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

` 2 HI(g) hArr H_(2) (g) + I_(2) (g), K= 1/54*8 `
At equilibrium , ` [HI] = 0*5 "mol" L^(-1) , [H_(2)] = [I_(2)] = x "mol" L^(-1) :. K= (x xx x)/(0*5)^(2)=1/(54*8).`
This gives ` x = 0*068 , i.e., [H_(2)] = [I_(2) = 0*068 "mol"L^(-1) .`