Introduction to Thermodynamics
Some useful constants in thermodynamics:
 1 eV = 9.6522E4 J/mol
 k, Boltzmann's constant = 1.38E23 J/K
 volume: 1 cm^{3} = 0.1 kJ/kbar = 0.1 J/bar
 mole: 1 mole of a substance contains Avogadro's number (N = 6.02E23) of molecules. Abbreviated as 'mol'.
 atomic weights are based around the definition that ^{12}C is exactly 12 g/mol
 R gas constant = Nk = 8.314 J mol^{1} K^{1}
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
zerothis 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 H_{2}O
has two phases and that the reaction transforming one to the other is
reversibleapparently 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 highpressure 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 supraatomic "blackbox" levelit 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.,  2Fe^{3+} + 3H_{2}O = Fe_{2}O_{3} + 6H^{+}
ReactionProduced Change in Mass, Density, Volume
The change in volume _{r}V of a reaction is the volume V of the products minus the volume of the reactants:
 _{r}V = Vproducts  Vreactants
Thus, if the products are smaller than the reactants, _{r}V < 0.
In a generalized reaction such as
 _{r}V = cV_{C} + dV_{D}  aV_{A}  bV_{B}
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 H_{2}O at 5°C and + 5°C and 1 atm. Left: at 5°C, solid H_{2}O has the lowest possible energy state. Right: at +5°C, liquid H_{2}O has the lowest possible energy state. When solid H_{2}O is actually present at +5°C, the difference between the free energy of solid H_{2}O and liquid H_{2}O is available to drive the reaction to form the stable solid H_{2}O 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 pressurevolume work, which, because mechanical
work is Fx, you can imagine might be
 PV
or
 VP
Because we noted that _{r}V < 0 if the products are smaller than the reactants, we choose to write the PV work term as
 PV
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=mc^{2}, but the presence of the c^{2}
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 Pointwithout 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 nonzero 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) = x^{3}y^{4}
is
= 4x^{3}y^{3}

First Law of Thermodynamics
Adding heat Q to a crystal increases its internal energy U:
 dU dQ
( 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 + PV work) is conserved.
Heat Capacity
Heat capacity C describes the amount of heat required to change the temperature of a substance:
 C =
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:
 C = 0
The heat capacity is written with a subscript P or V depending on whether it obtains for constant pressure C_{P} or constant volume C_{V}.
As an aside,
 C_{P} = C_{V} + TV^{2}/
where and are the expansivity and compressibilityfor solids the difference between C_{P} and C_{V} is minimal and can be ignored as a first approximation. For gases, C_{P} = C_{V} + R, and is quite significant.
Heat capacities are measured by calorimetry and often fit by a function of the form:
 C_{P} = a + bT + cT^{2} + dT^{0.5}
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 Ha 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 H_{0}°, plus the energy required to increase temperature:
 H = H_{0}° + C_{P}dT
i.e., we can find the enthalpy change H produced by changing temperature by integrating the heat capacity C_{P}:
 H = C_{P}dT
Integration Reminder 
How to integrate the heat capacity (to determine change in enthalpy H):
 C_{P} dT = (a + bT + cT^{2} + dT^{0.5})
 =aT + bT^{2}/2  c/T + 2dT^{0.5}
and is evaluated as
 =a(T_{2}  T_{1}) + b(T_{2}^{2}  T_{1}^{2})/2  c(T_{2}^{1}  T_{1}^{1}) + 2d(T_{2}^{0.5}  T_{1}^{0.5})
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 T_{2}  ln T_{1}) + b(T_{2}  T_{1})  c(T_{2}^{2}  T_{1}^{2})/2  2d(T_{2}^{0.5}  T_{1}^{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°_{2981000} that results from heating quartz from 298 K to 1000 K, given the following heat capacity expansion coefficients:
a = 104.35, b = 6.07E3, c = 3.4E+4, d = 1070
 (C_{P} dT = (a + bT + cT^{2} + dT^{0.5})
 =aT + bT^{2}/2  c/T + 2dT^{0.5}
evaluated from 298 to 1000K
 =a*(1000298) + b*(1000^{2}298^{2})/2  c*(1000^{1}298^{1}) + 2d*(1000^{0.5}298^{0.5})
 = 45.37 kJ/mol
Relation Among Enthalpy, Heat, and Heat Capacity (H_{P}=Q_{P})
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 C_{P}:
 = = C_{P}
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 C_{P} 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" _{f}H°
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, Cu^{2+}, Hg^{2+}, NO, Ag^{+}, and S^{2}. 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 _{r}H° we can measure the enthalpies of formation of the reactants and products _{f}H° and then take the difference between them as
 _{r}H° = _{f}H°_{products} _{f}H°_{reactants}
For example, we can compute the enthalpy of the reaction
 anhydrite + water = gypsum:
 CaSO_{4} + 2H_{2}O = CaSO_{4}2H_{2}O
from
Ca + S + 2O_{2} = CaSO_{4} 
_{f}H° = 1434.11 kJ/mol 
H_{2} + 0.5O_{2} = H_{2}O 
_{f}H° = 285.830 kJ/mol 
Ca + S + 3O_{2} + 2H_{2} = CaSO_{4}2H_{2}O 
_{f}H° = 2022.63 kJ/mol 
Thus,
 _{r}H° = _{f}H°gypsum  _{f}H°anhydrite  _{f}H°water = 16.86 kJ/mol.
