Copolymer Equation | MATSE 202: Introduction to Polymer Materials

A useful way of representing information from the copolymer equation is to plot F_A as a function of f_A. Let’s walk through how to do this using the reactivity ratios from the previous question, r_A = 0.9 and r_B = 1.2. First, we are going to just choose example values of f_A ranging from 0 to 1 (listed in the f_A column of the data Table 8.1 for Figure 8.2). Next, we can find f_B for each data point, because f_B = 1-f_A. Then, we plug the given values of r_A = 0.9 and r_B = 1.2 and the values of f_A and f_B into the expression:

$F_{A} = \frac{r_{A} f_{A}^{2} + f_{A} f_{B}}{r_{A} f_{A}^{2} + 2 f_{A} f_{B} + r_{B} f_{B}^{2}}$

The values we calculated are shown below in Table 8.1. Plotting F_A for each value of f_Ayields the graph shown below in Figure 8.2. Notice that the answer to the previous problem, which was F_A=0.46, in fact corresponds to f_A=0.5 as expected.

Table 8.1: Plotting F_A as a function of f_A for r_A = 0.9 and r_B = 1.2
f_A	f_B	F_A
0	1	0.00
0.05	0.95	0.04
0.1	0.9	0.09
0.15	0.85	0.13
0.2	0.8	0.17
0.25	0.75	0.22
0.3	0.7	0.27
0.35	0.65	0.31
0.4	0.6	0.36
0.45	0.55	0.41
0.5	0.5	0.46
0.55	0.45	0.51
0.6	0.4	0.57
0.65	0.35	0.62
0.7	0.3	0.67
0.75	0.25	0.73
0.8	0.2	0.78
0.85	0.15	0.83
0.9	0.1	0.89
0.95	0.05	0.94
1	0	1.00

Figure 8.2. Plotting F_A as a function of f_A for r_A = 0.9 and r_B = 1.2

Source: Lauren Zarzar

Looking at plots of F_A as a function of f_A is very helpful in visualizing the relative incorporation of A and B monomers into the polymer. Figure 9.1 in the textbook (and replicated as Figure 8.3 below) shows such plots for varying combination of reactivity ratios. Can you rationalize why each curve as the shape it does? We will spend a significant amount of time considering this graph!

Figure 8.3: Plotting F_A as a function of f_A

Source: Figure 9.1 from Young, Robert J., and Peter A. Lovell.
Introduction to Polymers, Third Edition, CRC Press, 2011.

Figure 8.3 is a plot of $F_{A} = \frac{r_{A} f_{A}^{2} + f_{A} f_{B}}{r_{A} f_{A}^{2} + 2 f_{A} f_{B} + r_{B} f_{B}^{2}}$ for various values of reactivity ratios and monomer ratios. This is a key figure illustrating how the reactivity ratios and mole fraction of monomers affect the incorporation of the monomer into the polymer, ultimately determining the polymer composition. The shapes of each of these curves is meaningful, and we will discuss how to interpret this plot.

Take a minute to use the interractive graph below to explore the copolymerization reaction. You can use the sliders to change the values of r_A and r_B. You can also ust the text entry box to directly enter values for r_A and r_B. Try adjusting the values to match some of the curves in Figure 8.3 above.
Note: If you hover your mouse over the plot, the X = N/A and Y = N/A boxes will show the coordinates of you cursor.

Note:

If you are unable to interract with the graph above, try opening the page in a different browser.

PROBLEM

“Random” copolymers are defined as having F_A=f_A for all values of f_A. To put that into words, it means that the fraction of A incorporated into the polymer is the same as the fraction of A in the monomer solution for any given monomer fraction of A. What reactivity ratios must we have in order to form a random copolymer?

Click here for the answer.

ANSWER

r_A and r_B both must equal 1. The random copolymerization conditions correspond to the solid line f_A=F_A in Figure 8.3. Looking at the key to this figure, we can find that this corresponds to reactivity ratio values of 1. Mathematically, we need values of reactivity ratios that would simplify $F_{A} = \frac{r_{A} f_{A}^{2} + f_{A} f_{B}}{r_{A} f_{A}^{2} + 2 f_{A} f_{B} + r_{B} f_{B}^{2}}$ to F_A=f_A. This happens when r_A=r_B=1. Shown below is the substitution and simplification math (don’t forget that f_A+f_B =1!)

$F_{A} = \frac{f_{A}^{2} + f_{A} f_{B}}{f_{A}^{2} + 2 f_{A} f_{B} + f_{B}^{2}} = \frac{f_{A} (f_{A} + f_{B})}{{(f_{A} + f_{B})}^{2}} = \frac{f_{A} (1)}{{(1)}^{2}} = f_{A}$

Intuitively, when r_B and r_A = 1 then both homopropagation and cross-propagation for ~A* and ~B* happen at equal rates. If k_AA=k_AB (which is the case for r_A=1) then ~A* has no preference for adding to A or B. Similarly, if k_BB=k_BA(which is true for r_B=1) then ~B* also has no preference for adding to A or B. Therefore, the relative amounts of A and B incorporated into the polymer simply reflect the distribution of A and B in the monomer solution. There are relatively few copolymerizations that fulfill this requirement. However, also realize that the reactivities of the 2 active centers (~A* and ~B*) don’t have to be the same for random copolymerization.

So we just learned that a random copolymer forms when f_A=F_A, which can happen when r_A=r_B=1. What happens when this is not the case? For instance, what if f_A<F_A, what does that mean? Look at the conditions highlighted below, where an exemplary line for r_A=4 and r_B=0.1 ( highlighted in blue ). We see that this curve always falls above the line f_A=F_A indicating that the fraction of A being incorporated into the polymer is greater than the fraction of A in the monomer mixture for all molar fractions of A. This is because both active centers (~A* and ~B*) prefer to add to A monomer, which we know because k_AA>k_AB and k_BA>k_BB, based on the given values of reactivity ratios (i.e. r_A>1 and r_B<1). We can use the same pattern of logic to understand why for the exemplary curve r_A=0.1 and r_B=10 ( highlighted in yellow ), f_A >F_A for all molar ratios of A; both active centers prefer to add to B monomer, so we always incorporate less A into the polymer than there is A in the monomer solution.

Enter image and alt text here. No sizes!

Figure 8.4: Plotting F_A as a function of f_A.
r_A=4 and r_B=0.1 ( blue plot )
r_A=0.1 and r_B=10 ( yellow plot )

Source: Modified from Figure 9.1 from Young, Robert J., and Peter A. Lovell.
Introduction to Polymers, Third Edition, CRC Press, 2011.