§6.1 Fundamentals of Step-Growth Polymerization: Mechanisms & Reversible Equilibrium
Step-growth polymerization occurs between bifunctional or polyfunctional monomers bearing complementary reactive functional groups. Unlike addition chain polymerizations, any two molecular species present in the reaction vessel—monomers, dimers, trimers, or long oligomers—can undergo condensation with each other at any stage during the synthesis.
Reaction Classifications: $A-B$ vs $A-A + B-B$
Step-growth systems are classified by their monomer stoichiometry:
1. $A-B$ Monomer Systems:
- A single monomer molecule possesses both complementary functional groups in equal stoichiometric 1:1 proportion within the same molecule.
- Examples: $\omega$-amino acids in polyamide synthesis (e.g., 6-aminocaproic acid forming Nylon 6), or $\omega$-hydroxy acids forming polyesters (e.g., glycolic acid forming polyglycolide).
- Strict equimolar stoichiometry is chemically guaranteed.
2. $A-A + B-B$ Monomer Systems:
- Polymerization occurs between two distinct bifunctional monomers, one bearing two $A$ functional groups and the other bearing two $B$ functional groups.
- Examples:
- Polyamides: Hexamethylenediamine ($H_2N-(CH_2)_6-NH_2$, $A-A$) reacting with adipic acid ($HOOC-(CH_2)_4-COOH$, $B-B$) to form Nylon 6,6.
- Polyesters: Ethylene glycol ($HO-CH_2CH_2-OH$, $A-A$) reacting with dimethyl terephthalate ($B-B$) to form poly(ethylene terephthalate) (PET).
- Polycarbonates: Bisphenol A reacting with phosgene ($COCl_2$) or diphenyl carbonate.
- Polyurethanes: Diisocyanates ($OCN-R-NCO$) reacting with diols ($HO-R'-OH$) without elimination of small molecule by-products (polyaddition).
Reversible Thermodynamic Equilibrium & Le Chatelier Removal
Most condensation reactions (esterification, amidation, transesterification) are reversible equilibria:
The equilibrium constant is:
For typical polyesterifications, $K$ is modest ($K \approx 1 - 10$). For polyamidations, $K \approx 100 - 400$. To drive the reaction to the high conversions ($p > 0.99$) required for engineering-grade mechanical properties, the low-molecular-weight condensation by-product ($\text{H}_2\text{O}, \text{CH}_3\text{OH}, \text{HCl}$) must be continuously removed from the melt under high vacuum ($< 1\text{ mbar}$) at elevated temperatures ($200 - 280^\circ\text{C}$) with vigorous mechanical agitation.
§6.2 Carothers Equation for Linear Systems: Conversion, $X_n$ & High-Conversion Imperative
In 1929, Wallace Carothers formulated the fundamental mathematical relationship between the fractional conversion of reactive functional groups and the resulting average degree of polymerization in step-growth systems.
Mathematical Derivation of the Carothers Equation
Consider an equimolar linear polymerization ($A-B$ or stoichiometric $A-A + B-B$). Let $N_0$ be the total number of monomer molecules initially present. Because each monomer molecule has 2 functional groups, the initial number of functional groups of each type is $N_0$. At reaction time $t$, let $N$ be the total number of macromolecular chains (molecules) remaining in the reaction mixture. Each reaction event between two functional groups forms one new covalent bond and reduces the total number of separate molecules by exactly one:
Since forming one inter-unit bond consumes two functional groups (one $A$ and one $B$), the total number of functional groups consumed is $2(N_0 - N)$. The fractional conversion $p$ of functional groups is defined as the fraction of initial functional groups that have reacted:
Rearranging gives the ratio of initial to remaining molecules:
The number-average degree of polymerization $X_n$ is defined as the average number of monomer units per macromolecule:
Substituting $N/N_0 = 1 - p$ yields the celebrated Carothers equation:
The High-Conversion Imperative
The Carothers equation demonstrates the mathematical difficulty of synthesizing high molecular weight polymers via step-growth:
- At $p = 0.50$ ($50\%$ conversion): $X_n = 1 / (1 - 0.50) = 2$ (only dimers on average).
- At $p = 0.80$ ($80\%$ conversion): $X_n = 1 / (1 - 0.80) = 5$ (short oligomers).
- At $p = 0.90$ ($90\%$ conversion): $X_n = 1 / (1 - 0.90) = 10$.
- At $p = 0.98$ ($98\%$ conversion): $X_n = 1 / (1 - 0.98) = 50$.
- At $p = 0.99$ ($99\%$ conversion): $X_n = 1 / (1 - 0.99) = 100$.
- At $p = 0.999$ ($99.9\%$ conversion): $X_n = 1 / (1 - 0.999) = 1,000$.
Polymers typically develop useful structural, mechanical, and tensile properties (fiber forming, impact resistance, film toughness) only when $X_n \ge 100$. Therefore, step-growth industrial processes must achieve conversions exceeding $99.0\%$ to $99.9\%$, which requires near-perfect monomer purity, precise stoichiometric balance, and total by-product removal.
§6.3 Stoichiometric Imbalance ($r$) & Monofunctional Chain Stoppers: Molecular Weight Control
In actual industrial synthesis, achieving $X_n = \infty$ is neither desirable nor processable; extremely high molecular weights lead to unworkable melt viscosities. Molecular weight is precisely controlled and limited through stoichiometric imbalance or the intentional addition of monofunctional chain stoppers.
Derivation of the Modified Carothers Equation with Imbalance
Consider an $A-A + B-B$ system where $A$ groups are in deficiency relative to $B$ groups. Define the stoichiometric ratio:
where $N_A$ and $N_B$ are the initial number of $A$ and $B$ functional groups, respectively ($N_A = 2 N_{AA}$ and $N_B = 2 N_{BB}$). The total number of monomer molecules initially present is:
Let $p$ be the fractional conversion of the minority group ($A$ groups). Number of reacted $A$ groups $= p N_A = p r N_B$. Because one $B$ group reacts for every $A$ group, the number of reacted $B$ groups is also $p r N_B$. Total number of unreacted functional groups remaining:
Since each remaining linear polymer chain terminates with two functional groups (one at each end), the total number of polymer molecules $N$ remaining is half the number of remaining functional groups:
The number-average degree of polymerization $X_n = N_0 / N$ is:
Asymptotic Limit at Complete Conversion ($p \to 1$)
When all minority groups are completely consumed ($p = 1.0$), all chain ends are capped with the excess $B$ functional group, terminating further polymerization:
- If $r = 1.0$ (perfect stoichiometry): $X_{n, \text{max}} \to \infty$.
