Introduction to Thermodynamics


Some useful constants in thermodynamics:

Units of Temperature: Degrees Celsius and Kelvin

The Celsius scale is based on defining 0 °C as the freezing point of water and 100°C as the boiling point.

The Kelvin scale is based on defining 0 K, "absolute zero," as the temperature at zero pressure where the volumes of all gases is zero--this turns out to be -273.15 °C. This definition means that the freezing temperature of water is 273.15 K. All thermodynamic calculations are done in Kelvin!

kilo and kelvin: write k for 1000's and K for kelvin. Never write °K.

Units of Energy: Joules and Calories

Joules and calories and kilocalories: A calorie is defined as the amount of energy required to raise the temperature of 1 g of water from 14.5 to 15.5°C at 1 atm.
4.184 J = 1 cal; all food 'calories' are really kcal.

Many times it is easiest to solve equations or problems by conducting "dimensional analysis," which just means using the same units throughout an equation, seeing that both sides of an equation contain balanced units, and that the answer is cast in terms of units that you want. As an example, consider the difference between temperature (units of K) and heat (units of J). Two bodies may have the same temperature, but contain different amounts of heat; likewise, two bodies may contain the same heat, but be at different temperatures. The quantity that links these two variables must have units of J/K or K/J. In fact, the heat capacity C describes the amount of heat dQ involved in changing one mole of a substance by a given temperature increment dT:

The heat capacity C is then and must have units of J K-1 mol-1. (The specific heat is essentially the same number, but is expressed per gram rather than per mole.)
Don't forget significant digits. 1*2=2; 1.1*2=2; 1.1*2.0=2.2; 1.0*2.0=2.0

Why Thermodynamics?

Think about some everyday experiences you have with chemical reactions.
Your ability to melt and refreeze ice shows you that H2O has two phases and that the reaction transforming one to the other is reversible--apparently the crystallization of ice requires removing some heat.
Frying an egg is an example of an irreversible reaction.
If you dissolve halite in water you can tell that the NaCl is still present in some form by tasting the water. Why does the NaCl dissolve? Does it give off heat? Does it require energy?
How is it that diamond, a high-pressure form of C, can coexist with the low pressure form, graphite, at Earth's surface? Do diamond and graphite both have the same energy? If you burn graphite and diamond, which gives you more energy?
When dynamite explodes, why does it change into a rapidly expanding gas, which provides the energy release, plus a few solids?
Chemical thermodynamics provides us with a means of answering these questions and more.

A Few Definitions

A system is any part of the universe we choose to consider.
Matter and energy can flow in or out of an open system but only energy can be added to or subtracted from a closed system. An isolated system is one in which matter and energy are conserved.
A phase is a homogeneous body of matter. The components of a system are defined by a set of chemical formula used to describe the system.
The phase rule: Extensive parameters are proportional to mass (e.g., V, mass, energy).
Intensive parameters are independent of mass (e.g., P, T); these are the "degrees of freedom" F contained in the phase rule.

Thermodynamics: Power and Limitations

Thermodynamics allows you to predict how chemical systems should behave from a supra-atomic "black-box" level--it says nothing about how chemical systems will behave. Thermodynamics also pertains to the state of a system, and says nothing about the path taken by the system in changing from one state to another.

Chemical Reactions and Equations

How to write chemical reactions; stoichiometry.
Mass and charge balance: e.g.,

Reaction-Produced Change in Mass, Density, Volume

The change in volume rV of a reaction is the volume V of the products minus the volume of the reactants: Thus, if the products are smaller than the reactants, rV < 0.
In a generalized reaction such as This sort of additive relationship is true for other state variables and is usually stated as where i are the stoichiometric coefficients, positive for products and negative for reactants.

What Actually Drives Reactions? Is it Energy? Can We Just Calculate or Measure the Energy Difference of Reactants and Products and Know Which Way the Reaction Will Go?

For many years people felt that chemical reactions occurred because the reactants had some kind of energy to give up (i.e., use to do work)--and that therefore the energy of the products would be less than the energy of the reactants. However, we all know that when ice melts it consumes rather than releases heat, so there must be more to the story behind why chemical reactions occur.

Le Chatelier's Principle

"If a change is made to a system, the system will respond so as to absorb the force causing the change."

Equilibrium

A mechanical analogy for chemical change is that of a ball rolling down a slope with multiple valleys; we explain the ball's behavior by saying that mechanical systems have a tendency to reduce their potential energy.
At equilibrium none of the properties of a system change with time. A system at equilibrium returns to equilibrium if disturbed.
"Stable" describes a system or phase in its lowest energy state.
"Metastable" describes a system or phase in any other energy state.


The figure above shows the mechanical analogy for H2O at -5°C and + 5°C and 1 atm. Left: at -5°C, solid H2O has the lowest possible energy state. Right: at +5°C, liquid H2O has the lowest possible energy state. When solid H2O is actually present at +5°C, the difference between the free energy of solid H2O and liquid H2O is available to drive the reaction to form the stable solid H2O phase, and the reaction will go to completion if kinetically possible.

