At temperature T a compound $A B_{2} (g)$ dissociates according to the reaction
$2 A B_{2} (g) \overset{}{⇌} 2 A B (g) + B_{2} (g)$
with a degree of dissociation ‘x’ which is very small compared to unity. The value of x is …… (consider ‘P’ at equilibrium position)

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$\sqrt{\frac{2 K_{p}}{P}}$

$\sqrt[3]{\frac{2 K_{p}}{P}}$

$\sqrt[3]{\frac{K_{p}}{P}}$

$\sqrt[3]{K_{p}}$

answer is B.

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Detailed Solution

$\underset{1 - x}{\underset{\begin{array}{l} 1 \\ I n i t i a l l y \end{array}}{2 A B_{2} (g)}} \overset{}{⇌} \underset{x}{\underset{\begin{array}{l} 0 \end{array}}{2 A B (g)}} + \underset{\frac{x}{2}}{\underset{\begin{array}{l} 0 \end{array}}{B_{2}}} (g)$
At equilibrium
Total moles at equilibrium $= 1 - x + x + \frac{x}{2} = 1 + \frac{x}{2} = 1$
$[∴ x is small in comparision to unity]$
$p_{A B_{2}} = (1 - x) P p_{A B} = x P {P_{B}}_{_{2}} = \frac{x P}{2}$
$K_{p} = \frac{x^{3} P^{3}}{2 {(1 - x)}^{2} P^{2}} = \frac{x^{3} P}{2} [∴ (1 - x) \approx 1]$