JEE Main 2025 — Chemical Equilibrium Question with Solution

$\begin{array}{rl}& 1\mathrm{mol}\\ & {\mathrm{X}}_2\mathrm{Y}(\mathrm{g})\to \underset{\mathrm{x}}{\mathrm{X}}{\mathrm{X}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{Y}}_2(\mathrm{g})\\ & \frac{\mathrm{x}}{2}\\ & \therefore P_{{\mathrm{x}}_2\mathrm{y}}\\ & =\frac{1-\mathrm{x}}{1+\frac{x}{2}}\times \mathrm{p}\\ & P_{x_2}=\frac{x}{1+\frac{x}{2}}\times p\end{array}$ $\begin{aligned} & P_{y_2}=\frac{x / 2}{1+\frac{x}{2}} \times p \\ \therefore \quad K_p= & \frac{\left(\frac{x}{1+\frac{x}{2}} p\right)\left(\frac{x}{2\left(1+\frac{x}{2}\right)} p\right)^{1 / 2}}{\left(\frac{1-x}{1+\frac{x}{2}}\right) \times p} \\ \therefore \quad & K_p=\left(\frac{x}{1-x}\right)\left(\frac{x}{2\left(1+\frac{x}{2}\right)}\right)^{1 / 2} \times p^{1 / 2} \end{aligned}$ $\therefore \mathrm{x}$ to be very small $\begin{aligned} \therefore & K_p=\frac{x^{3 / 2}}{2^{(1 / 2)}} \times p^{1 / 2} \\ \therefore & x^{3 / 2}=\frac{K_p \times 2^{1 / 2}}{p^{1 / 2}} \\ & x^3=\frac{K_p^2 \times 2}{p} \\ & x=\left(\frac{K_p^2 \times 2}{p}\right)^{1 / 3} \end{aligned}$