`1`mol of `Cl_(2)` and `3` mol of `PCl_(5)` are placed in a `100 L` vessel heated to `227^(@)C`. The equilibrium pressure is `2.05` atm. Assuming ideal behaviour, calculate the degree of dissociation for `PCl_(5)` and `K_(p)` for the reaction.
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`

Explanation

Dissociation of `PCl_(5)` is written as :
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`
Let x be the no. of moles of `PCl_(5)` decomposed at equilibrium.

Now total gaseous moles in the container `=n_(T)`
`rArr n_(T)="moles of" (PCl_(5)+PCl_(3)+Cl_(2))+"moles of"N_(2)`
`rArr n_(T)=3-x+x+x+1=4+x`
The mixture behaves ideally, hence `PV=n_(T)RT`
Let us calculate no. of moles by using gas equation
`rArr n_(T)=(PV)/(RT)=(2.05xx100)/(0.0821xx500) rArr n_(T)=5`
Now, equating the two values of `n_(T)`, we have:
`4+x=5 rArr x=1`.
`rArr` Degree of dissociation `=1//3=0.333`
Now `K_(P)=(P_(PCl_(3))xxP_(Cl_(2)))/(P_(PCl_(5)))`
`P_(PCl_(5))=(3-x)/(4+x)p" " P=` equilibrium pressure
`=(2)/(5)xx2.05=0.82`atm
`P_(Cl_(2))=P_(PCl_(3))=(x)/(4+x)P=(1)/(5)xx2.05=0.41` atm
`rArr K_(p)=(0.4xx0.4)/(0.8)=0.20` atm