BASIC THERMODYNAMIC PRINCIPLES

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

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.

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 = 298.15 K = 25 ° C 0.987 atm = 1 bar = 101.35 kPa Standard enthalpies of H2 and O2 are by definition zero. DH °R= the enthalpy change of a chemical reaction in the standard state

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

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 + 3 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 = 4.184 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

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

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 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 + 3 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 = 4.184 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

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

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

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

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

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 = -23.262 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).

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 + 3 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

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