At temperature T, compound AB2(g) dissociates as AB2(g) ⇄ AB(g) + 1/2 B2(g) having degree

Question

At temperature T , compound \(\mathrm{AB}_{2(\mathrm{~g})}\) dissociates as \(\mathrm{AB}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{AB}_{(\mathrm{g})}+\frac{1}{2} \mathrm{~B}_{2(\mathrm{~g})}\) having degree of dissociation x (small compared to unity). The correct expression for x in terms of \(\mathrm{K}_{\mathrm{p}}\) and p is

(1) \(\sqrt[3]{\frac{2 K_{p}}{p}}\)

(2) \(\sqrt[4]{\frac{2 K_{p}}{p}}\)

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

(4) \(\sqrt{\mathrm{K}_{\mathrm{p}}} \)

Answer 1

Correct option is (3) \(\sqrt[3]{\frac{2 K_{p}^{2}}{p}}\)  

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

\(t_{\text {eq. }} \frac{(1-x)}{1+\frac{x}{2}} P \frac{x P}{1+\frac{x}{2}} \frac{\left(\frac{x}{2}\right) P}{1+\frac{x}{2}}\)

\(\Rightarrow \mathrm{x} \ll 1 \Rightarrow 1+\frac{\mathrm{x}}{2} \simeq 1\) and \(1-\mathrm{x} \simeq 1\)

\(\Rightarrow k_{P}=\frac{(x p) \cdot\left(\frac{x p}{2}\right)^{\frac{1}{2}}}{P}\)

\(\Rightarrow \mathrm{k}_{\mathrm{P}}^{2}=\mathrm{x}^{2} \cdot \frac{\mathrm{xP}}{2}\)

\(x=\sqrt[3]{\frac{2 \mathrm{k}_{\mathrm{P}}^{2}}{\mathrm{P}}}\)