Ch.3. Thermodynamics for Geochemistry

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Presentation transcript:

Ch.3. Thermodynamics for Geochemistry The Law of Mass Action A mathematical model that explains and predicts behaviors of solutions in dynamic equilibrium (wikipedia) Defining the relation among the activities of the dissolved constituents at equilibrium in the (solution) system

The Gibbs free energy of the reaction at T & P becomes For a reaction aA + bB = cC + dD The Gibbs free energy of the reaction at T & P becomes ΔGrT,P = Σi=products ΔGfT,P(i) – Σj=reactants ΔGfT,Pj) ΔGrT,P = ΣiνiΔGfT,P(i) =(cΔGfT,P(C) + dΔGfT,P(D)) - (aΔGfT,P(A) + bΔGfT,P(B)). (10) The Gibbs free energy of individual species is given by ΔGfT,P(i) = ΔGfo,T,P(i) + RT ln X i (ideal solution). (11) ΔGfT,P(i) = ΔGfo,T,P(i) + RT ln a i (real solution). (12)

Keq = EXP(- ΔGro,T,P / RT ) (15) Substituting eqn (11) & (12) into (10) ΔGrT,P = ΣiνiΔGfo,T,P(i) +RTΣiνiln a i. (13) When it’s in equilibrium, ΔGrT,P = 0. Then, 0 = ΣiνiΔGfo,T,P(i) +RTΣiνiln a i ΔGro,T,P = - RT ln Keq (14) That is, Keq = EXP(- ΔGro,T,P / RT ) (15)

Ch.3. Gibbs Free Energy at given T & P From the definition of Gibbs free energy (G=H-TS), For a given T & P, ΔGrT,P = ΔHrT,P - TΔSrT,P If T’ & P’, what would be ΔG? ΔGrT',P = ΔHrT',P - T'ΔSrT',P. (16) Put eqn (7) & (9) into (16), ΔGrT',P = ΔHrT,P - T'ΔSrT,P + ∫TT'ΔcpdT - T'∫TT'ΔcpdT/T (17)

From dG=VdP -SdT If T=const, then dG=VdP That is, d(DG) = (DV)dP Integration gives ΔGrT',P' = ΔGrT',P + ∫PP'ΔVrdP (18) Combining eqn (17) & (18) ΔGrT',P' = ΔHrT,P - T'ΔSrT,P + ∫TT'ΔcpdT - T'∫TT'ΔcpdT/T + ∫PP'ΔVrdP (19)

Ch.3. Chemical Potential (m) Σinidμi = 0. (24) : Gibbs-Duhem equation The molal Gibbs free energy at a constant T & P G is a state function, so perfect differential dG = (∂G/∂T)P,ndT + (∂G/∂P)T,ndP + i(∂G/∂ni)T,Pdni (20) dG = -SdT + VdP + Σiμidni (21) For a constant T & P G = Σiμini (22), that is dG = Σidμini + Σiμidni. (23). Comparing (21) & (22) gives Σinidμi = 0. (24) : Gibbs-Duhem equation

If a reaction is in equilibrium (dG=0) at a constant T & P (dT=dP=0), from eqn (21) 0 = Σiμidni (3-40) Which means μi(1) = μi(2) = μi(3) = ........ μi(n).

Ch.3. Nernst Equation E = Eo +(RT/nF)Σiνiln a i. (25) ΔG = nFE, where F= Faraday constant & E=electrode potential E = Eo +(RT/nF)Σiνiln a i. (25)