In reaction $$A + 2B \rightleftharpoons 2C + D,$$     initial concentration of $$B$$  was 1.5 times of $$\left[ A \right],$$  but at equilibrium the concentrations of $$A$$ and $$B$$  became equal. The equilibrium constant for the reaction is :

Solution :
$$\eqalign{ & \,\,A\,\,\,\,\, + \,\,\,\,\,\,\,\,\,2B\,\,\,\,\,\,\,\,\, \rightleftharpoons \,\,\,\,2C + D \cr & \,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1.5a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,0 \cr & \left( {a - x} \right)\,\left( {1.5a - 2x} \right)\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,x \cr & {\text{Hence}}\,\,{K_C} = \frac{{\left( {2{x^2}} \right) \times x}}{{\left( {a - x} \right){{\left( {1.5a - 2x} \right)}^2}}} \cr & {\text{Given, at equilibrium}} \cr & \therefore \,\,\left( {a - x} \right) = \left( {1.5a - 2x} \right) \cr & \therefore \,\,a = 2x \cr & {\text{On solving}}\,\,{K_C} = 4 \cr} $$