Environmental Engineering Reference
In-Depth Information
E
XAMPLE
5.4 E
QUILIBRIUM
C
ONSTANT FROM
S
TANDARD
T
HERMODYNAMIC
D
ATA
Find the equilibrium constant at 298 K for the reaction: H
2
S(g)
⇔
H
2
S(aq) from
the standard free energy of formation for the compounds
G
f
(kJ/mol)
−
33.6
H
2
S(g)
−
27.9
H
2
S(aq)
The overall free energy change for the reaction is the difference between the
G
f
of the product and reactant.
Δ
G
f
=−
27.9
+
33.6
=+
5.7 kJ/mol. Hence,
K
eq
=
exp
[−
5.7
/(
8.314
×
10
−
3
)(
298
)
]=
0.1.Notethat
K
eq
=
k
f
/k
b
=[
H
2
S
]
w
/P
H
2
S
=
1
/K
aw
as defined previously in Chapter 3. Hence the Henry's constant
K
aw
=
9.98 L atm/mol.
From Appendix 1, the value is 8.3 L atm/mol.
5.3.3.2
Series Reactions and Steady-State Approximation
A large number of reactions in the environment occur either in series or in paral-
lel. Reaction in series is of particular relevance to us since it introduces both the
concepts of
steady state
and
rate-determining steps
that are of importance in envi-
ronmental chemical kinetics. Let us consider the conversion of A to C through an
intermediate B:
A
k
1
B
k
2
−→
−→
C.
(5.41)
The rates of production of A, B, and C are given by
[
]
d
t
=−
d
A
k
1
[
]
A
,
d
[
]
d
t
=
B
k
1
[
A
]−
k
2
[
B
]
,
(5.42)
d
[
]
d
t
=
C
k
2
[
B
]
.
Throughout the reaction, we have mass conservation such that
[
A
]+[
B
]+[
C
]=
[
0. Solving
the equations in succession and making use of the above mass balance and initial
conditions one obtains, after some manipulation, the values of [A], [B], and [C] as
given in Table 5.1. Figure 5.6 is a plot of [A], [B], and [C] with time for repre-
sentative values of
k
1
=
A
]
0
. The initial conditions are at
t
=
0,
[
A
]=[
A
]
0
, and
[
B
]=[
C
]=
0.2 min
−
1
. We note that the value of [A]
decreases to zero in an exponential manner. The value of [B] goes through a max-
imum and then falls off to zero. The value of [C], however, shows a slow initial
1 min
−
1
and
k
2
=
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