n [1 max]\n \n for “fingerprint regions will be different for both”.\n \n
\n\n This question which asked for an explanation of how IR spectroscopy can be used to distinguish aspirin from salicyclic acid was generally very well answered. The majority stated that salicyclic acid contains an absorption in the IR spectrum in the 3200-3600 cm\n \n -1\n \n range due to the phenolic OH group, which is not present in aspirin. A few stated that aspirin has a methyl group and hence the CH stretch will appear in the 2850-3090 cm\n \n -1\n \n region of the IR spectrum in aspirin (using Section 26 of the Data Booklet) which will not appear in the corresponding IR spectrum for salicyclic acid. This is somewhat incorrect as in salicyclic acid the benzene ring will also have CH bonds and the CH stretch for the benzene ring will occur in a similar region of the IR spectrum (as indicated in Section 26 of the Data Booklet) and hence cannot be used to distinguish fully between the two structures per se if using the Data Booklet range. Of course, in practice the alkyl CH stretch would be at a slightly lower wavenumber (e.g. 2850-2950 cm\n \n -1\n \n ) in the IR spectrum compared to the aromatic CH stretch (3030 cm\n \n -1\n \n ), but virtually no candidate gave this type of precise detail.\n
\n\n Which reagents and conditions are best for converting propan-1-ol into propanoic acid?\n
\n\n A. Reflux with acidified potassium dichromate (VI)\n
\n\n B. Reflux with aqueous sodium hydroxide\n
\n\n C. Distil with acidified potassium dichromate (VI)\n
\n\n D. Distil with aqueous sodium hydroxide\n
\n\n [1]\n
\n\n A\n
\n\n What is the conjugate acid of HS\n \n −\n \n ?\n
\n\n
\n A. H\n \n 2\n \n S\n
\n B. S\n \n 2−\n \n
\n\n C. H\n \n 2\n \n SO\n \n 3\n \n
\n\n D. H\n \n 2\n \n SO\n \n 4\n \n
\n\n [1]\n
\n\n A\n
\n\n What is the maximum number of electrons in energy level n = 4?\n
\n\n
\n A. 8\n
\n B. 18\n
\n\n C. 32\n
\n\n D. 50\n
\n\n [1]\n
\n\n C\n
\n\n \n MnO\n \n 2\n \n is another possible catalyst for the reaction. State the IUPAC name for MnO\n \n 2\n \n .\n \n
\n\n [1]\n
\n\n \n manganese(IV) oxide\n \n
\n\n \n \n \n OR\n \n \n \n
\n\n \n manganese dioxide\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “manganese(IV) dioxide”.\n \n \n
\n\n A well answered question. Very few candidates had problem with nomenclature.\n
\n\n What is the general formula of alkynes?\n
\n\n
\n A. C\n \n n\n \n H\n \n 2n+2\n \n
\n B. C\n \n n\n \n H\n \n 2n\n \n
\n\n C. C\n \n n\n \n H2\n \n n−2\n \n
\n\n D. C\n \n n\n \n H\n \n n\n \n
\n\n [1]\n
\n\n C\n
\n\n Which complex ion contains a central ion with an oxidation state of +3?\n
\n\n
\n A. [PtCl\n \n 6\n \n ]\n \n 2−\n \n
\n B. [Cu(H\n \n 2\n \n O)\n \n 4\n \n (OH)\n \n 2\n \n ]\n
\n\n C. [Ni(NH\n \n 3\n \n )\n \n 4\n \n (H\n \n 2\n \n O)\n \n 2\n \n ]\n \n 2+\n \n
\n\n D. [Co(NH\n \n 3\n \n )\n \n 4\n \n Cl\n \n 2\n \n ]\n \n +\n \n
\n\n [1]\n
\n\n D\n
\n\n \n Draw a best-fit line on the graph.\n \n
\n\n [1]\n
\n\n \n best-fit smooth curve ✔\n \n
\n\n \n \n NOTE: Do\n \n not\n \n accept a series of connected lines that pass through all points\n \n OR\n \n any straight line representation.\n \n \n
\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 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 Naming the organic compound using IUPAC rules was generally done well.\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 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 \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 ✔\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 Identify a metal, in the same period as magnesium, that does\n \n not\n \n form a basic oxide.\n
\n\n [1]\n
\n\n aluminium/Al ✔\n
\n\n Many students confused \"period\" and \"group\" and also many did not read metal, so aluminium was not chosen by the majority.\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 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 Another 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 SO\n \n 3\n \n 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 SO\n \n 3\n \n .\n
\n\n Calculate the enthalpy change of reaction,\n \n ΔH\n \n , in kJ, for the decomposition of calcium carbonate.\n
\n\n [2]\n
\n\n «\n \n ΔH\n \n =» (−635 «kJ» – 393.5 «kJ») – (−1207 «kJ») ✓\n
\n\n «\n \n ΔH\n \n = + » 179 «kJ» ✓\n
\n\n \n
\n Award\n \n [2]\n \n for correct final answer.\n \n
\n \n Award\n \n [1 max]\n \n for −179 kJ.\n \n
\n\n \n Ignore an extra step to determine total enthalpy change in kJ: 179 kJ mol\n \n -1\n \n x 5.55 mol = 993 kJ.\n \n
\n\n \n Award\n \n [2]\n \n for an answer in the range 990 - 993« kJ».\n \n
\n\n Calculate coefficients that balance the equation for the following reaction.\n
\n\n __ Mg\n \n 3\n \n N\n \n 2\n \n (s) + __ H\n \n 2\n \n O (l) → __ Mg(OH)\n \n 2\n \n (s) + __ NH\n \n 3\n \n (aq)\n
\n\n [1]\n
\n\n «1» Mg\n \n 3\n \n N\n \n 2\n \n (s) +\n \n 6\n \n H\n \n 2\n \n O (l) →\n \n 3\n \n Mg(OH)\n \n 2\n \n (s) +\n \n 2\n \n NH\n \n 3\n \n (aq)\n
\n\n This was generally very well done with almost all candidates being able to determine the correct coefficients.\n
\n\n Outline how\n \n one\n \n calcium compound in the lime cycle can reduce a problem caused by acid deposition.\n
\n\n [1]\n
\n\n «add» Ca(OH)\n \n 2\n \n /CaCO\n \n 3\n \n /CaO\n \n \n AND\n \n \n to «acidic» water/river/lake/soil\n
\n \n \n OR\n \n \n
\n «use» Ca(OH)\n \n 2\n \n /CaCO\n \n 3\n \n /CaO in scrubbers «to prevent release of acidic pollution» ✓\n
\n
\n\n \n Accept any correct name for any of the calcium compounds listed.\n \n
\n\n Which species is the oxidizing agent?\n
\n\n 14H\n \n +\n \n (aq) + 2Mn\n \n 2+\n \n (aq) + 5BiO\n \n 3\n \n \n −\n \n (aq) → 2MnO\n \n 4\n \n \n −\n \n (aq) + 5Bi\n \n 3+\n \n (aq) + 7H\n \n 2\n \n O (l)\n
\n\n
\n A. H\n \n +\n \n (aq)\n
\n B. Mn\n \n 2+\n \n (aq)\n
\n\n C. BiO\n \n 3\n \n \n −\n \n (aq)\n
\n\n D. MnO\n \n 4\n \n \n −\n \n (aq)\n
\n\n [1]\n
\n\n C\n
\n\n State, with a reason, the effect of an increase in temperature on the position of this equilibrium.\n
\n\n [1]\n
\n\n «shifts» left/towards reactants\n \n \n AND\n \n \n «forward reaction is» exothermic/ΔH is negative ✔\n
\n\n State the electron configuration of copper.\n
\n\n [1]\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
\n Sketch the first eight successive ionisation energies of sulfur.\n
\n\n \n
\n [2]\n
\n\n \n
\n two regions of small increases\n \n \n AND\n \n \n a large increase between them✔\n
\n\n large increase from 6th to 7th ✔\n
\n\n \n
\n Accept line/curve showing these trends.\n \n
\n What is the explanation for the malleability of metals?\n
\n\n
\n A. The bonds are strong.\n
\n B. The bonds are weak.\n
\n\n C. The bonds involve free electrons.\n
\n\n D. The bonds do not have a specific direction.\n
\n\n [1]\n
\n\n D\n
\n\n \n Nitrogen monoxide reacts with oxygen gas to form nitrogen dioxide.\n \n
\n\n \n The following experimental data was obtained.\n
\n \n
\n \n \n \n
\n \n Deduce the partial order of reaction with respect to nitrogen monoxide and oxygen.\n
\n \n
\n \n \n \n
\n
\n\n
\n\n [2]\n
\n\n \n : second ✔\n
\n \n : first ✔\n
\n Most candidates could correctly deduce the order of each reactant from rate experimental rate data.\n
\n\n Predict, giving a reason, how the value of the ΔS\n \n ⦵\n \n \n reaction\n \n would be affected if\n \n (g) were used as a reactant.\n
\n\n [1]\n
\n\n Δ\n \n lower/less positive\n \n \n AND\n \n \n same number of moles of gas\n
\n\n \n \n OR\n \n \n
\n\n Δ\n \n lower/less positive\n \n \n AND\n \n \n a solid has less entropy than a gas ✔\n
\n\n 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.\n
\n\n 4(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.\n
\n\n \n Aqueous sodium hydrogencarbonate has a pH of approximately 7 at 298 K.\n \n
\n\n \n Sketch a graph of pH against volume when 25.0cm\n \n 3\n \n of 0.100 mol dm\n \n −3\n \n NaOH (aq) is gradually added to 10.0cm\n \n 3\n \n of 0.0500 mol dm\n \n −3\n \n NaHCO\n \n 3\n \n (aq).\n \n
\n\n \n \n \n
\n [2]\n
\n\n \n
\n \n S-shaped curve from ~7 to between 12 and 14\n \n [✔]\n \n \n
\n\n \n equivalence point at 5 cm\n \n 3\n \n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept starting point >6~7.\n \n \n
\n\n Most 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.\n
\n\n State the oxidation state of C2 in the reactant and product in the following reaction and the type of the reaction.\n
\n\n \n
\n
\n\n Reactant: ..............................................................................................................................................\n
\n\n
\n Product: ................................................................................................................................................\n
\n
\n Type of reaction: ....................................................................................................................................\n
\n [2]\n
\n\n Reactant: −2\n
\n \n \n AND\n
\n \n \n Product: −4 ✔\n
\n Mechanism: reduction/hydrogenation ✔\n
\n\n Which elements are considered to be metalloids?\n
\n\n I. Gallium\n
\n\n II. Germanium\n
\n\n III. Arsenic\n
\n\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n C\n
\n\n The question about identifying metalloids in the Periodic Table was challenging for many candidates with only 40% of the candidates identifying the correct answer (germanium and arsenic). The most commonly chosen distractor was A (gallium and germanium).\n
\n\n \n State what the presence of alternative Lewis structures shows about the nature of the bonding in the molecule.\n \n
\n\n [1]\n
\n\n \n delocalization\n \n
\n\n \n OR\n \n
\n\n \n delocalized\n \n π\n \n -electrons\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “resonance”.\n \n \n
\n\n Most candidates identified resonance for this given Lewis representation.\n
\n\n \n To 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.\n \n
\n\n [1]\n
\n\n \n \n Any one of:\n \n
\n copper(II) sulfate\n \n \n AND\n \n \n mass/amount of zinc is independent variable/being changed.\n
\n \n \n OR\n \n \n
\n copper(II) sulfate\n \n \n AND\n \n \n with zinc in excess there is no independent variable «as amount of copper(II) sulfate is fixed»\n \n [✔]\n \n \n
\n \n copper(II) sulfate\n \n \n AND\n \n \n having excess zinc will not yield different results in each trial\n \n [✔]\n \n \n
\n\n \n zinc\n \n \n AND\n \n \n results can be used to see if amount of zinc affects temperature rise «so this can be allowed for»\n \n [✔]\n \n \n
\n\n \n zinc\n \n \n AND\n \n \n reduces variables/keeps the amount reacting constant\n \n [✔]\n \n \n
\n\n The 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.\n
\n\n \n How many carbon atoms are sp\n \n 3\n \n , sp\n \n 2\n \n and sp hybridized in the molecule?\n \n
\n\n \n \n \n
\n \n \n \n
\n [1]\n
\n\n A\n
\n\n Identify carbons with sp, sp\n \n 2\n \n , and sp\n \n 3\n \n hybridization was well answered.\n
\n\n State the type of reaction.\n
\n\n [1]\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 Very well answered, some mentioned halogenation which is a different reaction.\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 2CO(g) → C(s) + CO\n \n 2\n \n (g) ✓\n
\n\n
\n \n Accept reversible arrows.\n \n
\n \n Outline how alloys conduct electricity and why they are often harder than pure metals.\n \n
\n\n \n \n Conduct electricity:\n \n \n
\n\n \n \n \n Harder than pure metals:\n \n \n \n
\n\n [2]\n
\n\n \n \n Conduct electricity:\n \n
\n «delocalized/valence» electrons free to move «under potential difference»\n \n [✔]\n \n \n
\n \n \n Harder than pure metals:\n \n
\n atoms/ions of different sizes prevent layers «of atoms/ions» from sliding over one another\n \n [✔]\n \n \n
\n Most candidates were awarded one mark for how alloys conduct electricity. Some struggled with describing why they are harder than pure metals.\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 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 Only the very weak candidates were unable to calculate the enthalpy change correctly, eventually missing 1 mark for inverted calculations.\n
\n\n \n Explain the mechanism for the nitration of benzene, using curly arrows to indicate the movement of electron pairs.\n \n
\n\n [4]\n
\n\n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5945/1cii.PNG\")
\n
\n \n curly arrow going from benzene ring to N «of\n \n +\n \n NO\n \n 2\n \n /NO\n \n 2\n \n \n +\n \n »\n \n [\n \n ✔\n \n ]\n \n
\n carbocation with correct formula and positive charge on ring\n \n [\n \n ✔\n \n ]\n \n
\n curly arrow going from C–H bond to benzene ring of cation\n \n [\n \n ✔\n \n ]\n \n
\n formation of organic product nitrobenzene\n \n \n AND\n \n \n H\n \n +\n \n \n [\n \n ✔\n \n ]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept mechanism with corresponding Kekulé structures.\n
\n Do\n \n not\n \n accept a circle in M2 or M3.\n
\n Accept first arrow starting either inside the circle or on the circle.\n
\n If Kekulé structure used, first arrow must start on the double bond.\n
\n M2 may be awarded from correct diagram for M3.\n
\n M4: Accept “C\n \n 6\n \n H\n \n 5\n \n NO\n \n 2\n \n + H\n \n 2\n \n SO\n \n 4\n \n ” if HSO\n \n 4\n \n \n −\n \n used in M3.\n \n \n
\n Performance 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.\n
\n\n Draw the Lewis structure of NO\n \n 3\n \n \n −\n \n .\n
\n\n [1]\n
\n\n \n
\n
\n\n \n Do\n \n not\n \n accept the delocalised structure.\n \n
\n\n \n Accept any combination of dots, crosses and lines.\n \n
\n\n \n Coordinate/dative bond may be represented by an arrow.\n \n
\n\n Only 20% of the candidates scored the mark for the Lewis structure of NO\n \n 3\n \n \n -\n \n . Mistakes included: missing charge, missing lone pairs, 3 single bonds, 2 double bonds.\n
\n\n What is the molecular geometry of the central atom in SF\n \n 4\n \n Cl\n \n 2\n \n ?\n
\n\n
\n A. linear\n
\n B. tetrahedral\n
\n\n C. hexagonal\n
\n\n D. octahedral\n
\n\n [1]\n
\n\n D\n
\n\n Suggest\n \n two\n \n reasons why atoms are no longer regarded as the indivisible units of matter.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n
\n subatomic particles «discovered»\n
\n \n \n OR\n \n \n
\n particles smaller/with masses less than atoms «discovered»\n
\n \n \n OR\n \n \n
\n «existence of» isotopes «same number of protons, different number of neutrons» ✔\n
\n
\n charged particles obtained from «neutral» atoms\n
\n \n \n OR\n \n \n
\n atoms can gain or lose electrons «and become charged» ✔\n
\n
\n atom «discovered» to have structure ✔\n
\n
\n fission\n
\n \n \n OR\n \n \n
\n atoms can be split ✔\n
\n
\n\n \n Accept atoms can undergo fusion «to produce heavier atoms»\n \n
\n\n \n Accept specific examples of particles.\n \n
\n\n \n Award\n \n [2]\n \n for “atom shown to have a nucleus with electrons around it” as both M1 and M3.\n \n
\n\n The 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.\n
\n\n What is the correct ground state electron orbital configuration for 2s\n \n 2\n \n 2p\n \n 2\n \n ?\n
\n\n
\n \n
\n [1]\n
\n\n A\n
\n\n State the meaning of a strong Brønsted–Lowry acid.\n
\n\n [2]\n
\n\n fully ionizes/dissociates ✔\n
\n\n proton/H\n \n +\n \n «donor »✔\n
\n\n State the full electron configuration of the sulfide ion.\n
\n\n [1]\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 ✔\n
\n\n \n
\n Do\n \n not\n \n accept “[Ne] 3s\n \n 2\n \n 3p\n \n 6\n \n ”.\n \n
\n State the number of subatomic particles in this ion.\n
\n\n \n
\n [1]\n
\n\n \n Protons\n \n : 7\n \n \n AND\n \n Neutrons\n \n : 7\n \n \n AND\n \n Electrons\n \n : 10 ✔\n
\n\n Most candidates could answer the question about subatomic particles correctly.\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 \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 Drawing arrows in the boxes to represent the electron configuration of a nitrogen atom was done extremely well.\n
\n\n \n How do the following properties change down Group 17 of the periodic table?\n \n
\n\n \n \n \n
\n [1]\n
\n\n C\n
\n\n 85 % of the candidates know the trends in ionization energy and ionic radius down Group 17. The other distractors were equally chosen by the remainder of the candidates.\n
\n\n \n Alcohol levels in the breath can also be determined using IR spectroscopy.\n \n
\n\n \n Suggest, giving a reason, which bond’s absorbance is most useful for detecting ethanol in breath.\n \n
\n\n Bond:\n
\n\n Reason:\n
\n\n [2]\n
\n\n \n \n Bond:\n \n
\n C–O\n
\n \n \n OR\n \n \n
\n C–H\n \n [✔]\n \n \n
\n \n \n Reason\n \n :\n
\n cannot use O–H bonds as in water «found in breath»\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note\n \n : Accept “C–O/C–H «bonds in molecules in breath» most likely to be in ethanol”.\n \n \n
\n\n \n \n Do\n \n not\n \n apply ECF here.\n \n \n
\n\n Most candidates incorrectly identified O-H, failing to realise it is unsuitable due to its abundant presence in the breath.\n
\n\n \n Which statement is correct about a catalyst?\n \n
\n\n \n A. It decreases the activation energy of the forward reaction but not the reverse.\n \n
\n\n \n B. It increases the proportion of products to reactants in an equilibrium.\n \n
\n\n \n C. It decreases the enthalpy change of the reaction.\n \n
\n\n \n D. It changes the mechanism of the reaction.\n \n
\n\n [1]\n
\n\n D\n
\n\n This was surprisingly one of the most challenging questions on the paper and discriminated very well between high scoring and low scoring candidates. It tested understanding of the role of the catalyst. 52 % of candidates chose the correct answer (a catalyst changes the mechanism of the reaction). The most commonly chosen distractor was A (a catalyst decreases the activation energy of the forward reaction but not the reverse).\n
\n\n Calculate the equilibrium constant,\n \n K\n \n \n c\n \n , for this reaction at 298 K. Use your answer to (d)(iii) and sections 1 and 2 of the data booklet.\n
\n\n (If you did not obtain an answer to (d)(iii) use a value of 2.0 kJ mol\n \n −1\n \n , although this is not the correct answer).\n
\n\n [2]\n
\n\n «ln\n \n K\n \n \n c\n \n = – (3.4 × 10\n \n 3\n \n J mol\n \n –1\n \n /8.31 J K\n \n –1\n \n mol\n \n –1\n \n × 298 K)» = –1.37 ✔\n
\n\n «\n \n K\n \n \n c\n \n =» 0.25 ✔\n
\n\n \n Award\n \n [2]\n \n for “0.45” for the use of 2.0 kJ mol\n \n –1\n \n for ΔG\n \n \n ⦵\n \n \n .\n \n
\n\n Which combustion reaction releases the\n \n least\n \n energy per mole of C\n \n 3\n \n H\n \n 8\n \n ?\n
\n\n Approximate bond enthalpy / kJ mol\n \n −1\n \n
\n O=O 500\n
\n C=O 800\n
\n C≡O 1000\n
\n
\n A. C\n \n 3\n \n H\n \n 8\n \n (g) + 5O\n \n 2\n \n (g) → 3CO\n \n 2\n \n (g) + 4H\n \n 2\n \n O (g)\n
\n B. C\n \n 3\n \n H\n \n 8\n \n (g) +\n \n O\n \n 2\n \n (g) → 2CO\n \n 2\n \n (g) + CO (g) + 4H\n \n 2\n \n O (g)\n
\n\n C. C\n \n 3\n \n H\n \n 8\n \n (g) + 4O\n \n 2\n \n (g) → CO\n \n 2\n \n (g) + 2CO (g) + 4H\n \n 2\n \n O (g)\n
\n\n D. C\n \n 3\n \n H\n \n 8\n \n (g) +\n \n O\n \n 2\n \n (g) → 3CO (g) + 4H\n \n 2\n \n O (g)\n
\n\n
\n\n \n Chemistry: Atoms First 2e, https://openstax.org/books/chemistry-atoms-first-2e/pages/9-4-strengths-of-ionic-andcovalent-bonds © 1999–2021, Rice University. Except where otherwise noted, textbooks on this site are licensed under a Creative Commons Attribution 4.0 International License.\n \n
\n \n (CC BY 4.0) https://creativecommons.org/licenses/ by/4.0/.\n \n
\n [1]\n
\n\n D\n
\n\n Which is a homologous series?\n
\n\n A. C\n \n 2\n \n H\n \n 4\n \n , C\n \n 3\n \n H\n \n 5\n \n , C\n \n 4\n \n H\n \n 6\n \n
\n\n B. C\n \n 2\n \n H\n \n 2\n \n , C\n \n 3\n \n H\n \n 4\n \n , C\n \n 4\n \n H\n \n 6\n \n
\n\n C. C\n \n 2\n \n H\n \n 2\n \n , C\n \n 2\n \n H\n \n 4\n \n , C\n \n 2\n \n H\n \n 6\n \n
\n\n D. C\n \n 2\n \n H\n \n 2\n \n , C\n \n 4\n \n H\n \n 4\n \n , C\n \n 6\n \n H\n \n 6\n \n
\n\n [1]\n
\n\n B\n
\n\n 45% of the candidates identified the members of the same homologous series (ethyne, propyne and butyne). The most commonly chosen distractor included the compounds C\n \n 2\n \n H\n \n 2\n \n , C\n \n 4\n \n H\n \n 4\n \n and C\n \n 6\n \n H\n \n 6\n \n . The question had good discrimination between high-scoring and low-scoring candidates.\n
\n\n \n Describe 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.\n \n
\n\n [2]\n
\n\n \n place «pieces of» Re into each solution\n \n [✔]\n \n \n
\n\n \n if Re reacts/is coated with metal, that metal is less reactive «than Re»\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept other valid observations such as “colour of solution fades” or “solid/metal appears” for “reacts”.\n \n \n
\n\n This 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.\n
\n\n Deduce the hybridization of the central carbon atom in Compound A.\n
\n\n [1]\n
\n\n sp\n \n 2\n \n ✔\n
\n\n sp\n \n 2\n \n hybridization of the central carbon atom in the ketone was very done well.\n
\n\n The enthalpy of formation of ammonia gas is −46 kJ mol\n \n −1\n \n .\n
\n\n N\n \n 2\n \n (g) + 3H\n \n 2\n \n (g) → 2NH\n \n 3\n \n (g)\n
\n\n What is the energy released, in kJ, in the reaction?\n
\n\n
\n A. 23\n
\n B. 46\n
\n C. 69\n
\n D. 92\n
\n [1]\n
\n\n D\n
\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 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 Naming the organic compound using IUPAC rules was generally done well.\n
\n\n \n Describe, using a suitable equation, how the buffer solution formed during the titration resists pH changes when a small amount of acid is added.\n \n
\n\n [2]\n
\n\n \n \n \n ALTERNATIVE 1:\n \n \n
\n H\n \n +\n \n (aq) + CH\n \n 3\n \n COO\n \n –\n \n (aq) → CH\n \n 3\n \n COOH (aq) ✔\n \n
\n \n added acid neutralised by ethanoate ions\n
\n \n \n OR\n \n \n
\n «weak» CH\n \n 3\n \n COOH (aq)/ethanoic acid replaces H\n \n +\n \n (aq)\n
\n \n \n OR\n \n \n
\n CH\n \n 3\n \n COOH/CH\n \n 3\n \n COO\n \n –\n \n ratio virtually/mostly unchanged ✔\n \n
\n \n
\n \n \n ALTERNATIVE 2:\n \n \n
\n CH\n \n 3\n \n COOH (aq)\n \n \n \n H\n \n +\n \n (aq) + CH\n \n 3\n \n COO\n \n –\n \n (aq) ✔\n \n
\n \n equilibrium shifts to the ethanoic acid side\n
\n \n \n OR\n \n \n
\n CH\n \n 3\n \n COOH/CH\n \n 3\n \n COO\n \n −\n \n ratio virtually/mostly unchanged ✔\n \n
\n Which are the correct sequences of\n \n increasing\n \n bond strengths and bond lengths between two carbon atoms?\n
\n\n
\n| \n | \n\n \n Bond strength\n \n | \n\n \n Bond length\n \n | \n
| \n A.\n | \n\n \n | \n\n \n | \n
| \n B.\n | \n\n \n | \n\n \n | \n
| \n C.\n | \n\n \n | \n\n \n | \n
| \n D.\n | \n\n \n | \n\n \n | \n
\n [1]\n
\n\n C\n
\n\n \n Describe how this mixture is separated by fractional distillation.\n \n
\n\n [2]\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 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
\n\n Suggest what can be concluded about the gold atom from this experiment.\n
\n\n \n
\n [2]\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 \n Very few\n \n 4\n \n He\n \n 2+\n \n deviating largely from their path:\n \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 Comment on why peracetic acid, CH\n \n 3\n \n COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.\n \n
\n\n \n H\n \n 2\n \n O\n \n 2\n \n (aq) + CH\n \n 3\n \n COOH (aq)\n \n \n \n CH\n \n 3\n \n COOOH (aq) + H\n \n 2\n \n O (l)\n \n
\n\n [1]\n
\n\n \n move «position of» equilibrium to right/products\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “reactants are always present as the reaction is in equilibrium”.\n \n \n
\n\n Suggesting 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.\n
\n\n Which best explains why complexes of d-block elements are coloured?\n
\n\n
\n A. Light is absorbed when electrons are promoted between d orbitals.\n
\n B. Light is emitted when electrons are promoted between d orbitals.\n
\n\n C. Light is absorbed when electrons return to lower energy d orbitals.\n
\n\n D. Light is emitted when electrons return to lower energy d orbitals.\n
\n\n [1]\n
\n\n A\n
\n\n About 70% of candidates selected the best explanation for the colour of d-block elements. The most commonly chosen distractor (D) indicates that a significant number of students believe that the elements emit light.\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 \n Identify the initiation step of the reaction and its conditions.\n \n
\n\n [2]\n
\n\n \n Br\n \n 2\n \n 2Br•\n \n [\n \n ✔\n \n ]\n \n
\n \n
\n \n «sun»light/UV/\n \n hv\n \n
\n \n \n OR\n \n \n
\n high temperature\n \n [\n \n ✔\n \n ]\n \n \n
\n \n \n \n Note:\n \n Do not penalize missing radical symbol on Br.\n
\n Accept “homolytic fission of bromine” for M1.\n \n \n
\n Very well done, with a few making reference to a catalyst.\n
\n\n State the full electronic configuration of Fe\n \n 2+\n \n .\n
\n\n [1]\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 3d\n \n 6\n \n ✔\n
\n\n Mostly well done which was a pleasant surprise since this is not overly easy, predictably some gave [Ar] 4s\n \n 2\n \n 3d\n \n 4\n \n .\n
\n\n \n Explain why the first ionization energy of nitrogen is greater than both carbon and oxygen.\n \n
\n\n \n Nitrogen and carbon:\n \n
\n\n \n Nitrogen and oxygen:\n \n
\n\n [2]\n
\n\n \n \n Nitrogen and carbon:\n \n \n
\n\n \n N has greater nuclear charge/«one» more proton «and electrons both lost from singly filled p-orbitals»\n \n [✔]\n \n
\n \n
\n
\n\n \n \n Nitrogen and oxygen:\n \n \n
\n\n \n O has a doubly filled «p-»orbital\n
\n \n \n OR\n \n \n
\n N has only singly occupied «p-»orbitals\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “greater e–\n \n -\n \n e\n \n -\n \n repulsion in O” or “lower\n \n e–\n \n \n -\n \n \n e\n \n \n -\n \n repulsion in N”.\n \n \n
\n\n \n \n Accept box annotation of electrons for M2.\n \n \n
\n\n Mixed 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.\n
\n\n Evaluate whether this, rather than the loss of product, could explain the yield found in (b)(iii).\n
\n\n [1]\n
\n\n yes\n
\n \n \n AND\n \n \n
\n «each Mg combines with\n \n N, so» mass increase would be 14x\n \n which is less than expected increase of 16x\n
\n \n \n OR\n \n \n
\n 3 mol Mg would form 101g of Mg\n \n 3\n \n N\n \n 2\n \n but would form 3 x MgO = 121 g of MgO\n
\n \n \n OR\n \n \n
\n 0.2614 mol forms 10.536 g of MgO, but would form 8.796 g of Mg\n \n 3\n \n N\n \n 2\n \n ✔\n
\n
\n\n \n Accept Yes\n \n AND\n \n “the mass of N/N\n \n 2\n \n that combines with each g/mole of Mg is lower than that of O/O\n \n 2\n \n ”\n \n
\n\n \n Accept YES\n \n AND\n \n “molar mass of nitrogen less than of oxygen”.\n \n
\n\n This proved to be a very difficult question to answer in the quantitative manner required, with hardly any correct responses.\n
\n\n Determine the coefficients that balance the equation for the reaction of lithium with water.\n
\n\n \n
\n [1]\n
\n\n 2 Li (s) + 2 H\n \n 2\n \n O (l) → 2 LiOH (aq) + H\n \n 2\n \n (g) ✔\n
\n\n State the molecular formula of the next member of the homologous series to which ethene belongs.\n
\n\n [1]\n
\n\n C\n \n 3\n \n H\n \n 6\n \n ✔\n
\n\n \n Accept structural formula.\n \n
\n\n \n Which transition in the hydrogen atom emits visible light?\n \n
\n\n \n A. n = 1 to n = 2\n \n
\n\n \n B. n = 2 to n = 3\n \n
\n\n \n C. n = 2 to n = 1\n \n
\n\n \n D. n = 3 to n = 2\n \n
\n\n [1]\n
\n\n D\n
\n\n This question discriminated well between high scoring and low scoring candidates. 68% of the candidates chose the correct transition that emits visible light (n = 3 to n = 2). The most commonly chosen distractor C was the only other option that involved emission.\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 \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 Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.\n \n
\n\n
\n\n \n Molecular geometry:\n \n
\n\n \n H–N–H bond angle:\n \n
\n\n [2]\n
\n\n \n \n Molecular geometry:\n \n
\n «trigonal» pyramidal\n \n [✔]\n \n \n
\n \n \n H–N–H bond angle:\n \n
\n 107°\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept angles in the range of 100–109.\n \n \n
\n\n The 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.\n
\n\n \n What will be the major product in the reaction between but-1-ene and\n \n ?\n \n
\n\n \n A. 2-bromobut-1-ene\n
\n \n
\n \n B. 1-bromobut-1-ene\n
\n \n
\n \n C. 2-bromobutane\n
\n \n
\n \n D. 1-bromobutane\n \n
\n\n [1]\n
\n\n C\n
\n\n The vast majority selected the correct product for the reaction of but-1-ene and HBr although quite a few selected the minor product of 1-bromobutane.\n
\n\n Curve\n \n X\n \n on the following graph shows the volume of oxygen formed during the catalytic decomposition of a 1.0 mol dm\n \n −3\n \n solution of hydrogen peroxide.\n
\n\n \n
\n Which change would produce the curve\n \n Y\n \n ?\n
\n\n
\n A. Adding water.\n
\n B. Adding some 0.1 mol dm\n \n −3\n \n hydrogen peroxide solution.\n
\n\n C. Adding some 2.0 mol dm\n \n −3\n \n hydrogen peroxide solution.\n
\n\n D. Repeating the experiment without a catalyst.\n
\n\n [1]\n
\n\n B\n
\n\n The following diagram shows a chromatogram.\n
\n\n \n
\n
\n\n \n Identity of spot C:\n \n
\n leucine ✓\n
\n What is the enthalpy change,\n \n in J\n \n , when 5 g of water is heated from 10°C to 18°C?\n
\n\n Specific heat capacity of water: 4.18 kJ kg\n \n −1\n \n K\n \n −1\n \n
\n\n A. 5 × 4.18 × 8\n
\n\n B. 5 × 10\n \n −3\n \n × 4.18 × 8\n
\n\n C. 5 × 4.18 × (273 + 8)\n
\n\n D. 5 × 10\n \n −3\n \n × 4.18 × (273 + 8)\n
\n\n [1]\n
\n\n A\n
\n\n \n Iodine and bromine gases were mixed and allowed to reach equilibrium.\n \n
\n\n \n \n \n
\n \n \n What is the value of the equilibrium constant?\n \n \n
\n\n \n \n A. 0.05\n \n \n
\n\n \n \n B. 1\n \n \n
\n\n \n \n C. 4\n \n \n
\n\n \n \n D. 10\n \n \n
\n\n [1]\n
\n\n C\n
\n\n This was a challenging question with the highest discrimination index on the paper. 62 % of the candidates were able to deduce the equilibrium concentration of IBr and calculate the equilibrium constant correctly. The most commonly chosen distractor was B where the stoichiometric ratio was not taken into account when calculating the equilibrium concentration of IBr.\n
\n\n State the block of the periodic table in which magnesium is located.\n
\n\n [1]\n
\n\n s ✔\n
\n\n \n
\n Do not allow group 2\n \n
\n Many students did not know what \"block\" meant, and often guessed group 2 etc.\n
\n\n \n Draw the repeating unit of polyphenylethene.\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 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 Write the equation for this reaction.\n
\n\n [1]\n
\n\n 4FeS(s) + 7O\n \n 2\n \n (g) → 2Fe\n \n 2\n \n O\n \n 3\n \n (s) + 4SO\n \n 2\n \n (g) ✔\n
\n\n \n
\n Accept any correct ratio.\n \n
\n In which series are all carbon atoms sp\n \n 2\n \n hybridized?\n
\n\n A. C\n \n 2\n \n H\n \n 2\n \n H\n \n 2\n \n CO HCOOH\n
\n\n B. C\n \n 2\n \n H\n \n 4\n \n H\n \n 2\n \n CO HCOOH\n
\n\n C. C\n \n 2\n \n H\n \n 2\n \n CO HCN\n
\n\n D. C\n \n 2\n \n H\n \n 6\n \n CH\n \n 3\n \n OH CH\n \n 3\n \n OCH\n \n 3\n \n
\n\n [1]\n
\n\n B\n
\n\n \n Which is the species oxidized and the oxidizing agent in the reaction?\n \n
\n\n \n MnO\n \n 2\n \n (s) + 4HCl (aq) → MnCl\n \n 2\n \n (aq) + Cl\n \n 2\n \n (g) + 2H\n \n 2\n \n O (l)\n \n
\n\n \n \n \n
\n [1]\n
\n\n D\n
\n\n This question on identifying oxidizing agents was well answered by high scoring candidates, but not handled well by low scoring candidates.\n
\n\n What happens to the amount of hydroxide ions and hydroxide ion concentration when water is added to a solution of NH\n \n 3\n \n (aq)?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n 35% of the candidates deduced the change in amount of OH\n \n -\n \n and [OH\n \n -\n \n ] when water is added to NH\n \n 3\n \n (aq). The majority of candidates recognized the increase in the amount of OH\n \n -\n \n due to the shift in the equilibrium position but assumed that the concentration of OH\n \n -\n \n increased as well ignoring the effect of the dilution. Some teachers commented that this question was too complicated.\n
\n\n Which graphs show a first order reaction?\n
\n\n \n
\n A. V and X\n
\n\n B. V and Y\n
\n\n C. W and X\n
\n\n D. W and Y\n
\n\n [1]\n
\n\n B\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 \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 Which chemical process would produce a voltaic cell?\n
\n\n
\n A. spontaneous redox reaction\n
\n B. spontaneous non-redox reaction\n
\n\n C. non-spontaneous redox reaction\n
\n\n D. non-spontaneous non-redox reaction\n
\n\n [1]\n
\n\n A\n
\n\n State the full electron configuration of the sulfide ion.\n
\n\n [1]\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 ✔\n
\n\n \n
\n Do\n \n not\n \n accept “[Ne] 3s\n \n 2\n \n 3p\n \n 6\n \n ”.\n \n
\n Which compound has the shortest C to N bond?\n
\n\n A. HCN\n
\n\n B. CH\n \n 3\n \n CH\n \n 2\n \n NH\n \n 2\n \n
\n\n C. CH\n \n 3\n \n CHNH\n
\n\n D. (CH\n \n 3\n \n )\n \n 2\n \n NH\n
\n\n [1]\n
\n\n A\n
\n\n Outline the reason that sodium hydroxide is considered a Brønsted–Lowry base.\n
\n\n [1]\n
\n\n «OH\n \n −\n \n is a» proton acceptor ✔\n
\n\n Which of the following is the electron configuration of a metallic element?\n
\n\n A. [Ne] 3s\n \n 2\n \n 3p\n \n 2\n \n
\n\n B. [Ne] 3s\n \n 2\n \n 3p\n \n 4\n \n
\n\n C. [Ne] 3s\n \n 2\n \n 3p\n \n 6\n \n 3d\n \n 3\n \n 4s\n \n 2\n \n
\n\n D. [Ne] 3s\n \n 2\n \n 3p\n \n 6\n \n 3d\n \n 10\n \n 4s\n \n 2\n \n 4p\n \n 5\n \n
\n\n [1]\n
\n\n C\n
\n\n \n S\n \n \n tate the names of\n \n two\n \n functional groups present in all three molecules, using section 37 of the data booklet.\n \n
\n\n [2]\n
\n\n \n \n Any two of:\n \n
\n benzene/aromatic ring ✔\n
\n «tertiary» amino «group» ✔\n
\n ethenylene/1,2-ethenediyl «group» ✔\n
\n ether «group» ✔\n \n
\n \n \n NOTE: Accept “phenyl” for “benzene ring” although there are no phenyl groups as the benzene ring in this compound is a part of a polycyclic structure.\n
\n Do\n \n not\n \n accept “arene” or “benzene” alone.\n
\n Accept “amine” for “amino «group»”.\n
\n Accept “alkenyl/alkene/vinylene” for ethenylene/1,2-ethenediyl «group».\n \n \n
\n Consider the following equilibrium reaction:\n
\n\n 2SO\n \n 2\n \n (g) + O\n \n 2\n \n (g)\n \n 2SO\n \n 3\n \n (g)\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 pressure decrease «due to larger volume» ✓\n
\n\n reactant side has 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 is awarded.\n \n
\n \n What is the enthalpy change of reaction for the following equation?\n \n
\n\n \n \n \n
\n \n \n A.\n \n x + y + z\n \n \n \n
\n\n \n \n B.\n \n −x − y + z\n \n \n \n
\n\n \n \n C.\n \n x − y − z\n \n \n \n
\n\n \n \n D.\n \n x − y + z\n \n \n \n
\n\n
\n\n [1]\n
\n\n D\n
\n\n 83 % of candidates answered this Hess Law question correctly. This topic was obviously well covered.\n
\n\n \n Identify the independent and dependent variables in this experiment.\n \n
\n\n \n \n \n
\n [1]\n
\n\n \n Independent variable:\n \n
\n chain length\n \n \n OR\n \n \n number of carbon «atoms in alcohol»\n
\n \n \n AND\n \n \n
\n \n Dependent variable:\n \n
\n volume of\n \n \n \n OR\n \n \n \n /equilibrium constant\n \n \n OR\n \n \n \n equilibrium\n \n concentration/moles of\n \n ✔\n
\n Well answered. Students mostly identified (alcohol) chain length as the independent variable and K\n \n c\n \n at the dependent. For the latter [ethanoic acid] at equilibrium was another popular choice with some students neglecting to clarify \"equilibrium\" which was needed for the mark. This evidences an issue already identified in the Internal assessment that very often students only identify the processed variables. The proportion of students referring to volume of NaOH was too low for expectations.\n
\n\n \n Proteins are polymers of amino acids.\n \n
\n\n \n A paper chromatogram of two amino acids, A1 and A2, is obtained using a non-polar solvent.\n \n
\n\n \n
\n \n © International Baccalaureate Organization 2020.\n \n
\n\n \n Determine the\n \n \n \n value of A1.\n \n
\n\n [1]\n
\n\n \n ✔\n
\n\n \n Accept any value within the range “\n \n ”.\n \n
\n\n It was surprising that more did not manage to determine the Rf value of A1 within the acceptable range of \"0.67 to 0.73\". An out of range value of 0.75 was frequently seen.\n
\n\n What is the maximum number of electrons that can occupy a p-orbital?\n
\n\n A. 2\n
\n\n B. 3\n
\n\n C. 6\n
\n\n D. 8\n
\n\n [1]\n
\n\n A\n
\n\n Question 6 was poorly answered as it asked for the number of electrons in a p orbital. Very few students gave the correct answer of 2. The majority chose answer C (6) for the maximum number of electrons that can occupy a p-orbital, rather than A (2). It appears candidates reflexively conflated p-orbitals with the entire p subshell in any given period.\n
\n\n \n Which species does\n \n not\n \n have resonance structures?\n \n
\n\n \n A. C\n \n 6\n \n H\n \n 6\n \n \n
\n\n \n B. NH\n \n 4\n \n \n +\n \n \n
\n\n \n C. CO\n \n 3\n \n \n 2−\n \n \n
\n\n \n D. O\n \n 3\n \n \n
\n\n [1]\n
\n\n B\n
\n\n Identifying resonance structure had one of the highest discriminatory indices. Only the higher achieving candidates had much success with this.\n
\n\n Explain why Si has a smaller atomic radius than Al.\n
\n\n [2]\n
\n\n nuclear charge/number of protons/Z/Z\n \n eff\n \n increases «causing a stronger pull on the outer electrons» ✓\n
\n\n same number of shells/«outer» energy level/shielding ✓\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 \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 The 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 T\n \n 1\n \n and T\n \n 2\n \n , some made the two curves meet at high energy or even cross, and some did not have the correct relationship between the peaks of T\n \n 1\n \n and T\n \n 2\n \n .\n
\n\n Determine the volume, in dm\n \n 3\n \n , of 0.015 mol dm\n \n −3\n \n calcium hydroxide solution needed to neutralize 35.0 cm\n \n 3\n \n of 0.025 mol dm\n \n −3\n \n HCl (aq).\n
\n\n [2]\n
\n\n «\n \n n\n \n \n HCl\n \n = 0.0350 dm\n \n 3\n \n × 0.025 mol dm\n \n −3\n \n =» 0.00088 «mol»\n
\n\n \n \n OR\n \n \n
\n \n n\n \n \n Ca(OH)2\n \n =\n \n \n n\n \n \n HCl\n \n /0.00044 «mol» ✓\n
\n
\n «\n \n V\n \n =\n \n =» 0.029 «dm\n \n 3\n \n » ✓\n
\n \n
\n Award\n \n [2]\n \n for correct final answer.\n \n
\n \n Award\n \n [1 max]\n \n for 0.058 «dm\n \n 3\n \n ».\n \n
\n\n Curve\n \n X\n \n on the following graph shows the volume of oxygen formed during the catalytic decomposition of a 1.0 mol dm\n \n −3\n \n solution of hydrogen peroxide.\n
\n\n \n
\n Which change would produce the curve\n \n Y\n \n ?\n
\n\n
\n A. Adding water.\n
\n B. Adding some 0.1 mol dm\n \n −3\n \n hydrogen peroxide solution.\n
\n\n C. Adding some 2.0 mol dm\n \n −3\n \n hydrogen peroxide solution.\n
\n\n D. Repeating the experiment without a catalyst.\n
\n\n [1]\n
\n\n B\n
\n\n \n Which conditions are required for the reaction between two molecules?\n \n
\n\n \n I. a collision\n
\n II.\n \n E ≥ E\n \n a\n \n \n
\n III. proper orientation\n \n
\n \n A. I and II only\n \n
\n\n \n B. I and III only\n \n
\n\n \n C. II and III only\n \n
\n\n \n D. I, II and III\n \n
\n\n [1]\n
\n\n D\n
\n\n This question on collision theory was one of the best answered in the exam.\n
\n\n Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO\n \n 3\n \n , that obeys the octet rule, showing any non-zero formal charges on the atoms.\n
\n\n [2]\n
\n\n \n
\n bonds and non-bonding pairs correct ✔\n
\n\n formal charges correct ✔\n
\n\n
\n\n \n Accept dots, crosses or lines to represent electron pairs.\n \n
\n\n \n Do\n \n not\n \n accept resonance structures with delocalised bonds/electrons.\n \n
\n\n \n Accept + and – sign respectively.\n \n
\n\n \n Do not accept a bond between nitrogen and hydrogen.\n \n
\n\n \n For an incorrect Lewis structure, allow ECF for non-zero formal charges.\n \n
\n\n Drawing the Lewis structure of HNO\n \n 3\n \n 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.\n
\n\n Explain why Si has a smaller atomic radius than Al.\n
\n\n [2]\n
\n\n nuclear charge/number of protons/Z/Z\n \n eff\n \n increases «causing a stronger pull on the outer electrons» ✓\n
\n\n same number of shells/«outer» energy level/shielding ✓\n
\n\n Determine the rate expression for the reaction.\n
\n\n \n
\n [2]\n
\n\n \n BrO\n \n 3\n \n \n –\n \n :\n \n 1/first\n \n \n AND\n \n Br\n \n –\n \n :\n \n 1/first\n \n \n AND\n \n H\n \n +\n \n :\n \n 2/second ✓\n
\n\n «Rate =»\n \n k\n \n [BrO\n \n 3\n \n \n −\n \n ][Br\n \n −\n \n ][H\n \n +\n \n ]\n \n 2\n \n ✓\n
\n\n \n
\n M2: Square brackets required for the mark.\n \n
\n Which substance is likely to have an ionic lattice structure at 298 K and 100 kPa?\n
\n| \n | \n\n \n Melting point\n \n | \n\n \n Conducts electricity in a liquid state?\n \n | \n
| \n A.\n | \n\n low\n | \n\n yes\n | \n
| \n B.\n | \n\n low\n | \n\n no\n | \n
| \n C.\n | \n\n high\n | \n\n no\n | \n
| \n D.\n | \n\n high\n | \n\n yes\n | \n
\n [1]\n
\n\n D\n
\n\n The structure shows the repeating unit of a polymer found in some plastics.\n
\n\n \n
\n
\n Which monomer is used to form this plastic?\n
\n
\n A. H\n \n 2\n \n C=C(CH\n \n 3\n \n )\n \n 2\n \n
\n B. CH\n \n 3\n \n CH(CH\n \n 3\n \n )\n \n 2\n \n
\n\n C. (H\n \n 3\n \n C)\n \n 2\n \n C=C(CH\n \n 3\n \n )\n \n 2\n \n
\n\n D. (H\n \n 3\n \n C)\n \n 2\n \n C=CHCH(CH\n \n 3\n \n )\n \n 2\n \n
\n\n [1]\n
\n\n A\n
\n\n \n Predict, giving a reason, whether the reduction of ReO\n \n 4\n \n \n −\n \n to [Re(OH)\n \n 2\n \n ]\n \n 2+\n \n would oxidize Fe\n \n 2+\n \n to Fe\n \n 3+\n \n in aqueous solution. Use section 24 of the data booklet.\n \n
\n\n [1]\n
\n\n \n no\n \n \n AND\n \n \n ReO\n \n \n 4\n \n \n −\n \n \n is a weaker oxidizing agent than Fe\n \n 3+\n \n
\n \n \n OR\n \n \n
\n no\n \n \n AND\n \n \n Fe\n \n 3+\n \n is a stronger oxidizing agent than\n \n ReO\n \n \n 4\n \n \n −\n \n
\n \n \n OR\n \n \n
\n no\n \n \n AND\n \n \n Fe\n \n 2+\n \n is a weaker reducing agent than\n \n [Re(OH)\n \n \n 2\n \n \n ]\n \n \n 2+\n \n
\n \n \n OR\n \n \n
\n no\n \n \n AND\n \n \n [Re(OH)\n \n \n 2\n \n \n ]\n \n \n 2+\n \n \n is a stronger reducing agent than Fe\n \n 2+\n \n
\n \n \n OR\n \n \n
\n no\n \n \n AND\n \n \n cell emf would be negative/–0.41 V\n \n [✔]\n \n \n
\n Many students understood that the oxidation of Fe\n \n 2+\n \n was not viable but were unable to explain why in terms of oxidizing and reducing power; many students simply gave numerical values for\n \n E\n \n \n Θ\n \n often failing to realise that the oxidation of Fe\n \n 2+\n \n would have the inverse sign to the reduction reaction.\n
\n\n \n What is the hybridization of carbon and oxygen in methanol?\n \n
\n\n \n \n \n
\n [1]\n
\n\n A\n
\n\n 77 % of the candidates deduced the hydridization of carbon and oxygen in methanal. A teacher commented that it was unusual to ask for the hydridization of oxygen, however, this is a reasonable application of the concept.\n
\n\n Why is hydrated copper (II) sulfate blue?\n
\n\n A. Blue light is emitted when electrons return to lower d-orbitals.\n
\n\n B. Light complimentary to blue is absorbed when electrons return to lower d-orbitals.\n
\n\n C. Blue light is emitted when electrons are promoted between d-orbitals.\n
\n\n D. Light complimentary to blue is absorbed when electrons are promoted between d-orbitals.\n
\n\n [1]\n
\n\n D\n
\n\n Higher scoring candidates managed to identify why hydrated copper(II) sulfate is blue in colour.\n
