Chemistry · Chemical Equilibrium

JEE Main 2026 — 23 January, Morning Shift — Question 67

For the following gas phase equilibrium reaction at constant temperature, NH3( g)⇌12 N2( g)+32H2( g)\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) If the total pressure is 3 atm\sqrt{3} \mathrm{~atm} and the pressure equilibrium constant (Kp)\left(\mathrm{K}_{\mathrm{p}}\right) is 9 atm , then the degree of dissociation is given as (x×10−2)−1/2\left(\mathrm{x} \times 10^{-2}\right)^{-1 / 2}. The value of xx is ____\_\_\_\_ (Nearest integer)

Answer: 125

Numerical answer — enter this value.

Step-by-step solution

NH3(g)  ⇌  12 N2(g)+32 H2(g)\mathrm{NH_3(g)} \;\rightleftharpoons\; \frac{1}{2}\,\mathrm{N_2(g)} + \frac{3}{2}\,\mathrm{H_2(g)} NH3N2H2t=01 mole−−t=teq1−αα23α2\begin{array}{c|ccc} & \mathrm{NH_3} & \mathrm{N_2} & \mathrm{H_2} \\ \hline t = 0 & 1\,\text{mole} & - & - \\ t = t_{\mathrm{eq}} & 1-\alpha & \dfrac{\alpha}{2} & \dfrac{3\alpha}{2} \end{array} Kp=(α2)1/2(3α2)3/2(1−α)[PT1+α]−1[∴PT=3 atm]K_p = \frac{\left(\dfrac{\alpha}{2}\right)^{1/2} \left(\dfrac{3\alpha}{2}\right)^{3/2}} {(1-\alpha)} \left[\frac{P_T}{1+\alpha}\right]^{-1} \qquad \left[\therefore P_T = \sqrt{3}\,\text{atm}\right] 9=(α2)1/2(3α2)3/2(1−α)×(3)1/21+α9 = \frac{\left(\dfrac{\alpha}{2}\right)^{1/2} \left(\dfrac{3\alpha}{2}\right)^{3/2}} {(1-\alpha)} \times \frac{(3)^{1/2}}{1+\alpha} 9=9(α2)21−α29 = \frac{9\left(\dfrac{\alpha}{2}\right)^2}{1-\alpha^2} 1−α2=α241 - \alpha^2 = \frac{\alpha^2}{4} 5α24=1\frac{5\alpha^2}{4} = 1 α2=0.8\alpha^2 = 0.8 α=(0.8)1/2\alpha = (0.8)^{1/2} α=[10.8]−1/2\alpha = \left[\frac{1}{0.8}\right]^{-1/2} α=[125×10−2]1/2\alpha = \left[125 \times 10^{-2}\right]^{1/2} x=125x = 125

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Chemistry
Chapter
Chemical Equilibrium
Topic
Factors affecting equilibrium &Le Chatelier's Principle