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CHM 8304 Experimental Kinetics 1 Enzyme Kinetics Experimental Kinetics Outline: Experimental kinetics rate vs rate constant rate laws and molecularity practical kinetics kinetic analyses (simple equations) first order second order pseudo-first order kinetic analyses (complex reactions) consecutive first order reactions steady state changes in kinetic order saturation kinetics rapid pre-equilibrium see A&D sections 7.4-7.5 for review 2

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Page 1: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 1

Enzyme Kinetics

Experimental Kinetics

Outline: Experimental kinetics •  rate vs rate constant •  rate laws and molecularity •  practical kinetics •  kinetic analyses (simple equations)

–  first order –  second order –  pseudo-first order

•  kinetic analyses (complex reactions) –  consecutive first order reactions –  steady state –  changes in kinetic order –  saturation kinetics –  rapid pre-equilibrium

see A&D sections 7.4-7.5 for review

2

Page 2: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 2

Rate vs rate constant

•  the reaction rate depends on the activation barrier of the global reaction and the concentration of reactants, according to rate law for the reaction –  e.g. v = k[A]

•  the proportionality constant, k, is called the rate constant

3

Rate vs rate constant •  reaction rate at a given moment is the instantaneous slope of [P] vs time •  a rate constant derives from the integration of the rate law

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Temps

[Produit]

different initial rates

same half-lives; same rate constants

different concentrations of reactants; different end points

4

Page 3: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 3

Rate law and molecularity •  each reactant may or may not affect the reaction rate, according to the

rate law for a given reaction •  a rate law is an empirical observation of the variation of reaction rate as

a function of the concentration of each reactant –  procedure for determining a rate law:

•  measure the initial rate (<10% conversion) •  vary the concentration of each reactant, one after the other •  determine the order of the variation of rate as a function of the concentration of

each reactant •  e.g. v ∝ [A][B]2

•  the order of each reactant in the rate law indicates the stoichiometry of its involvement in the transition state of the rate-determining step

5

Integers in rate laws

•  integers indicate the number of equivalents of each reactant that are found in the activated complex at the rae-limiting transition state –  e.g.:

•  reaction: A + B à P –  mechanism: A combines with B to form P

•  rate law: v ∝ [A][B] –  one equivalent of each of A and B are present at the TS of the rds

6

Page 4: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 4

Fractions in rate laws

•  fractions signify the dissociation of a complex of reactants, leading up to the rds: –  e.g. :

•  elementary reactions: A à B + B; B + C à P –  mechanism: reactant A exists in the form of a dimer that must dissociate before

reacting with C to form P

•  rate law: v ∝ [A]½[C] –  true rate law is v ∝ [B][C], but B comes from the dissociation of dimer A –  observed rate law, written in terms of reactants A and C, reflects the dissociation of A –  it is therefore very important to know the nature of reactants in solution!

7

•  negative integers indicate the presence of an equilibrium that provides a reactive species: –  e.g. :

•  elementary reactions: A B + C; B + D à P –  mechanism: A dissociates to give B and C, before B reacts with D to give P

•  rate law: v ∝ [A][C]-1[D] –  true rate law is v ∝ [B][D], but B comes from the dissociation of A –  observed rate law, written in terms of reactants A and D, reflects the dissociation of A –  apparent inhibition by C reflects the displacement of the initial equilibrium

Negative integers in rate laws

8

Page 5: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 5

Practical kinetics

1.  development of a method of detection (analytical chemistry!) 2.  measurement of concentration of a product or of a reactant as a function

of time 3.  measurement of reaction rate (slope of conc/time; d[P]/dt or -d[A]/dt)

–  correlation with rate law and reaction order 4.  calculation of rate constant

–  correlation with structure-function studies

9

Kinetic assays

•  method used to measure the concentration of reactants or of products, as a function of time –  often involves the synthesis of chromogenic or fluorogenic reactants

•  can be continuous or non-continuous

10

Page 6: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 6

Continuous reaction assay

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Time

[Pro

duct

]

Continuous assay •  instantaneous detection of reactants or products as the reaction is

underway –  requires sensitive and rapid detection method

•  e.g.: UV/vis, fluorescence, IR, (NMR), calorimetry

Continuous reaction assay

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Time

[Pro

duct

]

11

Continuous reaction assay

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Time

[Pro

duct

]

Discontinuous assay •  involves taking aliquots of the reaction mixture at various time points,

quenching the reaction in those aliquots and measuring the concentration of reactants/products –  wide variety of detection methods applicable

•  e.g.: as above, plus HPLC, MS, etc.

12

Continuous reaction assay

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Time

[Pro

duct

]

Discontinuous

Page 7: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 7

Initial rates •  the first ~10 % of a reaction is almost linear, regardless of the

order of a reaction •  ΔC vs Δt gives the rate, but the rate constant depends on the

order of the reaction

0 10 20 30 40 50 60 70 80 90 100

Temps (min)

0

10

20

30

40

50

60

70

80

90

100[A

] (m

M)

k = 5 mM min- 1 k1 = 0.07 min- 1

k2 = 0.001 mM- 1 min- 1

tΩ = 10 min

Time

13

½

Calculation of a rate constant

•  now, how can a rate constant, k, be determined quantitatively? •  the mathematical equation to use to determine the value of k differs

according to the order of the reaction in question –  the equation must be derived from a kinetic scheme –  next, the data can be “fitted” to the resulting equation using a computer

(linear, or more likely, non-linear regression)

14

Page 8: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 8

Outline: Kinetic analyses (simple eqns)

–  first order reactions –  second order reactions –  pseudo-first order –  third order reactions –  zeroth order

15

The language of Nature •  “Nul ne saurait comprendre la nature si

celui-ci ne connaît son langage qui est le langage mathématique”

- Blaise Pascal –  (‘None can understand Nature if one does not know its

language, which is the language of mathematics’)

•  Natural order is revealed through special mathematical relationships

•  mathematics are our attempt to understand Nature –  exponential increase: the value of e –  volume of spherical forms: the value of π

Blaise Pascal, French mathematician

and philosopher, 1623-1662

16

Page 9: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 9

First order (simple)

A Pk1

= k1 [A]Vitesse = v = = -d[P]dt

d[A]dt

d[P] k ([P] [P])1dt= −∞

d[P] k ([A] [P])1 0dt= −

d dt[P][P] [P]

k1∞ −

=

∫∫ =−∞

t

01

P

0dtk

[P][P][P]d

ln ([P]∞ - [P]0) - ln ([P]∞ - [P]) = k1 t

ln t[A][A]

k01=

[A] = [A]0k1e− t

linear relation

mono-exponential decrease

ln t[A]½[A]

k0

01 ½=

t ln½

1

2k

=half-life

Rate

17

First order (simple)

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

kobs = k1

[A]∞

[A]0

0 10 20 30 40 50 60 70 80 90 100Temps

-7-6-5-4-3-2-101

ln ([

A]/[A

] 0)

pente = kobs = k1

mono-exponential decrease

Time

Time

slope

18

Page 10: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 10

First order (reversible)

A Pk1

k-1

= k1 [A] - k-1 [P]Vitesse = v = = -d[P]dt

d[A]dt

( )( )[P]kk[P]k[A]k

[P]k[P][P][A]kdt[P]d

dt[A]d

1-1éq1éq1

1-éqéq1

+−+=

−−+==−

À l’équilibre, k1 [A]éq = k-1 [P]éq ( )

( )( )− = = + − +

= + −

ddt

ddt

[A] [P]k [P] k [P] k k [P]

k k [P] [P]

-1 éq 1 éq 1 -1

1 -1 éq

( )ddtt[P]

[P] [P]k k

éq0

P

1 -1 0−= +∫ ∫

( )( ) ( )ln t[P] [P]

[P] [P]k k

éq 0

éq

1 -1

−= +linear relation kobs = (k1 + k-1)

Rate

At equilibrium,

19

First order (reversible)

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]Èq

[P]Èq

kobs = k1

kobs = k1 + k-1

[A]∞

[A]0

[P]0

( )t

ln½

1

2k k

=+ −1

Kéq 1

-1

kk

=

Time

20

Page 11: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 11

Second order (simple)

Vitesse = v = = k2[A][B]d[P]dt

A B P+k2

d[P] k ([A] [P])([B] [P])2 0 0dt= − −

d dt[P]([A] [P])([B] [P])

k0 0

2− −=

ddtt[P]

([A] [P])([B] [P])k

0 00

P

2 0− −=∫ ∫

1[B] [A]

