Ch 27 more Gibbs Free Ch 27 more Gibbs Free EnergyEnergy

Gibbs free energy is a measure of Gibbs free energy is a measure of chemicalchemical energyenergy

Gibbs free energy for a phase:Gibbs free energy for a phase:G = E + PV – TS => G = H - TSG = E + PV – TS => G = H - TS

Where:Where:G = Gibbs Free EnergyG = Gibbs Free EnergyE = Internal EnergyE = Internal EnergyH = Enthalpy (heat content) = E + PVH = Enthalpy (heat content) = E + PVT = Temperature in degrees Kelvin T = Temperature in degrees Kelvin ooKKP = Pressure, V = VolumeP = Pressure, V = VolumeS = Entropy (randomness, disorder)S = Entropy (randomness, disorder)

ChangesChanges Thermodynamics treats changesThermodynamics treats changes Regardless of path Regardless of path G = E + PV – TSG = E + PV – TS We should rewrite the equation for We should rewrite the equation for

Gibbs Free Energy in terms of Gibbs Free Energy in terms of changes, changes, GG

G = E + PG = E + PV – TV – TS S for P, T constantfor P, T constant G = G = H – TH – TSSpronounced “delta” means “the change in”

H can be measured in the laboratory with a calorimeter. S can also be measured with heat capacity measurements.Values are tabulated in books.

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.

ThermodynamicsThermodynamicsFor a reaction at other temperatures and pressuresFor a reaction at other temperatures and pressuresThe change in Gibbs Free Energy is The change in Gibbs Free Energy is ddG = G = VdP - VdP -

SdTSdT

We can use this equation to calculate G for any phase We can use this equation to calculate G for any phase at any T and P by integrating the above equation.at any T and P by integrating the above equation.

If V and S are ~constants, If V and S are ~constants, dG = V dP – S dTour equation reduces to:our equation reduces to:

GGT2 P2T2 P2 - G - GT1 P1T1 P1 = V(P = V(P22 - P - P11) - S (T) - S (T22 - T - T11))

FOR A SOLID_SOLID REACTIONFOR A SOLID_SOLID REACTION

Suppose 3A + 2B = 2C +1D reactants = products G = 2GC +1GD -3GA – 2GB

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

One small calorie cal ~ 4.2 Joules J

)reactants()( 000i

iii

iiR GnproductsGnG

Gibbs for a chemical reactionHess’s Law applied to Gibbs for a reaction 298.15K, 0.1 MPa

Same procedure for H, S, V

Which direction will the reaction go?Which direction will the reaction go?

G for a reaction of the type:G for a reaction of the type:2 A + 3 B = C + 4 D2 A + 3 B = C + 4 D

G = G = (n G) (n G)productsproducts - - (n G)(n G)reactantsreactants

= G= GCC + 4G + 4GDD - 2G - 2GAA - 3G - 3GBB

The reaction with negative The reaction with negative G will be more stable, G will be more stable, i.e. if i.e. if G G is negative for the reaction as written, the reaction will go to the is negative for the reaction as written, the reaction will go to the rightright“For chemical reactions, we say that a reaction proceeds to the right when G is negative and the reaction proceeds to the left when G is positive.” Brown, LeMay and Bursten (2006) Virtual Chemistry p 163

Same Same procedurprocedure for e for H, H, S, S, VV

)reactants()( 000i

iii

iiR GnproductsGnG

Since G = E + PV – TSAnd we saw the slope of a sum is the sum of the And we saw the slope of a sum is the sum of the

slopesslopes

Differentiating dG = dE +PdV +VdP -TdS – SdTWhat is dE? dE = dQ – dW First Law, and dQ =TdS

2nd lawSo dE = dQ - PdV => dE = TdS – PdV

Most of these terms cancel, sodG = VdP –SdT

And if we need the changes when moving to a new T,PAnd if we need the changes when moving to a new T,PddG = G = VdP - VdP - SdTSdT

