Transcript Geochemistry & Lab


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
 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).
ΔGfT,P(i) = ΔGfo,T,P(i) + RT ln a i (real solution).
(11)
(12)
 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)

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)

Chemical Potential (m)
 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).

Nernst Equation
▪ ΔG = nFE, where F= Faraday constant & E=electrode
potential
▪ E = Eo +(RT/nF)Σiνiln a i.
(25)