Thermodynamics

A system: Some portion of the
universe that you wish to study
•
The surroundings: The adjacent part
of the universe outside the system
Changes in a system are associated with the
transfer of energy
Natural systems tend toward states of minimum energy
Basic Thermodynamics
 The theoretical basis of phase equilibrium
 Three Laws of Thermodynamics
1. First Law:
Change in Energy
dE = dQ – dW
Q – heat energy
W – work = F * distance
(notice distance is a length)
Pressure has units Force/Area = (F/dist2)
so Work = F * dist = P * area * dist = P * V
At constant pressure dW = PdV
So if Pressure is constant:
(1) dE = dQ – PdV where dV is thermal expansion or contraction
Does not depend on path
The symbol d means “the small change in”
Energy E
 Kinetic + potential energy + Pressure energy + ….
 Kinetic energy due particles motion (vibration (heat),
rotation, translation)
 Potential energy: position, rest mass energy, charge of
atoms, energy of bonds.
 Changed by heating or by doing work on system
 Pressure energy = Pressure x unit Volume
 First law: dE = dQ – dW , change in energy is equal to
the total heat added/removed and work done by/on
the surroundings. If the system is isolated from its
surroundings, its energy cannot change.
Second and Third Laws of
Thermodynamics
2. All substances move to the greatest state of
disorder (highest Entropy ”S”) for a particular T
and P.
(2) dQ/T = dS
“The state of greatest order [lowest S] is at the lowest temperature. With increasing
temperature, disorder becomes more prevalent.”
Minerals with simple atomic structure and simple chemistry have lower entropy.
3. At absolute zero (0ºK), Entropy is zero
Does this suggest how to measure S?
Can we measure changes in S by keeping track of temperature as we add heat to a system?
“The entropy of a pure crystalline substance can therefore be obtained directly from heatcapacity measurements by assuming that S0 (at 0o K) is zero” Wood and Fraser
(1978) p38
 Stability of a phase (or mineral) is partly related to its
internal energy (here “E”), which strives to be as low as
possible under the external conditions.
 Metastability exists in a phase when its energy is higher
than P-T conditions indicate it should be. (1)
 Activation Energy is the energy necessary to push a phase
from its metastable state to its stable state. (2 minus 1)
 Equilibrium exists when the phase is at its lowest energy
level for the current P-T conditions. (3) (Two minerals that
are reactive with one another, may be found to be in
equilibrium at particular P-T conditions which on phase
diagrams are recognized as phase boundaries)
Note: most metamorphic and igneous minerals at the earth’s
surface are metastable.
Example: Exsolution of K-spar and Albite at low temperatures;
they were in Solid Solution at higher temperatures.
Moles (mol)
 A number ~6.022 x 1023 of anything
 A mole of a substance is equal to its
molecular mass in grams.
 Thus Aluminum has a molecular mass
of ~26.98, so a mole of aluminum
metal has a mass of 26.98 grams.
Enthalpy H
 Enthalpy (symbol H) is the sum of internal energy plus
work* related to changes in pressure and volume.
 H = E + PV
 The change in Enthalpy DH = Cp DT
 Cp is the Heat Capacity at constant pressure
P. Cp the amount of heat energy required to raise 1 mole of a
substance by 1oC. SI units Joules
 T is the Temperature in Kelvin
*The energy required to "make room for it" by displacing its surroundings.
Entropy S
 Entropy, S, is a measure of the
disorder of a system.
 Example: A liquid (maybe a melt) is
more disordered than a solid.
 SI units Joules/K
Gibbs Free Energy
 Define G
G = E + PV – TS
The internal energy E might be thought of as the energy
required to create a system in the absence of
changes in temperature or volume. An additional
amount of work PV must be done if the system is
created from a very small volume in order to "create
room" for the system. A system at constant
temperature T will contain an amount of energy TS
Gibbs is the energy in excess of that energy TS
Stability and Gibbs Free Energy G
G = E + PV – TS
and we just defined enthalpy
 G(p,T) = H − TS
H = E + PV
so
thermal energy minus Temp x disorder
 Stable forms of a chemical system have the minimum
possible Gibbs Free Energy
 H is the Enthalpy, a measure of Thermal Energy
 S is the Entropy, a measure of randomness, of
probability
Changes
 Thermodynamics treats changes
 Regardless of path G = E + PV – TS
 We should rewrite the equation for
Gibbs Free Energy in terms of
changes, D G
 DG = E + PDV – TDS
 DG =
DH
for P, T constant
– TDS
D = pronounced “delta” means “the
change in”
The change in Gibbs free energy, ΔG, in a reaction is
a very useful parameter. It can be thought of as the
maximum amount of work obtainable from a reaction.
DH can be measured in the
laboratory with a calorimeter.
DS can also be measured with
heat capacity measurements.
Values are tabulated in books.
Gibbs for a chemical reaction
Hess’s Law applied to Gibbs for a reaction
298.15K, 0.1 MPa
DG =  ni G ( products)   ni G (reactants )
0
R
0
i
i
0
i
i
 Suppose 3A + 2B = 2C +1D
reactants = products
 DG = 2GC +1GD -3GA – 2GB
 Gibbs Free Energy (G) is measured in
KJ/mol or Kcal/mol
 One small calorie cal ~ 4.2 Joules J
Molar Gibbs Free Energy of Formation
 G extensive, depends on amount
 Make intensive by using molar
free energy, Joules/mole
 Measure H with calorimetry
 and S (measure DQ/T from T = 0 to 298K and
define S0K = 0). More precision at low T’s
 Conditions 298K and 0.1 MPa = 1 bar are called “the
reference state”.
 For the reaction Si + O2 = SiO2
 Gf = Hf – TSf = -856.3 kJ/mol
298K and 0.1 MPa
Gibbs free energy of formation
 The Gibbs free energy of formation is Gf° = Hf° - TfS°
 For example, to calculate the Gibbs free energy of
formation of anhydrite, CaSO4 , we can use
Hf°CaSO4 = -1434.11 kJ/mole
 S°CaSO4 = 106.7 J mol-1 K-1
 S°Ca = 41.42 J mol-1 K-1
Hfo Ca = 0
 S°S = 31.80 J mol-1 K-1
Hfo S= small
 S°O2 = 205.138 J mol-1 K-1
Hfo O2= 0
 and we calculate the entropy of formation of anhydrite
Sf° = S°CaSO4 - S°Ca - S°S - 2 * S°O2 = -376.796 J mol-1 K-1
 and then use Gf° = Hf° - TfSf° = -1434,110 - 298.15 *
-376.796 = -1321.77 kJ/mol
G for reactions
 DGR = DHR – TDSR
 Again: For a reaction A  B
 That is, for reactants  products
 When DGR is negative the forward
reaction has excess energy and will
occur spontaneously A  B
 When DGR is positive  there is not
enough energy in the forward direction,
and the back reaction will occur B  A
 When DGR is zero  reaction is at
equilibrium, both reactions occur
equally.
Ex: Oxidation of Ferrous Iron, Fe+2
 Groundwater: iron in two oxidation states
Reduced ferrous iron (Fe+2) Oxidized ferric
iron (Fe+3).
 Modern atmosphere 21% oxygen, so
 most O2 in shallow soils ferric state, (Fe+3).
 With time, mineralized. Fe+2 => Fe+3
Fe+3 is less soluble, forms oxides, hydroxides
 Initially Ferric Hydroxide (Fe(OH)3)
 amorphous hydrous ferric oxide
(Fe2O3•xH2O),
 Hematite (Fe2O3), and Goethite FeO(OH).
Gibbs Calc 1: Fe+2 => Fe+3
Oxidation of ferrous ion to ferric ion
DG0R = DH0R – TDS0R
DGR0 =  ni Gi0 ( products)   ni Gi0 (reactants )
i
i
H2O(l)=-63.32 kcal/mol = -63320 cal/mol
You look these up in tables

Reactants
Products
Fe2+ + ¼ O2 + H+  Fe3+ + ½ H2O
=[-4120+(-63320*0.5)]-[-21870+(3954*0.25)]
=[-67440]-[-20887.5]=-46557.5
cal/mol
Negative, so forward (left to right) reaction will proceed
Gibbs details
 dE = dQ – PdV
 dQ/T = dS,
d(uv) = udv + vdu
1st Law
so dQ = TdS 2nd Law
G = E + PV – TS
Differentiating dG = dE +PdV +VdP -TdS - SdT
dE = dQ - PdV = TdS – PdV so
dG = TdS – PdV + PdV + VdP -TdS – SdT
dG = VdP –SdT
All other terms cancel.
Gibbs Calc 2: SiO2 at elevated P and T
 Suppose we want G for Quartz at 500oC
and 500MPa
 Doesn’t depend on path, can do Pressure
and Temperature separately
 dG = VdP –SdT we just saw.
