1. 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

2. 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)

3. 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)

4. 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)

5. 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)

6. 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

7. 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).

8. 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)