| Date | May 2017 | Marks available | 2 | Reference code | 17M.2.hl.TZ2.9 |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Predict | Question number | 9 | Adapted from | N/A |
Question
The Bombardier beetle sprays a mixture of hydroquinone and hydrogen peroxide to fight off predators. The reaction equation to produce the spray can be written as:
| C6H4(OH)2(aq) + H2O2(aq) | → | C6H4O2(aq) + 2H2O(l) |
| hydroquinone | quinone |
Hydrogenation of propene produces propane. Calculate the standard entropy change, ΔS θ, for the hydrogenation of propene.
The standard enthalpy change, ΔH θ, for the hydrogenation of propene is –124.4 kJ mol–1. Predict the temperature above which the hydrogenation reaction is not spontaneous.
Markscheme
«ΔS θ =» 270 «J K–1 mol–1» – 267 «J K–1 mol–1» – 131 «J K–1 mol–1»
«ΔS θ =» –128 «J K–1 mol–1»
Award [2] for correct final answer.
[2 marks]
«non spontaneous if» ΔG θ = ΔH θ – TΔS θ > 0
OR
ΔH θ > TΔS θ
«T above» \(\frac{{ - 124.4{\text{ }}\ll {\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - {\text{1}}}}\gg }}{{ - 0.128{\text{ }}\ll {\text{kJ}}\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - {\text{1}}}}\gg }} = \)» 972 «K»
Award [2] for correct final answer.
Accept 699 °C.
Do not award M2 for any negative T value.
[2 marks]
