The equilibrium constant for the reaction ${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right)\iff 2\mathrm{HI}\left(\mathrm{g}\right)$is 2 at a certain temperature .The equilibrium concentrations of ${\mathrm{H}}_2 \mathrm{and} \mathrm{HI} \mathrm{are} 2 \mathrm{mole} .{\mathrm{lit}}^{-1}.$The equilibrium concentrations of $(\mathrm{in} \mathrm{mole}.{\mathrm{lit}}^{-1}) \mathrm{of} {\mathrm{I}}_2?$

The chemical equilibrium reactions is

${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{HI}\left(\mathrm{g}\right)$

Equilibrium constant ${\mathrm{K}}_{\mathrm{C}}=2$

At equilibrium,

Concentration of HI, $\left[\mathrm{HI}\right]=2 mole.li{t}^{-1}$

Concentration of ${\mathrm{H}}_2,$$\left[{\mathrm{H}}_2\right]=2 mole.li{t}^{-1}$

${\mathrm{K}}_{\mathrm{C}}=\frac{{\left[\mathrm{HI}\right]}^2}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]}$

$\left[{\mathrm{I}}_2\right]=\frac{{\left[HI\right]}^2}{\left[H_2\right]K_C}=\frac{2\times 2}{2\times 2}$

$\therefore {\left[{\mathrm{I}}_2\right]}_{\mathrm{equilibrium}}=1.$