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At temperature T, a compound AB2(g) dissociates according to the reaction $2 {AB}_{2} (g) ⇌ 2 AB (g) + B_{2} (g)$ with a degree of dissociation ‘x’ which is small as compared to unity. The expression for $K_{P},$ in terms of ‘x’ and total pressure ‘P’ is

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$\frac{{Px}^{3}}{2}$

$\frac{{px}^{2}}{3}$

$\frac{{px}^{3}}{3}$

$\frac{{px}^{2}}{2}$

answer is A.

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Detailed Solution

$2 {AB}_{2} (g) ⇌ 2 AB (g) + B_{2} (g)$

Initial : P O O

At equli: $p - xp px \frac{xp}{2}$

$∴ K_{p} = \frac{P_{AB .} {}^{2}P_{B_{2}}}{P_{{AB}_{2}}}$

$= \frac{{(p_{2})}^{2} [\frac{xP}{2}]}{{(p - xp)}^{2}}$

$K_{P} = \frac{P^{2} x^{2} . X . P}{2} \times \frac{1}{P^{2}} = \frac{{px}^{2}}{2}$

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