Lesson 5: Free Radical Polymerization (Part II)

Lesson 5: Free Radical Polymerization (Part II) sxr133
The links below provide an outline of the material for this lesson. Be sure to carefully read through the entire lesson before submitting your assignments.

Overview/Checklist

Overview/Checklist mjg8

Overview

In this Lesson, we continue learning about free radical polymerization, with a focus on the kinetics of the polymerization. We describe the rate of reaction for initiation, propagation, and termination, and how those reactions play into the overall rate of polymerization. Introduction of inhibitors or retarders can be used to modulate the rate or onset of polymerization. We also learn about how to calculate the number average degree of polymerization for free radical polymerization and how it relates to factors such as concentration of monomer, initiator, and chain transfer.

Learning Outcomes

By the end of this lesson, you should be able to:

  • Calculate rates of initiation, propagation, and termination, and how these influence rate of polymerization
  • Describe the effect of monomer concentration, initiator concentration, radical concentration, and rate of chain transfer on rate of polymerization and degree of polymerization
  • Determine the polymer skeletal structure from monomers that can be polymerized by free radical pathways
  • Describe steady state conditions
  • Describe the effects of inhibitors and retarders on polymerization.

Lesson Checklist

Lesson 5 Checklist
ActivityContentAccess / Directions
To ReadRead all of the online material for Lesson 5.Continue navigating the online material.
To Read

Chapter 4 - Radical Polymerization

  • § 4.3 - 4.3.7
  • § 4.6.1
The chapter readings come from the textbook, Introduction to Polymers.
To DoHomework Assignment 5 (Practice)Registered students can access the homework assignment in the Lesson 5 module.

Please refer to the Canvas Calendar for specific timeframes.

Questions?

If you have questions, please feel free to post them to the General Questions and Discussion forum. While you are there, feel free to post your own responses if you, too, are able to help a classmate.

Polymerization Kinetics

Polymerization Kinetics jls164

Now that we have a working understanding of the reactions and processes that can happen during free radical polymerization, we can begin to discuss the polymerization kinetics for free radical polymerization. We start from the very beginning and define the rate of polymerization:

R p = d[ M ] dt
Eq 5.1

As shown in Eq. 5.1, we are defining the rate of polymerization as the rate at which monomer M is consumed. Note the negative sign, which is there because we are defining the rate as the disappearance or consumption of monomer.

You may want to refresh your memory of rate equations before going much further!

Rate of Initiation:

We should be able to write a rate equation for each "step" in free radical polymerization: initiation, propagation, and termination. Let's start with initiation; there are actually two parts to initiation: 1) the formation of the radical from decomposition of the initiator and 2) transfer of the radical to a monomer.

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Figure 5.1:
Source: Lauren Zarzar

The first step of initiation, generation of the radicals, is generally much slower than transfer of the active center to the monomer. Thus, the first step of initiation is the rate determining step, meaning that the rate of initiation is really only a function of the slow step, formation of radicals. Thus, we can represent the rate of initiation as the rate of formation of radicals:

R i = d[ R· ] dt
Eq 5.2

Notice that we don't have anything about monomer concentration in rate of initiation. Because we consume so little monomer during initiation, we neglect any changes in monomer concentration here. But we can make this more specific. Let's say we generate two radicals per initiator molecule, which is common, and that the rate of dissociation of the initiator has a rate constant of kd.

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Figure 5.2:
Source: Lauren Zarzar

We can thus re-write the rate for first step of initiation as:

R i = d[ R· ] dt =2 k d [ I ]
Eq 5.3

However, we also know that not all the radicals that are produced actually get transferred to monomer; only some fraction of them make it through step 2 of initiation. Radicals may combine, react with solvent, etc. To account for the fact that not all radicals produced in step 1 are actually used, we introduce the concept of initiator efficiency, f. For example, If half of our radicals produced actually get used and transferred to monomer, then f =0.5. We can update our description of the rate of initiation:

R i =2f k d [ I ]
Eq 5.4

Rate of Propagation:

Propagation involves the transfer of the radical from one monomer to another monomer, or to polymer. We are using up lots of monomer in this step.

