The following notes were written as a an introduction to Thermodynamics for geologists interested in '

** STANDARD GIBBS FREE ENERGIES
OF FORMATION IN KILOCALORIES**

Data from Faure unless otherwise stated

Albite -:
- 883.988 kcal/mole (Robie and Walbaum)

Andalusite -:
-584.25

Anorthite -:
-960.4

Aragonite -:
-269.55

Ca2+ -:
-132.3

Calcite -:
-269.8

Corundum -:
-378.2

Dolomite -:
-517.1

Enstatite (MgSiO3) -: -348.7

Ferrosilite -:
-267.16

Forsterite -:
-491.5

Fayalite -:
-329.6

Gibbsite -:
-276.1

H+ -: 0

Jadeite -:
-677.206 (Robie and Walbaum)

Kaolinite -:
-906.84

Kyanite -:
-583.38

Magnesite -:
-246.1

Mullite (Al6SiO11) -: -1341.2 (-1541.2 in Faure)

Nepheline -:
-472.8

Periclase -:
-136.04

Quartz amorph. -:
-203.33

Quartz -:
-204.646 (Robie and Walbaum)

Sillimanite -:
-582.81

Stishovite -:
-191.8

Tridymite -:
-204.42

Tschermak's mole. (CaAl2SiO6) -: -745.13

Water, liq -:
-56.687

Water, gas -:
-54.636

RULE: the stable phase has the lowest G_{0}
(Gibbs Free Energy of Formation) value (usually the largest negative value)
under standard conditions of P and T.

Quartz

Quartz amorph. -: -203.33
Quartz -: -204.75

Tridymite -:
-204.42 Stishovite -:
-191.8

We can predict that quartz is the most stable phase because it has the lowest free energy (largest negative number) - but which is also indicated by the abundance of quartz in common rocks.

DG is a function of mineral structure as well as composition, e.g. the carbonate and aluminosilicate polymorphs.

**Carbonates**

Calcite is stable relative to aragonite, viz:

Aragonite = Calcite

-269.55
-269.8

**Aluminosilicates**

Andalusite = Kyanite =
Sillimanite;

-584.25;
-583.38; -582.81

Note however that Al, Si and O could also exist as a mixture of corundum, Al2O3, and quartz, SiO2, or as the more complex aluminosilicate phase mullite, Al6SiO11, and the basis of the relative free energy values:

Al2O3 + SiO2
=
Al2SiO5;

-378.2 + -204.75

= -582.95
-584.25

3Al203 + SiO2
=
Al6SiO11

3 * -378.2 + -204.75

= -1339.35
-1341.2

Al2SiO5 + 2Al2O3
=
Al6SiO11

-584.25 + 2 * -378.2

= -1340.65
-1341.2

3Al2SiO5
=
Al6SiO11 + 2SiO2

3 * -584.25
-1341.2 + 2 * -204.75

= -1752.75
= -1750.72

we can predict that andalusite and mullite are more stable than
corundum in the presence of quartz; that mullite is more stable than andalusite
and corundum; and that andalusite is more stable than mullite in the presence
of quartz.

Mullite is not the natural aluminosilicate at low temperatures, and
the free energy value given by Faure is incorrect.

In the **MgO - SiO2
system**, olivine is more stable than MgO
and quartz, and enstatite is more stable than forsterite in the presence
of quartz:

2MgO + SiO2
= Mg2SiO4

2 x -136.04 + -204.75

= -476.83
-491.5

Mg2SiO4 + SiO2
= 2 * MgSiO3 (Mg2Si2O6)

-491.5 + -204.75
2 * -348.7

= -696.25
= -697.4

******************************************************************

However, fayalite and quartz are stable relative to ferrosilite:

Fayalite + Quartz
= 2 Ferrosilite

Fe2SiO4 + SiO2
2 * FeSiO3 (Fe2Si2O6)

-329.6 + -204.75
2* -267.16

= -534.35
= -534.32

******************************************************************

2MgO + 2SiO2
=
Mg2Si2O6

2 x -136.04 + 2 x -204.75
2 x -348.7

= -681.58
= -697.4

2MgO + Mg2Si2O6
=
2Mg2SiO4

-136.04 + -697.4
2 x -491.5

= -833.44
= -983

*****************************************************************

We can also predict that albite is more stable than nepheline and quartz

NaAlSi3O8
=
NaAlSiO4 + 2 SiO2

-884.8
-472.8 + 2 * -204.75

-884.8
= -882.3

*****************************************************************

but that albite and olivine are only very slightly more stable than nepheline and enstatite:

