RVHS Prelim_H2_Chemistry_P4_Soln

River Valley High School 9729/02/PRELIMS/25 2025 Preliminary Examination [Turn over River Valley High School 9729 H2 Chemistry Preliminary Examination Paper 4 Suggested solution 1 (a) (i) Results Titration number 1 2 Final burette reading /cm3 23.75 23.75 Initial burette reading /cm3 0.00 0.00 VFA2 /cm3 23.75 23.75 Subject supervisor’s reading: 23.75 cm3 2 marks if difference  0.20 cm3, 1 mark if difference  0.30 cm3 [5] (ii) average volume of FA 2 used = 23.75 23.75 2 + = 23.75 cm3 [1] (iii) [Ag+(aq)] = 23.75 100.011000 1000 = 0.02375 = 0.0238 mol dm–3 [1] (iv) Fe2+(aq) + Ag+(aq) = Fe3+(aq) + Ag(s) I/ mol dm−3 (0.1)/2 (0.1)/2 0 - C -x -x +x E 0.02375 0.02375 x Initial [Fe2+] or initial [Ag+] = 0.1 / 2 = 0.05 mol dm–3 [Fe3+(aq)] = 0.05 – 0.02375 = 0.02625 = 0.0263 mol dm–3 OR Initial amount of Fe2+ or initial amount of Ag+ = 0.1 × 100 1000 = 0.01 mol [2]

2 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination [Fe3+] = 2000.01 0.023751000 200 1000 − = 0.02625 = 0.0263 mol dm–3 (v) Kc = 2 0.0263 (0.0237) = 46.8 mol–1 dm3 [2] (b) Disagree. Either of the following: • The equilibrium is established in 4 hours and thus the position of equilibrium does not shift within minutes of titration. • Titration is carried out very quickly such that it will not shift the position of equilibrium/ no time to establish new equilibrium. • The value of Kc implies that position of equilibrium lies far right, the impact of titration does not affect position significantly. • FA 1 is left to stand for 4h, all Ag would have aggregated as solid at the bottom, which was left undisturbed. There is no Ag in the aliquot, so backward reaction will not occur, shifting of position of equilibrium will not occur. [1] (c) Use a lower concentration of FA1/ FA2 , increase volume used, lower percentage uncertainty. Use higher volume of FA 1 used, lower percentage uncertainty of Kc. Conduct the experiment in a thermostatically controlled water bath OR controlled at constant temperature. Filter the equilibrium mixture prior to titration to remove the solid so that the backward reaction will not occur during titration to affect the titre and value of Kc. [1] [Total: 13]

3 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination [Turn over 2 (a) Results V/cm3 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 T/ °C 31.0 35.0 38.0 40.0 41.5 42.0 41.0 40.0 39.0 (b) (i) [3] (ii) volume of FA 4 used = 23.00 …………………cm3 [1] (iii) Energy change = (25.0 + 23.0) (4.18) (42.20 – 31.0) = 2247.2 J = 2250 J [1] 23.00 42.20

4 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination (iv) Amount of Ba(OH)2 = 0.95 (0.025) = 0.02375 = 0.0238 mol [1] (v) ∆Hneut = (–2247.2J / 1000) / 2(0.02375) = – 47.3 kJmol−1 [1] [1] Accuracy −43 ≤ ∆Hneut