in a mixture if N2 and H2 in the reaction 1:3 at 30atm and 300°C , the% of NH3 at equilibrium is 17.8.Calculate kp for N2+H2=2NH3
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11th
Chemistry
Equilibrium
Solubility Equilibria
In a mixture of N2 and H2 i...
CHEMISTRY
In a mixture of N
2
and H
2
initially in a mole ration of 1:3 at 30 atm and 300
o
C, the percentage of ammonia by volume under the equilibrium is 17.8. Calculate the equilibrium constant (K
P
) of the mixture, for the reaction, N
2
(g)+3H
2
(g)⇌2NH
3
(g)
HARD
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ANSWER
Let the initial moles of N
2
and H
2
be 1 and 3.
N
2
+3H
2
⟶2NH
3
Initial 1 3 0
at Eqb. 1−x 3−3x 2x
% of volume is same as % by mole NH
3
4−2x
2x
=0.178
x=
2+(2×0.178)
4×0.178
x=0.302
mole fraction of H
2
at eqb.=
4−2x
3−3x
=0.6165
Mole fraction of N
2
at eqb.=1−0.6165=0.178=0.2055
K
p
=(X
NH
3
.P
T
)
2
/(X
N
2
)(X
H
2
.P
T
)
3
=
(0.2055×30)(0.6165×30)
3
(0.178×30)
2
=7.31×10
−4
atm
−2