JJC H2 Chemistry 9746 Prelim P3 Answer Scheme

Suggested Answers for 9746/H2 Chemistry Preliminary Examination Paper 3 1. (a) (i) • forward reaction is exothermic • low temperature ⇒ high yield of SO3 • but equilibrium reached at a lower rate • a moderately high temperature is used to achieve equilibrium more rapidly (ii) 2SO2 (g) + O2(g) = 2SO3(g) Initial partial pressure / atm 2 1 − Eqm partial pressure / atm 0.02(2) = 0.04 0.02(1) = 0.02 0.98(2) = 1.96 (0.02)(0.04) (1.96) 2 2 p =K = 1.20 × 105 atm−1 (iii) (effect on yield) • when volume is reduced, pressure is increased • equilibrium position shifts right to the side with fewer gaseous molecules • percentage conversion of SO 2 into SO3 increases (effect on Kp) • no change in temperature so Kp remain unchanged (b) (i) (high melting points of Na2O and Al2O3) • giant ionic structure • strong electrostatic attraction between oppositely charged ions in Na2O and Al2O3 (low melting point of SO2) • simple molecular structure • weak van der Waals’ forces (or weak permanent dipole-permanent dipole interactions) between molecules (Al2O3 has higher melting point than Na2O) • stronger ionic bond • A l3+ has higher charge density than Na+ Page 1 of 11

1. (b) (ii) (Na2O) • reacts vigorously with water • solution with pH = 13-14 • Na 2O + H2O → 2NaOH (Al2O3) • insoluble in water (SO2) • reacts with water • solution with pH = 1-3 • SO 2 + H2O → H2SO3 (c) (i) A Page 2 of 11 CH3 3 CH3 OCOCH B or C B or C CH3 OH CH3O2N CH3 OH CH3 NO2 E or D D or E CH3 CH3 CH2 CH3 (ii) an electron-deficient species which attacks electron-rich benzene ring (iii) acid HNO3 + H2SO4 = NO2 + + H2O+ HSO4 − (iv) phenol has an electron-donating −OH group which strongly activates the benzene ring

2. (a) (i) Ag(s) + 21 N2(g) + 23 O2(g) → AgNO3(aq) (ii) ΔHrxn = [4(+106) + 4(−207) + 90 + 33 + 3(−286)] − 6(−207) = +103 kJ mol−1 (iii) • ΔG = ΔH − TΔS • Since ΔH>0 and ΔG<0 (i.e. spontaneous), TΔS must be as positive as ΔH, so ΔS>0 • increase in disorder because disor der achieved in breaking the lattice into ions is greater than the orde ring of water molecules around the ions (b) Amount of silver produced = 108 1.44 = 1.33 × 10−2 mol Since Ag ≡ e, amount of e = 1.33 × 10−2 mol Q = I × time = n × F 10 × (t × 60) = 1.33 × 10−2 × 96500 ⇒ t = 2.14 mins Since X ≡ 3e, Amount of X produced = 1.33 × 10−2 ÷ 3 Ar of X = 3) 10 (1.33 0.120 2 ÷ ×− = 27.0 (c) (i) • white ppt of AgC l when AgNO3 is added • dissolves to form a colourless solution of Ag(NH 3)2 + when excess NH 3 is then added. AgCl(s) = Ag+(aq) + Cl−(aq) -------------(*) • due to formation of complex Ag(NH 3)2 +, [Ag+(aq)] decreases • equilibrium position in (*) shift right, increasing solubility of AgCl (ii) AgI Ksp(AgI) is smaller than Ksp(AgCl) 2. (d) (i) nucleophilic addition Page 3 of 11

(ii) optical isomerism C CH2COCH3 HHO C CH2COCH3 H OH non-superimposable mirror images (iii) O HBr CH CHCOCH3 + Br O δ+ δ− H C H C COCH3 H + Br OH−+ C CC O C H 3 OH Br HH (iv) CH2CH2CHCH3 OH acidified K2Cr2O7, heat under reflux 3. (a) (i) a solution which resists pH change when a small amount of acid or base is added Page 4 of 11

