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`.

Explanation

The reverse reaction may be written as :
`2HI(g) hArr H_(2)(g) +I_(2)(g) , K_(c) =1//54.8`
`K_(c) =[[H_(2)(g)][I_(2)(g)]]/[[HI(g)]^(2)] = [[x][x]]/[[0.5 " mol " L^(-1)]^(2)]`
`(1)/(54.8) = (x^(2))/((0.5" mol "L^(-1))) " or " x^(2) =(1)/(54.8) xx(0.5 " mol " L^(-1))^(2)`
`x= ((0.25)/(54.8) " mol"^(2)L^(-2))^(1//2) = 0.068 " mol" L^(-1)