The `K_(c)` for `A_(2(g))+B_(2(g))hArr2AB_((g))` at `100^(@)C` is `50`. If one litre flasks containing one mole of `A_(2)` is connected with a two litre flask containing `2 "mole"` of `B_(2)`, how many mole of `AB` will be formed at `100 ^(@)C`?

Explanation

`A_(2)(g)+B_(2)(g) hArr 2AB(g)`
As the two vessels are connected, the final volume for the contents is now 3.0 L. Let x mole each of `A_(2) and B_(2)` react to form 2x moles of AB (from stoichiometry of reaction)

`K_(C)=([AB]^(2))/([A_(2)][B_(2)])=50`
Concentration of species at equilibrium are : `[A_(2)]=(1-x)//3, [B_(2)]=(2-x)//3, [AB]=2x//3`
`K_(C)=(((2x)/(3))^(2))/(((1-x)/(3))((2-x)/(3)))=(4x^(2))/(2-3x+x^(2))=50`
`rArr 46x^(2)+150x+100=0 rArr x=0.93 or 2.33` (neglecting this value)
`rArr` moles of AB(g) formed at equilibrium `=2x=1.86`