- If $r = 0.99$ ($1\%$ molar excess of $B-B$): $X_{n, \text{max}} = \frac{1 + 0.99}{1 - 0.99} = \frac{1.99}{0.01} = 199$.
- If $r = 0.98$ ($2\%$ molar excess of $B-B$): $X_{n, \text{max}} = \frac{1 + 0.98}{1 - 0.98} = \frac{1.98}{0.02} = 99$.
- If $r = 0.95$ ($5\%$ molar excess): $X_{n, \text{max}} = \frac{1.95}{0.05} = 39$.
Monofunctional Chain Stoppers
Adding a monofunctional compound $R-B$ (such as acetic acid in Nylon 6,6 synthesis) has the exact same stoichiometric capping effect. Each molecule of $R-B$ provides one $B$ group. The effective stoichiometric ratio becomes:
where $N_{B'}$ is the number of monofunctional chain stopper molecules. The factor of 2 accounts for the fact that one monofunctional stopper eliminates one chain end without supplying a second reactive site for chain growth.
§6.4 Kinetics of Step-Growth: Self-Catalyzed vs Acid-Catalyzed Polyesterification
The kinetics of step-growth polymerization depend fundamentally on whether the reaction requires an added external catalyst or is self-catalyzed by one of the reacting monomer functional groups.
Case 1: Self-Catalyzed Polyesterification
In the absence of an added strong acid catalyst, polyesterification between a dicarboxylic acid and a diol is catalyzed by the carboxylic acid monomer itself:
The rate of disappearance of carboxylic acid groups is third-order overall:
For equimolar initial concentrations $c_0 = [\text{COOH}]_0 = [\text{OH}]_0$, at conversion $p$:
Substituting into the rate equation:
Separating variables and integrating from $t = 0$ ($c = c_0$) to $t$:
Substituting $c = c_0(1 - p)$:
Since $X_n = 1 / (1 - p)$:
Thus, for self-catalyzed polyesterification, a plot of $\frac{1}{(1 - p)^2}$ (or $X_n^2$) versus time $t$ is strictly linear with slope $2 c_0^2 k$. The degree of polymerization scales as the square root of time ($X_n \propto t^{1/2}$).
Case 2: Externally Acid-Catalyzed Polyesterification
When a catalytic amount of strong mineral or sulfonic acid (such as $p$-toluenesulfonic acid, $\text{H}_2\text{SO}_4$, or titanium alkoxide) is added, the proton concentration $[\text{H}^+]$ remains constant:
For stoichiometric concentrations $c = c_0(1 - p)$:
Integrating this second-order rate equation:
Since $X_n = 1 / (1 - p)$:
In externally catalyzed step-growth, $X_n$ grows linearly with time ($X_n \propto t$). This linear kinetics achieves target molecular weights dramatically faster than the self-catalyzed route.
§6.5 Flory's Principle of Equal Reactivity of Functional Groups
A cornerstone of macromolecular kinetic theory is Flory's principle of equal reactivity of functional groups, proposed by Paul Flory in 1939.
Formal Statement
The intrinsic chemical reactivity of a functional group in a step-growth reaction is independent of the size of the macromolecule to which it is attached.
That is:
where $k_{x,y}$ is the rate constant for the reaction between an $x$-mer and a $y$-mer.
Theoretical Justification: Diffusion in Condensed Phase
At first glance, Flory's principle seems counterintuitive: as polymer chains grow longer, their center-of-mass translational diffusion coefficient decreases dramatically ($D_{\text{cm}} \propto M^{-1}$ or $M^{-2}$). One might expect large coils to react much more slowly. However, in condensed liquids, reaction rate is governed by the collision rate inside a solvent cage:
- Although a large macromolecule diffuses into a new encounter volume slowly, once two coils overlap, their terminal functional groups remain trapped in close proximity inside the same solvent cage for an extended duration.
- The collision frequency between reactive end groups within the cage is determined entirely by local segmental mobility (bond rotation, segmental tumbling), not by whole-molecule translation.
- Because the local steric and electronic chemical environment of a carboxylic acid or hydroxyl group on a long flexible chain is virtually indistinguishable from that on a short chain, the intrinsic rate constant $k$ remains unchanged.
Experimental Evidence
Flory validated this principle by measuring the rate constants for esterification of homologous series of monocarboxylic and dicarboxylic acids:
- Formic acid ($n = 1$): $k = 4.70 \times 10^{-4}$
- Acetic acid ($n = 2$): $k = 2.45 \times 10^{-4}$
- Propionic acid ($n = 3$): $k = 1.95 \times 10^{-4}$
- Butyric acid ($n = 4$): $k = 1.98 \times 10^{-4}$
- Caproic acid ($n = 6$): $k = 2.01 \times 10^{-4}$
- Adipic acid to sebacic acid ($n = 6 \to 10$): $k = 2.00 \times 10^{-4}\text{ L/(mol s)}$
Beyond the first two or three carbons where electronic induction from adjacent groups is felt, the rate constants reach an exact invariant plateau.
§6.6 Molecular Weight Distributions: Flory-Schulz 'Most Probable' Distribution ($N_x, W_x$)
Because every functional group has identical reactivity throughout the reaction, the distribution of chain lengths in linear step-growth polymerization can be derived purely from probability theory.
Derivation of the Number Fraction Distribution ($N_x$)
Consider a linear polymer formed by step-growth with conversion $p$.
- The probability that a given functional group has reacted is $p$.
- The probability that a functional group has not reacted is $1 - p$.
An $x$-mer (a chain consisting of exactly $x$ monomer units) contains:
- Exactly $x - 1$ reacted bonds (each with probability $p$).
- Exactly one unreacted end group terminating the chain (with probability $1 - p$).
