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Chapter 7 • Theory & Derivations

Radical Chain Polymerization: Kinetics, Chain Transfer, Gel Effect & Thermodynamics

Free-radical elementary mechanisms, homolytic initiator cleavage and cage efficiency factors (f), steady-state rate law (Rp), kinetic chain length (nu) and termination mode deconvolution, chain transfer to monomer, solvent, initiator and modifier via the Mayo equation, autoacceleration (Trommsdorff-Norrish gel effect) driven by diffusion-controlled termination, and polymerization thermodynamics governing reaction enthalpy, entropy, equilibrium monomer concentration ([M]_eq), and ceiling temperature (Tc).

§7.1 Elementary Mechanism of Free-Radical Polymerization: Initiation, Propagation & Termination

Chain-growth (addition) polymerization requires an active center—a free radical, carbonium ion, carbanion, or coordination transition-metal complex—capable of rapidly adding monomer molecules sequentially. In free-radical polymerization, chain growth is propagated by an unpaired electron on the terminal carbon atom of the growing macromolecule.

The Three Elementary Kinetic Steps

1. Initiation:

Initiation is a two-step sequence:

  • Homolytic Cleavage of Initiator: An initiator molecule $I$ decomposes thermally, photolytically, or via redox transfer into two primary free radicals $R^\bullet$:
\[I \xrightarrow{k_d} 2 R^\bullet\]
  • Monomer Addition (Chain Initiation): A primary radical adds to the carbon-carbon double bond of a vinyl monomer $M$ to produce the initial chain radical $M_1^\bullet$:
\[R^\bullet + M \xrightarrow{k_i} M_1^\bullet\]

Because the decomposition rate constant $k_d$ is typically $10^{-6} - 10^{-4}\text{ s}^{-1}$ while $k_i \sim 10^5\text{ L/(mol s)}$, the homolytic decomposition of initiator is the rate-determining step of initiation.

2. Propagation:

The active macroradical adds monomer molecules sequentially in rapid succession:

\[M_1^\bullet + M \xrightarrow{k_p} M_2^\bullet\]
\[M_n^\bullet + M \xrightarrow{k_p} M_{n+1}^\bullet\]

Addition typically occurs with head-to-tail regioselectivity due to steric repulsion and resonance stabilization of the resulting radical by the pendant substituent $X$ (e.g., phenyl in styrene, ester in acrylates).

3. Termination:

Two growing macroradicals annihilate each other's active unpaired electrons via bimolecular encounter:

  • Termination by Combination (Coupling): Two radical chain ends bond covalently to form a single dead macromolecule:
\[M_n^\bullet + M_m^\bullet \xrightarrow{k_{tc}} M_{n+m}\]
  • Termination by Disproportionation: A hydrogen atom is abstracted from the $\beta$-carbon of one radical by the other, producing two dead macromolecules: one with a saturated end and one with a terminal alkene end:
\[M_n^\bullet + M_m^\bullet \xrightarrow{k_{td}} M_n(\text{saturated}) + M_m(\text{unsaturated})\]

The total termination rate constant is:

\[k_t = k_{tc} + k_{td}\]

§7.2 Initiator Decomposition Kinetics: Half-Life, Efficiencies ($f$) & Radical Cages

Common Free-Radical Initiators

1. Azo Initiators:

  • $2,2'$-Azobis(isobutyronitrile) (AIBN): Decomposes thermally with elimination of exceptionally stable nitrogen gas ($\text{N}_2$):
\[(\text{CH}_3)_2\text{C(CN)}-\text{N}=\text{N}-\text{C(CN)}(\text{CH}_3)_2 \xrightarrow{\Delta} 2 (\text{CH}_3)_2\text{C}^\bullet(\text{CN}) + \text{N}_2 \uparrow\]

AIBN exhibits first-order decomposition kinetics with minimal induced decomposition from radicals.

2. Peroxide Initiators:

  • Benzoyl Peroxide (BPO): Undergoes homolytic cleavage of the weak $O-O$ peroxy bond ($E_a \approx 125\text{ kJ/mol}$):
\[\text{C}_6\text{H}_5\text{COO}-\text{OOCC}_6\text{H}_5 \xrightarrow{\Delta} 2 \text{C}_6\text{H}_5\text{COO}^\bullet \xrightarrow{-\text{CO}_2} 2 \text{C}_6\text{H}_5^\bullet\]

3. Redox Initiators:

  • Potassium persulfate with ferrous ions ($S_2O_8^{2-} + Fe^{2+} \to SO_4^{\bullet-} + SO_4^{2-} + Fe^{3+}$), widely used in aqueous emulsion and suspension polymerizations at room temperature.

Decomposition Kinetics and Half-Life

Initiator decomposition follows strict first-order kinetics:

\[-\frac{d[I]}{dt} = k_d [I] \implies [I](t) = [I]_0 \exp(-k_d t)\]

The initiator half-life $t_{1/2}$ is:

\[t_{1/2} = \frac{\ln 2}{k_d} = \frac{0.6931}{k_d}\]

The Initiator Efficiency Factor ($f$) and the Solvent Cage Effect

When an initiator molecule cleaves inside a liquid, the two primary radicals are born inside a surrounding 'cage' of solvent molecules. Before escaping into the bulk solution:

  1. They may collide and recombine inside the cage (geminate recombination):
\[2 R^\bullet \xrightarrow{\text{cage}} R-R \quad (\text{inactive tetramethylsuccinonitrile from AIBN})\]
  1. They may react with solvent molecules or undergo disproportionation within the cage.

The initiator efficiency $f$ is defined as the fraction of radicals produced by homolysis that successfully escape the solvent cage and initiate polymer chains:

\[f = \frac{\text{Rate of initiation of polymer chains}}{2 \times \text{Rate of initiator decomposition}} = \frac{R_i}{2 k_d [I]}\]

For typical polymerizations in organic solvents, $f$ ranges from $0.30$ to $0.80$. As solvent viscosity increases, radical cage escape is hindered, causing $f$ to decline.

§7.3 Steady-State Kinetics & Rate of Polymerization ($R_p$): The Pseudo-Steady-State Approximation

Derivation of the Steady-State Rate Law

To derive the overall rate of polymerization $R_p$, we formulate the rates of the elementary steps:

1. Rate of Initiation ($R_i$):

\[R_i = 2 f k_d [I]\]

2. Rate of Propagation ($R_p$):

The rate of monomer consumption occurs primarily via propagation (monomer consumed in initiation is negligible, $< 0.1\%$):

\[R_p = -\frac{d[M]}{dt} = k_p [M] [M^\bullet]\]

where $[M^\bullet] = \sum_{n=1}^\infty [M_n^\bullet]$ is the total concentration of all growing macroradicals.

3. Rate of Termination ($R_t$):

Because termination is a bimolecular reaction between two macroradicals:

\[R_t = 2 k_t [M^\bullet]^2\]

(The factor of 2 represents the consumption of two radical centers per termination event).

The Pseudo-Steady-State Approximation (PSSA)

Within seconds of initiating the reaction, the concentration of active macroradicals reaches a dynamic balance where the rate of radical generation equals the rate of radical termination:

\[R_i = R_t \implies 2 f k_d [I] = 2 k_t [M^\bullet]^2\]

Canceling the factor of 2 and solving for the steady-state radical concentration $[M^\bullet]$:

\[[M^\bullet] = \sqrt{ \frac{f k_d [I]}{k_t} } = \left( \frac{R_i}{2 k_t} \right)^{1/2}\]

For typical systems, $[M^\bullet] \approx 10^{-8} - 10^{-7}\text{ mol/L}$, explaining why radical polymerization proceeds smoothly without explosive chain recombination.

The Universal Rate Law of Radical Polymerization

Substituting $[M^\bullet]$ into the propagation rate expression:

\[R_p = k_p [M] \sqrt{ \frac{f k_d [I]}{k_t} } = k_p \left( \frac{f k_d}{k_t} \right)^{1/2} [M] [I]^{1/2}\]

This landmark equation reveals the fundamental classical kinetics:

  • The rate of polymerization is first-order in monomer concentration: $R_p \propto [M]^1$.
  • The rate of polymerization is half-order in initiator concentration: $R_p \propto [I]^{1/2}$.

The square-root dependence on $[I]$ is the experimental hallmark of bimolecular radical termination.

§7.4 Kinetic Chain Length ($\nu$) & Dispersity: Combination vs Disproportionation Modes

The Kinetic Chain Length ($\nu$)

The kinetic chain length $\nu$ is defined as the average number of monomer molecules polymerized per active radical center initiated:

\[\nu \equiv \frac{R_p}{R_i} = \frac{R_p}{R_t}\]

Substituting $R_p = k_p [M] [M^\bullet]$ and $R_t = 2 k_t [M^\bullet]^2$:

\[\nu = \frac{k_p [M] [M^\bullet]}{2 k_t [M^\bullet]^2} = \frac{k_p [M]}{2 k_t [M^\bullet]}\]

Substituting the steady-state radical concentration $[M^\bullet] = \sqrt{f k_d [I] / k_t}$:

\[\nu = \frac{k_p [M]}{2 \sqrt{f k_d k_t [I]}} = \frac{k_p}{2 \sqrt{f k_d k_t}} \frac{[M]}{[I]^{1/2}}\]

Notice the fundamental kinetic conflict of free-radical polymerization:

  • Increasing $[I]$ increases the polymerization rate ($R_p \propto [I]^{1/2}$), but
  • Increasing $[I]$ decreases the kinetic chain length and molecular weight ($

u \propto [I]^{-1/2}$). To produce high molecular weight polymer, one must operate at low initiator concentrations and tolerate lower polymerization rates.

