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
Transcript
Chapter 12 Environmental Chemical Reactions and Transformations
There are more than 70,000 synthetic chemicals that are in daily use.
solventscomponents of detergentsdyes and varnishesadditives in plastics and textileschemicals used for constructionantifouling agentsherbicidesinsecticidesfungicides
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
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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
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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
[A]t= [A]0 e-kt
typically to get a 1st order fit, 70% of A needs to react ifthe rate constant is independent of concentration
-CH2- Cl + H2O---> -CH2- OH + H+ + Cl-
benzyl chloride benzyl alcohol
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Figure 12.1 page 470 Fig. 12.2
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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 differentlyln 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
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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
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For the special case of [A] reacting with [A]
A + A products
Since A is reacting with A
and
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
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More on second order reactions
if A + B----> productsand 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]
this has an exact solution
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
Rate Constants and Temperature
1850 - Wilhelmy related rate constant to temperature
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1862 - Bertholet proposed k = A eDT
1889 - Arrhenius showed that the rate constant increases exponentially with 1/temp.
1884 - van’t Hoff Equation
where Kc= “concentration equilibrium constant”U = ‘standard internal energy changeand
Kc = k1/k-1
so
van’t Hoff proposed two energy factors, so that
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U = E1 - E-1
It follows that: ln k1 = - E1/RT +const
or k1 = A e-E1 /RT (Arrhenius Equation)
There are other equations relating temp and rate constants
1898 van’t Hoff rate equation usually give very good empirical fits
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
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krate relative to 25oC x 100 (%)EakJ/mol
10oC 20oC 30oC Avg increase in krate
40 42 76 130 1.850 34 71 139 2.060 28 43 149 2.3
Theory of Arrhenius
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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
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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 const1.38 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
= kT/h
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[BC] ‡--> [D] + [E]
rate = kT/h [BC]‡
since we said
B + C --> <-- [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‡-TS‡
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for a bimolecular reaction H‡ = Ea - RT
this looks like k= A e-Ea / RT
the entropy term in A --> orientation probability
and temp in A may be related to the frequency of encounters; # collisions ~ (RT)1/2
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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
FeedbackA--> X--> Y---> Z
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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
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(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
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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] ‡
for the formation of product HOSO2
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HOSO2‡ + M ---> HOSO2 + M
d[HOSO2]/dt = +k3[HOSO2]‡
Substituting [HOSO2]‡
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In the atmosphere the formation of ozone can be represented as
hNO2 -------> NO + O. k1= 0.4xTSR
MO. + 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
plot your results for NO, NO2, and O3 and if this was a real atmosphere explain.
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Rate determining steps
k1 k3
A + B X Z k-1
if k3 >> k1
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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
R
COO-
R
+H+
Go= GoH + Go
i
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
Go = -2.303 RT log Ka
and Go
H = -2.303 RT log KaH
Goi = -2.303 RTi
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-2.303 RT log Ka=-2.303 RT log KaH +-2.303 RTi
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
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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‡o
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show that log(krate) = log krateH + m,p
or log(krate/krateH) = m,p
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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
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The Taft Relationship
Attempts to extend Hammett type LFERs to aliphatic compounds.
G‡= G‡ref + G‡
i,electronic+G‡i,steric
= 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
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
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directly from the rate constant kA for the acid-catalyzed 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)
