Reaction Order
Rate laws, Reaction Orders
Molecularity
The rate or velocity of a chemical reaction is
loss of reactant or appearance of product in
concentration units, per unit time
The order in each reactant is its exponent, ni
The
! overall order of the reaction is the sum of the n i
• The order of a reaction is determined by the
number of participants that must come
together before the rate-limiting-step (RLS)
of the reaction
• It is determined by the mechanism, not
stoichiometry of a reaction
• It must be determined (measured)
experimentally
Determining Reaction Order
Determining Reaction Order
Initial Rate Method
Initial Rate Method
• Choose initial concentrations of reactants
• Measure “Initial Rate”, before concentrations of reactants
change appreciably (tangents)
• Choose initial concentrations of reactants
• Measure “Initial Rate”, before concentrations of reactants
change appreciably
• Change initial concentrations, and measure initial rate
again
•Example: For A + B  C, assume Rate = k[A]n a [B]n b
d[P]
d[S]
="
dt
dt
The rate law for a reaction is of the form
d[P]
na
nb
nc
! Rate = dt = k[A] [B] [C] ...
Initial rate
Course of reaction
[A]o
1
2
1
[B]o
1
1
2
Rate
R1
R2
R3
if
then
R 2= R1
na = 0
R 2= 2R1 n a = 1
R 2= 4R1 n a = 2
R 3= R1
n b = 0, etc.
•Once you know the ni, plug in and solve for k
Determining Reaction Order
Determining Reaction Order
Integrated Rate Equation Method
Integrated Rate Equation, Zero Order
• Allow the rx to proceed and measure change in
concentration of reactant or product.
• By trial and error, determine which equation fits the data
• Choose initial concentrations of reactants
• Measure Rate over a substantial portion of the reaction
• The product and reactant concentrations change
“automatically”
• Fit to theory
•Example: zero order reaction A
B
d[B]
d[A]
•The rate law is
="
= k[A]0 = k
dt
dt
•Integrating from [A]0 at t = 0 to [A]t at t = t
[A ]
t
" d[A] =!# " kdt
[A ]o
[A]t " [A]0 = "kt
0
•So, a plot of [A]t - [A]0 vs t is linear, with a slope of -k
!
!
1
Integrated Rate Equation
Integrated Rate Equation
First Order Reaction
Second Order Reaction
A
B
• The rate law is
A
B
• The rate law is
• Rewrite as
d[B]
d[A]
="
= k[A]
dt
dt
d[A]
= "kdt
[A]
• Rewrite as
[A ]
• Integrate!from [A]0 at t = 0 to [A]t at t = t
t
d[A]
= # " kdt
"
[A ]o [A]
0
!
A
ln[A]t " ln[A]0 = ln( t ) = "kt
A0
!
d[B]
d[A]
="
= k[A]2
dt
dt
d[A]
= "kdt
[A]2
[A ]
• Integrate
! from [A]0 at t = 0 to [A]t at t = t
"2
d[A] = " # kdt
A
or ln( 0 ) = kt
At
!
• A plot of ln (A0/At) vs t is linear, with a slope of k
• If so, the reaction is 1st Order in A and zero order in all
other components
!
1
1
=
+ kt
[A] [A]0
!
• So, a plot of (1/[A]t) vs t is linear, with a slope of k
• If so, the reaction is 2nd Order in A and zero order in all
! other components
!
Determining Reaction Order
Units of Rate Constants
Multiple reactants, first order in each
A+B→C
d[C]
d[A]
d[B]
="
="
= k[A][B]
dt
dt
dt
