Ch 27 more Gibbs Free Energy
Gibbs free energy is a measure of chemical energy
Gibbs free energy for a phase:
G = E + PV – TS
=> G = H - TS
Where:
G = Gibbs Free Energy
E = Internal Energy
H = Enthalpy (heat content) = E + PV
T = Temperature in degrees Kelvin oK
P = Pressure, V = Volume
S = Entropy (randomness, disorder)
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
for P, T constant
 DG =
DH
– 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.
Thermodynamics
For a reaction at other temperatures and pressures
The change in Gibbs Free Energy is dDG = DVdP - DSdT
We can use this equation to calculate G for any phase at any
T and P by integrating the above equation.
FOR A SOLID_SOLID REACTION
If V and S are ~constants,
our equation reduces to:
dG = V dP – S dT
GT2 P2 - GT1 P1 = V(P2 - P1) - S (T2 - T1)
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
Same
procedure
for DH, DS,
DV
 Gibbs Free Energy (G) is measured in
KJ/mol or Kcal/mol
 One small calorie cal ~ 4.2 Joules J
Which direction will the reaction go?
DGR0 =  ni Gi0 ( products)   ni Gi0 (reactants )
i
i
DG for a reaction of the type:
2A + 3B =C +4D
DG = S (n G)products - S(n G)reactants
= GC + 4GD - 2GA - 3GB
Same
procedure
for DH, DS,
DV
The reaction with negative DG will be more stable, i.e. if DG is
negative for the reaction as written, the reaction will go to the right
“For chemical reactions, we say that a reaction proceeds to the right
when DG is negative and the reaction proceeds to the left when DG is
positive.” Brown, LeMay and Bursten (2006) Virtual Chemistry p 163
Since G = E + PV – TS
And we saw the slope of a sum is the sum of the slopes
Differentiating dG = dE +PdV +VdP -TdS – SdT
What is dE? dE = dQ – dW First Law, and dQ =TdS 2nd law
So dE = dQ - PdV => dE = TdS – PdV
Most of these terms cancel, so
dG = VdP –SdT
And if we need the changes when moving to a new T,P
dDG = DVdP - DSdT
To get an equilibrium curve for a phase diagram,
could use dDG = DVdP - DSdT
and G, S, V values for Albite, Jadeite and Quartz to
calculate the conditions for which DG of the reaction:
Ab = Jd + Q
is equal to 0
Method:



Table 27-1. Thermodynamic Data at 298K and
0.1 MPa from the SUPCRT Database
Mineral
S(J)
G (J)
V
3
(cm /mol)
Low Albite
Jadeite
Quartz
207.25
133.53
41.36
-3,710,085
-2,844,157
-856,648
100.07
60.04
22.688
From Helgeson et al. (1978).
From G values for each phase at 298K and 0.1 MPa list DG298, 0.1 for the
reaction, do the same for DV and DS
DG at equilibrium = 0, so we can calculate an isobaric change in T that
would be required to bring DG298, 0.1 to 0
0 - DG298, 0.1 = -DS (Teq - 298) (at constant P)
Similarly we could calculate an isothermal change
0 - DG298, 0.1 = -DV (Peq - 0.1)
(at constant T)
NaAlSi3O8 = NaAlSi2O6 + SiO2
Albite
=
Jadeite
+ Quartz
P - T phase diagram of the equilibrium curve
How do you know which side has which phases?
Calculate DG for products and reactant for pairs of P and T, spontaneous
reaction direction at that T P will have negative DG
When DG < 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
dG = VdP –SdT
• 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
To get the slope, at a boundary DG is 0
dDG = 0 = DVdP - DSdT
dP DS
=
solve
dT DV
Figure 27-1. Temperaturepressure 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 review
Gas Phases
Return to dG = VdP – SdT. For an isothermal
process dT is zero, so:
GP  GP =
2
1
P2
VdP
P1
For solids it was fine to assume V stays ~ constant
For gases this assumption is wrong
A gas compresses as P increases
How can we define the relationship between V and P for a
gas?
Gas Laws
• 1600’s to 1800’s
Pressure times Volume is a constant
Increase Temp, Volume increases
Increase Temp, Pressure increases
Increase moles of gas, Volume increases
• Combined as ideal gas law:
• n= # moles, and R is the universal gas constant
• R = 8.314472 N·m·K−1·mol−1
Gas Pressure-Volume
Relationships
Ideal Gas
– As P increases V decreases
– PV=nRT Ideal Gas Law
 P = pressure
 V = volume
 T = temperature
 n = # of moles of gas
 R = gas constant
= 8.3144 J mol-1 K-1
So P x V is a constant at constant T
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 Relationships
Since
GP  GP =
2
1
P2
VdP
P1
we can substitute RT/P for V (for a single mole of gas),
thus:
GP  GP =
2
1
P2
P1
RT
dP
P
and, since R and T are certainly independent of P:
G P  G P = RT
2
1
P2
P1
1
P
dP
Logarithms