NaAlSi3O8 + 2 Mg2SiO4
= NaAlSiO4
+ 2Mg2Si2O6

-884.8 + 2 * -491.5
-472.8 + 4 * -348.7

= -1867.8
= -1867.6

**Weathering**

Under conditions of natural weathering by acid rain anorthite is not stable
relative to kaolinite:

Primary material
Weathered material

Anorthite + 2H+ + H2O
= Al2Si2O5(OH)4 + Ca2+

-960.4 +
0 + -56.7
-906.84 + -132.2

= -1017.1
= -1039.04

Weathering is therefore a spontaneous process.

The difference between the products and the reactants is recorded as DG, which will have a negative value if the Gibbs free energy of the products is less than that of the reactants. In the case of enstatite as a product of the reactants forsterite and quartz, DG = -1.15 kcal, a negative value. Enstatite is stable relative to olivine and quartz.

If the reactants and products have the same value of Gibbs free energy, both the reactants and products will be equally stable, and it will be noted that Gr = Gp and that Gp-Gr = DG = 0. The reactants and products are said to be in equilibrium.

**When might the reactants
and products have the same Gibbs free energy?**

If we were to react calcium, carbon and oxygen
(=calcite or aragonite) at varying conditions of P and T, we would find
that aragonite is stable at higher pressures and lower temperatures and
calcite stable at lower pressures and higher temperatures. If a sufficiently
large number of experiments were carried out, we would find that the area
of a P-T graph would be subdivisible into an Aragonite field and a Calcite
field separated by a sharp boundary. We would also find that at each point
on the P-T graph the value of G for aragonite and calcite would be different,
but that along the field boundary the value of G for Aragonite and Calcite
would be the same. Along the field boundary therefore, DG
= 0. It is also implied that the G values of Aragonite and Calcite are
different from their values at the standard conditions of 298 K and 1 bar.

It is important to note that the free
energy G for individual minerals changes with P and T, and that the relative
sum of G of the phases on either side of the reaction boundary also changes.
DG
usually refers to the latter case. At 900 K, DG
for the reaction Forsterite + Quartz = 2 Enstatite is more negative than
at 298 K, implying that under these conditions Enstatite is even more stable.
Thermodynamics is very much concerned with calculating the variations of
G with T and P.

The variation in G is a reflection of the relative variation in two other parameters of state, H, the enthalpy, and S, the entropy, both of which also vary with T and P. In the case of a reaction such as Forsterite and Quartz -> Enstatite, there is a relative decrease in the values of both H, and S. What determines the stable assemblage is however the relative value of H and the function TS. If DH is more negative than TDS so that (DH - T DS), that is DG, is negative, the reaction will proceed. The reaction involves a loss of energy released as heat, and a decrease in entropy. Even if a reaction was endothermic, that is it takes in heat energy, it may still take place as long as the increase in TDS is greater than the positive change in DH.

At 900 K and 1 bar (kcals):

Reactants
Products

Forsterite + Quartz =
2Enstatite (MgSiO3)

DH = -(-519.207
+ -216.401) + (-738.988) (2*369.494; the enthalpy of MgSiO3 is wrongly
given by Faure as 364.9)

= (735.608) + (-738.988); Products - reactants = -3.380 kcal

DS
= -(.06188 + .02609) + .08612

= -(.08797) + .08612; Products - reactants = -.00185

DG
= -(-491.5 + -204.75) + 2 * -348.7

=
696.25 + -697.4; Products - reactants = -1.15

As a second example, at the standard state, 298 K, 1 bar (kcals); data from Robie and Walbaum:

reactants
products

Albite =
Jadeite + Quartz

**H**

-937.146 kcal -719.871
-217.65

DH = -(-937.146) + (-719.871
+ -217.65)

=
937.146 + (- 937.521) (diff = -.375)

** S**

.0502 kcal
.0319 .00988

DS = -(.0502)
+ (.0319 + .00988)

-.0502 +
.04178 (diff. = -.00842)

**G**

-884.8 kcal
-681.7 -204.75 Data from Faure

DG = -(-884.8)
+ (-681.7 + -204.75) Faure

= 884.8
+ (-886.45) (diff = -1.65)
Faure

- 883.988
-677.206 -204.646 ) Data from Robie and Waldbaum

DG = -(-883.988)
+ (-677.206 + -204.646) (R &W)

= 883.988
+ (-881.852)
(diff = +2.136) (R &W)

Calculated free energy change: DG
= DH - TDS

DG = (-937.521 - (-937.146))
- (298 * (.0319 + .00988 - .0502)

= -.375 - ( 298 * -.00842)

= -.375 - -2.509 (diff. = +2.134)

Both parameters of state H and S are less (more negative in the case of Enthalpy) in Jadeite + Quartz than in Albite, but the determining factor lies in the fact that DH is less negative than TDS (the internal energy of the system decreases more than does the entropy). The conversion of albite to jadeite and quartz therefore involves a net decrease in Gibbs free energy.

**The Free energy of
a mineral phase**

In the case of mineral phases, such as albite, the free energy of formation of the phase is a measure of the difference in enthalpy and entropy for the reaction:

Na + Al + 3Si + 8O = NaAlSi3O8

where the free energy, enthalpy and entropy of formation
of albite and of its constituent elements are related by the equation:

G^{0}f_{albite} - G^{0}_{elements} = H^{0}f_{albite}
- H^{0}_{elements} - T(S^{0}f _{albite}
- S^{0}_{elements}) [f = formation]

and since the free energy and enthalpy values of the elements are 0, but the sum of the entropies of formation of the elements is a finite value :

then
G^{0}f _{albite} =
H^{0}f _{albite} - T(S^{0}f _{albite}
- S^{0}_{elements})

- 883.988 = -937.146 - 298 * 50.2
+ 298 * (S_{Na} + S_{Al }+ 3S_{Si} + 8S_{O})

IMPORTANT: the
free energy of a compositionally identical assemblage to albite such as
jadeite and quartz (NaAlSi3O8 = NaAlSi2O6 + SiO2) would also contain the
term ‘ 298. (S_{Na} + S_{Al} + 3S_{Si} + 8S_{O})’,

but DG (where
DG
= G _{jadeite+quartz} - G_{albite} ) as a measure
of the difference in free energy between Albite and 'Jadeite + Quartz'
would not need to take into account the existence of this term, because,
with respect to the change in Free energy, Enthalpy and Entropy involved
in the reaction Albite -> Jadeite + Quartz, the term ‘ 298. (S_{Na}
+ S_{Al} + 3S_{Si} + 8S_{O})' would cancel out
when:

G^{0}f _{albite} = H^{0}f _{albite} - TS^{0}f
_{albite}
+ TS
_{elements} for albite

is subtracted from:

G^{0}f _{jadeite+quartz} = H^{0}f _{jadeite+quartz}
- TS^{0}f _{jadeite+quartz} + TS_{elements}
for jadeite + quartz.

**How do we determine
the variation of H, S, and G of Albite with T and P?**

If energy is added to a chemical system, one
of three changes to the physical state of the system may take place:

1) the phases of which the system is made up may
simply change temperature as a reflection of heat absorption.

2) the phases may undergo a change of state at constant
temperature.

3) there may be a change of phase as the result
of a chemical reaction.

** ENTHALPY**

Enthalpy, H, is a measure of the energy content of the system, as is the internal energy, E, and a raise in temperature reflects an increase in the translational, rotational and vibrational energy of individual molecules of the system. The difference between internal energy and enthalpy is that a change in enthalpy includes the energy that must be supplied to account for the work done should the system change volume during the change. The work done is an additional energy-storage mode. Consequently, H = E + PDV. For solids, changes in enthalpy and internal energy are effectively the same, because raising the temperature involves only negligible changes in volume and therefore little work is done.

The enthalpy of formation of a substance such
as Albite at some condition of P and T is the heat exchanged as a result
of the formation of Albite from its elemental components Na Al Si and O.
Since this usually involves loss of heat (a decrease in the potential energy
of the system), the value of H will be a negative term. If the Albite is
subsequently heated, the change in enthalpy is equal to the heat absorbed.

If the absorption of heat involves an increase
of temperature
DT at constant pressure, the
heat absorbed, q, can be measured as the quantity
DT.Cp,
where Cp is the heat capacity, the heat that needs to be absorbed to change
the temperature of the substance by one degree.

When a system changes via an infinite sequence
of infinitesimally close equilibrium states, the change is said to be reversible.
In this case, dq = Cp.dT. If the change involves a single step the change
is irreversible.

