What is degree of dissociation of PCl5 at 5 atmospheric pressure?
PCl5 PCl3 + Cl2 =2.4
Answers
Answer:
K
p
=
1−α
2
αP
2
PCl
5
(g)
⇌ PCl
3
(g)
+Cl
2
(g)
Initial t=0 1 0 0
At equilibrium 1−α α α
Total number of molus =1−α+α+α
=1+α
K
P
will known that partial pressure is the product of mole fraction and the total pressure role fraction is the number of moles of that component divided by the number of mole in mixture
P
PCl
5
=
1+α
(1−α)p
,P
PCl
3
=
1+α
αP
, P
PCl
2
=
1+α
αP
Kp=
P
PCl
5
P
PCl
3
.P
Cl
2
Put all these value:
kp=
(1+α)
(1−α)P
1+α
αP
.
1+α
αP
⇒
(1+α)
(1−α)
(1+α)
2
α
2
p
=
1−α
2
α
2
P
kp=
1−α
2
α
2
P
Explanation: