In a typical step-growth polymerization, the Anderson–Schulz–Flory distribution is the probability mass function that describes the number fraction or weight fraction of polymers of length $x$ at a given extent of reaction $p$. The extent of reaction, $p$, is a value between 0 and 1, and $x$ is the degree of polymerization for any given oligomer or polymer present in the system.

$$\begin{align} \text{mass fraction} &= p^{x-1} \cdot (p-1)^2 \\ \text{weight fraction} &= x \cdot p^{x-1} \cdot (p-1)^2 \end{align}$$

A nice derivation of these equations can be found in this video. Figure 1 illustrates a plot of the number fraction distribution as a function of the degree of polymerization for several extent of reactions. Figure 2 illustrates the plot of the weight fraction distribution as a function of the degree of polymerization for several extent of reactions.

enter image description here

Note that high extents of reaction in step-growth polymerizations result in very broad distributions, and are also required for high molecular weight polymers. Carothers' equation describes how stoichiometric imbalances necessarily limit the possible degree of polymerization:

$$x_\text{average} = \frac{1+r}{1 + r - 2rp}$$

where $x_\text{average}$ is the average degree of polymerization, $p$ is the extent of reaction, and $r$ is the mole fraction of the reacting functional groups. $r$ is always a number between 0 and 1. The excess functional group is always taken to be in the denominator. When $r$ is equal to 1, stoichiometric balance exists, and the equation reduces to:

$$x_\text{average} = \frac{1}{1-p}$$

For example, if I react a diol and a diacid under step-growth conditions:

enter image description here

My goal is to purposely limit the degree of polymerization of a chemical reaction by introducing a stoichiometric imbalance in order to form short chain oligomers. I'd like to predict/calculate the statistical distribution of molecular weights for a given extent of reaction, p and stoichiometric imbalance, r.

I'm having trouble convincing myself that the Flory–Schulz distribution is appropriate for modeling step-growth systems that contain stoichiometric imbalances. Is there a straightforward way to incorporate stoichiometric imbalance into these distribution functions? Any help would be appreciated.


1 Answer 1


I came up with a way to solve my own problem as I was writing my question, but I feel as though it is a bit hand-wavy. I would appreciate any input regarding the validity of this solution.

The following is a workaround to force the Flory distribution to have a maximum value that is consistent with the average degree of polymerization due to stoichiometric imbalance via Carothers' equation. Carothers' equation models how stoichiometric functional group imbalances limit the average degree of polymerization for the step-growth mechanism. This average degree of polymerization $(\text{DP}_\text{average})$ can be calculated very simply:

$$\begin{align} \text{DP}_\text{average} &= \frac{r+1}{r+1-2rp} \\ \end{align}$$

When there are no stoichiometric imbalances, and the Flory distribution is valid:

$$\text{DP}_\text{average} = \frac{1}{1-p}\\ f(\text{DP}_\text{average}) = x \cdot p^{x-1} \cdot (p-1)^2$$

Let $p_\text{eff}$ be the effective extent of reaction, which is related to $r$ and $p$ via:

$$\frac{1}{1-p_\text{eff}} = \text{DP}_\text{average} = \frac{1+r}{1+r-2rp}\\ p_\text{eff} = 1 - \frac{1+r-2rp}{1+r}$$

This $p_\text{eff}$ can be used to calculate the ideal Flory distribution according to

$$F(\text{DP}_\text{average}) = x \cdot p_\text{eff}^{x-1} \cdot (p_\text{eff}-1)^2$$

The function $F(\text{DP}_\text{average})$ will force the molecular weight distribution to be properly centered about the $\text{DP}_\text{average}$ that is consistent with Carothers' equation. I'm not sure if this is chemically sensible, or accurately describes a real-ish polymer system. Comments/alternative answers would be appreciated.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.