Consider a system composed of the mineral phase Albite which is being heated at constant pressure in infinitesimally small steps represented by dq, and whose heat capacity is Cp. Then:

dq = dH =Cp.dT

and therefore, assuming Cp does not change with temperature:

H2 - H1 = DH
= **f**Cp.dT = Cp.(T2 - T1) = Cp. DT
(=DQ)

if Cp changes with temperature then Cp is replaced by the empirical Cp - temperature relationship

Cp = a + bT + c/T^2, (^ means 'to the power of'; squared in this case)

where a, b, and c are experimentally determined constants. (Note: c may be a negative term.) In this case

H2 - H1 = DH
= **f**Cp. dT = **f**_{T1}^{T2}
(a + bT + c/T^2).dT = [aT + bT^2/2 - c/T]_{T1}^{T2}

**ENTROPY**

The absorption of heat also causes a change in another parameter of state known as the entropy of the system. This parameter can be envisaged as a measure of the chance that a given molecule will be found in a given element of space in the system. The entropy of a substance will evidently therefore be higher as a gas phase than a liquid phase, and higher as a liquid phase than a solid phase. At absolute zero, the entropy of the system is also 0.

In a reversible exchange at equilibrium dS
= dq/T = (Cp.dT)/T (where dq = Cp.dT)

and therefore, assuming Cp does not change with temperature:

Then S2 - S1 = DS
= **f**(Cp/T). dT = Cp.ln(T2/T1)

If Cp changes with temperature however, and
Cp = a + bT + c/T^2, then:

S2 - S1 = DS
= **f**Cp/T. dT = **f**(a/T
+ b + c/T^3).dT = [a.lnT + b.T - c/(2.T^2)]_{T1}^{T2}

The free energy, G2, of a single phase, e.g.
Albite, with an increase of temperature of DT from a temperature T1 to
a temperature T2 is the free energy at temperature T1, plus the change
in enthalpy, less the increase in entropy, less the function T.Cp.ln(T2/T1),
where with reference to the standard state T1 = 298 K:

G2 = G1 + DH
- (DT.S1 + T1.DS
+ DT.DS) + f(T.S_{elemental
components})

That is, DG
= G2 - G1 = DH
- f(T.S) + f(T.S_{elemental
components})

**NOTE:** G2 = H2 - T2S2 + f(T.Selemental components)

G2 = H1 + DT.Cp
- T2.(S1 + Cp.ln(T2/T1)) + f(T.S_{elemental components})

= H1 + DT.Cp - (T1+DT).(S1
+ Cp.ln(T2/T1)) + f(T.S_{elemental components})

= H1 + DT.Cp - T1.S1 - DT.S1
- T2.Cp.ln(T2/T1) + f(T.S_{elemental components})

and since G1 = H1 - T1.S1

= G1 + DT.Cp - DT.S1
- T2.Cp.ln(T2/T1) + f(T.S_{elemental components})

and
= G1 + DH - DT.S1
- T2.Cp.ln(T2/T1) + f(T.S_{elemental components})

= G1 + DH - DT.S1
- (T1+DT).Cp.ln(T2/T1) + f(T.S_{elemental components}s)

=
G1 + DH - DT.S1 -
(T1 + DT).DS + f(T.S_{elemental
components})

= G1 + DH - (DT.S1
+ T1.DS + DT.DS)
+ f(T.S_{elemental components})

=
G1 + DH - (T1.DS
+ (DT.(S1 + DS))
+ f(T.S_{elemental components})

= G1 + DH - (T1.DS
+ S2.DT) + f(T.S_{elemental components})

*************************************************************************************

** Example**:
The ENTHALPY of albite at 298K and 1 bar is -937,146 cal /mol, and the
heat required to increase the temperature of albite 1 degree (the heat
capacity CP) is 61.7 cal /mol/K; that is Cp = DH/DT.
The ENTROPY of Albite at 298 K and 1 bar is 50.2 cal/mol/degree.Supposing
the heat capacity does not vary with temperature, what would be the enthalpy
and entropy at 900K?

The ENTHALPY of Albite would be:

dH = Cp.dT; H^{900} =
H^{0} + Cp.DT

H^{900} = H^{0} + (900
- 298) * 61.7 = -900,002.6 cal

The ENTROPY of Albite at 298 K and 1 bar is
50.2 cal/mol/degree. Supposing the heat capacity does not vary with temperature,
what would it's entropy be at 900 K?

dS = Cp/T.dT; S^{900} = S0 +
Cp.lnT/298

S^{900} = 50.2 + 61.7*ln 900/298
= 118.39 e.u.

