Basics and Discovery of Atmospheric Chemicals - Air Pollution and Global Warming: History, Science, and Solutions - page 27

1.7. Summary
In this chapter, atoms, molecules, elements, and com-
pounds were defined. In addition, a history of the dis-
covery of elements and compounds of atmospheric
importance was given. Only a few elements, includ-
ing carbon, sulfur, and certain metals, and a few solid
compounds, including calcite, halite, and nitre, were
identified in the ancient world. An acceleration of the
discovery of elements and compounds, particularly of
gases such as oxygen and nitrogen, occurred near the
end of the eighteenth century. Several types of chem-
ical reactions take place in the air, including photoly-
sis, kinetic, thermal decomposition, isomerization, and
combination reactions. The rate of a reaction depends
on the reactivity and concentration of molecules. The
chemical e -folding lifetime of a substance is the time
required for its concentration to decrease to 1/ e its orig-
inal value and gives an indication of the reactivity of
the substance. Molecules with free electrons are called
free radicals and are highly reactive.

1.6. Chemical Units
Throughout this text, the atmospheric abundance of a
chemical will generally be quantified in terms of its
mixing ratio or concentration. A mixing ratio is the
number of molecules of a chemical per molecule of
dry air (total air minus water vapor). Mixing ratios can
be in units of parts per trillion volume (pptv), parts
per billion volume (ppbv), parts per million volume
(ppmv), parts per thousand volume, or fraction of air .
Forexample, 40 ppbv of ozone gas indicates that, of
every billion molecules of dry air in a given volume, 40
are ozone molecules. The mixing ratio of oxygen gas
is so large that it is expressed as a fraction (0.2095) or
percent (20.95 percent) of dry air.
Concentrations of a substance can be expressed in
terms of either number or mass concentration. Number
concentration is the number of molecules of a gas per
unit volume of air (e.g., molec. cm − 3 )orthenumber of
aerosol particles per unit volume of air (e.g., particles
cm − 3 ). An example of the conversion between mixing
ratio and number concentration is given in Example
3.6. Mass concentration of a gas or particle is usu-
ally expressed in units of micrograms per cubic meter
of air (

1.8. Problems
1.1. What are the main differences between gases and
aerosol particles?

gm − 3 ). The conversion from mixing ratio

( ,expressed as a fraction) to mass concentration ( m ,

expressed in

gm − 3 )is

=

mN d z
A

m (1.14) 1.2. What compound might you expect to form on the

surface of a statue made of marble or limestone (both
of which contain calcium carbonate) if aqueous sulfuric
acid deposits onto the statue?

where m is molecular weight of the substance of inter-
est (g mol − 1 ), A is Avogadro's number (6.02252 10 23

molec. mol − 1 ), N d is the number concentration of dry
air molecules (molec. cm − 3 ), and z

×

gcm 3 g − 1
m − 3 is a conversion constant. Some of these parame-
ters are defined more precisely in Section 3.4. Example
1.2 illustrates the conversion from mixing ratio to mass
concentration.

10 12

=

1.3. Describe one experiment you could devise to iso-
late molecular oxygen.

1.4. What was the fundamental flaw with the theory of
phlogiston?

1.5. Why did Lavoisier name oxygen as he did? Was
his definition correct? Why or why not?

Example 1.2
Convert 40 ppbv O 3 (g) to g-O 3 (g) m −3 under

typical near-surface conditions.
Solution
The number concentration of dry air molecules
under near-surface conditions is approximated
in Example 3.4 as N d =

1.6. Is a termolecular combination reaction the result of
the collision of three molecules simultaneously? Why
or why not?

10 19 molec. cm −3 .
The molecular weight of ozone is m

1.7. If the chemical e -folding lifetimes of the harmless
substances A , B , and C are 1 hour, 1 week, and 1 year,
respectively, and all three substances produce harm-
ful products when they break down, which substance
would you prefer to eliminate from urban air first?
Why?

2.55 ×

= 48.0 g

mol −1 . Thus, from Equation 1.14, m O 3 =

48.0 g mol −1 10 19 molec.

0.00000004 × × 2.55 ×

cm −3 10 12 gcm 3 g −1 m −3 10 23 × / 6.02252 ×

molec. mol −1 g-O 3 (g) m −3 . = 81.29