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Figure 5.3:
Source: Lauren Zarzar

The rate of disappearance of monomer (note the negative sign in the rate!) is a function of the rate constant, the monomer concentration, and concentration of active centers (M·)

R prop = d[ M ] dt  = k p [ M ][ M· ]
Eq 5.5

Note that this is the only place where we directly account for the rate as consumption of monomer (but NOT the radical species).

Rate of Termination:

Recall that we have two different pathways for termination: combination and disproportionation.

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Figure 5.4:
Source: Lauren Zarzar

We write the rate equations for these two termination pathways:

R tc = d[ M· ] dt  =2 k tc [ M· ][ M· ]
Eq 5.6
R td = d[ M· ] dt =2 k td [ M· ][ M· ]
Eq 5.7

The coefficient of two results from the fact that two growing polymers are consumed by either pathway.

Add up these two rates (ktc +ktc = kt ) to get an overall rate equation:

R t = d[ M· ] dt =2 k td [ M· ] 2 +2 k tc [ M· ] 2 =2 k t [ M· ] 2
Eq 5.8

Steady State

Steady State jls164

We have now written rate equations for initiation (generation of radicals), propagation (growth of polymer), and termination (consumption of radicals). At the very beginning of the reaction, the rate of formation of radicals is faster than consumption of monomer. But the number of radicals cannot keep growing forever! Eventually, the rate of formation of radicals and consumption of radical monomer/active centers reaches a steady state — radicals are formed as quickly as they are consumed/propagated in the reaction. Radicals are consumed just as fast as they are created. This is called steady state. We work under this assumption for the rest of our derivation of steady state polymerization rates.

The rate of production of radicals is the same as the consumption of radicals:

R i = R t
Eq 5.9

Substitution of our previous expressions:

d[ R· ] dt =2 k t [ M· ] 2
Eq 5.10

Rearrange to solve for the steady state concentration of radicals:

[ M· ]= ( R i 2 k t ) 1/2
Eq 5.11

OK great, why do we care? Well, we ultimately want to figure out the rate of polymerization, which recall that we defined at the very beginning of the section. To refresh your memory, we defined rate of polymerization as rate of consumption of monomer. Where else have we seen rate of consumption of monomer? We saw it in the expression for the rate of propagation!

R p = R prop = d[ M ] dt = k p [ M ][ M· ]
Eq 5.12

While we can measure the concentration of monomer, it's not easy to measure the concentration of radical. But if we are under steady state conditions, we can substitute for [ M· ]  using [ M· ]= ( R i 2 k t ) 1/2

R p = k p [ M ] ( R i 2 k t ) 1/2
Eq 5.13

Note: the upper limit for [ M ]  is pure monomer. When pure monomer is used, it's called bulk polymerization

PROBLEM 1

What happens to the total radical concentration at steady state if we double the concentration of initiator?

  1. [ M· ]  doesn't depend on [ I ]  so nothing changes
  2. [ M· ]  doubles
  3. [ M· ]  increases by a factor of 2
  4. [ M· ]  decreases by half

ANSWER 1

C. [ M· ]  increases by a factor of 2

We substitute for R i  in the expression for [ M· ]  and find [ M· ]= ( 2 k d [ I ] 2 k t ) 1/2  so  [ M· ] [ I ] 1/2

PROBLEM 2

What happens to the total rate of polymerization at steady state if we double the concentration of monomer?

  1. R p  doesn't depend on [ M ]  so nothing changes
  2. R p  doubles
  3. R p  increases by a factor of 2
  4. R p  decreases by half

ANSWER 2

B. R p  doubles.

R p [ M ]

Number average degree of polymerization

Number average degree of polymerization jls164

We have already introduced the idea of number average degree of polymerization. Recall:

x ¯ n = moles monomer consumed in time t moles of polymer formed in time t
Eq 5.14

