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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2026 · 23 January · Morning Shift · Chemistry · Chemical Equilibrium

For the following gas phase equilibrium reaction at constant...

For the following gas phase equilibrium reaction at constant temperature,
NH3( g)⇌1/2 N2( g)+3/2H2( g)\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons 1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})
if the total pressure is 3 atm\sqrt{3} \mathrm{~atm} and the pressure equilibrium constant \left(\mathrm{K}_{\mathrm{p}}\right)is9atm,thenthedegreeofdissociationisgivenas(x×10−2)−1/2is 9 atm, then the degree of dissociation is given as \left(x \times 10^{-2}\right)^{-1 / 2}. The value of xx is ____\_\_\_\_. (nearest integer)

Correct answer

0

Step-by-step solution

For the reaction: NH3(g)⇌12N2(g)+32H2(g)NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)

At t=0t = 0: 1 mole, –, –
At equilibrium: 1−α1-\alpha, α2\frac{\alpha}{2}, 3α2\frac{3\alpha}{2}
Total moles =1+α= 1 + \alpha

KP=(α2)1/2(3α2)3/2(1−α)⋅(PT1+α)1K_P = \frac{\left(\frac{\alpha}{2}\right)^{1/2}\left(\frac{3\alpha}{2}\right)^{3/2}}{(1-\alpha)} \cdot \left(\frac{P_T}{1+\alpha}\right)^1

Given PT=3P_T = \sqrt{3} atm and KP=9K_P = 9 atm:

9=(α2)1/2(3α2)3/2(1−α)×(3)1/21+α9 = \frac{\left(\frac{\alpha}{2}\right)^{1/2}\left(\frac{3\alpha}{2}\right)^{3/2}}{(1-\alpha)} \times \frac{(3)^{1/2}}{1+\alpha}

Simplifying: 9=9(α2)21−α29 = \frac{9\left(\frac{\alpha}{2}\right)^2}{1-\alpha^2}

1−α2=α241 - \alpha^2 = \frac{\alpha^2}{4}

5α24=1⇒α2=0.8\frac{5\alpha^2}{4} = 1 \Rightarrow \alpha^2 = 0.8

α=(0.8)1/2=(10.8)−1/2=(125×10−2)−1/2\alpha = (0.8)^{1/2} = \left(\frac{1}{0.8}\right)^{-1/2} = (125 \times 10^{-2})^{-1/2}

So x=125x = 125.
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