Exothermic vs. Endothermic
If _{r}H° < 0 the reaction
produces a reduction in enthalpy and is exothermic (heat is given up by
the rock and gained by the surroundings). If _{r}H°
> 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
U_{P} H_{P}
a decrease in H_{P} is also a decrease in U_{P}.
Calculating _{f}H° 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 _{r}C_{P} for the reaction that forms the mineral of interest:
 _{r}C_{P} = _{r}C_{P}_{products}  _{r}C_{P}_{reactants}
and then integrating. Thus, for example if we want to know _{f}H° for quartz at 1000 K, we get coefficients for the heat capacities of Si, O_{2} and SiO_{2}:
compound  a  b  c  d 
Si  31.778  5.3878E4  1.4654E5  1.7864E2 
O_{2}  48.318  6.9132E4  4.9923E5  4.2066E2 
SiO_{2}  104.35  6.07E3  3.4E4  1070 
for the reaction
and we calculate
a = 24.254
b = 6.2225E3
c = 3.5E5
d = 470.7
and thus,
 _{f}H°_{1000}  _{f}H°_{298} = C_{P} dT = a*(1000298) + b/2*(1000^{2}298^{2})  c*(1000^{1}298^{1}) + 2d*(1000^{0.5}298^{0.5}) = 5.511 kJ/mol
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:
 _{f}H°_{1000} = (_{f}H°1000  _{f}H°298) + _{f}H°298 = 5.511  910.700 = 905.2 kJ/mol
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°_{2981000} 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 heatsome, 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 Ti^{3+}, the single 3d electron can occupy one of three possible t_{2g} orbitals and S_{electronic} = 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 Thermobetter 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
 dS >
then
 >
and recalling that
 C_{P} =
then
 >
and
 S = dT
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:
 C = 0
However, because the rate of atomic diffusion also goes to zero at 0 K, all compounds have some zeropoint 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
 P(X=1) = P(X=2) = ... = P(X=6) = 1/6
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
 = N^{M}
Copyright 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
With the overcounting removed we have
 = N! / (N_{p1}! * N_{p2}! ...N_{pn}!)
where is the probability that a given number of atoms n in a given number of sites N will have a particular configuration.
Or
 ln = ln N!  ln(N_{pi})!
N is always large where moles of material are concerned, so we can simplify this (using Stirling's approximation: ln (N!) = NlnN  N) to
 ln = N _{pi}ln_{pi}
For N atomic sites that can contain fraction X_{A} A atoms and X_{B} 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
 S_{configurational} = k ln (This is engraved on Boltzmann's tomb in Vienna!)
