1 mole of `PCl_(5)` is placed in a closed vessel at 523K. At equilibrium, if it dissociates to an extent of 35%, calculate `K_(p)" for "PCl_(5)hArrPCl_(3)+Cl_(2)`. Equilibrium pressure is found to be `5xx10^(5)` Pa.

Explanation

Initial no. of moles: `underset(1)(PCl_(5))hArrunderset(0)(PCl_(3))+underset(0)(Cl_(2))`
`{:("No. of moles present"),("at equilibrium"):}}1-x=0.65" "x" "x" "(x=0.35)`
Total number of moles present at equilibrium = 0.65 + 0.35 + 0.35 = 1.35
We know that partial pressure = Mole fraction = Mole fraction `xx` total pressure
`thereforeP_(PCl_(5))=(0.65/1.35)xx5xx10^(5)Pa,P_(PCl_(3))=(0.35/1.35)xx5xx10^(5)Pa,P_(Cl_(2))=(0.35/1.35)xx5xx10^(5)Pa`
`K_(p)=(P_(PCl_(3))xxP_(Cl_(2)))/(P_(PCl_(5)))=((0.35/1.35)xx5xx10^(5)(0.35/1.35)xx5xx10^(5))/((0.65/1.35)xx5xx10^(5))=6.98xx10^(4)Pa`