JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Equilibrium

One mole each of He and

A (g)

are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium.

A (g) ⇌ B (g)

K_{C}

for this reaction at 400 K is 4.0. The partial pressures (in atm ) of He and

B (g)

are respectively (at equilibrium).
(Assume

He

A (g)

and

B (g)

behave as ideal gases)
(Given :

R = 0.082 L atm K^{- 1} mol^{- 1}

)

Answer: (A)

Solution

Let

x

mol of A convert to B in the

10 L

flask. Since the volumes cancel,

K_{C} = \frac{[ B ]}{[ A ]} = \frac{x /10}{( 1 - x ) /10} = \frac{x}{1 - x} = 4 \Rightarrow x = 0.8.

Helium is inert, so its amount stays

1 mol

P_{He} = \frac{n R T}{V} = \frac{1 \times 0.082 \times 400}{10} = 3.28 atm .

For B, with

n = 0.8 mol

P_{B} = \frac{0.8 \times 0.082 \times 400}{10} = 2.624 atm .

The partial pressures of helium and B are

3.28 atm

and

2.624 atm

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option