• Use stoichiometry to relate all reactant concentrations to
their initial values and to [A]
• If [A]!
0 = [B]0 and the stoichiometry is 1, then the
integrated rate equation is the same as for 2nd order in A
• If [A]0 ≠ [B]0 and Stoichiometry is n (A + nB → C)
Then substitute [B] = [B]0 -2([A]0 -[A]) before integrating
• Remember, stoichiometry doesn’t determine reaction
order, but it may be useful in guessing what the order
might be
Sometimes initial rate and integrated
rate measurements disagree - Why?
The integrated rate equation gives the concentrations of
components whose concentrations change during the reaction
consider: A + B → C
Rate = k2[A][B]
The reaction is second order overall, first order in each reactant
Rate decreases as [A] decreases and as [B] decreases.
What if [B] remains constant during a run?
because it is present in large excess (e.g., solvent),
because its concentration is buffered (e.g., H+, OH-), or
because it is a catalyst, and is regenerated.
Then the reaction is pseudo-zero order in B, and
pesudo-first order overall:
Rate = kapparent[A], where kapparent = k2[B]
!
1
1
"
= kt
[A] [A]0
t
# [A]
[A ]o
If rate =
0
or
d[A]
(concentration)
, it must have units of
dt
(time)
And the products on the right side of the equation must
have the same units, so the units of k must be different
!Zero order reactions
!
Rate (M•min-1) =
k (M•min-1)
First order reactions
Rate (M•min-1)
=
k (min-1) [A](M)
Second order reactions
Rate (M•min-1) =
k (M-1•min-1) [A]2(M2 )
Third order reactions
Rate (M•min-1) =
k (M-2•min-1) [A]3(M3 )
Model an Enzyme Reaction
The enzyme and substrate(s) get together
Transformations occur
The product(s) is (are) released, regenerating enzyme
k1
k2
""
#
""
#
E + S%
" " E • S %" " E + P
k$1
k$2
We have to worry about four concentrations
! [E], [S], [E•S], and [P]
and four reactions, some probably 1st order, some 2nd
k1[E][S], k2[E•S], k-1[E•S] and k-2[E][P]
Too complicated!
2
Simplifying Assumptions
Simplifying Assumptions
steady state
1. Measure initial rates V0 (<10% of reaction)
• then [P] = 0
• so we can neglect the reverse reaction, k-2[E][P]
• and we can ignore product inhibition (later)
• and [S]0 = [S] + [E•S]
2. Use catalytic amounts of enzyme, [E] << [S]
• Then [S]0 = [S] + 0
3. Enzyme exists in only two forms, E and E•S, which sum to ET
4. [E•S] reaches a constant level rapidly and remains constant
during the measurement
• Could happen in two ways
k1
k
E+S
E•S 2 E + P
• k-1 >> k2 , rapid equilibrium
k
-1
• k-1 ≤ k2 , steady state
5. V0 = k2 [E•S]
k1
E+S
E•S
k2
E+P
k-1
d[E •S]
= 0 = k1[E][S] " k -1[E • S] " k 2 [E • S]
dt
or
!
0 = k1[E][S] " [E • S]{k -1 + k 2 }
But we also know
that [S] = [S]0 and
! ET = [E]+[E•S], so that [E] = ET - [E•S], and
V0 = k2 [E•S] (assumes the catalytic step is the RLS)
So, let’s substitute for free enzyme [E] and solve for [E•S]
0 = k1{E T - [E • S]}[S]0 " [E • S]{k -1 + k 2 }
Collect terms in [E•S] (next slide)
!
