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Q.

At temperature T, compound AB_2(g) dissociates as having degree of dissociation x (small compared to unity). The correct expression for x in terms of K_p and p is

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a

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

b

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

c

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

d

$\sqrt{K_{p}}$

answer is C.

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Detailed Solution

$\begin{array}{l} {AB}_{2 (g)} ⇌ {AB}_{(g)} + \frac{1}{2} B_{2 (g)} \\ t_{eq.} \frac{(1 - x)}{1 + \frac{x}{2}} P \frac{xP}{1 + \frac{x}{2}} \frac{(\frac{x}{2}) P}{1 + \frac{x}{2}} \\ \Rightarrow x ≪ 1 \Rightarrow 1 + \frac{x}{2} ≃ 1 and 1 - x ≃ 1 \\ \begin{array}{l} \Rightarrow k_{P} = \frac{(xp) \cdot {(\frac{xp}{2})}^{\frac{1}{2}}}{P} \\ \Rightarrow k_{P}^{2} = x^{2} \cdot \frac{xP}{2} \\ x = \sqrt[3]{\frac{2 k_{P}^{2}}{P}} \end{array} \end{array}$

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