Energy: How Do We Calculate and Measure Energy and How Can We Use this Knowledge to Predict Reaction Behavior?

Thermodynamics works equally well to describe any kind of work or energy: magnetic, potential, kinetic, etc. For geological systems we typically talk about pressure-volume work, which, because mechanical work is Fx, you can imagine might be or Because we noted that rV < 0 if the products are smaller than the reactants, we choose to write the P-V work term as so that a decrease in volume -V is seen as positive work or that an increase in volume +V results in a decrease in crystal energy.
The absolute energy of a body can be calculated from Einstein's equation U=mc2, but the presence of the c2 term means that the energy of any system is quite large and that measuring this energy is impractical. It is more practical to measure differences in energy U, and we always discuss or measure differences relative to some arbitrary standard state. Analogous to this might be if someone in Namibia asked you to measure the elevation of the crests of waves at Campus Point--without agreement on some kind of standard, you wouldn't be able to do much more than measure the heights of individual waves. If however, you could both agree on an equivalent "sea level" at both localities, you could then compare the absolute elevations of the wave crests.

A typical thermodynamic standard state is normal laboratory conditions: 25°C (298.15 K) and 1 atm (often called STP for standard temperature and pressure).

The internal energy U of a mineral is the sum of the potential energy stored in the interatomic bonds and the kinetic energy of the atomic vibrations. Thus, you might expect that weakly bonded minerals have relatively low potential energy and thus low internal energy, and when a mineral is cold such that its atomic vibrations are slow it will have low kinetic energy and thus low internal energy. Internal energies are always defined relative to some non-zero standard state, so we typically talk about changes in internal energy dU.
An Aside on Differences and Differentials
What's the difference among , d, and ?
is used to indicate any kind of difference.
d is used to indicate a differential.
is used to indicate a partial differential. For example, the partial differential, with respect to y, of
f(x,y) = x3y4
is
= 4x3y3

First Law of Thermodynamics

Adding heat Q to a crystal increases its internal energy U: ( indicates 'proportional') but if the crystal is allowed to expand, some of the added energy will be consumed by expansion dV, so the total energy of the crystal is reduced: This is effectively the First Law of Thermo: that total energy (heat + P-V work) is conserved.

Heat Capacity

Heat capacity C describes the amount of heat required to change the temperature of a substance: By definition, the heat capacity of water at 15°C is 1 cal K-1 g-1 or 18 cal K-1 mol-1 (i.e., the heat required to heat 1 gram of water from 14.5 to 15.5°C is 1 calorie).
Heat capacities of solids approach zero as absolute zero is approached: The heat capacity is written with a subscript P or V depending on whether it obtains for constant pressure CP or constant volume CV.

As an aside, where and are the expansivity and compressibility--for solids the difference between CP and CV is minimal and can be ignored as a first approximation. For gases, CP = CV + R, and is quite significant.
Heat capacities are measured by calorimetry and often fit by a function of the form: but there are other expansions for the heat capacity involving more or fewer terms.
Below are some examples of heat capacities of minerals. Note how silicates have a nearly constant heat capacity of ~1 J K-1 g-1 above 400K.

Enthalpy

We have already talked about the familiar concept of heat as energy.
Let's define another measure of energy called enthalpy H--a kind of measure of the thermal energy of a crystal. As we will see below, Recall that we interpreted to mean that the internal energy change is the heat change minus the energy lost to relaxation of the crystal. Thus, means that the enthalpy change is the heat change plus the energy the crystal gains by virtue of not being allowed to expand.
Enthalpy includes the vibrational and bonding energy at absolute zero H0°, plus the energy required to increase temperature: i.e., we can find the enthalpy change H produced by changing temperature by integrating the heat capacity CP:

Integration Reminder
How to integrate the heat capacity (to determine change in enthalpy H):
  • CP dT = (a + bT + cT-2 + dT-0.5)
  • =aT + bT2/2 - c/T + 2dT0.5
and is evaluated as
  • =a(T2 - T1) + b(T22 - T12)/2 - c(T2-1 - T1-1) + 2d(T20.5 - T10.5)
How to integrate the heat capacity divided by T (to determine entropy S):
  • dT = (a/T + b + cT-3 + dT-1.5)
  • = a ln T + b T - c T-2/2 - 2 d T -0.5
and is evaluated as
  • a(ln T2 - ln T1) + b(T2 - T1) - c(T2-2 - T1-2)/2 - 2d(T2-0.5 - T1-0.5)
Nowadays this will all be done for you by the software you are using.

