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.

1 mole of PCl_(5) is placed in a closed vessel at 523K. At equilibrium

1.	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.
Answer» <P> Solution :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`