RI Prelim P2 (Ans)

1 © Raffles Institution 2020 9729/02/S/20 2020 RI H2 Chemistry Prelim Paper 2 – Suggested Solutions 1(a)(i) oxidation number of C(1): +1 oxidation number of C(2): –1 oxidation number of C(3): +3 1(a)(ii) Disproportionation 1(b)(i) H2(g) + ½O2(g)  H2O(g) ∆Hf(H2O(g)) = BE(H–H) + ½ BE(O=O) – 2BE(O–H) = 436 + ½(496) – 2(460) = –236 kJ mol–1 1(b)(ii) By Hess’ Law, ∆Hr = +44 – 2(–87.0) – (– 236) –160.7 – 385.2 = –91.9 kJ mol–1 1(b)(iii) equation 2 2C6H5CHO(l) + H2O(l)  C6H5CH2OH(l) + C6H5COOH(s) ∆S = 216.7 + 167.6 – 2(221.2) – 70 = –128.1 J K–1 mol–1 ∆G = ∆Hr – T∆S = –91.9 – 298(–128.1/1000) = –53.7 kJ mol–1 ∆G < 0, hence the reaction is feasible.

2 © Raffles Institution 2020 9729/02/S/20 2(a)(i) As the titration progresses, [Ag+(aq)] decreases. If AgCl and AgBr are not filtered away before titration, the ionic products of the silver halides fall below their respective Ksp values, causing the silver halide precipitates to dissolve and produce more Ag+ ions. OR the position of equilibria (I) and (II) will shift to the left, causing the silver halide precipitates to dissolve and produce more Ag+ ions. Ag+(aq) + Cl(aq) ⇌ AgCl(s) ----- (I) Ag+(aq) + Br(aq) ⇌ AgBr(s) ----- (II) More SCN ions will be required to react with these additional Ag+ ions (from the dissolution of silver halide precipitates) and the titre value will be larger than expected. 2(a)(ii) Amount of remaining Ag+ ions in 10.0 cm3 filtrate = Amount of SCN ions = 14.25/1000  0.100 = 1.425  103 = 1.43  103 mol 2(a)(iii) Amount of remaining Ag+ ions in 40 cm3 of mixture after precipitation = 40.0/10.0  1.425  103 = 5.70  103 mol Initial amount of Ag+ ions added = 0.700  15.0/1000 = 1.05  102 mol Amount of Ag+ ions that reacted with the halide ions = (10.5  5.70)  103 = 4.80  103 mol Total concentration of halide ions = 4.80  103 ÷ 25.0/1000 = 0.192 mol dm3 Since sample X contains equimolar amounts of Cl– and Br– ions, concentration of Cl– ions = 0.192 / 2 = 0.0960 mol dm3 (shown) 2(b)(i) At the end of the separation, a saturated solution of AgCl is formed. Hence, [Ag+][Cl–] = Ksp = 1.77  10–10 [Ag+] = 1.77  10–10 ÷ 0.0960 = 1.844  10–9 = 1.84  10–9 mol dm3 2(b)(ii) At the end of the separation, the solution is still a saturated solution of AgBr. Hence, [Ag+][Br–] = Ksp = 5.35  10–13 [Br–] = 5.35  10–13 ÷ (1.844  10–9) = 2.90  10–4 mol dm3 2(b)(iii) Amount of Ag+ ions in solution = 1.844  10–9  25.0/1000 = 4.61  10–11 mol Amount of Ag+ ions used to precipitate AgBr = (0.0960  2.90  10–4)  25.0/1000