Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications (2013)
13. Mixing Liquids: Polymeric Solutions
13.3. The Flory–Huggins Model
Let us now turn to the thermodynamics of polymer solutions. In 1941, Flory  and Huggins  published their (in the meantime) classic papers on the solutions of polymers. The principle of dealing with the problem remains the same as for molecular solutions: calculate the entropy of mixing ΔmixS, the enthalpy of mixing ΔmixH, and from these two quantities the Gibbs energy of mixing ΔmixG. If ΔmixG < 0, the polymer is soluble in the solvent. In this section we deal with this approach, using again the lattice model with coordination number z and having n sites, based on the original discussion of Flory  and the more recent version provided by Gedde . As usual, we refer to component 1 as the solvent, and component 2 as the polymer.
13.3.1 The Entropy
The basis of the thermodynamic estimates for polymeric solutions is similar to that of molecular solutions (Figure 13.6). For molecular solutions, we have seen that the entropy of mixing was given by (see Section 11.4)
where, as before, ni and xi represent the number of moles and the mole fraction of component i, respectively, while R represents the gas constant. This expression can also be written (because it is assumed that for the molar volumes we have ) as:
where ϕi represents the volume fraction of component i. In Chapter 11, arguments were advanced to show that the proper expression is indeed in volume fractions.
Figure 13.6 Lattice models for (a) molecular and (b) polymeric solutions.
Here, we derive the entropy expression for a polymer solution. In fact, in one set of considerations we can derive the relevant expressions not only for solutions but also for polymer blends and molecular solutions. To that purpose, we deal with solvent and solute on a symmetrical basis. We introduce first a reference volume V0 that represents the molar volume of a lattice site which can be chosen arbitrarily, for example, V0 = V1, V2 or (V1V2)1/2. Often, one takes for V0 the molar volume of the pure solvent . In the sequel we will also use the molecular reference volume v0 = V0/NA where, as usual, NA represents Avogadro's constant. Assuming no volume change, for a volume Vj of component j, the total volume of the solution is V = V1 + V2, and the number of sites5) is n = V/v0. Of these sites a volume fraction ϕj = Vj/V is occupied by component j, and therefore Vj/vref = nϕj.
Now, consider a molecule which in terms of the equivalent chain consists of r segments. If the molecular volume vj of component j is given by vj = Vj/NA, the degree of polymerization is defined by rj = vj/v0 = Vj/V0 = Mj/ρjV0, where Mj is the molar mass of the molecule that has density ρj (volume Vj) in the amorphous state at the solution temperature. Because vj = rjv, the number of molecules nj of component j is nj = Vj/vj = Vj/rjv0 = nϕj/rjv0. Hence, on the lattice r1n1 sites are occupied by solvent molecules and r2n2 sites by polymer molecules, and the total number of lattice sites is n = r1n1 + r2n2.
We now turn again to solutions and use r1 = 1 and r2 = r. The analysis starts with having already i polymer molecules on the lattice, so that the number of vacant sites is then n − ir. This number equals the number of different ways of placing the (i + 1)th molecule on the lattice, which equals the product of the coordination number z and the fraction of remaining vacancies (1 − fi), that is, z(1 − fi). The next segment can only occupy (z − 1) (1 − fi) positions. The number of different ways of arranging the (i + 1)th molecule is then
The number of ways to arrange all polymer molecules is accordingly
where the denominator n2! takes into account that the solute molecules are indistinguishable. The fraction vacant sites is (1 − fi) = (n − ir)/n which, upon substitution in Eq. (13.38), meanwhile replacing once z by z − 1, results in
Since we consider only diluted solutions, we have n >> r and therefore we can approximate Eq. (13.40) by
Substitution of Eq. (13.41) in Eq. (13.39) results in
where the last step can be made only for dilute solutions with n >> r.
