Number average degree of polymerization

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

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: $R_{prop}=-\frac{d\left[ M \right]}{dt}=k_p\left[ M \right]\left[ M· \right]$. 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:

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

Again, we can't do so much with radical concentrations in that expression. So let's substitute the steady state concentration of radicals, $\left[ M· \right]={\left( \frac{R_i}{2k_t}\right)}^{1/2}$ and we get:

$q$ 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 $R_i=2f{k}_d\left[ I \right]$:

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: ${\left( {\overline{x}}_n\right)}_0\propto \left[ M \right]{\left[ I \right]}^{-1/2}$

We call the equations for ${\text{R}}_{\text{p}}$ and ${\left( {\overline{x}}_n\right)}_0$ that we derived instantaneous equations because they apply to a specific concentration of $\left[ \text{M} \right]$ and $\left[ \text{I} \right]$ (which change over time). As a refresher: $R_p\propto \left[ M \right]{\left[ I \right]}^{1/2}$ and ${\left( {\overline{x}}_n\right)}_0\propto \left[ M \right]{\left[ I \right]}^{-1/2}$. Thus, both go up with increasing $\left[ \text{M} \right]$, and the expressions are oppositely correlated with $\left[ \text{I} \right]$. Thus, we can predict how changing $\left[ \text{M} \right]$ and $\left[ \text{I} \right]$, which we can control, will affect our resulting polymer. Remember $\left[ \text{M} \right]$ 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 ${\overline{x}}_n$? We find that experimental values of ${\overline{x}}_n$ 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 ${\overline{x}}_n$) 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).

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 ${\overline{x}}_n$. We thus update our previous expression, which was ${\left( {\overline{x}}_n\right)}_0=\frac{k_p\left[ M \right]\left[ M· \right]}{ 2k_{td}{\left[ M· \right]}^2+k_{tc}{\left[ M· \right]}^2}$, and we alter it:

Notice that the $k_{trM}\left[ M· \right]\left[ M \right]$ takes into account chain transfer to monomer, $k_{trI}\left[ M· \right]\left[ I \right]$ is for chain transfer to initiator, and $k_{trS}\left[ M· \right]\left[ S \right]$ 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 ${\left( {\overline{x}}_n\right)}_0$ (which we previously derived without chain transfer) to generate the Mayo Walling Equation:

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).

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, $k_p{ }^2/k_t=22Lmo{l}^{-1}{s}^{-1}$ 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: $t_{1/2}=\frac{\text{ln}\left( 2 \right)}{k_d}$)

Click here for the answer.

ANSWER

We are looking to solve for the initial steady state rate of polymerization, thus $R_p=k_p\left[ M \right]{\left( \frac{f{k}_d\left[ I \right] }{k_t}\right)}^{1/2}$

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:

$$\left[ M \right]=\left[ acrylamide \right]=\frac{100g}{L}\text{*}\frac{mol}{71g}=1.408 mol/L$$

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 $k_p{ }^2/k_t=22Lmo{l}^{-1}{s}^{-1}$ and we want to make sure all our units are going to be consistent throughout the problem.

$$k_d=\frac{\text{ln}\left( 2 \right)}{9hr*3600s/hr}=2.139x{10}^{-5}/s$$

In order to actually make sure of $k_p{ }^2/k_t=22Lmo{l}^{-1}{s}^{-1}$ we are going to have to rearrange our expression for $R_p$ a little.

$$R_p=k_p\left[ M \right]{\left( \frac{f{k}_d\left[ I \right] }{k_t}\right)}^{\frac{1}{2}}= \frac{k_p}{k_t^{1/2}}\left[ M \right]{\left( f{k}_d\left[ I \right] \right)}^{\frac{1}{2}}$$

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

$$R_p={\left( \frac{22 L}{mol*s}\right)}^{\frac{1}{2}}\left( \frac{1.408 mol}{L}\right)\left[ 0.3\left( \frac{2.139x{10}^{-5}}{s}\right)\left( \frac{0.1\text{ mol}}{\text{L}}\right) \right]{}^{\frac{1}{2}}=5.292x{10}^{-3}mol/\left( L*s \right)$$