Applying the multiplication rule of independent probabilities, the probability that a randomly chosen chain contains exactly $x$ units—which equals the mole fraction (number fraction) $N_x$ of $x$-mers—is:
This is the celebrated Flory-Schulz most probable distribution (geometric distribution). Checking normalization:
Derivation of the Weight Fraction Distribution ($W_x$)
The weight fraction $W_x$ of $x$-mers is the mass of all $x$-mers divided by total sample mass. Neglecting end-group mass differences, the mass of an $x$-mer is proportional to $x$:
Substituting $N_x = (1 - p) p^{x-1}$ and $X_n = 1 / (1 - p)$:
Checking normalization:
Molecular Weight Averages and Dispersity
1. Number-Average Degree of Polymerization:
2. Weight-Average Degree of Polymerization:
3. Polydispersity Index (Dispersity $\text{Đ}$):
As conversion approaches unity ($p \to 1.0$):
Thus, linear step-growth polycondensation has a theoretical limiting dispersity of exactly 2.0.
§6.7 Non-Linear Step-Growth & Cross-Linking: Branching, Polyfunctionality & Networks
When monomers with functionality greater than two ($f \ge 3$) are introduced into a step-growth mixture, the polymer chains branch in multiple spatial dimensions, leading to hyperbranched polymers or macroscopic three-dimensional cross-linked network gels.
Average Functionality of Monomer Mixtures
For a reaction mixture containing $N_i$ molecules of monomer $i$, each bearing functionality $f_i$: The average functionality $f_{\text{avg}}$ is defined as:
- If $f_{\text{avg}} = 2.0$: strictly linear polymers are formed.
- If $f_{\text{avg}} > 2.0$: branched structures form, which may reach a critical point of infinite molecular weight (the gel point).
For stoichiometric systems where $A$ and $B$ functional groups are present in stoichiometric balance ($N_A = N_B$):
Topological Regimes of Non-Linear Polymerization
1. Pre-Gel Regime ($p < p_c$):
- The reaction mixture consists of soluble branched molecules, dendrimer-like structures, and unreacted monomers.
- The mixture remains completely soluble in appropriate solvents and flows as a viscous liquid.
2. The Gel Point ($p = p_c$):
- At this precise critical conversion, the first continuous, macroscopic macromolecular cluster of infinite size (spanning the entire dimensions of the reactor) forms.
- Viscosity diverges to infinity ($\eta \to \infty$).
- The mixture suddenly transforms from a liquid to an elastic, insoluble gel.
3. Post-Gel Regime ($p > p_c$):
- The system partitions into two distinct phases:
- Gel Fraction ($w_{\text{gel}}$): The insoluble, infinite cross-linked macroscopic network.
- Sol Fraction ($w_{\text{sol}}$): Finite, extractable soluble branched oligomers trapped within the network interstices ($w_{\text{sol}} + w_{\text{gel}} = 1.0$).
- As conversion increases further, finite sol oligomers are progressively incorporated into the infinite gel network, so $w_{\text{gel}} \to 1.0$.
§6.8 Gelation Theory: Carothers vs Flory-Stockmayer Gel Point ($\alpha_c$) & Percolation
Predicting the critical conversion $p_c$ at which gelation occurs is one of the classic problems in theoretical physical chemistry, historically resolved by two distinct theories: the Carothers approach and the Flory-Stockmayer statistical percolation model.
1. The Carothers Gel Point Derivation
Carothers defined the gel point as the conversion at which the number-average degree of polymerization diverges to infinity:
For a monomer mixture with average functionality $f_{\text{avg}}$, initial total functional groups $= N_0 f_{\text{avg}}$. At conversion $p$, the number of reacted functional groups is $p N_0 f_{\text{avg}}$. Since each reaction forms one bond and reduces the number of molecules by one:
Therefore:
Setting the denominator to zero ($X_n \to \infty$) yields the Carothers gel point:
- For a trifunctional monomer ($f_{\text{avg}} = 3$): $p_c = 2/3 = 0.667$ ($66.7\%$ conversion).
- For a tetrafunctional monomer ($f_{\text{avg}} = 4$): $p_c = 2/4 = 0.500$ ($50.0\%$ conversion).
2. The Flory-Stockmayer Statistical Gelation Theory
Paul Flory and Walter Stockmayer pointed out a profound flaw in Carothers' reasoning: Gelation occurs when the first infinite macromolecular network forms—which corresponds to the divergence of the weight-average molecular weight ($X_w \to \infty$), while the number-average molecular weight ($X_n$) remains modest and finite!
Flory defined the branching coefficient $\alpha$ as the probability that a given functional group on a multifunctional branch unit leads, via a chain of bifunctional units, to another multifunctional branch unit. Consider an $A_f$ branch monomer reacting with bifunctional $A-A$ and $B-B$ monomers. The critical branching coefficient for gelation in a system where branch units have functionality $f$ is:
- If branch units are trifunctional ($f = 3$): $\alpha_c = 1 / (3 - 1) = 1/2 = 0.500$.
- If branch units are tetrafunctional ($f = 4$): $\alpha_c = 1 / (4 - 1) = 1/3 = 0.333$.
For a stoichiometric mixture of bifunctional $A-A$ and $B-B$ monomers with a fraction $\rho$ of $A$ groups belonging to $A_f$ branch units ($p_A = p_B = p$):
For pure $A_f + B-B$ stoichiometric systems where all $A$ groups are on branch units ($ ho = 1$):
For a trifunctional system ($f = 3$):
Carothers predicts $p_c = 2 / f_{\text{avg}} = 2 / 2.4 = 0.833$ ($83.3\%$).
Experimental Reality and Intramolecular Cyclization
Experimental gel points measured via rheology (where $\tan \delta = G'' / G'$ becomes frequency-independent) are typically slightly higher than Flory-Stockmayer predictions ($p_{c, \text{exp}} > p_{c, \text{Flory}}$) because a small fraction of bonds form unreactive intramolecular loops that waste functional groups without contributing to the infinite percolating network.
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.
An equimolar mixture of adipic acid ($HOOC-(CH_2)_4-COOH$, formula weight $146.14\text{ g/mol}$) and 1,4-butanediol ($HO-(CH_2)_4-OH$, formula weight $90.12\text{ g/mol}$) is polymerized to form poly(butylene adipate). During polycondensation, water ($18.02\text{ g/mol}$) is removed. The repeating unit is $-[O-(CH_2)_4-O-CO-(CH_2)_4-CO]-$ with formula weight $M_0 = 200.24\text{ g/mol}$. Calculate: (a) The number-average degree of polymerization $X_n$ at conversions $p = 0.900, 0.980, 0.990, 0.995$, and $0.999$. (b) The number-average molecular weight $M_n$ at each of these conversions (accounting for unreacted end groups). (c) The weight-average degree of polymerization $X_w$ and polydispersity index $\text{Đ}$ at $p = 0.990$ and $p = 0.999$.
