`0.5 mol` of `H_(2)` and `0.5` mol of `I_(2)` react in `10 L` flask at `448^(@)C`. The equilibrium constant `(K_(c ))` is `50` for
`H_(2)+I_(2) hArr 2HI`
a. What is the value of `K_(p)`?
b. Calculate the moles of `I_(2)` at equilibrium.

Explanation

`{:(,,H_(2),+,I_(2),hArr,2HI),(,"Mole at t=0",0.5,,0.5,,0),(,"Mole at equilibrium",(0.5-x),,(0.5-x),,(2x)):}`
`:. K_(p)=K_(c)=(4x^(2))/((0.5-x)^(2))`
Note: Volume term is eliminated, if `Delta n=0`
(`a`) `:' K_(p)=K_(c)` (`:'Deltan=0`)
`:. K_(p)=50`
(`b`) `50=(4x^(2))/((0.5-x)^(2))` or `(2x)/((0.5-x))=sqrt(50)`
`:. x=0.39`
`:. `Mole of `I_(2)` at equilibrium `=0.50-0.39=0.11 "mole"`