The equilibrium constant for the reversible reaction $2\mathrm{A(g)} \rightleftharpoons 2\mathrm{B(g)}+\mathrm{C(g)}$ is $\mathrm{K_1}$ and for the reaction $\frac{3}{2}\mathrm{A(g)} \rightleftharpoons \frac{3}{2}\mathrm{B(g)}+\frac{3}{4}\mathrm{C(g)}$ is $\mathrm{K_2}$. $\mathrm{K_1}$ and $\mathrm{K_2}$ are related as :

Option 3: $\mathrm{K_2}=\mathrm{K_1^{3/4}}$

$\mathrm{K_1}=\sqrt{\mathrm{K_2}}$

$\mathrm{K_2}=\sqrt{\mathrm{K_1}}$

$\mathrm{K_2}=\mathrm{K_1^{3/4}}$

$\mathrm{K_1}=\mathrm{K_2^{3/4}}$

JEE Main 2022 · ChemistryMultiple choiceSingle correctEasyApplication

JEE Main 30 June 2022, Shift 1, Chemistry Q64

Question 64 of 90 in this shift, Chemistry question 4 of 30, Section A.

The equilibrium constant for the reversible reaction

2\mathrm{A(g)} \rightleftharpoons 2\mathrm{B(g)}+\mathrm{C(g)}

\mathrm{K_1}

and for the reaction

\frac{3}{2}\mathrm{A(g)} \rightleftharpoons \frac{3}{2}\mathrm{B(g)}+\frac{3}{4}\mathrm{C(g)}

\mathrm{K_2}

\mathrm{K_1}

and

\mathrm{K_2}

are related as :

(1) $\mathrm{K_1}=\sqrt{\mathrm{K_2}}$
(2) $\mathrm{K_2}=\sqrt{\mathrm{K_1}}$
(3) $\mathrm{K_2}=\mathrm{K_1^{3/4}}$ Official answer
(4) $\mathrm{K_1}=\mathrm{K_2^{3/4}}$

Official answer

Option 3

NTA final key.

Chapter: Equilibrium
Topic: Chemical Equilibrium
Idea tested: Equilibrium Constant for Modified Equations

Same idea in other shifts

Asked 9× in all

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.