JEE Chemistry 2025 Question with Solution

Given:

• Reaction: $\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}$
• Initial moles of CO = $0.1 \, \text{mol}$
• Volume = $2 \, \text{dm}^3$
• Temperature = $500 \, \text{K}$
• Total pressure at equilibrium = $5 \, \text{bar}$
• Methanol formed = $0.04 \, \text{mol}$
• Gas constant = $0.08 \, \text{dm}^3 \, \text{bar} \, \text{K}^{-1} \, \text{mol}^{-1}$

Find: $K_p$ in the form $x \times 10^{-3}$.

Using the ideal gas law, total moles at equilibrium are

$$n_{\text{total}} = \frac{PV}{RT} = \frac{(5)(2)}{(0.08)(500)} = \frac{10}{40} = 0.25 \; \text{mol}$$

Methanol formed is $0.04 \, \text{mol}$, so CO reacted is also $0.04 \, \text{mol}$.

$$n_{\text{CO, eq}} = 0.1 - 0.04 = 0.06$$

Now use total moles to find hydrogen at equilibrium:

$$0.25 = 0.06 + n_{\text{H}_2} + 0.04$$ $$n_{\text{H}_2, eq} = 0.15$$

Partial pressures are

$$P_{\text{CO}} = \left(\frac{0.06}{0.25}\right) \times 5 = 1.2 \; \text{bar}$$ $$P_{\text{H}_2} = \left(\frac{0.15}{0.25}\right) \times 5 = 3.0 \; \text{bar}$$ $$P_{\text{CH}_3\text{OH}} = \left(\frac{0.04}{0.25}\right) \times 5 = 0.8 \; \text{bar}$$

For

$$\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}$$

we have

$$K_p = \frac{P_{\text{CH}_3\text{OH}}}{P_{\text{CO}}(P_{\text{H}_2})^2}$$

Substituting values,

$$K_p = \frac{0.8}{(1.2)(3.0)^2} = \frac{0.8}{10.8} = \frac{2}{27} \approx 0.074074$$

Now,

$$0.074074 = 74.074 \times 10^{-3}$$

So the nearest integer is $74$.

Therefore, the correct option is A.

Given: the equilibrium mixture contains CO, $\text{H}_2$ and $\text{CH}_3\text{OH}$ at $5 \, \text{bar}$, $500 \, \text{K}$ in $2 \, \text{dm}^3$.

Find: the integer $x$ in $K_p = x \times 10^{-3}$.

First determine total moles from the ideal gas law:

$$PV = nRT$$ $$n = \frac{PV}{RT} = \frac{5 \times 2}{0.08 \times 500} = 0.25$$

Next use stoichiometry. Formation of $0.04 \, \text{mol}$ methanol consumes:

• CO = $0.04 \, \text{mol}$
• $\text{H}_2$ = $0.08 \, \text{mol}$

Hence equilibrium moles are:

$$n_{\text{CO}} = 0.10 - 0.04 = 0.06$$ $$n_{\text{CH}_3\text{OH}} = 0.04$$

Since total moles are $0.25$,

$$n_{\text{H}_2} = 0.25 - 0.06 - 0.04 = 0.15$$

Now calculate mole fractions:

$$X_{\text{CO}} = \frac{0.06}{0.25} = 0.24$$ $$X_{\text{H}_2} = \frac{0.15}{0.25} = 0.60$$ $$X_{\text{CH}_3\text{OH}} = \frac{0.04}{0.25} = 0.16$$

Thus partial pressures are:

$$P_{\text{CO}} = 0.24 \times 5 = 1.2 \; \text{bar}$$ $$P_{\text{H}_2} = 0.60 \times 5 = 3.0 \; \text{bar}$$ $$P_{\text{CH}_3\text{OH}} = 0.16 \times 5 = 0.8 \; \text{bar}$$

Equilibrium constant:

$$K_p = \frac{0.8}{1.2 \times 9} = \frac{0.8}{10.8} = 0.074074$$

Expressing in the asked form,

$$K_p = 74.074 \times 10^{-3}$$

Therefore, the nearest integer is $74$, so the correct option is A.