JEE Main 2026 Apr 4 Shift 1, Chemistry Q5: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2026Apr 4, Shift 1Chemistry

At $T (K)$ , the equilibrium constant of $A_{2} (g) + B_{2} (g) ⇌ C (g)$ is $2.7 \times 1 0^{- 5}$ .

What is the equilibrium constant for $\frac{1}{3} A_{2} (g) + \frac{1}{3} B_{2} (g) ⇌ \frac{1}{3} C (g)$ at the same temperature?

A
$(2.7 \times 1 0^{- 5})^{3}$
B
$6 \times 1 0^{- 2}$
C
$2.7 \times 1 0^{- 5}$
D
$3 \times 1 0^{- 2}$

Solution

If a balanced equation is multiplied by a factor $n$ , the new equilibrium constant becomes $K^{n}$ . Here the stoichiometric coefficients are scaled by $\frac{1}{3}$ , so

$K_{e q}^{'} = K_{e q}^{1/3} = (2.7 \times 1 0^{- 5})^{1/3} .$

Compute: $2.7 \times 1 0^{- 5} = 27 \times 1 0^{- 6}$ , and $(27 \times 1 0^{- 6})^{1/3} = 3 \times 1 0^{- 2} .$

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JEE Main 2026 Apr 4 Shift 1, Chemistry Q5: Law of Chemical Equilibrium And Equilibrium Constant

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