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Lecture 6
Crystal Chemistry
Part 5:
 Rock Suites
 Applications of Thermodynamics
Rock Cart, Kyanite. Andalusite, Sillimanite
Serpentinite, Greenschist, Amphibolite
Mica Schist with Staurolite, Microcline with Perthitic Texture
Basaltic Magma Cooling - Bowen’s Reaction Series
Molten- VERY Hot
No solids
First mineral to crystallize out
Molten- Not so hot
100% Solid
Fine crystals
Need a microscope
Low silica, HOT, fluid
Course crystals
Easily seen
High silica, warm, viscous
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)
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)
Marble Demo
A melt will crystallize its mafic components first, and the remaining melt may be granitic
Assimilation and magmatic
Show Samples
0 km
Increasing depth
and temperature
10 km
50 km
The rocks don’t melt
200 and
800o C
Contact metamorphism
Produced by contact with heat source
Hydrothermal Metamorphism
Basaltic Magma
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.
Dynamothermal Metamorphism,
Before collision
Sediments are “unconsolidated”. They will fold if pushed.
Dynamothermal Metamorphism,
convergent margins (subduction, collision)
Stable temperature ranges depend on Pressures
Index Minerals in metamorphic rocks
Note Quartz and Feldspar are not index minerals: Why?
Thermometers and Pressure Gauges
Polymorphs of Al2SiO5  Al2SiO5
solid - solid
Kyanite in Andalucite
Role of Volatiles - (H2O & CO2)
 Catalyzes reactions
 Mobility during
metamorphism leads to
 Dehydration and
decarbonation during
prograde reactions
 Lack of volatiles slows
retrograde reactions
Example: Dehydration:
Muscovite + Qtz = K-spar + Sill. + H2O
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.
Greenschist Hand Sample
Greenschist Thin Section
Blueschist glaucophane
a sodic amphibole
Amphibolite hornblende + Plag.
Pleochroism, cleavage in hornblende
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
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
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
 Gibbs Free Energy (G) is measured in
KJ/mol or Kcal/mol
DG   ni G ( products)   ni G (reactants )
G is a measure of driving force
 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
Petrology Field Trip to Bemco Mining District
Supergene Enrichment, Bemco Mine
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
 Hematite (Fe2O3), and
 Goethite FeO(OH).
Gibbs Free Energy Example
Oxidation of ferrous ion to ferric ion
DGR0   ni Gi0 ( products)   ni Gi0 (reactants )
H2O(l)=-63.32 kcal/mol = -63320 cal/mol
(You look these up in these tables)
 Fe2+ + ¼ O2 + H+  Fe3+ + ½ H2O
=[-67440]-[-20887.5]=-46557.5 cal/mol
Negative, so forward (left to right) reaction will proceed
Al2SiO5 polymorphs
differ in coordination
of the Al+3
In Andalusite, one
Aluminum is in 5-fold
coordination, very
 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.
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
Here three Phases: Ky,
Sill, Andalucite
Degrees of Freedom f by Examples
 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
On a phase diagram f=0
corresponds to a point,
f = 1 to a reaction line,
f = 2 to a 1 phase area.
The Phase Rule
 The number of minerals (phases) that
may stably coexist is limited by the
number of chemical components
where P is the number of mineral phases, c
the number of chemical components, and f
is the number of degrees of freedom.
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
Second and Third Laws of
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
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
Gibbs Free Energy
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)
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
DS can also be measured with heat capacity
Values are tabulated in books.
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
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
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
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