One mole of `H_(2)` two moles of `I_(2)` and three moles of `HI` are injected in one litre flask. What will be the concentration of `H_(2), I_(2)` and HI at equilibrium at `500^(@)C. K_(c )` for reaction `H_(2)+I_(2) hArr 2HI` is `45.9`.

Explanation

`{:(,,H_(2),+,I_(2),hArr,2HI),(,"Initial mole",1,,2,,3),(,"Mole at equilibrium",(1-x),,(2-x),,(3+2x)):}`
Let volume of container be `V litre`.
`:. K_(c)=([HI]^(2))/([H_(2)][I_(2)])=([(3+2x)/(V)]^(2))/([(1-x)/(V)][(2-x)/(V)])`
`45.9=((3+2x)^(2))/((1-x)(2-x))` (`:' K_(c)=45.9`)
`45.9=(9+4x^(2)+12x)/(2-3x+x^(2))`
`:. x=0.684`
`:. [H_(2)]=1-0.684=0.316 mol litre^(-1)`
(`:' V= 1 litre`)
`[I_(2)]=2-0.684=1.316 mol litre^(-1)`
`[HI]=3+2(0.684)=4.368 mol litre^(-1)`