\n\n Identify a metal produced by reacting its oxide with carbon or carbon monoxide. Use section 25 of the data booklet.\n
\n\n [1]\n
\n\n \n Any one of:\n \n
\n\n Zn, Cr, Fe, Cd, Co, Ni, Sn, Pb, Sb, As, Bi, Cu, Ag, Pd, Hg, Pt ✓\n
\n\n
\n\n \n Accept “Au”.\n
\n Accept name or symbol of metal.\n \n
\n Which pair of statements about electrons in this molecule is correct?\n
\n\n \n
\n
\n \n
\n [1]\n
\n\n D\n
\n\n Deduce the half-equations for the reaction at each electrode.\n
\n\n \n
\n [2]\n
\n\n Cathode (negative electrode):\n
\n\n Zn\n \n 2+\n \n + 2e\n \n −\n \n → Zn (l) ✔\n
\n\n
\n\n Anode (positive electrode):\n
\n\n 2Cl\n \n −\n \n → Cl\n \n 2\n \n (g) + 2e\n \n −\n \n
\n\n \n \n OR\n \n \n
\n\n Cl\n \n −\n \n → ½ Cl\n \n 2\n \n (g) + e\n \n −\n \n ✔\n
\n\n The 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\n \n +\n \n instead of Zn\n \n 2+\n \n . Some candidates reversed the anode and cathode equations earning only 1 of the 2 marks.\n
\n\n Which observation would explain a systematic error for an experiment involving the combustion of magnesium to find the empirical formula of its oxide?\n
\n\n
\n A. The crucible lid was slightly ajar during heating.\n
\n B. The product was a white powdery substance.\n
\n\n C. The crucible had black soot on the bottom after heating.\n
\n\n D. The flame colour during heating was yellow.\n
\n\n [1]\n
\n\n C\n
\n\n Calcium carbonate is heated to produce calcium oxide, CaO.\n
\n\n CaCO\n \n 3\n \n (s) → CaO (s) + CO\n \n 2\n \n (g)\n
\n\n Calculate the volume of carbon dioxide produced at STP when 555 g of calcium carbonate decomposes. Use sections 2 and 6 of the data booklet.\n
\n\n [2]\n
\n\n «\n \n n\n \n \n CaCO3\n \n =\n \n =» 5.55 «mol» ✓\n
\n\n «\n \n V\n \n = 5.55 mol × 22.7 dm\n \n 3\n \n mol\n \n −1\n \n =» 126 «dm\n \n 3\n \n » ✓\n
\n\n \n
\n Award\n \n [2]\n \n for correct final answer.\n \n
\n \n Accept method using pV = nRT to obtain the volume with p as either 100 kPa (126 dm\n \n 3\n \n ) or 101.3 kPa (125 dm\n \n 3\n \n ).\n \n
\n\n \n Do not penalize use of 22.4 dm\n \n 3\n \n mol\n \n –1\n \n to obtain the volume (124 dm\n \n 3\n \n ).\n \n
\n\n \n How do the following properties change down Group 17 of the periodic table?\n \n
\n\n \n \n \n
\n
\n\n [1]\n
\n\n C\n
\n\n 74 % of the candidates chose the correct trends for ionization energy and ionic radius down Group 17.\n
\n\n Explain in terms of its structure, why β-carotene appears orange in visible white light. Refer to section 17 of the data booklet.\n
\n\n [2]\n
\n\n contains many/multiple conjugated «carbon–carbon/C=C» double bonds\n
\n \n \n OR\n \n \n
\n extended system of delocalized electrons ✓\n
\n absorbs blue\n
\n \n \n OR\n \n \n
\n complementary to orange «light» ✓\n
\n
\n\n \n M1 requires the concept of many or multiple conjugated double bonds in structure for mark.\n
\n Do\n \n not\n \n accept either “conjugation”\n \n OR\n \n “double bonds” alone for M1.\n \n
\n Explain the decrease in radius from Na to Na\n \n +\n \n .\n
\n\n [2]\n
\n\n Na\n \n +\n \n has one less energy level/shell\n
\n \n \n OR\n \n \n
\n Na\n \n +\n \n has 2 energy levels/shells\n \n \n AND\n \n \n Na has 3 ✓\n
\n less shielding «in Na\n \n +\n \n so valence electrons attracted more strongly to nucleus»\n
\n \n \n OR\n \n \n
\n effective nuclear charge/Z\n \n eff\n \n greater «in Na\n \n +\n \n so valence electrons attracted more strongly to nucleus» ✓\n
\n \n
\n Accept “more protons than electrons «in Na\n \n +\n \n »”\n \n OR\n \n “less electron-electron repulsion «in Na\n \n +\n \n »” for M2.\n \n
\n Explain the electron domain geometry of SO\n \n 3\n \n .\n
\n\n [2]\n
\n\n three electron domains «attached to the central atom» ✔\n
\n\n repel/as far away as possible /120° «apart» ✔\n
\n\n Most 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.\n
\n\n \n Suggest why the existence of salts containing an ion with this formula could be predicted. Refer to section 7 of the data booklet.\n \n
\n\n [1]\n
\n\n \n same group as Mn «which forms M\n \n n\n \n O\n \n 4\n \n \n -\n \n »\n
\n \n \n OR\n \n \n
\n in group 7/has 7 valence electrons, so its «highest» oxidation state is +7\n \n [✔]\n \n \n
\n This 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.\n
\n\n Sketch the Lewis (electron dot) structure of the P\n \n 4\n \n molecule, containing only single bonds.\n
\n\n
\n\n [1]\n
\n\n \n
\n \n
\n Accept any diagram with each P joined to the other three.\n
\n \n
\n \n Accept any combination of dots, crosses and lines.\n \n
\n\n \n Additional experiments were carried out at an elevated temperature. On the axes below, sketch Maxwell–Boltzmann energy distribution curves at two temperatures T\n \n 1\n \n and T\n \n 2\n \n , where T\n \n 2\n \n > T\n \n 1\n \n .\n \n
\n\n \n \n \n
\n [2]\n
\n\n \n
\n \n peak of T\n \n 2\n \n to right of\n \n \n AND\n \n \n lower than T\n \n 1\n \n \n [✔]\n \n \n
\n\n \n lines begin at origin\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 The drawing of the two curves at T1 and T2 was generally poorly done.\n
\n\n In which of the following situations is the forward reaction spontaneous?\n
\n\n
\n A. The equilibrium constant is greater than one under standard conditions.\n
\n B. The cell potential is negative.\n
\n\n C. The Gibbs free energy change of the reverse reaction is negative.\n
\n\n D. The entropy change of the universe for the forward reaction is negative.\n
\n\n [1]\n
\n\n A\n
\n\n \n Deduce the coefficients required to complete the half-equation.\n \n
\n\n \n \n ReO\n \n 4\n \n \n −\n \n (aq) + ____H\n \n +\n \n (aq) + ____e\n \n −\n \n ⇌ [Re(OH)\n \n 2\n \n ]\n \n 2+\n \n (aq) + ____H\n \n 2\n \n O (l) E\n \n θ\n \n = +0.36 V\n \n \n
\n\n [1]\n
\n\n \n ReO\n \n \n 4\n \n \n −\n \n \n (aq) + 6H\n \n \n +\n \n \n (aq) + 3e\n \n \n −\n \n \n ⇌\n \n \n [Re(OH)\n \n \n 2\n \n \n ]\n \n \n 2+\n \n \n (aq) + 2H\n \n \n 2\n \n \n O (l)\n \n [\n \n ✔]\n \n \n \n
\n\n Surprisingly, 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.\n
\n\n Deduce the ratio of Fe\n \n 2+\n \n :Fe\n \n 3+\n \n in Fe\n \n 3\n \n O\n \n 4\n \n .\n
\n\n [1]\n
\n\n 1:2 ✔\n
\n\n \n Accept 2 Fe\n \n 3+\n \n : 1 Fe\n \n 2+\n \n \n
\n \n Do\n \n not\n \n accept 2:1 only\n \n
\n Determine 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
\n\n (If you do not have answers for b(i) and b(ii), use Δ\n \n H\n \n = 190 kJ and Δ\n \n S\n \n = 180 J K\n \n −1\n \n , but these are not the correct answers.)\n
\n\n [2]\n
\n\n «spontaneous» if Δ\n \n G\n \n = Δ\n \n H\n \n −\n \n T\n \n Δ\n \n S\n \n < 0\n
\n \n \n OR\n \n \n
\n Δ\n \n H\n \n <\n \n T\n \n Δ\n \n S\n \n ✓\n
\n «\n \n T\n \n >\n \n =» 1112 «K» ✓\n
\n\n \n
\n Award\n \n [2]\n \n for correct final answer.\n \n
\n \n Accept “1056 K” if both of the incorrect values are used to solve the problem.\n \n
\n\n \n Do\n \n not\n \n award M2 for any negative T value.\n \n
\n\n The transition metal copper, Cu, is used in electrical circuits given its high electrical conductivity. Outline why it is preferred over s- and p-block metals.\n
\n\n [1]\n
\n\n «Cu has» additional delocalization of d-electrons ✓\n
\n\n \n What is the IUPAC name of the following molecule?\n \n
\n\n \n \n \n
\n \n \n A. 2-bromo-3-ethylbutane\n \n \n
\n\n \n \n B. 3-methyl-4-bromopentane\n \n \n
\n\n \n \n C. 2-ethyl-3-bromobutane\n \n \n
\n\n \n \n D. 2-bromo-3-methylpentane\n \n \n
\n\n [1]\n
\n\n D\n
\n\n 57 % of candidates could correctly apply IUPAC nomenclature.\n
\n\n \n Outline, with a reason, another property that could be monitored to measure the rate of this reaction.\n \n
\n\n [2]\n
\n\n \n \n \n ALTERNATIVE 1\n \n \n
\n colour\n \n [✔]\n \n \n
\n \n Br\n \n 2\n \n /reactant is coloured «Br\n \n –\n \n (aq)/product is not»\n \n [✔]\n \n \n
\n\n \n \n \n Note:\n \n Do\n \n not\n \n accept “changes in temperature” or “number of bubbles”.\n \n \n
\n\n
\n\n \n \n \n ALTERNATIVE 2\n \n \n
\n conductivity\n \n [✔]\n \n \n
\n \n greater/increased concentration of ions in products\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n ALTERNATIVE 3\n \n \n
\n mass/pressure\n \n [✔]\n \n \n
\n \n gas is evolved/produced\n \n [✔]\n \n \n
\n\n \n \n \n Note\n \n : Do\n \n not\n \n accept “mass of products is less than mass of reactants”.\n \n \n
\n\n \n
\n \n \n ALTERNATIVE 4\n \n \n
\n pH\n \n [✔]\n \n \n
\n \n methanoic acid is weak\n \n \n AND\n \n \n HBr is strong\n
\n \n \n OR\n \n \n
\n increase in [H\n \n +\n \n ]\n \n [✔]\n \n \n
\n The reaction rate was originally monitored by measuring the volume of CO\n \n 2\n \n produced. Students needed to propose another method for this reaction, with a reason, that could be used to measure the rate. There were several possible correct answers and most students received at least one mark with many receiving both marks. The most common incorrect answer involved changes in temperature.\n
\n\n \n What is the molar mass, in\n \n \n \n , of a compound if\n \n of the compound has a mass of\n \n ?\n \n
\n\n \n A.\n \n
\n \n
\n \n B.\n \n
\n \n
\n \n C.\n \n
\n \n
\n \n D.\n \n \n
\n\n [1]\n
\n\n A\n
\n\n 70% of candidates correctly found the molar mass of a substance given the mass of 0.200 mol. Good to see that very few were fooled by the same answer but incorrect significant figures.\n
\n\n Predict the shape of the hydrogen sulfide molecule.\n
\n\n [1]\n
\n\n bent/non-linear/angular/v-shaped✔\n
\n\n The activation energy of a reaction can be obtained from the rate constant,\n \n k\n \n , and the absolute temperature,\n \n . Which graph of these quantities produces a straight line?\n
\n\n
\n A.\n \n k\n \n against\n \n
\n B.\n \n k\n \n against\n \n
\n\n C. ln\n \n k\n \n against\n \n
\n\n D. ln\n \n k\n \n against\n \n
\n\n [1]\n
\n\n D\n
\n\n Over 80% of candidates correctly identified the variables that should be plotted to determine activation energy from rate data, with the other distractors being about equally favoured.\n
\n\n State, with a reason, the effect of an increase in temperature on the position of this equilibrium.\n
\n\n [1]\n
\n\n «shifts» left/towards reactants\n \n \n AND\n \n \n «forward reaction is» exothermic/ΔH is negative ✔\n
\n\n Which molecule produces this\n \n 1\n \n H-NMR spectrum?\n
\n\n \n
\n \n [Source: SDBS, National Institute of Advanced Industrial Science and Technology.]\n \n
\n\n
\n A. CH\n \n 3\n \n COOCH\n \n 3\n \n
\n B. CH\n \n 3\n \n COCH\n \n 3\n \n
\n\n C. CH\n \n 3\n \n CHO\n
\n\n D. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 3\n \n
\n\n [1]\n
\n\n B\n
\n\n Which two colliding species have the highest probability of having the proper orientation for a reaction to occur?\n
\n\n
\n A. Cl• + Cl•\n
\n B. Cl• + CH\n \n 3\n \n •\n
\n\n C. HCl + CH\n \n 2\n \n \n CH\n \n 2\n \n
\n\n D. CF\n \n 3\n \n Cl + O\n \n 3\n \n
\n\n [1]\n
\n\n A\n
\n\n The complete combustion of 20.0 cm\n \n 3\n \n of a gaseous hydrocarbon, C\n \n x\n \n H\n \n y\n \n , produces 80.0 cm\n \n 3\n \n of gaseous products. This volume reduces to 40.0 cm\n \n 3\n \n when the water vapour present condenses. All volumes are measured at the same temperature and pressure.\n
\n\n
\n What is the molecular formula of the hydrocarbon?\n
\n
\n A. CH\n \n 4\n \n
\n B. C\n \n 2\n \n H\n \n 2\n \n
\n\n C. C\n \n 2\n \n H\n \n 4\n \n
\n\n D. C\n \n 3\n \n H\n \n 6\n \n
\n\n [1]\n
\n\n C\n
\n\n Calculate the mass of CO\n \n 2\n \n produced per 1000 kJ of heat obtained from the combustion of ethanol. Use section 13 of the data booklet.\n
\n\n [2]\n
\n\n «1000/1367=» 0.7315 «mol ethanol» ✓\n
\n\n «C\n \n 2\n \n H\n \n 6\n \n O + 3O\n \n 2\n \n → 2CO\n \n 2\n \n + 3H\n \n 2\n \n O / each ethanol forms two carbon dioxides»\n
\n\n «0.7315 × 2 × 44.01 g mol\n \n −1\n \n = »64 «g» ✓\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n Suggest how the extent of decomposition could be measured.\n
\n\n [1]\n
\n\n use colorimeter\n
\n \n \n OR\n \n \n
\n change in colour\n
\n \n \n OR\n \n \n
\n change in volume\n
\n \n \n OR\n \n \n
\n change in pressure ✔\n
\n \n Accept suitable instruments, e.g. pressure probe/oxygen sensor.\n \n
\n\n Which reaction would be expected to have the largest Arrhenius (pre-exponential) factor, A, at constant temperature?\n
\n\n
\n A. H (g)\n \n I (g) → HI (g)\n
\n B. H\n \n 2\n \n (g)\n \n I\n \n 2\n \n (g) → 2HI (g)\n
\n\n C. 2HCl (g) → H\n \n 2\n \n (g)\n \n Cl\n \n 2\n \n (g)\n
\n\n D. H\n \n 2\n \n \n C\n \n 2\n \n H\n \n 4\n \n → C\n \n 2\n \n H\n \n 6\n \n
\n\n [1]\n
\n\n A\n
\n\n Suggest how the extent of decomposition could be measured.\n
\n\n [1]\n
\n\n use colorimeter\n
\n \n \n OR\n \n \n
\n change in colour\n
\n \n \n OR\n \n \n
\n change in volume\n
\n \n \n OR\n \n \n
\n change in pressure ✔\n
\n \n Accept suitable instruments, e.g. pressure probe/oxygen sensor.\n \n
\n\n \n Aqueous sodium hydrogencarbonate has a pH of approximately 7 at 298 K.\n \n
\n\n \n Sketch a graph of pH against volume when 25.0cm\n \n 3\n \n of 0.100 mol dm\n \n −3\n \n NaOH (aq) is gradually added to 10.0cm\n \n 3\n \n of 0.0500 mol dm\n \n −3\n \n NaHCO\n \n 3\n \n (aq).\n \n
\n\n \n \n \n
\n [2]\n
\n\n \n
\n \n S-shaped curve from ~7 to between 12 and 14\n \n [✔]\n \n \n
\n\n \n equivalence point at 5 cm\n \n 3\n \n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept starting point >6~7.\n \n \n
\n\n Most 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.\n
\n\n For the reaction\n \n \n 2\n \n (g) + 3Cl\n \n 2\n \n (g)\n \n 2\n \n Cl\n \n 3\n \n (g) at a certain temperature, the equilibrium concentrations are (in mol dm\n \n −3\n \n ):\n
\n\n [\n \n \n 2\n \n ] = 0.20, [Cl\n \n 2\n \n ] = 0.20, [\n \n Cl\n \n 3\n \n ] = 2.0\n
\n\n What is the value of\n \n K\n \n \n c\n \n ?\n
\n\n
\n A. 0.25\n
\n B. 50\n
\n\n C. 2500\n
\n\n D. 5000\n
\n\n [1]\n
\n\n C\n
\n\n Over 72% of the candidates were able to calculate value of\n \n K\n \n \n c\n \n from the concentrations given, with the other distractors being about equally favoured. Some teachers expressed concern at candidates having to do this without a calculator, but the figures were quite easy to manipulate and the number answering correctly indicates it did not hinder many candidates.\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 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 Which element has the highest metallic character in Group 14?\n
\n\n
\n A. C\n
\n B. Si\n
\n\n C. Ge\n
\n\n D. Sn\n
\n\n [1]\n
\n\n D\n
\n\n Draw\n \n one\n \n Lewis (electron dot) structure of the sulfate ion.\n
\n\n [1]\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 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 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 Which structure represents a repeating unit of a polymer formed from propene?\n
\n\n A. –CH\n \n 2\n \n –CH(CH\n \n 3\n \n )–\n
\n\n B. –CH\n \n 2\n \n –CH\n \n 2\n \n –CH\n \n 2\n \n –\n
\n\n C. –CH(CH\n \n 3\n \n )–CH(CH\n \n 3\n \n )–\n
\n\n D. –CH\n \n 2\n \n –CH\n \n 2\n \n –\n
\n\n [1]\n
\n\n A\n
\n\n 44% of the candidates identified the repeating unit of the polymer formed from propene. The most commonly chosen distractor was a straight chain.\n
\n\n \n Cl\n \n 2\n \n (g) +\n \n \n \n (g)\n \n \n Cl (g)\n \n K\n \n \n c\n \n = 454\n
\n\n What is the\n \n K\n \n \n c\n \n value for the reaction below?\n
\n\n 2\n \n Cl (g)\n \n Cl\n \n 2\n \n (g) +\n \n \n \n (g)\n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n D\n
\n\n Question 18 was identified as being mathematically challenging. However, as all answers were shown as calculations rather than as final answers, this does not seem a reasonable complaint.\n
\n\n \n What is the enthalpy change of the reaction?\n \n
\n\n \n C\n \n 6\n \n H\n \n 14\n \n (l) → C\n \n 2\n \n H\n \n 4\n \n (g) + C\n \n 4\n \n H\n \n 10\n \n (g)\n \n
\n\n \n \n \n
\n \n \n A. + 1411 + 2878 + 4163\n
\n \n \n
\n \n \n B. + 1411 − 2878 − 4163\n
\n \n \n
\n \n \n C. + 1411 + 2878 − 4163\n
\n \n \n
\n \n \n D. − 1411 − 2878 + 4163\n \n \n
\n\n [1]\n
\n\n C\n
\n\n \n Classify PVC and polyethene terephthalate, PET, as addition or condensation polymers and deduce the structural formulas.\n \n
\n\n \n \n \n
\n [3]\n
\n\n \n
\n \n PVC\n \n : addition\n \n \n AND\n \n PET\n \n : condensation ✔\n
\n\n structure of PVC monomer ✔\n
\n\n structure of PET monomers ✔\n
\n\n
\n \n Accept full\n \n OR\n \n condensed structural formulas.\n \n
\n This question on PVC and PET was very well answered. All three marks for their correct classification and the corresponding structures of their monomers was frequently scored.\n
\n\n Which compound is both volatile and soluble in water?\n
\n\n
\n A. NaCl\n
\n B. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 3\n \n
\n\n C. CH\n \n 3\n \n OH\n
\n\n D. C\n \n 12\n \n H\n \n 22\n \n O\n \n 11\n \n
\n\n [1]\n
\n\n C\n
\n\n \n Classify polybutadiene as either an addition or condensation polymer, giving a reason.\n \n
\n\n [1]\n
\n\n \n addition\n \n \n AND\n \n \n not two different functional groups reacting\n
\n \n \n OR\n \n \n
\n addition\n \n \n AND\n \n \n formed by breaking one bond of the carbon–carbon double bonds\n
\n \n OR\n \n
\n addition\n \n \n AND\n \n \n empirical formula of monomer equals empirical formula of polymer\n
\n \n OR\n \n
\n addition\n \n \n AND\n \n \n no atoms removed/all atoms accounted for/no loss of water/ammonia/inorganic by-product/small molecules\n
\n \n OR\n \n
\n addition\n \n \n AND\n \n \n atom economy/efficiency is 100 %\n
\n \n OR\n \n
\n addition\n \n \n AND\n \n \n there is only one «reaction» product\n \n [✔]\n \n \n
\n Some candidates failed to score the mark as they did not give a reason for classifying polybutadiene as an addition polymer.\n
\n\n Outline why metals, like iron, can conduct electricity.\n
\n\n [1]\n
\n\n mobile/delocalized «sea of» electrons\n
\n\n \n The graph below shows the Maxwell–Boltzmann distribution of molecular energies at a particular temperature.\n \n
\n\n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5983/2f.PNG\")
\n
\n \n \n The rate at which dinitrogen monoxide decomposes is significantly increased by a metal oxide catalyst.\n \n \n
\n\n \n \n Annotate and use the graph to outline why a catalyst has this effect.\n \n \n
\n\n [2]\n
\n\n \n
\n \n catalysed and uncatalysed E\n \n a\n \n marked on graph\n \n \n AND\n \n \n with the catalysed being at lower energy\n \n [✔]\n \n
\n \n
\n
\n\n \n «for catalysed reaction» greater proportion of/more molecules have E ≥ E\n \n a\n \n / E > E\n \n a\n \n
\n \n \n OR\n \n \n
\n «for catalysed reaction» greater area under curve to the right of the E\n \n a\n \n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “more molecules have the activation energy”.\n \n \n
\n\n The 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\n \n E\n \n (particles) >\n \n E\n \n \n a\n \n or did so too vaguely.\n
\n\n Determine the volume, in dm\n \n 3\n \n , of 0.015 mol dm\n \n −3\n \n calcium hydroxide solution needed to neutralize 35.0 cm\n \n 3\n \n of 0.025 mol dm\n \n −3\n \n HCl (aq).\n
\n\n [2]\n
\n\n «\n \n n\n \n \n HCl\n \n = 0.0350 dm\n \n 3\n \n × 0.025 mol dm\n \n −3\n \n =» 0.00088 «mol»\n
\n\n \n \n OR\n \n \n
\n \n n\n \n \n Ca(OH)2\n \n =\n \n \n n\n \n \n HCl\n \n /0.00044 «mol» ✓\n
\n
\n «\n \n V\n \n =\n \n =» 0.029 «dm\n \n 3\n \n » ✓\n
\n \n
\n Award\n \n [2]\n \n for correct final answer.\n \n
\n \n Award\n \n [1 max]\n \n for 0.058 «dm\n \n 3\n \n ».\n \n
\n\n Which conditions best favour oxidation of primary alcohols directly to carboxylic acids?\n
\n\n
\n A. Excess acidified potassium dichromate (VI) and distillation\n
\n B. Excess acidified potassium dichromate (VI) and reflux\n
\n\n C. Few drops of acidified potassium dichromate (VI) and distillation\n
\n\n D. Few drops of acidified potassium dichromate (VI) and reflux\n
\n\n [1]\n
\n\n B\n
\n\n 51% of the candidates identified the conditions that favour the oxidation of primary alcohols directly to carboxylic acids. The question had good discrimination between high-scoring and low-scoring candidates.\n
\n\n Draw the Lewis structure of SO\n \n 3\n \n .\n
\n\n [1]\n
\n\n \n ✔\n
\n
\n\n \n Note:\n \n
\n\n \n
\n \n Accept any of the above structures as formal charge is not being assessed.\n \n
\n\n Drawing the Lewis structure of SO\n \n 3\n \n 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.\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 \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 Identify the type of reaction.\n
\n\n [1]\n
\n\n «nucleophilic» substitution\n
\n \n \n OR\n \n \n
\n SN2 ✔\n
\n \n
\n Accept “hydrolysis”.\n \n
\n \n Accept SN1\n \n
\n\n Excellent performance on the type of reaction but with some incorrect answers such as alkane substitution, free radical substitution or electrophilic substitution.\n
\n\n State how the use of a catalyst affects the position of the equilibrium.\n
\n\n [1]\n
\n\n same/unaffected/unchanged ✔\n
\n\n Good performance; however, some misread the question as asking for the effect of a catalyst on equilibrium, rather than on the position of equilibrium.\n
\n\n \n What is the intercept on the\n \n y\n \n -axis when a graph of ln\n \n k\n \n is plotted against\n \n on the\n \n x\n \n -axis?\n \n
\n\n \n \n \n
\n\n \n \n A. ln\n \n A\n \n \n \n
\n\n \n \n B.\n \n \n \n
\n\n \n \n C.\n \n \n \n
\n\n \n \n D.\n \n \n \n \n \n
\n\n [1]\n
\n\n A\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 \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 Explain why there is a large increase from the 8th to the 9th ionization energy of iron.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n\n IE\n \n 9\n \n : electron in lower energy level\n
\n \n \n OR\n \n \n
\n IE\n \n 9\n \n : more stable/full electron level ✔\n
\n
\n IE\n \n 9\n \n : electron closer to nucleus\n
\n \n \n OR\n \n \n
\n IE\n \n 9\n \n : electron more tightly held by nucleus ✔\n
\n
\n IE\n \n 9\n \n : less shielding by «complete» inner levels ✔\n
\n Despite 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.\n
\n\n \n Suggest why it is surprising that dinitrogen monoxide dissolves in water to give a neutral solution.\n \n
\n\n [1]\n
\n\n \n oxides of nitrogen/non-metals are «usually» acidic\n \n [✔]\n \n \n
\n\n Only a quarter of the candidates answered correctly. Some simply stated that N\n \n 2\n \n O forms HNO\n \n 3\n \n with water which did not gain the mark.\n
\n\n A student heated a known mass of zinc powder in an open crucible until there was no further mass change and recorded the final mass.\n
\n\n What would the student be able to derive from this data?\n
\n\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\n
\n\n Outline the requirements for a collision between reactants to yield products.\n
\n\n [2]\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 Generally 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.\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 dilute the titrant to use larger volumes «of titrant» ✔\n
\n\n ensure spinach leaf fragments are the same size ✔\n
\n\n Outline how\n \n one\n \n calcium compound in the lime cycle can reduce a problem caused by acid deposition.\n
\n\n [1]\n
\n\n «add» Ca(OH)\n \n 2\n \n /CaCO\n \n 3\n \n /CaO\n \n \n AND\n \n \n to «acidic» water/river/lake/soil\n
\n \n \n OR\n \n \n
\n «use» Ca(OH)\n \n 2\n \n /CaCO\n \n 3\n \n /CaO in scrubbers «to prevent release of acidic pollution» ✓\n
\n
\n\n \n Accept any correct name for any of the calcium compounds listed.\n \n
\n\n 0.20 mol of magnesium is mixed with 0.10 mol of hydrochloric acid.\n
\n\n \n
\n\n Which is correct?\n
\n\n \n
\n [1]\n
\n\n C\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 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 Which of these changes would shift the equilibrium to the right?\n
\n\n [Co(H\n \n 2\n \n O)\n \n 6\n \n ]\n \n 2+\n \n (aq) + 4Cl\n \n −\n \n (aq)\n \n [CoCl\n \n 4\n \n ]\n \n 2−\n \n (aq) + 6H\n \n 2\n \n O (l)\n
\n\n I. Addition of 0.01 M HCl\n
\n II. Addition of concentrated HCl\n
\n III. Evaporation of water\n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n C\n
\n\n This proved to be the most difficult question on the paper with <30% of candidates deducing which changes would shift the equilibrium to the right and success rate seemed to be only weakly related to overall performance. The most popular answer was (D), indicating that most students did not realise that adding dilute HCl would add much more water than Cl\n \n −\n \n .\n
\n\n What are the most likely reactions ethene and benzene will undergo?\n
\n\n
\n| \n | \n\n \n Ethene\n \n | \n\n \n Benzene\n \n | \n
| \n A.\n | \n\n Addition\n | \n\n Substitution\n | \n
| \n B.\n | \n\n Addition\n | \n\n Addition\n | \n
| \n C.\n | \n\n Substitution\n | \n\n Substitution\n | \n
| \n D.\n | \n\n Substitution\n | \n\n Addition\n | \n
\n
\n\n [1]\n
\n\n A\n
\n\n \n MnO\n \n 2\n \n is another possible catalyst for the reaction. State the IUPAC name for MnO\n \n 2\n \n .\n \n
\n\n [1]\n
\n\n \n manganese(IV) oxide\n \n
\n\n \n \n \n OR\n \n \n \n
\n\n \n manganese dioxide\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “manganese(IV) dioxide”.\n \n \n
\n\n A well answered question. Very few candidates had problem with nomenclature.\n
\n\n Determine the coefficients that balance the equation for the reaction of lithium with water.\n
\n\n \n
\n [1]\n
\n\n 2 Li (s) + 2 H\n \n 2\n \n O (l) → 2 LiOH (aq) + H\n \n 2\n \n (g) ✔\n
\n\n This 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.\n
\n\n Determine the oxidation state of nitrogen in Mg\n \n 3\n \n N\n \n 2\n \n and in NH\n \n 3\n \n .\n
\n\n \n
\n [1]\n
\n\n \n Mg\n \n 3\n \n N\n \n 2\n \n : -3\n \n
\n \n \n AND\n \n \n
\n \n NH\n \n 3\n \n : -3 ✔\n \n
\n \n
\n Do not accept 3 or 3-\n \n
\n About 40% of students managed to correctly determine both the oxidation states, as -3, with errors being about equally divided between the two compounds.\n
\n\n Which enthalpy changes can be calculated using only bond enthalpy data?\n
\n\n
\n I. N\n \n 2\n \n (g)\n \n 2H\n \n 2\n \n (g) → N\n \n 2\n \n H\n \n 4\n \n (g)\n
\n II. CH\n \n 4\n \n (g)\n \n 2O\n \n 2\n \n (g) → 2H\n \n 2\n \n O (l)\n \n CO\n \n 2\n \n (g)\n
\n III. H\n \n 2\n \n (g)\n \n Cl\n \n 2\n \n (g) → 2HCl (g)\n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n B\n
\n\n State how the use of a catalyst affects the position of the equilibrium.\n
\n\n [1]\n
\n\n same/unaffected/unchanged ✔\n
\n\n Good performance; however, some misread the question as asking for the effect of a catalyst on equilibrium, rather than on the position of equilibrium.\n
\n\n A pH probe is placed in a small volume of 0.10 mol dm\n \n -3\n \n solution of hydrochloric acid. The pH is recorded while a steady stream of distilled water is added to the acid at constant temperature.\n
\n\n
\n\n \n
\n start at pH = 1 ✔\n
\n curve with decreasing gradient ✔\n
\n must finish below pH = 7 ✔\n
\n Determine the enthalpy change, Δ\n \n H\n \n , for the Haber–Bosch process, in kJ. Use Section 11 of the data booklet.\n
\n\n [3]\n
\n\n \n bonds broken\n \n : N≡N + 3(H-H) /«1 mol×»945 «kJ mol\n \n –1\n \n » + 3«mol»×436 «kJ mol\n \n –1\n \n » / 945 «kJ» + 1308 «kJ» / 2253 «kJ» ✔\n
\n\n \n bonds formed\n \n : 6(N-H) / 6«mol»×391 «kJ mol\n \n –1\n \n » / 2346 «kJ» ✔\n
\n\n Δ\n \n H\n \n = «2253 kJ - 2346 kJ = » -93 «kJ» ✔\n
\n\n
\n\n \n Award\n \n [2 max]\n \n for (+)93 «kJ».\n \n
\n\n Good 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.\n
\n\n \n Which is a major product of the electrophilic addition of hydrogen chloride to propene?\n \n
\n\n \n A. ClCH\n \n 2\n \n CH=CH\n \n 2\n \n \n
\n\n \n B. CH\n \n 3\n \n CH(Cl)CH\n \n 3\n \n \n
\n\n \n C. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n Cl\n \n
\n\n \n D. CH\n \n 3\n \n CH=CHCl\n \n
\n\n [1]\n
\n\n B\n
\n\n Markovnikov addition was handled much better by higher scoring candidates.\n
\n\n \n Which can be reduced to an aldehyde?\n \n
\n\n \n A. Butanone\n
\n \n
\n \n B. Butan-1-ol\n
\n \n
\n \n C. Butanoic acid\n
\n \n
\n \n D. Butan-2-ol\n \n
\n\n [1]\n
\n\n C\n
\n\n Which statement is correct for both voltaic and electrolytic cells?\n
\n\n A. The oxidation reaction releases electrons.\n
\n\n B. The oxidation reaction occurs at the positive electrode.\n
\n\n C. The cathode is negative.\n
\n\n D. Electrons flow through the electrolyte.\n
\n\n [1]\n
\n\n A\n
\n\n \n Data collected from a larger number of silicon samples could also be plotted to determine the density using the following axes.\n \n
\n\n \n \n \n
\n \n \n Which statements are correct?\n \n \n
\n\n \n \n I. The density is the slope of the graph.\n
\n II. The data will show that mass is proportional to volume.\n
\n III. The best-fit line should pass through the origin.\n \n \n
\n \n \n A. I and II only\n \n \n
\n\n \n \n B. I and III only\n \n \n
\n\n \n \n C. II and III only\n \n \n
\n\n \n \n D. I, II and III\n \n \n
\n\n [1]\n
\n\n D\n
\n\n This question was answered in two different ways. 33 % of the candidates chose the correct answer (D) considering what would be true about a mass-volume graph for silicon samples. However, 39 % chose distractor (B) as they considered the silicon samples in the previous question which did not show a directly proportional relationship. Both answers were accepted as we agree that the wording was ambiguous (it will be amended before publication).\n
\n\n \n The following data were recorded for determining the density of three samples of silicon, Si.\n \n
\n\n \n \n \n
\n \n \n Which average density value, in g cm\n \n −3\n \n , has been calculated to the correct number of significant figures?\n \n \n
\n\n \n \n A. 2\n \n \n
\n\n \n \n B. 2.3\n \n \n
\n\n \n \n C. 2.27\n \n \n
\n\n \n \n D. 2.273\n \n \n
\n\n [1]\n
\n\n B\n
\n\n Candidates found this question relatively challenging and only 67 % chose the answer with two significant figures. The most commonly chosen distractor was C which expressed the answer to three significant figures.\n
\n\n Which causes acid deposition?\n
\n\n A. SO\n \n 2\n \n
\n\n B. SiO\n \n 2\n \n
\n\n C. SrO\n
\n\n D. CO\n \n 2\n \n
\n\n [1]\n
\n\n A\n
\n\n In which group of ions and molecules are electrons delocalized in all species?\n
\n\n
\n A. CH\n \n 3\n \n COOH, O\n \n 3\n \n , C\n \n 60\n \n
\n B. CH\n \n 3\n \n COO\n \n −\n \n , NO\n \n 2\n \n \n −\n \n , C(graphite)\n
\n\n C. C\n \n 2\n \n H\n \n 2\n \n , (COOH)\n \n 2\n \n , C(diamond)\n
\n\n D. C\n \n 2\n \n H\n \n 4\n \n , NO\n \n 2\n \n \n +\n \n , SiO\n \n 2\n \n
\n\n [1]\n
\n\n B\n
\n\n Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.\n
\n\n [1]\n
\n\n allows them to explain the properties of different compounds/substances\n
\n \n \n OR\n \n \n
\n enables them to generalise about substances\n
\n \n \n OR\n \n \n
\n enables them to make predictions ✔\n
\n \n
\n Accept other valid answers.\n \n
\n \n An iron nail and a copper nail are inserted into a lemon.\n \n
\n\n \n \n \n
\n \n \n Explain why a potential is detected when the nails are connected through a voltmeter.\n \n \n
\n\n [2]\n
\n\n \n lemon juice is the electrolyte\n
\n \n \n OR\n \n \n
\n lemon juice allows flow of ions\n
\n \n \n OR\n \n \n
\n each nail/metal forms a half-cell with the lemon juice\n \n [✔]\n \n \n
\n \n \n \n Note:\n \n Accept “lemon juice acts as a salt bridge”.\n \n \n
\n\n \n \n Any one of\n \n :\n
\n iron is higher than copper in the activity series\n
\n \n \n OR\n \n \n
\n each half-cell/metal has a different redox/electrode potential\n \n [✔]\n \n \n
\n \n \n \n Note:\n \n Accept “iron is more reactive than copper”.\n \n \n
\n\n \n iron is oxidized\n
\n \n \n OR\n \n \n
\n Fe → Fe\n \n 2+\n \n + 2e\n \n –\n \n
\n \n \n OR\n \n \n
\n Fe → Fe\n \n 3+\n \n + 3e\n \n −\n \n
\n \n \n OR\n \n \n
\n iron is anode/negative electrode of cell\n \n [✔]\n \n \n
\n \n copper is cathode/positive electrode of cell\n
\n \n \n OR\n \n \n
\n reduction occurs at the cathode\n
\n \n \n OR\n \n \n
\n 2H\n \n +\n \n + 2e\n \n −\n \n → H\n \n 2\n \n \n [✔]\n \n \n
\n \n electrons flow from iron to copper\n \n [✔]\n \n \n
\n\n \n \n \n \n
\n\n \n \n Notes:\n
\n Accept “lemon juice acts as a salt bridge”.\n
\n Accept “iron is more reactive than copper”.\n \n \n
\n Very 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 Which attacking species is matched with its mechanism of reaction?\n
\n\n
\n \n
\n [1]\n
\n\n D\n
\n\n Explain the general increase in trend in the first ionization energies of the period 3 elements, Na to Ar.\n
\n\n [2]\n
\n\n increasing number of protons\n
\n \n \n OR\n \n \n
\n increasing nuclear charge ✔\n
\n «atomic» radius/size decreases\n
\n \n \n OR\n \n \n
\n same number of shells/electrons occupy same shell\n
\n \n \n OR\n \n \n
\n similar shielding «by inner electrons» ✔\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 (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 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
\n\n \n Which element is in group 13?\n \n
\n\n \n
\n [1]\n
\n\n B\n
\n\n The majority of candidates could identify the group an element belongs in based on its successive ionization energies.\n
\n\n \n Distinguish ultraviolet light from visible light in terms of wavelength and energy.\n \n
\n\n [1]\n
\n\n \n «UV» shorter wavelength\n \n \n AND\n \n \n higher energy «than visible» ✔\n \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 \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 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 \n Which alcohol would produce a carboxylic acid when heated with acidified potassium dichromate(VI)?\n \n
\n\n \n A. propan-2-ol\n \n
\n\n \n B. butan-1-ol\n \n
\n\n \n C. 2-methylpropan-2-ol\n \n
\n\n \n D. pentan-3-ol\n \n
\n\n [1]\n
\n\n B\n
\n\n 75 % of candidates could correctly identify the oxidation of a primary alcohol to form carboxylic acid.\n
\n\n \n Which compound with the molecular formula\n \n \n \n has this high resolution\n \n \n \n ?\n \n
\n\n \n
\n \n From: libretexts.org. Courtesy of Chris Schaller, Professor (Chemistry)\n \n
\n \n at College of Saint Benedict/Saint John’s University.\n \n
\n
\n\n \n A. but-3-en-2-ol,\n \n \n \n
\n \n
\n \n B. butanal,\n \n \n \n
\n \n
\n \n C. butanone,\n \n \n \n
\n \n
\n \n D. but-3-en-1-ol,\n \n \n
\n\n [1]\n
\n\n B\n
\n\n Just over 50% of candidates could identify the correct compound from high resolution 1H NMR data. The question had a relatively high discriminatory index.\n
\n\n \n Which technique is used to detect the isotopes of an element?\n \n
\n\n \n A. Mass spectrometry\n
\n \n
\n \n B. Infrared spectroscopy\n
\n \n
\n \n C. Titration\n
\n \n
\n \n D. Recrystallization\n \n
\n\n [1]\n
\n\n A\n
\n\n A weak base is titrated with a strong acid. Which value of p\n \n K\n \n \n b\n \n can be estimated from this titration curve?\n
\n\n \n
\n A. 11.3\n
\n\n B. 9.2\n
\n\n C. 4.8\n
\n\n D. 1.8\n
\n\n [1]\n
\n\n C\n
\n\n Only 29% of candidates determined p\n \n K\n \n \n b\n \n from the titration curve. The majority of the candidates made the mistake of using the pH without calculating the corresponding pOH.\n
\n\n Sketch the first eight successive ionisation energies of sulfur.\n
\n\n \n
\n [2]\n
\n\n \n
\n two regions of small increases\n \n \n AND\n \n \n a large increase between them✔\n
\n\n large increase from 6th to 7th ✔\n
\n\n \n
\n Accept line/curve showing these trends.\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 systematic error\n \n \n AND\n \n \n all values «equally» inaccurate ✔\n
\n\n \n Which of the following shows a general increase across period 3 from\n \n to\n \n \n \n ?\n \n
\n\n \n A. Ionic radius\n
\n \n
\n \n B. Atomic radius\n
\n \n
\n \n C. Ionization energy\n
\n \n
\n \n D. Melting point\n \n
\n\n [1]\n
\n\n C\n
\n\n Although 75% of candidates had the correct answer, this question about trends across period 3 was not a very good discriminator.\n
\n\n Explain which form of the anthocyanin,\n \n X\n \n or\n \n Y\n \n , predominates at a pH of 12.\n
\n\n [1]\n
\n\n X\n \n \n AND\n \n \n «high pH moves» equilibrium left\n
\n \n \n OR\n \n \n
\n X\n \n \n AND\n \n \n loss of proton/H\n \n +\n \n «at high pH/due to excess OH\n \n −\n \n (aq)»\n
\n \n \n OR\n \n \n
\n X\n \n \n AND\n \n \n positive ion unlikely «at high pH» ✓\n
\n What is the correct comparison of H–N–H bond angles in NH2-, NH3, and NH4+?\n
\n\n
\n A. NH\n \n 2\n \n \n −\n \n < NH\n \n 3\n \n < NH\n \n 4\n \n \n +\n \n
\n B. NH\n \n 4\n \n \n +\n \n < NH\n \n 3\n \n < NH\n \n 2\n \n \n −\n \n
\n\n C. NH\n \n 3\n \n < NH\n \n 2\n \n \n −\n \n < NH\n \n 4\n \n \n +\n \n
\n\n D. NH\n \n 3\n \n < NH\n \n 4\n \n \n +\n \n < NH\n \n 2\n \n \n −\n \n
\n\n [1]\n
\n\n A\n
\n\n 20 cm\n \n 3\n \n of gas A reacts with 20 cm\n \n 3\n \n of gas B to produce 10 cm\n \n 3\n \n of gas A\n \n x\n \n B\n \n y\n \n and 10 cm\n \n 3\n \n of excess gas A. What are the correct values for subscripts\n \n x\n \n and\n \n y\n \n in the empirical formula of the product A\n \n x\n \n B\n \n y\n \n (g)?\n
\n\n
\n| \n | \n\n \n x\n \n | \n\n \n y\n \n | \n
| \n A.\n | \n\n 2\n | \n\n 1\n | \n
| \n B.\n | \n\n 2\n | \n\n 2\n | \n
| \n C.\n | \n\n 1\n | \n\n 1\n | \n
| \n D.\n | \n\n 1\n | \n\n 2\n | \n
\n [1]\n
\n\n D\n
\n\n 8.8 g of an oxide of nitrogen contains 3.2 g of oxygen. What is the empirical formula of the compound?\n
\n\n A. N\n \n 2\n \n O\n \n 5\n \n
\n\n B. N\n \n 2\n \n O\n
\n\n C. NO\n \n 2\n \n
\n\n D. NO\n
\n\n [1]\n
\n\n B\n
\n\n \n Which contains the greatest number of moles of oxygen atoms?\n \n
\n\n \n A. 0.05 mol Mg(NO\n \n 3\n \n )\n \n 2\n \n
\n \n
\n \n B. 0.05 mol C\n \n 6\n \n H\n \n 4\n \n (NO\n \n 2\n \n )\n \n 2\n \n
\n \n
\n \n C. 0.1 mol H\n \n 2\n \n O\n
\n \n
\n \n D. 0.1 mol NO\n \n 2\n \n \n
\n\n [1]\n
\n\n A\n
\n\n \n Which is correct for all solid ionic compounds?\n \n
\n\n \n A. High volatility\n
\n \n
\n \n B. Poor electrical conductivity\n
\n \n
\n \n C. Low melting point\n
\n \n
\n \n D. Good solubility in water\n \n
\n\n [1]\n
\n\n B\n
\n\n \n Iron is a stronger reducing agent than silver.\n \n
\n\n \n What is correct when this voltaic cell is in operation?\n \n
\n\n \n
\n [1]\n
\n\n D\n
\n\n 53% of candidates identified the anode, cathode, AND direction of electron flow in a voltaic cell, with misconceptions being equally distributed. This was answered much better by higher scoring candidates.\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 sulfuric acid/H\n \n 2\n \n SO\n \n 4\n \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\n A very well answered question. 70% of the candidates stated H2SO4 as the product from the reaction of SO\n \n 3\n \n with water.\n
\n\n Which statement is correct about the ions in a cell assembled from these half-cells?\n
\n
\n
| \n \n Reaction\n \n | \n\n \n E\n \n ⦵\n \n \n | \n
| \n Ni\n \n 2+\n \n (aq) + 2e\n \n −\n \n ⇌ Ni (s)\n | \n\n −0.26 V\n | \n
| \n Zn\n \n 2+\n \n (aq) + 2e\n \n −\n \n ⇌ Zn (s)\n | \n\n −0.76 V\n | \n
\n
\n A. Negative ions flow into the zinc half-cell from the salt bridge.\n
\n B. Negative ions flow into the nickel half-cell from the salt bridge.\n
\n\n C. Zn\n \n 2+\n \n ions are reduced to Zn.\n
\n\n D. The concentration of Ni\n \n 2+\n \n ions increases.\n
\n\n [1]\n
\n\n A\n
\n\n \n Identify the chiral carbon atom using an asterisk, *.\n \n
\n\n [1]\n
\n\n \n \n [\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 A student heated a known mass of zinc powder in an open crucible until there was no further mass change and recorded the final mass.\n
\n\n What would the student be able to derive from this data?\n
\n\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\n
\n\n \n What is the equilibrium constant expression for the following equation?\n \n
\n\n \n 2NO\n \n 2\n \n (g) + F\n \n 2\n \n (g)\n \n \n \n 2NO\n \n 2\n \n F (g)\n \n
\n\n \n A.\n \n \n \n \n
\n\n \n B.\n \n \n \n \n
\n\n \n C.\n \n \n \n \n
\n\n \n D.\n \n \n \n \n
\n\n [1]\n
\n\n D\n
\n\n One of the easiest questions on the paper. 89 % of the candidates chose the correct\n \n K\n \n c expression.\n
\n\n Consider the Born–Haber cycle for the formation of sodium oxide:\n
\n\n \n
\n What is the lattice enthalpy, in kJ mol\n \n −1\n \n , of sodium oxide?\n
\n\n
\n A. 414 + 2(108) + 249 + 2(496) − 141 + 790\n
\n B. 414 + 2(108) + 249 + 2(496) + 141 + 790\n
\n\n C. −414 + 2(108) + 249 + 2(496) − 141 + 790\n
\n\n D. −414 − 2(108) − 249 − 2(496) + 141 − 790\n
\n\n [1]\n
\n\n A\n
\n\n What is the name of this substance using IUPAC rules?\n
\n\n \n
\n A. 2-ethyl-1-methylbutan-1-ol\n
\n\n B. 1-methyl-2-ethylbutan-1-ol\n
\n\n C. 3-ethylpentan-2-ol\n
\n\n D. 3-ethylpentan-4-ol\n
\n\n [1]\n
\n\n C\n
\n\n Which property increases down group 1?\n
\n\n A. atomic radius\n
\n\n B. electronegativity\n
\n\n C. first ionization energy\n
\n\n D. melting point\n
\n\n [1]\n
\n\n A\n
\n\n \n Suggest a modification to the procedure that would make the results more reliable.\n \n
\n\n [1]\n
\n\n \n repetition / take several samples «and average» ✔\n \n
\n\n Identify a metal produced by reacting its oxide with carbon or carbon monoxide. Use section 25 of the data booklet.\n
\n\n [1]\n
\n\n \n Any one of:\n \n
\n\n Zn, Cr, Fe, Cd, Co, Ni, Sn, Pb, Sb, As, Bi, Cu, Ag, Pd, Hg, Pt ✓\n
\n\n
\n\n \n Accept “Au”.\n
\n Accept name or symbol of metal.\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 «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 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 «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 Suggest, giving reasons, the relative volatilities of SCl\n \n 2\n \n and H\n \n 2\n \n O.\n
\n\n [3]\n
\n\n H\n \n 2\n \n O forms hydrogen bonding «while SCl\n \n 2\n \n does not» ✓\n
\n\n SCl\n \n 2\n \n «much» stronger London/dispersion/«instantaneous» induced dipole-induced dipole forces ✓\n
\n\n \n \n
\n Alternative 1:\n \n \n
\n H\n \n 2\n \n O less volatile\n \n \n AND\n \n \n hydrogen bonding stronger «than dipole–dipole and dispersion forces» ✓\n
\n \n \n
\n Alternative 2:\n \n \n
\n SCl\n \n 2\n \n less volatile\n \n \n AND\n \n \n effect of dispersion forces «could be» greater than hydrogen bonding ✓\\\n
\n
\n\n \n Ignore reference to Van der Waals.\n \n
\n\n \n Accept “SCl\n \n 2\n \n has «much» larger molar mass/electron density” for M2.\n \n
\n\n \n How is colour produced in transition metal complexes?\n \n
\n\n \n A. Light is absorbed when electrons are promoted between split d-orbitals.\n \n
\n\n \n B. Light is emitted when electrons fall between split d-orbitals.\n \n
\n\n \n C. Light is absorbed when electrons escape from the complex.\n \n
\n\n \n D. Light is emitted when the complex returns to ground state.\n \n
\n\n [1]\n
\n\n A\n
\n\n 69 % of the candidates understood how colour is produced in transition metal complexes. The most commonly chosen distractor was B, which also recognized the involvement of the split d-orbitals, however stated that colour is produced when light is emitted when electrons fall between split d-orbitals.\n
\n\n Write\n \n two\n \n equations showing how these antacids neutralize excess hydrochloric acid.\n
\n\n
\n\n Magnesium carbonate: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n Aluminium hydroxide: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n
\n\n
\n\n [1]\n
\n\n MgCO\n \n 3\n \n (s) + 2HCl(aq) → MgCl\n \n 2\n \n (aq) + H\n \n 2\n \n O(l) + CO\n \n 2\n \n (g)\n
\n \n \n AND\n \n \n
