JEE MAIN - Chemistry (2025 - 2nd April Morning Shift - No. 24)
Consider the following equilibrium,
$$\mathrm{CO}(\mathrm{~g})+2 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{~g})$$
0.1 mol of CO along with a catalyst is present in a $2 \mathrm{dm}^3$ flask maintained at 500 K . Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of $\mathrm{CH}_3 \mathrm{OH}$ is formed. The $K_p^\theta$ is __________ $\times 10^{-3}$ (nearest integer).
Given : $\mathrm{R}=0.08 \mathrm{~dm}^3$ bar $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$
Assume only methanol is formed as the product and the system follows ideal gas behaviour.
Explanation
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$$\begin{aligned} & V=2 L \\ & T=500 K \\ & P_{\text {total }}=5 \text { bar } \\ & n_{\text {Total }}=0.25=\frac{1}{4} \mathrm{~mol} \\ & P_{\text {total }}=n_{\text {total }} \times \frac{R T}{V} \\ & \Rightarrow 5=(0.06+a-0.08+0.04) \times \frac{0.08 \times 500}{2} \\ & \Rightarrow 10=(0.02+a) \times 0.08 \times 500 \end{aligned}$$
$$\begin{aligned} &\Rightarrow \mathrm{a}=0.25-0.02=0.23 \mathrm{~mol}\\ &\begin{aligned} & \mathrm{K}_{\mathrm{P}}=\frac{\mathrm{X}_{\mathrm{CH}_3 \mathrm{OH}}}{\mathrm{X}_{\mathrm{CO}} \times \mathrm{X}_{\mathrm{H}_2}^2} \times \frac{1}{\left(\mathrm{P}_{\mathrm{T}}\right)^2}=\frac{0.04}{0.06 \times(0.15)^2} \times\left[\frac{1 / 4}{5}\right]^2 \\ & =\frac{4}{6 \times(0.15)^2 \times 16} \times \frac{1}{25} \\ & =\frac{100 \times 100}{24 \times 225 \times 25}=\frac{100 \times 100}{135000} \\ & =0.074=74 \times 10^{-3} \end{aligned} \end{aligned}$$
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