Phreeqc教程2 -热力学数据库

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

2、om 21.0oC to 27.5oC. 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

3、 4.18 J/g-K.qrxn = - (cs solution J/g-K) (mass of solution g) (D DT K)= - (4.18 J/g-K) (1.0g/mL)(100 mL) (6.5 K)= - 2700 J or 2.7 kJD DH = 2.7 kJEnthalpy change per mole of HCl = (-2.7 kJ)/(0.050 mol) = - 54 kJ/molHesss LawKnown values of D DH for reactions can be used to determine D DHs for other r

4、eactions.D DH 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 D DH of the reaction will be the same regardless of

5、 the details of the process (single vs multi- step).CH4(g) + O2(g) - CO2(g) + 2H2O(l) D DH = -890 kJIf the same reaction was carried out in two steps:CH4(g) + O2(g) - CO2(g) + 2H2O(g) D DH = -802 kJ2H2O(g) - 2H2O(l) D DH = -88 kJCH4(g) + O2(g) - CO2(g) + 2H2O(l) D DH = -890 kJNet equationHesss law :

6、 if a reaction is carried out in a series of steps, D DH for the reaction will be equal to the sum of the enthalpy change for the individual steps.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

7、/from the surroundings measured from 0K (actually just close to it), where S0=0 for ANY substance (at 0 K, atoms do not MOVE!) S0 for water = 69.9 J/mol Entropy detemrination S0 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 v

8、ery small part of that energy! How to evaluate that small heat change CAREFULLY determine Cp over this range in incremental steps to subtract H componentTheoretical estimations In natural systems, there are many species, minerals, gases that are very difficult to impossible to determine with any acc

9、uracy 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!)Theoretical determinations Ab i

10、nitio (first principles) calculations based on electron energy (complicated rules for ESTIMATING this) can be used to determine enthalpy, entropy, Gibbs energy from a molecular basisDetermining K - Titrations Especially important in acid-base equilibrium constantszzaMHMOHK1AHHAKxxGreg Wed Oct 06 200

11、40510152025303540455023456789101112NaOH reacted (mmoles)pHCarbonate titrationVoltammetric titrations Can use voltammetry to meausure acid-base reactions for electroactive species HS- + Hg = HgS + H+ + 2e- H2S + Hg = HgS + 2 H+ + 2e- Where E is Ep from the analytical peak, E0 is the formal potential,

12、 n is # e-s, F is Faradays constant Plot of Ep vs. pH gives a slope proportional to H+ complexed to sulfideHHSnFRTEEln0Voltammetry for complexes DeHume and Ford formalism nM + L Mn(L) Stability constants can be fit from the relation: F0(X)=SBnXn=B0 + B1X + B2X2 + + BnXn where F0(X) is a polynomial f

13、unction representing Bn=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, Ip is the peak current, DEp=(Ep)s-(Ep)c, n=#e-s in rxn LMLMnnnDcpsppIIERTnFantiXF)()log(434. 0log)(0Voltammetric titrations Can also titrate su

14、lfide 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 - Error in thermodynamic data There can be significant er

15、ror in the thermodynamic data used in different databases. For example, DG0 Fe2+ 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

16、used (KNOBS)Calculating uncertainty Because so much of what we use in thermodynamic databases is additive, the general accumulation of error is estimated: x=(a2x2+ b2x2+)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

17、 error is not additive for the same species)Thermodynamic Database ConsistencyData consistent with thermodynamic relationships (appropriate basic laws and consequences)Common scales used for T, energy, mass, physical constantsConflicts between different reports for same data are resolvedSame mathema

18、tical model used to fit data from different setsSame chemical model is used to fit data from different setsAppropriate standard states are used, and consistently applied throughoutFrom Nordstrom and Munoz (1994; p. 370)log Keq CaCO3(calcite) = Ca2+ + CO32- -8.48 CO2(g) + H2O = H2CO30-1.47 H2CO30 = H+ + HCO3-6.35 H+ + CO32- = HCO3-+10.33CaCO3(calcite) + CO2(g) + H2O = Ca2+ + 2 HCO3- -5.97Another way to do this is to simply combine the Keq data algebraically:212KKKKKCOcalciteeqStill another way is recompute the D DG0R for the reaction of interest and calculate KeqWhat d