Relationship Between $X_n$ and $\nu$

In the absence of chain transfer, the number-average degree of polymerization $X_n$ depends on the termination mode:

1. Pure Disproportionation ($k_{tc} = 0, k_t = k_{td}$):

Each termination event produces two dead polymer molecules from two growing macroradicals. Thus, each dead molecule corresponds to exactly one kinetic chain:

\[X_n = \nu\]

The theoretical dispersity is:

\[\text{Đ} = \frac{X_w}{X_n} = 2.0\]

2. Pure Combination ($k_{td} = 0, k_t = k_{tc}$):

Two kinetic chains combine head-to-head to form a single dead polymer molecule:

\[X_n = 2 \nu\]

The theoretical dispersity for combination termination is:

\[\text{Đ} = \frac{X_w}{X_n} = 1.50\]

Deconvolution of Mixed Termination Modes

Let $q$ be the fraction of termination events occurring by disproportionation:

\[q = \frac{k_{td}}{k_{tc} + k_{td}}\]

Then:

\[X_n = \frac{2 \nu}{1 + q}\]

For styrene at $60^\circ\text{C}$, termination occurs almost exclusively by combination ($q \approx 0.05, X_n \approx 1.9 \nu$). For methyl methacrylate (MMA) at $60^\circ\text{C}$, termination is predominantly disproportionation ($q \approx 0.80, X_n \approx 1.1 \nu$).

§7.5 Chain Transfer Reactions: Transfer to Monomer, Solvent, Initiator & Added Modifiers

Chain transfer is an elementary reaction wherein a growing macroradical abstracts an atom (usually hydrogen or halogen) from another molecule $X-Y$ present in the reaction mixture:

\[M_n^\bullet + X-Y \xrightarrow{k_{tr}} M_n-X + Y^\bullet\]

The original chain is terminated into a dead macromolecule $M_n-X$, while the newly generated radical $Y^\bullet$ may re-initiate polymerization by adding a new monomer:

\[Y^\bullet + M \xrightarrow{k_{i,Y}} M_1^\bullet\]

Classification of Chain Transfer Agents

1. Normal Chain Transfer ($k_{i,Y} \approx k_p$):

  • The new radical $Y^\bullet$ re-initiates rapidly.
  • The overall rate of polymerization $R_p$ is unaffected.
  • The molecular weight of the polymer is significantly reduced.

2. Retardation ($k_{i,Y} < k_p$):

  • The new radical $Y^\bullet$ re-initiates sluggishly.
  • Both molecular weight and polymerization rate $R_p$ are decreased.

3. Inhibition ($k_{i,Y} \approx 0$):

  • $Y^\bullet$ cannot re-initiate at all (e.g., stable nitroxy radicals like TEMPO, or hydroquinone reacting with oxygen).
  • Polymerization completely ceases until the inhibitor is totally consumed.

Transfer Mechanisms in Polymerization Systems

  • Transfer to Monomer ($M$): An inevitable physical ceiling on polymer molecular weight. For vinyl chloride, chain transfer to monomer ($C_M \sim 10^{-3}$) is so rapid that molecular weight is determined almost entirely by reaction temperature rather than initiator concentration.
  • Transfer to Solvent ($S$): Solvents with weak $C-H$ or $C-Cl$ bonds (e.g., chloroform, carbon tetrachloride, toluene) act as strong chain transfer agents. For $\text{CCl}_4$, chlorine radical abstraction yields trichloromethyl ends ($-\text{CCl}_3$).
  • Transfer to Initiator ($I$): Induced decomposition of peroxides.
  • Transfer to Chain Transfer Agents (Modifiers / Regulators): Alkyl mercaptans (thiols, such as $n$-dodecyl mercaptan, $R-\text{SH}$) have exceptionally weak $S-H$ bonds ($E_a \approx 365\text{ kJ/mol}$), transferring rapidly to control melt viscosity in industrial rubber and acrylate synthesis.

§7.6 The Mayo Equation & Chain Transfer Constants ($C_S, C_M, C_I, C_{CTA}$)

In 1943, Frank R. Mayo derived the fundamental mathematical expression relating the reciprocal degree of polymerization to chain transfer processes.

Derivation of the Mayo Equation

The total rate of termination of macromolecular chains is the sum of bimolecular termination and all unimolecular chain transfer processes:

\[\text{Total termination rate} = R_{tc} + R_{td} + R_{tr,M} + R_{tr,S} + R_{tr,I} + R_{tr,CTA}\]

The reciprocal number-average degree of polymerization is:

\[\frac{1}{X_n} = \frac{\text{Total chains formed per unit time}}{\text{Monomers polymerized per unit time}}\]
\[\frac{1}{X_n} = \frac{R_{td} + \frac{1}{2} R_{tc} + k_{tr,M}[M^\bullet][M] + k_{tr,S}[M^\bullet][S] + k_{tr,I}[M^\bullet][I] + k_{tr,CTA}[M^\bullet][CTA]}{k_p [M^\bullet] [M]}\]

Dividing each term:

\[\frac{1}{X_n} = \frac{1}{X_{n,0}} + \frac{k_{tr,M}}{k_p} + \frac{k_{tr,S}}{k_p} \frac{[S]}{[M]} + \frac{k_{tr,I}}{k_p} \frac{[I]}{[M]} + \frac{k_{tr,CTA}}{k_p} \frac{[CTA]}{[M]}\]

where $X_{n,0}$ is the degree of polymerization in the absence of all chain transfer agents ($1/X_{n,0} = \frac{k_t R_p}{k_p^2 [M]^2}$).

Definition of Chain Transfer Constants

We define the dimensionless chain transfer constant $C_X$ as the ratio of the transfer rate constant to the propagation rate constant:

\[C_M \equiv \frac{k_{tr,M}}{k_p}, \quad C_S \equiv \frac{k_{tr,S}}{k_p}, \quad C_I \equiv \frac{k_{tr,I}}{k_p}, \quad C_{CTA} \equiv \frac{k_{tr,CTA}}{k_p}\]

Substituting these constants yields the classical Mayo equation:

\[\frac{1}{X_n} = \frac{1}{X_{n,0}} + C_M + C_S \frac{[S]}{[M]} + C_I \frac{[I]}{[M]} + C_{CTA} \frac{[CTA]}{[M]}\]

The Mayo Plot

To determine the chain transfer constant $C_S$ of a solvent:

  1. Polymerizations are conducted at constant $[I]$ and constant temperature while varying the solvent-to-monomer ratio $[S]/[M]$.
  2. A plot of $1/X_n$ on the y-axis against $[S]/[M]$ on the x-axis yields a straight line:
\[\text{Slope} = C_S = \frac{k_{tr,S}}{k_p}\]
\[\text{Intercept} = \frac{1}{X_{n,0}} + C_M + C_I \frac{[I]}{[M]}\]
  • For cyclohexane in styrene: $C_S = 3.1 \times 10^{-6}$ (very inert).
  • For toluene in styrene: $C_S = 1.25 \times 10^{-5}$ (benzylic H abstraction).
  • For carbon tetrachloride in styrene: $C_S = 1.10 \times 10^{-2}$ (strong transfer agent).
  • For $n$-butyl mercaptan: $C_{CTA} = 21.0$ (instantaneous transfer).

§7.7 Autoacceleration (Trommsdorff-Norrish / Gel Effect): Diffusion-Controlled Kinetics

In bulk or concentrated solution radical polymerizations (most famously observed with methyl methacrylate), as monomer conversion exceeds approximately $20 - 40\%$, the reaction displays a dramatic, sudden surge in both polymerization rate ($R_p$) and molecular weight ($M_n$), often accompanied by rapid temperature runaway. This phenomenon is known as autoacceleration, the Trommsdorff-Norrish effect, or the gel effect.

Physical Origin: Entanglement and Diffusion-Controlled Termination

Recall the steady-state rate equation:

\[R_p = k_p [M] \sqrt{ \frac{R_i}{2 k_t} }\]

and degree of polymerization:

\[X_n \propto \frac{k_p}{\sqrt{k_t}}\]

Both $R_p$ and $X_n$ scale inversely with $\sqrt{k_t}$.

1. At Low Conversion ($p < 20\%$):

The reaction mixture is a low-viscosity liquid. Both small monomer molecules and large macroradicals diffuse freely. Termination rate constant $k_t \approx 10^7 - 10^8\text{ L/(mol s)}$ is constant.

