Chemical Engineering 160/260Polymer Science and Engineering

Lecture 9 -Lecture 9 - Flory Flory--Huggins Huggins ModelModelfor Polymer Solutionsfor Polymer Solutions

February 5, 2001February 5, 2001

Read Sperling, Chapter 4

Objectives

!! To develop the classicalTo develop the classical FloryFlory--HugginsHuggins theory for the freetheory for the freeenergy of mixing of polymer solutions based on aenergy of mixing of polymer solutions based on astatistical approach on a regular lattice.statistical approach on a regular lattice.

!! To describe the criteria for phase stability and illustrateTo describe the criteria for phase stability and illustrate

typical phase diagrams for polymer blends and solutions. typical phase diagrams for polymer blends and solutions.

Outline

!! Lattice Theory for Solutions of Small MoleculesLattice Theory for Solutions of Small Molecules

"" Thermodynamic probability and the Thermodynamic probability and the Boltzmann Boltzmann EquationEquation

"" Ideal solutionIdeal solution

!! FloryFlory--Huggins Huggins Theory of Polymer SolutionsTheory of Polymer Solutions

"" Placement of a new polymer molecule on a partially filled latticePlacement of a new polymer molecule on a partially filled lattice

"" Entropy of mixingEntropy of mixing

"" Enthalpy of mixing (for Enthalpy of mixing (for dispersive dispersive or dipole-dipole interactions)or dipole-dipole interactions)

"" Cohesive energy density and solubility parameterCohesive energy density and solubility parameter

"" Free energy of mixingFree energy of mixing

Lattice Theory for Solutions ofSmall Molecules

Assume that a solution may be formed by distributing thepure components on the sites of a regular lattice. Furtherassume that there are N1 molecules of Type 1, N2 moleculesof Type 2, and that Type 1 and Type 2 molecules areindistinguishable but identical in size and interaction energy.

Small molecule of Type 1(e.g. solvent)

Small molecule of Type 2(e.g. solute)

Thermodynamic Probability

Place the molecules on the N = N1 + N2 sites of a three-dimensional lattice.

Ω is the thermodynamic probability, which counts the numberof ways that a particular state can come about.

Total no. different waysof arranging molecules ofTypes 1 and 2 on the lattice

Ω =N

N N

!! !1 2

≡

Total number of arrangementsof N molecules

Interchanging the 1’s or 2’s makes no difference.

Boltzmann Equation

The thermodynamic probability (or the number of ways thatthe system may come about )may be related to the entropyof the system through a fundamental equation from statisticalthermodynamics that is known as the Boltzmann Equation.

S k= lnΩ

Configurational Entropy

∆ΩΩ

S S S k= − =

2 1

2

1

ln∆S > 0

∆S < 0

Ω Ω2 1>

Ω Ω2 1<Consider the entropy of the mixture:

S k k N N Nmix = = − −( )ln ln ! ln ! ln !Ω 1 2

ln ! lny y y y≅ −Stirling’s approximation:

S k NN

NN

N

Nmix = −

+

1

12

2ln lnN

NA

A

Multiply r.h.s.by

S R x x x xmix = − +( )1 1 2 2ln ln

∆S S S S R x xm mix i i= − − = − ∑1 2 ln S Smix m= ∆

Smix is that part of the total entropy of the mixture arising fromthe mixing process itself. This is the configurational entropy.

Apply the Boltzmann Equation to the mixing process:

N NA≡

Ideal Solution of Small Molecules

∆S R x xm i i= − ∑ ln

Η∆Hm = 0

∆Vm = 0

∆G RT x xm i i= − ∑ ln

What entropy effects can you envision other than theconfigurational entropy?

If the 1-1, 2-2, and 1-2 interactions are equal, then

If the solute and the solvent molecules are the same size,

The thermodynamics of mixing will be governed by the Gibbs free energy of mixing.

Do you expect a polymersolution to be ideal?

(athermal mixing)

Lattice Approach to Polymer Solutions

To place a macromolecule on a lattice, it is necessary that thepolymer segments, which do not necessarily correspond to a single repeat unit, are situated in a contiguous string.

Small molecule of Type 1(solvent)

Macromoleculeof Type 2(solute)

Flory-Huggins Theory ofPolymer Solutions

Assume (for now) that the polymer-solvent system showsathermal mixing. Let the system consist of N1 solvent molecules,each occupying a single site and N2 polymer molecules, eachoccupying n lattice sites.

N nN N1 2+ =

What assumption about molecular weight distribution isimplicit in the system chosen?

Placement of a New Polymer Moleculeon a Partially Filled Lattice

Let (i) polymer molecules be initially placed on an empty latticeand determine the number of ways that the (i + 1)st polymermolecule can be placed on the lattice.

How can we get the (i+1)stmolecule to fit on the lattice?

Placement of Polymer Segments on a LatticePlacement of first segment of polymer (i + 1):

N ni− = Number of remaining sitesNumber of ways to add segment 1

Placement of second segment of polymer (i + 1):

Let Z = coordination number of the lattice

ZN ni

N

−

= Number of ways to add segment 2

Average fraction of vacant sites on the lattice as a whole (When is this most valid?)

Number of latticesites adjacent to thefirst segment

Number of sites occupiedby initial i polymer molecules

Probability of Placement of the (i)th MoleculePlacement of third segment (and all others) of polymer (i + 1):

ZN ni

N−( ) −

=1 Number of ways to add segment 3

One site on the coordination sphereis occupied by the second segment

Ignore contributions to theaverage site vacancy due to segments of molecule (i + 1)

Thus, the (i + 1)st polymer molecule may be placed on a latticealready containing (i) molecules in ωi+1 ways.

