JEE Main 2026 — Chemical Equilibrium Question with Solution

Dissociation of a gas ${\mathrm{A}}_2$ takes place according to the following chemical reaction. 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:
$\begin{array}{cc}\text{ Substance }& \Delta {\mathrm{G}}_{\mathrm{f}}^{\circ }/\mathrm{kJ}{\mathrm{mol}}^{-1}\\ {\mathrm{A}}_2& -100.00\\ \mathrm{A}& -50.832\end{array}$
The degree of dissociation of ${\mathrm{A}}_2(\mathrm{g})$ is given by ${\left(x\times 10^{-2}\right)}^{1/2}$ where $x=$ $\_\_\_\_$. (Nearest integer).
[Given: $\mathrm{R}=8\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1},\log 2=0.3010,\log 3=0.48$ ]
Assume degree of dissociation is not negligible.