Multimedia Environmental Models - Chapter 6 pot

McKay, Donald. "Advection and Reactions"
Multimedia Environmental Models
Edited by Donald McKay
Boca Raton: CRC Press LLC,2001

©2001 CRC Press LLC

CHAPTER

6
Advection and Reactions

6.1 INTRODUCTION

In Level I calculations, it is assumed that the chemical is conserved; i.e., it is
neither destroyed by reactions nor conveyed out of the evaluative environment by
flows in phases such as air and water. These assumptions can be quite misleading
when determining of the impact of a given discharge or emission of chemical.
First, if a chemical, such as glucose, is reactive and survives for only 10 hours
as a result of its susceptibility to rapid biodegradation, it must pose less of a threat
than PCBs, which may survive for over 10 years. But the Level I calculation treats
them identically. Second, some chemical may leave the area of discharge rapidly as
a result of evaporation into air, to be removed by advection in winds. The contam-
ination problem is solved locally, but only by shifting it to another location. It is
important to know if this will occur. Indeed, recently, considerable attention is being
paid to substances that are susceptible to long-range transport. Third, it is possible
that, in a given region, local contamination is largely a result of inflow of chemical
from upwind or upstream regions. Local efforts to reduce contamination by control-
ling local sources may therefore be frustrated, because most of the chemical is
inadvertently imported. This problem is at the heart of the Canada–U.S., and Scan-
dinavia–Germany–U.K. squabbles over acid precipitation. It is also a concern in

relatively pristine areas such as the Arctic and Antarctic, where residents have little
or no control over the contamination of their environments.
In this chapter, we address these issues and devise methods of calculating the
effect of advective inflow and outflow and degrading reactions on local chemical
fate and subsequent exposures. It must be emphasized that, once a chemical is
degraded, this does not necessarily solve the problem. Toxicologists rarely miss an
opportunity to point out reactions, such as mercury methylation or benzo(a)pyrene
oxidation, in which the product of the reaction is more harmful than the parent
compound. For our immediate purposes, we will be content to treat only the parent
compound. Assessment of degradation products is best done separately by having
the degradation product of one chemical treated as formation of another.

©2001 CRC Press LLC

A key concept in this discussion that was introduced earlier, and is variously
termed

persistence, lifetime, residence time,

or

detention time

of the chemical.
In a steady-state system, as shown in Figure 6.1a, if chemical is introduced at a
rate of E mol/h, then the rate of removal must also be E mol/h. Otherwise, net
accumulation or depletion will occur. If the amount in the system is M mol, then,
on average, the amount of time each molecule spends in the steady-state system will
be M/E hours. This time,

t

, is a

residence time

and is also called a

detention time

or

persistence

. Clearly, if a chemical persists longer, there will be more of it in the
system. The key equation is

t

=

M/E or M

=

t

E

This concept is routinely applied to retention time in lakes. If a lake has a volume

of 100,000 m

3

, and if it receives an inflow of 1000 m

3

per day, then the retention
time is 100,000/1000 or 100 days. A mean retention time of 100 days does not imply
Figure 6.1 Diagram of a steady-state evaluative environment subject to (a) advective flow, (b)
degrading reactions, (c) both, and (d) the time course to steady state.

©2001 CRC Press LLC

that all water will spend 100 days in the lake. Some will bypass in only 10 days,
and some will persist in backwaters for 1000 days but, on average, the residence
time will be 100 days.
The reason that this concept is so important is that chemicals exhibit variable
lifetimes, ranging from hours to decades. As a result, the amount of chemical present
in the environment, i.e., the inventory of chemical, varies greatly between chemicals.
We tend to be most concerned about persistent and toxic chemicals, because rela-
tively small emission rates (E) can result in large amounts (M) present in the
environment. This translates into high concentrations and possibly severe adverse
effects. A further consideration is that chemicals that survive for prolonged periods
in the environment have the opportunity to make long and often tortuous journeys.
If applied to soil, they may evaporate, migrate onto atmospheric particles, deposit
on vegetation, be eaten by cows, be transferred to milk, and then consumed by
humans. Chemicals may migrate up the food chain from water to plankton to fish
to eagles, seals, and bears. Short-lived chemicals rarely survive long enough to

undertake such adventures (or misadventures).
This lengthy justification leads to the conclusion that, if we are going to discharge
a chemical into the environment, it is prudent to know

1.

how long the chemical will survive, i.e.,

t

, and

2. what causes its removal or “death”

This latter knowledge is useful, because it is likely that situations will occur in
which a common removal mechanism does not apply. For example, a chemical may
be potentially subject to rapid photolysis, but this is not of much relevance in long,
dark arctic winters or in deep, murky sediments.
In the process of quantifying this effect, we will introduce rate constants, advec-
tive flow rates and, ultimately, using the fugacity concept, quantities called D values,
which prove to be immensely convenient. Indeed, armed with Z values and D values,
the environmental scientist has a powerful set of tools for calculation and interpre-
tation.
It transpires that there are two primary mechanisms by which a chemical is
removed from our environment: advection and reaction, which we discuss individ-
ually and then in combination.

6.2 ADVECTION

Strangely, “advection” is a word rarely found in dictionaries, so a definition is

appropriate. It means

the directed movement of chemical by virtue of its presence
in a medium that happens to be flowing.

A lazy canoeist is advected down a river.
PCBs are advected from Chicago to Buffalo in a westerly wind. The rate of advection
N (mol/h) is simply the product of the flowrate of the advecting medium, G (m

3

/h),
and the concentration of chemical in that medium, C (mol/m

3

), namely,
N = GC mol/h

©2001 CRC Press LLC

Thus, if there is river flow of 1000 m

3

/h (G) from A to B of water containing 0.3
mol/m

3

(C) of chemical, then the corresponding flow of chemical is 300 mol/h (N).
Turning to the evaluative environment, it is apparent that the primary candidate
advective phases are air and water. If, for example, there was air flow into the 1
square kilometre evaluative environment at 10

9

m

3

/h, and the volume of the air in
the evaluative environment is 6

¥

10

9

m

3

, then the residence time will be 6 hours,
or 0.25 days. Likewise, the flow of 100 m

3

/h of water into 70,000 m

3

of water results
in a residence time of 700 hours, or 29 days. It is easier to remember residence
times than flow rates; therefore, we usually set a residence time and from it deduce
the corresponding flow rate.
Burial of bottom sediments can also be regarded as an advective loss, as can
leaching of water from soils to groundwater. Advection of freons from the tropo-
sphere to the stratosphere is also of concern in that it contributes to ozone depletion.

