JEE Main 2025 — Chemistry Physical Chemistry

$$\begin{array}{llll} & \mathrm{CO}(\mathrm{g}) & + & 2 \mathrm{H}_2(\mathrm{g}) & \rightleftharpoons & \mathrm{CH}_3 \mathrm{OH}(\mathrm{g}) \\ \mathrm{t}=0 & 0.1 \mathrm{~mol} & & \mathrm{a} \mathrm{~mol} & & - \\ \mathrm{t}_{\text {eq }} & 0.1-\mathrm{x} & & \mathrm{a}-2 \mathrm{x} & & \mathrm{x}=0.04 \\ & =0.06 & & =\mathrm{a}-0.08 & \\ & & & =0.23-0.08 \\ & & & =0.15 \mathrm{~mole} \end{array}$$t=0teq ​​CO(g)0.1 mol0.1−x=0.06​+​2H2​(g)a mola−2x=a−0.08=0.23−0.08=0.15 mole​⇌​CH3​OH(g)−x=0.04

$$\begin{aligned} & \mathrm{V}=2 \mathrm{~L} \\ & \mathrm{~T}=500 \mathrm{~K} \\ & \mathrm{P}_{\text {total }}=5 \mathrm{bar} \\ & \mathrm{n}_{\text {Total }}=0.25=\frac{1}{4} \mathrm{~mol} \\ & \mathrm{P}_{\text {total }}=\mathrm{n}_{\text {total }} \times \frac{\mathrm{RT}}{\mathrm{V}} \\ & \Rightarrow 5=(0.06+\mathrm{a}-0.08+0.04) \times \frac{0.08 \times 500}{2} \\ & \Rightarrow 10=(0.02+\mathrm{a}) \times 0.08 \times 500\end{aligned}$$​V=2 L T=500 KPtotal ​=5barnTotal ​=0.25=41​ molPtotal ​=ntotal ​×VRT​⇒5=(0.06+a−0.08+0.04)×20.08×500​⇒10=(0.02+a)×0.08×500​

$$\Rightarrow \mathrm{a}=0.25-0.02=0.23 \mathrm{~mol}.$$⇒a=0.25−0.02=0.23 mol.

$$\begin{aligned} & \mathrm{K}_{\mathrm{P}}=\frac{\mathrm{X}_{\mathrm{CH}_3 \mathrm{OH}}}{\mathrm{X}_{\mathrm{CO}} \times \mathrm{X}_{\mathrm{H}_2}^2} \times \frac{1}{\left(\mathrm{P}_{\mathrm{T}}\right)^2}=\frac{0.04}{0.06 \times(0.15)^2} \times\left[\frac{1 / 4}{5}\right]^2 \\ & =\frac{4}{6 \times(0.15)^2 \times 16} \times \frac{1}{25} \\ & =\frac{100 \times 100}{24 \times 225 \times 25}=\frac{100 \times 100}{135000} \\ & =0.074=74 \times 10^{-3} \end{aligned}$$​KP​=XCO​×XH2​2​XCH3​OH​​×(PT​)21​=0.06×(0.15)20.04​×[51/4​]2=6×(0.15)2×164​×251​=24×225×25100×100​=135000100×100​=0.074=74×10−3​