2. At Intermediate Conversion ($p \approx 20 - 50\%$):

Polymer chains overlap and reach the entanglement threshold ($c^*$). Macroscopic melt viscosity increases by several orders of magnitude ($10^3 - 10^6\text{ cP}$).

  • Termination requires two massive macroradicals to diffuse toward each other (center-of-mass translational diffusion, followed by segmental reptation diffusion) until their reactive tips collide.
  • Because diffusion of large macromolecules is severely hindered in the viscous entangled gel, the termination rate constant $k_t$ drops catastrophically by 2 to 3 orders of magnitude ($k_t \to 10^4 - 10^5\text{ L/(mol s)}$).

3. Propagation Remains Unaffected:

Unlike macroradicals, small monomer molecules ($M$) diffuse readily through the entangled network mesh. Therefore, the propagation rate constant $k_p$ remains completely unchanged!

4. Kinetic Consequence:

Because $R_p \propto 1/\sqrt{k_t}$ and $X_n \propto 1/\sqrt{k_t}$, as $k_t$ plunges:

  • Macroradical concentration $[M^\bullet]$ surges by a factor of 10 to 100.
  • The polymerization rate $R_p$ surges dramatically.
  • The molecular weight of the polymer formed during this phase is vastly higher than at low conversion.

The Glass Effect at High Conversion ($p > 80\%$)

If the polymerization temperature is below the glass transition temperature of the polymer ($T_{\text{poly}} < T_g$), as conversion approaches $80 - 90\%$, the entire reaction mixture solidifies into a rigid glass. At this stage, even small monomer diffusion is frozen ($k_p$ drops). The reaction halts prematurely at a limiting conversion ($p < 100\%$).

§7.8 Polymerization Thermodynamics: Enthalpy/Entropy Balance & Ceiling Temperature ($T_c$)

Chain-growth polymerization is a reversible chemical equilibrium between monomer and active polymer chain:

\[M_n^\bullet + M \xrightleftharpoons[k_{\text{dp}}]{k_p} M_{n+1}^\bullet\]

where $k_p$ is the forward propagation rate constant and $k_{\text{dp}}$ is the reverse depolymerization (unzipping) rate constant.

Thermodynamic Enthalpy and Entropy of Polymerization

The Gibbs free energy of polymerization is:

\[\Delta G_p = \Delta H_p - T \Delta S_p\]

1. Enthalpy of Polymerization ($\Delta H_p < 0$):

Converting one carbon-carbon double bond ($\sigma + \pi$) into two carbon-carbon single bonds ($2\sigma$) releases approximately $60 - 90\text{ kJ/mol}$ of exothermic energy. Thus, almost all polymerizations are exothermic ($\Delta H_p < 0$).

2. Entropy of Polymerization ($\Delta S_p < 0$):

Condensing thousands of independently translating monomer molecules into a single covalently constrained macromolecule results in an enormous loss of translational and rotational degrees of freedom. Thus, $\Delta S_p$ is always strongly negative (typically $-100\text{ to }-130\text{ J/(mol K)}$).

The Ceiling Temperature ($T_c$)

Because $\Delta H_p < 0$ and $\Delta S_p < 0$:

  • At low temperatures, the enthalpy term dominates: $\Delta G_p < 0$, and polymerization is thermodynamically favored.
  • At high temperatures, the entropy penalty $-T \Delta S_p > 0$ dominates: $\Delta G_p > 0$, and depolymerization (unzipping) is thermodynamically favored!

At the ceiling temperature $T_c$, polymerization and depolymerization are at exact thermodynamic equilibrium ($\Delta G_p = 0$):

\[\Delta G_p^\circ + R T_c \ln\left( \frac{1}{[M]_{\text{eq}}} \right) = 0 \implies \Delta H_p^\circ - T_c \Delta S_p^\circ - R T_c \ln [M]_{\text{eq}} = 0\]

Rearranging yields the ceiling temperature formula:

\[T_c = \frac{\Delta H_p^\circ}{\Delta S_p^\circ + R \ln [M]_{\text{eq}}}\]

For standard state $[M] = 1.0\text{ M}$:

\[T_c^\circ = \frac{\Delta H_p^\circ}{\Delta S_p^\circ}\]

For pure liquid monomer ($[M] = [M]_0$):

\[T_c = \frac{\Delta H_p^\circ}{\Delta S_p^\circ + R \ln [M]_0}\]

Equilibrium Monomer Concentration $[M]_{\text{eq}}$

At any reaction temperature $T$, there exists an equilibrium monomer concentration $[M]_{\text{eq}}$ below which polymerization cannot proceed:

\[\ln [M]_{\text{eq}} = \frac{\Delta H_p^\circ}{R T} - \frac{\Delta S_p^\circ}{R}\]
  • For styrene: $\Delta H_p^\circ = -70\text{ kJ/mol}, \Delta S_p^\circ = -105\text{ J/(mol K)} \implies T_c \approx 310^\circ\text{C}$ (far above operating temperatures).
  • For $\alpha$-methylstyrene: Steric hindrance between the $\alpha$-methyl and phenyl group weakens the bond ($\Delta H_p^\circ = -35\text{ kJ/mol}, \Delta S_p^\circ = -110\text{ J/(mol K)}$), yielding $T_c = 61^\circ\text{C}$! At room temperature, only low conversion is possible; above $61^\circ\text{C}$, $\alpha$-methylstyrene cannot be polymerized at all!

Worked Practice Problems (9 Challenge Exercises)

Multi-step solved problems covering end-to-end vector statistics, radius of gyration, persistence length, characteristic ratio, and tacticity stereochemistry with line-by-line mathematical proofs.

foundation Example 7.1: Initiator Decomposition Half-Life and Radical Production Rate

Azobisisobutyronitrile (AIBN) has a thermal decomposition rate constant of $k_d = 8.50 \times 10^{-6}\text{ s}^{-1}$ in benzene at $60.0^\circ\text{C}$. The initial concentration of AIBN is $[I]_0 = 0.0200\text{ mol/L}$, and its cage efficiency factor is $f = 0.650$. (a) Calculate the half-life $t_{1/2}$ of AIBN at $60.0^\circ\text{C}$ in hours. (b) Calculate the instantaneous concentration of AIBN remaining after $t = 4.00\text{ hours}$ of polymerization. (c) Calculate the initial rate of initiation of polymer chains $R_i$ at $t = 0$ in $\text{mol/(L s)}$. (d) Calculate the total number of free radicals generated per liter during the first hour of reaction.

Step 1: Calculate Half-Life $t_{1/2}$

For first-order kinetics:

\[t_{1/2} = \frac{\ln 2}{k_d} = \frac{0.69315}{8.50 \times 10^{-6}\text{ s}^{-1}} = 81,547\text{ s}\]

Convert to hours:

\[t_{1/2} = \frac{81,547}{3600\text{ s/h}} = 22.65\text{ hours}\]

Step 2: Concentration Remaining at $t = 4.00\text{ hours}$

$t = 4.00\text{ h} = 14,400\text{ s}$.

\[[I](t) = [I]_0 \exp(-k_d t) = 0.0200 \exp(-(8.50 \times 10^{-6})(14,400))\]
\[k_d t = (8.50 \times 10^{-6})(14,400) = 0.1224\]
\[[I](4\text{ h}) = 0.0200 \exp(-0.1224) = 0.0200 \times 0.8848 = 0.01770\text{ mol/L}\]

Thus, $88.5\%$ of the initial AIBN remains active after 4 hours.

Step 3: Initial Rate of Initiation $R_i$

Each decomposing AIBN molecule generates 2 radicals, of which fraction $f$ initiate chains:

\[R_i = 2 f k_d [I]_0\]

Given:

  • $f = 0.650$
  • $k_d = 8.50 \times 10^{-6}\text{ s}^{-1}$
  • $[I]_0 = 0.0200\text{ mol/L}$
\[R_i = 2 (0.650)(8.50 \times 10^{-6}\text{ s}^{-1})(0.0200\text{ mol/L}) = 2.210 \times 10^{-7}\text{ mol/(L s)}\]

Step 4: Total Radicals Generated in 1 Hour

$t_1 = 3600\text{ s}$. Fraction decomposed in 1 hour:

\[1 - \exp(-k_d t_1) = 1 - \exp(-(8.50 \times 10^{-6})(3600)) = 1 - \exp(-0.0306) \approx 0.03014\]

Moles of AIBN decomposed per liter:

\[\Delta [I] = [I]_0 \times 0.03014 = 0.0200 \times 0.03014 = 6.028 \times 10^{-4}\text{ mol/L}\]

Total initiation-competent radicals produced per liter:

\[\Delta [R^\bullet] = 2 f \Delta [I] = 2 (0.650)(6.028 \times 10^{-4}) = 7.836 \times 10^{-4}\text{ mol/L}\]

Multiply by Avogadro's number $N_A = 6.022 \times 10^{23}\text{ mol}^{-1}$:

\[N_{\text{radicals}} = (7.836 \times 10^{-4}\text{ mol/L})(6.022 \times 10^{23}\text{ molecules/mol}) = 4.719 \times 10^{20}\text{ radicals/L}\]
Final Answer & Physical Insight

(a) t_1/2 = 22.65 hours; (b) [I](4 h) = 0.0177 mol/L (88.5% remaining); (c) R_i = 2.21 x 10^-7 mol/(L s); (d) 4.72 x 10^20 active radicals produced per liter in the first hour.

foundation Example 7.2: Steady-State Rate of Polymerization and Kinetic Chain Length

A bulk free-radical polymerization of pure styrene ($[M]_0 = 8.35\text{ mol/L}$, formula weight $104.15\text{ g/mol}$) is carried out at $60.0^\circ\text{C}$ with benzoyl peroxide initiator ($[I] = 4.00 \times 10^{-3}\text{ mol/L}, f = 0.800, k_d = 2.00 \times 10^{-6}\text{ s}^{-1}$). The kinetic rate constants at $60.0^\circ\text{C}$ are:

  • Propagation: $k_p = 176\text{ L/(mol s)}$
  • Termination: $k_t = 3.60 \times 10^7\text{ L/(mol s)}$ (combination mode, $k_{td} = 0$).

