JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Equilibrium

Observe the following equilibrium in a 1 L flask.

A (g) ⇌ B (g)

T (K)

, the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to

T (K)

to establish the equilibrium again. The new equilibrium concentrations (in M ) of A and B are respectively.

Answer: (D)

Solution

The equilibrium constant is

K = \frac{[ B ]}{[ A ]} = \frac{0.375}{0.5} = 0.75

. Adding 0.1 mol of A to the 1 L flask raises

[A]

to 0.6 M; letting

x

of it convert to B,

K = \frac{0.375 + x}{0.6 - x} = 0.75

gives

0.45 - 0.75 x = 0.375 + x

, so

x = 0.043

. The new equilibrium concentrations are

[A] = 0.6 - 0.043 = 0.557 M

and

[B] = 0.375 + 0.043 = 0.418 M

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option