\n Al(OH)\n \n 3\n \n (s) +3HCl(aq) → AlCl\n \n 3\n \n (aq) + 3H\n \n 2\n \n O(l) ✓\n
\n
\n\n \n Accept appropriate ionic equations.\n
\n \n
\n \n Do\n \n not\n \n accept H\n \n 2\n \n CO\n \n 3\n \n as a product of first reaction.\n
\n \n
\n \n Ignore equilibrium arrows.\n \n
\n\n Draw the Lewis structure of SO\n \n 3\n \n .\n
\n\n [1]\n
\n\n \n ✔\n
\n
\n\n \n Note:\n \n
\n\n \n
\n \n Accept any of the above structures as formal charge is not being assessed.\n \n
\n\n Drawing the Lewis structure of SO\n \n 3\n \n 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.\n
\n\n What is the change of state for a gas to a solid?\n
\n\n
\n A. Condensation\n
\n B. Deposition\n
\n\n C. Freezing\n
\n\n D. Sublimation\n
\n\n [1]\n
\n\n B\n
\n\n 52% of the candidates identified \"deposition\" as the change of state for a gas to a solid. The most commonly chosen distractor was sublimation.\n
\n\n SO\n \n 2\n \n (g), O\n \n 2\n \n (g) and SO\n \n 3\n \n (g) are mixed and allowed to reach equilibrium at 600 °C.\n
\n\n \n
\n Determine the value of\n \n K\n \n \n c\n \n at 600 °C.\n
\n\n [2]\n
\n\n [O\n \n 2\n \n ] = 1.25 «mol dm\n \n −3\n \n »\n \n \n AND\n \n \n [SO\n \n 3\n \n ] = 3.50 «mol dm\n \n −3\n \n » ✓\n
\n\n «\n \n K\n \n \n c\n \n =\n \n =» 4.36 ✓\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer\n \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 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 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 What is the correct interpretation of the following potential energy profile?\n
\n\n \n
\n
\n A. Endothermic reaction; products more stable than reactants.\n
\n B. Exothermic reaction; products more stable than reactants.\n
\n\n C. Endothermic reaction; products less stable than reactants.\n
\n\n D. Exothermic reaction; products less stable than reactants.\n
\n\n
\n\n [1]\n
\n\n C\n
\n\n A 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 cm\n \n 3\n \n to 525 cm\n \n 3\n \n .\n
\n\n Determine the mole percentage of hydrogen sulfide in the sample, stating one assumption you made.\n
\n\n \n
\n [3]\n
\n\n \n Mole percentage H\n \n 2\n \n S:\n \n
\n volume of H\n \n 2\n \n S = «550 − 525 = » 25 «cm\n \n 3\n \n » ✔\n
\n mol % H\n \n 2\n \n S = «\n \n = » 4.5 «%» ✔\n
\n \n Award [2] for correct final answer of 4.5 «%»\n \n
\n\n
\n\n \n Assumption:\n \n
\n «both» gases behave as ideal gases ✔\n
\n
\n \n Accept “volume of gas\n \n mol of gas”.\n \n
\n \n Accept “reaction goes to completion”.\n \n
\n \n Accept “nitrogen is insoluble/does not\n \n \n react with NaOH/only H\n \n 2\n \n S reacts with\n \n \n NaOH”.\n \n
\n Three elements, X, Y, and Z are in the same period of the periodic table. The relative sizes of their atoms are represented by the diagram.\n
\n\n \n
\n Which general trends are correct?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n The question involved the identification of three trends. 43% of the candidates selected the correct trends in ionization energy, effective nuclear charge and acidity of the oxides. The most commonly chosen distractor reversed the trend in the acidity of the oxides.\n
\n\n Which compound is an aromatic ester?\n
\n\n
\n \n
\n [1]\n
\n\n B\n
\n\n \n In which compound is the halogen substituted the most rapidly by aqueous hydroxide ions?\n \n
\n\n \n A. (CH\n \n 3\n \n )\n \n 3\n \n CCl\n
\n \n
\n \n B. (CH\n \n 3\n \n )\n \n 3\n \n CI\n
\n \n
\n \n C. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n Cl\n
\n \n
\n \n D. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n I\n \n
\n\n [1]\n
\n\n B\n
\n\n Determine the mass, in g, of CaCO\n \n 3\n \n (s) produced by reacting 2.41 dm\n \n 3\n \n of 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n of Ca(OH)\n \n 2\n \n (aq) with 0.750 dm\n \n 3\n \n of CO\n \n 2\n \n (g) at STP.\n
\n\n [2]\n
\n\n «\n \n n\n \n \n Ca(OH)2\n \n = 2.41 dm\n \n 3\n \n × 2.33 × 10\n \n −2\n \n mol dm\n \n −3\n \n =» 0.0562 «mol»\n \n \n AND\n \n \n
\n\n «\n \n n\n \n \n CO2\n \n =\n \n =» 0.0330 «mol» ✓\n
\n\n «CO\n \n 2\n \n is the limiting reactant»\n
\n\n «\n \n m\n \n \n CaCO3\n \n = 0.0330 mol × 100.09 g mol\n \n −1\n \n =» 3.30 «g» ✓\n
\n\n
\n\n \n Only award ECF for M2 if limiting reagent is used.\n \n
\n\n \n Accept answers in the range 3.30 - 3.35 «g».\n \n
\n\n State the meaning of a strong Brønsted–Lowry acid.\n
\n\n [2]\n
\n\n fully ionizes/dissociates ✔\n
\n\n proton/H\n \n +\n \n «donor » ✔\n
\n\n While 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.\n
\n\n How many signals are observed in the\n \n 1\n \n H NMR spectrum of this compound?\n
\n \n
\n A. 1\n
\n B. 2\n
\n\n C. 3\n
\n\n D. 4\n
\n\n [1]\n
\n\n A\n
\n\n The compound used for this\n \n 1\n \n H NMR question was unfamiliar to the candidates and it was a skeletal formula. They were expected to notice its symmetry and that all H’s had the same chemical environment. The majority of candidates saw only part of this symmetry and decided on two chemical environments. 29% of the candidates obtained the correct answer. Some teachers commented on the difficulty of this question.\n
\n\n \n Which species has a square planar molecular geometry?\n \n
\n\n \n \n A. SF\n \n 4\n \n \n \n
\n\n \n \n B. XeF\n \n 4\n \n \n \n
\n\n \n \n C. CF\n \n 4\n \n \n \n
\n\n \n \n D. PF\n \n 4\n \n \n +\n \n \n \n
\n\n [1]\n
\n\n B\n
\n\n This was a highly discriminating question. 67 % of the candidates identified the species that has a square planar molecular geometry. One teacher commented that this was a difficult recall question. However, the intention here is not to memorize shapes but rather to use VSEPR theory to deduce the shape.\n
\n\n Deduce, giving reasons, whether the reaction of magnesium nitride with water is an acid–base reaction, a redox reaction, neither or both.\n
\n\n \n
\n [2]\n
\n\n \n Acid–base:\n \n
\n yes\n \n AND\n \n N\n \n 3-\n \n accepts H\n \n +\n \n /donates electron pair«s»\n
\n \n \n OR\n \n \n
\n yes\n \n AND\n \n H\n \n 2\n \n O loses H\n \n +\n \n «to form OH\n \n -\n \n »/accepts electron pair«s» ✔\n
\n \n Redox:\n \n
\n no\n \n AND\n \n no oxidation states change ✔\n
\n
\n\n \n Accept “yes\n \n AND\n \n proton transfer takes place”\n \n
\n\n \n Accept reference to the oxidation state of specific elements not changing.\n \n
\n\n \n Accept “not redox as no electrons gained/lost”.\n \n
\n\n \n Award\n \n [1 max]\n \n for Acid–base: yes\n \n AND\n \n Redox: no without correct reasons, if no other mark has been awarded\n \n
\n\n Probably 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.\n
\n\n Suggest an experiment that shows that magnesium is more reactive than zinc, giving the observation that would confirm this.\n
\n\n [2]\n
\n\n \n \n Alternative 1\n \n \n
\n\n put Mg in Zn\n \n 2+\n \n (aq) ✔\n
\n\n Zn/«black» layer forms «on surface of Mg» ✔\n
\n\n \n
\n Award\n \n [1 max]\n \n for “no reaction when Zn placed in Mg\n \n 2+\n \n (aq)”.\n \n
\n
\n\n \n \n Alternative 2\n \n \n
\n\n place both metals in acid ✔\n
\n\n bubbles evolve more rapidly from Mg\n
\n \n \n OR\n \n \n
\n Mg dissolves faster ✔\n
\n
\n\n \n \n Alternative 3\n \n \n
\n\n construct a cell with Mg and Zn electrodes ✔\n
\n\n bulb lights up\n
\n \n \n OR\n \n \n
\n shows (+) voltage\n
\n \n \n OR\n \n \n
\n size/mass of Mg(s) decreases «over time»\n
\n \n \n OR\n \n \n
\n size/mass of Zn increases «over time»\n
\n \n
\n \n \n Accept “electrons flow from Mg to Zn”.\n \n
\n \n Accept Mg is negative electrode/anode\n \n
\n \n \n OR\n \n \n
\n \n Zn is positive electrode/cathode\n \n
\n \n
\n Accept other correct methods.\n \n
\n Many 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\".\n
\n\n \n Which combination corresponds to a strong metallic bond?\n \n
\n\n \n \n \n
\n
\n\n [1]\n
\n\n B\n
\n\n The concept of charge and ionic radii size on the strength of the bond was answered well, but had the lowest discriminatory index on the paper.\n
\n\n Deduce the number of signals that you would expect in the\n \n 1\n \n H NMR spectrum of nitrobenzene and the relative areas of these.\n
\n\n \n
\n [2]\n
\n\n \n Number of signals\n \n : three/3 ✔\n
\n\n \n Relative areas\n \n : 2 : 2 : 1 ✔\n
\n\n Deducing 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.\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 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 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
\n\n Why does benzene undergo substitution more readily than addition?\n
\n\n
\n A. Benzene is unsaturated.\n
\n B. Addition could produce an alkane.\n
\n\n C. Resonance makes carbon–carbon bonds too strong to break.\n
\n\n D. A benzene molecule is planar.\n
\n\n [1]\n
\n\n C\n
\n\n State\n \n two\n \n different drug administration methods.\n
\n\n [1]\n
\n\n Any\n \n two\n \n of:\n
\n oral ✓\n
\n inhalation ✓\n
\n topical/applied to the skin ✓\n
\n parenteral/injection ✓\n
\n suppositories ✓\n
\n eye/ear drops ✓\n
\n
\n\n \n One mark for any\n \n two\n \n correct methods.\n
\n \n
\n \n Count multiple methods of injection (intramuscular, subcutaneous, intravenous) as just one method.\n \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 dipole-dipole ✔\n
\n\n
\n\n \n Do not accept van der Waals’ forces.\n \n
\n\n How are emission spectra formed?\n
\n\n A. Photons are absorbed when promoted electrons return to a lower energy level.\n
\n\n B. Photons are absorbed when electrons are promoted to a higher energy level.\n
\n\n C. Photons are emitted when electrons are promoted to a higher energy level.\n
\n\n D. Photons are emitted when promoted electrons return to a lower energy level.\n
\n\n [1]\n
\n\n D\n
\n\n What is the equilibrium constant expression for the following reaction?\n
\n\n 2SO\n \n 3\n \n (g)\n \n 2SO\n \n 2\n \n (g) + O\n \n 2\n \n (g)\n
\n\n
\n A.\n \n
\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n The enthalpy of combustion of a fuel was determined using the calorimeter shown. The final result was lower than the literature value.\n
\n\n \n
\n Which factors could have contributed to this error?\n
\n\n I. Not all heat from the combustion was transferred to the calorimeter.\n
\n II. Incomplete combustion occurred.\n
\n III. The temperature probe touched the bottom of the calorimeter.\n
\n A. I and II only\n
\n\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\n
\n\n Write the equation for this reaction.\n
\n\n [1]\n
\n\n 4FeS(s) + 7O\n \n 2\n \n (g) → 2Fe\n \n 2\n \n O\n \n 3\n \n (s) + 4SO\n \n 2\n \n (g) ✔\n
\n\n \n
\n Accept any correct ratio.\n \n
\n \n Calculate the percentage uncertainty of the day 5 titre.\n \n
\n\n [1]\n
\n\n \n «\n \n \n \n «%»✔\n \n
\n\n \n Predict, giving a reason, whether the reduction of ReO\n \n 4\n \n \n −\n \n to [Re(OH)\n \n 2\n \n ]\n \n 2+\n \n would oxidize Fe\n \n 2+\n \n to Fe\n \n 3+\n \n in aqueous solution. Use section 24 of the data booklet.\n \n
\n\n [1]\n
\n\n \n no\n \n \n AND\n \n \n ReO\n \n \n 4\n \n \n −\n \n \n is a weaker oxidizing agent than Fe\n \n 3+\n \n
\n \n \n OR\n \n \n
\n no\n \n \n AND\n \n \n Fe\n \n 3+\n \n is a stronger oxidizing agent than\n \n ReO\n \n \n 4\n \n \n −\n \n
\n \n \n OR\n \n \n
\n no\n \n \n AND\n \n \n Fe\n \n 2+\n \n is a weaker reducing agent than\n \n [Re(OH)\n \n \n 2\n \n \n ]\n \n \n 2+\n \n
\n \n \n OR\n \n \n
\n no\n \n \n AND\n \n \n [Re(OH)\n \n \n 2\n \n \n ]\n \n \n 2+\n \n \n is a stronger reducing agent than Fe\n \n 2+\n \n
\n \n \n OR\n \n \n
\n no\n \n \n AND\n \n \n cell emf would be negative/–0.41 V\n \n [✔]\n \n \n
\n Many students understood that the oxidation of Fe\n \n 2+\n \n was not viable but were unable to explain why in terms of oxidizing and reducing power; many students simply gave numerical values for\n \n E\n \n \n Θ\n \n often failing to realise that the oxidation of Fe\n \n 2+\n \n would have the inverse sign to the reduction reaction.\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 \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 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 \n Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.\n \n
\n\n
\n\n \n Molecular geometry:\n \n
\n\n \n H–N–H bond angle:\n \n
\n\n [2]\n
\n\n \n \n Molecular geometry:\n \n
\n «trigonal» pyramidal\n \n [✔]\n \n \n
\n \n \n H–N–H bond angle:\n \n
\n 107°\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept angles in the range of 100–109.\n \n \n
\n\n The 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.\n
\n\n What is the relative molecular mass of bromine, according to the following mass spectrum?\n
\n\n \n
\n NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce\n
\n on behalf of the United States of America. All rights reserved.\n
\n
\n A.\n \n
\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n What are the signs of ΔH and ΔS for a reaction that is non-spontaneous at low temperatures but spontaneous at high temperatures?\n
\n\n \n
\n [1]\n
\n\n D\n
\n\n Two-thirds of the candidates related information about the spontaneity of a reaction at different temperatures to the signs of ΔH and ΔS correctly. This question discriminated well between high-scoring and low-scoring candidates.\n
\n\n \n Which statement is correct when a zinc spoon is electroplated with silver?\n \n
\n\n \n A. The cathode (negative electrode) is made of silver.\n
\n \n
\n \n B. The anode (positive electrode) is the zinc spoon.\n
\n \n
\n \n C. The anode (positive electrode) is made of silver.\n
\n \n
\n \n D. The electrolyte is zinc sulfate solution.\n \n
\n\n [1]\n
\n\n C\n
\n\n Very pleasing to see that nearly 80% could identify the material best suited as anode in an electrolytic cell. Many candidates, however, thought that the material being electroplated should be the anode.\n
\n\n \n State one reason why the mixture was not heated strongly.\n \n
\n\n [1]\n
\n\n solvent/oil is flammable\n
\n \n \n OR\n \n \n
\n solvent/oil must be kept below its flash point\n
\n \n \n OR\n \n \n
\n oxidation/decomposition of oil\n
\n \n \n OR\n \n \n
\n mixture has a low boiling point ✔\n
\n \n Accept “to prevent evaporation of oil”.\n \n
\n\n Another 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.\n
\n\n \n Identify a conjugate acid–base pair in the equation.\n \n
\n\n [1]\n
\n\n \n C\n \n 6\n \n H\n \n 8\n \n O\n \n 7\n \n \n \n AND\n \n \n C\n \n 6\n \n H\n \n 7\n \n O\n \n 7\n \n \n −\n \n
\n \n \n OR\n \n \n
\n H\n \n 2\n \n O\n \n \n AND\n \n \n H\n \n 3\n \n O\n \n +\n \n ✔\n \n
\n \n Deduce the hybridization of the central nitrogen atom in the molecule.\n \n
\n\n [1]\n
\n\n \n sp\n \n [✔]\n \n \n
\n\n Hybridisation of the N atom was correct in most cases.\n
\n\n \n 100.0 cm\n \n 3\n \n of soda water contains 3.0 × 10\n \n −2\n \n g NaHCO\n \n 3\n \n .\n \n
\n\n \n Calculate the concentration of NaHCO\n \n 3\n \n in mol dm\n \n −3\n \n .\n \n
\n\n [2]\n
\n\n \n «molar mass of NaHCO\n \n 3\n \n =» 84.01 «g mol\n \n –1\n \n »\n \n [✔]\n \n \n
\n\n \n «concentration =\n \n \n \n » 3.6 × 10\n \n –3\n \n «mol dm\n \n –3\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 Very well answered. Most candidates calculated the molar concentration correctly.\n
\n\n \n The 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.\n \n
\n\n [2]\n
\n\n \n \n \n ALTERNATIVE 1:\n \n \n
\n [H\n \n +\n \n ] «= 10\n \n −2.95\n \n » = 1.122 × 10\n \n −3\n \n «mol dm\n \n −3\n \n »\n \n [✔]\n \n \n
\n \n «[OH\n \n −\n \n ] =\n \n \n \n =» 8.91 × 10\n \n −12\n \n «mol dm\n \n −3\n \n »\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n ALTERNATIVE 2:\n \n \n
\n pOH = «14 − 2.95 =» 11.05\n \n [✔]\n \n \n
\n \n «[OH\n \n −\n \n ] = 10\n \n −11.05\n \n =» 8.91 × 10\n \n −12\n \n «moldm\n \n −3\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 \n \n Accept other methods.\n \n \n
\n\n Many students could correctly calculate the hydroxide concentration, but some weaker students calculated hydrogen ion concentration only.\n
\n\n What is the correct order of reaction types in the following sequence?\n
\n\n \n
\n
\n
| \n | \n\n \n I\n \n | \n\n \n II\n \n | \n\n \n III\n \n | \n
| \n A.\n | \n\n substitution\n | \n\n oxidation\n | \n\n reduction\n | \n
| \n B.\n | \n\n addition\n | \n\n substitution\n | \n\n reduction\n | \n
| \n C.\n | \n\n oxidation\n | \n\n substitution\n | \n\n reduction\n | \n
| \n D.\n | \n\n substitution\n | \n\n oxidation\n | \n\n substitution\n | \n
\n [1]\n
\n\n A\n
\n\n \n A sample of gas was enriched to contain 2 % by mass of\n \n 15\n \n N with the remainder being\n \n 14\n \n N.\n \n
\n\n \n Calculate the relative molecular mass of the resulting N\n \n 2\n \n O.\n \n
\n\n [2]\n
\n\n \n «\n \n \n \n =» 14.02\n \n [✔]\n \n \n
\n\n \n «M\n \n r\n \n = (14.02 × 2) + 16.00 =» 44.04\n \n [✔]\n \n \n
\n\n Most candidates were able to calculate the accurate mass of N\n \n 2\n \n O, though quite a few candidates just calculated the mass of N and didn’t apply it to N\n \n 2\n \n O, losing an accessible mark.\n
\n\n Draw an arrow, labelled\n \n Z\n \n , to represent the lowest energy electron transition in the visible spectrum.\n
\n\n [1]\n
\n\n \n
\n downward or upward arrow between n = 3 and n = 2 ✔\n
\n\n 1.0 mol each of sulfur dioxide, oxygen, and sulfur trioxide are in equilibrium.\n
\n\n \n
\n\n Which change in the molar ratio of reactants will cause the greatest increase in the amount of sulfur trioxide?\n
\n\n Assume volume and temperature of the reaction mixture remain constant.\n
\n\n \n
\n [1]\n
\n\n D\n
\n\n Deduce the half-equations for the reaction at each electrode.\n
\n\n \n
\n [2]\n
\n\n Cathode (negative electrode):\n
\n\n Zn\n \n 2+\n \n + 2e\n \n −\n \n → Zn (l) ✔\n
\n\n
\n\n Anode (positive electrode):\n
\n\n 2Cl\n \n −\n \n → Cl\n \n 2\n \n (g) + 2e\n \n −\n \n
\n\n \n \n OR\n \n \n
\n\n Cl\n \n −\n \n → ½ Cl\n \n 2\n \n (g) + e\n \n −\n \n ✔\n
\n\n The 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\n \n +\n \n instead of Zn\n \n 2+\n \n . Some candidates reversed the anode and cathode equations earning only 1 of the 2 marks.\n
\n\n \n What are the products when concentrated KBr (aq) is electrolyzed?\n \n
\n\n \n \n \n
\n [1]\n
\n\n D\n
\n\n Another challenging question with a high discrimination index. 57 % of the candidates were able to identify the electrode products during the electrolysis of concentration KBr (aq). The most commonly chosen distractor was C where K was the product at the cathode (instead of H\n \n 2\n \n ). Some teachers commented that the data booklet was needed to solve this question and others said “concentrated” was vague. But the effect of concentration is clearly stated in the syllabus and does not need the data booklet to be determined. As for the cation, potassium is known as a reactive metal according to the periodic trends and should have been easy to recognize as more reactive than hydrogen. Similar questions have appeared in past papers.\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 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 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 \n Predict\n \n two\n \n other chemical properties you would expect rhenium to have, given its position in the periodic table.\n \n
\n\n [2]\n
\n\n \n \n Any two of:\n \n
\n variable oxidation states\n \n \n \n [✔]\n \n \n
\n \n forms complex ions/compounds\n \n \n \n [✔]\n \n \n
\n\n \n coloured compounds/ions\n \n \n \n [✔]\n \n \n
\n\n \n «para»magnetic compounds/ions\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept other valid responses related to its\n \n chemical\n \n metallic properties.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “catalytic properties”.\n \n \n
\n\n Generally well answered though some students suggested physical properties rather than chemical ones.\n
\n\n \n State the number of 1H NMR signals for this isomer of xylene and the ratio in which they appear.\n \n
\n\n \n Number of signals:\n \n \n
\n \n
\n \n Ratio:\n \n
\n\n [2]\n
\n\n \n Number of signals:\n
\n 2\n \n [✔]\n \n \n
\n \n Ratio:\n
\n 3 : 2\n
\n \n \n OR\n \n \n
\n 6 : 4\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept any correct integer or fractional ratio.\n \n \n
\n\n \n \n Accept ratios in reverse order.\n \n \n
\n\n Most students gained M1 but very few gained M2, suggesting that the correct answer of 2 signals may have been a guess.\n
\n\n \n What is correct when temperature increases in this reaction at equilibrium?\n \n
\n\n \n \n
\n\n \n
\n [1]\n
\n\n D\n
\n\n A good majority of candidates understood how position of equilibrium and rate constant of an endothermic reaction are affected by temperature increase.\n
\n\n \n Which equation represents the N–H bond enthalpy in NH\n \n 3\n \n ?\n \n
\n\n A. NH\n \n 3\n \n (g) → N (g) + 3H (g)\n
\n\n \n B.\n \n NH\n \n 3\n \n (g) →\n \n N (g) + H (g)\n
\n \n
\n \n C. NH\n \n 3\n \n (g) →\n \n N\n \n 2\n \n (g) +\n \n H\n \n 2\n \n (g)\n
\n \n
\n \n D. NH\n \n 3\n \n (g) → •NH\n \n 2\n \n (g) + •H (g)\n \n
\n\n [1]\n
\n\n B\n
\n\n Deduce the change in the oxidation state of sulfur.\n
\n\n [1]\n
\n\n +6\n
\n \n \n OR\n \n \n
\n −2 to +4 ✔\n
\n \n Accept “6/VI”.\n \n
\n \n Accept “−II, 4//IV”.\n \n
\n Do\n \n not\n \n accept 2− to 4+.\n
\n Describe the bonding in this type of solid.\n
\n\n [2]\n
\n\n electrostatic attraction ✔\n
\n\n between oppositely charged ions/between Fe\n \n 2+\n \n and S\n \n 2−\n \n ✔\n
\n\n \n Which combination correctly describes the geometry of\n \n \n \n ?\n \n
\n\n \n
\n [1]\n
\n\n C\n
\n\n Identifying correct electron domain and molecular geometries around a central atom was answered well by the better candidates. This had the highest discrimination index of all the questions.\n
\n\n Which terms describe the nitronium ion in the nitration of benzene?\n
\n\n \n
\n
\n| \n | \n\n \n Type of reactant\n \n | \n\n \n Acid-base nature\n \n | \n
| \n A.\n | \n\n nucleophile\n | \n\n Lewis base\n | \n
| \n B.\n | \n\n nucleophile\n | \n\n Lewis acid\n | \n
| \n C.\n | \n\n electrophile\n | \n\n Lewis base\n | \n
| \n D.\n | \n\n electrophile\n | \n\n Lewis acid\n | \n
\n [1]\n
\n\n D\n
\n\n \n Comment on the spontaneity of the reaction at 298 K.\n \n
\n\n [1]\n
\n\n \n non-spontaneous\n \n \n AND\n \n \n \n \n \n positive ✔\n \n
\n\n Deduce the change in the oxidation state of sulfur.\n
\n\n [1]\n
\n\n +6\n
\n \n \n OR\n \n \n
\n −2 to +4 ✔\n
\n \n Accept “6/VI”.\n \n
\n \n Accept “−II, 4//IV”.\n \n
\n Do\n \n not\n \n accept 2− to 4+.\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 \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 What is the enthalpy change of the reaction, in kJ?\n
\n\n 2C (graphite) + O\n \n 2\n \n (g) → 2CO (g)\n
\n\n \n
\n A. −394 − 283\n
\n\n B. 2(−394) + 2(−283)\n
\n\n C. −394 + 283\n
\n\n D. 2(−394) + 2(283)\n
\n\n
\n\n [1]\n
\n\n D\n
\n\n Suggest\n \n two\n \n differences in the\n \n 1\n \n H NMR of but-2-ene and the organic product from (d)(ii).\n
\n\n [2]\n
\n\n \n \n ALTERNATIVE 1:\n \n Any two of:\n \n
\n\n but-2-ene: 2 signals\n \n \n AND\n \n \n product: 4 signals ✔\n
\n\n but-2-ene: «area ratio» 3:1/6:2\n \n \n AND\n \n \n product: «area ratio» 3:3:2:1 ✔\n
\n\n product: «has signal at» 3.5-4.4 ppm «and but-2-ene: does not» ✔\n
\n\n but-2-ene: «has signal at» 4.5-6.0 ppm «and product: does not» ✔\n
\n\n
\n\n \n \n ALTERNATIVE 2:\n \n \n
\n\n but-2-ene: doublet\n \n \n AND\n \n \n quartet/multiplet/4 ✔\n
\n\n product: doublet\n \n \n AND\n \n \n triplet\n \n \n AND\n \n \n quintet/5/multiplet\n \n \n AND\n \n \n sextet/6/multiplet ✔\n
\n\n
\n\n \n Accept “product «has signal at» 1.3–1.4 ppm «and but-2-ene: does not»”.\n \n
\n\n A 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.\n
\n\n Calculate the entropy change of reaction, Δ\n \n S\n \n \n ⦵\n \n , in J K\n \n −1\n \n mol\n \n −1\n \n .\n
\n\n \n
\n [1]\n
\n\n «Δ\n \n = 2 × 206.6 – (130.6 + 116.1) =» 166.5 «J K\n \n –1\n \n mol\n \n –1\n \n » ✔\n
\n\n ΔS was well calculated in general except for some inverted calculations or failure to consider the ratios of the reactants.\n
\n
\n
\n \n Where does oxidation occur in a voltaic cell?\n \n
\n\n \n A. positive electrode and anode\n \n
\n\n \n B. negative electrode and anode\n \n
\n\n \n C. positive electrode and cathode\n \n
\n\n \n D. negative electrode and cathode\n \n
\n\n [1]\n
\n\n B\n
\n\n One G2 form thought the question was awkwardly worded, implying oxidation is occurring in two places rather than one. This question also had a relatively high discriminatory index.\n
\n\n A sample of a compound contains approximately 24.0 g C, 3.0 g H, and 1.6 g O. What is the empirical formula of the compound?\n
\n\n A. C\n \n 20\n \n H\n \n 30\n \n O\n
\n\n B. C\n \n 84\n \n H\n \n 10\n \n O\n \n 6\n \n
\n\n C. C\n \n 2\n \n H\n \n 3\n \n O\n
\n\n D. C\n \n 24\n \n H\n \n 30\n \n O\n \n 2\n \n
\n\n [1]\n
\n\n A\n
\n\n The two containers shown are connected by a valve. What is the total pressure after the valve is opened and the two gas samples are allowed to mix at constant temperature?\n
\n\n \n
\n
\n A. 1.5 × 10\n \n 5\n \n Pa\n
\n B. 2.3 × 10\n \n 5\n \n Pa\n
\n\n C. 2.5 × 10\n \n 5\n \n Pa\n
\n\n D. 5.0 × 10\n \n 5\n \n Pa\n
\n\n [1]\n
\n\n B\n
\n\n \n Which can be identified using infrared (IR) spectroscopy?\n \n
\n\n \n A. functional groups\n \n
\n\n \n B. molar mass\n \n
\n\n \n C. 3-D configuration\n \n
\n\n \n D. bond angle\n \n
\n\n [1]\n
\n\n A\n
\n\n The vast majority of candidates knew that IR spectroscopy can be used to identify functional groups.\n
\n\n Outline why many elements have atomic volumes greater than 10 000 cm\n \n 3\n \n mol\n \n −1\n \n .\n
\n\n [1]\n
\n\n gases «and others are solids» ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “lower density” alone.\n \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 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 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
\n\n How many chiral centres are there in the following molecule?\n
\n\n \n
\n A. 2\n
\n\n B. 3\n
\n\n C. 4\n
\n\n D. 6\n
\n\n [1]\n
\n\n C\n
\n\n Which statement is correct when 2-chloro-2-methylpentane reacts with water to form 2-methylpentan-2-ol?\n
\n\n
\n A. Water acts as a nucleophile and attacks the chlorine atom.\n
\n B. The reaction occurs in a single step.\n
\n\n C. A carbocation intermediate is formed.\n
\n\n D. Homolytic bond fission occurs.\n
\n\n [1]\n
\n\n C\n
\n\n Which condition will cause the given equilibrium to shift to the right?\n
\n\n Ag\n \n +\n \n (aq) + Cl\n \n −\n \n (aq) ⇌ AgCl (s)\n
\n\n
\n A. One half of solid AgCl is removed.\n
\n B. Water is added.\n
\n\n C. Solid NaCl is added.\n
\n\n D. The system is subjected to increased pressure.\n
\n\n [1]\n
\n\n C\n
\n\n \n Describe the effect of infrared (IR) radiation on carbon dioxide molecules.\n \n
\n\n [2]\n
\n\n \n bond length/C=O distance changes\n
\n \n \n OR\n \n \n
\n «asymmetric» stretching «of bonds»\n
\n \n \n OR\n \n \n
\n bond angle/OCO changes\n \n [✔]\n \n \n
\n
\n\n \n polarity/dipole «moment» changes\n
\n \n \n OR\n \n \n
\n dipole «moment» created «when molecule absorbs IR»\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept appropriate diagrams.\n \n \n
\n\n This question was fairly well answered with most students receiving one of the two marks. There were many students who mentioned the information in M1 (asymmetric stretching and bonds vibrate) or M2 (polarity and dipole changes) more than one time but could only receive one mark. Teachers need to remind students each mark is a different topic or concept.\n
\n\n An\n \n -particle is a helium-4 nucleus. In an experiment,\n \n -particles are accelerated towards a thin sheet of gold and their resulting paths are detected, giving evidence of the positive charge of the nucleus.\n
\n\n \n
\n Angle of detection\n \n increase from 0° to 180°\n
\n
\n
\n The number of\n \n -particles detected at different angles of deflection\n \n are shown.\n
\n\n \n
\n
\n\n \n Gold\n \n :\n \n 79\n \n \n AND\n \n \n \n Helium\n \n :\n \n 2 ✔\n
\n\n \n Suggest how benzoic acid,\n \n M\n \n r\n \n \n = 122.13, forms an apparent dimer,\n \n M\n \n r\n \n \n = 244.26, when dissolved in a non-polar solvent such as hexane.\n \n
\n\n [1]\n
\n\n \n «intermolecular» hydrogen bonding\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept diagram showing hydrogen bonding.\n \n \n
\n\n Very few students answered this question correctly, thinking benzoic would bond with the hexane even though it was a non-polar solvent. It was very rare for a student to realize there was intermolecular hydrogen bonding.\n
\n\n \n Proteins are polymers of amino acids.\n \n \n A paper chromatogram of two amino acids, A1 and A2, is obtained using a non-polar solvent.\n \n
\n\n \n
\n \n © International Baccalaureate Organization 2020.\n \n
\n\n \n Determine the\n \n \n \n value of A1.\n \n
\n\n [1]\n
\n\n \n ✔\n
\n\n \n Accept any value within the range “\n \n ”.\n \n
\n\n Many students scored this mark. The students who missed this mark that were close either measured from the top or bottom of the spot rather than the middle. A few students had answers that were greater than 1 which indicated a clear lack of understanding of this concept.\n
\n\n \n Outline a green chemistry solution for problems generated by the use of organic solvents.\n \n
\n\n [1]\n
\n\n «use of» alternative solvents such as supercritical/liquid\n \n
\n \n Do\n \n not\n \n accept political or regulatory solutions.\n \n
\n \n \n OR\n \n \n
\n use of water «as solvent»\n
\n \n \n OR\n \n \n
\n solvent-free reactions «for example, polymerization of propene»\n
\n \n \n OR\n \n \n
\n solid-state chemistry\n
\n \n \n OR\n \n \n
\n recycle «waste» solvents\n
\n \n \n OR\n \n \n
\n catalysis that leads to better/higher yield\n
\n \n “catalysis” alone not sufficient for mark.\n \n
\n \n \n OR\n \n \n
\n reducing number of steps ✔\n
\n
\n
\n
\n Many correct answers. The most common correct answer was using water as a solvent. Some students lost the mark with responses like \" less dangerous organic solvents\", \"alternative solvents (but not giving a specific example)\" or \"proper disposal of organic solvents\".\n
\n\n How many moles of carbon dioxide are produced by the complete combustion of 7.0 g of ethene, C\n \n 2\n \n H\n \n 4\n \n (g)?\n
\n\n \n M\n \n \n r\n \n = 28\n
\n\n
\n A. 0.25\n
\n B. 0.5\n
\n\n C. 0.75\n
\n\n D. 1.0\n
\n\n [1]\n
\n\n B\n
\n\n 54% of the candidates deduced the number of CO\n \n 2\n \n moles produced from 7g of C\n \n 2\n \n H\n \n 4\n \n . The most commonly chosen distractor was A which ignored the stoichiometric ratio.\n
\n\n \n Which product will be obtained at the anode (positive electrode) when molten NaCl is electrolysed?\n \n
\n\n \n A. Na (l)\n \n
\n\n \n B. Cl (g)\n \n
\n\n \n C. Cl\n \n 2\n \n (g)\n \n
\n\n \n D. Na (s)\n \n
\n\n [1]\n
\n\n C\n
\n\n 46 % of candidates correctly identified products of electrolysis at the anode with the incorrect answers being split by the remaining 54 %\n
\n\n \n Draw the structure of the conjugate base of benzoic acid showing\n \n all\n \n the atoms and\n \n all\n \n the bonds.\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 Accept Kekulé structures.\n \n \n
\n\n \n \n Negative sign must be shown in correct position- on the O or delocalised over the carboxylate.\n \n \n
\n\n Most failed to score a mark for the conjugate base of benzoic acid as either they didn’t show all bonds and atoms in the ring and/or they did not put the minus sign in the correct place. Some didn't read the question carefully so gave the structure of the acid form.\n
\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
\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 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 \n Suggest why the titration must be conducted quickly even though a low temperature is maintained.\n \n
\n\n [1]\n
\n\n equilibrium shifts to left\n
\n \n \n OR\n \n \n
\n more ethanoic acid is produced «as ethanoic acid is neutralized»\n
\n \n \n OR\n \n \n
\n prevents/slows down ester hydrolysis ✔\n
\n \n Accept “prevents equilibrium shift” if described correctly without direction.\n \n
\n\n Not well answered and of the few students that replied correctly most referred to preventing equilibrium shift and few candidates identified the direction of the shift. It was rather common to see answers where Le Chatelier's principle was stated without any attempt in adapting it to the context. Very few students described the specific equilibrium shift that could occur during the titration, changing the results.\n
\n\n What is the correct comparison of H–N–H bond angles in NH2-, NH3, and NH4+?\n
\n\n
\n A. NH\n \n 2\n \n \n −\n \n < NH\n \n 3\n \n < NH\n \n 4\n \n \n +\n \n
\n B. NH\n \n 4\n \n \n +\n \n < NH\n \n 3\n \n < NH\n \n 2\n \n \n −\n \n
\n\n C. NH\n \n 3\n \n < NH\n \n 2\n \n \n −\n \n < NH\n \n 4\n \n \n +\n \n
\n\n D. NH\n \n 3\n \n < NH\n \n 4\n \n \n +\n \n < NH\n \n 2\n \n \n −\n \n
\n\n [1]\n
\n\n A\n
\n\n \n \n \n
\n \n Which shows the first ionization energies of successive elements across period 2, from left to right?\n \n
\n\n \n \n
\n\n [1]\n
\n\n C\n
\n\n State the number of protons, neutrons and electrons in each species.\n
\n\n \n
\n [2]\n
\n\n \n
\n \n Award\n \n [1 max]\n \n for 4 correct values.\n \n
\n\n Explain why the addition of small amounts of carbon to iron makes the metal harder.\n
\n\n [2]\n
\n\n disrupts the regular arrangement «of iron atoms/ions»\n
\n \n \n OR\n \n \n
\n carbon different size «to iron atoms/ions» ✔\n
\n prevents layers/atoms sliding over each other ✔\n
\n\n Explain why transition metal cyanide complexes are coloured.\n
\n\n [3]\n
\n\n \n Any three from:\n \n
\n\n partially filled d-orbitals ✔\n
\n\n «CN- causes» d-orbitals «to» split ✔\n
\n\n light is absorbed as electrons transit to a higher energy level «in d–d transitions»\n
\n \n \n OR\n \n \n
\n light is absorbed as electrons are promoted ✔\n
\n energy gap corresponds to light in the visible region of the spectrum ✔\n
\n\n
\n\n \n Do\n \n not\n \n accept “colour observed is the complementary colour” for M4.\n \n
\n\n This 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\n
\n\n Calculate the value of the rate constant stating its units.\n
\n\n [2]\n
\n\n \n ✔\n
\n\n mol\n \n –1\n \n dm\n \n 3\n \n s\n \n –1\n \n ✔\n
\n\n \n Calculate the standard Gibbs free energy change,\n \n , in kJ mol\n \n −1\n \n , for the first dissociation of citric acid at 298 K, using section 1 of the data booklet.\n \n
\n\n [1]\n
\n\n \n \n «\n \n \n \n \n = −\n \n RT\n \n ln\n \n K\n \n = −8.31 J K\n \n –1\n \n mol\n \n –1\n \n × 298 K × ln(5.01 × 10\n \n –4\n \n ) ÷ 1000 =» 18.8 «kJ mol\n \n –1\n \n » ✔\n \n
\n\n Equal volumes of 0.10 mol dm\n \n −3\n \n weak acid and strong acid are titrated with 0.10 mol dm\n \n −3\n \n NaOH solution. Which of these is the same for the two acids?\n
\n\n
\n A. Initial pH\n
\n B. Heat evolved in the neutralization\n
\n\n C. Volume of NaOH for complete neutralization\n
\n\n D. Initial electrical conductivity\n
\n\n [1]\n
\n\n C\n
\n\n 48% of the candidates deduced that the volume of NaOH required to neutralize the strong and weak acid samples would be the same. The most commonly chosen distractor was that the heat evolved would be the same. Surprisingly this question did not discriminate very well between high-scoring and low-scoring candidates. Some teachers commented that the question should have specified that the acids were monoprotic.\n
\n\n \n What are typical characteristics of metals?\n \n
\n\n \n \n \n
\n [1]\n
\n\n A\n
\n\n Very disappointing results with a large percentage of the candidates thinking metals have a high electron affinity.\n
\n\n What is the formula of the compound formed between magnesium ions and hydrogencarbonate ions?\n
\n\n
\n A. MgHCO\n \n 3\n \n
\n B. Mg(HCO\n \n 3\n \n )\n \n 2\n \n
\n\n C. Mg(HCO\n \n 3\n \n )\n \n 3\n \n
\n\n D. Mg\n \n 3\n \n (HCO\n \n 3\n \n )\n \n 2\n \n
\n\n [1]\n
\n\n B\n
\n\n Explain why transition metal cyanide complexes are coloured.\n
\n\n [3]\n
\n\n \n Any three from:\n \n
\n\n partially filled d-orbitals ✔\n
\n\n «CN- causes» d-orbitals «to» split ✔\n
\n\n light is absorbed as electrons transit to a higher energy level «in d–d transitions»\n
\n \n \n OR\n \n \n
\n light is absorbed as electrons are promoted ✔\n
\n energy gap corresponds to light in the visible region of the spectrum ✔\n
\n\n
\n\n \n Do\n \n not\n \n accept “colour observed is the complementary colour” for M4.\n \n
\n\n This 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\n
\n\n State the equilibrium constant expression,\n \n K\n \n \n c\n \n , for this reaction.\n
\n\n [1]\n
\n\n «\n \n K\n \n \n c\n \n =»\n \n ✔\n
\n\n \n Which will react with a halogen by an electrophilic substitution mechanism?\n \n
\n\n \n \n \n
\n [1]\n
\n\n A\n
\n\n \n Distinguish between a weak and strong acid.\n \n
\n\n \n Weak acid:\n \n
\n\n \n Strong acid:\n \n
\n\n [1]\n
\n\n \n \n Weak acid:\n \n partially dissociated/ionized «in aqueous solution/water»\n
\n \n \n AND\n \n \n
\n \n Strong acid\n \n : «assumed to be almost» completely/100 % dissociated/ionized «in aqueous solution/water»\n \n [✔]\n \n \n
\n As expected, many candidates were able to distinguish between strong and weak acids; some candidates referred to “dissolve” rather than dissociate.\n
\n\n \n Aspirin crystals are rinsed with water after recrystallization to remove impurities.\n
\n Suggest why\n \n \n \n cold\n \n \n \n water is used.\n \n
\n [1]\n
\n\n to avoid dissolving the crystals/aspirin ✔\n
\n\n \n Accept “to avoid loss of product”\n \n OR\n \n “aspirin is less soluble in cold water”.\n \n
\n\n While there were many good answers it was worrying to correct as many where the student clearly didn't establish a connection between solubility and purification. Many incorrect responses indicated the cold water was to \"stay below the melting point of the aspirin\" rather than relate it to the solubility of the final product.\n
\n\n Magnesium can be produced by the electrolysis of molten magnesium chloride.\n
\n\n Write the half-equation for the formation of magnesium.\n
\n\n [1]\n
\n\n Mg\n \n 2+\n \n + 2 e\n \n -\n \n → Mg ✔\n
\n\n
\n\n \n Do\n \n not\n \n penalize missing charge on electron.\n \n
\n\n \n Accept equation with equilibrium arrows.\n \n
\n\n Unfortunately, only 40% of the students could write this quite straightforward half equation.\n
\n\n Outline why increasing the concentration of N\n \n 2\n \n O\n \n 5\n \n increases the rate of reaction.\n
\n\n [1]\n
\n\n greater frequency of collisions «as concentration increases»\n
\n \n \n OR\n \n \n
\n more collisions per unit time «as concentration increases» ✔\n
\n \n
\n Accept “rate/chance/probability/likelihood” instead of “frequency”.\n \n
\n \n Do\n \n not\n \n accept just “more collisions”.\n \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 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 Good performance on how the polarity of C-X bond changes going down group 17.\n
\n\n Calculate the enthalpy change of reaction,\n \n ΔH\n \n , in kJ, for the decomposition of calcium carbonate.\n
\n\n [2]\n
\n\n «\n \n ΔH\n \n =» (−635 «kJ» – 393.5 «kJ») – (−1207 «kJ») ✓\n
\n\n «\n \n ΔH\n \n = + » 179 «kJ» ✓\n
\n\n \n
\n Award\n \n [2]\n \n for correct final answer.\n \n
\n \n Award\n \n [1 max]\n \n for −179 kJ.\n \n
\n\n \n Ignore an extra step to determine total enthalpy change in kJ: 179 kJ mol\n \n -1\n \n x 5.55 mol = 993 kJ.\n \n
\n\n \n Award\n \n [2]\n \n for an answer in the range 990 - 993« kJ».\n \n
\n\n \n Predict, giving\n \n two\n \n reasons, how the first ionization energy of\n \n 15\n \n N compares with that of\n \n 14\n \n N.\n \n
\n\n [2]\n
\n\n \n \n Any two\n \n :\n
\n same\n \n \n AND\n \n \n have same nuclear charge/number of protons/Z\n \n eff\n \n \n [✔]\n \n \n
\n \n same\n \n \n AND\n \n \n neutrons do not affect attraction/ionization energy/Z\n \n eff\n \n
\n \n \n OR\n \n \n
\n same\n \n \n AND\n \n \n neutrons have no charge\n \n [✔]\n \n \n
\n \n same\n \n \n AND\n \n \n same attraction for «outer» electrons\n \n [✔]\n \n \n
\n\n \n same\n \n \n AND\n \n \n have same electronic configuration/shielding\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n \n Note:\n \n Accept “almost the same”.\n
\n “same” only needs to be stated once.\n \n \n \n
\n This 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.\n
\n\n What is the enthalpy change for the reaction in kJ mol\n \n −1\n \n ?\n
\n\n CO\n \n 2\n \n (g)\n \n H\n \n 2\n \n (g) → CO (g)\n \n H\n \n 2\n \n O (g)\n
\n\n 2CO (g) + O\n \n 2\n \n (g) → 2CO\n \n 2\n \n (g)\n \n \n H\n \n =\n \n 566 kJ mol\n \n \n 1\n \n
\n\n 2H\n \n 2\n \n (g) + O\n \n 2\n \n (g) → 2H\n \n 2\n \n O (l)\n \n \n H\n \n =\n \n 572 kJ mol\n \n \n 1\n \n
\n\n H\n \n 2\n \n O (g) → H\n \n 2\n \n O (l)\n \n \n H\n \n =\n \n 44 kJ mol\n \n \n 1\n \n
\n\n
\n A.\n \n 1182\n
\n B.\n \n 899\n
\n\n C.\n \n 41\n
\n\n D.\n \n 41\n
\n\n [1]\n
\n\n D\n
\n\n Combustion of coal emits particulates into the atmosphere.\n
\n\n
\n\n reflects «sun» light ✓\n
\n\n
\n\n \n Accept “results in global dimming”\n
\n \n OR\n \n
\n “reduces the amount of energy reaching the Earth”\n
\n \n OR\n \n
\n “acts as nucleation points for cloud formation”.\n \n
\n \n Do\n \n not\n \n accept answers that only indicate increases in global temperatures.\n \n
\n\n Which substance, made from two elements with electronegativities E\n \n X\n \n and E\n \n Y\n \n , is an alloy?\n
\n\n
\n \n
\n [1]\n
\n\n D\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 Incorrect\n \n \n AND\n \n \n two readings, uncertainty is ±0.1 ✔\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 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 Most could suggest one environmental disadvantage of producing biodiesel from renewable resources.\n
\n\n Identify the type of bond and by-product when monosaccharides combine.\n
\n\n
\n\n Bond: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n By-product: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n [2]\n
\n\n \n Bond:\n \n
\n glycosidic ✓\n
\n \n By-product:\n \n
\n water/H\n \n 2\n \n O ✓\n
\n
\n\n \n Accept “ether/C-O-C”\n
\n \n OR\n \n
\n “covalent/polar covalent” for M1.\n \n
\n Explain why the reaction produces more (CH\n \n 3\n \n )\n \n 3\n \n COH than (CH\n \n 3\n \n )\n \n 2\n \n CHCH\n \n 2\n \n OH.\n
\n\n [2]\n
\n\n carbocation formed from (CH\n \n 3\n \n )\n \n 3\n \n COH is more stable / (CH\n \n 3\n \n )\n \n 3\n \n C\n \n +\n \n is more stable than (CH\n \n 3\n \n )\n \n 2\n \n CHCH\n \n 2\n \n \n +\n \n ✔\n
\n\n
\n «because carbocation has» greater number of alkyl groups/lower charge on the atom/higher e\n \n -\n \n density\n
\n \n \n OR\n \n \n
\n «greater number of alkyl groups» are more electron releasing\n
\n \n \n OR\n \n \n
\n «greater number of alkyl groups creates» greater inductive/+I effect ✔\n
\n
\n\n \n Do\n \n not\n \n award any marks for simply quoting Markovnikov’s rule.\n \n
\n\n Poor performance, particularly in light of past feedback provided in similar questions since there was repeated reference simply to Markovnikov's rule, without any explanation.\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 (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 \n Note:\n \n Accept “(n-1)H\n \n 2\n \n O”.\n \n \n
\n\n \n \n Do\n \n not\n \n award mark if “n” not included.\n \n \n
