Dissociation of a gas $\mathrm{A}_{2}$ takes place according to the following chemical reactions. At equilibrium, the total pressure is 1 bar at 300 K. $\mathrm{A}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g})$ The standard Gibbs energy of formation of the involved substances has been provided below: | Substance | $\Delta \mathrm{G}_{\mathrm{f}}^{\circ} / \mathrm{kJ} \mathrm{mol}^{-1}$ | |-----------|-------------| | $\mathrm{A}_{2}$ | -100.00 | | A | -50.832 | The degree of dissociation of $\mathrm{A}_{2}(\mathrm{~g})$ is given by $(\mathrm{x} \times 10^{-2})^{1 / 2}$ where $\mathrm{x}=$ ______. (Nearest integer). [Given: $\mathrm{R}=8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, $\log 2=0.3010$, $\log 3=0.48$]

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Chemical Thermodynamics

Dissociation of a gas

A_{2}

takes place according to the following chemical reactions. At equilibrium, the total pressure is 1 bar at 300 K.

A_{2} (g) ⇌ 2 A (g)

The standard Gibbs energy of formation of the involved substances has been provided below:

Substance	$Δ G_{f}^{\circ} / kJ mol^{- 1}$
$A_{2}$	-100.00
A	-50.832

The degree of dissociation of

A_{2} (g)

is given by

(x \times 1 0^{- 2})^{1/2}

where

x =

______.
(Nearest integer).
[Given:

R = 8 J mol^{- 1} K^{- 1}

lo g 2 = 0.3010

lo g 3 = 0.48

]

Answer: 33

The standard free-energy change of

A_{2} ⇌ 2A

Δ G^{\circ} = 2 (- 50.832) - (- 100.00) = - 1.664 kJ mo l^{- 1} = - 1664 J mo l^{- 1}

. From

Δ G^{\circ} = - R T ln K_{p}

ln K_{p} = \frac{1664}{8 \times 300} = 0.693

, so

K_{p} = 2

. With a degree of dissociation

α

at a total pressure of 1 bar,

K_{p} = \frac{4 α ^{2}}{1 - α ^{2}}

; putting

K_{p} = 2

gives

2 (1 - α^{2}) = 4 α^{2}

, hence

6 α^{2} = 2

and

α^{2} = \frac{1}{3}

. Writing

α = (x \times 1 0^{- 2})^{1/2}

means

x \times 1 0^{- 2} = \frac{1}{3} = 33.3 \times 1 0^{- 2}

, so

x = 33

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer