Thermodynamics Internal Energy (E) of a mineral = Sum of energy stored A system, e.g. a mineral, is an isolated portion of the Universe. Internal Energy (E) of a mineral = Sum of energy stored in the bonding and in the kinetic energy of the atomic vibration Adding heat to a mineral dQ increasing kinetic energy increasing internal energy (dE) dE = dQ - pdV pdV = work at constant pressure, volume expansion This is a statement of the First Law of Thermodynamics

Thermodynamics Added heat (Q) to a system versus its Entropy (S) At constant temperature: dQ/T = dS Second law of thermodynamics: In any reversible process the change in the Entropy of the system (dS) is equal to the heat received by the system (dQ) divided by the absolute temperature. Adding heat to ice = increase in entropy through Vibration Breaking hydrogen bonds between (H2O) groups (Partial Melting)

Thermodynamics Combinations of first and second law of thermodynamics: E = Q – PV, Q = H at constant P H = E + PV dQ = TdS dE = TdS - PdV G = H – TS G = E + PV – TS dG = dE + PdV + VdP – TdS – SdT dG = TdS – PdV + PdV + VdP – TdS – SdT To integrate properly we must know how V and S vary with P and T (hence the calculus), but we shall simplify the math and assume V and S are constant This simplifies our math considerably (but may lead to some errors) We can check the validity of our assumptions by comparing to experiments or more rigorous calculations (performed by computer)

Free Energy vs. Temperature Slope of GLiq > Gsol since Ssolid < Sliquid A: Solid more stable than liquid (low T) B: Liquid more stable than solid (high T) Slope dG/dT = -S Slope S < Slope L Equilibrium at Teq GLiq = GSol dG = VdP – SdT

Free Energy vs. Pressure Slope of GLiq > Gsol since Vsolid < Vliquid A: Solid more stable than liquid (high P) B: Liquid more stable than solid (low P) Slope dG/dP = V Slope S < Slope L Equilibrium at Peq GLiq = GSol dG = VdP – SdT

Thermodynamics dGB = VBdP – SBdT dG = VdP – SdT In a phase diagram with the environmental variables temperatures and pressures we can relate the change of the free energy of a phase A as dGA = VAdP – SAdT and of the phase B dGB = VBdP – SBdT To integrate properly we must know how V and S vary with P and T (hence the calculus), but we shall simplify the math and assume V and S are constant This simplifies our math considerably (but may lead to some errors) We can check the validity of our assumptions by comparing to experiments or more rigorous calculations (performed by computer)

Thermodynamics VBdP – SBdT = VAdP – SAdT At equilibrium or = dGB = dGA or VBdP – SBdT = VAdP – SAdT = (VA- VB)dP = (SA– SB)dT dP/dT = Δ S/ Δ V Clapeyron equation To integrate properly we must know how V and S vary with P and T (hence the calculus), but we shall simplify the math and assume V and S are constant This simplifies our math considerably (but may lead to some errors) We can check the validity of our assumptions by comparing to experiments or more rigorous calculations (performed by computer)

Thermodynamics dG = VdP – SdT, at constant T: dG = VdP Ideal gas law: V = RT/P dG = RT(dP/P) Integrating from the standard pressure P° to a pressure P : Gp – G° = RT(lnP - lnP°) (n G°R)products – S(nG°R)reactants = RT(S(nlnP)products – S(nlnP)reactants)

Thermodynamics RT(S(nlnP)products – S(nlnP)reactants) = RT ln K - DG°R = RT ln K R = 1.987cal/deg mol T = 298.15 K ln K = 2.3025 log K DG°R = -1.364 log K K = 10-DG°R/1.364 To integrate properly we must know how V and S vary with P and T (hence the calculus), but we shall simplify the math and assume V and S are constant This simplifies our math considerably (but may lead to some errors) We can check the validity of our assumptions by comparing to experiments or more rigorous calculations (performed by computer) Note: R = 8.314 J/deg mol

The Equilibrium Constant For a one-component system, the Second Law of Thermodynamics is dG = -SdT + VdP but dG = ndG = ndµ where µ = (∂G/∂n) At constant T and for simplicity an ideal gas, dG = VdP From the gas law, V = nRT/P Integrating, µ P n∫µ◦ dµ = nRT∫P◦ dP µ - µº = RT ln P/P◦ If P◦ is the standard state of the ideal gas at 1 atm pressure and T, P◦=1 µ = µº + RT ln P (Note that P is unitless as it is a ratio as defined above.)

