Chapter 12 Environmental Chemical
Reactions and Transformations
There are more than 70,000 synthetic chemicals that are
in daily use.
solvents
components of detergents
dyes and varnishes
additives in plastics and textiles
chemicals used for construction
antifouling agents
herbicides
insecticides
fungicides
The US EPA in the 1990 Clean Air Amendments
estimates that there are ~2000 excess deaths in the US
each year due to exposures to hazardous chemicals
1
Chapter 12 Environmental Chemical Reactions and
Transformations
Photochemical
Biological-Microbial
Dark reactions
1. kinetics
 order of a reaction
 Arrhnenius temperature -rate const
 rate limiting steps
 steady-state approximation
 Hammett relationships and rate constants
 Langmiurian rate constants
 kinetic simulations
2. mechanisms
 hydrolysis
 organics in the environment
2
Kinetics: the rate law of a reaction must be verified by
experimentation
1st order reactions
A ---> B
-d [A] /dt
=
krate [A]
- d [A]/[A] = kratedt
,t  t
ln[ A ] A
A, t  0   k t
[A]t= [A]0 e-kt
typically to get a 1st order fit, 70% of A needs to react if
the rate constant is independent of concentration
-CH - Cl
+ H2O--->
benzyl chloride
2
-CH - OH + H
2
+
+ Cl-
benzyl alcohol
3
Figure 12.1 page 470 Fig. 12.2
4
t1/2 = time it takes for [A] to decrease by a
factor of 2
ln [A]/[A]o=-k t1/2 ; ln2 /k =t1/2
life times are define slightly differently
ln e = -k 
1/k = 
1/k = time scale for the reaction
Pseudo first order rate constants
A. The reaction of benzyl chloride to produce benzyl
alcohol water reacts with benzyl chloride
CH2 -Cl + H2O
CH2-OH + H+ + Cl-
5
d[benz-Cl]/dt= -k [benz-Cl] [H2O]
k here is a second order rate const. in L mol-1sec-1
If we assume that the reaction is run in dilute
solutions as we could say that the [H2O] is constant
kpseudo = k[H2O]
d[benz-Cl]/dt= -kpseudo [benz-Cl]
B. Consider the reaction of methyl mercaptan in
water to produce dimethyl disulfide
2CH3 SH + ½ O2 ----> H3 C-S-S-CH3
d[CH3 SH]/dt = k [CH3 SH]2 [O2]1/2
if we assume that O2 is constant
d[CH3 SH]/dt = kpseudo [CH3 SH]2
6
For the special case of [A] reacting with [A]
A + A  products
Since A is reacting with A
d[ A ]
  k [ A]2
dt
and

d[ A]
[ A]
2
 kdt
1 At
 k t
Ao
[ A]
1
1

k t
[ A] t [ A] o
for this type of simple second order reaction, plot of
1/[A]t vs. t gives a straight line with a slope of...??? and
an intercept of....?
the half-life is when [A]t = ½ [A]o
1
  k t1 / 2
[A]o
7
More on second order reactions
if A + B----> products
and x is the amount of A and B reacted
the differential equation that describes a second
order rate law for the change in x with respect to
time
dx/dt = - k[A] [B]
or
dx/dt = - k [Ao-x] [Bo-x]
x
t
1
dx
dx
(

)k  dt
0
[ A o ]  [Bo ] 0 [ A o ]  x [Bo ]  x

this has an exact solution
[B ]( [ A o ] x )
1
ln o
kt
[ A o ][Bo ] [ A o ]( [Bo ] x )
so if we measure the amount reacted, x, over time and
plot the left side of the equation vs time, the rate
constant, k can be measured
8
Rate Constants and Temperature
1850 - Wilhelmy related rate constant to temperature
1862 - Bertholet proposed
k = A eDT
1889 - Arrhenius showed that the rate constant
increases exponentially with 1/temp.


