Question

For the reaction, CO(g) + 2H2 (g)= CH3 OH (g) derive an expression for the equilibrium constant, Kp in terms of the extent of reaction, € and the total pressure pt, if initially 2 mol of CO and 3 mol of H2 are mixed.

Answer 1

For the reaction, $\mathrm{CO(g)} + 2\mathrm{H2(g)} = \mathrm{CH3OH(g)}$ derive an expression for the equilibrium constant, Kp in terms of the extent of reaction, € and the total pressure pt, if initially 2 mol of CO and 3 mol of H2 are mixed.

**Solution.** The general expression for the equilibrium constant is written as: $K_{\mathrm{p}} = \frac{p_{\mathrm{CH}_3\mathrm{OH}}}{p_{\mathrm{H}_2}^2 \times p_{\mathrm{CO}}}$

We have a degree of reaction €. Then reacted: 2€ moles CO, 4€ moles H₂, and formed: 2€ moles CH₃OH. Then the equilibrium quantities of substances:

- for CO: (2-2€) moles;

- for H₂: (3-4€) moles;

- for CH₃OH: 2€ moles.

We have a total pressure pt, then partial pressures:

- for CO: $p_{\mathrm{CO}} = \frac{2 - 2\epsilon}{2 - 2\epsilon + 3 - 4\epsilon + 2\epsilon} pt = \frac{2 - 2\epsilon}{5 - 4\epsilon} pt;$

- for H₂: $p_{\mathrm{H_2}} = \frac{3 - 4\epsilon}{5 - 4\epsilon} pt;$

- for CH₃OH: $p_{\mathrm{CH}_3\mathrm{OH}} = \frac{2\epsilon}{5 - 4\epsilon} pt.$

Then the expression of the equilibrium constant takes the form:

$$K_{\mathrm{p}} = \frac{\epsilon (5 - 4\epsilon)^2}{(1 - \epsilon)(3 - 4\epsilon)^2 \mathrm{pt}^2}.$$

**Answer:** $K_{\mathrm{p}} = \frac{\epsilon (5 - 4\epsilon)^2}{(1 - \epsilon)(3 - 4\epsilon)^2 \mathrm{pt}^2}$.

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