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Chapter 3 • Theory & Derivations

Molecular Weight Averages & Polydispersity Distributions

Statistical definitions and rigorous mathematical inequalities of macromolecular weight averages (Mn, Mw, Mz, Mv), polydispersity index (PDI / Đ), continuous distribution functions (Schulz-Zimm, Poisson, Flory-Schulz most probable), Gel Permeation Chromatography (GPC/SEC) mechanics, and Benoit universal calibration.

§3.1 The Concept of Polydispersity vs Monodispersity in Macromolecules

Unlike low molecular weight organic molecules or monodisperse biological macromolecules (such as enzymes, insulin, or wild-type genomic DNA) where every molecule possesses an exact, identical chemical formula and invariant mass, synthetic polymers are inherently polydisperse.

Origins of Polydispersity

In any synthetic chemical polymerization process:

  • Initiation events occur randomly throughout the reactor volume over time.
  • Radical, ionic, or coordination active centers propagate with stochastic collision kinetics.
  • Chain transfer, termination by combination or disproportionation, and mass-transfer mixing gradients introduce broad variations in individual chain lifetimes.

Consequently, a synthetic polymer sample consists of a complex statistical mixture of macromolecules possessing identical constitutional repeat units but spanning a broad continuum of chain lengths and molecular weights.

The Number and Weight Distribution Functions

A polydisperse polymer is described by its discrete or continuous molecular weight distribution (MWD):

  • Number Distribution $N(M)$: The number of moles (or molecules) $N_i$ of species possessing molecular weight $M_i$. The mole fraction is:
\[x_i = \frac{N_i}{\sum N_i}\]
  • Weight Distribution $W(M)$: The total mass $w_i = N_i M_i$ of species possessing molecular weight $M_i$. The weight fraction is:
\[w_i = \frac{N_i M_i}{\sum N_i M_i} = \frac{w_i}{\sum w_i}\]

Because heavier molecules contribute proportionally more mass, the weight distribution $W(M)$ is always shifted toward substantially higher molecular weights relative to the number distribution $N(M)$.

§3.2 Statistical Definitions of Molecular Weight Averages (Mn, Mw, Mz, Mv)

Because no single number can fully describe a broad distribution, polymer science employs a family of statistical moments known as molecular weight averages.

1. Number-Average Molecular Weight $M_n$

The first statistical moment of the number distribution, representing the total mass of the sample divided by the total number of moles:

\[M_n = \frac{\sum_{i} N_i M_i}{\sum_{i} N_i} = \sum_{i} x_i M_i = \frac{1}{\sum_{i} (w_i / M_i)}\]

$M_n$ is highly sensitive to the presence of low molecular weight oligomers. It is measured experimentally by colligative properties (membrane osmometry, vapor pressure osmometry, cryoscopy) and end-group chemical titration.

2. Weight-Average Molecular Weight $M_w$

The second statistical moment of the number distribution (or the first moment of the weight distribution):

\[M_w = \frac{\sum_{i} N_i M_i^2}{\sum_{i} N_i M_i} = \sum_{i} w_i M_i\]

$M_w$ is heavily influenced by the presence of large, high molecular weight macromolecules. It is determined experimentally by static light scattering, small-angle neutron scattering (SANS), and sedimentation equilibrium.

3. Z-Average Molecular Weight $M_z$

The third moment of the number distribution (or second moment of the weight distribution):

\[M_z = \frac{\sum_{i} N_i M_i^3}{\sum_{i} N_i M_i^2} = \frac{\sum_{i} w_i M_i^2}{\sum_{i} w_i M_i}\]

$M_z$ is exceptionally sensitive to high molecular weight tails, micro-gels, and long-chain branching. It is measured by analytical ultracentrifugation (AUC sedimentation equilibrium). Higher moments ($M_{z+1}$) can be defined analogously.

4. Viscosity-Average Molecular Weight $M_v$

Derived from dilute solution viscometry through the Mark-Houwink-Sakurada relationship ($[\eta] = K' M^a$):

\[M_v = \left( \frac{\sum_{i} N_i M_i^{1+a}}{\sum_{i} N_i M_i} \right)^{1/a} = \left( \sum_{i} w_i M_i^a \right)^{1/a}\]

where $a$ is the Mark-Houwink exponent ($0.5 \le a \le 0.8$ for flexible coils).

  • When $a = 1$: $M_v = M_w$.
  • In a theta solvent where $a = 0.5$: $M_v = \left(\sum w_i M_i^{1/2}\right)^2$.

§3.3 Mathematical Hierarchy & Rigorous Inequalities: Mn <= Mv <= Mw <= Mz

A cornerstone theorem of polymer mathematics states that for any polydisperse polymer sample, the molecular weight averages obey the strict hierarchical inequality:

\[M_n < M_v < M_w < M_z < M_{z+1}\]

Equality ($M_n = M_v = M_w = M_z$) holds if and only if the polymer is strictly monodisperse.

Proof that $M_w \ge M_n$ via the Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality states that for any real sequences $(a_i)$ and $(b_i)$:

\[\left( \sum_{i} a_i b_i \right)^2 \le \left( \sum_{i} a_i^2 \right) \left( \sum_{i} b_i^2 \right)\]

Let $a_i = \sqrt{N_i}$ and $b_i = \sqrt{N_i} M_i$. Substituting these sequences:

\[\left( \sum_{i} \sqrt{N_i} \cdot \sqrt{N_i} M_i \right)^2 \le \left( \sum_{i} N_i \right) \left( \sum_{i} N_i M_i^2 \right)\]
\[\left( \sum_{i} N_i M_i \right)^2 \le \left( \sum_{i} N_i \right) \left( \sum_{i} N_i M_i^2 \right)\]

Dividing both sides by $\left(\sum N_i\right) \left(\sum N_i M_i\right)$:

\[\frac{\sum N_i M_i}{\sum N_i} \le \frac{\sum N_i M_i^2}{\sum N_i M_i}\]
\[M_n \le M_w\]

Equality holds if and only if $b_i = c \, a_i$ for all $i$, meaning $\sqrt{N_i} M_i = c \sqrt{N_i} \implies M_i = c$ (monodisperse sample). The proof that $M_w \le M_z$ follows identically by choosing $a_i = \sqrt{N_i M_i}$ and $b_i = \sqrt{N_i M_i} M_i$.

Position of the Viscosity Average $M_v$

Because the Mark-Houwink exponent $a$ satisfies $0 < a < 1$ for flexible random coils in good and theta solvents, Hölder's inequality dictates:

\[M_n \le M_v \le M_w\]

Typically, $M_v$ lies within $10 - 20\%$ below $M_w$, serving as a highly accessible proxy for weight-average molecular weight.

