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)`

Answer

A, B

Explanation

`{:(,2HI,hArr,1/2 H_(2),+,1/2 I_(2)),("Initial",1,,0,,0),("moles at Eq",(1-alpha),,alpha//2,,alpha//2):}`
where `alpha` is degree of dissociation and volume of container is VL.
`K_(p)=K_(c)((alpha/(2V))^(1//2)(alpha/(2V))^(1//2))/(((1-alpha))/V)`
`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`