If the heat capacity varies with temperature according to Cp = 61.7 + 13.9 x 10^-3 T - 15.01 x 10^5/T^2 what will be the values of enthalpy?

H^{900} = H^{0} + **f**_{298}^{900}
(61.7 + 13.9 x 10^-3 T - 15.01 x 10^5/T^2) dT

H^{0 }+
(61.7T + 13.9 x 10^-3 T^2/2 + 15.01 x 10^5/T^2)_{298}^{900}

= -937146
+ 37143 + 5012 +
(-3369) = -898,360 **cal**

** What would be the
value of the entropy under the same conditions?**

S^{900} = S0 + **f**_{298}^{900}
(61.7/T + 13.9 x 10^-3 - 15.01 x 10^5/T^3) dT

= S0 + [61.7*lnT + 13.9
x 10^-3*T + 15.01 x 10^5/(2.T^2)]_{298}^{900}

= S0 + [61.7*ln(900/298)
+ 13.9 x 10^-3*(900-298) + (15.01 x 10^5/2)*(1/900^2 - 1/298^2)]

= 50.2
+ 68.2 + 8.368 - 7.5247 = 119.24 e.u.

*************************************************************************************

** Energy changes in the case
of a change of state.**

In the the conversion of ice to water or water
to steam the addition of heat energy does not cause a change in temperature
but will nevertheless cause a change in enthalpy, DH
and entropy,
DS. The change is reversible since
subtraction of a small amount of heat will convert a small amount of steam
to water but the intensive parameters T and P will not change. The heat
absorbed in this case, Dq, is the latent heat
of fusion or vaporization, and Dq is equal to
both DH and T.DS.
The change in entropy, DS, equals Dq/T,
where T is the absolute temperature in degrees Kelvin..

Since the water and steam are in equilibrium
it is evident that
DH = T.DS,
and that
DH - T.DS
= 0. The relative difference between these two terms is known as the Gibbs
free energy, DG, and is a measure of how close
the system is to equilibrium, that is, when
DG
= 0. Note that the heat absorbed is not partitioned between H and S; the
terms DH and T.DS
are simply different means of expressing the quantity Dq.
DD
compares the difference between DH and T.DS
resulting from the absorption of Dq.

**Note: **DH
= Dq and T.DS = T*(Dq/T)
= Dq; and therefore DH
= T.DS

**************************************************************************************************
**Example**: Calculate the work
done and the value of the change in internal energy, DE,
and entropy DS, involved in the conversion at
373 K and 1 bar of 1 mole of water into steam, where the latent heat of
vaporization = 9,720 cal, the volume of 1 mol of water is 18cc, the gas
constant is 83.1441 cc bars / deg mol (8.31441 J / mol / K); and 1 calorie
/ bar unit = 41.84 cc.

**************************************************************************************************

Conversion of SI units to cm bar units

1 bar = 10^5 pascals; 1 metre^3 = 10^6 cm^3; 1 calorie (heat energy units) = 4.814 joules

Energy/Pressure = ms^2t^-2 / ms^-1t^-2 = s^3 = Volume

1 metre^{3} = 1 joule / 1pascal;
1 metre^{3} = 1 joule / pascal

**10^6 cm^3 = 1 joule / 10^-5 bars;**
**and 1 cm^3 = 1 joule / (10^-5 * 10^6) = 0.1 joule /
bar = 1 / 48.14 calories**

**1 bar cc = .1 joules = 1/41.84 calories**

*****************************************************************************

Cm Bar units
SI units (metres, joules)

Q = 9720 calories 9720 * 4.184 = 4.0668 x 10^4 joules

V1 = 18cc 18 x 10^-6 metre3

V2 = RT/P = 83.1441 * 373 cc
8.31441 * 373 / 10^5 metre^3

= 3.1011854 x 104 cc
3.1011.854 x 10-2 metre3

V2 -V1 = 3.0993854 x 104 cc 3.0993854 x 10^-2 metre^3

P(V2-V1) = 1 * 3.0993854 x 10^4 bar cc
10^5 * 3.0993854 x 10^-2 joules

= 3.0993854 x 10^4 bar cc
3.0993854 x 10^3 joules

= 3.0993854 x 10^4 / 41.84 cals

= 740.771 cals

Q - P(V2-V1) = 9720 - 740.77 cals
4.0668 x 10^4 - 3.0993854 x 10^3 joules

= 8979 cals
37568.615 joules

8979 cals