We are going to try to describe number average degree of polymerization using the concentration of monomer and initiator, and for now let's neglect chain transfer. Can we substitute for the numerator, moles of monomer consumed in time t ? What's the rate of monomer consumption? We know that, it's the rate of propagation: R prop = d[ M ] dt = k p [ M ][ M· ] . What about the denominator, moles of polymer formed in time t ? Think about in what step we actually form a complete polymer — it's the termination step. In Figure 5.4, and we noticed that for termination by combination we form one polymer, and for termination by disproportionation we form 2 polymers. Thus the total rate of polymer formation is the sum of formation from combination and disproportionation. We write rate equations for this:

rate of polymer formation= 2 k td [ M· ] 2 + k tc [ M· ] 2
Eq 5.15

Let's substitute for the numerator and denominator in our number average degree of polymerization. We are going to use the subscript '0' for x ¯ n  to denote that this is the degree of polymerization in the absence of chain transfer.

( x ¯ n ) 0 = k p [ M ][ M· ]  2 k td [ M· ] 2 + k tc [ M· ] 2
Eq 5.16

Again, we can't do so much with radical concentrations in that expression. So let's substitute the steady state concentration of radicals, [ M· ]= ( R i 2 k t ) 1/2  and we get:

( x ¯ n ) 0 = k p [ M ] ( 1+q ) k t  1/2 ( R i 2 ) 1/2  where  q= k td k t
Eq 5.17

q  is the fraction of termination happening by disproportionation. If there is only termination by combination then q=0 . If there is only termination by disproportionation then q=1 . We can further substitute with R i =2f k d [ I ] :

( x ¯ n ) 0 = k p [ M ] ( 1+q ) k t  1/2 ( f k d [ I ] ) 1/2
Eq 5.18

This equation looks a little complicated. But importantly, we want to gain insight as to how the number average degree of polymerization is affected by inputs we can control, namely the concentration of monomer and concentration of initiator. We find: ( x ¯ n ) 0 [ M ] [ I ] 1/2

We call the equations for R p  and ( x ¯ n ) 0  that we derived instantaneous equations because they apply to a specific concentration of [ M ]  and [ I ]  (which change over time). As a refresher: R p [ M ] [ I ] 1/2  and ( x ¯ n ) 0 [ M ] [ I ] 1/2 . Thus, both go up with increasing [ M ] , and the expressions are oppositely correlated with [ I ] . Thus, we can predict how changing [ M ]  and [ I ] , which we can control, will affect our resulting polymer. Remember [ M ]  is limited to pure monomer in bulk polymerization. So there are constraints on the degrees of polymerization that you can achieve.

How accurate are these expressions as compared to experimental values of x ¯ n ? We find that experimental values of x ¯ n  tend to be lower than expected based on these derived relationships. Why? What have we neglected in our derivations? Chain transfer! Chain transfer can contribute significantly to the termination of growing polymer chains. Even though the transferred radical can still continue to react, the chain from which it was transferred is now shorter (lower x ¯ n ) than would have been expected (Figure 5.5). The rate of polymerization may not be so greatly affected if A·  reacts quickly with monomer (Figure 5.5).

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Figure 5.5: Chain transfer effect on rate of polymerization and degree of polymerization
Source: Lauren Zarzar

We can account for chain transfer, including chain transfer to solvent, monomer, and initiator, in addition to termination by combination and disproportionation by building this into the "moles of polymer formed at time t" part of the denominator in our expression for  x ¯ n . We thus update our previous expression, which was ( x ¯ n ) 0 = k p [ M ][ M· ]  2 k td [ M· ] 2 + k tc [ M· ] 2 , and we alter it:

x ¯ n = k p [ M ][ M· ]  2 k td [ M· ] 2 + k tc [ M· ] 2 + k trM [ M· ][ M ]+ k trI [ M· ][ I ]+ k trS [ M· ][ S ]
Eq 5.19

Notice that the k trM [ M· ][ M ]  takes into account chain transfer to monomer, k trI [ M· ][ I ]  is for chain transfer to initiator, and k trS [ M· ][ S ]  is for chain transfer to solvent. This is a more accurate description of degree of polymerization. Notice that we didn't include a term for chain transfer to polymer in our updated description of degree of polymerization because if the active center is transferred between polymers, then there is actually no effect on "moles of polymer formed at time t". Rearrange, substitute with ( x ¯ n ) 0  (which we previously derived without chain transfer) to generate the Mayo Walling Equation:

1 x ¯ n = 1 ( x ¯ n ) 0 + C M + C I [ I ] [ M ] + C s [ S ] [ M ]
Eq 5.20

We have thus introduced transfer constants C M C I C S  are k tr / k p  for each type of chain transfer reaction (transfer to monomer, initiator, and solvent, respectively). Notice that while chain transfer to polymer can greatly affect skeletal structure, it doesn't affect degree of polymerization because the number of polymers before and after chain transfer are the same.

