lecture 6 crystal chemistry part 5: rock suites applications of thermodynamics rock cart,...

Lecture 6 Crystal Chemistry

Part 5:

Rock Suites Applications of Thermodynamics

Rock Cart, Kyanite. Andalusite, SillimaniteSerpentinite, Greenschist, AmphiboliteMica Schist with Staurolite, Microcline with Perthitic Texture

MARBLE DEMO

Basaltic Magma Cooling - Bowen’s Reaction Series

Molten- VERY HotNo solids

Molten- Not so hot

100% Solid

First mineral to crystallize out

Fine crystalsNeed a microscope

Course crystalsEasily seen

Low silica, HOT, fluid High silica, warm, viscousIntermediateFractionationminerals

Zoned feldspar (plagioclase) showing change in composition with time in magma chamber (calcium-rich in core to sodium-rich at rim)

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

Crystals can react with the melt if they touch it

FRACTIONATION: if early crystals are removed, the melt becomes richer in Silica

A melt will crystallize its mafic components first, and the remaining melt may be granitic

Remove Fe, Mg, CaSome Si

Left withK and AlMost of Si

You can start with aMafic (silica-poor) magmaand end up with some Felsic (silica-rich)Granites. Marble Demo

Assimilation and magmatic differentiation

Show Samples

0 kmSedimentaryrock

Metamorphicrock

Igneousrock

50 km

10 km

~200ºC

~800ºC

Incr

easi

ng d

epth

and

tem

pera

ture

Melting

Metamorphism

Sedimentaryrock

Sediment

Most metamor-phism occurs between about 200 and 800o C

The rocks don’t melt

Metamorphism

Contact metamorphismProduced 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)

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.

http://collections.ic.gc.ca/geoscience/images/detail/F92S0220.jpg

Copper plus Tin makes Bronze

Iron plus Carbon makes steel.

Dynamothermal Metamorphism, Before collision

Sediments are “unconsolidated”. They will fold if pushed.

Dynamothermal Metamorphism, in convergent margins (subduction, collision)

Index Minerals in metamorphic rocks

Note Quartz and Feldspar are not index minerals: Why?

Note Temperature gradient

Stable temperature ranges depend on Pressures

Thermometers and Pressure GaugesPolymorphs of Al2SiO5 Al2SiO5

Kyanite

Sillimanite

AndalusiteKyanite in Andalucite

solid - solid

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

Prograde/

Prograde/

Dehydration

Dehydration

Retrograde

Retrograde

Example: Dehydration: Muscovite + Qtz = K-spar + Sill. + H2O

Metamorphic Facies near Subduction Zones and Arcs

We can note mineral facies in Metamorphic Rocks and determine where they formed.

More Basaltic Magmaformed by dehydration of ocean lithosphere and hydration of mantle

Greenschist Hand Sample

Greenschist Thin SectionChl-Ep

Blueschist glaucophanea 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

G = 0. If G<0, a phase can spontaneously transform

to another phase; e.g., solid to liquid If G > 0, a phase transformation will not

occur. Multiple phases can occur simultaneously.

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 G0R = H0

R – TS0R

Gibbs Free Energy (G) is measured in KJ/mol or Kcal/mol

)reactants()( 000i

iii

iiR GnproductsGnG

E + PV is Enthalpy H

G is a measure of driving force GR = HR – TSR

Again: For a reaction A B That is, for reactants products When GR is negative the forward reaction

has excess energy and will occur spontaneously A B

When GR is positive there is not enough energy in the forward direction, and the back reaction will occur B A

When GR is zero reaction is at equilibrium, both reactions occur equally.

Petrology Field Trip to Bemco Mining District

Supergene Enrichment, Bemco MineSupergene 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 (Fe2O3•xH2O),

Hematite (Fe2O3), and Goethite FeO(OH).

Gibbs Free Energy Example

G0R = H0

R – TS0R

H2O(l)=-63.32 kcal/mol = -63320 cal/mol (You look these up in these tables)

Fe2+ + ¼ O2 + H+ Fe3+ + ½ H2O=[-4120+(-63320*0.5)]-[-21870+(3954*0.25)]

=[-67440]-[-20887.5]=-46557.5 cal/molNegative, so forward (left to right) reaction will proceed

)reactants()( 000i

iii

iiR GnproductsGnG

Oxidation of ferrous ion to ferric ion

http://www.uwgb.edu/DutchS/PETROLGY/coordinationx.htm

Andalusite

Al2SiO5 polymorphsdiffer in coordinationof the Al+3In Andalusite, oneAluminum is in 5-foldcoordination, very unusual.

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 fluidsHere three Phases: Ky, Sill, Andalucite

And

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 freedom f = 1

Ky.

And.

Sill.

On a phase diagram f=0 corresponds to a point,

f = 1 to a reaction line,

f = 2 to a 1 phase area.

p + f = c + 2

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.

Basic Thermodynamics

The theoretical basis of phase equilibrium Three Laws of Thermodynamics

1. First Law: Change in Internal Energy (E)= dE = dQ – dWQ – 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 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

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 - SdTdE = dQ - PdV = TdS – PdV sodG = 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)

Enthalpy

Earlier we saw G(T,p) = E + pV − TS

• But the Enthalpy H(S,p) = E + pV

• So G = H –TS

H can be measured in the laboratory with a calorimeter.

S can also be measured with heat capacity measurements.

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 (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

Constructing Phase DiagramsOne 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 (G = 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, G = 0 in the reaction Andalucite Sillimanite

Nesse fig 5.3

Of Nesse

The KSA Phase diagram allows us to assign PT conditions to various Plate Tectonic settings