To get an equilibrium curve for a phase diagram, To get an equilibrium curve for a phase diagram, could use could use ddG = G = VdP - VdP - SdTSdT and G, S, V values for Albite, Jadeite and Quartz and G, S, V values for Albite, Jadeite and Quartz to calculate the conditions to calculate the conditions for which for which G G of the of the reaction: reaction:

Ab = Jd + Q Ab = Jd + Q is equal to 0is equal to 0

From G values for each phase at 298K and 0.1 MPa list From G values for each phase at 298K and 0.1 MPa list GG298, 298,

0.10.1 for the reaction, do the same for for the reaction, do the same for V and V and SS G at equilibrium = 0G at equilibrium = 0, so we can calculate an isobaric change , so we can calculate an isobaric change

in T that would be required to bring in T that would be required to bring GG298, 0.1298, 0.1 to 0 to 00 - 0 - GG298, 0.1298, 0.1 = - = -S (S (TTeqeq - 298) - 298) (at constant P)(at constant P)

Similarly we could calculate an isothermal changeSimilarly we could calculate an isothermal change0 - 0 - GG298, 0.1298, 0.1 = - = -V (V (PPeqeq - 0.1) - 0.1) (at constant T)(at constant T)

Mineral S(J) G (J) V (cm3/mol)

Low Albite 207.25 -3,710,085 100.07 Jadeite 133.53 -2,844,157 60.04 Quartz 41.36 -856,648 22.688From Helgeson et al. (1978).

Table 27-1. Thermodynamic Data at 298K and0.1 MPa from the SUPCRT Database

Method:Method:

NaAlSiNaAlSi33OO88 = NaAlSi = NaAlSi22OO66 + SiO + SiO22

Albite = Jadeite + Quartz Albite = Jadeite + QuartzP - T phase diagram of the equilibrium curveP - T phase diagram of the equilibrium curve

How do you know which side has which phases?How do you know which side has which phases?Calculate Calculate G for products and reactant for pairs of P and T, spontaneous G for products and reactant for pairs of P and T, spontaneous

reaction direction at that T P will have negative reaction direction at that T P will have negative GGWhen When G < 0 the product is stableG < 0 the product is stable

Figure 27-1. Temperature-pressure phase diagram for the reaction: Albite = Jadeite + Quartz calculated using the program TWQ of Berman (1988, 1990, 1991).

Clausius -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 slopeslope of the equilibrium curve will be positive if S and V both decrease or increase

with increased T and P

dG = VdP –SdT

To get the slope, at a boundary To get the slope, at a boundary G is 0G is 0ddG = 0 = G = 0 = VdP - VdP - SdTSdT

solvedPdT

SV

Figure 27-1. Temperature-pressure phase diagram for the reaction: Albite = Jadeite + Quartz calculated using the program TWQ of Berman (1988, 1990, 1991). Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

gives us the slope

End of reviewEnd of review

Return to dG = VdP – SdT. For an isothermal Return to dG = VdP – SdT. For an isothermal process dT is zero, so:process dT is zero, so:

G G VdPP P P

P

2 11

2

Gas PhasesGas Phases

For solids it was fine to assume V stays ~ constantFor solids it was fine to assume V stays ~ constant

For gases this assumption is wrongFor gases this assumption is wrong

A gas compresses as P increasesA gas compresses as P increases

How can we define the relationship between V and P for a How can we define the relationship between V and P for a gas?gas?

Gas Laws

• 1600’s to 1800’s

• Combined as ideal gas law:• n= # moles, and R is the universal gas constant• R = 8.314472 N·m·K−1·mol−1

Pressure times Volume is a constant

Increase Temp, Volume increases

Increase Temp, Pressure increases

Increase moles of gas, Volume increases

Ideal GasIdeal Gas– As P increases V As P increases V

decreasesdecreases– PV=nRTPV=nRT Ideal Gas LawIdeal Gas Law

P = pressureP = pressure V = volumeV = volume T = temperatureT = temperature n = # of moles of gasn = # of moles of gas R = gas R = gas constantconstant

