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 = 2.39  10–3 mol Total amount of Ag+ ions added = (2.39  10–3) + (4.61  10–11) = 2.39  10–3 mol Mass of AgNO3 added = 2.39  10–3  169.9 = 0.407 g

3 © Raffles Institution 2020 9729/02/S/20 2(c)(i) Let the solubility of Ag2CrO4 in water be s mol dm–3. Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42−(aq) eqm conc 2s s Ksp = 1.12  10–12 = [Ag+]2 [CrO42−] = 4s3 s = 6.54  10–5 mol dm–3 2(c)(ii) Let the solubility of Ag2CrO4 in K2CrO4 solution be s’ mol dm–3. [K2CrO4] = 0.100 / 0.500 = 0.200 mol dm–3 Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42−(aq) eqm conc 2s’ s’ + 0.2 Ksp = 1.12  10−12 = (2s’)2(s’ + 0.200) Due to common ion effect, the solubility is reduced. Hence, assume (s’ + 0.200) ≈ 0.200 1.12  10−12 = (2s’)2(0.200) s’ = 1.18  10−6 mol dm−3 2(c)(iii) Difference in solubility = (6.54  10–5) – (1.18 x 10−6) = 6.42  10−5 mol dm–3 Amount of Ag2CrO4 precipitated = 6.42  10−5  0.500 = 3.21  10–5 mol Mass of Ag2CrO4 precipitated = 3.21  10–5  331.8 = 0.0107 g 3(a)(i) CO(NH2)2 + 2OH−  2NH3 + CO32– Since the initial concentration of CO(NH2)2 (limiting reagent) is 0.015 mol dm–3 for experiment A, [NH3] when the reaction goes to completion = 2  0.015 = 0.0300 mol dm–3

4 © Raffles Institution 2020 9729/02/S/20 3(a)(ii) From the graph of expt A, First t1/2 (from 0.0  0.015 mol dm–3) = 29.5 min; Second t1/2 (from 0.015  0.0225 mol dm–3) = 29.5 min  t1/2 is constant  reaction is first order with respect to CO(NH2)2. (Note: Show all construction lines and label at least two t½ on the graph) For expt A, initial rate = ½ x 0 - 0.016 0 - 24 = 3.333 x 10–4 ≈ 3.33 x 10–4 mol dm–3 min–1 For expt B, initial rate = ½ x 0 - 0.015 0 - 45 = 1.666 x 10–4 ≈ 1.67 x 10–4 mol dm–3 min–1 (Note: On the graph, label coordinates used to obtain the initial rates.) Comparing expt A and expt B, when [OH]  2, initial rate  2.  reaction is first order with respect to OH– ions.

5 © Raffles Institution 2020 9729/02/S/20 [Alternative method to find order of reaction with respect to OH– ions] t1/2 of expt A = 29.5 min t1/2 of expt B = 59.0 min Since OH– is in large excess, [OH–] remains effectively constant during the reaction. Hence the reaction follows pseudo first-order kinetics. rate = k’[CO(NH2)2], where k’ = k[OH–]x t1/2= In 2 k' = ln 2 k[OH–]x By comparing expt A and expt B and their t1/2 values, when [OH] is halved then t1/2 is doubled.  x = 1 and reaction is first order with respect to OH– ions. 3(a)(iii) From expt A, 3.333  10–4 = k(0.015)(2.0)  k = 0.01111 ≈ 0.0111 mol–1 dm3 min–1 OR t½ = ln 2 k [OH–] From expt A, 29.5 = ln 2 k (2.0) k = 0.0117 mol–1 dm3 min–1 3(a)(iv) Since OH– is in large excess, [OH–] remains effectively constant during the reaction. The reaction follows pseudo first-order kinetics. rate = k’[CO(NH2)2], where k’ = k[OH–] t1/2= In 2 k' = ln 2 k[OH–] Since [OH–] in expt C is twice that of expt A (or four times that of expt B), t1/2 = 29.5  ½ = 14.8 min (or t1/2 = 59  ¼ = 14.8 min) OR t1/2 = ln 2 ÷ 0.0111(4.00) = 15.6 min

6 © Raffles Institution 2020 9729/02/S/20 3(a)(v) As temperature increases from T 1 to T2 K, the average kinetic energy of the reactant particles also increases. As such, significantly more reactant particles have energy greater than or equal to the activation energy of the reaction. Consequently, the frequency of effective collisions increases accordingly, and hence the reaction rate increases. Also, an increase in temperature results in a larger rate constant, and hence an increase in the reaction rate. 3(b)(i) 3(b)(ii) z = 13.6 + (38.7 – 30.3) = +22.0 3(b)(iii) At low [CO(NH 2)2], not all of the active sites on urease are occupied . In this case, rate  [CO(NH2)2] or reaction is first-order with respect to CO(NH2)2. At high [ CO(NH2)2], all the active sites on urease become saturated / are fully occupied by molecules of CO(NH 2)2. The reaction is zero order with respect to CO(NH 2)2 or rate is independent of [CO(NH2)2]. 4(a) Nucleophilic addition 4(b)(i) sp3

7 © Raffles Institution 2020 9729/02/S/20 4(b)(ii) The O−Ca−Cb bond angle in the intermediate is 90° and deviates greatly from the ideal bond angle of 109.5°. Hence, the orbital overlap is less effective / the bonds are weakened / reaction in step 2 will open the ring and relieve ring strain. Thus the reaction happens readily. 4(c) An alkene does not have an electron-deficient carbon atom for the nucleophilic ylide to attack. 4(d) 4(e)(i) 4(e)(ii) OR 4(f)(i) K2Cr2O7(aq), H2SO4(aq), heat (under reflux) 4(f)(ii) Oxidation would lead to two ketone functional groups , which will lead to the formation of a compound with two alke