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*75)]`
D. `[(0*75)^(3) (0*25)] div [(0*75)^(2) (0*25)]`

Answer

B

Explanation

` 2 A (g) + B (g) hArr 3 C (g) + D (g)`
`{:(,2A(g),+,B(g),hArr,3 C(g),+,D(g)),("Intial",1.0 M,,1.0 M,,0,,0),("conc.",,,,,,,),("Eqm.",1-(2xx0.25),,1-0.25,,3xx0.25,,0.25),("conc".,0.5 M,,0.75 M,,0.75 M,,):}`
` K=([C]^(3) [D])/([A]^(2) [B])= ((0.75)^(3) (0.25))/((0.5)^(2) (0.75))`