Geochemistry. William M. White

−1459.92 31.28 102.72 0.01983 2627552 Wollatonite CaSiO₃ −1632.0 82.03 −1656.45 39.93 139.58 0.00236 1401200 Forsterite Mg₂SiO₄ −2175.68 95.19 −2056.70 43.79 149.83 0.02736 3564768 Clinozoisite Ca₂Al₃Si₃O₁₂(OH) −68798.42 295.56 −6482.02 136.2 787.52 0.10550 11357468 Tremolite Ca₂MgSi₈O₂₂(OH)₂ −12319.70 548.90 −11590.71 272.92 188.22 0.05729 4482200 Chlorite MgAl(AlSi₃)O₁₀(OH)₈ −8857.38 465.26 −8207.77 207.11 696.64 0.17614 15677448 Pargasite NaCa₂Mg₄Al₃Si₈O₂₂(OH)₂ −12623.40 669.44 −11950.58 273.5 861.07 0.17431 21007864 Phlogopite KMg₃AlSi₃O₁₀(OH)₂ −6226.07 287.86 −5841.65 149.66 420.95 0.01204 8995600 Muscovite KAl₃Si₃O₁₀(OH)₂ −5972.28 287.86 −5591.08 140.71 408.19 0.110374 10644096 Gibbsite Al(OH)₃ −1293.13 70.08 −1155.49 31.96 36.19 0.19079 0 Boehmite AlO(OH) −983.57 48.45 −908.97 19.54 60.40 0.01757 0 Brucite Mg(OH)₂ −926.30 63.14 −835.32 24.63 101.03 0.01678 2556424

      Data for the standard state of 298.15 K and 0.1 MPa. ΔH_f is the molar heat (enthalpy) of formation from the elements; S° is the standard state entropy; V is the molar volume; a, b and c are constants for the heat capacity (C_p) computed as: C_p = a + bT − cT^–2 J/K-mol.

where S₀ is the entropy at 0 K (the configurational, or third law entropy) and ΔSφ is the entropy change associated with any phase change. Compilations for S₂₉₈ are available for many minerals. Table 2.2 lists some heat capacity constants for the power series formula as well as other thermodynamic data for a few geologically important minerals. Example 2.6 illustrates how entropy and enthalpy changes are calculated.

Example 2.6 Calculating enthalpy and entropy changes

If the heat capacity of steam can be represented by a three-term power series:

      with constants , , and , and the enthalpy of vaporization at 100°C is 40.6 kJ/mol, calculate the S and H changes when 1 mol of liquid water at 100°C and 1 atm is converted to steam and brought to 200°C and 3 atm. Assume that with respect to volume, steam behaves as an ideal gas (which, in reality, it is certainly not).

      Answer: We need to calculate entropy and enthalpy associated with three changes: the conversion of water to steam, raising the steam from 100°C to 200°C, and increasing the pressure from 1 atm to 3 atm. Since both S and H are state variables, we can treat these three processes separately; our answer will be the sum of the result for each of these processes and will be independent of the order in which we do these calculations.

      1 Conversion of water to steam. This process will result in ΔH of 40.6 kJ. For entropy, . We converted centigrade to Kelvin, or absolute, temperature.

      2 Raising the steam from 100°C to 200°C (from 373 K to 473 K) isobarically. Since heat capacity is a function of temperature, we will have to integrate eqn. 2.112 over the temperature interval:Evaluating this, we find that . The entropy change is given by:Evaluating this, we find that .

      3 Increasing pressure from 1 atm to 3 atm (0.1 MPa to 0.3 MPa) isothermally. We can use eqn. 2.114 to determine the enthalpy change associated with the pressure change. On the assumption of ideal gas behavior, we can substitute 1/T for α. Doing so, we find the equation