Geoscience Reference
In-Depth Information
1.5. Chemical Lifetimes
Some gases are important in the atmosphere because
they persist in high concentrations since they have either
ahigh production rate or a low removal rate. Gases
are produced by either emissions or chemical reaction.
Gases are removed by chemical reaction, scavenging
by rain or aerosol particles, or deposition to a surface.
Some gases that persist in the air for a long time due
to a slow removal rate include N 2 (g), CO 2 (g), CH 4 (g),
N 2 O(g), and H 2 (g).
Some gases are important in the atmosphere because
they chemically react quickly with other gases to form
one or more harmful products. For example, OH(g)
reacts quickly with many organic gases to produce
organic radicals that ultimately contribute to ozone
formation in polluted urban air.
The rate at which a chemical species is lost from the
air is characterized by its e -folding lifetime .Thisisthe
time required for the concentration of a gas to decrease
to 1/ e its original concentration as a result of chemi-
cal reaction. It is similar to the half-lifetime of a gas,
which is the time required for the gas's concentration
to decrease to one-half its original value. The e -folding
lifetime is a lifetime against loss only and is indepen-
dent of a gas's emission rate or rates of production from
other sources. It is possible to estimate the e -folding
lifetimes of a gas against loss by an individual reaction
or against all reactions.
The e -folding lifetime of a gas against loss by a sin-
gle chemical reaction is calculated as follows. Suppose
species A is lost by the photolysis reaction A
In the case of a collision reaction, B
products,
which has a rate coefficient of k in units of cm 3 molec. -1
s 1 ,the rate of change of concentration of [B] is first
estimated by writing
d [B]
dt
+
C
=−
k [B] [C] 0
(1.11)
where [C] is held constant (as denoted by the subscript
0). Integrating Equation 1.11 gives [B]
[B] 0 e k [C] 0 h .
The resulting e -folding lifetime of B occurs when
[B]/[B] 0 =
=
1/ e ,which occurs when
1
k [C] 0
B =
h
=
(1.12)
The overall chemical lifetime of a species is deter-
mined by calculating the lifetime of the species against
loss due to each individual chemical reaction or other
loss process that it is involved in, and then applying
1
A =
(1.13)
1
A 1 +
1
A 2 + ···
1
An
In this equation,
A is the overall chemical lifetime of
species A and
An are the lifetimes of A due
to loss from chemical reactions 1, ..., n ,respectively.
Equation 1.13 can account for losses due not only to
chemistry, but also to rainfall and removal by biological
consumption of the species at the ground.
A 1 , ...,
Example 1.1
The main loss of methane in the atmosphere is
due to the chemical reaction CH 4 (g)
products (e.g., Reaction 1.1). This reaction, which has
one reactant, has rate coefficient J ,inunits of s 1 .The
rate of change of the concentration of A, denoted by
[A], which has units of molecules per cubic centimeter
of air (molec. cm 3 ), is
d [A]
dt
+
h
+
OH(g)
CH 3 (g)
H 2 O(g). Calculate the e -folding lifetime
of methane against loss by this reaction when
the rate coefficient for this reaction is k
+
×
10 −15 cm 3 molec. −1 s −1 at 298 K and the concen-
tration of OH in the air is [OH]
=
6.2
10 5 molec.
cm −3 .Ifthe e -folding lifetime of methane against
loss by soil bacteria metabolism at the surface is
160 years, calculate the overall lifetime of CH 4 (g)
against loss by both processes combined.
=
5.0
×
=−
J [A]
(1.8)
Integrating Equation 1.8 from concentration [A] 0 at
time t
=
0toconcentration [A] at time t
=
h gives
Solution
From Equation 1.12,
[A] 0 e Jh .The e -folding lifetime of A (denoted
[A]
=
CH4
=
1
/
(6.2
×
10 −15
×
by
A )isthetime h at which
[A]
[A] 0 =
5.0
10.2 years due to
chemical reaction alone. From Equation 1.13,
the overall
×
10 5 )
=
3.23
×
10 8
s
=
1
e =
e Jh
(1.9)
lifetime of methane is thus
CH4
=
1
9.6 years. The long lifetime
of methane combined with its ability to warm the
atmosphere per molecule make it an important
contributor to global warming (Chapter 12).
/
(1
/
10.2
+
1
/
160)
=
This occurs when the lifetime of A against loss by the
reaction is
1
J
A =
h
=
(1.10)
 
 
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