1. Lecture 6 Crystal Chemistry
Part 5:
Rock Suites
Applications of Thermodynamics
MARBLE DEMO
Rock Cart, Kyanite. Andalusite, Sillimanite
Serpentinite, Greenschist, Amphibolite
Mica Schist with Staurolite, Microcline with Perthitic Texture

2. Basaltic Magma Cooling - Bowen’s Reaction Series
Molten- VERY Hot
No solids
First mineral to crystallize out
Molten- Not so hot
100% Solid

3. Fine crystals
Need a microscope
Fractionation
minerals
Low silica, HOT, fluid
Intermediate
High silica, warm, viscous
Course crystals
Easily seen

4. Crystals can react with the melt if they touch it
If the first formed crystals of Calcium-rich (Ca) Plagioclase touch the melt they will react with it,
and will become more sodium-rich on their outer rims
Zoned feldspar (plagioclase) showing change in
composition with time in magma chamber
(calcium-rich in core to sodium-rich at rim)

5. FRACTIONATION: if early crystals are removed, the melt becomes richer in Silica
Fe, Mg, Ca
Some Si
Left with
K and Al
Most of Si
You can start with a
Mafic (silica-poor) magma
and end up with some
Felsic (silica-rich)
Granites.
Marble Demo
A melt will crystallize its mafic components first, and the remaining melt may be granitic

6. Assimilation and magmatic differentiation
Show Samples

7. Metamorphism Most metamor-phism occurs between about 200 and 800o C
Sedimentary
0 km
rock
Metamorphic
rock
Igneous
Sediment
rock
10 km
Most metamor-phism occurs between about 200 and 800o C
~200ºC
Sedimentary
rock
The rocks don’t melt
Metamorphism
Increasing depth
and temperature
50 km
Melting
~800ºC

8. Contact metamorphism
Produced by contact with heat source

9. Hydrothermal Metamorphism
Basaltic Magma

10. MORs,Black Smokers, Cyprus & Bronze Age (Cu), Rome’s conquest of Britain (Sn), Sterling Zinc (and Manganese)
Copper plus Tin makes Bronze
Iron plus Carbon makes steel.
Circulation of hot water in cracks at mid-ocean ridge dissolves metals (Copper, Tin, Iron, Zinc, Lead, Barium) which are re-precipitated as sulfide ores. Hydrothermal waters are capable of metamorphism.

11. Dynamothermal Metamorphism, Before collision
Sediments are “unconsolidated”. They will fold if pushed.

12. Dynamothermal Metamorphism, in convergent margins (subduction, collision)

13. Index Minerals in metamorphic rocks
Note Temperature gradient
Stable temperature ranges depend on Pressures
Index Minerals in metamorphic rocks
Note Quartz and Feldspar are not index minerals: Why?

14. Thermometers and Pressure Gauges
Polymorphs of Al2SiO5  Al2SiO5
solid - solid
Sillimanite
Kyanite
Andalusite
Kyanite in Andalucite

15. Role of Volatiles - (H2O & CO2)
Catalyzes reactions
Mobility during metamorphism leads to non-isochemical reactions
Dehydration and decarbonation during prograde reactions
Lack of volatiles slows retrograde reactions
Retrograde
Prograde/
Dehydration
Example: Dehydration:
Muscovite + Qtz = K-spar + Sill. + H2O

16. Metamorphic Facies near Subduction Zones and Arcs
More Basaltic Magma
formed by dehydration of ocean lithosphere and hydration of mantle
We can note mineral facies in Metamorphic Rocks and determine where they formed.

17. Greenschist Hand Sample
Greenschist Thin Section
Chl-Ep

18. Blueschist glaucophane Amphibolite hornblende + Plag.
a sodic amphibole
Amphibolite hornblende + Plag.
Pleochroism, cleavage in hornblende

19. Mineral Stability/Equilibrium
Phase Stability The stability of a phase is determined by the Gibbs free energy, G.
A Mineral of constant composition is considered a solid phase
Mineral stability is commonly portrayed on a Phase Diagram

20. Stability and Gibbs Free Energy G
G(p,T) = E + PV − TS
For a reaction or change of phase:
System in equilibrium if no unbalanced forces DG = 0.
If DG<0, a phase can spontaneously transform to another phase; e.g., solid to liquid
If DG > 0, a phase transformation will not occur. Multiple phases can occur simultaneously.
E is the internal energy
P the pressure, V the volume, T the Temperature, and S the Entropy (disorder)

