`8` mol of gas `AB_(3)` are introduced into a `1.0 dm^(3)` vessel. It dissociates as `2AB_(3)(g) hArr A_(2)(g)+3B_(2)(g)`
At equilibrium, 2 mol of `A_(2)` is found to be present. The equilibrium constant for the reaction is

Options

A. `72 mol^(2)L^(-2)`
B. `36 mol^(2)L^(-2)`
C. `3 mol^(2)L^(-2)`
D. `27 mol^(2)L^(-2)`

Answer

D

Explanation

`{:(,2AB_(3(g)),hArr,A_(2(g)),+,3B_(2(g))),("At t=0",8,,0,,0),("At eqm.",8-a,,a//2,,3a//2):}`
Thus `K_(c)=([A_(2)][B_(2)]^(3))/([AB_(3)]^(2))`, Also `(a)/(2)=2`
`:. A=4`
and `[AB_(3)]=(4)/(1)`, `[A_(2)]=(2)/(1)` and `[B_(2)]=(6)/(1)`
`=(2xx6^(3))/(4^(2))=27 mol^(2)litre^(2-)`