As an example, let's calculate the change in enthalpy H°298-1000 that results from heating quartz from 298 K to 1000 K, given the following heat capacity expansion coefficients:

a = 104.35, b = 6.07E-3, c = 3.4E+4, d = -1070

evaluated from 298 to 1000K

Relation Among Enthalpy, Heat, and Heat Capacity (HP=QP)

An important relationship between enthalpy change H and heat change Q is revealed by differentiating H = U + PV to obtain the total differential substituting dU = dQ - PdV we get dividing by dT gives at constant pressure, = 0, leaving which is equal to CP:

Determining Enthalpies

Thus, if we want to measure how the internal energy U of a crystal changes U with increasing temperature at constant pressure, we want to know H, and we can get that by integrating the heat capacity CP over the temperature range of interest.
There's another way to measure H, though: calorimetry. By dissolving a mineral in acid and measuring the heat produced by the dissolution, we get a heat of dissolution (usually positive). The enthalpy of "formation" fH° of the mineral is then just the opposite of the heat of dissolution (usually negative). Exceptions to the "usually positive/negative" rule include CN, HCN, Cu2+, Hg2+, NO, Ag+, and S2-. Enthalpies of formation appear in tables of thermodynamic data and are usually referenced to 298 K and 1 atm.

Enthalpy of Reaction

To get an enthalpy of reaction rH° we can measure the enthalpies of formation of the reactants and products fH° and then take the difference between them as For example, we can compute the enthalpy of the reaction from
Ca + S + 2O2 = CaSO4 fH° = -1434.11 kJ/mol
H2 + 0.5O2 = H2O fH° = -285.830 kJ/mol
Ca + S + 3O2 + 2H2 = CaSO42H2O fH° = -2022.63 kJ/mol

Thus,

Exothermic vs. Endothermic

If rH° < 0 the reaction produces a reduction in enthalpy and is exothermic (heat is given up by the rock and gained by the surroundings). If rH° > 0 the reaction produces an increase in enthalpy and is endothermic (heat from the surroundings is consumed by the rock). An easy way to remember this is that spontaneous reactions produce a decrease in internal energy, and because we know that
UP HP
a decrease in HP is also a decrease in UP.

Calculating fH° at Temperatures Other Than 298 K

So far we know how to calculate the change in enthalpy caused by heating and we know that we can get enthalpies of formation from tables. What if we want to know the enthalpy of formation of a mineral at a temperature other than 298 K?
We do this by calculating rCP for the reaction that forms the mineral of interest: and then integrating. Thus, for example if we want to know fH° for quartz at 1000 K, we get coefficients for the heat capacities of Si, O2 and SiO2:
compoundabcd
Si31.7785.3878E-4-1.4654E5-1.7864E2
O248.318-6.9132E-44.9923E5-4.2066E2
SiO2104.356.07E-33.4E-4-1070

for the reaction and we calculate
a = 24.254
b = 6.2225E-3
c = -3.5E5
d = -470.7
and thus, This is the change in the enthalpy of formation that results from heating. We add this to the enthalpy of formation at 298 K to get the enthalpy of formation at 1000 K: In other words, forming quartz from the elements at 1000 K yields slightly less heat than at 298 K.
Compare this with the change in enthalpy H°298-1000 that results from heating quartz from 298 K to 1000 K, which we calculated is 45.37 kJ/mol.

Entropy

We have discussed the intuitive statement that reactions probably proceed because the reactants can decrease their internal energy by reacting. We also noted that internal energy scales with enthalpy, suggesting that reactions might 'go' because of a decrease in enthalpy. However, we then noted that not all reactions give off heat--some, such as the melting of ice, proceed in spite of consuming heat. Moreover, there are other processes that proceed in the apparent absence of any heat change: e.g., mixing of gases or the spreading of dye in water. What is it that causes these reactions to proceed spontaneously even if the heat change is zero or endothermic?
The answer is entropy S, which is a measure of the order or disorder.
Entropy has three sources: configurational, electronic, and vibrational.


Electronic entropy arises when an electron in an unfilled orbital can occupy more than one orbital; e.g., for Ti3+, the single 3d electron can occupy one of three possible t2g orbitals and Selectronic = 9 J mol-1 K-1.

Vibrational (or calorimetric) entropy arises because the energy of lattice vibrations can only increase or decrease in discrete steps and the energy quanta (phonons) can be distributed within the possible energy steps in different ways. Vibrational entropy is very difficult to calculate from statistical mechanics but can be calculated easily from heat capacity. Here's why:
The entropy of a system always increases during irreversible processes; i.e., for a reversible process, dS = 0, whereas for irreversible processes dS >0. This is the Second Law of Thermo--better known as "You can't feed s**t into the rear of a horse and get hay out the front."
If a mineral becomes more ordered during a reaction, reducing its entropy, the heat liberated must increase the entropy of the surroundings by an even greater amount. Thus, we write

then and recalling that then and In other words, the vibrational entropy can be found by integrating the heat capacity divided by temperature.
In a perfectly ordered, pure crystalline material the entropy is zero. This is a simple statement of the Third Law of Thermo, which follows from the fact that heat capacities approach zero at zero K: However, because the rate of atomic diffusion also goes to zero at 0 K, all compounds have some zero-point entropy S°0.
Entropy is thus the only thermodynamic potential for which we can calculate an absolute value. What we typically do is determine the heat capacity from near absolute zero to ambient conditions and then integrate it to get the (absolute) entropy (in fact this gives us only the vibrational entropy and ignores configurational and electronic contributions to entropy).

