Under what pressure must an equimolar mixture of `PCl_(5)` and `Cl_(2)` be placed at `250^(@)C` in order to obtain `PCl_(5)` at `1` atm? `(K_(p) "for dissociation of" PCl_(5)=1.78)`.

Answer

A, C

Explanation

Suppose partial pressure of `PCl_(3)` at equilibrium `=p "atm"`
Then partial pressure of `Cl_(2)` at equilibrium `=p "atm"`
Partial pressure of `PCl_(5)` at equilibrium `=1 "atm"`
for dissociation of `PCl_(5)`
`PCl_(5) hArr PCl_(3)+Cl_(2)`
`K_(p)=1.78=(p_(PCl_(3))xxp_(Cl_(2)))/(p_(PCl_(5)))=(pxxp)/(1)=p^(2)`
`:. p=sqrt(1.78)=1.33 "atm"`
`:.` Total pressure `=p_(PCl_(5))+p_(PCl_(3))+p_(PCl_(2))`
`=1+1.33+1.33=3.66 "atm"`