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

Answer

C

Explanation

`{:(,2AB_(3)(g),hArr,A_(2)(g),+,3B_(2)(g)),("Initial",8,,0,,0),("At eq.",(8-2x)/1,,x/1,,(3x)/1):}`
Since volume `=1 dm^(3)= 1 L`
At eq. `[A_(2)]=2 "mol"=x`
`:. [AB_(3)]=8-2xx2=4 M`
`[A_(2)]=2 M`
`[B_(2)]=3xx2=6 M`
`K=([B_(2)]^(3)[A_(2)])/([AB_(3)]^(2))=(6xx6xx6xx2)/(4xx4)=27 "mol"^(-2) L^(-2)`