Chapter 12 Environmental Chemical Reactions and Transformations

Environmental Chemical Reactions and Kinetics

1.kinetics  order of a reaction  Arrhnenius temperature -rate const  rate limiting steps  steady-state approximation  Hammett relationships and rate constants  kinetic simulations

1 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  t ln[A] A, t 0   k t

-kt [A]t= [A]0 e typically to get a 1st order fit, 70% of A needs to react if the rate constant is independent of concentration

2 Some time vs conc. data

Hr Conc [A] Ln[A] 0 2.718 1 0.3 2.117 0.75 0.6 1.649 0.50 0.9 1.284 0.25 1.2 1.000 0.00 1.5 0.779 -0.25

Time vs. concentration

2 m p p

. c n o

0 0 0.5 1 1.5 2 tim e in hours

3 1st order plot

0.6 ] A [ 0.4 n l 0.2

0 0 0.5 1 1.5 2 -0.2

-0.4 time in hours

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 = 

4 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

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

5 2CH3 SH + ½ O2 ----> H3 C-S-S-CH3

2 1/2 d[CH3 SH]/dt = k [CH3 SH] [O2]

if we assume that O2 is constant

2 d[CH3 SH]/dt = kpseudo [CH3 SH]

For the special case of [A] reacting with [A]

A + A  products

Since A is reacting with A d[A] d[A]   k[A]2 and kdt dt [A]2

1  At  kt [A] Ao

1 1  kt [A]t [A]o

6 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  kt1/ 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]

1 x dx dx t (  )k dt 0 [Ao ]  [Bo ] 0 [Ao ]  x [Bo ]  x this has an exact solution

1 [B ]([A ]x) ln o o   k t [Ao][Bo] [Ao]([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 Time -pinene O3 (hh.hh) (ppm) (ppm) 9.15 0.820 0.600 9.23 0.406 0.435 9.50 0.158 0.223 9.83 0.097 0.149

1 [Bo]([Ao]x) ln   k t ???? [Ao][Bo] [Ao]([Bo]x)

9 Rate Constants and Temperature

1850 - Wilhelmy related rate constant to temperature

DT 1862 - Bertholet proposed k = A e

1889 - Arrhenius showed that the rate constant increases exponentially with 1/temp.

 lnK  U 1884 - van’t Hoff Equation c    2  T  p RT where

Kc= “concentration equilibrium constant” U = ‘standard internal energy change and

Kc = k1/k-1

10 dlnk dlnk U 1  1  dT dT RT2

dlnk dlnk U 1  1  dT dT RT2 van’t Hoff proposed two energy factors, so that

 U = E1 - E-1

It follows that: ln k1 = - E1/RT +const

-E /RT or k1 = A e 1 (Arrhenius Equation)

There are other equations relating temp and rate constants

1898 van’t Hoff rate equation usually give very good empirical fits 2 k  ATme(BDT )/T

11 The Arrhenius equation is most widely used because it provides insight into how reactions proceed

-Ea/RT k1 = A e ln k1 = ln A - Ea/RT

Ea (1/T2-1/T1)/R krate (Ti/T2) = e

o krate relative to 25 C x 100 (%) Ea 10oC 20oC 30oC Avg KJ/mol increase in krate 40 42 76 130 1.8 50 34 71 139 2.0 60 28 43 149 2.3

Theory of Arrhenius

12 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

13 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

14  = kT/h [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

15 and G‡ = H‡-TS‡

16 kT   rate    eS /R eH /RT[B][[C]  h 

kT   k    eS /R eH /RT  h 

‡ for a bimolecular reaction H = Ea - RT

kT  k    eS /R e1 eEa /RT  h  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

17 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

18 Consecutive Reactions

k1 k2 A ---> X---> Z

-k1t -d[A]/dt = k1 [A] [A] = [A]o e

+d[X]/dt = k1[A]

d[X]/dt = +k1[A] - k2[X]

-k1t d[X]/dt = +k1[A]oe +- k2[X] solution

-k1t -k2t [X] =[A]ok1/(k2-k1) (e - e )

-k1t -k2t [Z] =[A]o/(k2-k1) [k1(1-e ) -k1(1- e )

These types of expressions are cumbersome

19 (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

k3 ‡ + HOSO2 M ---> HOSO2 + M

‡ for the rate of formation of HOSO2 , we would write

20 ‡ 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]

 k1[OH][SO2] [HOSO2 ]  k1  k3

for the formation of product HOSO2

21 ‡ + HOSO2 M ---> HOSO2 + M

‡ d[HOSO2]/dt = +k3[HOSO2]

‡ Substituting [HOSO2]

[HOSO ] k k [OH][SO ] 2  1 3 2 dt k1  k3

22 In the atmosphere the formation of ozone can be represented as

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 +03 conditions; assume k3 = 2.64310 exp(-1370/ T) time 6:00 7:30 9:00 12:00 15:00

NO (ppm) 0.06 0.11 0.08 0.03 0.005 NO2 (ppm) 0.03 0.07 0.14 0.15 0.10 Temp (oC) 25 26 28 35 34 TSR(cal cm-2min-1) 0.05 0.12 0.25 0.5 0.3 *TSR= total solar radiation flux in cal /cm2/min plot your results for NO, NO2, and O3 and if this was a real atmosphere explain.

23 Rate determining steps

k1 k3 A + B X Z k-1

[HOSO ] k k [OH][SO ] 2  1 3 2 dt k1  k3

[Z] k k [A][B]  1 3 dt k1  k3

if k3 >> k1

[Z]  k [A] [B] dt 1

24 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-

o o o G = G H +  G 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.

-RTln Ka = -RTln KaH - RTln Ka,i

Effect on the free energy change from dissociation could be represented as the sum of the free energy

25 change by the unsubstituted benzoic acid and the contributions from the various R groups.

Hammett then says that the effects of the substituent group i, on the free energy can be represented o as:G i = -2.303 RTi; where i =ln Ka,i we know that

o G = -2.303 RT log Ka and o G H = -2.303 RT log KaH

o G i = -2.303 RTi so

-2.303 RT log Ka=-2.303 RT log KaH +-2.303 RTi going back to:

-RTln Ka = -RTln Ka,H - RTln Ka,i

log Ka= log KaH +i

so, log (Ka / KaH )= I

and pKa = pKaH - I 26 27 Phenols and Hammet Constants anilines

28 Hammet also observes that when considering other compound classes, like phenyl acetic acids the  values developed for benzoic acid can be used

log (Ka / KaH) = 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

29 Rate constants and Hammet sigmas 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

‡o ‡o ‡o G = G H +  G i show that

log(krate) = log krateH +  m,p

30 or log(krate/krateH) =  m,p

Different rates are obtained in different solvents, so reaction rates are not directly applicable to water if in another solvent