[A] ([B] [P])[B] ([A] [P])

k0 0

0 0

0 02−

⎝⎜

⎠⎟ =ln t

a lot of error is introduced when [B]0 and [A]0 are similar

Rate

21

Second order (simplified)

If [A]0 = [B]0 :

Vitesse = v = = k2[A][B]d[P]dt

A B P+k2

ddtt[P]

([A] [P])k

020

P

2 0−=∫ ∫

1 1[A] [P] [A]

k0 0

2−− = t

1 1[A] [A]

k0

2− = tlinear relation

t½2 0

1k [A]

=half-life

1 1[A]2

[A]k

0 02 ½⎛

⎝⎜

⎞⎠⎟

− = t

Rate

22

Page 12: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 12

Second vs first order

•  one must often follow the progress of a reaction for several (3-5) half-lives, in order to be able to distinguish between a first order reaction and a second order reaction :

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

premier ordre, k1

k2 = k1 ˜ 70

k2 = k1

first order,

23

Réactions cinétiques

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20 25

Temps

[Pro

duit]

Calcul de constante de vitesse

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

16.00

18.00

0 20 40 60 80 100 120

[A]

vite

sse

initi

ale

Example: first or second order? •  measure of [P] as a function of time •  measure of v0 as a function of time [A]

Reaction kinetics

first order [A] = [A]0 e –k1t

Time

[Pro

duct

]

Rate constant calculation

v0 = k1 [A]

In

itial

rate

24

Page 13: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 13

Réactions cinétiques

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20 25

Temps

[Pro

duit]

Calcul de constante de vitesse

0.00

5.00

10.00

15.00

20.00

25.00

30.00

0 20 40 60 80 100 120

[A] = [B]

vite

sse

initi

ale

second order

•  measure of [P] as a function of time •  measure of v0 as a function of time [A]

Time

Reaction kinetics

[Pro

duct

]

Rate constant calculation

In

itial

rate

1/[A] – 1/[A]0 = k2t v0 = k2 [A][B]

Example: first or second order?

25

Pseudo-first order

•  in the case where the initial concentration of one of the reactants is much larger than that of the other, one can simplify the treatment of the experimental data

Vitesse = v = = k2[A][B]d[P]dt

A B P+k2

1[B] [A]

[A] ([B] [P])[B] ([A] [P])

k0 0

0 0

0 02−

⎝⎜

⎠⎟ =ln t general solution:

second order:

26

Page 14: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 14

Pseudo-first order •  consider the case where [B]0 >> [A]0 (>10 times larger)

–  the concentration of B will not change much over the course of the reaction; [B] = [B]0 (quasi-constant)

1[B]

[A] [B][B] ([A] [P])

k0

0 0

0 02ln

⎝⎜

⎠⎟ = t

1[B] [A]

[A] ([B] [P])[B] ([A] [P])

k0 0

0 0

0 02−

⎝⎜

⎠⎟ =ln t

ln [A]([A] [P])

k [B]0

02 0−

= t

ln [A]([A] [P])

k où k k [B]0

01'

1'

2 0−= =t

ln t[A][A]

k01'=

[A] = [A]0k1'

e− tmono-exponential decrease

where

27

Pseudo-first order •  from a practical point of view, it is more reliable to determine a

second order rate constant by measuring a series of first order rate constants as a function of the concentration of B (provided that [B]0 >> [A]0)

0 10 20 30 40 50 60 70 80 90 100

Temps (min)

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]∞

[A]0

0.15 min- 10.10 min- 1

0.07 min- 1

0.035 min- 1

ko b s = k1'

[B]0 = 0.033 M[B]0 = 0.065 M[B]0 = 0.098 M[B]0 = 0.145 M

Time 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16

[B]0 (M)

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

k obs

(min

-1)

pente = k2 = (1.02 ± 0.02) M- 1min- 1slope

28

Page 15: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 15

Third order

•  reactions that take place in one termolecular step are rare in the gas phase and do not exist in solution –  the entropic barrier associated with the simultaneous collision of three

molecules is too high

A B P+k3

C+

Vitesse = v = = k3[A][B][C]d[P]dt

Rate

29

Third order, revisited

•  however, a reaction that takes place in two consecutive bimolecular steps (where the second step is rate-limiting) would have a third order rate law!