Steady State
E+S
k1
E•S
k2
Simplifying Assumptions
E+P
rapid equilibrium
k-1
E+S
0 = k1[S]0 E T - [E • S]{k1[S]0 + (k -1 + k 2 )}
solve for [E•S]
!
[E • S] =
KS =
k 2 [S]0 E T
(k + k 2 )
[S]0 + -1
k1
V0 =
Vmax [S]0
K M + [S]0
KM =
[E][S]0
[E • S]
Vmax = k 2E T
!
[S]
E [S]
[E • S]{1 + 0 } = T 0
KS
! KS
!
!
V0 = k 2 [E • S] =
1000
The same equation
Vmax (units of velocity)
k"1
k1
Vo
time"1
= concentration
time -1conc"1
!
units:
k 2E T [S]0 Vmax [S]0
=
KS +[S]!0 KS +[S]0
k "1 + k 2
k1
Rapid Equilibrium KS =
!
{E T - [E •S]}[S]0
KS
E T [S]0
E [S]
KS
[E • S] =
= T 0
[S]
1 + 0 KS +[S]0
KS
[E • S] =
Michaelis-Menten Equation
steady state vs rapid equilibrium
units:
[E][S]0
KS
!
Comparison
KM =
[E • S] =
(k -1 + k 2 )
k1
!
Steady State
E+P
!
!
!
k2
[E•S] is controlled by the equilibrium and is slowly bled
off toward product
k1[S]0 E T
k1[S]0 + (k -1 + k 2 )
multiply by k2
V0 =
E•S
k-1
k1[S]0 E T
V0 = k 2 [E • S] = k 2
!
k1[S]0 + (k -1 + k 2 )
!
k1
so
time"1
= concentration
time -1conc"1
!
So, they really only differ in the
k"1 [E][S]0
=
k1 [E • S]
KM (units of concentration)
!
!relative magnitude of k and k
-1
2
500
Because at equilibrium k1[E][S]0 = k"1[E • S]
0
!
E+S
k1
k-1
E•S
k2
0
E+P
20
40
60
80
100
120
140
160
[S]o
3
What Data are Required?
Order of the Reaction
If [S]0 is too high
poor estimation of KM - looks zero order
If [S]0 is too low
poor estimation of KM -no evidence of saturation
poor estimation of V max - looks first order
Need data with [S]0 in the vicinity of KM,
Need data with [S]0 both above and below KM
1000
[S]0 >> K M
V0 =
!
Vmax [S]0
X
K M + [S]0
Vmax [S]0
= Vmax
[S]0
zero order
Mixed order
Vo
"
!
500
Vo
1000
500
[S]0 << K M
So… how can it be both
first order and zero order?
0
0
20
40
60
80
100
120
140
160
0
!
[S]o
[S]0 >> K M
[S ]t
"
[S]0 " [S] = Vmax t
max
[S ]0
[S ]
d[S] Vmax
=
#
KM
[S ]0 [S]
"
!
and
t
#
ln
dt
0
[S]0 Vmax
=
t
[S] K M
first order
100
120
140
160
!
!
[S]0
+ {[S] " [S]0 }
[S]
[S]0
0.2
=
= 0.25
K M 1" 0.2
V0
[S]0
= 0.2 =
Vmax
K M + [S]0
so, S0 = 0.25KM
Or, if [S]0 = 5KM!, how close is the rate to V max?
In Between -can’t! simplify … invert before integrating:
Vmax t = K M ln
80
Vmax [S]0
V
[S]0
" 0 =
K M + [S]0
Vmax K M + [S]0
So, if V0 is 20% of V max,
!
So, the semilog plot vs t is linear, with slope = V max/KM
!
60
0
First Order Region! [S]0 << K M
d[S] Vmax
=
[S]
dt
KM
!
40
[S]o
V0 =
t
# d[S] = V # dt
So, a plot of [S]0 -[S] or [P] vs t is linear, with slope = Vmax
V0 = "
20
Vo and [S]o Relations
d[P]
d[S]
V0 =
="
= Vmax
dt
dt
!
!
Vmax [S]0
V [S] V
" max 0 = max [S]0
X0
K M + [S]
KM
KM
!
More Order
Zero Order Region
V0 =
0
V0
5K M
=
= 0.83
Vmax K M + 5K M
V0
10K M
=
= 0.91
Vmax K M + 10K M
How about [S]0 = 10KM?
The sum of the limiting cases
!
!
!
Complications -
Lineweaver-Burk
Substrate Inhibition
Invert both sides:
It’s actually hard to get to V max
At [S]0 = 10KM, you observe only 91% of Vmax
The example actually had V max = 1000
