At 700K, equilibrium constant for the reaction: `H_(2)(g)+I_(2)(g)hArr 2HI(g)` is 54.8. if 0.5 `mol*L^(-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

At 700K, `K_(c)` for the following reaction is 54.8.
`H_(2)(g)+I_(2)(g)hArr 2HI(g)` . . . [1]
At the same temperature, for the reaction,
`2HI(g)hArr H_(2)(g)+I_(2)(g)` . . [2]
`K_(c)=(1)/(54.8)=0.0182`
If the reaction is initially started with HI, then the equilibrium represented by eq. [2] will be established. in reaction [2], equal number of moles of `H_(2) and I_(2)` are produced. let at equilibrium the concentration of each of `H_(2) and I_(2)` be x `mol*L^(-1)`. at equilibrium `[HI]=0.5mol*L^(-1)`.
Therefore, for reaction [2],
`K_(c)=([H_(2)][I_(2)])/([HI]^(2))=(x xx x)/((0.5)^(2))=0.0182 or, x=0.0675`
Hence, at equilibrium `[H_(2)]=[I_(2)]=0.0675mol*L^(-1)`