`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.

Explanation

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

P = 5 atm and `K_(p)= 0.82`
`P_(PCl_(5))=((1-x)/(1+x))P, P_(PCl_(3))=(xP)/(1+x) and P_(Cl_(2))=(xP)/(1+x)`
Now, `K_(p)= ((P_(PCl_(5)))(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. i.e. proceeding in the same manner, we get
`K_(p)=((y)^(2)P)/(1-(y)^(2)) rArr 0.82 = ((y)^(2)10)/(1-(y)^(2)) rArr y = sqrt((0.82)/(10.82)) or y=0.275 (or 27.5%)`