This document provides an outline and overview of step-growth and chain-growth polymerization mechanisms and kinetics. It discusses the step-growth mechanism, kinetics of step-growth polymerization using Carother's equation, and controlling molecular weight. It then covers the chain-growth mechanism, kinetics of chain-growth polymerization using steady-state kinetics, and examples of free radical polymerization initiation, propagation and termination reactions. Major classes of natural and commercial polymers are also briefly mentioned.

1. CHE 305
Introduction to Polymer Chemistry
Chapter 9: Kinetics of chain and step
growth polymerization
• Name: Winnie Hung
• E-mail: r61042i@gmail.com
1

2. Outline
1. Step-growth mechanism
2. Kinetics of step-growth polymerization
3. Chain-growth mechanism
4. Kinetics of chain-growth polymerization
2

3. Polymer synthesis
3
• Polymers may be formed by two major kinetic
schemes.
1.Step polymerization
2.Chain polymerization (Faster)
• Free radical polymerization
Initiation
Propagation
Termination
• Anionic polymerization
• Cationic polymerization

4. Step-growth mechanism
4

5. Step-growth polymerization
5
+𝐻𝐶𝑙

6. 6
Chain growth
1. Uncontrolled (Popcorn)
2. Controlled/ Living (Grass)
Controlled
chain-growth

7. Polyamide-I
7
A B A BA BA B

8. Polyamide-II
8
Diacid Diamina Polyamide
A BA B A BA BA A

9. Assume all functional
groups equally reactive
9
• Xn: Number average chain length
• Imagine 16 monomers (or 32 functional groups)
Xn=
8×1+4×2
12
= 1.3
25% reaction
Xn=
4×1+1×2+2×3+1×4
8
= 2
50% reaction
Xn=?
75% reaction
4
Still very low

10. Step-growth
10
For high molecular weight
• the reaction must go toward completion!
• ( >99% reaction rate)

11. Carother's Equation
11
• N0 ≡ # of monomers originally present in system
• N ≡ # of molecules present in system at any time t
• (N0-N) ≡ Total # of functional groups of either A or B that
have reacted at t
• 𝑝 ≡ 𝐸𝑥𝑡𝑒𝑛𝑡 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
=
𝑁0 − 𝑁
𝑁0
→ 𝑁 = 𝑁0 1 − 𝑝
• Since 𝑋 𝑛 =
𝑁0
𝑁
Xn =
1
1 − 𝑝
Carother's Equation

12. Carother's Equation
12
p Xn
0.95 20
0.99 100
0.999 1000
Example:
Good fibers of nylon 6-6 (fishing line),
Mn=12,000 g/mol, Xn~ 110, so p should be more
than?
Ans: >0.99

13. How to control the MW
13
If the Mw
• Too low→ Poor properties
• Too high→ Difficult to process (Melt/ solubilize)
Nylon rope trick
• A.Q. phase (Hexine diamine)+Organic phase (Adipoyl chloride)

14. How to control the MW
14
Control
• Heat+Pressure

15. How to control the MW
15
Control
• Stoichiometric imbalance
• Excess of one reactant in A-A/B-B system to limit MW
𝑋 𝑛 =
1 + 𝑟
1 + 𝑟 − 2𝑟𝑝
, r =
𝑁0𝐴－𝐴
𝑁0𝐵－B
=
𝑁0𝐴
𝑁0𝐵
# of
unreacted
functional
groups
Excess goes to denominator, r<1

16. How to control the MW
16
For “quantitative” reaction, P=0.999
N0AA N0BB r Xn
1 1 1 1000
1 1.05 0.952 39
𝑋 𝑛 =
1 + 𝑟
1 + 𝑟 − 2𝑟𝑝
, r =
𝑁0𝐴－𝐴
𝑁0𝐵－B
=
𝑁0𝐴
𝑁0𝐵

17. Kinetics of step-growth
17

18. Polyesterification
18
A-B type Polyester
RCOOH + OH → COOR + H2O
H+ (cat)
Self-catalyzed
−𝑑[𝐶𝑂𝑂𝐻]
𝑑𝑡
= 𝑘 𝐶𝑂𝑂𝐻 2[𝑂𝐻]

19. Polyesterification
19
Self-catalyzed
−𝑑[𝐶𝑂𝑂𝐻]
𝑑𝑡
= 𝑘 𝐶𝑂𝑂𝐻 2[𝑂𝐻]
If [COOH]=[OH] → C
Then
−𝑑𝐶
𝑑𝑡
= 𝑘𝐶3
→− 𝐶0
𝐶 𝑑𝐶
𝐶3 = 𝑘 0
𝑡
𝑑𝑡
→2kt=
1
𝐶2 −
1
𝐶0
2
𝑋 𝑛 =
𝐶0
𝐶
=
1
1 − 𝑝
→C=C0(1-P)
Then,
2C0
2kt=[
1
1−𝑃 2] − 1
Xn
2