Simplified with Stirling's approximation (and recalling that R = Nk) to
 S =  n R (X_{A} ln X_{A} + X_{B} ln X_{B})
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
 S =  9 R (4/9 ln 4/9 + 5/9 ln 5/9) = 51.39 J mol^{1} K^{1}
Note that the form of the configurational entropy equation (and electronic entropy as well) indicates that if X_{A} or X_{B} are 0 or 1, S_{config} is zero:
Entropy Change of Reaction
Just like _{r}H and _{r}V, we can calculate entropies of reactions by using absolute entropies S and calculating a difference in entropy _{r}S. For example, if we know that
 S°_{CaSO4} = 106.7 J mol^{1} K^{1}
 S°_{Ca} = 41.42 J mol^{1} K^{1}
 S°_{S} = 31.80 J mol^{1} K^{1}
 S°_{O2} = 205.138 J mol^{1} K^{1}
then the entropy of the reaction
 Ca + S + 2O_{2} = CaSO_{4}
is _{r}S° = 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:
 dU TS
(The minus sign is there for reasons similar to the PV we encountered earlier.)
The energy associated with configurational entropy in the Al_{4}Si_{5} 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
conditionsneither 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
 dG = dU + PdV + VdP  TdS  SdT
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
 dG = dP  dT
which means that
 = V
and
 = S
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 _{f}G° is
 _{f}G° = _{f}H°  T_{f}S°
For example, to calculate the Gibbs free energy of formation of anhydrite, we can use
 _{f}H°_{CaSO4} = 1434.11 kJ/mole
 S°_{CaSO4} = 106.7 J mol^{1} K^{1}
 S°_{Ca} = 41.42 J mol^{1} K^{1}
 S°_{S} = 31.80 J mol^{1} K^{1}
 S°_{O2} = 205.138 J mol^{1} K^{1}
and we calculate the entropy of formation of anhydrite
 _{f}S° = S°_{CaSO4}  S°_{Ca}  S°_{S}  2 * S°_{O2} = 376.796 J mol^{1} K^{1}
and then use
 _{f}G° = _{f}H°  T_{f}S° = 1434,110  298.15 * 376.796 = 1321.77 kJ/mol
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
 _{r}H = T_{r}S
then
 _{r}G = 0
and there is no reaction. Finally we have come to a
satisfying pointwe 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 _{r}G < 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 _{r}G
> 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 _{r}G° at STP for the reaction
we use
 _{r}H° = 370 J
 _{r}S° = 3.7 J mol^{1} K^{1}
to calculate
 _{r}G° = 370  298.15 * 3.7 = 733 J/mol
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  _{r}G = V_{r}dP  S_{r}dT
At equilibrium G = 0, such that
 _{r}VdP = _{r}SdT
or
 =
So, the PT 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 _{r}G, 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
 _{r}G° = _{f}G°_{products} _{f}G°_{reactants}
Instead of using _{f}H° and _{f}S°, it is often possible to obtain _{f}G°
values for most compounds from electronic data bases. For example, if
the following Gibbs free energies of formation are known:
 _{f}G°_{CaSO42H2O} = 1707.280 kJ/mol
 _{f}G°_{CaSO4} = 1321.790 kJ/mol
 _{f}G°_{H2O} = 237.129 kJ/mol
then for
 CaSO4 + 2H_{2}O = CaSO_{4}2H_{2}O
 _{r}G° = 1.232 kJ/mol
Gibbs Free Energy at Any Pressure and Temperature
We know many ways to determine _{r}G at STPbut how do we calculate _{r}G
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
 _{r}GdP =
_{r}G_{P}  _{r}G_{Pref} =
_{r}VdP
or
 _{r}G_{P} = _{r}G_{Pref} + _{r}VdP
and
 _{r}GdT =
_{r}G_{T}  _{r}G_{Tref} = 
_{r}SdT
or
 Δ_{r}G_{T} = Δ_{r}G_{Tref}  Δ_{r}SdT
thus
 _{r}G_{PT} = _{r}G_{PrefTref} + _{r}VdP  _{r}SdT
Solving the Pressure Integral at Constant Temperature
