Environmental Engineering Reference
In-Depth Information
For a 1 L container:
2
K
c
=
NO
½
L
−1
=0
:
05mol
N
2
O½
2
ðÞ
2
x
0
:
05 =
ð
0
:
5
−
x
Þ
4
x
2
+5×10
−2
x
5×10
−2
=0
−
2
:
x
1
<0;
x
2
=0
:
073
Answers
: The amount of NO
2
present is 2
0.073 = 0.146 mol = 6.72 g, and the
dissociation grade is (0.073/0.5)
100% = 14.6%.
Example 5.5 Calculation of the equilibrium constant based
on H
0
and S
0
data
Calculate the equilibrium constant at 25
C for the following reaction:
ðÞ
!
CH
3
OH g
ðÞ
CO g
ðÞ
+2H
2
g
Data:
S
0
= 236
mol
−1
K
−1
H
0
=
mol
−1
CH
3
OH g
ðÞ
:
:
8J
−
202
:
1kJ
:
S
0
= 197
mol
−1
K
−1
H
0
=
mol
−1
CO
:
9J
−
110
:
5kJ
S
0
= 130
mol
−1
K
−1
H
0
=0kJ
mol
−1
H
2
:
:
6J
Solution
Step 1: Calculate the enthalpy and entropy changes:
mol
−1
H
0
=
Δ
ð
−
202, 100
Þ−−
ð
110, 500
Þ
=
−
91, 600J
S
0
= 236
mol
−1
K
−1
Δ
:
8
−
ð
261
:
2 + 197
:
9
Þ
=
−
222
:
3J
Step 2: Considering that the free energy is a function of state, we can use Equation
(5.15) to calculate the standard free energy change:
G
0
=
mol
−1
Δ
−
91, 600
−
298
:
15
ð
−
222
:
3
Þ
=
−
91,600 + 66,245 =
−
25,355J
Answer
: The value of
K
p
follows from the relation Equation (5.26):
K
p
=
e
−
Δ
G
0
25
,
355
8×10
4
R
u
T
=
e
15
=2
:
8
:
314 × 298
:
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