Reaction: $2\text{A} + \text{B} \rightleftharpoons 3\text{C} + \text{D}$

Initial concentrations: $[\text{A}]_0 = 1.00 \text{ M}$ $[\text{B}]_0 = 1.00 \text{ M}$ $[\text{C}]_0 = 0 \text{ M}$ $[\text{D}]_0 = 0 \text{ M}$

At equilibrium, $[\text{D}] = 0.25 \text{ M}$. From the stoichiometry of the reaction, if $x$ is the concentration of D formed at equilibrium, then $x = 0.25 \text{ M}$. The change in concentrations will be: Change in $[\text{D}] = +x = +0.25 \text{ M}$ Change in $[\text{C}] = +3x = +3(0.25) = +0.75 \text{ M}$ Change in $[\text{A}] = -2x = -2(0.25) = -0.50 \text{ M}$ Change in $[\text{B}] = -x = -0.25 \text{ M}$

Therefore, the equilibrium concentrations are: $[\text{A}] = 1.00 - 0.50 = 0.50 \text{ M}$ $[\text{B}] = 1.00 - 0.25 = 0.75 \text{ M}$ $[\text{C}] = 0 + 0.75 = 0.75 \text{ M}$ $[\text{D}] = 0.25 \text{ M}$

The equilibrium constant $K_c$ is given by the expression: $K_c = \frac{[\text{C}]^3[\text{D}]}{[\text{A}]^2[\text{B}]}$

Substituting the equilibrium values into the expression: $K_c = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)}$