To a first approximation, we can ignore the expansivity and compressibility of solids and use
 _{r}V_{s}dP = _{r}V_{s}(P  1)
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
 _{nepheline} = 54.16 cm^{3}
 _{albite} = 100.43 cm^{3}
 _{jadeite} = 60.40 cm^{3}
First we calculate _{r}V and find
 _{r} = _{nepheline} + _{albite}  2_{jadeite} = 33.79 cm^{3} = 3.379 J/bar
and thus
 _{r}G_{PT}  _{r}G_{1,T} = _{r}V_{s}P = 33.79 kJ/mol
Solving the Temperature Integral at Constant Pressure
Recall that the effect of temperature on the entropy change of reaction _{r}S depends on the heat capacity change of reaction _{r}C_{P}:
Thus
 _{r}G_{T} = _{r}G_{Tref}  _{r}SdT
expands to
 _{r}G_{T} = _{r}G_{Tref}
 S_{Tref}
+ dTdT
If the form of the heat capacity expansion is
 C_{P} = a + bT + cT^{2} + dT^{0.5}
then the above double integral is
 a(T  T ln T)
 bT^{2}/2
 cT^{1}/2 + 4dT^{0.5}

aT_{ref}
 bT_{ref}^{2}/2
+ cT_{ref}^{1}
 2dT_{ref}^{0.5}
+
aTlnT_{ref}
+ bTT_{ref}
 cTT_{ref}^{2}/2
 2dTT_{ref}^{0.5}
 T_{r}S_{Tref}
+ T_{ref}_{r}S_{Tref}
Note that this considers only vibrational entropy and ignores configurational entropy. This means of solving for _{r}G requires that you know _{r}G at the reference temperature.
An alternative path that requires that you know the enthalpy change _{r}H at the reference temperature is
 _{r}G_{T} =
_{r}H_{Tref}
+ C_{P}dT  T_{r}S_{Tref}
+ dT
Solving the Temperature and Pressure Integrals for _{r}GP,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 _{r}H or _{r}G
 _{r}G_{P,T} = _{r}G_{1,Tref}
 _{r}S_{Tref}
+ dTdT
+ _{r}V_{s}P
 _{r}G_{P,T} =
_{r}H_{1,Tref}
+ C_{P}dT  T_{r}S_{Tref}
+ dT
+ _{r}V_{s}P
If you don't have heat capacity data for the reaction of interest, these equations can be roughly approximated as
 _{r}G_{P,T} = _{r}G_{1,Tref}
 _{r}S_{1,Tref}(T  T_{ref})
+ _{r}V_{s}P
 _{r}G_{P,T} =
_{r}H_{1,Tref}
 T_{r}S_{1,Tref}
+ _{r}V_{s}P
For example, calculate _{r}G for jadeite + quartz = albite at 800 K and 20 kbar. The data at 298 K and 1 bar are
_{r}H° = 15.86 kJ/mol
_{r}S° = 51.47 J K^{1} mol^{1}
_{r}V_{s}° = 1.7342 J/bar = 17.342 cm^{3}/mol
Using
 _{r}G_{P,T} =
_{r}H_{1,Tref}
 T_{r}S_{1,Tref}
+ _{r}V_{s}P
 = 15,860  800 * 51.47 + 1.7342 * 20,000 = 9.37 kJ/mol
If we had used the complete equation for solids, integrating
the heat capacities, we would have obtained an answer of 9.86
kJ/molnot horrifically different.
Calculating the PT Position of a Reaction
If we say that _{r}G = 0 at equilibrium, then we can write our solidsonly and constantheatcapacity approximations as
 0 = _{r}G_{1,T}
 _{r}S_{1,Tref}(T  T_{ref})
+ _{r}V_{s}P
 0 = _{r}H_{1,Tref}
 T_{r}S_{1,Tref}
+ _{r}V_{s}P
and thus we can calculate the pressure of a reaction at different temperatures by
 P = _{r}G_{1,Tref}
 _{r}S_{1,Tref}(T  T_{ref})
/ _{r}V_{s}
 P = _{r}H_{1,Tref}
 T_{r}S_{1,Tref}
/ _{r}V_{s}
and we can calculate the temperature of a reaction at different pressures by
 T = T_{ref} + _{r}G_{1,Tref} + _{r}V_{s}P
/ _{r}S_{1,Tref}
 T = T_{ref} + _{r}H_{1,Tref}
+ _{r}V_{s}P
/ _{r}S_{1,Tref}
Let's do this for the albite = jadeite + quartz reaction at T = 400 K and T = 1000 K:
 P = (15,860  5147 * 400) / 1.7342 = 2.7 kbar
 P = (15,860  5147 * 1000) / 1.7342 = 20.6 kbar
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 Mg_{2}SiO_{4} to pure fayalite Fe_{2}SiO_{4}, and can have any composition in betweenit 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 (Mg_{0.25}Fe_{0.75})_{2}SiO_{4} or fo_{25}fa_{75}, etc. Mole fractions are denoted as X_{i}.
