In a closed container, when initially 60mL of $H_2$ and 42 mL of $I_2$ are heated, the equilibrium volume of HI formed was found to be 28mL. Find the degree of dissociation of HI. All measurements are made at the same temperature and pressure conditions.

                       $H_{2_{\left(g\right)}}+I_{2_{\left(g\right)}}\underset{ }{\rightleftarrows }2H{I}_{\left(g\right)}$

Vol initial:    60          42              0
Vol eqbm    60-x       42-x           2x
Under same $P\&T$, $V\alpha$ No. of moles. Also since  $\Delta n_g=0,$
$K_c=\frac{{\left(2x\right)}^2}{(60-x)(42-x)}=\frac{{\left(28\right)}^2}{(60-14)(42-14)}=\frac{28}{46}$
For dissociation
$\begin{array}{l}2H{I}_{\left(g\right)}\underset{ }{\rightleftarrows }{H}_{2_{\left(g\right)}}+I_{2_{\left(g\right)}}{K}_c^1=\frac{46}{28}\\ 1-\alpha \frac{\alpha }{2}\frac{\alpha }{2}\end{array}$

$\therefore K_c=\frac{46}{28}=\frac{{\alpha }^2}{4{(1-\alpha )}^2}or\frac{\alpha }{2(1-\alpha )}=\sqrt{\frac{46}{28}}$

$\alpha =0.719$