From WolfWikis

Phase diagrams

Diagrams

A good map will take you to your destination with ease, provided you know how to read it. A map is an example of a diagram, a pictorial representation of a body of knowledge. In science they play a considerable role. Next to plots and tables diagrams are an important means of making information and/or theoretical knowledge accessible.

Constructing them takes quite a bit of thought. You want to represent as much of what you know and give as accurate a picture of it without conveying anything incorrect. If the drawing can be made to scale that makes it quite a bit more powerful, but this is not strictly necessary. A remark like not to scale or schematically does need to be given if applicable. A good caption or description is essential.

Thermodynamic stability

A schematic G curve showing various types of stability
A: Stable
B: Metastable
C: Indifferent
D: Labile
E: Instable
animated version

The concept stability is in a sense the obverse side of the concept spontaneity. A system is stable if no spontaneous changes are possible any more (without changing variables like pressure, temperature or composition). This again means that the Gibbs free energy must be at a minimum and that we have allowed the system to equilibrate. (How much time that takes is not part of the diagram),

There are different kinds of equilibrium, besides the stable equilibrium that represents an absolute minimum in the G function. Of course G is potentially a function of a great number of variables but let us look at a diagram in which G is shown as a function of only one unspecified variable. You could think of the density, the mole fraction of one of the components of a mixture or an applied electrical field or whatever, but the argument is general.

The diagram is helpful to point out that besides a stable equilibrium (A) there can also be an metastable equilibrium (B) or an indifferent equilibrium (C). The local derivatives of G (versus all variables of which we only show one) are zero in all three cases, which means that changes in the variables are not spontaneous.

For a labile equilibrium (D) the opposite is true. Any small deviation will make the system role down hill. (Note that the second derivative has the opposite sign compared to cas A) and B)) A labile equilibrium is seldom or never observed except in a circus where artists delight in balancing objects on their heads (because you pay for it).. This usually requires continuous small corrections to maintain the precarious balance.

All other points in our diagram represent state of instability because locally dG is not zero and a spontaneous process can take place.

Fluctuations

The fact that dG=0 in the equilibrium points does not mean small deviations from the minimum cannot happen at times. We have seen e.g. that the Boltzmann distribution was simply the most likely distribution. The most likely one is the one that has the highest number of realizations W. Another way of saying that is that it is the one with the highest entropy S. A slightly less likely distribution may occur from time to time by chance. It will have a little less entropy but the same <E>. That means it will have a slightly higher G (G=H-TS). From time to time therefore G will fluctuate a bit. Such fluctuations are very small for large systems, but they are of greater relative importance for small systems (like a nanoparticle). (Statistical averaging works best on large ensembles.)

The fluctuations in G mean that small fluctuations in its variables like density etc. can also occur. They are usually kept in check, because dG is no longer zero when moving away from the equilibrium state. This drives the system back to the minimum spontaneously. You could picture the system wobbling a bit around in its G-well. This holds for stable and metastable equilibria alike.

In the indifferent case (C) however the derivative is zero (or very close to zero) for a range of neighboring values of some variable. In contrast to A) and B) also the second derivative is zero. This means that there is little penalty to much larger deviations in the variable. If this variable is the density the system becomes milky and shows opalescence a strong scattering of light because the refractive index depends on the strongly fluctuating density. This is observed near critical points and is called critical opalescence.

Unary phase diagrams

A unary phase diagram summarizes the equilibrium states of a single pure substance. We will see that we can also look at mixtures of two components (binary diagrams) or more (ternary, quaternary, quinary, senary etc.) . Usually a phase diagram only maps out stable equilibriums but occasionally metastable ones may be given too (e.g. with a dashed line).

Liquid-vapor equilibrium

We have seen that the Gibbs function G depends strongly (logarithmically) on pressure for a gas but only slightly (and linearly) for a liquid. The two curves intersect in a point representing the equilibrium vapor pressure of the liquid. At lower pressures the vapor is more stable, at higher ones the liquid. (For a solid the same holds as for the liquid).

This means that except at the intersection point we only will observe one phase.

It is important to stress that this holds in the absence of other matter, e.g. when we put only water into an evacuated cylinder. We may get three cases:

• we compress the cylinder until it only contains the liquid under hydrostatic pressure (P>P_eq)
• we expand the cylinder until all water has vaporized (P<P_eq)
• we let part of the water evaporate, just enough so that the space above the liquid is filled with an equilibrium vapor pressure (P=P_eq)

At room temperature P_eq for water is only about 15 Torr. If we apply 1 bar -or let the atmosphere do the job- we will only have liquid water.

A cylinder with one substance animated version

If other gases are present ,e.g. air, we must distinguish between the total pressure (say 1 bar) and the equilibrium vapor pressure which will now be the partial pressure.

In a cylinder with water and one bar of air just enough water will evaporate to establish equilibrium. The evaporation will be limited to the gas-liquid interface unless the partial pressure equals the total pressure. Then the liquid will boil.