We need a way of splitting up the Gibbs free energy of a phase
among the various components of the phasehow 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
compositionis 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
 μ_{i} =
where n is the amount of substance. For olivine solid
solution composed of fayalite and forsterite components or endmembers,
we write
 dG = dn_{fayalite} + dn_{forsterite}
Volume of Mixing
Imagine that mole fractions of phase A and phase B with molar volumes V_{A} and V_{B}, are mixed together. We can describe the volume of the mixture as
 V = X_{A}V_{A}° + X_{B}V_{B}°
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 V_{A}^{o} and V_{B}^{o}, and the partial molar volumes are _{A} and _{B}._{}^{}
Partial Molar Volume
The partial molar volume is defined as
 _{i}
If you mix two compounds A and B together and find a volume
of mixing that is nonideal, 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:
 V_{mix} = X_{A}_{A} + X_{B}_{B}
or
 V_{mix} = X_{i}_{i}
The behavior of this function is such that when X_{A} is 1, V_{mix} = V_{A} and when X_{A} is 0, V_{mix} = V_{B} = V_{B}^{o}. Alternatively,

_{A} = (V_{mix}  X_{B}_{B}) / X_{A}
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
 S_{mix} = R (X_{i} ln X_{i})
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  H_{mix} = 0.5 * N z X_{A}X_{B} [2_{AB}  _{AA}  _{BB}]
where _{AB} is the interaction energy among AB atoms, _{AA} is the interaction energy among AA atoms, and _{BB} is the interaction energy among BB 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 negativeotherwise
mixing would not occur. The fact that μ_{A} < G°_{A} and μ_{B} < G°_{B} illustrates why compounds combine spontaneouslyeach 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
 _{A}  G°_{A} = _{A}  °_{A} = RT lnX_{A}
 _{B}  G°_{B} = _{B}  °_{B} = RT lnX_{B}
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 twocomponent mineral if X_{A} = 0.4, at T = 298 K,
 μ_{A}  μ°A = 8.314 * 298 ln 0.4 = 2271 J
 μ_{B}  μ°B = 8.314 * 298 ln 0.6 = 1266 J
The equation of the G_{mix} line is the sum of the chemical potentials of the endmembers:
 G_{mix} = RT (X_{A} ln X_{A} + X_{B} ln X_{B}) or
 G_{mix} = RT (X_{i} ln X_{i})
which looks like this for two components:
Actually, all this discussion has been predicated on the assumption that H_{mix} = 0. If this is not true, G_{mix} is not a simple function of composition, but has the general form:
Depending on the relative values of H_{mix} and TS_{mix},
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 H_{mix} > 0, i.e., if 2_{AB} > _{AA} + _{BB},
which makes sense because it means that the AB 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 ideallythat is, their chemical
potentials are never simple logarithmic functions of composition as
 μ_{A}  μ°_{A} = RT lnX_{A}
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 Mg_{2}SiO_{4}), and is the activity coefficient that describes the nonideal behavior. Thus we write
 μ_{A}  μ°_{A} = RT lna_{A}
For pure compounds a=1 because X=1. For ideal compounds =1.
As a specific example, the chemical potential of the almandine (Fe_{3}Al_{2}Si_{3}O_{12}) component of a garnet solid solution ((Fe, Mg, Ca, Mn)_{3}Al_{2}Si_{3}O_{12}) is
 μalm = μ°alm + RT lnX_{alm}
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  SiO_{2} + 2H_{2}O = H_{4}SiO_{4}
at equilibrium,
 μH_{4}SiO_{4} = μSiO_{2} + 2μH_{2}O
More generally, for
then
or, at equilibrium
 _{r}μ = 0 = cμC + dμD  aμA  bμB
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
 _{r}μ = 0 = c(μ_{C°} + RT lna_{C}) + d(μ_{D°} + RT lna_{D})  a(μ_{A°} + RT lna_{A})  b(μ_{B°} + RT lna_{B})
which looks nicer as
 _{r}μ = _{r}μ° + RT ln (a_{C}^{c} a_{D}^{d}/a_{A}^{a} a_{B}^{b})
To be completely general we write
 _{r}μ = _{r}μ° + RT ln a_{i} ( means to multiply all i terms)
This equation is invariably simplified to
 _{r}μ = _{r}μ° + RT lnQ
and Q is the activity product ratio. The activities in the Q term change as the reaction progresses toward equilibrium.
