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THERMODYNAMICS

THERMODYNAMICS. WHO NEEDS IT?. Kaolinite. Andalusite. Al 2 Si 2 O 5 (OH) 4 = Al 2 SiO 5 + SiO 2 +2H 2 O. Magma. Olivine. Mg 2+ + SiO 4 4- = Mg 2 SiO 4. WHY THERMODYNAMICS?. At what conditions does one mineral transform into another?.

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THERMODYNAMICS

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  1. THERMODYNAMICS WHO NEEDS IT?

  2. Kaolinite Andalusite Al2Si2O5(OH)4 = Al2SiO5 + SiO2 +2H2O Magma Olivine Mg2+ + SiO44-= Mg2SiO4 WHY THERMODYNAMICS? At what conditions does one mineral transform into another? At what conditions does a magma crystallize a particular mineral? The conditions of interest are P, T and the activities or Thermodynamic concentrations of chemical species

  3. THE GIBBS FUNCTION Recall the fundamental equation: dU = dq + dw or dU = TdS – PdV and (U/S)v = T; (U/V)s = – P w q As U is a function of S, V we can transform the fundamental equation and define Gibbs energy as : G = U – S(U/S)v – V(U/V)s = U –TS +PV dG = dU– TdS – SdT + PdV + VdP = TdS – PdV – TdS – SdT + PdV + VdP dG= -SdT + VdP

  4. G/P = V G GAndalusite GKyanite 1 Bar Peq G G/T = -S GAndalusite GKyanite 298 K T eq IMPLICATIONS FOR EQUILIBRIUM Consider the reaction Kyanite = Andalusite

  5. STABLE AND METASTABLE EQUILIBRIUM Metastable equilibrium Stable equilibrium GT,P dGP,T = 0 dGP,T = 0

  6. THE GIBBS FUNCTION (Continued) Since dG = (G/T)pdT + (G/P)TdP At equilibrium G Andalustite = G Kyanite dGT,P = 0 Whereas for the Gibbs function we transformed the fundamental equation on both S and V we can define enthalpy by transforming it only on S and thus H = U + V(U/V)s or H = U + PV but G = U –TS +PV or U = G + TS – PV and thus G = H –TS or GT,P = H - TS

  7. C.P. 220 Water Ice Pressure (bar) 1 Steam 25 374 Temperature (oC) APPLICATION OF THE GIBBS FUNCTION • We cannot measure energy parameters of the phase of interest in absolute terms • We need a standard set of conditions to which the parameters can be related  We need to relate energies of phases to those of their elements and define a standard state, e.g., 298 K and 1 bar Consider the reaction H2O(V)=H2O (L) at 25 C, (298 K, 1 bar) fGoH2O = GoH2O(L) – GoH2 - 0.5GoO2 fGoH2O = GoH2O(V) – GoH2 - 0.5GoO2 rGoH2O = GoH2O(L) – GoH2O(V)

  8. 10000 HT – H298 6000 dH/dT = Cp J mol-1 = a + bT –cT-2 2000 298 500 1000 Temperature (K) T rHT = rH298 +  rCpdT 298 T rST = rS298 +  (rCp/T)dT 298 P T T rGP,T = rH298 +  rCpdT – T(rS298 +  (rCp/T)dT) +  rV 298 1 298 HEAT CAPACITY AND PHASE EQUILIBRIA Consider the reaction Kyanite =Andalusite At equilibrium, GKyanite =GAndalusite and rG = 0 rG = rH - TrS and therefore rS = rH/T Kyanite Andalusite

  9. T (rG/T)p = -rS and rGT - rG298 =  - rSdT 298 CLAPEYRON EQUATION Consider the reaction Kyanite =Andalusite At equilibrium, GKyanite =GAndalusite and rG = 0 but dG = -SdT + VdP and dG =0  VdP = SdT i.e., dP/dT = rS/rV Calculate the equilibrium temperature at 1 bar If rST = rS298 then rGT - rG298 = - rS (T – 298.15)

  10. APPLY THE CLAPEYRON EQUATION i.e., dP/dT = rS/rV rGT - rG298 = - rS (T – 298.15) dP/dT Pressure (Kbar) Kyanite Andalusite Temperature (oC)

  11. CLAPEYRON EQUATION (Continued) rG/T)p = - rS T rGT - rG298 =  - rS 298 rGT - rG298 = - rS (T – 298.15) 0 – 1220 = -9.41 (T - 298.15) = 427.8 K or 154.6 C dP/dT = rS/rV = 9.41/0.744 (1 cm3 = 0.01Jbar-1) = 12.65 bar/ K fGo So Vo kJ mol-1 J mol-1K-1 cm3 mol-1 Kyanite -2443.88 83.81 44.09 Andalusite -2442.66 93.22 51.53 Sillimanite -2440.99 96.11 49.90

  12. ALUMINOSILICATE STABILITY Kyanite = Andalusite Teq @ 1 bar = 154.6 dP/dT = 12.65 bar K-1 Andalusite = Sillimanite Teq @ 1 bar = 602.9 C; dP/dT = -17.73 bar K-1 10 Andalusite = Sillimanite 8 dP/dT = 21.17 bar K-1 6 Pressure (Kbar) Kyanite Triple point 4 Sillimanite 416 C; 3310 bar 2 Andalusite 0 100 300 700 200 400 600 500 Temperature (oC)