The entropy becomes S = klnW2 which, upon some straightforward evaluation, using Stirling's approximation (lnx! = xlnx − x) and the fact that n = n1 + n2r, becomes
The entropy difference upon dissolution is calculated as ΔS = S(n1,n2) − S(0,n2) because S(0,n2) = kn2lnr + kn2(z − 1)ln[(z − 1)/e] represents the entropy of the amorphous polymer. Considering that the volume fractions of the solvent and solute are ϕ1 = n1/(n1 + rn2) and ϕ1 = rn2/(n1 + rn2), respectively, the result is (after some algebra)
Using n = n1 + rn2, the entropy change per site becomes ΔmixS/n (note that ΔmixS/n is an intensive quantity), so that ΔmixS/nv0 represents the entropy change per unit volume
The entropy expression also applies if both components are polymers, that is, for a blend. In this case we use again rj = Mj/ρjV0 leading to volume fractions given by
The entropy change per unit volume then becomes
and this quantity, since ϕ ≤ 1, is always positive. From these equations we conclude that, in the case of a polymer blend, the entropy gain per unit volume is relatively small as compared to that of the polymer solution because, in the former case n1 = (ϕ1Σjnjrj/r1) is much smaller. Note that, if r1 = r2 = 1, we regain the ideal solution entropy. We included here only the mixing entropy, due to the translational degrees of freedom, and assumed that the configurational contribution would remain the same as for the pure polymer. Although this assumption is appropriate for blends, we discussed that excluded volume interactions for solutions have a significant influence. Moreover, we neglected volume changes which, although small, are actually present.
Estimate for which size ratio the volume fraction expression for the entropy deviates by more than 5% from the mole fraction expression.
13.3.2 The Energy
For the energy we must calculate the interaction between the solvent and solute segments, and we do this in a rather similar fashion to that used for molecular solutions. In particular, we neglect (again) volume changes and assume that the segments are placed at random over the lattice, thereby neglecting all correlation between segments. We again use a lattice of n sites with coordination number z and n1 solvent molecules and n2 polymer molecules, so that n = n1 + rn2. Each solute segment is surrounded by zϕ1 solvent molecules and zϕ2 solute segments, and therefore the interaction energy for that segment reads zϕ1w12 + zϕ2w22. Because the lattice contains nϕ2 segments, each interacting in this way, the energy U2 for the solute becomes
where the factor ½ is introduced to avoid counting interactions twice. Similarly, for the solvent molecules we obtain the energy
The energy for the pure solvent and solute are U01 = ½nϕ1zw11 and U02 = ½nϕ2zw22, respectively, so that the energy difference upon dissolution is calculated as
where w ≡ w12 − ½(w11 + w22) is the interchange energy, the increase in energy when a solvent–polymer contact is formed from molecules that were originally contacting molecules of the same type.
In Flory–Huggins theory it is conventional to render the interaction parameter w dimensionless according to χ = zw/kT, where χ (chi) is the dimensionless interaction parameter. Values for many polymers are listed in handbooks . Since ϕ1 = n1/(n1 + n2r2), we can also write
The total volume of the solution is V = nv0 = (n1 + n2r)v0, where v0 is the reference volume. The energy of mixing per unit volume then becomes
13.3.3 The Helmholtz Energy
From the entropy and energy, we obtain the Helmholtz energy
For the case of mixture of two polymers we obtain the result
with, as before, the DP rj = Mj/ρjV0. Note that the value of χ depends on the choice of V0. Remembering that χ contributes to the Helmholtz energy, one should not be surprised that χ is temperature-dependent. Often, one writes
using the “entropic” and “enthalpic” parameters A and B, respectively, comparable to zw = a + bT, Eq. (11.75), for regular solutions. A typical example is the system polystyrene (PS)/polymethyl methacrylate (PMMA) with A = 0.0129 and B = 1.96 K using v0 = 100 Å3, resulting in χ ≅ 0.01, so that only low-molar mass components form a miscible blend. Empirically, this relationship is not generally observed, that is, the graph of χ versus 1/T may show curvature, local minima or even change sign, and is somewhat dependent on composition and chain length. Finally, we note that, as expected, for r1 = r2 = 1, we regain the regular solution description.
13.3.4 Phase Behavior
From the Helmholtz energy expression one can calculate the stability of the polymer solution via the conventional route, as used before for molecular (regular) solutions (see Eqs 11.67 to 11.70) which led in that case to kTcri = ½zw. Here, the main result for a binary polymer mixture with ϕ1 = ϕ and ϕ2 = 1 − ϕ, is
Equating this result to zero, one obtains the expression for the inflection points of the Helmholtz curve, dependent on composition and separating unstable and metastable regions. It is conventionally labeled as the spinodal (line) (see Sections 16.1 and 16.2), and reads for this case
Using χ = A + B/T one obtains
The critical point corresponds to the minimum in this curve and is obtained from
which corresponds to the critical composition
and upon substitution this result in the spinodal to
Since n2 is usually large this implies for solutions (n1 = 1) that, in the limit of infinite molar mass, χcri = ½ or kTcri = 2zw, in contrast to the result for regular solutions kTcri = ½zw. For the most common case where B > 0, χdecreases with increasing T and the solubility increases. The highest temperature for a two-phase system to exist is the upper critical solution temperature (UCST) and for T > Tcri the homogeneous solution is stable. However, B < 0 also occurs and χ increases with increasing T so that the solubility decreases. The lowest temperature for a homogeneous solution to exist is the lower critical solution temperature (LCST). If B changes sign as a function of T, the system will show both an UCST and a LCST.