ω i

n

N ni ZN ni

NZ

N ni

N+

−

= −( ) −

−( ) −

1

2

1

ω in

n

Z Z NN ni

N+−= −( ) −

1

21

For the (i)th molecule: ω in

n

Z Z NN n i

N= −( ) − −( )

−112

Total Number of Ways of Placing N2

Polymer Molecules on a Lattice

Ω = =

=∏

ω ω ω ωω1 2

2 2 1

2

21L Li N

i

N

iN N! !

Apply Boltzmann’s Equation:S kN imix

i

N

=

=

∏ln!

1

2 1

2

ωSubstitute for ωi to obtain:

Ω =−( )

− −( )[ ]−( )

−=∏

Z Z

N NN n i

N N n

N ni

Nn

2 2

2

211

2

21

1! ( )

Examine the product:

i

Nn nN

i

N n

N n i nN

ni

= =∏ ∏− −( )[ ] = + −

1 1

2

2

2

1 1nN

nN=

Entropy of the MixtureWrite out several terms in the product expression:

∏ = + −

+ −

+ −

+ −

N

n

N

n

N

n

N

nN

n n n n

1 1 1 2 1 3 1 2L

Note that:

N

nN

nN

N

nN

N

nN

N

nN

nN

−

=( )( )( ) −

− +

( )( )( ) −

!

!2

2 2

2

1 2 3 1

1 2 3

L L

L

Thus

∏ =

−

N

nN

nN

n

!

!2

Ω =−( )

−

−( )

−( )Z Z n

N N

N

nN

nN

N N n nN

N n

n

2 2 2

2

1 2

21

2!

!

!

Apply Stirling’s approximation to obtain:S

kN

nN

NN

N

N

N Z n Z n n

mix = −

−

+ + −( ) −( ) + −( ) +[ ]

22

11

2 2 1 1

ln ln

ln ln ln

Flory-Huggins Entropy of MixingCalculate entropy of pure solvent and pure polymer:

Pure solvent: N2 0= S1 0=

Pure polymer: N1 0=

S kN Z n Z n n2 2 2 1 1= + −( ) −( ) + −( ) +[ ]ln ln ln

Entropy of the disordered polymer when it fills the lattice

∆S S S Sm mix= − −2 1 ∆S k NN

NN

nN

Nm = −

+

1

12

2ln ln

Multiply and divide r.h.s. by N1+N2 and assume N1+N2=NA

∆S R xN

Nx

nN

Nm = −

+

1

12

2ln lnN

N1

1= ϕnN

N2

2= ϕ

Calculate entropy of mixing:

Flory-Huggins Theory for anAthermal Solution

∆G RT x xm = − +[ ]1 1 2 2ln lnϕ ϕ

∆S R x xm = − +[ ]1 1 2 2ln lnϕ ϕ

Entropy of mixing:

Gibbs free energy of mixing:

Enthalpy of mixing:

∆Hm = 0

Concentration Conversions

ϕϕ

2

1

2

1

=nN

NN

N n2

1

2

1

1=

ϕϕ

xN

N N

N

NN

N

n

n

22

1 2

2

1

2

1

2

1

2

1

1

1

11

=+

=+

=

+

ϕϕ

ϕϕ

x x1 2 1+ = ϕ ϕ1 2 1+ =

xn

n

2

2

2

2

2

11

11

1

=

−

+ −

ϕϕ

ϕϕ

ϕ22

2

11

1=+ −

x

nx

n

Flory-Huggins Enthalpy of MixingUse the same lattice model as for the entropy of mixing, andconsider a quasi-chemical reaction:

(1,1) + (2,2) 2(1,2)

1 represents a solvent and 2 represents a polymer repeat unit

The interaction energy is then given by:

∆w w w w= − −2 12 11 22

∆w

2= Change in interaction energy per (1,2) pair

Define the system to be a filled lattice with Z nearest neighbors.Each polymer segment is then surrounded by Zϕ2 polymer segments and Zϕ1 solvent molecules.

Contributions to the Interaction EnergyContributions of polymer segments

Interaction of a polymer segment with its neighbors yields

Z w Z wϕ ϕ2 22 1 12+

The total contribution is 12

12 1 22 1 12

−( ) +[ ]Z N w wϕ ϕ ϕ

Contributions of solvent molecules

Each solvent molecule is surrounded by Zϕ2 polymer segmentsand Zϕ1 solvent molecules. Interaction of the solvent with itsneighbors then yields Z w Z wϕ ϕ2 12 1 11+

The total contribution is 12

11 2 12 2 11

+ −( )[ ]Z N w wϕ ϕ ϕ

Remove double counting

Remove double counting

Flory-Huggins Enthalpy of Mixing

∆H ZN w w wm =

− −( )12

2 1 2 12 1 2 11 1 2 22ϕ ϕ ϕ ϕ ϕ ϕ

∆ ∆H ZN wm =

12 1 2ϕ ϕ

Let12

=Z w RT∆ χ Flory-Huggins interactionparameter (the “chi” parameter)

χ

χ

χ

=

>

<

0

0

0

∆H N RT N RTm = =ϕ ϕ χ ϕ χ1 2 1 2

For athermal mixturesFor endothermic mixingFor exothermic mixing

Flory-Huggins Free Energy of Mixing:General Case

∆ ∆ ∆G H T Sm m m= −

∆G RT N x xm = + +( )[ ]ϕ ϕ χ ϕ ϕ1 2 1 1 2 2ln ln

∆H N RTm = ϕ ϕ χ1 2

∆S R x xm = − +[ ]1 1 2 2ln lnφ φ

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