A container whose volume is V contains an equilibrium mixture that consits of 2mol each of `PCl_(5) , PCl_(3)` and `Cl_(2)` ( all as gases ). The pressure is 30.3975 kPa and temperature is T. A certain amount of `Cl_(2)` ( g) is now introduced keeping the pressure and temperature constant until the equilibrium volume is 2V. Calculate the amount of `Cl_(2)` that was added and the value of `K_(p)`.

Explanation

`PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g))`
At equilibrium 2mol 2 mol 2 mol
Total pressure = 30.3975 Kpa = 3 atm = P ( say )
`K _(p) = ( P_(PCl_(3)) xx P_(Cl_(2)))/(P_(Cl_(5))) = (( P)/( 3) xx ( P)/( 3))/((P)/( 3))`
`K_(p ) = ( P)/( 3) = 1` .....(1)
When chlorine is added to the system. The system will behave to nullify the effect and hence formation of `PCl_(5)` will be preferred.
Since P and T are constant are
`( V_(1))/( V_(2)) = ( n _(1))/( n _(2)) = n _(1) = 6" " V_(1) = V " " V_(2) = 2V `
`:. n _(2) = 12 ` moles

Say a moles of `Cl_(2)` were added
`PCl_(5) hArr PCl_(3) = Cl_(2)`
Initial 2 2 2
FInal 2`+x` 2-x `2+a-x`
`n_(T )- 12 = 6+a -x `
`K_(p) = ( (( 2+ax) - x)/( 12) xx ( 2-x)/( 12))/((2+x)/(12)) xxP=((2+a-x)(2-x))/(4( 2+x)) =1 ` .....(2)
Solving equation ( 1) and ( 2) we get
`a = ( 20)/(3)` moles
Hence `( 20)/( 3)` moles of `Cl_(2)` was added