Step 1: Calculate $X_n$ via Carothers Equation
For an equimolar linear step-growth system:
- $p = 0.900 \implies X_n = \frac{1}{1 - 0.900} = \frac{1}{0.100} = 10.0$
- $p = 0.980 \implies X_n = \frac{1}{1 - 0.980} = \frac{1}{0.020} = 50.0$
- $p = 0.990 \implies X_n = \frac{1}{1 - 0.990} = \frac{1}{0.010} = 100.0$
- $p = 0.995 \implies X_n = \frac{1}{1 - 0.995} = \frac{1}{0.005} = 200.0$
- $p = 0.999 \implies X_n = \frac{1}{1 - 0.999} = \frac{1}{0.001} = 1,000.0$
Step 2: Calculate $M_n$
The chain structure is $H-[O-(CH_2)_4-O-CO-(CH_2)_4-CO]_{X_n/2}-OH$ (where each esterification unit pair is one adipate + one butanediol, $M_0 = 200.24\text{ g/mol}$ per repeat unit, corresponding to $X_n$ monomer residues, or $X_n/2$ repeat units). Let $M_{\text{monomer, avg}} = (146.14 + 90.12) / 2 = 118.13\text{ g/mol}$. Loss of water ($18.02\text{ g/mol}$) occurs per bond formed ($X_n - 1$ bonds for $X_n$ monomer residues):
- $p = 0.900$: $M_n = 10(100.11) + 18 = 1,019\text{ g/mol}$
- $p = 0.980$: $M_n = 50(100.11) + 18 = 5,024\text{ g/mol}$
- $p = 0.990$: $M_n = 100(100.11) + 18 = 10,029\text{ g/mol}$
- $p = 0.995$: $M_n = 200(100.11) + 18 = 20,040\text{ g/mol}$
- $p = 0.999$: $M_n = 1000(100.11) + 18 = 100,128\text{ g/mol}$
Step 3: Calculate $X_w$ and Dispersity $\text{Đ}$
For the Flory-Schulz most probable distribution:
- At $p = 0.990$:
- At $p = 0.999$:
(a) X_n = 10.0 (p=0.90), 50.0 (p=0.98), 100.0 (p=0.99), 200.0 (p=0.995), 1000.0 (p=0.999); (b) M_n: 1,019 g/mol (0.90), 5,024 g/mol (0.98), 10,029 g/mol (0.99), 20,040 g/mol (0.995), 100,130 g/mol (0.999); (c) At p=0.99: X_w = 199.0, PDI = 1.990; At p=0.999: X_w = 1,999.0, PDI = 1.999.
Nylon 6,6 is synthesized by polycondensation of hexamethylenediamine ($H_2N-(CH_2)_6-NH_2$) and adipic acid ($HOOC-(CH_2)_4-COOH$). To prevent unworkably high melt viscosity during fiber spinning, the number-average molecular weight at complete conversion ($p = 1.0$) must be limited to exactly $M_n = 15,000\text{ g/mol}$. The repeat unit formula weight is $M_0 = 226.32\text{ g/mol}$ (corresponding to two monomer residues, so average monomer mass is $M_0 / 2 = 113.16\text{ g/mol}$). (a) Determine the required number-average degree of polymerization $X_n$ at complete conversion. (b) Calculate the required stoichiometric imbalance ratio $r = N_A / N_B$ if excess adipic acid is used to control molecular weight. (c) Alternatively, if an equimolar mixture of diamine and diacid is used, calculate the mole percent of acetic acid (monofunctional chain stopper, $CH_3COOH$) that must be added relative to adipic acid.
Step 1: Calculate Target $X_n$
The repeat unit contains 2 monomer residues ($X_n = 2$ corresponds to one repeat unit). Average monomer residue mass in the chain:
Target $M_n = 15,000\text{ g/mol}$:
Step 2: Calculate Required Stoichiometric Ratio $r$
At complete conversion ($p = 1.0$), the modified Carothers equation is:
Solve for $r$:
Substitute $X_n = 132.56$:
This means there must be a $1.50\%$ stoichiometric deficit of diamine relative to diacid ($r = 0.9850$).
Step 3: Mole Percent of Monofunctional Acetic Acid
When adding acetic acid ($B'$) to an equimolar mixture ($N_A = N_B$):
Set $r_{\text{eff}} = 0.98503$:
Adding just $0.76\text{ mol}\%$ of acetic acid relative to adipic acid precisely caps the polymer at $M_n = 15,000\text{ g/mol}$.
(a) X_n = 132.6 monomer residues; (b) r = 0.9850 (1.50% stoichiometric excess of adipic acid); (c) 0.760 mol% acetic acid required relative to adipic acid.
The polyesterification of an equimolar mixture of diethylene glycol and adipic acid ($c_0 = 4.00\text{ mol/L}$) is investigated at $160.0^\circ\text{C}$ under two experimental conditions:
- Condition 1 (Self-Catalyzed): No external catalyst is added.
- Condition 2 (Externally Catalyzed): $0.10\text{ mol}\%$ $p$-toluenesulfonic acid catalyst is added.
The following reaction times are required to reach specific conversions:
- Condition 1 reaches $p = 0.800$ in $t = 50.0\text{ min}$, and $p = 0.900$ in $t = 194.0\text{ min}$.
- Condition 2 reaches $p = 0.800$ in $t = 8.00\text{ min}$, and $p = 0.900$ in $t = 18.00\text{ min}$.
(a) Verify the kinetic order for both conditions using the integrated rate equations. (b) Calculate the rate constants $k_1$ (in $\text{L}^2\text{ mol}^{-2}\text{ min}^{-1}$) and $k_2$ (in $\text{L mol}^{-1}\text{ min}^{-1}$). (c) Calculate the time required for each condition to reach an engineering conversion of $p = 0.990$ ($X_n = 100$).
Step 1: Kinetic Verification for Condition 1 (Self-Catalyzed)
For self-catalyzed polyesterification, the integrated rate equation is:
Calculate $[1/(1-p)^2 - 1]$:
- At $p = 0.800$: $\frac{1}{(1 - 0.800)^2} - 1 = \frac{1}{0.040} - 1 = 25.0 - 1 = 24.0$
- At $p = 0.900$: $\frac{1}{(1 - 0.900)^2} - 1 = \frac{1}{0.010} - 1 = 100.0 - 1 = 99.0$
Calculate rate constant $k_1$ from both data points:
- At $t = 50.0\text{ min}$:
- At $t = 194.0\text{ min}$:
The rate constant is consistent within experimental precision ($k_1 = 0.0155\text{ L}^2\text{ mol}^{-2}\text{ min}^{-1}$), confirming third-order self-catalyzed kinetics.
Step 2: Kinetic Verification for Condition 2 (Externally Catalyzed)
For acid-catalyzed polyesterification, the integrated rate equation is second-order:
Calculate $[1/(1-p) - 1]$:
- At $p = 0.800$: $\frac{1}{1 - 0.800} - 1 = 5.0 - 1 = 4.0$
- At $p = 0.900$: $\frac{1}{1 - 0.900} - 1 = 10.0 - 1 = 9.0$
Calculate rate constant $k_2$:
- At $t = 8.00\text{ min}$:
- At $t = 18.00\text{ min}$:
The rate constant is identical ($k_2 = 0.125\text{ L mol}^{-1}\text{ min}^{-1}$), confirming second-order kinetics.
Step 3: Time Required to Reach $p = 0.990$ ($X_n = 100$)
1. Condition 1 (Self-Catalyzed):
2. Condition 2 (Externally Catalyzed):
This striking comparison ($3.3\text{ hours}$ vs $14\text{ days}$) demonstrates why commercial polyester reactors universally employ acid catalysts!
(a) Verified: Condition 1 is 3rd order (linear 1/(1-p)^2 vs t); Condition 2 is 2nd order (linear 1/(1-p) vs t); (b) k_1 = 0.0155 L^2 mol^-2 min^-1, k_2 = 0.125 L mol^-1 min^-1; (c) Time to reach p=0.99: Self-catalyzed = 20,160 min (336 hours); Acid-catalyzed = 198 min (3.3 hours).
A stoichiometric linear step-growth polymerization has reached a conversion of $p = 0.980$. (a) Calculate the number-average degree of polymerization $X_n$ and weight-average degree of polymerization $X_w$. (b) Calculate the mole fraction $N_x$ and weight fraction $W_x$ for monomer ($x = 1$), dimer ($x = 2$), pentamer ($x = 5$), and 50-mer ($x = 50$). (c) Find the chain length $x_{\text{max}}$ at which the weight fraction distribution $W_x$ achieves its maximum value. (d) Calculate the fraction of the total polymer sample mass that consists of chains with lengths greater than $X_n$.
Step 1: Calculate $X_n, X_w$, and Dispersity
Given $p = 0.980$:
Step 2: Compute $N_x$ and $W_x$
Formulas:
1. Monomer ($x = 1$):
- $N_1 = 0.020 \times (0.980)^0 = 0.0200\ (2.00\%)$
- $W_1 = 1 \times (4.00 \times 10^{-4}) \times 1 = 0.000400\ (0.040\%)$
2. Dimer ($x = 2$):
- $N_2 = 0.020 \times (0.980)^1 = 0.0196\ (1.96\%)$
- $W_2 = 2 \times (4.00 \times 10^{-4}) \times 0.980 = 0.000784\ (0.0784\%)$
3. Pentamer ($x = 5$):
- $N_5 = 0.020 \times (0.980)^4 = 0.020 \times 0.92237 = 0.01845\ (1.845\%)$
- $W_5 = 5 \times (4.00 \times 10^{-4}) \times 0.92237 = 0.001845\ (0.185\%)$
4. 50-mer ($x = 50 = X_n$):
- $N_{50} = 0.020 \times (0.980)^{49} = 0.020 \times 0.3716 = 0.00743\ (0.743\%)$
- $W_{50} = 50 \times (4.00 \times 10^{-4}) \times 0.3716 = 0.007432\ (0.743\%)$
Notice that on a mole basis, monomer ($x = 1$) is the single most abundant species in the mixture ($N_1 > N_2 > N_3 \dots$), whereas on a mass basis, monomer accounts for only $0.04\%$ of the total polymer!
Step 3: Chain Length at Maximum Weight Fraction $x_{\text{max}}$
To find the maximum of $W_x$, treat $x$ as continuous and differentiate $\ln W_x$:
Since $p = 0.980$, $\ln(0.980) = -0.0202027$:
Because $\ln p = \ln(1 - (1-p)) \approx -(1-p)$, for high conversion:
The weight distribution achieves its maximum at precisely the number-average degree of polymerization!
Step 4: Mass Fraction of Chains with $x > X_n$
The cumulative weight fraction of chains with length up to $X_n$ is:
Given $X_n = 50$ and $p = 0.980$:
The fraction of mass with chain length greater than $X_n$ is:
Nearly three-quarters ($72.8\%$) of the sample mass resides in chains longer than the number-average length $X_n$.
(a) X_n = 50.0, X_w = 99.0, PDI = 1.980; (b) Monomer: N_1 = 2.00%, W_1 = 0.040%; Dimer: N_2 = 1.96%, W_2 = 0.078%; Pentamer: N_5 = 1.85%, W_5 = 0.185%; 50-mer: N_50 = 0.743%, W_50 = 0.743%; (c) x_max = 50 (= X_n); (d) 72.83% of total mass resides in chains with x > X_n.
A melt polycondensation between dimethyl terephthalate and ethylene glycol reaches an equilibrium constant of $K = 4.00$ at $280^\circ\text{C}$ for the transesterification equilibrium:
where ethylene glycol (EG) is the volatile condensation by-product. The total concentration of repeating ester units in the melt is $[\text{Ester}]_0 = 5.50\text{ mol/L}$. (a) Derive the equilibrium degree of polymerization $X_n$ as a function of the equilibrium constant $K$ and the mole fraction of residual by-product $n_{\text{EG}} / n_{\text{polymer}}$. (b) If the reaction is carried out in a closed autoclave without removing EG, calculate the maximum achievable conversion $p_{\text{eq}}$ and degree of polymerization $X_n$. (c) To achieve an engineering fiber-grade $X_n = 120$ ($M_n \approx 23,000\text{ g/mol}$), calculate the maximum permissible concentration of residual ethylene glycol $[\text{EG}]$ in the melt (in $\text{mol/L}$). (d) Given Henry's law constant for ethylene glycol in PET melt $H = 1.20 \times 10^4\text{ Pa L/mol}$, calculate the required vacuum pressure $P_{\text{vac}}$ (in Pascals and millibars).
