2024 Re-NEET EquilibriumRelation between equilibrium constants of related reactions Medium Question 13 of 104 in Equilibrium

At a given temperature and pressure, the equilibrium constant values for the equilibria are given below: $3A_2+B_2\rightleftharpoons2A_3B,\ K_1$ and $A_3B\rightleftharpoons\dfrac{3}{2}A_2+\dfrac{1}{2}B_2,\ K_2$. The relation between $K_1$ and $K_2$ is:

A $K_1^2=2K_2$
B $K_2=\dfrac{K_1}{2}$
C $K_1=\dfrac{1}{\sqrt{K_2}}$
D $K_2=\dfrac{1}{\sqrt{K_1}}$

The answer is D.

Solution

Reversing and halving the first reaction: $\dfrac{1}{2}(3A_2+B_2\rightleftharpoons2A_3B)$ reversed gives $A_3B\rightleftharpoons\dfrac{3}{2}A_2+\dfrac{1}{2}B_2$, with equilibrium constant $K_2=\sqrt{\dfrac{1}{K_1}}=\dfrac{1}{\sqrt{K_1}}$.

Concept Relation between equilibrium constants of related reactions

Equilibrium in past papers

104 questions from this chapter have appeared across 27 exam years.