At a certain temperature the equilibrium constant ${\text{K}}_{\text{c}}$ is 0.25 for the reaction ${\text{A}}_{2\left(g\right)}+B_{2\left(g\right)}\rightleftharpoons C_{2\left(g\right)}+D_{2\left(g\right)}$ . If we take 1 mole of each of the four gases in a 10 litre container, what would be equilibrium concentration of  ${\text{A}}_{\text{2}\left(\text{g}\right)}$ ?

$QC=\frac{1\times 1}{1\times 1}=1$ 
        $Q_C>K_C\to \text{backward\hspace{0.17em}\hspace{0.17em}reaction}$
Given that  $A_2+B_2\rightleftharpoons C_2+D_2$
                   $C_2+D_2\rightleftharpoons A_2+B_2$
                   $\frac{(1-x)}{10}\frac{(1-x)}{10}\frac{(1+x)}{10}\frac{(1+x)}{10}$
                   $1<C=\frac{\left[A_2\right]\left[B_2\right]}{\left[C_2\right]\left[D_2\right]}=\frac{{\left(\frac{1-x}{10}\right)}^2}{{\left(\frac{1+x}{10}\right)}^2}$
                   $0.25=\frac{{\left(\frac{1-x}{10}\right)}^2}{{\left(\frac{1+x}{10}\right)}^2}\Rightarrow x=0.333$
                  $\left(A_2\right)=\frac{1+x}{10}=\frac{1+0.333}{10}=0.133$