JEE Main 2026 — Chemistry Physical Chemistry

2026integerhard

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.

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Held on 22 Jan 2026 · Verified 6 Jul 2026.

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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 $x = 33.3 \approx 33$ .

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One mole each of He and $A(g)$ are taken in a $10$ L closed flask and heated to $400$ K to establish the following equilibrium. $A(g) \rightleftharpoons B(g)$. $K_c$ for this reaction at $400$ K is $4.0$. The partial pressures (in atm) of He and $B(g)$ are respectively (at equilibrium) (Assume He, $A(g)$ and $B(g)$ behave as ideal gases) (Given: $R = 0.082$ L atm K$^{-1}$ mol$^{-1}$)

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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 $x = 33.3 \approx 33$ .

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main Physical Chemistry in other years

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