The compounds A and B are mixed in equimolar proportion to form the products.
$A+B\rightleftharpoons C+D$
At equilibrium, one-third of A and B are consumed. The equilibrium constant for the reaction is:
(A) 0.5
(B) 4.0
(C) 2.5
(D) 0.25

Hint: The term "chemical equilibrium state" refers to reactant and product concentrations in a reaction occurring in a closed system that are no longer changing over time. Chemical equilibrium is reached when a reversible reaction proceeds both backwards and forwards by the same amount at the same time. In the presence of a catalyst, this equilibrium can be reached more quickly. Chemical equilibrium cannot be changed by a catalyst since it only changes the rate of both forward and backward reactions to the same degree. Its nature is dynamic.

Formula Used: For the reaction,
$A+B\rightleftharpoons C+D$
The equilibrium state is calculated as: $K=\frac{\left[ C \right]\left[ D \right]}{\left[ A \right]\left[ B \right]}$; where A and B are the reactants, C and D are the products.

Complete step by step solution:
The compounds A and B are mixed in equal proportions to form the products C and D.
The initial concentrations of the reactants A and B are taken as 1M. It is given that at equilibrium, one-third of A and B are consumed.

The equilibrium constant is given by:
$K=\frac{\left[ C \right]\left[ D \right]}{\left[ A \right]\left[ B \right]}$
$K=\frac{\frac{1}{3}\times \frac{1}{3}}{\frac{2}{3}\times \frac{2}{3}}$
$K=\frac{1}{4}$
$K=0.25$
Thus, the equilibrium constant for this reaction is 0.25.
Correct Option: (D) 0.25.

Note: The factors which affect the equilibrium constant are change in pressure, change in temperature, change in concentration, addition of catalyst, etc. The value of the equilibrium constant is definite for each chemical reaction.