Configurational entropy refers to the entropy resulting from imperfect mixing of different atoms in the same site in a crystal.

When you roll a die, there are six possible outcomes, with the following probability

    P(X=1) + P(X=2) + ... + P(X=6) = 1

The probability that any particular number will come up is

The uncertainty associated with that probability is 0 if one outcome has a probability of 1, and a maximum when the outcomes are equally likely.

Imagine a material with M atoms that can occupy N sites. If the mixing is ideal, the atoms have no interaction with each other, so more than one can occupy the same site. The total number of configurations, , in this system is

config entropCopyright c 2009 by C.H. Mak

But, this is slightly incorrect (in what is known as Gibbs' paradox) because there is overcounting of identical configurations like these

config entrop 2

With the overcounting removed we have

where is the probability that a given number of atoms n in a given number of sites N will have a particular configuration. Or
N is always large where moles of material are concerned, so we can simplify this (using Stirling's approximation: ln (N!) = NlnN - N) to
For N atomic sites that can contain fraction XA A atoms and XB B atoms, where is the probability that a given number of atoms M in a given number of sites N will have a particular configuration.

Boltzmann defined configurational entropy as

Simplified with Stirling's approximation (and recalling that R = Nk) to where n is the number of sites per mole. For example in cordierite there are 4 Al atoms and 5 Si atoms distributed over 9 tetrahedral sites. For a random distribution the entropy is Note that the form of the configurational entropy equation (and electronic entropy as well) indicates that if XA or XB are 0 or 1, Sconfig is zero:


Entropy Change of Reaction

Just like rH and rV, we can calculate entropies of reactions by using absolute entropies S and calculating a difference in entropy rS. For example, if we know that then the entropy of the reaction is rS° = 106.7 - 41.42 - 31.80 - 2 * 205.138 = -376.8 J mol-1 K-1

Energy Associated With Entropy

The units of entropy suggest that the energy associated with S scales with temperature: (The minus sign is there for reasons similar to the -PV we encountered earlier.)
The energy associated with configurational entropy in the Al4Si5 cordierite we talked about earlier looks like this:

The energy associated with vibrational entropy in tremolite, quartz, and chalcopyrite looks like this:

(Josiah Willard) Gibbs Free Energy of a Phase

The Gibbs free energy G is the thermodynamic potential that tells us which way a reaction goes at a given set of physical conditions--neither the enthalpy change nor the entropy change for a reaction alone can provide us with this information. The two measures of energy (enthalpy H and entropic energy TS) are brought together in the Gibbs free energy equation: (the chemical potential is the equivalent for a component) which says that the Gibbs free energy G is the internal energy of the crystal U plus the energy the crystal gains by virtue of not being allowed to expand minus the entropic energy TS. Recalling that we can write this in a more understandable way which says that G is the difference between the heat energy and the entropic energy.

Relationship Among G, S, and V

If we differentiate to obtain and substitute (this comes from dS = dQ/T and dU = dQ - PdV); we are left with meaning that changes in Gibbs free energy are produced by changes in pressure and temperature acting on the volume and entropy of a phase.
Realize that when we write we are implicitly writing which means that and These relations indicate that the change in Gibbs free energy with respect to pressure is the molar volume V and the change in Gibbs free energy with respect to temperature is minus the entropy S.

Gibbs Free Energy of Formation

The defining equation for Gibbs free energy can be written as such that the Gibbs free energy of formation fG° is For example, to calculate the Gibbs free energy of formation of anhydrite, we can use and we calculate the entropy of formation of anhydrite and then use

Gibbs Free Energy of Reaction

We can write the Gibbs free energy of reaction as the enthalpy change of reaction minus the entropic energy change of reaction If the heat energy equals the entropic energy then and there is no reaction. Finally we have come to a satisfying point--we can now determine whether a given reaction will occur if we know H and S, and both of these are measurable or can be calculated.

If rG < 0, the Gibbs free energy of the products is lower than the Gibbs free energy of the reactants and the reaction moves to produce more products. If rG > 0, the Gibbs free energy of the products is greater than the Gibbs free energy of the reactants and the reaction moves to produce more reactants.
For example, to calculate rG° at STP for the reaction we use to calculate The negative value of G tells us that calcite has lower Gibbs free energy and that the reaction runs forward (aragonite calcite).

Clapeyron Relation

There is a useful relation between the slope of a reaction in PT space (i.e., dP/dT) and the entropy and volume changes of the reaction that follows from At equilibrium G = 0, such that or So, the P-T slope of a reaction is equal to the ratio of the entropy change to the volume change. Alternatively, along the equilibrium curve, the changes in pressure times the volume change are equal to changes in temperature times the entropy change. This is the Clapeyron Equation.

So, a phase diagram is a kind of free energy map. = along an equilibrium, < at high P and low T, and > at high T and low P. Along the equilibrium boundary the Gibbs Free energies of the reactants and products are equal and the Gibbs Free energy of reaction rG, is zero.