And, sometimes substrate binds incorrectly and inhibits
Never reaches Vmax
Vo vs [S]o
Can’t get KM
y - intercept =
1000
!
1
Vmax
1 K M + [S]0 K M 1
1
=
=
+
V0 Vmax [S]0 Vmax [S]0 Vmax
1
V0
slope =
Vo
750
KM
Vmax
500
!
250
x - intercept =
0
0
2
4
6
8
10
12
[S]o
-1
KM
!
1
[S]0
!
4
Eadie-Hofstee Plot
Lineweaver-Burk
alternative linear transform
“Double-reciprocal” plot
1/V0 vs 1/[S]0 has linear region, even with substrate inhibition
V0 =
Vmax [S]0
K M +[S]0
V0K M + V0 [S]0 = Vmax [S]0
V0
V
V
+ 0 = max
[S]0 K M K M
V0
K M + V0 =!Vmax
[S]0
!
-
V0 Vmax
1
=
"
V0!
[S]0 K M K M
!
V0
[S]0
!
!
!
Vmax
KM
-
1
KM
Vmax
!
V0
!
!
Non-Michaelis-Menten Behavior
Induced Fit
cooperativity
Michaelis-Menten
Enzymes and Ideas are Flexible
Conformational Change
Paradigm change
Allostery
Energy Coupling (e.g., F1 F 0)
Negative Cooperativity
Vo
Hexokinase
Bound Glc
Hexokinase
Positive Cooperativity
[So]
How Cooperative is it?
Cooperativity
S
+S
Kd1
S
Kd2
S
Koshland’s Cooperativity factor: RS =
+S
For normal MM kinetics
V
[S]0
= 0.9 =
Vmax
K M + [S]0
S
Higher [S] always shifts the equilibrium to the right
(from cubes to cylinders)
If Kd1 > Kd2, cooperativity is positive.
If Kd1 < Kd2, cooperativity is negative.
!
[S]0!= 9K M
RS!=
[S]0 giving V = 0.9Vmax
[S]0 giving V = 0.1Vmax
V
[S]0
= 0.1 =
Vmax
K M + [S]0
[S]0 = 19 K M
9
= 81
1/9
For positive cooperativity, Rs < 81
For negative !cooperativity, Rs > 81
5
Dependence of Vo on ET
Vmax = k2E T, and V0 ∝ V max, always
At low [S]0, V0 = V max[S]0/KM
At high [S]0 , V0 = Vmax
So, V0 ∝ Vmax, always, always, always
Dependence of Vo on ET
exceptions - irreversible inhibitor
The irreversible inhibitor inactivates the enzyme
stoichiometrically
When you have added more enzyme than there is
inhibitor, the reaction rate is proportional to ET
Except
If an irreversible inhibitor is present
If a second substrate is limiting
If the assay system becomes limiting
If an Enzyme-cofactor or other equilibrium is involved
V0
no
l
ma
vio
ha
be
r
r
ET
Enzyme
is inactivited until irreversible inhibitor is used up
Dependence of Vo on ET
exceptions - second substrate becomes limiting
Example: glucose oxidase
β-D-glucose + O2 + H2O  δ-gluconolactone + H2O2
But O2 is not very soluble in water (2.4 x 10-4M @ 25°C)
So, rate can’t continue to increase with ET forever.
Dependence of Vo on ET
exceptions - coupled enzyme assay becomes limiting
A common way to assay an enzyme is to use a second enzyme that
uses the product of the first enzyme to make something that’s easy
to detect, such as NADH, which absorbs at 340nm
NAD+
A
E1
[O2] starts to become limiting, here
V0
n
b
al
orm
ior
av
eh
NADH + H+
B
C
E2
Coupling enzyme can’t keep up
V0
ior
av
eh
lb
a
rm
no
ET
ET
More than one Substrate
Except for isomerases, most enzymes have more than one substrate
How do we deal with this fact?
A + E  E•A + B… is awkward notation
A better notation is due to W. W. Cleland (Madison,WI):
Rewrite normal MM sequence E + S  E•S  E + P as
S

