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

1 Answer

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Best answer
Correct Answer - A
`{:(,,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`

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