Overview

In this Lesson, we will take a close look at step polymerization. If we know the number of reactive groups in our monomer mixture, and we can control how many groups react, then can we predict the number average degree of polymerization? We can! First we consider if we have stoichiometric balance of reactive groups, then we consider if we have a non-stoichiometric balance and what effects that has on the polymer produced. Finally, we end with a statistical analysis of step polymerization, which gives us insight into the distributions of the molar masses produced.

Learning Outcomes

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

• Define extent of reaction and reactant ratio
• Apply Carothers theory to step polymerization reactions
• Calculate the probability of finding a polymer of specific degree of polymerization, and the weight fraction or number fraction of polymers of a specific degree of polymerization

Lesson Checklist

Please refer to the Canvas Calendar for specific time frames.

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.

We will now consider in greater depth the theoretical treatment of linear step-growth polymerization. Specifically, we will examine the Carothers equation, which relates degree of polymerization to extent of reaction. We learned about degree of polymerization in Lesson 2, but what is extent of reaction?

If we assume equal amounts of mutually reactive groups (e.g., $A$ and $B$ groups), then we can define the extent of reaction $\left( p \right)$as

$$p=\frac{\#functionalgroupsreacted}{\#functionalgroupsinitially}$$

Extent of reaction is really just the probability that any single functional group that was present at the start has reacted. Consider the mixture of bifunctional monomers in Figure 3.1.

Figure 3.1: Stoichiometric balance of A and B

Source: Lauren Zarzar

What is the extent of reaction shown in Figure 3.1? $A$ and $B$ each represent a functional group, so there were 16 functional groups (2 on each of 8 molecules) initially. Looking at the products, we find that 6 of those functional groups are now part of new bonds, so they reacted. We plug these values our equation for extent of reaction:

$$p=\frac{\#functionalgroupsreacted}{\#functionalgroupsinitially}=\frac{6}{16}=0.375$$

But notice, there is another way to represent extent of reaction in terms of number of molecules, where $N$ is the number of molecules left after the reaction and $N_O$ is the number of molecules initially. Again, using Figure 3.1, where our initial number of molecules is 8 and our final number of molecules is 5:

$$p=\frac{N_O-N}{N_O}=\frac{8-5}{8}=0.375$$

Let’s rearrange this new expression for extent of reaction:

$$\frac{N_O}{N}=\frac{1}{1-p}$$

This value, $N_O/N$, should be familiar, because it’s the same as the number average degree of polymerization!

$${\overline{x}}_n=\frac{N_O}{N}$$

We combine the equations for extent of reaction with number average degree of polymerization to yield the Carothers equation for stoichiometric balance of reactive groups:

$${\overline{x}}_n=\frac{1}{1-p}$$

So what does this equation tell us?

If you want polymers of any significant length (i.e., high degree of polymerization, ${\overline{x}}_n$) you need very high extent of reaction $\left( p \right)$!

If there is a stoichiometric imbalance, we consider the reactant ratio, r:

$$r=\frac{N_A}{N_B}$$

We always have a choice of which groups are $A$ and which groups are $B$; here, we must define the reactant ratio such that it is equal to or less than 1. The way we have defined $r$, then the $A$ groups would be in the minority.

Now because we are still confining our analysis to linear polymerization, each monomer has only 2 reactive groups. So, the total number of molecules initially present can be represented as:

$$N_O=\frac{N_A+N_B}{2}$$

Let’s simplify this expression, and try to represent $N_O$ as a function of only the number of $B$ groups for instance, but substituting in $r$:

$$N_O=\frac{N_B\left( 1+r \right)}{2}$$

Consider the following example, which is a modification of Figure 3.1. Notice, we now have a different number of $A$ and $B$ groups:

Figure 3.4: Stoichiometric imbalance of A and B

Source: Lauren Zarzar

Our minority functional group is $A$. So our reactant ratio is:

$$r=\frac{N_A}{N_B}=\frac{7}{9}$$

We can still define an extent of reaction:

$$p=\frac{\text{\#}minorityfunctionalgroupsreacted}{\text{\#}minorityfunctionalgroupsinitially}=\frac{3}{7}$$

Let’s keep our eye on the prize: we want to be able to define ${\overline{x}}_n$ as a function of $r$ and $p$. How can we do this?  Similarly to our derivation with the case of stoichiometric balance, we need to have an expression for $N$, the number of molecules remaining. The number of molecules remaining is:

$$N=\frac{\#unreactedfunctionalgroups}{\text{\#}functionalgroupspermolecule}$$

Convince yourself with our example that this is true. There are 10 unreacted $A$ and $B$ groups, and 2 functional groups per molecule. Thus, we have 5 molecules remaining; and indeed we do! But how can we write $N$ as a function of $r$ and $p$? Let’s rewrite our expression for $N$, taking into account that we already know there are 2 functional groups per molecule, and we only have $A$ and $B$ unreacted groups:

$$N=\frac{\left( \#unreactedA \right)+\left( \#unreactedB \right)}{2}$$

We can define $\#unreactedA$ as:

$$\#unreactedA\text{ = \# }Apresentinitially-\text{ \# }Areacted$$

$$\#unreactedA\text{ = r}{N}_B\left( 1-p \right)$$

We can define $\#unreactedB$ as:

$$\#unreactedB\text{ = }{N}_B-p{N}_A\text{= }{N}_B\left( 1-rp \right)$$

Notice that the because $A$ only reacts with $B$, the number of $A$ groups that react $\left( p{N}_A\right)$ is the same as the number of $B$ groups that react.