= 8.3144 J mol= 8.3144 J mol-1-1 K K-1-1

So P x V is a constant at constant T

Gas Pressure-Volume Gas Pressure-Volume RelationshipsRelationships

Figure 5-5. Piston-and-cylinder apparatus to compress a gas. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

Gas Pressure-Volume Gas Pressure-Volume RelationshipsRelationships

SinceSince

we can substitute RT/P for V (for a single mole of we can substitute RT/P for V (for a single mole of gas), thus:gas), thus:

and, since R and T are certainly independent of and, since R and T are certainly independent of P:P:

G G VdPP P P

P

2 11

2

G G RT P

dPP P P

P

2 11

2

G G RTP

dPP PP

P

2 11

2

1

LogarithmsLogarithms Logarithms (Logs) are just exponentsLogarithms (Logs) are just exponents if bif byy = x then y = log = x then y = logbb x x loglog10 10 (100) = 2 because 10(100) = 2 because 1022 = 100 = 100

Natural logs (ln) use e = 2.718 as a baseNatural logs (ln) use e = 2.718 as a base For example ln(1) = logFor example ln(1) = logee(1) = 0(1) = 0 because ebecause e00 = (2.718) = (2.718)00 = 1 = 1 Anything to the zero power is one.Anything to the zero power is one. bbxx /b /byy = b = bx-y x-y so log so logbbx - logx - logbb y = log y = logbb(x/y)(x/y)

Early on we looked at slopes and areas, and defined derivatives and integrals.

We can just look these up in tables.

Here is another slope

d ln u = 1 du dx u dx

The area under the curve is the reverse operation

Gas Pressure-Volume RelationshipsGas Pressure-Volume Relationships

bx /by = bx-y so logbx - logb y = logb(x/y)

Gas Pressure-Volume Gas Pressure-Volume RelationshipsRelationships

The form of this equation is very usefulThe form of this equation is very useful

GGP, TP, T - G - GTT = RT = RT lnln (P/P (P/Poo))

For a For a non-ideal gasnon-ideal gas (more geologically appropriate) the same (more geologically appropriate) the same form is used, but we substitute form is used, but we substitute fugacity ( fugacity ( f f )) for P for P

wherewhere f f = = PP is the fugacity coefficient is the fugacity coefficient GGP, TP, T - G - Goo

TT = RT = RT lnln ( (f f /P/Poo) so ) so H2OH2O ranges 0.1 – 1.5, ranges 0.1 – 1.5, CO2CO2 ranges 2 – 50 ranges 2 – 50

At low pressures most gases are ideal, but at high P they are notAt low pressures most gases are ideal, but at high P they are not

o

Solid Solutions: T-X Solid Solutions: T-X relationshipsrelationships

Ab = Jd + Q was calculated for Ab = Jd + Q was calculated for purepure phases phasesWhen solid solution results in impure phases When solid solution results in impure phases

the activity of each phase is reducedthe activity of each phase is reducedUse the same form as for gases (RT Use the same form as for gases (RT ln ln P or RT P or RT ln ln

f f ))Instead of fugacity f, we can use Instead of fugacity f, we can use activity aactivity a

Ideal solution: Ideal solution: aaii = X = Xii y = # of crystallographic sites y = # of crystallographic sites in in which mixing takes placewhich mixing takes place

Non-ideal: Non-ideal: aaii = = ii X Xi i

where gamma where gamma ii is the is the activity coefficient activity coefficient

y

y

Dehydration ReactionsDehydration Reactions Ms + Qtz = Kspar + Sillimanite + HMs + Qtz = Kspar + Sillimanite + H22OO We can treat the solids and gases separatelyWe can treat the solids and gases separately

GGP, TP, T - G - GTT = = VVsolidssolids ( (PP - 0.1) + RT - 0.1) + RT ln ln ( (PP/0.1) /0.1) (isothermal) (isothermal) The treatment is then quite similar to solid-solid The treatment is then quite similar to solid-solid

reactions, but you have to solve for the equilibrium reactions, but you have to solve for the equilibrium pressure pressure PP by iteration. by iteration.

