2022 HCI Prelim P1 Worked Solutions

2022 HCI C2 H2 Chemistry Prelims / Paper 4 HWA CHONG INSTITUTION 2022 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS (PAPER 1) Paper 1 1 2 3 4 5 6 7 8 9 10 C B B D D D C A B A 11 12 13 14 15 16 17 18 19 20 B A A B B C C D B A 21 22 23 24 25 26 27 28 29 30 A C A D C A D D C D 1 C Option D can be eliminated immediately as He contains 2 protons. Option No. of protons No. of neutrons No. of electrons A 1 0 2 B 1 1 3 C 1 2 3 2 B The large drop in the graph suggests that A + ion (which has already lost its first electron) has electronic configuration of ns 1. Hence A originally has electronic configuration of ns 2 and is a Group 2 element. Since the graph is showing consecutive elements in the Period Table, B is a Group 13 element. 3 B Option Explanation A Methanol forms favourable hydrogen bonding between its –OH group and that in water. B The hydrogen atom is not bonded to F, O or N atom, hence there is no intermolecular hydrogen bonding for HBr and HCl. C For 2-aminophenol, the –NH2 and the –OH groups are in close proximity which allows for intramolecular hydrogen bonding. Hence there is less extensive intermolecular hydrogen bonding for 2- aminophenol and less energy is required to overcome the intermolecular forces, leading to a lower boiling point. D When ethanoic acid, CH3CO2H (Mr = 60), is dissolved in a non-polar solvent (hexane), the molecules form hydrogen-bonded dimers.

4 D Option Explanation 1 about 107 109.5 2 about 104.5 90 3 90 180 5 D Recall that real gases would deviate from ideal behaviour at high pressures and low temperatures. Option Explanation A HCl is a polar molecule, which will deviate less at higher temperature since the molecules have more kinetic energy to overcome the intermolecular permanent dipole -permanent dipole (pd -pd) interactions. B CH4 is a non-polar molecule with only weak intermolecular dispersion forces. CH 4 will also deviate less at higher temperature since the molecules have more kinetic energy to overcome the intermolecular forces. C O2 is a non-polar molecule with only weak intermolecular dispersion forces, hence should deviate less from ideal behaviour than HCl. D HF is a polar molecule with stronger hydrogen bonding interactions than the permanent dipole-permanent dipole interactions of HCl. Hence HF deviates more from ideal behaviour than HCl at the same temperature of 300 K. 6 D An acid -base indicator is a weak aqueous acid whose acid form, HIn, is a different colour from its ionised form (or conjugate base form), In –. As such, it shows different colours in solutions of different pH. HIn(aq) ⇌ H+(aq) + In–(aq) red yellow

2022 HCI C2 H2 Chemistry Prelims / Paper 4 7 C When the metal nitrate decomposes, metal oxide, nitrogen dioxide and oxygen are formed. M(NO3)2 d MO + 2NO2 + ½O2 The metal oxide then reacts with the added acid to form salt and water. MO + 2HCl d MCl2 + H2O The excess HCl will undergo neutralisation with NaOH. HCl + NaOH d NaCl + H2O Amount of NaOH = 0.0780  0.0216 = 0.001685 mol = amount of excess HCl Amount of HCl used to dissolve residue = 0.0350  0.080 = 0.00280 mol Amount of HCl reacted with MO = 0.00280 − 0.001685 = 0.001115 mol Amount of MO = 0.001115 / 2 = 0.0005575 mol = Amount of M(NO3)2 Mr of M(NO3)2 = 0.118 / 0.0005575 = 211.7 Ar of M = 211.7 – 2(14 + 16  3) = 87.7 Checking the periodic table, M is Sr. 8 A Statement Explanation 1 Down the group, the size of electron cloud and hence polarisability of the halogen molecule increases. More energy is required to overcome stronger intermolecular dispersion forces. Hence, the volatility of halogens decreases down Group 17. 2 Large anions have high polarisability as their electron cloud is less attracted by the nucleus and can be more easily distorted by a cation. Hence, there is a higher degree of covalent character in the silver halide ionic compounds down Group 17. 3 As the atomic radius increases down the group, the bond length of H−X increases and thus bond strength decreases. Hence, the H−X bond energy decreases down Group 17 and less energy is required to break the bond and HX decomposes more readily. 9 B In the experiment with benzoic acid, No. of moles of benzoic acid combusted = 0.986 / 122 = 0.008082 q = −Hc × no. of moles of benzoic acid = 3054 × 0.008082 = 24.68 kJ Heat capacity of the bomb calorimeter = 24.68 / 2.14 kJ C−1 In the experiment with glycine, No. of moles of glycine combusted = 0.483 / 75 = 0.00644 q = (24.68 / 2.14)  0.54 = 6.228 kJ Enthalpy change of combustion of glycine = −6.228/ 0.00644 = −967 kJ mol−1

