For a reaction `2HI hArr H_(2)+I_(2)`, at equilibrium `7.8 g, 203.2 g`, and `1638.4 g` of `H_(2), I_(2)`, and HI, respectively were found. Calculate `K_(c)`.

Explanation

`{:(,2HI,hArr,H_(2),+,I_(2)),("Moles at equilibrium",(1638.4)/(128),,(7.8)/(2),,(203.2)/(25.4)),(,=12.8,,3.9,,0.8):}`
Let volume of container by `V` litre.
`[H_(2)]=(3.9)/(V)`, `[HI]=(12.8)/(V)`, `[I_(2)]=(0.8)/(V)`
`:. K_(c)=([H_(2)][I_(2)])/([HI]^(2))=(3.9xx0.8)/(VxxVxx((12.8)/(2))^(2))=0.019`
`:. K_(c)=0.019`