Consider the following equilibrium, \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2(\mathrm{g}) \rightleftharpoons…
\(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})\)
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 \(\mathrm{K}_{\mathrm{p}}^0\) 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.
Solution
& \mathrm{CO}(\mathrm{g}) & + & 2 \mathrm{H}_2(\mathrm{g}) & \rightleftharpoons & \mathrm{CH}_3 \mathrm{OH}(\mathrm{g}) \\
\mathrm{t}=0 & 0.1 \mathrm{~mol} & & \mathrm{a} \mathrm{~mol} & & - \\
\mathrm{t}_{\text {eq }} & 0.1-\mathrm{x} & & \mathrm{a}-2 \mathrm{x} & & \mathrm{x}=0.04 \\
& =0.06 & & =\mathrm{a}-0.08 & \\
& & & =0.23-0.08 \\
& & & =0.15 \mathrm{~mole}
\end{array}\)
\(\begin{aligned} & \mathrm{V}=2 \mathrm{~L} \\ & \mathrm{~T}=500 \mathrm{~K} \\ & \mathrm{P}_{\text {total }}=5 \mathrm{bar} \\ & \mathrm{n}_{\text {Total }}=0.25=\frac{1}{4} \mathrm{~mol} \\ & \mathrm{P}_{\text {total }}=\mathrm{n}_{\text {total }} \times \frac{\mathrm{RT}}{\mathrm{V}} \\ & \Rightarrow 5=(0.06+\mathrm{a}-0.08+0.04) \times \frac{0.08 \times 500}{2} \\ & \Rightarrow 10=(0.02+\mathrm{a}) \times 0.08 \times 500\end{aligned}\)
\(\Rightarrow \mathrm{a}=0.25-0.02=0.23 \mathrm{~mol}.\)
\(\begin{aligned}
& \mathrm{K}_{\mathrm{P}}=\frac{\mathrm{X}_{\mathrm{CH}_3 \mathrm{OH}}}{\mathrm{X}_{\mathrm{CO}} \times \mathrm{X}_{\mathrm{H}_2}^2} \times \frac{1}{\left(\mathrm{P}_{\mathrm{T}}\right)^2}=\frac{0.04}{0.06 \times(0.15)^2} \times\left[\frac{1 / 4}{5}\right]^2 \\
& =\frac{4}{6 \times(0.15)^2 \times 16} \times \frac{1}{25} \\
& =\frac{100 \times 100}{24 \times 225 \times 25}=\frac{100 \times 100}{135000} \\
& =0.074=74 \times 10^{-3}
\end{aligned}\)
Asked in: JEE Main 2025 (02 Apr Shift 1)