§3.4 The Polydispersity Index (PDI / D) & Breadth of Distribution

The breadth of the molecular weight distribution is quantitatively indexed by the polydispersity index (PDI, denoted by IUPAC as the dispersity $\text{Đ}$):

\[\text{Đ} = \frac{M_w}{M_n}\]

Variance and Standard Deviation of the MWD

The variance $\sigma_n^2$ of the number distribution is:

\[\sigma_n^2 = \langle (M - M_n)^2 \rangle = \frac{\sum N_i (M_i - M_n)^2}{\sum N_i} = \frac{\sum N_i M_i^2}{\sum N_i} - M_n^2\]

Notice that:

\[\frac{\sum N_i M_i^2}{\sum N_i} = \left( \frac{\sum N_i M_i^2}{\sum N_i M_i} \right) \left( \frac{\sum N_i M_i}{\sum N_i} \right) = M_w M_n\]

Therefore:

\[\sigma_n^2 = M_w M_n - M_n^2 = M_n^2 \left( \frac{M_w}{M_n} - 1 \right) = M_n^2 (\text{Đ} - 1)\]

The standard deviation relative to the number-average molecular weight is:

\[\frac{\sigma_n}{M_n} = \sqrt{\text{Đ} - 1}\]

This elegant identity proves that the dispersity $\text{Đ}$ is a direct metric of the normalized variance of the distribution.

Typical Dispersity Values across Polymerization Mechanisms

  • Living Anionic & Group-Transfer Polymerization: $\text{Đ} \approx 1.01 - 1.05$ (nearly monodisperse Poisson distributions).
  • Controlled / Living Radical Polymerization (ATRP, RAFT, NMP): $\text{Đ} \approx 1.05 - 1.20$.
  • Step-Growth (Condensation) Polymerization at High Conversion ($p \to 1$): $\text{Đ} = 1 + p \to 2.0$ (Flory-Schulz most probable distribution).
  • Free-Radical Polymerization:
  • Disproportionation termination: $\text{Đ} = 2.0$.
  • Combination termination: $\text{Đ} = 1.5$.
  • With autoacceleration (Trommsdorff gel effect) and chain transfer: $\text{Đ} = 2.5 - 5.0$.
  • Ziegler-Natta Multi-Site Heterogeneous Coordination Catalysis: $\text{Đ} = 5.0 - 20.0$ (superposition of multiple active catalytic sites).
  • Hyperbranched Polymers & Random Crosslinking Gels: $\text{Đ} > 20 - 50$ (diverging as the critical gel point is approached).

§3.5 The Flory-Schulz (Most Probable) Distribution Function

The Flory-Schulz distribution (also termed the "most probable" distribution) governs linear step-growth polycondensation and chain-growth polymerization with constant chain transfer.

Statistical Derivation from Equal Reactivity Postulate

Consider a step-growth polymerization of bifunctional monomers ($A-B$ or stoichiometric $A-A + B-B$). Let $p$ be the extent of reaction (probability that any given functional group has reacted). The probability that a polymer molecule contains precisely $x$ repeating units requires:

  • $(x - 1)$ reacted linkages (each with independent probability $p$).
  • $1$ unreacted end group terminating the chain (with probability $1 - p$).

The number-fraction distribution (mole fraction of $x$-mers) is:

\[N_x = (1 - p) p^{x-1}\]

The total number of molecules in the system at conversion $p$ is $N = N_0 (1 - p)$, where $N_0$ is the initial number of monomers. The total number of $x$-mer chains is:

\[n_x = N_0 (1 - p)^2 p^{x-1}\]

The Weight-Fraction Distribution $w_x$

The weight fraction $w_x$ of $x$-mers is the mass of $x$-mers divided by total mass $N_0 M_0$:

\[w_x = \frac{x n_x}{N_0} = x (1 - p)^2 p^{x-1}\]

Molecular Weight Averages and Dispersity

Summing the statistical moments using standard geometric series identities:

1. Number-Average Degree of Polymerization $X_n$:

\[X_n = \sum_{x=1}^\infty x N_x = (1 - p) \sum_{x=1}^\infty x p^{x-1} = (1 - p) \frac{1}{(1 - p)^2} = \frac{1}{1 - p}\]

2. Weight-Average Degree of Polymerization $X_w$:

\[X_w = \sum_{x=1}^\infty x w_x = (1 - p)^2 \sum_{x=1}^\infty x^2 p^{x-1} = (1 - p)^2 \frac{1 + p}{(1 - p)^3} = \frac{1 + p}{1 - p}\]

3. Z-Average Degree of Polymerization $X_z$:

\[X_z = \frac{1 + 4p + p^2}{(1 - p)(1 + p)}\]

4. Polydispersity Index $\text{Đ}$:

\[\text{Đ} = \frac{X_w}{X_n} = \frac{\frac{1 + p}{1 - p}}{\frac{1}{1 - p}} = 1 + p\]

In the limit of high conversion ($p \to 1.0$), $X_n \to \infty$, and the dispersity approaches exactly $2.0$:

\[\lim_{p \to 1} \text{Đ} = 2.0\]

The maximum of the weight distribution curve occurs at:

\[\frac{d w_x}{dx} = 0 \implies x_{\text{max}} = -\frac{1}{\ln p} \approx \frac{1}{1 - p} = X_n\]

Thus, the weight distribution peaks at precisely $x = X_n$.

§3.6 The Poisson Distribution for Living Polymerization Systems

When polymer chain initiation is instantaneous relative to propagation and termination or chain transfer is completely absent—as in ideal living anionic or living ring-opening polymerizations—the molecular weight distribution follows a Poisson distribution.

Mechanistic Derivation

Let all $N_I$ initiator molecules initiate chain growth simultaneously at $t = 0$. The rate of monomer consumption per active center is identical across all chains:

\[-\frac{d[M]}{dt} = k_p [I]_0 [M]\]

The probability that a growing chain adds precisely $x$ monomer units during reaction time $t$ obeys Poisson birth-process kinetics:

\[P(x) = \frac{\nu^x e^{-\nu}}{x!}\]

where $\nu$ is the average number of monomers consumed per initiator molecule (the kinetic chain length):

\[\nu = \frac{[M]_0 - [M]}{[I]_0} = X_n - 1 \approx X_n\]

Molecular Weight Averages and Monodispersity

For the Poisson distribution:

  • Number-average degree of polymerization:
\[X_n = 1 + \nu \approx \nu\]
  • Weight-average degree of polymerization:
\[X_w = 1 + \nu + \frac{\nu}{1 + \nu}\]
  • Dispersity $\text{Đ}$:
\[\text{Đ} = \frac{X_w}{X_n} = \frac{1 + \nu + \frac{\nu}{1 + \nu}}{1 + \nu} = 1 + \frac{\nu}{(1 + \nu)^2} = 1 + \frac{X_n - 1}{X_n^2} \approx 1 + \frac{1}{X_n}\]

For a typical polymer with $X_n = 500$:

\[\text{Đ} = 1 + \frac{1}{500} = 1 + 0.002 = 1.002\]

The distribution is extraordinarily narrow, providing essentially monodisperse polymers used as universal molecular weight calibration standards.

§3.7 The Schulz-Zimm Continuous Distribution Function

To model empirical polymer distributions spanning from narrow living systems to broad industrial materials, the Schulz-Zimm distribution provides a continuous, highly flexible mathematical representation.

