1. Activity II For solids or liquid solutions: a i =X i  i For gases: a i =P i  i = f i For aqueous solutions: a i =m i  i X i =mole fraction of component i P i = partial pressure of component i m i = molal concentration of component i

2. Activity Coefficients Debye-Huckel approximation: Where A and B are constants (depending on T, see table 10.3 in your book), and a is a measure of the effective diameter of the ion (table 10.4)

3. Equilibrium Constant  G R –  G 0 R = RT ln K AT equilibrium,  G R =0, therefore:  G 0 R = -RT ln K eq where K eq is the equilibrium constant

4. Equilibrium constants  G 0 R = -RT ln K Rearrange: ln K = -  G 0 R / RT Find K from thermodynamic data for any reaction Q is also found from the activities of the specific minerals, gases, and species involved in a reaction (in turn affected by the solution they are in)

5. Solubility Product Constant For mineral dissolution, the K is K sp, the solubility product constant Use it for a quick look at how soluble a mineral is, often presented as pK (table 10.1)  G 0 R = -RT ln K sp Higher values  more soluble CaCO 3(calcite)  Ca 2+ + CO 3 2- Fe 3 (PO 4 ) 2 *8H 2 O  3 Fe 2+ + 2 PO 4 3- + 8 H 2 O

6. Ion Activity Product For reaction aA + bB  cC + dD: For simple mineral dissolution, this is only the product of the products  activity of a solid phase is equal to one CaCO 3  Ca 2+ + CO 3 2- IAP = [Ca 2+ ][CO 3 2- ]

7. Saturation Index When  G R =0, then ln Q/K eq =0, therefore Q=K eq. For minerals dissolving in water: Saturation Index (SI) = log Q/K or IAP/K eq When SI=0, mineral is at equilibrium, when SI<0 (i.e. negative), mineral is undersaturated

8. Calculating K eq  G 0 R = -RT ln K eq Look up G 0 i for each component in data tables (such as Appendix F3-F5 in your book) Examples: CaCO 3(calcite) + 2 H +  Ca 2+ + H 2 CO 3(aq) CaCO 3(aragonite) + 2 H +  Ca 2+ + H 2 CO 3(aq) H 2 CO 3(aq)  H 2 O + CO 2(aq) NaAlSiO 4(nepheline) + SiO 2(quartz)  NaAlSi 3 O 8(albite)

9. Using K eq to define equilibrium concentrations  G 0 R = -RT ln K eq K eq sets the amount of ions present relative to one another for any equilibrium condition AT Equilibrium

10. If the system is at equilibrium, then the ratio is a constant Example: CaCO 3(calcite)  Ca 2+ + CO 3 2- pK=8.4, T=25, Ca 2+ = CO 3 2-  what’s the concentration of Ca 2+ ?? What if there is already some CO 3 2- there? Ca 2+ ≠ CO 3 2- This is the law of mass action!

11. Summation of reaction- thermodynamic properties Can sum a set of reactions (cancelling out equivalent terms on opposite sides) to form a new reaction, and derive that reaction’s K eq from it’s constituents… Consider rxn: CaCO 3(calcite) + CO 2(g) + H 2 O = Ca 2+ + 2 HCO 3 - –SUM of reactions: CaCO 3(calcite) = Ca 2+ + CO 3 2- CO 2(g) + H 2 O = H 2 CO 3 0 H 2 CO 3 0 = H + + HCO 3 - H + + CO 3 2- = HCO 3 -

12. log K eq CaCO 3(calcite) = Ca 2+ + CO 3 2- -8.48 CO 2(g) + H 2 O = H 2 CO 3 0 -1.47 H 2 CO 3 0 = H + + HCO 3 - -6.35 H + + CO 3 2- = HCO 3 - +10.33 CaCO 3(calcite) + CO 2(g) + H 2 O = Ca 2+ + 2 HCO 3 - -5.97 Another way to do this is to simply combine the K eq data algebraically: Still another way is recompute the  G 0 R for the reaction of interest and calculate K eq

13. Mineral Chemistry How could we describe the complete chemistry of a mineral or rock? What processes affect how this complete chemistry might change?

14. Pauling’s Rules for ionic structures 1.Radius Ratio Principle – cation-anion distance can be calculated from their effective ionic radii cation coordination depends on relative radii between cations and surrounding anions Geometrical calculations reveal ideal R c /R a ratios for selected coordination numbers Larger cation/anion ratio yields higher C.N.  as C.N. increases, space between anions increases and larger cations can fit Stretching a polyhedra to fit a larger cation is possible

15. Pauling’s Rules for ionic structures 2. Electrostatic Valency Principle –Bond strength = ion valence / C.N. –Sum of bonds to an ion = charge on that ion –Relative bond strengths in a mineral containing >2 different ions: Isodesmic – all bonds have same relative strength Anisodesmic – strength of one bond much stronger than others – simplify much stronger part to be an anionic entity (SO 4 2-, NO 3 -, CO 3 2- ) Mesodesmic – cation-anion bond strength = ½ charge, meaning identical bond strength available for further bonding to cation or other anion

16. Pauling’s Rules for ionic structures 3. Sharing of edges or faces by coordinating polyhedra is inherently unstable –This puts cations closer together and they will repel each other

17. Pauling’s Rules for ionic structures 4. Cations of high charge do not share anions easily with other cations due to high degree of repulsion 5. Principle of Parsimony – Atomic structures tend to be composed of only a few distinct components – they are simple, with only a few types of ions and bonds.

18. Isoelectronic series Ions of different elements with equal numbers of electrons are said to be isoelectronic Example: Al 3+, Si 4+, P 5+, S 6+, Cl 7+ each have how many e-?? Size patterns can be related to the concept of shielding vs. electrostatic interaction For any element – how would charge and size be related??

20. Goldschmidt’s rules of Substitution 1.The ions of one element can extensively replace those of another in ionic crystals if their radii differ by less than about 15% 2.Ions whose charges differ by one may substitute readily if electrical neutrality is maintained – if charge differs by more than one, substitution is minimal

21. 3.When 2 ions can occupy a particular position in a lattice, the ion with the higher ionic potential forms a stronger bond with the anions surrounding the site 4.Substitution may be limited when the electronegativities of competing ions are different, forming bonds of different ionic character Goldschmidt’s rules of Substitution

22. Goldschmidt’s rules updated… After Wood (2003): Equilibrium partitioning depends primarily on 2 energies of substitution into the crystal (a) the energy of elastic strain generated by inserting an ion which is either too large or too small for the site. (b) the electrostatic work done in inserting an ion which is either more or less highly charged than the major ion normally occupying the site. The theory requires modifications to Goldschmidt’s rules (2) and (3). Rule (2) should now be: The site has a preferred radius of ion (r O ) which enters most easily. For ions of the same charge, those which are closest in radius to r O enter most easily. Ions which are smaller or larger are discriminated against. Rule (3): The site has a preferred charge Z O. For ions of similar size, but different charge the one whose charge is closest to Z O enters most easily.

23. FeS 2 What ions would substitute nicely into pyrite?? S - radius=219 pm Fe 2+ radius=70 pm

24. Coupled Substitution When the ion of a major element in a mineral is replaced with something having a different charge, the charge imbalance created must be neutralized by addition of a counter ion Example  addition of Al 3+ in a silicate structure (replacing Si 4+ ) requires addition of a Na + or K + (Key to understanding feldspar chemistry…). When 2 Al 3+ are added for Si 4+, this then can be balanced by adding a Ca 2+ ion

26. Goldschmidt’s Classifications Principally for how trace elements/ transition metals are distributed: Atmophile elements are generally extremely volatile (i.e., they form gases or liquids at the surface of the Earth) and are concentrated in the atmosphere and hydrosphere. Lithophile elements are those showing an affinity for silicate phases and are concentrated in the silicate portion (crust and mantle) of the earth. Siderophile elements have an affinity for a metallic liquid phase. They are depleted in the silicate portion of the earth and presumably concentrated in the core. Chalcophile elements have an affinity for a sulfide liquid phase. They are also depleted in the silicate earth and may be concentrated in the core. They are also often associated with ore deposits formed via aqueous processes.

27. Atmophile elements are generally extremely volatile Lithophile elements are those showing an affinity for silicate phases Siderophile elements have an affinity for a metallic liquid phase. Chalcophile elements have an affinity for a sulfide liquid phase.