Consider the following gaseous equilibrium in a closed container of volume "V" at T(K). P 2 ( g )+ Q 2 ( g ) rightleftharpoons 2 PQ ( g ) 2 moles each of P 2 ( g ), Q 2 ( g ) and PQ ( g ) are present at equilibrium. Now one mole each of ' P 2 ' and ' Q 2 ' are added to the equilibrium keeping the temperature at T ( K ) . The number of moles of P 2 , Q 2 and PQ at the new equilibrium, respectively, are -

Chemistry · Chemical Equilibrium

JEE Main 2026 — 24 January, Evening Shift — Question 52

Consider the following gaseous equilibrium in a closed container of volume "V" at T(K). $\mathrm{P}_{2}(\mathrm{~g})+\mathrm{Q}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{PQ}(\mathrm{g})$ 2 moles each of $\mathrm{P}_{2}(\mathrm{~g}), \mathrm{Q}_{2}(\mathrm{~g})$ and $\mathrm{PQ}(\mathrm{g})$ are present at equilibrium. Now one mole each of ' $\mathrm{P}_{2}$ ' and ' $\mathrm{Q}_{2}$ ' are added to the equilibrium keeping the temperature at $\mathrm{T}(\mathrm{K})$ . The number of moles of $\mathrm{P}_{2}$ , $\mathrm{Q}_{2}$ and PQ at the new equilibrium, respectively, are -

Option A:
$2.67,2.67,2.67$
Correct
Option B:
$1.21,2.24,1.56$
Option C:
1.66, 1.66, 1.66
Option D:
2.56, 1.62, 2.24

Answer: A

Step-by-step solution

Initial equilibrium: P₂(g) + Q₂(g) ⇌ 2PQ(g) At equilibrium: moles of P₂ = 2, Q₂ = 2, PQ = 2. Equilibrium constant: $K_c = \frac{[PQ]^2}{[P₂][Q₂]} = \frac{(2/V)^2}{(2/V)(2/V)} = 1$ . After adding 1 mole each of P₂ and Q₂, initial moles for new equilibrium: P₂ = 3, Q₂ = 3, PQ = 2. Let x moles of P₂ react. Then: P₂ = 3 - x, Q₂ = 3 - x, PQ = 2 + 2x. New equilibrium constant expression: $K_c = \frac{(2+2x)^2/V^2}{(3-x)(3-x)/V^2} = 1$ . Simplify: $\frac{4(1+x)^2}{(3-x)^2} = 1$ . Take square root: $\frac{2(1+x)}{3-x} = 1$ (positive root since moles increase). Solve: 2 + 2x = 3 - x → 3x = 1 → x = 1/3. New moles: P₂ = 3 - 1/3 = 8/3 ≈ 2.67, Q₂ = 8/3 ≈ 2.67, PQ = 2 + 2/3 = 8/3 ≈ 2.67.

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient