Chemical Engineering 160/260 Polymer Science and Engineering
LectureLecture 99 -- FloryFlory--HugginsHuggins ModelModel forfor PolymerPolymer SolutionsSolutions FebruaryFebruary 5,5, 20012001
Read Sperling, Chapter 4 Objectives
! To develop the classical Flory-Huggins theory for the free energy of mixing of polymer solutions based on a statistical approach on a regular lattice. ! To describe the criteria for phase stability and illustrate typical phase diagrams for polymer blends and solutions. Outline
! Lattice Theory for Solutions of Small Molecules " Thermodynamic probability and the Boltzmann Equation " Ideal solution ! Flory-Huggins Theory of Polymer Solutions " Placement of a new polymer molecule on a partially filled lattice " Entropy of mixing " Enthalpy of mixing (for dispersive or dipole-dipole interactions) " Cohesive energy density and solubility parameter " Free energy of mixing Lattice Theory for Solutions of Small Molecules Assume that a solution may be formed by distributing the pure components on the sites of a regular lattice. Further assume that there are N1 molecules of Type 1, N2 molecules of Type 2, and that Type 1 and Type 2 molecules are indistinguishable 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. Total number of arrangements of N molecules Total no. different ways N! of arranging molecules of Ω = ≡ NN!! Types 1 and 2 on the lattice 12 Interchanging the 1’s or 2’s makes no difference.
Ω is the thermodynamic probability, which counts the number of ways that a particular state can come about. Boltzmann Equation
The thermodynamic probability (or the number of ways that the system may come about )may be related to the entropy of the system through a fundamental equation from statistical thermodynamics that is known as the Boltzmann Equation. Sk= lnΩ Configurational Entropy Apply the Boltzmann Equation to the mixing process:
Ω 2 ΩΩ21> ∆S > 0 ∆SS= 21− S= kln Ω 1 ΩΩ21< ∆S < 0 Consider the entropy of the mixture:
Skmix ==lnΩ kNN() ln !−− ln12 ! ln N ! NN≡ Stirling’s approximation: lnyyyy !≅− ln A
N1 N2 NA SkNln N ln Multiply r.h.s.by mix = − 1 + 2 N N NA
SRxxxxmix = − ()112ln+ ln 2 ln ∆SSm= mix−− SS12= − Rxx∑ i i SSmix= ∆ m
Smix is that part of the total entropy of the mixture arising from the mixing process itself. This is the configurational entropy. Ideal Solution of Small Molecules What entropy effects can you envision other than the configurational entropy? ln ∆SRxxmii= − ∑ If the 1-1, 2-2, and 1-2 interactions are equal, then
Η ∆Hm = 0 (athermal mixing) If the solute and the solvent molecules are the same size,
∆Vm = 0 The thermodynamics of mixing will be governed by the Gibbs free energy of mixing. GRTxxln Do you expect a polymer ∆ mii= − ∑ solution to be ideal? Lattice Approach to Polymer Solutions
To place a macromolecule on a lattice, it is necessary that the polymer segments, which do not necessarily correspond to a single repeat unit, are situated in a contiguous string.
Small molecule of Type 1 (solvent)
Macromolecule of Type 2 (solute) Flory-Huggins Theory of Polymer Solutions Assume (for now) that the polymer-solvent system shows athermal mixing. Let the system consist of N1 solvent molecules, each occupying a single site and N2 polymer molecules, each occupying n lattice sites.
NnNN12+=
What assumption about molecular weight distribution is implicit in the system chosen? Placement of a New Polymer Molecule on a Partially Filled Lattice Let (i) polymer molecules be initially placed on an empty lattice and determine the number of ways that the (i + 1)st polymer molecule can be placed on the lattice.
