2020 H2 Chem P1 Q14 — Chemical equilibrium Solution

The answer is C (the forward reaction is endothermic).

\(\text{P} + \text{Q} \rightleftharpoons \text{R} + 2\text{S}\) \(K_c = \frac{[\text{R}][\text{S}]^2}{[\text{P}][\text{Q}]}\)

\(K_c(1) = \frac{0.075 \times 0.040^2}{0.050 \times 0.064} = 0.0375\text{ mol dm}^{-3}\) (300 K, 100 kPa)
\(K_c(2) = \frac{0.070 \times 0.035^2}{0.040 \times 0.042} = 0.0510\text{ mol dm}^{-3}\) (400 K, 100 kPa)
\(K_c(3) = \frac{0.015 \times 0.080^2}{0.080 \times 0.032} = 0.0375\text{ mol dm}^{-3}\) (300 K, 200 kPa)

\(K_c\) increases when the temperature is raised from 300 K to 400 K. A rise in temperature favours the endothermic direction, so the forward reaction is endothermic.

• A and B: experiments 1 and 3 are at the same temperature and give the same \(K_c\). \(K_c\) depends only on temperature whether the species are gases or in solution, and the experiments start from different concentrations, so the data give no way of telling which state the substances are in.
• D: for an exothermic forward reaction, \(K_c\) would decrease as the temperature rises, but it increases from 0.0375 to 0.0510.