For the reaction $$ 2 \mathrm{HI}(\mathrm{g}) ightleftharpoons…
$$
2 \mathrm{HI}(\mathrm{g}) ightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})
$$
The degree of dissociation $(\alpha)$ of $\mathrm{HI}(\mathrm{g})$ is related to equilibrium constant, $K_{p}$ by the expression
- $\frac{1+2 \sqrt{\mathrm{K}_{\mathrm{p}}}}{2}$
- $\sqrt{\frac{1+2 \mathrm{~K}_{\mathrm{p}}}{2}}$
- $\sqrt{\frac{2 \mathrm{~K}_{\mathrm{p}}}{1+2 \mathrm{~K}_{\mathrm{p}}}}$
- $\frac{2 \sqrt{\mathrm{K}_{\mathrm{p}}}}{1+2 \sqrt{\mathrm{K}_{\mathrm{p}}}}$
Solution
$\mathrm{K}_{\mathrm{p}}=\frac{\left(\frac{\alpha}{2} \mathrm{P}_{\mathrm{T}}ight)^{2}}{(1-\alpha)^{2} \mathrm{P}_{\mathrm{T}}^{2}}$
$\frac{\alpha}{1-\alpha}=2 \sqrt{\mathrm{K}_{\mathrm{p}}}$
$\alpha=\frac{2 \sqrt{\mathrm{K}_{\mathrm{p}}}}{1+2 \sqrt{\mathrm{K}_{\mathrm{p}}}}$
Asked in: JEE-TOPICTESTS-CHEMISTRY