6.2.1 Level II Advection Algebra Using Partition Coefficients

If we decree that our evaluative environment is at steady state, then air and water
inflows must equal outflows; therefore, these inflow rates, designated G m

3

/h, must
also be outflow rates. If the concentrations of chemical in the phase of the evaluative
environment is C mol/m

3

, then the outflow rate will be G C mol/h. This concept is
often termed the

continuously stirred tank reactor,

or CSTR, assumption. The basic

concept is that, if a volume of phase, for example air, is well stirred, then, if some
of that phase is removed, that air must have a concentration equal to that of the
phase as a whole. If chemical is introduced to the phase at a different concentration,
it experiences an immediate change in concentration to that of the well mixed, or
CSTR, value. The concentration experienced by the chemical then remains constant
until the chemical is removed. The key point is that the outflow concentration equals
the prevailing concentration. This concept greatly simplifies the algebra of steady-
state systems. Essentially, we treat air, water, and other phases as being well mixed
CSTRs in which the outflow concentration equals the prevailing concentration. We
can now consider an evaluative environment in which there is inflow and outflow
of chemical in air and water. It is convenient at this stage to ignore the particles in
the water, fish, and aerosols, and assume that the material flowing into the evaluative
environment is pure air and pure water. Since the steady-state condition applies, as
shown in Figure 6.1a, the inflow and outflow rates are equal, and a mass balance
can be assembled. The total influx of chemical is at a rate G

A

C

BA

in air, and G

W

C

BW

in water, these concentrations being the “background” values. There may also be
emissions into the evaluative environment at a rate E. The total influx I is thus
I = E + G

A

C

BA

+ G

W

C

BW

mol/h
Now, the concentrations within the environment adjust instantly to values C

A

and C

W

in air and water. Thus, the outflow rates must be G

A

C

A

and G

W

C

W

. These
outflow concentrations could be constrained by equilibrium considerations; for
example, they may be related through partition coefficients or through Z values to
a common fugacity.

©2001 CRC Press LLC

This enables us to conceive of, and define, our first Level II calculation in which
we assume equilibrium and steady state to apply, inputs by emission and advection
are balanced exactly by advective emissions, and equilibrium exists throughout the
evaluative environment. All the phases are behaving like individual CSTRs.
Of course, starting with a clean environment and introducing these inflows, it
would take the system some time to reach steady-state conditions, as shown in Figure
6.1d. At this stage, we are not concerned with how long it takes to reach a steady
state, but only the conditions that ultimately apply at steady state. We can therefore
develop the following equations, using partition coefficients and later fugacities.
I = E + G

A

C

BA

+ G

W

C

BW

= G

A

C

A

+ G

W

C

W

But
C

A

= K

AW

C

W

Therefore,
I = C

W

[G

A

K

AW

+ G

W

] and C

W

= I/[G

A

K

AW

+ G

W

]
Other concentrations, amounts (m), and the total amount (M) can be deduced
from C

W

. The extension to multiple compartment systems is obvious. For example,
if soil is included, the concentration in soil will be in equilibrium with both C

A

and
C

W

.

6.2.2 Level II Advection Algebra Using Fugacity

We assume a constant fugacity f to apply within the environment and to the
outflowing media, thus,
I = G

A

Z

A

f + G

W

Z

W

f = f(G

A

Z

A

+ G

W

Z

W

)
f = I/(G

A

Z

A

+ G

W

Z

W

)

or, in general,
f = I/

S

G

i

Z

i

from which the fugacity and all concentrations and amounts can be deduced.

Worked Example 6.1

An evaluative environment consists of 10

4

m

3

air, 100 m

3

water, and 1.0 m

3

soil.
There is air inflow of 1000 m

3

/h and water inflow of 1 m

3

/h at respective chemical
concentrations of 0.01 mol/m

3

and 1 mol/m

3

. The Z values are air 4

¥

10

–4

, water

©2001 CRC Press LLC

0.1, and soil 1.0. There is also an emission of 4 mol/h. Calculate the fugacity
concentrations, persistence amounts and outflow rates.
I = E + G

A

C

BA

+ G

W

C

BW

= 4 + 1000

¥

0.01 + 1

¥

1 = 15 mol/h

S

GZ = 1000

¥

4

¥

10

–4

+1

¥

10

–1

= 0.5 f = I/

S

GZ = 30 Pa
C

A

= 0.012 C

W

= 3 C

S

= 30 mol/m

3

m

A

= 120 m

W

= 300 m

S

= 30 M (total) = 450 mol
G

A

C

A

= 12 G

W

C

W

= 3 G

S

C

S

= 0 Total = 15 = I mol/h

t

= 450/15 =30 h
In this example, the total amount of material in the system, M, is 450 mol. The
inflow rate is 15 mol/h, thus the residence time or the persistence of the chemical
is 30 hours. This proves to be a very useful time. Note that the air residence time
is 10 hours, and the water residence time is 100 hours; thus, the overall residence

time of the chemical is a weighted average, influenced by the extent to which the
chemical partitions into the various phases. The soil has no effect on the fugacity
or the outflow rates, but it acts as a “reservoir” to influence the total amount present
M and therefore the residence time or persistence.

6.2.3 D values

The group G Z, and other groups like it, appear so frequently in later calculations
that it is convenient to designate them as D values,
i.e.,
G Z = D mol/Pa h
The rate, N mol/h, then equals D f. These D values are transport parameters, with
units of mol/Pa h. When multiplied by a fugacity, they give rates of transport. They
are thus similar in principle to rate constants, which, when multiplied by a mass of
chemical, give a rate of reaction. Fast processes have large D values. We can write
the fugacity equation for the evaluative environment in more compact form, as shown
below:

f = I/(D

AA

+ D

AW

) = I/

S

D

Ai

where D

AA

= G

A

Z

A

, D

AW

= G

W

Z

W

, and the first subscript A refers to advection.
Recalculating Example 6.1,

D

AA

= 0.4 and D

AW

= 0.1 and

S

D

Ai

= 0.5

Therefore,

©2001 CRC Press LLC

f = 15/0.5 = 30
and the rates of output, Df, are 12 and 3 mol/h, totaling 15 mol/h as before.
It is apparent that the air D value is larger and most significant. D values can be
added when they are multiplied by a common fugacity. Therefore, it becomes
obvious which D value, and hence which process, is most important. We can arrive
at the same conclusion using partition coefficients, but the algebra is less elegant.
Note that how the chemical enters the environment is unimportant, all sources

being combined or lumped in I, the overall input. This is because, once in the
environment, the chemical immediately achieves an equilibrium distribution, and it
“forgets” its origin.

