JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

Consider the following gaseous equilibrium in a closed container of volume "V" at T(K).

$P_{2} (g) + Q_{2} (g) ⇌ 2 P Q (g)$

2 moles each of $P_{2} (g)$ , $Q_{2} (g)$ and $P Q (g)$ 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 -

A
2.67, 2.67, 2.67
B
1.21, 2.24, 1.56
C
1.66, 1.66, 1.66
D
2.56, 1.62, 2.24

Solution

Initial equilibrium has $[P_{2}] = [Q_{2}] = [P Q] = 2$ moles in volume $V$ , so

$K_{c} = \frac{( 2/ V ) ^{2}}{( 2/ V ) ( 2/ V )} = 1$ .

After adding 1 mole each of $P_{2}$ and $Q_{2}$ , the system has 3, 3, 2 moles. Let $x$ moles of each reactant convert further:

Moles: $P_{2} = 3 - x$ , $Q_{2} = 3 - x$ , $P Q = 2 + 2 x$ .

Applying $K_{c} = 1$ (volume cancels):

$\frac{( 2 + 2 x ) ^{2}}{( 3 - x ) ^{2}} = 1 \Rightarrow \frac{2 + 2 x}{3 - x} = 1 \Rightarrow x = \frac{1}{3}$ .

Therefore: $P_{2} = Q_{2} = \frac{8}{3} = 2.67$ , $P Q = \frac{8}{3} = 2.67$ .

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