1. Calorimetry Measurement of heat flow (through temperature) associated with a reaction Because dH = q / dT, measuring Temperature change at constant P yields enthalpy

3. Problem When 50.mL of 1.0M HCl and 50.mL of 1.0M NaOH are mixed in a calorimeter, the temperature of the resultant solution increases from 21.0 o C to 27.5 o C. Calculate the enthalpy change per mole of HCl for the reaction carried out at constant pressure, assuming that the calorimeter absorbs only a negligible quantity of heat, the total volume of the solution is 100. mL, the density of the solution is 1.0g/mL and its specific heat is 4.18 J/g-K. q rxn = - (c s solution J/g-K) (mass of solution g) (  T K) = - (4.18 J/g-K) [(1.0g/mL)(100 mL)] (6.5 K) = - 2700 J or 2.7 kJ  H = 2.7 kJ Enthalpy change per mole of HCl = (-2.7 kJ)/(0.050 mol) = - 54 kJ/mol

4. Hess’s Law Known values of  H for reactions can be used to determine  H’s for other reactions.  H is a state function, and hence depends only on the amount of matter undergoing a change and on the initial state of the reactants and final state of the products. If a reaction can be carried out in a single step or multiple steps, the  H of the reaction will be the same regardless of the details of the process (single vs multi- step).

5. CH 4 (g) + O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ If the same reaction was carried out in two steps: CH 4 (g) + O 2 (g) --> CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g) --> 2H 2 O(l)  H = -88 kJ CH 4 (g) + O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ Net equation Hess’s law : if a reaction is carried out in a series of steps,  H for the reaction will be equal to the sum of the enthalpy change for the individual steps.

6. Determining Entropy As for H, dS=q/dT  can be measured as heat energy (q) Another way to think of entropic energy – for any reaction, energy is ‘dispersed’ to/from the surroundings –measured from 0K (actually just close to it), where S 0 =0 for ANY substance (at 0 K, atoms do not MOVE!) S 0 for water = 69.9 J/mol

7. Entropy detemrination S 0 for water = 69.9 J/mol –0 K to 298 K what happens to water? –Heats up, changes phase (ice-ice  liquid) –69.9 joules/mol is a very small part of that energy! How to evaluate that small heat change – CAREFULLY determine C p over this range in incremental steps to subtract H component

8. Theoretical estimations In natural systems, there are many species, minerals, gases that are very difficult to impossible to determine with any accuracy by experiment Correlation methods based on isostructural- isovalent analogues, electrosatic models, ligand field models exist, but are based on empirical evidence and have little grounding in theory – thus these often suffer from innaccuracy (if that is even known!)

9. Theoretical determinations Ab initio (first principles) calculations based on electron energy (complicated rules for ESTIMATING this) can be used to determine enthalpy, entropy, Gibbs energy from a molecular basis

10. Determining K - Titrations Especially important in acid-base equilibrium constants Carbonate titration

11. Voltammetric titrations Can use voltammetry to meausure acid-base reactions for electroactive species –HS - + Hg = HgS + H + + 2e - –H 2 S + Hg = HgS + 2 H + + 2e - –Where E is Ep from the analytical peak, E 0 is the formal potential, n is # e-’s, F is Faraday’s constant –Plot of Ep vs. pH gives a slope proportional to H+ complexed to sulfide

12. Voltammetry for complexes DeHume and Ford formalism –nM + L  [M n (L)] Stability constants can be fit from the relation: –F 0 (X)=  n [X] n =B 0 + B 1 [X] + B 2 [X] 2 + … + B n [X] n where F 0 (X) is a polynomial function representing B n =overall stability constant of the nth complex, [X] is the added species (such as M) Where c is the complexed ion and s is the free ion, I p is the peak current,  E p =(E p ) s -(E p ) c, n=#e - s in rxn

13. Voltammetric titrations Can also titrate sulfide or metal into an electrochemical cell and measure the changes in free species associated with complexation Competetive coomplexation approach – where one species is displaced from a weaker complex as a titrant is added Mole ratio approach -

14. Error in thermodynamic data There can be significant error in the thermodynamic data used in different databases. For example,  G 0 Fe 2+ data was evaluated at - 78.8 KJ/ml for a long time, recently re-evaluated at -89.9 KJ/mol… One program, PHREEQC, has a function built in to evaluate equilibrium values for minerals using +/- 10% on the thermodynamic data used (KNOBS…)

15. Calculating uncertainty Because so much of what we use in thermodynamic databases is additive, the general accumulation of error is estimated: σ x =(a 2 σ x 2+ + b 2 σ x 2+ +…) 1/2 BUT – that assumes none of the values are directly related, which reduces the error (i.e. if 2 equations share the same data the error is not additive for the same species…)

16. Thermodynamic Database Consistency 1.Data consistent with thermodynamic relationships (appropriate basic laws and consequences) 2.Common scales used for T, energy, mass, physical constants 3.Conflicts between different reports for same data are resolved 4.Same mathematical model used to fit data from different sets 5.Same chemical model is used to fit data from different sets 6.Appropriate standard states are used, and consistently applied throughout From Nordstrom and Munoz (1994; p. 370)

17. log K eq CaCO 3(calcite) = Ca 2+ + CO 3 2- -8.48 CO 2(g) + H 2 O = H 2 CO 3 0 -1.47 H 2 CO 3 0 = H + + HCO 3 - -6.35 H + + CO 3 2- = HCO 3 - +10.33 CaCO 3(calcite) + CO 2(g) + H 2 O = Ca 2+ + 2 HCO 3 - -5.97 Another way to do this is to simply combine the K eq data algebraically: Still another way is recompute the  G 0 R for the reaction of interest and calculate K eq

18. What does a database look like? 2.0000 H2O + 1.0000 O2 + 1.0000 Mn++ = MnO4-- +4.0000 H+ -llnl_gamma 4.0 log_k -32.4146 -delta_H151.703kJ/mol# Calculated enthalpy of reactionMnO4-2 #Enthalpy of formation:-156 kcal/mol -analytic -1.0407e+001 -4.6464e-002 -1.0515e+004 1.0943e+001 -1.6408e+002 # -Range: 0-300