lnK

U


c
1884 - van’t Hoff Equation 
 
 T  p RT 2
where
Kc= “concentration equilibrium constant”
U = ‘standard internal energy change
and
Kc = k1/k-1
so
9
d ln k1 d ln k 1 U


dT
dT
RT 2
d ln k1 d ln k 1 U


dT
dT
RT 2
van’t Hoff proposed two energy factors, so that
U = E1 - E-1
It follows that:
or
ln k1 = - E1/RT +const
k1 = A e-E
1
/RT
(Arrhenius Equation)
There are other equations relating temp and rate
constants
1898 van’t Hoff rate equation usually give very good
empirical fits
m (B DT 2 )/ T
k  AT e
10
The Arrhenius equation is most widely used because it
provides insight into how reactions proceed
k1 = A e-Ea/RT
ln k1 = ln A - Ea/RT
(see Table 12.2 page 349- old book) effect of temp on
rate constants (new book does not have this table)
krate (Ti/T2) = eEa (1/T2-1/T1)/R
Ea
kJ/mol
40
50
60
krate relative to 25oC x 100 (%)
10oC
20oC
30oC
Avg
increase
in krate
42
76
130
1.8
34
71
139
2.0
28
43
149
2.3
Theory of Arrhenius
11
Consider the reaction of
B +C ---> D + E
dB/dt = -k [B] [C]
to react B and C have to collide, and the rate should
depend
1. on the frequency of encounters of B and C which is
proportional to the conc. of B and C and how fast B
and C move (diffuse) toward each other.
2. the orientation of B and C
3. the fraction of the collisions that will have sufficient
energy to break the bonds of B and C.
The fraction of reaction species with an energy greater
than the activation energy is given by the Botlzmann
distribution of energies
e-Ea/RT
hence in
rate = A e-Ea/RT[B] [C]
the coef A must include frequency and orientation
factors
12
Activated Complex or transition state theory
the reaction of B + C ---> on its way to products goes
though an activated intermediate called [BC]
B + C ---> [BC]‡--> [D] + [E]

it is assumed that there is an equilibrium between
the reactants and the activated complex [B-C]
 and that [B-C] decomposes to products with a
rate constant of
kT/h
where k = Boltzmann const
1.38 x10-23 J K-1
h= Plancks const,
1.63 x10-34 J/sec
k T/h assumes that the rate constant is directly
proportional to the vibrating frequency of the transition
state and the energy associated with this is
proportional to kT
E  kT  h
 = kT/h
13
[BC] ‡--> [D] + [E]
rate = kT/h [BC]‡
since we said
B+C
-->
<--
[B-C]‡

[
BC
]
K 
[B][C]
substituting for [BC]* in: rate = kT/h [BC]‡
rate = kT/h K‡ [B] [C]
The equilibrium const. is related to the free energy of
activation by
K‡= e-G‡/RT
and
G‡ = H‡-TS‡
14
 kT  S /R  H /RT
rate    e
e
[B][[C]
 h
 kT  S /R  H /RT
k   e
e
 h
for a bimolecular reaction H‡ = Ea - RT
 kT  S /R 1  Ea /RT
k   e
e e
 h
this looks like k= A e-Ea / RT
the entropy term in A --> orientation probability
of
and temp in A may be related to the frequency
encounters; # collisions ~ (RT)1/2
15
Composite Reactions
Simultaneous Reactions
A--> Y
A---> Z
Competition reactions
A + B --> Y
A + C --> Z
Opposing Reactions
A+B
Z
Consecutive reactions
A--> X--> Y-->Z
Feedback
A--> X--> Y---> Z
16
Consecutive Reactions
k1
k2
A ---> X---> Z
-d[A]/dt = k1 [A]
[A] = [A]o e-k1t
+d[X]/dt = k1[A]
d[X]/dt = +k1[A] - k2[X]
d[X]/dt = +k1[A]oe-k1t +- k2[X]
solution
[X] =[A]ok1/(k2-k1) (e-k1t- e-k2t)
[Z] =[A]o/(k2-k1) [k1(1-e-k1t) -k1(1- e-k2t)
These types of expressions are cumbersome
17
(Steady- state approximation)
“The rate of change of the concentration of an
intermediate, to a good approximation, can be
set equal to zero whenever the intermediate is
formed slowly and disappears rapidly” -D.L.
Chapman and L.K. Underhill, 1913
Use of the Steady-State Assumption
(Consecutive Reactions with an Opposing Reaction as
the 1st step)
Consider the reaction of OH radicals in the
atmosphere with SO2 to form sulfuric acid
particles.
k1
OH + SO2
HOSO2‡
k-1
k3
HOSO2‡ + M ---> HOSO2 + M
for the rate of formation of HOSO2‡, we would write
18
d [HOSO2]‡/dt = +k1[OH][SO2]
for the loss we have
-k-1[HOSO2] ‡
and
-k3[HOSO2]‡
so the total rate expression is
d [HOSO2]‡/dt = +k1[OH][SO2]
-k-1 [HOSO2]‡ -k3[HOSO2]‡
at steady state d[HOSO2]‡/dt = zero
0=+k1[OH][SO2] -k-1[HOSO2]‡ -k3[HOSO2] ‡
k [OH][SO2 ]
[HOSO2 ]  1
k 1  k 3
for the formation of product HOSO2
19
HOSO2‡ + M ---> HOSO2 + M
d[HOSO2]/dt = +k3[HOSO2]‡
Substituting [HOSO2]‡
[HOSO 2 ] k1k 3[ OH][ SO 2 ]