Step 1: Derivation of $X_n$ at Reversible Equilibrium
Let $c_0$ be the initial concentration of functional groups. At conversion $p$:
- Unreacted end groups: $c_{\text{end}} = c_0(1 - p)$
- Formed ester linkages: $c_{\text{ester}} = c_0 p$
- Residual volatile by-product: $[\text{EG}]$
The equilibrium expression is:
Since $X_n = 1 / (1 - p)$, we have $1 - p = 1 / X_n$ and $p = 1 - 1/X_n \approx 1.0$ for high molecular weights. Substituting:
Solving for $X_n$:
Step 2: Closed System without By-Product Removal
In a closed vessel, all EG formed remains in the melt. Each bond formed generates one molecule of EG: $[\text{EG}] = c_0 p / 2$ (since 2 end groups yield 1 by-product).
Taking the square root:
Given $K = 4.00$:
Conversion:
In a closed system, equilibrium stops the reaction at $X_n < 4$ (short oligomers)!
Step 3: Maximum Permissible Residual By-Product $[\text{EG}]$ for $X_n = 120$
Rearranging the formula from Step 1:
Given:
- $K = 4.00$
- $c_0 = 5.50\text{ mol/L}$
- $X_n = 120$
Step 4: Required Reactor Vacuum Pressure $P_{\text{vac}}$
Applying Henry's law:
Given $H = 1.20 \times 10^4\text{ Pa L/mol}$:
Convert to millibars ($1\text{ mbar} = 100\text{ Pa}$):
To produce PET fiber, the finishing finisher reactor must operate under a high vacuum of less than $0.2\text{ mbar}$!
(a) X_n = sqrt(K * c_0 / [EG]); (b) Closed system: p_eq = 0.739, X_n = 3.83 (reaction stops at oligomers); (c) Residual [EG] <= 1.53 x 10^-3 mol/L; (d) P_vac = 18.3 Pa = 0.183 mbar.
Determine the critical gel point conversion $p_c$ according to the Carothers theory for each of the following reaction mixtures: (a) Pure glycerol ($f = 3$) reacting with phthalic anhydride ($f = 2$) in stoichiometric proportions ($2\text{ moles of glycerol to }3\text{ moles of phthalic anhydride}$, forming glyptal resin). (b) Pentaerythritol ($f = 4$) reacting with adipic acid ($f = 2$) in exact stoichiometric proportions ($1\text{ mole of pentaerythritol to }2\text{ moles of adipic acid}$). (c) A ternary mixture consisting of $2.0\text{ moles of adipic acid } (f = 2), 1.6\text{ moles of ethylene glycol } (f = 2)$, and $0.267\text{ moles of glycerol } (f = 3)$. Verify stoichiometric balance and compute the Carothers gel point $p_c$.
Step 1: Glyptal Resin ($A_3 + B_2$, 2:3 Moles)
- Initial molecules: $N_{\text{glycerol}} = 2$, $N_{\text{phthalic}} = 3$.
- Total molecules: $N_0 = 2 + 3 = 5\text{ moles}$.
- Hydroxyl groups ($A$): $2 \times 3 = 6\text{ moles}$.
- Carboxyl groups ($B$): $3 \times 2 = 6\text{ moles}$.
Stoichiometry is exact ($N_A = N_B = 6$). Total functional groups $= 6 + 6 = 12\text{ moles}$. Average functionality:
Carothers gel point:
Step 2: Pentaerythritol + Adipic Acid ($A_4 + B_2$, 1:2 Moles)
- Initial molecules: $N_{\text{penta}} = 1$, $N_{\text{adipic}} = 2$.
- Total molecules: $N_0 = 1 + 2 = 3\text{ moles}$.
- Hydroxyl groups ($A$): $1 \times 4 = 4\text{ moles}$.
- Carboxyl groups ($B$): $2 \times 2 = 4\text{ moles}$.
Stoichiometry is exact ($N_A = N_B = 4$). Total functional groups $= 4 + 4 = 8\text{ moles}$. Average functionality:
Carothers gel point:
Step 3: Ternary System ($2.0\text{ Adipic} + 1.6\text{ EG} + 0.267\text{ Glycerol}$)
Count functional groups:
- Carboxyl groups ($B$): $2.0\text{ mol} \times 2 = 4.00\text{ moles}$.
- Hydroxyl groups from EG ($A$): $1.6\text{ mol} \times 2 = 3.20\text{ moles}$.
- Hydroxyl groups from Glycerol ($A$): $0.267\text{ mol} \times 3 = 0.80\text{ moles}$.
Total hydroxyl groups $= 3.20 + 0.80 = 4.00\text{ moles}$. Stoichiometry is exact ($N_A = N_B = 4.00\text{ moles}$). Total monomer molecules:
Total functional groups:
Average functionality:
Carothers gel point:
Because the trifunctional brancher comprises only a small fraction of the diol component, gelation is postponed until $96.7\%$ conversion, allowing significant linear chain growth before network formation.
(a) Glyptal (f_avg = 2.40): p_c = 0.833 (83.3%); (b) Pentaerythritol + Adipic (f_avg = 2.67): p_c = 0.750 (75.0%); (c) Ternary system (f_avg = 2.069): p_c = 0.967 (96.7%).
Consider a polymerization system containing multifunctional branch units $A_f$ with functionality $f \ge 3$, along with bifunctional monomers $A-A$ and $B-B$. Let $\alpha$ be the branching coefficient, defined as the probability that a given functional group on a branch unit connects through a chain of bifunctional units to another branch unit. (a) Using Cayley tree (Bethe lattice) branching analysis, show that the expected number of new chains branching out from generation $n$ to generation $n+1$ is multiplied by the branching factor $(f - 1) \alpha$. (b) Prove that the condition for an infinite network (percolation) to form with non-zero probability requires:
(c) For a stoichiometric mixture of $A_f$ and $B-B$ ($r = 1, \rho = 1$), express $\alpha$ in terms of fractional conversion $p$, and derive the Flory-Stockmayer critical conversion $p_c$. (d) For a trifunctional monomer ($f = 3$), compare the Flory-Stockmayer critical conversion $p_c$ with the Carothers prediction and explain physically why $p_{c, \text{Flory}} < p_{c, \text{Carothers}}$.
