Chemistry of the Atmosphere - Principles of Environmental Chemistry - page 25

Example lifetime calculation
For the reaction, CH 4 þ OH

CH 3 þ H 2 O that is the key loss process
for CH 4 (a greenhouse gas), the rate coefficient for the reaction at
atmospheric temperatures is given by k ¼ 8.4 10 15 cm 3 molecule 1
s 1 . Given the mean atmospheric concentration is [OH] ¼ 5 10 5
molecule cm 3 , what is the atmospheric lifetime of CH 4 ?
Answer: For a reaction of the type A þ B

-

- P(i.e. second-order) the

atmospheric lifetime is given by

1
k ½ OH ¼

1

t CH 4

ð 6 : 3 10 15 Þð 5 10 5 Þ ¼ 3 : 1 10 8 s ¼ 9 : 9 years
The type of calculations are useful as they give an indication of the
likely chemical lifetime (i.e. the amount of time it will take before a
molecule is reacted away) of a molecule in the atmosphere. Clearly,
this form of calculation does not take into account any other chemical
loss routes other than reaction with OH or any other physical process
that may remove a molecule.

OH ¼

Temperature dependence of a reaction
The lifetime of a compound in the atmosphere will depend on how
fast it reacts with main atmospheric oxidants. Many reaction rates
vary with temperature, therefore in the atmosphere the lifetime will
vary with altitude. For the reaction between OH and CH 4 , the
temperature dependence of the reaction is given by k ¼ 1.85
10 12 exp( 1690/T). How does the reaction rate vary between 0.1
and 10 km and therefore effect the lifetime? For the lifetime calcula-
tion see previous description.
Answer:

k (cm 3
molecule 1 s 1 )

Altitude
(km)

t OH
(years)

Region T (1C)

6.3 10 15 Boundary layer 0.1 25 9.9

4.6 10 15 Lower troposphere 1 9 13.8

2.0 10 15 Middle troposphere 6 24 30.3

7.6 10 16 Lower stratosphere 10 56 82.7