Two moles of `PCl_(5)` were heated to `327^(@)C` in a closed two-litre vessel, and when equilibrium was achieved, `PCl_(5)` was found to be `40%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant `K_(p)` and `K_(c)` for this reaction.

Explanation

`PCl_(5)hArrPCl_(3)+Cl_(2)" ""Here concentration of" PCl_(5)=n/V=2/2=1`
`{:(1,0,0),(1-x,x,x):}`
`x=40%=0.4`
`1-x=1-0.4=0.6`
`K_(c)([PCl_(3)][Cl_(2)])/([PCl_(5)])`
`=(0.4xx0.4)/(0.6)=(1.6)/(6)=(0.8)/(3)=0.267 ` moles/L