JEE MAIN - Chemistry (2025 - 29th January Morning Shift - No. 13)
At temperature T, compound $AB_{2(g)}$ dissociates as $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$ having degree of dissociation $ x $ (small compared to unity). The correct expression for $ x $ in terms of $ K_p $ and $ p $ is:
$ \sqrt{K_p} $
$\sqrt[3]{\frac{2 K_{\mathrm{p}}^2}{\mathrm{p}}}$
$\sqrt[3]{\frac{2 K_p}{p}}$
$\sqrt[4]{\frac{2 K_p}{p}}$
Explanation
$$\begin{aligned}
& \mathrm{AB}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{AB}_{(\mathrm{g})}+\frac{1}{2} \mathrm{~B}_{2(\mathrm{~g})} \\
& \mathrm{t}_{\mathrm{cq}} \cdot \frac{(1-\mathrm{x})}{1+\frac{\mathrm{x}}{2}} \mathrm{P} \frac{\mathrm{xP}}{1+\frac{\mathrm{x}}{2}} \frac{\left(\frac{\mathrm{x}}{2}\right) \mathrm{P}}{1+\frac{\mathrm{x}}{2}} \\
& \Rightarrow \mathrm{x} \ll 1 \Rightarrow 1+\frac{\mathrm{x}}{2} \simeq 1 \text { and } 1-\mathrm{x} \simeq 1 \\
& \Rightarrow \mathrm{k}_{\mathrm{P}}=\frac{(\mathrm{xp}) \cdot\left(\frac{\mathrm{xp}}{2}\right)^{\frac{1}{2}}}{\mathrm{P}} \\
& \Rightarrow \mathrm{k}_{\mathrm{P}}^2=\mathrm{x}^2 \cdot \frac{\mathrm{xP}}{2} \\
& \mathrm{x}=\sqrt[3]{\frac{2 \mathrm{k}_{\mathrm{P}}^2}{\mathrm{P}}}
\end{aligned}$$
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