The gas `A_(2)` in the left flask allowed to react with gas `B_(2)` present in right flask as `A_(2)(g)+B_(2)(g)hArr2AB(g),K_(c)=4` at `27^(@)C.` What is the concentration of AB when equilibrium is established ?

Answer

`0.66`

Explanation

`A_(2)(g)+B_(2)(g)hArr2AB(g)`
Moles at eqm `2-x 4-x 2x`
`K_(C)=(4x_(2))/((2-x)(4-x))`
`rArr x=(32)/(24)=1.33` mole
`[AB(g)]=(2x1.33)/(4)=0.66` M