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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2025 · 2 April · Morning Shift · Chemistry · Chemical Equilibrium

Consider the following equilibrium, (CO(g)+ 2 H _ 2 ( g)...

Consider the following equilibrium,
(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}}_{2}( \mathrm{g}) \rightleftharpoons \mathrm{CH}}_{3} OH(g)\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 ______ (\times10^{-3}) (nearest integer).
Given : (\mathrm{R}=0.08 \mathrm{dm}^3) bar (\mathrm{K}^{-1} mol−1\mathrm{ mol}^{-1})
Assume only methanol is formed as the product and the system follows ideal gas behaviour.

Correct answer

0

Step-by-step solution

(CO(g)+2H2(g)⇌CH3OH(g)t=00.1molamol−teq 0.1−xa−2xx=0.04&=0.06=a−0.08&=0.23−0.08&=0.15mole\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})(V=2LT=500KPtotal =5barnTotal =0.25=14molPtotal =ntotal ×RTV⇒5=(0.06+a−0.08+0.04)×0.08×5002⇒10=(0.02+a)×0.08×500\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}.)(KP=XCH3OHXCO×XH22×1(PT)2=0.040.06×(0.15)2×[1/45]2&=46×(0.15)2×16×125&=100×10024×225×25=100×100135000&=0.074=74×10−3\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})
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