Initiator concentration for a target free-radical conversion: 23-Chem-B8 Polymer Engineering, December 2015, Q2

Question 2: Initiator concentration for a target free-radical conversion (25 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. Benzoyl peroxide (BPO) initiator: $t_{1/2}=7.3$ h at 70 °C, $E_{a,d}=29.7$ kcal/mol. Target: 50% monomer conversion in $t=6$ h at 60 °C. Efficiency $f=0.4$; rate-constant group $k_p^2/k_t=1.04\times10^{-2}$ L·mol$^{-1}$·s$^{-1}$ at 60 °C.

Find. The required initiator concentration $[I]$ (mol/L).

Approach. Get $k_d$ at 60 °C from the 70 °C half-life via Arrhenius, then integrate the standard free-radical rate law $R_p=k_p[M](f k_d[I]/k_t)^{1/2}$ over the batch to relate conversion to $[I]$, and solve for $[I]$.

Decomposition constant at 60 °C. From $k_d=\ln2/t_{1/2}$, $k_d(70^\circ)=\ln2/(7.3\times3600)=2.64\times10^{-5}\ \text{s}^{-1}$. Shifting with Arrhenius ($E_a=29\,700$ cal/mol, $R=1.987$): $$k_d(60^\circ)=k_d(70^\circ)\exp\!\left[-\frac{E_a}{R}\!\left(\frac{1}{333.15}-\frac{1}{343.15}\right)\right] =2.64\times10^{-5}(0.270)=7.13\times10^{-6}\ \text{s}^{-1}$$ (so $t_{1/2}\approx27$ h at 60 °C — the initiator decays only $\sim15\%$ over the 6-h run).
Rate law and integration. The polymerization rate is $R_p=-d[M]/dt=k_p[M]\,(f k_d[I]/k_t)^{1/2}$. Grouping the given constant $k_p/\sqrt{k_t}=\sqrt{k_p^2/k_t}$ and treating $[I]$ as sensibly constant: $$\ln\frac{[M]_0}{[M]}=\sqrt{\tfrac{k_p^2}{k_t}}\,\sqrt{f k_d[I]}\;t$$ For 50% conversion $\ln(1/0.5)=0.693$.
Solve for the initiator concentration. Rearranging, $$[I]=\frac{1}{f k_d}\!\left[\frac{\ln 2}{\sqrt{k_p^2/k_t}\;t}\right]^2 =\frac{1}{(0.4)(7.13\times10^{-6})}\!\left[\frac{0.693}{\sqrt{1.04\times10^{-2}}\,(21600)}\right]^2$$ $$[I]=\boxed{3.5\times10^{-2}\ \text{mol/L}}$$

Quantity	Value
$k_d$ at 60 °C	$7.13\times10^{-6}\ \text{s}^{-1}$ ($t_{1/2}\approx27$ h)
Required initiator concentration $[I]$	$\approx0.035$ mol/L

Check

The constant-$[I]$ assumption is justified because the BPO half-life at 60 °C (27 h) far exceeds the 6-h reaction; over the run $[I]$ falls only to $e^{-k_dt}=0.86$ of its start value. Carrying the decay exactly (replace $t$ by the effective time $(2/k_d)(1-e^{-k_dt/2})=2.08\times10^4$ s) raises the requirement modestly to $[I]_0\approx0.037$ mol/L. Either value rounds to $\sim0.035\!-\!0.04$ mol/L.

Topic: Free-radical chain polymerization — steady-state rate and initiator sizing.

Key relations: $k_d=\ln2/t_{1/2}$; Arrhenius shift; rate of initiation $R_i=2fk_d[I]$; steady-state $R_p=k_p[M](f k_d[I]/k_t)^{1/2}$; half-order dependence on $[I]$.

Why this works: Under the steady-state approximation, radical generation ($2fk_d[I]$) equals termination ($2k_t[M\!\cdot]^2$), fixing the radical population at $[M\!\cdot]=(fk_d[I]/k_t)^{1/2}$. Propagation then consumes monomer at $k_p[M][M\!\cdot]$, giving the characteristic $R_p\propto[I]^{1/2}[M]$. Integrating at nearly constant $[I]$ turns the conversion target directly into an $[I]$ requirement.

Common pitfall: Forgetting the efficiency $f$ and the factor structure of $k_p^2/k_t$ (the group already is $k_p^2/k_t$, so use $\sqrt{\;}$ once). Also, the half-order in $[I]$ means a 4× error in $[I]$ only doubles the rate — but it still matters here.

Source: Odian, Principles of Polymerization 4th ed., §3.4 (rate of radical polymerization); Rudin & Choi, Elements of Polymer Science 3rd ed., Ch. 6.

Quantity	Value
\(k_d\) at 60 °C	\(7.13\times10^{-6}\ \text{s}^{-1}\) (\(t_{1/2}\approx27\) h)
Required initiator concentration \([I]\)	\(\approx0.035\) mol/L

Question 2 of 6: Initiator concentration for a target free-radical conversion

Question 2: Initiator concentration for a target free-radical conversion (25 marks)