For a reaction `A_((g)) + B_((g))hArrC_((g)) + D_((g))` the intial concentration of A and B are equals but the equilibrium constant of C is twice that of equilibrium concentration of A. Then K is

Options

A. 4
B. 9
C. `1//4`
D. `1//9`

Answer

A

Explanation

`{:(,,A(g),+,B(g),hArr,C(g),+,D(g)),("Initial Conc".,,a,,a,,0,,0),("Eg.Conc",,a-x,,a-x,,x,,x):}`
Given `= 2(a - x)` or `x = (2a)/(3)`
`K_(c) = (x^(2))/((a - x)^(2))= ((2a//3)^(2))/((a - 2a//3)^(2)) = 4`