CHM 8304

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

Experimental Kinetics

1

CHM 8304

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 100

different concentrations of reactants; different end points

90 80

different initial rates

[Produit]

70 60 50 40 30 20 10 0 0

5

10

15

20

Temps

same half-lives; same rate constants 4

Experimental Kinetics

2

CHM 8304

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 (> [A]0 (>10 times larger) –  the concentration of B will not change much over the course of the reaction; [B] = [B]0 (quasi-constant) ⎛ [A] 0 ([B] 0 − [P]) ⎞ 1 ⎜ ln ⎟ = k 2 t [B] 0 − [A] 0 ⎝ [B] 0 ([A] 0 − [P]) ⎠

⎞ [A] 0 [B] 0 1 ⎛ ⎜ ln ⎟ = k 2 t [B] 0 ⎝ [B] 0 ([A] 0 − [P]) ⎠ ln

ln

[A]0 = k 2 [B]0 t ([A]0 − [P])

[A]0 = k 1't where

où k 1' = k 2 [B]0 ([A]0 − [P])

ln

mono-exponential decrease

[A]0 = k 1' t [A]

[A] = [A]0 e

− k '1 t

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)

[A]0

100

0.16

90 80

0.14

k o b s = k 1'

70

kobs (min-1)

% [A]0

60

0.035 min- 1

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

40 30

0.07 min- 1

20

10

[A]∞ 20

30

40

0.08 0.06

0.02

0.15 min- 1 0

0.10

0.04

0.10 min- 1

10 0

slope pente = k 2 = (1.02 ± 0.02) M- 1min- 1

0.12

50

60

Temps (min) Time

70

80

90

100

0.00 0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

[B]0 (M)

28

Experimental Kinetics

14

CHM 8304

Third order A

+

B

+

Rate

= v = Vitesse

k3

C

P

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

•  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

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

Rate

= v = Vitesse

d[P] dt

k1 k2

AB P

= k3[A][B][C]

30

Experimental Kinetics

15

CHM 8304

Zeroth order k

catalyseur catalyst

+

A

P

catalyseur catalyst

+

d[P] - d[A] = = k dt dt

Vitesse Rate

= v =

•  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 A

t

A0

0

− ∫ d[A] = k ∫ dt

[A]0 - [A] = k t



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

[A]0 - ½ [A]0 = k t½

t½ =



linear relation



[A]0 2k

half-life

31

Zeroth order

100 90 80

pente = -k slope

70

% [A]0

60

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

50 40 30 20 10 0 0

10

20

30

40

50

60

70

80

90

100

Temps Time

32

Experimental Kinetics

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CHM 8304

Summary of observed parameters Reaction order

Rate law

Explicit equation

Linear equation

zero

d[P] =k dt

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

[A]0 - [A] = k t

first

d[P] = k1[A] dt

[A] = [A]0 e − k1 t

second

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

ln

1 1 − = k 2t [A] [A]0

1

[A] =

k 2t +

[A]0 = k1 t [A]

1 [A]0

complex

⎛ [A]0 ([B]0 − [P]) ⎞ 1 ⎜ ln ⎟ = k 2 t [B]0 − [A]0 ⎝ [B]0 ([A]0 − [P]) ⎠

non-linear regression

linear regression

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

Half-life t½ =

[A]0 2k

t½ =

ln 2 k1

t½ =

1 k 2 [A]0

complex

33

Exercise A: Data fitting •  use conc vs time data to determine order of reaction and its rate constant 100.0

80.0

[A] (mM)

60.0

40.0

20.0

0.0 0

10

20

30

40

50

60

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

70

Tim e (m in)

34

Experimental Kinetics

17

CHM 8304

Exercise A: Data fitting •  use conc vs time data to determine order of reaction and its rate constant 0.060

2.50

non-linear

2.00

linear

0.050

0.040 1/[A] - 1[A]0

ln[A]0/[A]