  13. P T T rGP,T = rH298+  rCpdT – T(rS298 +  (rCp/T)dT) +  rVdP 1 298 298 T T = rH298+  rCpdT – T(rS298 +  (Cp/T)dT) + Vs(P-1) 298 298 P 1 REACTIONS INVOLVING A GAS An assumption implicit in evaluating equilibria involving solids is that volume does not change with P and T This is not true for gases, e.g., PV = RT Consider the reaction Calcite + Quartz= Wollastonite + CO2 +  r(RT/P)dPg T T = rH298+  rCpdT – T(rS298 +  (rCp/T)dT) + rVs(P-1) 298 298 + RTlnPg

  14. 2000 Wollastonite 1500 Calcite 1000 + Pressure (bar) CO2 Quartz + 500 Wollastonite CO2 300 400 600 700 500 Temperature (oC) T = rH298+  rCpdT – T(rS298 +  (rCp/T)dT) + rVs(P-1) 298 298 + RTlnf CO2 REACTIONS INVOLVING A GAS (Continued) Calcite + Quartz = Wollastonite + CO2 CaCO3 + SiO2=CaSiO3 + CO2 Calcite dp/dt large dp/dt small Quartz T

  15. - The partial G of component i, Gi = (G/ni)P,T,ni (I = 1,2…) = μi G = niμi and dG =  nidμi + μidni i i i dG =-SdT + VdP +  μi dni i 0 = -SdT + PdV + nidμi  nidμi + μidni = -SdT + PdV + μidni i i i i At constant P and T  nidμi = 0 i CHEMICAL POTENTIAL If the system is open then we need to consider the contribution of components that act as intensive parameters Rearrange

  16. PRACTICAL SIGNIFICANCE OF CHEMICAL POTENTIAL Aragonite Z P Y Calcite X T X Y Z Calcite Aragonite Aragonite μ = CaCO3 Calcite Calcite Aragonite Consider the reaction Aragonite = Calcite (CaCO3)

  17. At constant P and T  nidμi = 0 i GIBBS DUHEM EQUATION Consider two components to which the system is open nadμa+ nbdμb = 0 dμa/dμb= -nb/na Consider phase relations in the system Fe-S-O Magnetite Slope = 3/2 FeS + 1/2S2 = FeS2 μO2 3FeS +2O2 = Fe3O4 + 3/2S2 Slope = 3/4 Pyrite Fe3O4 + 3S2 = 3FeS2 +2O2 Pyrrhotite μS2

  18. P P P  dG = GP – Go =  VdP =  (RT/P)dP 1 1 1 FUGACITY AND ACTIVITY Consider the change in G of an ideal gas with P G – Go = RT lnP Most gases are not ideal. We therefore define an effective thermodynamic pressure, fugacity, which is related to pressure as follows f= P G – Go = RT ln f Therefore μi–μio = RT lnfi/fio for component i in a gas mixture Extending the concept to solutions generally we define: ai= fi/fio and ai = Xii μi – μio = RT ln ai

  19. V =XAVAo+XBVBo VAo VMixing VA VBo VB B A XB μBo XAμAo + XBμAo μAo μB GMixing μA B A XB THE EFFECT OF MIXING Ideal mixing Non-ideal mixing V ideal > Vnon-ideal No change of V or H with Ideal mixing. S increases with both ideal and non ideal mixing. S = k ln 

  20. EFFECT OF MIXING ON ACTIVITY Ideal solid solutions For a pure phase a = 1 but for a solid solution ai = (Xii)nwhere i is a component and n is the number of crystallographic sites of mixing If mixing ideal i = 1 and ai = (Xi)n Consider olivine comprising forsterite (Mg2SiO4) + fayalite (Fe2SiO4) in reaction Mg2SiO4 + 2H2O = 2Mg(OH)2 + SiO2 Olivine Brucite Quartz If Olivine contains 70 mol% forsterite then aForsterite = (0.7)2 = 0.49 and μForsterite = μForsteriteo + RT ln(0.49)

  21. LAW OF MASS ACTION Rate of reaction is directly proportional to the concentration of each reacting substance Consider reaction aA + bB  cC + dD k1[A]a[B]b At equilibrium aA + bB= cC + dD k1[A]a[B]b = k2[C]c[D]d or k1/k2 = K The equilibrium constant, K = [C]c[D]d/[A]a[B]b ln K =cln[C] + dln[D] – aln[A] – bln[B] Ln K = ni  lnXi More generally

  22. THE EQUILIBRIUM CONSTANT Consider reaction aA + bB = cC + dD rG = rμi = cμC + dμD – aμA - bμB butμi – μio = RT ln ai rG = rμa-d= rμa-do + RTln (aCcaDd/aAaaBb) rG = r μo +RT ln K Ln K = ni  ln ai At equilibrium rG = 0 and r μo = -RT lnK rG = - RT lnK and ln K = - rG/RT

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