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 -

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Equilibrium

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 PQ at the new equilibrium, respectively, are -

Answer: (A)

Because

P_{2} + Q_{2} ⇌ 2 P Q

has equal moles of gas on both sides, the volume cancels and

K_{c}

may be written in moles. At the first equilibrium

K_{c} = \frac{n _{P Q}^{2}}{n _{P_{2}} n _{Q_{2}}} = \frac{2 ^{2}}{2 \times 2} = 1

. Adding one mole each of

P_{2}

and

Q_{2}

gives starting amounts

3, 3, 2

. Let

x

moles of each reactant be consumed on re-equilibration:

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

. Taking the square root,

\frac{2 + 2 x}{3 - x} = 1

, so

2 + 2 x = 3 - x

and

x = \frac{1}{3}

. Hence

n_{P_{2}} = n_{Q_{2}} = 3 - \frac{1}{3} = \frac{8}{3} = 2.67

and

n_{P Q} = 2 + \frac{2}{3} = \frac{8}{3} = 2.67

. The new equilibrium amounts are

2.67, 2.67, 2.67

mol.

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option