QJEEM25Ch5f04: JEE Main 2025 Chemistry - Class 11 JEEM25Ch5f04 Solution: Consider the reaction The equation representing correct…

$\begin{aligned}& 1 \text { mole } \\& \mathrm{X}_2 \mathrm{Y}_{(\mathrm{g})} \rightleftharpoons \mathrm{X}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{Y}_{2(\mathrm{~g})} \\& \text { 1-x mole } \quad \mathrm{x} \text { mole } \quad \frac{\mathrm{x}}{2} \text { mole } \\& \therefore \mathrm{P}_{\mathrm{X}_2 \mathrm{Y}}=\frac{1-\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P} \\& \mathrm{P}_{\mathrm{X}_2}=\frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P} \\& \mathrm{P}_{\mathrm{Y}_2}=\frac{\mathrm{x} / 2}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P}\end{aligned}$

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

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

$\because \mathrm{x}$ to be very very small

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

(2) ${ }^{\frac{1}{2}}$

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

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

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