1. Lecture 7 Chemical Equilibrium Define equilibrium constant – K Define the free energies:  G f ,  G r  and  G r Calculate K from  G r  Define Q – How is it different from K Goal – Learn how to do equilibrium calculations! Sounds hard but the tools to use are actually very straightforward

2. Why do we need to study chemical equilibria? Two questions are asked 1.Is a geochemical system at chemical equilibrium? 2. If not, what reaction (s) are most likely to occur?

3. Example 1: Solubility - Diatoms (with opal shells) exist in surface seawater SiO 2 2H 2 O (opal)  H 4 SiO 4  (diatom shell material) (silicic acid) SiO 2 (Opal)  SiO 2 (quartz) opal is undersaturated yet diatoms grow quartz is more stable why doesn’t it form?

4. Example 2: redox and complexation – Iron speciation and plankton growth Oxidation Reaction Fe 2+ = Fe 3+ + e - Hydrolysis reactions Fe 3+ + H 2 O  FeOH 2+ + H + Organic complexation (e.g. with siderophores) reactions Fe 3+ + H 3 CL  FeCL + 3H + FeSO 4 is added to SW but need to know Fe 3+ one of several reactions can have complexation with inorganic and organic ligands

5. Example 4: Carbonate System – CaCO 3 in marine sediments CaCO 3  Ca 2+ + CO 3 2- Or CaCO 3 + CO 2 + H 2 O  Ca 2+ + 2 HCO 3 - important for lysocline solubility reaction written in terms of the main species in SW

6. Chemical reactions can be characterized by an equilibrium constant, K. This constant expresses the ratio of the product of the concentrations of the reaction products (right side) to the product of the concentration of the reactants (left side). ** Always check that a reaction is balanced for elements and charge** Example: The solubility of gypsum is written as: CaSO 4.2H 2 O ===== Ca 2+ + SO 4 2- + 2 H 2 O Gypsum forms when SW is evaporated 5x. There are geological formations with gypsum. Used as dry wall in houses!

7. The solubility of gypsum is written as: CaSO 4. 2H 2 O = Ca 2+ + SO 4 2- + 2H 2 O The equilibrium constant is written as: K = aCa 2+. aSO 4 2-. aH 2 O 2 / aCaSO 4. 2H 2 O The magnitude of the equilibrium constant is: K = 10 -4.58 pK = 4.58 (pK = -logK = -log (10 -4.58 ) = 4.58) K = 2.63 x 10 -5 (10 -4.58 = 10 +0.42. 10 -5.00 = 2.63 x 10 -5 ) written in terms of a = activity = effective conc. note: 2H 2 O = a 2 H2O log 1.5 = 0.2 log 2 = 0.3 log 3 = 0.5 log 5 = 0.7 log 8 = 0.9

8. What does this K mean? K s = a Ca2+. a SO42- = (Ca 2+ ) (SO 4 2- ) For Ideal Solutions (means activity = concentration) = [Ca 2+ ][SO 4 2- ] If [Ca 2+ ] = [SO 4 2- ] = 10 -3 M If [Ca 2+ ] = [SO 4 2- ] = 10 -2.29 M If [Ca 2+ ] = [SO 4 2- ] = 10 -2 M activity of solvent = 1 thus a H2O = 1 activity of pure solids = 1 thus a CaCO3.2H2O = 1 K = 10 -4.58 solution is: undersaturated at saturation (equilibrium) supersaturated

9. How do we calculate equilibrium constants? Answer: From Gibbs Free Energy of reaction  G r  We need to define three types of free energy.

10. 1.  G  f ---- Standard Free Energy of Formation Definition: the energy content of one mole of a substance at standard temperature and pressure (STP = 1atm, 25  C). See Appendix I from Drever (1988) It is the energy to form one mole of substance from stable elements under standard state conditions. By definition  G f  = 0 for all elements in their stable form at STP. For example, the most stable form of elemental Ca and O at STP are elemental Ca  and O 2 gas respectively. Example: the free energy of formation of CaO is given from the following reaction where the elements Ca and O (in their most stable form) react to form CaO:  G  r 2Ca  + O 2 = 2CaO  G  f = 0 0 2(-604.2) kJ/mol the free energy of formation of 2CaO can be obtained from this reaction. Thus:  G  r = 2(-604.2) = -1208.4 kJ/mol  G  f CaO = -1208.4  2 = -604.2 kJ/mol

11. 2.  G  r ---- Standard Free Energy of Reaction Definition: Free energy released or absorbed by a chemical reaction with all substances in their standard states. This means P = 1 atm, T = 25  C, activity = 1 (pure solids = 1, solvent = 1, ideal gas = 1, ions = 1M)  G  r = (  G  f ) products - (  G  f ) reactants For a generalized reaction where B and C are reactant compounds with stoichiometric coefficients b and c and D and E are product compounds with coefficients d and e: bB + cC = dD + eE Then:  G  r = {d  G  f D + e  G  fE } - { b  G  f B + c  G  fC }

12. Example: for CaSO 4.2H 2 O ===== Ca 2+ + SO 4 2- + 2 H 2 O  G  r =  G  fCa2+ +  G  fSO42- + 2  G  fH2O -  G  fCaSO4.2H2O  G  r = (-553.58) + (-744.53) + 2 (-237.13) – (-1797.28) (from Drever, 1988)  G  r = (-1773.37) + (1797.28)  G  r = + 24.91 kJ/mol

13. The equilibrium constant (K) is calculated from  G  r as follows:  G  r = - RT ln K = -2.3 RT logK(note: ln K = 2.3log K) where R = gas constant = 8.314 J deg -1 mol -1 or 1.987 cal deg -1 mol -1 T = absolute temperature = C  + 273  = 298° for 25° C at 25  C, K can be simply calculated from:  G  r = -5.708 log K (for G in kJ/mol) = -1.364 log K (for G in kcal/mol) Example: For the solubility of gypsum +24.91 = -5.708 log K K = 10 -4.364

14. The equilibrium constant K gives the ratio of the products to the reactants at equilibrium, with all species raised to the power that corresponds to their stoichiometry. For the generalized reaction, K would be written as: (where a = activity) K = aD d aE e / aB b aC c Thus  G r  and K tell you the equilibrium ratio of products to reactants but not the concentration of any one species!

15. 3.  G r ---- Free Energy of Reaction There is a whole class of problems where we know the concentrations for a specific set of conditions and we want to ask the question a slightly different way. Is a specific reaction at equilibrium, and if not, which way will it spontaneously want to proceed for these conditions. The way we do this is to calculate the reaction quotient using the observed concentrations (converted to activities). We call this ratio Q to distinguish it from K, but they are in the same form. Comparison of the magnitude of Q with K tells us the direction a reaction should proceed. The free energy change for a given set of conditions is written as  G r =  G  r + 2.3 RT log (aD d aE e / aB b aC c ) in SW  G r =  G  r + 2.3 RT log Q where Q = aD d. aE e / aB b. aC c = Ion Activity Product (IAP) in seawater

16. We can replace  G r  with its log K expression. Then:  G r = -2.3 RTlogK + 2.3 RT log Q or  G r = 2.3 RT log Q/K = 2.3 RT log  where  = Q / K if K = Q  = 1  G r = 0 reaction is at equilibrium system's free energy is at a minimum Q > K  >1  G r > 0 reaction goes to left system has too many products and shifts to the left. Q < K  <1  G r < 0 reaction goes to right system has too many reactants and shifts to the right

17. G = Total Energy versus extent of reaction G minimum at equilibrium  G = Change in Energy versus extent of reaction  G = 0 at equilibrium only B + C at 0% only D + E at 100%