TJC 2020 JC2 Prelim H2 Paper 4 Solutions

1 9729/04/TJC Prelims/2020 Answer all the questions in the spaces provided. 1 (a) Initial burette reading/ cm3 0.00 0.00 Final burette reading/ cm3 22.00 21.90 Volume of FA 1 used/ cm3 22.00 21.90 [4] (b) volume of FA 1 used = ½ (22.00 + 21.90)  = 21.95 cm3 (c) (i) No. of moles of KMnO4 = 21.95/1000 x 0.0300  = 6.59 x 10-4 mol (ii) No. of moles of H2O2 = 5/2 x 6.59 x 10-4  = 1.65 x 10-3 mol Concentration of H2O2 in FA 4 = 1.65 x 10-3 / 0.025  = 0.0659 mol dm-3 (iii) Concentration of H2O2 in FA 2 = 0.0659 x 250 25  = 0.659 mol dm-3 (iv) Volume of O2 = 0.659 x ½ x 24 = 7.90  Volume strength = 7.90 OR  Volume strength = 1.02 x ½ x 24 = 12.2 [Total: 10] 2 (b) Table of results Expt VFA 4/ cm3 Volume of water/ cm3 t/ s 1/t /s-1 log(1/t) log(VFA 4) 1 20.0 0.0 23.8 0.0420 -1.38 1.30 2 10.0 10.0 48.0 0.0208 -1.68 1.00 3 13.0 7.0 37.0 0.0270 -1.57 1.11 4 17.0 3.0 27.9 0.0358 -1.45 1.23 [4]

2 9729/04/TJC Prelims/2020 (c) (i) [3] log(1/t) -1.30 -1.35 -1.40 -1.45 -1.50 -1.55 -1.60 -1.65 -1.70 0.95 1.0 1.05 1.10 1.15 1.20 1.25 1.30 X X X X (1.28, -1.40) (1.0, -1.68)

3 9729/04/TJC Prelims/2020 (ii) Gradient = -1.40+1.68 1.28-1.0 = 1.00 (iii) log VFA 4 = log 16.50 = 1.22 From graph, log (1/t) = -1.46 t = 28.8 s [1] (d) Number of moles of S2O32- used = 0.025 x 10.0/1000 = 2.50 x 10-5 mol Total volume of reacting mixture = 51.0 cm3 [S2O32-]initial = (2.50 x 10-5) / (51.0 x 10-3) = 4.90 x 10-3 mol dm-3 [S2O32-]final = 0 Rate of change of [S2O32-] = (4.90 x 10-3 – 0)/ 23.8 = 2.10 x10-4 mol dm-3 s-1 [3] (e) (i) Higher conc. of thiosulfate means greater reaction time (allow reaction will be slower) and so a smaller percentage error. [1] (ii) Reason: change of temperature Use thermostatically-controlled water bath to maintain constant temperature Reason: decomposition of hydrogen peroxide Store H 2O2(aq) at low temperature, make up fresh H 2O2(aq), keep H 2O2(aq) in dark/dim light [2] (f)  Increase number of experiments carried out and hence data points plotted. OR  Have a greater spread of data points, e.g. VFA4 less than 10cm3 [1] (g) For Experiment 1 and 2, same amount of H2O2 will be reacted with as the same fixed amount of S2O32- is added into each reacting mixture. Since experiment 2 has the lower initial amount or conc of H2O2 added, it will have the larger percentage drop/ decrease in concentration of H2O2. Hence, experiment 2 shows a greater difference than experiment 1. [1] [Total: 18] 3 (a) Test Observation (i) To a 1 cm depth of FA 7 in a test- tube, add aqueous sodium hydroxide until it is in excess.