| Date | May 2018 | Marks available | 2 | Reference code | 18M.2.hl.TZ2.6 |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Deduce | Question number | 6 | Adapted from | N/A |
Question
A mixture of 1.00 mol SO2(g), 2.00 mol O2(g) and 1.00 mol SO3(g) is placed in a 1.00 dm3 container and allowed to reach equilibrium.
2SO2(g) + O2(g) \( \rightleftharpoons \) 2SO3(g)
Nitrogen oxide is in equilibrium with dinitrogen dioxide.
2NO(g) \( \rightleftharpoons \) N2O2(g) ΔHΘ < 0
Deduce, giving a reason, the effect of increasing the temperature on the concentration of N2O2.
A two-step mechanism is proposed for the formation of NO2(g) from NO(g) that involves an exothermic equilibrium process.
First step: 2NO(g) \( \rightleftharpoons \) N2O2(g) fast
Second step: N2O2(g) + O2 (g) → 2NO2(g) slow
Deduce the rate expression for the mechanism.
The rate constant for a reaction doubles when the temperature is increased from 25.0 °C to 35 °C.
Calculate the activation energy, Ea, in kJ mol−1 for the reaction using section 1 and 2 of the data booklet.
Markscheme
[N2O2] decreases AND exothermic «thus reverse reaction favoured»
Accept “product” for [N2O2].
Do not accept just “reverse reaction favoured/shift to left” for “[N2O2] decreases”.
[1 mark]
ALTERNATIVE 1:
«from equilibrium, step 1»
\({K_c} = \frac{{{\text{[}}{{\text{N}}_2}{{\text{O}}_2}{\text{]}}}}{{{{{\text{[NO]}}}^2}}}\)
OR
[N2O2] = Kc[NO]2
«from step 2, rate «= k1[N2O2][O2] = k2K[NO]2[O2]»
rate = k[NO]2[O2]
ALTERNATIVE 2:
«from step 2» rate = k2[N2O2][O2]
«from step 1, rate(1) = k1[NO]2 = k–1[N2O2], [N2O2] = \(\frac{{{k_1}}}{{{k_{ - 1}}}}\) [NO]2»
«rate = \(\frac{{{k_1}}}{{{k_{ - 1}}}}\) k2[NO]2[O2]»
rate = k[NO]2[O2]
Award [2] for correct rate expression.
[2 marks]
«\(\ln \frac{{{k_1}}}{{{k_2}}} = \frac{{{E_a}}}{R}\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right)\)»
T2 = «273 + 35 =» 308 K AND T1 = «273 + 25 =» 298 K
Ea = 52.9 «kJ mol–1»
Award [2] for correct final answer.
[2 marks]
