Q68 / 199 · NEET 2008 (Mains) · Chemistry
The dissociation equilibrium of a gas $\mathrm{AB}_2$ can be represented as, $2 \mathrm{AB}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_2(\mathrm{~g})$. The degree of dissociation is ' $\mathrm{x}$ ' and is small compared to 1. The expression relating the degree of dissociation $(\mathrm{x})$ with equilibrium constant $K_P$ and total pressure $\mathrm{P}$ is
- A$\left(\frac{K_p}{P}\right)$
- B$\left(\frac{2 K_P}{P}\right)$
- Correct answerC$\left(\frac{2 K_P}{P}\right)^{1 / 3}$
- D$\left(\frac{2 K_P}{P}\right)^{1 / 2}$
Solution
$$
\begin{aligned}
& 2 \mathrm{AB}_2 \rightleftharpoons 2 \mathrm{AB}+\mathrm{B}_2 \\
& 1 \\
& 00 \\
& 1-x \\
& \mathrm{x} \quad \mathrm{x} / 2 \\
&
\end{aligned}
$$
Total mole at equi. $=1+\frac{x}{2}$
$$
\therefore K_P=\frac{\left(\frac{x}{1+x / 2} \times P\right)^2\left(\frac{x / 2}{1+x / 2} \times P\right)}{\left(\frac{1-x}{1+x / 2} \times P\right)^2}
$$
(Here $x$ is degree of dissociation)
or $K_P=\frac{x^3 P}{2}$
or $x^3=\frac{2 K_P}{P}$
or $x=\left(\frac{2 K_P}{P}\right)^{1 / 3}$
Hence, the correct answer is option: 'C'.