Question 2 of 5: Combining Two Iron-Oxide Equilibria to get K and ΔG° for the Water–Gas-Shift Reaction

Question 2: Combining Two Iron-Oxide Equilibria to get K and ΔG° for the Water–Gas-Shift Reaction (15%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. K₁, K₂ at 600, 800, 1000 °C from the table above.

Find. K and ΔG° for CO + H₂O ↔ CO₂ + H₂ at the same three temperatures.

Approach. Combine the two given equilibria by Hess’s law: adding Reaction 1 (forward) to Reaction 2 (reversed) cancels FeO(s) and Fe(s) and gives exactly the target reaction, so K_target = K₁/K₂, then ΔG° = −RT ln K.

1. Combine the two equilibria. Reaction 1 forward + Reaction 2 reversed: [FeO+CO→Fe+CO₂] + [Fe+H₂O→FeO+H₂] = CO+H₂O→CO₂+H₂, the target reaction. Reversing Reaction 2 inverts its equilibrium constant, so $$K_{target}=K_1\times\frac{1}{K_2}=\frac{K_1}{K_2}$$
2. T = 600°C (873 K). $$K_{target}=\frac{0.900}{0.332}=2.711\qquad \Delta G^{\circ}=-RT\ln K=-(8.314)(873)\ln(2.711)$$ $$\boxed{\Delta G^{\circ}=-7238\text{ J}=-7.24\text{ kJ}}$$
3. T = 800°C (1073 K). $$K_{target}=\frac{0.535}{0.499}=1.072\qquad \Delta G^{\circ}=-(8.314)(1073)\ln(1.072)$$ $$\boxed{\Delta G^{\circ}=-621\text{ J}=-0.62\text{ kJ}}$$
4. T = 1000°C (1273 K). $$K_{target}=\frac{0.396}{0.668}=0.593\qquad \Delta G^{\circ}=-(8.314)(1273)\ln(0.593)$$ $$\boxed{\Delta G^{\circ}=+5534\text{ J}=+5.53\text{ kJ}}$$ Note the sign change between 800°C and 1000°C: the water–gas-shift reaction switches from mildly product-favoured to mildly reactant-favoured as K₁ falls faster than K₂ rises with temperature.