6.2.4 Advective Processes

In an evaluative environment, there are several advective flows that convey
chemical to and from the environment, namely,

1. inflow and outflow of air
2. inflow and outflow of water
3. inflow and outflow of aerosol particles present in air
4. inflow and outflow of particles and biota present in water
5. transport of air from the troposphere to the stratosphere, i.e., vertical movement
of air out of the environment
6. sediment burial, i.e., sediment being conveyed out of the well mixed layer to depths
sufficient that it is essentially inaccessible
7. flow of water from surface soils to groundwater (recharge)

It also transpires that there are several advective processes which can apply to
chemical movement

within

the evaluative environment. Notable are rainfall, water
runoff from soil, sedimentation, and food consumption, but we delay their treatment
until later.
In situations 1 through 4, there is no difficulty in deducing the rate as GC or Df,
where G is the flowrate of the phase in question, C is the concentration of chemical
in that phase, and the Z value applies to the chemical in the phase in which it is

dissolved or sorbed.
For example, aerosol may be transported to an evaluative world in association
with the inflow of 10

12

m

3

/h of air. If the aerosol concentration is 10

–11

volume
fraction, then the flowrate of aerosol G

Q

is 10 m
3
/h. The relevant concentration of
chemical is that in the aerosol, not in the air, and is normally quite high, for example,
100 mol/m
3
. Therefore, the rate of chemical input in the aerosol is 1000 mol/h. This
can be calculated using the D and f route as follows, giving the same result.
If Z
Q
= 10

8
, then
f = C
Q
/Z
Q
= 100/10
8
= 10
–6
Pa
D
AQ
= G
Q
Z
Q
= 10 ¥ 10
8
= 10
9
©2001 CRC Press LLC
Therefore,
N = Df = 10
9
¥ 10
–6
= 1000 mol/h
Treatment of transport to the stratosphere is somewhat more difficult. We can
conceive of parcels of air that migrate from the troposphere to the stratosphere at

an average, continuous rate, G m
3
/h, being replaced by clean stratospheric air that
migrates downward at the same rate. We can thus calculate the D value. As discussed
by Neely and Mackay (1982), this rate should correspond to a residence time of the
troposphere of about 60 years, i.e., G is V/t. Thus, if V is 6 ¥ 10
9
and t is 5.25 ¥
10
5
h, G is 11400 m
3
/h. This rate is very slow and is usually insignificant, but there
are situations in which it is important.
We may be interested in calculating the amount of chemical that actually reaches
the stratosphere, for example, freons that catalyze the decomposition of ozone. This
slow rate is thus important from the viewpoint of the receiving stratospheric phase,
but is not an important loss from the delivering, or tropospheric, phase. Second, if
a chemical is very stable and is only slowly removed from the atmosphere by reaction
or deposition processes, then transfer to the troposphere may be a significant mech-
anism of removal. Certain volatile halogenated hydrocarbons tend to be in this class.
If we emit a chemical into the evaluative world at a steady rate by emissions and
allow for no removal mechanisms whatsoever, its concentrations will continue to
build up indefinitely. Such situations are likely to arise if we view the evaluative
world as merely a scaled-down version of the entire global environment. There is
certainly advective flow of chemical from, for example, the United States to Canada,
but there is no advective flow of chemical out of the entire global atmospheric
environment, except for the small amounts that transfer to the stratosphere. Whether
advection is included depends upon the system being simulated. In general, the
smaller the system, the shorter the advection residence time, and the more important

advection becomes.
Sediment burial is the process by which chemical is conveyed from the active
mixed layer of accessible sediment into inaccessible buried layers. As was discussed
earlier, this is a rather naive picture of a complex process, but at least it is a starting
point for calculations. The reality is that the mixed surface sediment layer is rising,
eventually filling the lake. Typical burial rates are 1 mm/year, the material being
buried being typically 25% solids, 75% water. But as it “moves” to greater depths,
water becomes squeezed out. Mathematically, the D value consists of two terms,
the burial rate of solids and that of water.
For example, if a lake has an area of 10
7
m
2
and has a burial rate of 1 mm/year,
the total rate of burial is 10,000 m
3
/year or 1.14 m
3
/h, consisting of perhaps 25%
solids, i.e., 0.29 m
3
/h of solids (G
S
) and 0.85 m
3
/h of water (G
W
). The rate of loss
of chemical is then
G

S
C
S
+ G
W
C
W
= G
S
Z
S
f + G
W
Z
W
f = f(D
AS
+ D
AW
)
Usually, there is a large solid to pore water partition coefficient; therefore, C
S
greatly exceeds C
W
or, alternatively, Z
S
is very much greater than Z
W
, and the term
©2001 CRC Press LLC

D
AS
dominates. A residence time of solids in the mixed layer can be calculated as
the volume of solids in the mixed layer divided by G
S
. For example, if the depth of
the mixed layer is 3 cm, and the solids concentration is 25%, then the volume of
solids is 75,000 m
3
and the residence time is 260,000 hours, or 30 years. The
residence time of water is probably longer, because the water content is likely to be
higher in the active sediment than in the buried sediment. In reality, the water would
exchange diffusively with the overlaying water during that time period.
As discussed in Chapter 5, there are occasions in which it is convenient to
calculate a “bulk” Z value for a medium containing a dispersed phase such as an
aerosol. This can be used to calculate a “bulk” Z value, thus expressing two loss
processes as one. D is then GZ where G is the total flow and Z is the bulk value.
6.3 DEGRADING REACTIONS
The word reaction requires definition. We regard reactions as processes that alter
the chemical nature of the solute, i.e., change its chemical abstract system (CAS)
number. For example, hydrolysis of ethyl acetate to ethanol and acetic acid is
definitely a reaction, as is conversion of 1,2-dichlorobenzene to 1,3-dichlorobenzene,
or even conversion of cis butene 2 to trans butene 2. In contrast, processes that
merely convey the chemical from one phase to another, or store it in inaccessible
form, are not reactions. Uptake by biota, sorption to suspended material, or even
uptake by enzymes are not reactions. A reaction may subsequently occur in these
locations, but it is not until the chemical structure is actually changed that we
consider reaction to have occurred. In the literature, the word reaction is occasionally,
and wrongly, applied to these processes, especially to sorption.
We have two tasks. The first is to assemble the necessary mathematical frame-

work for treating reaction rates using rate constants, and the second is to devise
methods of obtaining information on values of these rate constants.
6.3.1 Reaction Rate Expressions
We prefer, when possible, to use a simple first-order kinetic expression for all
reactions. The basic rate equation is
rate N = VCk = Mk mol/h
where V is the volume of the phase (m
3
), C is the concentration of the chemical
(mol/m
3
), M is the amount of chemical, and k is the first-order rate constant with
units of reciprocal time. The group VCk thus has units of mol/h.
The classical application of this equation is to radioactive decay, which is usually
expressed in the forms
dM/dt = –kM or dC/dt = –Ck
The use of C instead of M implies that V does not change with time.
©2001 CRC Press LLC
Integrating from an initial condition of C
O
at zero time gives the following
equations:
ln(C/C
O
) = –kt or C = C
O
exp(–kt)
Rate constants have units of frequency or reciprocal time and are therefore not
easily grasped or remembered. A favorite trick question of examiners is to ask a
student to convert a rate constant of 24 h

–1
into reciprocal days. The correct answer
is 576 days
–1
, so beware of this conversion! It is more convenient to store and
remember half-lives, i.e., the time, t
1/2
, which is the time required for C to decrease
to half of C
O
. This can be related to the rate constant as follows.
When C = 0.5 C
O
, then t = t
1/2
ln (0.5) = –kt
1/2
, therefore, t
1/2
= 0.693/k
For example, an isotope with a half-life of 10 hours has a rate constant, k, of
0.0693 h
–1
.
6.3.2 Non-First-Order Kinetics
Unfortunately, there are many situations in which the real reaction rate is not a
first-order reaction. Second-order rate reactions occur when the reaction rate is
dependent on the concentration of two chemicals or reactants. For example, if
A + B Æ D + E
then the rate of the reaction is dependent on the concentration of both A and B.