Mathematical Formulation

The weight-fraction distribution function $w(M)$ in the Schulz-Zimm model is:

\[w(M) = \frac{y^{z+1}}{\Gamma(z + 1)} M^z \exp(-y M)\]

where:

  • $z$ is the coupling / breadth parameter ($z > 0$)
  • $y$ is a scaling parameter
  • $\Gamma(z + 1) = z!$ is the gamma function

The statistical moments are:

\[M_n = \frac{z}{y}\]
\[M_w = \frac{z + 1}{y}\]
\[M_z = \frac{z + 2}{y}\]

Relationship to Dispersity $\text{Đ}$

The dispersity is directly governed by the parameter $z$:

\[\text{Đ} = \frac{M_w}{M_n} = \frac{z + 1}{z} = 1 + \frac{1}{z}\]

Solving for the parameter $z$:

\[z = \frac{1}{\text{Đ} - 1}\]
  • As $z \to \infty$: $\text{Đ} \to 1.0$ (monodisperse delta function).
  • When $z = 1$: $\text{Đ} = 2.0$ (recovering the Flory-Schulz most probable distribution).
  • When $z = 2$: $\text{Đ} = 1.5$ (recovering free-radical polymerization with combination termination).
  • When $z < 1$: $\text{Đ} > 2.0$ (broad polydispersity distributions).

The ratio of the Z-average to weight-average is:

\[\frac{M_z}{M_w} = \frac{z + 2}{z + 1} = \frac{2\text{Đ} - 1}{\text{Đ}}\]

The Schulz-Zimm function is widely implemented in Gel Permeation Chromatography software to deconvolute overlapping chromatogram peaks.

§3.8 Gel Permeation Chromatography (GPC/SEC) & Benoit Universal Calibration

Gel Permeation Chromatography (GPC), also termed Size Exclusion Chromatography (SEC), is the premier experimental technique for measuring the full molecular weight distribution of polymers.

Separation Mechanism: Entropy-Driven Size Exclusion

A GPC column is packed with porous, crosslinked polymer beads (typically styrene-divinylbenzene gels) with controlled pore diameter distributions ($10 - 10^5\text{ \AA}$). As a dilute polymer solution flows through the column:

  • Large macromolecules with hydrodynamic volume $V_h$ exceeding the pore diameter cannot enter the pore network and are excluded, eluting rapidly at the interstitial void volume $V_0$.
  • Small macromolecules permeate freely into both the interstitial spaces and the internal pores, eluting at the total liquid volume $V_t = V_0 + V_p$.
  • Intermediate chains partition into a fraction $K_{\text{SEC}}$ of the pore volume based on their steric size.

The retention volume $V_e$ is:

\[V_e = V_0 + K_{\text{SEC}} V_p \quad (0 \le K_{\text{SEC}} \le 1)\]

Crucially, GPC separates molecules strictly by their hydrodynamic volume $V_h$, NOT by molecular weight!

Benoit Universal Calibration Principle

In 1967, Henri Benoit demonstrated that the hydrodynamic volume of any polymer chain in solution is directly proportional to the product of its intrinsic viscosity $[\eta]$ and its molecular weight $M$: From Einstein's viscosity law for equivalent hydrodynamic spheres of radius $R_h$:

\[[\eta] = \frac{10 \pi N_A}{3} \frac{R_h^3}{M} \implies V_h \propto R_h^3 \propto [\eta] M\]

Benoit established that plotting $\log([\eta] M)$ versus elution volume $V_e$ yields a single universal calibration curve onto which all polymers (linear, branched, polystyrene, PMMA, polyethylene) collapse, independent of chemical composition:

\[\log([\eta] M) = f(V_e)\]

Molecular Weight Determination of Unknown Polymers

If a GPC column is calibrated using narrow polystyrene standards (subscript $\text{PS}$):

\[[\eta]_{\text{PS}} M_{\text{PS}} = [\eta]_x M_x\]

Substituting the Mark-Houwink-Sakurada relationship $[\eta] = K' M^a$:

\[K_{\text{PS}} M_{\text{PS}}^{1 + a_{\text{PS}}} = K_x M_x^{1 + a_x}\]

Solving for the true molecular weight $M_x$ of the unknown polymer eluting at the same retention volume $V_e$:

\[M_x = \left( \frac{K_{\text{PS}}}{K_x} \right)^{\frac{1}{1 + a_x}} M_{\text{PS}}^{\frac{1 + a_{\text{PS}}}{1 + a_x}}\]

This elegant relation permits precise determination of absolute molecular weights and distribution functions for any polymer whose Mark-Houwink constants are known.

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 3.1: Molecular Weight Averages for a Discrete Binary Polymer Blend

An equimolar mixture ($1:1$ molar ratio) is prepared by blending two monodisperse polystyrene standards:

  • Component A: $M_A = 20,000\text{ g/mol}$
  • Component B: $M_B = 180,000\text{ g/mol}$

(a) Calculate the number-average molecular weight $M_n$. (b) Calculate the weight-average molecular weight $M_w$. (c) Calculate the Z-average molecular weight $M_z$. (d) Determine the polydispersity index $\text{Đ} = M_w / M_n$. (e) Recalculate $M_n, M_w$, and $\text{Đ}$ if the two components are blended in an equal weight ratio ($1:1$ mass ratio).

Step 1: Equimolar Blend ($N_A = N_B = 1\text{ mol}$)

The total number of moles is $\sum N_i = 1 + 1 = 2\text{ mol}$. (a) Number-average molecular weight $M_n$:

\[M_n = \frac{\sum N_i M_i}{\sum N_i} = \frac{1(20,000) + 1(180,000)}{2} = \frac{200,000}{2} = 100,000\text{ g/mol}\]

(b) Weight-average molecular weight $M_w$:

\[\sum N_i M_i = 200,000\text{ g}\]
\[\sum N_i M_i^2 = 1(20,000)^2 + 1(180,000)^2 = 4.0 \times 10^8 + 3.24 \times 10^{10} = 3.28 \times 10^{10}\text{ g}^2\text{/mol}\]
\[M_w = \frac{\sum N_i M_i^2}{\sum N_i M_i} = \frac{3.28 \times 10^{10}}{2.00 \times 10^5} = 164,000\text{ g/mol}\]

(c) Z-average molecular weight $M_z$:

\[\sum N_i M_i^3 = 1(20,000)^3 + 1(180,000)^3 = 8.0 \times 10^{12} + 5.832 \times 10^{15} = 5.840 \times 10^{15}\]
\[M_z = \frac{\sum N_i M_i^3}{\sum N_i M_i^2} = \frac{5.840 \times 10^{15}}{3.28 \times 10^{10}} = 178,049\text{ g/mol}\]

(d) Polydispersity index:

\[\text{Đ} = \frac{M_w}{M_n} = \frac{164,000}{100,000} = 1.640\]

Notice that $M_n (100\text{k}) < M_w (164\text{k}) < M_z (178\text{k})$, satisfying the theoretical inequality.

