Chapter 4: Flory Huggins

Flory Huggins Theory:

We are now going to get a bit more complex and not only talk about a single polymer but talk about what happens when we mix polymers with other polymers, polymer blends, or when we mix polymers with solvents, polymer solutions. This requires using Flory-Huggins which is an extension of Bragg-Williams which should have been covered if you have taken a thermodynamics course but don’t panic we will be going over this again for review (yes same Bragg as diffraction, those old scientists always had multiple breakthrough discoveries). The Flory-Huggins framework will allow us to generate phase diagrams, understand fractionization by molecular weight, melting point depression in spherulites, and rubber elasticity. Additionally it also again will harken back to this idea of entropic and enthalpic competition.

Now why are polymer blends important? Well there are a number of critically important polymer blends that you might utilize every day and one such example of this is High Impact Polystyrene or HIPS. HIPS is a mixture of a very stiff yet brittle PS with a more compliant but more ductile polybutadine (PB). This combination leads to a material, HIPS, that exhibits a much larger toughness as cracks are blunted by the rubbery PB regions as seen here

So if we are dealing with polymer blends and polymer solution this involves mixing which is simply starting from a state with two pure phases and then these two pure phases must spontaneously mix. We know from our supplemental thermodynamics lecture that spontaneous reactions can only occur when the change in free energy of the system upon mixing decreases and you can guess what are going to be the two competing factors yes once again, say it with me

1. Entropy: Two mixed species will have more available configurations than the separated pure phases which increases entropy. Thus we have a entropic driving force that is configuration, combinatorial, or translational in nature as opposed to vibrational, rotational, etc.
2. Enthalpic: Two mixed species have some interaction between them. If positive this will avoid mixing, which will often be the case.

We combine these terms to get

$$\Delta G_{mix} = \Delta H_{mix} - T\Delta S_{mix} \nonumber$$

We will determine if the increase in entropy due to more configurations is enough to overcome the energy penalty associated with mixing two species with unfavorable enthalpic interactions. The key idea that drives mixing is that the increase in entropy due to many more available configurations associated with two mixed species is sufficient to overcome the energy associated with bringing two species with unfavorable interactions into contact.

Entropy of Mixing Bragg-Williams:

Before we deal with polymers, which hopefully you appreciate by now is always more complicated than hard or typical materials, we will start off by thinking about the entropy of mixing for small molecules or even gasses. Let’s consider we have two equally sized molecules, species 1 and 2, that we mix on a lattice or grid with a fixed number of sites N₀ where the number of species 1, N₁, and the number of species 2, N₂, will fill all the points on the lattice or alternatively N₀ = N₁ + N₂ as seen schematically here

Well we know that the number of configurations we can arrange species 1 and 2 is

$$\Omega_{12} = \frac{N_0!}{N_1!N_2!} \nonumber$$

and we know entropy is

$$S= k_b \ln(\Omega) \nonumber$$

Now for the initial state of the system prior to mixing how many configurations can we arrange the system? Just 1! all species 1 on one side and 2 on the other. So in this case Ω = 1 and thus S = 0. So the change in mixing is

$$\Delta S_{mix} = S_{12} - S_{11} - S_{22} = k_b \ln(\Omega_{12}) \nonumber$$

We can use Sterling’s approximation lnN! ≈ N lnN − N to get

$$\Delta S_{mix}= k_b (-N_1 \ln \Phi_1 - N_2 \ln \Phi_2) \nonumber$$

where $\Phi_1 = \frac{N_1}{N_0}$ and Φ₁ is both the mole fraction and volume fraction of species 1. We have just derived the Bragg-Williams entropy change for small molecules and we can divide the full expression by N₀ to give a molar quantity

$$\frac{\Delta S_{mix}}{N_0} = k(-\Phi _1 \ln \Phi _1 - \Phi _2 \ln \Phi _2) \nonumber$$

Let’s take a quick look at this equation.

Is it every negative?

No!

This means that mixing always increases the entropy of the system (for this ideal case).

Also when is entropy maximized?

Looking at the equation the curve is symmetric around Φ₁ = 0.5 and maximized at this point.