\n\n It was very unusual to find a candidate who could give a correct equation for starch hydrolysis.\n
\n\n \n The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.\n \n
\n\n \n State the formula of its conjugate base.\n \n
\n\n [1]\n
\n\n \n CO\n \n 3\n \n \n 2–\n \n \n [✔]\n \n \n
\n\n A 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 HCO\n \n 3\n \n \n -\n \n . Wrong answers included water, the hydroxide ion and carbon dioxide.\n
\n\n What is the correct order for\n \n increasing\n \n first ionization energy?\n
\n\n A. Na < Mg < Al\n
\n\n B. Na < Al < Mg\n
\n\n C. Al < Mg < Na\n
\n\n D. Al < Na < Mg\n
\n\n [1]\n
\n\n B\n
\n\n The majority of candidates focussed on the general trend of ionization energy increasing across the period and missed out the finer details due to sub-levels. The correct order for increasing first ionization energy was Na < Al < Mg rather than Na < Mg < Al.\n
\n\n Suggest what should be used as a blank for spectrophotometric reading.\n
\n\n [1]\n
\n\n water\n \n \n AND\n \n \n all samples dissolved «in water» ✔\n
\n\n \n Identify the initiation step of the reaction and its conditions.\n \n
\n\n [2]\n
\n\n \n Br\n \n 2\n \n 2Br•\n \n [\n \n ✔\n \n ]\n \n
\n \n
\n \n «sun»light/UV/\n \n hv\n \n
\n \n \n OR\n \n \n
\n high temperature\n \n [\n \n ✔\n \n ]\n \n \n
\n \n \n \n Note:\n \n Do not penalize missing radical symbol on Br.\n
\n Accept “homolytic fission of bromine” for M1.\n \n \n
\n Very well done, with a few making reference to a catalyst.\n
\n\n What is formed in a propagation step of the substitution reaction between bromine and ethane?\n
\n\n A. CH\n \n 3\n \n CH\n \n 2\n \n •\n
\n\n B. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n
\n\n C. H•\n
\n\n D. Br\n \n −\n \n
\n\n [1]\n
\n\n A\n
\n\n Describe two observations that indicate the reaction of lithium with water is exothermic.\n
\n\n [2]\n
\n\n \n Any two:\n \n
\n\n temperature of the water increases ✔\n
\n\n lithium melts ✔\n
\n\n pop sound is heard ✔\n
\n\n
\n\n \n Accept “lithium/hydrogen catches fire”.\n \n
\n\n \n Do not accept “smoke is observed”.\n \n
\n\n Outline why increasing the concentration of N\n \n 2\n \n O\n \n 5\n \n increases the rate of reaction.\n
\n\n [1]\n
\n\n greater frequency of collisions «as concentration increases»\n
\n \n \n OR\n \n \n
\n more collisions per unit time «as concentration increases» ✔\n
\n \n
\n Accept “rate/chance/probability/likelihood” instead of “frequency”.\n \n
\n \n Do\n \n not\n \n accept just “more collisions”.\n \n
\n\n A compound consists of the ions Ca\n \n 2+\n \n and PO\n \n 4\n \n \n 3–\n \n . What are the name and formula of the compound?\n
\n\n \n
\n [1]\n
\n\n D\n
\n\n In which molecule does the central atom have an incomplete octet of electrons?\n
\n\n A. H\n \n 2\n \n Se\n
\n\n B. PH\n \n 3\n \n
\n\n C. OF\n \n 2\n \n
\n\n D. BF\n \n 3\n \n
\n\n [1]\n
\n\n D\n
\n\n Deduce the ratio of Fe\n \n 2+\n \n :Fe\n \n 3+\n \n in Fe\n \n 3\n \n O\n \n 4\n \n .\n
\n\n [1]\n
\n\n 1:2 ✔\n
\n\n \n Accept 2 Fe\n \n 3+\n \n : 1 Fe\n \n 2+\n \n \n
\n \n Do\n \n not\n \n accept 2:1 only\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 0.050−1.000 ✔\n
\n\n Suggest how the extent of decomposition could be measured.\n
\n\n [1]\n
\n\n use colorimeter\n
\n \n \n OR\n \n \n
\n change in colour\n
\n \n \n OR\n \n \n
\n change in volume\n
\n \n \n OR\n \n \n
\n change in pressure ✔\n
\n \n Accept suitable instruments, e.g. pressure probe/oxygen sensor.\n \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 \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 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 Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.\n
\n\n \n
\n [1]\n
\n\n HCN has stronger dipole–dipole attraction ✔\n
\n\n
\n\n \n Do\n \n not\n \n accept reference to H-bonds.\n \n
\n\n This 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.\n
\n\n \n The dotted line represents the formation of oxygen, O\n \n 2\n \n (g), from the uncatalysed complete decomposition of hydrogen peroxide, H\n \n 2\n \n O\n \n 2\n \n (aq).\n \n
\n\n \n \n \n
\n \n \n Which curve represents a catalysed reaction under the same conditions?\n \n \n
\n\n
\n\n [1]\n
\n\n B\n
\n\n Graphical representation of catalysis was also well answered.\n
\n\n \n Describe the effect of infrared (IR) radiation on carbon dioxide molecules.\n \n
\n\n [2]\n
\n\n \n bond length/C=O distance changes\n
\n \n \n OR\n \n \n
\n «asymmetric» stretching «of bonds»\n
\n \n \n OR\n \n \n
\n bond angle/OCO changes\n \n [✔]\n \n \n
\n \n polarity/dipole «moment» changes\n
\n \n \n OR\n \n \n
\n dipole «moment» created «when molecule absorbs IR»\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept appropriate diagrams.\n \n \n
\n\n This part was fairly well answered with most candidates receiving one of the two marks. There were many candidates who stated asymmetric stretching and bonds vibrate but missed writing polarity and dipole changes, which deprived them of the second mark.\n
\n\n \n Pure magnesium needed for making alloys can be obtained by electrolysis of molten magnesium chloride.\n \n
\n\n \n \n \n
\n \n © International Baccalaureate Organization 2020.\n
\n \n
\n \n Write the half-equations for the reactions occurring in this electrolysis.\n \n
\n\n \n \n \n
\n [2]\n
\n\n Anode:\n \n ✔\n
\n\n Cathode:\n \n ✔\n
\n\n \n
\n Accept\n \n \n .\n
\n \n Award\n \n [1 max]\n \n for correct equations at incorrect electrodes.\n \n
\n\n This was a very popular option with approximately 34% of candidates attempting Option B. Many students appeared well prepared for this option. Some candidates continue to provide answers with a heavy Biology bias that often make them lose valuable points.\n
\n\n What is the heat change, in kJ, when 100.0 g of aluminium is heated from 19.0 °C to 32.0 °C?\n
\n\n Specific heat capacity of aluminium: 0.90 J g\n \n −1\n \n K\n \n −1\n \n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n C\n
\n\n What is the main reason for an increase in rate of reaction when the temperature is raised?\n
\n\n
\n A. A greater proportion of collisions are successful.\n
\n B. Particles collide more frequently.\n
\n\n C. The bonds in the reactants are weakened.\n
\n\n D. The activation energy of the reaction decreases.\n
\n\n [1]\n
\n\n A\n
\n\n Alloying a metal with a metal of smaller atomic radius can disrupt the lattice and make it more difficult for atoms to slide over each other. Which property will increase as a result?\n
\n\n
\n A. Electrical conductivity\n
\n B. Ductility\n
\n\n C. Malleability\n
\n\n D. Strength\n
\n\n [1]\n
\n\n D\n
\n\n A well-answered question. 62% of the candidates were able to identify strength as the property that increases when a metal is alloyed. Good performance on this question correlated well with candidates who scored well overall.\n
\n\n \n Which represents the shape of an s atomic orbital?\n \n
\n\n \n
\n [1]\n
\n\n A\n
\n\n Which molecule is most polar?\n
\n\n A. CF\n \n 4\n \n
\n\n B. CCl\n \n 4\n \n
\n\n C. CHF\n \n 3\n \n
\n\n D. CClF\n \n 3\n \n
\n\n [1]\n
\n\n C\n
\n\n Question 11 was identified in the G2s as a tricky question, because students could have difficulty resolving the conflict between the greater polarity of the C-Cl bond compared with the C-H bond, and their contrast with the polarity of the C-F bond.) However, 56% of student did identify the correct answer of C.\n
\n\n State the number of\n \n (sigma) and\n \n (pi) bonds in Compound A.\n
\n\n \n
\n [1]\n
\n\n \n : 9\n
\n \n \n AND\n \n \n
\n \n : 1 ✔\n
\n Mediocre 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.\n
\n\n Calculate the standard enthalpy change (Δ\n \n H\n \n \n ⦵\n \n ) for the forward reaction in kJ mol\n \n −1\n \n .\n
\n\n Δ\n \n H\n \n \n ⦵\n \n \n f\n \n PCl\n \n 3\n \n (g) = −306.4 kJ mol\n \n −1\n \n
\n\n Δ\n \n H\n \n \n ⦵\n \n \n f\n \n PCl\n \n 5\n \n (g) = −398.9 kJ mol\n \n −1\n \n
\n\n [1]\n
\n\n «−398.9 kJ mol\n \n −1\n \n − (−306.4 kJ mol\n \n −1\n \n ) =» −92.5 «kJ mol\n \n −1\n \n » ✔\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 capture where produced «and store»\n \n [✔]\n \n \n
\n \n use scrubbers to remove\n \n [✔]\n \n \n
\n\n \n use as feedstock for synthesising 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 and reduce energy use\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Do\n \n not\n \n accept “planting more trees”.\n \n \n
\n\n \n \n Accept specific correct examples.\n \n \n
\n\n This part was reasonably answered although there were many candidates who gave vague answers that did not receive marks.\n
\n\n Suggest\n \n two\n \n reasons why atoms are no longer regarded as the indivisible units of matter.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n
\n subatomic particles «discovered»\n
\n \n \n OR\n \n \n
\n particles smaller/with masses less than atoms «discovered»\n
\n \n \n OR\n \n \n
\n «existence of» isotopes «same number of protons, different number of neutrons» ✔\n
\n
\n charged particles obtained from «neutral» atoms\n
\n \n \n OR\n \n \n
\n atoms can gain or lose electrons «and become charged» ✔\n
\n
\n atom «discovered» to have structure ✔\n
\n
\n fission\n
\n \n \n OR\n \n \n
\n atoms can be split ✔\n
\n
\n\n \n Accept atoms can undergo fusion «to produce heavier atoms»\n \n
\n\n \n Accept specific examples of particles.\n \n
\n\n \n Award\n \n [2]\n \n for “atom shown to have a nucleus with electrons around it” as both M1 and M3.\n \n
\n\n The 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.\n
\n\n The concentration of excess sodium hydroxide was 0.362 mol dm\n \n −3\n \n . Calculate the pH of the solution at the end of the experiment.\n
\n\n [1]\n
\n\n «pOH = –log(0.362) = 0.441»\n
\n\n «pH = 14.000 – 0.441 =» 13.559 ✔\n
\n\n Determine the equilibrium constant,\n \n K\n \n , for this reaction at 25 °C, referring to section 1 of the data booklet.\n
\n\n If you did not obtain an answer in (c)(iii), use Δ\n \n G\n \n = –43.5 kJ mol\n \n −1\n \n , but this is not the correct answer.\n
\n\n [2]\n
\n\n «ΔG = –41.8 kJ mol\n \n –1\n \n =\n \n × 298 K × ln\n \n K\n \n »\n
\n \n \n OR\n \n \n
\n «ΔG = –41800 J mol\n \n –1\n \n = –8.31 J mol\n \n –1\n \n K\n \n –1\n \n × 298 K × ln\n \n K\n \n »\n
\n
\n «ln\n \n K\n \n = =» 16.9 ✔\n
\n «\n \n K\n \n = e\n \n 16.9\n \n =» 2.19 × 10\n \n 7\n \n ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Accept range of 1.80 × 10\n \n 6\n \n –2.60 × 10\n \n 7\n \n .\n \n
\n\n \n If –43.5 is used then 4.25 × 10\n \n 7\n \n .\n \n
\n\n Outline how these drug administration methods affect bioavailability.\n
\n\n
\n\n Oral: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n
\n\n Intravenous: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n
\n\n [2]\n
\n\n \n Oral:\n \n
\n low/lower «bioavailability»\n \n \n AND\n \n \n drugs pass through digestive system «and breakdown» ✓\n
\n \n Intravenous:\n \n
\n high/higher «bioavailability»\n \n \n AND\n \n \n «more» direct route to bloodstream ✓\n
\n
\n\n \n Accept “low/lower\n \n AND\n \n drugs not easily absorbed from digestive system”\n
\n \n \n OR\n \n \n
\n “low/lower\n \n AND\n \n drugs broken down in digestive system”\n
\n \n OR\n \n
\n “low/lower\n \n AND\n \n drugs affected by acid” for M1.\n \n
\n \n Do not penalize use of “slow” for “low/lower” or “fast” for “high/higher”.\n \n
\n\n \n Accept “100 % bioavailability” for “high/higher” within Intravenous answer in M2.\n \n
\n\n \n Award\n \n [1 max]\n \n for “oral drugs have slower absorption/distribution than intravenous drugs”\n
\n \n OR\n \n
\n “Oral: low/lower «bioavailability»\n \n AND\n \n intraveneous: high/higher «bioavailability»”.\n \n
\n Outline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.\n
\n\n [1]\n
\n\n «valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔\n
\n\n \n
\n Accept 2,8 (for O\n \n 2–\n \n ) and 2,8,8 (for S\n \n 2–\n \n )\n \n
\n Justify why sulfur is classified as a non-metal by giving\n \n two\n \n of its chemical properties.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n\n forms acidic oxides «rather than basic oxides» ✔\n
\n\n forms covalent/bonds compounds «with other non-metals» ✔\n
\n\n forms anions «rather than cations» ✔\n
\n\n behaves as an oxidizing agent «rather than a reducing agent» ✔\n
\n\n \n
\n Award\n \n [1 max]\n \n 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.\n \n
\n Discuss the relative length of the two O−O bonds in ozone.\n
\n\n [2]\n
\n\n both equal ✔\n
\n\n delocalization/resonance ✔\n
\n\n \n
\n Accept bond length between 121 and 148 pm/ that of single O−O bond and double O=O bond for M1.\n \n
\n Which equilibrium constant corresponds to the spontaneous reaction with the most negative value of Δ\n \n G\n \n \n ⦵\n \n ?\n
\n
\n
\n A. 4.9 × 10\n \n −3\n \n
\n\n B. 8.2 × 10\n \n −3\n \n
\n\n C. 4.9 × 10\n \n 2\n \n
\n\n D. 8.2 × 10\n \n 2\n \n
\n\n [1]\n
\n\n D\n
\n\n \n Which reaction has the greatest increase in entropy of the system?\n \n
\n\n \n A. HCl (g) + NH\n \n 3\n \n (g) → NH\n \n 4\n \n Cl (s)\n
\n \n
\n \n B. (NH\n \n 4\n \n )\n \n 2\n \n Cr\n \n 2\n \n O\n \n 7\n \n (s) → Cr\n \n 2\n \n O\n \n 3\n \n (s) + N\n \n 2\n \n (g) + 4H\n \n 2\n \n O (g)\n
\n \n
\n \n C. CaCO\n \n 3\n \n (s) → CaO (s) + CO\n \n 2\n \n (g)\n
\n \n
\n \n D. I\n \n 2\n \n (g) → I\n \n 2\n \n (s)\n \n
\n\n [1]\n
\n\n B\n
\n\n Calculate the standard enthalpy change (Δ\n \n H\n \n \n ⦵\n \n ) for the forward reaction in kJ mol\n \n −1\n \n .\n
\n\n Δ\n \n H\n \n \n ⦵\n \n \n f\n \n PCl\n \n 3\n \n (g) = −306.4 kJ mol\n \n −1\n \n
\n\n Δ\n \n H\n \n \n ⦵\n \n \n f\n \n PCl\n \n 5\n \n (g) = −398.9 kJ mol\n \n −1\n \n
\n\n [1]\n
\n\n «−398.9 kJ mol\n \n −1\n \n − (−306.4 kJ mol\n \n −1\n \n ) =» −92.5 «kJ mol\n \n −1\n \n » ✔\n
\n\n Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.\n
\n\n [1]\n
\n\n allows them to explain the properties of different compounds/substances\n
\n \n \n OR\n \n \n
\n enables them to generalise about substances\n
\n \n \n OR\n \n \n
\n enables them to make predictions ✔\n
\n \n
\n Accept other valid answers.\n \n
\n What volume of oxygen, in dm\n \n 3\n \n at STP, is needed when 5.8 g of butane undergoes complete combustion?\n
\n\n \n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n B\n
\n\n State one advantage of the use of hydrogen, H\n \n 2\n \n , as an alternative greener fuel to fossil fuels.\n
\n\n [1]\n
\n\n it does not produce CO\n \n 2\n \n «when combusted» ✓\n
\n\n \n Determine the initial rate of reaction of limestone with nitric acid from the graph.\n \n
\n\n \n Show your working on the graph and include the units of the initial rate.\n \n
\n\n [3]\n
\n\n \n tangent drawn at time zero ✔\n
\n g day\n \n −1\n \n ✔\n
\n 0.16 ✔\n \n
\n
\n\n \n \n NOTE: Accept other reasonable units for initial rate eg, mol dm\n \n −3\n \n s\n \n −1\n \n , mol dm\n \n −3\n \n min\n \n −1\n \n , g s\n \n −1\n \n \n OR\n \n g min\n \n −1\n \n .\n \n \n
\n\n \n \n M3 can only be awarded if the value corresponds to the correct unit given in M2.\n
\n Accept values for the initial rate for M3 in the range: 0.13 − 0.20 g day\n \n −1\n \n \n OR\n \n 1.5 × 10\n \n −6\n \n g s\n \n −1\n \n − 2.3 × 10\n \n −6\n \n g s\n \n −1\n \n \n OR\n \n 7.5 × 10\n \n −8\n \n − 1.2 × 10\n \n −7\n \n mol dm\n \n −3\n \n s\n \n −1\n \n \n OR\n \n 4.5 × 10\n \n −6\n \n − 6.9 × 10\n \n −6\n \n mol dm\n \n −3\n \n min\n \n −1\n \n \n OR\n \n 9.0 × 10\n \n −5\n \n − 1.4 × 10\n \n −4\n \n g min\n \n −1\n \n \n OR\n \n a range based on any other reasonable unit for rate.\n \n \n
\n \n \n Ignore any negative rate value.\n
\n Award\n \n [2 max]\n \n for answers such as 0.12/0.11 g day\n \n −1\n \n , incorrectly obtained by using the first two points on the graph (the average rate between t = 0 and 1 day).\n
\n Award\n \n [1 max]\n \n for correctly calculating any other average rate.\n \n \n
\n Which is a Lewis acid, but not a Brønsted-Lowry acid?\n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n \n Which is the\n \n 1\n \n H NMR spectrum of tetramethylsilane, TMS, (CH\n \n 3\n \n )\n \n 4\n \n Si?\n \n
\n\n \n \n \n
\n [1]\n
\n\n C\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 \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 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 3.00 mol of C\n \n 3\n \n H\n \n 8\n \n is mixed with 20.00 mol of O\n \n 2\n \n . Which quantity is present at the end of the reaction?\n
\n\n C\n \n 3\n \n H\n \n 8\n \n (g) + 5O\n \n 2\n \n (g) → 3CO\n \n 2\n \n (g) + 4H\n \n 2\n \n O (g)\n
\n\n
\n A. 1.00 mol of C\n \n 3\n \n H\n \n 8\n \n
\n B. 5.00 mol of O\n \n 2\n \n
\n\n C. 12.00 mol of CO\n \n 2\n \n
\n\n D. 16.00 mol of H\n \n 2\n \n O\n
\n\n [1]\n
\n\n B\n
\n\n \n Identify\n \n one\n \n error associated with the use of an accurate stopwatch.\n \n
\n\n [1]\n
\n\n \n human reaction time/delay «starting/stopping the stopwatch»\n \n [✔]\n \n \n
\n\n \n \n Note\n \n :\n \n Do\n \n not\n \n accept “inaccurate stopwatch”.\n \n \n
\n\n This question was well answered by most candidates although some students did not read the question clearly and commented on the stopwatch having problems or not being accurate.\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 brown/orange/red/yellow to colourless ✔\n
\n\n
\n \n Do\n \n not\n \n accept clear for colourless.\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 \n Sodium percarbonate, 2Na\n \n 2\n \n CO\n \n 3\n \n •3H\n \n 2\n \n O\n \n 2\n \n , is an adduct of sodium carbonate and hydrogen peroxide and is used as a cleaning agent.\n \n
\n\n \n M\n \n r\n \n (2Na\n \n 2\n \n CO\n \n 3\n \n •3H\n \n 2\n \n O\n \n 2\n \n ) = 314.04\n \n
\n\n \n Calculate the percentage by mass of hydrogen peroxide in sodium percarbonate, giving your answer to two decimal places.\n \n
\n\n [2]\n
\n\n \n M( H\n \n 2\n \n O\n \n 2\n \n ) «= 2 × 1.01 + 2 × 16.00» = 34.02 «g»\n \n [✔]\n \n \n
\n\n \n «% H\n \n 2\n \n O\n \n 2\n \n = 3 ×\n \n \n \n × 100 =» 32.50 «%»\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 It is recommended that candidates use the relative atomic masses given in the periodic table.\n
\n\n Which molecule has a carbonyl functional group?\n
\n\n
\n A. CH\n \n 3\n \n OCH\n \n 3\n \n
\n B. CH\n \n 3\n \n COCH\n \n 3\n \n
\n\n C. CH\n \n 3\n \n CH\n \n 2\n \n OH\n
\n\n D. CH\n \n 3\n \n CH\n \n 2\n \n NH\n \n 2\n \n
\n\n [1]\n
\n\n B\n
\n\n Determine the value and unit of the rate constant using the rate expression in (a).\n
\n\n [2]\n
\n\n «\n \n k\n \n =\n \n =» 8.0 ✓\n
\n\n mol\n \n −3\n \n dm\n \n 9\n \n s\n \n −1\n \n ✓\n
\n\n Which compound has the greatest volatility under the same conditions?\n
\n\n A. SO\n \n 2\n \n
\n\n B. SiO\n \n 2\n \n
\n\n C. SnO\n \n 2\n \n
\n\n D. SrO\n
\n\n [1]\n
\n\n A\n
\n\n \n Which properties can be monitored to determine the rate of the reaction?\n \n
\n\n \n Fe (s) + CuSO\n \n 4\n \n (aq) → Cu (s) + FeSO\n \n 4\n \n (aq)\n \n
\n\n \n I. change in volume\n
\n II. change in temperature\n
\n III. change in colour\n \n
\n \n A. I and II only\n \n
\n\n \n B. I and III only\n \n
\n\n \n C. II and III only\n \n
\n\n \n D. I, II and III\n \n
\n\n [1]\n
\n\n C\n
\n\n One G2 form queried if visible spectroscopy is on the core syllabus and therefore should candidates be aware of monitoring a reaction via colour change. However, 6.1 clearly states that following change in colour is one method of following reactions.\n
\n\n Outline the reasons that sodium hydroxide is considered a Brønsted–Lowry and Lewis base.\n
\n\n \n
\n [1]\n
\n\n \n Brønsted–Lowry base:\n \n
\n proton acceptor\n
\n \n \n AND\n \n \n
\n\n \n Lewis Base:\n \n
\n e\n \n –\n \n pair donor/nucleophile ✔\n
\n Explain the mechanism of the reaction between 1-bromopropane, CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n Br, and aqueous sodium hydroxide, NaOH (aq), using curly arrows to represent the movement of electron pairs.\n
\n\n [4]\n
\n\n \n
\n curly arrow going from lone pair/negative charge on O in HO\n \n –\n \n to C ✔\n
\n\n curly arrow showing Br breaking ✔\n
\n\n representation of transition state showing negative charge, square brackets and partial bonds ✔\n
\n\n formation of organic product CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n OH\n \n \n AND\n \n \n Br– ✔\n
\n\n
\n\n \n Do not allow curly arrow originating on H in HO\n \n –\n \n .\n \n
\n\n \n Accept curly arrow either going from bond between C and Br to Br in 1-bromopropane or in the transition\n \n
\n \n state.\n \n
\n \n Do\n \n not\n \n penalize if HO and Br are not at 180° to each other.\n \n
\n\n \n Award\n \n [3 max]\n \n for S\n \n N\n \n 1 mechanism.\n \n
\n\n As 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.\n
\n\n State the class of compound to which ethene belongs.\n
\n\n [1]\n
\n\n alkene ✔\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 \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 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 Which combination describes a strong Brønsted–Lowry acid?\n
\n\n
\n| \n | \n\n \n Proton donor\n \n | \n\n \n Conjugate base\n \n | \n
| \n A.\n | \n\n good\n | \n\n strong\n | \n
| \n B.\n | \n\n good\n | \n\n weak\n | \n
| \n C.\n | \n\n poor\n | \n\n strong\n | \n
| \n D.\n | \n\n poor\n | \n\n weak\n | \n
\n [1]\n
\n\n B\n
\n\n \n Which corresponds to a system at equilibrium?\n \n
\n\n \n
\n [1]\n
\n\n B\n
\n\n State the class of compound to which ethene belongs.\n
\n\n [1]\n
\n\n alkene ✔\n
\n\n Draw an arrow, labelled\n \n X\n \n , to represent the electron transition for the ionization of a hydrogen atom in the ground state.\n
\n\n [1]\n
\n\n \n
\n upward arrow X\n \n \n AND\n \n \n starting at n = 1 extending to n = ∞ ✔\n
\n\n Only 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.\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 2CO(g) → C(s) + CO\n \n 2\n \n (g) ✓\n
\n\n
\n \n Accept reversible arrows.\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 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 Which series is in order of increasing boiling point?\n
\n\n A. CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n OH CH\n \n 3\n \n COCH\n \n 3\n \n CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 3\n \n
\n\n B. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 3\n \n CH\n \n 3\n \n COCH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n OH\n
\n\n C. CH\n \n 3\n \n COCH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n OH CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 3\n \n
\n\n D. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n OH CH\n \n 3\n \n COCH\n \n 3\n \n
\n\n [1]\n
\n\n B\n
\n\n Which combination describes a strong Brønsted–Lowry acid?\n
\n\n
\n| \n | \n\n \n Proton donor\n \n | \n\n \n Conjugate base\n \n | \n
| \n A.\n | \n\n good\n | \n\n strong\n | \n
| \n B.\n | \n\n good\n | \n\n weak\n | \n
| \n C.\n | \n\n poor\n | \n\n strong\n | \n
| \n D.\n | \n\n poor\n | \n\n weak\n | \n
\n [1]\n
\n\n B\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 \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 Some antacids contain carbonates.\n \n
\n\n \n Determine the pH of a buffer solution which contains 0.160 mol dm\n \n −3\n \n CO\n \n 3\n \n \n 2−\n \n and 0.200 mol dm\n \n −3\n \n HCO\n \n 3\n \n \n −\n \n , using section 1 of the data booklet.\n \n
\n\n \n p\n \n K\n \n \n a\n \n (HCO\n \n 3\n \n \n −\n \n ) = 10.32\n \n
\n\n [1]\n
\n\n \n «pH = p\n \n K\n \n \n a\n \n \n \n \n »\n
\n «pH =»10.22\n \n [✔]\n \n \n
\n This was generally a well-answered question. Most candidates who did not receive the mark inverted the concentration of the conjugate base/concentration of the acid in the calculation.\n
\n\n Suggest\n \n two\n \n differences in the\n \n 1\n \n H NMR of but-2-ene and the organic product from (d)(ii).\n
\n\n [2]\n
\n\n \n \n ALTERNATIVE 1:\n \n Any two of:\n \n
\n\n but-2-ene: 2 signals\n \n \n AND\n \n \n product: 4 signals ✔\n
\n\n but-2-ene: «area ratio» 3:1/6:2\n \n \n AND\n \n \n product: «area ratio» 3:3:2:1 ✔\n
\n\n product: «has signal at» 3.5-4.4 ppm «and but-2-ene: does not» ✔\n
\n\n but-2-ene: «has signal at» 4.5-6.0 ppm «and product: does not» ✔\n
\n\n
\n\n \n \n ALTERNATIVE 2:\n \n \n
\n\n but-2-ene: doublet\n \n \n AND\n \n \n quartet/multiplet/4 ✔\n
\n\n product: doublet\n \n \n AND\n \n \n triplet\n \n \n AND\n \n \n quintet/5/multiplet\n \n \n AND\n \n \n sextet/6/multiplet ✔\n
\n\n
\n\n \n Accept “product «has signal at» 1.3–1.4 ppm «and but-2-ene: does not»”.\n \n
\n\n A 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.\n
\n\n Which change results in the largest negative value of Δ\n \n S\n \n ?\n
\n\n A. C\n \n 2\n \n H\n \n 5\n \n OH (l) + SOCl\n \n 2\n \n (l) → C\n \n 2\n \n H\n \n 5\n \n Cl (l) + SO\n \n 2\n \n (g) + HCl (g)\n
\n\n B. CaCO\n \n 3\n \n (s) → CaO (s) + CO\n \n 2\n \n (g)\n
\n\n C. H\n \n 2\n \n O (l) → H\n \n 2\n \n O (s)\n
\n\n D. NH\n \n 3\n \n (g) + HCl (g) → NH\n \n 4\n \n Cl (s)\n
\n\n [1]\n
\n\n D\n
\n\n A container holds 30 g of argon and 60 g of neon.\n
\n\n
\n What is the ratio of number of atoms of argon to number of atoms of neon in the container?\n
\n
\n A. 0.25\n
\n B. 0.50\n
\n\n C. 2.0\n
\n\n D. 4.0\n
\n\n [1]\n
\n\n A\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 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 \n Which compound has the shortest C to O bond?\n \n
\n\n \n A. CH\n \n 3\n \n CHO\n
\n \n
\n \n B. CO\n
\n \n
\n \n C. CO\n \n 2\n \n
\n \n
\n \n D. C\n \n 2\n \n H\n \n 5\n \n OC\n \n 2\n \n H\n \n 5\n \n \n
\n\n [1]\n
\n\n B\n
\n\n Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl bond angle in PCl\n \n 3\n \n .\n
\n\n \n
\n [3]\n
\n\n \n Electron domain geometry\n \n : tetrahedral ✔\n
\n\n \n Molecular geometry\n \n : trigonal pyramidal ✔\n
\n\n \n Bond angle\n \n : 100«°» ✔\n
\n\n \n
\n Accept any value or range within the range 91−108«°» for\n \n M3\n \n .\n \n
\n Identify the type of reaction.\n
\n\n [1]\n
\n\n «nucleophilic» substitution\n
\n \n \n OR\n \n \n
\n SN2 ✔\n
\n \n
\n Accept “hydrolysis”.\n \n
\n \n Accept SN1\n \n
\n\n Excellent performance on the type of reaction but with some incorrect answers such as alkane substitution, free radical substitution or electrophilic substitution.\n
\n\n Which compound produces this mass spectrum?\n
\n\n \n
\n \n [Spectral Database for Organic Compounds, SDBS. SDBS Compounds and Spectral Search. [graph] Available at:\n \n
\n \n https://sdbs.db.aist.go.jp [Accessed 3 January 2019].]\n \n
\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n Question 29 contained a mass spectrum and 44% of students chose the correct answer of C - ethanoic acid. There were comments that this was unfair without access to the data booklet. However, as the other molecules are clearly wrong this did not seem so difficult to do without the data booklet as they did have the periodic table.\n
\n\n Why does benzene undergo substitution more readily than addition?\n
\n\n A. Benzene is unsaturated.\n
\n\n B. Addition could produce an alkane.\n
\n\n C. Resonance makes carbon–carbon bonds too strong to break.\n
\n\n D. A benzene molecule is planar.\n
\n\n [1]\n
\n\n C\n
\n\n Which spectra would show the difference between propan-2-ol, CH\n \n 3\n \n CH(OH)CH\n \n 3\n \n , and propanal, CH\n \n 3\n \n CH\n \n 2\n \n CHO?\n
\n\n I. mass\n
\n II. infrared\n
\n III.\n \n 1\n \n H NMR\n
\n A. I and II only\n
\n\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n D\n
\n\n Which reaction involves homolytic fission?\n
\n\n A. CH\n \n 4\n \n + Cl\n \n 2\n \n
\n\n B. CH\n \n 3\n \n Br + NaOH\n
\n\n C. (CH\n \n 3\n \n )\n \n 3\n \n CBr + NaOH\n
\n\n D. C\n \n 6\n \n H\n \n 6\n \n + HNO\n \n 3\n \n + H\n \n 2\n \n SO\n \n 4\n \n
\n\n [1]\n
\n\n A\n
\n\n A well answered question. 66% of the candidates identified the chlorination of methane as the reaction involving homolytic bond fission.\n
\n\n What is the concentration of chloride ions, in mol dm\n \n −3\n \n , in a solution formed by mixing 200 cm\n \n 3\n \n of 1 mol dm\n \n −3\n \n HCl with 200 cm\n \n 3\n \n of 5 mol dm\n \n −3\n \n NaCl?\n
\n\n A. 1\n
\n\n B. 2\n
\n\n C. 3\n
\n\n D. 6\n
\n\n [1]\n
\n\n C\n
\n\n 52% of the candidates calculated the concentration of chloride ions in the titration. The distractors were chosen almost equally.\n
\n\n Which of the following elements yields a basic oxide when combusted?\n
\n\n
\n A. Au\n
\n B. P\n
\n\n C. Ca\n
\n\n D. N\n
\n\n [1]\n
\n\n C\n
\n\n 2.85 g of CaCO\n \n 3\n \n was collected in the experiment in d(i). Calculate the percentage yield of CaCO\n \n 3\n \n .\n
\n\n (If you did not obtain an answer to d(i), use 4.00 g, but this is not the correct value.)\n
\n\n [1]\n
\n\n «\n \n × 100 =» 86.4 «%» ✓\n
\n\n
\n\n \n Accept answers in the range 86.1-86.4 «%».\n \n
\n\n \n Accept “71.3 %” for using the incorrect given value of 4.00 g.\n \n
\n\n Write the equation for this reaction.\n
\n\n [1]\n
\n\n 4FeS(s) + 7O\n \n 2\n \n (g) → 2Fe\n \n 2\n \n O\n \n 3\n \n (s) + 4SO\n \n 2\n \n (g) ✔\n
\n\n \n
\n Accept any correct ratio.\n \n
\n Which is the product when but-1-yne reacts with excess hydrogen gas?\n
\n\n
\n A. But-1-ene\n
\n B. Butane\n
\n\n C. But-2-ene\n
\n\n D. No reaction\n
\n\n [1]\n
\n\n B\n
\n\n What is the mass of one molecule of C\n \n 60\n \n ?\n
\n\n \n N\n \n \n A\n \n = 6.0 × 10\n \n 23\n \n
\n\n
\n A. 1.0 × 10\n \n −22\n \n g\n
\n B. 2.0 × 10\n \n −23\n \n g\n
\n\n C. 8.3 × 10\n \n −24\n \n g\n
\n\n D. 1.2 × 10\n \n −21\n \n g\n
\n\n [1]\n
\n\n D\n
\n\n Which allotrope, oxygen or ozone, has the stronger bond between O atoms, and which absorbs higher frequency UV radiation in the atmosphere?\n
\n\n
\n| \n | \n\n \n Stronger bond between O atoms\n \n | \n\n \n Absorbs higher frequency UV\n \n | \n
| \n A.\n | \n\n ozone\n | \n\n ozone\n | \n
| \n B.\n | \n\n ozone\n | \n\n oxygen\n | \n
| \n C.\n | \n\n oxygen\n | \n\n oxygen\n | \n
| \n D.\n | \n\n oxygen\n | \n\n ozone\n | \n
\n [1]\n
\n\n C\n
\n\n \n Explain why the rate of reaction of limestone with nitric acid decreases and reaches zero over the period of five days.\n \n
\n\n [2]\n
\n\n \n acid used up\n
\n \n OR\n \n
\n acid is the limiting reactant ✔\n \n
\n \n concentration of acid decreases\n
\n \n OR\n \n
\n less frequent collisions ✔\n \n
\n \n \n NOTE: Award\n \n [1 max]\n \n for \"surface area decreases\" if the idea that CaCO\n \n 3\n \n is used up/acts as the limiting reactant” is conveyed for M1.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “reaction reaches equilibrium” for M2.\n \n \n
\n\n Formulate an equation for the reaction of one mole of phosphoric acid with one mole of sodium hydroxide.\n
\n\n [1]\n
\n\n H\n \n 3\n \n PO\n \n 4\n \n (aq) + NaOH (aq) → NaH\n \n 2\n \n PO\n \n 4\n \n (aq) + H\n \n 2\n \n O (l) ✔\n
\n\n \n
\n Accept net ionic equation.\n \n
\n A reversible reaction has a reaction quotient,\n \n Q\n \n , of 4.5 and equilibrium constant,\n \n K\n \n \n c\n \n , of 6.2.\n
\n\n 2A (g)\n \n A\n \n 2\n \n (g)\n
\n\n Which statement describes the reaction at this time?\n
\n\n
\n A. The system has reached equilibrium.\n
\n B. The rate of the forward reaction is greater than the rate of the reverse reaction.\n
\n\n C. The concentration of reactant is greater than the concentration of product.\n
\n\n D. At equilibrium, the concentration of reactant is greater than the concentration of product.\n
\n\n [1]\n
\n\n B\n
\n\n \n What type of reaction occurs when C\n \n 6\n \n H\n \n 13\n \n Br becomes C\n \n 6\n \n H\n \n 13\n \n OH?\n
\n \n
\n \n A. Nucleophilic substitution\n
\n \n
\n \n B. Electrophilic substitution\n
\n \n
\n \n C. Radical substitution\n
\n \n
\n \n D. Addition\n \n
\n\n [1]\n
\n\n A\n
\n\n State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.\n
\n\n \n
\n [4]\n
\n\n \n
\n \n
\n Award\n \n [1]\n \n for all bonding types correct.\n \n
\n \n Award\n \n [1]\n \n for\n \n each\n \n correct description.\n \n
\n\n \n Apply ECF for M2 only once.\n \n
\n\n About 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.\n
\n\n Calculate the oxidation state of sulfur in iron(II) disulfide, FeS\n \n 2\n \n .\n
\n\n [1]\n
\n\n –1 ✔\n
\n\n
\n\n \n Accept “– I”.\n \n
\n\n This 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.\n
\n\n What is correct about energy changes during bond breaking and bond formation?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n 55% of the candidates selected the correct statements about bond breaking and bond formation. The most commonly chosen distractor reversed them. The question discriminated well between high-scoring and low-scoring candidates.\n
\n\n \n The\n \n 1\n \n H NMR spectrum of one of the products has four signals. The integration trace shows a ratio of the areas under the signals of 9 : 3 : 2 : 2.\n \n
\n\n \n Deduce the structural formula of the product.\n \n
\n\n [1]\n
\n\n \n
\n \n NOTE:\n \n Accept a skeletal formula or a full or condensed structural formula.\n
\n Penalise missing hydrogens or bond connectivities once only in Option C.\n \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 \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 Sketch the shape of one sigma (\n \n ) and one pi (\n \n ) bond.\n
\n\n \n
\n [2]\n
\n\n \n Sigma (\n \n ):\n \n
\n\n \n
\n
\n\n \n Pi (\n \n ):\n \n
\n\n \n
\n
\n\n \n Accept overlapping p-orbital(s) with both lobes of equal size/shape.\n \n
\n\n \n Shaded areas are not required in either diagram.\n \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 \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 Which statement is correct about points X and Y on the energy profile diagram?\n
\n\n \n
\n
\n A. X is a transition state and Y is an intermediate.\n
\n B. X is an intermediate and Y is a transition state.\n
\n\n C. X and Y are transition states.\n
\n\n D. X and Y are intermediates.\n
\n\n [1]\n
\n\n C\n
\n\n Calculate the entropy change for the Haber–Bosch process, in J mol\n \n –1\n \n K\n \n –1\n \n at 298 K. Use your answer to (b)(i) and section 1 of the data booklet.\n
\n\n [2]\n
\n\n Δ\n \n G\n \n = «Δ\n \n H\n \n –\n \n T\n \n Δ\n \n S\n \n =» –93000 «J» – 298«K» × Δ\n \n S\n \n = –33000 ✔\n
\n\n Δ\n \n S\n \n = 〈〈\n \n 〉〉 = –201 «J mol\n \n –1\n \n K\n \n –1\n \n » ✔\n
\n\n
\n\n \n Do\n \n not\n \n penalize failure to convert kJ to J in\n \n both\n \n (c)(ii) and (c)(iii).\n \n
\n\n \n Award\n \n [2]\n \n for correct final answer\n \n
\n\n \n Award\n \n [1 max]\n \n for (+) 201 «J mol\n \n –1\n \n K\n \n –1\n \n ».\n \n
\n\n \n Award [2] for –101 or –100.5 «J mol\n \n –1\n \n K\n \n –1\n \n ».\n \n
\n\n Very good performance; since the unit for\n \n S\n \n is J mol\n \n ˗1\n \n K\n \n ˗1\n \n , Δ\n \n G\n \n and Δ\n \n H\n \n needed to be converted from kJ to J, but that was not done in some cases.\n
\n\n \n What is formed at the electrodes during the electrolysis of molten sodium bromide?\n \n
\n\n \n
\n
\n\n [1]\n
\n\n D\n
\n\n Outline how these drug administration methods affect bioavailability.\n
\n\n
\n\n Oral: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n
\n\n Intravenous: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n
\n\n [2]\n
\n\n \n Oral:\n \n
\n low/lower «bioavailability»\n \n \n AND\n \n \n drugs pass through digestive system «and breakdown» ✓\n
\n \n Intravenous:\n \n
\n high/higher «bioavailability»\n \n \n AND\n \n \n «more» direct route to bloodstream ✓\n
\n
\n\n \n Accept “low/lower\n \n AND\n \n drugs not easily absorbed from digestive system”\n
\n \n \n OR\n \n \n
\n “low/lower\n \n AND\n \n drugs broken down in digestive system”\n
\n \n OR\n \n
\n “low/lower\n \n AND\n \n drugs affected by acid” for M1.\n \n
\n \n Do not penalize use of “slow” for “low/lower” or “fast” for “high/higher”.\n \n
\n\n \n Accept “100 % bioavailability” for “high/higher” within Intravenous answer in M2.\n \n
\n\n \n Award\n \n [1 max]\n \n for “oral drugs have slower absorption/distribution than intravenous drugs”\n
\n \n OR\n \n
\n “Oral: low/lower «bioavailability»\n \n AND\n \n intraveneous: high/higher «bioavailability»”.\n \n
\n \n How does a lithium atom form the most stable ion?\n \n
\n\n \n A. The atom gains a proton to form a positive ion.\n \n
\n\n \n B. The atom loses a proton to form a negative ion.\n \n
\n\n \n C. The atom loses an electron to form a positive ion.\n \n
\n\n \n D. The atom gains an electron to form a negative ion.\n \n
\n\n [1]\n
\n\n C\n
\n\n One of the most straight-forward questions on the paper about how lithium forms its most stable ion.\n
\n\n \n What is the sum of the coefficients when the equation is balanced with whole numbers?\n \n
\n\n \n __MnO\n \n 2\n \n (s) + __HCl (aq) → __MnCl\n \n 2\n \n (aq) + __H\n \n 2\n \n O (l) + __Cl\n \n 2\n \n (g)\n
\n \n
\n \n A. 6\n
\n \n
\n \n B. 7\n
\n \n
\n \n C. 8\n
\n \n
\n \n D. 9\n \n
\n\n [1]\n
\n\n D\n
\n\n \n Proteins are polymers of amino acids.\n \n \n A paper chromatogram of two amino acids, A1 and A2, is obtained using a non-polar solvent.\n \n
\n\n \n
\n \n © International Baccalaureate Organization 2020.\n \n
\n\n \n Determine the\n \n \n \n value of A1.\n \n
\n\n [1]\n
\n\n \n ✔\n
\n\n \n Accept any value within the range “\n \n ”.\n \n
\n\n Many students scored this mark. The students who missed this mark that were close either measured from the top or bottom of the spot rather than the middle. A few students had answers that were greater than 1 which indicated a clear lack of understanding of this concept.\n
\n\n \n Suggest a modification to the procedure that would make the results more reliable.\n \n
\n\n [1]\n
\n\n \n repetition / take several samples «and average» ✔\n \n
\n\n What is ΔH, in kJ, for the reaction N\n \n 2\n \n H\n \n 4\n \n (l) + H\n \n 2\n \n (g) → 2NH\n \n 3\n \n (g)?\n
\n| \n \n Reaction\n \n | \n\n \n ΔH\n \n | \n
| \n N\n \n 2\n \n H\n \n 4\n \n (l) + CH\n \n 3\n \n OH (l) → CH\n \n 2\n \n O (g) + N\n \n 2\n \n (g) + 3H\n \n 2\n \n (g)\n | \n\n −37 kJ\n | \n
| \n N\n \n 2\n \n (g) + 3H\n \n 2\n \n (g) → 2NH\n \n 3\n \n (g)\n | \n\n −46 kJ\n | \n
| \n CH\n \n 3\n \n OH (l) → CH\n \n 2\n \n O (g) + H\n \n 2\n \n (g)\n | \n\n −65 kJ\n | \n
\n
\n A. −18\n
\n B. 18\n
\n\n C. −83\n
\n\n D. −148\n
\n\n [1]\n
\n\n A\n
\n\n Calculate the Gibbs free energy change (Δ\n \n G\n \n ), in kJ mol\n \n −1\n \n , for this reaction at 25 °C. Use section 1 of the data booklet.\n
\n\n If you did not obtain an answer in c(i) or c(ii) use −87.6 kJ mol\n \n −1\n \n and −150.5 J mol\n \n −1\n \n K\n \n −1\n \n respectively, but these are not the correct answers.\n
\n\n [2]\n
\n\n «ΔS =» –0.1702 «kJ mol\n \n –1\n \n K\n \n –1\n \n »\n
\n \n \n OR\n \n \n
\n 298 «K» ✔\n
\n «ΔG = –92.5 kJ mol\n \n –1\n \n – (298 K × –0.1702 kJ mol\n \n –1\n \n K\n \n –1\n \n ) =» –41.8 «kJ mol\n \n –1\n \n » ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n If –87.6 and -150.5 are used then –42.8.\n \n
\n\n Which is correct when benzene undergoes electrophilic substitution by chlorine, Cl\n \n 2\n \n , in presence of the catalyst, Cl\n \n 3\n \n Al?\n
\n\n
\n A. C\n \n 6\n \n H\n \n 6\n \n + Cl\n \n 2\n \n → C\n \n 6\n \n H\n \n 5\n \n Cl + HCl\n
\n B. C\n \n 6\n \n H\n \n 6\n \n + Cl\n \n 2\n \n → C\n \n 6\n \n H\n \n 4\n \n Cl\n \n 2\n \n + H\n \n 2\n \n
\n\n C. C\n \n 6\n \n H\n \n 6\n \n + Cl\n \n 2\n \n → C\n \n 6\n \n H\n \n 6\n \n Cl\n \n 2\n \n
\n\n D. C\n \n 6\n \n H\n \n 6\n \n + Cl\n \n 2\n \n → C\n \n 6\n \n H\n \n 6\n \n Cl + Cl\n \n −\n \n
\n\n [1]\n
\n\n A\n
\n\n The unit cell of lead (II) sulfide is shown:\n
\n\n \n
\n
\n\n 6 ✓\n
\n\n Outline why ionization energies have positive values but most electron affinities have negative values\n
\n\n [1]\n
\n\n ionization energy breaks bond/attractive force between nucleus and electron\n
\n \n \n AND\n \n \n
\n electron affinity forms bond/attractive force between nucleus and electron ✓\n
\n
\n \n Accept for ionization energy “energy needed/endothermic to remove an electron\".\n
\n \n \n AND\n \n \n
\n for electron affinity “energy released/exothermic adding an electron”.\n \n
\n What is the activation energy according to the following plot of the linear form of the Arrhenius equation?\n
\n\n Arrhenius equation:\n \n .\n
\n\n \n
\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n A discriminating question in which high scoring candidates had much greater success in transforming the Arrhenius equation into the units used in the graph by taking its natural log and using the dimensions of the slope to arrive at the correct answer.\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 systematic error\n \n \n AND\n \n \n lower accuracy ✔\n
\n\n overestimation of [Fe(II)] ✔\n
\n\n Determine the enthalpy change, Δ\n \n H\n \n , for the Haber–Bosch process, in kJ. Use Section 11 of the data booklet.\n
\n\n [3]\n
\n\n \n bonds broken\n \n : N≡N + 3(H-H) /«1 mol×»945 «kJ mol\n \n –1\n \n » + 3«mol»×436 «kJ mol\n \n –1\n \n » / 945 «kJ» + 1308 «kJ» / 2253 «kJ» ✔\n
\n\n \n bonds formed\n \n : 6(N-H) / 6«mol»×391 «kJ mol\n \n –1\n \n » / 2346 «kJ» ✔\n
\n\n Δ\n \n H\n \n = «2253 kJ - 2346 kJ = » -93 «kJ» ✔\n
\n\n
\n\n \n Award\n \n [2 max]\n \n for (+)93 «kJ».\n \n
\n\n Good 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.\n
\n\n Which element is found in the 4th group, 6th period of the periodic table?\n
\n\n A. Selenium\n
\n\n B. Lead\n
\n\n C. Chromium\n
\n\n D. Hafnium\n
\n\n [1]\n
\n\n D\n
\n\n A very well answered question. 76% of the candidates were able to identify the element in the fourth group and sixth period of the periodic table as hafnium. The most commonly chosen distractor was lead which is in group 14 not 4.\n
\n\n Lipids and carbohydrates both release energy in the body.\n
\n\n
\n\n C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n + 6O\n \n 2\n \n → 6CO\n \n 2\n \n + 6H\n \n 2\n \n O ✓\n
\n\n C\n \n 57\n \n H\n \n 110\n \n O\n \n 6\n \n + 81.5O\n \n 2\n \n → 57CO\n \n 2\n \n + 55H\n \n 2\n \n O\n
\n \n \n OR\n \n \n
\n 2C\n \n 57\n \n H\n \n 110\n \n O\n \n 6\n \n + 163O\n \n 2\n \n →114CO\n \n 2\n \n + 110H\n \n 2\n \n O ✓\n
\n What is a possible value of pH at the equivalence point in the titration of a strong acid with a weak base?\n
\n\n
\n A. 5\n
\n B. 7\n
\n\n C. 9\n
\n\n D. 11\n
\n\n [1]\n
\n\n A\n
\n\n Which equation represents the deposition of iodine?\n
\n\n A. I\n \n 2\n \n (g) → I\n \n 2\n \n (l)\n
\n\n B. I\n \n 2\n \n (g) → I\n \n 2\n \n (s)\n
\n\n C. I\n \n 2\n \n (l) → I\n \n 2\n \n (g)\n
\n\n D. I\n \n 2\n \n (s) → I\n \n 2\n \n (g)\n
\n\n [1]\n
\n\n B\n
\n\n 58% of the candidates identified the equation that represented the deposition of iodine. The most commonly chosen distractor was the condensation of gaseous iodine. Sublimation was also quite commonly chosen.\n
\n\n Suggest what can be concluded about the gold atom from this experiment.\n
\n\n \n
\n [2]\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 Calculate the number of moles of oxygen in the day 0 sample.\n \n
\n\n [2]\n
\n\n \n \n \n \n ✔\n \n
\n\n \n \n «\n \n \n \n ✔\n \n \n
\n\n \n \n \n NOTE: Award\n \n [2]\n \n for correct final answer.\n \n \n \n
\n\n The standard electrode potential of zinc can be measured using a standard hydrogen electrode (SHE).\n
\n\n Draw and annotate the diagram to show the complete apparatus required to measure the standard electrode potential of zinc.\n
\n\n \n
\n [4]\n
\n\n \n
\n H\n \n 2\n \n (g) entering at «298 K and» 100 kPa ✔\n
\n\n platinum electrode on left ✔\n
