Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Lecture 7 Chemical EquilibriumDefine equilibrium constant – KDefine the free energies: Gf, Gr and GrCalculate K from GrDefine Q – How is it different from KGoal – Learn how to do equilibrium calculations!Sounds hard but the tools to use are actually very straightforwardWhy do we need to study chemical equilibria?Two questions are asked1. Is a geochemical system at chemical equilibrium?2. If not, what reaction (s) are most likely to occur?Example 1: Solubility - Diatoms (with opal shells) exist in surface seawater SiO2 • 2H2O (opal)  H4SiO4(diatom shell material) (silicic acid) SiO2 (Opal)  SiO2 (quartz)opal is undersaturated yet diatoms growquartz is more stablewhy doesn’t it form?Example 2: redox and complexation – Iron speciation and plankton growth Oxidation ReactionFe2+ = Fe3+ + e-Hydrolysis reactionsFe3+ + H2O  FeOH2+ + H+Organic complexation (e.g. with siderophores) reactionsFe3+ + H3CL  FeCL + 3H+FeSO4 is added to SWbut need to know Fe3+one of several reactionscan have complexationwith inorganic andorganic ligandsExample 4: Carbonate System – CaCO3 in marine sedimentsCaCO3  Ca2+ + CO32-OrCaCO3 + CO2 + H2O  Ca2+ + 2 HCO3-important for lysoclinesolubility reactionwritten in terms of the main species in SWChemical 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:CaSO4.2H2O ===== Ca2+ + SO42- + 2 H2OGypsum forms when SWis evaporated 5x. There aregeological formations withgypsum.Used as dry wall in houses!The solubility of gypsum is written as:CaSO4.2H2O = Ca2+ + SO42- + 2H2OThe equilibrium constant is written as:K = aCa2+ . aSO42- . aH2O2 / aCaSO4.2H2OThe 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:2H2O = a2H2O log 1.5 = 0.2log 2 = 0.3log 3 = 0.5log 5 = 0.7log 8 = 0.9What does this K mean?Ks = aCa2+ . a SO42- = (Ca2+) (SO42-) For Ideal Solutions (means activity = concentration) = [Ca2+ ][SO42-]If [Ca2+] = [SO42-] = 10-3 M If [Ca2+] = [SO42-] = 10-2.29 M If [Ca2+] = [SO42-] = 10-2 M activity of solvent = 1 thus aH2O = 1activity of pure solids = 1 thus aCaCO3.2H2O = 1K = 10-4.58solution is:undersaturatedat saturation (equilibrium)supersaturatedHow do we calculate equilibrium constants?Answer: From Gibbs Free Energy of reaction GrWe need to define three types of free energy.1. Gf ---- Standard Free Energy of FormationDefinition: 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 Gf = 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 O2 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r2Ca + O2 = 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/mol2. Gr ---- Standard Free Energy of ReactionDefinition: 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 }Example: for CaSO4.2H2O ===== Ca2+ + SO42- + 2 H2O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/molThe 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 = 10-4.364The 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 = aDd aEe / aBb aCcThus Gr and K tell you the equilibrium ratio of products to reactants but not the concentration of any one species!3. Gr ---- 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