(ii) ] CO [H ] ][H [HCO 3 2 3 a +− =K pH = 7.4 ⇒ [H+] = 10 −7.4 mol dm−3 = 3.98 x 10 −8 mol dm−3 ] CO [H ]( [HCO 3 2 3 )-7.40 7 - 1010 7.90 − = × 19.810 3.98 10 7.90 ] CO [H ] [HCO 8 7 3 2 3 =× ×= − −− • ratio is high • HCO 3 − is needed to remove lactic acid in the blood (iii) H+(aq) + HCO3 −(aq) → H2CO3(aq) (iv) When lactic acid is removed, H2CO3(aq) = CO2(aq) + H2O(l) ----- (1) [H2CO3] increases ⇒ equilibrium position in (1) shifts right, forming more CO2(aq) CO 2(aq) = CO2(g) ----- (2) [CO2(aq)] increases ⇒ equilibrium position in (2) shifts right, forming more CO2(g) and thus resulting in higher rate of breathing (b) (i) Ksp(CaCO3) = [Ca2+][CO3 2−] (ii) CaCO3(s) = Ca2+(aq) + CO3 2− (aq) ------- (3) At eqm: − x x where x mol dm−3 is the solubility of CaCO3. [Ca 2+] = [CO 3 2−] = 9.35 x 10−5 mol dm−3, Ksp(CaCO3) = (9.35 x 10 −5)2 = 8.74 x 10−9 mol2 dm−6 (b) (iii) BeCO3 is not stable due to • high charge density of Be 2+ • Be 2+ has a high polarising power Be(OH)2 is relatively more stable as • OH − is smaller than CO3 2− • OH − is less polarisable by Be2+ Page 5 of 11

3. (c) (i) CH3CHCO2H OH HBr, heat under reflux CH3CHC O OHCl conc./excess NH3, heat under reflux in ethanol CH3CHCOOH NH2 OR HCl, ZnCl2, heat under reflux CH3CHC O OHBr intermediate G (ii) CH3CHCO2H CH3CHCH2OH ⏐ ⏐ Page 6 of 11 H H C = C H CH2OH Q OH OH P hot Al2O3 LiAlH4 in dry ether (reduction) • On passing P over hot Al2O3 ⇒ P undergoes dehydration/elimination to give Q • Possible structures of Q are CH2=CHCH2OH, CH3CH=CH(OH) and CH3(OH)C=CH2 • Q decolourises Br2 ⇒ Q contains a C=C bond/is an alkene and Q undergoes electrophilic addition • Q does not react with 2,4-DNPH ⇒ Q is not a carbonyl compound/ not an aldehyde or ketone. (or not an enol, CH3CH=CH(OH) or CH3(OH)C=CH2 which can be rearranged to give carbonyl group) • Q undergoes vigorous oxidation with hot conc. acidified KMnO4 to gives only CO2 ⇒ Q is CH2=CHCH2OH. Q is oxidised to CO2 and (COOH)2 which then further oxidises to CO2 and H2O

4. (a) (i) A homogeneous catalyst • is in the same phase as the reactants and products • increases the rate of reaction by providing an alternative reaction pathway/mechanism of a lower activation energy than that of the uncatalysed reaction (ii) iron is able to vary its oxidation stat es in its compounds or Fe can exist as Fe2+ or Fe3+ (iii) Step 1: 2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq) E, cell = (+0.77) – (+0.54) = +0.23 V > 0 (energetically feasible) Step 2: S2O8 2− (aq) + 2Fe2+(aq) → 2SO4 2− (aq) + 2Fe3+(aq) E, cell = (+2.01) – (+0.77) = +1.24 V > 0 (energetically feasible) (iv) [Fe(CN)6]3− + e− = [Fe(CN)6]4− E, = +0.36 V E, value not within +0.54 V and +2.01 V so the intermediate compound, [Fe(CN)6]4−, cannot be formed [Fe(CN)6]3− is negatively charged so ac tivation energy is high due to electrostatic repulsion between like charges (v) • ligands causes the d orbitals to sp lit into 2 sets of different energies • difference in energies between these 2 sets of d orbitals is so small • such that radiation from the visible region of the electromagnetic spectrum • when an electron moves from a d or bital of lower energy to another unfilled/partially filled d orbital of higher energy • Hence, complexes are coloured as the colour seen is the complement of the colours absorbed (b) I2 + 2e = 2I− E, = +0.54 V Cr2O7 2− + 14H+ + 6e = 2Cr3+ + 7H2O E, = +1.33 V E, cell = (+1.33) – (+0.54) = +0.79 V > 0 (energetically feasible) Overall eqn: Cr 2O7 2− + 6I− + 14H+ → 2Cr3+ + 3I2 + 7H2O (c) (i) brown I2(aq) decolourises and pale yellow ppt forms CHI3 and O CH3CHCH2C O− CH3 Page 7 of 11

CH3C = CHCHCH3 CH3 OH 4. (c) (ii) • R undergoes substitution with PCl5 ⇒ R contains –OH group / is an alcohol • R undergoes ester formation/esterification