20. Polyesterification
20
Acid-catalyzed
−𝑑[𝐶𝑂𝑂𝐻]
𝑑𝑡
= 𝑘′ 𝐶𝑂𝑂𝐻 [𝑂𝐻]
If [COOH]=[OH] → C
Then
−𝑑𝐶
𝑑𝑡
= 𝑘′𝐶2
→− 𝐶0
𝐶 𝑑𝐶
𝐶2 = 𝑘′ 0
𝑡
𝑑𝑡
→𝐶0 𝑘′ 𝑡 =
1
1−𝑝
− 1

21. Polyesterification
21
Self-catalyzed Acid-catalyzed
1
1 − 𝑝 2
= Xn2
t
1
1 − 𝑝
= Xn
t
𝐶0 𝑘′ 𝑡 =
1
1 − 𝑝
− 12C0
2kt=[
1
1−𝑃 2] − 1

22. Practice
22
Q:
80 moles of monomers react to prepare Nylon 12.
After completion of 8 h, 4 moles of monomers are
still left. What is the number average molecular
weight of polymer system?
(the molecular weights of the repeating units of
polymer Nylon 12 is 197)
The extent of reaction is p= (80-4)/80 = 0.95
Xn=1/(1-p) = 1/(1-0.95)= 20
Mn=20*197=3940

23. Practice
23
Q:
If the value of C0 and k are 10 mol L-1 and 10-3 L mol s-1,
respectively, how long would it take to obtain a Xn of 37?
Assume it’s self-catalyzed
2C0
2kt=[
1
1−𝑃 2] − 1
→2C0
2kt=Xn
2−1
→t=68400 s

24. Flory distribution
24
Molar mass/ Degree of polymerization distribution
→ Calculate the probability P(x) of finding a chain
comprising x units also know as the mole fraction P(x)
P(x)=
𝑁 𝑥
𝑁
=
# 𝑋−𝑚𝑒𝑟𝑠
𝑇𝑜𝑡𝑎𝑙 # 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
(X=1 for monomer, 2 for dimer, 3 for trimer)

25. Flory distribution
25
 Step 1: Probability of random molecule being a monomer
P(x=1)=(1-p)
 Step 2: Probability of random molecule being a dimer
P(x=2)=p(1-p)
The first molecule reacted The adjacent is unreacted
At extent of reaction p for either
1. x A-A + x B-B → A-A[B-BA-A]x-1B-B
2. x A-B → A[BA]x-1B
Note:
• A, B are functional groups
• “Molecules” are either monomers or polymer chains

26. Flory distribution
26
P(x=2) =p(1-p), P(x=3) =p2(1-p)
→P(x)=px-1(1-p)
Since probability of x=P(x)=Mol. Fraction=
𝑁 𝑥
𝑁
𝑃 𝑥 =
𝑁 𝑥
𝑁
= 𝑝 𝑥−1 1 − 𝑝
→𝑁 𝑥 = 𝑁𝑝 𝑥−1
1 − 𝑝

27. Flory distribution
27
Since Xn=
𝑁0
𝑁
→ N =
𝑁0
𝑋 𝑛
= 𝑁0(1 − 𝑝)
→ 𝑁 𝑥 = 𝑁0 𝑝 𝑥−1
1 − 𝑝 2
In terms of mass fraction of x-mers, Wx
→ 𝑊 𝑥 = 𝑥𝑝 𝑥−1 1 − 𝑝 2
Where Wx=
𝑥𝑁 𝑥
𝑁0
=
𝑇𝑜𝑡𝑎𝑙 # 𝑜𝑓 𝑚𝑜𝑛𝑜𝑚𝑒𝑟𝑠 𝑖𝑛𝑐𝑜𝑟𝑝𝑎𝑟𝑎𝑡𝑒𝑑 𝑖𝑛𝑡𝑜 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑤𝑖𝑡ℎ 𝑙𝑒𝑛𝑔𝑡ℎ 𝑥
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑛𝑜𝑚𝑒𝑟𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
𝑁 𝑥 = 𝑁𝑝 𝑥−1 1 − 𝑝

28. Flory distribution here
28
Number-fraction distribution Weight-fraction distribution
𝑁 𝑥
𝑁
X
50 100 150 200
Wx=
𝑥𝑁 𝑥
𝑁0
X
50 100 150 200
p=0.96
p=0.9875
p=0.9950
p=0.96
p=0.9875
p=0.9950

