Question

At `540 K, 0.10 mol` of `PCl_(5)` is heated in a `8L` flask. The pressure of equilibrium mixture is found to be `1.0 atm`. Calculate `K_(p)` and `K_(c )` for the reaction.

Answer 1

`PCl_(5)hArrPCl_(3)+Cl_(2)`

`1xx8=(0.1+x)xx0.082xx540`
or `x=0.08`
`K_(c )=([PCl_(3)][Cl_(2)])/([PCl_(5)])`
`=x^(2)/((0.1-x)xx8)`
`=(0.08xx0.08)/((0.1-0.08)8)=4xx10^(-2) "mol" L^(-1)`
`K_(p)=K_(c )(RT)^(Deltan)`
`=K_(c )RT(Deltan=+1)`
`=4xx10^(-2)xx0.082xx540=1.77 "atm"`