JEE Main 2022 · 30 Jun · Shift 1 · Chemistry Q64

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ChemistrySingle correct+4 / -1Question 64 of 90

The equilibrium constant for the reversible reaction $\mathrm{2A(g)\rightleftharpoons2B(g)+C(g)}$ is $K_1$ and for the reaction $\mathrm{\frac32A(g)\rightleftharpoons\frac32B(g)+\frac34C(g)}$ is $K_2$ . $K_1$ and $K_2$ are related as :

K_1=\sqrt{K_2}

K_2=\sqrt{K_1}

K_2=K_1^{3/4}

K_1=K_2^{3/4}

Answer and worked solution

Official answer: C

Scaling. The second reaction is $\frac34$ of the first, so $K_2=K_1^{3/4}$ .

(C)

Topics:Law of mass action and Kc, Kp

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JEE Main 2022 · 30 Jun · Shift 1, Chemistry question 64