Shortcutting H and S and Finding G Directly

Like other thermodynamic potentials, we can write the change in Gibbs free energy of reaction as Instead of using fH° and fS°, it is often possible to obtain fG° values for most compounds from electronic data bases. For example, if the following Gibbs free energies of formation are known: then for

Gibbs Free Energy at Any Pressure and Temperature

We know many ways to determine rG at STP--but how do we calculate rG for other pressures and temperatures? Recall that the changes in Gibbs free energy with pressure and temperature are given by two of Maxwell's relations If we recast these as and
and integrate, we get or
and or
thus

Solving the Pressure Integral at Constant Temperature

To a first approximation, we can ignore the expansivity and compressibility of solids and use as a simplification. Don't forget that this approximation is valid for solids only! An even more common assumption for P>>1 is For example, calculate the change in Gibbs free energy for the reaction
2 jadeite = albite + nepheline
if pressure increases from 1 bar to 10 kbar, given First we calculate rV and find and thus

Solving the Temperature Integral at Constant Pressure

Recall that the effect of temperature on the entropy change of reaction rS depends on the heat capacity change of reaction rCP: Thus expands to If the form of the heat capacity expansion is then the above double integral is Note that this considers only vibrational entropy and ignores configurational entropy. This means of solving for rG requires that you know rG at the reference temperature.
An alternative path that requires that you know the enthalpy change rH at the reference temperature is

Solving the Temperature and Pressure Integrals for rGP,T

To calculate the Gibbs free energy change of a reaction at any pressure and temperature, we can use either of the following equations, depending on whether we know rH or rG If you don't have heat capacity data for the reaction of interest, these equations can be roughly approximated as For example, calculate rG for jadeite + quartz = albite at 800 K and 20 kbar. The data at 298 K and 1 bar are
rH° = 15.86 kJ/mol
rS° = 51.47 J K-1 mol-1
rVs° = 1.7342 J/bar = 17.342 cm3/mol
Using If we had used the complete equation for solids, integrating the heat capacities, we would have obtained an answer of 9.86 kJ/mol--not horrifically different.

Calculating the PT Position of a Reaction

If we say that rG = 0 at equilibrium, then we can write our solids-only and constant-heat-capacity approximations as and thus we can calculate the pressure of a reaction at different temperatures by and we can calculate the temperature of a reaction at different pressures by Let's do this for the albite = jadeite + quartz reaction at T = 400 K and T = 1000 K: Assuming that dP/dT is constant (a bad assumption, we know), the reaction looks like this


Solutions

Almost no phases are pure, but typically are mixtures of components. For example, olivine varies from pure forsterite Mg2SiO4 to pure fayalite Fe2SiO4, and can have any composition in between--it is a solid solution. We need a way to calculate the thermodynamic properties of such solutions.
As a measure of convenience, we use mole fraction to describe the compositions of phases that are solid solutions. For example, a mix of 1 part forsterite and 3 parts fayalite yields an olivine with 25 mol% forsterite and 75 mol% fayalite, which can be written as (Mg0.25Fe0.75)2SiO4 or fo25fa75, etc. Mole fractions are denoted as Xi.
We need a way of splitting up the Gibbs free energy of a phase among the various components of the phase--how for example do we decide how much of the Gibbs free energy of an olivine is related to the forsterite component and how much derives from the fayalite component? Likewise, how does the Gibbs free energy of a phase vary with composition--is the relationship linear between endmembers?? We address these issues by defining a partial Gibbs free energy for each component at constant pressure and temperature and constant composition of other components, called the partial molar Gibbs free energy or chemical potential where n is the amount of substance. For olivine solid solution composed of fayalite and forsterite components or endmembers, we write

Volume of Mixing

Imagine that mole fractions of phase A and phase B with molar volumes VA and VB, are mixed together. We can describe the volume of the mixture as and it is a linear mixing of the two endmember volumes. We call this ideal mixing or mechanical mixing. Real solutions, however, do not behave this way, and the mixing is always non ideal, although sometimes only weakly so. The figure shows mixing that produces a smaller volume than expected, but it is not possible to predict the shapes and positions of such mixing curves.

in this figure, the molar volumes of the endmembers are VAo and VBo, and the partial molar volumes are A and B.

Partial Molar Volume

The partial molar volume is defined as If you mix two compounds A and B together and find a volume of mixing that is non-ideal, how can you determine the contribution that A and B each make to the volume? That is, what are the partial molar volumes of A and B, A and B? Graphically, the partial molar volumes are the A and B axis intercepts of the tangent to the mixing curve, and can be described by the simple relationship: or
The behavior of this function is such that when XA is 1, Vmix = VA and when XA is 0, Vmix = VB = VBo. Alternatively,

Entropy of Mixing

The entropy of mixing is never zero because mixing increases entropy. As we discussed days ago, the entropy of mixing (i.e., the configurational entropy) is where i = 1..n is the number of sites over which mixing is occurring.