Now for polymers the mole fraction and volume fractions will typically not be equivalent so from here on out we will say that Φ₁ and Φ₂ refer to the volume fraction of each species.

Enthalpy of Mixing Bragg-Williams:

Now let’s take a look at the enthalpic contribution to free energy and again we have to think about interactions between molecules on the lattice. We can make things a bit easier by using a mean field theory. Instead of considering each lattice site discretely and thinking about it’s specific interactions with species, we can instead think of the neighbors of each lattice site taking on the average properties of the entire lattice based on the volume fraction of the two species. This is the mean field approximation.

We can calculate the enthalpy of mixing by thinking that the probability of finding any species i on any given lattice site is given the by the volume fraction Φ_i and thus enthalpy of mixing is

$$H_{12} = \text{(\# 1-2 interactions)}(\epsilon_{12}) + \text{(\# 1-1)}(\epsilon_{11}) + \text{(\# 2-2)}(\epsilon_{22}) \nonumber$$

Let’s dissect this equation step by step. Well let’s think about the number of interactions between species i and j

$$\nu_{ij} \nonumber$$

And the number of species 1 and 2 interactions will be

$$\nu _{12} = \text{(\# num of 1 sites)(\# nearest neighbors)(Prob. of adj. site being species 2)} \nonumber$$

To keep this general each lattice site will have z number of nearest neighbors. The total number of species 1 lattice site is N₁ and the probability of the nearest neighbor lattice site being species 2 will be the volume fraction of species 2 so

$$\nu _{12} = N_1 z \Phi _2 \nonumber$$

The same logic can be used to find the number of 1-1 and 2-2 interactions but we have to divide by 2 to avoid overcounting.

$$\nu_{11} = \frac{N_1 z \Phi_1}{2} \nonumber$$

$$\nu_{22} = \frac{N_2 z \Phi_2}{2} \nonumber$$

Then to find total enthalpy of mixing H₁₂ is just the number of interactions multiplied by the energy/strength of the interaction, $\epsilon_{ij}$,

$$H_{12} = \nu_{12} \epsilon_{12} + \nu_{11} \epsilon_{11} + \nu_{22} \epsilon_{22} \nonumber$$
$$H_{12} = N_1 z \Phi_2 \epsilon_{12} + \frac{N_1 z \Phi_1}{2} \epsilon_{11} + \frac{N_2 z \Phi_2}{2} \epsilon_{22} \nonumber$$

So this is the enthalpy upon mixing but we also need the initial state which is fairly simple to calculate from the discussion above and doesn’t require a mean-field approximation

$$H_{11} = \frac{N_1 z}{2} \epsilon_{11} \nonumber$$
$$H_{22} = \frac{N_2 z}{2} \epsilon_{22} \nonumber$$

So with these expression we can combine everything to obtain the expression for the enthalpy of mixing, remember that N₀ = N₁ + N₂.

$$\Delta H_{mix} = H_{12} - H_{11} - H_{22} \nonumber$$
$$\Delta H_{mix} = N_1 z \Phi_2 \epsilon_{12} + \frac{N_1 z \Phi_1}{2} \epsilon_{11} + \frac{N_2 z \Phi_2}{2} \epsilon_{22} - \frac{N_1 z}{2} \epsilon_{11} - \frac{N_2 z}{2} \epsilon_{22} \nonumber$$
$$\Delta H_{mix} = z \left[ N_1 \Phi_2 \epsilon_{12} + \frac{N_1 \epsilon_{11}}{2} (\Phi_1 - 1) + \frac{N_2 \epsilon_{22}}{2} (\Phi_2 - 1) \right] \nonumber$$
$$\Delta H_{mix} = z N_0 \left[ \epsilon_{12} - \frac{1}{2} (\epsilon_{11} + \epsilon_{22}) \right] \Phi_1 \Phi_2 \nonumber$$

To clean up this expression we will introduce a critical new parameter, the Flory χ parameter which is

$$\chi = \frac{z}{kT} (\epsilon _{12} - \frac{1}{2} (\epsilon _{11} + \epsilon _{22})) \nonumber$$