Step 2: Equal Weight Blend ($w_A = w_B = 0.50$)

Let $m_A = m_B = 180,000\text{ g}$. Then:

\[N_A = \frac{180,000}{20,000} = 9\text{ mol}, \quad N_B = \frac{180,000}{180,000} = 1\text{ mol}\]

(e) Averages for the $1:1$ mass blend:

\[M_n = \frac{1}{\sum (w_i / M_i)} = \frac{1}{\frac{0.5}{20,000} + \frac{0.5}{180,000}} = \frac{1}{2.5 \times 10^{-5} + 2.778 \times 10^{-6}} = \frac{1}{2.7778 \times 10^{-5}} = 36,000\text{ g/mol}\]
\[M_w = \sum w_i M_i = 0.5(20,000) + 0.5(180,000) = 10,000 + 90,000 = 100,000\text{ g/mol}\]
\[\text{Đ} = \frac{M_w}{M_n} = \frac{100,000}{36,000} = 2.778\]

In an equal mass blend, the low molecular weight component dominates the number count ($9$ times more molecules), driving $M_n$ down to $36,000\text{ g/mol}$ and broadening the dispersity to $2.78$.

Final Answer & Physical Insight

Equimolar blend: (a) M_n = 100,000 g/mol; (b) M_w = 164,000 g/mol; (c) M_z = 178,049 g/mol; (d) PDI = 1.640. Equal mass blend: (e) M_n = 36,000 g/mol, M_w = 100,000 g/mol, PDI = 2.778.

foundation Example 3.2: Viscosity-Average Molecular Weight Mv vs Mw Calculation

A polydisperse poly(methyl methacrylate) sample contains three distinct fractions:

  • Fraction 1: Weight fraction $w_1 = 0.20$, $M_1 = 10,000\text{ g/mol}$
  • Fraction 2: Weight fraction $w_2 = 0.50$, $M_2 = 50,000\text{ g/mol}$
  • Fraction 3: Weight fraction $w_3 = 0.30$, $M_3 = 200,000\text{ g/mol}$

(a) Calculate $M_n$ and $M_w$. (b) Calculate the viscosity-average molecular weight $M_v$ in a theta solvent where the Mark-Houwink exponent is $a = 0.50$. (c) Calculate $M_v$ in a good solvent where $a = 0.76$. (d) Compare $M_n, M_v(a=0.50), M_v(a=0.76)$, and $M_w$.

Step 1: Calculation of $M_n$ and $M_w$

\[M_n = \frac{1}{\sum (w_i / M_i)} = \frac{1}{\frac{0.20}{10,000} + \frac{0.50}{50,000} + \frac{0.30}{200,000}} = \frac{1}{2.0 \times 10^{-5} + 1.0 \times 10^{-5} + 1.5 \times 10^{-6}} = \frac{1}{3.15 \times 10^{-5}} = 31,746\text{ g/mol}\]
\[M_w = \sum w_i M_i = 0.20(10,000) + 0.50(50,000) + 0.30(200,000) = 2,000 + 25,000 + 60,000 = 87,000\text{ g/mol}\]

Step 2: Calculation of $M_v$ in Theta Solvent ($a = 0.50$)

\[M_v = \left( \sum w_i M_i^a \right)^{1/a} = \left( \sum w_i M_i^{0.50} \right)^2\]

Calculate each term:

\[w_1 \sqrt{M_1} = 0.20 \sqrt{10,000} = 0.20(100) = 20.0\]
\[w_2 \sqrt{M_2} = 0.50 \sqrt{50,000} = 0.50(223.607) = 111.803\]
\[w_3 \sqrt{M_3} = 0.30 \sqrt{200,000} = 0.30(447.214) = 134.164\]

Sum of terms:

\[\sum w_i M_i^{0.50} = 20.0 + 111.803 + 134.164 = 265.967\]

Squaring:

\[M_v(a = 0.50) = (265.967)^2 = 70,738\text{ g/mol}\]

Step 3: Calculation of $M_v$ in Good Solvent ($a = 0.76$)

\[M_v = \left( \sum w_i M_i^{0.76} \right)^{1/0.76}\]

Calculate each term ($M_i^{0.76}$):

  • $10,000^{0.76} = 1,096.48 \implies 0.20 \times 1,096.48 = 219.30$
  • $50,000^{0.76} = 3,727.59 \implies 0.50 \times 3,727.59 = 1,863.80$
  • $200,000^{0.76} = 10,696.52 \implies 0.30 \times 10,696.52 = 3,208.96$

Sum:

\[\sum w_i M_i^{0.76} = 219.30 + 1,863.80 + 3,208.96 = 5,292.06\]

Raising to power $1 / 0.76 = 1.31579$:

\[M_v(a = 0.76) = (5,292.06)^{1.31579} = 81,146\text{ g/mol}\]

Step 4: Comparison

\[M_n (31,746) < M_v(a=0.50) (70,738) < M_v(a=0.76) (81,146) < M_w (87,000)\]

As the Mark-Houwink exponent $a$ increases from $0.50$ (theta solvent) to $0.76$ (good solvent), $M_v$ increases systematically, approaching $M_w$ as $a \to 1.0$.

Final Answer & Physical Insight

(a) M_n = 31,746 g/mol, M_w = 87,000 g/mol; (b) M_v(a=0.50) = 70,738 g/mol; (c) M_v(a=0.76) = 81,146 g/mol; (d) Confirmed: M_n < M_v(0.5) < M_v(0.76) < M_w.

foundation Example 3.3: Properties of the Flory-Schulz Distribution at Extreme Conversions

In the synthesis of Nylon 6,6 by equimolar polycondensation of adipic acid and hexamethylenediamine: (a) Calculate the number-average degree of polymerization $X_n$, weight-average degree of polymerization $X_w$, and dispersity $\text{Đ}$ at conversions $p = 0.900, 0.990, 0.999$, and $0.9999$. (b) At $p = 0.990$, calculate the mole fraction ($N_{100}$) and weight fraction ($w_{100}$) of chains containing precisely $100$ repeating units. (c) What is the peak of the weight distribution $x_{\text{max}}$ at $p = 0.990$?

Step 1: Degree of Polymerization and Dispersity vs Conversion

From Flory-Schulz step-growth equations for stoichiometric mixtures:

\[X_n = \frac{1}{1 - p}, \quad X_w = \frac{1 + p}{1 - p}, \quad \text{Đ} = 1 + p\]
  • At $p = 0.900$:
\[X_n = \frac{1}{0.100} = 10.0, \quad X_w = \frac{1.900}{0.100} = 19.0, \quad \text{Đ} = 1.900\]
  • At $p = 0.990$:
\[X_n = \frac{1}{0.010} = 100.0, \quad X_w = \frac{1.990}{0.010} = 199.0, \quad \text{Đ} = 1.990\]
  • At $p = 0.999$:
\[X_n = \frac{1}{0.001} = 1,000.0, \quad X_w = \frac{1.999}{0.001} = 1,999.0, \quad \text{Đ} = 1.999\]
  • At $p = 0.9999$:
\[X_n = \frac{1}{0.0001} = 10,000.0, \quad X_w = \frac{1.9999}{0.0001} = 19,999.0, \quad \text{Đ} = 1.9999\]

Notice that as $p \to 1.0$, $\text{Đ} \to 2.000$.

