The equilibrium constant for the reaction, ${\mathrm{H}}_2+{\mathrm{I}}_2\underset{ }{\rightleftharpoons } 2\mathrm{HI}$ at 700K is 56. If 0.5 mole of hydrogen and 1.0 mole of iodine are added in the system at equilibrium, the value of equilibrium constant will be.

                              H₂              +I₂        →      2HI
At t=0                   a                a                 0
At equilibrium:- (a-x)           (a-x)            2x
$\begin{array}{l}\mathrm{Equilibrium} \mathrm{constant}(\mathrm{k}) =\frac{2{\mathrm{x}}^2}{(\mathrm{a}-\mathrm{x}{)}^2}\\ \therefore 56=\frac{4{\mathrm{x}}^2}{{\mathrm{a}}^2-2\mathrm{x}+{\mathrm{x}}^2}\\ \therefore 56{\mathrm{a}}^2-112\mathrm{x}+56{\mathrm{x}}^2-4{\mathrm{x}}^2=0\\ \therefore 52{\mathrm{x}}^2-112\mathrm{x}+56{\mathrm{a}}^2=0\\ \therefore 13{\mathrm{x}}^2-28\mathrm{x}+14{\mathrm{a}}^2=0\\ \mathrm{Assuming}, \mathrm{a}=1\\ \therefore 13{\mathrm{x}}^2-28\mathrm{x}+14=0\\ \therefore \mathrm{x}=0.789\end{array}$
Assuming, a=1
$\begin{array}{l}\therefore 13{\mathrm{x}}^2-28\mathrm{x}+14=0\\ \therefore \mathrm{x}=0.789\end{array}$
$\therefore$If 0.5 mole of Hydrogen of 1 mole of Iodine is added at equilibrium then,
$\begin{array}{l}{\mathrm{k}}^{\mathrm{\prime }}=\frac{2\mathrm{x}}{(\mathrm{a}-\mathrm{x}+0.5)(\mathrm{a}-\mathrm{x}+1)}\\ {\mathrm{k}}^{\mathrm{\prime }}=\frac{2\times 0.789}{(1-\mathrm{x}+0.5)(-\mathrm{x})}\\ {\mathrm{k}}^{\mathrm{\prime }}=\frac{2\times 0.789}{(1-0.789+0.5)(2-0.789)}\\ \mathrm{k}=25\end{array}$