The reaction,
`2A(g) + B(g)hArr3C(g) + D(g)`
is begun with the concentration of A and B both at an intial 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:

Options

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)]`

Answer

B

Explanation

`2A +B hArr 3C +D`
`{:("Initial",1,1,0,0),("Eq.",1–0.50,1–0.25,0.75, 0.25):}`
`K=((0.75)^3(0.25))/((0.50)^2(0.75))`