Iterative methods are those which are used to are those which are used to produce approximate numerical solutions to problems. produce approximate numerical solutions to problems. Newton's method is an example of an iterative is an example of an iterative method.method.

o

Newton’s MethodNewton’s Method

Dehydration ReactionsDehydration ReactionsdPdT

SV

Figure 27-2. Pressure-temperature phase diagram for the reaction muscovite + quartz = Al2SiO5 + K-feldspar + H2O, calculated using SUPCRT (Helgeson et al., 1978). Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

•Muscovite is unstable at High T while Qtz present, dehydrates by reacting w Qtz, forms K-spar and Al-silicate + water.

• V high at low P so high Vgas -> S/V low (gentle slope)• V low at high P (already near limit of compressibility) so -> S/V high (steep slope)

• Result: Characteristic concave shape;decarbonation and other devolitilazation reactions are similar

Ch 27b Ch 27b GeothermobarometryGeothermobarometry For any reaction with one or more variable For any reaction with one or more variable

components, at any given P,T ,we can components, at any given P,T ,we can solve for the equilibrium curve usingsolve for the equilibrium curve using

G=0= G=0= GG00 + RT ln K (27-17) + RT ln K (27-17) So ln K = - So ln K = - GG00/RT/RT

Equilibrium Constant KEquilibrium Constant K GGP, TP, T = G = Goo

TT + RT + RT ln ln ( (PP/0.1 /0.1 MPaMPa)) At equilibrium the ratio in the At equilibrium the ratio in the

parentheses, regardless of how it is parentheses, regardless of how it is expressed (Pressures, chemical expressed (Pressures, chemical potentials, activities), is a constant, called potentials, activities), is a constant, called the equilibrium constant, Kthe equilibrium constant, K

GGP, TP, T = G = GooTT + RT + RT ln ln ( (KK))

Calculating an Equilibrium Constant for a Calculating an Equilibrium Constant for a ReactionReaction

The units M (molar) are moles per liter

A mixture of gasses in an inclusion was allowed to reach equilibrium. 0.10 M NO, 0.10 M H2, 0.05 M N2 and 0.10 M H2 O was measured. Calculate the Equilibrium Constant for the equation:

K for an example reactionK for an example reaction

For a reaction 2A + 3B = C + 4DFor a reaction 2A + 3B = C + 4D

K = XK = XCCXX44DD . . CC 44

DD

XX22AAXX33

BB . . AA 33

BB where Xi is the mole fraction and i is the correction, i.e. the activity coefficient, so i.e. K = KD . KWe will define the Distribution Coefficient, KD, again below. We saw it earlier in Chapter 9.

GGP, TP, T - G - GooTT = RT = RT ln ln ( (KK) ) and at equilibrium Gand at equilibrium GP, TP, T = =

00

ln K = - ln K = - GG00/RT/RT

but but GGoo = = HHoo –T –TSSoo + + V dPV dP

So So

ln K = - ln K = - HHoo/RT +/RT +SSoo/R - (/R - (V/RT) dP V/RT) dP (27-26)(27-26)

A Garnet-Biotite ReactionA Garnet-Biotite Reaction Below is the stoichiometric equation for the Fe-Mg exchange in the reaction between the biotites and Ca-free garnets:Below is the stoichiometric equation for the Fe-Mg exchange in the reaction between the biotites and Ca-free garnets:

FeFe33AlAl22SiSi33OO12 12 + KMg+ KMg33SiSi33AlOAlO1010(OH)(OH)22 = Mg = Mg33AlAl22SiSi33OO1212 + KFe + KFe33SiSi33AlOAlO1010(OH)(OH)22

Almandine + Phlogopite = Pyrope + AnniteAlmandine + Phlogopite = Pyrope + Annite

This false color image of a garnet crystal in equilibrium with biotites. The garnet passed from an initial composition of Magnesium-rich Pyrope in its core to Fe-rich Almandine on its rim.

