JEE Main 2025 — Chemical Equilibrium Question with Solution

$A{B}_2(g)\rightleftharpoons A{B}_{(g)}+\frac{1}{2}{B}_2(g)$

Degree of dissociation $\to x$ (small compared to unity)

Equilibrium constant interms of pressure $\to k_p$

Partial pressure of each gas $\to P_{A{B}_2},P_{AB}$ and $P_{B_2}$

Total pressure $\to P$

ICE Table:

$\begin{array}{rl}& \text{ partial pressure }\\ & \text{ at equilibrium }\end{array}=\frac{\text{ moles at equilibrium }}{\begin{array}{l}\text{ total moles at }\\ \text{ equilibrium }\end{array}}\times$ total pressure

Partial pressure of $A{B}_2(g)$

$$P_{A{B}_2}=\left(\frac{1-x}{1+\frac{x}{2}}\right)P$$

Partial pressure of $AB(g)$

$$P_{AB}=\left(\frac{x}{1+\frac{x}{2}}\right)P$$

Partial pressure of $B_2(g)$

$$P_{B_2}={\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}\right)}^P$$

$K_p$ for the equation $A_2(g)\rightleftharpoons A{B}_{(g)}+\frac{1}{2}{B}_2(g)$ can be written as

$k_p=\frac{P_{AB}{P}_{B_2}^{1/2}}{P_{A{B}_2}}$

$\begin{array}{rl}& =\frac{\frac{x}{1+\frac{x}{2}}P1\times {\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}P\right)}^{1/2}}{\frac{1-x}{1+\frac{x}{2}}P}\\ & =\frac{x\times {\left(\frac{x}{1+\frac{x}{2}}P\right)}^{1/2}}{1-x}\end{array}$

$$\begin{array}{rl}& \text{ given, }x⋘1\\ & \begin{array}{rl}& \text{ So, }1-x\approx 1\\ & K_p=\frac{x\times {\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}p\right)}^{1/2}}{1}\end{array}\end{array}$$

$=x\times \frac{{\left(\frac{x}{2}\right)}^{1/2}}{{\left(\frac{2+x}{2}\right)}^{1/2}}{p}^{1/2}$

$=\frac{x\times \frac{x^{1/2}}{2^{1/2}}\times p^{1/2}}{\frac{(2+x{)}^{1/2}}{2^{1/2}}}$

$=\frac{x^{3/2}\times p^{1/2}}{2^{1/2}}\times \frac{2^{1/2}}{(2+x{)}^{1/2}}$

$$\begin{array}{c}x⋘1\\ 2+x\approx 2\end{array}$$

$$\begin{array}{rl}& =\frac{x^{3/2}\times p^{1/2}}{2^{1/2}}\\ & ={\left(\frac{x^{3}p}{2}\right)}^{1/2}\end{array}$$

$$\begin{array}{rl}& k_p={\left(\frac{x^{3}p}{2}\right)}^{1/2}\\ & \text{ Square on both sides, }\end{array}$$

$$\begin{array}{rl}{k}_p^2& =\frac{x^{3}p}{2}\\ x^{3}p& =2k_p^2\\ x^3& =\frac{2k_p^2}{p}\\ x& =\sqrt[3]{\frac{2k_p^2}{p}}\end{array}$$

Correct answer option (2) $\sqrt[3]{\frac{2K_p^2}{p}}$