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.