21. E + PV is Enthalpy H
A certain amount of energy goes to an increase in entropy of a system and a certain amount goes to a heat exchange for a reaction.
G = H –TS or DG0R = DH0R – TDS0R
Gibbs Free Energy (G) is measured in KJ/mol or Kcal/mol

22. G is a measure of driving force
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.

23. Petrology Field Trip to Bemco Mining District

24. Supergene Enrichment, Bemco Mine

25. Oxidation of Ferrous Iron, Fe+2
Groundwater: iron in two oxidation states Reduced soluble ferrous iron (Fe+2) Oxidized insoluble ferric iron (Fe+3).
Modern atmosphere 21% oxygen, so
most in shallow soils ferric state, (Fe+3).
Initially Ferric hydroxide (Fe(OH)3)
With time, mineralized. Decreasing solubility
amorphous hydrous ferric oxide (Fe2O3•xH2O),
Hematite (Fe2O3), and
Goethite FeO(OH).

26. Gibbs Free Energy Example
Oxidation of ferrous ion to ferric ion
DG0R = DH0R – TDS0R
H2O(l)= kcal/mol = cal/mol
(You look these up in these tables)
Fe2+ + ¼ O2 + H+  Fe3+ + ½ H2O
=[-4120+(-63320*0.5)]-[ (3954*0.25)]
=[-67440]-[ ]= cal/mol
Negative, so forward (left to right) reaction will proceed

27. Al2SiO5 polymorphs
differ in coordination
of the Al+3
In Andalusite, one
Aluminum is in 5-fold
coordination, very
unusual.
Andalusite

28. 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.

29. Components and Phases
Components are the chemical entities necessary to define all the potential phases in a system of interest
Here one Component, Al2SiO5
Phases: number of mineral species plus fluids
Here three Phases: Ky, Sill, Andalucite
And

30. Degrees of Freedom f by Examples
p + f = c + 2
If T and P can change without changing the mineral assemblage, the system has two degrees of freedom f = 2
If neither T or P can change without changing the mineral assemblage, the system has zero degrees of freedom f = 0
If T and P must change together to maintain the same mineral, the system has one degree of freedom f = 1
The number of thins you can change before everything is determined
Sill.
Ky.
On a phase diagram f=0 corresponds to a point,
f = 1 to a reaction line,
f = 2 to a 1 phase area.
And.

31. The Phase Rule
The number of minerals (phases) that may stably coexist is limited by the number of chemical components
p + f = c + 2
where P is the number of mineral phases, c the number of chemical components, and f is the number of degrees of freedom.

32. Basic Thermodynamics The theoretical basis of phase equilibrium
Three Laws of Thermodynamics
1. First Law: Change in Internal Energy (E)= dE = dQ – dW
Q – heat energy
W – work = F * distance (notice distance is a length)
Pressure has units Force/Area = (F/dist2)
so 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

33. 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 heat-capacity measurements by assuming that S0 (at 0o K) is zero” Wood and Fraser (1978) p38 ss

34. Gibbs Free Energy Differentiating dG = dE +PdV +VdP -TdS - SdT
Define G – the energy of a system in excess of its internal energy E. This is the energy necessary for a reaction to proceed
G = E + PV – TS
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 (3) dG = VdP –SdT
If T = constant dT = 0, then dG = VdP, if V decreases, P can increase without increasing G
(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, if T increases then S can increase without increasing G
(3) at constant P (dG/dT)P = -S (disordered phases (high S) are favored at high T)

35. Enthalpy Earlier we saw G(T,p) = E + pV − TS
But the Enthalpy H(S,p) = E + pV
So DG = DH –TDS
DH can be measured in the laboratory with a calorimeter.
DS can also be measured with heat capacity measurements.
Values are tabulated in books.
Enthalpy is a measure of the total energy of a thermodynamic system. It includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure.

36. Clapeyron Equation From Eqn.3, if dG =0, dP/dT = ΔS / ΔV (eqn.4)
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 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

37. Constructing Phase Diagrams
One Component: H2O
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
Pressure-Temperature Phase Diagram

38. 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

39. The KSA Phase diagram allows us to assign PT conditions
to various Plate Tectonic settings
Of Nesse