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>

2][][

Akdt

Ad

0

0

0

0

1 ][

][

][

][][

Akt

Akt

A

AAp

0

0

1 ][

][][

Akt

AA

pA

An

1

10

][

][

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+]).

]][[

][

OHOH

OHK

322

23

]][[][ BrOHk

dt

Od23

2

]][][[][ BrOHOHKk

dt

Od322

2

• 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

111

][maxmax Sv

K

vvM

03 ][

max

E

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

1032

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