Transcript Lecture 20 - University of Windsor

• Midterm Exam II:
March 11: between 1 PM and 2:10PM.
Students whose family name starts with
letter A, B, C, or D will write the exam in
DH354.
23.5 Features of homogeneous catalysis
• A Catalyst is a substance that
accelerates a reaction but
undergoes no net chemical
change.
• Enzymes are biological
catalysts and are very specific.
• Homogeneous catalyst: a
catalyst in the same phase as
the reaction mixture.
• heterogeneous catalysts: a
catalyst exists in a different
phase from the reaction mixture.
Example: Bromide-catalyzed decomposition of hydrogen peroxide:
2H2O2(aq) →
2H2O(l) + O2(g)
is believed to proceed through the following pre-equilibrium:
H3O+
+ H2O2 ↔
H3O2+ + H2O
[ H 3O2 ]
K
[ H 2O2 ][ H 3O  ]
H3O2+ + Br- → HOBr + H2O
v = k[H3O2+][Br-]
HOBr + H2O2 → H3O+ + O2 + Br-
(fast)
The second step is the rate-determining step. Thus the production rate
of O2 can be expressed by the rate of the second step.
d [O2 ]
 k[ H 3O2 ][ Br  ]
dt
The concentration of [H3O2+] can be solved
[H3O2+] = K[H2O2][H3O+]
Thus
d[O2 ]
 Kk[ H 2O2 ][ H 3O  ][ Br  ]
dt
The rate depends on the concentration of Br- and on the pH of the solution
(i.e. [H3O+]).
• Exercise 23.4b: Consider the acid-catalysed reaction
(1) HA + H+ ↔ HAH+
k1, k1’ , both fast
(2) HAH+ + B → BH+ + AH
k2, slow
Deduce the rate law and show that it can be made
independent of the specific term [H+]
Solution:
23.6 Enzymes
Three principal features of enzyme-catalyzed reactions:
1.
For a given initial concentration of substrate, [S]0, the initial rate
of product formation is proportional to the total concentration of
enzyme, [E]0.
2.
For a given [E]0 and low values of [S]0, the rate of product
formation is proportional to [S]0.
3.
For a given [E]0 and high values of [S]0, the rate of product
formation becomes independent of [S]0, reaching a maximum
value known as the maximum velocity, vmax.
• Michaelis-Menten mechanism
E + S → ES
ES
→E + S
ES
→P + E
The rate of product formation:
k1
k2
k3
d[ P ]
 k 3 [ ES ]
dt
To get a solution for the above equation, one needs to know the
value of [ES]
d [ ES ]
 k1[ E ][ S ]  k 2 [ ES ]  k 3 [ ES ]
dt
Applying steady-state approximation
k1[ E ][S ]  k2[ ES ]  k3[ ES ]  0
[ ES ] 
k1
[ E ][S ]
k2  k3
Because [E]0 = [E] + [ES], and [S] ≈ [S]0
[ ES ] 
[ E ]0
 k  k3  1

1   2
 k1  [ S ]0
• Michaelis-Menten equation can be obtained by plug the value of
[ES] into the rate law of P:
dP

dt
k3 [ E ]0
 k  k3  1

1   2
k
1

 [ S ]0
• Michaelis-Menten constant:
KM 
k2  k3
k1
KM can also be expressed as [E][S]/[ES].
•
Analysis:
1. When [S]0 << KM, the rate of product formation is proportional to [S]0:
v
k1k3
[ S ]0 [ E ]0
k2  k3
2. When [S]0 >> KM, the rate of product formation reaches its maximum
value, which is independent of [S]0:
v = vmax = k3[E]0
With the definition of KM and vmax, we get
v
The above Equation can be rearranged into:
v m ax
K
1 M
[ S ]0
K  1
1
1

  M 
v v max  v max  [ S ]0
Therefore, a straight line is expected with the slope of KM/vmax, and a yintercept at 1/vmax when plotting 1/v versus 1/[S]0. Such a plot is
called Lineweaver-Burk plot,
• The catalytic efficiency of enzymes
Catalytic constant (or, turnover number) of an enzyme, kcat, is the
number of catalytic cycles (turnovers) performed by the active site in
a given interval divided by the duration of the interval.
vmax
kcat  k3 
• Catalytic efficiency, ε, of an enzyme is the ratio kcat/KM,
k
kk
  cat  1 3
k M k2  k3
[ E ]0
Example: The enzyme carbonic anhydrase catalyses the hydration of CO2 in red blood
cells to give bicarbonate ion:
CO2 + H2O →
HCO3- + H+
The following data were obtained for the reaction at pH = 7.1, 273.5K, and an enzyme
concentration of 2.3 nmol L-1.
[CO2]/(mmol L-1)
1.25
2.5
5.0
20.0
rate/(mol L-1 s-1)
2.78x10-5
5.00x10-5
8.33x10-5 1.67x10-4
Determine the catalytic efficiency of carbonic anhydrase at 273.5K
Answer:
Make a Lineweaver-Burk plot and determine
the values of KM and vmax from the graph.
The slope is 40s and y-intercept is 4.0x103 L
mol-1s
1
vmax = 4.0 10 3 Lmol 1 s
= 2.5 x10-4 mol L-1s-1
KM = (2.5 x10-4 mol L-1s-1)(40s) = 1.0 x 10-2
mol L-1
4
1 1
s
kcat = 2.5  10 molL
= 1.1 x 105s-1
9
1
2.3  10 molL
ε=
k cat
KM
= 1.1 x 107 L mol-1 s-1
Mechanisms of enzyme inhibition
• Competitive inhibition: the inhibitor (I) binds only to the
active site.
EI ↔ E + I
• Non-competitive inhibition: binds to a site away from the
active site. It can take place on E and ES
EI ↔ E + I
ESI ↔ ES + I
• Uncompetitive inhibition: binds to a site of the enzyme
that is removed from the active site, but only if the
substrate us already present.
ESI ↔ ES + I
• The efficiency of the inhibitor (as well as the type of
inhibition) can be determined with controlled experiments
Autocatalysis
• Autocatalysis: the catalysis of a reaction by its products
A + P →
2P
d[ P ]
The rate law is
= k[A][P]
dt
To find the integrated solution for the above differential equation, it is
convenient to use the following notations
[A] = [A]0 - x; [P] = [P]0 + x
One gets
d[ P ]
dt
= k([A]0 - x)( [P]0 + x)
integrating the above ODE by using the following relation
1
1
1
1

(

)
([ A]0  x )([ P ]0  x ) [ A]0  [ P ]0 [ A]0  x [ P ]0  x
gives
or rearrange into
 ([P ]0  x )[ A]0 
1
  kt
ln
[ A]0  [ P ]0  [ P ]0 ([ A]0  x ) 
x
e at  1

[ P ]0 1  be at
with a=([A]0 + [P]0)k and b = [P]0/[A]0