How can we get the (i+1)st molecule to fit on the lattice? Placement of Polymer Segments on a Lattice Placement of first segment of polymer (i + 1): Nni− = Number of remaining sites Number of ways to add segment 1 Number of sites occupied by initial i polymer molecules
Placement of second segment of polymer (i + 1): Let Z = coordination number of the lattice Nni− Z = Number of ways to add segment 2 N Number of lattice Average fraction of vacant sites on sites adjacent to the the lattice as a whole (When is this first segment most valid?) Probability of Placement of the (i)th Molecule Placement of third segment (and all others) of polymer (i + 1): Nni− ()Z −1 = Number of ways to add segment 3 N Ignore contributions to the One site on the coordination sphere average site vacancy due to is occupied by the second segment segments of molecule (i + 1) Thus, the (i + 1)st polymer molecule may be placed on a lattice already containing (i) molecules in ωi+1 ways. n−2 Nni− Nni− ω i +1 = ()NniZ− ()Z − 1 N N n n−2 Nni− ω = ZZ()−1 N i+1 N n n−2 Nni−−()1 For the (i)th molecule: ω = ZZ()−1 N i N Total Number of Ways of Placing N2 Polymer Molecules on a Lattice
N2 ωω12 ωLL ω iN2 1 Ω == NN!!∏ i 22i= 1 1 N2ω Apply Boltzmann’s Equation:Sk= ln mix !∏ω i N2 i=1 Substitute for ωi to obtain:
N2 Nn2 ()−2 N2 ZZ()− 1 n Ω = Nni−−()1 Nn2 ()−1 ∏[] NN2! i=1 Examine the product:
N2 N 2 n nN n N N Nni−−()11=+ nnN2 − i = ∏∏[] n i==11i n Entropy of the Mixture Write out several terms in the product expression: N nnnN N N n ∏ =+ 11− + 12− + 13− + 1− N L 2 n n n n N N N N ! ()()()123LL− N22− N + 1 n n n n Note that: = N N − N2 ! ()()()123L − N2 n n Thus n n N N ! N Nn()−2 nN ! n ZZ2 ()−1 2 n2 n ∏ = Ω = 1 N Nn2 ()− NN2! N − N2 ! − N2 ! n n Apply Stirling’s approximation to obtain: S nN N mix = −N ln 2 − N ln 1 k 2 N 1 N
++NZn2[]ln()− 211 ln() Z− + ()− n+ ln n Flory-Huggins Entropy of Mixing Calculate entropy of pure solvent and pure polymer: Pure solvent: N2 = 0 S1 = 0 Entropy of the disordered polymer Pure polymer: N = 0 1 when it fills the lattice
SkNZn22=+[]ln()− 211 ln() Z− + ()− n+ ln n
N nN ∆SS= −− SS SkNln 1 N ln 2 m mix 21∆ m = − 1 + 2 N N
Multiply and divide r.h.s. by N1+N2 and assume N1+N2=NA Calculate entropy of mixing: N nN N1 nN2 1 2 ∆SRxm = − 1 ln+ x2 ln = ϕ1 = ϕ 2 N N N N Flory-Huggins Theory for an Athermal Solution
Entropy of mixing:
∆SRxxm = − []1122lnϕϕ+ ln Enthalpy of mixing:
∆Hm = 0
Gibbs free energy of mixing:
∆GRTxxm = − []1122lnϕϕ+ ln Concentration Conversions
ϕ N2 1 2 ϕ ϕ2 nN2 N2 1 ϕ2 N2 N1 n 1 = = x2= = = N2 ϕ1 N1 Nn ϕ NN12+ 1 ϕ 2 1 1 1 + 1 + N1 n ϕ1
xx12+=1 ϕϕ12+=1 1 ϕ 2 x2 n1 − ϕ ϕ2 = = 2 1 1 x2 1 1 ϕ + x2 − 1 + 2 n n 1 − n ϕ2 Flory-Huggins Enthalpy of Mixing Use the same lattice model as for the entropy of mixing, and consider 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:
∆wwww= 2 12−− 11 22 ∆w = Change in interaction energy per (1,2) pair 2
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 Energy Contributions of polymer segments Interaction of a polymer segment with its neighbors yields
Zwϕϕ222+ Zw 112
The total contribution is 1 ZNϕϕϕ2[]()1 − 1 w 22+ 1 w 12 Remove double counting 2 Contributions of solvent molecules
Each solvent molecule is surrounded by Zϕ2 polymer segments and Zϕ1 solvent molecules. Interaction of the solvent with its neighbors then yields Zwϕϕ212+ Zw 111 The total contribution is 1 ZNϕϕ1[] 2 w ϕ 12+ ()1 − 2 w 11 Remove double counting 2 Flory-Huggins Enthalpy of Mixing 1 ∆HZNwww= ()2ϕϕ ϕϕ ϕϕ−− m 2 1212 1211 12 22 1 ∆∆HZNw= ϕϕ m 2 12 1 Flory-Huggins interaction Let Zw∆ = χ RT 2 parameter (the “chi” parameter)
χ = 0 For athermal mixtures χ> 0 For endothermic mixing For exothermic mixing χ< 0
∆Hm == Nϕϕ12 χ ϕ RT χ N 12 RT Flory-Huggins Free Energy of Mixing: General Case
∆∆GHTSmm= − ∆ m
∆HNm = ϕϕ12 χ RT
∆SRxxm = − []112lnφφ+ ln 2
∆GRTNm =++[]ϕϕ12 χ ϕ ϕ () x 1ln 1 x 2 ln 2 >0 0 <0 <0