$4$ moles of $\mathrm{A}$ are mixed with $4$ moles of $\mathrm{B}$. At equilibrium for the reaction $\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}, 2$ moles of $\mathrm{C}$ and $\mathrm{D}$ are formed, the equilibrium constant for the reaction will be,

The initial concentrations of A, B, C, and D are 4M, 4M, 0M, and 0M respectively.

The equilibrium concentrations of A,B,C and D are 4−2=2M, 4−2=2M, 2M and 2M respectively.

$$\begin{gathered} \mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D} \\ \mathrm{initial} \mathrm{conc} 4 4 0 0 \\ \mathrm{at} \mathrm{equilibrium} \left(4-2\right) \left(4-2\right) 2 2 \end{gathered}$$

The expression for the equilibrium constant is $=\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}$

$\therefore {\mathrm{K}}_{\left(\mathrm{eq}\right)}=\frac{2\times 2}{2\times 2}=1$