Phlogopite is the magnesium end-member of the biotite solid solution seriesAnnite is the iron end-member of the biotite solid solution series

ln K = - H/RT +S/R - (V/RT) dP

Garnet-Biotite Garnet-Biotite GeothermometerGeothermometer

The Distribution Coefficient The Distribution Coefficient KKDD

The Garnet - Biotite Fe –Mg exchange reactionThe Garnet - Biotite Fe –Mg exchange reaction

Figure 27-5. Graph of lnK vs. 1/T (in Kelvins) for the Ferry and Spear (1978) garnet-biotite exchange equilibrium at 0.2 GPa from Table 27-2. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

Application to Application to H and H and S S determinationdetermination

lnK = - H/RT +S/R - (V/RT) dP

This is a line!From (27-26) we can extractH from the slope and S from the intercept!

y = slope . x + b

The GASP geobarometerThe GASP geobarometerGarnet-aluminosilicate-silica-plagioclaseGarnet-aluminosilicate-silica-plagioclase

Figure 27-8. P-T phase diagram showing the experimental results of Koziol and Newton (1988), and the equilibrium curve for reaction (27-37). Open triangles (yellow) indicate runs in which An grew, closed triangles (red) indicate runs in which Grs + Ky + Qtz grew, and half-filled triangles (yellow/red) indicate no significant reaction. The univariant equilibrium curve is a best-fit regression of the data brackets. The line at 650oC is Koziol and Newton’s estimate of the reaction location based on reactions involving zoisite. The shaded area is the uncertainty envelope. After Koziol and Newton (1988) Amer. Mineral., 73, 216-233

GeothermobarometryGeothermobarometry

Assessment of reaction Assessment of reaction texturestextures

Identify which minerals are early, which are Identify which minerals are early, which are late, and which are part of a stable late, and which are part of a stable assemblage. assemblage.

Early minerals are likely to be inclusions or Early minerals are likely to be inclusions or broken. broken.

Late minerals may be in cracks or strain Late minerals may be in cracks or strain shadows.shadows.

Minerals that are in textural equilibrium Minerals that are in textural equilibrium should not be separated by reaction zones. should not be separated by reaction zones.

The GASP geobarometerThe GASP geobarometer 3CaAl3CaAl22SiSi22OO88 = Ca = Ca33AlAl22SiSi33OO1212 + 2Al + 2Al22SiOSiO55 + +

SiOSiO22 3 Anorthite = Grossular + 2 Al3 Anorthite = Grossular + 2 Al22SiOSiO55 + +

Quartz Quartz

Garnet-aluminosilicate-silica-plagioclase

These Grossular garnets (in association with SiO2 and Al2SiO5) have Anorthite plagioclase rims. They tell us only that the rock passed somewhere through this equilibrium line.

However …However … if we have another mineral if we have another mineral

equilibrium, we may get a crossing equilibrium, we may get a crossing line on our PT diagramline on our PT diagram

Pyrophyllite is Al2Si4O10(OH)2

Determining P-T-t HistoryDetermining P-T-t History

Zoning in Pl Zoning in Pl gives gives successive successive stages in P-T stages in P-T history;history;

if we can if we can date these date these different different stages, then stages, then we canwe can

get P-T-t get P-T-t path.path.

How is this done?

GASP

Gar-Bt

Spear’s Classic Paper

1 bar = 100000 pascal1 mb [mbar, millibar] = 100 Pascals

1 atmosphere [atm, standard] = 1.01 barSpears 15-47

Calculate KD then draw in a Garnet-Biotite Garnet-Biotite linelineCalculate Pressures in Kilobars for 400 and 700C1000 bar = 1 kilobarDraw in the GASP LineGASP LineCrossing Point gives the P-T conditions

You have a thick section of a metamorphic rock containing Plagioclase, Biotites, Garnets and aluminosilicates (Al2SiO5) , so you run electron microprobe scans across interesting areas. In a scan where garnet contacts biotite, you find XMg = 0.310, XFe = 0.690 for Garnets; and XMg = 0.606, XFe = 0.324 for Biotite.

Find the Pressure and Temperature