In order to be able to describe the UCST, both the “entropic” part A and “enthalpic” part B can be considered to be temperature-independent. From Problem 13.5 we have
so that for the DP r2 ≡ r → ∞, χcri = ½. Combining with χ = A + B/T and solving for Tcri(r = ∞) ≡ θ, we obtain ½ = A + B/θ and therefore
describing 1/Tcri as a function of θ and r. Hence, given a set of experimental values of Tcri as a function of r, the slope of Eq. (13.63) determines B and the intercept determines θ, and therefore A. An example is shown in Figure 13.7.
Figure 13.7 Experimental and calculated phase diagrams for three polyisobutylene fractions in diisobutyl ketone of the molecular weights indicated. The dashed curves are the calculated Flory–Huggins binodals, while the dotted curves are the spinodals. The model description is due to Casassa , using the data of Schultz and Flory .
The temperature θ can be shown to be indeed the theta temperature. From Problem 13.6 we have, remembering that a1 = γ1x1 and that ϕ1 = 1 − ϕ2,
where in the third line r → ∞ is used. In the last step we apply an expansion similar to that used for the osmotic pressure (see Section 13.2), but now for lna1 instead of the osmotic factor g. The concentration of the polymer is given by c2 = n2M2/V, where n2 is the number of moles of polymer of molecular mass M2 in a volume V. If we assume that we have no volume change upon mixing, the volume of the polymer V2 is given by , with the specific volume of the pure polymer as defined by , so that . The Flory–Huggins first and second osmotic virial coefficients A1 and A2 become, respectively,
Similarly to what is discussed in Section 11.3, the first osmotic virial coefficient A1 is determined entirely by the molecular mass M2. From Eq. (13.65) we see further that A2 vanishes if χ = ½ and the solution behaves to third order as an ideal solution, that is, the chains are “unperturbed.” This behavior is comparable to the pseudo-ideal pressure behavior of gases at the Boyle temperature, where the second virial coefficient vanishes. We have already indicated that for r → ∞, χcri = ½, and therefore it is appropriate to consider θ as the theta temperature.
In the spirit of the osmotic expansion, we expect that the second osmotic virial coefficient A2 characterizes the onset of intermolecular interactions between the polymer solute molecules, and therefore depends on M2. However, we predict that A2 is independent of M2, which experimentally is not the case. This shows one of the deficiencies of the Flory–Huggins model, namely that it is not valid for very dilute systems because the assumption that a segment experiences the interaction for a random distribution of segments and solvent molecules cannot be true for such a solution.
Let us now consider the LCST. Whereas, for the UCST the parameters A and B could be considered as temperature-independent, when describing the LCST they should be considered as temperature-dependent. This can be done, for example, by writing
with ΔCP the exchange heat capacity. Using the Gibbs–Helmholtz equation, we obtain
If ΔCP is considered as constant, we easily evaluate this expression to
If we accept that it is possible that ΔCP < 0, χ as a function of T shows a minimum and two critical temperatures exist: T = θ, representing the UCST, and another temperature representing the LCST. Typical values of ΔCP required are ΔCP ≅ −k to −2k. One interpretation for ΔCP < 0 is that the interchange energy between solute segments and solvent molecules becomes lower as the system expands with increasing temperature.
Some further aspects of the FH theory are discussed by Boyd and Phillips .
Problem 13.4: Asymmetry
Plot ΔmixF as resulting from the Flory–Huggins theory with χ = 1, 2, and 3 for:
a) r1 = 1 and r2 = 1, 10, 102, and 104.
b) r1 = 102 and r2 = 102 and 104.
c) r1 = 104 and r2 = 104.
Problem 13.5: Phase separation
Using the Flory–Huggins expression Eq. (13.54) as well as r1 = 1 and r2 = r:
a) Show that the critical values of ϕ2 and χ for phase separation to occur in polymer solutions are given by, respectively,
Check that χcri = ½ for infinite molecular mass of the polymer.
b) Show that the critical temperature Tcri can be described by
Problem 13.6: Activity coefficient
Show by using Eq. (11.9) that the activity coefficients of the solvent (1) and the polymer (2) containing r monomers per polymer in the Flory–Huggins model are given by, respectively,
Problem 13.7: LCST
Derive Eq. (13.68).