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

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