At 250C temperature and 1 atm pressure PCL5 is 70percent dissociated . Calculate kp?
Answers
Explanation:
At 250
o
C and 1 atmospheric pressure, the vapour density of PCl
5
is 57.9.
Calculate:
(i) K
p
for the reaction, PCl
5
⇌PCl
3
(g)+Cl
2
(g) at 250
o
C
(ii) the percentage dissociation when pressure is doubled.
Hard
Solution
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(i) Molecular mass of PCl
5
=208.5
Vapour density, D=
2
208.5
=104.25
Observed vapour density, d=57.9
Degree of dissociation, α=
d
D−d
=0.80
PCl
5
⇌PCl
3
(g)+Cl
2
(g)
At equilibrium: 1−α α α
(1−0.8) 0.80 0.80
Total number of moles=1+α=1.80
Partial pressure of PCl
5
=
1.80
0.2
×1=
9
1
Partial pressure of PCl
3
=
1.80
0.80
×1=
9
4
Partial pressure of Cl
2
=
1.80
0.80
×1=
9
4
So, K
p
=
p
PCl
5
p
PCl
3
×p
Cl
2
=1.78
(ii) Let the degree of dissociation be α.
At equilibrium.
p
PCl
5
=(
1+α
1−α
).P=
1+α
1−α
×2
p
PCl
3
=(
1+α
α
).P=
1+α
α
×2
p
Cl
2
==(
1+α
α
).P=
1+α
α
×2
K
p
=
1+α
1−α
×2
1+α
α
×2×
1+α
α
×2
=1.78
or α
2
=
1.89
0.89
α=0.686
Thus, PCl
5
is 68.6% dissociated.