Enthalpy of Mixing

Enthalpies also do not combined ideally (linearly) in mixtures because the mixture may have stronger bonds than were present in either of the unmixed phases. The excess enthalpy is where AB is the interaction energy among A-B atoms, AA is the interaction energy among A-A atoms, and BB is the interaction energy among B-B atoms.

Gibbs Free Energy of Mixing

Recall that all spontaneous processes/reactions occur because of a decrease in Gibbs free energy. It should therefore not surprise you that the Gibbs free energy of mixing is always negative--otherwise mixing would not occur. The fact that μA < G°A and μB < G°B illustrates why compounds combine spontaneously--each compound is able to lower its free energy.

The above figure is hypothetical because we cannot measure or calculate the absolute Gibbs free energy of phases. For this reason, μ is always expressed as a difference from some standard state measurement, as μ, μ - μ°, or μ - G°.
The difference between the absolute Gibbs free energy G° per mole μ° of a pure compound and the chemical potential per mole μ of dissolved compound (i.e., the difference between the blue and red curves in the above figure) is

This function has the following shape:

implying that when the mineral is pure (X = 1) then μ = 0, and when the mineral is infinitely dilute (X = 0), the chemical potential is undefined. For example, in a two-component mineral if XA = 0.4, at T = 298 K,

The equation of the Gmix line is the sum of the chemical potentials of the endmembers: which looks like this for two components:

Actually, all this discussion has been predicated on the assumption that Hmix = 0. If this is not true, Gmix is not a simple function of composition, but has the general form:

Depending on the relative values of Hmix and -TSmix, the free energy of mixing may be negative throughout the whole composition range if the entropic energy contribution outweighs the enthalpy increase; this is more likely at higher temperature.
The two free energy minima in the above figure indicate that minerals of intermediate compositions can reduce their free energy by unmixing into two phases. This explains the appearance and driving force for exsolution. Note that this can only be true if Hmix > 0, i.e., if 2AB > AA + BB, which makes sense because it means that the A-B bonds have a higher free energy than the sum of the free energies of separate AA and BB bonds.

Activities

In reality, no phases behave ideally--that is, their chemical potentials are never simple logarithmic functions of composition as implies. Instead, we say that the chemical potential is a simple logarithmic function of activity and define activity as
a = (X)
where a is the activity of a compound, is the "site occupancy coefficient" (e.g., = 2 for Mg in Mg2SiO4), and is the activity coefficient that describes the non-ideal behavior. Thus we write For pure compounds a=1 because X=1. For ideal compounds =1. As a specific example, the chemical potential of the almandine (Fe3Al2Si3O12) component of a garnet solid solution ((Fe, Mg, Ca, Mn)3Al2Si3O12) is To be clear, μ° is the chemical potential of the component in its pure reference state and varies as a function of pressure and temperature; this we measure with calorimetry. μ is the chemical potential as it actually occurs and varies as a function of phase composition; this we measure with an electron microprobe. The activity forms a bridge between idealized behavior and real behavior.

The Equilibrium Constant

At equilibrium the sum of the Gibbs free energies of the reactants equals the sum of the Gibbs free energies of the products. Equally, the sum of the partial molar Gibbs free energies (chemical potentials) of the reactants equals the sum of the partial molar Gibbs free energies (chemical potentials) of the products. In other words, for at equilibrium, More generally, for then or, at equilibrium which we can reformat as where i is the stochiometric coefficient of a product or reactant and is positive if for a product and negative if for a reactant. If we then remember that and rewrite it as we can reformat the earlier equation as which looks nicer as To be completely general we write This equation is invariably simplified to and Q is the activity product ratio. The activities in the Q term change as the reaction progresses toward equilibrium.
To be clear again, rμ° is the difference in the Gibbs free energies of the products and reactants when each is in its pure reference state and varies as a function of pressure and temperature. rμ is the difference in the Gibbs free energies of the products and reactants as they actually occur and varies as a function of phase composition.
At equilibrium, the product and reactant activities have adjusted themselves such that rμ = 0. We write this (with K instead of Q, to signify equilibrium) as K is called the equilibrium constant. If K is very large (ln K positive), the combined activities of the reaction products are enormous relative to the combined activities of the reactants and the reaction will likely progress. On the other hand, if K is small (ln K negative), there is unlikely to be any reaction.
The utility of K is that it tells us for any reaction and any pressure and temperature, what the activity ratios of the phases will be at equilibrium. For example, for the reaction let's say that at a particular P and T, Using we calculate This means that at equilibrium, Where ajadeite is the activity of NaAlSi2O6 in clinopyroxene and aalbite is the activity of NaAlSi3O8 in plagioclase.