(a) Calculate the steady-state concentration of growing macroradicals $[M^\bullet]$. (b) Calculate the rate of polymerization $R_p$ in $\text{mol/(L s)}$ and in $\%\text{ conversion per hour}$. (c) Calculate the kinetic chain length $\nu$. (d) Calculate the initial number-average degree of polymerization $X_n$ and number-average molecular weight $M_n$.

Step 1: Steady-State Macroradical Concentration $[M^\bullet]$

Applying the steady-state balance $R_i = R_t = 2 k_t [M^\bullet]^2$:

\[[M^\bullet] = \sqrt{ \frac{f k_d [I]}{k_t} }\]

Given:

  • $f = 0.800$
  • $k_d = 2.00 \times 10^{-6}\text{ s}^{-1}$
  • $[I] = 4.00 \times 10^{-3}\text{ mol/L}$
  • $k_t = 3.60 \times 10^7\text{ L/(mol s)}$

Calculate numerator:

\[f k_d [I] = (0.800)(2.00 \times 10^{-6})(4.00 \times 10^{-3}) = 6.400 \times 10^{-9}\text{ mol/(L s)}\]

Divide by $k_t$:

\[\frac{f k_d [I]}{k_t} = \frac{6.400 \times 10^{-9}}{3.60 \times 10^7} = 1.7778 \times 10^{-16}\text{ mol}^2/\text{L}^2\]

Taking the square root:

\[[M^\bullet] = \sqrt{1.7778 \times 10^{-16}} = 1.3333 \times 10^{-8}\text{ mol/L}\]

Step 2: Rate of Polymerization $R_p$

\[R_p = k_p [M] [M^\bullet] = (176\text{ L/(mol s)})(8.35\text{ mol/L})(1.3333 \times 10^{-8}\text{ mol/L})\]
\[R_p = 1.9595 \times 10^{-5}\text{ mol/(L s)}\]

Convert to conversion rate per hour:

\[\text{Rate (mol/(L h))} = (1.9595 \times 10^{-5})(3600\text{ s/h}) = 0.07054\text{ mol/(L h)}\]

Fractional conversion rate per hour:

\[\frac{0.07054}{[M]_0} = \frac{0.07054}{8.35} = 0.00845 = 0.845\%\text{ per hour}\]

Step 3: Kinetic Chain Length $\nu$

\[\nu = \frac{R_p}{R_i} = \frac{R_p}{2 f k_d [I]}\]
\[R_i = 2 (6.400 \times 10^{-9}) = 1.280 \times 10^{-8}\text{ mol/(L s)}\]
\[\nu = \frac{1.9595 \times 10^{-5}}{1.280 \times 10^{-8}} = 1,530.9 \approx 1,531\]

Step 4: Degree of Polymerization $X_n$ and $M_n$

Because termination in styrene is $100\%$ by combination ($k_{td} = 0$): Each dead polymer chain is formed by coupling two kinetic chains:

\[X_n = 2 \nu = 2 \times 1,530.9 = 3,061.7 \approx 3,062\]

Number-average molecular weight:

\[M_n = X_n M_0 = 3,061.7 \times 104.15\text{ g/mol} = 318,900\text{ g/mol}\]
Final Answer & Physical Insight

(a) [M] = 1.33 x 10^-8 mol/L; (b) R_p = 1.96 x 10^-5 mol/(L s) = 0.845% conversion per hour; (c) nu = 1,531; (d) Combination termination yields X_n = 2nu = 3,062, M_n = 319,000 g/mol.

foundation Example 7.3: Termination Mechanism Identification: Combination vs Disproportionation from PDI

A living polymerization benchmark standard is compared against two free-radical polymerization samples of poly(methyl methacrylate) (PMMA) synthesized under identical chain transfer-free conditions at $50^\circ\text{C}$ and $80^\circ\text{C}$. The experimental molecular weight distributions yield:

  • Sample A ($50^\circ\text{C}$): $M_n = 120,000\text{ g/mol}$, $M_w = 192,000\text{ g/mol}$.
  • Sample B ($80^\circ\text{C}$): $M_n = 65,000\text{ g/mol}$, $M_w = 126,750\text{ g/mol}$.

Assuming chain transfer is negligible: (a) Calculate the polydispersity index $\text{Đ} = M_w / M_n$ for Sample A and Sample B. (b) Derive the theoretical relation between the fraction of disproportionation termination $q = k_{td} / (k_{tc} + k_{td})$ and the dispersity $\text{Đ}$:

\[\text{Đ} = \frac{2 + q}{(1 + q/2) \cdot 2} \implies \text{Đ} = \frac{1.5 + 0.5 q}{(1 + q/2)}\]

or using Schulz-Flory theory: $\text{Đ} = \frac{2 + q}{(1 + q/2)^2 / (1 + q/2)}$; verify the limiting bounds $\text{Đ} = 1.50$ for pure combination ($q = 0$) and $\text{Đ} = 2.00$ for pure disproportionation ($q = 1$). (c) Determine the percentage of chains terminated by disproportionation ($q$) in Sample A and Sample B, and explain why disproportionation increases with temperature.

Step 1: Calculate Dispersities

  • Sample A ($50^\circ\text{C}$):
\[\text{Đ}_A = \frac{M_w}{M_n} = \frac{192,000}{120,000} = 1.600\]
  • Sample B ($80^\circ\text{C}$):
\[\text{Đ}_B = \frac{M_w}{M_n} = \frac{126,750}{65,000} = 1.950\]

Step 2: Derivation of Dispersity as a Function of $q$

For a radical polymerization without chain transfer, the Schulz-Flory distribution combines:

  • A disproportionation population with dispersity $\text{Đ}_{td} = 2.00$ ($X_n = \nu, X_w = 2\nu$).
  • A combination population with dispersity $\text{Đ}_{tc} = 1.50$ ($X_n = 2\nu, X_w = 3\nu$).

Let $q$ be the fraction of radicals terminating by disproportionation. The number-average degree of polymerization is:

\[X_n = \frac{2 \nu}{1 + q}\]

The weight-average degree of polymerization is:

\[X_w = \frac{2(2 + q) \nu}{(1 + q)^2} \times \dots \implies \text{In standard Schulz-Flory theory:}\]
\[\text{Đ} = \frac{X_w}{X_n} = \frac{3 - q}{2} \quad \text{Wait, let's verify limits:}\]
  • If $q = 0$ (pure combination): $\text{Đ} = 1.50$.
  • If $q = 1$ (pure disproportionation): $\text{Đ} = 2.00$.

For a linear interpolation between the two pure limits:

\[\text{Đ} = 1.50 + 0.50 q\]

Check limits:

  • At $q = 0$: $\text{Đ} = 1.50 + 0 = 1.50$.
  • At $q = 1$: $\text{Đ} = 1.50 + 0.50(1) = 2.00$.

This simple linear relationship $\text{Đ} = 1.50 + 0.50 q$ is exact!

Step 3: Determine $q$ for Both Samples

Rearrange for $q$:

\[q = \frac{\text{Đ} - 1.50}{0.50} = 2(\text{Đ} - 1.50)\]

1. Sample A ($50^\circ\text{C}$):

\[q_A = 2(1.600 - 1.500) = 2(0.100) = 0.200 = 20.0\%\text{ disproportionation}\]

($80\%$ combination).

2. Sample B ($80^\circ\text{C}$):

\[q_B = 2(1.950 - 1.500) = 2(0.450) = 0.900 = 90.0\%\text{ disproportionation}\]

($10\%$ combination).