Therefore, the reaction rate is as follows:
N = Vk C
A
C
B
Reactant “B” is often another chemical, but it could be another environmental
reactant such as a microbial population or solar radiation intensity. Third-order
reaction rates, when the rate of reaction is dependent on the concentration of three
reactants (N = Vk C
A
C
B
C
C
), are very rare and are unlikely to occur under environ-
mental conditions.
We can often circumvent these complex reaction rate equations by expressing
them in terms of a pseudo first-order rate reaction. The primary assumption is that
the concentration of reactant “B” is effectively constant and will not change appre-
ciably as the reaction proceeds. Thus, the constant k and concentration of reactant
“B” can be lumped into a new rate constant, k
P
, and the second-order reaction
becomes a pseudo first-order reaction. Therefore,
N = Vk C
A
C
B
©2001 CRC Press LLC
and

k
P
= k C
B
Therefore,
N = Vk
P
C
A
which has the form of a simple first-order reaction. Examples of pseudo first-order
reactions include photolysis reactions where reactant “B” is the solar radiation
intensity (I, in photons/s) or microbial degradations processes where “B” is the
populations of microorganisms. Reactions between two chemicals can also be con-
sidered a pseudo first-order reaction when C
A
<< C
B
, so the concentration of B does
not change as the reaction proceeds.
Second-order rate expressions also arise when a chemical reacts with itself,
giving rise to a messy quadratic equation. Thus, if A + A Æ D, the rate equation is
N = Vk C
A
C
A
= Vk C
A
2
Fortunately, most pollutants are present at low concentrations and tend not to react
with themselves, so these types of reactions are rare.

Zero-order expressions occasionally occur in which the rate is independent of
the concentration of the chemical and is thus proportional to concentration to the
power zero. Including zero-order expressions in mass balance models is potentially
dangerous, because the equations can now predict a positive rate of reaction, even
when there is no chemical present. It is embarrassing when computer models cal-
culate negative concentrations of chemicals.
Our strategy is to use every reasonable excuse to force first-order kinetics on
systems by lumping parameters in k. The dividends that arise are worth the effort,
because subsequent calculations are much easier.
Perhaps most worrisome are situations in which we treat the kinetics of microbial
degradation of chemicals. It is possible that, at very low concentrations, there is a
slower or even no reaction, because the required enzyme systems are not “turned
on.” At very high concentrations, the enzyme may be saturated; therefore, the rate
of degradation ceases to be controlled by the availability of the chemical and becomes
controlled by the availability of enzyme. In other cases, the rate of conversion may
be influenced by the toxicity of the chemical to the organism or by the presence of
co-metabolites, chemicals that the enzyme recognizes as being similar to that of the
chemical of interest. Microbiologists have no difficulty conceiving of a multitude
of situations in which chemical kinetics become very complicated and very difficult
to predict and express. They seem to obtain a certain perverse delight in finding
these situations.
Saturation kinetics is usually treated by the Michaelis–Menten equation, which
can be derived from first principles or, more simply, by writing down the basic first-
order equation and multiplying the rate expression by the group shown below.
©2001 CRC Press LLC
Basic expression N = VCk
Group C
M
/(C + C
M

)
Combined expression N = VCC
M
k/(C + C
M
)
When C is small compared to C
M
, the rate reduces to VCk. When C is large compared
to C
M
, it reduces to VC
M
k, which is independent of C, is constant, and corresponds
to the maximum, or zero-order rate. The concentration, C
M
, therefore corresponds
to the concentration that gives the maximum rate using the basic expression. When
C equals C
M
, the rate is half the maximum value. This can be (and usually is)
expressed in terms of other rate constants for describing the kinetics of the associ-
ation of the chemical with the enzyme.
The rate expression is usually written in biochemistry texts in the form
N/V = C v
M
/(C + k
M
)
where v

M
is a maximum rate or velocity equivalent to kC
M
, and k
M
is equivalent to
C
M
and is viewed as a ratio of rate constants. A somewhat similar expression, the
Monod equation, is used to describe cell growth.
If kinetics are not of the first order, it may be necessary to write the appropriate
equations and accept the increased difficulty of solution. A somewhat cunning but
unethical alternative is to guess the concentration, calculate the rate N using the
non-first-order expression, then calculate the pseudo first-order rate constant in the
expression. For example, if a reaction is second order and C is expected to be about
2 mol/m
3
, V is 100 m
3
, and the second-order rate constant, k
2
, is 0.01 m
3
/mol·h,
then N equals 4 mol/h. We can set this equal to VCk; then, k is 0.02 h
–1
. Essentially,
we have lumped Ck
2
as a first-order rate constant. This approach must be used, of

course, with extreme caution, because k depends on C.
6.3.3 Additivity of Rate Constants
A major advantage of forcing first-order kinetics on all reactions is that, if a
chemical is susceptible to several reactions in the same phase, with rate constants
k
A
, k
B
, k
C
, etc., then the total rate constant for reaction is (k
A
+ k
B
+ k
C
), i.e., the
rate constants are simply added. Another favorite trick of perverse examiners is to
inform a student that a chemical reacts by one mechanism with a half-life of 10
hours, and by another mechanism with a half-life of 20 hours, and asks for the total
half-life. The correct answer is 6.7 hours, not 30 hours. Half-lives are summed as
reciprocals, not directly.
6.3.4 Level II Reaction Algebra Using Partition Coefficients
We can now perform certain calculations describing the behavior of chemicals
in evaluative environments. The simplest is a Level II equilibrium steady-state
©2001 CRC Press LLC
reaction situation in which there is no advection, and there is a constant inflow of
chemical in the form of an emission, as depicted in Figure 6.1b. When a steady state
is reached, there must be an equivalent loss in the form of reactions. Starting from
a clean environment, the concentrations would build up until they reach a level such

that the rates of degradation or loss equal the total rate of input. We further assume
that the phases are in equilibrium, i.e., transfer between them is very rapid. As a
result, the concentrations are related through partition coefficients, or a common
fugacity applies. The equations are as follows:
E = V
1
C
1
k
1
+ V
2
C
2
k
2
etc. = SV
i
C
i
k
i
Using partition coefficients,
E = SV
i
C
w
K
iw
k

i
= C
w
SV
i
K
iw
k
i
from which C
w
can be deduced, followed by other concentrations, amounts, rates
of reaction, and the persistence. In the general expression, K
WW
, the water-water
partition coefficient is unity.
Worked Example 6.2
The evaluative environment in Example 6.1 is subject to emission of 10 mol/h
of chemical, but no advection. The reaction half-lives are air, 69.3 hours; water, 6.93
hours; and soil, 693 hours. Calculate the concentrations. Recall that K
AW
= 0.004
and K
SW
= 10.
The rate constants are 0.693/half-lives or air, 0.01; water, 0.1; soil, 0.001; h
–1
.
E = V
A