Step 1: Branching Tree Analysis
Consider an $A_f$ branch unit selected as the root of a tree graph (Generation 0). The root has $f$ reactive arms. Pick one arm and trace outward:
- It reacts with probability $p_A$.
- Through alternating $B-B$ and $A-A$ linkages, it reaches another $A_f$ branch unit with overall probability $\alpha$.
Once this new branch unit is reached (Generation 1):
- One of its $f$ functional groups is used by the incoming bond from Generation 0.
- The remaining number of outgoing arms capable of continuing the tree outward is strictly $f - 1$.
Each of these $f - 1$ outgoing arms independently has probability $\alpha$ of reaching a further branch unit. Therefore, the expected number of branch connections in Generation 1 is $(f - 1) \alpha$. By induction, the expected number of active branching paths in Generation $n$ is:
Step 2: Critical Condition for Infinite Network
- If $(f - 1) \alpha < 1$:
As $n \to \infty$, $[(f - 1) \alpha]^n \to 0$. The probability of an infinite path is zero; all molecular clusters are finite.
- If $(f - 1) \alpha > 1$:
As $n \to \infty$, $[(f - 1) \alpha]^n \to \infty$. The branching tree diverges exponentially; there is a non-zero probability of an infinite percolating cluster.
- The transition occurs precisely when the propagation factor equals unity:
Step 3: Critical Conversion for Pure $A_f + B-B$
For a stoichiometric mixture of $A_f$ and $B-B$: All $A$ groups reside on branch units ($\rho = 1$), and stoichiometry is balanced ($p_A = p_B = p$). Tracing from an $A$ group on branch unit 1:
- The $A$ group must react with a $B$ group (probability $p$).
- The second $B$ group on the $B-B$ monomer must react with an $A$ group on another branch unit (probability $p$).
Therefore:
Equating $\alpha$ to $\alpha_c$:
Step 4: Comparison for Trifunctional Monomer ($f = 3$)
1. Flory-Stockmayer Prediction:
2. Carothers Prediction:
A stoichiometric mixture of $A_3$ and $B_2$ requires 2 moles of $A_3$ per 3 moles of $B_2$ (6 $A$ groups, 6 $B$ groups). Total molecules $N_0 = 2 + 3 = 5$. Average functionality $f_{\text{avg}} = 12 / 5 = 2.40$.
Physical Explanation of the Discrepancy
- Carothers' condition requires the number-average molecular weight to diverge: $X_n = N_0 / N \to \infty$.
For $N \to 0$, virtually all molecules in the reactor would have to be linked into a single macroscopic super-molecule.
- Flory-Stockmayer recognizes that gelation begins when the first infinite network appears, which corresponds to the divergence of the weight-average molecular weight ($X_w \to \infty$).
- At $p = 70.7\%$, a tiny infinitesimal weight fraction of the system forms the first infinite percolating network spanning the container, causing the liquid to lose fluidity and gel. Meanwhile, millions of finite soluble oligomers remain ($X_n$ is only $\approx 3.4$!).
- Therefore, gelation occurs much earlier than Carothers predicted ($70.7\%$ vs $83.3\%$). Flory's statistical approach is experimentally confirmed.
(a) Generation scaling:
Starting from the Flory-Schulz most probable mole fraction distribution:
and weight fraction distribution:
(a) Using generating functions or summation series formulas, prove analytically that:
(b) Derive the analytical expression for the z-average degree of polymerization:
(c) For conversions $p = 0.900, 0.990$, and $0.999$, compute $X_n, X_w, X_z$, and the ratio $X_z / X_w$. (d) Show that in the limit $p \to 1.0$, $X_n : X_w : X_z \to 1 : 2 : 3$.
Step 1: Analytical Proof for $X_w$
By definition:
Recall the standard geometric series for $|p| < 1$:
Differentiating with respect to $p$:
Multiplying by $p$:
Differentiating again with respect to $p$:
Substitute into the expression for $X_w$:
This completes the exact proof.
Step 2: Derivation of $X_z$
The z-average degree of polymerization is:
Let us evaluate $S_3 = \sum_{x=1}^\infty x^3 p^{x-1}$: Multiply the sum for $x^2 p^{x-1}$ by $p$:
Differentiating with respect to $p$:
Factor out $(1 - p)^2$:
Now substitute into $X_z$:
Dividing by $X_w = \frac{1 + p}{1 - p}$:
Step 3: Tabulate Moments at $p = 0.900, 0.990, 0.999$
1. $p = 0.900$:
- $X_n = 1 / (1 - 0.900) = 10.0$
- $X_w = (1 + 0.900) / (1 - 0.900) = 1.900 / 0.100 = 19.0$
- $X_z = [1 + 4(0.900) + (0.900)^2] / [(0.100)(1.900)] = [1 + 3.60 + 0.81] / 0.190 = 5.41 / 0.190 = 28.47$
- Ratio $X_z / X_w = 28.47 / 19.0 = 1.498$
2. $p = 0.990$:
- $X_n = 1 / 0.010 = 100.0$
- $X_w = 1.990 / 0.010 = 199.0$
- $X_z = [1 + 4(0.990) + (0.990)^2] / [(0.010)(1.990)] = [1 + 3.96 + 0.9801] / 0.01990 = 5.9401 / 0.01990 = 298.50$
- Ratio $X_z / X_w = 298.50 / 199.0 = 1.500$
3. $p = 0.999$:
- $X_n = 1 / 0.001 = 1,000.0$
- $X_w = 1.999 / 0.001 = 1,999.0$
- $X_z = [1 + 4(0.999) + (0.999)^2] / [(0.001)(1.999)] = [5.994] / 0.001999 = 2,998.5$
- Ratio $X_z / X_w = 2,998.5 / 1,999.0 = 1.500$
Step 4: Asymptotic Ratio at $p \to 1.0$
Let $\epsilon = 1 - p \to 0$:
- $X_n = 1 / \epsilon$
- $X_w = (2 - \epsilon) / \epsilon \approx 2 / \epsilon = 2 X_n$
- $X_z = (1 + 4(1) + 1) / (\epsilon \cdot 2) = 6 / (2 \epsilon) = 3 / \epsilon = 3 X_n$
Therefore:
(a) Proved: X_w = (1+p)/(1-p); (b) Proved: X_z = (1 + 4p + p^2) / [(1-p)(1+p)]; (c) p=0.90: X_n=10, X_w=19, X_z=28.5 (X_z/X_w=1.50); p=0.99: X_n=100, X_w=199, X_z=298.5; p=0.999: X_n=1000, X_w=1999, X_z=2998.5; (d) As p -> 1, X_n : X_w : X_z -> 1 : 2 : 3.