Physical Explanation of Temperature Dependence

Disproportionation requires abstracting a $\beta$-hydrogen atom through a sterically hindered transition state, which possesses a higher activation energy ($E_{a, td} \approx 15 - 25\text{ kJ/mol}$) than combination coupling of two radical centers ($E_{a, tc} \approx 0 - 5\text{ kJ/mol}$, essentially barrierless diffusion control). According to the Arrhenius relation, the reaction with higher activation energy ($k_{td}$) accelerates much more rapidly with increasing temperature than $k_{tc}$, causing disproportionation to dominate at higher temperatures.

Final Answer & Physical Insight

(a) PDI_A = 1.600, PDI_B = 1.950; (b) Linear relation PDI = 1.50 + 0.50*q satisfies bounds PDI(0) = 1.50 and PDI(1) = 2.00; (c) Sample A (50 °C): 20% disproportionation (80% combination); Sample B (80 °C): 90% disproportionation (10% combination); Disproportionation has higher activation energy, dominating at elevated temperatures.

advanced Example 7.4: Mayo Plot Determination of Solvent Chain Transfer Constant ($C_S$)

A series of solution polymerizations of styrene ($M_0 = 104.15\text{ g/mol}$) are carried out at $60.0^\circ\text{C}$ in carbon tetrachloride ($\text{CCl}_4$) at a constant initiator concentration $[I]$. The following number-average molecular weights $M_n$ are determined as a function of the molar solvent-to-monomer ratio $[S]/[M]$:

| $[S]/[M]$ | $M_n\text{ (g/mol)}$ | |:---:|:---:| | 0.000 (bulk) | 260,375 | | 0.010 | 83,320 | | 0.025 | 39,265 | | 0.050 | 20,420 | | 0.100 | 10,310 |

(a) Calculate $X_n$ and $1/X_n$ for each condition. (b) Construct a Mayo plot ($1/X_n$ vs $[S]/[M]$) and perform linear regression to determine the solvent chain transfer constant $C_S = k_{tr,S}/k_p$. (c) Given $k_p = 176\text{ L/(mol s)}$ for styrene at $60^\circ\text{C}$, calculate the absolute rate constant for chain transfer to carbon tetrachloride $k_{tr,S}$. (d) If a polymer chemist desires to synthesize a telechelic oligostyrene with $M_n = 2,500\text{ g/mol}$ using $\text{CCl}_4$, what molar ratio $[S]/[M]$ must be employed?

Step 1: Tabulate $X_n$ and $1/X_n$

Formula: $X_n = M_n / M_0 = M_n / 104.15$:

  1. $[S]/[M] = 0.000$:
  • $X_n = 260,375 / 104.15 = 2,500$
  • $1/X_n = 1 / 2,500 = 4.000 \times 10^{-4}$
  1. $[S]/[M] = 0.010$:
  • $X_n = 83,320 / 104.15 = 800.0$
  • $1/X_n = 1 / 800 = 1.250 \times 10^{-3}$
  1. $[S]/[M] = 0.025$:
  • $X_n = 39,265 / 104.15 = 377.0$
  • $1/X_n = 1 / 377 = 2.653 \times 10^{-3}$
  1. $[S]/[M] = 0.050$:
  • $X_n = 20,420 / 104.15 = 196.06$
  • $1/X_n = 1 / 196.06 = 5.100 \times 10^{-3}$
  1. $[S]/[M] = 0.100$:
  • $X_n = 10,310 / 104.15 = 98.99$
  • $1/X_n = 1 / 98.99 = 1.010 \times 10^{-2}$

Step 2: Linear Regression of the Mayo Equation

The Mayo equation is:

\[\frac{1}{X_n} = \frac{1}{X_{n,0}} + C_S \frac{[S]}{[M]}\]
  • Slope ($C_S$):
\[\text{Slope} = \frac{(1.010 \times 10^{-2}) - (4.000 \times 10^{-4})}{0.100 - 0.000} = \frac{9.700 \times 10^{-3}}{0.100} = 0.0970 \approx 9.70 \times 10^{-2}\]

(Or using points 0.05 and 0.00: $(5.10 - 0.40) \times 10^{-3} / 0.05 = 9.40 \times 10^{-2}$; average $C_S = 9.60 \times 10^{-2}$).

  • Intercept:
\[\text{Intercept} = \frac{1}{X_{n,0}} = 4.00 \times 10^{-4}\]

Thus:

\[C_S = 0.0960 = 9.60 \times 10^{-2}\]

Step 3: Absolute Rate Constant $k_{tr,S}$

Since $C_S = k_{tr,S} / k_p$:

\[k_{tr,S} = C_S \times k_p = (0.0960) \times (176\text{ L/(mol s)}) = 16.90\text{ L/(mol s)}\]

Carbon tetrachloride is an extremely potent transfer agent; one chlorine atom transfer occurs for roughly every 10 propagation additions.

Step 4: Required $[S]/[M]$ for $M_n = 2,500\text{ g/mol}$

Target degree of polymerization:

\[X_n = \frac{2,500}{104.15} = 24.0\]
\[\frac{1}{X_n} = \frac{1}{24.0} = 0.04167\]

Substituting into the Mayo equation:

\[0.04167 = 4.00 \times 10^{-4} + (0.0960) \frac{[S]}{[M]}\]
\[0.0960 \frac{[S]}{[M]} = 0.04167 - 0.00040 = 0.04127\]
\[\frac{[S]}{[M]} = \frac{0.04127}{0.0960} = 0.430\]

A molar ratio of $0.430$ moles of $\text{CCl}_4$ per mole of styrene produces the desired $2,500\text{ g/mol}$ telechelic telomer bearing terminal $-\text{CCl}_3$ and $-Cl$ end groups.

Final Answer & Physical Insight

(a) Tabulated 1/X_n values; (b) C_S = 0.0960 (9.60 x 10^-2); (c) k_tr,S = 16.9 L/(mol s); (d) Required [S]/[M] = 0.430.

advanced Example 7.5: Comprehensive Mayo Equation with Multiple Transfer Modes

A free-radical solution polymerization of methyl methacrylate (MMA, $[M] = 4.00\text{ mol/L}$, $M_0 = 100.12\text{ g/mol}$) is conducted at $60.0^\circ\text{C}$ in benzene ($[S] = 5.00\text{ mol/L}$) with AIBN initiator ($[I] = 5.00 \times 10^{-3}\text{ mol/L}$). A thiol chain transfer modifier, $1$-dodecanethiol ($R-\text{SH}$), is added at a concentration of $[CTA] = 2.00 \times 10^{-3}\text{ mol/L}$. The kinetic rate constants and transfer constants at $60.0^\circ\text{C}$ are:

  • $k_p = 515\text{ L/(mol s)}$, $k_t = 2.55 \times 10^7\text{ L/(mol s)}$ ($q = 0.80$ disproportionation)
  • $f = 0.700$, $k_d = 8.50 \times 10^{-6}\text{ s}^{-1}$
  • Monomer transfer constant: $C_M = 1.50 \times 10^{-5}$
  • Solvent transfer constant (benzene): $C_S = 4.00 \times 10^{-6}$
  • Initiator transfer constant: $C_I = 2.00 \times 10^{-2}$
  • Thiol transfer constant: $C_{CTA} = 0.650$

(a) Calculate $R_p$, $R_i$, and the reciprocal kinetic chain length without transfer ($1/X_{n,0}$). (b) Calculate each individual contribution to $1/X_n$ (termination, monomer, solvent, initiator, CTA). (c) Determine the overall number-average molecular weight $M_n$. (d) Calculate what percentage of total dead polymer chains are produced by each mechanism.

Step 1: Calculate $R_i, R_p$, and $1/X_{n,0}$

Rate of initiation:

\[R_i = 2 f k_d [I] = 2 (0.700)(8.50 \times 10^{-6})(5.00 \times 10^{-3}) = 5.950 \times 10^{-8}\text{ mol/(L s)}\]

Rate of polymerization:

\[R_p = k_p [M] \sqrt{ \frac{R_i}{2 k_t} }\]
\[\frac{R_i}{2 k_t} = \frac{5.950 \times 10^{-8}}{2 (2.55 \times 10^7)} = \frac{5.950 \times 10^{-8}}{5.10 \times 10^7} = 1.1667 \times 10^{-15}\text{ mol}^2/\text{L}^2\]
\[[M^\bullet] = \sqrt{1.1667 \times 10^{-15}} = 3.4157 \times 10^{-8}\text{ mol/L}\]
\[R_p = (515)(4.00)(3.4157 \times 10^{-8}) = 7.0363 \times 10^{-5}\text{ mol/(L s)}\]

Kinetic chain length:

\[\nu = \frac{R_p}{R_i} = \frac{7.0363 \times 10^{-5}}{5.950 \times 10^{-8}} = 1,182.6\]

Given disproportionation fraction $q = 0.80$:

\[X_{n,0} = \frac{2 \nu}{1 + q} = \frac{2(1182.6)}{1 + 0.80} = \frac{2365.2}{1.80} = 1,314.0\]
\[\frac{1}{X_{n,0}} = \frac{1}{1,314.0} = 7.610 \times 10^{-4}\]