Logarithms (Logs) are just exponents
if by = x then y = logb x
log10 (100) = 2 because 102 = 100
Natural logs (ln) use e = 2.718 as a base
For example ln(1) = loge(1) = 0
because e0 = (2.718)0 = 1
Anything to the zero power is one.
 bx /by = bx-y so logbx - logb y = logb(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 Relationships
bx /by = bx-y so logbx - logb y = logb(x/y)
Gas Pressure-Volume Relationships
The form of this equation is very useful
o
GP, T - GT = RT ln (P/Po)
For a non-ideal gas (more geologically appropriate) the same
form is used, but we substitute fugacity ( f ) for P
where
f = gP
g is the fugacity coefficient
GP, T - GoT = RT ln (f /Po) so
g
H2O
ranges 0.1 – 1.5, g
CO2
ranges 2 – 50
At low pressures most gases are ideal, but at high P they are not
Solid Solutions: T-X relationships
Ab = Jd + Q was calculated for pure phases
When solid solution results in impure phases the
activity of each phase is reduced
Use the same form as for gases (RT ln P or RT ln f )
Instead of fugacity f, we can use activity a
Ideal solution: ai = Xi
y
y = # of crystallographic sites in
which mixing takes place
Non-ideal: ai = gi Xi
y
where gamma gi is the activity coefficient
Dehydration Reactions

Ms + Qtz = Kspar + Sillimanite + H2O

We can treat the solids and gases separately
o
GP, T - GT = DVsolids (P - 0.1) + RT ln (P/0.1)
(isothermal)

The treatment is then quite similar to solid-solid reactions,
but you have to solve for the equilibrium pressure P by
iteration.

Iterative methods are those which are used to produce
approximate numerical solutions to problems. Newton's
method is an example of an iterative method.
Newton’s Method
Dehydration Reactions
dP = DS
dT DV
•Muscovite is unstable at
High T while Qtz present,
dehydrates by reacting w
Qtz, forms K-spar and Alsilicate + water.
• DV high at low P so high
DVgas -> DS/DV low (gentle
slope)
• DV low at high P (already
near limit of compressibility)
so -> DS/DV high (steep
slope)
• Result: Characteristic
concave shape;
decarbonation and other
devolitilazation reactions are
similar
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.
Ch 27b Geothermobarometry

For any reaction with one or more variable
components, at any given P,T ,we can solve for
the equilibrium curve using

DG=0= DG0 + RT ln K

So ln K = - DG0/RT
(27-17)
Equilibrium Constant K
GP, T = GoT + RT ln (P/0.1 MPa)
 At equilibrium the ratio in the parentheses,
regardless of how it is expressed (Pressures,
chemical potentials, activities), is a constant,
called the equilibrium constant, K


GP, T = GoT + RT ln (K)
The units M (molar) are moles per liter
Calculating an Equilibrium Constant for a Reaction
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 reaction

For a reaction 2A + 3B = C + 4D
K = XCX4D . gC g4D
X2AX3B . g2A g3B
where Xi is the mole fraction and gi is the correction, i.e.
the activity coefficient, so i.e. K = KD . Kg
We will define the Distribution Coefficient, KD, again
below. We saw it earlier in Chapter 9.

GP, T - GoT = RT ln (K)

ln K = - DG0/RT

but DGo = DHo –TDSo + DV dP
and at equilibrium GP, T = 0
So

ln K = - DHo/RT +DSo/R - (DV/RT) dP
(27-26)
Phlogopite is the magnesium end-member of the biotite solid solution series
Annite is the iron end-member of the biotite solid solution series
A Garnet-Biotite Reaction

Below is the stoichiometric equation for the Fe-Mg exchange in the
reaction between the biotites and Ca-free garnets:

Fe3Al2Si3O12

Almandine + Phlogopite = Pyrope +
+ KMg3Si3AlO10(OH)2 = Mg3Al2Si3O12 + KFe3Si3AlO10(OH)2
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.
ln K = - DH/RT +DS/R - (DV/RT) dP
Annite
Garnet-Biotite Geothermometer
The Distribution Coefficient KD
Application to DH and DS determination
The Garnet - Biotite Fe –Mg exchange reaction
lnK = - DH/RT +DS/R - (DV/RT) dP
y
= slope . x
+b
This is a line!
From (27-26) we can extract
DH from the slope and DS from
the intercept!
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.
Geothermobarometry
The GASP geobarometer
Garnet-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 halffilled 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, 216233
Assessment of reaction textures
Identify which minerals are early, which
are late, and which are part of a stable
assemblage.
 Early minerals are likely to be inclusions or
broken.
 Late minerals may be in cracks or strain
shadows.
 Minerals that are in textural equilibrium
should not be separated by reaction
zones.

The GASP geobarometer
Garnet-aluminosilicate-silica-plagioclase
3CaAl2Si2O8 = Ca3Al2Si3O12 + 2Al2SiO5 + SiO2
 3 Anorthite = Grossular + 2 Al2SiO5 + Quartz

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 …

if we have another mineral equilibrium, we
may get a crossing line on our PT diagram
Pyrophyllite is Al2Si4O10(OH)2
Determining P-T-t History


Zoning in Pl
gives
successive
stages in P-T
history;
if we can date
these different
stages, then
we can
get P-T-t path.
How is this
done?
GASP
1 bar = 100000 pascal
1 mb [mbar, millibar] = 100 Pascals
Spears 15-47
1 atmosphere [atm, standard] = 1.01 bar
Spear’s Classic Paper
Gar-Bt
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
Calculate KD then draw in a Garnet-Biotite line
Calculate Pressures in Kilobars for 400 and 700C
1000 bar = 1 kilobar
Draw in the GASP Line
Crossing Point gives the P-T conditions