For A + B $\rightleftharpoons$C + D, if initially the concentrations of A and B are equal but at equilibrium, concentration of D is twice that of A.  The equilibrium constant of the reaction is                         

$A + B \rightleftharpoons C + D$

1          1            0         0    (Given conditions)

1-x       1-x         x         x    (Equilibrium conditions)

Given , 

          $$\begin{gathered} \left[\mathrm{D}\right]=2\left[\mathrm{A}\right] \\ \mathrm{x} = 2(1-\mathrm{x}) \\ \mathrm{x}=2-2\mathrm{x} \\ \mathrm{x}=\frac{2}{3} \end{gathered}$$

$\left[\mathrm{A}\right]=1-\frac{2}{3}=\frac{1}{3}=\left[\mathrm{B}\right] : \left[\mathrm{C}\right] = \left[\mathrm{D}\right] =\mathrm{x}=\frac{2}{3}$

${\mathrm{k}}_{\mathrm{C}}=\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}=\frac{\frac{2}{3}\times \frac{2}{3}}{\frac{1}{3}\times \frac{1}{3}}=4$