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Polymerization kinetics. Stepwise polymerization: any two monomers present in the reaction mixture can link together at any time. The growth of the polymer is not confined to chains that are already formed. - PowerPoint PPT Presentation
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Polymerization kinetics
• Stepwise polymerization: any two monomers present in the reaction mixture can link together at any time. The growth of the polymer is not confined to chains that are already formed.
• Chain polymerization: an activated monomer attacks another monomer, links to it, then that unit attacks another monomer, and so on.
23.3 Stepwise polymerization
• Commonly proceeds through a condensation reaction, in which a small molecule is eliminated in each step.
• The formation of nylon-66 H2N(CH2)6NH2 + HOOC(CH2)4COOH →
H2N(CH2)6NHOC(CH2)4COOH
• HO-M-COOH + HO-M-COOH → HO-M-COO-M-COOH
• Because the condensation reaction can occur between molecules containing any number of monomer units, chains of many different lengths can grow in the reaction mixture.
Stepwise polymerization
• The rate law can be expressed as
• Assuming that the rate constant k is independent
of the chain length, then k remains constant throughout the reaction.
• The degree of polymerization: The average number of monomers per polymer molecule, <n>
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23.4 Chain polymerization
• Occurs by addition of monomers to a growing polymer, often by a radical chain process.
• Rapid growth of an individual polymer chain for each activated monomer.
• The addition polymerizations of ethene, methyl methacrylate, and styrene.
• The rate of polymerization is proportional to the square root of the initiator concentration.
The three basic types of reaction step in a chain polymerization
(a) Initiation: I → R. + R. vi = ki[I] M + R. → .M1 (fast)
(b) Propagation: M + .M1→ .M2 M + .M2→ .M3
░ vp = kp[M][.M]
M + .Mn-1→ .Mn (c) Termination:
Mutual termination: .Mn + .Mm→ Mn+m
Disproportionation: .Mn + .Mm→ Mn + Mm
Chain transfer: M + .Mn→ Mn + .M
Influences of termination step on the polymerization
• Mutual termination: two growing radical chains combine. vt = kt ([.M])2
• Disproportionation: Such as the transfer of a hydrogen atom from one chain to another, which corresponds to the oxidation of the donor and the reduction of acceptor.
vt = kt ([.M])2
• Chain transfer: vt = ?
• the net rate of change of radical concentration is calculated as
• Using steady-state approximation (the rate of production of radicals equals the termination rate)
• The rate of polymerization
vp = kp[.M][M] = kp[M]
• The above equation states that the rate of polymerization is proportional to the square root of the concentration of the initiator.
• Kinetic chain length, v,
• <n> = 2v (for mutual termination)
222 ][][][ .
.
MkIfkdt
Mdti
production
2121
//
. ][][ Ik
fkM
t
i
21
21
/
/
][Ik
fk
t
i
2/1
2/1
)(2
1
]][[
tip kfkkkwhere
IMkproducedcentresactivatedofnumber
consumedunitsmonomerofnumberv
• Example: For a free radical addition polymerization with ki = 5.0x10-5
s-1 , f = 0.5, kt = 2.0 x107 dm3 mol-1 s-1, and kp = 2640 dm3 mol-1 s-1 , and with initial concentrations of [M] = 2.0 M and [I] = 8x10-3 M. Assume the termination is by combination.
(a) The steady-state concentration of free radicals.
(b) The average kinetic chain length.
(c) The production rate of polymer.
Solution: (a)
(b)
(c) The production rate of polymer corresponds to the rate of polymerization is vp:
2121
//
. ][][ Ik
fkM
t
i
2/12/1 )(2
1]][[ tip kfkkkwhereIMkv
vp = kp[.M][M]
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
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.
The concentration of [H3O2+] can be solved
[H3O2+] = K[H2O2][H3O+]
Thus
The rate depends on the concentration of Br- and on the pH of the solution (i.e. [H3O+]).
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322
23
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Od23
2
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• 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 k1
ES → E + S k2
ES → P + E k3
The rate of product formation:
To get a solution for the above equation, one needs to know the value of [ES]
Applying steady-state approximation
Because [E]0 = [E] + [ES], and [S] ≈ [S]0
][][
ESkdt
Pd3
][][]][[][
ESkESkSEkdt
ESd321
0321 ][][]][[ ESkESkSEk
]][[][ SEkk
kES
32
1
01
32
0
11
][
][][
Sk
kk
EES
• Michaelis-Menten equation can be obtained by plug the value of [ES] into the rate law of P:
• Michaelis-Menten constant:
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:
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
01
32
03
11
][
][
Skkk
Ek
dt
dP
1
32
k
kkKM
0032
31 ][][ ESkk
kkv
With the definition of KM and vmax, we get
The above Equation can be rearranged into:
Therefore, a straight line is expected with the slope of KM/vmax, and a y-intercept 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.
• Catalytic efficiency, ε, of an enzyme is the ratio kcat/KM,
0
1][
max
S
Kv
vM
0
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][maxmax Sv
K
vvM
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max
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vkkcat
32
31
kk
kk
k
k
M
cat
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 20.0rate/(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
vmax = = 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
kcat = = 1.1 x 105s-1
ε = = 1.1 x 107 L mol-1 s-1
sLmol 131004
1.
19
114
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1052
molL
smolL
.
.
M
cat
K
k
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 ESEI ↔ E + IESI ↔ 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