Question 1 of 6: Emulsion Polymerization of MMA — Number-Average Molecular Weight
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Open-book, 3 hours; six numbered problems of equal value (20 marks each), of which five constitute a complete paper (only the first five in the answer book are marked). All six problems are solved below so the set is complete for study.
Reference texts: Odian, Principles of Polymerization (4th ed., Wiley) — chain- and step-growth kinetics, emulsion polymerization, molecular-weight distributions; Rudin & Choi, The Elements of Polymer Science and Engineering (3rd ed., Academic Press) — dilute-solution properties, osmometry, viscoelasticity; Sperling, Introduction to Physical Polymer Science (4th ed., Wiley) — linear viscoelasticity, terminal-zone moduli; Bird, Armstrong & Hassager, Dynamics of Polymeric Liquids, Vol. 1 — non-Newtonian tube flow; supporting polyolefin process detail from Young & Lovell, Introduction to Polymers (3rd ed.).
Find. The number-average molecular weight M̄n of the polymer produced under Smith–Ewart Case 2 conditions (n̄ = 0.5).
Approach. Count the particles from the volume fraction and single-particle volume, form the total radical concentration and hence the rate of propagation Rp; divide by the rate of radical generation Ri (which sets the rate of chain stopping in a zero–one particle) to get the kinetic chain length, then multiply by the monomer molar mass.
Check
With no chain-transfer data given, chain transfer is neglected and every chain is assumed to be stopped by the entry of the next radical (the Smith–Ewart Case 2, n̄ = 0.5 limit), so the number-average degree of polymerization equals the kinetic chain length Rp/Ri. If chain transfer to monomer were included, M̄n would be lower.
Volume of one particle and the particle number density. A particle is a sphere of diameter dp, and the particles occupy a fraction φ of the mixture:
$$v_p=\frac{\pi}{6}d_p^3=\frac{\pi}{6}(80\times10^{-9})^3=2.681\times10^{-22}\ \text{m}^3$$
$$N_p=\frac{\phi}{v_p}=\frac{0.4}{2.681\times10^{-22}}=\boxed{1.492\times10^{21}\ \text{particles}\cdot\text{m}^{-3}}$$
Total radical concentration. Each particle holds n̄ radicals on average; dividing the radical number density by Avogadro's number puts it in molar units:
$$[R^\bullet]=\frac{\bar n\,N_p}{N_A}=\frac{0.5\times1.492\times10^{21}}{6.023\times10^{26}}=1.239\times10^{-6}\ \text{kmol}\cdot\text{m}^{-3}$$
Rate of propagation. Propagation consumes monomer at the particle concentration [M]p:
$$R_p=k_p[M]_p[R^\bullet]=830\times4\times1.239\times10^{-6}=4.11\times10^{-3}\ \text{kmol}\cdot\text{m}^{-3}\text{s}^{-1}$$
Rate of radical generation. Persulfate dissociates into two radicals, of which a fraction f are effective:
$$R_i=2fk_d[I]=2(0.5)(4\times10^{-3})(1\times10^{-3})=4.0\times10^{-6}\ \text{kmol}\cdot\text{m}^{-3}\text{s}^{-1}$$
Kinetic chain length and molecular weight. In a zero–one particle each entering radical terminates the resident chain, so the number-average degree of polymerization is Rp/Ri:
$$\bar X_n=\frac{R_p}{R_i}=\frac{4.11\times10^{-3}}{4.0\times10^{-6}}=1028$$
$$\bar M_n=\bar X_n\,M_0=1028\times100.121=\boxed{1.03\times10^{5}\ \text{g}\cdot\text{mol}^{-1}}$$
