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

Answer

D

Explanation

`K_(c)=([A_(2)][B_(2)]^(3))/([AB_(3)]^(2)),`
`{:(,2AB_(3),hArr,A_(2),+,3B_(2)),("Eqbm. conc.",2-2alpha,,alpha,,3alpha),("For 8 moles the conc. is ",4(2-2 alpha),,4(alpha),,4(3alpha)):}`
`4 alpha=2 implies alpha=1//2`
`:. [AB_(3)]=4(2-1)=4 mol^(-1)L,[A_(2)]=2 mol L^(-1)`
`[B_(2)]=4 (3 alpha)=4(3 xx 1//2)=6 mol L^(-1)`
`:. K=(2 xx (6)^(3))/((4)^(2))=(2xx216)/(16)=27`