Q21 / 75 · JEE Main 2026 (23 January Shift 1) · Chemistry
For the following gas phase equilibrium reaction at constant temperature,
$\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons 1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})$
if the total pressure is $\sqrt{3} \mathrm{~atm}$ and the pressure equilibrium constant $\left(\mathrm{K}_{\mathrm{p}}\right)$ is 9 atm, then the degree of dissociation is given as $\left(x \times 10^{-2}\right)^{-1 / 2}$. The value of $x$ is $\_\_\_\_$. (nearest integer)
$\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons 1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})$
if the total pressure is $\sqrt{3} \mathrm{~atm}$ and the pressure equilibrium constant $\left(\mathrm{K}_{\mathrm{p}}\right)$ is 9 atm, then the degree of dissociation is given as $\left(x \times 10^{-2}\right)^{-1 / 2}$. The value of $x$ is $\_\_\_\_$. (nearest integer)
Correct answer: 125
Solution
For the reaction: $NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$
At $t = 0$: 1 mole, –, –
At equilibrium: $1-\alpha$, $\frac{\alpha}{2}$, $\frac{3\alpha}{2}$
Total moles $= 1 + \alpha$
$K_P = \frac{\left(\frac{\alpha}{2}\right)^{1/2}\left(\frac{3\alpha}{2}\right)^{3/2}}{(1-\alpha)} \cdot \left(\frac{P_T}{1+\alpha}\right)^1$
Given $P_T = \sqrt{3}$ atm and $K_P = 9$ atm:
$9 = \frac{\left(\frac{\alpha}{2}\right)^{1/2}\left(\frac{3\alpha}{2}\right)^{3/2}}{(1-\alpha)} \times \frac{(3)^{1/2}}{1+\alpha}$
Simplifying: $9 = \frac{9\left(\frac{\alpha}{2}\right)^2}{1-\alpha^2}$
$1 - \alpha^2 = \frac{\alpha^2}{4}$
$\frac{5\alpha^2}{4} = 1 \Rightarrow \alpha^2 = 0.8$
$\alpha = (0.8)^{1/2} = \left(\frac{1}{0.8}\right)^{-1/2} = (125 \times 10^{-2})^{-1/2}$
So $x = 125$.
At $t = 0$: 1 mole, –, –
At equilibrium: $1-\alpha$, $\frac{\alpha}{2}$, $\frac{3\alpha}{2}$
Total moles $= 1 + \alpha$
$K_P = \frac{\left(\frac{\alpha}{2}\right)^{1/2}\left(\frac{3\alpha}{2}\right)^{3/2}}{(1-\alpha)} \cdot \left(\frac{P_T}{1+\alpha}\right)^1$
Given $P_T = \sqrt{3}$ atm and $K_P = 9$ atm:
$9 = \frac{\left(\frac{\alpha}{2}\right)^{1/2}\left(\frac{3\alpha}{2}\right)^{3/2}}{(1-\alpha)} \times \frac{(3)^{1/2}}{1+\alpha}$
Simplifying: $9 = \frac{9\left(\frac{\alpha}{2}\right)^2}{1-\alpha^2}$
$1 - \alpha^2 = \frac{\alpha^2}{4}$
$\frac{5\alpha^2}{4} = 1 \Rightarrow \alpha^2 = 0.8$
$\alpha = (0.8)^{1/2} = \left(\frac{1}{0.8}\right)^{-1/2} = (125 \times 10^{-2})^{-1/2}$
So $x = 125$.
Hence, the correct answer is 125.