Non ideal conditions Ionic strength (I or ) Already talked about - PDF document

CEE 680 Lecture #15 2/18/2020 Print version Updated: 18 February 2020 Lecture #15 Kinetics and Thermodynamics: Fundamentals and Temperature effects (Stumm & Morgan, Chapt.2 ‐ 3 ) (Benjamin, Chapt 3,4) David Reckhow CEE 680 #15 1 Non ‐ ideal conditions  Ionic strength (I or µ) Already talked about this (ionic strength corrections)  Not zero (or infinite dilution)  Temperature (T) Focus of this section  Not 25°C  Concentration (C ) Mostly of concern for G  Not 1 M David Reckhow CEE 680 #15 2 1

CEE 680 Lecture #15 2/18/2020 Ionic Strength Effects  Ideal or infinite dilution constants (K) are in terms of activity quotients  These can be factored into molar concentrations and activity coefficients:       { H } f [ B ] H B   K B K   f [ HB ] HB HB     f [ H ] f [ B ] f { H }[ B ]      B B H   f [ HB ] f [ HB ]   HB HB  f f   [ H ][ B ]     B H   K’ f [ HB ]   HB c K David Reckhow CEE 680 #15 3 Operational Acidity Constants  The common practice of using molar concentrations results in a conditional constant:    [ H ][ B ] f   c   K K HB   [ HB ] f f    B H  And pH measurements generally give us H + activity, so it’s often convenient to leave H + in these terms, which results in the mixed acidity constant:  { H }[ B ] f    C  K Kf K HB  H [ HB ] f B David Reckhow CEE 680 #15 4 2

CEE 680 Lecture #15 2/18/2020 Temperature Effects on rates  Chemist's Approach: Arrhenius Equation Pre-exponential Factor Activation energy or frequency factor (ln )  d k E a   E / RT k Ae a a 2 dT RT T a a a R = universal gas constant   E ( T 293 )/ RT 293 k k e a a a = 1.987 cal/ o K/mole T o 293 K a T a = absolute temp ( o K)  Engineer's Approach: Or more generally where    T T  o   T 20 C k k k k o T o is any “baseline” T T T o 20 C o temperature Typical values:  =1.02 to 1.15 David Reckhow 5 Determination of E a and A   E / RT k Ae a a T  Use Arrhenius equation a     E 1  Take natural log of both       ln k ln A a   sides T  R  T a   a  Evaluate slope and intercept See: equation 3.4 (pg83) in Benjamin, 2015 David Reckhow CEE 680 #15 6 3

CEE 680 Lecture #15 2/18/2020  G  Catalysis  G  A Catalyst enhances rates reactants by providing alternative pathways with lower activation products energies Reaction coordinate  It is not “consumed” in the reaction  Homogeneous  Acid/base catalysis  Trace metal catalysis  Heterogeneous  Reactions on particle surfaces  Reactions mediated by microorganisms (enzymes)  Engineered surface catalysis  Catalytic converters, activated carbon David Reckhow CEE 680 #15 7 Chapt 2: Basics  Thermodynamics  Will tell you  which reactions are favorable or “Possible” Sets limits  composition of systems at equilibrium  Won’t tell you  how quickly the reactions proceed  good for systems with constant P & T  Air: 1.0 atm at sea level  Water: 1.0 atm for each additional 10.7 m of water  Earth: wt of overlying rock and soil  Governing property @ const. T&P is the Gibbs Free Energy  for constant T&V, it is the Helmholtz Free Energy David Reckhow CEE 680 #15 8 4

CEE 680 Lecture #15 2/18/2020 Gibbs Free Energy  Combines enthalpy and entropy  1st and 2nd 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      o o o G H T S David Reckhow CEE 680 #15 9 Standard State About:  Standard State 1 mole/L for dissolved substances 1 atmosphere for gases  Conditions:  Unit activity (a=1)  Rarely encountered in practice; but easier to base calculations on  State variables:  G o , µ o  Non ‐ standard State  Conditions:  Non ‐ unit activity, often quite low  This is the “real world”  State Variables:  G , µ David Reckhow CEE 680 #15 10 5

CEE 680 Lecture #15 2/18/2020 Enthalpy  A fundamental thermodynamic variable  enthalpy change is equal to heat of reaction (for systems at constant pressure)   H<0, heat is given off  exothermic   H>0, heat is absorbed  endothermic   H can be calculated from standard enthalpies of formation (  H o f )  available in many texts    o   o H H  e.g., Snoeyink & Jenkins, Table 3 ‐ 1 i f David Reckhow CEE 680 #15 11 Example H 2 O (g )  Evaporation of water H 2 O (l)  H 2 O (l) = H 2 O (g)   H 2 O (g) H 2 O (l)    o o H H i f              1 mole 57 . 80 kcal 1 mole 68 . 32 kcal mole mole   10 . 52 kcal   H>0, heat is absorbed  endothermic  However, this does not tell us if the reaction is favorable, or proceeds spontaneously  to answer this we need to know the entropy change David Reckhow CEE 680 #15 12 6

