RI 2018 Prelim P1 Ans

1 2018 Y6 H2 Chemistry Preliminary Examinations Paper 1 (Suggested Solutions) Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C A C C C A B B D C A C A D B Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer A C D B D A A B D D C B D D A Q1(C) Since the Ar of Fe is closest to 55.94, the most abundant isotope is 56Fe   abundance of 56Fe = 91.8  abundance of 54Fe & 58Fe = 100 – 91.8 = 8.2% Let the  abundance of 54Fe be x%  abundance of 58Fe = 8.2 – x % = z% Ar of Fe = (53.94)(x) + (55.94)(91.8) + (8.2−x)(57.93) 100 = 55.849  abundance of 54Fe = x = 6.37 %  abundance of 58Fe = 8.2 – x = 1.83 % Q2(A) S2Clx disproportionates completely in water to give S:SO2 = 3:1 = 6:2 Writing and balancing half-equations: [R] 3S2Clx + (3x)e− ⎯→ 6S + (3x)Cl− [O] S2Clx + 4H2O ⎯→ 2SO2 + (8−x)e− + 8H+ + xCl− electrons loss = electrons gain 3x = 8−x x = 2 Q3(C) Angle of deflection ∝ charge mass A Angle of deflection ∝ 1/(15 + 15) = 0.0333 B Angle of deflection ∝ 3/(12 + 16 + 16) = 0.0682 C Angle of deflection ∝ 2/(12 +16) = 0.0714 (greatest value) D Angle of deflection ∝ 2/(16 + 18) = 0.0588 Q4(C) Mass number of E = 241 − 4 = 237 Number of protons in E = 95 − 2 = 93 Number of neutrons in E = 237 − 93 = 144 Q5(C) A Only G has an aldehyde functional group that can be oxidised by potassium dichromate (VI). B Both F and G exhibit intramolecular hydrogen bonding C Both F and G have 8 𝜋 electrons (6 in benzene ring and 2 in C=O). D The carbon on the methyl side chain of F is sp 3 hybridised and tetrahedral in shape, hence F is not planar. G is planar as all the carbon atoms are sp2 hybridised (trigonal planar) and the O atom has a bent shape. Q6(A) Manipulate the ideal gas equation, pV = nRT A p T = nR V = nR (1 V) Since p T ∝ ( 1 V), A is correct. B V T = nR P = constant The graph should be a vertical straight line. C pV = m M RT, where n = m M p = (m V) RT M Since 𝜌 = m V , p = 𝜌 RT M 𝜌 = ( M RT) p Since M RT is constant, gradient is constant and it should be a straight line through origin. D pV = nRT = (nR)T The y-intercept should be 0, i.e. a straight line through origin, as temperature is measured in Kelvins. Q7(B) There are 2 propagation steps that lead to the formation of bromomethane, which explains the presence of 2 peaks (i.e. 2 transition states). The 2 propagation steps are: (a) CH4 + Cl• ⎯→ •CH3 + HCl ∆Ha = [BE(C−H)−BE(H−Cl)] =410−431= −21 kJ mol−1