 Or approximately
 DG = VDP –SDT
 Gp2 – Gp1 = V(P2 – P1) if T, V const
T constant means DT = 0, V constant is a good approximation for solids
Variation in Gibbs of a phase with T and P
Doesn’t depend on path, do Pressure and Temperature separately
PRESSURE calc, assume volume const for solids
 Gp2 – Gp1 = V(P2 – P1) if T, V const
For P2 = 500 MPa, P1 = 0.1 MPa,
V = 22.69 x 10-6 m3,
Gp1 = - 856,300 J (lo-qtz looked up in a table)
Gp2 = Gp1 + V(P2 – P1)
= -856,300J + 22.69 x 10-6m3(500x106 – 0.1x106
= - 844,957 Joules ~ -845 KJ
Data from Robie and Waldbaum (1970), not Table 5.1 in the text
SiO2 at Elevated P and T
Pa)
Variations Gibbs of a phase with T and P
Doesn’t depend on path, do Pressure and Temperature separately
Assume S constant over change temperature
error is small
 DG = VDP –SDT
 GT2 – GT1 = -S(T2 – T1) if P = 500 MPa const
G500MPa, 773K = GT1 –S0.1MPa(773K – 298K)
= -844957 J – 41.46 J/K (773 -298K)
= -864650.5 Joules
G SiO2 at 500MPa and 773K ~ -864.7 KJ
P constant means DP = 0, S constant is a rough approximation provided DT not too big
Gibbs Free Energy

We defined G – the energy of a system in excess of its internal energy E.
This is the energy necessary for a reaction to proceed. H = E + PV
G = E + PV – TS = H – TS
Now let’s derive some useful relationships.
Differentiating dG = dE +PdV +VdP -TdS - SdT
dE = dQ - PdV = TdS – PdV so
dG = TdS – PdV + PdV + VdP -TdS – SdT = VdP –SdT
This is equation
dQ =TdS is the 2nd law
(3) dG = VdP –SdT
Reactions go to the phase with lower G
If T = constant dT = 0, then dG = VdP
(2) at constant T (dG/dP)T = V
(dense (low V) phases are favored at high P)
If P = constant dP = 0, then dG = -SdT
(3) at constant P (dG/dT)P = -S
(disordered phases (high S) are favored at high T)
We saw (equation 3) that
dG = VdP –SdT
The Clapeyron Equation
• defines the state of Equilibrium between reactants
and products in terms of S and V
From Eqn.3, if dG =0,
dP/dT = ΔS / ΔV
(eqn.4)
The slope of the equilibrium curve on a P T
diagram will be positive if S and V both
decrease or increase with increased T and P
The function G can be represented graphically on P T diagrams
Constructing Phase Diagrams
One Component: H2O
Pressure-Temperature Phase Diagram
The Gibbs and Clapeyron
Equations allow us to
estimate phase diagrams
with extrapolations from
laboratory measurements.
The lines show where
equilibrium conditions
(DG = 0) occur. Clapeyron
tells us the slope
Dense (low V) phases are favored at high P
Disordered phases (high S) are favored at high T
dP/dT = ΔS / ΔV
Clapeyron
1.Does the liquid or solid
have the larger volume/unit
mass? Usually liquid. (except H2O)
Low V usually solid.
2. High pressure favors
low volume, so which
phase should be stable
at high P? Solid
5. Both DV and DS increase to right. We
can thus predict that the slope of solidliquid equilibrium should be positive
and that increased pressure raises the
melting point.
3.Does liquid or solid have
a higher entropy? liquid
4. High temperature favors
randomness, so which
phase should be stable at
higher T? Liquid is more
random, expect at high T.
Does the liquid or solid
have the lowest G at
point A? Solid, pt. A is
in the solid field.
What about at point B?
Figure 5-2. Schematic P-T phase diagram of a melting reaction.
Winter (2001) An Introduction to Igneous and Metamorphic
Petrology. Prentice Hall.
Liquid, same argument
The phase assemblage with the lowest G under a specific set of
conditions is the most stable
The Payoff
• Our experiments and
calculations allow us
to construct the 3-D
plot in (a), and to
project the mineral
with the lowest free
energy at each PT
onto the graph in (b).
Notice the points a,b and c at right.
At c, DG = 0 in the reaction
Andalucite  Sillimanite
Nesse fig 5.3
The KSA Phase diagram allows us to assign PT conditions
to various Plate Tectonic settings
Of Nesse