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Mean Field Flory Huggins Lattice Theory Mean Field Flory Huggins Lattice Theory • Mean field: the interactions between molecules are assumed to be due to the interaction of a given molecule and an average field due to all the other molecules in the system. To aid in modeling, the solution is imagined to be divided into a set of cells within which molecules or parts of molecules can be placed (lattice theory). • The total volume, V, is divided into a lattice of No cells, each cell of volume v. The molecules occupy the sites randomly according to a probability based on their respective volume fractions. To model a polymer chain, one occupies xi adjacent cells. No =N1 + N2 = n1x1 + n2x2 V =Nov • Following the standard treatment for small molecules (x1 = x2 = 1) N0! Ω1,2 = N1!N2! using Stirling’s approximation: !ln = − >>1MforMMlnMM φ ΔSm = k(−N1 lnφ1 −N2 lnφ2) 1 = ? Entropy Change on Mixing ΔSm ΔSm =+k(−φ1 lnφ1 −φ2 lnφ2) Remember this is No for small molecules x1 = x2 = 1 0.8 0.7 0.6 0.5 0.4 0.3 0.2 Note the symmetry 0.1 0 0 0.2 0.4 0.6 0.8 1 φ2 The entropic contribution to ΔGm is thus seen to always favor mixing if the random mixing approximation is used. Enthalpy of Mixing ΔHm • Assume lattice has z nearest-neighbor cells. • To calculate the enthalpic interactions we consider the number of pairwise interactions. The probability of finding adjacent cells filled by component i, and j is given by assuming the probability that a given cell is occupied by species i is equal to the volume fraction of that species: φi. ΔHm cont’d υij = # of i,j interactions υ12 = N1z φ2 υ11 = N1 zφ1 /2 υ22 = N2 zφ2 /2 Mixed state enthalpic H1, 2 = υ12 ε12 + υ11ε11 + υ22 ε22 interactions z z H = zN φ ε + N φ ε + N φ ε 1, 2 1 2 12 2 1 1 11 2 2 2 22 z N z N Pure state enthalpic 1 2 H1 = ε11 H2 = ε22 interactions 2 2 ⎡ ⎤ N1ε11 N2ε22 ΔHM = zN⎢ 1φ2ε12 + ()φ1 −1 + ()φ2 −1 ⎥ ⎣ 2 2 ⎦ recall N1 + N2 = N0 ⎡ 1 ⎤ HΔ =M zε N − ε() ε + φ φ Note the symmetry Some math 0⎣⎢ 12 211 22⎦⎥ 1 2 χ Parameter z ⎡ 1 ⎤ • Define χ: χ= ε − ε + ε 12 ()11 22⎥ kT ⎣⎢ 2 ⎦ χ represents the chemical interaction between the components ΔH Ned: add eqn for M = kTχ φ φ Computing Chi from 1 2 V (delta-delta)2/RT N0 seg ΔG = ΔH −T ΔS • ΔGM : M M M ΔGM =χφkT φ1 − 2 kT[]1 − φln 1 φ 2 − ln φ 2 φ N0 ΔGM χφ= kT1[] φ 2+ φln 1 1 φ + 2 ln φ 2 φ N0 Note: ΔGM is symmetric in φ1 and φ2. This is the Bragg-Williams result for the change ng of binary metal alloys.in free energy for the mixi ΔGM(T, φ, χ) • lory showed how to pack chains onto a lattice and correctly evaluate F Ω1,2 for polymer-solvent and polymer-polymer systems. Flory made a complex derivation but got a very simple and intuitive result, namely that ΔSM is decreased by factor (1/x) due to connectivity of x segments into a single molecule: ΔSM ⎡ φ1 φ2 ⎤ ersFor polym = −k⎢ lnφ1 + lnφ2 ⎥ N0 ⎣x1 x2 ⎦ • Systems of Interest - solvent – solvent x1 = 1 x2 = 1 - solvent – polymer x1 = 1 x2 = large - polymer – polymer x1 = large, x2 = large ΔG ⎡ φ φ ⎤ M 1 2 Note possible huge = kT χφ1+ 2 φ lnφ1 + lnφ2 ⎢ ⎥ asymmetry in x1, x2 N 0 ⎣ x1 x2 ⎦ z ⎡ 1 ⎤ χ= ε − ε + ε ⎢ 12 ()11 22⎥ kT ⎣ 2 ⎦ Need to examine variation of ΔGM with T, χ, φi, and xi to determine phase behavior. Flory Huggins Theory • Many Important Applications 1. Phase diagrams 2. Fractionation by molecular weight, fractionation by composition nd 3. Tm depression in a semicrystalline polymer by 2 component 4. Swelling behavior of networks (when combined with the theory of rubber elasticity) ΔG The two parts of free energy per site M N0 kT φ1 φ2 S ΔSM =− ln(φ1) − ln(φ2) x1 x2 H ΔHM = χ φ1 φ2 • ΔHM can be measured for low molar mass liquids and estimated for nonpolar, noncrystalline polymers by the Hildebrand solubility approach. Solubility Parameter cohesive 1/2 • Liquids ⎛ ΔE⎞ 1/2 ⎡ ΔH − RT⎤ δ=⎜ ⎟ = energy = ⎢ ν ⎥ ⎝ V ⎠ ⎣⎢ V ⎦⎥ density m ΔHν = molar enthalpy of vaporization Hildebrand proposed that compatibility between components 1 and 2 arises as their solubility parameters approach one another δ1→δ2. • δp for polymers Take δP as equal to δ solvent for which there is: (1) maximum in intrinsic viscosity for soluble polymers (2) maximum in swelling of the polymer network or (3) calculate an approximate value of δP by chemical group contributions for a particular monomeric repeat unit. Estimating the Heat of Mixing Hildebrand equation: 2 ΔHM = Vm φ1 φ2 (δ1 − δ2 ) ≥ 0 Vm = average molar volume of solvent/monomers δ1, δ2 = solubility parameters of components Inspection of solubility parameters can be used to estimate possible compatibility (miscibility) of solvent-polymer or polymer-polymer pairs. This approach works well for non-polar solvents with non-polar amorphous polymers. Think: usual phase behavior for a pair of polymers? Note: this approach is not appropriate for systems with specific interactions, for which ΔHM can be negative. Influence of χ on Phase Behavior Assume kT = 1 and x1 =x2 Expect ΔH symmetry m 0.8 0.6 ΔG 0.2 0.4 0.2 m 0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -0.2 0 0.2 0.4 0.6 0.8 1 -0.4 χ = -1,0,1,2,3 -0.2 -0.4 -T ΔSm -0.6 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -0.1 χ = -1,0,1,2,3 -0.2 -0.3 -0.8 -0.4 -0.5 -0.6 -1 -0.7 -0.8 At what value of chi does the system go biphasic? What happens to entropy for a pair of polymers? Polymer-Solvent Solutions • Equilibrium: Equal Chemical potentials: so need partial molar quantities pure mixed ⎡∂G ⎤ μi = ⎢ ⎥ o o ∂n μ1 μ2 μ1, μ2 ⎣ iT⎦ ,,, P n j 0 ⎡∂ΔGm ⎤ ∂ΔGm ∂φi μi−where μ i RT ≡ ln i a is the ≡ ⎢ activity⎥ = ∂n ∂φ ∂n ⎣ i T⎦ ,,, P n j i i 0 and is the chemicalμi potential in the standard state ⎡ ⎛ ⎞ ⎤ 0 1 2 μ1 − μ1 = RT ⎢ lnφ1 + ⎜1 − ⎟ φ2 + χφ2 ⎥ solvent P.S. #1 ⎣ ⎝ x2 ⎠ ⎦ μ − μ 0 = RT φ − x − φ + x χφ 2 2 2 []ln 2 ()2 1 1 2 1 polymer Note: For a polydisperse system of chains, use x2 = <x2> the number average α β Recall at equilibrium μi = μi etc Construction of Phase Diagrams ΔGm ⎡ φ1 φ2 ⎤ = kT χφ⎢ 1+ 2 φ lnφ1 + lnφ2 ⎥ N0 ⎣ x1 x2 ⎦ • Chemical Potential ⎡∂G ⎤ μi = ⎢ ⎥ ∂n ⎣ iT⎦ ,,, P n j • Binodal - curve denoting the region of 2 distinct coexisting phases μ1' = μ1" μ2' = μ2" o o emPhases called pri or equivalently μ1' – μ1 = μ1" – μ1 o o and double prime μ2' – μ2 = μ2" – μ2 ⎡∂ΔGm ⎤ ⎡∂ΔGm ⎤ ⎡∂ΔGm ⎤ ⎡∂φ2 ⎤ ⎢ ⎥ = Δμ2 and since ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ∂n ∂n ∂φ ∂n ⎣ 2 ⎦TP, ⎣ 2 ⎦ ⎣ 2 ⎦ ⎣ 2 ⎦ Binodal curve is given by finding common tangent to ΔGm(φ) curve for each φ, T combination. Note with lattice model (x1 = x2) (can use volume fraction of component in place of moles of component Construction of Phase Diagrams cont’d 6 T low Two phases • Spinodal (Inflection Points) 5 α β Equilibrium 2 2 ⎡ ∂ ΔGm ⎤ ⎡ ∂ ΔGm ⎤ Stability = 0 = 0 4 ⎢ 2 ⎥ ⎢ 2 ⎥ ⎣ ∂φ2 ⎦ ⎣ ∂φ2 ⎦ N χ 3 A' B' •Critical Point 2 2 3 Critical point ⎡ ∂ ΔG ⎤ ⎡ ∂ ΔG ⎤ hot m m A B χ c ⎢ ⎥ = 0 and ⎢ ⎥ = 0 1 2 3 φ φ" ⎣ ∂φ ⎦ ⎣ ∂φ ⎦ 'A One phase A 2 2 T high 0 φ 0.0 0.2 0.4 0.6 0.8 1.0 1,c A B φ B Figure by MIT OCW. Dilute Polymer Solution • # moles of solvent (n1) >> polymer (n2) and n1 >> n2x2 n1v1 n2 x2v2 n2 x2 φ1 = φ2 = ~ n1v1 + n2 x2v2 n1v1 + n2 x2v2 n1 Useful Approximations n x 2 2 Careful… φ2 ≡ ~ X 2 x2 n1 x 2 x 3 ln(1+ x) ≅ x − + − ..... 2 3 x 2 x 3 ln(1− x) ≅−x − − − ..... 2 3 • For a dilute solution: 0 ⎡ φ ⎛ 1 ⎞ 2 ⎤ −μ = μ RT −2 ⎜χ +⎟ − φ Let’s do the math: 1 1 ⎢ 2 ⎥ ⎣ x2 ⎝ 2 ⎠ ⎦ Dilute Polymer Solution cont’d • Recall for an Ideal Solution the chemical potential is proportional to the activity which is equal to the mole fraction of the species, Xi 0 φ2 μ1−RT μ 1 = =ln X1 ln( RT − 12 ≅ ) X −RTX 2 RT = − x2 0 • Comparing to theμ1 − μ1 expression we have for the dilute solution, we see the first term corresponds to that of an ideal solution. The 2nd term is called the excess chemical potential ⎡ ⎤ 0 φ ⎛ 1⎞ 2 μ − μ = RT⎢− 2 + ⎜ χ − ⎟φ ⎥ 1 1 ⎣ x ⎝ 2⎠ 2 ⎦ – This term has 2 parts due to 2 2 • Contact interactions (solvent quality) χ φ2 RT 2 • Chain connectivity (excluded volume) -(1/2) φ2 RT • Notice that for the special case of χ = 1/2, the entire 2nd term disappears! This implies that in this special situation, the dilute solution acts as an Ideal solution. The excluded volume effect is precisely compensated by the solvent quality effect.
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