Consider the following gaseous equilibrium in a closed container of volume ' V ' at T ( K ). P _ 2 ( ~g )+ Q _ 2 ( ~g ) 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

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Consider the following gaseous equilibrium in a closed container of volume ' V ' at T ( K ). P _ 2 ( ~g )+ Q _ 2 ( ~g ) 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 P Q at the new equilibrium, respectively, are

Quantrex Academy · Accepted Answer

Correct answer: C. 2.67,2.67,2.67

Initial equilibrium: 2 moles each of P_2, Q_2, PQ with K = [PQ]^2 [P_2][Q_2] = 4 4 = 1. After adding 1 mole each of P_2 and Q_2: initial composition is (3, 3, 2). Reaction quotient Q = 4 9 Let x moles of reactants combine: (2+2x)^2 (3-x)^2 = 1 2+2x = 3-x 3x = 1 → x = 1 3 Final moles: P_2 = 3 - 1 3 = 2.67, Q_2 = 2.67, PQ = 2 + 2 3 = 2.67

JEE Main Chemistry Question (2026) — Solution

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