JEE Main 2025 — 23 January, Evening Shift — Question 30

1 mole

$\mathrm{X}_{2} \mathrm{Y}_{(\mathrm{g})} \rightleftharpoons \mathrm{X}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{Y}_{2(\mathrm{~g})}$

1 -x mole $\quad x$ mole $\quad \frac{x}{2}$ mole

$\therefore \mathrm{P}_{\mathrm{X}_{2} \mathrm{Y}}=\frac{1-\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P}$ $P_{X_{2}}=\frac{x}{1+\frac{x}{2}} \times P$ $\mathrm{P}_{\mathrm{Y}_{2}}=\frac{\mathrm{x} / 2}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P}$

$\therefore \mathrm{K}_{\mathrm{p}}=\left(\frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \mathrm{P}\right)\left(\frac{\mathrm{x}}{2\left(1+\frac{\mathrm{x}}{2}\right)} \mathrm{P}\right)^{\frac{1}{2}} /\left(\frac{1-\mathrm{x}}{1+\frac{\mathrm{x}}{2}}\right) \times \mathrm{P}$

$\therefore K_{p}=\left(\frac{x}{1-x}\right)\left(\frac{x}{2\left(1+\frac{x}{2}\right)}\right)^{\frac{1}{2}} \times \mathrm{p}^{\frac{1}{2}}$

$\because \mathrm{x}$ to be very very small $\therefore \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{x}^{3 / 2}}{1} \times \mathrm{P}^{\frac{1}{2}}$ $(2)^{\frac{1}{2}}$

$\therefore \mathrm{x}^{\frac{3}{2}}=\frac{\mathrm{K}_{\mathrm{p}} \times 2^{\frac{1}{2}}}{\mathrm{P}^{\frac{1}{2}}}$

$\therefore \mathrm{x}^{3}=\frac{\mathrm{K}_{\mathrm{p}}^{2} \times 2}{\mathrm{P}}$

$\mathrm{x}=\left(\frac{\mathrm{K}_{\mathrm{p}}^{2} \times 2}{\mathrm{P}}\right)^{\frac{1}{3}}$