0.5 mol of `H_(2)` and 0.5 mole of `I_(2)` react in 10 litre flast at `448^(@)C`. The equilibrium constant `K_(c)` is 50 for
`H_(2)(g)+I_(2)(g)hArr2HI(g)`
a. What is the value of `K_(p)`
b. Calculate mole of `I_(2)` at equilibrium.

Explanation

a. `K_(p)=K_(c)[RT]^(Deltan)`
`H_(2)+I_(2)hArr2HI`
For this reaction `DeltaN=0`
`:.K_(p)=K_(c)=50`
b. `underset(0.5-x)underset(0.5)(H_(2))+underset(0.5-x)underset(0.5)(I_(2))hArr underset(2x "equi conc.") underset(- "initial colnc")(2HI)`
`K_(c)=([HI]^(2))/([H_(2)][I_(2)])`
`50=((2x)/(0.5-x))^(2)`
`:.` Moles of `I_(2)=0.111`