` PCI_5` is 47.1% dissociated at 18°C and one atmospheric pressure. Calculate the value of `K_p` .

Explanation

Let the initial number of moles of `PCl_(5)` be 1 mole.
Number of moles dissociated = 47.1% of 1 `= (1 xx 47.1)/(100+ 0.471`
Number of moles left at equilibrium `= 1-0.471= 0.529`
`PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g)`
`{:("Initial moles",1,0,0),("Number of moles at Equilibrium",0.529,0.471,0.471):}`
As per equation, 0.529 mole of `PCl_(5)` will dissociate to give 0.471 mole of `PCl_(3)` and 0.471 mole of `Cl_(2)`. Therefore, total number of moles at equilibrium `= 0.529+ 0.471 + 0.471= 1.471` mole.
Total pressure at equilibrium = 1atm
Since Partial pressure= Mole fraction `xx` Total pressure, we have
Partial pressure of `PCl_(5)= P_(PCl_(5))= (1- 0.471)/(1.471)= 0.3596` atm
Partial pressure of `PCl_(3)= P_(PCl_(3)) = (0.471)/(1.471)= 0.320` atm
Partial pressure of `PCl_(5)= P_(PCl_(5)) = (0.471)/(1.471)= 0.320` atm.
According to the law of mass action and law of equilibrium ,
`K_(p)= (P_(PCl_(3)) xx P_(Cl_(2)))/(P_(PCl_(5)))= (0.320 xx 0.320)/(0.3596)= 0.285`