Try plugging some numbers in from our example in Figure 3 to convince yourself these expressions hold.

$$\#unreactedA\mathrm{groups}\text{ = }r{N}_B\left( 1-p \right)=\frac{7}{9}\left( 9 \right)\left( 1-\frac{3}{7}\right)=4$$

$$\#unreactedB\mathrm{groups}\text{ = }{N}_B\left( 1-rp \right)=9\left( 1-\frac{7}{9}\ast \frac{3}{7}\right)=6$$

Ultimately, we wanted to write our expression for $N$ in terms of $r$ and $p$. So:

$$N=\frac{\left( \#unreactedA \right)+\left( \#unreactedB \right)}{2}=\frac{r{N}_B\left( 1-p \right)+N_B\left( 1-rp \right)}{2}$$

We can simplify:

$$N=\frac{N_B\left( 1+r-2rp \right)}{2}$$

Now, finally, we have expressions for $N$ and $N_O$ which we substitute into our general expression for ${\overline{x}}_n$

$${\overline{x}}_n=\frac{N_o}{N}=\frac{\frac{N_B\left( 1+r \right)}{2}}{\frac{N_B\left( 1+r-2rp \right)}{2}}=\frac{1+r}{1+r-2rp}$$

Finally, we have a more general Carothers equation that is now applicable to polymerization with non-stoichiometric balance of reactive groups. And what does this equation tell us? If you want polymers of any significant length (high degree of polymerization) you need VERY HIGH extent of reaction AND to control $r$ as close to 1 as possible!

Table 3.1 (Table 3.3 in the text) calculates values of degree of polymerization for varying values of extent of reaction and reactant ratio. Take a look; you may be surprised really how high extent of reaction needs to be in order to get polymers of any significant length, especially as your reactant ratio deviates more greatly from 1.

$${\overline{x}}_n=\frac{1+r}{1+r-2rp}$$

These short chains are barely polymers!

PROBLEM

You’re polymerizing the two monomers shown below. You want to limit your number average degree of polymerization to 39. Assuming you can achieve a quantitative reaction (p approaching 1), and you start with 2 moles of phenolphthalein as your limiting reagent, then how many moles of terephthaloyl chloride should you use?

Phenolphthalein (left) and terephthaloyl chloride (TCL) (right)

Source: Lauren Zarzar

Click here for the answer.

ANSWER

We are told degree of polymerization, x_n is 39. We can therefore use the Carothers equation to solve for reactant ratio:

$${\overline{x}}_n=39=\frac{1+r}{1+r-2r\left( 1 \right)}$$

$$r=\frac{38}{40}=0.95$$

Once we know the reactant ratio, that tells us the mole ratio of reactive groups that we should use. Since the reactant ratio always has to be 1 or less, and we are told that phenolphthalein is the limiting reagent (and we have 2 moles of it, with 2 functional groups per molecule), we define the reactant ratio as:

$$r=0.95=\frac{N_A}{N_B}$$

$$=\frac{molphenolphthalein\ast 2}{molterephthaloylchloride\ast 2}$$

$$=\frac{2}{molterephthaloylchloride}$$

We solve for moles of terephthaloyl chloride, which equals 2.1 moles.

With Carothers theory, we were dealing with number-average quantities. But how do we predict the molar mass distributions? If we treat each reaction between A and B groups in a step of polymerization as a random event, we can start to apply probability and statistics to the reaction and get an estimation of the distribution of the degrees of polymerization of the polymers formed for a given extent of reaction. Essentially, we are trying to find $P\left( x \right)$, which is the probability of existence of a molecule containing exactly x units at time t with extent of reaction p. Without going into the derivation here, we find the probability of finding a sequence of x units:

$$P\left( x \right)=\left( 1-p \right)p^{\left( x-1 \right)}$$

Since $P\left( x \right)$ is the probability of finding an “x-mer”, it must also be representative of the mole fraction of x-mers:

$$molfraction\text{ x-mer}=\frac{N_x}{N}=\left( 1-p \right)p^{\left( x-1 \right)}$$

