JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

Observe the following equilibrium in a 1 L flask.

$A (g) ⇌ B (g)$

At 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.

A
0.367, 0.275
B
0.53, 0.4
C
0.742, 0.557
D
0.557, 0.418

Solution

$K_{e q} = \frac{[ B ] _{e q}}{[ A ] _{e q}} = \frac{0.375}{0.5} = 0.75$ .

After adding 0.1 mol of A (1 L flask), $[A] = 0.6$ M, $[B] = 0.375$ M. Let $x$ mol of A convert to B at new equilibrium:

$K = \frac{0.375 + x}{0.6 - x} = 0.75 \Rightarrow 0.45 - 0.75 x = 0.375 + x \Rightarrow 1.75 x = 0.075 \Rightarrow x = \frac{3}{70} \approx 0.043$ .

$[A]_{n e w} = 0.6 - 0.043 \approx 0.557$ M; $[B]_{n e w} = 0.375 + 0.043 \approx 0.418$ M.

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