If we consider the set of equilibrium pressures as a function of temperature and plot that in a P versus T diagram we have one component of our phase diagram.

Gas-solid

For solids the situation is similar as the G(P) curve is once again an almost flat straight line. The intersection with the logarithmic curve for the gas will define an equilibrium pressure for gas-solid co-existence. Generally vapor pressures above solids are quite small but not negligible. As for liquids we can construct a line representing the equilibrium pressures for sublimation as function of temperature and add it to the phase diagram.

Liquid-solid

The transition solid <=> liquid known as melting or freezing is not very dependent on pressure. Usually melting points increase a little bit with pressure. Water is a peculiar exception. It expands upon freezing and the melting point goes down (a bit) with pressure. In our diagram this will represent an almost vertical line leaning a little forwards for most substances, but backwards for water and a few others.

Putting the lines together

A typical unary diagram

The three lines come together in the triple point, the only point where all three phases are at equilibrium with each other. For water its temperature is only .01K different from the normal melting point (273.16K) and its pressure is only 4.58 Torr.

The intersection points with a line representing atmospheric pressure give the melting and boiling points at that pressure.

If the triple point lies above the line that represents atmospheric pressure this implies that a liquid is never observed. On earth CO₂ is such a substance. The intersection of the solid-vapor equilibrium line with the 1 bar line represents a state where the solid will 'boil' (evaporate from inside out). This is known as the sublimation point.

The melting points at P=1 bar are known as the standard melting point, the only slightly different one at 760 Torr = 1 atm is called normal melting point. The same goes for boiling and sublimation points.

We should realize that there is nothing magical about P=1 bar. It just happens to be the pressure of our home planet. On a planet with higher atmospheric pressures CO₂ may well be a liquid. On such a planet all boiling points will be quite different (higher than on earth). The melting points will also differ but only slightly so. On Mars where atmospheric pressure is much lower water can not occur in liquid form, much like carbondioxide on earth. It sublimes.

We should also realize that in a closed container (glass ampoule, hermetically sealed DSC pan), we can observe melting points at only very slightly different temperature values but we will not see a boiling effect.

The liquid evaporation line ends in a point that we have encountered before: the critical point T_C. As temperature increases the liquid and vapor phases in equilibrium with each other start to resemble each other more and more and at T_c they coalesce. At this point the liquid-gas equilibrium becomes indifferent with respect to density and large fluctuations occur leading to critical opalescence.

Notice that there is a relationship of dimensionality between the objects in the diagram and the number of phases present:

• 2D planes: one phase
• 1D lines : two phases
• 0D point: three phases

As you see the sum is always 3.

Number of moles

So far we have typically considered one substance at the time but for chemists it is imperative to deal with more than one because we are typically changing one into the other in our reactions. This means that the number of moles n, that we often simply set equal to one now becomes an important variable in its own right. Besides we will actually have two (or more) of them: the number of moles of component one and the one for the other component. This makes n a much less trivial variable.

This is already the case at a simple melting point, say when ice melts, because we are dealing with changing quantities of ice and water:

n_ice + n_water = n_total

If all we do is turn water into ice or vice versa, we have dn_total=0, so that:

dn_ice =- dn_water

To deal with changing n's, we need to expand our mathematical notation a bit.

Partial variables

So far we have simply divided our thermodynamic functions if they were extensive by the number of moles and arrived at intensive molar values:

G_molar = G / n

V_molar = V / n

We have written such intensive molar values by writing a bar over the symbol G or V. We should note that scaling the function this way departs from the assumption that the function G depends on the variable n as a straight line that passes through the origin.

If we have the same pure compound in two phases, like ice and water we can still apply this principle and write:

G_molar^ice = G^ice / n^ice

V_molar^ice = V^ice / n^ice

G_molar^water = G^water / n^water

V_molar^water = V^water / n^water

If we have a mixture of two substances present as n₁ and n₂ moles the dependency need not be linear on either if the two substances interact with each other. This is also true for function like the volume of a liquid mixture. In the presence of interactions volumes do not have to be linearly additive. We can define a partial molar value of e.g. for the volume:

V_{partial molar,1} = ∂V /∂ n₁ at n₂ = constant

V_{partial molar,2} = ∂V /∂ n₂ at n₁ = constant

The notation of putting a bar over the V symbol is used for these partial quantities as well. Partial molar volumes have been measured for many binary systems. They are functions of the composition (mole fraction) as well as the temperature and to a lesser extent the pressure.

Thermodynamic potentials

When numbers of moles can change we can write the corresponding change in G as:

dG = -SdT + VdP + (∂G/∂n₁)_P,T,njdn₁ + (∂G/∂n₂)_P,T,njdn₂ + etc.

dG = -SdT + VdP + μ₁dn₁ + μ₂dn₂ + etc.

(As you can see we are adding a set of conjugate variables μ_in_i for each phase i.)

