Thermodynamics for Geologists

     The following notes were written as a an introduction to Thermodynamics for geologists interested in 'hard rock' geology as distinct from hydrous solution chemistry. It purports to provide some understanding of the relationship between Free energy, Enthalpy, and Entropy, the variation of these parameters with Pressure and Temperature,

    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 G0 (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
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     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
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     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

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    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:
                      G0falbite - G0elements = H0falbite - H0elements - T(S0f albite - S0elements)  [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                              G0albite   =      H0f albite  -  T(S0f albite   -  S0elements)
                                        - 883.988  =  -937.146    -  298 * 50.2  +  298 * (SNa + SAl + 3SSi + 8SO)

     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. (SNa + SAl + 3SSi + 8SO),
     but DG  (where DG = G jadeite+quartz - Galbite )  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. (SNa + SAl + 3SSi + 8SO)' would cancel out when:

                                     G0f albite = H0f albite - TS0f albite + TS elements for albite
    is subtracted from:
                           G0f jadeite+quartz  = H0f jadeite+quartz  - TS0f jadeite+quartz + TSelements  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 = fCp.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 = fCp. dT =  fT1T2 (a + bT + c/T^2).dT = [aT + bT^2/2 - c/T]T1T2

     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 = fCp/T. dT = f(a/T + b + c/T^3).dT = [a.lnT + b.T - c/(2.T^2)]T1T2

     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.Selemental components)
     That is,   DG = G2 - G1 =      DH  -         f(T.S) + f(T.Selemental components)

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

      G2 = H1 + DT.Cp - T2.(S1 + Cp.ln(T2/T1)) + f(T.Selemental components)
            = H1 + DT.Cp - (T1+DT).(S1 + Cp.ln(T2/T1)) + f(T.Selemental components)
            = H1 + DT.Cp - T1.S1 - DT.S1 - T2.Cp.ln(T2/T1) + f(T.Selemental components)
    and since    G1 = H1 - T1.S1
            = G1 + DT.Cp - DT.S1 - T2.Cp.ln(T2/T1) + f(T.Selemental components)
    and         = G1 + DH - DT.S1 - T2.Cp.ln(T2/T1) + f(T.Selemental components)
            = G1 + DH - DT.S1 - (T1+DT).Cp.ln(T2/T1) + f(T.Selemental componentss)

            = G1 + DH - DT.S1 - (T1 + DT).DS + f(T.Selemental components)
            = G1 + DH - (DT.S1 + T1.DS + DT.DS) + f(T.Selemental components)

            = G1 + DH - (T1.DS + (DT.(S1 + DS)) + f(T.Selemental components)
            = G1 + DH - (T1.DS + S2.DT) + f(T.Selemental components)

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    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;   H900 = H0 + Cp.DT
      H900 = H0 + (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;  S900 = S0 + Cp.lnT/298
      S900 = 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?

     H900 = H0 + f298900 (61.7 + 13.9 x 10^-3 T - 15.01 x 10^5/T^2) dT

          H0 + (61.7T + 13.9 x 10^-3 T^2/2 + 15.01 x 10^5/T^2)298900

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

     What would be the value of the entropy under the same conditions?
      S900 = S0 + f298900 (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)]298900
        = 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

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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.
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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 metre3  = 1 joule / 1pascal;        1 metre3 = 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