At temperature T, compound \(\mathrm{AB}_{2(\mathrm{~g})}\) dissociates as \(\mathrm{AB}_{2(\mathrm{~g})}…

1. \(\sqrt[4]{\frac{2 K_p}{p}}\)
2. \(\sqrt[3]{\frac{2 K_{\mathrm{p}}}{\mathrm{p}}}\)
3. \(\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}^2}{\mathrm{p}}}\)
4. \(\sqrt{K_p}\)

Solution

$\mathrm{AB}_2(\mathrm{~g}) \rightleftharpoons \mathrm{AB}(\mathrm{g})+\frac{1}{2} \mathrm{~B}_2(\mathrm{~g})$

$\begin{gathered}p=p_0\left(1+\frac{x}{2}\right) \\ p_0=\frac{p}{\left(1+\frac{x}{2}\right)} \\ K_p=\frac{\left(p_{A B}\right)\left(p_{B_2}\right)^{1 / 2}}{\left(p_{\mathrm{AB}_2}\right)} \\ K_p=\frac{\left(p_0 x\right)\left(\frac{p_0 x}{2}\right)^{1 / 2}}{p_0(1-x)}\end{gathered}$
$\mathrm{K}_{\mathrm{p}}=\frac{\frac{\mathrm{px}}{\left(1+\frac{x}{2}\right)}\left(\frac{\mathrm{p}}{1+\frac{x}{2}} \times \frac{\mathrm{x}}{2}\right)^{1 / 2}}{\frac{\mathrm{p}(1-\mathrm{x})}{\left(1+\frac{x}{2}\right)}}$
Since $x \ll 1$
$\begin{aligned}
& K_p=\frac{\mathrm{p}^{1 / 2} \mathrm{x}^{3 / 2}}{2^{1 / 2}} \\
& \mathrm{x}=\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}^2}{\mathrm{p}}}
\end{aligned}$

Asked in: JEE Main 2025 (29 Jan Shift 1)

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