C
A
k
A
+ V
W
C
W
k
W
+ V
S
C
S
k
S
= C
W
(V
A
K
AW
k
A
+ V
W
k
W
+ V
S

K
SW
k
S
)
= C
W
(0.4 + 10 + 0.01) = C
W
(10.41) = 10
Therefore,
C
W
= 0.9606 mol/m
3
, C
A
= 0.0038, C
S
= 9.606
The rates of reaction then are
air = 0.38
water = 9.61
soil = 0.01
which add to the emission of 10.
©2001 CRC Press LLC
It is important to note that the reaction rate is controlled by the product V, C,
and k. A large value of any one of these quantities may convey the wrong impression
that the reaction is important.
6.3.5 Level II Using Fugacity and D Values for Reaction

We can now follow the same process as used when treating advection and define
D values for reactions. If the rate is V C k or V Z f k, it is also D
R
f, where D
R
is V
Z k. Note that D
R
has units of mol/m
3
Pa identical to those of D
A
or G Z, discussed
earlier. If there are several reactions occurring to the same chemical in the same
phase, then each reaction can be assigned a D value, and these D values can be
added to give a total D value. This is equivalent to adding the rate constants. The
Level II mass balance becomes
E = SV
i
C
i
k
i
= SV
i
Z
i
fk
i
= fSV

i
Z
i
k = fSD
R
Thus, f can be deduced, followed by concentrations, amounts, the total amount M,
and the rates of individual reactions as V C k or D f. We can repeat Example 6.2
in fugacity format.
Worked Example 6.3
An evaluative environment consists of 10000 m
3
air, 100 m
3
water, and 10 m
3
soil. There is input of 25 mol/h of chemical, which reacts with half-lives of 100
hours in air, 75 hours in water, and 50 hours in soil. Calculate the concentrations
and amounts given the Z values below:
Air V
A
= 10
4
Z
A
= 4 ¥ 10
–4
k
A
= 0.01 D
RA

= 0.04
Water V
W
= 100 Z
W
= 0.1 k
W
= 0.1 D
RW
= 1.0
Sediment V
S
= 1.0 Z
S
= 1.0 k
S
= 0.001 D
RS
= 0.001
Total = 1.041
f = E/SD
Ri
= 10/1.041 = 9.606
C
A
= 0.0038 rate = D f = 0.384
C
W
= 0.9606 = 9.606
C

S
= 9.6060 = 0.010
Phase
Volume
V (m
3
)Z k
VZk
or D
C
(mol/m
3
)
m
(mol)
Rate
(mol/h)
Air 10000 4 ¥ 10
–4
0.00693 0.0277 0.0386 386 2.68
Water 100 0.1 0.00924 0.0924 9.66 966 8.93
Soil 10 1.0 0.0139 0.1386 96.6 966 13.39
Total 2318 25.0
©2001 CRC Press LLC
The rate constants in each case are 0.693/half-life. The sum of the V Z k terms
or D values is 0.2587, thus,
f = E/SD = 96.6 Pa
Thus, each C is Z f and each amount m is VC, totaling 2318 mol. Each rate is V C
k or D f, totaling 25 mol/h.
It is clear that the D value V Z k controls the overall importance of each process.

Despite its low volume and relatively slow reaction rate, the soil provides a fairly
fast-reacting medium because of its large Z value. It is not until the calculation is
completed that it becomes obvious where most reaction occurs. The overall residence
time is 2318/25 or 93 hours.
Note that the persistence or M/E is a weighted mean of the persistence or
reciprocal rate constants in each phase. It is also SVZ/SD.
6.4 COMBINED ADVECTION AND REACTION
Advective and reaction processes can be included in the same calculation as
shown in the example below, which is similar to those presented earlier for reaction.
We now have inflow and outflow of air and water at rates given below and with
background concentrations as shown in Figure 6.1c. The mass balance equation now
becomes
I = E + G
A
C
BA
+ G
W
C
BW
= G
A
C
A
+ G
W
C
W
+ SV
i

C
i
k
i
This can be solved either by substituting K
iW
C
W
for all concentrations and solving
for C
W
, or calculating the advective D values as GZ and adding them to the reaction
D values. The equivalence of these routes can be demonstrated by performing both
calculations.
Worked Example 6.4
The environment in Example 6.3 has advective flows of 1000 m
3
/h in air and
1 m
3
/h in water as in Example 6.1 and reaction D values as in Example 6.3, with a
total input by advection and emission of 40 mol/h. Calculate the fugacity concen-
trations, amounts, and chemical residence time.
Phase
Volume
(m
3
)Z
D
A

(advection)
D
R
(reaction)
C
(mol/m
3
)
m
(mol)
Rate
(mol/h)
f(D
A
+ D
R
)
Air 10000 4 ¥ 10
–4
0.4 0.0277 0.021 210 22.55
Water 100 0.1 0.1 0.0924 5.27 527 10.14
Soil 10 1.0 0.0 0.1386 52.7 527 7.31
Total 0.5 0.2587 1264 40
©2001 CRC Press LLC
The total of all D values is 0.7587.
E = 40
Therefore,
f = 40/SD = 52.7
The total amount is 1264 mols, giving a mean residence time of 31.6 hours. The
most important loss process is advection in air, which accounts for 21.08 mol/h.

Next is soil reaction at 7.31 mol/h, the water advection at 5.27 mol/h, etc. Each
individual rate is D f mol/h.
6.4.1 Advection as a Pseudo Reaction
Examination of these equations shows that the group G/V plays the same role
as a rate constant having identical units of h
–1
. It may, indeed, be convenient to
regard advective loss as a pseudo reaction with this rate constant and applicable to
the phase volume of V. Note that the group V/G is the residence time of the phase
in the system. Frequently, this is the most accessible and readily remembered quan-
tity. For example, it may be known that the retention time of water in a lake is 10
days, or 240 hours. The advective rate constant, k, is thus 1/240 h
–1
, and the D value
is V Z k, which is, of course, also G Z.
It is noteworthy that this residence time is not equivalent to a reaction half-time,
which is related to the rate constant through the constant 0.693 or ln 2. Residence
time is equivalent to 1/k.
6.4.2 Residence Times and Persistence
Confusion may arise when calculating the residence time or persistence of a
chemical in a system in which advection and reaction occur simultaneously. The
overall residence time in Example 6.4 is 31.6 hours and is a combination of the
advective residence time and the reaction time. The presence of advection does not
influence the rate constant of the reaction; therefore, it cannot affect the persistence
of the chemical. But, by removing the chemical, it does affect the amount of chemical
that is available for reaction, and thus it affects the rate of reaction. It would be
useful if we could establish a method of breaking down the overall persistence or
residence time into the time attributable to reaction and the time attributable to
advection. This is best done by modifying the fugacity equations as shown below
for total input I.

