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1、1,THE GEOCHEMISTRY OF NATURAL WATERS,CHEMICAL BACKGROUND Readings： CHAPTER 2,11 - Drever(1997),2,Case studies,Become acquainted with equilibrium and the equilibrium constant. Become acquainted with activity and activity coefficients.,3,WHAT CAN THERMODYNAMICS TELL US?,In the context of the geochemis
2、try of natural waters, thermodynamics can tell us: Whether a mineral should dissolve in or precipitate from a solution of a given composition. What types of other reactions that control water chemistry (e.g., acid-base, oxidation-reduction) might occur.,4,THE MEANING OF EQUILIBRIUM,A system at equil
3、ibrium has none of its properties changing with time, no matter how long it is observed. A system at equilibrium will return to that state after being disturbed, i.e., after having one or more of its parameters slightly changed, then changed back to the original values. Thermodynamically speaking, a
4、 system is at equilibrium when rG = 0,5,THE GIBBS FREE ENERGY CHANGE OF REACTION,Consider the reaction: aA + bB cC + dD Where a,b, c and d are the molar amounts of compounds A, B, C and D, respectively. The Gibbs free energy change of reaction is written as: rG = cf GC + df GD - af GA - bf GB or mos
5、t generallyIf rG = 0, the reaction is at equilibrium; if rG 0, the reaction will proceed to the left.,6,STABLE VS. METASTABLE EQUILIBRIUM,Stable equilibrium - System is at its lowest possible energy level. Metastable equilibrium - System satisfies criteria for equilibrium, but is not at lowest possi
6、ble energy.,G,7,ACTIVITY AND ACTIVITY COEFFICIENTS,In thermodynamic expressions, the activity takes the place of concentration. Activity and concentration are related: ai = iMi where ai is the activity, Mi is the concentration and i is the activity coefficient. In dilute solutions, i 1, so ai Mi. Ho
7、wever, in concentrated solutions activity and concentration may be far from equal.,8,THE LAW OF MASS ACTION,Consider the reaction: aA + bB cC + dD Where a,b, c and d are the molar amounts of compounds A, B, C and D, respectively. At equilibrium it must be true that:Thus, if we, for example, increase
8、 aA, then to maintain equilibrium, the reaction must shift to the right so that the activities of the reactants decrease and the activities of the products increase, keeping K constant. This is an example of Le Chatliers Principle.,9,HOW DO WE CALCULATE K?,For the above reaction, we can write:rG = c
9、fGC + dfGD - afGA - bfGBAnd we also write:where R is the gas constant and equals 8.314 J K-1 mol-1 or 1.987 cal K-1 mol-1.,10,AN EXAMPLE,Suppose we have the reaction: CaSO4(s) Ca2+ + SO42- For which we can write:Note that, the activity of most pure solids can be taken equal to unity (i.e., aCaSO4(s)
10、 = 1). If we increase the activity of Ca2+, Le Chatliers Principle tells us that the reaction will shift to the left (anhydrite will precipitate) so that K will remain constant.,11,SOLUBILITY PRODUCT,The equilibrium constant for a reaction of the type: CaSO4(s) Ca2+ + SO42- is called a solubility pr
11、oduct (KSP). The KSP can be calculated according to: rG = f GCa2+ + f GSO42- - f GCaSO4(s) and,12,AN EXAMPLE CALCULATION OF THE SOLUBILITY PRODUCT,From the Appendix of Kehew (2001) we obtain: f GCa2+ = -553.6 kJ mol-1 f GSO42- = -744.0 kJ mol-1 f GCaSO4(s) = -1321.8 kJ mol-1 so: rG = -553.6 + (-744.
12、0) - (-1321.8) = 24.2 kJ mol-1,13,A SECOND EXAMPLE CALCULATION OF KSP,Calculate the solubility product of mackinawite at 25C: FeS(s) Fe2+ + S2-rG = f GFe2+ + f GS2- - f GFeS(s) rG = -90.0 + (85.8) - (-93.0) = 88.8 kJ mol-1,14,A THIRD EXAMPLE CALCULATION OF KSP,Calculate the solubility product of dolomite at 25C: CaMg(CO3)2(s) Mg2+ + Ca2+ + 2CO32-rG = f GMg2+ + f GCa2+ + 2f GCO32- - f GCaMg(CO3)2(s) rG = -455.5 + (-553.6) + 2(-527.0) - (-2161.3) = 98.2 kJ mol-1,