A + B AB

C P

k1

+ABk2

Vitesse = v =d[P]dt

= k3[A][B][C]Rate

30

Page 16: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 16

Zeroth order

•  in the presence of a catalyst (organo-metallic or enzyme, for example) and a large excess of reactant, the rate of a reaction can appear to be constant

Vitesse = v = = kd[P]dt

= - d[A]dt

A Pk

catalyseur+ catalyseur+

− =∫ ∫d dtt

[A] kA

A

00[A]0 - [A] = k t [A] = -k t + [A]0 linear relation

[A]0 - ½ [A]0 = k t½

t½ 0[A]2 k

= half-life

catalyst catalyst

Rate

31

Zeroth order

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

pente = -k

not realistic that rate would be constant all the way to end of reaction...

Time

slope

32

Page 17: Enzyme Kinetics - Faculty of Sciencemysite.science.uottawa.ca/jkeillor/English/Teaching_files... · Enzyme Kinetics Experimental Kinetics ... see A&D sections 7.4-7.5 for review 2

CHM 8304

Experimental Kinetics 17

Summary of observed parameters

Reaction order

Rate law Explicit equation Linear equation Half-life

zero d[P] kdt

= [A] = -k t + [A]0 [A]0 - [A] = k t t½ 0[A]2 k

=

first d[P] k [A]1dt= [A] = [A]0

k1e− t ln t[A][A]

k01= t ln

½1

2k

=

second d[P] k [A]22

dt= [A]

k[A]2

0

=+

11t

1 1[A] [A]

k0

2− = t t½2 0

1k [A]

=

d[P] k [A][B]2dt=

complex 1[B] [A]

[A] ([B] [P])[B] ([A] [P])

k0 0

0 0

0 02−

⎝⎜

⎠⎟ =ln t

complex

linear regression

non-linear regression

33

Exercise A: Data fitting •  use conc vs time data to determine order of reaction and its rate

constant

0.0

20.0

40.0

60.0

80.0

100.0

0 10 20 30 40 50 60 70

Time (min)

[A] (

mM

)

Time (min) [A] (mM) ln ([A]0/[A]) 1/[A] – 1/[A]0 0 100.0 0.00 0.000 10 55.6 0.59 0.008 20 38.5 0.96 0.016 30 29.4 1.22 0.024 40 23.8 1.44 0.032 50 20.0 1.61 0.040 60 17.2 1.76 0.048

34

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CHM 8304

Experimental Kinetics 18

0.00

0.50

1.00

1.50

2.00

2.50

0 10 20 30 40 50 60 70

Time (min)

ln[A

]0/[A

] y = 0.0008x

0.000

0.010

0.020

0.030

0.040

0.050

0.060

0 10 20 30 40 50 60 70

Time (min)

1/[A

] - 1

[A]0

Exercise A: Data fitting •  use conc vs time data to determine order of reaction and its rate

constant

linear non-linear

second order; kobs = 0.0008 mM-1min-1

35

Exercise B: Data fitting •  use conc vs time data to determine order of reaction and its rate

constant

0.0

20.0

40.0

60.0

80.0

100.0

0 10 20 30 40 50 60 70

Time (min)

[A] (

mM

)

Time (min) [A] (mM) ln ([A]0/[A]) 1/[A] – 1/[A]0 0 100.0 0.00 0.00 10 60.7 0.50 0.006 20 36.8 1.00 0.017 30 22.3 1.50 0.035 40 13.5 2.00 0.064 50 8.2 2.50 0.112 60 5.0 3.00 0.190

36

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CHM 8304

Experimental Kinetics 19

Exercise B: Data fitting •  use conc vs time data to determine order of reaction and its rate

constant

y = 0.05x

0

0.5

1

1.5

2

2.5

3

3.5

0 10 20 30 40 50 60 70

Time (min)

ln ([

A]0/

[A]) y = 0.003x - 0.0283

-0.05

0

0.05

0.1

0.15

0.2

0.25

0 10 20 30 40 50 60 70

Time (min)

1/[A

] - 1

/[A]0

linear non-linear

first order; kobs = 0.05 min-1

37

Outline: Kinetic analyses (complex rxns)

–  consecutive first order reactions –  steady state –  changes in kinetic order –  saturation kinetics –  rapid pre-equilibrium

38

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CHM 8304

Experimental Kinetics 20

First order (consecutive)