As shown above, as long as we have only one pure component (but in different modifications, like ice and steam), we can still write:

μ_i = (∂G/∂n_i)_P,T,nj =G_i/n_i

As soon as we are dealing with mixtures we really do have derivatives.

The thermodynamic potential at the melting point. Clapeyron

For an equilibrium between e.g. ice and water at constant T and P we get

dG = -SdT + VdP +μ_icedn_ice+μ_waterdn_water

As dn_ice = -dn_water, we get

0 = [μ_ice-μ_water]dn_water

As dn_water is not zero, this means that Δμ must be zero! This must hold true for any set of points where ice and water are in equilibrium. That is the almost vertical line in the diagram. Its points are not at the same P and T, but we can find out where they should be by considering the thermodynamic potential μ as a function of T and P:

dμ = (∂μ/∂P)_TdT + (∂μ/∂T)_PdP

Because μ is simply G/n, it is not hard to identify the partial derivatives. :(∂μ/∂P) = (∂G/n/∂P) =V/n = V_molar

(∂μ/∂T) = (∂G/n/∂T) =-S/n = -S_molar

This is true for both water and ice. As the Δμ=0 we can equate the dμ expressions for both water and ice:

(∂μ_ice/∂P)_TdT + (∂μ_ice/∂T)_P dP=(∂μ_water/∂P)_TdT + (∂μ_water/∂T)_PdP

Rearranging and identifying the partials gives:

V_molaricedP -S_molaricedT= V_molarwaterdP -S_molarwaterdT

Solving for dP/dT we get:

dP/dT = ΔS_molar/ΔV_molar

As ΔG= ΔH-TΔS = 0, we have:

ΔS_molar = ΔH_molar /T_m

So:

dP/dT = ΔH_molar/TΔV_molar

This expression should be valid for all points on the melt line, in fact it tells use that this line is defined by ΔH_molar/TΔV_molar. We immediately see why for water the line runs a little to the left: exceptionally ΔV_molar is negative for ice, because water is actually a little denser that ice.

The above expression(s) are named after Clapeyron

The values of ΔH_molar and ΔV_molar do not change much with pressure and can often be considered constants for the melting line. When gases are involved that is not really true.

Evaporation. Clausius-Clapeyron

We derived the Clapeyron equation for melting points.

dP/dT = ΔH_molar/TΔV_molar

However, our argument is actually quite general and should hold for vapor equilibria as well. The only problem is that the molar volume of gases are by no means so nicely constant as they are for condensed phases. (For the latter both α and κ are pretty small)

We can write:

dP/dT = ΔH_molar/TΔV_molar = ΔH_molar/T[V_molar^g-V_molar^l]

As V_molar^g>>V_molar^l, we can approximate V_molar^g-V_molar^l by just taking V_molar^g. Further more if the vapor is taken as an ideal gas V_molar^g = RT/P

We get

1/P.dP/dT = dlnP/dT = ΔH_molar^vap/RT²

This euation is known as the Clausius-Clapeyron equation.

We can further work our the integration and find the how the equilibrium vapor pressure changes with temperature:

ln[P₂/P₁] = -ΔH_molar^vap/R [1/T₂-1/T₁]

Thus if we know the molar enthalpy of vaporization we can predict the vapor lines in the diagram.

Of course the approximations made are likely to lead to deviations if the vapor is not ideal and very dense (approaching the critical point).

Thermodynamic and electrochemical potentials

Let's do a bit of chemistry and consider a chemical reaction that transforms one pure compound into the other. We'll assume they do not mix (odd assumption that).

Consider the change in Gibbs free energy when a bit of compound one is reacted to form an identical amount of compound two in moles, i.e. dn₁= -dn₂ :

dG = -SdT + VdP +μ₁dn₁+μ₂dn₂

If the reaction is done in an electrochemical cell at constant pressure and temperature, we get:

dG = -SdT + VdP +μ₁dn₁+μ₂dn₂ +ℰde

At equilibrium dG=0, this gives:

0 =μ₁dn₁-μ₂dn₁ +ℰde

[μ₂-μ₁]dn=ℰde

[Δμ]=[de/dn]ℰ

The factor de/dn is determined by the stoichiometric coefficients of the electrochemical reaction (the number of electrons involved in the balanced reaction). It has a dimension Coulomb/mole. Clearly the open cell voltage (the electromotive force ℰ) of the cell is directly related to the difference in thermodynamic potentials between reactants and products. Notice that the righthand side has dimensions [Coulomb*Volt/mole]. This indeed equals the [Joules/mole] on the left

As you can see the term thermodynamic potential is not a joke. It really gives you what work the system can potentially do and electrical potentials are a closely related quantity.

We have treated the reactant and the product as two pure compounds here and that is a bit of a fib. In reality we are usually dealing with mixtures (solutions in which one species replaces the other e.g.). In such cases μ is a function of the composition of the mixture and therefore so is ℰ.

We will need to consider the thermodynamics of mixtures