In a closed container for the reaction ${\mathrm{H}}_{2_{\left(\mathrm{g}\right)}}+{\mathrm{I}}_{2_{\left(\mathrm{g}\right)}}\rightleftharpoons 2{\mathrm{HI}}_{\left(\mathrm{g}\right)}$ the equilibrium concentrations $\left[{\mathrm{H}}_2\right]=0.5\mathrm{M},\left[{\mathrm{I}}_2\right]=0.5\mathrm{M}\text{ and }\left[\mathrm{HI}\right]=1.25\mathrm{M}$. If 0.5 moles of HI was added to 1 L of equilibrium mixture, then $\left[{\mathrm{H}}_2\right]$ at new equilibrium mixture is -------------M

Equilibrium constant 

$$\begin{gathered} \left({\mathrm{K}}_{\mathrm{C}}\right)=\frac{[\mathrm{HI}{]}^2}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]} \\ =\frac{[1.25{]}^2}{[0.5][0.5]} \\ =6.25 \\ {\mathrm{H}}_2 + {\mathrm{I}}_{2\left(\mathrm{g}\right)} \rightleftharpoons 2\mathrm{HI} \\ \mathrm{at} \mathrm{old} \mathrm{equilibrium} :0.5 0.5 1.25 \\ \mathrm{at} \mathrm{new} \mathrm{equilibrium} :=0.5+\frac{\mathrm{x}}{2} =0.5+\frac{\mathrm{x}}{2} 1.25+0.5-\mathrm{x} \\ \frac{(1.75-\mathrm{x}{)}^2}{\left(0.5+\frac{\mathrm{x}}{2}\right)\left(0.5+\frac{\mathrm{x}}{2}\right)}=6.25 \\ \frac{1.75-\mathrm{x}}{0.5+\frac{\mathrm{x}}{2}}=2.5 \\ 1.75-\mathrm{x}=1.25+1.25 \mathrm{x} \\ 2.25 \mathrm{x}=0.5 \\ \mathrm{x}=0.222 \\ \mathrm{Equilibrium} \mathrm{concentration} \mathrm{of} {\mathrm{H}}_2=0.61\mathrm{M} \end{gathered}$$