χ is essentially a measure of the energy of the interaction between the components mixed. If the value of χ is large and positive then we know the $\epsilon _{12} >> \frac{1}{2} (\epsilon _{11} + \epsilon _{22})$ and since higher energy is always less favorable this will push towards a phase separated state as this is more energetically favorable. This is typically the more common case but we can have negative values of χ like if we have a very good solvent. So re-writing our expression in terms of χ

$$\frac{\Delta H_{mix}}{N_0} = kT\chi \Phi_1 \Phi_2 \nonumber$$

Full Free Energy Expression: Bragg Williams

Now with this we can write the full free energy expression

$$\frac{\Delta G_{mix}}{N_0} = \Delta H_{mix} - T \Delta S_{mix} \nonumber$$
$$\frac{\Delta G_{mix}}{N_0} = kT \chi \Phi_1 \Phi_2 - kT (-\Phi_1 \ln \Phi_1 - \Phi_2 \ln \Phi_2) \nonumber$$
$$\frac{\Delta G_{mix}}{N_0} = kT \left[ \chi \Phi_1 \Phi_2 + \Phi_1 \ln \Phi_1 + \Phi_2 \ln \Phi_2 \right] \nonumber$$

Again we can see the for the Bragg-Williams is symmetric around Φ₁ and Φ₂.

Flory-Huggins Free Energy of Mixing for Polymers:

Now polymers are obviously a bit more complicated because they are not just simple small molecules on a lattice. Polymers are composed of connected monomers and this is a complication we must address. Let’s see if we can adjust some parameters and work from there

$$n_1 = \text{Number of molecules/polymers of species 1} \nonumber$$
$$n_2 = \text{Number of molecules/polymers of species 2} \nonumber$$
$$x_1 = \text{Degree of polymerization } \propto \text{ molecular weight of 1} \nonumber$$
$$x_2 = \text{Degree of polymerization } \propto \text{ molecular weight of 2} \nonumber$$
$$v_1 = \text{Volume of each monomer of species 1} \nonumber$$
$$v_2 = \text{Volume of each monomer of species 2} \nonumber$$
$$V_1 = \text{Total volume of 1} \nonumber$$
$$V_2 = \text{Total volume of 2} \nonumber$$

With these variables redefined let’s redefine our lattice model. Let’s now assume our lattice sites have a size commensurate with the monomer size of species 1 and 2 and thus we assume the monomers of each species has the same size v₁ = v₂ = v₀.

With this we know now the volume of any species will be

$$V_i = n_i x_i v_0 \nonumber$$

And with that the volume fraction of a species is simply defined as

$$\Phi_i = \frac{n_i x_i}{n_1x_1 + n_2 x_2} \nonumber$$

notice that the total number of lattice sites is N₀ = n₁x₁+n₂x₂ and is equivalent to the total number of monomers. Because we assume monomers of each species occupy the same volume, we can see that Φ_i is a volume fraction (and is not equivalent to the mole fraction, given by $\frac{n_i}{n_1 + n_2}$.

Flory Entropy of Mixing for Polymers

Before we get into the math let’s take a step back and think about the qualitative, physical picture of what will happen when we mix polymers. So far we have focused on the configurational entropy associated with polymer, i.e. when we stretch polymers we reduce the number of available configurations. This contribution to entropy should not change significantly upon mixing, assuming chains behave ideally, i.e. if they don’t feel the excluded volume from other chains. Whether mixed or separated you can imagine that the fluctuations around the center of mass will be essentially the same as long as the center of mass remains constant. And this brings us the major component of entropy that will change significantly for polymers which is translational entropy. This refers to the fact that when we mix polymers the entire chain can access more microstates by intermixing with polymers, via translation of the center of mass. Basically you can treat the entire polymer as a small molecule as was done for Bragg-Williams.

