2017 H2 Chem P1 Q14 — Thermodynamics Solution

The answer is D (3 and 4 only).
Starting with \(x\) mol each of \(\text{H}_2\text{O}\) and CO, if \(\alpha\) mol reacts, at equilibrium \(n(\text{H}_2) = n(\text{CO}_2) = \alpha\) and \(n(\text{H}_2\text{O}) = n(\text{CO}) = x - \alpha\) (see the table). So \(K_\text{c} = \dfrac{\alpha^2}{(x - \alpha)^2}\), and \([\text{H}_2] > [\text{H}_2\text{O}]\) means \(\alpha > x - \alpha\), which is true only when \(K_\text{c} > 1\).
Using \(\Delta G^\circ = -RT\ln K\): \(K > 1\) means \(\ln K > 0\), so \(\Delta G^\circ < 0\). From the graph, \(\Delta G^\circ\) is negative at point 3 (about \(-10 \text{ kJ mol}^{-1}\) at 770 K) and point 4 (about \(-20 \text{ kJ mol}^{-1}\) at 500 K).
• A is wrong: at point 1, \(\Delta G^\circ\) is positive (about \(+10 \text{ kJ mol}^{-1}\)), so \(K \lt 1\) and \([\text{H}_2] \lt [\text{H}_2\text{O}]\); at point 2, \(\Delta G^\circ = 0\), so \(K = 1\) and \([\text{H}_2] = [\text{H}_2\text{O}]\), not greater.
• B is wrong: at point 1, \(\Delta G^\circ > 0\) and \(K \lt 1\), so there is less \(\text{H}_2\) than \(\text{H}_2\text{O}\).
• C is wrong: it includes point 2, where \(\Delta G^\circ = 0\) and \(K = 1\), so \([\text{H}_2]\) equals \([\text{H}_2\text{O}]\) rather than exceeding it.