At a given temperature and pressure, the equilibrium constant values for the equilibria are given below: $3…

At a given temperature and pressure, the equilibrium constant values for the equilibria are given below: $3 \mathrm{~A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{~A}_3 \mathrm{~B}, \mathrm{~K}_1$ $\mathrm{A}_3 \mathrm{~B} \rightleftharpoons \frac{3}{2} \mathrm{~A}_2+\frac{1}{2} \mathrm{~B}_2, \mathrm{~K}_2$ The relation between $\mathrm{K}_1$ and $\mathrm{K}_2$ is :
  1. $\mathrm{K}_1^2=2 \mathrm{~K}_2$
  2. $\mathrm{K}_2=\frac{\mathrm{K}_1}{2}$
  3. $\mathrm{K}_1=\frac{1}{\sqrt{\mathrm{K}_2}}$
  4. $\mathrm{K}_2=\frac{1}{\sqrt{\mathrm{K}_1}}$

Solution

$3 \mathrm{~A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{~A}_3 \mathrm{~B} ; \mathrm{K}_1$ $2 \mathrm{~A}_3 \mathrm{~B} \rightleftharpoons 3 \mathrm{~A}_2+\mathrm{B}_2 ; \mathrm{K}^{\prime}=\frac{1}{\mathrm{~K}_1}$ $\frac{1}{2} A_3 B \rightleftharpoons \frac{3}{2} A_2+\frac{1}{2} B_2 ; K_2=\sqrt{K^{\prime}}$ $\mathrm{K}_2=\sqrt{\mathrm{K}^{\prime}}$ $=\frac{1}{\sqrt{K_1}}$

Asked in: NEET 2024 (Re-NEET)

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