10 A This decomposition is endothermic since energy is required for breaking the covalent bonds in the OH− anion. S is expected to be positive since there is formation of gaseous H 2O, thus statement 3 is wrong. G can be positive or negative depending on the temperature. For reactions to be spontaneous, G needs to be negative. As G = H − TS, T has to be high so that |TS| > H. Thus statement 2 is correct. Similar to Group 2 carbonates, the thermal decomposition of Group 2 hydroxides is dependent on the charge density of the cation. Since Ba2+ is larger in cationic radius as compared to Mg2+, the charge density of Ba2+ is smaller. Hence its polarising power is weaker and is less able to distort the electron cloud of the hydroxide ion, weakening the O -H covalent bond to a smaller extent. Thus, barium hydroxide will decompose at a higher temperature than magnesium hydroxide. 11 B Comparing the first two sets of experiments, when concentration of bromine doubles while the concentrations of other reactants remain unchanged, the rate of reaction remains the same. Hence, it should be zero order with respect to Br2. Comparing the first and third sets of experiments, when concentration of H + doubles while the concentrations of other reactants remain unchanged, rate doubles. Hence it is first order with respect to H+. Comparing the third and fourth sets of experiments, when the concentration of H+ doubles and the concentration of propanone increases 4/3 times, the rate of reaction increases by (2 × 4/3) times. Hence it is first order with respect to propanone. Option Explanation A Rate equation should be rate = k[CH3COCH3][H+]. B The rate constant value may be obtained by substituting the rate and concentrations from any set of experiments. C Rate constant is only affected by temperature or addition of catalyst. D Overall order is not 1. Hence half lives may not be the same across different experiments.

2022 HCI C2 H2 Chemistry Prelims / Paper 4 12 A Recall Haber process: N2(g) + 3H2 ⇌ 2NH3(g) The forward reaction is exothermic. Option Explanation A High temperature will favour the endothermic backward reaction. Hence the yield will decrease but the rate of the reaction will increase. B At lower temperature, Kp increases. There will be more NH3 formed and less N2 and H2 present. C Higher pressure will favour the forward reaction to produce less number of gaseous molecules. Hence the yield will increase. D The presence of catalyst does not affect the yield. 13 A CH3CO2H + CH3CH2OH ⇌ CH3CO2CH2CH3 + H2O initial amt/ mol 0.50 0.50 − − change/ mol −x −x +x +x eqm amt/ mol 0.50 − x 0.50 − x x x Kc = (x/V)2 / ((0.50 – x)/V)2 = 4.0 Taking square root on both sides, x = 0.33 Hence amount of CH3CO2H = 0.50 – 0.33 = 0.17 14 B pH = −lg[H+] = 7 pKw = −lg[H+][OH−] = 14 pKa = −lg [𝐻+][𝑂𝐻−] [𝐻2𝑂] = −lg((10−7)2/(55.6)) = 15.7 15 B Recall that a buffer should be made up of a weak acid and its conjugate base or a weak base and its conjugate acid. Option Explanation A This mixture consists of NaCl(aq) (a salt) and HCl(aq) (a strong acid). Thus this is not a buffer solution. B HCl + NH3 → NH4C