Beyond the gel point conversion ($p > p_c$), a polymerizing network partitions into an extractable, soluble branched fraction (sol, weight fraction $w_{\text{sol}}$) and an insoluble infinite network (gel, weight fraction $w_{\text{gel}} = 1 - w_{\text{sol}}$). Consider a trifunctional monomer $A_3$ reacting with itself ($f = 3$) or a stoichiometric $A_3 + B_2$ system with branching coefficient $\alpha > \alpha_c = 0.50$. Let $\beta$ be the extinction probability that a randomly chosen bond attached to a branch unit does not lead to an infinite network path. (a) Show that $\beta$ satisfies the recursive algebraic relation:
(b) Solve this quadratic equation for $\beta$ and identify the two roots: the trivial root $\beta_1 = 1.0$ (pre-gel) and the non-trivial root $\beta_2(\alpha)$ (post-gel). (c) Prove that the sol weight fraction is given by:
(d) Calculate $w_{\text{sol}}$ and $w_{\text{gel}}$ for branching coefficients $\alpha = 0.55, 0.60, 0.70, 0.80$, and $0.90$.
Step 1: Derivation of the Self-Consistent Relation for $\beta$
Let $\beta$ be the probability that an outgoing path from a branch unit does NOT connect to the infinite gel network. There are two mutually exclusive possibilities for an outgoing arm:
- The arm is unreacted: this occurs with probability $1 - \alpha$. An unreacted arm definitely does not reach the infinite network.
- The arm has reacted: this occurs with probability $\alpha$.
If it reacts, it connects to another trifunctional branch unit with $f - 1 = 2$ new outgoing arms. For this path to still remain finite, both of these 2 new outgoing arms must fail to connect to the infinite network. Since the arms branch independently, the probability that both fail is $\beta \times \beta = \beta^2$.
Summing the two probabilities yields the fundamental recursive equation:
Step 2: Solve the Quadratic Equation
Rearranging into standard quadratic form:
Factoring the quadratic: Notice that $\beta = 1$ is always a root:
Dividing $(\alpha \beta^2 - \beta + 1 - \alpha)$ by $(\beta - 1)$:
The two roots are:
1. $\beta_1 = 1.0$: For $\alpha \le \alpha_c = 0.50$, all paths are strictly finite, so the extinction probability is identically $1.0$.
2. $\beta_2 = \frac{1 - \alpha}{\alpha}$: For $\alpha > 0.50$, $\frac{1 - \alpha}{\alpha} < 1.0$, representing the true physical extinction probability in the post-gel regime.
Step 3: Derivation of the Sol Fraction $w_{\text{sol}}$
A trifunctional monomer molecule belongs to the sol fraction if and only if all three of its functional arms lead to finite paths. Since each of its 3 arms independently has extinction probability $\beta$:
The gel fraction is the remainder of the total sample:
Step 4: Numerical Calculation of $w_{\text{sol}}$ and $w_{\text{gel}}$
1. $\alpha = 0.55$:
- $\beta = (1 - 0.55) / 0.55 = 0.45 / 0.55 = 0.81818$
- $w_{\text{sol}} = (0.81818)^3 = 0.5477\ (54.77\%)$
- $w_{\text{gel}} = 1 - 0.5477 = 0.4523\ (45.23\%)$
2. $\alpha = 0.60$:
- $\beta = (1 - 0.60) / 0.60 = 0.40 / 0.60 = 0.66667$
- $w_{\text{sol}} = (0.66667)^3 = 0.2963\ (29.63\%)$
- $w_{\text{gel}} = 1 - 0.2963 = 0.7037\ (70.37\%)$
3. $\alpha = 0.70$:
- $\beta = (1 - 0.70) / 0.70 = 0.30 / 0.70 = 0.42857$
- $w_{\text{sol}} = (0.42857)^3 = 0.0787\ (7.87\%)$
- $w_{\text{gel}} = 1 - 0.0787 = 0.9213\ (92.13\%)$
4. $\alpha = 0.80$:
- $\beta = (1 - 0.80) / 0.80 = 0.20 / 0.80 = 0.2500$
- $w_{\text{sol}} = (0.2500)^3 = 0.0156\ (1.56\%)$
- $w_{\text{gel}} = 1 - 0.0156 = 0.9844\ (98.44\%)$
5. $\alpha = 0.90$:
- $\beta = (1 - 0.90) / 0.90 = 0.10 / 0.90 = 0.11111$
- $w_{\text{sol}} = (0.11111)^3 = 0.00137\ (0.14\%)$
- $w_{\text{gel}} = 1 - 0.00137 = 0.99863\ (99.86\%)$
Notice how rapidly the gel fraction consumes the reaction mixture: by $\alpha = 0.70$, over $92\%$ of the sample mass is locked into the macroscopic gel network!
(a) Self-consistent equation proved: beta = (1-alpha) + alpha * beta^2; (b) Roots: beta_1 = 1.0 (pre-gel), beta_2 = (1-alpha)/alpha (post-gel); (c) w_sol = beta^3 = [(1-alpha)/alpha]^3, w_gel = 1 - w_sol; (d) alpha=0.55: w_sol=54.8%, w_gel=45.2%; alpha=0.60: w_sol=29.6%, w_gel=70.4%; alpha=0.70: w_sol=7.9%, w_gel=92.1%; alpha=0.80: w_sol=1.6%, w_gel=98.4%; alpha=0.90: w_sol=0.14%, w_gel=99.86%.