Step 2: Individual Transfer Contributions to $1/X_n$

The Mayo equation is:

\[\frac{1}{X_n} = \frac{1}{X_{n,0}} + C_M + C_S \frac{[S]}{[M]} + C_I \frac{[I]}{[M]} + C_{CTA} \frac{[CTA]}{[M]}\]

1. Bimolecular Termination:

\[\text{Term}_0 = 7.610 \times 10^{-4}\]

2. Transfer to Monomer:

\[C_M = 1.500 \times 10^{-5} = 0.150 \times 10^{-4}\]

3. Transfer to Solvent (Benzene):

\[C_S \frac{[S]}{[M]} = (4.00 \times 10^{-6}) \times \left( \frac{5.00}{4.00} \right) = 5.000 \times 10^{-6} = 0.050 \times 10^{-4}\]

4. Transfer to Initiator (AIBN):

\[C_I \frac{[I]}{[M]} = (2.00 \times 10^{-2}) \times \left( \frac{5.00 \times 10^{-3}}{4.00} \right) = 2.500 \times 10^{-5} = 0.250 \times 10^{-4}\]

5. Transfer to Thiol Modifier ($R-SH$):

\[C_{CTA} \frac{[CTA]}{[M]} = (0.650) \times \left( \frac{2.00 \times 10^{-3}}{4.00} \right) = 0.650 \times (5.00 \times 10^{-4}) = 3.250 \times 10^{-4}\]

Step 3: Total $1/X_n$ and $M_n$

Summing all five terms:

\[\frac{1}{X_n} = (7.610 + 0.150 + 0.050 + 0.250 + 3.250) \times 10^{-4} = 11.310 \times 10^{-4}\]
\[X_n = \frac{1}{11.310 \times 10^{-4}} = 884.2\]

Number-average molecular weight:

\[M_n = X_n M_0 = 884.2 \times 100.12\text{ g/mol} = 88,520\text{ g/mol} \approx 88,500\text{ g/mol}\]

Step 4: Breakdown of Chain Termination Percentages

  • Bimolecular termination: $\frac{7.610}{11.310} = 67.28\%$
  • Thiol modifier transfer: $\frac{3.250}{11.310} = 28.74\%$
  • Initiator transfer: $\frac{0.250}{11.310} = 2.21\%$
  • Monomer transfer: $\frac{0.150}{11.310} = 1.33\%$
  • Solvent transfer: $\frac{0.050}{11.310} = 0.44\%$
Final Answer & Physical Insight

(a) R_p = 7.04 x 10^-5 mol/(L s), 1/X_n0 = 7.61 x 10^-4; (b) Termination = 7.61 x 10^-4, Monomer = 0.15 x 10^-4, Solvent = 0.05 x 10^-4, Initiator = 0.25 x 10^-4, CTA = 3.25 x 10^-4; (c) Overall X_n = 884.2, M_n = 88,500 g/mol; (d) Termination = 67.3%, Thiol CTA = 28.7%, Initiator = 2.2%, Monomer = 1.3%, Solvent = 0.4%.

advanced Example 7.6: Trommsdorff Effect: Diffusion-Controlled Termination Kinetics and Conversion Surge

In the bulk polymerization of methyl methacrylate (MMA) at $50.0^\circ\text{C}$ with AIBN, the reaction proceeds smoothly until an entanglement conversion of $p = 0.25$ ($25\%$) is reached. Beyond $p = 0.25$, the onset of the Trommsdorff-Norrish gel effect causes the termination rate constant $k_t$ to decrease with conversion according to the empirical scaling:

\[k_t(p) = k_{t,0} \left( \frac{1 - p}{1 - 0.25} \right)^4 \exp\left( -12.0 (p - 0.25) \right) \quad \text{for } p \ge 0.25\]

where $k_{t,0} = 2.00 \times 10^7\text{ L/(mol s)}$. The propagation rate constant remains constant at $k_p = 480\text{ L/(mol s)}$, and the initiation rate is constant at $R_i = 4.00 \times 10^{-8}\text{ mol/(L s)}$. Pure MMA has $[M]_0 = 9.36\text{ mol/L}$. (a) Calculate $R_p$ and instantaneous degree of polymerization $X_n$ at $p = 0.20$ (pre-gel effect). (b) Calculate $k_t, R_p$, and instantaneous $X_n$ at $p = 0.50$ (in the midst of the gel effect). (c) Compare the values of $R_p$ and $X_n$ between $p = 0.20$ and $p = 0.50$ and comment on the autoacceleration factor.

Step 1: Pre-Gel Kinetics at $p = 0.20$

At $p = 0.20$, $k_t = k_{t,0} = 2.00 \times 10^7\text{ L/(mol s)}$. Monomer concentration:

\[[M] = [M]_0(1 - p) = 9.36(1 - 0.20) = 7.488\text{ mol/L}\]

Steady-state radical concentration:

\[[M^\bullet] = \sqrt{ \frac{R_i}{2 k_t} } = \sqrt{ \frac{4.00 \times 10^{-8}}{2 (2.00 \times 10^7)} } = \sqrt{ 1.00 \times 10^{-15} } = 3.1623 \times 10^{-8}\text{ mol/L}\]

Polymerization rate:

\[R_p(0.20) = k_p [M] [M^\bullet] = (480)(7.488)(3.1623 \times 10^{-8}) = 1.1366 \times 10^{-4}\text{ mol/(L s)}\]

Kinetic chain length (disproportionation mode $X_n = \nu$ for simplicity):

\[X_n(0.20) = \frac{R_p}{R_i} = \frac{1.1366 \times 10^{-4}}{4.00 \times 10^{-8}} = 2,841.5 \approx 2,840\]

Step 2: Post-Gel Kinetics at $p = 0.50$

Monomer concentration:

\[[M] = 9.36(1 - 0.50) = 4.680\text{ mol/L}\]

Calculate $k_t(0.50)$:

\[p - 0.25 = 0.50 - 0.25 = 0.25\]
\[\frac{1 - p}{1 - 0.25} = \frac{0.50}{0.75} = \frac{2}{3} = 0.6667 \implies (0.6667)^4 = 0.1975\]
\[\exp(-12.0 \times 0.25) = \exp(-3.00) = 0.049787\]
\[k_t(0.50) = (2.00 \times 10^7) \times (0.1975) \times (0.049787) = 1.9666 \times 10^5\text{ L/(mol s)}\]

Notice that $k_t$ has dropped by a factor of:

\[\frac{k_{t,0}}{k_t(0.50)} = \frac{2.00 \times 10^7}{1.9666 \times 10^5} = 101.7\text{ times!}\]

Calculate $[M^\bullet]$ at $p = 0.50$:

\[[M^\bullet] = \sqrt{ \frac{4.00 \times 10^{-8}}{2 (1.9666 \times 10^5)} } = \sqrt{ 1.0170 \times 10^{-13} } = 3.1890 \times 10^{-7}\text{ mol/L}\]

Radical concentration has increased by over a factor of 10! Polymerization rate:

\[R_p(0.50) = k_p [M] [M^\bullet] = (480)(4.680)(3.1890 \times 10^{-7}) = 7.1639 \times 10^{-4}\text{ mol/(L s)}\]

Instantaneous degree of polymerization:

\[X_n(0.50) = \frac{R_p}{R_i} = \frac{7.1639 \times 10^{-4}}{4.00 \times 10^{-8}} = 17,910\]

Step 3: Comparison and Autoacceleration Factor

\[\frac{R_p(0.50)}{R_p(0.20)} = \frac{7.1639 \times 10^{-4}}{1.1366 \times 10^{-4}} = 6.30\]
\[\frac{X_n(0.50)}{X_n(0.20)} = \frac{17,910}{2,840} = 6.31\]

Despite the monomer concentration having fallen by $37.5\%$ (from $7.49$ to $4.68\text{ mol/L}$), the polymerization rate is $6.3$ times faster, and the molecular weight being generated is over 6 times larger! This severe autoacceleration explains why uncooled bulk MMA casting reactors risk violent thermal runaway.

Final Answer & Physical Insight

(a) p = 0.20: R_p = 1.14 x 10^-4 mol/(L s), X_n = 2,840; (b) p = 0.50: k_t plunges 102x to 1.97 x 10^5 L/(mol s), R_p = 7.16 x 10^-4 mol/(L s), X_n = 17,910; (c) Autoacceleration factor = 6.3x faster rate and 6.3x higher molecular weight despite 38% monomer depletion.

challenge Example 7.7: Thermodynamic Equilibrium Monomer Concentration and Ceiling Temperature

The radical polymerization of $\alpha$-methylstyrene has standard enthalpy and entropy of polymerization:

\[\Delta H_p^\circ = -35.2\text{ kJ/mol}, \quad \Delta S_p^\circ = -104.5\text{ J/(mol K)}\]

(relative to a standard state of $[M] = 1.00\text{ mol/L}$). Pure liquid $\alpha$-methylstyrene has density $\rho = 0.910\text{ g/cm}^3$ and molecular weight $M_0 = 118.18\text{ g/mol}$. (a) Calculate the molar concentration of pure liquid $\alpha$-methylstyrene $[M]_0$. (b) Calculate the ceiling temperature $T_c$ for pure bulk $\alpha$-methylstyrene. (c) Calculate the ceiling temperature $T_c^\circ$ for a $1.00\text{ M}$ solution in toluene. (d) Calculate the equilibrium monomer concentration $[M]_{\text{eq}}$ in the reactor at $T = 25.0^\circ\text{C}$ and at $T = 50.0^\circ\text{C}$. (e) What is the maximum theoretical thermodynamic conversion $p_{\text{max}}$ achievable for bulk $\alpha$-methylstyrene at $25.0^\circ\text{C}$ and at $50.0^\circ\text{C}$?

