MATS3004 · Polymer Science and Engineering 1
Radical Polymerization Kinetics
The Lecture-3 kinetics handout derives the rate expressions that dominate the calculation questions: the rates of initiation, propagation and termination, the steady-state assumption giving [M•] = (Ri/kt)^½, the polymerization rate Rp = kp(f·kd/kt)^½[M][I]^½, and the number-average degree of polymerization (Xn)₀ ∝ [M][I]^−½. Because both exams carry the shared hurdle and the course supplies calculus/gas-constant reference sheets, this is the most calculation-heavy chapter — expect a full worked Rp/Xn or initiator problem with units required for full marks.
What this chapter covers
- 01Rate of initiation Ri = f·kd[I] (thermolysis) and initiator efficiency f (typically 0.3-0.8)
- 02Rate of propagation Rp = kp[M•][M]; rate of termination Rt = kt[M•]²
- 03Steady-state assumption Ri = Rt ⇒ [M•] = (Ri/kt)^½ = (f·kd[I]/kt)^½
- 04The signature result Rp = kp(f·kd/kt)^½[M][I]^½ ⇒ Rp ∝ [M][I]^½ (first order in monomer, half order in initiator)
- 05Number-average degree of polymerization (Xn)₀ = kp[M] / {(1+q)(f·kd·kt)^½[I]^½}, with q = ktd/kt ⇒ (Xn)₀ ∝ [M][I]^−½
- 06Effect of scaling [M] or [I]: [M•] depends on [I] only, Rp and Xn scale with [M], Rp rises but Xn falls with [I]
- 07Initiator efficiency from a rate dataset: f = Rp²·kt / (kp²·kd·[M]²·[I])
- 08Photolysis and redox initiation give analogous Rp forms; temperature raises Rp but lowers Xn
Effect of raising monomer concentration on Rp, radical concentration and chain length
- +1Write the dependences from the steady-state results. [M•] = (f·kd[I]/kt)^½ depends on [I] only; Rp = kp(f·kd/kt)^½[M][I]^½ ∝ [M] at fixed [I]; (Xn)₀ = kp[M]/{(1+q)(f·kd·kt)^½[I]^½} ∝ [M] at fixed [I].
- +1(a) Radical concentration. Because [M•] ∝ [I]^½ and does not contain [M], raising [M] at constant [I] leaves [M•] unchanged.
- +1(b) New rate. Rp ∝ [M], so scaling [M] from 4.0 to 6.0 mol·L⁻¹ multiplies Rp by 6.0/4.0 = 1.5: new Rp = 1.5 × 2.0 × 10⁻⁴ = 3.0 × 10⁻⁴ mol·L⁻¹·s⁻¹.
- +1(c) Chain length. (Xn)₀ ∝ [M] at constant [I], so it also rises by the factor 1.5 — the chains get 50% longer. (Raising [I] instead would lift Rp but shorten the chains, since Xn ∝ [I]^−½.)
Key terms
- Rate of initiation (Ri)
- The rate at which chain radicals are generated; for thermal decomposition Ri = f·kd[I], where kd is the decomposition rate constant and f is the initiator efficiency. Initiation's slow sub-step (decomposition) sets Ri.
- Initiator efficiency (f)
- The fraction of primary radicals that actually start chains rather than being lost to cage recombination or side reactions; typically 0.3-0.8. It can be found from a rate dataset via f = Rp²·kt / (kp²·kd·[M]²·[I]).
- Steady-state assumption
- The approximation that the radical concentration is constant because radicals are generated and destroyed at equal rates, Ri = Rt. It gives [M•] = (Ri/kt)^½ and is the pivot that turns the mechanism into a usable rate law.
- Rate of polymerization (Rp)
- Rp = kp(f·kd/kt)^½[M][I]^½, obtained by substituting the steady-state [M•] into Rp = kp[M•][M]. It is first order in monomer and half order in initiator, so Rp ∝ [M][I]^½ — the classic free-radical signature.
- Number-average degree of polymerization (Xn)₀
- (Xn)₀ = kp[M] / {(1+q)(f·kd·kt)^½[I]^½}, with q = ktd/kt the fraction of termination by disproportionation. It scales as [M][I]^−½, so raising initiator shortens chains.
- q (termination-mode parameter)
- q = ktd/kt, the fraction of termination that proceeds by disproportionation. q = 0 for pure combination and q = 1 for pure disproportionation; it enters the (1+q) factor in the (Xn)₀ expression.
Radical Polymerization Kinetics FAQ
Why is the rate of polymerization half order in initiator?
Because radicals are made one-at-a-time by initiation but destroyed two-at-a-time by termination. Setting Ri = Rt (steady state) gives f·kd[I] = kt[M•]², so [M•] = (f·kd[I]/kt)^½ — the radical concentration goes as the square root of [I]. Since Rp = kp[M•][M], that square-root dependence carries straight through, giving Rp ∝ [M][I]^½. The half order is a direct fingerprint of bimolecular termination.
Why does adding more initiator make the chains shorter?
More initiator means more radicals, so more chains start at once and monomer is shared among a larger radical population. Each chain therefore grows for less time before it meets another radical and terminates. Quantitatively (Xn)₀ ∝ [I]^−½, the opposite sign to Rp ∝ [I]^½: you speed the reaction up but at the cost of molar mass. To make long chains you keep [I] low (and, from the temperature rules, avoid running too hot).
What does the steady-state assumption actually assume?
That after a very short start-up the radical concentration is essentially constant because radicals are created (initiation) and destroyed (termination) at equal rates, so d[M•]/dt ≈ 0. This is justified because initiator decomposes slowly (its concentration barely changes over a run) while radicals are highly reactive and short-lived. The assumption converts an intractable set of rate equations into the compact Rp and Xn expressions you actually use.
How should I present a kinetics calculation for full marks?
Start from the quoted rate law, show the algebraic rearrangement, substitute values with consistent units, and give the answer with units and a sanity check (for example, initiator efficiency should land in 0.3-0.8). The Assignment brief is explicit that problem-solving must show a full worked solution and include units for full marks, and the course provides reference sheets so you are expected to manipulate the equations rather than recall numbers. Confirm the permitted calculator and any formula sheet on the UNSW course outline / Moodle.
Exam move
Memorise the three scaling laws as a single card — [M•] ∝ [I]^½; Rp ∝ [M][I]^½; (Xn)₀ ∝ [M][I]^−½ — because most kinetics questions are 'change one variable, predict the three responses', and the initiator sign flip (faster but shorter) is the recurring trap. Practise the full initiator-efficiency rearrangement f = Rp²·kt / (kp²·kd·[M]²·[I]) with a unit check and the 0.3-0.8 sanity range, since that algebra is the flagship worked skill. Rehearse first-order initiator decay too (half-life t½ = ln2/kd, fraction remaining e^(−kd·t)), which justifies treating [I] as constant. Because both the Mid-term and Final reward clean, unit-carrying working under the shared hurdle, drill for speed and tidy layout, not just the right final number. Confirm the reference sheets and calculator policy on the UNSW course outline / Moodle.
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