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

`2HI(g) hArr H_(2)(g)+I_(2)(g), K=1//54.8`
At equilibrium, `[HI]=0.5 mol L^(-1)`,
`[H_(2)]=[I_(2)]=xmol L^(-1)`
`:. K=(x xx x)/((0.5)^(2))=1/54.8 (Given)`
This gives `x=0.068, i.e., [H_(2)]=[I_(2)]=0.068 mol L^(-1)`.