dt
k 1  k 3
20
In the atmosphere the formation of ozone can be
represented as
h
NO2 -------> NO + O.
k1= 0.4xTSR
M
O. + O2 ----> O3
k2 = fast,fast
O3 + NO -----> NO2
k3
a. derive a steady state relationship for O3 as a function
of NO2, NO, k1 and k3; assume that dNO2/dt is at ss..
b. calculate the equilibrium ozone for the following
conditions
time
7:00
9:00
12:00
15:00
NO (ppm)
0.1
NO2 (ppm)
0.03
Temp (oC)
25
TSR(cal cm-2min-1) 0.05
0.05
0.14
27
0.2
0.03
0.15
35
0.5
0.005
0.10
34
0.3
plot your results for NO, NO2, and O3 and if this was a
real atmosphere explain.
21
Rate determining steps
k1
A + B
k3
X
Z
k-1
[HOSO 2 ] k1k 3[ OH][ SO 2 ]

dt
k 1  k 3
[ Z] k1k 3[ A][B]

dt
k 1  k 3
if k3 >> k1
[ Z]
 k1[ A] [B]
dt
22
The Hammett Equation and rates constants
In 1940 Hammett recognized for substituted benzoic
acids the effects of substituent groups on the
dissociation of the acid group
COOH
COO-
+H+
R
R
Go= GoH +  Goi
Effect on the free energy change from dissociation could
be represented as the sum of the free energy change by
the unsubstituted benzoic acid and the contributions
from the various R groups.
we know that


and
Go = -2.303 RT log Ka
GoH = -2.303 RT log KaH
23
Goi = -2.303 RTi
-2.303 RT log Ka=-2.303 RT log KaH +-2.303 RTi
log (Ka / KaH )= i
when considering other compound classes, like phenyl
acetic acids the  values developed for benzoic acid can
be used
log (Ka / KaH) = i
Figure 8.7 page 174
24
if an aromatic reaction is going through a transition
state
B + C ---> [BC]‡--> [D] + [E]
we said that the rate
rate = (kT/h) K‡ [B] [C]
the rate constant is (kT/h) K‡
log krate= log (kT/h)+ log K‡
since -2.303 RT log K‡= G‡o
Using the Hammett argument that
G‡o= G‡oH +  G‡oi
show that
or
log(krate) = log krateH +  m,p
log(krate/krateH) =  m,p
25
Figure 12.2 page 354 (old book) new book
does not have this figure
Different rates are obtained in different solvents, so
reaction rates are not directly applicable to water if in
another solvent
26
The Taft Relationship
Attempts to extend Hammett type LFERs to aliphatic
compounds.

G‡= G‡ref + G‡i,electronic+G‡i,steric
log
k
k ref
 *  *   E s

 = polar effects
Es= steric effects
 and  are fitting parameters to a reference system
Taft chose the hydrolysis of carboxylic acid ester system
because he could use different R groups with different
steric and inductive effects
27
By varying R1 but keeping R2, solvent and temp.
constant, Taft proposed that the steric effects of R1
as compared to a methyl group and can be derived
directly from the rate constant kA for the acidcatalyzed hydrolysis reaction.
Es = log (ka/ kA, ref)
This implies that the acid-catalyzed reaction
compared to the base catalyzed reaction does not
have inductive effects (when we look at this we will
see why in about two lectures).
To determine inductive effects (*)the reaction is run
under a basic catalyzed regime and when both
inductive and steric effects are operative.
* = log (kB/kB,ref) – log (ka/ kA, ref )
In the literature, you will sometimes see
*’ = * /2.48 ,to put it on the same scale as the
Hammett  values.
The direction and tends of * values, is similar to
Hammett  values; ie withdrawing groups (Fl, Cl,
NO2) are positive and donating slightly positive or
negative (C2H5)
28
page 356 Table 12.4
29