The value of `K_(c)` for the reaction
`H_(2)(g)+I_(2)(g) hArr 2HI(g)`
is `64` at `773K`. If one "mole" of `H_(2)`, one mole of `I_(2)`, and three moles of HI are taken in a `1 L` flask, find the concentrations of `I_(2)` and HI at equilibrium at `773 K`.

Explanation

For the reaction `H_(2)(g)+I_(2)(g) hArr 2HI(g)`
`Q=([HI]^(2))/([H_(2)][I_(2)])=3^(2)/(1xx1)=9 [V=1L]`
Note: When `Deltan_(g)=0` not only `K_(p)=K_(c )` but volume terms always get cancelled in the expression of `K`.
`rArr Q lt K_(eq)(=64)`. Hence, the reaction proceeds to forward direction to acchieve equilibrium.
Let `x` mol of `H_(2)` and `I_(2)` combine to produce `2x` mol of `HI`.

`K_(c )=([HI]^(2))/([H_(2)][I_(2)])=64`
Concentration of species at equilibrium are:
`[H_(2)]=(1-x)//1, [I_(2)]=(1-x)//1, [AB]=(3+2x)//1`
`K_(c )=(((3+2x)/1)^(2))/(((1-x)/1)((1-x)/1))=((3+2x)^(2))/((1-x))=64`
`rArr x=0.5`
`[I_(2)]=(1-x)/1=1-0.5=0.5 M`
`[HI]=(3+2x)/(1)=3+1.0=4.0 M`