In the dissociation of HI, `20%` of HI is dissociated at equilibrium. Calculate `K_(p)` for
`HI(g) hArr 1//2 H_(2)(g)+1//2 I_(2)(g)`

Explanation

`{:(,HI,hArr,(1)/(2)H_(2),+,(1)/(2)I_(2)),("Initial moles",1,,0,,0),("Moles at equilibrium",(1-alpha),,alpha//2,,alpha//2):}`
Where `alpha` is the degree of dissociation and volume of container is `V` litre.
`K_(p)=K_(c)=(((alpha)/(2V))^(2)((alpha)/(2V))^(1//2))/((1-alpha)/(V))` (`:' Deltan=0`)
`K_(p)=K_(c)=(alpha)/(2(1-alpha))` (`:'alpha=0.2`)
`K_(p)=K_(c)=(0.2)/(2(1-0.2))`
`:. K_(p)=K_(c)=0.125`