Q25 / 75 · JEE Main 2026 (22 January Shift 1) · Chemistry
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:
\(\begin{array}{|c|c|} \hline \text { Substance } & \Delta \mathrm{G}_{\mathrm{f}}^{\circ} / \mathrm{kJ} \mathrm{~mol}^{-1} \\ \hline \mathrm{~A}_2 & -100.00 \\ \hline \mathrm{~A} & -50.832 \\ \hline \end{array}\)
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.
$\mathrm{A}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g})$
The standard Gibbs energy of formation of the involved substances has been provided below:
\(\begin{array}{|c|c|} \hline \text { Substance } & \Delta \mathrm{G}_{\mathrm{f}}^{\circ} / \mathrm{kJ} \mathrm{~mol}^{-1} \\ \hline \mathrm{~A}_2 & -100.00 \\ \hline \mathrm{~A} & -50.832 \\ \hline \end{array}\)
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.
Correct answer: 33
Solution
For the dissociation reaction $A_2(g) \rightleftharpoons 2A(g)$, first calculate the standard Gibbs free energy:
$\Delta G°_{rxn} = 2(-50.832) - (-100.00) = -1.664$ kJ/mol.
Using $\Delta G° = -RT \ln K_p$:
$-1664 = -8 \times 300 \times \ln K_p$, giving $K_p = e^{0.6933} = 2$.
For dissociation with degree of dissociation $\alpha$, starting with 1 mole of $A_2$ at 1 bar total pressure:
At equilibrium, moles are $(1-\alpha)$ for $A_2$ and $2\alpha$ for $A$, with total $(1+\alpha)$ moles.
$K_p = \frac{4\alpha^2}{1-\alpha^2} = 2$.
Solving: $2(1-\alpha^2) = 4\alpha^2$ gives $\alpha^2 = 1/3$ or $\alpha = 0.577$.
Since degree of dissociation is $(x \times 10^{-2})^{1/2}$:
$(x \times 10^{-2})^{1/2} = 0.577$ gives $x \times 10^{-2} = 0.333$, so $x = 33.3 \approx 33$.
$\Delta G°_{rxn} = 2(-50.832) - (-100.00) = -1.664$ kJ/mol.
Using $\Delta G° = -RT \ln K_p$:
$-1664 = -8 \times 300 \times \ln K_p$, giving $K_p = e^{0.6933} = 2$.
For dissociation with degree of dissociation $\alpha$, starting with 1 mole of $A_2$ at 1 bar total pressure:
At equilibrium, moles are $(1-\alpha)$ for $A_2$ and $2\alpha$ for $A$, with total $(1+\alpha)$ moles.
$K_p = \frac{4\alpha^2}{1-\alpha^2} = 2$.
Solving: $2(1-\alpha^2) = 4\alpha^2$ gives $\alpha^2 = 1/3$ or $\alpha = 0.577$.
Since degree of dissociation is $(x \times 10^{-2})^{1/2}$:
$(x \times 10^{-2})^{1/2} = 0.577$ gives $x \times 10^{-2} = 0.333$, so $x = 33.3 \approx 33$.
Hence, the correct answer is 33.