Question

`K_(p)` for the reaction
`PCl_(5)(g)ghArrPCl_(3)(g)+Cl_(2)(g)`
at `250^(@)C` is `0.82`. Calculate the degree of dissociation at given temperature under a total pressure of `5 atm`. What will be the degree of dissociation if the equilibrium pressure is `10 atm`, at same temperature.

Answer 1

Let 1 mol of `PCl_(5)` be taken initially. If x moles of `PCl_(5)` dissociate at equilibrium its degree of dissociation `=x`

Total moles `=1-x+x+x=1+x`
`P=5` atm and `K_(p)=0.82`
`p_(PCl_(5))=((1-x)/(1+x))P, p_(PCl_(3))=x/(1+x)P`
and `p_(Cl_(2))=x/(1+x)P`
Now, `K_(p)=((p_(PCl_(3)))(p_(Cl_(2))))/((p_(PCl_(5))))rArr K_(p)=x^(2)/(1-x^(2))P=0.82`
or `(x^(2)(5))/(1-x^(2))=0.82 rArr x=sqrt(0.82/5.82)`
`x=0.375 (or 37.5 %)`
Now the new pressure `P=10` atm.
Let y be the new degree of dissociation. as the temperature is same `(250^(@)C)`, the value of `K_(p)` will remain same.
Proceeding in the same manner.
`K_(p)=(y^(2)P)/(1-y^(2))rArr 0.82=y^(2)/(1-y^(2))xx10`
`rArr y=sqrt(0.82/10.82)` or `y=0.275 (or 27.5 %)`