Where $N_x$ is number of x-mers and $N$ is total number of molecules at time $t$. Often, it’s hard to measure the number of molecules present at any given time $\left( N \right)$, but it’s easy to find $N_0$, which is the number of molecules we initially start with . Can we substitute in $N_0$? In order to relate $N$ to $N_0$ we use a familiar equation:

$${\overline{x}}_n=\frac{1}{1-p}=\frac{N_0}{N}$$

If we solve this above equation for $N$, we find $N=N_0\left( 1-p \right)$. We substitute this relationship into $N_x=N\left( 1-p \right)p^{\left( x-1 \right)}$ and find an expression for the number of “x-mers” at a given extent of reaction:

$$N_x=N_0{\left( 1-p \right)}^2{p}^{\left( x-1 \right)}$$

We may also be interested in the weight fraction of x-mers, in addition to the number fraction. How do we solve for that? We start with the standard definition of weight fraction:

$$w_x=\frac{massofmoleculeswithdegreeofpolymerizationx}{totalmassofallmolecules}$$

$$=\frac{N_x\left( x{\overline{M}}_0\right)}{N_0{\overline{M}}_0}=\frac{x{N}_x}{N_0}$$

We substitute into this with $\frac{N_x}{N_0}={\left( 1-p \right)}^2{p}^{\left( x-1 \right)}$ and get the weight fraction of x-mers at a given extent of reaction:

$$w_x=x{\left( 1-p \right)}^2{p}^{\left( x-1 \right)}$$

PROBLEM

You are polymerizing the below monomers (1 mole of each) by step growth polymerization. When you stop the reaction, there are 0.1 moles of acid chloride groups left. What weight fraction of monomer do you have left?

Phenolphthalein (left) and terephthaloyl chloride (TCL) (right)

Source: Lauren Zarzar

Click here for the answer.

ANSWER

First, we find the extent of reaction:

$$p=\frac{N_A}{N_B}=\frac{\left( 2mol-0.1mol \right)}{\left( 2mol \right)}=0.95$$

Then we can find the weight fraction of monomer (because x=1 is monomer!)

$$w_x=x{\left( 1-p \right)}^2{p}^{\left( x-1 \right)}$$

$$=1{\left( 1-0.95 \right)}^2{0.95}^{\left( 1-1 \right)}$$

$$=0.0025$$

PROBLEM 2

You are polymerizing the below monomers (1 mole of each) by step growth polymerization. You stop the reaction when it's 99% complete. What fraction of the mixture is monomer, on a per mole basis?

Phenolphthalein (left) and terephthaloyl chloride (TCL) (right)

Source: Lauren Zarzar

Click here for the answer.

ANSWER 2

We can plug directly into the $P\left( x \right)$ equation:

$$P\left( x \right)=\left( 1-p \right)p^{\left( x-1 \right)}$$

$$=\left( 1-0.99 \right){0.99}^{\left( 1-1 \right)}$$

We can use these relationships we have learned from the statistical theory to plot the most probable distributions for $P\left( x \right)$ and $w_x$:

Figure 3.5: Probable distributions for $P\left( x \right)$ and $w_x$

Source: Figure 3.1 from Young, Robert J., and Peter A. Lovell.
Introduction to Polymers, Third Edition, CRC Press, 2011.

What can we learn from these plots? First, consider the plot of $P\left( x \right)$ for a few different values of extent of reaction (which is the left graph). Notice that $P\left( x \right)$, even for the high extents of reaction, is always larger for the lower values of $x$. This is telling us that even if we go to high extents of reaction, polymers with low degrees of polymerization are still more likely to be found than those with high degrees of polymerization. Now look at the second graph, at right, which is a plot of $w_x$. In contrast, this plot has a peak, which is actually very close to $M_n$. As extent of reaction increases, we see that the peak weight fraction moves to high values of $x$– this makes sense, because the polymers with higher degrees of polymerization are formed at higher extents of reaction. We also notice that the breadth of the curve increases as extent of reaction increases, which tells us that the dispersity of the polymer is increasing.

From these distributions, we can solve for the number average molar mass using this relationship:

$${\overline{M}}_n={\displaystyle \sum P\left( x \right) }{M}_x$$

Doing a series of substitutions (i.e. substitute in $P\left( x \right)=\left( 1-p \right)p^{\left( x-1 \right)}$, $M_x=x{\overline{M}}_0$, and then simplify) we find:

$${\overline{M}}_n=\frac{{\overline{M}}_0}{1-p}$$

Does this look familiar? It’s the same relationship as we got from Carothers theory!

Without going through the derivation, we can apply a similar analysis to get information about the weight average molar mass as well. We find:

${\overline{M}}_w={\overline{M}}_0\frac{1+p}{1-p}$ and ${\overline{x}}_w=\frac{1+p}{1-p}$ and $dispersity=1+p$

Summary

In Lesson 3, we got a more in depth look at step polymerization. Using the Carothers equation, we now understand the relationship between extent of reaction, reactant ratio, and the degree of polymerization. Out of these relationships, we learned that in order to achieve high degrees of polymerization, we need to go to high extents of reaction and have a reactant ratio as close to 1 as possible. We also used statistical theory to derive expressions for the probability of finding an x-mer, the weight fraction of x-mers, and the number fraction of x-mers. From these, we can learn about the distribution of the polymer molar mass for different extents of reaction.

Reminder - Complete all of the Lesson 3 tasks!

You have reached the end of Lesson 3! Review the checklist on the Lesson 3 Overview / Checklist page to make sure you have completed all of the activities listed there before you begin Lesson 4.