The Equilibrium Constant For a multi-component gas mixture at equilibrium: vj+1Aj+1+ . . . + vi-1Ai-1+ viAi = v1A1 + v2A2 + . . . + vjAj where v refers to stoichiometric coefficients and A to the gas species. The subscripts j+1 to I refer to the reactants and 1 to j refer to the products.

The Equilibrium Constant µi = µi° + RT ln Pi. where Pi = P (ni /n) = PXi ∆Gr = ∑viµi = ∑vi (µiº + RT ln Pi) = ∑viµiº + ∑viRT ln Pi = ∆G◦ + RT ∑ln Pivi = ∆G◦ + RT ln K where P1v1P2v2 . . . Pj v j K = Pj+1vj+1Pj+2vj+2 . . . Pivi the equilibrium constant (note that the value of K is dependent on the quanitites used in the expression and the standard states selected for them. In other words, K is not a universal constant for a reaction, but only for the conditions considered in deriving it.)

Thermodynamics DG°R = RT ln K DG°R = DH°R – TDS°R ln K = - DG°R / RT lnK = -(DH°R/T – DS°R)/ R dlnK/dT = DH°R / RT2 dln K = DH°R/R (dT/T2) dln K = DH°R/R (dT/T2)

TEMPERATURE DEPENDENCE OF THE EQUILIBRIUM CONSTANT Van’t Hoff Equation

EXAMPLE VAN’T HOFF PLOT

Application: Solubility of silica SiO2(am) + 2H2O(l)  H4SiO40 G°f(H4SiO40) = -312.66 kcal G°f(SiO2(am)) = -203.33 kcal nG°f(2H2O(l)) = 2(-56.687) kcal DG°R = -312.66 – (-203.33 + 2(-56.687) = 4.044 kcal K = 10-2.96 DH°R = +3.47 kcal, endothermic reaction Log KT = -2.96 – [3.47x103(cal) / (2.3025 x 1.987) x (1/T – 1/298.15) Log KT = -2.96 – 758.4(1/T – 0.00335)

Application: Solubility of silica Log KT = -2.96 – 758.4(1/T – 0.00335) With increasing T, the term (1/T – 0.00335) becomes negative, the term – 758.4(1/T – 0.00335) becomes positive and log KT > -2.96 If the enthalpy change is negative (exothermic), the last term will be negative with increasing T and log KT < log K(298.15) Endothermic reaction: K increases with T Exothermic reactions: K decreases with T

Application: Solubility of silica Endothermic reaction: K increases with T Exothermic reactions: K decreases with T Endothermic dissolution reaction = higher ratio of products versus reactants = increasing solubility with increasing T Exothermic dissolution reactions = lower ratio of products versus reactants = decreasing solubility with increasing T

Second Example: dissolution of feldspar 2KAlSi3O8 + 9H2O + 2H+  Al2Si2O5(OH)4 + 2K+ + 4H4SiO4 Equilibrium constant K at 25 and 100°C??? Compound G°f (kcal) H°f (kcal) KAlSi3O8 -894.9 -948.7 H2O -56.687 -68.315 Al2Si2O5(OH)4 -906.84 -983.5 K+ -67.70 -60.32 H4SiO4 -312.66 -348.3

DH °R = S (n H°f)products - S(n H°f)reactants= (-983.5 + 2(-60.32) + 4(-348.3)) – (2(-948.7) + 9(-68.315)) = -2497.34 – -2512.235 = 14.895 kcal Compound G°f (kcal) H°f (kcal) KAlSi3O8 -894.9 -948.7 H2O -56.687 -68.315 Al2Si2O5(OH)4 -906.84 -983.5 K+ -67.70 -60.32 H4SiO4 -312.66 -348.3

Log KT = -5.207 – 3275.36(-0.000674) Log KT = -5.207 – -2.208 DG°R = 7.103 kcal DH °R = 14.895 kcal K = 10-DG°R/1.364 = 10-5.207 Log K = -5.207 at 298.15 K Log K at 373.15K? R = 1.987cal/deg mol Log KT = -5.207 – 3275.36(-0.000674) Log KT = -5.207 – -2.208 Log KT = -3.002

DG°R = 7.103 kcal DH °R = 14.895 kcal K = 10-DG°R/1.364 = 10-5.207 Log K = -5.207 at 298.15 K Log K at 373.15K = -3.002 Endothermic dissolution reaction = higher ratio of products versus reactants at higher T= increasing solubility with increasing T