JEE Main 2025 — Chemical Equilibrium Question with Solution

To determine the relationship between the degree of dissociation ($x$) of ${\mathrm{X}}_2\mathrm{Y}(\mathrm{g})$ and its equilibrium constant $K_p$, let's analyze the reaction:

${\mathrm{X}}_2\mathrm{Y}(\mathrm{g})\rightleftharpoons {\mathrm{X}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{Y}}_2(\mathrm{g})$

Initial and Change in Moles

Initially, we start with 1 mole of ${\mathrm{X}}_2\mathrm{Y}$.

At equilibrium:

${\mathrm{X}}_2\mathrm{Y}$ is $(1-x)$ moles.

${\mathrm{X}}_2$ is $x$ moles.

$\frac{1}{2}{\mathrm{Y}}_2$ is $\frac{x}{2}$ moles.

Partial Pressures

The total pressure ${\mathrm{P}}_{\text{total}}$ is given by:

${\mathrm{P}}_{{\mathrm{X}}_2\mathrm{Y}}=\frac{1-x}{1+\frac{x}{2}}\times \mathrm{P}$

${\mathrm{P}}_{{\mathrm{X}}_2}=\frac{x}{1+\frac{x}{2}}\times \mathrm{P}$

${\mathrm{P}}_{{\mathrm{Y}}_2}=\frac{x/2}{1+\frac{x}{2}}\times \mathrm{P}$

Equilibrium Constant Expression

The equilibrium constant $K_p$ can be written in terms of partial pressures:

$K_p=\frac{\left(\frac{x}{1+\frac{x}{2}}\cdot \mathrm{P}\right)\cdot {\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}\cdot \mathrm{P}\right)}^{1/2}}{\frac{1-x}{1+\frac{x}{2}}\cdot \mathrm{P}}$

Simplifying the expression, considering $x$ to be very small, results in:

$K_p=\left(\frac{x}{1-x}\right){\left(\frac{x}{2\left(1+\frac{x}{2}\right)}\right)}^{1/2}\cdot {\mathrm{P}}^{\frac{1}{2}}$

For small $x$, $(1-x)\approx 1$, thus:

$K_p=\frac{x^{3/2}}{\frac{1}{3}}\cdot {\mathrm{P}}^{\frac{1}{2}}$

$x^{3/2}=\frac{K_p\cdot 2^{1/2}}{{\mathrm{P}}^{1/2}}$

Finally, by cubing both sides:

$x^3=\frac{2\cdot K_p^2}{\mathrm{P}}$

$x={\left(\frac{2\cdot K_p^2}{\mathrm{P}}\right)}^{1/3}$

Thus, the degree of dissociation $x$ relates to $K_p$ by the equation:

$x={\left(\frac{2\cdot K_p^2}{\mathrm{P}}\right)}^{1/3}$