1. Write the reaction and assume 1 mole initially

Given: $\mathrm{X}_2\mathrm{Y}(g) \rightleftharpoons \mathrm{X}_2(g)+\frac12\mathrm{Y}_2(g)$

Let the initial moles of $\mathrm{X}_2\mathrm{Y}$ be $1$. If the degree of dissociation is $x$, then at equilibrium:

• $\mathrm{X}_2\mathrm{Y}$: $1-x$
• $\mathrm{X}_2$: $x$
• $\mathrm{Y}_2$: $\dfrac{x}{2}$

So total moles at equilibrium are: $n_{\text{total}}=(1-x)+x+\frac{x}{2}=1+\frac{x}{2}$

2. Write partial pressures

If total pressure is $p$, then partial pressures are: $P_{\mathrm{X}_2\mathrm{Y}}=\frac{1-x}{1+x/2}p$ $P_{\mathrm{X}_2}=\frac{x}{1+x/2}p$ $P_{\mathrm{Y}_2}=\frac{x/2}{1+x/2}p$

3. Expression for $K_p$

For the reaction $\mathrm{X}_2\mathrm{Y}(g) \rightleftharpoons \mathrm{X}_2(g)+\frac12\mathrm{Y}_2(g)$ we have $K_p=\frac{P_{\mathrm{X}_2}\,(P_{\mathrm{Y}_2})^{1/2}}{P_{\mathrm{X}_2\mathrm{Y}}}$

Substitute the partial pressures: $K_p=\frac{\left(\frac{x}{1+x/2}p\right)\left(\frac{x/2}{1+x/2}p\right)^{1/2}}{\left(\frac{1-x}{1+x/2}p\right)}$

4. Simplify

First cancel the common factor $\dfrac{1}{1+x/2}$ appropriately: $K_p=\frac{x p\cdot \sqrt{\frac{x p}{2(1+x/2)}}}{(1-x)p}$

The $p$ cancels partly, giving: $K_p=\frac{x}{1-x}\sqrt{\frac{x p}{2(1+x/2)}}$

5. Use the approximation: $x$ is very very small

Since $x\ll 1$, $1-x\approx 1, \qquad 1+\frac{x}{2}\approx 1$

Hence, $K_p\approx x\sqrt{\frac{x p}{2}}$

So, $K_p\approx \sqrt{\frac{p}{2}}\;x^{3/2}$

6. Solve for $x$

$x^{3/2}=K_p\sqrt{\frac{2}{p}}$

Squaring both sides: $x^3=\frac{2K_p^2}{p}$

Therefore, $x=\sqrt[3]{\frac{2K_p^2}{p}}$

7. Match with options

This matches Option C.

Comparison with stored answer: Stored correct answer is C, which matches the derived result.