This means that we might expect that the change in entropy for mixing depends only on the number of polymer molecules and not on the total number of monomers. But we see above that the volume fractions are defined by the number of monomers. So we should expect that the entropy of mixing will decrease by a factor of where x is the degree of polymerization and x will in general be different for each species. Let’s see if this our conceptual understanding hold true for the derivation below

Now the original Flory derivation is extremely complicated as he explicitly counted the number of conformations of monomer on a lattice assuming each monomer was restricted by the position of the previous monomer so we will look at a simpler derivation developed by Hildebrand. Let’s start by assuming that polymers are completely ideal and that their entropy is equivalent to the entropy of an ideal gas

$$S_i = k\ln V_i^{n_i} = n_i k\ln V_i \nonumber$$

So for our two polymers we have

$$S_1 = n_1 k \ln V_1 = n_1 k \ln(n_1 x_1 v_0) \nonumber$$
$$S_2 = n_2 k \ln V_2 = n_2 k \ln(n_2 x_2 v_0) \nonumber$$
$$S_{12} = (n_1+n_2) k \ln V_{12} = (n_1+n_2) k \ln \left[ (n_1 x_1 + n_2 x_2) v_0 \right] \nonumber$$

then for the change in entropy upon mixing is

$$\Delta S_M = S_{12} - S_1 - S_2 \nonumber$$
$$\Delta S_M = -k \left[ n_1 \ln \left( \frac{n_1 x_1}{n_1 x_1 + n_2 x_2} \right) + n_2 \ln \left( \frac{n_2 x_2}{n_1 x_1 + n_2 x_2} \right) \right] \nonumber$$
$$\Delta S_M = -k \left[ n_1 \ln (\Phi_1) + n_2 \ln (\Phi_2) \right] \nonumber$$
$$\frac{\Delta S_M}{N_0} = -k \left[ \frac{n_1}{n_1 x_1 + n_2 x_2} \ln (\Phi_1) + \frac{n_2}{n_1 x_1 + n_2 x_2} \ln (\Phi_2) \right] \nonumber$$
$$\frac{\Delta S_M}{N_0} = -k \left[ \frac{\Phi_1}{x_1} \ln (\Phi_1) + \frac{\Phi_2}{x_2} \ln (\Phi_2) \right] \nonumber$$

Notice here that we used N₀ = n₁x₁+n₂x₂ and the assumption of v₁ = v₂ = v₀ which allowed the terms to cancel out nicely and we see that the term is scaled by the degree of polymerization as we expected. Also if the degree of polymerization decreases to 1 we recover our small molecule Bragg-Williams theory.

Flory Enthalpy of Mixing for Polymers:

During our last discussion with Bragg-Williams enthalpy we introduced the critical χ parameter for the small molecules. Luckily our derivation for enthalpy won’t change too much for polymers because we are assuming ideal chains. If we want to get more complicated and include details about chain connectivity this will only change zx_i the number of sites around each monomer but again this will be a very small change. So the enthalpic part of the free energy is identical to the small molecule case, thank goodness!

$$\frac{\Delta H_{mix}}{N_0} = kT\chi \Phi_1 \Phi_2 \nonumber$$

Total Flory Free Energy of Mixing for Polymers:

So this gives us a total Flory Free Energy of

$$\frac{\Delta G_{mix}}{N_0} = kT \lbrack \chi \Phi_1 \Phi_2 + \frac{\Phi _1}{x_1} \ln \Phi _1 + \frac{\Phi _2}{x_2} \ln \Phi _2) \rbrack \nonumber$$

where χ is still

$$\chi = \frac{z}{kT} (\epsilon _{12} - \frac{1}{2} (\epsilon _{11} + \epsilon _{22})) \nonumber$$

One quick aside we have defined x₁ is the degree of polymerization of polymer species 1, and x₂ is the degree of polymerization of polymer species 2. This allows us to define several systems that we will be working with quite a bit

1. solvent-solvent x₁ = x₂ = 1
2. solvent-polymer x₁ = 1,x₂ = large
3. polymer-polymer x₁ = large,x₂ = large

Looking at these systems we see that solvent-solvent is our small molecule system and the free energy is the same as Bragg-Williams and symmetric around a volume fraction of 0.5. However for the solvent-polymer case

we will no longer be symmetric around this value and it will lead to some interesting phase behavior which we will get into into right now.

