Ozone (
) is the form of oxygen consisting of three bound oxygen atoms. It is stable only in high-energy environments (such as near electric discharges). It can be formed by the reactions,
and depleted by
The Chapman stratospheric ozone theory assumes pure oxygen chemistry (no surface chemistry or catalysis) and photochemical equilibrium (no transport). Consider the following reactions, where M is an unspecified third species,
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where
,
,
and are rate constants and 
and 
are photodissociation constants given at 40 km altitude mid-latitude (250 K, 2.9 mbar). At this location, the concentrations of M and O2 are
The destruction rates are
In equilibrium, [O2] and [M] are constant. Therefore, define the following constants
In equilibrium, (9) and (10) must be zero. Substituting the constants in then gives
Define odd oxygen by
then
If we assume that recycling is much more rapid than net production and loss of O and O3 in photochemical equilibrium
then (15) becomes
We know that
. If we know assume that the major loss of Ox is by the second (and not third reaction), we can approximate (17) by
Dividing (19) by (20)
Plugging the solution for [O3] into (19) gives
) is the form of oxygen consisting of three bound oxygen atoms. It is stable only in high-energy environments (such as near electric discharges). It can be formed by the reactions, (1)
(2)
(3)
and depleted by
(4)
(5)
(6)
The Chapman stratospheric ozone theory assumes pure oxygen chemistry (no surface chemistry or catalysis) and photochemical equilibrium (no transport). Consider the following reactions, where M is an unspecified third species,
[Ctrl+A 全部选择 提示：你可先修改部分代码，再按运行]
where
, , and are rate constants and and are photodissociation constants given at 40 km altitude mid-latitude (250 K, 2.9 mbar). At this location, the concentrations of M and O2 are (7)
(8)
The destruction rates are
(9)
(10)
In equilibrium, [O2] and [M] are constant. Therefore, define the following constants
(11)
(12)
(13)
In equilibrium, (9) and (10) must be zero. Substituting the constants in then gives
(14)
(15)
Define odd oxygen by
(6)
then
(17)
If we assume that recycling is much more rapid than net production and loss of O and O3 in photochemical equilibrium
(18)
then (15) becomes
(19)
We know that
. If we know assume that the major loss of Ox is by the second (and not third reaction), we can approximate (17) by(20)
Dividing (19) by (20)
(21)
(22)
Plugging the solution for [O3] into (19) gives
(23)
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