Chemistry · 2022
JEE Main · 30 June 2022, Shift 1 · Q64
The equilibrium constant for the reversible reaction 2 A ( g ) ⇌ 2 B ( g )+ C ( g ) is K_1 and for the reaction 3/2 A ( g ) ⇌ 3/2 B ( g )+3/4 C ( g )…
The equilibrium constant for the reversible reaction $\displaystyle 2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$ is $\displaystyle \mathrm{K}_1$
and for the reaction
$\displaystyle \frac{3}{2} \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \frac{3}{2} \mathrm{~B}(\mathrm{~g})+\frac{3}{4} \mathrm{C}(\mathrm{g})$ is $\displaystyle \mathrm{K}_2$.
$\displaystyle \mathrm{K}_1$ and $\displaystyle \mathrm{K}_2$ are related as :
Official answer
(3)
$\displaystyle \mathrm{K}_2=\mathrm{K}_1^{3 / 4}$
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JEE Main 2022 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.