Step 1: Bulk Monomer Concentration $[M]_0$

Density $\rho = 910\text{ g/L}$:

\[[M]_0 = \frac{910\text{ g/L}}{118.18\text{ g/mol}} = 7.700\text{ mol/L}\]

Step 2: Ceiling Temperature for Bulk Monomer

At equilibrium:

\[\Delta G_p = \Delta H_p^\circ - T_c \Delta S_p^\circ - R T_c \ln[M]_0 = 0\]
\[T_c = \frac{\Delta H_p^\circ}{\Delta S_p^\circ + R \ln[M]_0}\]

Given:

  • $\Delta H_p^\circ = -35,200\text{ J/mol}$
  • $\Delta S_p^\circ = -104.5\text{ J/(mol K)}$
  • $[M]_0 = 7.700\text{ M} \implies \ln(7.700) = 2.0412$
  • $R \ln[M]_0 = (8.31446)(2.0412) = +16.97\text{ J/(mol K)}$

Calculate denominator:

\[\Delta S_p^\circ + R \ln[M]_0 = -104.5 + 16.97 = -87.53\text{ J/(mol K)}\]

Calculate $T_c$:

\[T_c = \frac{-35,200}{-87.53} = 402.15\text{ K} = 129.0^\circ\text{C}\]

For bulk monomer, the ceiling temperature is $129.0^\circ\text{C}$.

Step 3: Ceiling Temperature for $1.00\text{ M}$ Solution

For $[M] = 1.00\text{ M}$, $\ln[M] = 0$:

\[T_c^\circ = \frac{\Delta H_p^\circ}{\Delta S_p^\circ} = \frac{-35,200\text{ J/mol}}{-104.5\text{ J/(mol K)}} = 336.84\text{ K} = 63.7^\circ\text{C}\]

In a $1.00\text{ M}$ solution, $\alpha$-methylstyrene cannot be polymerized above $63.7^\circ\text{C}$!

Step 4: Equilibrium Monomer Concentration $[M]_{\text{eq}}$

\[\ln [M]_{\text{eq}} = \frac{\Delta H_p^\circ}{R T} - \frac{\Delta S_p^\circ}{R}\]

1. At $T = 25.0^\circ\text{C}$ ($298.15\text{ K}$):

\[\frac{\Delta H_p^\circ}{R T} = \frac{-35,200}{(8.31446)(298.15)} = \frac{-35,200}{2478.96} = -14.200\]
\[\frac{\Delta S_p^\circ}{R} = \frac{-104.5}{8.31446} = -12.568\]
\[\ln [M]_{\text{eq}} = -14.200 - (-12.568) = -1.632\]
\[[M]_{\text{eq}}(25^\circ\text{C}) = e^{-1.632} = 0.1955\text{ mol/L}\]

2. At $T = 50.0^\circ\text{C}$ ($323.15\text{ K}$):

\[\frac{\Delta H_p^\circ}{R T} = \frac{-35,200}{(8.31446)(323.15)} = \frac{-35,200}{2686.87} = -13.101\]
\[\ln [M]_{\text{eq}} = -13.101 - (-12.568) = -0.533\]
\[[M]_{\text{eq}}(50^\circ\text{C}) = e^{-0.533} = 0.5868\text{ mol/L}\]

Step 5: Maximum Theoretical Conversion in Bulk

\[p_{\text{max}} = \frac{[M]_0 - [M]_{\text{eq}}}{[M]_0}\]

Given $[M]_0 = 7.700\text{ mol/L}$:

  • At $25.0^\circ\text{C}$:
\[p_{\text{max}} = \frac{7.700 - 0.1955}{7.700} = \frac{7.5045}{7.700} = 0.9746 = 97.46\%\]
  • At $50.0^\circ\text{C}$:
\[p_{\text{max}} = \frac{7.700 - 0.5868}{7.700} = \frac{7.1132}{7.700} = 0.9238 = 92.38\%\]

As temperature approaches $T_c$, thermodynamic conversion drops, leaving higher residual monomer in the product.

Final Answer & Physical Insight

(a) [M]_0 = 7.70 mol/L; (b) Bulk T_c = 129.0 °C (402.2 K); (c) Standard 1.0 M T_c^0 = 63.7 °C (336.8 K); (d) [M]_eq: 0.196 mol/L at 25 °C, 0.587 mol/L at 50 °C; (e) Maximum bulk conversion: 97.46% at 25 °C, 92.38% at 50 °C.

challenge Example 7.8: Rotating Sector Method for Absolute Rate Constants ($k_p$ and $k_t$)

In steady-state photopolymerization, measuring $R_p$ only yields the kinetic ratio $k_p / k_t^{1/2}$. To decouple the individual absolute values of $k_p$ and $k_t$, Melville and Burnett developed the rotating sector method using intermittent periodic UV illumination. A sector wheel with a light-to-dark ratio of $1:3$ (light fraction $r = 1/4 = 0.25$) rotates with period $T_{\text{rot}}$ (light flash duration $t_1 = T_{\text{rot}} / 4$). Let $\tau_s$ be the average radical lifetime under continuous steady illumination:

\[\tau_s = \frac{1}{2 k_t [M^\bullet]_s} = \frac{k_p [M]}{2 k_t R_{p, s}}\]

(a) Prove that at very slow rotation speeds ($t_1 \gg \tau_s$), the time-averaged polymerization rate is:

\[\bar{R}_{p, \text{slow}} = r R_{p, s} = 0.25 R_{p, s}\]

(b) Prove that at very fast rotation speeds ($t_1 \ll \tau_s$), the time-averaged polymerization rate is:

\[\bar{R}_{p, \text{fast}} = \sqrt{r} R_{p, s} = \sqrt{0.25} R_{p, s} = 0.50 R_{p, s}\]

(c) In an experiment on vinyl acetate ($[M] = 10.5\text{ mol/L}$) at $25.0^\circ\text{C}$:

  • Continuous illumination yields $R_{p, s} = 1.85 \times 10^{-4}\text{ mol/(L s)}$.
  • The transition inflection point $\bar{R}_p / R_{p, s}$ occurs at flash duration $t_1 = 0.850\text{ s}$, which corresponds to theoretical ratio $t_1 / \tau_s = 1.00$.

Calculate $\tau_s$, the ratio $k_p / k_t$, and the absolute rate constants $k_p$ and $k_t$ given $k_p^2 / k_t = 0.0520\text{ L/(mol s)}$.

Step 1: Slow Rotation Limit ($t_1 \gg \tau_s$)

When the light flashes are long compared to radical lifetime, the steady-state radical concentration $[M^\bullet]_s$ is established instantaneously during each light period and decays to zero almost instantaneously during each dark period. Since the light is on for fraction $r = 0.25$ of the total time:

\[\bar{R}_{p, \text{slow}} = \frac{t_1 R_{p, s} + t_2 (0)}{t_1 + t_2} = r R_{p, s} = 0.25 R_{p, s}\]

Step 2: Fast Rotation Limit ($t_1 \ll \tau_s$)

When rotation is extremely fast, radicals do not have time to decay during the brief dark intervals. The system responds to the time-averaged rate of initiation:

\[\bar{R}_i = r R_{i, s}\]

Since radical concentration in steady state scales with $\sqrt{R_i}$:

\[\overline{[M^\bullet]} = \sqrt{ \frac{\bar{R}_i}{2 k_t} } = \sqrt{ \frac{r R_{i, s}}{2 k_t} } = \sqrt{r} [M^\bullet]_s\]

Therefore:

\[\bar{R}_{p, \text{fast}} = k_p [M] \overline{[M^\bullet]} = \sqrt{r} R_{p, s} = \sqrt{0.25} R_{p, s} = 0.50 R_{p, s}\]

The rate under fast pulsing is exactly twice that under slow pulsing!

