1Chpt. 5 - Chemical Equilibrium James W. Murray(10/03/01) Univ. WashingtonWhy do we need to study chemical equilibria?The material in this lecture comes from the field of of chemical thermodynamics.A rigorous presentation is included in an appendix for those interested. For the purpose ofthis class there are only a few basic concepts you need to know in order to conductequilibrium calculations. With these calculations we can predict chemical compositionusing chemical models. The main questions we ask are:1. Is a geochemical system at chemical equilibrium?2. If not, what reaction (s) are most likely to occur?Here are some examples of problems where equilibrium calculations are useful.Example 1: Diatoms exist in surface seawaterThe solubility of diatom shell material (called opal) is written as:SiO2. 2H2O (Opal) ↔ H4SiO4(diatom shell material) (silicic acid)There are two chemical problems here:1) The surface ocean is everywhere undersaturated withrespect to Opal, yet diatoms grow. And furthermore they arepreserved in sediments over long time periods (millions of years).2) The favored form of solid SiO2(s) at earth surface conditions isquartz not opal! Why does opal form instead? This reaction would be written as:SiO2. H2O (Opal) ↔ SiO2 (quartz)Example 2: Acid mine drainageImagine a stream in Idaho that has been grossly contaminated by waste material fromdecades of silver mining activity but has been left to recover for the past 10 years. As anenvironmental chemist you are interested in the pollutant fluxes and the form of thepollutant in the sediments. Are solid phases still dissolving? Are harmless chemicalsbeing transformed into deadly ones in the sediments, or visa versa? You could makeinformed guesses about these questions using simple equilibrium calculations. Forexample: arsenic (As) has two main oxidation states, As (V) and As (III). As (III), orarsenite, is considerably more mobile and toxic than As (V) or arsenate. The reductionreaction of arsenate to arsenite is written as :H2AsO4- + 3 H+ + 2 e- ↔ H3AsO3 + H2OExample 3: Iron speciation and plankton growthIron has been shown to limit plankton growth in some ocean areas. Experiments havebeen conducted where iron solutions are spread from a ship across the sea surface. We2know from lab experiments that the form of iron taken up by phytoplankton is the Fe3+cation. In order to understand these iron fertilization experiments we have to be able tocalculate the activity of free Fe3+ from the total iron concentration. Such calculations needto include hydrolysis reactions of iron with water as well as speciation reactions withinorganic and organic ligands.The hydrolysis reactions are written in the form of:Fe3+ + H20 ↔ FeOH2+ + H+The organic complexation reactions may be written as:Fe3+ + H3CL ↔ FeCL + 3H+Example 4: CaCO3 in marine sedimentsCaCO3 is present in shallow sediments and disappears in sediments below a certain depthin the water column. Is this depth controlled by mineral solubility? The saturation statevaries as a function of temperature and pressure and can be calculated fromthermodynamics. The solubility reactions can be written as either:CaCO3 ↔ Ca2+ + CO32-OrCaCO3 + CO2 + H2O ↔ Ca2+ + 2 HCO3-The purpose of these lectures is to show you how to do these calculations.There are two main types of calculations that we do in chemical oceanography. 1. We can calculate chemical distributions at equilibrium. For example we can comparethe actual composition of seawater with that calculated from equilibrium. Can the oceanbe described by an equilibrium model? We will look at this question in detail next week. 2. We can determine the potential for a given chemical reaction to occur for a given setof conditions. What reactions are likely to occur? For example we can determine whatsolids will precipitate or dissolve in seawater or we can determine if a metal in seawaterwill react with specific organic compounds to form complexes.The main points of this lecture are: define the equilibrium constant - K define the free energies: ∆Gf°, ∆Gr° and ∆Gr show how to calculate K from ∆Gr° explain the difference between Q and K and how they are used to determine which direction a reaction will proceed.3The Equilibrium Constant (K)Chemical reactions can be characterized by an equilibrium constant, K. Thisconstant 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**For example: The solubility of gypsum is written as:CaSO4.2H2O ===== Ca2+ + SO42- + 2 H2OK = 10-4.58pK = 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)For this reaction the equilibrium constant can be defined as:K = aCa2+ . aSO42- . a2H2O / aCaSO4.2H2OThis type of equilibrium constant for a reaction where a solid goes into solution as freeions, is often called the solubility product (Ks) where the "a" values stand for activity.For now think of activity as the effective concentration and we will discuss it in moredetail in the next lecture. Note that activities are raised to the power of theirstoichiometric coefficient (e.g. 2H2O = a2H2O). If ions are dilute enough (this is calledan ideal solution) the activity "a" or "( )" and absolute concentration "c" or "[ ]" areessentially equal (i.e. aCa2+ = cCa2+ ). The thermodynamic convention is to set the activityof solvent (e.g. H2O) and pure solids (e.g. CaSO4 . 2H2O) equal to 1 thus H2O andCaSO4 do not appear in this version of the expression. After these simplifications, theequilibrium constant, for this case, can be written as:Ks = aCa2+ . aSO42- = (Ca2+) (SO4 2-) = [Ca2+][SO42-]Assuming ideal solutions we can imagine hypothetical situations where:If [Ca2+] = [SO42-] = 10-3 M, The product of [Ca2+][SO42-] is 10-6 which is less than the value of Ks = 10-4.58. The solution is undersaturated. The reaction will shift to the right, which means any solid present will dissolve.4If [Ca2+] = [SO42-] = 10-2.29 M, The product of [Ca2+][SO42-] is 10-4.58 which is exactly equal to the value of Ks. The