See Table 5.1 below for a list of example transfer constants of various solvents for the free radical polymerization of styrene. Solvents with high transfer constants would prevent propagation and growth of long polymers, but can be added in small quantities to produce smaller polymers if they are desired (i.e., used as a chain transfer agent).

Table 5.1 Typical Magnitudes of Transfer Constants Observed in Free-Radical Polymerization of Styrene at 60 °C
CompoundBond Cleaved
(T-A)
Transfer Constant
CTA=(KtrTA/Kp)
StyreneH—C(Ph)=CH27x10-5
Benzoyl peroxidePhCOO—OOCPh5x10-2
BenzeneH—Ph2x10-6
TolueneH—CH2Ph12x10-6
ChloroformH—CCl35x10-5
Carbon tetrachlorideCl—CCl31x10-2
Carbon tetrabromideBr—CBr32
Dodecyl mercaptanH—SC12H2515

PROBLEM

A solution of 100 g/L acrylamide in methanol is polymerized at 25°C with 0.1 mol/L diisobutyryl peroxide whose half life is 9.0 h at this temperature and efficiency in methanol is 0.3. For acrylamide, k p  2 / k t = 22 L mo l 1 s 1  at 25°C and termination is by coupling alone.

What is the initial steady state rate of polymerization?
(half life for a first order reaction: t 1/2 = ln( 2 ) k d )

EnterimageandalttexthereNosizes

ANSWER

We are looking to solve for the initial steady state rate of polymerization, thus R p = k p [ M ] ( f k d [ I ] k t ) 1/2

To solve this equation, we need to substitute for many variables. Can we pull all of that necessary information out of the question? Let's start with the concentration of the monomer, acrylamide. It's density is given, but we need the concentration in mol/L:

[ M ]=[ acrylamide ]= 100g L * mol 71g =1.408 mol/L

Let's move on to k d . We can get this rate constant for decomposition of the initiator from the half life. We are going to solve for this with the time units of seconds, because in the problem we are given k p  2 / k t = 22 L mo l 1 s 1  and we want to make sure all our units are going to be consistent throughout the problem.

k d = ln( 2 ) 9hr*3600s/hr =2.139x 10 5 /s

In order to actually make sure of k p  2 / k t = 22 L mo l 1 s 1  we are going to have to rearrange our expression for R p  a little.

R p = k p [ M ] ( f k d [ I ] k t ) 1 2 = k p k t 1/2 [ M ] ( f k d [ I ] ) 1 2

Now it looks like we have all the numbers we need to plug into this equation!

R p = ( 22 L mol*s ) 1 2 ( 1.408 mol L )[ 0.3( 2.139x 10 5 s )( 0.1 mol L ) ]  1 2 =5.292x 10 3 mol/( L*s )

Inhibitors and Retarders

Inhibitors and Retarders jls164

We can add specific molecules to our reaction if we want to prevent or slow down polymerization. For example, when you purchase a commercially available monomer from a supplier, you may read on the bottle that the monomer is stabilized, perhaps by a molecule such as MEHQ (monomethyl ether hydroquinone). Such inhibitors are added to extend the shelf life of the monomer by preventing unwanted, premature polymerization. Inhibitors work by "trapping" any radicals that are generated in a very stable molecule that does not further react. If a "normal" polymerization follows a trajectory such as in Figure 5.6 (Figure 4.7 in the textbook), then that same polymerization with an inhibitor present would follow a trajectory like Figure 5.6 (c); the inhibitor prevents polymerization for a certain amount of time (the induction time), which depends on how much inhibitor is present. As soon as all the inhibitor is used up, then the reaction proceeds as normal.