Graphical Portrayal of K

Another way to write the activity product ratio is
or At equilibrium, where ΔrG°= 0, ln K = 0 and K = 1. Let's see what K looks like for jadeite + quartz = albite at 800 K and 20 kbar: If we do this for all of PT space, we can contour PT space in terms of lnK:


Alternative Route to the Equilibrium Constant

When we think of mass balance in a reaction, we can explicitly write where i are the stoichiometric coefficients and Mi are the masses or the phase components. Analogously, we can explicitly write a similar balance among the chemical potentials: For each chemical potential we can write Combining these two equations we find and eventually or The equation is the law of mass action (which actually discusses the action of chemical potential rather than mass). We can also write for 298 K and 1 atm and for any P and T of interest: This has been called "the most important equation in thermodynamics," so you'd better like it(!) The equilibrium constant K is a function of 1/T Which looks like

Activity Models (Activity-Composition Relations) for Crystalline Solutions

Garnets are solid solutions of
component abbrev.formula
pyropeprpMg3Al2Si3O12
almandinealmFe3Al2Si3O12
grossulargrsCa3Al2Si3O12
spessartinespsMn3Al2Si3O12
andraditeandCa3Fe23+Si3O12

Mixing models derive from entropy considerations. In particular the relation although we will not go through the derivation.

Mixing on a Single Site

The simplest type of useful activity model is the ionic model, wherein we assume that mixing occurs on crystallographic sites. For a Mg-Fe-Ca-Mn garnet with mixing on one site, which we can idealize as (A,B,C,D)Al2Si3O12, the activities are


The form of a cubic ideal activity model is shown in the above figure.
In general, for ideal mixing in a mineral with a single crystallographic site that can contain ions,

where a, the activity of component i, is the mole fraction of element j raised to the power. For non-ideal mixing, we include an activity coefficient

Mixing on a Several Sites

For minerals with two distinct sites and the general formula there are four possible end member components AY, AZ, BY, and BZ.
The ideal activities of these components are For non-ideal garnet activities we write where the X3 term describes mixing on the 8-fold trivalent site and the X2 term describes mixing on the octahedral divalent site.


The pyrope activity is shown in the above figure for Mg from 0 3 and Al from 0 2.
It is common to modify such models--that are based on completely random mixing of elements--with models that consider local charge balance on certain sites or the Al-avoidance principle.

Geothermometry and Geobarometry

Exchange Reactions

Many thermometers are based on exchange reactions, which are reactions that exchange elements but preserve reactant and product phases. For example:
  • Fe3Al2Si3O12
 + KMg3AlSi3O10(OH)2 = Mg3Al2Si3O12 + KFe3AlSi3O10(OH)2
  • almandine
+ phlogopite = pyrope + annite

We can reduce this reaction to a simple exchange vector: Popular thermometers include garnet-biotite (GARB), garnet-clinopyroxene, garnet-hornblende, and clinopyroxene-orthopyroxene; all of these are based on the exchange of Fe and Mg, and are excellent thermometers because rV is small, such that is large (i.e., the reactions have steep slopes and are little influenced by pressure). Let's write the equilibrium constant for the GARB exchange reaction thus
This equation implies that the activities of the Fe and Mg components of biotite and garnet are a function of Gibbs free energy change and thus are functions of pressure and temperature.
If we assume ideal behavior ( = 1) in garnet and biotite and assume that there is mixing on only 1 site Thus the equilibrium constant is When discussing element partitioning it is common to define a distribution coefficient KD, which is just the equilibrium constant without the exponent (this just describes the partitioning of elements and not the partitioning of chemical potential): Long before most of you were playground bullies (1978) a couple of deities named John Ferry and Frank Spear measured experimentally the distribution of Fe and Mg between biotite and garnet at 2 kbar and found the following relationship:



If you compare their empirical equation

this immediately reminds you of and you realize that for this reaction (the three comes from the site occupancy coefficient; i.e., K = KD3) and or Molar volume measurements show that for this exchange reaction rV = 0.238 J/bar, thus The full equation is then To plot the KD lines in PT space


Net-Transfer Reactions

Net-transfer reactions are those that cause phases to appear or disappear. Geobarometers are often based on net-transfer reactions because rV is large and relatively insensitive to temperature. A popular one is GASP:
3CaAl2Si2O8 = Ca3Al2Si3O12 + 2Al2SiO5 + SiO2
anorthite = grossular + kyanite + quartz

which describes the high-pressure breakdown of anorthite.


For this reaction

(the activities of quartz and kyanite are one because they are pure phases). A best fit through the experimental data for this reaction by Andrea Koziol and Bob Newton yields for rV = -6.608 J/bar. Again, if we use and set ln K = 0 to calculate values at equilibrium, we can rewrite the above as or if TrS / rV = 22.8 then rS = -150.66 J/mol K
if rH / rV = 7317 then rH = -48.357 kJ/mol
So, we can write the whole shmear as Contours of ln K on a PT diagram for GASP look like this:


Kinetics

Thermodynamics places no constraints on the rate or mechanism of reaction--that is the realm of kinetics. A popular method for describing the rate at which reactions proceed is to talk of an activated state through which the reaction must pass:

When a system passes from an initial to a final state it must overcome an activation energy barrier G*.
The advantages of this activation energy barrier paradigm are that it qualitatively explains the i) persistence of metastable states; ii) effect of catalysts in lowering G*; iii) temperature dependence of transformation. We can draw a similar diagram for a change in enthalpy induced by the reaction rH and an activation enthalpy barrier H* (usually called an activation energy Q*). Of course, unlike rG, rH can be positive or negative:

It is not easy to generalize about the activation entropy S*, however, in general, reactions with positive entropy change rS are faster. For example, evaporation is faster than condensation, melting is faster than crystallization, and disordering is faster than ordering.
Because thermal energy dictates whether an atom has sufficient energy to overcome an activation energy barrier, we will write that the fraction of atoms with thermal energy greater than H* is

i.e., if f << 1, few atoms have enough thermal energy to overcome the activation energy barrier, whereas if f = 1 all atoms can overcome the activation energy barrier.