To be clear again, _{r}μ°
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. _{r}μ 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 _{r}μ = 0. We write this (with K instead of Q, to signify equilibrium) as
 0 = _{r}G° = RT ln K
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
 albite = jadeite + quartz
let's say that at a particular P and T,
 _{r}G° = 20.12 kJ/mol
Using
 _{r}G° = RT lnK
we calculate
This means that at equilibrium,
 (a_{jadeite} a_{quartz} / a_{NaAlSi3O8}) = e^{3.52}
Where a_{jadeite} is the activity of NaAlSi_{2}O_{6} in clinopyroxene and aalbite is the activity of NaAlSi_{3}O_{8} in plagioclase.
Graphical Portrayal of K
Another way to write the activity product ratio is
or
At equilibrium, where Δ_{r}G°= 0, ln K = 0 and K = 1. Let's see what K looks like for jadeite + quartz = albite at 800 K and 20 kbar:
 ln K =  (Δ_{r}H_{1,Tref}
 TΔ_{r}S_{Tref}
+ Δ_{r}V_{s}P)/ RT
 = (15,860  800 * 51.47 + 1.7342 * 20,000)/(8.314*800) = 1.4
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
 0 = _{i}M_{i}
where _{i} are the stoichiometric coefficients and M_{i} are the masses or the phase components. Analogously, we can explicitly write a similar balance among the chemical potentials:
 0 = _{i}μ_{i}
For each chemical potential we can write
Combining these two equations we find
 0 = _{i}μ°_{i} + _{i}RTln a_{i}
 0 = _{i}μ°_{i} + RTln (a_{i})
 0 = _{i}μ°_{i} + RTln a_{i}
 0 = _{i}μ°_{i} + RTln K
and eventually
 0 = _{r}G° + RT ln K
or
 _{r}G° = RT ln K
The equation
 K = a_{i}
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
 _{r}H°  T_{r}S° =  RT ln K
and for any P and T of interest:
 _{r}H_{1,Tref}
+ C_{P}dT  T_{r}S_{Tref}
+ dT
+ _{r}VP =  RT ln K
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
 ln K = (_{r}G° / RT) = [(_{r}H / RT)  (_{r}S / R)]
Which looks like
Activity Models (ActivityComposition Relations) for Crystalline Solutions
Garnets are solid solutions of
component  abbrev.  formula 
pyrope  prp  Mg_{3}Al_{2}Si_{3}O_{12} 
almandine  alm  Fe_{3}Al_{2}Si_{3}O_{12} 
grossular  grs  Ca_{3}Al_{2}Si_{3}O_{12} 
spessartine  sps  Mn_{3}Al_{2}Si_{3}O_{12} 
andradite  and  Ca_{3}Fe_{2}^{3+}Si_{3}O_{12} 
Mixing models derive from entropy considerations. In particular the relation
 S_{mix} = R X_{i} ln X_{i}
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
MgFeCaMn garnet with mixing on one site, which we can idealize as
(A,B,C,D)Al_{2}Si_{3}O_{12}, the activities are
 a_{prp} = _{Mg}^{3}X_{Mg}^{3}
 a_{alm} = _{Fe}^{3}X_{Fe}^{3}
 a_{grs} = _{Ca}^{3}X_{Ca}^{3}
 a_{sps} = _{Mn}^{3}X_{Mn}^{3}
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,
 a_{i} = X_{j}
where a, the activity of component i, is the mole fraction of element j raised to the power. For nonideal mixing, we include an activity coefficient
Mixing on a Several Sites
For minerals with two distinct sites and the general formula
 (A,B)(Y,Z)
there are four possible end member components AY, AZ, BY, and BZ.
