CEE 370 Environmental Engineering Principles. Equilibrium Chemistry

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1 Updated: 9 September 015 Print version CEE 370 Environmental Engineering Principles Lecture #6 Environmental Chemistry IV: Thermodynamics, Equilibria, Acids-bases I Reading: Mihelcic & Zimmerman, Chapter 3 Davis & Masten, Chapter Mihelcic, Chapt 3 David Reckhow CEE 370 L#6 1 Equilibrium Chemistry Tells us what direction the reaction is headed in Doesn t tell us how fast the reaction is going (kinetics) Solving equilibrium problems identify reactants and products formulate equations equilibrium equations mass balance equations electroneutrality equation solve equations David Reckhow CEE 370 L#6 Lecture #6 Dave Reckhow 1

2 Temperature Effects on K Need H (enthalpy change) H < 0, exothermic (heat evolved) H > 0, endothermic (heat absorbed) The Van t Hoff Equation: log K K 1 recall that: o H T T.303RT T 1 1 Log K o H H i o f 1/T David Reckhow CEE 370 L#6 3 Acids & Bases ph of most mineral-bearing waters is 6 to 9. (fairly constant) ph and composition of natural waters is regulated by reactions of acids & bases chemical reactions; mostly with minerals carbonate rocks: react with CO (an acid) CaCO 3 + CO = Ca + + HCO - 3 other bases are also formed: NH 3, silicates, borate, phosphate acids from volcanic activity: HCl, SO Biological reactions: photosynthesis & resp. Sillen: Ocean is result of global acid/base titration David Reckhow CEE 370 L#6 4 Lecture #6 Dave Reckhow

3 Acids & Bases (cont.) Equilibrium is rapidly established proton transfer is very fast we call [H + ] the Master Variable because Protons react with so many chemical species, affect equilibria and rates Strength of acids & bases strong acids have a substantial tendency to donate a proton. This depends on the nature of the acid as well as the base accepting the proton (often water). David Reckhow CEE 370 L#6 5 ph: the intensity factor Alkalinity: a capacity factor David Reckhow CEE 370 L#6 6 Lecture #6 Dave Reckhow 3

4 Mathematical Expression of Acid/Base Strength Equilibrium constant acids: HA = H + + A - HCl + H O = H 3 O + + Cl - H Cl HCl = H + + Cl - K a HCl Bases: B + H O = BH + + OH - NH 3 + H O = NH 4+ + OH - K b NH OH NH 3 3 David Reckhow CEE 370 L#6 7 Thermodynamics & Equilibrium Classical Thermodynamics Stoichiometry Equilibrium Chemistry Working with equations Phase Transfer Acids/Bases The Carbonate System Precipitation/Dissolution Adsorption David Reckhow CEE 370 L#7 8 Lecture #6 Dave Reckhow 4

5 Enthalpy or heat of reaction For a generic reaction: aa + bb pp + rr H = (ph + rh ) + (ah + bh ) rxn P R A B where, A,B = reactant species P,R = product species a,b = stoichiometric coefficients of the reactants p,r = stoichiometric coefficients of the products H A,B = enthalpy of reactants A and B H P,Q = enthalpy of products P and Q David Reckhow CEE 370 L#7 9 Enthalpy (cont.) Or more generally: n H rxn = aih prod - a jh i=1 where, a i = stoichiometric coefficient of product species i a j = stoichiometric coefficient of reactant species j H i = enthalpy of product species i, [kcal/mol] H j = enthalpy of reactant species j, [kcal/mol] David Reckhow CEE 370 L#7 10 m j=1 rxtnts Lecture #6 Dave Reckhow 5

6 Species G, kcal/mol H, kcal/mol Thermodynamic Constants Table 4., pg. 57 Ca + (aq) CaCO3(s), calcite Ca(OH)(s), lime CO(g) CO(aq) CO - (aq) Cl - (aq) Cl(aq) OCl - (aq) H + (aq) 0 0 HNO3(aq) NO(aq) HOCl(aq) OH - (aq) O(aq) HO(l) H(g) 0 0 O(g) 0 0 David Reckhow CEE 370 L#7 11 Thermodynamic Constants for Species of Importance in Water Chemistry (Table 3-1 from Snoeyink & Jenkins) Part I Species o H f o G f Species o H f o G f kcal/mole kcal/mole kcal/mole kcal/mole Ca + (aq) CO3 - (aq) CaC03(s), calcite CH3COO -, acetate CaO (s) H + (aq) 0 0 C(s), graphite 0 0 H (g) 0 0 CO(g) Fe + (aq) CO(aq) Fe +3 (aq) CH4 (g) Fe(OH)3 (s) HCO3 (aq) Mn + (aq) HCO3 - (aq) MnO (s) Conversion: 1kcal = kj David Reckhow CEE 370 L#7 1 Lecture #6 Dave Reckhow 6

7 Thermodynamic Constants for Species of Importance in Water Chemistry (Table 3-1 from Snoeyink & Jenkins) Part II Species o H f o G f Species o H f o G f kcal/mole kcal/mole kcal/mole kcal/mole Mg + (aq) O (g) 0 0 Mg(OH) (s) OH - (aq) NO - 3 (aq) H O (g) NH 3 (g) H O (l) NH 3 (aq) SO NH + 4 (aq) HS (aq) HNO 3 (aq) H S(g) O (aq) H S(aq) Conversion: 1kcal = kj David Reckhow CEE 370 L#7 13 Gibbs Free Energy G = H - TS where, G = Gibbs free energy, [kcal/mol] H = enthalpy, [kcal/mol] S = entropy, [kcal/k-mol] T = temperature, [K] David Reckhow CEE 370 L#7 14 Lecture #6 Dave Reckhow 7