A retarder works slightly differently than an inhibitor, and serves to slow down a polymerization rather than prevent it, as shown in Figure 5.6 (b). Retarders act in a similar way as inhibitors in that they stabilize radicals, but are not quite as good as preventing further reactions of the radicals as inhibitors. Hence, polymerization can still proceed in the presence of active retarder.

Figure 4.7 from the textbook
FIGURE 5.6: Schematic examples of inhibition and retardation: (a) normal polymerization; (b) polymerization in the presence of a retarder; and (c) polymerization in the presence of an inhibitor (where Δt is the induction period).
Source: Figure 4.7 from Young, Robert J., and Peter A. Lovell.
Introduction to Polymers, Third Edition, CRC Press, 2011.

Effects of Temperature

Effects of Temperature jls164

Reaction rates typically depend on temperature, so what about polymerization reactions? More specifically, how are rate of polymerization and degree of polymerization dependent on temperature? As with any other reaction, temperature changes the rate constants associated with the polymerization. But there are many rate constants that affect polymerization, so the effect of temperature is not necessarily straightforward. If the temperature increased, we expect there to be more radicals generated in a shorter time frame, so the concentration of active centers will go up. Because rate of polymerization is directly correlated with concentration of radicals, the rate of polymerization will also go up with temperature. Can we simply keep increasing the temperature to increase R p indefinitely? That's not really an effective strategy, because depolymerization, which is the polymer reverting to monomer, is favored at elevated temperatures. At the ceiling temperature, the rate of polymerization and depolymerization are equal. Also consider that at higher temperatures, with higher concentration of radicals, that also means that termination rates will be higher. As the rate of termination increases, the degree of polymerization decreases, and the polymers will be shorter. Thus, the temperature at which you run a polymerization reaction can have significant impact on the resulting polymer formed and should be considered carefully.

Nonlinear Radical Polymerization

Nonlinear Radical Polymerization jls164

Thus far, we have primarily considered free radical linear polymerizations of monomers containing a functionality of 2. With the exception of branching that can occur from chain transfer to polymer, a monomer with a functionality of 2 can only lead to linear polymers. But what if we use a monomer that has a functionality greater than 2? Much like what we discussed for step polymerization, we will get branching and possible formation of crosslinks leading to a network structure. In fact, many monomers for free radical polymerization that have a functionality higher than 2 are often called crosslinkers. Examples of common crosslinkers are shown in Figure 5.7 and Figure 4.2 in the textbook.

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Figure 5.7: Examples of crosslinkers for free radical polymerization
Source: Lauren Zarzar

PROBLEM

What is the functionality of divinylbenzene?

  1. 1
  2. 2
  3. 3
  4. 4

ANSWER

  1. 4

Polymerization of a monomer and crosslinker would create a network polymer, such as diagramed in Figure 5.8. The monomers that have a functionality of 2 form linear polymers, that are linked by the crosslinker.

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Figure 5.8: Schematic diagram of a network polymer formed from monomer 'X' and crosslinker 'Z'.
Source: Replicated from Figure 4.19 from Young, Robert J., and Peter A. Lovell.
Introduction to Polymers, Third Edition, CRC Press, 2011

Summary and Final Tasks

Summary and Final Tasks sxr133

Summary

We have now completed our lessons focused on free radical polymerization. The reactions, kinetics, and mechanisms you learned in Lessons 4 and 5 will be found elsewhere in the course, however, as radical polymerization is one of the most ubiquitous polymerization mechanisms. Next, we will be moving on to ionic polymerization, which as you will see has quite a few similarities in terms of the reaction mechanism to free radical. A key difference, however, is that the active center is no longer a radical, but an ion - and the fact that it is charged is what is going to give ionic polymerization distinguishing characteristics from free radical polymerization. Pay attention to similarities and difference between these reaction pathways as we move forward in the course.

Reminder - Complete all of the Lesson 5 tasks!

You have reached the end of Lesson 5! Review the checklist on the Lesson 5 Overview / Checklist page to make sure you have completed all of the activities listed there before you begin Lesson 6.