YJC H2 CHEM P3 Answer Prelim

1 9647 / YJC / 2012 / JC2 Preliminary Examinations / Paper 3 answers YJC 2014 Prelim Paper 3 answers 1 (a) Na2O : dissolves readily in H 2O to form a strongly alkaline solution with pH = 13 (accept 11 – 14) . Na2O + H2O → 2NaOH SO3 : dissolves in H 2O forming a strongly acidic solution with pH = 2 (accept 0 – 3) . SO3 + H2O → H2SO4 (b) (i) Buffer solutions resist pH changes when a small amount of acid or alkali is added. (ii) Small amount of H+: CO3 2− + H+ (or H3O+) → HCO3 − ( + H2O) The H+ added is removed by the large reservoir of CO3 2−. Small amount of OH−: HCO3 − + OH− → CO3 2− + H2O The OH− added is removed by the large reservoir of HCO3 − (iii) Buffer most effective when [HCO 3 −] = [CO 3 2−] (maximum buffering capacity), using Hasselbalch equation, pH = pKa = −lg (5.61 × 10−11) = 10.3 Range of effectiveness = pKa ±1 = 9.3 to 11.3 (iv) By Hasselbalch equation, pH = pKa + lg ][ ]2   3 3 HCO [CO When ][ ]2   3 3 HCO [CO increases from 0.50 to 0.85, pH change = lg (0.85) – lg (0.50) = −0.0706 – (−0.301) = +0.230 OR ratio 0.5, pH = 9.95 ratio 0.85, pH = 10.18 Difference = 10.18 – 9.95 = 0.230 (c) (i) Mr AlCl3 = [27 + 3(35.5)] = 133.5 Mr (AlCl3)n = 267, n = 2 Molecular formula is Al2Cl6. Cl Cl Cl Cl Cl Cl Al Al x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x xx xxxx xx xx xx

2 9647 / YJC / 2012 / JC2 Preliminary Examinations / Paper 3 answers (ii) Al2O3 (iii) When large amount of water added, aluminium chloride forms Al(H2O)6 3+. Al(H2O)6 3+ (+ H2O) ⇌ Al(H2O)5(OH)2+ + H+ (or H3O+) pH = 3.0 (d) (i) Electrophilic substitution (ii) Step 1: Cl2 + AlCl3 → Cl+ + AlCl4 − Step 2: Step 3: (iii) 2 (a) Shape: Tetrahedral Angle: 109.5° (b) (i) (ii) (iii) (iv) CH3 Cl + CH3 Cl H + CH3 Cl H + AlCl4 - CH3 Cl + HCl + AlCl3 ClCl Cl S O O OO xx xx H H x x x x x x x x xx xx xx xx x x x x

3 9647 / YJC / 2012 / JC2 Preliminary Examinations / Paper 3 answers For (iii) and (iv), accept if Br/OH is in 3rd position of cyclic ring instead of 2nd. (c) (i) Eᶱcell = EPbO2/Pb2+ − EPb2+/Pb = +1.47 – (−0.13) = +1.60 V (ii) 2 PbSO4(s) + 2 H2O(l) → Pb(s) + PbO2(s) + H2SO4(aq) + HSO4 −(aq) + H+(aq) OR 2 PbSO4(s) + 2 H2O(l) → Pb(s) + PbO2(s) + 2 H2SO4(aq) OR 2 PbSO4(s) + 2 H2O(l) → Pb(s) + PbO2(s) + 2 HSO4 −(aq) + 2 H+(aq) (iii) As time passes, more of the reactants are being used up, causing the concentration of the reactants (H2SO4) to decrease. As a result, Eᶱcell decreases. (iv) Hydrogen fuel cell . The reactants are constantly replenished as it is supplied from the air, thus e.m.f. remains constant. . (d) (i) Ksp = [Pb2+][CrO4 2−] = s2 = 1.69 × 10−14 mol2 dm−6 (where s is the solubility of PbCrO4.) s = 1.30 × 10−7 mol dm−3 (ii) Precipitation occurs when ionic product > Ksp, so the maximum ionic product before precipitation = Ksp. [Pb2+] [0.01] = 1.69 × 10−14 [Pb2+]saturation = 1.69 × 10−12 mol dm−3 (iii) 2CrO4 2− + 2H+ ⇌ Cr2O7 2− + H2O In the presence of acid, the H+ concentration increases, causing the equilibrium to shift to the right by Le Chatelier’s Principle, with CrO4 2− forming Cr2O7 2−. In the presence of alkali, [H+] decreases as it is neutralized by OH −, resulting in the equilibrium shifting left to replenish the H+ by LCP, thus more CrO4 2− is formed. (e) (i) Pb(OH)2.PbCO3(s) → 2PbO(s) + H2O(g) + CO2(g) (ii) Mr Pb(OH)2.PbCO3 = [2(207) + 5(16.0) + 2(1) + 12.0] = 508 1 mole white lead produces 1 mole CO2 and 1 mole H2O, so… Mass (in g) loss per mole white lead = Mr CO2 + Mr H2O = 62.0 Hence, % loss in mass = (62.0 / 508) × 100% = 12.2% CH3 CH3 CH2 CH3 CH3 O CH3 O CH3 CH3 Br Br CH3 CH3 OH OH