CEE 680 Lecture #15 2/18/2020 Thermodynamic Constants for Species of Importance in Water Chemistry (Table 3 ‐ 1 from Snoeyink & Jenkins) Part I   o  o o  o Species H G Species H G f f f f kcal/mole kcal/mole kcal/mole kcal/mole Ca +2 (aq) -2 (aq) CO 3 -129.77 -132.18 -161.63 -126.22 CH 3 COO - , CaC0 3 (s), calcite -288.45 -269.78 -116.84 -89.0 acetate H + (aq) CaO (s) -151.9 -144.4 0 0 C(s), graphite H 2 (g) 0 0 0 0 Fe +2 (aq) CO 2 (g) -94.05 -94.26 -21.0 -20.30 Fe +3 (aq) CO 2 (aq) -98.69 -92.31 -11.4 -2.52 CH 4 (g) Fe(OH) 3 (s) -17.889 -12.140 -197.0 -166.0 Mn +2 (aq) H 2 CO 3 (aq) -167.0 -149.00 -53.3 -54.4 - (aq) HCO 3 MnO 2 (s) -165.18 -140.31 -124.2 -111.1 Conversion: 1kcal = 4.184 kJ David Reckhow CEE 680 #15 13 Thermodynamic Constants for Species of Importance in Water Chemistry (Table 3 ‐ 1 from Snoeyink & Jenkins) Part II     o o o o Species Species H G H G f f f f kcal/mole kcal/mole kcal/mole kcal/mole Mg +2 (aq) -110.41 -108.99 O 2 (g) 0 0 OH - (aq) Mg(OH) 2 (s) -221.00 -199.27 -54.957 -37.595 - (aq) NO 3 -49.372 -26.43 H 2 O (g) -57.7979 -54.6357 NH 3 (g) -11.04 -3.976 H 2 O (l) -68.3174 -56.690 -2 NH 3 (aq) -19.32 -6.37 SO 4 -216.90 -177.34 + (aq) NH 4 -31.74 -19.00 HS (aq) -4.22 3.01 HNO 3 (aq) -49.372 -26.41 H 2 S(g) -4.815 -7.892 O 2 (aq) -3.9 3.93 H 2 S(aq) -9.4 -6.54 Conversion: 1kcal = 4.184 kJ David Reckhow CEE 680 #15 14 7

CEE 680 Lecture #15 2/18/2020  A measure of a system’s Entropy randomness  remove the partition and randomness T=0 V 1 , c 1 V 2 , c 2 increases  2nd law of Thermo.      o o S S T=1 i f  Spontaneous in isolated T=2 system  Like water running downhill  Or hot objects heating colder ones T=large David Reckhow CEE 680 #15 15 Gibbs Energy of a System Fig. 2.5  G Changes as reaction Pg. 45 progresses due to changing concentrations  G reaches a minimum at the point of equilibrium dG  G   d Extent of reaction David Reckhow CEE 680 #15 16 8

CEE 680 Lecture #15 2/18/2020 o Convention The G f o values are essentially  G’s for the  Since the G f formation of chemical substances from the “most stable” (reference) forms of their constituent elements o values for those most stable elemental forms are  The G f zero, by definition  Examples  Zero ‐ valent, Metallic Ag, Al, Fe, Mn, Pb, Zn  graphite ‐ C, white ‐ P, rhombic ‐ S  diatomic H 2 , I 2 , N 2 , O 2 David Reckhow CEE 680 #15 17 Simple examples Reaction  o G H 2 (g) + S (s) = H 2 S (aq) -27.87 H 2 (g) + S (s) = H 2 S (g) -33.56 O 2 (g) + S (s) = SO 2 (g) -300.2 Hg (l) + S (s) = HgS (s) -43.3 H 2 (g) + ½O 2 (g) = H 2 O (l) -237.18 O 2(g) = O 2 (aq) 16.32 o equal to zero, i.e., they are  In all of these cases reactants have a G f the reference forms of the elements  so the  G o is simply equal to the G f o of the product compound David Reckhow CEE 680 #15 18 9

CEE 680 Lecture #15 2/18/2020 Solving problems with G   o    o G Recall, at any T & P: i i   G can be calculated at 25 o C from But for STP, we use: standard gibbs free energies of   o   o G G formation (G o f ) i f  These are essentially  G’s for the formation of chemical substances from the most stable forms of their constituent elements  available in many texts  e.g., Stumm & Morgan, Appendix 3 In: kJ/mole  e.g., Benjamin, Table 2.1 In: kcal/mole  e.g., Snoeyink & Jenkins, Table 3 ‐ 1 Conversion: 1kcal = 4.184 kJ David Reckhow CEE 680 #15 19 Ammonia Problem (1/7)  Determine  G o for dissolution of ammonia in water at 25 o C  NH ( g ) NH ( aq ) 3 3  Based on example 2.5 in text  Two approaches  A. Determine  G o directly from individual G o f ’s  The easiest way  B. Determine  G o from  H o and  S o    o  o  o G H T S David Reckhow CEE 680 #15 20 10