1. BASIC THERMODYNAMIC PRINCIPLES

2. Thermodynamics Internal Energy (E) of a mineral = Sum of energy stored
Mineral structure = System
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 the First Law of Thermodynamics

3. Enthalpy
dE = dQ – pdV
The amount of heat added Q to the system at constant pressure = enthalpy H = E + PV
The change of the enthalpy of a system
ΔH = (E2 + PV2) – (E1 + PV1)
of a reversible process is equal to the heat absorbed by the system during that change in state. The amount of heat absorbed or released from the system during a chemical reaction can be measured in a calorimeter.

4. Enthalpy
H°f = the enthalpy of formation of compounds and their ions and molecules in aqueous solution is the heat absorbed or given off by chemical reactions in which the compounds, ions and molecules form from the elements in the standard state.
Standard state = K = 25 ° C
0.987 atm = 1 bar = kPa
Standard enthalpies of H2 and O2 are by definition zero.
DH °R= the enthalpy change of a chemical reaction in the standard state

5. Thermodynamics Reaction enthalpy DHR = 0 = equilibrium
DH °R cannot in general be zero because it refers to a reference state in which equilibrium does not occur.
DH °R> 0 = endothermic, heat flows from the surroundings to the system
DH °R< 0 = exothermic, heat flows from the system
to the surroundings
The side of the reaction with lower G will be more stable
How do we go about determining this for a reaction?
First we must be able to determine G for the phases in the reaction at any P and T

6. Thermodynamics
a Reaction: some change in the nature or types of phases in a system
reactions are written in the form:
reactants products
The reaction enthalpy DHr of a reaction of the type:
2 A B C + 4 D
is: DH °R = S (n H°f)products - S(n H°f)reactants
= H°f C + 4 H°f D - 2 H°f A - 3 H°f B
Units: calories or joules per mol
1 kcal mol-1 = kJ mol-1
The side of the reaction with lower G will be more stable
How do we go about determining this for a reaction?
First we must be able to determine G for the phases in the reaction at any P and T

7. Thermodynamics Entropy (S): Degree of disorder
High entropy in a solid = strong vibration or diffusion
of atoms, high disorder of atoms in the structure
Entropy increase: solid < melt < liquid < vapor
Increase of entropy in water: adding salt to the solution

8. 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)

9. Thermodynamics
a Reaction: some change in the nature or types of phases in a system
reactions are written in the form:
reactants products
The reaction enthalpy DSr of a reaction of the type:
2 A B C + 4 D
is: DS°R = S (n S°f)products - S(n S°f)reactants
= S°f C + 4 S°f D - 2 S°f A - 3 S°f B
Units: calories or joules per degree
1 kcal degree-1 = kJ degree-1
The side of the reaction with lower G will be more stable
How do we go about determining this for a reaction?
First we must be able to determine G for the phases in the reaction at any P and T

10. Gibbs Free Energy DG °R = DH °R – TDS°R
Second law of thermodynamics implies:
Increase in enthalpy in the system is diminished because a certain amount of the enthalpy is consumed by an increase in the entropy of the system.
Change of the free energy during a chemical reaction DG °R = difference between heat flow at constant P and entropy change at constant T
G = H – TS =
DG °R = DH °R – TDS°R

11. Gibbs Free Energy
Gibbs free energy (G) is a measure of the chemical energy of a mineral, melt, liquid ..
Gibbs free energy of a chemical reaction is the excess of energy to drive the chemical reaction
All chemical systems tend naturally toward states of minimum Gibbs free energy

12. Energy States Unstable: falling or rolling high free energy
Stable: at rest in lowest energy state, lowest possible free energy
Metastable: in low- energy perch with lower free energy than energy barrier

13. 1000 2000 3000 20 40 60 80 Diamond Graphite T e mperature (K)
Pressure (kbars)

15. Free Energy of Formation
The free energy of formation ΔGºf is defined as zero for the elements
and for the hydrogen ion H+.
The free energy of formation of all other material is the free energy of
formation of the material from the elements in their standard states.
For example, consider:
Pb + S = PbS, ΔGºf = kcal/mole
The free energy of formation can be evaluated from the expression for
free energy, ΔGºf = ΔH – TΔS
where ΔH is the enthalpy of formation of PbS
ΔS is the entropy of formation of PbS minus the sum of the
entropies of formation of Pb and S
T is the absolute temperature in Kelvins
The usual standard state is 298K and 1 atm (101.35kPa).

16. Thermodynamics
a Reaction: some change in the nature or types of phases in a system
reactions are written in the form:
reactants products
The free energy DG°R of a reaction of the type:
2 A B C + 4 D
is: DG°R = S (n G°f)products - S(n G °f)reactants
= G °f C + 4G °f D - 2G °f A - 3G °f B
The side of the reaction with lower G will be more stable
How do we go about determining this for a reaction?
First we must be able to determine G for the phases in the reaction at any P and T

17. Thermodynamics 2 A + 3 B C + 4 D
DG° = S (n G°f)products - S(nG°f)reactants
= GfC + 4GfD - 2GfA - 3GRB
If DGR = negative, S(n Gf)reactants > S (n Gf)products
Reaction from left to right
If DG°R = positive, S(n G°f)reactants < S (n G°f)products
The side of the reaction with lower G will be more stable
How do we go about determining this for a reaction?
First we must be able to determine G for the phases in the reaction at any P and T
Reaction from right to left
If DG°R = 0, S(n G°f)reactants = S (n G°f)products
Equilibrium