8 Gibbs Free Energy Combines enthalpy and entropy 1st and nd laws of thermodynamics Determines whether a reaction is favorable or spontaneous Practical form is based on an arbitrary datum the pure and most stable form of each element at standard state G o H o T S o David Reckhow CEE 370 L#7 15 Reactions under Standard State Conditions aa bb pp rr G = (pg + rg ) - (ag + bg ) rxn P R A B where, A,B = reactant species P,R = product species a,b = stoichiometric coefficients of the reactants p,r = stoichiometric coefficients of the reactants G A,B = Gibbs free energy of reactants A and B G P,Q = Gibbs free energy of reactants P and Q David Reckhow CEE 370 L#7 16 Lecture #6 Dave Reckhow 8

9 General Equation n G rxn = a igi - a jg i=1 prod m j=1 j rxtnts where, a i = stoichiometric coefficient of product species i a j = stoichiometric coefficient of reactant species j G i = standard Gibbs free energy of product species i, [kcal/mol] G j = standard Gibbs free energy of reactant species j, [kcal/mol] David Reckhow CEE 370 L#7 17 Non-standard state conditions And we define: G = G + RT ln Q = P p R a A B r b p P R a A B r b At equilibrium, G=0, and Q=K eq G = -RT ln K eq End equilibria David Reckhow CEE 370 L#7 18 Lecture #6 Dave Reckhow 9

10 What are standard state conditions? Parameter Temperature Gas Solid Liquid Solution Standard State or Condition 5C 1 atm Pure solid Pure liquid 1 M Element * 0 David Reckhow CEE 370 L#7 19 Roller Coaster Analogy Fig. 3-8 in Mihelcic G o react A reaction with a negative G will proceed from reactants to products. A reaction with a positive G will proceed from products to reactants (backwards). A reaction in which G is zero is at equilibrium David Reckhow CEE 370 L#7 0 Lecture #6 Dave Reckhow 10

11 Temperature effects Increases in temperature cause: A. Increases in solubility of salts B. Decreases in solubility of salts C. Depends on the salt David Reckhow CEE 370 L#6 1 Temperature effects Increases in temperature cause: A. Increases in solubility of dissolved gases B. Decreases in solubility of dissolved gases C. Depends on the gas David Reckhow CEE 370 L#6 Lecture #6 Dave Reckhow 11

12 Lime Example Lime, or calcium hydroxide, is commonly used to improve sedimentation processes or to precipitate toxic metals. A 500 L reaction vessel contains an excess amount of solid calcium hydroxide. The solution contains 10-8 M of Ca + and 10-8 M of OH. Is the lime still dissolving, is the solution at equilibrium, or is lime precipitating? Ca(OH ) (s) Ca + OH + - Example 4.6 from Ray David Reckhow CEE 370 L#7 3 Solution to Lime example First calculate the Gibbs Free Energy under standard state conditions G = ( G + G ) - G Ca(OH) rxn Ca OH G rxn = [(-13.3 kcal/ mol) + x (-37.6 kcal/ mol) - (-14.7 kcal/ mol)] G rxn = + 7. kcal/ mol David Reckhow CEE 370 L#7 4 Lecture #6 Dave Reckhow 1

13 Solution (cont.) Now we need to adapt to actual conditions: G = G + RT ln p P R a A B r b Ca(OH ) (s) Ca + OH + - G G = G + RT ln [Ca ][OH ] rxn kcal K- mol cal K- mol kcal 1000 cal rxn = (7. ) + (1.987 ) x ( ) x (98 K) x ) G rxn = -13. kcal/ mol ln (10 )(10 Since G is negative, the reaction will proceed as written David Reckhow CEE 370 L#7 5 Kinetics & Equilibrium Kinetics Rates of reactions Dynamic, complex Best approach for slow reactions Oxidation reactions, biochemical transformations, photochemical reactions, radioactive decay Equilibrium: thermodynamics Final stopping place Static, simple Best approach for fast reactions Acid/base reaction, complexation, some phase transfer David Reckhow CEE 370 L#7 6 Lecture #6 Dave Reckhow 13

14 DO Example A polluted stream with a temperature of 5C has a dissolved oxygen concentration of 4 mg/l. Use Gibbs free energy to determine if oxygen from the atmosphere is dissolving into the water, the oxygen is at equilibrium, or oxygen from the stream is going into the atmosphere. O (aq) O (g) Example 4.7 from Ray David Reckhow CEE 370 L#7 7 Solution to DO example G = G -G = (0 kcal mol ) - (-3.9 kcal rxn O (g) O (aq) mol ) G = G = 3.9 kcal rxn mol po G + RT ln [O (aq)] [O (aq)] = 4 mg O L g O x 1000 mg O x mol O 3 g O -4 = 1.5 x 10 M David Reckhow CEE 370 L#7 8 Lecture #6 Dave Reckhow 14

15 Solution (cont.) G = 3.9 kcal mol x 1000 cal kcal cal 0.09 atm (98 K) ln -4 K- mol 1.5 x 10 M G = 491 cal mol > 0 Since G is positive, the reaction will proceed in the reverse direction as written. From the atmosphere to the water. David Reckhow CEE 370 L#7 9 To next lecture David Reckhow CEE 370 L#6 30 Lecture #6 Dave Reckhow 15