From WolfWikis

LECTURE 14

Thermodynamics of solutions

A lot of chemistry takes place in solution and therefore this topic is of prime interest for chemistry.

Thermodynamic potentials of solutions

The Gibbs free energy of an ideal gas depends logaritmically on pressure

G = G^o + RT ln [P/P^o]

Usually P^o is taken to be 1 bar and is often dropped out of the formula. and we write.

G = G^o + RT ln [P]

If we have a gas mixture we can hold the same logarithmic argument for each partial pressure as the gases do not notice each other. We do need to take into acount the number of moles of each and work with (partial) molar values, i.e. the thermodynamic potential:

μ_j = μ_j^o + RT ln [P_j/P^o] (B)

(Again P^o is often dropped to confuse the enemy).

If we are dealing with an equilibrium over an ideal liquid solution the situation in the gas phase gives us a probe for the situation in the liquid. The equilibrium must hold for each of all components j (say two in binary mixture). That means that for each of them the thermodynamic potential in the liquid and in the gas must be equal:

μ_j^sln = μ_j^gas for all j.

Consider what happens to a pure component, e.g. j=1 in equilibrium with its vapor We can write:

μ₁^{pure liq}= μ₁^{pure vapor}=μ₁^o + RT ln [P*₁/P^o]

The asterisks P*₁ denotes the equilibrium vapor pressure of pure component 1 and we will use that to indicate the thermodynamic potential of pure compounds too:

μ₁*^liq= μ₁^o + RT ln [P*₁/P^o] (A)

Combining (A) and (B) we find a relationship between the solution and the pure liquid :

μ_j^sln=μ*_j + RT ln [P_j/P*_j]

Notice that the gas and its pressure is used to link the mixture and the pure compound.

Thermodynamic potentials of ideal solutions

As we have seen, an ideal solution obeys Raoult's law:

P_j = x_jP*_j

When we substitute we find a logarithmic relationship between thermodynamic potential and mole fraction:

μ_j^sln=μ*_j + RT ln [x_j]

Thermodynamics of ideal mixing

When we ad n₁ moles of component 1 and n₂ moles of component 2 to form an ideal liquid solution this is generally a spontaneous process. Let us consider the Gibbs free energy change of that process:

ΔG_mix= G(T,P,n₁,n₂)^sln - G*₁(T,P,n₁)* - G*₂(T,P,n₂)

G^sln = n₁μ^sln₁ + n₂μ^sln₂

G*₁ =n₁μ*₁

G*₂ =n₂μ*₂

ΔG_mix= n₁μ^sln₁ + n₂μ^sln₂-n₁μ*₁-n₂μ*₂

ΔG_mix= = RT(n₁ln[x₁]+n₂ln[x₂])

And yes this is always negative, so mixing is spontaneous. However the expression looks suspiciously familiar. Apart from a factor -T is is equal to the entropy of mixing!

ΔG_mix= = -T.{-R(n₁lnx₁+n₂lnx₂)}= -T{ΔS_mix}

This leaves no room at all for an enthalpy effect:

ΔG_mix =ΔH_mix-TΔS_mix

ΔG_mix^ideal = 0 -TΔS_mix^ideal

ΔH_mix^ideal =0

Even though in the liquid there are stong interactions between neighboring particles there is no enthalpy change because it does not matter what this neighboring molecule is.

If we represent the average interaction energy between molecule i and j by U_ij we are essentially assuming that

2U₁₂=U₁₁+U₂₂

In practice this is seldom the case. It usually does matter and then the enthalpy term is not zero. As this affects the thermodynamics of the liquid solution it should als affect the vapor pressures that are in equilibrium with it.

Molar volumes

From the change of G in its natural variables we know that:

∂G/∂P |_T = V

This means that if we take

∂ΔG_mix/∂P |_T = ΔV_mix

In the ideal case we get:

∂ΔG_mix^ideal/∂P |_T = ΔV_mix^ideal

∂RT(n₁ln[x₁]+n₂ln[x₂])/∂P |_T = ΔV_mix^ideal= 0

In the ideal case volumes are simply additive and we need not worry about how the partial molar volumes change with composition.