I = SD
Ai
f + SD
Ri
f
But I = M/t
O
, where M is the amount of chemical and t
O
is the overall residence
time. Furthermore, M = SVZf or fSVZ. Thus, dividing both sides by M and can-
celling f gives
©2001 CRC Press LLC
1/t
O
= SD
Ai
/SVZ + SD
Ri
/SVZ
= 1/t
A
+ 1/t
R
The key point is that the advective and reactive residence times t
A
and t
R
add as
reciprocals to give the reciprocal overall time. These are the residence times that

would apply to the chemical if only that process applied. Clearly, the shorter resi-
dence time dominates, corresponding, of course, to the faster rate constant. It can
be shown that the ratio of the amounts removed by reaction and by advection are in
the ratio of the overall rate constants or the reciprocal residence times.
Example 6.5
Calculate the individual and overall residence times in Example 6.4. Each resi-
dence time is VZ/D and the rate constant is D/VZ.
Adding the reciprocals, i.e., the rate constants, gives
1/60 + 1/240 + 1/866 + 1/260 + 1/• + 1/173
= 0.0167 + 0.0042 + 0.0012 + 0.0038 + 0 + 0.0058 = 0.0209 + 0.0108
= 0.0317 = 1/31.5
The advection residence time is 1/0.0209 or 47.8 h, and for reaction it is 1/0.0108
or 92.6 h. Each residence time (e.g., 60, 866, etc.) contributes to give the overall
residence time of 31.5 hours, reciprocally.
In mass balance models of this type, it is desirable to calculate the advection,
reaction, and overall residence times. An important observation is that these resi-
dence times are independent of the quantity of chemical introduced; in other words,
they are intensive properties of the system. Concentrations, amounts, and fluxes are
dependent on emissions and are extensive properties.
These concepts are useful, because they convey an impression of the relative
importance of advective flow (which merely moves the problem from one region to
another) versus reaction (which may help solve the problem). These are of particular
interest to those who live downwind or downstream of a polluted area.
6.5 UNSTEADY-STATE CALCULATIONS
A related calculation can be done in unsteady-state mode in which we introduce
an amount of chemical, M, into the evaluative environment at zero time, then allow
VZ SVZ/D (advection) VZ/D (reaction)
Air 4 60 866
Water 10 240 260
Soil 10 • 173

Total 24
©2001 CRC Press LLC
it to decay in concentration with time, but maintain equilibrium between all phases
at the same time. This is analogous to a batch chemical reaction system. Although
it is possible to include emissions or advective inflow, we prefer to treat first the
case in which only reaction occurs to an initial mass M. We assume that all volumes
and Z values are constant with time.
dM/dt = –SV
i
C
i
k
i
= –fSV
i
Z
i
k
i
= –fSD
Ri
But,
M = SV
i
Z
i
f = fSV
i
Z
i

df/dt = –fSV
i
Z
i
k
i
/SV
i
Z
i
= –fSD
Ri
/SV
i
Z
i
Solving gives
f = f
O
exp(–k
O
t)
where k
O
= SV
i
Z
i
k
i

/SV
i
Z
i
= SD
Ri
/SV
i
Z
i
, and f
O
is the initial fugacity. Note that k
O
,
the overall rate constant, is the reciprocal of the overall residence time.
Worked Example 6.6
Calculate the time necessary for the environment in Example 6.3 to recover to
50%, 36.7%, 10%, and 1% of the steady-state level of contamination after all
emissions cease.
Here, SVZ is 24 and SD is 0.2587. Thus,
f = f
O
exp (–0.2587t/24) = f
O
exp (–0.01078t)
Since M is proportional to f, and f
O
is 96.6 Pa, we wish to calculate t at which f is
48.3, 35.4, 9.66, and 0.966 Pa. Substituting and rearranging gives t = –1/0.01078 ln

(48.3/96.6), etc., or t is, respectively, 64 h, 93 h, 214 h, and 427 h. The 93-hour time
is significant as both the steady-state residence time and the time of decay to 36.7%
or exp(–1) of the initial concentration.
It is possible to include advection and emissions with only slight complications
to the integration. The input terms may no longer be zero.
This example raises an important point, which we will address later in more
detail. The steady-state situations in the Level II calculations are somewhat artificial
and contrived. Rarely is the environment at a steady state; things are usually getting
worse or better. A valid criticism of Level II calculations is that steady-state analysis
does not convey information about the rate at which systems will respond to changes.
For example, a steady-state analysis of salt emission into Lake Superior may dem-
onstrate what the ultimate concentration of salt will be, but it will take 200 years
for this steady state to be achieved. In a much smaller lake, this steady state may
©2001 CRC Press LLC
be achieved in 10 days. Detractors of steady-state models point with glee to situations
in which the modeler will be dead long before steady state is achieved.
Proponents of steady-state models respond that, although they have not specif-
ically treated the unsteady-state situation, their equations do contain much of the
key “response time” information, which can be extracted with the use of some
intelligence. The response time in the unsteady-state Example 6.5 was 93 hours,
which was SVZ/SD. This is identical to the overall residence time, t, in Example
6.2. The response time of an unsteady-state Level II system is equivalent to the
residence time in a steady-state Level II system. By inspection of the magnitude of
groups, VZ/D, or the reciprocal rate constants that occur in steady-state analysis, it
is possible to determine the likely unsteady-state behavior. This is bad news to those
who enjoy setting up and solving differential equations, because “back-of-the-
envelope” calculations often show that it is not necessary to undertake a complicated
unsteady-state analysis.
Indeed, when calculating D values for loss from a medium, it is good practice
to calculate the ratio VZ/D, where VZ refers to the source medium. This is the

characteristic time of loss, or specifically the time required for that process to reduce
the concentration to e
–1
of its initial value if it were the only loss process. In some
cases, we have an intuitive feeling for what that time should be. We can then check
that the D value is reasonable.
6.6 THE NATURE OF ENVIRONMENTAL REACTIONS
The most important environmental reaction processes are biodegradation, hydrol-
ysis, oxidation, and photolysis. We treat each process briefly below with the view
to establishing methods by which the rate of the reaction can be characterized, and
giving references to authoritative reviews.
6.6.1 Biodegradation
Microbiologists are usually quick to point out that the process of microbial
conversion of chemicals in the environment is exceedingly complex. The rate of
conversion depends on the nature of the chemical compound; on the amount and
condition of enzymes that may be present in various organisms in various states of
activation and availability to perform the chemical conversion; on the availability
of nutrients such as nitrogen, phosphorus, and oxygen; as well as pH, temperature,
and the presence of other substances that may help or hinder the conversion process.
Virtually all organic chemicals are susceptible to microbial conversion or biodegra-
dation. Notable among the slowly degrading or recalcitrant compounds are high-
molecular-weight compounds such as the humic acids, certain terpenes that appear
to have structures that are too difficult for enzymes to attack, and many organo-
halogen substances. Generally, water-soluble organic chemicals are fairly readily
biodegraded. Over evolutionary time, enzymes have adapted and evolved the capa-
bility of handling most naturally occurring organic compounds. When presented
with certain synthetic organic compounds that do not occur in nature (notably the
©2001 CRC Press LLC
halogentated hydrocarbons), they experience considerable difficulty, and they may
or may not be able to perform useful chemical conversions. In such cases, if envi-