E
P

E•S  E•P
E
Then, more complex reactions can be written simply
Note also that
V [S]
V
V0 = max
= max
K M + [S] K M + 1
[S]
Models
two substrates on, two products released
E + A + B  P + Q + E
Substrates could add together, one after the other
A
B
P
Q




E
E•A
E•A•B  E•P•Q
EQ
E
This is an ordered bi bi reaction
Kinetic equation can be generated as before
We’ll assume rapid equilibrium
!
6
Random bi bi
Ordered bi bi
either substrate can add first, either product
can leave first
assume rapid equilibrium
KA
E + A  EA
+
B
KB  kcat
EAB  E + P + Q
1.
2.
3.
4.
V0 = kcat[EAB]
A

ET = [E] + [EA] + [EAB]
KA =
[A][E]
[B][EA]
KB =
[EA]
[EAB]
B

P

E•A
E

B
Substitute [E] = ET - [EA] - [EAB] into eq for KA
! that into eq
! for K
Solve for [EA]; substitute
B
Solve for [EAB]
Vmax
Multiply by kcat:
V0 =
K
K K
1+ B + A B
[B] [A][B]
E•B

A
Q

E•Q
E•A•B  E•P•Q
E•P

Q
E

P
Binding of A and B may or may not be independent
!
Random bi bi
KA
E + A  EA
+
+
B
B
KB 
αK A  αK B
Ping-Pong
One product leaves before second substrate adds
V0 = kcat[EAB]
A
P B
Q

 

E E•A  E•P F
F•B  EQ
ET = [E] + [EA] + [EB] + [EAB]
EB + A  EAB  E + P + Q
kcat
Same steps - just more of them
!
solving
substituting
! V0 =
multiply by kcat
KA =
[A][E]
[EA]
"K A =
[A][EB]
[EAB]
KB =
[B][E]
[EB]
!
"K B =
[B][EA]
[EAB]
1+
V
! max
"K A "K B "K AK B
+
+
[A]
[B]
[A][B]
V0 =
1+
E
Vmax
KB KA
+
[B] [A]
!
!
Double reciprocal plots
Vary one substrate at different fixed levels of other substrate
1
K "K % 1
1 " KB %
= A $ B'
+
$1+
'
V0 Vmax # [B] & [A] Vmax # [B] &
!
[B]
Double reciprocal plots
ordered bi bi, continued
1
K "K % 1
1 " KB %
= A $ B'
+
$1+
'
V0 Vmax # [B] & [A] Vmax # [B] &
!Note that plot of same
data vs second substrate
looks different
(rearrange to see why)
[A]
7
Random bi
Double reciprocal plots
ping pong
1 "K A # K B & 1
1 # "K B &
=
+
%1+
(
%1+
(
V0 Vmax $ [B] ' [A] Vmax $
[B] '
α=1
1
K 1
1 " KB %
= A
+
$1+
'
V0 Vmax [A] Vmax # [B] &
α = .5
!
Constant
slope
!
[B]
[B]
[B]
Decreasing
intercept
Reversible Inhibition
Getting KI
mixed type
slope or intercept replots
If slope changes (competitive or noncompetitive)
Slope = Slope[I]= 0 (1+
[I]
)
KI
Substrate & inhibitor binding are neither independent nor
mutually exclusive. Binding of one alters K of the other by
a factor of α.
Slope[I]= 0
Slope = Slope[I]= 0 +
[I]
KI
If 1/V-intercept changes, (noncompetitive or uncompetitive)
[I]
Intercept = Intercept [I]= 0 (1+ )
! KI
!
Intercept = Intercept [I]= 0 +
Intercept [I]= 0
[I]
KI
E + S
For one inhibitor concentration, solve the approprite equation
For more than one [I], plot the parameter (slope or intercept) vs [I]
KI
!
Y = Slope or 1/V-intercept
!
Slope =
To get KI,
divide the intercept of the replot
by the slope of the replot
Y[I]= 0
KI
Intercept = Y[I]= 0
KI =
!
+I
EI + S
KM
ES
+I
k2
[EI] =
E + P
αK I
[ESI] =
ESI X
!
αK M
!
Two inhibitory forms of the enzyme
Two dissociation constants to account for
! them
Replot Intercept
Replot Slope
[E][I]
KI
[ES][I]
KI
E T = [E] + [ES] + [EI] + [ESI]
Vo =
K M (1+
Vmax [S]0
[I]
[I]
) + [S]0 (1+
)
KI
"K I
[I]
!
!
!
Double Reciprocal Plots
Mixed Type
Inhibition
α=2
Vmax = 100
KM = 2
slope =
K M " [I] %
$1+ '
Vmax # K I &
[I]
1/V0
!
1
1
="
[S]0
#K M
$ 1'
1
1 !
=
&1" )
V0 Vmax % # (
intercept =
1 #
[I] &
%1+
(
Vmax $ "K I '
1/Vmax
!
-1/KM
!
1/[S]0
8