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)])/([(1.00)^(2) (1.00)])`
B. `([(0.75)^(3) (0.25)])/([(0.50)^(2) (0.75)])`
C. `([(0.75)^(3) (0.25)])/([(0.50)^(2) (0.25)])`
D. `([(0.75)^(3) (0.25)])/([(0.75)^(2) (0.25)])`

Answer

B

Explanation

`{:(,,2A,+,B,hArr,3C,+,D),("Initial Conc",,1,,1,,,,),("Reacted",,.5,,.25,,,,),("Produced",,,,,,.75,,.25),("At.equilibrium",,.5,,.75,,.75,,.25):}`
`K_(c) = ((0.25)xx(0.75)^(3))/((0.5)^(2)xx(0.757))`