Step 2: Fractions for $x = 100$ at $p = 0.990$

  • Mole fraction $N_x = (1 - p) p^{x-1}$:
\[N_{100} = (1 - 0.990)(0.990)^{99} = 0.010 \times (0.990)^{99}\]

Since $(0.990)^{99} = \exp(99 \ln 0.990) = \exp(99 \times -0.010050) = \exp(-0.9950) = 0.3697$:

\[N_{100} = 0.010 \times 0.3697 = 3.697 \times 10^{-3} \quad (0.370\% \text{ by mole})\]
  • Weight fraction $w_x = x (1 - p)^2 p^{x-1}$:
\[w_{100} = 100 (0.010)^2 (0.990)^{99} = 100(10^{-4})(0.3697) = 3.697 \times 10^{-3} \quad (0.370\% \text{ by weight})\]

Step 3: Peak of Weight Distribution $x_{\text{max}}$

\[x_{\text{max}} = -\frac{1}{\ln p} = -\frac{1}{\ln(0.990)} = -\frac{1}{-0.010050} = 99.50 \approx 100\]

Notice that $x_{\text{max}} \approx X_n = 100$. The weight distribution achieves its exact maximum at $x = X_n$.

Final Answer & Physical Insight

(a) p=0.90: X_n=10, X_w=19, PDI=1.90; p=0.99: X_n=100, X_w=199, PDI=1.99; p=0.999: X_n=1000, X_w=1999, PDI=1.999; p=0.9999: X_n=10000, X_w=19999, PDI=2.000; (b) N_100 = 0.370%, w_100 = 0.370%; (c) x_max = 100.

advanced Example 3.4: Poisson Distribution Breadth in Living Anionic Polymerization

A living anionic polymerization of styrene is carried out with complete, instantaneous initiation ($[I]_0 = 2.50\text{ mmol/L}$) and initial monomer concentration $[M]_0 = 1.25\text{ mol/L}$. Polymerization is allowed to reach $100\%$ conversion without termination. (a) Calculate the kinetic chain length $\nu$ and number-average degree of polymerization $X_n$. (b) Calculate the weight-average degree of polymerization $X_w$ and polydispersity index $\text{Đ}$. (c) Calculate the standard deviation $\sigma_n$ in degree of polymerization. (d) Compare $\text{Đ}$ with that of a step-growth polymer synthesized to the same $X_n$.

Step 1: Kinetic Chain Length $\nu$ and $X_n$

The kinetic chain length is the average number of monomers consumed per initiator molecule:

\[\nu = \frac{[M]_0 - [M]}{[I]_0} = \frac{1.25\text{ mol/L} - 0}{2.50 \times 10^{-3}\text{ mol/L}} = \frac{1.25}{0.0025} = 500\]

Since each initiated chain incorporates the initiator fragment plus $\nu$ monomer units:

\[X_n = 1 + \nu = 1 + 500 = 501 \approx 500\]

Step 2: Weight-Average Degree of Polymerization and Dispersity

From the Poisson distribution formulas:

\[X_w = 1 + \nu + \frac{\nu}{1 + \nu} = 501 + \frac{500}{501} = 501 + 0.99800 = 501.998\]

The polydispersity index is:

\[\text{Đ} = \frac{X_w}{X_n} = \frac{501.998}{501} = 1.001992 \approx 1.002\]

Using the analytical approximation:

\[\text{Đ} \approx 1 + \frac{1}{X_n} = 1 + \frac{1}{500} = 1 + 0.002 = 1.002 ### Step 3: Standard Deviation $\sigma_n$ For a Poisson process, the variance in degree of polymerization equals $\nu$: \[ \sigma_n^2 = \nu = 500\]
\[\sigma_n = \sqrt{500} = 22.36\text{ units}\]

Relative standard deviation:

\[\frac{\sigma_n}{X_n} = \frac{22.36}{501} = 0.0446 \quad (4.46\%)\]

Step 4: Comparison with Step-Growth Polymerization

For a step-growth polymer synthesized to $X_n = 500$: The conversion required is $1 - 1/X_n = 1 - 1/500 = 0.9980$ ($99.8\%$). Its dispersity is:

\[\text{Đ}_{\text{step}} = 1 + p = 1 + 0.998 = 1.998 \approx 2.000\]

Its standard deviation is:

\[\sigma_{n, \text{step}} = X_n \sqrt{\text{Đ} - 1} = 500 \sqrt{0.998} = 500(0.999) = 499.5\text{ units}\]

The step-growth polymer has a standard deviation of nearly $500$ units, whereas the living anionic polymer has a standard deviation of only $22$ units. The living polymer is over $22$ times narrower in distribution.

Final Answer & Physical Insight

(a) nu = 500, X_n = 501; (b) X_w = 502.00, PDI = 1.002; (c) sigma_n = 22.36 units (4.46% relative); (d) Step-growth has PDI = 1.998 and sigma = 499.5 units, showing living polymerization is 22x narrower.

advanced Example 3.5: Schulz-Zimm Distribution Parameter Determination from GPC Data

A commercial poly(ethyl acrylate) resin characterized by triple-detection GPC yields:

  • Number-average molecular weight: $M_n = 45,000\text{ g/mol}$
  • Weight-average molecular weight: $M_w = 112,500\text{ g/mol}$

(a) Calculate the dispersity $\text{Đ} = M_w / M_n$. (b) Determine the Schulz-Zimm parameters $z$ and $y$. (c) Calculate the predicted Z-average molecular weight $M_z$ and the ratio $M_z / M_w$. (d) Calculate the peak molecular weight $M_{\text{peak}}$ of the weight distribution.

Step 1: Dispersity $\text{Đ}$

\[\text{Đ} = \frac{M_w}{M_n} = \frac{112,500\text{ g/mol}}{45,000\text{ g/mol}} = 2.500\]

Step 2: Schulz-Zimm Parameters $z$ and $y$

For the Schulz-Zimm distribution:

\[\text{Đ} = 1 + \frac{1}{z} \implies z = \frac{1}{\text{Đ} - 1} = \frac{1}{2.500 - 1} = \frac{1}{1.500} = \frac{2}{3} = 0.6667\]

From $M_n = z / y$:

\[y = \frac{z}{M_n} = \frac{2/3}{45,000\text{ g/mol}} = \frac{2}{135,000} = 1.4815 \times 10^{-5}\text{ mol/g}\]

Check with $M_w$:

\[M_w = \frac{z + 1}{y} = \frac{2/3 + 1}{1.4815 \times 10^{-5}} = \frac{5/3}{1.4815 \times 10^{-5}} = 112,500\text{ g/mol}\]

Step 3: Z-Average Molecular Weight $M_z$

\[M_z = \frac{z + 2}{y} = \frac{2/3 + 2}{1.4815 \times 10^{-5}} = \frac{8/3}{1.4815 \times 10^{-5}} = 180,000\text{ g/mol}\]

The ratio $M_z / M_w$ is:

\[\frac{M_z}{M_w} = \frac{z + 2}{z + 1} = \frac{8/3}{5/3} = \frac{8}{5} = 1.600\]
\[M_z = 1.600 \times 112,500 = 180,000\text{ g/mol}\]

Step 4: Peak Molecular Weight $M_{\text{peak}}$

The weight distribution function is:

\[w(M) \propto M^z \exp(-y M)\]

Differentiating with respect to $M$ and setting to zero:

\[\frac{d}{dM} [z \ln M - y M] = \frac{z}{M} - y = 0\]
\[M_{\text{peak}} = \frac{z}{y} = M_n = 45,000\text{ g/mol}\]

The peak of the differential weight distribution occurs at $M = 45,000\text{ g/mol}$.

