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) 𝜂𝐻2𝑃𝑂4− = 2.00 𝑔 136.1 𝑔/𝑚𝑜𝑙 = 0.0147 mol hence [H2PO4–] = 0.0147 mol dm–3 [1] 𝜂𝐻𝑃𝑂42− = 1.42𝑔 142.0 𝑔/𝑚𝑜𝑙 = 0.0100 mol hence [HPO42–] = 0.0100 mol dm–3 [1] Ka(H2PO4–) = [𝐻+][𝐻𝑃𝑂42−] [𝐻2𝑃𝑂4−] = 10–7.20 [𝐻+](0.0100) (0.0147) = 10−7.20 hence [H+] = 9.275 × 10–8 mol dm–3 hence pH = 7.03 [1] (iv) H2PO4- + OH- → HPO42- + H2O (v) I: 10.0 cm3 pH = 7.20 + log (0.0011 V⁄ ) (0.00137 V⁄ ) = 7.10 [1] II: 200 cm3 [OH-] = 0.000530 0.3 = 0.001768 [1] pOH = 2.75 pH = 14 – 2.75 = 11.2 [1] H2PO4- + OH- → HPO42- + H2O Before rxn /mol 0.00147 0.0001 0.001 [1] Change -0.0001 -0.0001 +0.0001 After rxn /mol 0.00137 0 0.0011 [1] H2PO4- + OH- → HPO42- + H2O Before rxn /mol 0.00147 0.002 0.001 Change -0.00147 -0.00147 +0.00147 After rxn /mol 0 0.000530 [1] 0.00247

5 (b) (i) 2Na2HPO4 + NaH2PO4 ⟶ Na5P3O10 + 2H2O (ii) tetrahedral (iii) Ca 2+ O – P O OO P O – P O – O O – O O – overall 3– charge okay (iv) [Ca2+] present = 5.06 × 10−5 𝑔/𝑑𝑚3 40.1 𝑔/𝑚𝑜𝑙 = 1.2618 × 10–6 mol dm–3 [1] Ksp(Ca3(PO4)2) = [Ca2+]3[PO43–]2 = 2.07 × 10–33 mol5 dm–15 [𝐶𝑎2+]3(6.53 × 10−8)2 = 2.07 × 10−33 hence maximum [Ca2+] allowed in solution = 7.859254 × 10–7 mol dm–3 [1] hence 𝜂𝐶𝑎2+ to remove = 𝜂𝑁𝑎5𝑃3𝑂10 to add = 1.2618 × 10–6 – 7.859254 × 10–7 = 4.7587 × 10–7 mol mass of Na5P3O10 to add = 4.7587 × 10–7 mol × 368.0 g/mol = 1.75 × 10–4 g [1]

6 4 (a) I O OH O OH H J O O K OH L OH O (b) Information Deduction A, C6H10O5, is a chiral compound A has a carbon with 4 different groups / atoms attached to it When 1 mol of A is reacted with excess solid sodium carbonate at room temperature and pressure, 24 dm 3 of CO2 is formed. Acid-carbonate reaction A has 2 -COOH groups A does not react with hot acidifed potassium dichromate(VI) A cannot be oxidised A reacts with excess hot concentrated H2SO4 to form B, C6H8O4 elimination A has a tertiary alcohol (since cannot be oxidised but can undergo elimination) B has an alkene group 1 mol of B reacts completely with 2 mol of NaOH(aq) B has 2 -COOH groups acid-base reaction When B is heated with acidified KMnO4(aq), C, C 3H4O3, and D, C 3H4O4, are formed. oxidative cleavage / oxidation C gives a yellow precipitate with warm aqueous alkaline iodine C has CH3CO- oxidation Both C and D reacts with magnesium to give effervescence Both C and D have -COOH group

7 O OH CH3 OH O OH A O OH CH3 O OH B CH3 O O OH C O OH O OH D Section B 5 (a) (i) pTVT = p1V1 + p2V2 pT(200) = 155(150) + 80(120) [1] sub pT = 164 kPa [1] ans (ii) nucleophilic substitution, SN1 3 electron-donating methyl groups disperse positive charge stabilising the carbocation 3 methyl groups sterically hinder nucleophile approach from opposite C–Cl [2] 4 / 4 [1] 2 / 4 (iii) SN1: C Cl CH3 CH3 CH3 slow C + CH3 CH3 CH3 Cl – NH3 C N + CH3 CH3 CH3 H H H Cl – C NH2 CH3 CH3 CH3 ClH no need to name again [3] 10 / 11 [2] 7 / 11 [1] 4 / 11 - 3D tetrahedral reactant molecule - 𝛿+ and 𝛿 − - arrow from C–Cl bond to Cl - slow - trigonal planar intermediate with C+ - arrow from N lone pair to C+ - C–NH3+ intermediate - arrow from Cl– ion lone pair to H - arrow from N–H bond to N+ - 3D tetrahedral product molecule - HCl by-product molecule

8 SN2 (accept only if (ii) is SN2): C Cl CH3 CH3 CH3 H3N C CH3 CH3 CH3 Cl C CH3 CH3CH3 N + HH H Cl – C CH3 CH3CH3 NH2 ClH no need to name again [3] 10 / 11 [2] 7 / 11 [1] 4 / 11 (iv) NH N N + Cl – use limiting amount of (CH3)3Cl / use excess NH3 OR HCl (if HCl given as answer, no possible method to limit the formation) any one (b) (i) [1] correct 2 species [1] dipoles on HF, correctly labelled (ii) HF is a Brønsted acid as it lost / released H+ (iii) anode: 2F– ⟶ F2 + 2e– cathode: 2H+ + 2e– ⟶ H2 electrode labels required (iv) 𝜂𝐾𝐻𝐹2 = 355 𝑔 78.1 𝑔/𝑚𝑜𝑙 = 4.545 mol 𝜂𝐻𝐹 = 145 𝑔 20.0 𝑔/𝑚𝑜𝑙 = 7.25 mol [1] from KHF2: H3N ǂ - 3D tetrahedral reactant molecule - 𝛿+ and 𝛿 − - arrow from N lone pair to C - arrow from C–Cl bond to Cl - trigonal bipyramidal transition state with dotted lines - square brackets, dagger - C–NH3+ intermediate - arrow from Cl– ion lone pair to H - arrow from N–H bond to N+ - 3D tetrahedral product molecule - HCl by-product molecule 𝛿+ 𝛿– F H F – ion-dipole

9 𝜂𝐻+ discharged = 𝜂𝐹− discharged = 4.545 mol from HF: 𝜂𝐻+ discharged = 𝜂𝐹− discharged = 7.25 mol hence total 𝜂𝐻+ discharged = total 𝜂𝐹− discharged = 4.545 mol + 7.25 mol = 11.795 mol [1] since 𝜂𝐻2 𝜂𝐻+ = 𝜂𝐹2 𝜂𝐹− = 1 2, 𝜂𝐻2 = 𝜂𝐹2 = 11.795 𝑚𝑜𝑙 2 = 5.8977 mol pV = nRT volume of each gas H2