In a closed container for the reaction $H_{2_{\left(g\right)}}+I_{2_{\left(g\right)}}\rightleftharpoons 2H{I}_{\left(g\right)}$ the equilibrium concentrations $\left[H_2\right]=0.5M,\left[I_2\right]=0.5M\text{ and }\left[HI\right]=1.25M$. If 0.5 moles of HI was added to 1 L of equilibrium mixture, then $\left[H_2\right]$ at new equilibrium mixture is

Equilibrium constant 

$$\begin{gathered} \left(K_C\right)=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]} \\ =\frac{[1.25{]}^2}{[0.5][0.5]} \\ =6.25 \\ H_{2_2,}+I_{2\left(g\right)}\rightleftharpoons 2HI \\ at old equilibrium 0.5 0.5 1.25 \\ at new equilibrium \frac{+\frac{x}{2}}{0.5+\frac{x}{2}}\frac{+\frac{x}{2}}{0.5+\frac{x}{2}} \frac{-x}{1.75-x} \\ \frac{(1.75-x{)}^2}{\left(0.5+\frac{x}{2}\right)\left(0.5+\frac{x}{2}\right)}=6.25 \\ \frac{1.75-x}{0.5+\frac{x}{2}}=2.5 \\ 1.75-x=1.25+1.25 x \\ 2.25 x=0.5 \\ x=0.222 \\ Equilibrium concentration of H_2=0.61M \end{gathered}$$