JEE Main 2026 — Chemistry Physical Chemistry

2026integerhard

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)

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

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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 $x = 125$ .

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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 $x = 125$ .

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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