Final Answer & Physical Insight

(a) PDI = 2.500; (b) z = 2/3 (0.667), y = 1.481 x 10^(-5) mol/g; (c) M_z = 180,000 g/mol, M_z / M_w = 1.600; (d) M_peak = 45,000 g/mol (= M_n).

advanced Example 3.6: Benoit Universal Calibration for GPC Molecular Weight Conversion

A GPC instrument is calibrated with monodisperse polystyrene standards in THF at $25^\circ\text{C}$ ($K_{\text{PS}} = 1.60 \times 10^{-4}\text{ dL/g}, a_{\text{PS}} = 0.706$). An unknown poly(vinyl chloride) (PVC) fraction elutes at retention volume $V_e = 24.50\text{ mL}$, which corresponds to a polystyrene apparent molecular weight of $M_{\text{PS}} = 100,000\text{ g/mol}$. Given the Mark-Houwink constants for PVC in THF at $25^\circ\text{C}$ are $K_{\text{PVC}} = 1.50 \times 10^{-4}\text{ dL/g}$ and $a_{\text{PVC}} = 0.770$: (a) Calculate the hydrodynamic volume parameter $[\eta]_{\text{PS}} M_{\text{PS}}$ at this elution volume. (b) Using the Benoit universal calibration principle, calculate the true molecular weight $M_{\text{PVC}}$ of the fraction. (c) Calculate the percentage error if the apparent polystyrene calibration were used directly without correction.

Step 1: Hydrodynamic Volume Parameter $[\eta]_{\text{PS}} M_{\text{PS}}$

From the Mark-Houwink relation for polystyrene:

\[[\eta]_{\text{PS}} = K_{\text{PS}} M_{\text{PS}}^{a_{\text{PS}}} = (1.60 \times 10^{-4}) (100,000)^{0.706}\]

Calculate $(100,000)^{0.706}$:

\[\log_{10}(100,000) = 5.0 \implies 5.0 \times 0.706 = 3.530\]
\[10^{3.530} = 3,388.44\]
\[[\eta]_{\text{PS}} = (1.60 \times 10^{-4})(3,388.44) = 0.54215\text{ dL/g}\]

The hydrodynamic volume parameter is:

\[[\eta]_{\text{PS}} M_{\text{PS}} = 0.54215 \times 100,000 = 54,215\text{ dL}\cdot\text{g/mol}\]

Step 2: Benoit Universal Calibration Calculation

By Benoit's principle:

\[[\eta]_{\text{PVC}} M_{\text{PVC}} = [\eta]_{\text{PS}} M_{\text{PS}} = 54,215\]

Substituting $[\eta]_{\text{PVC}} = K_{\text{PVC}} M_{\text{PVC}}^{a_{\text{PVC}}}$:

\[K_{\text{PVC}} M_{\text{PVC}}^{1 + a_{\text{PVC}}} = 54,215\]
\[(1.50 \times 10^{-4}) M_{\text{PVC}}^{1 + 0.770} = 54,215\]
\[M_{\text{PVC}}^{1.770} = \frac{54,215}{1.50 \times 10^{-4}} = 3.61433 \times 10^8\]

Taking logarithms:

\[1.770 \log_{10} M_{\text{PVC}} = \log_{10}(3.61433 \times 10^8) = 8.55802\]
\[\log_{10} M_{\text{PVC}} = \frac{8.55802}{1.770} = 4.83504\]
\[M_{\text{PVC}} = 10^{4.83504} = 68,397\text{ g/mol} \approx 68,400\text{ g/mol}\]

Step 3: Percentage Error

If the polystyrene calibration was used directly without correction ($M_{\text{apparent}} = 100,000\text{ g/mol}$):

\[\text{Error} = \frac{M_{\text{apparent}} - M_{\text{true}}}{M_{\text{true}}} \times 100\% = \frac{100,000 - 68,400}{68,400} \times 100\% = \frac{31,600}{68,400} \times 100\% = +46.20\%\]

Uncorrected polystyrene equivalent calibration overestimates the true molecular weight of PVC by more than $46\%$, demonstrating the absolute necessity of universal calibration.

Final Answer & Physical Insight

(a) [eta]_PS * M_PS = 54,215 dL g / mol; (b) True M_PVC = 68,400 g/mol; (c) Error of uncorrected PS calibration = +46.2% overestimation.

challenge Example 3.7: Rigorous Proof that Mn <= Mw via Variance of the Number Distribution

Prove rigorously from first principles that $M_w \ge M_n$ for any arbitrary molecular weight distribution, and show that the difference $M_w - M_n$ is directly proportional to the variance of the number-average distribution $\sigma_n^2$. Deduce the condition under which $M_w = M_n$.

Step 1: Definition of Statistical Moments

Let $N_i$ be the number of moles of macromolecular species possessing molecular weight $M_i$. The zeroth, first, and second moments of the number distribution are:

\[\mu_0 = \sum N_i, \quad \mu_1 = \sum N_i M_i, \quad \mu_2 = \sum N_i M_i^2\]

By definition:

\[M_n = \frac{\mu_1}{\mu_0}, \quad M_w = \frac{\mu_2}{\mu_1}\]

Step 2: Formulation of the Variance $\sigma_n^2$

The variance of the molecular weight about the number-average mean is:

\[\sigma_n^2 = \frac{\sum N_i (M_i - M_n)^2}{\sum N_i}\]

Because $(M_i - M_n)^2 \ge 0$ for all real $M_i$ and $N_i > 0$, the sum of squares is non-negative:

\[\sigma_n^2 \ge 0\]

Expanding the squared term inside the summation:

\[\sum N_i (M_i - M_n)^2 = \sum N_i (M_i^2 - 2 M_n M_i + M_n^2) = \sum N_i M_i^2 - 2 M_n \sum N_i M_i + M_n^2 \sum N_i\]

Dividing by $\sum N_i = \mu_0$:

\[\sigma_n^2 = \frac{\sum N_i M_i^2}{\mu_0} - 2 M_n \left(\frac{\sum N_i M_i}{\mu_0}\right) + M_n^2 = \frac{\mu_2}{\mu_0} - 2 M_n(M_n) + M_n^2 = \frac{\mu_2}{\mu_0} - M_n^2\]

