Prove that the pressure necessary to obtain 50% dissociation of PCl_(5)at 500 K is numerically equal to three times the value of the equilibrium constant, K_(p).

Prove that the pressure necessary to obtain 50% dissociation of PCl_(5

1.	Prove that the pressure necessary to obtain 50% dissociation of PCl_(5)at 500 K is numerically equal to three times the value of the equilibrium constant, K_(p).
Answer» Solution :` {: (,PCl_(5),hArr,PCl_(3),+,Cl_(2),) , (" Intial moles",1,,0,,0,) , ("Moles at EQM".,1-05=05,,05,,05,"TOTAL " 15 "moles") :}` IfP isthe total requiredpressure, then ` p_(PCl_(5)) = (05)/(15) XX P=P/3, p_(PCl_(3)) = (05)/(15) xx P = P/3, p_(Cl_(2)) = (05)/(1*5) xx P= P/3` ` K_(p) = (p_(pCl_(3)) xx p_(cl_(2)))/(p_(PCl_(5)))= ((P//3)(P//3))/(P//3) = P/3 or P = 3 K_(P).`