With our free energy of mixing equation complete we can now create phase diagrams of polymer blends!. This will look a bit different from the phase diagrams in materials science because we see in the equation above that the entropy of mixing decreases as the degree of polymerization increases so high molecular weight polymers are less likely to mix. So we will develop phase diagrams that are a function of polymer molecular weight and χ which has temperature built in as we know that

$$\chi \propto \frac{1}{T} \nonumber$$

So you can see that to increase the probability of mixing the χ value will have to decrease. χ or more typically χN will be a critical parameter that will be used to determine when mixing will occur and when the solution will phase separate.

Polymer-Solvent Solutions:

To begin our discussion of polymer phase behavior we will discuss Scenario 2, which was the polymer-solvent scenario where x₁ = 1. Thus, the solvent is species 1. And to build up our phase diagram we should remember our conditions from Gibbs Phase Rule that the chemical potential µ of every species in every possible phase must be equivalent at equilibrium where µ is

$$\mu_i = \left [ \frac{\partial G}{\partial n_i}\right ]_{T,P,n_j} \nonumber$$

Again physical interpretation time, chemical potential is the change in free energy when we add or remove a molecule of a species and at equilibrium if we add or remove a molecule of species the change in free energy is 0. Alternatively if we can change the number of species and lower free energy then we are not at equilibrium, non-equilibrium.

Now in the equation we derived the change in free energy upon mixing so this is different from the absolute free energy but we can still work with this equation by simply changing how we express the chemical potential as a change as well

$$\mu_i - \mu_i^0 = \left ( \frac{\partial \Delta G}{\partial n_i} \right ) _{T,P, n_j} = RT\ln a_i \nonumber$$

where µ_i is the chemical potential of species i in the final state, µ⁰_i is the chemical potential in the initial state, and a is the activity.

We can simplify the equation by relating n to volume fraction using the chain

rule

$$\left ( \frac{\partial \Delta G}{\partial n_i} \right ) _{T,P, n_j} = \left ( \frac{\partial \Delta G}{\partial \Phi_i} \frac{\partial \Phi_i}{\partial n_i} \right ) _{T,P, n_j} \nonumber$$

We can now take our free energy equation and plug into the derivation above to get

$$\mu_1 - \mu_1^0 = RT \left[ \ln \Phi_1 + \left( 1 - \frac{1}{x_2} \right) \Phi_2 + \chi \Phi_2^2 \right] \nonumber$$
$$\mu_2 - \mu_2^0 = RT \left[ \ln \Phi_2 - (x_2 - 1) \Phi_1 + x_2 \chi \Phi_1^2 \right] \nonumber$$

where we assumed that x₁ = 1 since this species is our solvent.

Now as we have stated previously to build our phase diagram the we have to set the chemical potential of all the species in the co-existing phases equal to one another or the change must be equal (assuming same initial state)

$$\mu_1^\alpha = \mu_1^\beta \nonumber$$
$$\mu_1^\alpha - \mu_0 = \mu_1^\beta - \mu_0 \nonumber$$

This requirement for equilibrium is also described by the common tangent rule where you find you equilibrium volume fraction of species in each phase via the common tangents in free energy curves, i.e.

$$\bigg (\frac{\partial \Delta G_{mix}}{\partial \phi_2}\bigg)_{\phi_2 = \phi^\alpha} = \bigg (\frac{\partial \Delta G_{mix}}{\partial \phi_2}\bigg)_{\phi_2 = \phi^\beta} \nonumber$$

You have most likely seen this before in materials science but just as a quick reminder (also read the supplementary notes as well)

For polymers it is a little different since for mixing we already have our full free energy equation so we will typically develop our phase diagram as such

here if we see multiple minimum that indicates a two phase region as we can construct a common tangent. If we only have a single minimum then we are in a single phase region.