4 9647 / YJC / 2012 / JC2 Preliminary Examinations / Paper 3 answers (iii) Decomposition temperature of Ca(OH) 2.CaCO3 would be lower than that of white lead, as…  Ionic radius of Ca 2+ (0.099 nm) < Ionic radius of Pb 2+ (0.120 nm) (students need not state values)  Charge density of Ca2+ > charge density of Pb2+  Ca2+ polarizes OH− and CO3 2− electrons cloud to a greater extent  Bond energy of O−H and O−C bond weakened  Less energy needed to overcome the bond energies to decompose. Penalised if OH−/CO3 2− not mentioned in decomposition. 3 (a) (i) k = (ln 2) / t½ = (ln 2) / 5 = 0.139 min−1 (ii) Final [R-Br] = (6.25 / 100) × 1.6 = 0.1 mol dm−3 1.6 → 0.8 → 0.4 → 0.2 → 0.1 4 t½ must be drawn. . (b) (i) SN2 reaction occurs because the 2 methyl groups do not pose much stearic hindrance, while SN1 may occur as the 2 electron-donating methyl groups can stabilize the carbocation. OR SN1 reaction occurs because the 2 methyl groups pose significant stearic hindrance , while SN2 may occur as the 2 electron -donating methyl groups may not stab ilize the carbocation. (ii) (iii) For [OH−] = 0.01 mol dm−3, % rate = %10024.001.07.4 01.07.4   = 16.4% CH3 Br H CH3 OH- δ+ δ- CH3 H CH3 BrHO - CH3 OH CH3 H + Br -

5 9647 / YJC / 2012 / JC2 Preliminary Examinations / Paper 3 answers For [OH−] = 0.1 mol dm−3, % rate = %10024.01.07.4 1.07.4   = 66.2% For [OH−] = 1.0 mol dm−3, % rate = %10024.07.4 7.4  = 95.1% (iv) The higher the concentration of OH −, the higher the percentage rate for S N2, as the second order kinetics rate is dependent / proportional to [OH−]. (v) (vi) (c) (i) Boiling point increases from C2H5Cl → C2H5Br → C2H5I . As size of molecule (HX) increases, number of electrons increases, leading to greater tendency to polarize and instantaneous-dipole-induced-dipole attraction strength increases . Thus more energy is required to overcome to id-id. (ii) Atomic radius: I > Br > Cl. Thus bond energy: C – I < C – Br < C – Cl Thus C2H5I would react with nucleophile more readily than C2H5Br (than C 2H5Cl) as the bond is more easily overcome. (d) (i) Electrode I. (ii) Tix+ + xe− → Ti m M It zF  (CH3)2CHBr (CH3)2CHCOOH (CH3)2CHCN alcoholic KCN heat dilute HCl heat energy Ea2 Number of molecules Ea1

6 9647 / YJC / 2012 / JC2 Preliminary Examinations / Paper 3 answers 45.0 9.47 )5400)(5.0( )96500( x x = 2.98 ≈ 3 OR nTi = 31039.99.47 45.0  mol Q = It = (0.5A)(90 × 60s) = 2700C If 9.39 × 10−3 mol Ti requires 2700C, 1 mol Ti would require 31039.9 2700  = 287 400 C x = Amount of electrons per mole of Ti = 96500 287400 = 2.98 ≈ 3 4 (a) (i) ∆Hf = 6(−394) + 3(−286) +3054 = −168 kJ mol−1 (ii) ∆G = ∆H − T∆S = −168 – (298)(−0.384) = −53.6 kJ mol−1 (b) (i) 2,4,6-trichlorophenol has a larger Ka value than phenol. 3 electron withdrawing Cl further disperses the negative charge of the phenoxide ion, hence stabilizing the phenoxide ion to a larger extent, thus acidity (and K a) increases. (ii) Add aqueous Br 2 (or conc HNO 3). Phenol will decolourise brown Br 2 to colourless while forming a white precipitate (or for conc HNO 3, yellow precipitate) , while 2,4,6 - tricholorophenol would not. Full mark given only for correct test and results. (c) (i) (ii) Sn, conc HCl, heat, followed by NaOH(aq). (iii) Q may form intramolecular hydrogen bonding between OH and NH 2, which would make unavailable the lone pairs or hydrogen for hydrogen bonding with water. 1-napthol will only form intermolecular bonding with H2O. 6C(s) + 3H2(g) + 15/2 O2(g) Hf OH 6CO2(g) + 3H2O(l) 6(-394) 3(-286) -3054 OH NO2

7 9647 / YJC / 2012 / JC2 Preliminary Examinations / Paper 3 answers (d) (i) More accurate if ionic bond indicated between O− and Cu2+. Penalised if phenol instead