JEE Main 2026 · 4 Apr · Shift 1 · Chemistry Q55

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ChemistrySingle correct+4 / -1Question 55 of 75

At T(K), the equilibrium constant of $\ce{A2(g) + B2(g) <=> C(g)}$ is $2.7\times10^{-5}$ . What is the equilibrium constant for $\ce{1/3A2(g) + 1/3B2(g) <=> 1/3C(g)}$ at the same temperature?

\left(2.7\times10^{-5}\right)^3

6\times10^{-2}

\sqrt{2.7\times10^{-5}}

3\times10^{-2}

Answer and worked solution

Official answer: D

Scaling an equation. Multiplying an equation by $\frac13$ raises K to the power $\frac13$ :
$K'=(2.7\times10^{-5})^{1/3}=3\times10^{-2}.$

(D)

Topics:Law of mass action and Kc, Kp

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JEE Main 2026 · 4 Apr · Shift 1, Chemistry question 55