Ak1

B Pk2

= k2 [B]Vitesse = v = d[P]dt ≠ -

d[A]dt

− =d[A] k [A]1dt

[A] = [A]0k1e− t

d[B] k [A] k [B]1 2dt= −

d t[B] k [A] k [B]1 0k

21

dte= −− d t[B] k [B] k [A]2 1 0

k1

dte+ = −

solved by using the technique of partial derivatives

[B] k [A](k k )

( )1

2 1

k k=−

−− −0 1 2e et t

[P] [A] [A] k [A](k k )

( )k 1

2 1

k k= − −−

−− − −0 0

01 1 2e e et t t [P] [A]k

(k k )( )k 1

2 1

k k= − −−

−⎛

⎝⎜

⎠⎟− − −

0 1 1 1 2e e et t t

Rate

39

Induction period

•  in consecutive reactions, there is an induction period in the production of P, during which the concentration of B increases to its maximum before decreasing

A à B à P

•  the length of this period and the maximal concentration of B varies as a function of the relative values k1 and k2 –  consider three representative cases :

•  k1 = k2 •  k2 < k1

•  k2 > k1

40

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Experimental Kinetics 21

Consecutive reactions, k1 = k2

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]0

[B]0, [P]0

[P]∞

[A]∞ , [B]∞

[B]max

pÈriode d'induction

k2 ≈ k1

0 10 20 30 40 50 60 70 80 90 100

Temps

-1

0

1

2

3

4

5

ln([P

] ∞ -

[P] 0)

- ln

([P] ∞

- [P

])

pente = kobs = k2 ≈ k1

k2 ≈ k1

pÈriode d'induction

induction period

induction period slope

Time

Time

41

Consecutive reactions, k2 = 0.2 × k1

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]0

[B]max

[A]∞

[P]

[B]

k2 = 0.2 ◊ k1

0 10 20 30 40 50 60 70 80 90 100

Temps

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

ln([

P] ∞

- [

P] 0

) -

ln([

P] ∞

- [

P])

pÈriode d'induction

pente = kobs = k2

Time

induction period slope

Time

42

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CHM 8304

Experimental Kinetics 22

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]0 [P]∞

[A]∞ , [B]∞

[B]max

k2 = 5 ◊ k1

0 10 20 30 40 50 60 70 80 90 100

Temps

-1

0

1

2

3

4

5

6

7

ln([P

] ∞ -

[P] 0)

- ln

([P] ∞

- [P

])

pÈriode d'induction

pente = kobs = k1

k2 = 5 ◊ k1

Consecutive reactions, k2 = 5 × k1

Time

induction period

slope

Time

43

Steady state •  often multi-step reactions involve the formation of a reactive intermediate

that does not accumulate but reacts as rapidly as it is formed •  the concentration of this intermediate can be treated as though it is

constant •  this is called the steady state approximation (SSA)

44

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Experimental Kinetics 23

Steady State Approximation

•  consider a typical example (in bio-org and organometallic chem) of a two step reaction: –  kinetic scheme:

–  rate law:

–  SSA :

•  expression of [I] :

–  rate equation:

A I Pk1

k-1

k2[B]

[ ] [ ][ ]BItP

2kdd

=

[ ] [ ] [ ] [ ][ ] 0BIIAtI

211 =−−= − kkkdd

[ ] [ ][ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BAI21

1

kkk

[ ] [ ][ ][ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BBA

tP

21

21

kkkk

dd first order in A;

less than first order in B

45

Steady State Approximation

•  consider another example of a two-step reaction: –  kinetic scheme:

–  rate law:

–  SSA:

•  expression of [I] :

–  rate equation:

[ ] [ ]ItP

2kdd

=

[ ] [ ][ ] [ ] [ ] 0IIBAtI

211 =−−= − kkkdd

[ ] [ ][ ]⎟⎟⎠

⎞⎜⎜⎝

+=

− 21

1 BAIkk

k

[ ] [ ][ ] [ ][ ]BABAtP

obs21

21 kkk

kkdd

=⎟⎟⎠

⎞⎜⎜⎝

+=

A + B I Pk1

k-1

k2

kinetically indistinguishable from the mechanism with no intermediate!

