JEE Main 2026 Jan 22 Shift 1, Chemistry Q25: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2026Jan 22, Shift 1Chemistry

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:

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 $^{- 1}$ , $lo g 2 = 0.3010$ , $lo g 3 = 0.48$ ]

Your answer

Solution

Reaction Gibbs energy: $Δ G_{r x n}^{\circ} = 2Δ G_{f}^{\circ} (A) - Δ G_{f}^{\circ} (A_{2}) = 2 (- 50.832) - (- 100) = - 101.664 + 100 = - 1.664$ kJ/mol $= - 1664$ J/mol.

From $Δ G^{\circ} = - R T ln K_{p}$ : $ln K_{p} = \frac{1664}{8.3 \times 300} \approx 0.693$ , so $K_{p} \approx 2$ .

Set up the equilibrium for $A_{2} ⇌ 2 A$ . Starting with 1 mole of $A_{2}$ , at equilibrium: $A_{2} = 1 - α$ , $A = 2 α$ , total $= 1 + α$ . Mole fractions give partial pressures (with $P_{t o t a l} = 1$ bar):

$K_{p} = \frac{p _{A}^{2}}{p _{A_{2}}} = \frac{[ 2 α / ( 1 + α ) ] ^{2} \cdot P}{( 1 - α ) / ( 1 + α ) \cdot P} = \frac{4 α ^{2} \cdot P}{1 - α ^{2}}$ .

With $K_{p} = 2$ and $P = 1$ : $2 = \frac{4 α ^{2}}{1 - α ^{2}} \Rightarrow 2 - 2 α^{2} = 4 α^{2} \Rightarrow α^{2} = \frac{1}{3}$ .

So $α^{2} = \frac{1}{3} = \frac{100}{3} \times 1 0^{- 2} \approx 33.33 \times 1 0^{- 2}$ .

Comparing with $α = (x \times 1 0^{- 2})^{1/2}$ : $x = 33.33 \approx 33$ .

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JEE Main 2026 Jan 22 Shift 1, Chemistry Q25: Law of Chemical Equilibrium And Equilibrium Constant

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