2017 Paper 4 Ans

Universal Indicator paper turns red / orange, indicating pH = 3 (match with chart provided) When yellow FA 2 is added to colourless FA 1, the solution turns brown. Vigorous effervescence of colourless, odourless gas. Gas relights a glowing splint. Gas is oxygen. After a few minutes, the reaction mixture turns yellow. 2017 A level H2 Chem Paper 4 Answer

30.4 30.5 30.5 38.9 +8.4 30.4 30.4 30.4 38.8 +8.4 Average ΔT1max = (8.4 + 8.4) / 2 = +8.4 °C +8.4 °C Note: must include + or – sign for “change”

Average heat change, q = maq c ΔT = 100 × 4.18 × 8.4 = 3511.2 = 3510 J 3510 J No of mol of H2O2 = 50 1000 × 0.85 = 0.0425 mol ΔHrxn = −3511.2 0.0425 = −82.62 kJ mol−1 (per mol of H2O2) ΔH1decomp = −82.62 × 2 = −165 kJ mol –1 (for 2 mol of H2O2 as shown in eqn 1) −165 kJ mol−1 [Fe(H2O)6]3+ does not appear in the overall equation of the decomposition of H2O2. it is consumed in step 1 and regenerated in step 3 of the mechanism, this shows that it acts as a catalyst in the reaction. The starting species [Fe(H2O)6]3+ is yellow. During the reaction, the mixture turned brown, however, the yellow colour is obtained again at the end of the reaction. Before [Fe(H2O)6]3+ was added, there was no effervescence seen in H2O2. However upon adding [Fe(H2O)6]3+, vigorous effervescence is observed, implying that [Fe(H2O)6]3+ acts as a catalyst and speed up the decomposition of H2O2. Note: must include – sign as reaction is exothermic (T is +ve)

[Fe(H2O)5(OH)]2+ + H3O+ From test 1, colour of Universal Indicator paper is orange-red which shows that the solution is acidic (pH = 3). Fe3+ has high charge over size ratio, polarises and weakens the O−H bond of H2O, causing one of the six H2O to undergo hydrolysis to give [Fe(H2O)5(OH)]2+ and H3O+, resulting in an acidic solution. [Fe(H2O)5(OH)]2+ is yellow solution while [Fe(H2O)5O]3+ is brown solution. After [Fe(H2O)6]3+ is regenerated in step 3 of the mechanism, it reacts with H2O and undergoes hydrolysis to give [Fe(H2O)5(OH)]2+

At t = 4 min, Tmax = 33.0°C At t = 4 min, Tinitial = 21.5°C ΔT2max = 33.0 – 21.5 = +11.5°C +11.5 heat change, q = maqcΔT = 100 × 4.18 × 11.5 = 4807 J ΔH2decomp = −4807 0.0425 × 2 = – 226211 J mol−1 = – 226 kJ mol –1 (for 2 mol of H2O2 as shown in eqn 1) −226 kJ mol−1 ΔT2max is likely to be more accurate as the heat lost to the environment is accounted for through the extrapolation temperature readings to the time of mixing (t = 4 min).

Time of transfer, t (time at which reaction mixture is quenched) 1 min 40 s 4 min 50 s 7 min 54 s 11 min 17 s 15 min 6 s Decimal time, td /min 1.7 4.8 7.9 11.3 15.1 Final burette reading / cm3 40.00 47.50 14.55 24.80 30.80 Initial burette reading / cm3 12.40 27.75 0.75 14.60 24.80 Volume of FA 3 added / cm3 27.60 19.75 13.80 10.20 6.00

Gradient = 33.00−20.00 0.0−4.0 = −3.25 cm3 min–1 Note: Must indicate it is a negative gradient −3.25 Amt of MnO4– in 3.25 cm3 = 3.25 1000 × 0.0200 = 6.50 × 10–5 Rate of change of amount of MnO4– ions = −6.50 × 10–5 mol min–1 −6.50 × 10–5 Amount of H2O2 decomposed per minute = (6.50 × 10–5) × 5 2 = 1.625 × 10–4 mol Rate of change of [H2O2] at td = 0.0 min = 1.625 × 10−4 (101000⁄ ) = −0.01625 mol dm3 min–1 (Only 10.0 cm3 of reaction mixture is used for each titration with KMnO4) −0.0163 Time taken for vol of FA 3 to decrease from 25.00 cm3 to 12.50 cm3 = 6.0 min Time taken for vol of FA 3 to decrease from 20.00 cm3 to 10.00 cm3 = 6.0 min Since vol of KMnO4 used for each titration is directly proportional to the concentration of H2O2 in the reaction mixture, the shape of the [H2O2] vs time graph will be similar to VKMnO4 vs time graph. As [H2O2] decreases with time, rate of reaction (gradient) also decreases with a constant half-life of 6.0 min. Hence, decomposition of H2O2 is first order w.r.t. [H2O2]