Environmental Engineering Reference
In-Depth Information
where
[
B
]
eq
is the concentration of B at equilibrium. If the pH of a sample of
water is 5.7, then
[
H
+
]=[
C
]
eq
=
2
×
10
−
6
mol/L. By stoichiometry
[
HCO
3
]
eq
=
[
B
]
eq
=[
C
]
eq
=
2
×
10
−
6
mol/L. If the closed system considered has an initial CO
2
of
5
×
10
−
5
mol/L, then
[
A
]
0
=[
CO
2
(
aq
)
]=
5
×
10
−
5
mol/L. Then we have
[
B
]=
2
×
10
−
6
−
0.96
)
]
and
[
A
]=[
A
]
0
−[
B
]=
5
×
10
−
5
[
(
e
1.5
t
−
1
)/(
e
1.5
t
−[
B
]
. If the equi-
librium pH is 5, then
[
B
]
eq
=
1
×
10
−
5
mol/L, and hence
[
B
]=
1
×
10
−
5
[
(
e
0.27
t
−
1
)/(
e
0.27
t
−
0.8
)
]
and
[
A
]=
5
×
10
−
5
−[
B
]
. Figure 5.7 gives the concentration of
aqueous CO
2
with time as it is being converted to HCO
3
in the system. The func-
tional dependence on pH is shown. Equilibrium values of CO
2
(aq) is pH dependent
and reaches 48
μ
Min
≈
0.7 s at a pH of 5.7 and 40
μ
Min
≈
14satapHof5.Ifthe
final equilibrium pH is to increase, more CO
2
(aq) has to be consumed and hence the
concentration falls to a lower equilibrium value.
In the case of SO
2
solution in water, we have
k
f
k
b
HSO
3
(
aq
)
+
H
+
(
aq
)
,
SO
2
(
aq
)
+
H
2
O
(5.55)
where
k
f
=
3.4
×
10
−
6
s
−
1
and
k
b
=
2
×
10
8
mol/L/s. Notice first of all that
k
f
in this
case is much larger than for CO
2
, and hence a virtually instantaneous reaction can be
expected. Let the initial SO
2
concentration be 5
×
10
−
5
mol/L. A similar analysis as
above for CO
2
can be carried out to determine the approach to equilibrium for SO
2
.
Figure 5.8 displays the result at two pH values of 5.7 and 5.0. The striking differences
in time to equilibrium from those for CO
2
dissolution reactions are evident. At a pH of
5.7 the equilibrium value is reached in
≈
6 ns, whereas at a pH of 5.0 the characteristic
time is
≈
0.1
μ
s.
5.2 10
-5
5 10
-5
4.8 10
-5
4.6 10
-5
pH = 5.7
pH = 5.0
4.4 10
-5
4.2 10
-5
4 10
-5
0
5
10
15
20
25
t
/s
FIGURE 5.7
Kinetics of solution of CO
2
in water at different pH values.
continued
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