29. Note
29
• Average molecular weight Mn and Mw as function of p for
step-growth
𝑀0 ≡ 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑟𝑒𝑝𝑒𝑎𝑡 𝑢𝑛𝑖𝑡
𝑋 𝑛 =
𝑀 𝑛
𝑀0
=
1
1 − 𝑝
→ 𝑀 𝑛 =
𝑀0
1 − 𝑝
→Mw=
𝑀0
(1+𝑝)
1−𝑝
• Dispersity (PDI)
PDI=Đ=
𝑀 𝑊
𝑀 𝑛
= 1 + 𝑝
As p→1 (reaction ↑)
Đ→2

30. Major stepwise polymer classes
30

31. Nature product polymers
31
• Living organisms make many polymers enzymatically.
• Most such natural polymers strongly resemble step-
polymerized materials.
• The structure ultimately being controlled by DNA.

32. Commercialization dates
32

33. Chain-growth
By radical polymerization by unpaired electrons traveling
down the chain and add new monomers as they go! Or made
by ionic polymerization (anionic or cationic)
33

34. Free radical polymerization
34
1. Initiation
2. Propagation

35. Free Radical Polymerization
Ｉ + C=C
Ｉ + C － C
Ｉ－C－C
35

36. Free radical polymerization
36
3. Termination

37. Radical initiators
37
benzoyl peroxide

38. Radical intiators
38
𝐻2 𝑂2 + 𝐹𝑒2 +
→ 𝐹𝑒3 +
+𝑂𝐻
−
+𝑂𝐻●
Redox

39. Initiation + Propagation
39

40. Termination I
40
(a) Combination (Two chain ends couple)
 Double molecular weight
 2 initiator fragments

41. Termination II
41
(b) Disproportionation
 One initiator fragment per chain
 Predicted molecular weight

42. Kinetic of chain-growth
42
• Steady-state kinetics
• Assume rate of radical formation = rate of radical destruction
Initiation, 2-step
1. Decombination of initiator
𝐼 →
𝑘 𝑑
2𝑅●
2. Attack on the monomer
𝑅● + M →
𝑘𝑖
𝑅𝑀●
→Rate of thermal initiation
𝑣𝑖 =
𝑑[𝑅𝑀●]
𝑑𝑡
= 2𝑘𝑑𝑓[𝐼]
Slow-rate determining step
2 initiators per I
Initiators
efficiency, the
ability of 𝑅● to
propagate chains
0.3-0.8

43. Kinetic of chain-growth
43
Propagation
𝑅𝑀 𝑛● + 𝑀 →
𝑘 𝑝
𝑅𝑀 𝑛 + 1●
𝑣 𝑝 = 𝑘𝑝 𝑀 [𝑀●]
• [M] : concentration of monomers
• [𝑀●]: concentration of the growing chains (active centers)

44. Kinetic of chain-growth
44
Termiation
𝑣𝑡 = 2𝑘𝑡 𝑀● 𝑀●
2 radicals cost for each termination step
kt=ktc+ktd
• ktc: combination
• ktd: disproportionation

45. Kinetic of chain-growth
45
Steady state≡ 𝑣𝑖 = 𝑣𝑡
2𝑘𝑑𝑓 𝐼 = 2kt M● 2
M● =
𝑘 𝑑 𝑓[𝐼]
𝑘𝑡
𝑣 𝑝 = 𝑘𝑝[𝑀]
𝑘 𝑑 𝑓[𝐼]
𝑘𝑡
Too low to measure
experimentally

46. Kinetic of chain-growth
46
• 𝐷𝑒𝑓𝑖𝑛𝑒 𝑣 ≡ 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑐ℎ𝑎𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ
• Average # of monomers reacting with active
center during its lifetime
→ 𝑣＝
𝑚𝑜𝑛𝑜𝑚𝑒𝑟𝑠 𝑎𝑑𝑑𝑒𝑑/𝑠𝑒𝑐
# 𝑜𝑓 𝑐ℎ𝑎𝑖𝑛𝑠 𝑓𝑜𝑟𝑚𝑒𝑑/𝑠𝑒𝑐
=
𝑣 𝑝
𝑣𝑡
1. For combination: Xn=2 𝑣
2. For disproportionation: Xn= 𝑣

47. Kinetic of chain-growth
47
• Under steady state
𝑣＝
𝑣 𝑝
𝑣𝑡
=
𝑣 𝑝
𝑣𝑖
=
𝑘 𝑝 𝑀 [𝑀●]
2𝑘𝑡 𝑀● 2
=
𝑘 𝑝 𝑀
2𝑘𝑡 𝑀●
=
𝑘 𝑝
2 𝑀 2
2𝑘𝑡𝑣𝑝
★ 𝑀● =
𝑣 𝑝
𝑘𝑝[𝑀]

