Question

In the reaction,
`H_(2)(g)+I_(2)(g)hArr2HI(g)`
The concentration of `H_(2), I_(2)`, and `HI` at equilibrium are `8.0, 3.0` and `28.0` mol per `L` respectively. Determine the equilibrium constant.

Answer 1

`H_(2)(g)+I_(2)(g)hArr2HI(g)`
Applying the law of mass action
`K_(c )=([HI]^(2))/([H_(2)][I_(2)])`
Given `[H_(2)]=8.0 "mol" L^(-1)`
`[I_(2)]=3.0 "mol" L^(-1)`
`[HI]=2.8 "mol" L^(-1)`
So, `K_(c )=((28.0)^(2))/((8.0)xx(3.0))=32.66`