In a closed container 3 mol H2, 1 mol N2 and 2 mol ammonia is present at equilibrium condition What is the value of equilibrium constant
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Answer:
N
2
+
(3−3x)
3
3H
2
⟶
2x
0
2NH
3
Initially-
Total no. of moles, n
i
=1+3=4
Pressue, P
i
=4 atm
At equillibrium-
Total no. of moles, n
f
=(1−x)+(3−3x)+2x=4−2x
Pressure, P
f
=3 atm
From ideal gas law,
PV=nRT
At constat temperature,
P∝n
P
f
P
i
=
n
f
n
i
⇒
3
4
=
4−2x
4
⇒4−2x=3
⇒x=0.5
Mole fraction =
total no. of moles
no. of moles
∴ At equillibrium-
Partial pressure of N
2
,P
N
2
=X
N
2
×P=0.5
Partial pressure of H
2
,P
H
2
=X
H
2
×P=1.5
Partial pressure of NH
3
,P
NH
3
=X
NH
3
×P=1
K
P
for the above reaction is given by-
K
P
=
P
N
2
(P
H
2
)
3
(P
NH
3
)
2
=
0.5×(1.5)
3
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