The ideal activities of these components are
 a_{AY} = X_{A}X_{Y}
 a_{AZ} = X_{A}X_{Z}
 a_{BY} = X_{B}X_{Y}
 a_{BZ} = X_{B}X_{Z}
For nonideal garnet activities we write
 a_{prp} = X_{Mg}^{3} X_{Al}^{2}
or _{Mg}^{3} X_{Mg}^{3}
_{Al}^{2} X_{Al}^{2}
 a_{alm} = X_{Fe}^{3} X_{Al}^{2}
or _{Fe}^{3} X_{Fe}^{3}
_{Al}^{2} X_{Al}^{2}
 a_{grs} = X_{Ca}^{3} X_{Al}^{2}
or _{Ca}^{3} X_{Ca}^{3}
_{Al}^{2} X_{Al}^{2}
 a_{sps} = X_{Mn}^{3} X_{Al}^{2}
or _{Mn}^{3} X_{Mn}^{3}
_{Al}^{2} X_{Al}^{2}
 a_{and} = X_{Ca}^{3} X_{Fe3+}^{2}
or _{ Ca}^{3} X_{Mn}^{3}
_{Fe3+}^{2} X_{Fe3+}^{2}
where the X^{3} term describes mixing on the 8fold trivalent site and the X^{2} 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 modelsthat are based on completely
random mixing of elementswith models that consider local charge
balance on certain sites or the Alavoidance 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:

+ KMg_{3}AlSi_{3}O_{10}(OH)_{2} 
= Mg_{3}Al_{2}Si_{3}O_{12} 
+ KFe_{3}AlSi_{3}O_{10}(OH)_{2} 

+ phlogopite 
= pyrope 
+ annite 
We can reduce this reaction to a simple exchange vector:
 (FeMg)^{gar}_{+1} = (FeMg)^{bio}_{1}
Popular thermometers include garnetbiotite (GARB),
garnetclinopyroxene, garnethornblende, and
clinopyroxeneorthopyroxene; all of these are based on the exchange of
Fe and Mg, and are excellent thermometers because _{r}V 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  K = (a_{prp}a_{ann})/(a_{alm}a_{phl})
thus
 _{r}G = RT ln (a_{prp}a_{ann})/(a_{alm}a_{phl})
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
 a_{alm} = X_{alm}^{3} = [Fe/(Fe + Mg + Ca + Mn)]^{3}
 a_{prp} = X_{prp}^{3} = [Mg/(Fe + Mg)]^{3}
 a_{ann} = X_{ann}^{3} = [Fe/(Fe + Mg)]^{3}
 a_{phl} = X_{phl}^{3} = [Mg/(Fe + Mg)]^{3}
Thus the equilibrium constant is
 K = (X_{Mg}^{gar} X_{Fe}^{bio})/(X_{Fe}^{gar} X_{Mg}^{bio})
When discussing element partitioning it is common to define a distribution coefficient K_{D},
which is just the equilibrium constant without the exponent (this just
describes the partitioning of elements and not the partitioning of
chemical potential):
 K_{D} = (X_{Mg}^{gar} X_{Fe}^{bio})/(X_{Fe}^{gar} X_{Mg}^{bio}) = (Mg/Fe)^{gar} /(Mg/Fe)^{bio} = K^{1/3}
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
 ln K_{D} = 2109 / T + 0.782
this immediately reminds you of
 ln K =  (_{r}G° / RT) = (_{r}H / RT)  (P_{r}V / RT) + (_{r}S / R)
and you realize that for this reaction
 _{r}S = 3*0.782*R = 19.51 J/mol K
(the three comes from the site occupancy coefficient; i.e., K = K_{D}^{3}) and
 (_{r}H / R)  (P_{r}V / R) = 2109
or
 _{r}H = 3*2109*R 2070*_{r}V
Molar volume measurements show that for this exchange reaction _{r}V = 0.238 J/bar, thus
 _{r}H = 52.11 kJ/mol
The full equation is then
 52,110  19.51*T(K) + 0.238*P(bar) + 3RT ln K_{D} = 0
To plot the K_{D} lines in PT space
NetTransfer Reactions
Nettransfer reactions are those that cause phases to appear or
disappear. Geobarometers are often based on nettransfer reactions
because _{r}V is large and relatively insensitive to temperature. A popular one is GASP:
3CaAl_{2}Si_{2}O_{8} 
= Ca3Al_{2}Si_{3}O_{12} 
+ 2Al_{2}SiO_{5} 
+ SiO_{2}

anorthite 
= grossular 
+ kyanite 
+ quartz 
which describes the highpressure breakdown of anorthite.
