Chemistry · Chemical Equilibrium

JEE Main 2025 — 2 April, Morning Shift — Question 20

Consider the following equilibrium, CO(g)+2H2( g)⇌CH3OH(g)\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 2dm32 \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 CH3OH\mathrm{CH}_{3} \mathrm{OH} is formed. The Kpθ\mathrm{K}_{\mathrm{p}}^{\theta} is _____\_\_\_\_\_ 10−310^{-3} (nearest

integer). Given : R=0.08dm3\mathrm{R}=0.08 \mathrm{dm}^{3} bar K−1 mol−1\mathrm{K}^{-1} \mathrm{~mol}^{-1} Assume only methanol is formed as the product and the

system follows ideal gas behaviour.

Answer: 74

Numerical answer — enter this value.

Step-by-step solution

ntotal n_{\text {total }} at equilibrium : 5×2(0.08)×500\frac{5 \times 2}{(0.08) \times 500}

=0.250=0.250

Moles of CH3OH=0.04\mathrm{CH}_{3} \mathrm{OH}=0.04

PCH3OH=(0.04)×(0.08)×5002=0.8bar\mathrm{P}_{\mathrm{CH}_{3} \mathrm{OH}}=\frac{(0.04) \times(0.08) \times 500}{2}=0.8 \mathrm{bar}

PH2=3\mathrm{P}_{\mathrm{H}_{2}}=3 bar

Pco =1.2=1.2 bar

KP=0.8(1.2)(3)2=0.07407=74.07×10−3K_{P}=\frac{0.8}{(1.2)(3)^{2}}=0.07407=74.07 \times 10^{-3}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Chemistry
Chapter
Chemical Equilibrium
Topic
Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient
Consider the following equilibrium, CO ( g )+2 H 2 ( g )… | JEE Main 2025 PYQ with Solution · DhiX AI