Figure 9. Phase Diagram Construction for Polymers Plotted as χ

This is the same as our equilibrium statement as tangent lines gives the chemical potential of both species in both phases and if the slope is the same then the chemical potentials are equal. We can use the common tangent rule on the plot of ∆G_mix vs Φ to define the binodal curve. Binodal curves/points are denoted by

$$\frac{\partial\Delta G}{\partial \Phi_i} = 0 \nonumber$$

and we can also identify spinodal curves and points as well. The spinodal curve which determines the kinetics of phase separation (spontaneous or nucleation/growth) via the inflection point, which denotes the spinodal point.

$$\frac{\partial^2 \Delta G}{\partial \Phi_i ^2} = 0 \nonumber$$

You will also notice that the kinetics and mechanism by which the solution phase separates is different in the spinodal and binodal regions. We see that Spontaneous/Spinodal Decomposition will occur when one is locally unstable or equivalently when

$$\frac{\partial^2 \Delta G}{\partial \Phi_i ^2} < 0 \nonumber$$

Whereas nucleation and growth will occur where one is locally stable or equivalently when

$$\frac{\partial^2 \Delta G}{\partial \Phi_i ^2} > 0 \nonumber$$

You can see here that nucleation and growth will occur in the binodal regions while spinodal decomposition will occur in the spinodal regions. With all these key points we can now build our phase diagram. We can then take these key points and build our phase diagram which is plotted as χ vs. Φ. These points and the interpolated line between them denotes the region of coexistence between phases, back once again to good old Gibbs Phase Rule

$$D + P = C + 1 \nonumber$$

where D is the degrees of freedom, P is the number of phases, and C is the number of components. As χ increases, phase separation is the preferred energetic state while at low values mixing is preferred. This should make sense knowing $\chi ~ \frac{1}{T}$.

Finally, we can define several critical points, the first begin χ_c, where the spinodal obtains its lowest value in terms of χ also the point where the binodal and spinodal intersect.

$$\frac{\partial^2 \Delta G}{\partial \Phi_i ^2} = \frac{\partial \Delta G}{\partial \Phi_i} = 0 \nonumber$$

You can also find the critical composition, φ_c associated with χ_c by solving

$$\frac{\partial \chi}{\partial \phi} = 0 \nonumber$$

There is also a temperature associated with both χ_c and φ_c which is either the Upper Critical Solution Temperature (UCST) which denotes the temperature above which the solution will mix or the Lower Critical Solution Temperature (LCST) which denotes the temperature above which the solution will de-mix. Atomistically you can understand the appearance of a LCST as there is a temperature at which you start breaking weak intermolecular interactions that may have promoted mixing which leads to phase separation. You can see some of these critical points here

You will proves this in the homework but we can actually solve for these critical values here for all the typical scenarios that you may come across

Flory Huggins Polymer Blends Summary:

(1) Solvent-Solvent Scenario x₁ = x₂ = 1:

• φ_c = 0.5

• χ_c = 2

(2) Solvent-Polymer Scenario x₁ = 1;x₂ = N:

$\phi_c = \frac{1}{1+\sqrt{x_2}}$

$\chi_c = \frac{1}{2} \bigg( 1 + \frac{1}{\sqrt{x_2}}\bigg)^2$

(3) Symmetric Polymer-Polymer Scenario x₁ = N;x₂ = N:

$\phi_c = 0.5$

$\chi_c = \frac{2}{N}$

(4) Most General Scenario x₁,x₂:

$\phi_c = \frac{\sqrt{x_1}}{\sqrt{x_1}+\sqrt{x_2}}$

$\chi_c = \frac{1}{2} \bigg( \frac{1}{\sqrt{x_1}} + \frac{1}{\sqrt{x_2}}\bigg)^2$

Dilute Polymer-Solvent Solutions

You may often encounter scenarios where you are working with a dilute solution where the moles of solvent is much greater than the number of moles of polymer (i.e. really anytime working with biological concentrations of proteins/peptides). When you have such a scenario there are a couple of simplifications that simplify this derivation the first concerns the volume fraction of polymer

$$\Phi _2 = \frac{n_2 x_2}{n_1 x_1 + n_2 x_2} \approx \frac{n_2 x_2}{n_1} \nonumber$$

Also since Φ₁ = 1 − Φ₂ and Φ₂ is small in dilute solution we can do an expansion of lnΦ₁ from our chemical potential above

$$\ln (1-x) \approx -x - \frac{x^2}{2} - ... \nonumber$$

We can then use these expression and re-write the change in chemical potential of the solvent

$$\mu _1 - \mu _1^0 = RT \left [ -\frac{\Phi_2}{x_2} +(\chi -\frac{1}{2})\Phi_2^2 \right ] \nonumber$$