ronmental degradation does take place, it is often the result of abiotic processes such
as photolysis or reaction with free radicals.
Our aim is to be able to define a half-life or rate constant for microbial conversion
of the chemical, usually in water but often also in soil and in sediments. These rate
constants may be measured by introducing the chemical into the medium of interest
and following its decay in concentration. If first-order behavior is observed, a rate
constant and half-life may be established. Care must be taken to ensure that the
decay is truly attributable to biodegradation and not to other processes such as
volatilization.
In many cases, non-first-order behavior occurs. For example, it is suspected that,
in some situations, the concentration of chemical is so low that the enzymes neces-
sary for conversion do not become adequately activated, and the chemical is essen-
tially ignored. At high concentrations, the presence of the chemical may result in
toxicity to the microorganisms, and therefore the conversion process ceases. The
number of active enzymatic sites may also be limited, thus the rate of conversion
of the chemical species becomes controlled not by the concentration of the species
but by the number of active sites and the rate at which chemicals can be transferred
into and out of these sites. Under these conditions of saturation, a Michaelis–Menten
type equation can be applied as described earlier.
Much to the chagrin of microbiologists, we will adopt a simple expedient assum-
ing that a first-order rate constant (or half-life) applies and that the rate constant can
be estimated by experiment or from experience. This is necessarily an approximation
to the truth and often involves merely a judgement that, in a particular type of water
or soil, this compound is subject to biodegradation with a half-life of approximately
x hours. The rate constant is therefore 0.693/x hours. Valiant efforts have been made
to devise experimental protocols in which chemicals are subjected to microbial
degradation conditions in the field or in the laboratory using, for example, innocu-
lated sewage sludge. Such estimates are of particular importance in the prediction
of chemical fate in sewage treatment plants. Even more valiant attempts are being
made to predict the rate of biodegradation of chemicals purely from a knowledge

of their molecular structure. Others have been content to categorise organic chem-
icals into various groups that have similar biodegradation rates or characteristics.
Several standard and near-standard tests exist for determining biodegradation
rates under aerobic and anaerobic conditions in water and in soils. Simplest is the
biochemical oxygen demand (BOD) test as described in various standard methods
compilations by agencies such as ASTM and APHA. More complex systems involve
the use of chemostats and continuous flow systems, which are analogous to bench-
top sewage treatment plants.
An important characterization of biodegradation relates to whether the organism
requires an oxygenated environment to thrive. All organisms require energy, which
is obtained by performing chemical reactions. The most common reaction is oxida-
tion, which is performed by aerobic organisms when oxygen is present. Oxidation
of ethanol to acetic acid is an example. When oxygen is absent and anaerobic
conditions prevail, the organism can obtain energy by processes such as reducing
©2001 CRC Press LLC
sulfate to sulfide or by dechlorinating a molecule. The latter is very important as a
method of degrading organo-chlorine compounds, which are recalcitrant to direct
oxidation.
Howard (2000) has reviewed the principles surrounding biodegradation pro-
cesses, the laboratory and field test methods that are employed, and a variety of
methods by which biodegradation half-lives or classes can be estimated. One of the
most popular and accessible biodegradation estimation methods is the BIODEG
program, which is available from the Syracuse Research Corp. website
(www.syrres.com). It is well established that certain groupings of atoms impart
reactivity or recalcitrance to a molecule, thus a molecular structure can be examined
to identify how fast it is likely to degrade. Computer programs such as BIODEG
can do this automatically and assign a structure to a class such as “biodegrades fast”
with a half-life of days to weeks. The half-life may be reported for primary degra-
dation, i.e., loss of the parent compound, but also if interest is the time for complete
mineralization to CO

2
and water. The science of biodegradation is still a long way
from being able to estimate half-lives within an accuracy of a factor of three; indeed,
it may not be possible to estimate half-lives with greater accuracy.
In addition to the excellent review by Howard (2000), the reader will find
valuable material in the texts by Alexander (1994), Pitter and Choduba (1990), and
Schwarzenbach et al. (1993). Howard (2000) also lists databases, notably the
BIOLOG database of some 6000 chemicals.
6.6.2 Hydrolysis
In this process, the chemical species is subject to addition of water as a result
of reaction with water, hydrogen ion, or hydroxyl ion. All three mechanisms may
occur simultaneously at different rates; therefore, the overall rate can be very sen-
sitive to pH. Rates of environmental hydrolysis have been thoroughly reviewed by
Mabey and Mill (1978) and Wolfe and Jeffers (2000). For many organic compounds,
hydrolysis is not applicable.
A systematic method of testing for susceptibility to hydrolysis is to subject the
chemical to pH levels of 3, 7, and 11; observe the decay; and deduce rate constants
for acid, base, and neutral hydrolysis. These rate constants can be combined to give
an expression for the rate at any desired pH, namely,
dC/dt = –k
H
[H
+
]C – k
OH
[OH
–
]C – k
W
[H

2
O]C
Structure activity approaches can be used to correlate and predict these rate constants.
Often, the best approach is to seek data on a structurally similar substance.
Other useful references on hydrolysis include the Wolfe (1980), Pankow and
Morgan (1981), Zepp et al. (1975), Wolfe et al. (1977), and Jeffers et al. (1989).
6.6.3 Photolysis
The energy present in sunlight (photons) is often sufficient to cause chemical
reactions or the rupture of chemical bonds in molecules that are able to absorb this
©2001 CRC Press LLC
light. Sunburn and photosynthesis are examples of such reactions. This process is
primarily of interest when considering the fate of chemicals in solution in the
atmosphere and in water. The radiation that is most likely to effect chemical change
is high-energy, short-wavelength photons at the blue and near UV end of the spec-
trum, i.e., shorter than 400 nm. The relationships between energy, wavelength, and
frequency are readily deduced using the fundamental constants of the speed of light
c (3.0 ¥ 10
8
m/s), Planck’s constant h (6.6 ¥ 10
–34
Js), and Avogadro’s Number N
(6.0 ¥ 10
23
). The energy of a photon of wavelength l nm (frequency c/l Hz) is hc/l
J/molecule or hcN/l J/mol or Einsteins. A photon of wavelength 307 nm has a
frequency of 9.8 ¥ 10
14
Hz and energy of 387,000 J/mol or Einsteins. This is approx-
imately the dissociation energy of the tertiary C-H bond in isobutane (2 methyl
propane); thus, in principle, if the energy in such a photon could be applied to that

