Experimental kinetics.pptx

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 (<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
Experimental Kinetics
3
CHM 8304
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 in rate laws
•  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
8
Experimental Kinetics
4
CHM 8304
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
Experimental Kinetics
5
CHM 8304
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
100
90
80
[Product]
70
60
50
40
30
20
10
0
0
5
10
15
20
20
Time
11
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.
Continuous reaction assay
Discontinuous
100
90
80
[Product]
70
60
50
40
30
20
10
0
0
5
10
15
20
Time
12
Experimental Kinetics
6
CHM 8304
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
100
90
80
[A] (mM)
70
60
50
tΩ = 10 min
½
40
30
20
10
0
k 2 = 0.001 mM- 1 min- 1
k 1 = 0.07 min- 1
k = 5 mM min- 1
0
10
20
30
40
50
60
70
80
90
100
Temps
Time (min)
13
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
Experimental Kinetics
7
CHM 8304
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
Blaise Pascal,
French mathematician
and philosopher,
1623-1662
–  exponential increase: the value of e
–  volume of spherical forms: the value of π
16
Experimental Kinetics
8
CHM 8304
First order (simple)
k1
A
Vitesse
Rate
= v =
P
0
d[P]
d[A]
= = k1 [A]
dt
dt
d [P]
= k 1 dt
[P]∞ − [P]
d[P]
= k1 ([P]∞ − [P])
dt
d[P]
= k1 ([A]0 − [P])
dt
∫
P
t
d[P]
= k 1 ∫ dt
0
[P]∞ − [P]
ln ([P]∞ - [P]0) - ln ([P]∞ - [P]) = k1 t ln
linear relation
[A]0
= k1 t
[A]
ln
[A]0
= k 1 t½
½[A]0
[A] = [A]0 e − k1 t
mono-exponential decrease
t½ =
half-life
ln 2
k1
17
First order (simple)
1
0
[A]0
-1
ln ([A]/[A]0)
100
90
mono-exponential
decrease
80
% [A]0
70
-2
pente = kobs = k1
slope
-3
-4
-5
60
-6
50
-7
0
40
10 20 30 40 50 60 70 80 90 100
Temps
Time
30
20
kobs = k1
10
ttΩ½ 0
0
10
20
[A]∞
30
40
50
60
70
80
90
100
Temps
Time
18
Experimental Kinetics
9
CHM 8304
First order (reversible)
k1
A
Vitesse
Rate
= v =
d[P]
d[A]
= = k1 [A] - k-1 [P]
dt
dt
−
l’équilibre, k1 [A]éq = k-1 [P]éq
At À
equilibrium,
P
k-1
d[A] d[P]
=
= k 1 ([A]éq + [P]éq − [P]) − k -1[P]
dt
dt
= k 1[A]éq + k 1[P]éq − (k 1 + k -1 )[P]
−
d[A] d[P]
=
= k -1 [P] éq + k 1 [P] éq − (k 1 + k -1 )[P]
dt
dt
(
)
= (k 1 + k -1 ) [P] éq − [P]
∫
P
0
linear relation
ln
t
d[P]
= (k 1 + k -1 )∫ dt
0
[P] éq − [P]
([P]
([P]
éq
éq
) = (k
− [P])
− [P] 0
1
+ k -1 ) t
kobs = (k1 + k-1) 19
First order (reversible)
[A] 0
100
90
t½ =
80
ln 2
(k
1
+ k −1 )
70
[P] Èq
% [A]0
60
Kéq =
k obs = k 1 + k -1
50
[A] Èq
40
k1
k -1
30
20
k obs = k 1
10
[A] ∞
[P] 0
0
0
10
20
30
40
50
60
70
80
90
100
Temps
Time
20
Experimental Kinetics
10
CHM 8304
Second order (simple)
+
A
B
Vitesse
Rate
= v =
k2
P
d[P]
= k2[A][B]
dt
d [P]
= k 2 dt
([A]0 − [P])([B]0 − [P])
d[P]
= k 2 ([A]0 − [P])([B]0 − [P])
dt
∫
P
0
t
d[P]
= k 2 ∫ dt
0
