Environmental Engineering Reference
In-Depth Information
To simplify the above expression, we should have some idea about the relative
magnitudes of the various rate constants. Generally, in the stratosphere
[
Z
]=[
N 2 ]
10 24 molecules/m 3 , J 1
10 10 min 1 , and J 3
0.019 min 1 . k 2
1.5
×
1.2
×
10 43 m 6 /molecule 2 min, k 4
10 20 m 3 /molecule min, [O 2 ]∼
10 23.5 mole-
7.5
×
cules/m 3 . Hence,
) 1 / 2
10 18.8 molecules/m 3 and
[
O 3 ]=[
O 2 ]
((J 1 /J 3 )(k 2 /k 4 )
·[
N 2 ]
10 13.4 molecules/m 3 .
The steady-state concentration of [O 3 ] given by the equation earlier can be
rearranged to obtain the expression
[
]=
O
k 2 [
O 2 ][
Z
]
[
O 3 ]=
(6.176)
k 4 ( 1
+
(J 3 /k 4 [
O
]
))
10 20
10 13.4 )
10 4 . Hence,
with J 3 /k 4 [
O
]≈
( 0.019 / 7.5
×
×
=
1
×
[
O 3 ]≈
(k 2 /J 3 )
[
O 2 ][
Z
][
O
]
. Thus we have the ratio
[
O 3 ]
[
k 2
J 3 [
O 2 ][
Z
]
.
(6.177)
O
]
10 5 )
if [O 2 ] remains constant. This ratio can vary if either [O] or [O 3 ] varies due to other
reactions. Since the production of O atoms by photolysis of O 2 is a slow process
( J 1 is very small), any competing reaction that decomposes O 3 faster will reduce the
ratio rapidly. These competing reactions involve trace species such as OH , NO, and
CFCs. The general catalytic cycle follows the scheme
The steady-state concentration ratio [O 3 ]/[O] should remain constant (
2.5
×
X
+
O 3 −→
XO
+
O 2 ,
(6.178)
XO
+
O
−→
O 2 +
X
with the net reaction
O
+
O 3 −→
2O 2 .
(6.179)
Note that the species X is neither formed nor removed from the system, and thus acts
as a catalyst.
ThereactionwithOH isestimatedtoaccountforthedecompositionofonly
15%
of the stratospheric ozone. Excited ozone molecules (
< 310 nm) give rise to excited
oxygen atoms, which in turn decomposes H 2 OorCH 4 to provide OH :
λ
ν
−→
h
O +
O 3
O 2 ,
O +
HO ,
(6.180)
CH 4 −→
CH 3 +
O +
2HO .
H 4 O
−→
The OH radicals react with ozone in a self-propagating series of reactions to give two
O 2 molecules:
HO +
O 3 −→
O 2 +
HO 2 ,
(6.181)
HO .
HO 2 +
O 3 −→
O 2 +
 
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