We previously described that for an ideal solution, the chemical potential is proportional the log of the activity, which is in turn proportional to the mole fraction of that species in solution ( Henry’s law). We can use this to

$$\mu _1 - \mu _1^0 = RT \ln a_i = RT\ln X_1 = RT \ln(1-X_2) \approx -RTX_2 = -RT\frac{\Phi_2}{x_2} \nonumber$$

where X_i is the mole fraction of the species in solution. As you see in the previous expression the first term matches with the ideal solution. But the second term which has χ and Φ²₂ describes the interaction between polymers and is the correction to the purely ideal assumption. This is sometimes referred to as the excess chemical potential because it is an additive term which has two components

(1) Contact interactions (solvent quality) : $\chi \Phi_2^2RT$

(2) Chain connectivity (excluded volume effects): $-\frac{1}{2}\Phi_2^2RT$

But there is a an extremely important fact in the expression where we see that the excess chemical potential disappears if $\chi = \frac{1}{2}$. There is a balance between the contact interaction and chain connectivity or the solvent quality and excluded volume effects that can allow for the chain to behave ideally (θ conditions). So you can change the value of χ to obtain ideal solutions, most typically via changing temperature!. We can summarize when/where we obtain θ conditions for some different scenarios you might encounter

Limitations of Theory

As we have mentioned before we are typically assuming only van der Waals interactions and have ignored hydrogen bonding. So this framework will only hold for reasonable temperature ranges. We can predict UCST, or upper critical solution temperature, behavior where at high temperature we switch from phase-separated solutions to mixed. This is typical when χ and the solubility parameters are positive.

But there are some systems which exhibit LCST, or lower critical solution temperature, where at a certain temperatures the opposite behavior occurs and the solutions demix. This is counterintuitive because at high temperatures the entropy of mixing should dominate and encourage mixing however, χ is also temperature dependent and can switch sign and become negative and encourages phase separation. Atomistically there is a temperature at which you start breaking any weak intermolecular interactions that may have promoted mixing which leads to phase separation. There are also scenarios where you can have both a LCST and UCST.

Determining χ Parameter

As we see above χ is a critical parameter for determining the phase behavior of polymer blends so it would be great to calculate it. Luckily Hildebrand developed a method to estimate χ. Imagine a liquid on a surface. The surface will have some interaction with the solvent in the liquid and to get an idea of the strength of these interactions you can heat up and evaporate the liquid. The amount of energy is the heat of evaporation, ∆H_V and if you divide this by the volume for a mole of molecules you get a cohesive energy density. And we can relate this to the solubility parameter δ which is

$$\delta _i = \sqrt{\frac{\Delta H_V - RT}{V_m}} \nonumber$$

This is essentially the energy required to remove a mole of moles from the bulk to an infinite distance. Hildebrand theorized that the compatibility between two components in a blend could be measured by measuring the difference in the solubility parameters. We can then relate this to χ. If the intermolecular forces are non-specific in nature like van der Waals interactions, the interaction could act on one species with itself or with the other species and if these interaction strengths are about the same they should be compatible. We can write the energies as we have done previously:

$$-\frac{z\epsilon _{11}}{2} = v_0\frac{\Delta E_1}{V_1} = v_0 \delta _1 ^2 \nonumber$$
$$-\frac{z\epsilon _{22}}{2} = v_0\frac{\Delta E_2}{V_2} = v_0 \delta _2 ^2 \nonumber$$
$$-z\epsilon_{12} = v_0 \delta _1 \delta _2 \nonumber$$

where ∆E is the energy change in moving that molecule to infinity. We can then re-write χ as

$$\chi = \frac{v_0}{kT}(\delta _1 - \delta _2)^2 \nonumber$$

Now there are some problems here as we have discussed previously it is possible to have negative values of χ but here we can only have positive values. Additionally, polymers will also have other specific intermolecular interactions like hydrogen bonds.