46

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Experimental Kinetics 24

Steady State Approximation

•  consider a third example of a two-step reaction: –  kinetic scheme:

–  rate law:

–  SSA:

•  expression of [I] :

–  rate equation:

[ ] [ ][ ]BItP

22 kdd

=

[ ] [ ] [ ][ ] [ ][ ] 0BIPIAtI

2111 =−−= − kkkdd

[ ] [ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BPAI

211

1

kkk

[ ] [ ][ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BPBA

tP

211

212

kkkk

dd

A I + P1k1

k -1k2[B]

P2

first order in A, less than first order in B; slowed by P1

47

SSA Rate equations

•  useful generalisations: 1.  the numerator is the product of the rate constants and concentrations

necessary to form the product; the denominator is the sum of the rates of the different reaction pathways of the intermediate

2.  terms involving concentrations can be controlled by varying reaction conditions

3.  reaction conditions can be modified to make one term in the denominator much larger than another, thereby simplifying the equation as zero order in a given reactant

48

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Experimental Kinetics 25

Change of reaction order

•  by adding an excess of a reactant or a product, the order of a rate law can be modified, thereby verifying the rate law equation

•  for example, consider the preceding equation:

–  in the case where k-1 >> k2 , in the presence of B and excess (added) P1, the equation can be simplified as follows:

•  now it is first order in A and in B

[ ] [ ][ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BPBA

tP

211

212

kkkk

dd

[ ] [ ][ ][ ] ⎟⎟

⎞⎜⎜⎝

⎛=

− 11

212

PBA

tP

kkk

dd

49

Change of reaction order

•  however, if one considers the same equation:

–  but reaction conditions are modified such that k-1[P1] << k2[B] , the equation is simplified very differently:

•  now the equation is only first order in A

•  in this way a kinetic equation can be tested, by modifying reaction conditions

[ ] [ ][ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BPBA

tP

211

212

kkkk

dd

[ ] [ ][ ][ ] ⎟⎟

⎞⎜⎜⎝

⎛=

BBA

tP

2

212

kkk

dd

[ ] [ ]AtP

12 kdd

=

50

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Experimental Kinetics 26

Saturation kinetics

•  on variation of the concentration of reactants, the order of a reactant may change from first to zeroth order –  the observed rate becomes “saturated” with respect to a reactant –  e.g., for the scheme

having the rate law

one observes:

A I Pk1

k-1

k2[B]

[ ] [ ][ ][ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BBA

tP

21

21

kkkk

dd

v

[B]0

vmax = k1[A]

first order in B

~zero order in B

51

Example of saturation kinetics

•  for a SN1 reaction, we are taught that the rate does not depend on [Nuc], but this is obviously false at very low [Nuc], where vobsà0 all the same

•  in reality, for

one can show that •  but normally k2 >> k-1 and [-CN] >> [Br-]

so the rate law becomes:

•  i.e., it is typically already saturated with respect to [-CN]

Brk1

k -1+ Br-

k2 [-CN] CN

[ ] [ ][ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− CNBrCNR

tP

21

21

kkkk

dd

[ ] [ ][ ][ ]

[ ]RCNCNR

tP

12

21 kkkk

dd

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

52

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CHM 8304

Experimental Kinetics 27

Rapid pre-equilibrium

•  in the case where a reactant is in rapid equilibrium before its reaction, one can replace its concentration by that given by the equilibrium constant

•  for example, for the protonated alcohol is always in rapid eq’m with the alcohol and therefore

•  normally, k2[I-] >> k-1[H2O] and therefore

OHk1

k -1+ H2O

k2 [-I] IH

OHKeq[H+]

[ ] [ ][ ][ ][ ] [ ] ⎟⎟

⎞⎜⎜⎝

+

+=

− IOHIBuOHH

tP

221

21

kktKkk

dd eq

[ ] [ ][ ][ ][ ]

[ ][ ]BuOHHI

IBuOHHtP

12

21 tKkktKkk

dd

eqeq +=⎟⎟

⎞⎜⎜⎝

⎛ += first order in tBuOH,

zero order in I-, pH dependent

53

Summary: Kinetic approach to mechanisms

•  kinetic measurements provide rate laws –  ‘molecularity’ of a reaction

•  rate laws limit what mechanisms are consistent with reaction order –  several hypothetical mechanisms may be proposed

•  detailed studies (of substituent effects, etc) are then necessary in order to eliminate all mechanisms – except one! –  one mechanism is retained that is consistent with all data –  in this way, the scientific method is used to refute inconsistent

mechanisms (and support consistent mechanisms)

54