([A] 0 − [P])([B] 0 − [P])
⎛ [A] 0 ([B] 0 − [P]) ⎞
1
⎜ ln
⎟ = k 2 t
[B] 0 − [A] 0 ⎝ [B] 0 ([A] 0 − [P]) ⎠
a lot of error is
introduced when [B]0
and [A]0 are similar
21
Second order (simplified)
A
+
B
Vitesse
Rate
= v =
If [A]0 = [B]0 :
∫
P
0
k2
P
d[P]
= k2[A][B]
dt
t
d[P]
= k 2 ∫ dt
0
([A] 0 − [P]) 2
1
1
−
= k 2t
[A]0 − [P] [A]0
linear relation
half-life
1
1
−
= k 2t
[A] [A]0
t½ =
1
k 2 [A]0
1
1
−
= k2 t½
⎛ [A] 0 ⎞ [A] 0
⎜
⎟
⎝ 2 ⎠
22
Experimental Kinetics
11
CHM 8304
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 :
100
90
80
70
% [A]0
60
50
premier
ordre, k 1
first order,
40
30
k 2 = k 1 ˜ 70
20
10
k2 = k1
0
0
10
20
30
40
50
60
70
80
90
100
Temps
23
Example: first or second order?
•  measure of [P] as a function of time
•  measure of v0 as a function of time [A]
[A] = [A]0 e
–k1t
v0 = k1 [A]
first order
Réactions
Reactioncinétiques
kinetics
Calcul
constante
de vitesse
Ratede
constant
calculation
18.00
90
16.00
80
14.00
vitesse
initiale
Initial rate
100
[Produit]
[Product]
70
60
50
40
30
20
12.00
10.00
8.00
6.00
4.00
2.00
10
0.00
0
0
5
10
15
Time
Temps
20
25
0
20
40
60
80
100
120
[A]
24
Experimental Kinetics
12
CHM 8304
Example: first or second order?
•  measure of [P] as a function of time
•  measure of v0 as a function of time [A]
1/[A] – 1/[A]0 = k2t
v0 = k2 [A][B]
second order
Réactions
Reactioncinétiques
kinetics
Calcul
constante
de vitesse
Ratede
constant
calculation
100
30.00
90
25.00
Initial rate
vitesse
initiale
80
[Produit]
[Product]
70
60
50
40
30
20
20.00
15.00
10.00
5.00
10
0.00
0
0
5
10
15
20
0
25
20
Temps
Time
40
60
80
100
120
[A] = [B]
25
Pseudo-first order
A
+
B
second order:
Vitesse = v =
general solution:
k2
P
d[P]
= k2[A][B]
dt
⎛ [A] 0 ([B] 0 − [P]) ⎞
1
⎜ ln
⎟ = k 2 t
[B] 0 − [A] 0 ⎝ [B] 0 ([A] 0 − [P]) ⎠
•  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
26
Experimental Kinetics
13
CHM 8304
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)
⎛ [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
16
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
18
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] << k2[B] , the
equation is simplified very differently:
d [P2 ] ⎛ k1k2 [A][B] ⎞
⎟⎟
= ⎜⎜
dt
⎝ k2 [B] ⎠
d [P2 ]
= k1 [A ]
dt
•  now the equation is only first order in A
•  in this way a kinetic equation can be tested, by modifying reaction
conditions
50
Experimental Kinetics
25
CHM 8304
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
k1
k [B]
A
having the rate law
k-1
I
2
P
d [P] ⎛ k1k2 [A][B] ⎞
⎟
= ⎜
dt ⎜⎝ k−1 + k2 [B] ⎟⎠
vmax = k1[A]
one observes:
~zero order in B
v
first order in B
[B]0
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 vobsà0 all the same
k1
k 2 [-CN]
•  in reality, for
Br
CN
k -1
one can show that
+ Br-
d [P] ⎛ k1k2 [R ][CN] ⎞
⎟
= ⎜
dt ⎜⎝ k−1 [Br ]+ k2 [CN] ⎟⎠
•  but normally k2 >> 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