At `700 K` equilibrium constant for the reaction, `H_(2(g))+I_(2(g))hArr2HI_((g))`
is `54.8`. If `0.5 mol litre^(-1)` of `HI_((g))` is present at equilibrium at `700 K`, what are the concentrations of `H_(2(g))` and `I_(2(g))`, assuming that we initially started with `HI_((g))` and allowed it to reach equilibrium at `700 K`.

Answer

`0.068 "molL"^(-1)` each of `H_(2)` and `I_(2)`

Explanation

It is given `K_(c)` for the forward reaction
`H_(2(g)) + I_(2(g)) harr 2HI_((g))` is `54.8`.
Therefore , at equilibrium, the equilibrium constant `K'_(c)` for the reaction
`2HI_((g)) harr H_(2(g)) + I_(2(g))` will be `(1)/(54.8)`
`[HI] = 0.5 "molL"^(-1)`
Let the concentrations of hydrogen and iodine at equilibrium be `x "molL"^(-1)`
`[H_(2)]= [I_(2)] = x "mol L"^(-1)`
Therefore `([H_(2)][I_(2)])/([HI]^(2)) = K'_(c)`
`rArr (x xx x)/((0.5)^(2)) = (1)/(54.8)`
`rArr x^(2) = (0.25)/(54.8)`
`rArr x = 0.06754`
`x = 0.068 "molL"^(-1)` (approximately)
Hence, at equilibrium, `[H_(2)] = [I_(2)] = 0.068 "molL^(-1)`.