For reactions that involve a single step characterized by a single activation energy, the rate of the reaction depends on the i) frequency with which atoms attempt to jump from one site to the next ; ii) fraction of atoms with thermal energy greater than Q*; and iii) probability p that the atom jumping satisfies some geometrical consideration (this is a fancy name for a fudge factor):

Often this is rewritten as rate (kT/h) exp(-Q*/RT) where k is Boltzmann's constant, and h is Planck's constant. Thus, the rate at which atoms jump is related to temperature and to atom-scale processes described by k and h.

Nucleation

Most transformations take place by nucleation and growth. The driving force for nucleation results from the fact that the formation of the new phase lowers the total free energy by VG, the Gibbs free energy change of reaction per unit volume. Nucleation is opposed by an interfacial energy and a strain energy . The overall change in nucleation energy for a spherical nucleus of radius r is then
nG = VG + Ginterfacial + Gstrain = (4/3)r3VG + 4 r2 + (4/3)r3
These different energy contributions yield an increase in total free energy up to a certain radius termed the critical radius, beyond which the total free energy decreases.

This change from increasing to decreasing nucleation free energy happens because the surface:volume ratio decreases with increasing radius, so the negative volume free energy term that favors nucleation eventually overwhelms the positive surface and strain free energy terms that oppose nucleation. We can re-order the terms in the above equation to calculate the critical size of the nucleus rc

If, at the critical radius, nG = G*, the activation energy for nucleation is

Growth

Following nucleation, grains of the product phase(s) grow to replace the parent phase(s). Either the reaction is polymorphic, or atoms of the parent phase must dissolve, be transported, and then attach themselves to the product phase. The rate of growth is thus controlled by the rate of the slowest of these three steps, and is described as either interface controlled or diffusion controlled.
Turnbull's (1956) formulation to quantify interface-controlled growth rate is: where is an "interface jump distance" and the rest you will recognize from previous equations. The [( -rG/RT)] term indicates that the rate of growth depends on the Gibbs free energy change of reaction. As rG 0, this term also approaches zero. As rG , this term approaches 1.

Transformation: Nucleation + Growth

A complete transformation involves nucleation and growth--the product phases must form and grow and the parent phases must be consumed. The letter _ is often used to indicate 'reaction progress' or 'degree of transformation,' and varies from 0 (no reaction) to 1 (complete reaction). The transformation is the time-integrated result of ongoing nucleation and growth, which can vary in time and space: Just as an example, the general rate equation for interface-controlled (i.e., not diffusion-controlled) growth of nuclei formed on grain boundaries is In their simplest form, these equations have the form Flip this around to linearize by means of logarithms change signs and cleverly insert ln 1 rewrite the two logarithms as one, recalling that ln (a/b) = ln a - ln b and linearize once more with logarithms to get the final glorious equation

The Last Supper: A Still Life of Thermodynamics & Kinetics

Zeolites, like laumontite and wairakite, form in oil fields as alteration products of plagioclase and other Ca-bearing phases. The thermodynamic properties (H, S, V, CP) of zeolites have been measured, and we can use those properties to calculate a phase diagram for the CASH system:


Laumontite is the phase that is stable at the lowest P and T. Wairakite is the next phase to form, and it does so via the reaction laumontite _ wairakite + H2O. At higher temperature wairakite + H2O decompose to anorthite + quartz + H2O.



Kinetic experiments conducted on the reaction laumontite _ wairakite + H2O, using crystals of laumontite suspended in H2O, reveal the transformation vs. time data in the figure above.


One could use the equation

to interpret the two sub-figures to reveal different n and k values for each temperature. This is the way rate data were interpreted in the bad old days, but this approach lacks any physical or mechanistic basis. Instead, we can measure the rate of growth at each temperature and relate the rate of growth to the Turnbull equation. Above is an example of grain growth data collected for 450°C. The rate of growth is interpreted to be a constant 2.0 ± 0.3 E-10 m/s. If growth rate data like this can be collected for every temperature of interest, an activation energy for growth can be calculated:


The activation energy for growth apparently varies from ~196 kJ/mol to ~ 72 kJ/mol depending on the reaction mechanism. The growth rate data can be combined with similar nucleation rate data to yield a complete transformation rate equation that can be extrapolated to geologic conditions of interest. The figure below shows extrapolated growth rates (m/s) for the laumontite _ wairakite + H2O reaction.