For this reaction
 _{r}G
= RT ln [(a_{qtz}a_{ky}^{2}a_{grs}) / a_{an}^{3}]
= RT ln a_{grs} / a_{an}^{3}
(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
 P(bar) = 22.80 T(K)  7317
for _{r}V = 6.608 J/bar. Again, if we use
 ln K = (_{r}H / RT)  (P_{r}V / RT) + (_{r}S / R)
and set ln K = 0 to calculate values at equilibrium, we can rewrite the above as
 (P_{r}V / R) = (_{r}H / R) + (T_{r}S / R)
or
 P = T_{r}S / _{r}V  _{r}H / _{r}V
if T_{r}S / _{r}V = 22.8 then _{r}S = 150.66 J/mol K
if _{r}H / _{r}V = 7317 then _{r}H = 48.357 kJ/mol
So, we can write the whole shmear as
 0 = 48,357 + 150.66 T(K) 6.608 P (bar) + RT ln K
Contours of ln K on a PT diagram for GASP look like this:
Kinetics
Thermodynamics places no constraints on the rate or mechanism of
reactionthat 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 _{r}H and an activation enthalpy barrier H* (usually called an activation energy Q*). Of course, unlike _{r}G, _{r}H can be positive or negative:
It is not easy to generalize about the activation entropy S*, however, in general, reactions with positive entropy change _{r}S
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
 f = exp ( H*/RT)
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):

rate p exp(Q*/RT)
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 atomscale 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
_{n}G = _{V}G + G_{interfacial} + G_{strain} = (4/3)r^{3}_{V}G + 4 r^{2} + (4/3)r^{3}
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 reorder the terms in the above equation
to calculate the critical size of the nucleus r_{c}
If, at the critical radius, _{n}G = G*, the activation energy for nucleation is
 G* = 16 ^{3} / 3(_{V}G + )^{2}
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 interfacecontrolled growth rate is:
 = (kT/h) exp( Q*/RT)[( _{r}G/RT)]
where is an "interface jump distance" and the rest you will recognize from previous equations. The [( _{r}G/RT)] term indicates that the rate of growth depends on the Gibbs free energy change of reaction. As _{r}G 0, this term also approaches zero. As _{r}G , this term approaches 1.
Transformation: Nucleation + Growth
A complete transformation involves nucleation and growththe
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 timeintegrated result
of ongoing nucleation and growth, which can vary in time and space:
Just as an example, the general rate equation for
interfacecontrolled (i.e., not diffusioncontrolled) growth of nuclei
formed on grain boundaries is
 = 1  exp{(6.7/d)[1exp([^{2}(t  )^{2}  y^{2}] d)] dy}
In their simplest form, these equations have the form
 = 1  exp(kt^{n})
Flip this around to
 exp(kt^{n}) = 1 
linearize by means of logarithms
 kt^{n} = ln (1  )
change signs and cleverly insert ln 1
 kt^{n} = ln 1  ln (1  )
rewrite the two logarithms as one, recalling that ln (a/b) = ln a  ln b
 kt^{n} = ln [1 / (1  )]
and linearize once more with logarithms to get the final glorious equation
 n ln t + ln k = ln [ ln (1  )]
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 Cabearing phases. The
thermodynamic properties (H, S, V, C_{P}) 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 + H_{2}O. At higher temperature wairakite + H_{2}O decompose to anorthite + quartz + H_{2}O.
Kinetic experiments conducted on the reaction laumontite _ wairakite + H_{2}O, using crystals of laumontite suspended in H_{2}O, reveal the transformation vs. time data in the figure above.
One could use the equation
to interpret the two subfigures 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 E10 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 + H_{2}O reaction.