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The reaction 2A+B(g)⇌3C(g)+D(g)2\text{A} + \text{B}(g) \rightleftharpoons 3\text{C}(g) + \text{D}(g) begins with the concentrations of A and B both at an initial value of 1.00 M1.00 \text{ M}. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M0.25 \text{ M}. The value for the equilibrium constant for this reaction is given by the expression:

A

[(0.75)3(0.25)][(0.50)2(0.75)]\frac{[(0.75)^3(0.25)]}{[(0.50)^2(0.75)]}

B

[(0.75)3(0.25)][(0.50)2(0.25)]\frac{[(0.75)^3(0.25)]}{[(0.50)^2(0.25)]}

C

[(0.75)3(0.25)][(0.75)2(0.25)]\frac{[(0.75)^3(0.25)]}{[(0.75)^2(0.25)]}

D

[(0.75)3(0.25)][(1.00)2(1.00)]\frac{[(0.75)^3(0.25)]}{[(1.00)^2(1.00)]}

Step-by-Step Solution

Reaction: 2A+B⇌3C+D2\text{A} + \text{B} \rightleftharpoons 3\text{C} + \text{D}

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

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

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

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

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

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