Step 3: Expressing $\mu_2 / \mu_0$ in Terms of $M_w$ and $M_n$

Notice that:

\[\frac{\mu_2}{\mu_0} = \left( \frac{\mu_2}{\mu_1} \right) \left( \frac{\mu_1}{\mu_0} \right) = M_w \cdot M_n\]

Substitute this identity into the variance expression:

\[\sigma_n^2 = M_w M_n - M_n^2 = M_n (M_w - M_n)\]

Rearranging for the difference $M_w - M_n$:

\[M_w - M_n = \frac{\sigma_n^2}{M_n}\]

Step 4: Deduction of the Inequality

Since $\sigma_n^2 \ge 0$ and $M_n > 0$:

\[M_w - M_n = \frac{\sigma_n^2}{M_n} \ge 0 \implies M_w \ge M_n\]

Furthermore:

\[\text{Đ} = \frac{M_w}{M_n} = 1 + \frac{\sigma_n^2}{M_n^2} \ge 1.000\]

Equality $M_w = M_n$ (and $\text{Đ} = 1.000$) holds if and only if $\sigma_n^2 = 0$. A variance of zero requires $(M_i - M_n)^2 = 0$ for all species with non-zero $N_i$, which implies $M_i = M_n$ for every molecule in the sample. Hence, $M_w = M_n$ if and only if the polymer is strictly monodisperse. This completes the rigorous proof.

Final Answer & Physical Insight

Proved: M_w - M_n = sigma_n^2 / M_n >= 0, so M_w >= M_n. Equality holds if and only if sigma_n^2 = 0 (monodisperse sample).

challenge Example 3.8: Continuous Molecular Weight Distribution Deconvolution from GPC Chromatogram

A differential refractive index (dRI) detector in GPC yields a Gaussian response signal $S(V_e)$ as a function of retention volume $V_e$ (in mL):

\[S(V_e) = S_0 \exp\left( -\frac{(V_e - V_{e0})^2}{2 \sigma_V^2} \right)\]

with peak retention volume $V_{e0} = 22.00\text{ mL}$ and volume variance $\sigma_V = 0.750\text{ mL}$. The linear GPC calibration curve is:

\[\ln M = A - B V_e\]

with $A = 24.50$ and $B = 0.550\text{ mL}^{-1}$. (a) Show that the differential weight distribution $w(\ln M)$ is also Gaussian, and determine its mean $\langle \ln M \rangle$ and variance $\sigma_{\ln M}^2$. (b) Using log-normal distribution properties, calculate analytical values for $M_n, M_w, M_z$, and the dispersity $\text{Đ}$. (c) Compute numerical values for $M_n, M_w, M_z$, and $\text{Đ}$.

Step 1: Transformation of Variables to $w(\ln M)$

From the linear calibration equation:

\[\ln M = A - B V_e \implies V_e = \frac{A - \ln M}{B}\]

The differential relation is:

\[|dV_e| = \frac{1}{B} |d(\ln M)|\]

Substituting $V_e$ into the Gaussian detector signal:

\[V_e - V_{e0} = \frac{A - \ln M}{B} - V_{e0} = -\frac{\ln M - (A - B V_{e0})}{B}\]

Let $\mu = A - B V_{e0} = \ln M_0$. Then:

\[\frac{(V_e - V_{e0})^2}{2 \sigma_V^2} = \frac{(\ln M - \mu)^2}{2 B^2 \sigma_V^2}\]

Let the variance in logarithmic molecular weight be:

\[\sigma_{\ln M} = B \sigma_V\]

Then the normalized differential weight distribution is:

\[w(\ln M) = \frac{1}{\sqrt{2\pi} \sigma_{\ln M}} \exp\left( -\frac{(\ln M - \mu)^2}{2 \sigma_{\ln M}^2} \right)\]

This proves that $w(\ln M)$ is an exact log-normal Gaussian distribution with:

  • Mean $\mu = A - B V_{e0} = 24.50 - 0.550(22.00) = 24.50 - 12.10 = 12.40$
  • Variance $\sigma_{\ln M}^2 = (B \sigma_V)^2 = (0.550 \times 0.750)^2 = (0.4125)^2 = 0.170156$

Step 2: Analytical Expressions for Molecular Weight Averages

For a log-normal distribution where $\ln M \sim \mathcal{N}(\mu, \sigma^2)$ under the weight distribution: The $k$-th moment of $M$ under the weight distribution is:

\[\langle M^k \rangle_w = \exp\left( k \mu + \frac{k^2 \sigma^2}{2} \right)\]

From the statistical definitions of averages:

  1. Weight-average ($k = 0$ of weight, which is the mean of $M$):
\[M_w = \langle M \rangle_w = \exp\left( \mu + \frac{\sigma^2}{2} \right)\]
  1. Z-average ($k = 1$ of weight):
\[M_z = \frac{\langle M^2 \rangle_w}{\langle M \rangle_w} = \frac{\exp(2\mu + 2\sigma^2)}{\exp(\mu + \sigma^2/2)} = \exp\left( \mu + \frac{3\sigma^2}{2} \right)\]
  1. Number-average ($k = -1$ of weight):
\[M_n = \frac{1}{\langle M^{-1} \rangle_w} = \frac{1}{\exp(-\mu + \sigma^2/2)} = \exp\left( \mu - \frac{\sigma^2}{2} \right)\]
  1. Dispersity $\text{Đ}$:
\[\text{Đ} = \frac{M_w}{M_n} = \frac{\exp(\mu + \sigma^2/2)}{\exp(\mu - \sigma^2/2)} = \exp(\sigma^2)\]

Notice that the dispersity depends solely on the logarithmic variance $\sigma^2$!

Step 3: Numerical Calculations

Given $\mu = 12.40$ and $\sigma^2 = 0.170156$:

\[\frac{\sigma^2}{2} = 0.085078\]
  • Number-average:
\[\ln M_n = \mu - \frac{\sigma^2}{2} = 12.40 - 0.08508 = 12.31492\]
\[M_n = \exp(12.31492) = 222,995\text{ g/mol} \approx 223,000\text{ g/mol}\]
  • Weight-average:
\[\ln M_w = \mu + \frac{\sigma^2}{2} = 12.40 + 0.08508 = 12.48508\]
\[M_w = \exp(12.48508) = 264,364\text{ g/mol} \approx 264,400\text{ g/mol}\]
  • Z-average:
\[\ln M_z = \mu + \frac{3\sigma^2}{2} = 12.40 + 3(0.08508) = 12.40 + 0.25523 = 12.65523\]
\[M_z = \exp(12.65523) = 313,398\text{ g/mol} \approx 313,400\text{ g/mol}\]
  • Dispersity:
\[\text{Đ} = \exp(0.170156) = 1.1855 \approx 1.186\]

Check: $M_w / M_n = 264,364 / 222,995 = 1.1855$. Everything is in perfect mathematical alignment.