Vapor pressures on non-ideal solutions

Henry's law

If we plot the partial pressure of one component say P₁ above a mixture with a mole fraction x₁ we should get a staight line with a slope of P*₁ (Raoult's law). Above non-ideal solutions the graph will no longer be a straight line but a curve. However towards x₁=1 the curve typically approaches the Raoult line. On the other extreme there often is a more or less linear region as well but with a different slope. This means that we have two limiting laws.

vapor pressure above an ideal and a non-ideal solution

For x-> 0: Henry's law

P₁ = K_H.x₁

For x-> 1: Raoult's law

P₁ = P*₁.x₁

Note that this implies that the straight line that indicates the Henry expression will intersect the y-axis at x=1 (pure compound) at a different point than P*.

For x-> 0 (low concentrations) we can speak of component 1 being the solute (the minority component. At the other end x-> 1 it plays the role of the solvent (majority component)

Another thing to note is that P* is a property of one pure component, the value of K_H by contrast is a property of the combination of two components, so it needst to be measures for each solute-solvent combination.

The other component

As you can see we have a descriptiom for both the high and the low end but not in the middle. In general, the more modest the deviations from ideality the larger the range of validity of the two limiting laws.

The way to determine K_H would be to actually determine vapor pressures. (Tedious that..) How about the other component? Do we need to measure them too? Fortunately we can use thermodynamics to answer this question with no. There is a handy expression that saves us the trouble.

Gibbs-Duhem

Consider a Gibbs free energy that only includes μn conjugate variables as we obtained it from our scaling experiment at T and P constant:

G = μ₁n₁ + μ₂n₂

Consider a change in G:

dG = d(μ₁n₁) + d(μ₂n₂)

dG = n₁dμ₁+μ₁dn₁ + n₂dμ₂+μ₂dn₂

But if we simply write out a change in G due to the number of moles we have:

dG = μ₁dn₁ +μ₂dn₂

Consequently the other terms must add up to zero:

0 = n₁dμ₁+ n₂dμ₂

dμ₁= - [n₂/n₁]dμ₂

dμ₁= - [x₂/x₁]dμ₂

In the last step we have simply divided both denominator and numerator by the total number of moles.

This expression is known as the Gibbs-Duhem equation. As you see it relates the change in one thermodynamic potential (dμ₁) to the other (dμ₂).

Gibbs-Duhem in the ideal case

In the ideal case we have:

μ₂^sln=μ*₂ + RT ln [x₂]

Gibbs-Duhem gives:

dμ₁= - [x₂/x₁]dμ₂

As dμ₂ = 0 + RT /x₂ (x₂ being the only active variable at constant temperature) we get:

dμ₁= - [x₂/x₁]RT /x₂ = RT/x₁

If we now wish to find μ₁ we need to integrate dμ₁, e.g. form pure 1 to x₁.

This produces:

μ₁^sln=μ*₁ + RT ln [x₁]

This demonstrates that Raoult's law can only hold over the whole range for one component if it also holds for the other over the whole range.

Gibbs-Duhem and Henry's law

What happens is Raoult does not hold over the whole range?

Recall that in a gas:

μ_j = μ_j^o + RT ln [P_j/P^o] (B)

or

μ_j = μ_j^o + RT ln [P_j] (B)

(After dropping P^o=1 bar out of the notation. Note that numerically this does not matter, but that P_j is now assumed to be dimensionless)

Let's look at dμ₁ at constant temperature:

dμ₁ = RT[∂lnP₁/∂x₁]dx₁

likewise:

dμ₂ = RT[∂lnP₂/∂x₂]dx₂

If we substitute into the Gibbs-Duhem expression we get:

x₁[∂lnP₁/∂x₁]dx₁+x₂[∂lnP₂/∂x₂]dx₂=0

Because dx₁= -dx₂:

x₁[∂lnP₁/∂x₁]=x₂[∂lnP₂/∂x₂]

(This is an alternative way of writing Gibbs-Duhem).

If in the limit for x₁-> 1 Raoult holds then P₁->x₁P*₁.