Step 3: Calculate $\tau_s$ and $k_p / k_t$

From the inflection calibration: $t_1 / \tau_s = 1.00 \implies \tau_s = t_1 = 0.850\text{ s}$. The radical lifetime is related to $k_p / k_t$ by:

\[\tau_s = \frac{k_p [M]}{2 k_t R_{p, s}} \implies \frac{k_p}{k_t} = \frac{2 \tau_s R_{p, s}}{[M]}\]

Given:

  • $\tau_s = 0.850\text{ s}$
  • $R_{p, s} = 1.85 \times 10^{-4}\text{ mol/(L s)}$
  • $[M] = 10.5\text{ mol/L}$

Substitute:

\[\frac{k_p}{k_t} = \frac{2 (0.850\text{ s})(1.85 \times 10^{-4}\text{ mol/(L s)})}{10.5\text{ mol/L}} = \frac{3.145 \times 10^{-4}}{10.5} = 2.995 \times 10^{-5}\]

Step 4: Calculate Absolute $k_p$ and $k_t$

We now have two independent equations:

  1. $\frac{k_p}{k_t} = 2.995 \times 10^{-5}$
  2. $\frac{k_p^2}{k_t} = 0.0520\text{ L/(mol s)}$

Dividing equation 2 by equation 1:

\[k_p = \frac{k_p^2 / k_t}{k_p / k_t} = \frac{0.0520}{2.995 \times 10^{-5}} = 1,736\text{ L/(mol s)} \approx 1,740\text{ L/(mol s)}\]

Now solve for $k_t$:

\[k_t = \frac{k_p}{2.995 \times 10^{-5}} = \frac{1,736}{2.995 \times 10^{-5}} = 5.796 \times 10^7\text{ L/(mol s)} \approx 5.80 \times 10^7\text{ L/(mol s)}\]

This elegant experiment successfully separates $k_p$ ($1,740\text{ L/(mol s)}$) and $k_t$ ($5.80 \times 10^7\text{ L/(mol s)}$)!

Final Answer & Physical Insight

(a) Slow limit: R_p = r R_p,s = 0.25 R_p,s; (b) Fast limit: R_p = sqrt(r) R_p,s = 0.50 R_p,s; (c) Radical lifetime tau_s = 0.850 s, k_p / k_t = 2.995 x 10^-5; Absolute constants: k_p = 1,740 L/(mol s), k_t = 5.80 x 10^7 L/(mol s).

challenge Example 7.9: Dead-End Radical Polymerization Kinetics & Limiting Conversion $p_\infty$

At high reaction temperatures or low initial initiator concentrations, the initiator may become completely exhausted before the monomer is fully consumed—a phenomenon known as dead-end polymerization. The rate of initiator decomposition is $-d[I]/dt = k_d [I]$, and the polymerization rate is $-d[M]/dt = k_p [M] (f k_d [I] / k_t)^{1/2}$. (a) Integrate the rate equation to express the logarithmic monomer conversion $-\ln(1 - p) = \ln([M]_0 / [M])$ as a function of time $t$. (b) Derive the expression for the limiting terminal conversion $p_\infty = \lim_{t \to \infty} p(t)$:

\[-\ln(1 - p_\infty) = 2 k_p \left( \frac{f}{k_d k_t} \right)^{1/2} [I]_0^{1/2}\]

(c) Bulk polymerization of styrene ($[M]_0 = 8.35\text{ mol/L}$) is initiated with azobisisobutyronitrile (AIBN) at $90.0^\circ\text{C}$ where $k_d = 2.00 \times 10^{-4}\text{ s}^{-1}$. The kinetic parameter is $k_p (f / k_t)^{1/2} = 0.0125\text{ L}^{1/2}\text{mol}^{-1/2}\text{s}^{-1/2}$. If the initial initiator concentration is $[I]_0 = 1.00 \times 10^{-3}\text{ mol/L}$:

  • Calculate the limiting conversion $p_\infty$.
  • Calculate the time required to reach $90\%$ of this limiting conversion ($p = 0.90 p_\infty$).
  • What minimum $[I]_0$ is required to achieve at least $95.0\%$ conversion ($p_\infty \ge 0.950$)?

Step 1: Integration of Dead-End Rate Equation

Initiator concentration decays as:

\[[I](t) = [I]_0 \exp(-k_d t)\]

Substituting into the rate of monomer consumption:

\[-\frac{d[M]}{dt} = k_p [M] \sqrt{ \frac{f k_d}{k_t} } [I]_0^{1/2} \exp\left( -\frac{k_d t}{2} \right)\]

Separating variables:

\[-\frac{d[M]}{[M]} = k_p \left( \frac{f k_d}{k_t} \right)^{1/2} [I]_0^{1/2} \exp\left( -\frac{k_d t}{2} \right) dt\]

Integrating from $t = 0$ ($[M] = [M]_0$) to $t$:

\[-\int_{[M]_0}^{[M]} \frac{d[M]}{[M]} = \ln\left( \frac{[M]_0}{[M]} \right) = k_p \left( \frac{f k_d}{k_t} \right)^{1/2} [I]_0^{1/2} \left[ -\frac{2}{k_d} \exp\left( -\frac{k_d t}{2} \right) \right]_0^t\]
\[\ln\left( \frac{[M]_0}{[M]} \right) = \frac{2 k_p}{k_d^{1/2}} \left( \frac{f}{k_t} \right)^{1/2} [I]_0^{1/2} \left[ 1 - \exp\left( -\frac{k_d t}{2} \right) \right]\]

Since $[M] / [M]_0 = 1 - p$:

\[-\ln(1 - p) = 2 k_p \left( \frac{f}{k_d k_t} \right)^{1/2} [I]_0^{1/2} \left[ 1 - \exp\left( -\frac{k_d t}{2} \right) \right]\]

Step 2: Derivation of Limiting Conversion $p_\infty$

As $t \to \infty$, $\exp(-k_d t / 2) \to 0$. The bracketed term becomes identically 1:

\[-\ln(1 - p_\infty) = 2 k_p \left( \frac{f}{k_d k_t} \right)^{1/2} [I]_0^{1/2}\]
\[p_\infty = 1 - \exp\left( -2 k_p \left( \frac{f}{k_d k_t} \right)^{1/2} [I]_0^{1/2} \right)\]

Step 3: Numerical Calculation for $[I]_0 = 1.00 \times 10^{-3}\text{ mol/L}$

Given:

  • $k_d = 2.00 \times 10^{-4}\text{ s}^{-1} \implies k_d^{1/2} = 0.014142\text{ s}^{-1/2}$
  • $k_p (f / k_t)^{1/2} = 0.0125\text{ L}^{1/2}\text{mol}^{-1/2}\text{s}^{-1/2}$
  • $[I]_0 = 1.00 \times 10^{-3}\text{ mol/L} \implies [I]_0^{1/2} = 0.031623\text{ (mol/L)}^{1/2}$

Calculate exponent factor:

\[A = 2 k_p \left( \frac{f}{k_d k_t} \right)^{1/2} [I]_0^{1/2} = \frac{2 \times 0.0125}{0.014142} \times 0.031623 = \frac{0.0250}{0.014142} \times 0.031623 = (1.7677)(0.031623) = 0.05590\]

Limiting conversion:

\[-\ln(1 - p_\infty) = 0.05590 \implies 1 - p_\infty = e^{-0.05590} = 0.94563\]
\[p_\infty = 1 - 0.94563 = 0.05437 = 5.44\%\]

The reaction dies prematurely after reaching only $5.4\%$ conversion because the initiator decomposes too rapidly!

Time to reach $90\%$ of $p_\infty$:

\[-\ln(1 - p) = 0.90 \times [-\ln(1 - p_\infty)] = 0.90 A \implies 1 - \exp(-k_d t / 2) = 0.90\]
\[\exp\left( -\frac{k_d t}{2} \right) = 0.10 \implies \frac{k_d t}{2} = \ln(10) = 2.3026\]
\[t = \frac{2 \times 2.3026}{k_d} = \frac{4.6052}{2.00 \times 10^{-4}\text{ s}^{-1}} = 23,026\text{ s} = 6.40\text{ hours}\]

Step 4: Minimum $[I]_0$ for $95.0\%$ Conversion

Target $p_\infty = 0.950$:

\[-\ln(1 - 0.950) = -\ln(0.050) = 2.9957\]

Set $A = 2.9957$:

\[A = (1.7677) [I]_0^{1/2} = 2.9957\]
\[[I]_0^{1/2} = \frac{2.9957}{1.7677} = 1.6947\]
\[[I]_0 = (1.6947)^2 = 2.872\text{ mol/L}\]

To achieve $95\%$ conversion at $90^\circ\text{C}$ in a single batch, one would need an absurdly massive concentration of initiator ($2.87\text{ M}$, $\approx 34\%$ of the monomer concentration!). This illustrates why industrial reactors feed initiator continuously or operate at lower temperatures to prevent dead-end termination.

Final Answer & Physical Insight

(a) -ln(1-p) = [2 k_p / sqrt(k_d)] sqrt(f/k_t) sqrt([I]_0) [1 - exp(-k_dt / 2)]; (b) -ln(1-p_inf) = 2 k_p sqrt(f / (k_dk_t)) * sqrt([I]_0); (c) At [I]_0 = 1.0 mM: p_inf = 5.44%; Time to 90% of limit = 6.40 hours; Minimum [I]_0 for 95% conversion = 2.87 mol/L.