At temperature T, compound $\mathrm{AB}_{2(\mathrm{~g})}$ dissociates as $\mathrm{AB}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{AB}_{(\mathrm{g})}+\frac{1}{2} \mathrm{~B}_{2(\mathrm{~g})}$ having degree of dissociation x (small compared to unity). The correct expression for x in terms of $\mathrm{K}_{\mathrm{p}}$ and p is

(C) $\sqrt[3]{\frac{2 K_{p}^{2}}{p}}$

(A) $\sqrt[3]{\frac{2 K_{p}}{p}}$

(B) $\sqrt[4]{\frac{2 K_{p}}{p}}$

(D) $\sqrt{\mathrm{K}_{\mathrm{p}}}$

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Equilibrium

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

Answer: (C)

Starting with 1 mole of

AB_{2 (g)}

at total pressure p, at equilibrium the amounts are

AB_{2} = (1 - x)

AB = x

B_{2} = x /2

, giving total moles

1 + x /2

. Since

x ≪ 1

1 - x \approx 1

and

1 + x /2 \approx 1

, so the partial pressures simplify to

p_{AB_{2}} \approx p

p_{AB} \approx x p

p_{B_{2}} \approx (x /2) p

. Then

K_{p} = \frac{p _{AB} ( p _{B_{2}} ) ^{1/2}}{p _{AB_{2}}} = \frac{( x p ) ( x p /2 ) ^{1/2}}{p} = x \frac{x p}{2}

Squaring both sides,

K_{p}^{2} = \frac{x ^{3} p}{2}

, so

x = 3 \frac{2 K _{p}^{2}}{p}

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option