If 50% of N204, converts into NO2, at the following equilibrium:
N204(g) = 2 NO2(g)
and the equilibrium pressure is 9 atm. Then Kp:
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Answer:
1−α
1
N
2
O
4
⇌
2α
0
2NO
2
[Total moles =1−α+2α=1+α]
Partial pressure
1+α
1−α
×P
1+α
2α
×P
Given, α=0.2,P=1
Partial pressure of N
2
O
4
=
1+0.2
1−0.2
×1=
1.2
0.8
=
3
2
p
NO
2
=
1+0.2
2×0.2
×1=
1.2
0.4
=
3
1
K
p
=
[p
N
2
O
4
]
2
[p
NO
2
]
2
=
3×3×2
1×1×3
=
6
1
Hence, the correct answer is C
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