\n\n voltmeter connecting electrodes\n \n \n AND\n \n \n salt bridge connecting electrolytes ✔\n
\n\n 1 mol dm\n \n –3\n \n H\n \n +\n \n on the left\n \n \n AND\n \n \n 1 mol dm\n \n –3\n \n Zn\n \n 2+\n \n on the right ✔\n
\n\n
\n\n \n Voltmeter and salt bridge need to be drawn but not necessarily annotated for\n \n M3\n \n .\n \n
\n\n \n Concentrations, but not state symbols, required for\n \n M4\n \n .\n \n
\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 ✔\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 Identify the strongest force between the molecules of Compound B.\n
\n\n [1]\n
\n\n hydrogen bonds ✔\n
\n\n Probably 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.\n
\n\n Calculate the molar concentration of the resulting solution of lithium hydroxide.\n
\n\n [2]\n
\n\n \n ✔\n
\n\n «n\n \n LiOH\n \n = n\n \n Li\n \n »\n
\n\n \n ✔\n
\n\n
\n\n Award\n \n [2]\n \n for correct final answer.\n
\n\n The graph shows Gibbs free energy of a mixture of N\n \n 2\n \n O\n \n 4\n \n (g) and NO\n \n 2\n \n (g) in different proportions.\n
\n\n N\n \n 2\n \n O\n \n 4\n \n (g)\n \n 2NO\n \n 2\n \n (g)\n
\n\n Which point shows the system at equilibrium?\n
\n\n \n
\n [1]\n
\n\n B\n
\n\n Identify the isomer of Compound B that exists as optical isomers (enantiomers).\n
\n\n [1]\n
\n\n butan-2-ol/CH\n \n 3\n \n CH(OH)C\n \n 2\n \n H\n \n 5\n \n ✔\n
\n\n Mediocre performance; some identified 2-methylpropan-1-ol or -2-ol, instead butan-2-ol/CH\n \n 3\n \n CH(OH)C\n \n 2\n \n H\n \n 5\n \n as the isomer that exists as an optical isomer.\n
\n\n \n What is the number of carbon atoms in\n \n of ethanoic acid\n \n \n \n ,\n \n \n \n ?\n \n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n D\n
\n\n Approximately 57% of candidates could correctly apply Avogadro's number in calculating number of carbon atoms in 12 g of ethanoic acid. Many candidates did not identify that there were 2 carbon atoms per molecule.\n
\n\n Demonstrate 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.\n
\n\n [2]\n
\n\n increased temperature decreases yield «as shown on graph» ✔\n
\n\n shifts equilibrium in endothermic/reverse direction ✔\n
\n\n Good 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.\n
\n\n Outline the requirements for a collision between reactants to yield products.\n
\n\n [2]\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 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 \n What is the major reason why the pH of unpolluted rain is less than 7?\n \n
\n\n \n A. methane\n \n
\n\n \n B. carbon dioxide\n \n
\n\n \n C. nitrogen oxides\n \n
\n\n \n D. sulfur dioxide\n \n
\n\n [1]\n
\n\n B\n
\n\n 69 % of the candidates identified CO2 as the gas responsible for the acidity of unpolluted rain. The majority of the candidates that answered incorrectly chose nitrogen oxides (C) or sulfur dioxide (D).\n
\n\n \n Deduce, giving a reason, the group of elements in the periodic table most likely to undergo sublimation.\n \n
\n\n [2]\n
\n\n \n group 18/noble gases\n \n [✔]\n \n \n
\n\n
\n\n \n smallest difference between melting and boiling points\n
\n \n \n OR\n \n \n
\n weakest intermolecular forces «in that period»\n \n \n [✔]\n \n
\n
\n\n \n \n \n Note\n \n : Accept “group 17/halogens”.\n \n \n
\n\n Most candidates correctly identified the group of elements most likely to undergo sublimation but did not score for the reason as they referred to low melting and boiling points, rather than the smallest difference between these temperatures. There were several G2 comments that “the group of elements” was a confusing requirement as elements could be grouped in many ways, including for instance, from B to Ne. The Chemistry Guide clearly states that a group on the periodic table refers to a vertical column of elements. A few complaints were received about the inclusion of a question on sublimation, but the question was designed to make candidates think, and did not require knowledge of phase diagrams.\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 systematic error\n \n \n AND\n \n \n lower accuracy ✔\n
\n\n overestimation of [Fe(II)] ✔\n
\n\n \n Draw a best-fit line on the graph.\n \n
\n\n [1]\n
\n\n \n best-fit smooth curve ✔\n \n
\n\n \n \n NOTE: Do\n \n not\n \n accept a series of connected lines that pass through all points\n \n OR\n \n any straight line representation.\n \n \n
\n\n State the name of Compound A, applying International Union of Pure and Applied Chemistry (IUPAC) rules.\n
\n\n [1]\n
\n\n propanone ✔\n
\n\n \n
\n Accept 2-propanone and propan-2-one.\n \n
\n Even though superfluous numbers (2-propanone, propan-2-one) were overlooked, only about half of the students could correctly name this simple molecule.\n
\n\n \n Which can show optical activity?\n \n
\n\n \n A. CHBrCHCl\n
\n \n
\n \n B. CH\n \n 3\n \n CH\n \n 2\n \n CHBrCH\n \n 2\n \n CH\n \n 3\n \n
\n \n
\n \n C. (CH\n \n 3\n \n )\n \n 2\n \n CBrCl\n
\n \n
\n \n D. CH\n \n 3\n \n CH\n \n 2\n \n CH(CH\n \n 3\n \n )Br\n \n
\n\n [1]\n
\n\n D\n
\n\n Sketch the Lewis (electron dot) structure of the P\n \n 4\n \n molecule, containing only single bonds.\n
\n\n
\n\n [1]\n
\n\n \n
\n \n Accept any diagram with each P joined to the other three.\n
\n Accept any combination of dots, crosses and lines.\n \n
\n \n Explain why the first ionization energy of nitrogen is greater than both carbon and oxygen.\n \n
\n\n \n Nitrogen and carbon:\n \n
\n\n \n Nitrogen and oxygen:\n \n
\n\n [2]\n
\n\n \n \n Nitrogen and carbon:\n \n \n
\n\n \n N has greater nuclear charge/«one» more proton «and electrons both lost from singly filled p-orbitals»\n \n [✔]\n \n
\n \n
\n
\n\n \n \n Nitrogen and oxygen:\n \n \n
\n\n \n O has a doubly filled «p-»orbital\n
\n \n \n OR\n \n \n
\n N has only singly occupied «p-»orbitals\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “greater e–\n \n -\n \n e\n \n -\n \n repulsion in O” or “lower\n \n e–\n \n \n -\n \n \n e\n \n \n -\n \n repulsion in N”.\n \n \n
\n\n \n \n Accept box annotation of electrons for M2.\n \n \n
\n\n Mixed 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.\n
\n\n Identify\n \n one\n \n similarity and\n \n one\n \n difference between the structures of starch and cellulose.\n
\n\n
\n Similarity: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n
\n\n Difference: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n [2]\n
\n\n \n Similarity:\n \n
\n polymers of glucose\n
\n \n \n OR\n \n \n
\n «1-4» glycosidic «links» ✓\n
\n \n Difference:\n \n
\n starch contains\n \n -glucose\n \n \n AND\n \n \n cellulose contains\n \n -glucose\n
\n \n \n OR\n \n \n
\n starch contains\n \n \n \n AND\n \n \n cellulose contains\n \n «1-4» glycosidic links\n
\n \n \n OR\n \n \n
\n starch «may contain» 1-6 glycosidic links AND cellulose does not\n
\n \n \n OR\n \n \n
\n starch «may be» branched AND cellulose unbranched ✓\n
\n
\n\n \n Do\n \n not\n \n accept “both are polysaccharides” for M1.\n
\n Accept alpha or beta for symbols in M2.\n \n
\n What is the preferred IUPAC name of the structure shown?\n
\n\n \n
\n
\n A. 2-ethyl-3-methylbutan-1-ol\n
\n B. 2,3-dimethylbutan-2-ol\n
\n\n C. 1-ethyl-2-methylpropan-1-ol\n
\n\n D. 1,1,2-trimethylpropan-1-ol\n
\n\n [1]\n
\n\n B\n
\n\n \n 1,4-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.\n \n
\n\n [2]\n
\n\n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5946/1dii_m.PNG\")
\n \n [\n \n \n ✔\n \n \n ]\n \n
\n \n HBr\n \n [\n \n ✔\n \n ]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept condensed formulae, such as CH\n \n 3\n \n C\n \n 6\n \n H\n \n 4\n \n CH\n \n 2\n \n Br.\n \n \n
\n\n Some candidates lost one mark for the bond originated from H in CH\n \n 3\n \n instead of C. Some teachers thought the use of the word “substituted alkane” made the question more difficult than it should have been.\n
\n\n Determine the rate expression for the reaction.\n
\n\n \n
\n [2]\n
\n\n \n BrO\n \n 3\n \n \n –\n \n :\n \n 1/first\n \n \n AND\n \n Br\n \n –\n \n :\n \n 1/first\n \n \n AND\n \n H\n \n +\n \n :\n \n 2/second ✓\n
\n\n «Rate =»\n \n k\n \n [BrO\n \n 3\n \n \n −\n \n ][Br\n \n −\n \n ][H\n \n +\n \n ]\n \n 2\n \n ✓\n
\n\n \n
\n M2: Square brackets required for the mark.\n \n
\n Which reactions involve the transfer of a proton?\n
\n\n
\n I. 2HCl (aq) + Mg (s) → MgCl\n \n 2\n \n (aq) + H\n \n 2\n \n (g)\n
\n II. 2HCl (aq) + MgO (s) → MgCl\n \n 2\n \n (aq) + H\n \n 2\n \n O (l)\n
\n III. 2HCl (aq) + MgCO\n \n 3\n \n (s) → MgCl\n \n 2\n \n (aq) + H\n \n 2\n \n O (l) + CO\n \n 2\n \n (g)\n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n C\n
\n\n Which contain delocalised electrons?\n
\n\n I. C\n \n 6\n \n H\n \n 5\n \n OH\n
\n II. CH\n \n 3\n \n COO\n \n −\n \n
\n III. CO\n \n 3\n \n \n 2−\n \n
\n A. I and II only\n
\n\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n D\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 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 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 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 Another 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\n Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO\n \n 3\n \n , that obeys the octet rule, showing any non-zero formal charges on the atoms.\n
\n\n [2]\n
\n\n \n
\n bonds and non-bonding pairs correct ✔\n
\n\n formal charges correct ✔\n
\n\n
\n\n \n Accept dots, crosses or lines to represent electron pairs.\n \n
\n\n \n Do\n \n not\n \n accept resonance structures with delocalised bonds/electrons.\n \n
\n\n \n Accept + and – sign respectively.\n \n
\n\n \n Do not accept a bond between nitrogen and hydrogen.\n \n
\n\n \n For an incorrect Lewis structure, allow ECF for non-zero formal charges.\n \n
\n\n Drawing the Lewis structure of HNO\n \n 3\n \n 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.\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 \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 Which statements explain the following reactions occurring in the upper atmosphere?\n
\n\n
\n \n
\n [1]\n
\n\n D\n
\n\n What is the order of increasing conductivity for aqueous solutions of these acids and bases at equal concentrations?\n
\n| \n | \n\n \n p\n \n K\n \n \n b\n \n \n | \n
| \n \n Methylamine\n \n | \n\n 3.34\n | \n
| \n \n Ethanol\n \n | \n\n 15.5\n | \n
| \n \n Phenylamine\n \n | \n\n 9.13\n | \n
\n
\n A. methylamine < ethanol < phenylamine\n
\n B. ethanol < phenylamine < methylamine\n
\n\n C. methylamine < phenylamine < ethanol\n
\n\n D. ethanol < methylamine < phenylamine\n
\n\n [1]\n
\n\n B\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 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 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 Which solutions will form a buffer when mixed?\n
\n\n
\n A. 50 cm\n \n 3\n \n of 1.0 mol dm\n \n −3\n \n HCl and 50 cm\n \n 3\n \n of 1.0 mol dm\n \n −3\n \n NaOH\n
\n B. 50 cm\n \n 3\n \n of 1.0 mol dm\n \n −3\n \n CH\n \n 3\n \n COOH and 50 cm\n \n 3\n \n of 1.0 mol dm\n \n −3\n \n NaOH\n
\n\n C. 50 cm\n \n 3\n \n of 1.0 mol dm\n \n −3\n \n CH\n \n 3\n \n COOH and 100 cm\n \n 3\n \n of 1.0 mol dm\n \n −3\n \n NaOH\n
\n\n D. 100 cm\n \n 3\n \n of 1.0 mol dm\n \n −3\n \n CH\n \n 3\n \n COOH and 50 cm\n \n 3\n \n of 1.0 mol dm\n \n −3\n \n NaOH\n
\n\n [1]\n
\n\n D\n
\n\n Over 70% of candidates answered correctly, realising that partial neutralization of a weak acid by a strong base produces a buffer solution and there was quite a strong link with overall performance. Complete neutralization to give the salt (B) was the most frequently chosen distractor.\n
\n\n Which gases are acidic?\n
\n\n I. nitrogen dioxide\n
\n II. carbon dioxide\n
\n III. sulfur dioxide\n
\n A. I and II only\n
\n\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n D\n
\n\n Magnesium can be produced by the electrolysis of molten magnesium chloride.\n
\n\n Write the half-equation for the formation of magnesium.\n
\n\n [1]\n
\n\n Mg\n \n 2+\n \n + 2 e\n \n -\n \n → Mg ✔\n
\n\n
\n\n \n Do\n \n not\n \n penalize missing charge on electron.\n \n
\n\n \n Accept equation with equilibrium arrows.\n \n
\n\n Unfortunately, only 40% of the students could write this quite straightforward half equation.\n
\n\n What is the molecular formula of a compound with an empirical formula of CHO\n \n 2\n \n and a relative molecular mass of 90?\n
\n\n A. CHO\n \n 2\n \n
\n\n B. C\n \n 2\n \n H\n \n 2\n \n O\n \n 4\n \n
\n\n C. C\n \n 3\n \n H\n \n 6\n \n O\n \n 3\n \n
\n\n D. C\n \n 4\n \n H\n \n 10\n \n O\n \n 2\n \n
\n\n [1]\n
\n\n B\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 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 Only the very weak candidates were unable to calculate the enthalpy change correctly, eventually missing 1 mark for inverted calculations.\n
\n\n \n Calculate the percentage uncertainty and percentage error in the experimentally determined value of\n \n \n \n for methanol.\n \n
\n\n \n
\n [2]\n
\n\n \n Percentage uncertainty:\n \n
\n \n ✔\n
\n \n Percentage error:\n \n
\n \n ✔\n
\n \n Award\n \n [1 max]\n \n if calculations are reversed\n \n OR\n \n if incorrect alcohol is used.\n \n
\n\n Many students scored both points and others at least one. Weaker students inverted the calculations.\n
\n\n \n 0.10 mol of hydrochloric acid is mixed with 0.10 mol of calcium carbonate.\n \n
\n\n \n 2HCl (aq) + CaCO\n \n 3\n \n (s) → CaCl\n \n 2\n \n (aq) + H\n \n 2\n \n O (l) + CO\n \n 2\n \n (g)\n \n
\n\n \n Which is correct?\n \n
\n\n \n
\n [1]\n
\n\n C\n
\n\n State the equilibrium constant expression,\n \n K\n \n \n c\n \n , for this reaction.\n
\n\n [1]\n
\n\n «\n \n K\n \n \n c\n \n =»\n \n ✔\n
\n\n \n When a bottle of carbonated water is opened, these equilibria are disturbed.\n \n
\n\n \n State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).\n \n
\n\n [1]\n
\n\n \n shifts to left/reactants\n \n \n AND\n \n \n to increase amount/number of moles/molecules of gas/CO\n \n 2\n \n (g)\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “shifts to left/reactants\n \n AND\n \n to increase pressure”.\n \n \n
\n\n Many 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.\n
\n\n \n The 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.\n \n
\n\n [2]\n
\n\n \n \n \n ALTERNATIVE 1:\n \n \n
\n [H\n \n +\n \n ] «= 10\n \n −2.95\n \n » = 1.122 × 10\n \n −3\n \n «mol dm\n \n −3\n \n »\n \n [✔]\n \n \n
\n \n «[OH\n \n −\n \n ] =\n \n \n \n =» 8.91 × 10\n \n −12\n \n «mol dm\n \n −3\n \n »\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n ALTERNATIVE 2:\n \n \n
\n pOH = «14 − 2.95 =» 11.05\n \n [✔]\n \n \n
\n \n «[OH\n \n −\n \n ] = 10\n \n −11.05\n \n =» 8.91 × 10\n \n −12\n \n «moldm\n \n −3\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 \n \n Accept other methods.\n \n \n
\n\n Many students could correctly calculate the hydroxide concentration, but some weaker students calculated hydrogen ion concentration only.\n
\n\n Benzene, C\n \n 6\n \n H\n \n 6\n \n , can undergo electrophilic substitution reactions that proceeds slowly. Suggest why methylbenzene, C\n \n 7\n \n H\n \n 6\n \n , proceeds much faster.\n
\n\n [2]\n
\n\n methyl/alkyl «group» positive inductive effect ✓\n
\n\n
\n\n stabilises intermediate carbocation\n
\n\n \n \n OR\n \n \n
\n\n increased electron density in ring ✓\n
\n\n
\n\n \n Allow appropriate diagrams for M1\n \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 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 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 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 \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 100.0cm\n \n 3\n \n of soda water contains 3.0 × 10\n \n −2\n \n g NaHCO\n \n 3\n \n .\n \n
\n\n \n Calculate the concentration of NaHCO\n \n 3\n \n in mol dm\n \n −3\n \n .\n \n
\n\n [2]\n
\n\n \n «molar mass of NaHCO\n \n 3\n \n =» 84.01 «g mol\n \n -1\n \n »\n \n [✔]\n \n \n
\n\n \n «concentration =\n \n \n \n =» 3.6 × 10\n \n –3\n \n «mol dm\n \n -3\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 This is another stoichiometry question that most candidates were able to solve well, with occasional errors when calculating\n \n M\n \n \n r\n \n of hydrogen carbonate.\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 «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 Sketch an energy profile for the decomposition of calcium carbonate based on your answer to b(i), labelling the axes and activation energy,\n \n E\n \n \n a\n \n .\n
\n\n \n
\n [3]\n
\n\n endothermic sketch ✓\n
\n\n x-axis labelled “extent of reaction/progress of reaction/reaction coordinate/reaction pathway”\n \n \n AND\n \n \n y-axis labelled “potential energy/energy/enthalpy✓\n
\n\n activation energy/\n \n E\n \n \n a\n \n ✓\n
\n\n \n
\n \n
\n Do\n \n not\n \n accept “time” for x-axis.\n \n
\n Which species are acids in the equilibrium below?\n
\n\n CH\n \n 3\n \n NH\n \n 2\n \n + H\n \n 2\n \n O\n \n CH\n \n 3\n \n NH\n \n 3\n \n \n +\n \n + OH\n \n –\n \n
\n\n A. CH\n \n 3\n \n NH\n \n 2\n \n and H\n \n 2\n \n O\n
\n\n B. H\n \n 2\n \n O and CH\n \n 3\n \n NH\n \n 3\n \n \n +\n \n
\n\n C. H\n \n 2\n \n O and OH\n \n –\n \n
\n\n D. CH\n \n 3\n \n NH\n \n 2\n \n and CH\n \n 3\n \n NH\n \n 3\n \n \n +\n \n
\n\n [1]\n
\n\n B\n
\n\n What is the explanation for the malleability of metals?\n
\n\n
\n A. The bonds are strong.\n
\n B. The bonds are weak.\n
\n\n C. The bonds involve free electrons.\n
\n\n D. The bonds do not have a specific direction.\n
\n\n [1]\n
\n\n D\n
\n\n \n What is the IUPAC name of the following molecule?\n \n
\n\n \n \n \n
\n \n \n A. 2-bromo-3-ethylbutane\n \n \n
\n\n \n \n B. 3-methyl-4-bromopentane\n \n \n
\n\n \n \n C. 2-ethyl-3-bromobutane\n \n \n
\n\n \n \n D. 2-bromo-3-methylpentane\n \n \n
\n\n [1]\n
\n\n D\n
\n\n Another relatively high scoring question on IUPAC nomenclature.\n
\n\n Draw the Lewis (electron dot) structure of hydrogen sulfide.\n
\n\n [1]\n
\n\n \n \n \n OR\n
\n ✔\n \n \n
\n \n Accept any combination of lines, dots or crosses to represent electrons.\n \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 +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 Determine the value and unit of the rate constant using the rate expression in (a).\n
\n\n [2]\n
\n\n «\n \n k\n \n =\n \n =» 8.0 ✓\n
\n\n mol\n \n −3\n \n dm\n \n 9\n \n s\n \n −1\n \n ✓\n
\n\n Deduce the relationship between the concentration of N\n \n 2\n \n O\n \n 5\n \n and the rate of reaction.\n
\n\n [1]\n
\n\n « rate of reaction is directly» proportional to/∝[N\n \n 2\n \n O\n \n 5\n \n ]\n
\n \n \n OR\n \n \n
\n doubling concentration doubles rate ✔\n
\n \n
\n Do\n \n not\n \n accept “rate increases as concentration increases”/ positive correlation\n \n
\n \n Accept linear\n \n
\n\n \n When a bottle of carbonated water is opened, these equilibria are disturbed.\n \n
\n\n \n State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).\n \n
\n\n [1]\n
\n\n \n shifts to left/reactants\n \n \n AND\n \n \n to increase amount/number of moles/molecules of gas/CO\n \n 2\n \n (g)\n \n [✔]\n \n \n
\n\n This 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.\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 ✔\n
\n\n Almost all were able to calculate the energy released from the complete combustion of ethanol.\n
\n\n \n Estimate the time at which the powdered zinc was placed in the beaker.\n \n
\n\n [1]\n
\n\n \n 100 «s»\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept 90 to 100 s.\n \n \n
\n\n Almost all candidates identified 100 s as the time at which the reaction was initiated.\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 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 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 Suggest why this process might raise environmental concerns.\n
\n\n [1]\n
\n\n sulfur dioxide/SO\n \n 2\n \n causes acid rain ✔\n
\n\n \n Accept sulfur dioxide/SO\n \n 2\n \n /dust causes respiratory problems\n \n
\n \n Do\n \n not\n \n accept just “causes respiratory problems” or “causes acid rain”.\n \n
\n What is the hybridization of nitrogen and chlorine in NCl\n \n 3\n \n ?\n
\n\n \n
\n [1]\n
\n\n D\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 \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 At 298 K the concentration of aqueous carbon dioxide in carbonated water is 0.200 mol dm\n \n −3\n \n and the pK\n \n a\n \n for Equilibrium (2) is 6.36.\n \n
\n\n \n Calculate the pH of carbonated water.\n \n
\n\n [3]\n
\n\n \n «K\n \n a\n \n =» 10\n \n –6.36\n \n /4.37 × 10\n \n –7\n \n =\n \n \n \n
\n \n \n OR\n \n \n
\n «K\n \n a\n \n =» 10\n \n –6.36\n \n /4.37 × 10\n \n –7\n \n =\n \n \n \n \n [✔]\n \n \n
\n
\n\n \n [H\n \n +\n \n ] «\n \n \n \n » = 2.95 × 10\n \n –4\n \n «mol dm\n \n –3\n \n »\n \n \n \n [✔]\n \n
\n «pH =» 3.53\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 Most candidates calculated the pH of the aqueous CO\n \n 2\n \n . Some candidates attempted to use the Henderson-Hasselback equation and others used the quadratic expression to calculate [H\n \n +\n \n ] (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.\n
\n\n Explain the electron domain geometry of NO\n \n 3\n \n \n −\n \n .\n
\n\n [2]\n
\n\n three electron domains repel\n
\n\n \n \n OR\n \n \n
\n\n three electron domains as far away as possible ✔\n
\n\n
\n\n trigonal planar\n
\n\n \n \n OR\n \n \n
\n\n «all» angles are 120° ✔\n
\n\n The 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.\n
\n\n Outline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.\n
\n\n [1]\n
\n\n «valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔\n
\n\n \n
\n Accept 2,8 (for O\n \n 2–\n \n ) and 2,8,8 (for S\n \n 2–\n \n )\n \n
\n \n Identify the initiation step of the reaction and its conditions.\n \n
\n\n [2]\n
\n\n \n Br2 → 2Br•\n \n [✔]\n \n \n
\n\n \n «sun»light/UV/hv\n
\n \n \n OR\n \n \n
\n high temperature\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Do\n \n not\n \n penalize missing radical symbol on Br.\n \n \n
\n\n \n \n Accept “homolytic fission of bromine” for M1.\n \n \n
\n\n Drawing or describing the homolytic fission of bromine was generally done well.\n
\n\n \n Identify the most suitable indicator for the titration using section 22 of the data booklet.\n \n
\n\n [1]\n
\n\n \n phenolphthalein ✔\n \n
\n\n \n \n NOTE: Accept phenol red.\n \n \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 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\n \n The formula\n \n q = mcΔT\n \n was used to calculate the energy released. The values used in the calculation were\n \n m\n \n = 25.00 g,\n \n c\n \n = 4.18 J g\n \n −1\n \n K\n \n −1\n \n .\n \n
\n\n \n State an assumption made when using these values for\n \n m\n \n and\n \n c\n \n .\n \n
\n\n \n \n \n
\n [2]\n
\n\n \n
\n
\n\n \n \n Note:\n \n \n Accept “copper(II) sulfate/zinc sulfate” for “solution”.\n \n \n
\n\n Explaining the assumptions made when using values for m and c was challenging in. Many referred to the accuracy of the data when using m = 25.00g or said that no mass was lost during the reaction. Most knew that the value of c used was for water and suggested that the water was pure, but did not say that the specific heat of solution was assumed to be the same as that of water.\n
\n\n The overall reaction occurring at the electrodes of a rechargeable metal hydride battery can be summarized as:\n
\n\n MH\n \n NiO(OH)\n \n M\n \n Ni(OH)\n \n 2\n \n
\n\n
\n Which statement is correct?\n
\n
\n A. The oxidation state of Ni does not change.\n
\n B. M is oxidized by loss of hydrogen.\n
\n\n C. The oxidation state of one H atom changes from\n \n 1 to\n \n 1.\n
\n\n D. The oxidation state of one O atom changes from\n \n 1 to\n \n 2.\n
\n\n [1]\n
\n\n C\n
\n\n Deduce the expression for the equilibrium constant,\n \n K\n \n \n c\n \n , for this equation.\n
\n\n [1]\n
\n\n \n ✔\n
\n\n Deducing the equilibrium constant expression for the given equation was done very well.\n
\n\n Calculate the percentage of oxygen present in the double salt.\n
\n\n [1]\n
\n\n «100 − (7.09 + 5.11 + 16.22 + 14.91) =» 56.67 «%» ✓\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 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 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
\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 dilute the titrant to use larger volumes «of titrant» ✔\n
\n\n ensure spinach leaf fragments are the same size ✔\n
\n\n The following graph shows the concentration of C\n \n 4\n \n H\n \n 9\n \n Cl versus time.\n
\n\n \n
\n
\n What is the average rate of reaction over the first 800 seconds?\n
\n
\n A. 1 × 10\n \n −3\n \n mol dm\n \n −3\n \n s\n \n −1\n \n
\n B. 1 × 10\n \n −4\n \n mol dm\n \n −3\n \n s\n \n −1\n \n
\n\n C. 2 × 10\n \n −3\n \n mol dm\n \n −3\n \n s\n \n −1\n \n
\n\n D. 2 × 10\n \n −4\n \n mol dm\n \n −3\n \n s\n \n −1\n \n
\n\n [1]\n
\n\n B\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 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 \n Sodium percarbonate, 2Na\n \n 2\n \n CO\n \n 3\n \n •3H\n \n 2\n \n O\n \n 2\n \n , is an adduct of sodium carbonate and hydrogen peroxide and is used as a cleaning agent.\n \n
\n\n \n M\n \n r\n \n (2Na\n \n 2\n \n CO\n \n 3\n \n •3H\n \n 2\n \n O\n \n 2\n \n ) = 314.04\n \n
\n\n \n Calculate the percentage by mass of hydrogen peroxide in sodium percarbonate, giving your answer to two decimal places.\n \n
\n\n [2]\n
\n\n \n M( H\n \n 2\n \n O\n \n 2\n \n ) «= 2 × 1.01 + 2 × 16.00» = 34.02 «g»\n \n [✔]\n \n \n
\n\n \n «% H\n \n 2\n \n O\n \n 2\n \n = 3 ×\n \n \n \n × 100 =» 32.50 «%»\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 It is recommended that candidates use the relative atomic masses given in the periodic table.\n
\n\n Which combination best describes what is happening to chloromethane, CH\n \n 3\n \n Cl, in the equation below?\n
\n\n CH\n \n 3\n \n Cl (g) + H\n \n 2\n \n (g)\n \n CH\n \n 4\n \n (g) + HCl (g)\n
\n\n A. Oxidation and addition\n
\n\n B. Oxidation and substitution\n
\n\n C. Reduction and addition\n
\n\n D. Reduction and substitution\n
\n\n [1]\n
\n\n D\n
\n\n 57% of the candidates were able to identify the type of reaction (oxidation or reduction & addition or substitution) correctly.\n
\n\n In a redox titration, manganate(VII) ions are reduced to manganese(II) ions and iron(II) ions are oxidized to iron(III) ions.\n
\n\n MnO\n \n 4\n \n \n \n \n (aq) reduced to Mn\n \n 2+\n \n (aq)\n
\n Fe\n \n 2+\n \n (aq) oxidized to Fe\n \n 3+\n \n (aq)\n
\n
\n What volume, in cm\n \n 3\n \n , of 0.1 mol dm\n \n \n 3\n \n MnO\n \n 4\n \n \n \n \n (aq) is required to reach the equivalence point in the titration of 20.00 cm\n \n 3\n \n of 0.1 mol dm\n \n \n 3\n \n Fe\n \n 2+\n \n (aq)?\n
\n
\n A. 2.00\n
\n B. 4.00\n
\n\n C. 20.00\n
\n\n D. 100.00\n
\n\n [1]\n
\n\n B\n
\n\n What are the electron domain and molecular geometries of the XeF\n \n 4\n \n molecule?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n Higher scoring candidates had greater success in correctly identifying electron domain and molecular geometry of the XeF\n \n 4\n \n molecule.\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 ✔\n
\n\n Even rather weak candidates answered this one correctly.\n
\n\n Which structure shows the repeating unit of the polymer formed by but-1-ene?\n
\n\n \n
\n [1]\n
\n\n D\n
\n\n Why does the melting point of the elements decrease down group 1?\n
\n
\n
\n A. Atomic mass increases\n
\n\n B. Number of electrons increases\n
\n\n C. Radius of metal ion increases\n
\n\n D. First ionization energy decreases\n
\n\n [1]\n
\n\n C\n
\n\n The energy from burning 0.250 g of ethanol causes the temperature of 150 cm\n \n 3\n \n of water to rise by 10.5 °C. What is the enthalpy of combustion of ethanol, in kJ mol\n \n –1\n \n ?\n
\n\n Specific heat capacity of water: 4.18 J g\n \n –1\n \n K\n \n –1\n \n .\n
\n
\n
\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n
\n\n [1]\n
\n\n B\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 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 \n An additional experiment was conducted in which only the sulfuric acid catalyst was titrated with\n \n \n \n . Outline why this experiment was necessary.\n \n
\n\n [1]\n
\n\n to determine volume/moles of\n \n used up by the catalyst/sulfuric acid «in the titration»\n
\n \n \n OR\n \n \n
\n to eliminate/reduce «systematic» error caused by acid catalyst ✔\n
\n \n
\n Do not accept “control”\n \n OR\n \n “standard” alone.\n \n
\n Some students achieved on mark. Many answers referred simple to \"control\" or \"standard\" underlining the lack of some skills as also identified in the Internal assessment. Once again very few students had the specific details necessary to explain why this separate titration was needed in their response to receive a mark.\n
\n\n Outline why increasing the concentration of N\n \n 2\n \n O\n \n 5\n \n increases the rate of reaction.\n
\n\n [1]\n
\n\n greater frequency of collisions «as concentration increases»\n
\n \n \n OR\n \n \n
\n more collisions per unit time «as concentration increases» ✔\n
\n \n
\n Accept “rate/chance/probability/likelihood” instead of “frequency”.\n \n
\n \n Do\n \n not\n \n accept just “more collisions”.\n \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 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 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 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 Which of the following is most likely to be a transition metal?\n
\n\n
\n \n
\n [1]\n
\n\n C\n
\n\n Describe the original source of Taxol and the disadvantages of obtaining the medication from this source.\n
\n\n [2]\n
\n\n bark of yew tree ✓\n
\n\n kills tree\n
\n \n \n OR\n \n \n
\n tree grows slowly\n
\n \n \n OR\n \n \n
\n low yield ✓\n
\n
\n\n \n Accept “Taxus brevifolia” for “yew tree”.\n \n
\n\n \n Do\n \n not\n \n accept “yew tree” alone for M1.\n \n
\n\n Which aqueous solutions produce oxygen gas during electrolysis?\n
\n\n I. Dilute CuCl\n \n 2\n \n (aq) with inert electrodes\n
\n II. Dilute FeSO\n \n 4\n \n (aq) with inert electrodes\n
\n III. Dilute CuCl\n \n 2\n \n (aq) with copper electrodes\n
\n The standard electrode potentials are provided in the table:\n
\n\n \n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\n
\n\n Which is a renewable energy source?\n
\n\n
\n A. natural gas\n
\n B. uranium\n
\n\n C. coal\n
\n\n D. wood\n
\n\n [1]\n
\n\n D\n
\n\n Determine the empirical formula of the compound HX.\n
\n\n [2]\n
\n\n n\n \n C\n \n = «\n \n = » 3.33 «mol»\n
\n\n n\n \n H\n \n = «\n \n = » 6.66 «mol»\n
\n\n n\n \n O\n \n = «\n \n =» 3.33 «mol» ✔\n
\n\n CH\n \n 2\n \n O ✔\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 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 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 \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 exist in different conformational forms (isomers).\n \n
\n\n \n Outline what this means.\n \n
\n\n [1]\n
\n\n \n benzene ring «of the C\n \n 6\n \n H\n \n 5\n \n –CH\n \n 2\n \n » and the bromine «on the CH\n \n 2\n \n –Br» can take up different relative positions by rotating about the «C–C,\n \n σ\n \n –»bond\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “different parts of the molecule can rotate relative to each other”.\n
\n \n \n
\n \n \n Accept “rotation around σ\n \n –\n \n bond”.\n \n \n
\n\n If 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.\n
\n\n What is the slope of the graph?\n
\n\n \n
\n
\n A. −0.0025 mol dm\n \n −3\n \n s\n \n −1\n \n
\n B. −0.0025 mol dm\n \n −3\n \n s\n
\n\n C. −0.0033 mol dm\n \n −3\n \n s\n \n −1\n \n
\n\n D. −0.0033 mol dm\n \n −3\n \n s\n
\n\n [1]\n
\n\n A\n
\n\n Determine 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
\n\n (If you do not have answers for b(i) and b(ii), use Δ\n \n H\n \n = 190 kJ and Δ\n \n S\n \n = 180 J K\n \n −1\n \n , but these are not the correct answers.)\n
\n\n [2]\n
\n\n «spontaneous» if Δ\n \n G\n \n = Δ\n \n H\n \n −\n \n T\n \n Δ\n \n S\n \n < 0\n
\n \n \n OR\n \n \n
\n Δ\n \n H\n \n <\n \n T\n \n Δ\n \n S\n \n ✓\n
\n «\n \n T\n \n >\n \n =» 1112 «K» ✓\n
\n\n \n
\n Award\n \n [2]\n \n for correct final answer.\n \n
\n \n Accept “1056 K” if both of the incorrect values are used to solve the problem.\n \n
\n\n \n Do\n \n not\n \n award M2 for any negative T value.\n \n
\n\n What are the type of reaction and role of the nitronium ion, NO\n \n 2\n \n \n +\n \n , in the following reaction?\n
\n\n C\n \n 6\n \n H\n \n 6\n \n + NO\n \n 2\n \n \n +\n \n → C\n \n 6\n \n H\n \n 5\n \n NO\n \n 2\n \n + H\n \n +\n \n
\n\n \n
\n [1]\n
\n\n A\n
\n\n Good performance on a discriminating question in the type of reaction and role of the nitronium ion, NO\n \n 2\n \n \n +\n \n , in the nitration of benzene.\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 \n Two more trials (2 and 3) were carried out. The results are given below.\n \n
\n\n \n \n \n
\n \n \n Determine the rate equation for the reaction and its overall order, using your answer from (b)(i).\n \n \n
\n\n Rate equation:\n
\n\n Overall order:\n
\n\n [2]\n
\n\n \n \n Rate equation\n \n :\n
\n Rate =\n \n k\n \n [H\n \n 2\n \n O\n \n 2\n \n ] × [KI]\n \n [✔]\n \n
\n \n
\n \n \n Overall order\n \n :\n
\n 2\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Rate constant must be included.\n \n \n
\n\n Good performance but with answers that either typically included only [H\n \n 2\n \n O\n \n 2\n \n ] with first or second order equation or even suggesting zero order rate equation.\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 Sketch the neutralisation curve obtained\n \n and\n \n label the equivalence point.\n
\n\n \n
\n [3]\n
\n\n \n
\n \n \n OR\n \n \n
\n\n \n
\n starts at 1.6\n \n \n AND\n \n \n finishes at 13.6 ✔\n
\n\n approximately vertical at the correct volume of alkali added ✔\n
\n\n equivalence point labelled\n \n \n AND\n \n \n above pH 7 ✔\n
\n\n
\n\n \n Accept any range from 1.1-1.9\n \n AND\n \n 13.1-13.9 for\n \n M1\n \n or ECF from 11c(i) and 11c(ii).\n \n
\n\n \n Award\n \n M2\n \n for vertical climb at 28 cm\n \n 3\n \n \n OR\n \n 15 cm\n \n 3\n \n .\n \n
\n\n \n Equivalence point must be labelled for\n \n M3\n \n .\n \n
\n\n Which reaction mechanisms involve heterolytic fission of chlorine?\n
\n\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n D\n
\n\n Which term cannot characterize ammonia, NH\n \n 3\n \n ?\n
\n\n
\n A. Lewis acid\n
\n B. Brønsted–Lowry acid\n
\n\n C. ligand\n
\n\n D. nucleophile\n
\n\n [1]\n
\n\n A\n
\n\n In which case would repetition produce an improvement in accuracy?\n
\n\n
\n A. A thermometer always gives low readings.\n
\n B. An electronic balance has not been zeroed.\n
\n\n C. A student always reads the burette from a seated position.\n
\n\n D. Judgement of the end-point of a titration.\n
\n\n [1]\n
\n\n D\n
\n\n Explain why Si has a smaller atomic radius than Al.\n
\n\n [2]\n
\n\n nuclear charge/number of protons/Z/Z\n \n eff\n \n increases «causing a stronger pull on the outer electrons» ✓\n
\n\n same number of shells/«outer» energy level/shielding ✓\n
\n\n What is the percentage error if the enthalpy of combustion of a substance is determined experimentally to be −2100 kJ mol\n \n -1\n \n , but the literature value is −3500 kJ mol\n \n -1\n \n ?\n
\n\n A. 80 %\n
\n\n B. 60 %\n
\n\n C. 40 %\n
\n\n D. 20 %\n
\n\n [1]\n
\n\n C\n
\n\n \n Formulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.\n \n
\n\n [2]\n
\n\n \n 2C\n \n 6\n \n H\n \n 5\n \n COOH(s) + 15O\n \n 2\n \n (g) → 14CO\n \n 2\n \n (g) + 6H\n \n 2\n \n O(l)\n \n
\n\n \n correct products\n \n [✔]\n \n \n
\n\n \n correct balancing\n \n [✔]\n \n \n
\n\n Most students earned at least one mark for writing the correct products of the combustion of benzoic acid but the balancing appeared to be difficult for some.\n
\n\n Which ion is a better leaving group in nucleophilic substitutions?\n
\n\n
\n A. bromide ion\n
\n B. chloride ion\n
\n\n C. fluoride ion\n
\n\n D. iodide ion\n
\n\n [1]\n
\n\n D\n
\n\n Explain why the addition of small amounts of carbon to iron makes the metal harder.\n
\n\n [2]\n
\n\n disrupts the regular arrangement «of iron atoms/ions»\n
\n \n \n OR\n \n \n
\n carbon different size «to iron atoms/ions» ✔\n
\n prevents layers/atoms sliding over each other ✔\n
\n\n Distinguish between the hazards of high-level and low-level nuclear waste.\n
\n\n [2]\n
\n\n level has large amounts of «ionizing» radiation ✓\n
\n\n high-level has long half-lives\n
\n \n \n OR\n \n \n
\n high-level last longer/persists ✓\n
\n
\n\n \n Accept converse statements for low-level.\n \n
\n\n \n Accept “high radioactivity for high-level\" for M1.\n \n
\n\n \n Do\n \n not\n \n accept “high-level has ionizing radiation” alone for M1.\n \n
\n\n \n Do\n \n not\n \n accept answers based on storage or disposal differences alone.\n \n
\n\n \n Accept “high-level releases heat” for M2.\n \n
\n\n \n Do\n \n not\n \n accept “high-level has more penetrating radiation”\n
\n \n OR\n \n
\n “high-level has higher frequency radiation” for M1.\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 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 Which of these factors explains why NiCl\n \n 4\n \n \n 2−\n \n and CoCl\n \n 4\n \n \n 2−\n \n have different colours?\n
\n\n
\n A. Identity of the metal ion\n
\n B. Charge on the metal ion\n
\n\n C. Identity of the ligand in the complex\n
\n\n D. Spectrochemical series\n
\n\n [1]\n
\n\n A\n
\n\n Which products are formed from the neutralization of nitric acid by calcium hydroxide?\n
\n\n
\n A. Calcium oxide and ammonia\n
\n B. Calcium nitrate and water\n
\n\n C. Calcium nitrate and ammonia\n
\n\n D. Calcium nitrate and hydrogen\n
\n\n [1]\n
\n\n B\n
\n\n \n Bromine consists of two stable isotopes that exist in approximately a 1 : 1 ratio. The relative atomic mass,\n \n A\n \n \n r\n \n , of bromine is 79.90. Which are the stable isotopes of bromine?\n \n
\n\n \n A.\n \n 79\n \n Br and\n \n 81\n \n Br\n \n
\n\n \n B.\n \n 80\n \n Br and\n \n 81\n \n Br\n \n
\n\n \n C.\n \n 78\n \n Br and\n \n 80\n \n Br\n \n
\n\n \n D.\n \n 79\n \n Br and\n \n 80\n \n Br\n \n
\n\n [1]\n
\n\n A\n
\n\n Almost 56 % of candidates could find a 1:1 ratio as an average of 2 items, however many had\n \n 79\n \n Br and\n \n 80\n \n Br with an average mass of 79.90\n
\n\n What is the relative atomic mass of a sample of chlorine containing 70 % of the\n \n 35\n \n Cl isotope and 30 % of the\n \n 37\n \n Cl isotope?\n
\n
\n
\n A. 35.4\n
\n\n B. 35.5\n
\n\n C. 35.6\n
\n\n D. 35.7\n
\n\n [1]\n
\n\n C\n
\n\n 73% of the candidates were able to calculate the relative atomic mass of the sample of chlorine correctly (which was 35.6) given the masses and percentage abundances of the isotopes without the use of a calculator.\n
\n\n Which homologous series has the general formula C\n \n n\n \n H\n \n 2n\n \n O (n > 2)?\n
\n\n
\n A. Alcohols\n
\n B. Carboxylic acids\n
\n\n C. Ethers\n
\n\n D. Ketones\n
\n\n [1]\n
\n\n D\n
\n\n 49% of the candidates deduced that ketones had the general formula CnH2nO. The distractors were chosen with almost equal frequency. The question had good discrimination between high-scoring and low-scoring candidates.\n
\n\n \n State the nuclear symbol notation,\n \n \n \n , for iron-54.\n \n
\n\n [1]\n
\n\n \n \n \n \n \n \n [\n \n ✔\n \n ]\n \n
\n\n The nuclear symbol notation was generally correct. However, some students swapped atomic and mass numbers and hence lost the mark.\n
\n\n Which combination is correct regarding the anode and electron flow in an electrolytic cell?\n
\n\n
\n \n
\n [1]\n
\n\n A\n
\n\n 51% of the candidates deduced the polarity of the anode and the direction of the flow of electrons in an electrolytic cell correctly. The question had good discrimination between high-scoring and low-scoring candidates.\n
\n\n \n Draw the Lewis structures of oxygen, O\n \n 2\n \n , and ozone, O\n \n 3\n \n .\n \n
\n\n [2]\n
\n\n \n
\n \n NOTES: Coordinate bond may be represented by an arrow.\n \n
\n\n \n Do\n \n not\n \n accept delocalized structure for ozone.\n \n
\n\n Which compound produces the following\n \n 1\n \n H NMR spectrum?\n
\n\n \n
\n SDBS, National Institute of Advanced Industrial Science and Technology (AIST).\n
\n
\n A. Propane\n
\n B. Propanone\n
\n\n C. Propanal\n
\n\n D. 2,2-dimethylpropane\n
\n\n [1]\n
\n\n A\n
\n\n Which trend is correct, going down group 1?\n
\n\n A. Melting point increases\n
\n\n B. Reactivity decreases\n
\n\n C. First ionisation energy increases\n
\n\n D. Electronegativity decreases\n
\n\n [1]\n
\n\n D\n
\n\n The following diagram shows a chromatogram.\n
\n\n \n
\n
\n\n \n Identity of spot C:\n \n
\n leucine ✓\n
\n The block structure of the periodic table groups elements according to which characteristic?\n
\n\n
\n A. atomic number\n
\n B. atomic mass\n
\n\n C. electron configuration\n
\n\n D. reactivity\n
\n\n [1]\n
\n\n C\n
\n\n \n What is the value of the temperature change?\n \n
\n\n \n Initial temperature: 2.0 ± 0.1 °C\n
\n \n
\n \n Final temperature: 15.0 ± 1.0 °C\n \n
\n\n \n A. 13.0 ± 0.1 °C\n
\n \n
\n \n B. 13.0 ± 0.9 °C\n
\n \n
\n \n C. 13.0 ± 1.0 °C\n
\n \n
\n \n D. 13.0 ± 1.1 °C\n \n
\n\n [1]\n
\n\n D\n
\n\n Calculate the entropy change, Δ\n \n S\n \n , in J K\n \n −1\n \n mol\n \n −1\n \n , for this reaction.\n
\n\n \n
\n
\n\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/.\n \n
\n\n [1]\n
\n\n «ΔS = 364.5 J K\n \n –1\n \n mol\n \n –1\n \n – (311.7 J K\n \n –1\n \n mol\n \n –1\n \n + 223.0 J K\n \n –1\n \n mol\n \n –1\n \n )=» –170.2 «J K\n \n –1\n \n mol\n \n –1\n \n » ✔\n
\n\n The equation for the reaction between two gases, A and B, is:\n
\n\n \n
\n\n
\n When the reaction is at equilibrium at 600 K, the concentrations of A, B, and C are 2, 1, and 2 mol dm\n \n −3\n \n respectively. What is the value of the equilibrium constant at 600 K?\n
\n
\n A. 0.25\n
\n B. 1\n
\n\n C. 2\n
\n\n D. 4\n
\n\n [1]\n
\n\n C\n
\n\n \n Which combination will give you the enthalpy change for the hydrogenation of ethene to ethane,\n \n \n \n ?\n \n
\n\n \n
\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n A good majority of candidates could select the Hess' Law cycle for hydrogenation of ethene to ethane.\n
\n\n What happens to the mass of each copper electrode when aqueous copper(II) sulfate solution is electrolysed?\n
\n\n \n
\n \n
\n [1]\n
\n\n C\n
\n\n Chlorofluorocarbons (CFCs) contain bonds of the following lengths:\n
\n\n C—C = 1.54 × 10\n \n −10\n \n m\n
\n\n C—F = 1.38 × 10\n \n −10\n \n m\n
\n\n C—Cl = 1.77 × 10\n \n −10\n \n m\n
\n\n What is the order of\n \n increasing\n \n bond strength in the CFC molecule?\n
\n\n
\n\n A. C—C < C—F < C—Cl\n
\n\n B. C—C < C—Cl < C—F\n
\n\n C. C—Cl < C—C < C—F\n
\n\n D. C—F < C—C < C—Cl\n
\n\n [1]\n
\n\n C\n
\n\n 63% of the candidates were able to deduce the order of increasing bond strength from the bond lengths. The most commonly chosen distractor was D which had the reverse order. The question discriminated well between high-scoring and low-scoring candidates.\n
\n\n Deduce the structural and empirical formulas of\n \n B\n \n .\n
\n\n \n
\n [3]\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 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 State an equation for aerobic respiration.\n
\n\n [1]\n
\n\n C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n (aq) + 6O\n \n 2\n \n (g) → 6CO\n \n 2\n \n (g) + 6H\n \n 2\n \n O (l) ✓\n