48. Practice
48
The following are data for the polymerization of styrene in
benzene at 60°C with benzoyl peroxide as the initiator.
[M] = 3.34 × 103 mol/m3
[I] = 4.0 mol/m3
kp
2/kt = 0.95 × 10–6 m3/mol-s
If the spontaneous decomposition rate of benzoyl peroxide is
3.2 × 10–6 m3/mol-s, calculate the initial rate of polymerization.
Assume the initiator efficiency f=1

49. Answer
49
𝑣 𝑝 = 𝑘𝑝 𝑀
𝑘 𝑑 𝑓 𝐼
𝑘𝑡
→ 𝑣 𝑝
2 = 𝑘𝑑
𝑘 𝑝
2
𝑘𝑡
𝐼 𝑀 2
→ 𝑣 𝑝
2 = (3.2 × 10–6)×( 0.95 × 10–6) × 4 × (3.34 × 103)2
→𝑣 𝑝
2 = 1.357 × 10–4
→𝑣 𝑝=0.0116 mol/m3-s

50. Practice
50
For vinyl acetate polymerized at 50 ᵒC, the value of the
ratio kp
2/kt is 0.0138 l/mol-s. What is the value of chain
length, when monomer concentration is 6.53 mol/l and
rate of polymerization is 2.0×10-4 mol/l-s?
The expression for kinetic chain length is
𝑣 =
𝑘 𝑝
2 𝑀 2
2𝑘𝑡𝑣𝑝
=
0.0138 × 6.532
2 × (2.0 × 10−4)
= 1471

51. Thermodynamics
51
∆𝐺 𝑝 = ∆H 𝑝 − T∆𝑆 𝑝
• ∆H 𝑝: negative, because from 𝜋 𝑏𝑜𝑛𝑑 𝑡𝑜 𝜎 𝑏𝑜𝑛𝑑 𝑖𝑠 𝑒𝑥𝑡ℎ𝑜𝑡ℎ𝑒𝑟𝑚𝑖𝑐
• −∆𝐻 𝑝 ≈ 30 − 150𝐾𝐽/𝑚𝑜𝑙
• ∆𝑆 𝑝: negative, because we are confining monomers to the chain
• −∆𝑆 𝑝 ≈ 100 −
130𝐽
𝑚𝑜𝑙𝐾
∆𝐺 𝑝 < 0 at normal temp.

52. Polymerization processes
52
1. Block (monomers only, no solvent)
-Efficient, eco-friendly, optically transparent
-Susceptible to auto acceleration, explosion
2. Solution (In a solvent)
-Heat dissipation, but susceptible to reaction with solvent
-Ungreen
3. Suspension (of monomers in AQ phase)
-Effectively bulk, Droplets 0.1-5mm
-Reaction must be stable to water
4. Emulsion (Much smaller particles 50nm-5𝜇m)
-Use Micelles with a surfactant to control Mw

53. Features of free radical polymerization
53
1. High Mw formed immediately
The average of molecular weight in the beginning is low
2. Steady decrease in [M] thru out the reaction
3. Only the active center (the growing chain ) is reactive toward
other monomers
4. Long reaction time increase the yield of polymer produce but
not Mw
5. Increasing temp. increases the rate but decreases Mw

54. Chain-growth
54
1. The rate of propagation is proportional to the concentration of the monomer and the
square root of the concentration of the initiator.
2. The rate of termination is proportional to the concentration of the initiator.
3. The average molecular weight is proportional to the concentration of the monomer
and inversely proportional to the square root of the concentration of initiator.
4. The first chain that is initiated rapidly produces a high molecular weight polymer.
5. The monomer concentration decreases steadily throughout the reaction and
approaches zero at the end.
6. The increases in rates of initiation, propagation, and termination with increases in
temperature are in accord with the Arrhenius equation. The energies of activation of
initiation, propagation, and termination are approximately 35, 5, and 3kcal/mol,
respectively. Data for typical energies of activation are given in Tables8.3 and 8.4.
7. Increasing the temperature increases the concentration of free radicals and thus the
rate of reactions, but it decreases the average molecular weight.
8. If the temperature exceeds the ceiling temperature (Tc), the polymer will decompose
and no propagation will take place at temperatures above the ceiling temperature.

55. Chain growth V.S. Step growth
55
Step growth Chain growth
Reactive
sites
All molecules (monomer,
oligomer, polymer)
Only monomer react to the
“active site”
Reaction
process
No termination 3 steps
1. Initiation
2. Propagation
3. Termination
Reaction
speed
Slower Faster (Initiator↑)
Final
product
Oligomers of many sizes Polymer+monomer+very
few growing chains

56. 56

57. Chain polymer strucutre
59

58. Chain polymer structure-ABS
60
3D printing, Tg? m.p.? Acrylonitrile Butadiene Styrene

59. Thanks for your attention
62