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.

Explanation

`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Mole before dissociation",0.1,,0,,0),("Mole after dissociation",(0.1-x),,x,,x):}`
Given volume of container `=8 litre`
Now `K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=((x)/(8)*(x)/(8))/(((0.1-x))/(8))=((x^(2)))/(8(0.1-x))` .....(`1`)
Also `PV=nRT` for the equilibrium muxture at `540 K`.
`1xx8=(0.1+x)xx0.082xx540`
`:. x=0.08` ...(`2`)
Thus, from eqs. (`1`) and (`2`),
`K_(c)=(0.08xx0.08)/(8(0.1-0.08))=4xx10^(-2)mol litre^(-1)`
Also `K_(p)=K_(c)(RT)^(Deltan)`, (`:' Deltan=+1`)
`=4x10^(-2)(0.082xx540)=1.77 atm`