Question

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.

Answer 1

Correct Answer - A::B::C::D
`PCl_(5)` dissociates as `PCl_(5) hArr PCl_(3)+Cl_(2)`
Initial amount of `PCl_(5)=2 "mol"` (given)
`%` age dissociation at equilibrium `=40%`
Therefore `PCl_(5)` dissociated at equilibrium
`=40//100xx2=0.8 "mol"`
Therefore Amounts of `PCl_(5), PCl_(3)` and `Cl_(2)` at equilibrium will be
`PCl_(5)=2-0.08=1.2 "mol"`
`(1 "mol of" PCl_(5) "on dissociation gives" 1 "mol of" PCl_(3) and 1 "mol of" Cl_(2)`]
Since the volume of the vessel is `2 L`, therefore, the molar concentration at equilibrium will be
`[PCl_(5)]=1.2//2=0.6 "mol" L^(-1)`
`[PCl_(3)]=0.8//2=0.4 "mol" L^(-1)`
and `[Cl_(2)]=0.8//2=0.4 "mol" L^(-1)`
Applying the law of chemical equilibrium to the dissociation equilibrium
`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.4xx0.4)/0.6=0.267 "mol" L^(-1)`
`K_(p)=K_(c)(RT)^(Deltan)`
Here, `Deltan=n_(p)-n_(r)=2-1=1 rArr K_(p)=K_(c)(RT)`
But `T=327+273=600 K` (given),
`R=0.0821 L-"atm" deg^(-1) "mol"^(-1)`
`:. K_(p)=0.267xx0.0821xx600=13.15 "atm"`