Under what pressure must an equimolar mixture of `PCl_(3)` 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

`3.66 atm`

Explanation

Let the partial pressure of `PCI_(3)` at equilibrium =p atm
`:.` The partial pressure of `CI_(2)` at equilibrium =p atm
The partial pressure of `PCI_(5)` at equilibrium =1 atm
The dissociation equation is :
`PCI_(5) hArr PCI_(3) +CI_(2)`
`K_(p) =(pPCI_(3)xxpCI_(2))/(pPCI_(5)) " or " 1.78 =((P)xx(P))/(1) =P^(2) ' or " P=(1.78)^(1//2) =1.33 atm`
Total pressure `=pPCI_(5) +pPCI_(3)+ pCI_(2)=1+1.33 + 1.33 =3.66 atm`