2022 MI H2 Chem Prelim Paper 3 - Answer

2022 Preliminary Examination Pre-University 3 H2 CHEMISTRY 9729/03 Paper 3 Free Response Section A 1 (a) (i) [3] 7 / 7 [2] 5 / 7 [1] 3 / 7 max [1]: battery or mixing of solutions salt bridge (ii) anode: Zn ⟶ Zn2+ + 2e– cathode: NO3– + 2H+ + e– ⟶ NO2 + H2O [1] sum: Zn + 2NO3– + 4H+ ⟶ Zn2+ + 2NO2 + 2H2O or Zn + 2HNO3 + H2SO4 ⟶ ZnSO4 + 2NO2 + 2H2O [1] (iii) Ecell = + 0.81 – (- 0.76) = +1.57 V (iv) concentration of H+ / NO3– higher than 1.00 mol dm–3 / temperature higher and reaction is endothermic / NO2 gas formed escapes the solution [1] equilibrium position of cathode half-equation shifts right, higher Ecell [1]

2 (b) (i) (ii) ΔHreaction = –2(33.2) + 9.2 = –57.2 kJ mol –1 (iii) bond is formed between two NO2 energy released / exothermic (iv) decrease in number of moles of gases [1] decreased ways to arrange molecules (1) and distribute energy (2) decreased disorder (3) hence decreased entropy (4) [1] (v) ΔG = ΔH – TΔS at high T, –TΔS (positive) has larger magnitude than ΔH (negative) ΔG > 0 / more positive at low T, –TΔS (positive) has smaller magnitude than ΔH (negative) ΔG < 0 / more negative (vi) 2NO2(g) ⇌ N2O4(g) pinitial / bar 1.00 0 pchange / bar –0.702 +0.351 pfinal / bar 0.298 0.351 Kp = 𝑝𝑁𝑂2 𝑝𝑁2𝑂4 2 = 0.351 (0.298)2 = 3.95 bar–1 N O O O N O 2NO2(g) N2O4(g) the elements +2(33.2) +9.2 ΔHreaction

3 2 (a) (i) octahedral, 90° [1] 6 electron domains maximise distance from one another to minimise repulsion [1] (ii) ligands approach at 3 axes () z2 and x2-y2 point at 3 axes, greater electronic repulsion, higher energy xy, yz and xz point away from the 3 axes / diagonal, less electronic repulsion, lower energy [2] 6 / 6 [1] 3 / 6 (iii) less electronic repulsion as z2 is further away from the ligand / z2 points directly at axial ligands (iv) Co2+: [Ar] 3d7 yes, decreased energy of z2 electron Ni2+: [Ar] 3d8 no net energy change, decreased z2 electron energy, increased x2-y2 electron energy or even number of electrons [2] 4 / 4 [1] 2 / 4 (v) Cu+: [Ar] 3d10 Since the d orbitals are fully -filled, there is no d-d transition. Hence, the complex ions are not coloured. (b) Cu2+ + 2e– ⟶ Cu E = +0.34 V Cu + 4NH3 ⟶ [Cu(NH3)4]2+ + 2e– E = +0.05 V [1] Ecell = +0.34 + 0.05 = +0.39 V [1] ΔG = –nFΔEcell = −(2)(96500)(+0.39) = –75.3 kJ mol–1 < 0, hence feasible [1]

4 3 (a) (i) Buffer solutions are solutions which resist pH changes when small amounts of acids or bases are added to it. (ii) H+ (that tries to leave) are more strongly attracted to the more negatively-charged ion more difficult dissociation (iii