\n\n \n Aspirin crystals are rinsed with water after recrystallization to remove impurities.\n
\n Suggest why\n \n \n \n cold\n \n \n \n water is used.\n \n
\n [1]\n
\n\n to avoid dissolving the crystals/aspirin ✔\n
\n\n \n Accept “to avoid loss of product”\n \n OR\n \n “aspirin is less soluble in cold water”.\n \n
\n\n Many were unable to explain why aspirin should be washed with cold water, namely, to avoid dissolving crystals. Surprisingly, the incorrect term \"melt\" was frequently used instead of \"dissolve\".\n
\n\n \n S\n \n \n tate the names of\n \n two\n \n functional groups present in all three molecules, using section 37 of the data booklet.\n \n
\n\n [2]\n
\n\n \n \n Any two of:\n \n
\n benzene/aromatic ring ✔\n
\n «tertiary» amino «group» ✔\n
\n ethenylene/1,2-ethenediyl «group» ✔\n
\n ether «group» ✔\n \n
\n \n \n NOTE: Accept “phenyl” for “benzene ring” although there are no phenyl groups as the benzene ring in this compound is a part of a polycyclic structure.\n
\n Do\n \n not\n \n accept “arene” or “benzene” alone.\n
\n Accept “amine” for “amino «group»”.\n
\n Accept “alkenyl/alkene/vinylene” for ethenylene/1,2-ethenediyl «group».\n \n \n
\n \n 1,4-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.\n \n
\n\n [2]\n
\n\n \n \n [\n \n \n ✔\n \n \n ]\n \n
\n \n HBr\n \n [\n \n ✔\n \n ]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept condensed formulae, such as CH\n \n 3\n \n C\n \n 6\n \n H\n \n 4\n \n CH\n \n 2\n \n Br.\n \n \n
\n\n Some candidates lost one mark for the bond originated from H in CH\n \n 3\n \n instead of C. Some teachers thought the use of the word “substituted alkane” made the question more difficult than it should have been.\n
\n\n Explain the general increase in trend in the first ionization energies of the period 3 elements, Na to Ar.\n
\n\n [2]\n
\n\n increasing number of protons\n
\n\n \n \n OR\n \n \n
\n\n increasing nuclear charge ✔\n
\n\n
\n\n «atomic» radius/size decreases\n
\n\n \n \n OR\n \n \n
\n\n same number of shells/electrons occupy same shell\n
\n\n \n \n OR\n \n \n
\n\n similar shielding «by inner electrons» ✔\n
\n\n Outline why vitamins usually need to be obtained from food sources.\n
\n\n [1]\n
\n\n cannot be synthesized «by the human body» ✓\n
\n\n The indicator, HInd, is used in an acid–base titration.\n
\n\n HInd (aq)\n \n H\n \n +\n \n (aq) + Ind− (aq)\n
\n\n colour A colour B\n
\n\n
\n Which statements are correct?\n
\n
\n I. In a strongly alkaline solution, colour B is observed.\n
\n II. Colour A is observed when [HInd] < [Ind\n \n −\n \n ].\n
\n III. [Ind\n \n −\n \n ] approximately equals [HInd] at the end point.\n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n B\n
\n\n \n Describe the effect of infrared (IR) radiation on carbon dioxide molecules.\n \n
\n\n [2]\n
\n\n \n bond length/C=O distance changes\n
\n \n \n OR\n \n \n
\n «asymmetric» stretching «of bonds»\n
\n \n \n OR\n \n \n
\n bond angle/OCO changes\n \n [✔]\n \n \n
\n \n polarity/dipole «moment» changes\n
\n \n \n OR\n \n \n
\n dipole «moment» created «when molecule absorbs IR»\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept appropriate diagrams.\n \n \n
\n\n This part was fairly well answered with most candidates receiving one of the two marks. There were many candidates who stated asymmetric stretching and bonds vibrate but missed writing polarity and dipole changes, which deprived them of the second mark.\n
\n\n Which elements are capable of forming expanded octets?\n
\n\n I. Nitrogen\n
\n II. Phosphorus\n
\n III. Arsenic\n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n C\n
\n\n Only about 60% of candidates were aware that nitrogen cannot expand its octet, unlike other elements in Group 15. There were quite a lot of teacher comments mainly regarding the inclusion of arsenic, but the critical factor did not require any knowledge of this.\n
\n\n What occurs during the operation of a voltaic cell based on the given reaction?\n
\n\n 2Cr (s) + 3Fe\n \n 2+\n \n (aq) → 2Cr\n \n 3+\n \n (aq) + 3Fe (s)\n
\n\n
\n \n
\n
\n\n [1]\n
\n\n B\n
\n\n 48% of the candidates deduced the direction of electron flow and the direction of ion flow correctly based on the equation of the overall reaction in the voltaic cell. The most commonly chosen distractor was A which had the correct direction of electron flow but reversed the ion flow.\n
\n\n Outline the meaning of homologous series.\n
\n\n [1]\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 Which compound rotates the plane of plane-polarized light?\n
\n\n A. CH\n \n 3\n \n C(CH\n \n 3\n \n )ClCH\n \n 3\n \n
\n\n B. CH\n \n 3\n \n CH\n \n 2\n \n CHClCH\n \n 3\n \n
\n\n C. CH\n \n 3\n \n C(Cl)\n \n 2\n \n CH\n \n 3\n \n
\n\n D. CH\n \n 3\n \n CClBrCH\n \n 3\n \n
\n\n [1]\n
\n\n B\n
\n\n \n Additional experiments were carried out at an elevated temperature. On the axes below, sketch Maxwell–Boltzmann energy distribution curves at two temperatures T\n \n 1\n \n and T\n \n 2\n \n , where T\n \n 2\n \n > T\n \n 1\n \n .\n \n
\n\n \n \n \n
\n [2]\n
\n\n \n
\n \n peak of T\n \n 2\n \n to right of\n \n \n AND\n \n \n lower than T\n \n 1\n \n \n [✔]\n \n \n
\n\n \n lines begin at origin\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 The drawing of the two curves at T1 and T2 was generally poorly done.\n
\n\n Calculate the Gibbs free energy change, Δ\n \n G\n \n \n ⦵\n \n , in kJ mol\n \n −1\n \n , for the reaction at 298 K. Use section 1 of the data booklet.\n
\n\n [1]\n
\n\n «Δ\n \n G\n \n \n ⦵\n \n = 53.0 kJ mol\n \n –1\n \n – (298K × 0.1665 kJ K\n \n –1\n \n mol\n \n –1\n \n ) =» 3.4 «kJ mol\n \n –1\n \n » ✔\n
\n\n \n Outline why ethanoic acid is classified as a weak acid.\n \n
\n\n [1]\n
\n\n \n partial dissociation «in aqueous solution»\n \n [✔]\n \n \n
\n\n The definition of a weak acid was generally correct.\n
\n\n \n Dinitrogen monoxide has a positive standard enthalpy of formation, Δ\n \n H\n \n \n f\n \n \n \n θ\n \n \n .\n \n
\n\n \n Deduce, giving reasons, whether altering the temperature would change the\n \n \n spontaneity of the\n \n decomposition\n \n reaction.\n \n
\n\n [3]\n
\n\n \n exothermic decomposition\n
\n \n \n OR\n \n \n
\n Δ\n \n H\n \n \n (decomposition)\n \n < 0\n \n [✔]\n \n \n
\n \n \n TΔS\n \n \n θ\n \n > Δ\n \n H\n \n \n \n \n θ\n
\n \n \n \n \n \n \n \n OR\n
\n \n \n \n \n Δ\n \n G\n \n \n \n \n θ\n \n \n \n «= Δ\n \n H\n \n \n \n \n θ\n \n \n \n –\n \n TΔS\n \n \n \n \n θ\n \n \n \n » < 0 «at all temperatures»\n \n [\n \n \n ✔]\n \n \n
\n \n reaction spontaneous at all temperatures\n \n [\n \n \n ✔]\n \n \n
\n\n Though the question asked for decomposition (in bold), most candidates ignored this and worked on the basis of a the Δ\n \n H\n \n 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 Δ\n \n H\n \n (probably taking it for granted) nor explicitly relate Δ\n \n G\n \n and spontaneity, which left the examiner with no possibility of evaluating their reasoning.\n
\n\n Deduce the structural and empirical formulas of\n \n B\n \n .\n
\n\n \n
\n [3]\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 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 What is the formal charge of the oxygen atom in H\n \n 3\n \n O\n \n +\n \n ?\n
\n\n A. −2\n
\n\n B. −1\n
\n\n C. 0\n
\n\n D. +1\n
\n\n [1]\n
\n\n D\n
\n\n 44% of candidates were able to determine the formal charge of the oxygen atom in H\n \n 3\n \n O\n \n +\n \n . The most commonly chosen distractor was -2 (the oxidation state of the oxygen).\n
\n\n Which ions are present in an aqueous solution of Na\n \n 2\n \n CO\n \n 3\n \n ?\n
\n\n I. HCO\n \n 3\n \n \n −\n \n
\n II. OH\n \n −\n \n
\n III. CO\n \n 3\n \n \n 2−\n \n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n D\n
\n\n \n Where does oxidation occur in a voltaic cell?\n \n
\n\n \n A. positive electrode and anode\n \n
\n\n \n B. negative electrode and anode\n \n
\n\n \n C. positive electrode and cathode\n \n
\n\n \n D. negative electrode and cathode\n \n
\n\n [1]\n
\n\n B\n
\n\n One G2 form thought the question was awkwardly worded, implying oxidation is occurring in two places rather than one. This was one of the better answered question with a relatively low discriminatory index.\n
\n\n \n Which arrow shows the activation energy of the uncatalysed forward reaction for this equilibrium?\n \n
\n\n \n \n
\n\n \n
\n [1]\n
\n\n C\n
\n\n Identifying the activation energy of an uncatalyzed reaction from an energy profile diagram was the best answered question in the exam.\n
\n\n Describe the bonding in iron, Fe (s).\n
\n\n [1]\n
\n\n electrostatic attraction/hold between «lattice of» positive ions/cations\n \n \n AND\n \n \n delocalized «valence» electrons ✔\n
\n\n Most 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 FeS\n \n 2\n \n (not reading the question well). Overall, only 30% answered satisfactorily.\n
\n\n State the equilibrium constant expression,\n \n K\n \n \n c\n \n , for this reaction.\n
\n\n [1]\n
\n\n «\n \n K\n \n \n c\n \n =»\n \n ✔\n
\n\n \n Explain the mechanism for the nitration of benzene, using curly arrows to indicate the movement of electron pairs.\n \n
\n\n [4]\n
\n\n \n
\n \n curly arrow going from benzene ring to N «of\n \n +\n \n NO\n \n 2\n \n /NO\n \n 2\n \n \n +\n \n »\n \n [\n \n ✔\n \n ]\n \n
\n carbocation with correct formula and positive charge on ring\n \n [\n \n ✔\n \n ]\n \n
\n curly arrow going from C–H bond to benzene ring of cation\n \n [\n \n ✔\n \n ]\n \n
\n formation of organic product nitrobenzene\n \n \n AND\n \n \n H\n \n +\n \n \n [\n \n ✔\n \n ]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept mechanism with corresponding Kekulé structures.\n
\n Do\n \n not\n \n accept a circle in M2 or M3.\n
\n Accept first arrow starting either inside the circle or on the circle.\n
\n If Kekulé structure used, first arrow must start on the double bond.\n
\n M2 may be awarded from correct diagram for M3.\n
\n M4: Accept “C\n \n 6\n \n H\n \n 5\n \n NO\n \n 2\n \n + H\n \n 2\n \n SO\n \n 4\n \n ” if HSO\n \n 4\n \n \n −\n \n used in M3.\n \n \n
\n Performance 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.\n
\n\n \n Which property shows a general increase from left to right across period 2, Li to F?\n \n
\n\n \n A. Melting point\n
\n \n
\n \n B. Electronegativity\n
\n \n
\n \n C. Ionic radius\n
\n \n
\n \n D. Electrical conductivity\n \n
\n\n [1]\n
\n\n B\n
\n\n How many oxygen atoms are present in 0.0500 mol Ba(OH)\n \n 2\n \n •8H\n \n 2\n \n O?\n
\n\n \n N\n \n \n A\n \n = 6.02 × 10\n \n 23\n \n
\n\n
\n\n A. 3.01 × 10\n \n 23\n \n
\n\n B. 6.02 × 10\n \n 23\n \n
\n\n C. 3.01 × 10\n \n 24\n \n
\n\n D. 6.02 × 10\n \n 24\n \n
\n\n [1]\n
\n\n A\n
\n\n 57% of the candidates calculated the number of oxygen atoms correctly. The most commonly chosen distractor was C which would be achieved if a power of ten error was made.\n
\n\n Which set of conditions describe a reaction in which the reactants are more stable than the products?\n
\n\n
\n A. endothermic and\n \n \n H\n \n negative\n
\n B. endothermic and\n \n \n H\n \n positive\n
\n\n C. exothermic and\n \n \n H\n \n negative\n
\n\n D. exothermic and\n \n \n H\n \n positive\n
\n\n [1]\n
\n\n B\n
\n\n Which electron transition in the hydrogen atom emits radiation with the highest energy?\n
\n\n
\n A. n = 1 to n = 2\n
\n B. n = 2 to n = 3\n
\n\n C. n = 2 to n = 1\n
\n\n D. n = 3 to n = 2\n
\n\n [1]\n
\n\n C\n
\n\n Which are the most reactive elements of the alkali metals and halogens?\n
\n\n A. Lithium and fluorine\n
\n\n B. Lithium and iodine\n
\n\n C. Caesium and fluorine\n
\n\n D. Caesium and iodine\n
\n\n [1]\n
\n\n C\n
\n\n 70% of candidates selected the most reactive alkali metal and halogen. The most commonly chosen distractor was lithium and fluorine.\n
\n\n Deduce the change in the oxidation state of sulfur.\n
\n\n [1]\n
\n\n +6\n
\n \n \n OR\n \n \n
\n −2 to +4 ✔\n
\n \n Accept “6/VI”.\n \n
\n \n Accept “−II, 4//IV”.\n \n
\n Do\n \n not\n \n accept 2- to 4+.\n
\n In which molecule does the central atom have an incomplete octet of electrons?\n
\n\n A. H\n \n 2\n \n Se\n
\n\n B. PH\n \n 3\n \n
\n\n C. OF\n \n 2\n \n
\n\n D. BF\n \n 3\n \n
\n\n [1]\n
\n\n D\n
\n\n \n A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.\n \n
\n\n \n Cl2 (g) + 2NaOH (aq)\n \n \n \n NaOCl (aq) + NaCl (aq) + H\n \n 2\n \n O (l)\n \n
\n\n \n Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.\n \n
\n\n [3]\n
\n\n \n ethanoic acid/vinegar reacts with NaOH\n \n [✔]\n \n \n
\n\n \n moves equilibrium to left/reactant side\n \n [✔]\n \n \n
\n\n \n releases Cl\n \n 2\n \n (g)/chlorine gas\n
\n \n \n OR\n \n \n
\n Cl\n \n 2\n \n (g)/chlorine\n \n gas\n \n is toxic\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “ethanoic acid produces H\n \n +\n \n ions”.\n \n \n
\n\n \n \n Accept “ethanoic acid/vinegar reacts with NaOCl”.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “2CH\n \n 3\n \n COOH + NaOCl + NaCl → 2CH\n \n 3\n \n COONa + Cl\n \n 2\n \n + H\n \n 2\n \n O” as it\n \n \n \n \n does not refer to equilibrium.\n
\n \n \n
\n \n \n Accept suitable molecular or ionic equations for M1 and M3.\n \n \n
\n\n Explaining 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.\n
\n\n State why NH\n \n 3\n \n is a Lewis base.\n
\n\n [1]\n
\n\n donates «lone/non-bonding» pair of electrons ✔\n
\n\n The 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.\n
\n\n Outline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.\n
\n\n [1]\n
\n\n «valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔\n
\n\n \n
\n Accept 2,8 (for O\n \n 2–\n \n ) and 2,8,8 (for S\n \n 2–\n \n )\n \n
\n The nitride ion and the magnesium ion are isoelectronic (they have the same electron configuration). Determine, giving a reason, which has the greater ionic radius.\n
\n\n [1]\n
\n\n nitride\n \n \n AND\n \n \n smaller nuclear charge/number of protons/atomic number ✔\n
\n\n In 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.\n
\n\n State an equation for aerobic respiration.\n
\n\n [1]\n
\n\n C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n (aq) + 6O\n \n 2\n \n (g) → 6CO\n \n 2\n \n (g) + 6H\n \n 2\n \n O (l) ✓\n
\n\n Which species has the weakest conjugate base?\n
\n\n
\n A. HCl\n
\n B. NH\n \n 4\n \n \n +\n \n
\n\n C. HCO\n \n 3\n \n \n −\n \n
\n\n D. H\n \n 2\n \n O\n
\n\n [1]\n
\n\n A\n
\n\n 48% of the candidates chose HCl as the species with the weakest conjugate base. This question had good discrimination between high-scoring and low-scoring candidates.\n
\n\n Outline the reasons that sodium hydroxide is considered a Brønsted–Lowry and Lewis base.\n
\n\n \n
\n [1]\n
\n\n \n Brønsted–Lowry base:\n \n
\n proton acceptor\n
\n \n \n AND\n \n \n
\n\n \n Lewis Base:\n \n
\n e\n \n –\n \n pair donor/nucleophile ✔\n
\n State the number of\n \n (sigma) and\n \n (pi) bonds in Compound A.\n
\n\n \n
\n [1]\n
\n\n \n : 9\n
\n \n \n AND\n \n \n
\n \n : 1 ✔\n
\n Mediocre 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.\n
\n\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 «\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 State, giving a reason, whether the reaction is spontaneous or not at 298 K.\n
\n\n [1]\n
\n\n spontaneous\n \n \n AND\n \n \n Δ\n \n G\n \n < 0 ✔\n
\n\n Reason for the reaction being spontaneous was generally very done well indeed.\n
\n\n Draw and label an enthalpy level diagram for this reaction.\n
\n\n \n
\n [2]\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 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 \n Comment on the spontaneity of the reaction at 298 K.\n \n
\n\n [1]\n
\n\n \n non-spontaneous\n \n \n AND\n \n \n \n \n \n positive ✔\n \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 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 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 Incorrect\n \n \n AND\n \n \n two readings, uncertainty is ±0.1 ✔\n
\n\n Outline why we need vitamins/micronutrients in our diets.\n
\n\n [1]\n
\n\n «mostly» not synthesized by body «and needed for proper growth/metabolism» ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “needed for proper growth/metabolism” alone.\n \n
\n\n Explain\n \n two\n \n different ways antiviral medications work.\n
\n\n [2]\n
\n\n Any two of:\n
\n\n alters «viral» enzyme\n \n \n AND\n \n \n prevent virus from entering the cell ✓.\n
\n\n alter the cell DNA\n \n \n AND\n \n \n virus cannot multiply ✓\n
\n\n block «cell» enzyme activity\n \n \n AND\n \n \n prevent virus multiplication ✓\n
\n\n alters «viral» enzyme\n \n \n AND\n \n \n prevents release of «new» viral particles «from the cell» ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “just interferes with viral reproductive cycle”.\n \n
\n\n \n Award\n \n [1 max]\n \n for two partial answers.\n \n
\n\n \n Which equation shows the enthalpy of formation,\n \n \n \n , of ethanol?\n \n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n C\n
\n\n Less than half of the candidates selected the correct equation for the enthalpy of formation of ethanol. The most common wrong answer involved ethanol formed as a gas.\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 \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 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 \n Which is correct for the reaction mechanism shown?\n \n
\n\n \n \n \n
\n [1]\n
\n\n D\n
\n\n Another well answered question on rate mechanisms.\n
\n\n Identify the isomer of Compound B that exists as optical isomers (enantiomers).\n
\n\n [1]\n
\n\n butan-2-ol/CH\n \n 3\n \n CH(OH)C\n \n 2\n \n H\n \n 5\n \n ✔\n
\n\n Mediocre performance; some identified 2-methylpropan-1-ol or -2-ol, instead butan-2-ol/CH\n \n 3\n \n CH(OH)C\n \n 2\n \n H\n \n 5\n \n as the isomer that exists as an optical isomer.\n
\n\n The diagram shows the first ionisation energies of consecutive elements in the same period of the periodic table.\n
\n\n \n
\n Which factor explains why element X has a higher first ionisation energy than element Y?\n
\n\n A. Element Y loses an electron from a different sub-level.\n
\n\n B. Element Y has a smaller atomic radius.\n
\n\n C. Element X has a full octet.\n
\n\n D. Element Y has a greater nuclear charge.\n
\n\n [1]\n
\n\n A\n
\n\n Which of the following describes the role of benzene in the reaction?\n
\n\n \n
\n
\n A. an electrophile\n
\n B. a nucleophile\n
\n\n C. a reducing agent\n
\n\n D. an acid\n
\n\n [1]\n
\n\n A\n
\n\n The graph shows the Maxwell–Boltzmann energy distribution curve for a given gas at a certain temperature.\n
\n\n \n
\n How would the curve change if the temperature of the gas decreases while the other conditions remain constant?\n
\n\n A. The maximum would be lower and to the left of M.\n
\n\n B. The maximum would be lower and to the right of M.\n
\n\n C. The maximum would be higher and to the left of M.\n
\n\n D. The maximum would be higher and to the right of M.\n
\n\n [1]\n
\n\n C\n
\n\n What do all greenhouse gases have in common?\n
\n\n
\n A. They are emitted by the burning of fossil fuels.\n
\n B. They absorb ultraviolet radiation.\n
\n\n C. They are symmetrical molecules with no polar bonds.\n
\n\n D. They absorb infrared radiation.\n
\n\n [1]\n
\n\n D\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 \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 Data is given for the reaction 2X\n \n 2\n \n (g) + Y\n \n 2\n \n (g) → 2X\n \n 2\n \n Y (g).\n
\n
\n
\n \n
\n What rate equation can be inferred from the data?\n
\n
\n
\n A. Rate =\n \n k\n \n [X\n \n 2\n \n ] [Y\n \n 2\n \n ]\n
\n\n B. Rate =\n \n k\n \n [X\n \n 2\n \n ]\n \n 2\n \n [Y\n \n 2\n \n ]\n
\n\n C. Rate =\n \n k\n \n [X\n \n 2\n \n ]\n \n 2\n \n [Y\n \n 2\n \n ]\n \n 0\n \n
\n\n D. Rate =\n \n k\n \n [X\n \n 2\n \n ]\n \n 2\n \n [Y\n \n 2\n \n ]\n \n 2\n \n
\n\n [1]\n
\n\n C\n
\n\n Well over 80% of candidates were able to deduce the rate equation from the reaction rate data provided. The most common distractor (B) indicates that a few students confused zero and first order dependence for Y\n \n 2\n \n .\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 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 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 State the equilibrium constant expression,\n \n K\n \n \n c\n \n , for this reaction.\n
\n\n [1]\n
\n\n «\n \n K\n \n \n c\n \n =»\n \n ✔\n
\n\n \n Suggest how benzoic acid,\n \n M\n \n r\n \n \n = 122.13, forms an apparent dimer,\n \n M\n \n r\n \n \n = 244.26, when dissolved in a non-polar solvent such as hexane.\n \n
\n\n [1]\n
\n\n \n «intermolecular» hydrogen bonding\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept diagram showing hydrogen bonding.\n \n \n
\n\n Very few students answered this question correctly, thinking benzoic would bond with the hexane even though it was a non-polar solvent. It was very rare for a student to realize there was intermolecular hydrogen bonding.\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 «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 \n A sample of ethanoic acid was titrated with sodium hydroxide solution, and the following pH curve obtained.\n \n
\n\n \n \n \n
\n \n \n Annotate the graph to show the buffer region and the volume of sodium hydroxide at the equivalence point.\n \n \n
\n\n [2]\n
\n\n \n
\n \n buffer region on graph ✔\n
\n equivalence point/V\n \n eq\n \n on graph ✔\n \n
\n \n \n NOTE: Construction lines not required.\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 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 State the class of compound to which ethene belongs.\n
\n\n [1]\n
\n\n alkene ✔\n
\n\n \n Which region of the electromagnetic spectrum is used to identify hydrogen environments in a molecule?\n \n
\n\n \n
\n \n A. X-ray\n
\n \n
\n \n B. UV\n
\n \n
\n \n C. IR\n
\n \n
\n \n D. radio waves\n \n
\n\n [1]\n
\n\n D\n
\n\n Not even a quarter of candidates knew the region of the electromagnetic spectrum associated with identifying hydrogen environments. This was the lowest scoring question on the exam with the most popular answer\n
\n being IR region.\n
\n Which graph shows a first order reaction?\n
\n\n \n
\n [1]\n
\n\n A\n
\n\n What is the uncertainty, in cm\n \n 3\n \n , of this measurement?\n
\n \n
\n A. ±0.01\n
\n B. ±0.1\n
\n\n C. ±0.15\n
\n\n D. ±1\n
\n\n [1]\n
\n\n C\n
\n\n This question generated a lot of debate among teachers and examiners. The answer B is the best answer but it is a rather low value. It was selected by 34% of the candidates. Many teachers argued that the correct answer should be 0.5 cm\n \n 3\n \n .\n
\n\n \n Suggest one source of error in the experiment, excluding faulty apparatus and human error, that would lead to the following:\n \n
\n\n \n
\n [2]\n
\n\n \n Experimental mass greater than actual mass of oil in crisps:\n \n
\n other substances «in the crisps» are soluble in the solvent\n
\n \n \n OR\n \n \n
\n not all the solvent evaporates ✔\n
\n \n Experimental mass less than actual mass of oil in crisps:\n \n
\n not all oil dissolved/extracted ✔\n
\n \n Accept “oil evaporated”\n \n OR\n \n “oil burned/decomposed”\n \n OR\n \n “oil absorbed by the filter”\n \n OR\n \n “assumption «all oil dissolved» was wrong” for M2.\n \n
\n\n \n Do\n \n not\n \n accept examples of human errors\n \n OR\n \n faulty apparatus.\n \n
\n\n Even 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 Outline the meaning of the term\n \n xenobiotic\n \n .\n
\n\n [1]\n
\n\n chemicals found in an organism that are not normally present ✓\n
\n\n Deduce the splitting pattern in the\n \n 1\n \n H NMR spectrum for 1-bromopropane.\n
\n\n [1]\n
\n\n triplet/3\n \n \n AND\n \n \n multiplet/6\n \n \n AND\n \n \n triplet/3 ✔\n
\n\n Another 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.\n
\n\n What is the product of the reaction of but-2-ene with bromine?\n
\n\n A. 1,2-dibromobutane\n
\n\n B. 2,2-dibromobutane\n
\n\n C. 2,3-dibromobutane\n
\n\n D. 3,3-dibromobutane\n
\n\n [1]\n
\n\n C\n
\n\n What is the organic product of the reaction of 1-chloropentane with aqueous sodium hydroxide?\n
\n\n
\n A. pentan-1-ol\n
\n B. 1-chloropentan-1-ol\n
\n\n C. 1-chloropent-1-ene\n
\n\n D. 1-chloropent-2-ene\n
\n\n [1]\n
\n\n A\n
\n\n What is the molecular geometry of SF\n \n 4\n \n ?\n
\n\n A. Tetrahedral\n
\n\n B. Trigonal bipyramidal\n
\n\n C. See-saw\n
\n\n D. Square planar\n
\n\n [1]\n
\n\n C\n
\n\n 54% of the candidates deduced the molecular geometry of SF\n \n 4\n \n . The distractors were chosen with equal frequency by the rest of the candidates. This question discriminated well between high-scoring and low-scoring candidates.\n
\n\n Which structure of CF\n \n 2\n \n Cl\n \n 2\n \n is shown with correct bond and molecular dipoles?\n
\n\n
\n \n
\n [1]\n
\n\n D\n
\n\n 41% of candidates were able to deduce the bond polarity and molecular polarity correctly. The most commonly chosen distractor was A which had the correct bond polarities but did not include an overall dipole.\n
\n\n Sodium emits yellow light with a frequency of 5.09 × 10\n \n 14\n \n Hz when electrons transition from 3p to 3s orbitals.\n
\n\n Calculate the energy difference, in J, between these two orbitals using sections 1 and 2 of the data booklet.\n
\n\n
\n\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].\n \n
\n\n [1]\n
\n\n «ΔE = hν = 6.63 × 10\n \n –34\n \n J s × 5.09 × 10\n \n 14\n \n s\n \n –1\n \n =» 3.37 × 10\n \n –19\n \n «J» ✔\n
\n\n \n Which series shows the correct order of metallic bond strength from strongest to weakest?\n \n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n D\n
\n\n A small majority of candidates could identify the relative strength of metallic bonding amongst group 1 and 2 metals.\n
\n\n \n Outline why ozone in the stratosphere is important.\n \n
\n\n [1]\n
\n\n \n absorbs\n \n UV/ultraviolet\n \n light «of longer wavelength than absorbed by O\n \n 2\n \n »\n \n [✔]\n \n \n
\n\n 60 % 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.\n
\n\n \n What is represented by A in\n \n \n \n ?\n \n
\n\n \n A. Number of electrons\n
\n \n
\n \n B. Number of neutrons\n
\n \n
\n \n C. Number of nucleons\n
\n \n
\n \n D. Number of protons\n \n
\n\n [1]\n
\n\n C\n
\n\n Calculate the amount of magnesium, in mol, that was used.\n
\n\n [1]\n
\n\n \n «mol» ✔\n
\n\n A number of students were not able to interpret the results and hence find the gain in mass and calculate the moles correctly.\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 sulfuric acid/H\n \n 2\n \n SO\n \n 4\n \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\n A very well answered question. 70% of the candidates stated H2SO4 as the product from the reaction of SO\n \n 3\n \n with water.\n
\n\n Explain how strong analgesics like morphine work.\n
\n\n [2]\n
\n\n «temporarily» binding to receptors in the nervous system/spinal cord/brain ✓\n
\n\n preventing transmission of pain impulses ✓\n
\n\n
\n\n \n Accept “bonding” for “binding” in M1.\n \n
\n\n \n Accept “without depressing the central nervous system” for M2.\n \n
\n\n Annotate the diagram to show the charge separation at the p-n junction and flow of electrons through the external circuit.\n
\n\n [2]\n
\n\n negative above junction\n \n \n AND\n \n \n positive below junction ✓\n
\n\n e- flow from n-type through wire to light globe\n
\n \n \n OR\n \n \n
\n from light globe to p-type through wire.✓\n
\n \n The graph shows the Maxwell–Boltzmann energy distribution curve for a given gas at a certain temperature.\n \n
\n\n \n \n \n
\n \n \n How will the curve change if the temperature of the gas is increased, while other conditions remain constant?\n \n \n
\n\n \n \n A. The maximum is higher and to the left of A.\n
\n \n \n
\n \n \n B. The maximum is higher and to the right of A.\n
\n \n \n
\n \n \n C. The maximum is lower and to the right of A.\n
\n \n \n
\n \n \n D. The maximum is lower and to the left of A.\n \n \n
\n\n [1]\n
\n\n C\n
\n\n \n Draw the repeating unit of polyphenylethene.\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 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 \n Classify PVC and polyethene terephthalate, PET, as addition or condensation polymers and deduce the structural formulas.\n \n
\n\n \n \n \n
\n [3]\n
\n\n \n
\n \n PVC\n \n : addition\n \n \n AND\n \n PET\n \n : condensation ✔\n
\n\n structure of PVC monomer ✔\n
\n\n structure of PET monomers ✔\n
\n\n
\n \n Accept full\n \n OR\n \n condensed structural formulas.\n \n
\n This question on PVC and PET was very well answered. All three marks for their correct classification and the corresponding structures of their monomers was frequently scored.\n
\n\n Which property of elements increases down a group but decreases across a period?\n
\n\n
\n A. Atomic radius\n
\n B. Electronegativity\n
\n\n C. Ionic radius\n
\n\n D. Ionization energy\n
\n\n [1]\n
\n\n A\n
\n\n A very well answered question. 70% of the candidates identified atomic radius as the property that increases down a group and decreases across a period.\n
\n\n \n Outline the operation of a polarimeter used to distinguish between enantiomers.\n \n
\n\n [2]\n
\n\n \n \n Any two of:\n \n
\n «plane-» polarized light\n
\n \n \n OR\n \n \n
\n light passes through polarizer/polarizing filter\n \n [✔]\n \n \n
\n \n enantiomers rotate plane of «plane-» polarized light «by equal angles» in opposite directions\n \n [✔]\n \n \n
\n\n \n measure angle/direction of rotation\n \n [✔]\n \n \n
\n\n The operation of a polarimeter to distinguish between enantiomers was generally well handled by the candidates while some missed stating to measure the angle/direction of rotation.\n
\n\n \n Which will increase the rate of reaction between calcium carbonate and hydrochloric acid?\n \n
\n\n \n I. an increase in temperature\n
\n II. an increase in concentration of hydrochloric acid\n
\n III. an increase in particle size of calcium carbonate\n \n
\n \n A. I and II only\n \n
\n\n \n B. I and III only\n \n
\n\n \n C. II and III only\n \n
\n\n \n D. I, II and III\n \n
\n\n [1]\n
\n\n A\n
\n\n One teacher pointed out that the question should have stated for a fixed mass of calcium carbonate. However, this question was one of the highest scoring so it did not cause confusion.\n
\n\n Sketch the Lewis (electron dot) structure of the P\n \n 4\n \n molecule, containing only single bonds.\n
\n\n
\n\n [1]\n
\n\n \n
\n \n
\n Accept any diagram with each P joined to the other three.\n
\n \n
\n \n Accept any combination of dots, crosses and lines.\n \n
\n\n What is the name of this compound, applying IUPAC rules?\n
\n\n \n
\n A. 4-methylhex-2-ene\n
\n\n B. 4-ethylpent-2-ene\n
\n\n C. 2-ethylpent-3-ene\n
\n\n D. 3-methylhex-4-ene\n
\n\n [1]\n
\n\n A\n
\n\n \n Suggest a risk of using sulfuric acid as the catalyst.\n \n
\n\n [1]\n
\n\n corrosive/burns/irritant/strong oxidizing agent/carcinogenic\n
\n \n \n OR\n \n \n
\n disposal is an environmental issue\n
\n \n \n OR\n \n \n
\n causes other side reactions/dehydration/decomposition ✔\n
\n
\n \n Do\n \n not\n \n accept just “risk of accidents”\n \n OR\n \n “health risks”\n \n OR\n \n “hazardous”.\n \n
\n Most students received a mark for this question base on specific hazards. Very few students related disposal to environmental issues which isn't surprising as this is often missed in the Internal Assessment. Weaker students provided vague answers related to health issues which did not receive a mark. Some students misunderstood the question.\n
\n\n \n What is the order, in increasing pH, of the following solutions of equal concentration?\n \n
\n\n \n \n \n
\n \n \n A. H\n \n 3\n \n BO\n \n 3\n \n < CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n COOH < CH\n \n 3\n \n CH\n \n 2\n \n COOH < CHCl\n \n 2\n \n COOH\n
\n \n \n
\n \n \n B. H\n \n 3\n \n BO\n \n 3\n \n < CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n COOH < CHCl\n \n 2\n \n COOH < CH\n \n 3\n \n CH\n \n 2\n \n COOH\n
\n \n \n
\n \n \n C. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n COOH < CH\n \n 3\n \n CH\n \n 2\n \n COOH < CHCl\n \n 2\n \n COOH < H\n \n 3\n \n BO\n \n 3\n \n
\n \n \n
\n \n \n D. CHCl\n \n 2\n \n COOH < CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n COOH < CH\n \n 3\n \n CH\n \n 2\n \n COOH < H\n \n 3\n \n BO\n \n 3\n \n \n \n
\n\n [1]\n
\n\n D\n
\n\n Calculate the value of the rate constant,\n \n k\n \n , giving its units.\n
\n\n [3]\n
\n\n \n k\n \n =\n \n ✔\n
\n\n «\n \n k\n \n =\n \n = » 0.030 «min\n \n –1\n \n » ✔\n
\n\n min\n \n –1\n \n ✔\n
\n\n \n
\n M1 can be awarded from correct M2 if not explicitly stated.\n \n
\n \n Accept k = gradient.\n \n
\n\n \n Accept values in the range 0.028–0.032.\n \n
\n\n \n Award\n \n [3]\n \n for correct final answer.\n \n
\n\n \n Which is a Lewis acid but not a Brønsted−Lowry acid?\n \n
\n\n \n A. AlCl\n \n 3\n \n \n
\n\n \n B. CH\n \n 3\n \n CO\n \n 2\n \n H\n \n
\n\n \n C. HF\n \n
\n\n \n D. CCl\n \n 4\n \n \n
\n\n [1]\n
\n\n A\n
\n\n Identifying a Lewis acid seemed relatively easy for the majority of candidates.\n
\n\n Successive ionization energies of an element,\n \n X\n \n , are shown.\n
\n\n \n
\n What energy, in kJ mol\n \n −1\n \n , is required for element\n \n X\n \n to reach its most stable oxidation state in ionic compounds?\n
\n\n
\n A. 740\n
\n B. 1450\n
\n\n C. 2190\n
\n\n D. 7730\n
\n\n [1]\n
\n\n C\n
\n\n Only just over 50% of candidates correctly calculated the energy required to convert an atom to its most stable ion, given successive ionisation energies. The ionisation energy corresponding to the loss of the final electron (B) was the distractor most commonly chosen.\n
\n\n \n State the name of this compound, applying IUPAC rules.\n \n
\n\n [1]\n
\n\n \n rhenium(III) chloride\n
\n \n \n OR\n \n \n
\n rhenium trichloride\n \n [✔]\n \n \n
\n Almost all candidates were able to name the compound according to IUPAC.\n
\n\n Draw a Lewis (electron dot) structure for ozone.\n
\n\n [1]\n
\n\n \n ✔\n
\n \n Accept any combination of lines, dots or crosses to represent electrons.\n \n
\n\n \n Do\n \n not\n \n accept structures that represent 1.5 bonds.\n \n
\n\n \n Suggest why a non-polar solvent was needed.\n \n
\n\n [1]\n
\n\n oil is non-polar «and dissolves best in non-polar solvents»\n
\n \n \n OR\n \n \n
\n oil does not dissolve in polar solvents ✔\n
\n \n Do\n \n not\n \n accept “like dissolves like” only.\n \n
\n\n A 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.\n
\n\n Consider this voltaic cell, where Cu is a more reactive metal than Ag:\n
\n\n \n
\n Which combination describes the movement of charge in this cell?\n
\n\n \n
\n [1]\n
\n\n D\n
\n\n \n A scientist wants to investigate the catalytic properties of a thin layer of rhenium\n \n \n metal on a graphite surface.\n
\n \n
\n \n Describe an electrochemical process to produce a layer of rhenium on graphite.\n \n
\n\n [2]\n
\n\n \n electrolyze «a solution of /molten» rhenium salt/Re\n \n n+\n \n \n [✔]\n \n \n
\n\n \n graphite as cathode/negative electrode\n
\n OR\n
\n rhenium forms at cathode/negative electrode\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “using rhenium anode” for M1.\n \n \n
\n\n Generally well done; most described the cell identifying the two electrodes correctly and a few did mention the need for Re salt/ion electrolyte.\n
\n\n The complete combustion of 20.0 cm\n \n 3\n \n of a gaseous hydrocarbon, C\n \n x\n \n H\n \n y\n \n , produces 80.0 cm\n \n 3\n \n of gaseous products. This volume reduces to 40.0 cm\n \n 3\n \n when the water vapour present condenses. All volumes are measured at the same temperature and pressure.\n
\n\n
\n What is the molecular formula of the hydrocarbon?\n
\n
\n A. CH\n \n 4\n \n
\n B. C\n \n 2\n \n H\n \n 2\n \n
\n\n C. C\n \n 2\n \n H\n \n 4\n \n
\n\n D. C\n \n 3\n \n H\n \n 6\n \n
\n\n [1]\n
\n\n C\n
\n\n Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl\n \n 2\n \n .\n
\n\n \n
\n [2]\n
\n\n \n
\n \n Deduce the equation for the decomposition of guanidinium nitrate.\n \n
\n\n [1]\n
\n\n \n C(NH\n \n 2\n \n )\n \n 3\n \n NO\n \n 3\n \n (s) → 2N\n \n 2\n \n (g) + 3H\n \n 2\n \n O (g) + C (s) ✔\n \n
\n\n Suggest what should be used as a blank for spectrophotometric reading.\n
\n\n [1]\n
\n\n water\n \n \n AND\n \n \n all samples dissolved «in water» ✔\n
\n\n Which amount, in mol, of sodium chloride is needed to make 250 cm\n \n 3\n \n of 0.10 mol dm\n \n −3\n \n solution?\n
\n\n A. 4.0 × 10\n \n −4\n \n
\n\n B. 0.025\n
\n\n C. 0.40\n
\n\n D. 25\n
\n\n [1]\n
\n\n B\n
\n\n \n The organic product is not optically active. Discuss whether or not the organic product is a racemic mixture.\n \n
\n\n [1]\n
\n\n \n no\n \n \n AND\n \n \n there is no chiral carbon\n \n
\n\n \n \n \n OR\n \n \n \n
\n\n \n no\n \n \n AND\n \n \n there is no carbon with four different substituents/groups\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “no\n \n AND\n \n no asymmetric carbon\n
\n atom”.\n \n \n
\n One 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.\n
\n\n The block structure of the periodic table groups elements according to which characteristic?\n
\n\n
\n A. atomic number\n
\n B. atomic mass\n
\n\n C. electron configuration\n
\n\n D. reactivity\n
\n\n [1]\n
\n\n C\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 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 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 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 «\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 Assume 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\n
\n\n [2]\n
\n\n \n ✔\n
\n\n \n ✔\n
\n\n
\n\n \n Award «0.2614 mol x 40.31 g mol\n \n –1\n \n »\n \n
\n\n \n Accept alternative methods to arrive at the correct answer.\n \n
\n\n \n Accept final answers in the range 91-92%\n \n
\n\n \n \n [2]\n \n for correct final answer.\n \n
\n\n This was not well answered, but definitely better than the previous question with quite a few gaining some credit for correctly determining the theoretical yield.\n
\n\n Draw a Lewis (electron dot) structure for ozone.\n
\n\n [1]\n
\n\n \n ✔\n
\n \n Accept any combination of lines, dots or crosses to represent electrons.\n \n
\n\n \n Do\n \n not\n \n accept structures that represent 1.5 bonds.\n \n
\n\n What is the formula of copper (I) sulfide?\n
\n\n
\n A. CuS\n
\n B. Cu\n \n 2\n \n S\n
\n\n C. CuSO\n \n 3\n \n
\n\n D. Cu\n \n 2\n \n SO\n \n 3\n \n
\n\n [1]\n
\n\n B\n
\n\n The diagram represents a cell in such a battery delivering a current. Complete the half-equations on the diagram and identify the species moving between the electrodes.\n
\n\n \n
\n [3]\n
\n\n \n
\n Suggest an experiment that shows that magnesium is more reactive than zinc, giving the observation that would confirm this.\n
\n\n [2]\n
\n\n \n \n Alternative 1\n \n \n
\n\n put Mg in Zn\n \n 2+\n \n (aq) ✔\n
\n\n Zn/«black» layer forms «on surface of Mg» ✔\n
\n\n \n
\n Award\n \n [1 max]\n \n for “no reaction when Zn placed in Mg\n \n 2+\n \n (aq)”.\n \n
\n
\n\n \n \n Alternative 2\n \n \n
\n\n place both metals in acid ✔\n
\n\n bubbles evolve more rapidly from Mg\n
\n \n \n OR\n \n \n
\n Mg dissolves faster ✔\n
\n
\n\n \n \n Alternative 3\n \n \n
\n\n construct a cell with Mg and Zn electrodes ✔\n
\n\n bulb lights up\n
\n \n \n OR\n \n \n
\n shows (+) voltage\n
\n \n \n OR\n \n \n
\n size/mass of Mg(s) decreases «over time»\n
\n \n \n OR\n \n \n
\n size/mass of Zn increases «over time»\n
\n \n
\n \n \n Accept “electrons flow from Mg to Zn”.\n \n
\n \n Accept Mg is negative electrode/anode\n \n
\n \n \n OR\n \n \n
\n \n Zn is positive electrode/cathode\n \n
\n \n
\n Accept other correct methods.\n \n
\n Many 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\".\n
\n\n \n Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled, while nitrogen does not.\n \n
\n\n [2]\n
\n\n \n \n Any\n \n two\n \n of:\n \n
\n nitrogen non-polar/London/dispersion forces\n \n \n AND\n \n \n water polar/H-bonding ✔\n
\n water has «much» stronger intermolecular forces ✔\n
\n water molecules attract/condense/occupy smaller volume «and therefore deviate from ideal behaviour» ✔\n \n
\n How many electrons are needed when the following half-equation is balanced using the lowest possible whole numbers?\n
\n\n __ NO\n \n 3\n \n \n –\n \n (aq) + __ H\n \n +\n \n (aq) + __ e\n \n –\n \n → __ NO (g) + __ H\n \n 2\n \n O (l)\n
\n\n
\n A. 1\n
\n B. 2\n
\n\n C. 3\n
\n\n D. 5\n
\n\n [1]\n
\n\n C\n
\n\n \n What is the order with respect to each reactant?\n \n
\n\n \n 2NO (g) + Cl\n \n 2\n \n (g) → 2NOCl (g)\n \n
\n\n \n \n \n
\n
\n\n [1]\n
\n\n C\n
\n\n 88 % of the candidates deduced the order of the reaction with respect to each reactant based on the experimental data of initial rate and concentrations of reactants.\n
\n\n Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.\n
\n\n [1]\n
\n\n allows them to explain the properties of different compounds/substances\n
\n \n \n OR\n \n \n
\n enables them to generalise about substances\n
\n \n \n OR\n \n \n
\n enables them to make predictions ✔\n
\n \n
\n Accept other valid answers.\n \n
\n A container holds 30 g of argon and 60 g of neon.\n
\n\n
\n What is the ratio of number of atoms of argon to number of atoms of neon in the container?\n
\n
\n A. 0.25\n
\n B. 0.50\n
\n\n C. 2.0\n
\n\n D. 4.0\n
\n\n [1]\n
\n\n A\n
\n\n Which of the following statements is correct when a 1.0 M NH\n \n 4\n \n +/NH\n \n 3\n \n buffer (pH = 9.2) is diluted to 0.5 M with water?\n
\n\n I. The ability of the buffer to resist changes in pH when acids are added will decrease.\n
\n\n II. The ability of the buffer to resist changes in pH when bases are added will decrease.\n
\n\n III. The pH of the buffer will be equal to 7.\n
\n\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n A\n
\n\n Determine the percentage uncertainty of the mass of product after heating.\n
\n\n [2]\n
\n\n mass of product\n \n ✔\n
\n\n \n ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer\n \n
\n\n \n Accept 0.021%\n \n
\n\n Only 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\".\n
\n\n \n Calculate the percentage uncertainty and percentage error in the experimentally determined value of\n \n \n \n for methanol.\n \n
\n\n \n
\n [2]\n
\n\n \n Percentage uncertainty:\n \n
\n \n ✔\n
\n \n Percentage error:\n \n
\n \n ✔\n
\n \n Award\n \n [1 max]\n \n if calculations are reversed\n \n OR\n \n if incorrect alcohol is used.\n \n
\n\n Many students scored both points and others at least one. Weaker students inverted the calculations.\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 «[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 \n Enzymatic activity is studied in buffered aqueous solutions.\n \n
\n\n \n Calculate 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 (aq) and 0.1 mol dm\n \n −3\n \n Na\n \n 2\n \n HPO\n \n 4\n \n (aq) should be mixed to obtain a buffer with pH = 6.10. Use section 1 of the data booklet.\n \n
\n\n \n pK\n \n a\n \n (NaH\n \n 2\n \n PO\n \n 4\n \n ) = 7.20\n \n
\n\n [2]\n
\n\n \n «pH = pK\n \n a\n \n + log\n \n \n \n / 6.10 = 7.20 + log\n \n \n \n \n \n »\n \n
\n\n \n log\n \n \n \n \n \n = «6.10 – 7.20 =» –1.10\n
\n \n \n OR\n \n \n
\n \n \n \n \n \n = «10\n \n –1.10\n \n =» 0.079\n \n [✔]\n \n \n
