JEE Main 2025 Jan 29 Shift 1, Chemistry Q4: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2025Jan 29, Shift 1Chemistry

At temperature $T$ , compound $AB_{2 (g)}$ dissociates as $AB_{2 (g)} ⇌ AB_{(g)} + \frac{1}{2} B_{2 (g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_{p}$ and $p$ is

A
$3 \frac{2 K _{p}}{p}$
B
$4 \frac{2 K _{p}}{p}$
C
$3 \frac{2 K _{p}^{2}}{p}$
D
$K_{p}$

Solution

Starting with 1 mole of $AB_{2}$ and degree of dissociation $x$ , equilibrium moles are $(1 - x)$ , $x$ , $x /2$ with total $n_{T} = 1 + x /2$ .

The partial pressures are $p_{i} = \frac{n _{i}}{n _{T}} p$ , so

$K_{p} = \frac{p _{AB} \cdot p _{B_{2}}^{1/2}}{p _{A B_{2}}} = \frac{( \frac{x}{1 + x /2} p ) ( \frac{x /2}{1 + x /2} p ) ^{1/2}}{( \frac{1 - x}{1 + x /2} p )} .$

Since $x ≪ 1$ , $1 - x \approx 1$ and $1 + x /2 \approx 1$ . The expression collapses to

$K_{p} \approx x p \cdot (\frac{x p}{2})^{1/2} = x^{3/2} \frac{p}{2} \cdot p^{1/2} \cdot \frac{1}{1} \Rightarrow K_{p}^{2} = \frac{x ^{3} p}{2} .$

Solving for $x$ : $x = 3 \frac{2 K _{p}^{2}}{p} .$

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JEE Main 2025 Jan 29 Shift 1, Chemistry Q4: Law of Chemical Equilibrium And Equilibrium Constant

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