Q5 / 75 · JEE Main 2026 (04 April Shift 1) · Chemistry
At $T(K)$, the equilibrium constant of $A_2(g) + B_2(g) \rightleftharpoons C(g)$ is $2.7 \times 10^{-5}$. What is the equilibrium constant for $\dfrac{1}{3}A_2(g) + \dfrac{1}{3}B_2(g) \rightleftharpoons \dfrac{1}{3}C(g)$ at the same temperature?
- A$(2.7 \times 10^{-5})^3$
- B$6 \times 10^{-2}$
- C$\sqrt{2.7 \times 10^{-5}}$
- Correct answerD$3 \times 10^{-2}$
Solution
For the reaction $A_2(g) + B_2(g) \rightleftharpoons C(g)$, the equilibrium constant is $K_1 = 2.7 \times 10^{-5}$.
When a reaction is multiplied by a factor $n$, the new equilibrium constant becomes $K^n$.
The given reaction is multiplied by $\dfrac{1}{3}$ to obtain the reaction $\dfrac{1}{3}A_2(g) + \dfrac{1}{3}B_2(g) \rightleftharpoons \dfrac{1}{3}C(g)$.
The new equilibrium constant $K_2$ is given by:
$K_2 = (K_1)^{1/3}$
$K_2 = (2.7 \times 10^{-5})^{1/3}$
$K_2 = (27 \times 10^{-6})^{1/3}$
$K_2 = 3 \times 10^{-2}$
Answer: $3 \times 10^{-2}$
When a reaction is multiplied by a factor $n$, the new equilibrium constant becomes $K^n$.
The given reaction is multiplied by $\dfrac{1}{3}$ to obtain the reaction $\dfrac{1}{3}A_2(g) + \dfrac{1}{3}B_2(g) \rightleftharpoons \dfrac{1}{3}C(g)$.
The new equilibrium constant $K_2$ is given by:
$K_2 = (K_1)^{1/3}$
$K_2 = (2.7 \times 10^{-5})^{1/3}$
$K_2 = (27 \times 10^{-6})^{1/3}$
$K_2 = 3 \times 10^{-2}$
Answer: $3 \times 10^{-2}$
Hence, the correct answer is option: 'D'.