\n \n NaH\n \n 2\n \n PO\n \n 4\n \n : Na\n \n 2\n \n HPO\n \n 4\n \n = 12.6 : 1\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 The calculation of the ratio of a conjugate acid-base pair to create a buffer with a specific pH was poorly done. Some candidates wrote a ratio without indicating which compound each value referred to and thus could not score. Many candidates used the concentrations of 0.1 mol dm\n \n −3\n \n for both compounds and could not proceed.\n
\n\n Write the equation for the fermentation of glucose, C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n , to produce ethanol.\n
\n\n [1]\n
\n\n C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n → 2C\n \n 2\n \n H\n \n 5\n \n OH + 2CO\n \n 2\n \n ✓\n
\n\n Discuss the relative length of the two O−O bonds in ozone.\n
\n\n [2]\n
\n\n both equal ✔\n
\n\n delocalization/resonance ✔\n
\n\n \n
\n Accept bond length between 121 and 148 pm/ that of single O−O bond and double O=O bond for M1.\n \n
\n What is the formula of the compound formed between magnesium ions and hydrogencarbonate ions?\n
\n\n
\n A. MgHCO\n \n 3\n \n
\n B. Mg(HCO\n \n 3\n \n )\n \n 2\n \n
\n\n C. Mg(HCO\n \n 3\n \n )\n \n 3\n \n
\n\n D. Mg\n \n 3\n \n (HCO\n \n 3\n \n )\n \n 2\n \n
\n\n [1]\n
\n\n B\n
\n\n Ice containing only the isotope\n \n 2\n \n H sinks and does not melt when dropped into ordinary distilled water maintained at 3 °C.\n
\n\n
\n Which statement is correct?\n
\n
\n A. The isotope\n \n 2\n \n H has a high natural abundance.\n
\n B.\n \n 2\n \n H\n \n 2\n \n O (s) has a higher melting point than normal ice.\n
\n\n C.\n \n 2\n \n H\n \n 2\n \n O (s) has a lower density than normal ice-cold water.\n
\n\n D.\n \n 2\n \n H\n \n 2\n \n O has different chemical properties from normal water.\n
\n\n [1]\n
\n\n B\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 \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 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 \n Determine the standard entropy change, in J K−1, for the decomposition of dinitrogen monoxide.\n \n
\n\n \n 2N\n \n 2\n \n O (g) → 2N\n \n 2\n \n (g) + O\n \n 2\n \n (g)\n \n
\n\n \n ![\"\"](\"https://ib-questionbank-attachments.s3.amazonaws.com/uploads/tinymce_asset/asset/5984/2gi.PNG\")
\n \n
\n [2]\n
\n\n Δ\n \n S\n \n θ\n \n = 2(S\n \n θ\n \n (N\n \n 2\n \n )) + S\n \n θ\n \n (O\n \n 2\n \n ) – 2(S\n \n \n \n θ\n \n \n \n (N\n \n 2\n \n O))\n
\n \n \n \n \n OR\n
\n \n \n \n Δ\n \n S\n \n \n \n θ\n \n \n \n = 2 × 193 «J mol\n \n -1\n \n K\n \n -1\n \n » + 205 «J mol\n \n -1\n \n K\n \n -1\n \n » – 2 × 220 «J mol\n \n -1\n \n K\n \n -1\n \n »\n \n [✔]\n \n \n
\n \n «ΔS\n \n \n \n θ\n \n \n \n = +»151 «J K\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 Candidates were able to calculate the ΔS of the reaction, though in some cases they failed to multiply by the number of moles.\n
\n\n What is correct as a system approaches equilibrium?\n
\n\n
\n A. Q remains constant.\n
\n B.\n \n K\n \n \n c\n \n increases.\n
\n C.\n \n \n G\n \n \n ⦵\n \n becomes more negative.\n
\n D.\n \n \n G\n \n approaches zero.\n
\n [1]\n
\n\n D\n
\n\n \n Which molecule will decolorize bromine water in the dark?\n \n
\n\n \n A. cyclohexane\n
\n \n
\n \n B. hexane\n
\n \n
\n \n C. hex-1-ene\n
\n \n
\n \n D. hexan-1-ol\n \n
\n\n [1]\n
\n\n C\n
\n\n Two-thirds of candidates knew that hex-1-ene would decolourize bromine water in the dark.\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 «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 \n Determine the mole ratio of S\n \n 2\n \n O\n \n 3\n \n \n 2−\n \n to O\n \n 2\n \n , using the balanced equations.\n \n
\n\n [1]\n
\n\n \n 4 : 1 ✔\n \n
\n\n \n What is the IUPAC name of this molecule?\n \n
\n\n \n
\n \n A. 1,1,2,4-tetramethylpent-1-ene\n
\n \n
\n \n B. 2,4,5-trimethylhex-4-ene\n
\n \n
\n \n C. 2,4,5,5-tetramethylpent-4-ene\n
\n \n
\n \n D. 2,3,5-trimethylhex-2-ene\n \n
\n\n [1]\n
\n\n D\n
\n\n It was good to see that 2/3 of candidates could pick a preferred IUPAC name from a skeletal structure.\n
\n\n State\n \n two\n \n conditions necessary for a successful collision between reactants.\n
\n\n [1]\n
\n\n \n E\n \n ≥\n \n E\n \n \n a\n \n \n \n AND\n \n \n appropriate «collision» geometry/correct orientation ✔\n
\n\n Generally 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\n \n E\n \n a\n \n \n .\n
\n\n Explain why the first ionization energy of calcium is greater than that of potassium.\n
\n\n [2]\n
\n\n increasing number of protons/nuclear charge/Z\n \n eff\n \n ✔\n
\n\n
\n «atomic» radius/size decreases\n
\n \n \n OR\n \n \n
\n same number of energy levels\n
\n \n \n OR\n \n \n
\n similar shielding «by inner electrons» ✔\n
\n It 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.\n
\n\n Which is in the same homologous series as CH\n \n 3\n \n OCH\n \n 3\n \n ?\n
\n\n A. CH\n \n 3\n \n COCH\n \n 3\n \n
\n\n B. CH\n \n 3\n \n COOCH\n \n 3\n \n
\n\n C. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n OH\n
\n\n D. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n OCH\n \n 3\n \n
\n\n [1]\n
\n\n D\n
\n\n Which combination describes the acid–base nature of aluminium and phosphorus oxides?\n
\n\n \n
\n [1]\n
\n\n A\n
\n\n \n The 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\n \n \n \n both\n \n \n \n reactions.\n \n
\n\n [2]\n
\n\n dilution/lower concentrations ✔\n
\n\n less frequent collisions «per unit volume» ✔\n
\n\n \n Accept “lowers concentration of acid catalyst” for M1. M2 must refer to increase in activation energy or different pathway.\n \n
\n\n \n Do\n \n not\n \n accept responses referring to equilibrium.\n \n
\n\n A significant number of students scored at least one mark, usually the first and many both. Weaker students lost the second mark by referring to less collisions instead of less frequent collision or other words to this effect. Very few students referred to more diluted catalyst and of those even less were able to provide an adequate explanation in terms of the increased Ea. Many students tried to answer this question in terms of equilibrium instead of kinetics. There were also several responses that replied as if the dilution would only occur for part of the reaction or individual reactants instead of the entire solution.\n
\n\n \n State\n \n one\n \n factor considered when making green chemistry polymers.\n \n
\n\n [1]\n
\n\n \n \n Any one of:\n \n
\n high content of raw materials in product/high atom economy\n \n [✔]\n \n \n
\n \n use of low toxic chemicals/catalysts/materials/solvents\n \n \n \n [✔]\n \n \n
\n\n \n renewable feedstock/raw materials\n \n \n \n [✔]\n \n \n
\n\n \n use of renewable/clean/low carbon energy source\n \n \n \n [✔]\n \n \n
\n\n \n high safety standards\n \n \n \n [✔]\n \n \n
\n\n \n increase energy efficiency\n \n \n \n [✔]\n \n \n
\n\n \n waste recycling\n \n \n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept other reasonable answers.\n \n \n
\n\n Most candidates were able to state a factor considered when making green chemistry polymers.\n
\n\n \n The cell potential for the spontaneous reaction when standard magnesium and silver half-cells are connected is +3.17 V.\n \n
\n\n \n Determine the cell potential at 298 K when:\n \n
\n\n \n [Mg\n \n 2+\n \n ] = 0.0500 mol dm\n \n −3\n \n
\n [Ag\n \n +\n \n ] = 0.100 mol dm\n \n −3\n \n \n
\n \n Use sections 1 and 2 of the data booklet.\n \n
\n\n
\n\n [2]\n
\n\n \n «\n \n E = E\n \n ᶱ\n \n \n \n × ln\n \n Q\n \n »\n \n
\n\n \n ln\n \n Q\n \n = «\n \n \n \n » 1.61\n \n [✔]\n \n \n
\n\n \n E = «3.17 V\n \n \n \n »3.15 «V»\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 The cell potential was correctly calculated by several candidates with some candidates managed an ECF mark for an error in the calculation. Unfortunately, the ln\n \n Q\n \n part was frequently wrong due to candidates forgetting to square the denominator.\n
\n\n The concentration of excess sodium hydroxide was 0.362 mol dm\n \n −3\n \n . Calculate the pH of the solution at the end of the experiment.\n
\n\n [1]\n
\n\n «pOH = –log(0.362) = 0.441»\n
\n\n «pH = 14.000 – 0.441 =» 13.559 ✔\n
\n\n Identify the type of reaction.\n
\n\n [1]\n
\n\n «nucleophilic» substitution\n
\n \n \n OR\n \n \n
\n SN2 ✔\n
\n \n
\n Accept “hydrolysis”.\n \n
\n \n Accept SN1\n \n
\n\n About 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\n
\n\n Explain the electron domain geometry of SO\n \n 3\n \n .\n
\n\n [2]\n
\n\n three electron domains «attached to the central atom» ✔\n
\n\n repel/as far away as possible /120° «apart» ✔\n
\n\n Most 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.\n
\n\n \n Outline\n \n one\n \n difference between a primary and a secondary cell.\n \n
\n\n [1]\n
\n\n \n Any one of:\n \n
\n\n \n \n \n
\n \n \n Note:\n \n \n Accept “primary cannot be recharged\n \n AND\n \n “secondary can be recharged”.\n \n \n
\n\n Most candidates were able to state one difference between a primary and a secondary cell.\n
\n\n Ice containing only the isotope\n \n 2\n \n H sinks and does not melt when dropped into ordinary distilled water maintained at 3 °C.\n
\n\n
\n Which statement is correct?\n
\n
\n A. The isotope\n \n 2\n \n H has a high natural abundance.\n
\n B.\n \n 2\n \n H\n \n 2\n \n O (s) has a higher melting point than normal ice.\n
\n\n C.\n \n 2\n \n H\n \n 2\n \n O (s) has a lower density than normal ice-cold water.\n
\n\n D.\n \n 2\n \n H\n \n 2\n \n O has different chemical properties from normal water.\n
\n\n [1]\n
\n\n B\n
\n\n A sample of calcium carbonate reacts with excess hydrochloric acid in a beaker. The solid line shows how the mass of the beaker changes with time.\n
\n\n Which dashed line represents the results obtained when the acid concentration is doubled?\n
\n\n \n
\n [1]\n
\n\n B\n
\n\n \n Consider the following electrochemical cell.\n \n
\n\n \n \n \n
\n \n \n What happens to the ions in the salt bridge when a current flows?\n \n \n
\n\n \n \n A. Na\n \n +\n \n ions flow to the zinc half-cell and SO\n \n 4\n \n \n 2−\n \n ions flow to the copper half-cell.\n \n \n
\n\n \n \n B. Na\n \n +\n \n ions flow to the copper half-cell and SO\n \n 4\n \n \n 2−\n \n ions flow to the zinc half-cell.\n \n \n
\n\n \n \n C. Na\n \n +\n \n and SO\n \n 4\n \n \n 2−\n \n ions flow to the copper half-cell.\n \n \n
\n\n \n \n D. Na\n \n +\n \n and SO\n \n 4\n \n \n 2−\n \n ions flow to the zinc half-cell.\n \n \n
\n\n [1]\n
\n\n B\n
\n\n A challenging question about direction of ion flow in a salt bridge. 60 % answered correctly and the option with the opposite directions of flow of ions (A) was the most commonly chosen distractor.\n
\n\n \n Which compound has the lowest boiling point?\n \n
\n\n \n A. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n \n
\n\n \n B. CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 2\n \n CH\n \n 3\n \n \n
\n\n \n C. CH\n \n 3\n \n CH(CH\n \n 3\n \n )CH\n \n 2\n \n CH\n \n 3\n \n \n
\n\n \n D. CH\n \n 3\n \n C(CH\n \n 3\n \n )\n \n 2\n \n CH\n \n 3\n \n \n
\n\n [1]\n
\n\n D\n
\n\n Surprisingly, this was one of the challenging questions on the paper. Only 57 % of the candidates chose dimethylpropane as the compound having the lowest boiling point. The most commonly chosen distractor was pentane (B) which did not take into account the effect of branching on the strength of London dispersion forces.\n
\n\n Outline, giving a reason, the effect of a catalyst on a reaction.\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 A 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≥E\n \n a\n \n . 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).\n
\n\n Suggest what can be concluded about the gold atom from this experiment.\n
\n\n \n
\n [2]\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 \n Very few\n \n 4\n \n He\n \n 2+\n \n deviating largely from their path:\n \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 Outline what is meant by the degradation of energy.\n \n
\n\n [1]\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
\n\n \n The following occurs when metal\n \n X\n \n is added to\n \n Y\n \n sulfate solution and\n \n Z\n \n sulfate solution. (\n \n X\n \n ,\n \n Y\n \n and\n \n Z\n \n represent metal elements but not their symbols.)\n \n
\n\n \n \n X\n \n (s) +\n \n Y\n \n SO\n \n 4\n \n (aq) →\n \n X\n \n SO\n \n 4\n \n (aq) +\n \n Y\n \n (s)\n
\n \n X\n \n (s) +\n \n Z\n \n SO\n \n 4\n \n (aq): no reaction\n \n
\n \n What is the order of increasing reactivity?\n \n
\n\n \n A. X < Y < Z\n
\n \n
\n \n B. Y < X < Z\n
\n \n
\n \n C. Z < Y < X\n
\n \n
\n \n D. Z < X < Y\n \n
\n\n [1]\n
\n\n B\n
\n\n \n \n K\n \n \n c\n \n for 2N\n \n 2\n \n O (g)\n \n \n \n 2N\n \n 2\n \n (g) + O\n \n 2\n \n (g) is 7.3 × 10\n \n 34\n \n .\n \n
\n\n \n What is\n \n K\n \n \n \n c\n \n \n for the following reaction, at the same temperature?\n \n
\n\n \n N\n \n 2\n \n (g) +\n \n \n \n O\n \n 2\n \n (g)\n \n \n \n \n \n N\n \n 2\n \n O (g)\n \n
\n\n \n \n A. 7.3 × 10\n \n 34\n \n \n \n
\n\n B.\n \n \n \n \n \n
\n\n \n C.\n \n \n \n \n
\n\n \n D.\n \n \n \n \n
\n\n [1]\n
\n\n B\n
\n\n Not scored well and a couple of teachers thought it was a trick question and particularly tough. Relationship between K values on reversed equations should be understood.\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 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 Which term in the expression ΔG\n \n ⦵\n \n = ΔH\n \n ⦵\n \n − TΔS\n \n ⦵\n \n is an indirect measure of the entropy change of the surroundings when divided by T?\n
\n\n A. ΔG\n \n ⦵\n \n
\n\n B. ΔH\n \n ⦵\n \n
\n\n C. ΔS\n \n ⦵\n \n
\n\n D. −TΔS\n \n ⦵\n \n
\n\n [1]\n
\n\n B\n
\n\n This challenging question raised some debate and several teachers did not think it was suitable for this level. Most candidates chose -TΔS as “an indirect measure of the entropy change of the surroundings when divided by T”. While candidates are not expected to be familiar with the equation ΔS surroundings = -ΔHsystem/T, they could solve the question by recognizing that ΔH is the value that affects the surroundings while ΔS relates to the system. The wording of the question could have been simplified.\n
\n\n Calcium carbonate is heated to produce calcium oxide, CaO.\n
\n\n CaCO\n \n 3\n \n (s) → CaO (s) + CO\n \n 2\n \n (g)\n
\n\n Calculate the volume of carbon dioxide produced at STP when 555 g of calcium carbonate decomposes. Use sections 2 and 6 of the data booklet.\n
\n\n [2]\n
\n\n «\n \n n\n \n \n CaCO3\n \n =\n \n =» 5.55 «mol» ✓\n
\n\n «\n \n V\n \n = 5.55 mol × 22.7 dm\n \n 3\n \n mol\n \n −1\n \n =» 126 «dm\n \n 3\n \n » ✓\n
\n\n \n
\n Award\n \n [2]\n \n for correct final answer.\n \n
\n \n Accept method using pV = nRT to obtain the volume with p as either 100 kPa (126 dm\n \n 3\n \n ) or 101.3 kPa (125 dm\n \n 3\n \n ).\n \n
\n\n \n Do not penalize use of 22.4 dm\n \n 3\n \n mol\n \n –1\n \n to obtain the volume (124 dm\n \n 3\n \n ).\n \n
\n\n \n What effect does increasing both pressure and temperature have on the equilibrium constant,\n \n K\n \n \n c\n \n ?\n \n
\n\n \n N\n \n 2\n \n (g) + 3H\n \n 2\n \n (g)\n \n 2NH\n \n 3\n \n (g) Δ\n \n H\n \n = −45.9 kJ\n \n
\n\n \n A. Decreases\n
\n \n
\n \n B. Increases\n
\n \n
\n \n C. Remains constant\n
\n \n
\n \n D. Cannot be predicted as effects are opposite\n \n
\n\n [1]\n
\n\n A\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 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 The standard electrode potentials for three half-cells involving chromium are shown.\n
\n\n Cr\n \n 3+\n \n (aq) + e\n \n −\n \n \n Cr\n \n 2+\n \n (aq)\n \n E\n \n \n ⦵\n \n \n \n = −0.407 V\n
\n\n Cr\n \n 3+\n \n (aq) + 3e\n \n −\n \n \n Cr (s)\n \n E\n \n \n ⦵\n \n = −0.744 V\n
\n\n Cr\n \n 2+\n \n (aq) + 2e\n \n −\n \n \n Cr (s)\n \n E\n \n \n ⦵\n \n = −0.914 V\n
\n\n Which statement is correct?\n
\n
\n
\n A. Cr\n \n 3+\n \n (aq) can oxidize Cr\n \n 2+\n \n (aq) but not Cr (s).\n
\n\n B. Cr\n \n 3+\n \n (aq) can oxidize Cr (s) but not Cr\n \n 2+\n \n (aq).\n
\n\n C. Cr\n \n 3+\n \n (aq) can oxidize both Cr\n \n 2+\n \n (aq) and Cr (s).\n
\n\n D. Cr\n \n 3+\n \n (aq) can oxidize Cr (s) and reduce Cr\n \n 2+\n \n (aq).\n
\n\n [1]\n
\n\n B\n
\n\n A poorly answered question with only 33% of candidates being able to deduce the relative strengths of oxidizing and reducing agents based on standard electrode potential data. All distractors were quite effective, indicating significant confusion.\n
\n\n The overall reaction occurring at the electrodes of a rechargeable metal hydride battery can be summarized as:\n
\n\n MH + NiO(OH)\n \n M + Ni(OH)\n \n 2\n \n
\n\n
\n Which statement is correct?\n
\n
\n A. The oxidation state of Ni does not change.\n
\n B. M is oxidized by loss of hydrogen.\n
\n\n C. The oxidation state of one H atom changes from\n \n 1 to +1.\n
\n\n D. The oxidation state of one O atom changes from\n \n 1 to\n \n 2.\n
\n\n [1]\n
\n\n C\n
\n\n State\n \n one\n \n problem associated with chlorinated organic solvents as chemical waste.\n
\n\n [1]\n
\n\n \n Any one of:\n \n
\n\n «most are» toxic «to living organisms»\n
\n \n \n OR\n \n \n
\n incomplete combustion/incineration can produce toxic products/dioxins/phosgene\n
\n \n \n OR\n \n \n
\n carcinogenic ✓\n
\n «some can be» greenhouse gases ✓\n
\n\n ozone-depleting ✓\n
\n\n can contribute to formation of «photochemical» smog ✓\n
\n\n accumulate in groundwater\n
\n \n \n OR\n \n \n
\n have limited biodegradability ✓\n
\n cost/hazards of disposal ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept “harmful to the environment”.\n \n
\n\n \n Do\n \n not\n \n accept just “pollutes water”.\n \n
\n\n \n Do\n \n not\n \n accept “increases acid rain/acidity/acid deposition”.\n \n
\n\n \n Several reactions of calcium carbonate with dilute hydrochloric acid are carried out at the same temperature.\n \n
\n\n \n CaCO\n \n 3\n \n (s) + 2HCl (aq) → CaCl\n \n 2\n \n (aq) + H\n \n 2\n \n O (l) + CO\n \n 2\n \n (g)\n \n
\n\n \n Which reaction has the greatest rate?\n \n
\n\n \n \n \n
\n
\n\n [1]\n
\n\n A\n
\n\n 65 % of the candidates chose the correct combination to give the greatest rate of reaction. The most commonly chosen distractor was D where “smaller surface area of same mass of CaCO\n \n 3\n \n (s)” was chosen. It seems these candidates confused “surface area” with “particle size”.\n
\n\n Suggest\n \n two\n \n reasons why the penicillin side-chain is modified.\n
\n\n [2]\n
\n\n bacterial resistance «to older penicillin’s/antibiotics» ✓\n
\n\n prevent penicillinase/beta-lactamase/enzyme in bacterium to deactivate/open penicillin/beta-lactam ring ✓\n
\n\n
\n\n \n Accept “antibiotic resistant bacteria” but\n \n not\n \n “antibiotic resistance” for M1.\n \n
\n\n \n Accept “reduce allergic reactions from penicillin” for M2.\n \n
\n\n \n Award\n \n [1 max]\n \n for \"increased efficiency/bioavailability\"\n
\n \n OR\n \n
\n \"increased stability in GIT\".\n \n
\n \n Do\n \n not\n \n accept \"bacterial tolerance\".\n \n
\n\n \n Which equation represents the standard enthalpy of atomization of bromine, Br\n \n 2\n \n ?\n \n
\n\n \n A.\n \n \n \n Br\n \n 2\n \n (l) → Br (g)\n \n
\n\n \n B. Br\n \n 2\n \n (l) → 2Br (g)\n \n
\n\n \n C. Br\n \n 2\n \n (l) → 2Br (l)\n \n
\n\n \n D.\n \n \n \n \n \n Br\n \n 2\n \n (l) → Br (l)\n \n
\n\n [1]\n
\n\n A\n
\n\n One G2 comment asked if enthalpy of atomization was on the syllabus. This Topic 15.1 question was answered correctly by 54 % of candidates but did not differentiate well between higher and lower scoring candidates.\n
\n\n \n Suggest why many chemicals, including hydrogen peroxide, are kept in brown bottles instead of clear colourless bottles.\n \n
\n\n [1]\n
\n\n \n decomposes in light\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “sensitive to light”.\n \n \n
\n\n The 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\n \n e.g.\n \n acid/water/peroxide.\n
\n\n \n Explain why, as the reaction proceeds, the pressure increases by the amount shown.\n \n
\n\n [2]\n
\n\n \n increase in the amount/number of moles/molecules «of gas»\n \n [✔]\n \n \n
\n\n \n from 2 to 3/by 50 %\n \n [✔]\n \n \n
\n\n Students were able in general to relate more moles of gas to increase in pressure.\n
\n\n \n Distinguish ultraviolet light from visible light in terms of wavelength and energy.\n \n
\n\n [1]\n
\n\n \n «UV» shorter wavelength\n \n \n AND\n \n \n higher energy «than visible» ✔\n \n
\n\n \n Some antacids contain carbonates.\n \n
\n\n \n Determine the pH of a buffer solution which contains 0.160 mol dm\n \n −3\n \n CO\n \n 3\n \n \n 2−\n \n and 0.200 mol dm\n \n −3\n \n HCO\n \n 3\n \n \n −\n \n , using section 1 of the data booklet.\n \n
\n\n \n p\n \n K\n \n a\n \n \n (HCO\n \n 3\n \n \n −\n \n ) = 10.32\n \n
\n\n [1]\n
\n\n \n «pH = p\n \n K\n \n a\n \n \n + log\n \n \n \n = 10.32 + log\n \n \n \n = 10.32 – 0.097»\n \n
\n\n \n «pH =»10.22\n \n [✔]\n \n \n
\n\n This was in general a well answered question. Most candidates who did not receive the mark inverted the concentration of the conjugate base/concentration of the acid in the calculation.\n
\n\n What is the electron configuration for an element in group 4 period 5?\n
\n\n
\n A. [Kr] 5s\n \n 2\n \n 4d\n \n 2\n \n
\n B. [Ar] 4s\n \n 2\n \n 3d\n \n 3\n \n
\n\n C. [Ar] 4s\n \n 2\n \n 3d\n \n 10\n \n 4p\n \n 3\n \n
\n\n D. [Kr] 5s\n \n 2\n \n 4d\n \n 10\n \n 5p\n \n 2\n \n
\n\n [1]\n
\n\n A\n
\n\n \n Label with an asterisk, *, the chiral carbon atom.\n \n
\n\n [1]\n
\n\n \n \n [\n \n ✔]\n \n \n
\n Identifying a chiral carbon atom was answered reasonably well.\n
\n\n Which are endothermic processes in a Born–Haber cycle for the formation of an ionic compound?\n
\n\n
\n I. Enthalpy of atomization\n
\n II. First electron affinity\n
\n III. First ionization energy\n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n B\n
\n\n Deduce the structural and empirical formulas of\n \n B\n \n .\n
\n\n \n
\n [3]\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 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 In which block of the periodic table would element 119 be placed, if it is found in the future?\n
\n\n
\n A. s\n
\n B. p\n
\n\n C. d\n
\n\n D. f\n
\n\n [1]\n
\n\n A\n
\n\n Describe which cancers are treated by TAT and BNCT and the particles used in each treatment.\n
\n\n
\n\n TAT:\n
\n\n
\n\n
\n\n
\n BNCT:\n
\n
\n\n
\n\n
\n\n [2]\n
\n\n TAT:\n
\n spread to multiple sites\n \n \n AND\n \n \n alpha particles\n
\n \n \n OR\n \n \n
\n «cancers of the» blood/leukaemia\n \n \n AND\n \n \n alpha particles ✓\n
\n BNCT:\n
\n head/neck/brain «cancers»\n \n \n AND\n \n \n neutrons ✓\n
\n
\n\n \n Award\n \n [1]\n \n for 2 correct cancers or 2 correct particles.\n \n
\n\n With reference to the reaction quotient, Q, explain why the percentage yield increases as the pressure is increased at constant temperature.\n
\n\n [3]\n
\n\n increasing pressure increases «all» concentrations\n
\n \n \n OR\n \n \n
\n increasing pressure decreases volume ✔\n
\n \n
\n Q\n \n becomes less than\n \n K\n \n \n c\n \n
\n \n \n OR\n \n \n
\n affects the lower line/denominator of Q expression more than upper line/numerator ✔\n
\n
\n «for\n \n Q\n \n to once again equal\n \n K\n \n \n c\n \n ,» ratio of products to reactants increases\n
\n \n \n OR\n \n \n
\n «for\n \n Q\n \n to once again equal\n \n K\n \n \n c\n \n ,» equilibrium shifts to right/products ✔\n
\n
\n\n \n Award\n \n [2 max]\n \n for answers that do not refer to Q.\n \n
\n\n Mediocre performance; very few identified the effect of increasing pressure on all concentrations. Consequently,\n \n Q\n \n becomes less than\n \n K\n \n \n c\n \n (it affects the denominator of\n \n Q\n \n 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.\n
\n\n \n The organic product is not optically active. Discuss whether or not the organic product is a racemic mixture.\n \n
\n\n [1]\n
\n\n \n no\n \n \n AND\n \n \n there is no chiral carbon\n \n
\n\n \n \n \n OR\n \n \n \n
\n\n \n no\n \n \n AND\n \n \n there is no carbon with four different substituents/groups\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “no\n \n AND\n \n no asymmetric carbon\n
\n atom”.\n \n \n
\n One 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.\n
\n\n A gas storage tank of fixed volume V contains N molecules of an ideal gas at 300 K with a pressure of 40 kPa.\n \n molecules are removed, and the temperature is changed to 450 K.\n
\n\n What is the new pressure of the gas in kPa?\n
\n\n
\n A. 15\n
\n B. 30\n
\n\n C. 45\n
\n\n D. 60\n
\n\n [1]\n
\n\n C\n
\n\n Suggest an explanation, other than product being lost from the crucible or reacting with nitrogen, that could explain the yield found in (b)(iii).\n
\n\n [1]\n
\n\n incomplete reaction\n
\n \n \n OR\n \n \n
\n Mg was partially oxidised already\n
\n \n \n OR\n \n \n
\n impurity present that evaporated/did not react ✔\n
\n
\n\n \n Accept “crucible weighed before fully cooled”.\n \n
\n\n \n Accept answers relating to a higher atomic mass impurity consuming less O/O\n \n 2\n \n .\n \n
\n\n \n Accept “non-stoichiometric compounds formed”.\n \n
\n\n \n Do\n \n not\n \n accept \"human error\", \"wrongly calibrated balance\" or other non-chemical reasons.\n \n
\n\n \n If answer to (b)(iii) is >100%, accept appropriate reasons, such as product absorbed moisture before being weighed.\n \n
\n\n Quite 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\".\n
\n\n A white solid was formed when ethene was subjected to high pressure.\n
\n\n Deduce the type of reaction that occurred.\n
\n\n [1]\n
\n\n «addition» polymerization ✔\n
\n\n \n Which oxide will dissolve in water to give the solution with the lowest pH?\n \n
\n\n A.\n \n
\n\n B.\n \n
\n\n C.\n \n
\n\n D.\n \n
\n\n [1]\n
\n\n A\n
\n\n Fairly well answered. Almost 70% of candidates could identify that non-metallic oxides would dissolve in water to yield a substance of low pH.\n
\n\n An\n \n -particle is a helium-4 nucleus. In an experiment,\n \n -particles are accelerated towards a thin sheet of gold and their resulting paths are detected, giving evidence of the positive charge of the nucleus.\n
\n\n \n
\n Angle of detection\n \n increase from 0° to 180°\n
\n\n The number of\n \n -particles detected at different angles of deflection\n \n are shown.\n
\n\n \n
\n
\n\n repelled by «hitting/close contact with» gold nucleus ✔\n
\n\n \n Describe, using a suitable equation, how the buffer solution formed during the titration resists pH changes when a small amount of acid is added.\n \n
\n\n [2]\n
\n\n \n \n \n ALTERNATIVE 1:\n \n \n
\n H\n \n +\n \n (aq) + CH\n \n 3\n \n COO\n \n –\n \n (aq) → CH\n \n 3\n \n COOH (aq) ✔\n \n
\n \n added acid neutralised by ethanoate ions\n
\n \n \n OR\n \n \n
\n «weak» CH\n \n 3\n \n COOH (aq)/ethanoic acid replaces H\n \n +\n \n (aq)\n
\n \n \n OR\n \n \n
\n CH\n \n 3\n \n COOH/CH\n \n 3\n \n COO\n \n –\n \n ratio virtually/mostly unchanged ✔\n \n
\n \n
\n \n \n ALTERNATIVE 2:\n \n \n
\n CH\n \n 3\n \n COOH (aq)\n \n \n \n H\n \n +\n \n (aq) + CH\n \n 3\n \n COO\n \n –\n \n (aq) ✔\n \n
\n \n equilibrium shifts to the ethanoic acid side\n
\n \n \n OR\n \n \n
\n CH\n \n 3\n \n COOH/CH\n \n 3\n \n COO\n \n −\n \n ratio virtually/mostly unchanged ✔\n \n
\n A student performed displacement reactions using metals W and X and solutions of salts of metals W, X, Y and Z. The results are summarized in the table.\n
\n\n \n
\n Which of the four metals is most reactive?\n
\n\n A. W\n
\n\n B. X\n
\n\n C. Y\n
\n\n D. Z\n
\n\n [1]\n
\n\n D\n
\n\n Deduce the splitting pattern in the\n \n 1\n \n H NMR spectrum for 1-bromopropane.\n
\n\n [1]\n
\n\n triplet/3\n \n \n AND\n \n \n multiplet/6\n \n \n AND\n \n \n triplet/3 ✔\n
\n\n Another 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.\n
\n\n \n The reaction of the hydroxide ion with carbon dioxide and with the hydrogencarbonate ion can be represented by Equations 3 and 4.\n \n
\n\n \n Equation (3) OH\n \n −\n \n (aq) + CO\n \n 2\n \n (g) → HCO\n \n 3\n \n \n −\n \n (aq)\n
\n Equation (4) OH\n \n −\n \n (aq) + HCO\n \n \n 3\n \n \n −\n \n \n (aq) → H\n \n 2\n \n O (l) + CO\n \n 3\n \n \n 2−\n \n (aq)\n
\n \n
\n \n Discuss how these equations show the difference between a Lewis base and a Brønsted–Lowry base.\n \n
\n\n
\n\n \n Equation (3):\n \n
\n\n \n Equation (4):\n \n
\n\n [2]\n
\n\n \n \n Equation (3):\n \n
\n OH\n \n -\n \n donates an electron pair\n \n \n AND\n \n \n acts as a Lewis base\n \n [✔]\n \n \n
\n \n \n Equation (4):\n \n
\n OH\n \n -\n \n accepts a proton/H\n \n +\n \n /hydrogen ion\n \n \n AND\n \n \n acts as a Brønsted–Lowry base\n \n \n \n [✔]\n \n \n
\n This 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\n \n -\n \n or also lost marks referring just to electrons, an electron or H instead of hydrogen ions or H\n \n +\n \n for example.\n
\n\n \n Write the equation for the production of the active nitrating agent from concentrated sulfuric and nitric acids.\n \n
\n\n [1]\n
\n\n \n 2H\n \n 2\n \n SO\n \n 4\n \n + HNO\n \n 3\n \n ⇌ NO\n \n 2\n \n \n +\n \n + 2HSO\n \n 4\n \n \n \n \n −\n \n + H\n \n 3\n \n O\n \n +\n \n \n \n [\n \n ✔\n \n ]\n \n
\n\n \n \n Note\n \n :\n \n Accept a single arrow instead of an equilibrium sign.\n
\n Accept “H\n \n 2\n \n SO\n \n 4\n \n + HNO\n \n 3\n \n ⇌ NO\n \n 2\n \n \n +\n \n + HSO\n \n 4\n \n \n −\n \n + H\n \n 2\n \n O”.\n
\n Accept “H\n \n 2\n \n SO\n \n 4\n \n + HNO\n \n 3\n \n ⇌ H\n \n 2\n \n NO\n \n 3\n \n \n +\n \n + HSO\n \n 4\n \n \n −\n \n ”.\n
\n 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.\n \n \n
\n The production of NO\n \n 3\n \n \n −\n \n was a common answer.\n
\n\n The table shows the variation of standard Gibbs energy with temperature for a reversible reaction.\n
\n\n \n
\n\n \n
\n\n \n
\n What can be concluded about the reaction?\n
\n\n A. Equilibrium shifts left as temperature increases.\n
\n\n B. The forward reaction is more spontaneous below 300 K.\n
\n\n C. Entropy is higher in the products than in the reactants.\n
\n\n D.\n \n K\n \n \n c\n \n decreases as temperature increases.\n
\n\n [1]\n
\n\n C\n
\n\n State the type of spectroscopy that could be used to determine their relative abundances.\n
\n\n [1]\n
\n\n mass «spectroscopy»/MS ✔\n
\n\n \n A scientist wants to investigate the catalytic properties of a thin layer of rhenium\n \n \n metal on a graphite surface.\n
\n \n
\n \n Describe an electrochemical process to produce a layer of rhenium on graphite.\n \n
\n\n [2]\n
\n\n \n electrolyze «a solution of /molten» rhenium salt/Re\n \n n+\n \n \n [✔]\n \n \n
\n\n \n graphite as cathode/negative electrode\n
\n OR\n
\n rhenium forms at cathode/negative electrode\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “using rhenium anode” for M1.\n \n \n
\n\n Generally well done; most described the cell identifying the two electrodes correctly and a few did mention the need for Re salt/ion electrolyte.\n
\n\n Write a balanced equation for the reaction that occurs.\n
\n\n [1]\n
\n\n 2 Mg(s) + O\n \n 2\n \n (g) → 2 MgO(s) ✔\n
\n\n \n
\n Do not accept equilibrium arrows. Ignore state symbols\n \n
\n This was not as well done as one might have expected with the most common errors being O instead of O\n \n 2\n \n oxygen and MgO rather than MgO\n \n 2\n \n .\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 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 Another 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\n \n Determine the initial rate of reaction of limestone with nitric acid from the graph.\n \n
\n\n \n Show your working on the graph and include the units of the initial rate.\n \n
\n\n [3]\n
\n\n \n tangent drawn at time zero ✔\n
\n g day\n \n −1\n \n ✔\n
\n 0.16 ✔\n \n
\n
\n\n \n \n NOTE: Accept other reasonable units for initial rate eg, mol dm\n \n −3\n \n s\n \n −1\n \n , mol dm\n \n −3\n \n min\n \n −1\n \n , g s\n \n −1\n \n \n OR\n \n g min\n \n −1\n \n .\n \n \n
\n\n \n \n M3 can only be awarded if the value corresponds to the correct unit given in M2.\n
\n Accept values for the initial rate for M3 in the range: 0.13 − 0.20 g day\n \n −1\n \n \n OR\n \n 1.5 × 10\n \n −6\n \n g s\n \n −1\n \n − 2.3 × 10\n \n −6\n \n g s\n \n −1\n \n \n OR\n \n 7.5 × 10\n \n −8\n \n − 1.2 × 10\n \n −7\n \n mol dm\n \n −3\n \n s\n \n −1\n \n \n OR\n \n 4.5 × 10\n \n −6\n \n − 6.9 × 10\n \n −6\n \n mol dm\n \n −3\n \n min\n \n −1\n \n \n OR\n \n 9.0 × 10\n \n −5\n \n − 1.4 × 10\n \n −4\n \n g min\n \n −1\n \n \n OR\n \n a range based on any other reasonable unit for rate.\n \n \n
\n \n \n Ignore any negative rate value.\n
\n Award\n \n [2 max]\n \n for answers such as 0.12/0.11 g day\n \n −1\n \n , incorrectly obtained by using the first two points on the graph (the average rate between t = 0 and 1 day).\n
\n Award\n \n [1 max]\n \n for correctly calculating any other average rate.\n \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 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 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 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 Estimate the time at which the powdered zinc was placed in the beaker.\n \n
\n\n [1]\n
\n\n \n 100 «s»\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept 90 to 100 s.\n \n \n
\n\n Almost all candidates identified 100 s as the time at which the reaction was initiated.\n
\n\n \n State one analytical technique that could be used to determine the ratio of\n \n 14\n \n N:\n \n 15\n \n N.\n \n
\n\n [1]\n
\n\n \n mass spectrometry/MS\n \n [✔]\n \n \n
\n\n Well answered. More than half of the candidates stated mass spectrometry is used to determine the ratio of the isotopes.\n
\n\n What is the index of hydrogen deficiency of adenine?\n
\n\n \n
\n
\n A. 3\n
\n B. 4\n
\n\n C. 5\n
\n\n D. 6\n
\n\n [1]\n
\n\n D\n
\n\n \n Determine the change in the average oxidation state of carbon.\n \n
\n\n \n From ethanol to ethanal:\n \n
\n\n \n From ethanol to carbon dioxide:\n \n
\n\n [2]\n
\n\n \n \n From ethanol to ethanal:\n \n
\n −2 to −1\n
\n \n \n OR\n \n \n
\n +1/increases by 1 ✔\n \n
\n \n \n NOTE: Do\n \n not\n \n accept “2− to 1−”.\n \n \n
\n\n \n \n From ethanol to carbon dioxide:\n \n
\n −2 to +4\n
\n \n \n OR\n \n \n
\n +6/increases by 6 ✔\n \n
\n \n \n NOTE: Do\n \n not\n \n accept “2− to 4+”.\n
\n \n \n
\n \n \n Do\n \n not\n \n penalize incorrect notation twice.\n \n \n
\n\n \n \n Penalize incorrect oxidation state value of carbon in ethanol once only.\n \n \n
\n\n The table lists successive ionization energies of an element\n \n Z\n \n .\n
\n| \n \n Ionization number\n \n | \n\n \n 1st\n \n | \n\n \n 2nd\n \n | \n\n \n 3rd\n \n | \n\n \n 4th\n \n | \n\n \n 5th\n \n | \n\n \n 6th\n \n | \n
| \n \n Ionization energy / kJ mol−1\n \n | \n\n 577.54\n | \n\n 1816.68\n | \n\n 2744.78\n | \n\n 11 577.5\n | \n\n 14 841.9\n | \n\n 18 379.0\n | \n
\n
\n Which is the formula of the stable oxide of the element\n \n Z\n \n ?\n
\n
\n A. Z\n \n 2\n \n O\n
\n B. ZO\n
\n\n C. Z\n \n 2\n \n O\n \n 3\n \n
\n\n D. ZO\n \n 2\n \n
\n\n [1]\n
\n\n C\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 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\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 \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 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 Which organic compound has the\n \n 1\n \n H NMR shown?\n
\n\n \n
\n \n Source: Spectral Database for Organic Compounds, SDBS, n.d. [online] Available at:\n
\n https://sdbs.db.aist.go.jp/sdbs/cgi-\n \n \n bin/direct_frame_top.cgi [Accessed 6 October 2021].\n \n
\n
\n A. Methanal\n
\n B. Ethanoic acid\n
\n\n C. Methyl ethanoate\n
\n\n D. Propanoic acid\n
\n\n [1]\n
\n\n D\n
\n\n Just under 70% of the candidates identified the\n \n 1\n \n H NMR spectrum as being that of propanoic acid. The most commonly mistakes were thinking it was ethanoic acid (B) or methyl ethanoate (C).\n
\n\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 «\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 \n Which describes a resonance structure?\n \n
\n\n \n A. Double bond can be drawn in alternative positions.\n
\n \n
\n \n B. Bonds vibrate by absorbing IR radiation.\n
\n \n
\n \n C. A double and a single bond in the molecule\n
\n \n
\n \n D. A Lewis structure\n \n
\n\n [1]\n
\n\n A\n
\n\n Describe metallic bonding and how it contributes to electrical conductivity.\n
\n\n [3]\n
\n\n electrostatic attraction ✓\n
\n\n between «a lattice of» cations/positive «metal» ions\n \n \n AND\n \n \n «a sea of» delocalized electrons ✓\n
\n\n
\n mobile electrons responsible for conductivity\n
\n \n \n OR\n \n \n
\n electrons move when a voltage/potential difference/electric field is applied ✓\n
\n
\n\n \n Do\n \n not\n \n accept “nuclei” for “cations/positive ions” in M2.\n \n
\n\n \n Accept “mobile/free” for “delocalized” electrons in M2.\n \n
\n\n \n Accept “electrons move when connected to a cell/battery/power supply”\n \n OR\n \n “electrons move when connected in a circuit” for M3.\n \n
\n\n \n Which apparatus can be used to monitor the rate of this reaction?\n \n
\n\n \n \n \n
\n\n \n A. I and II only\n
\n \n
\n \n B. I and III only\n
\n \n
\n \n C. II and III only\n
\n \n
\n \n D. I, II and III\n \n
\n\n [1]\n
\n\n B\n
\n\n It was pleasing to see that a vast majority of candidates could select equipment necessary to monitor rates of reaction. This was one of the best answered questions in the exam.\n
\n\n Which can be reduced to a secondary alcohol?\n
\n\n A. C\n \n 2\n \n H\n \n 5\n \n COOH\n
\n\n B. CH\n \n 3\n \n CH\n \n 2\n \n OCH\n \n 3\n \n
\n\n C. (CH\n \n 3\n \n )\n \n 2\n \n CHCHO\n
\n\n D. CH\n \n 3\n \n COC\n \n 2\n \n H\n \n 5\n \n
\n\n [1]\n
\n\n D\n
\n\n Calculate the value of the rate constant,\n \n k\n \n , giving its units.\n
\n\n [3]\n
\n\n \n k\n \n =\n \n ✔\n
\n\n «\n \n k\n \n =\n \n = » 0.030 «min\n \n –1\n \n » ✔\n
\n\n min\n \n –1\n \n ✔\n
\n\n \n
\n M1 can be awarded from correct M2 if not explicitly stated.\n \n
\n \n Accept k = gradient.\n \n
\n\n \n Accept values in the range 0.028–0.032.\n \n
\n\n \n Award\n \n [3]\n \n for correct final answer.\n \n
\n\n State the meaning of a strong Brønsted–Lowry acid.\n
\n\n [2]\n
\n\n fully ionizes/dissociates ✔\n
\n\n proton/H\n \n +\n \n «donor » ✔\n
\n\n While 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.\n
\n\n State the condensed electron configurations for Cr and Cr3\n \n +\n \n .\n
\n\n \n
\n [2]\n
\n\n \n Cr:\n \n
\n [Ar] 4s\n \n 1\n \n 3d\n \n 5\n \n ✓\n
\n \n
\n Cr\n \n 3+\n \n :\n \n
\n [Ar] 3d\n \n 3\n \n ✓\n
\n \n
\n Accept “[Ar] 3d\n \n 5\n \n 4s\n \n 1\n \n ”.\n \n
\n \n Accept “[Ar] 3d\n \n 3\n \n 4s\n \n 0\n \n ”.\n \n
\n\n \n Award\n \n [1 max]\n \n for two correct full electron configurations “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 5\n \n \n AND\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 3d\n \n 3\n \n ”.\n \n
\n\n \n Award\n \n [1 max]\n \n for 4s\n \n 1\n \n 3d\n \n 5\n \n \n AND\n \n 3d\n \n 3\n \n .\n \n
\n\n Calculate the value of the equilibrium constant,\n \n K\n \n , at 298 K. Use sections 1 and 2 of the data booklet.\n
\n\n [2]\n
\n\n \n ✔\n
\n\n \n ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Accept answers in the range 4.4×10\n \n 5\n \n to 6.2×10\n \n 5\n \n (arises from rounding of ln K).\n \n
\n\n Good performance; for ln\n \n K\n \n calculation in the equation ΔG = RTln\n \n K\n \n , ΔG unit had to be converted from kJ to J. This led to an error of 1000 in the value of ln\n \n K\n \n for some.\n
\n\n \n Draw the structure of the conjugate base of benzoic acid showing\n \n all\n \n the atoms and\n \n all\n \n the bonds.\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 Accept Kekulé structures.\n \n \n
\n\n \n \n Negative sign must be shown in correct position- on the O or delocalised over the carboxylate.\n \n \n
\n\n Most failed to score a mark for the conjugate base of benzoic acid as either they didn’t show all bonds and atoms in the ring and/or they did not put the minus sign in the correct place. Some didn't read the question carefully so gave the structure of the acid form.\n
\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 \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 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 Calculate the value of the rate constant stating its units.\n
\n\n [2]\n
\n\n \n ✔\n
\n\n mol\n \n –1\n \n dm\n \n 3\n \n s\n \n –1\n \n ✔\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 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 efficiency of a nuclear power plant is approximately 33 %.\n
\n\n
\n\n 33 % of energy input/released «in fission» is converted to electricity ✓\n
\n\n
\n\n \n Do\n \n not\n \n accept generalizations, such as “Only produces 33 % useful energy.”\n \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 exist in different conformational forms (isomers).\n \n
\n\n \n Outline what this means.\n \n
\n\n [1]\n
\n\n \n benzene ring «of the C\n \n 6\n \n H\n \n 5\n \n –CH\n \n 2\n \n » and the bromine «on the CH\n \n 2\n \n –Br» can take up different relative positions by rotating about the «C–C,\n \n σ\n \n –»bond\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n Accept “different parts of the molecule can rotate relative to each other”.\n
\n \n \n
\n \n \n Accept “rotation around σ\n \n –\n \n bond”.\n \n \n
\n\n If 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.\n
\n\n \n Deduce the coefficients required to complete the half-equation.\n \n
\n\n \n \n ReO\n \n 4\n \n \n −\n \n (aq) + ____H\n \n +\n \n (aq) + ____e\n \n −\n \n ⇌ [Re(OH)\n \n 2\n \n ]\n \n 2+\n \n (aq) + ____H\n \n 2\n \n O (l) E\n \n θ\n \n = +0.36 V\n \n \n
\n\n [1]\n
\n\n \n ReO\n \n \n 4\n \n \n −\n \n \n (aq) + 6H\n \n \n +\n \n \n (aq) + 3e\n \n \n −\n \n \n ⇌\n \n \n [Re(OH)\n \n \n 2\n \n \n ]\n \n \n 2+\n \n \n (aq) + 2H\n \n \n 2\n \n \n O (l)\n \n [\n \n ✔]\n \n \n \n
\n\n Surprisingly, 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.\n
\n\n The successive ionization energies for an element in period three are shown.\n
\n\n \n
\n Which element in period 3 has these successive ionization energies?\n
\n\n
\n A. Na\n
\n B. Mg\n
\n\n C. Al\n
\n\n D. Si\n
\n\n [1]\n
\n\n C\n
\n\n \n Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)\n \n
\n\n [2]\n
\n\n \n «additional HCO\n \n 3\n \n \n –\n \n » shifts position of equilibrium to left\n \n [✔]\n \n \n
\n\n \n pH increases\n \n [✔]\n \n \n
\n\n
\n\n \n \n \n Note:\n \n \n \n \n \n \n \n \n \n Do\n \n not\n \n award M2 without any justification in terms of equilibrium shift in M1.\n \n \n
\n\n This 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 NaHCO\n \n 3\n \n was an acid and decreased pH. Some answers had contradictions that showed poor understanding of the pH concept.\n
\n\n \n Which statement is\n \n not\n \n correct regarding benzene?\n \n
\n\n \n A. It is planar.\n \n
\n\n \n B. The ring contains delocalized electrons.\n \n
\n\n \n C. It always reacts in the same way as alkenes.\n \n
\n\n \n D. The carbon–carbon bond has a bond order of 1.5.\n \n
\n\n [1]\n
\n\n C\n
\n\n 73 % of candidates knew that benzene did not react the same way as alkenes.\n
\n\n Calculate the entropy change for the Haber–Bosch process, in J mol\n \n –1\n \n K\n \n –1\n \n at 298 K. Use your answer to (b)(i) and section 1 of the data booklet.\n
\n\n [2]\n
\n\n Δ\n \n G\n \n = «Δ\n \n H\n \n –\n \n T\n \n Δ\n \n S\n \n =» –93000 «J» – 298«K» × Δ\n \n S\n \n = –33000 ✔\n
\n\n Δ\n \n S\n \n = 〈〈\n \n 〉〉 = –201 «J mol\n \n –1\n \n K\n \n –1\n \n » ✔\n
\n\n
\n\n \n Do\n \n not\n \n penalize failure to convert kJ to J in\n \n both\n \n (c)(ii) and (c)(iii).\n \n