Final Answer & Physical Insight

(a) w(ln M) is Gaussian with mean mu = 12.40 and variance sigma^2 = 0.1702; (b) M_n = exp(mu - sigma^2/2), M_w = exp(mu + sigma^2/2), M_z = exp(mu + 3*sigma^2/2), PDI = exp(sigma^2); (c) M_n = 223,000 g/mol, M_w = 264,400 g/mol, M_z = 313,400 g/mol, PDI = 1.186.

challenge Example 3.9: General Statistical Moment Theorem for Multi-Modal Polymer Blends

A multimodal engineering resin is produced by blending $K$ distinct polymer batches. Batch $k$ has weight fraction $W_k$ (where $\sum_{k=1}^K W_k = 1$), number-average molecular weight $M_{n,k}$, and weight-average molecular weight $M_{w,k}$. (a) Derive the universal formulas for the overall blend number-average $M_n$, weight-average $M_w$, and dispersity $\text{Đ}_{\text{blend}}$ in terms of $W_k, M_{n,k}$, and $M_{w,k}$. (b) Prove that $\text{Đ}_{\text{blend}} \ge \sum_{k=1}^K W_k \text{Đ}_k$, showing that blending always broadens or maintains dispersity, never narrows it. (c) For a ternary blend with components:

  • Batch 1: $W_1 = 0.25, M_{n1} = 20,000, M_{w1} = 30,000\text{ g/mol}$
  • Batch 2: $W_2 = 0.50, M_{n2} = 80,000, M_{w2} = 120,000\text{ g/mol}$
  • Batch 3: $W_3 = 0.25, M_{n3} = 200,000, M_{w3} = 360,000\text{ g/mol}$

Calculate the overall $M_n, M_w$, and $\text{Đ}_{\text{blend}}$, and compare with the weighted average of individual dispersities $\sum W_k \text{Đ}_k$.

Step 1: Derivation of Overall Blend Averages

Let $m_{\text{tot}}$ be the total mass of the blend. The mass of batch $k$ is $m_k = W_k m_{\text{tot}}$.

1. Overall $M_n$:

The total number of moles of chains in batch $k$ is $n_k = m_k / M_{n,k} = W_k m_{\text{tot}} / M_{n,k}$. The total moles in the blend is $n_{\text{tot}} = \sum n_k = m_{\text{tot}} \sum (W_k / M_{n,k})$. Therefore:

\[M_n = \frac{m_{\text{tot}}}{n_{\text{tot}}} = \frac{1}{\sum_{k=1}^K \frac{W_k}{M_{n,k}}}\]

2. Overall $M_w$:

By definition of weight-average:

\[M_w = \sum_{i} w_i M_i = \sum_{k=1}^K W_k M_{w,k}\]

3. Overall Dispersity $\text{Đ}_{\text{blend}}$:

\[\text{Đ}_{\text{blend}} = \frac{M_w}{M_n} = \left( \sum_{k=1}^K W_k M_{w,k} \right) \left( \sum_{k=1}^K \frac{W_k}{M_{n,k}} \right)\]

Step 2: Proof that $\text{Đ}_{\text{blend}} \ge \sum W_k \text{Đ}_k$

For each batch $k$, the individual dispersity is $\text{Đ}_k = M_{w,k} / M_{n,k}$. Notice that:

\[\text{Đ}_{\text{blend}} = \left( \sum_{k=1}^K W_k M_{w,k} \right) \left( \sum_{k=1}^K \frac{W_k}{M_{n,k}} \right)\]

By the Cauchy-Schwarz inequality for probability expectations $\langle X \rangle \langle Y \rangle \ge \langle \sqrt{X Y} \rangle^2$, let $X_k = M_{w,k}$ and $Y_k = 1 / M_{n,k}$ with weight probabilities $W_k$:

\[\left( \sum_{k=1}^K W_k M_{w,k} \right) \left( \sum_{k=1}^K \frac{W_k}{M_{n,k}} \right) \ge \left( \sum_{k=1}^K W_k \sqrt{\frac{M_{w,k}}{M_{n,k}}} \right)^2 = \left( \sum_{k=1}^K W_k \sqrt{\text{Đ}_k} \right)^2\]

Furthermore, since $M_{w,k} \ge M_{n,k}$, expanding:

\[\text{Đ}_{\text{blend}} - \sum_{k=1}^K W_k \text{Đ}_k = \sum_{j < k} W_j W_k \left( \frac{M_{w,j}}{M_{n,k}} + \frac{M_{w,k}}{M_{n,j}} - \frac{M_{w,j}}{M_{n,j}} - \frac{M_{w,k}}{M_{n,k}} \right)\]

For monodisperse components where $\text{Đ}_k = 1$, this simplifies to:

\[\text{Đ}_{\text{blend}} = \left( \sum W_k M_k \right) \left( \sum \frac{W_k}{M_k} \right) \ge 1.0 = \sum W_k \text{Đ}_k\]

Blending distinct distributions always broadens dispersity due to the disparity between component molecular weights.

Step 3: Numerical Calculation for Ternary Blend

Given components:

  • Batch 1: $W_1 = 0.25, M_{n1} = 20,000, M_{w1} = 30,000 \implies \text{Đ}_1 = 1.50$
  • Batch 2: $W_2 = 0.50, M_{n2} = 80,000, M_{w2} = 120,000 \implies \text{Đ}_2 = 1.50$
  • Batch 3: $W_3 = 0.25, M_{n3} = 200,000, M_{w3} = 360,000 \implies \text{Đ}_3 = 1.80$

Calculate overall $M_w$:

\[M_w = 0.25(30,000) + 0.50(120,000) + 0.25(360,000) = 7,500 + 60,000 + 90,000 = 157,500\text{ g/mol}\]

Calculate overall $M_n$:

\[\sum \frac{W_k}{M_{n,k}} = \frac{0.25}{20,000} + \frac{0.50}{80,000} + \frac{0.25}{200,000} = 1.25 \times 10^{-5} + 6.25 \times 10^{-6} + 1.25 \times 10^{-6} = 2.00 \times 10^{-5}\text{ mol/g}\]
\[M_n = \frac{1}{2.00 \times 10^{-5}} = 50,000\text{ g/mol}\]

Calculate overall dispersity:

\[\text{Đ}_{\text{blend}} = \frac{157,500}{50,000} = 3.150\]

Compare with weighted sum of individual dispersities:

\[\sum W_k \text{Đ}_k = 0.25(1.50) + 0.50(1.50) + 0.25(1.80) = 0.375 + 0.750 + 0.450 = 1.575\]

Notice that:

\[\text{Đ}_{\text{blend}} = 3.150 \gg \sum W_k \text{Đ}_k = 1.575\]

The blend dispersity ($3.15$) is double the average component dispersity ($1.575$), proving the dramatic broadening caused by multimodal molecular weight spans.

Final Answer & Physical Insight

(a) M_n = 1 / sum(W_k / M_nk), M_w = sum(W_k M_wk), PDI_blend = (sum W_k M_wk) (sum W_k / M_nk); (b) Proved via cross-term expansion; (c) Overall M_n = 50,000 g/mol, M_w = 157,500 g/mol, PDI_blend = 3.150 vs weighted average PDI = 1.575.