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. Warm the tube, gently and carefully. Then, add a 1 cm depth of FA 2.  green ppt,  insoluble in excess, ppt turns brown on standing  on warming, NH3 gas evolves, turns moist red litmus blue  Brown ppt. Effervescence. Gas evolved relights glowing splint.

4 9729/04/TJC Prelims/2020 (ii) Place about 2 cm3 of FA 7 in a test- tube. Add 3-4 pieces of magnesium ribbon. Leave it on test tube rack for 5 minutes.  Effervescence  Gas evolved causes lighted splint to ‘pop’  Green ppt formed  Brown/black deposit on Mg ribbon OR brown/black ppt [3] (b) (i) FA 7: cations are are NH4+ and Fe2+ (ii) Green ppt is Fe(OH)2 which darkens/ turns brown as it is oxidised to Fe(OH)3. OH- is produced from the reaction between Mg and H2O. Mg is more reactive than Fe, so Fe is coated on Mg Mg + 2H2O → Mg(OH)2 + H2 Fe2+ + 2OH- → Fe(OH)2 2Fe(OH)2 + [O] + H2O → 2Fe(OH)3 Fe2+ + Mg → Fe + Mg2+ [5] (c) Test Observation (i) To a 1 cm depth of FA 8 in a test- tube, add a 1 cm depth of FA 5 and leave it on test tube rack for 1-2 minutes, then add FA 6. Solution turns brown/yellow brown or black ppt Solution turns blue (ii) Transfer about 5 cm3 of FA 8 into a boiling tube. Add all the sample of zinc powder provided to the solution. Stir the mixture. Record all the observations. When no further changes are seen, filter the reaction mixture into a test tube. This is solution FA 9. Effervescence, gas evolved pops with lighted splint Solution turns yellow → green → blue → green → violet Violet filtrate (can award mark for the 3rd colour here if violet is not mentioned above) (iii) To a 1 cm depth of FA 9 in a test- tube, add 2 cm of H2SO4 followed by FA 1 dropwise until in excess. Record all the observations. When FA 1 is in excess, the solution will be pink. Solution turns violet/final colour given in (a)(ii) →green →blue→ green →yellow(orange) → pink/purple (iv) To a 1 cm depth of FA 9 in a test- tube, add a 1 cm depth of FA 8. Solution turns blue/green [4]

5 9729/04/TJC Prelims/2020 (d) FA 9 undergoes oxidation. MnO4- is an oxidising agent/is reduced/changes from purple to colourless [1] [Total:14] 4 (a) Add an excess of aqueous Na2CO3 to precipitate MnCO3. Filter the mixture to separate the precipitate from the solution using a pre-weighed filter paper. Wash the precipitate to remove any impurities. Dry the precipitate together with the filter paper by heating in an oven to remove the water. [2] (b) (i) Pre-calculations Assume O2 gas collected in a 100 cm3 gas syringe to be 80 cm3. No of moles of O2 gas in 80 cm3 = 80 / 24000 = 3.33 x 10-3 mol No of moles of H2O2 required = 2(3.33 x 10-3) = 6.67 x 10-3 mol  Vol of H2O2 required = 6.67 x 10-3 / 0.1 = 66.7 cm3 Set-up for gas collection Experimental Procedure 1. Using a 50 cm3 measuring cylinder, measure approximately 70 cm3 (or any other vol in excess) of H2O2 and transfer it to a conical flask. 2. Assemble the set-up as shown in the diagram. 3. Note the initial volume reading of a 100 cm3 gas syringe before removing the string to start the reaction. Start stopwatch immediately. 4. Record volume of gas produced at 30-second interval (or any logical time intervals) 5. Record the final volume reading of the gas syringe approx 20 minutes after the reaction has completed and there is no movement observed of the plunger in the syringe (to allow the temperature and pressure to equilibrate with the surroundings). [6] 250 cm3 conical flask Stopper String Rubber tubing Graduated 100 cm3 syringe 0.100 mol dm–3 H2O2 MnO2 Glass tubing Retort stand

6 9729/04/TJC Prelims/2020 (ii) [1] (iii) Determine two t ½ values from the graph.  If both t ½ is a constant, order of reaction wrt [H2O2] is 1st order. [2] (c) H2O2(aq) H2O(l) + ½O2(g) H2O2(l) Energy cycle By Hess’ Law,  ∆Hr = -2.1 + (-93.6) = -95.7 kJ mol-1 [2] [Total: 13] –2.1 kJ mol-1 –93.6 kJ mol-1 ∆Hr 60 80 40 1st t½ 2nd t½ (iii) Volume of oxygen/ cm3 Time/ min