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The reaction $2A(g) + B(g) \rightleftharpoons 3C(g) + D(g)$ 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
A
$[(0.75)^3(0.25)] \div [(1.00)^2(1.00)]$
B
$[(0.75)^3(0.25)] \div [(0.50)^2(0.75)]$
C
$[(0.75)^3(0.25)] \div [(0.50)^2(0.25)]$
D
$[(0.75)^3(0.25)] \div [(0.75)^2(0.25)]$
Detailed Solution
$2A + B \rightleftharpoons 3C + D$; initial concentrations: [A] = 1.00, [B] = 1.00, [C] = 0, [D] = 0
At equilibrium [D] = 0.25 M, so the extent of reaction is x = 0.25 M.
[C] = 3x = 0.75 M
[A] = 1.00 − 2x = 1.00 − 0.50 = 0.50 M
[B] = 1.00 − x = 1.00 − 0.25 = 0.75 M
$K = \frac{[C]^3[D]}{[A]^2[B]} = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)}$
At equilibrium [D] = 0.25 M, so the extent of reaction is x = 0.25 M.
[C] = 3x = 0.75 M
[A] = 1.00 − 2x = 1.00 − 0.50 = 0.50 M
[B] = 1.00 − x = 1.00 − 0.25 = 0.75 M
$K = \frac{[C]^3[D]}{[A]^2[B]} = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)}$