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ramuramu

Q. In an evacuated vessel of capacity 110 Lt., 4 moles of argon and 5 moles of PCl₅ were introduced and equilibrated at a temperature at 250⁰C . At equilibrium, the total pressure of the mixture was found to be 4.678 atm. Calculate the degree of dissociation, of PCl₅ and K_p for the reaction

PCl₅ --à PCl₃ + Cl₂

at this temperature.

ANS 0.6 , 1.75 atm

nitneh

please solve this as early as possible

rishikesh_anshu

let degree of dissociation be

PCl₅ ---> PCl₃ + Cl₂
5 0 0
5(1-

) 5

5

total moles in container = 5(1+

) + 4 ( 4 for argon)
we know
PV=nRT
so
4.678*110 = (9+5

)*.0821*(250+273)

solving

=.5968

0.6

K_p=p(PCl₃)p(Cl₂)
      ----------------
            p(PCl₅)

by dalton law of partial pressure     partial pressure = molefraction(total pressute)
total mole =5(1+

) + 4 =12
mole fraction of PCl₃=mole fraction of Cl₂= 5

/(9+5

) = .25
similarly of PCl₅= .167

hence K_{p = (}.25_*4.678₎₍0.25*4.678₎
----------------------------
0.167*4.678

= 1.75 atm

iitkgp_bipin

Let the degree of dissociation be x.

At equilibrium PCl₅ remaining is 5(1-x) mole and PCl₃ formed is 5x mole and Cl₂ formes is 5x mole.

Hence total no. of moles = 4 + 5(1-x) + 5x + 5x = 9 + 5x

Now apply PV=nRT

(4.678)(110) = (9+5x)().0821)(523)

x

0.6

Mole fraction of PCl₅ = 5(1-x)/(5+9x) = 0.17
Partial pressure of PCl₅ = (0.17)(4.678)

Mole fraction of PCl₃ = 5x/(9+5x) = 0.25
Partial pressure of PCl₃ = (0.25)(4.678)

Mole fraction of Cl₂ = 5x/(9+5x) = 0.25
Partial pressure of Cl₂ = (0.25)(4.678)

Hence K_p = (0.25)(4.678)(0.25)(4.678)/{(0.17)(4.678)}

= 1.75 atm

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