Thus:

[∂lnP₁/∂x₁] = 1/x₁

and

x₁/x₁=x₂[∂lnP₂/∂x₂]

1=x₂[∂lnP₂/∂x₂]

1/x₂=[∂lnP₂/∂x₂]

We can integrate this to form a logarithmic impression but it will have an integration constant:

lnP₂ =lnx₂ + constant

lnP₂ =lnK.x₂

So for x₁-> 1, i.e x₂-> 0 we get that P₂=K.x₂ where K is some constant (not necessarily P*!).

What this shows is that when one component follows Raoult the other must follow Henry and vice versa.

(Note that the ideal case is a subset of this case, in that the value of K then becomes P* and the linearity must hold over the whole range.)

Margules functions

Of course a big drawback of the Henry law is that it only describes what happens at the two extremes of the phase diagram and not in the middle. In many cases of moderate non-ideality it is possible to describe the whole range (at least in good approximation) using a Margules function.:

P₁=x₁P*₁.f_Mar

The function f_Mar has the shape:

f_Mar= exp[ αx₂²+βx₂³+δx₂³+....]

Notice that the function involves the mole fraction of the opposite component. It is an exponential with a series expansion. The constant and linear term are missing. As you can see the function has a number of parameters α,β,δ etc. that need to be determined by experiment. In general, the more the system diverges from ideality, the more parameters you need.

Using Gibbs-Duhem is is possible to translate the expression for P₁ into the corresponding one for P₂. The book shows a good example of that for a Marugles with two terms (See example 24-7).

(Note that I skipped γ: it is usually given a different meaning, see below).

Activity and activity coefficients

In the ideal case we have seen that the thermodynamic potential can be written as:

μ₂^sln=μ*₂ + RT ln [x₂]=μ*₂ + RT ln [P₂/P*₂]

One approach to non-ideality is to simply redefine the problem away and say:

μ₂^sln ≡ μ*₂ + RT ln [a₂]

I am using ≡ here to indicate this is actually a definition. The newly defined variable a₂ is known as the activity. Alternatively we can define it as:

[a₂] ≡ [P₂/P*₂]

As at high enough values of the mole fraction we know that we can still apply Raoult. So a₂ must approach x₂ in this limit, but for other concentrations this will no longer hold.

Often this is expressed in terms of an activity coefficient γ:

[a₂] = γ₂.x₂

For high values of x₂, γ₂ will approach unity.

If we model the non-ideality with a Margules function we see that:

P₂=x₂P*₂.f_Mar

[a₂] = [P₂/P*₂]=[x₂P*₂.f_Mar/P*₂]=[x₂.f_Mar]

Obviously the activity coefficient and the Margules function are the same thing in this descripion.

Regular solutions

A special -and simplest- case of a Margules function is the case where all but one Margules parameters (α) can be negelected. Such a system is called a regular solution

In this case we can write

a₁= x₁. exp[αx₂²]

We can use Gibbs-Duhem to show that this implies (See example 24-8):

a₂= x₂. exp[αx₁²]

Gibbs free energy of regular solutions

Consider the change in Gibbs free energy when we mix two components to form a regular solution:

Δ_mixG = n₁μ₁^sln+n₂μ₂^sln-[n₁μ₁*+n₂μ₂*]

Using:

μ_j^sln ≡ μ*_j + RT ln [a_j]

and:

[a_j] = γ_j.x_j

we get:

Δ_mixG/RT = n₁lnx₁+n₂lnx₂+n₁lnγ₁+n₂lnγ₂

If we divide by the total number of moles we get

Δ_{mix, molar}G/RT = x₁lnx₁+x₂lnx₂+x₁lnγ₁+x₂lnγ₂

For a regular solution:

lnγ₁= lnf_Mar = αx₂²

lnγ₂= lnf_Mar = αx₁²

This gives:

Δ_{mix, molar}G/RT = x₁lnx₁+x₂lnx₂+x₁αx₂²+x₂αx₁²

Δ_{mix, molar}G/RT = x₁lnx₁+x₂lnx₂+α[x₁+x₂]x₁.x₂

Δ_{mix, molar}G/RT = x₁lnx₁+x₂lnx₂+α.x₁.x₂

([x₁+x₂] = 1 by definition)

In this expression we see that we have an additional term to the entropy of mixing term we had seen before. Its coefficient α is dimensionless but represents the fact that the (strong!) interactions between the molecules are different depending on who is the neighbor. In general α can be written as W/RT where W represents an energy (actually enthalpy) that brings the difference in interaction energies into account. W does not depend strongly on temperature. We could look at W as the difference in average interaction energies:

The comprimise of the entropy and enthalpy term can lead to two minima

W= 2U₁₂-U₁₁-U₂₂

Rearranging we get:

Δ_{mix, molar}G/W = RT/W*[x₁lnx₁+x₂lnx₂]+ x₁.x₂

As you see the two terms will compete as a function of temperature. The mixing entropy will be more important at high temperatures, the interaction enthalpy at low temperatures. The entropy term has a minimum at x₁=0.5, the enthalpy term a maximum if W is positive. So, one tends to favor mixing, the other segregation and we will get a compromise between the two.

Depending on the value of RT/W (read: temperature) we can either get one or two minima. This means that at low temperatures there will be a solubility limit of 1 into 2 and vice versa. At higher temperatures the two components can mix completely. At the transition between these two regimes we will have critical or consolute point

A binary diagram showing an upper consolute point

Notice that even though we used the vapor pressures of the gas to develop our theory, they are conspicuously absent from the final result. The same thing we said about melting points hold true here. Because we are dealing with the miscibility behavior of two condensed phases, the outcome should not depend very strongly on the total pressure of our experiment.

Although in regular solutions the consolute point is predicted to be a maximum in temperature, we can find them as minima as well in practice. The nicotine-water system even has two consolute points, an upper and a lower one. When heating up a mixture of these we first observe mixing, then segregation and then mixing again. Obviously this behavior is far more complicated than we can describe with just one Margules parameter.

Partial molar volumes

What we said above about volumes simply being additive in the ideal case is no longer true here.

∂ΔG_mix^regular/∂P |_T = ΔV_regular^ideal

∂ΔH_mix+RT(n₁ln[x₁]+n₂ln[x₂])/∂P |_T = ΔV_mix^ideal

∂ΔH_mix/∂P |_T = ΔV_mix^ideal

In general the enthalpy of mixing does depend on pressure as it is related to the interactions between the molecules in solution. (W depends on the distance between them). This means that partial molar volumes now become a function of composition and volume is no longer simply aditive

Real solutions

Notice that the curves are symmetrical around x=0.5. This implies that it is as easy (or not) dissolving A into B as vice versa. In many cases this is not realistic. Many systems diverge more seriously from ideal behavior that the regular one. Up to a point we can model that by adding more terms to the Margules function. For example, adding a β-term undoes the symmetry (see example 24-7). However, many systems are so non-ideal that the Margules expression become unwieldy with too many parameters.

Boiling non-ideal solutions

Azeotropes

For ideal solutions we have seen that there is a lense shaped two-phase region between the gas and the liquid phase. For non-ideal systems the two-phase region can attain different shapes. In many cases there is either a minimum or a maximum. As such a point the phase gap closes to a point that is known as an azeotrope. It represents a composition of the liquid that boils congruently. That means that the vapor and the liquid have the same composition for a change. Azeotropes impose an important limitation unto distillation: they represent the end point of a distillation beyond which we can not purify by this method.

Eutectics

Another point to be made is that in the diagram with the consolute point we are assuming the pressure to be constant. If we lower the pressure this would affect the boiling points strongly: the whole gas-liquid gap would come down in temperature (see animation). The mixing behavior is only weakly affected. (The reason is that one involves the volume term of the gas, the other only of the liquid(s)). At lower pressures it is possible therefore that the consolute point is above the gas-liquid gap. In other words: the mixtures will boil before they get a chance to mix.

a simple binary eutectic system

In such a case the two phase gaps will become adjacent and we can get a eutectic. In the middle of the diagram the gas phase becomes an attractive alternative to a combination of two liquids that do not mix. The boiling points will be lower there than for the pure compounds. There will be a composition for which the boiling point is at a minimum and where the mixture boils congruently (i.e. to a vapor with the same (overall) composition).

Notice that the mutual solubility limits increase as temperature increases, just as happens in the critical mixing case, but that due to the competition from the vapor phase this process comes to an end at the eutectic temperature.

At this temperature one liquid boils aways completely, the other one in part. At the eutectic composition they both boil away simultaneously.