Environmental Engineering Reference
In-Depth Information
5.26
3
Sedlak and Andren (1994) showed that the aqueous-phase oxidation of
polychlorinated biphenyls (PCBs) by OH
X
is influenced by the presence
of
A
−
P
k
1
−
k
−
1
A
+
P,
k
2
−−→
A
−
OH
A
+
OH
•
particles such as diatomaceous earth (represented as P). The mecha-
nism representing this is given below, where A-OH is the hydroxylated
PCB. Using appropriate assumptions derive the following equation for the
concentration of A in the aqueous phase:
[
A
]
[
A
]
0
=
k
1
(
β − α
)
(
e
−α
t
−
e
−β
t
)
,
where
αβ =
k
1
k
2
[
OH
•
]
and
α + β =
k
1
+
k
−
1
[
P
]+
k
2
[
OH
•
]
.For
2,2
,4,5,5
-pentachlorobiphenyl,
k
1
=
4.53
×
10
−
4
s
−
1
,
k
2
=
1.41 L/g s,
and
k
2
=
10
−
9
cm
3
/mol s.At an average concentration of [OH
X
]=
1
×
10
−
14
mol/L, and
[
P
]
/
[
Fe
]=
10
μ
M, sketch the progress of hydroxy-
lationofPCB.Repeatyourcalculationfor
[
P
]=
50
μ
M.Whatconclusions
can your draw from your plots?
5.27
2
The photochemical reaction in the upper atmosphere that leads to the
formation and dissipation of O
2
follows the following pathway:
4.6
×
h
ν
−−→
O
(
1
D
)
+
O,
O
2
k
2
−−→
O
+
M.
O
(
1
D
)
+
M
O(
1
D) is a transient state (singlet) for O atom and M is a third body
such as N
2
. The rate of the first reaction is represented as
R
1
.If
[
M
]=
10
14
molecules/cm
3
and
k
2
=
1
×
10
−
25
cm
6
/(molecule)
2
s, determine
the steady-state concentration of singlet oxygen as a function of
R
1
.
5.28
2
Atmospheric photo-oxidation of propane to acetone is said to occur by the
following pathway:
CH
3
CH
2
CH
3
+
OH
•
−→
CH
3
CH
•
CH
3
+
H
2
O,
fast
CH
3
CH
•
CH
3
+
O
2
−−→
CH
3
CH
2
(
O
2
)
CH
3
,
CH
3
CH
2
(
O
2
)
CH
3
+
NO
−→
NO
2
+
CH
3
CH
(
O
•
)
CH
3
,
CH
3
CH
(
O
•
)
CH
3
+
O
2
fast
−−→
CH
3
COCH
3
+
HO
2
.
Obtain an expression for the rate of production of NO
2
due to the
dissipation of propane. Make appropriate assumptions.
5.29
3
Atmospheric chemists consider the oxidation of CH
4
by OH
•
to be a
convenient way of estimating the concentration of hydroxyl ion in the
atmosphere: CH
4
+
OH
•
k
→
CH
3
+
H
2
O. In so doing, it is assumed that
the hydroxyl radical concentration is constant, since it is much larger than
the methane concentration. Given that
k
=
107 m
3
/mol s at 300 K and that
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