The reaction, 2A(g) + B(g) 3C(g) + D(g) is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression
- [(0.75)3(0.25)] [(1.00)2(1.00)]
- [(0.75)3(0.25)] [(0.50)2(0.75)]
- [(0.75)3(0.25)] [(0.50)2(0.25)]
- [(0.75)3(0.25)] [(0.75)2(0.25)]
Answer
(2) [(0.75)3(0.25)] [(0.50)2(0.75)]
Solution
Let x M of B react: [D]eq = x = 0.25 M
[A]eq = 1.00 - 2x = 0.50 M
[B]eq = 1.00 - x = 0.75 M
[C]eq = 3x = 0.75 M
Kc = [C]3[D][A]2[B]
∴ Kc = (0.75)3(0.25)(0.50)2(0.75)
Answer: (2)