1.50

1.00

y = 0.0008x 0.030

0.020

0.50

0.010

0.000

0.00 0

10

20

30

40

50

60

0

70

10

20

30

40

50

60

70

Tim e (m in)

Tim e (m in)

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

100.0

80.0

[A] (mM)

60.0

40.0

20.0

Time (min)

[A] (mM)

ln ([A]0/[A])

0

100.0

0.00

1/[A] – 1/[A]0 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

0.0 0

10

20

30

40

50

60

70

Tim e (m in)

36

Experimental Kinetics

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CHM 8304

Exercise B: Data fitting •  use conc vs time data to determine order of reaction and its rate constant 0.25

3.5 y = 0.05x 3

linear

0.2

non-linear

0.15

2

1/[A] - 1/[A]0

ln ([A]0/[A])

2.5

1.5

y = 0.003x - 0.0283 0.1

0.05

1 0.5

0 0

0 0

10

20

30

40

50

60

70

Time (min)

10

20

30

40

50

60

70

-0.05 Time (min)

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

Experimental Kinetics

19

CHM 8304

First order (consecutive) k1

A

Rate

= v = Vitesse −

d[A] = k 1[A] dt

B

k2

P

d[P] d[A] = k2 [B] ≠ dt dt

[A] = [A]0 e − k1 t

d[B] = k1[A] − k 2 [B] dt

d [B] = k 1[A]0 e − k1 t − k 2 [B] dt

d [B] + k 2 [B] = k 1[A]0 e − k1 t dt

solved by using the technique of partial derivatives

[B] =

[P] = [A]0 − [A]0 e − k1t −

k 1[A]0 (e − k1t − e − k 2t ) (k 2 − k 1 )

k 1[A]0 (e − k1t − e − k 2 t ) (k 2 − k 1 )

⎛ ⎞ k1 [P] = [A] 0 ⎜1 − e − k1t − (e − k1t − e − k 2 t )⎟ (k 2 − k 1 ) ⎝ ⎠ 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

Experimental Kinetics

20

CHM 8304

Consecutive reactions, k1 = k2 100

[A]0

90

[P]∞

induction period pÈriode d'induction

80

5

% [A]0

60

ln([P]∞ - [P]0) - ln([P]∞ - [P])

70

k2 ≈ k1

50 40

[B]max

30

k2 ≈ k1

4 3

induction period pÈriode d'induction

2

pente = k obs = k 2 ≈ k 1 slope

1 0

-1

0

10

20

30

40

50

60

70

80

90

100

Temps

Time

20 10

[A]∞ , [B]∞

[B]0, [P]0

0 0

10

20

30

40

50

60

70

80

90

100

Temps Time 41

Consecutive reactions, k2 = 0.2 × k1

[A]0

1.0

90

0.8

k 2 = 0.2 ◊ k 1

0.6

80

0.4

[B]max

70

[P]

60

% [A]0

ln([P]∞ - [P]0) - ln([P]∞ - [P])

1.2

100

pente = kobs = k2 slope pÈriode d'induction induction period

0.2 0.0

-0.2 0

50

10

20

30

40

50

60

70

80

90

100

Temps Time

40 30

[B]

20 10

[A]∞

0 0

10

20

30

40

50

60

70

80

90

100

Temps Time

42

Experimental Kinetics

21

CHM 8304

Consecutive reactions, k2 = 5 × k1

100

[A]0

[P]∞

90 80 ln([P]∞ - [P]0) - ln([P]∞ - [P])

70 60

% [A]0

k2 = 5 ◊ k1

7

50

k2 = 5 ◊ k1

40 30

6

4

pente = k obs = k 1 slope

3 2 1 0

-1

20

induction period pÈriode d'induction

5

0

10

[B]max

20

30

40

50

60

70

80

90

100

Temps

Time

10

[A]∞ , [B]∞

0 0

10

20

30

40

50

60

70

80

90

100

Temps 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

Experimental Kinetics

22

CHM 8304

Steady State Approximation •  consider a typical example (in bio-org and organometallic chem) of a two step reaction: k1 k2[B] –  kinetic scheme: I A P k-1

