Question

In the dissociation of $HI,20\%$ of $HI$ is dissociated at equilibrium. Calculate $K_p$ for

$H{I}_{\left(g\right)}\to \frac{1}{2}{H}_{2\left(g\right)}+\frac{1}{2}{I}_{2\left(g\right)}$.

Answer 1

$HI\rightleftharpoons \frac{1}{2}{H}_2+\frac{1}{2}{I}_2$
$1$ $0$ $0$ --- Initial moles
$(1-\alpha )$ $\frac{\alpha }{2}$ $\frac{\alpha }{2}$ ---- At Eqbm.

where $\alpha$ is the degree of dissociation and volume of the container is $V$ litre.

$K_p=K_c=\frac{{\left(\frac{\alpha }{2V}\right)}^{1/2}{\left(\frac{\alpha }{2V}\right)}^{1/2}}{\frac{(1-\alpha )}{V}}(\because △n=0)$

$K_p=K_c=\frac{\alpha }{2(1-\alpha )}(\alpha =0.2)$

$K_p=K_c=\frac{0.2}{2(1-0.2)}$

$K_p=K_c=0.125$.