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`.

Answer

A, B, C, D

Explanation

`{:(H_(2),+,I_(2),hArr,2HI,),(1,,2,,3,"Initial conc"),(1-x,,2-x,,3+2x,"At Eq."):}`
`:. K_(c)=([H])/([H_(2)][I_(2)])=((3+2x)^(2))/((1-x)(2-x))`
`=(9+4x^(2)+12x)/(2+x^(2)-3x)=45.9` (given)
or `9+4x^(2)+12x=91.8+45.9 x^(2)-137.7 x`
`rArr 41.9x^(2)-149.7x+82.8=0`
`rArr x=(149.7 +- sqrt((149.7)^(2)-4xx41.9xx82.8))/(2xx41.9)`
On solving we get, `x=2.89` and `0.68`
But `x=2.89` is impossible, hence `x=0.684`
`:. [H_(2)]=1-x=1-0.684=0.316 M`
`[I_(2)]=2-x=2-0.684=1.316 M`
`[HI]=3+2x=3+2xx0.684=4.368 M`