For the reversible reaction, $\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$taking place in one litre vessel at 450 K, initial moles of A and B are equal. If the number of moles of A at equilibrium are double the number of moles of C at equilibrium then the equilibrium constant of the reaction is equal to

$$\begin{gathered} \mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}; \\ \mathrm{Initial} \mathrm{moles} 1 1 - -; \\ \mathrm{Moles} \mathrm{at} \mathrm{Eqbm} \left(1-\mathrm{x}\right) \left(1-\mathrm{x}\right) \mathrm{x} \mathrm{x}; \\ \mathrm{Given} \\ 1-\mathrm{x}=2\mathrm{x}; \\ \mathrm{x}=\frac{1}{3}; \\ ∆\mathrm{n}=0 \mathrm{for} \mathrm{the} \mathrm{process}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{vessel} \mathrm{has} \mathrm{no} \mathrm{effect} \mathrm{in} \mathrm{the} \mathrm{calculation} \mathrm{of} {\mathrm{K}}_{\mathrm{c}}; \\ {\mathrm{K}}_{\mathrm{c}}=\frac{{\mathrm{x}}^2}{{\left(1-\mathrm{x}\right)}^2}=0.25 \end{gathered}$$