1. Reaction and initial data

Given equilibrium: $\mathrm{CO(g)} + 2\mathrm{H_2(g)} \rightleftharpoons \mathrm{CH_3OH(g)}$

Initial moles:

• CO = $0.10$ mol
• $\mathrm{H_2}$ is added until total pressure becomes $5$ bar
• Volume = $2\,\mathrm{dm^3}$
• Temperature = $500\,\mathrm{K}$
• At equilibrium, methanol formed = $0.04$ mol

Let initial moles of hydrogen be $n$.

2. Find initial moles of hydrogen using ideal gas equation

Before reaction proceeds, total moles are: $n_{\text{total, initial}} = 0.10 + n$

Using $PV = nRT$ we get $5 \times 2 = (0.10+n)(0.08)(500)$

$10 = (0.10+n)(40)$

$0.10+n = \frac{10}{40} = 0.25$

So, $n = 0.25 - 0.10 = 0.15$

Hence initial hydrogen moles = $0.15$ mol.

3. Equilibrium moles

Reaction stoichiometry: $\mathrm{CO} + 2\mathrm{H_2} \rightleftharpoons \mathrm{CH_3OH}$

If $0.04$ mol methanol is formed, then:

• CO consumed = $0.04$ mol
• $\mathrm{H_2}$ consumed = $0.08$ mol

Thus equilibrium moles are:

• CO: $0.10 - 0.04 = 0.06$
• $\mathrm{H_2}$: $0.15 - 0.08 = 0.07$
• $\mathrm{CH_3OH}$: $0.04$

Total equilibrium moles: $n_{\text{total, eq}} = 0.06 + 0.07 + 0.04 = 0.17$

4. Equilibrium total pressure

Again using ideal gas equation: $P_{\text{eq}} = \frac{nRT}{V}$

$P_{\text{eq}} = \frac{(0.17)(0.08)(500)}{2}$

$P_{\text{eq}} = \frac{6.8}{2} = 3.4\,\text{bar}$

5. Partial pressures at equilibrium

Using mole fraction $\times$ total pressure:

$p_{\mathrm{CO}} = \frac{0.06}{0.17}\times 3.4 = 1.2\,\text{bar}$

$p_{\mathrm{H_2}} = \frac{0.07}{0.17}\times 3.4 = 1.4\,\text{bar}$

$p_{\mathrm{CH_3OH}} = \frac{0.04}{0.17}\times 3.4 = 0.8\,\text{bar}$

6. Expression for $K_p^\theta$

For the reaction, $K_p^\theta = \frac{(p_{\mathrm{CH_3OH}}/p^\theta)}{(p_{\mathrm{CO}}/p^\theta)(p_{\mathrm{H_2}}/p^\theta)^2}$

Taking standard pressure $p^\theta = 1\,\text{bar}$, $K_p^\theta = \frac{0.8}{(1.2)(1.4)^2}$

$K_p^\theta = \frac{0.8}{1.2 \times 1.96}$

$K_p^\theta = \frac{0.8}{2.352} \approx 0.340$

But note that because the reaction reduces moles by 2, with the dimensionless standard-state form: $K_p^\theta = \frac{p_{\mathrm{CH_3OH}}/1}{(p_{\mathrm{CO}}/1)(p_{\mathrm{H_2}}/1)^2}$ this is indeed $0.340$.

Now express it as: $K_p^\theta = 340 \times 10^{-3}$

Nearest integer = 340.

7. Comparison with stored answer

Stored correct answer = $74$

My derived answer is $340$, which does not match the stored answer. The stoichiometric and ideal-gas calculations are straightforward, and the result $340 \times 10^{-3}$ follows consistently.