Molecular weight vs. monomer conversion (constant-initiator free-radical polymerization): 23-Chem-B8 Polymer Engineering, December 2014, Q3

Question 3: Molecular weight vs. monomer conversion (constant-initiator free-radical polymerization) (Part B — 30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Free-radical batch polymerization with $[\text{I}]$ held constant. Instantaneous $\overline{M}_n = 200{,}000$ when $[\text{M}]=0.20[\text{M}]_0$. Termination by coupling (combination) only; constant $k_p, k_t, k_d, f$; equal polydispersity at both instants.

Find. Instantaneous $\overline{M}_n$ of the polymer forming when $[\text{M}]=0.10[\text{M}]_0$.

Approach. Write the instantaneous kinetic chain length $\nu$, relate the instantaneous $\overline{X}_n$ to $\nu$ for coupling termination, note its dependence on $[\text{M}]$ and $[\text{I}]$, then take the ratio at the two monomer concentrations.

Kinetic chain length. For steady-state radical polymerization the kinetic chain length (monomers consumed per radical) is $$\nu=\frac{R_p}{R_t}=\frac{k_p[\text{M}]}{2\,(f k_d k_t[\text{I}])^{1/2}} \;\propto\;\frac{[\text{M}]}{[\text{I}]^{1/2}}.$$
Termination by coupling. When two growing chains combine, each dead chain contains two kinetic chains, so the instantaneous number-average degree of polymerization is $$\overline{X}_n = 2\nu \;\propto\; \frac{[\text{M}]}{[\text{I}]^{1/2}}.$$ Because $[\text{I}]$ is deliberately held constant, at fixed temperature $\overline{X}_n$ (hence $\overline{M}_n$) is directly proportional to the instantaneous monomer concentration: $\;\overline{M}_n \propto [\text{M}]$.
Take the ratio. With identical polydispersity at the two instants the $\overline{M}_n$ ratio equals the $[\text{M}]$ ratio: $$\frac{\overline{M}_{n,10\%}}{\overline{M}_{n,20\%}} = \frac{0.10[\text{M}]_0}{0.20[\text{M}]_0}=\tfrac12 \;\Rightarrow\; \overline{M}_{n,10\%} = \tfrac12(200{,}000)=\boxed{1.0\times10^{5}}.$$

Instant	Instantaneous $\overline{M}_n$
$[\text{M}] = 0.20[\text{M}]_0$	200,000 (given)
$[\text{M}] = 0.10[\text{M}]_0$	100,000

Topic: Instantaneous molecular weight in free-radical chain polymerization; kinetic chain length and mode of termination.

Key relations: $\nu = k_p[\text{M}]/2(fk_dk_t[\text{I}])^{1/2}$; $\overline{X}_n = 2\nu$ (coupling) or $\overline{X}_n = \nu$ (disproportionation); $\overline{M}_n=\overline{X}_n M_0$.

Why this works: The degree of polymerization is set by the competition between propagation ($\propto[\text{M}]$) and termination ($\propto[\text{I}]^{1/2}$ through the radical concentration). Normally, as monomer is consumed in a batch, $[\text{M}]$ drops and $[\text{I}]$ drops, so the analysis is messy; holding $[\text{I}]$ constant by make-up feed removes the initiator dependence and leaves a clean linear proportionality to $[\text{M}]$. Equal polydispersity (assumption c) lets us equate the $\overline{M}_n$ ratio to the $\overline{X}_n$ (chain-length) ratio directly.

Common pitfall: Using $\overline{X}_n=\nu$ (correct only for disproportionation) or forgetting the $[\text{I}]^{1/2}$ dependence — here it is inert only because $[\text{I}]$ is held fixed. If initiator were not replenished, $[\text{I}]$ would fall as well and the molecular weight would not simply halve.

Source: Odian, Principles of Polymerization 4th ed., §3.6–3.8; Rudin & Choi 3rd ed., Ch. 6.

Instant	Instantaneous \(\overline{M}_n\)
\([\text{M}] = 0.20[\text{M}]_0\)	200,000 (given)
\([\text{M}] = 0.10[\text{M}]_0\)	100,000

Question 3 of 7: Molecular weight vs. monomer conversion (constant-initiator free-radical polymerization)

Question 3: Molecular weight vs. monomer conversion (constant-initiator free-radical polymerization) (Part B — 30 marks)