\n\n \n Award\n \n [2]\n \n for correct final answer\n \n
\n\n \n Award\n \n [1 max]\n \n for (+) 201 «J mol\n \n –1\n \n K\n \n –1\n \n ».\n \n
\n\n \n Award [2] for –101 or –100.5 «J mol\n \n –1\n \n K\n \n –1\n \n ».\n \n
\n\n Very good performance; since the unit for\n \n S\n \n is J mol\n \n ˗1\n \n K\n \n ˗1\n \n , Δ\n \n G\n \n and Δ\n \n H\n \n needed to be converted from kJ to J, but that was not done in some cases.\n
\n\n Describe two observations that indicate the reaction of lithium with water is exothermic.\n
\n\n [2]\n
\n\n \n Any two:\n \n
\n\n temperature of the water increases ✔\n
\n\n lithium melts ✔\n
\n\n pop sound is heard ✔\n
\n\n
\n\n \n Accept “lithium/hydrogen catches fire”.\n \n
\n\n \n Do not accept “smoke is observed”.\n \n
\n\n \n The maximum temperature used to calculate the enthalpy of reaction was chosen at a point on the extrapolated (dotted) line.\n \n
\n\n \n State the maximum temperature which should be used and outline\n \n one\n \n assumption made in choosing this temperature on the extrapolated line.\n \n
\n\n
\n\n \n Maximum temperature:\n \n
\n\n Assumption:\n
\n\n [2]\n
\n\n \n \n Maximum temperature:\n \n
\n 73 «°C»\n \n [✔]\n \n
\n \n
\n \n \n Assumption\n \n :\n
\n \n \n «temperature reached if» reaction instantaneous\n
\n \n \n \n \n OR\n
\n \n \n \n \n «temperature reached if reaction occurred» without heat loss\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “rate of heat loss is constant”\n \n OR\n \n “rate of temperature decrease is constant”.\n \n \n
\n\n Stating the maximum temperature that should be used in calculations was less well answered, with answers between 63 and 65, or 78 commonly given instead of the correct answer of 73°C. Most candidates managed to score the second mark for stating “no heat loss”.\n
\n\n \n In 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 \n
\n\n \n \n \n
\n \n \n The data for the first trial is given below.\n \n \n
\n\n \n \n \n \n \n
\n \n \n \n Plot a graph on the axes below and from it determine the average rate of\n
\n formation of oxygen gas in cm\n \n 3\n \n O\n \n 2\n \n (g) s\n \n −1\n \n .\n \n \n \n
\n \n \n \n \n \n \n \n
\n \n \n \n \n Average rate of reaction:\n \n \n \n \n
\n\n [3]\n
\n\n \n
\n \n points correctly plotted\n \n [✔]\n \n \n
\n\n \n best fit line\n \n \n AND\n \n \n extended through (to) the origin\n \n [✔]\n \n \n
\n\n \n \n Average rate of reaction:\n \n
\n «slope (gradient) of line =» 0.022 «cm\n \n 3\n \n O\n \n 2\n \n (g) s\n \n −1\n \n »\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note\n \n : Accept range 0.020–0.024cm\n \n 3\n \n O\n \n 2\n \n (g) s\n \n −1\n \n .\n \n \n
\n\n Most candidates could plot a best fit line and find the slope to calculate an average rate of reaction.\n
\n\n Identify the type of reaction.\n
\n\n [1]\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 The graph below shows the Maxwell–Boltzmann distribution of molecular energies at a particular temperature.\n \n
\n\n \n
\n \n \n The rate at which dinitrogen monoxide decomposes is significantly increased by a metal oxide catalyst.\n \n \n
\n\n \n \n Annotate and use the graph to outline why a catalyst has this effect.\n \n \n
\n\n [2]\n
\n\n \n
\n \n catalysed and uncatalysed E\n \n a\n \n marked on graph\n \n \n AND\n \n \n with the catalysed being at lower energy\n \n [✔]\n \n
\n \n
\n
\n\n \n «for catalysed reaction» greater proportion of/more molecules have E ≥ E\n \n a\n \n / E > E\n \n a\n \n
\n \n \n OR\n \n \n
\n «for catalysed reaction» greater area under curve to the right of the E\n \n a\n \n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “more molecules have the activation energy”.\n \n \n
\n\n The 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\n \n E\n \n (particles) >\n \n E\n \n \n a\n \n or did so too vaguely.\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 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 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 \n Consider the following equilibrium reaction.\n \n
\n\n \n 2N\n \n 2\n \n O (g) + O\n \n 2\n \n (g)\n \n \n \n 4NO (g)\n \n ΔH\n \n = +16 kJ\n \n
\n\n \n Which change will move the equilibrium to the right?\n \n
\n\n \n A. Decrease in pressure\n \n
\n\n \n B. Decrease in temperature\n \n
\n\n \n C. Increase in [NO]\n \n
\n\n \n D. Decrease in [O\n \n 2\n \n ]\n \n
\n\n [1]\n
\n\n A\n
\n\n 77 % of the candidates applied Le Chatelier’s Principle correctly. The question had a high discrimination index.\n
\n\n State the oxidation state of sulfur in copper (II) sulfate.\n
\n\n [1]\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 State\n \n two\n \n different drug administration methods.\n
\n\n [1]\n
\n\n Any\n \n two\n \n of:\n
\n oral ✓\n
\n inhalation ✓\n
\n topical/applied to the skin ✓\n
\n parenteral/injection ✓\n
\n suppositories ✓\n
\n eye/ear drops ✓\n
\n
\n\n \n One mark for any\n \n two\n \n correct methods.\n
\n \n
\n \n Count multiple methods of injection (intramuscular, subcutaneous, intravenous) as just one method.\n \n
\n\n What is the coefficient for H\n \n +\n \n when the equation below is balanced?\n
\n\n __Pb (s) + __NO\n \n 3\n \n \n −\n \n (aq) + __H\n \n +\n \n (aq) → __Pb\n \n 2+\n \n (aq) + __NO (g) + __H\n \n 2\n \n O (l)\n
\n\n
\n\n A. 2\n
\n\n B. 4\n
\n\n C. 6\n
\n\n D. 8\n
\n\n [1]\n
\n\n D\n
\n\n Only 43% of candidates could balance the equation and determine. Most students failed to realise that the coefficients needed to be doubled, because of the change in the oxidation state of Pb, and answered (B).\n
\n\n Calculate coefficients that balance the equation for the following reaction.\n
\n\n __ Mg\n \n 3\n \n N\n \n 2\n \n (s) + __ H\n \n 2\n \n O (l) → __ Mg(OH)\n \n 2\n \n (s) + __ NH\n \n 3\n \n (aq)\n
\n\n [1]\n
\n\n «1» Mg\n \n 3\n \n N\n \n 2\n \n (s) +\n \n 6\n \n H\n \n 2\n \n O (l) →\n \n 3\n \n Mg(OH)\n \n 2\n \n (s) +\n \n 2\n \n NH\n \n 3\n \n (aq)\n
\n\n This was generally very well done with almost all candidates being able to determine the correct coefficients.\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 \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 Explain why the reaction produces more (CH\n \n 3\n \n )\n \n 3\n \n COH than (CH\n \n 3\n \n )\n \n 2\n \n CHCH\n \n 2\n \n OH.\n
\n\n [2]\n
\n\n carbocation formed from (CH\n \n 3\n \n )\n \n 3\n \n COH is more stable / (CH\n \n 3\n \n )\n \n 3\n \n C\n \n +\n \n is more stable than (CH\n \n 3\n \n )\n \n 2\n \n CHCH\n \n 2\n \n \n +\n \n ✔\n
\n\n
\n «because carbocation has» greater number of alkyl groups/lower charge on the atom/higher e\n \n -\n \n density\n
\n \n \n OR\n \n \n
\n «greater number of alkyl groups» are more electron releasing\n
\n \n \n OR\n \n \n
\n «greater number of alkyl groups creates» greater inductive/+I effect ✔\n
\n
\n\n \n Do\n \n not\n \n award any marks for simply quoting Markovnikov’s rule.\n \n
\n\n Poor performance, particularly in light of past feedback provided in similar questions since there was repeated reference simply to Markovnikov's rule, without any explanation.\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 «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 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
\n\n An example of a steroid is testosterone.\n
\n\n \n
\n
\n\n chromatography ✓\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 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 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
\n\n \n List the three products at the anode from the least to the most oxidized.\n \n
\n\n [1]\n
\n\n \n ethanal < ethanoic acid < carbon dioxide ✔\n \n
\n\n \n \n NOTE: Accept formulas.\n \n
\n \n No ECF from 2aii calculations.\n \n \n
\n Explain why the addition of small amounts of carbon to iron makes the metal harder.\n
\n\n [2]\n
\n\n disrupts the regular arrangement «of iron atoms/ions»\n
\n \n \n OR\n \n \n
\n carbon different size «to iron atoms/ions» ✔\n
\n prevents layers/atoms sliding over each other ✔\n
\n\n The reaction of lithium with water is a redox reaction. Identify the oxidizing agent in the reaction giving a reason.\n
\n\n [1]\n
\n\n H\n \n 2\n \n O\n \n \n AND\n \n \n hydrogen gains electrons «to form H\n \n 2\n \n »\n
\n\n \n \n OR\n \n \n
\n\n H\n \n 2\n \n O\n \n \n AND\n \n \n H oxidation state changed from +1 to 0 ✔\n
\n\n
\n\n \n Accept “H\n \n 2\n \n O\n \n AND\n \n H/H\n \n 2\n \n O is reduced”.\n \n
\n\n State how adding a catalyst to the reaction would impact the enthalpy change of reaction, Δ\n \n H\n \n , and the activation energy,\n \n E\n \n \n a\n \n .\n
\n\n \n
\n [1]\n
\n\n Δ\n \n H\n \n same\n \n \n AND\n \n \n lower\n \n E\n \n \n a\n \n ✓\n
\n\n Which molecules contain two pi (\n \n ) bonds?\n
\n\n I. HCN\n
\n II. H\n \n 2\n \n CO\n \n 3\n \n
\n III. H\n \n 2\n \n C\n \n 2\n \n O\n \n 4\n \n
\n
\n A. I and II only\n
\n B. I and III only\n
\n\n C. II and III only\n
\n\n D. I, II and III\n
\n\n [1]\n
\n\n B\n
\n\n Deduce the order of reaction with respect to hydrogen.\n
\n\n [1]\n
\n\n first order ✔\n
\n\n 4(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.\n
\n\n Which 0.01 mol dm\n \n –3\n \n aqueous solution has the highest pH?\n
\n\n A. HCl\n
\n\n B. H\n \n 2\n \n SO\n \n 4\n \n
\n\n C. NaOH\n
\n\n D. NH\n \n 3\n \n
\n\n [1]\n
\n\n C\n
\n\n Calculate the value of the equilibrium constant,\n \n K\n \n , at 298 K. Use sections 1 and 2 of the data booklet.\n
\n\n [2]\n
\n\n \n ✔\n
\n\n \n ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Accept answers in the range 4.4×10\n \n 5\n \n to 6.2×10\n \n 5\n \n (arises from rounding of ln K).\n \n
\n\n Good performance; for ln\n \n K\n \n calculation in the equation ΔG = RTln\n \n K\n \n , ΔG unit had to be converted from kJ to J. This led to an error of 1000 in the value of ln\n \n K\n \n for some.\n
\n\n State the oxidation state of sulfur in copper (II) sulfate.\n
\n\n [1]\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 \n Outline why the major product, C\n \n 6\n \n H\n \n 5\n \n –CHBr–CH\n \n 3\n \n , can exist in two forms and state the relationship between these forms.\n \n
\n\n
\n\n \n Two forms:\n \n
\n\n \n Relationship:\n \n
\n\n [2]\n
\n\n \n \n Two forms:\n \n
\n chiral/asymmetric carbon\n
\n \n \n OR\n \n \n
\n carbon atom attached to 4 different groups\n \n [✔]\n \n
\n \n
\n
\n\n \n \n Relationship\n \n :\n
\n mirror images\n
\n \n \n OR\n \n \n
\n enantiomers/optical isomers\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept appropriate diagrams for either or both marking points.\n \n \n
\n\n Candidates 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.\n
\n\n Which represents a\n \n p\n \n orbital?\n
\n\n \n
\n [1]\n
\n\n C\n
\n\n A chiral auxiliary used in the synthesis of Taxol is trans-2-phenylcyclohexanol. Deduce and label with asterisks (*) the positions of the\n \n two\n \n chiral carbon atoms in the molecule.\n
\n\n \n
\n [2]\n
\n\n \n
\n one for each carbon correctly identified\n
\n\n
\n\n \n Apply ‘list principle’ if any answer has more than two carbon atoms marked.\n \n
\n\n \n Methane reacts with chlorine in sunlight.\n \n
\n\n \n CH\n \n 4\n \n (g) + Cl\n \n 2\n \n (g) → CH\n \n 3\n \n Cl (g) + HCl (g)\n \n
\n\n \n Which type of reaction occurs?\n \n
\n\n \n A. free-radical substitution\n \n
\n\n \n B. electrophilic substitution\n \n
\n\n \n C. nucleophilic substitution\n \n
\n\n \n D. electrophilic addition\n \n
\n\n [1]\n
\n\n A\n
\n\n 74 % of the candidates chose free-radical substitution as the type of reaction occurring between methane and chlorine in sunlight.\n
\n\n Which species could be reduced to form SO\n \n 2\n \n ?\n
\n\n
\n A. S\n
\n B. H\n \n 2\n \n SO\n \n 3\n \n
\n\n C. H\n \n 2\n \n SO\n \n 4\n \n
\n\n D. (CH\n \n 3\n \n )\n \n 2\n \n S\n
\n\n [1]\n
\n\n C\n
\n\n Photosynthesis enables green plants to store energy from sunlight as glucose.\n
\n\n
\n\n 6CO\n \n 2\n \n (g) + 6H\n \n 2\n \n O (l) → C\n \n 6\n \n H\n \n 12\n \n O\n \n 6\n \n (aq) + 6O\n \n 2\n \n (g) ✓\n
\n\n Which statement describes an endothermic reaction?\n
\n\n
\n A. The bonds broken are stronger than the bonds formed.\n
\n B. The enthalpy of the reactants is higher than the enthalpy of the products.\n
\n\n C. The temperature of the surroundings increases.\n
\n\n D. The products are more stable than the reactants.\n
\n\n [1]\n
\n\n A\n
\n\n Large deposits of methane hydrate, CH\n \n 4\n \n ⋅6H\n \n 2\n \n O (\n \n M\n \n \n r\n \n = 124), have been discovered under the ocean floor. What mass of carbon dioxide would be produced by the complete combustion of 12.4 g of the methane hydrate?\n
\n\n
\n A. 4.40 g\n
\n B. 26.4 g\n
\n\n C. 34.1 g\n
\n\n D. 44.0 g\n
\n\n [1]\n
\n\n A\n
\n\n [Cr(OH\n \n 2\n \n )\n \n 6\n \n ]\n \n 3+\n \n is violet and [Cr(NH\n \n 3\n \n )\n \n 6\n \n ]\n \n 3+\n \n is yellow. What is correct?\n
\n\n The Colour Wheel\n
\n\n \n
\n \n
\n [1]\n
\n\n B\n
\n\n 61% of the candidates were able to determine the relative d-level splitting and the wavelength of light absorbed by complex ions with different ligands, given the colours of the complex ions and a colour wheel labelled with the wavelengths of light.\n
\n\n \n Which volume of ethane gas, in\n \n \n \n , will produce\n \n \n \n of carbon dioxide gas when mixed with\n \n \n \n of oxygen gas, assuming the reaction goes to completion?\n \n
\n\n \n \n \n
\n\n \n A.\n \n
\n \n
\n \n B.\n \n
\n \n
\n \n C.\n \n
\n \n
\n \n D.\n \n \n
\n\n [1]\n
\n\n B\n
\n\n More than 65% of candidates were able to apply coefficient ratios in a stoichiometry question involving only gaseous reactants and products.\n
\n\n \n Suggest a concern about the disposal of solvents from drug manufacturing.\n \n
\n\n [1]\n
\n\n \n \n Any one of:\n \n
\n «most are» toxic «to living organisms»\n
\n \n \n OR\n \n \n
\n incomplete combustion/incineration can produce toxic products/dioxins/phosgene\n
\n \n \n OR\n \n \n
\n carcinogenic/can cause cancer ✔\n
\n \n NOTE: Do\n \n not\n \n accept “harmful to the environment”.\n \n
\n \n
\n \n accumulate in groundwater\n
\n \n \n OR\n \n \n
\n have limited biodegradability ✔\n
\n \n NOTE: Do\n \n not\n \n accept just “pollutes water”.\n \n
\n \n
\n \n cost of disposal ✔\n
\n \n NOTE: Do\n \n not\n \n accept “hazard of disposal”.\n \n
\n \n
\n \n \n NOTE: Accept “ozone depletion” only if there is some reference to chlorinated solvents.\n \n \n
\n\n Determine the equilibrium constant,\n \n K\n \n , for this reaction at 25 °C, referring to section 1 of the data booklet.\n
\n\n If you did not obtain an answer in (c)(iii), use Δ\n \n G\n \n = –43.5 kJ mol\n \n −1\n \n , but this is not the correct answer.\n
\n\n [2]\n
\n\n «ΔG = –41.8 kJ mol\n \n –1\n \n =\n \n × 298 K × ln\n \n K\n \n »\n
\n \n \n OR\n \n \n
\n «ΔG = –41800 J mol\n \n –1\n \n = –8.31 J mol\n \n –1\n \n K\n \n –1\n \n × 298 K × ln\n \n K\n \n »\n
\n
\n «ln\n \n K\n \n = =» 16.9 ✔\n
\n «\n \n K\n \n = e\n \n 16.9\n \n =» 2.19 × 10\n \n 7\n \n ✔\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n \n Accept range of 1.80 × 10\n \n 6\n \n –2.60 × 10\n \n 7\n \n .\n \n
\n\n \n If –43.5 is used then 4.25 × 10\n \n 7\n \n .\n \n
\n\n \n Suggest a source of error in the procedure, assuming no human errors occurred and the balance was accurate.\n \n
\n\n [1]\n
\n\n \n surface area not uniform\n
\n \n \n \n NOTE: Accept “acids impure.\n \n
\n \n
\n \n \n \n OR\n \n \n
\n limestone pieces do not have same composition/source\n
\n \n NOTE: Accept “«limestone» contains impurities”.\n \n \n \n
\n \n
\n \n \n \n OR\n \n \n
\n limestone absorbed water «which increased mass»\n \n
\n \n \n \n OR\n \n \n
\n acid removed from solution when limestone removed\n
\n \n NOTE: Accept “loss of limestone when dried\"\n \n OR\n \n \"loss of limestone due to crumbling when removed from beaker”.\n \n
\n \n
\n \n \n \n OR\n \n \n
\n «some» calcium sulfate deposited on limestone lost\n \n
\n \n \n \n OR\n \n \n
\n pieces of paper towel may have stuck to limestone\n \n
\n \n \n \n OR\n \n \n
\n beakers not covered/evaporation\n \n
\n \n \n \n OR\n \n \n
\n temperature was not controlled ✔\n \n
\n \n Suggest a risk of using sulfuric acid as the catalyst.\n \n
\n\n [1]\n
\n\n corrosive/burns/irritant/strong oxidizing agent/carcinogenic\n
\n \n \n OR\n \n \n
\n disposal is an environmental issue\n
\n \n \n OR\n \n \n
\n causes other side reactions/dehydration/decomposition ✔\n
\n
\n \n Do\n \n not\n \n accept just “risk of accidents”\n \n OR\n \n “health risks”\n \n OR\n \n “hazardous”.\n \n
\n Most students received a mark for this question base on specific hazards. Very few students related disposal to environmental issues which isn't surprising as this is often missed in the Internal Assessment. Weaker students provided vague answers related to health issues which did not receive a mark. Some students misunderstood the question.\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 «\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 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 What are nucleophiles most likely to react with?\n
\n\n
\n A. Alkenes\n
\n B. Alcohol\n
\n\n C. Alkanes\n
\n\n D. Halogenoalkanes\n
\n\n [1]\n
\n\n D\n
\n\n 45% of the candidates knew that nucleophiles were most likely to react with halogenoalkanes. The distractors were chosen with almost equal frequency.\n
\n\n Suggest\n \n one\n \n piece of evidence, other than temperature rise, that shows correlation between greenhouse gas emission and global warming.\n
\n\n [1]\n
\n\n glacier/ice sheets/ retreat/melting\n
\n \n \n OR\n \n \n
\n rising sea levels ✓\n
\n
\n\n \n Accept other reasonable answers, e.g. ocean acidification, severe weather patterns.\n
\n Do not accept just ‘climate change’.\n \n
\n \n Explain how IR spectroscopy can be used to distinguish aspirin from salicylic acid.\n \n
\n\n \n \n \n
\n [2]\n
\n\n salicylic acid contains absorption in the range\n \n ✔\n
\n\n due to phenol/hydroxyl/\n \n group not present in aspirin ✔\n
\n\n
\n \n Award\n \n [2]\n \n for “additional\n \n «stretch» in IR for salicylic acid at higher wavenumber than corresponding\n \n «stretch» in aspirin”\n \n OR\n \n “aspirin has two absorption bands/one stronger absorption band in\n \n while salicylic acid has one/weaker absorption band in that region”.\n \n
\n \n Award\n \n [1 max]\n \n for “fingerprint regions will be different for both”.\n \n
\n\n This question which asked for an explanation of how IR spectroscopy can be used to distinguish aspirin from salicyclic acid was generally very well answered. The majority stated that salicyclic acid contains an absorption in the IR spectrum in the 3200-3600 cm\n \n -1\n \n range due to the phenolic OH group, which is not present in aspirin. A few stated that aspirin has a methyl group and hence the CH stretch will appear in the 2850-3090 cm\n \n -1\n \n region of the IR spectrum in aspirin (using Section 26 of the Data Booklet) which will not appear in the corresponding IR spectrum for salicyclic acid. This is somewhat incorrect as in salicyclic acid the benzene ring will also have CH bonds and the CH stretch for the benzene ring will occur in a similar region of the IR spectrum (as indicated in Section 26 of the Data Booklet) and hence cannot be used to distinguish fully between the two structures per se if using the Data Booklet range. Of course, in practice the alkyl CH stretch would be at a slightly lower wavenumber (e.g. 2850-2950 cm\n \n -1\n \n ) in the IR spectrum compared to the aromatic CH stretch (3030 cm\n \n -1\n \n ), but virtually no candidate gave this type of precise detail.\n
\n\n \n Which graph is obtained from a first order reaction?\n \n
\n\n \n \n \n
\n [1]\n
\n\n B\n
\n\n More than 78 % of candidates correctly identified a first order reaction graph.\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 \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 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 What is the type of bonding in a compound that has high boiling and melting points, poor electrical conductivity, and low solubility in water?\n
\n\n A. Ionic\n
\n\n B. Molecular covalent\n
\n\n C. Metallic\n
\n\n D. Giant covalent\n
\n\n [1]\n
\n\n D\n
\n\n 58% of the candidates identified giant covalent bonding as the bonding in compounds with high melting and boiling points, poor electrical conductivity and low solubility in water. The most commonly selected distractor was ionic bonding. The question discriminated well between high-scoring and low-scoring candidates.\n
\n\n \n What are the products of electrolysis of concentrated aqueous sodium bromide?\n \n
\n\n \n
\n [1]\n
\n\n D\n
\n\n Outline, with reference to the reaction equation, why this sign for the entropy change is expected.\n
\n\n [1]\n
\n\n «forward reaction involves» decrease in number of moles «of gas» ✔\n
\n\n Average performance for sign of the entropy change expected for the reaction. Some answers were based on Δ\n \n G\n \n value rather than in terms of decrease in number of moles of gas or had no idea how to address the question.\n
\n\n \n List the three products at the anode from the least to the most oxidized.\n \n
\n\n [1]\n
\n\n \n ethanal < ethanoic acid < carbon dioxide ✔\n \n
\n\n \n \n NOTE: Accept formulas.\n \n
\n \n No ECF from 2aii calculations.\n \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 \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 This question was well answered.\n
\n\n \n Pure magnesium needed for making alloys can be obtained by electrolysis of molten magnesium chloride.\n \n
\n\n \n \n \n
\n \n © International Baccalaureate Organization 2020.\n
\n \n
\n \n Write the half-equations for the reactions occurring in this electrolysis.\n \n
\n\n \n \n \n
\n [2]\n
\n\n Anode:\n \n ✔\n
\n\n Cathode:\n \n ✔\n
\n\n \n
\n Accept\n \n \n .\n
\n \n Award\n \n [1 max]\n \n for correct equations at incorrect electrodes.\n \n
\n\n This was a very popular option with approximately 34% of candidates attempting Option B. Many students appeared well prepared for this option. Some candidates continue to provide answers with a heavy Biology bias that often make them lose valuable points.\n
\n\n In a redox titration, manganate(VII) ions are reduced to manganese(II) ions and iron(II) ions are oxidized to iron(III) ions.\n
\n\n MnO\n \n 4\n \n \n −\n \n (aq) reduced to Mn\n \n 2+\n \n (aq)\n
\n Fe\n \n 2+\n \n (aq) oxidized to Fe\n \n 3+\n \n (aq)\n
\n
\n What volume, in cm\n \n 3\n \n , of 0.1 mol dm\n \n −3\n \n MnO\n \n 4\n \n \n −\n \n (aq) is required to reach the equivalence point in the titration of 20.00 cm\n \n 3\n \n of 0.1 mol dm\n \n −3\n \n Fe\n \n 2+\n \n (aq)?\n
\n
\n A. 2.00\n
\n B. 4.00\n
\n\n C. 20.00\n
\n\n D. 100.00\n
\n\n [1]\n
\n\n B\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 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
\n \n Nitrogen monoxide reacts with oxygen gas to form nitrogen dioxide.\n \n
\n\n \n The following experimental data was obtained.\n
\n \n
\n \n \n \n
\n \n Deduce the partial order of reaction with respect to nitrogen monoxide and oxygen.\n
\n \n
\n \n \n \n
\n
\n\n
\n\n [2]\n
\n\n \n : second ✔\n
\n \n : first ✔\n
\n Most candidates could correctly deduce the order of each reactant from rate experimental rate data.\n
\n\n Write an equation for the complete combustion of C\n \n 17\n \n H\n \n 36\n \n .\n
\n\n [1]\n
\n\n C\n \n 17\n \n H\n \n 36\n \n (l) + 26O\n \n 2\n \n (g) → 17CO\n \n 2\n \n (g) + 18H\n \n 2\n \n O(l) ✔\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 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 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 Determine the rate expression for the reaction.\n
\n\n \n
\n [2]\n
\n\n \n BrO\n \n 3\n \n \n –\n \n :\n \n 1/first\n \n \n AND\n \n Br\n \n –\n \n :\n \n 1/first\n \n \n AND\n \n H\n \n +\n \n :\n \n 2/second ✓\n
\n\n «Rate =»\n \n k\n \n [BrO\n \n 3\n \n \n −\n \n ][Br\n \n −\n \n ][H\n \n +\n \n ]\n \n 2\n \n ✓\n
\n\n \n
\n M2: Square brackets required for the mark.\n \n
\n Write\n \n two\n \n equations showing how these antacids neutralize excess hydrochloric acid.\n
\n\n
\n\n Magnesium carbonate: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n Aluminium hydroxide: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n
\n\n
\n\n
\n\n [1]\n
\n\n MgCO\n \n 3\n \n (s) + 2HCl(aq) → MgCl\n \n 2\n \n (aq) + H\n \n 2\n \n O(l) + CO\n \n 2\n \n (g)\n
\n \n \n AND\n \n \n
\n Al(OH)\n \n 3\n \n (s) +3HCl(aq) → AlCl\n \n 3\n \n (aq) + 3H\n \n 2\n \n O(l) ✓\n
\n
\n\n \n Accept appropriate ionic equations.\n
\n \n
\n \n Do\n \n not\n \n accept H\n \n 2\n \n CO\n \n 3\n \n as a product of first reaction.\n
\n \n
\n \n Ignore equilibrium arrows.\n \n
\n\n \n Calculate the percentage uncertainty of the day 5 titre.\n \n
\n\n [1]\n
\n\n \n «\n \n \n \n «%»✔\n \n
\n\n Predict, giving a reason, the major product of reaction between but-1-ene and steam.\n
\n\n [2]\n
\n\n CH\n \n 3\n \n CH\n \n 2\n \n CH(OH)CH\n \n 3\n \n ✔\n
\n\n «secondary» carbocation/CH\n \n 3\n \n CH\n \n 2\n \n CH\n \n +\n \n CH\n \n 3\n \n more stable ✔\n
\n\n
\n\n \n Do\n \n not\n \n accept “Markovnikov’s rule” without reference to carbocation stability.\n \n
\n\n Product 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.\n
\n\n State how adding a catalyst to the reaction would impact the enthalpy change of reaction, Δ\n \n H\n \n , and the activation energy,\n \n E\n \n \n a\n \n .\n
\n\n \n
\n [1]\n
\n\n Δ\n \n H\n \n same\n \n \n AND\n \n \n lower\n \n E\n \n \n a\n \n ✓\n
\n\n In which of the following species would sulfur be reduced if converted to SCl\n \n 2\n \n ?\n
\n\n A. S\n \n 2\n \n O\n \n 3\n \n \n 2–\n \n
\n\n B. H\n \n 2\n \n S\n
\n\n C. S\n
\n\n D. SO\n \n 2\n \n
\n\n [1]\n
\n\n D\n
\n\n What are the products when dilute aqueous copper (II) nitrate is electrolysed using platinum electrodes?\n
\n\n E\n \n ⦵\n \n (Cu | Cu\n \n 2+\n \n ) = –0.34 V.\n
\n\n \n
\n [1]\n
\n\n A\n
\n\n Average performance with a lower discrimination index and no clear misconception based on the incorrect choices.\n
\n\n What is the preferred IUPAC name of the structure shown?\n
\n\n \n
\n
\n A. 2-ethyl-3-methylbutan-1-ol\n
\n B. 2,3-dimethylbutan-2-ol\n
\n\n C. 1-ethyl-2-methylpropan-1-ol\n
\n\n D. 1,1,2-trimethylpropan-1-ol\n
\n\n [1]\n
\n\n B\n
\n\n Which mechanism does the nitration of benzene proceed by?\n
\n\n
\n A. electrophilic addition\n
\n B. electrophilic substitution\n
\n\n C. nucleophilic addition\n
\n\n D. nucleophilic substitution\n
\n\n [1]\n
\n\n B\n
\n\n Describe how viruses differ from bacteria.\n
\n\n [1]\n
\n\n viruses have no nucleus but bacteria do\n
\n \n \n OR\n \n \n
\n viruses have no organelles but bacteria do\n
\n \n \n OR\n \n \n
\n viruses cannot perform biological functions/reproduce without a host but bacteria can\n
\n \n \n OR\n \n \n
\n viruses are not cells but bacteria are\n
\n \n \n OR\n \n \n
\n viruses are smaller than bacteria ✓\n
\n
\n\n \n Accept specific structural differences. Such as “bacteria have cytoplasm/ribosomes but viruses do not\" or \"viruses can have double stranded RNA/single stranded DNA whereas bacteria always have double stranded DNA\".\n \n
\n\n \n Do\n \n not\n \n accept “Bacteria are living and viruses are not”.\n \n
\n\n Calculate the initial pH before any sodium hydroxide was added, using section 21 of the data booklet.\n
\n\n [2]\n
\n\n «\n \n K\n \n a = 10\n \n –2.87\n \n = 1.35 × 10\n \n –3\n \n »\n
\n\n «1.35 × 10\n \n –3\n \n =\n \n »\n
\n\n «x = [H\n \n +\n \n ] =\n \n =» 2.6 × 10\n \n –2\n \n «mol dm\n \n –3\n \n » ✔\n
\n\n
\n «pH = –log[H\n \n +\n \n ] = –log(2.6 × 10\n \n –2\n \n ) =» 1.59 ✔\n
\n
\n\n \n Accept final answer in range 1.58–1.60.\n \n
\n\n \n Award\n \n [2]\n \n for correct final answer.\n \n
\n\n Deduce the overall rate equation.\n
\n\n [2]\n
\n\n H\n \n +\n \n : second order\n
\n \n \n AND\n \n \n
\n BrO\n \n 3\n \n \n −\n \n : first order\n
\n \n \n AND\n \n \n
\n Br\n \n −\n \n : first order ✓\n
\n «rate =» k[Br\n \n −\n \n ][BrO\n \n 3\n \n \n −\n \n ][H\n \n +\n \n ]\n \n 2\n \n ✓\n
\n\n
\n\n \n Award\n \n [2]\n \n for correct rate equation.\n \n
\n\n Sketch an energy profile for the decomposition of calcium carbonate based on your answer to b(i), labelling the axes and activation energy,\n \n E\n \n \n a\n \n .\n
\n\n \n
\n [3]\n
\n\n endothermic sketch ✓\n
\n\n x-axis labelled “extent of reaction/progress of reaction/reaction coordinate/reaction pathway”\n \n \n AND\n \n \n y-axis labelled “potential energy/energy/enthalpy✓\n
\n\n activation energy/\n \n E\n \n \n a\n \n ✓\n
\n\n \n
\n \n
\n Do\n \n not\n \n accept “time” for x-axis.\n \n
\n Which is a possible empirical formula for a substance with\n \n M\n \n \n r\n \n = 42?\n
\n\n
\n A. CH\n
\n B. CH\n \n 2\n \n
\n\n C. C\n \n 3\n \n H\n \n 6\n \n
\n\n D. C\n \n 3\n \n H\n \n 8\n \n
\n\n [1]\n
\n\n B\n
\n\n This question needed careful reading. The majority of candidates chose distractor C (C\n \n 3\n \n H\n \n 6\n \n ) which had an\n \n M\n \n \n r\n \n of 42. But what the question asked for was the empirical formula of the compound. Only 40% of the candidates chose the correct answer (CH\n \n 2\n \n ). The question discriminated well between high-scoring and low-scoring candidates. However, it was the only question on this paper for which the correct answer was not the most commonly chosen answer.\n
\n\n Aspirin is most commonly used as a mild analgesic. State\n \n two\n \n other common medical uses for aspirin.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n\n reduce fever/antipyretic ✓\n
\n\n anti-inflammatory ✓\n
\n\n anti-coagulant/reduces blood clotting/blood thinner\n
\n \n \n OR\n \n \n
\n prevent cardiovascular disease/stroke ✓\n
\n
\n\n \n Accept \"prevents/reduces «risk of» heart attack\" for M3.\n
\n \n
\n \n Accept \"prevents heart disease\" for M3.\n
\n \n
\n \n Accept “may reduce colon/colorectal cancer” for M3.\n \n
\n\n Aspirin is most commonly used as a mild analgesic. State\n \n two\n \n other common medical uses for aspirin.\n
\n\n [2]\n
\n\n \n Any two of:\n \n
\n\n reduce fever/antipyretic ✓\n
\n\n anti-inflammatory ✓\n
\n\n anti-coagulant/reduces blood clotting/blood thinner\n
\n \n \n OR\n \n \n
\n prevent cardiovascular disease/stroke ✓\n
\n
\n\n \n Accept \"prevents/reduces «risk of» heart attack\" for M3.\n
\n \n
\n \n Accept \"prevents heart disease\" for M3.\n
\n \n
\n \n Accept “may reduce colon/colorectal cancer” for M3.\n \n
\n\n \n Identify a conjugate acid–base pair in the equation.\n \n
\n\n [1]\n
\n\n \n C\n \n 6\n \n H\n \n 8\n \n O\n \n 7\n \n \n \n AND\n \n \n C\n \n 6\n \n H\n \n 7\n \n O\n \n 7\n \n \n −\n \n
\n \n \n OR\n \n \n
\n H\n \n 2\n \n O\n \n \n AND\n \n \n H\n \n 3\n \n O\n \n +\n \n ✔\n \n
\n Outline how the toxicity of xenobiotics is reduced using host–guest chemistry.\n
\n\n [1]\n
\n\n host molecule/super molecule forms complex/bond with guest/xenobiotic «facilitating their removal»\n
\n \n \n OR\n \n \n
\n shape/size of host matches guest/xenobiotic\n \n \n AND\n \n \n binds ✓\n
\n \n Comment on the magnitudes of random and systematic errors in this experiment using the answers in (e).\n \n
\n\n [2]\n
\n\n \n Any two:\n \n
\n\n large percentage error means large systematic error «in procedure» ✔\n
\n\n small percentage uncertainty means small random errors ✔\n
\n\n random errors smaller than systematic error ✔\n
\n\n \n
\n \n \n Award\n \n [2]\n \n for “both random and systematic errors are significant.”\n \n
\n Of the many students that obtained the mark most did through the first alternative and a lesser percentage through the third. Many students were unable to relate their calculations from 2e (percentage error and percentage uncertainty) to systematic error and random error. They either compared the calculations to incorrect errors or in some cases did not discuss the errors at all. Once again this points to a general lack on experimental understanding.\n
\n\n The diagram shows the energy profile of a reaction.\n
\n\n \n
\n
\n Which combination is correct?\n
\n
\n \n
\n [1]\n
\n\n B\n
\n\n \n State the number of\n \n 1\n \n H NMR signals for this isomer of xylene and the ratio in which they appear.\n \n
\n\n
\n\n [2]\n
\n\n \n \n Number of signals\n \n : 2\n \n [\n \n \n ✔\n \n \n ]\n \n \n
\n\n \n \n \n Ratio\n \n :\n \n \n
\n\n \n \n 3 : 2\n \n \n
\n\n \n \n \n OR\n \n \n \n
\n\n \n \n 6 : 4\n \n [\n \n \n ✔\n \n \n ]\n \n \n \n
\n\n \n \n \n \n Note\n \n : Accept any correct integer or fractional ratio. Accept ratios in reverse order.\n \n \n \n
\n\n Many identified two correct peaks but quite a few less the correct ratio.\n
\n\n When sodium carbonate powder is added to ethanoic acid, the beaker becomes cooler.\n
\n\n Possible enthalpy diagrams are shown.\n
\n\n \n
\n Which correctly describes the reaction?\n
\n\n \n
\n [1]\n
\n\n C\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 (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 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
\n\n What is the amount, in mol, of H\n \n 2\n \n O produced for a reaction between 10.0 mol of C\n \n 2\n \n H\n \n 3\n \n Cl and 10.0 mol of O\n \n 2\n \n if the yield is 90 %?\n
\n\n 2C\n \n 2\n \n H\n \n 3\n \n Cl (g) + 5O\n \n 2\n \n (g) → 4CO\n \n 2\n \n (g) + 2H\n \n 2\n \n O (g) + 2HCl (g)\n
\n\n
\n A. 3.60\n
\n B. 4.00\n
\n\n C. 9.00\n
\n\n D. 10.00\n
\n\n [1]\n
\n\n A\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 Incorrect\n \n \n AND\n \n \n two readings, uncertainty is ±0.1 ✔\n
\n\n \n \n Justify this hypothesis.\n \n \n
\n\n [1]\n
\n\n \n sulfuric acid is diprotic/contains two H\n \n +\n \n «while nitric acid contains one H\n \n +\n \n »/releases more H\n \n +\n \n «so reacts with more limestone»\n
\n \n \n OR\n \n
\n \n higher concentration of protons/H\n \n +\n \n ✔\n \n
\n \n \n NOTE: Ignore any reference to the relative strengths of sulfuric acid and nitric acid.\n
\n Accept “sulfuric acid has two hydrogens «whereas nitric has one»”.\n
\n Accept \"dibasic\" for \"diprotic\".\n \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 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 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 the melting points of saturated and unsaturated fats.\n \n \n
\n\n A classic instance of candidates answering the question they thought (or hoped?) they had been asked rather than the one that was asked. Almost all answers referred to the differing amounts of saturated and unsaturated fatty acids present, totally ignoring the fact that the question clearly stated “\n \n their saturated fatty acid composition\n \n ”, where the relative lengths of the chains was the key point. Nevertheless some who went on to discuss the nature of the intermolecular forces between the chains gained some credit.\n
\n\n Write a balanced equation for the reaction that occurs.\n
\n\n [1]\n
\n\n 2 Mg(s) + O\n \n 2\n \n (g) → 2 MgO(s) ✔\n
\n\n \n
\n Do not accept equilibrium arrows. Ignore state symbols\n \n
\n This was not as well done as one might have expected with the most common errors being O instead of O\n \n 2\n \n oxygen and MgO rather than MgO\n \n 2\n \n .\n
\n\n Sketch the graphs on the axes to show the expected results of this experiment.\n
\n\n \n
\n [2]\n
\n\n separate curves/lines for Li and K sketched\n \n \n AND\n \n \n both increasing ✔\n
\n steeper gradient for Li\n
\n \n \n OR\n \n \n
\n curve/line for Li higher ✔\n
\n What is the name of the compound with formula Ti\n \n 3\n \n (PO\n \n 4\n \n )\n \n 2\n \n ?\n
\n\n A. Titanium phosphate\n
\n\n B. Titanium(II) phosphate\n
\n\n C. Titanium(III) phosphate\n
\n\n D. Titanium(IV) phosphate\n
\n\n [1]\n
\n\n B\n
\n\n 71% of candidates were able to name Ti\n \n 3\n \n (PO\n \n 4\n \n )\n \n 2\n \n correctly. The most commonly chosen distractor was titanium(III) phosphate.\n
\n\n \n Dinitrogen monoxide has a positive standard enthalpy of formation, Δ\n \n H\n \n \n f\n \n \n \n θ\n \n \n .\n \n
\n\n \n Deduce, giving reasons, whether altering the temperature would change the\n \n \n spontaneity of the\n \n decomposition\n \n reaction.\n \n
\n\n [3]\n
\n\n \n exothermic decomposition\n
\n \n \n OR\n \n \n
\n Δ\n \n H\n \n \n (decomposition)\n \n < 0\n \n [✔]\n \n \n
\n \n \n TΔS\n \n \n θ\n \n > Δ\n \n H\n \n \n \n \n θ\n
\n \n \n \n \n \n \n \n OR\n
\n \n \n \n \n Δ\n \n G\n \n \n \n \n θ\n \n \n \n «= Δ\n \n H\n \n \n \n \n θ\n \n \n \n –\n \n TΔS\n \n \n \n \n θ\n \n \n \n » < 0 «at all temperatures»\n \n [\n \n \n ✔]\n \n \n
\n \n reaction spontaneous at all temperatures\n \n [\n \n \n ✔]\n \n \n
\n\n Though the question asked for decomposition (in bold), most candidates ignored this and worked on the basis of a the Δ\n \n H\n \n 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 Δ\n \n H\n \n (probably taking it for granted) nor explicitly relate Δ\n \n G\n \n and spontaneity, which left the examiner with no possibility of evaluating their reasoning.\n
\n\n Write the equation for the reaction of Ca(OH)\n \n 2\n \n (aq) with hydrochloric acid, HCl (aq).\n
\n\n [1]\n
\n\n Ca(OH)\n \n 2\n \n (aq) + 2HCl (aq) → 2H\n \n 2\n \n O (l) + CaCl\n \n 2\n \n (aq) ✓\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 Outline, giving a reason, the effect of a catalyst on a reaction.\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 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 \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 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 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 Outline the requirements for a collision between reactants to yield products.\n
\n\n [2]\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 Generally 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.\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 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 Good performance on how the polarity of C-X bond changes going down group 17.\n
\n\n Which will eventually yield the greatest mass of deposited copper in the electrolysis of a fixed volume of 1 mol dm\n \n −3\n \n CuSO\n \n 4\n \n (aq)?\n
\n\n
\n A. copper anode and inert cathode\n
\n B. inert anode and copper cathode\n
\n\n C. inert anode and inert cathode\n
\n\n D. zinc anode and zinc cathode\n
\n\n [1]\n
\n\n A\n
\n\n Some nitride ions are\n \n 15\n \n N\n \n 3–\n \n . State the term that describes the relationship between\n \n 14\n \n N\n \n 3–\n \n and\n \n 15\n \n N\n \n 3–\n \n .\n
\n\n [1]\n
\n\n \n isotope\n \n «s» ✔\n
\n\n Identification of isotopes was answered correctly by most students.\n
\n\n \n Explain why the rate of reaction of limestone with nitric acid decreases and reaches zero over the period of five days.\n \n
\n\n [2]\n
\n\n \n acid used up\n
\n \n OR\n \n
\n acid is the limiting reactant ✔\n \n
\n \n concentration of acid decreases\n
\n \n OR\n \n
\n less frequent collisions ✔\n \n
\n \n \n NOTE: Award\n \n [1 max]\n \n for \"surface area decreases\" if the idea that CaCO\n \n 3\n \n is used up/acts as the limiting reactant” is conveyed for M1.\n \n \n
\n\n \n \n Do\n \n not\n \n accept “reaction reaches equilibrium” for M2.\n \n \n
\n\n \n Lecithin aids the body’s absorption of vitamin E.\n \n
\n\n \n \n \n
\n \n \n Suggest why vitamin E is fat-soluble.\n \n \n
\n\n [1]\n
\n\n \n long non-polar/hydrocarbon chain «and only one hydroxyl group»\n
\n \n \n OR\n \n \n
\n forms London/dispersion/van der Waals/vdW interactions with fat\n \n [✔]\n \n \n
\n
\n\n \n \n \n Note:\n \n Accept “alcohol/hydroxy/OH” for “hydroxyl” but\n \n not\n \n “hydroxide”.\n \n \n
\n\n This question was also not answered well even though it has appeared on previous tests. Many students missed the idea of a long or large non-polar chain when describing the structure.\n
\n\n \n Distinguish between a weak and strong acid.\n \n
\n\n
\n\n \n Weak acid:\n \n
\n\n \n Strong acid:\n \n
\n\n [1]\n
\n\n \n \n Weak acid\n \n : partially dissociated/ionized «in solution/water»\n
\n \n \n AND\n \n \n
\n \n Strong acid\n \n : «assumed to be almost» completely/100 % dissociated/ionized «in solution/water»\n \n [✔]\n \n \n
\n It was rather disappointing that less than 70 % of the candidates could distinguish between weak and strong acids. Many candidates referred to pH differences.\n
\n\n \n Which atom is sp\n \n 2\n \n hybridized?\n
\n \n
\n \n A. C in H\n \n 2\n \n CO\n
\n \n
\n \n B. C in CO\n \n 2\n \n
\n \n
\n \n C. N in CH\n \n 3\n \n NH\n \n 2\n \n
\n \n
\n \n D. O in H\n \n 2\n \n O\n \n
\n\n [1]\n
\n\n A\n
\n\n \n Identify the most suitable indicator for the titration using section 22 of the data booklet.\n \n
\n\n [1]\n
\n\n \n phenolphthalein ✔\n \n
\n\n \n \n NOTE: Accept phenol red.\n \n \n
\n\n What is thin-layer chromatography best used for separating?\n
\n\n
\n A. molecules of varying polarity\n
\n B. molecules of similar polarity\n
\n\n C. metals in an alloy\n
\n\n D. water of crystallization from hydrated salts\n
\n\n [1]\n
\n\n A\n
\n\n Outline why ionization energies have positive values but most electron affinities have negative values\n
\n\n [1]\n
\n\n ionization energy breaks bond/attractive force between nucleus and electron\n
\n \n \n AND\n \n \n
\n electron affinity forms bond/attractive force between nucleus and electron ✓\n
\n
\n \n Accept for ionization energy “energy needed/endothermic to remove an electron\".\n
\n \n \n AND\n \n \n
\n for electron affinity “energy released/exothermic adding an electron”.\n \n
\n Draw a Lewis (electron dot) structure for ozone.\n
\n\n [1]\n
\n\n \n ✔\n
\n \n Accept any combination of lines, dots or crosses to represent electrons.\n \n
\n\n \n Do\n \n not\n \n accept structures that represent 1.5 bonds.\n \n
\n\n Write the rate expression for this reaction.\n
\n\n [1]\n
\n\n rate =\n \n k\n \n [N\n \n 2\n \n O\n \n 5\n \n ] ✔\n
\n\n \n Suggest why a non-polar solvent was needed.\n \n
\n\n [1]\n
\n\n oil is non-polar «and dissolves best in non-polar solvents»\n
\n \n \n OR\n \n \n
\n oil does not dissolve in polar solvents ✔\n
\n \n Do\n \n not\n \n accept “like dissolves like” only.\n \n
\n\n A 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.\n
\n\n \n Determine the standard entropy change, in J K−1, for the decomposition of dinitrogen monoxide.\n \n
\n\n \n 2N\n \n 2\n \n O (g) → 2N\n \n 2\n \n (g) + O\n \n 2\n \n (g)\n \n
\n\n \n \n \n
\n [2]\n
\n\n Δ\n \n S\n \n θ\n \n = 2(S\n \n θ\n \n (N\n \n 2\n \n )) + S\n \n θ\n \n (O\n \n 2\n \n ) – 2(S\n \n \n \n θ\n \n \n \n (N\n \n 2\n \n O))\n
\n \n \n \n \n OR\n
\n \n \n \n Δ\n \n S\n \n \n \n θ\n \n \n \n = 2 × 193 «J mol\n \n -1\n \n K\n \n -1\n \n » + 205 «J mol\n \n -1\n \n K\n \n -1\n \n » – 2 × 220 «J mol\n \n -1\n \n K\n \n -1\n \n »\n \n [✔]\n \n \n
\n \n «ΔS\n \n \n \n θ\n \n \n \n = +»151 «J K\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 Candidates were able to calculate the ΔS of the reaction, though in some cases they failed to multiply by the number of moles.\n
\n\n State the full electronic configuration of Fe\n \n 2+\n \n .\n
\n\n [1]\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 3d\n \n 6\n \n ✔\n
\n\n Mostly well done which was a pleasant surprise since this is not overly easy, predictably some gave [Ar] 4s\n \n 2\n \n 3d\n \n 4\n \n .\n
\n\n Which statements explain the following reactions occurring in the upper atmosphere?\n
\n\n
\n\n \n
\n [1]\n
\n\n D\n
\n