At temperature T, compound dissociates as having degree of dissociation (small compared to unity). The correct expression for in terms of and is [2025]

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Solution

Q.
At temperature T, compound $A B_{2} (g)$ dissociates as $A B_{2} (g) ⇌ A B (g) + \frac{1}{2} B_{2 (g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_{p}$ and $p$ is [2025]

1 $\sqrt[3]{\frac{2 K_{p}^{2}}{p}}$

2 $\sqrt{K_{p}}$

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

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

Ans.
(1)

$\begin{matrix} {AB}_{2} (g) & ⇌ & AB (g) & + & \frac{1}{2} B_{2} (g) \\ t=0 (moles) & a & 0 & 0 \\ t =∞ (moles) & a (1 - x) & a x & a x / 2 \\ t =∞ (mole fraction) & \frac{a (1 - x)}{a (1 + \frac{x}{2})} & \frac{a x}{a (1 + \frac{x}{2})} & \frac{a x}{a (1 + \frac{x}{2})} \\ t =∞ (partial pressure) & \frac{a (1 - x)}{a (1 + \frac{x}{2})} p & \frac{a x}{a (1 + \frac{x}{2})} p & \frac{a x / 2}{a (1 + \frac{x}{2})} p \\ \approx p & \approx x p & \approx \frac{x p}{2} \end{matrix} \begin{matrix} Total = a (1 + x / 2) \end{matrix}$
$K_{p} = \frac{P_{AB} \times {(P_{B_{2}})}^{1 / 2}}{P_{{AB}_{2}}} = \frac{x p \times {(\frac{x p}{2})}^{1 / 2}}{p}$

${(K_{p})}^{2} = x^{3} p / 2$

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

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