The degree of dissociation of `PCl_(5)` at `1` atm pressure is `0.2`. Calculate the pressure at which `PCl_(5)` is dissociated to `50%`?

Answer

A, B, D

Explanation

`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("moles before",1,,0,,0),("dissociation",,,,,),("moles after",1-alpha,,alpha,,alpha),("dissociation",,,,,):}`
Given `=0.2` at `1` atm temperature
`:. K_(p)=(n_(PCl_(3))xxn_(Cl_(2)))/n_(PCl_(5))xx[P/(Sigman)]^(Deltan)`
`=(alpha.alpha)/((1-alpha))[P/(1+alpha)]=(Palpha^(2))/(1-alpha^(2))=(1xx(0.2)^(2))/(1-(0.2)^(2))`
`K_(p)=0.0416 "atm"`
Again when `alpha` is desired at `0.5, K_(p)` remains constant and thus,
`K_(SP)=(Palpha^(2))/(1-alpha^(2))`
`0.0416=(Pxx(0.5)^(2))/(1-(0.5)^(2)), :. P=0.1248 "atm"`