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

–  rate law: –  SSA : d [I] dt

= k1 [A] − k−1 [I] − k 2 [I][B] = 0

•  expression of [I] :

–  rate equation:

k1 [A] ⎞ ⎟⎟ k ⎝ −1 + k 2 [B] ⎠ ⎛

[I] = ⎜⎜

first order in A; less than first order in B

d [P] ⎛ k1k2 [A][B] ⎞ ⎟ = ⎜ dt ⎜⎝ k−1 + k2 [B] ⎟⎠

45

Steady State Approximation •  consider another example of a two-step reaction: k1 –  kinetic scheme: k2 A+B

–  rate law:

k-1

I

P

d [P] = k 2 [I] dt

d [I] = k1 [A][B] − k −1 [I] − k2 [I] = 0 dt ⎛ k [A][B] ⎞ ⎟⎟ •  expression of [I] : [I] = ⎜⎜ 1 ⎝ k −1 + k2 ⎠

–  SSA:

–  rate equation:

d [P] ⎛ k1k2 [A][B] ⎞ ⎟ = kobs [A][B] = ⎜ dt ⎜⎝ k−1 + k2 ⎟⎠

kinetically indistinguishable from the mechanism with no intermediate!

46

Experimental Kinetics

23

CHM 8304

Steady State Approximation •  consider a third example of a two-step reaction: k1 –  kinetic scheme: A

–  rate law: d [P2 ] dt

k -1

I + P1 k 2[B]

= k 2 [I][B]

P2

–  SSA: d [I] = k1 [A] − k−1 [I][P1 ] − k2 [I][B] = 0 dt

•  expression of [I] :

⎞ k1 [A] ⎟⎟ [ ] [ ] k P + k B 2 ⎝ −1 1 ⎠ ⎛

[I] = ⎜⎜

–  rate equation: d [P ] ⎛ k k [A][B] ⎞ 2 1 2 ⎟⎟ = ⎜⎜ dt ⎝ k−1 [P1 ] + k2 [B] ⎠

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

Experimental Kinetics

24

CHM 8304

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: d [P2 ] ⎛ k1k2 [A][B] ⎞ ⎟⎟ = ⎜⎜ dt ⎝ k−1 [P1 ] + k2 [B] ⎠

–  in the case where k-1 >> k2 , in the presence of B and excess (added) P1, the equation can be simplified as follows: d [P2 ] ⎛ k1k2 [A][B] ⎞ ⎟⎟ = ⎜⎜ dt ⎝ k−1 [P1 ] ⎠ •  now it is first order in A and in B

49

Change of reaction order •  however, if one considers the same equation: d [P2 ] ⎛ k1k2 [A][B] ⎞ ⎟⎟ = ⎜⎜ dt ⎝ k−1 [P1 ] + k2 [B] ⎠

–  but reaction conditions are modified such that k-1[P1] > k-1 and [-CN] >> [Br-] so the rate law becomes:

d [P] ⎛ k1k2 [R ][CN ] ⎞ ⎟ = k1 [R ] = ⎜ dt ⎜⎝ k2 [CN ] ⎟⎠

•  i.e., it is typically already saturated with respect to [-CN] 52

Experimental Kinetics

26

CHM 8304

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 OH

•  for example, for

K eq[H +]

H OH

k1

k 2 [-I]

I

k -1 + H2 O

the protonated alcohol is always in rapid eq’m with the alcohol and therefore d [P] ⎛ k1k2 K eq [H +][tBuOH][I] ⎞ dt

= ⎜⎜ ⎝

k−1 [H 2O] + k2 [I]

⎟⎟ ⎠

•  normally, k2[I-] >> k-1[H2O] and therefore d [P] ⎛ k1k2 K eq [H +][tBuOH][I] ⎞ ⎟⎟ = k1K eq [H +][tBuOH] = ⎜ dt ⎜⎝ k2 [I] ⎠

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

Experimental Kinetics

27