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MCQEasyJEE 2026Equilibrium Basics

JEE Chemistry 2026 Question with Solution

Observe the following equilibrium in a 1 L1 \, \text{L} flask: \ceA(g)<=>B(g)\ce{A(g) <=> B(g)} At T(K)T(\text{K}), the equilibrium concentrations of A and B are 0.5 M0.5 \, \text{M} and 0.375 M0.375 \, \text{M} respectively. 0.10.1 moles of A is added into the flask and heated to T(K)T(\text{K}) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively:

  • A

    0.7420.742, 0.5570.557

  • B

    0.3670.367, 0.2750.275

  • C

    0.530.53, 0.40.4

  • D

    0.5570.557, 0.4180.418

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: In a 1 L1 \, \text{L} flask, the equilibrium is \ceA(g)<=>B(g)\ce{A(g) <=> B(g)}. Initial equilibrium concentrations are [A]=0.5 M[A] = 0.5 \, \text{M} and [B]=0.375 M[B] = 0.375 \, \text{M}. Then 0.10.1 mol of A is added, so the concentration of A becomes 0.6 M0.6 \, \text{M} before re-establishing equilibrium.

Find: The new equilibrium concentrations of A and B, and hence the correct option.

First calculate the equilibrium constant from the original equilibrium state:

Kc=[B][A]=0.3750.5=0.75K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75

After adding 0.10.1 mol of A in a 1 L1 \, \text{L} flask:

[A]=0.5+0.1=0.6,[B]=0.375[A] = 0.5 + 0.1 = 0.6, \qquad [B] = 0.375

Let the system shift forward by xx to attain equilibrium again. Then:

[A]=0.6−x,[B]=0.375+x[A] = 0.6 - x, \qquad [B] = 0.375 + x

Using the same equilibrium constant:

Kc=0.375+x0.6−x=0.75K_c = \frac{0.375 + x}{0.6 - x} = 0.75

Now solve for xx:

0.375+x=0.75(0.6−x)0.375 + x = 0.75(0.6 - x) 0.375+x=0.45−0.75x0.375 + x = 0.45 - 0.75x 1.75x=0.0751.75x = 0.075 x=0.043x = 0.043

Substitute this value into the equilibrium concentrations:

[A]=0.6−0.043=0.557[A] = 0.6 - 0.043 = 0.557 [B]=0.375+0.043=0.418[B] = 0.375 + 0.043 = 0.418

Therefore, the new equilibrium concentrations are 0.557 M0.557 \, \text{M} and 0.418 M0.418 \, \text{M} respectively, so the correct option is D.

The solution explicitly notes that the given key suggests A, but the calculation shows option (4).

Common mistakes

  • Using the old equilibrium concentrations directly as the new answer is incorrect because adding 0.10.1 mol of A disturbs equilibrium. Recalculate the concentrations after perturbation using an ICE-type setup.

  • Writing Kc=[A][B]K_c = \frac{[A]}{[B]} is incorrect for the equilibrium \ceA(g)<=>B(g)\ce{A(g) <=> B(g)} as written. Here, Kc=[B][A]K_c = \frac{[B]}{[A]}. Reversing the ratio gives the wrong shift and wrong concentrations.

  • Forgetting that the flask volume is 1 L1 \, \text{L} leads to an incorrect concentration change. Since the volume is 1 L1 \, \text{L}, adding 0.10.1 mol of A increases its concentration by exactly 0.1 M0.1 \, \text{M}.

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