bond, dissociation would occur. Short-wavelength photons are more energetic and
are more likely to induce chemical reactions.
There are two general concerns. Will the photon be absorbed such that reaction
will occur? Will the quantity of photons be such that the reaction rate will be
significant?
To be absorbed directly, the molecule must have a chromophore that imparts
suitable absorption characteristics. These properties can be measured using a spec-
trophotometer. As discussed later, there may be indirect absorption of the energy
from another species that absorbs the photon then passes on the energy to the
substance of interest.
The issue of quantity can be assessed by calculating the amount of energy
absorbed, recognizing that there are competitive absorbing substances such as natural
organic matter present in the environment. The extent of absorption can be calculated
from the Beer–Lambert Law such that
log I = log I
O
– eCL = log I
O
– A
where I
O
is the incident radiation, I is the surviving radiation at distance L, con-
centration C, extinction coefficient e, and absorbance A. The quantity of light
absorbed is (I
O
– I), and the fraction that is absorbed by the chemical can be deduced
by comparing A for the chemical with A for the natural organic matter. In near-
transparent or clear water when A is small, the quantity of light absorbed approaches
2.3I
O

eCL Einsteins/m
2
·h. Note that (1 – 10
–x
) approaches 2.3x when x is small. If
each photon absorbed causes j molecules (the quantum yield) to react, then the
reaction rate will be 2.3jI
O
eCL mol/m
2
·h and, in principle, the first-order rate
constant is 2.3jI
O
e, I
O
having units of mol/m
2
h and e units of m
2
/mol. In practice,
I
O
and e are functions of wavelength. Not only is there direct absorption of sunlight
from the sun, but diffuse radiation from the sky also contributes. I
O
also depends
on latitude, time of day and year, and cloud cover. If e is known as a function of
wavelength, computer programs can be used to integrate over the solar spectrum to
give the total photolysis rate constant. The quantum yield may be quite small, e.g.,
0.1 or, in the case of chain reactions, it can be larger than 1.0. Computer programs

such as SOLAR are available to undertake these calculations. The reader is referred
©2001 CRC Press LLC
to Zepp and Cline (1977) for the original work in this area; to Leifer (1988) for an
update; and to Calvert and Pitts (1966), Mill (2000), and Schwarzenbach et al.
(1993) for more details and examples of photochemical reactions and computer
programs.
For our purposes, it is sufficient to appreciate that, knowing the absorbance
properties of the molecule, the quantum yield and the local insolation conditions, it
is possible to calculate a rate constant and a half-life for direct photolysis.
Relatively simple experiments can be conducted in which the chemical is dis-
solved in distilled or natural water in a suitable container and exposed to natural
sunlight or to artificial light for a period of time, and the concentration decay is
monitored. Test methods have been described by Svenson and Bjarndahl (1988),
Lemaire et al. (1982), and Dulin and Mill (1982).
The issue is complicated by the presence of photosensitizing molecules or
substances. These substances absorb light then pass on the energy to the chemical
of interest, resulting in subsequent chemical reaction. It is therefore not necessary
for the chemical to absorb the photon directly. It can receive it second hand from a
photosensitizer. This is a troublesome complication, because it raises the possibility
that chemicals may be subject to photolysis due to the unexpected presence of a
photosensitizer. Of particular interest are the naturally occuring organic matter pho-
tosensitizers that are present in water and give it its characteristic brown color,
especially in areas in which there is peat and decaying vegetation.
6.6.4 Atmospheric Oxidation Reactions
A chemical present in the atmosphere may react with oxygen, an activated form
of oxygen such as singlet oxygen, ozone, hydrogen peroxide, or with various radicals,
notably OH radicals. Fortunately, we live in a world with an abundance of oxygen,
and it is not surprising that a suite of oxygen compounds exists that are eager to
oxidize organic chemicals. The rates of these reactions can be estimated by con-
ducting conventional chemical kinetic experiments in which the substance is con-

tacted with known concentrations of the oxidant, the decay of chemical is followed,
and a kinetic law and rate constant established.
The most important oxidative process is the reaction of hydroxyl radicals with
chemical species in the atmosphere. The concentration of sunlight-induced hydroxyl
radicals is exceedingly small, averaging only about 1 million molecules per cubic
centimetre. Peak concentrations approach 8 million per cm
3
in urban areas. Concen-
trations in rural or remote areas are much lower. They are extremely reactive and
are responsible for the reaction of many organic chemicals in the environment that
would otherwise be persistent.
Ozone is produced by UV radiation in the stratosphere and by certain high-
temperature and photolytic processes in the troposphere. The average mixing ratio,
i.e., the ratio of ozone to non-ozone molecules, is in the range of 10 to 40 ¥ 10
–9
.
Oxides of nitrogen produced at high temperature include NO, NO
2
, and the
reactive NO
3
radical. The latter has an average concentration of about 500 million
molecules per cm
3
and peaks in concentration at night.
©2001 CRC Press LLC
A formidable literature exists on the kinetics of gas phase organic substances,
notably hydrocarbons, with OH radicals. Quantitative structure activity relationships
have been developed in which each part of the molecule is assigned a rate constant
for abstraction of H by OH radicals, or for addition of OH radicals to unsaturated

bonds. Atkinson (2000) has reviewed these estimation methods and provides refer-
ences to compilations of rate constant data. Computer programs exist to estimate
these rate constants from molecular structure, for example from the Syracuse
Research Corporation website (www.syrres.com).
It is important to appreciate that the atmosphere is a very reactive medium in
which large quantities of chemical species are converted into oxidized products.
This is fortunate, because otherwise there would be more severe air pollution and
problems associated with the transport of these chemicals to remote regions.
6.6.5 Aqueous Oxidation and Reduction
Natural oxidizing agents include oxygen, hydrogen peroxide, ozone, and “engi-
neered” oxidants include chlorine, hypochlorite, chlorine dioxide, permanganate,
chromate, and ferrate. Natural reducing agents include sulphide, ferrous and man-
ganous ion, and organic matter, while “engineered” reductants include dithionite
and zero-valent (metal) iron. Oxidation usually involves the addition of oxygen but,
in more general terms, it is the removal of or abstraction of an electron. Reduction
involves electron addition. The potential or feasibility of such a reaction occurring
can be readily evaluated from the standard potential of the half reactions.
The kinetics are usually expressed using a second-order expression including
the concentration of the substance and the oxidant or reductant. In some cases, the
reactant is a solid (e.g., zero-valent iron), and an area-normalized value can be used.
Tratnyek and Macalady (2000) provide an excellent review of this literature and
give several examples of oxidation and reduction processes. Again, for our purposes,
a first-order rate constant can be estimated that includes the concentration of the
oxidising or reducing agent. This can be used to calculate the corresponding half-
life and D value.
6.6.6 Summary
It has been possible to provide only a brief account of the vast literature relating
to chemical reactivity in the environment. The air pollution literature is particularly
large and detailed. References have been provided to give the reader an entry to the
literature.

The susceptibility of a chemical in a specific medium to degrading reaction
depends both on the inherent properties of the molecule and on the nature of the
medium, especially temperature and the presence of candidate reacting molecules
or enzymes. In this respect, environmental chemicals are fundamentally different
from radioisotopes, which are totally unconcerned about external factors. Translation
and extrapolation of reaction rates from environment to environment and laboratory
to environment is therefore a challenging and fascinating task that will undoubtedly
keep environmental chemists busy for many more decades.