JEE MAIN - Chemistry (2025 - 23rd January Evening Shift - No. 15)
Consider the reaction
$$\mathrm{X}_2 \mathrm{Y}(\mathrm{~g}) \rightleftharpoons \mathrm{X}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Y}_2(\mathrm{~g})$$
The equation representing correct relationship between the degree of dissociation (x) of $\mathrm{X}_2 \mathrm{Y}(\mathrm{g})$ with its equilibrium constant Kp is __________.
Assume $x$ to be very very small.
Explanation
$$\begin{aligned} & 1 \text { mole } \\ & \mathrm{X}_2 \mathrm{Y}_{(\mathrm{g})} \rightleftharpoons \mathrm{X}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{Y}_{2(\mathrm{~g})} \\ & \text { 1-x mole } \quad \mathrm{x} \text { mole } \quad \frac{\mathrm{x}}{2} \text { mole } \\ & \therefore \mathrm{P}_{\mathrm{X}_2 \mathrm{Y}}=\frac{1-\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P} \\ & \mathrm{P}_{\mathrm{X}_2}=\frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P} \\ & \mathrm{P}_{\mathrm{Y}_2}=\frac{\mathrm{x} / 2}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P} \end{aligned}$$
$\therefore \mathrm{K}_{\mathrm{p}}=\left(\frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \mathrm{P}\right)\left(\frac{\mathrm{x}}{2\left(1+\frac{\mathrm{x}}{2}\right)} \mathrm{P}\right)^{\frac{1}{2}} /\left(\frac{1-\mathrm{x}}{1+\frac{\mathrm{x}}{2}}\right) \times \mathrm{P}$
$\therefore \mathrm{K}_{\mathrm{p}}=\left(\frac{\mathrm{x}}{1-\mathrm{x}}\right)\left(\frac{\mathrm{x}}{2\left(1+\frac{\mathrm{x}}{2}\right)}\right)^{\frac{1}{2}} \times \mathrm{p}^{\frac{1}{2}}$
$\because \mathrm{x}$ 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) ${ }^{\frac{1}{2}}$
$$\therefore \mathrm{x}^{\frac{3}{2}}=\frac{\mathrm{K}_{\mathrm{p}} \times 2^{\frac{1}{2}}}{\mathrm{P}^{\frac{1}{2}}}$$
$$\therefore \mathrm{x}^3=\frac{\mathrm{K}_{\mathrm{p}}^2 \times 2}{\mathrm{P}}$$
$$\mathrm{x}=\left(\frac{\mathrm{K}_{\mathrm{p}}^2 \times 2}{\mathrm{P}}\right)^{\frac{1}{3}}$$
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