2 (b) •CH3 + Cl2 ⎯→ CH3Cl + Cl• ∆Hb= [BE(Cl−Cl)−BE(C−Cl)] = 244−340= −96 kJ mol−1 ∆Hprop = ∆Ha + ∆Hb= −117 kJ mol−1 Hence, the propagation step is exothermic. The products formed are at a lower energy level than the reactants. Q8(B) A The maximum attainable oxidation number of each element corresponds to the number of valence electron(s) in each atom of the element, i.e. all the valence electrons can be used for bonding. B MgO has a higher melting point than Na2O. Recall that qqstrength of ionic bonds L.E. r +r +− +− . The lattice energy of MgO will be more exothermic than Na 2O as Mg 2+ has a higher charge and smaller cationic radius than Na+. C The pH of the aqueous solutions of the chlori des decreases across the Period (refer to lecture notes on ‘The Periodic Table (I)’). D NaCl and MgCl2 are ionic compounds while AlCl3, SiCl4 and PCl5 are simple molecules. Q9(D) A Cl2 molecule has less electrons than P 4 molecule. Hence C l2 has weaker instantaneous dipole - induced dipole interactions and has a lower melting point. B Na has a lower melting point than Mg due to weaker metallic bonding in Na. C S8 molecule has more electrons than P4 molecule . Hence S8 has stronger instantaneous dipole - induced dipole interactions and has a highe r melting point. D S8 molecule has more electrons than P4 molecule . Hence S8 has stronger instantaneous dipole - induced dipole interactions and has a highe r melting point. SiO2 is also the only insoluble oxide amongst the three elements. Q10(C) By inspection, rate = k[NO2][SO2]. Since [SO2] >> [NO 2] in all 3 experiments, this is a pseudo first -order reaction and rate = k’[NO2], where k’ = k[SO2]. t1 2⁄ = ln 2 k' = ln 2 k[SO2] Since the [SO2] in expt 1 and 2 are the same, the half - life should be the same at 48 s. Since [SO 2] in expt 3 is twice that in expt 1, the half-life would be halved to 24 s. Q11(A) The time taken for the pink colour to fade also depends on “how pink” the solution was, i.e. [ M], at the start of each experiment. So rate is directly proportional to [M]/t. 1 Comparing expt 1 & 2, when [M] doubled, the rate also doubled.  order of rxn wrt M is 1. 2 Comparing expt 1 & 3, when [OH−] halved, the rate also halved.  order of rxn wrt OH− is 1. 3 In expt 4, the volumes were all double that of expt 1, so the concentrations of all reactants in expt 4 were the same as expt 1. So the time taken should be 100 s. 12(C) A A catalyst will increase the rate of reaction but does not affect the composition of the mixture at equilibrium, i.e. [S] remain unchanged. B Kp changes with temperature. C Compression of reaction mixture causes the partial pressures of all gases to increase. Since there are fewer gas particles on the product side, the equilibrium position will shift right to produce more S and [S] will increase. Kp only changes with temperature. The rates of the forward and backward reactions increase because the particles are closer to one another and the freq uency of effective collisions increase. D At constant volume, the partial pressure s (and hence concentration s) of all species remain unchanged and the equilibrium position will not shift, , i.e. [S] remain unchanged. Q13(A) From both graphs , the yield of NH3 decreases as the temperature increases. This means that as temperature increases, the backward reaction is favoured. Hence the backward reaction is endothermic and the forward reaction must be exothermic. (x < 0) Comparing the two graphs at any given t emperature, p1 gives a higher yield of NH3 than p2. Since there are fewer gas particles on the product side , an increase in pressure will favour the forward reaction which decreases the total pressure of the system. Hence, p1 > p2. Q14(D) [OH−] = √C0 × Kb = √C0 × Kw Ka = √0.2 × 1.0 × 10−14 5.6 × 10−10 pH = 14 − pOH = 14 − (− lg √0.2 × 1.0 × 10−14 5.6 × 10−10) = 14 + lg √0.2 × 1.0 × 10−14 5.6 × 10−10

3 Q15(B) A buffer solution is produced when there is a weak acid/base and its conjugate base/acid in solution. 1 Excess of PO43− reacts with limited H+, resulting in HPO42− and PO43− in solution. (Buffer) 2 Excess OH− reacts with limited CH3CO2H, resulting in CH3COO− and OH− in solution. (Not a buffer) 3 Excess HO 2C–CO2H reacts with limited OH −, resulting in HO2C–CO2− and HO2C–CO2H. (Buffer) Q16(A) A The ligand, edta, has a charge of 4−. Since overall charge is 2−, oxidation number of Fe = +2. B The ligand, CO, has a charge of 0. Since overall charge is 0, oxidation number of Fe = 0. C The ligand, phenoxide ion, has a charge of 1−. Since overall charge is 3−, oxidation number of Fe = +3. D The ligand, NCS, has a charge of 1−, while H2O has a charge of 0. Since overall charge is 2+, oxidation number of Fe = +3. Q17(C) The number of d-electrons in copper remains the same because there is no change in oxidation state of copper. Since there is a colour change, the wavelength of light absorbed must be different and that must be due to a change in energy gap between the d-orbitals. Q18(D) H2, Ni reduc es alkene and carbonyl functional group s, but not carboxylic acid functional group. The product contains 9 chiral centres and has 2 9 stereoisomers. Q19(B) U V 1 2 3 Options (1) and (3) are correct.

4 Q20(D