Consider the reaction $\mathrm{X}_2 \mathrm{Y}(\mathrm{~g})=\mathrm{X}_2(\mathrm{~g})+\frac{1}{2}…

Consider the reaction
$\mathrm{X}_2 \mathrm{Y}(\mathrm{~g})=\mathrm{X}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Y}_2(\mathrm{~g})$
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.
  1. $x=\sqrt[3]{\frac{2 \mathrm{Kp}}{\mathrm{p}}}$
  2. $x=\sqrt[3]{\frac{2 \mathrm{Kp}^2}{\mathrm{p}}}$
  3. $x=\sqrt[3]{\frac{\mathrm{Kp}}{\mathrm{p}}}$
  4. $x=\sqrt[3]{\frac{K \mathrm{p}}{2 \mathrm{p}}}$

Solution

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

Asked in: JEE Main 2025 (23 Jan Shift 2)

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