Question

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 1

` {: (,PCl_(5),hArr,PCl_(3),+,Cl_(2),) , (" Intial moles",1,,0,,0,) , ("Moles at eqm".,1-0*5=0*5,,0*5,,0*5,"Total " 1*5 "moles") :}`
If P is the total required pressure, then
` p_(PCl_(5)) = (0*5)/(1*5) xx P=P/3, p_(PCl_(3)) = (0*5)/(1*5) xx P = P/3, p_(Cl_(2)) = (0*5)/(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).`