Question

Calculate the degree of dissociation of $HI$ as $2HI\rightleftharpoons H_2+I_2$ at $450{}^{o}C$ , if the equilibrium constant for the dissociation reaction is 0.263.

• A. $0.688$
• B. $0.326$
• C. $0.454$
• D. $0.253$

Answer 1

The correct option is D $0.253$

Let degree of dissociation of $HI$ = $\alpha$
Let initial moles of the $HI$ be $1$ and the volume be V
$2HI\rightleftharpoons H_2+I_2$
initial moles $1$ $0$ $0$
At equili. $1-2\alpha$ $\alpha$ $\alpha$
$K_c$ = $\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}$
$K_c$ = $\frac{\frac{\alpha }{V}\times \frac{\alpha }{V}}{[\frac{1-2\alpha }{V}{]}^2}$
$K_c$ = $\frac{(\alpha {)}^2}{(1-2\alpha {)}^2}$
$\sqrt{0.263}=\frac{\alpha }{1-2\alpha }$
Solving for $\alpha$, we get 0.253