1. Volume Changes (Equation of State) Volume is related to energy changes: Mineral volume changes as a function of T: , coefficient of thermal expansion Mineral volume changes as a function of P: , coefficient of isothermal expansion For Minerals:

2. Volume Changes (Equation of State) Gases and liquids undergo significant volume changes with T and P changes Number of empirically based EOS solns.. For metamorphic environments: –Redlich and Kwong equation: V-bar denotes a molar quatity, a Rw and b RK are constants

3. Hess’s Law Known values of  H for reactions can be used to determine  H’s for other reactions.  H is a state function, and hence depends only on the amount of matter undergoing a change and on the initial state of the reactants and final state of the products. If a reaction can be carried out in a single step or multiple steps, the  H of the reaction will be the same regardless of the details of the process (single vs multi- step).

4. CH 4 (g) + O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ If the same reaction was carried out in two steps: CH 4 (g) + O 2 (g) --> CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g) --> 2H 2 O(l)  H = -88 kJ CH 4 (g) + O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ Net equation Hess’s law : if a reaction is carried out in a series of steps,  H for the reaction will be equal to the sum of the enthalpy change for the individual steps.

5. Reference States We recall that we do not know absolute energies!!! We can describe any reaction or description of reaction relative to another  this is all we need to describe equilibrium and predict reaction direction, just need an anchor… Reference States: –Standard state: 1 atm pressure, 25°C –Absolute states – where can a value be defined?  entropy at 0 Kelvin

6. Heat of reaction  H 0 R  H 0 R is positive  exothermic  H 0 R is negative  endothermic Example: 2A + 3B  A 2 B 3  H 0 R =H 0 f (A 2 B 3 )-[2H 0 f (A) + 3H 0 f (B)] Heat of Reaction

7. Entropy of reaction Just as was done with enthalpies: Entropy of reaction S 0 R : When  S 0 R is positive  entropy increases as a result of a change in state When  S 0 R is negative  entropy decreases as a result of a change in state

8. J. Willard Gibbs Gibbs realized that for a reaction, a certain amount of energy goes to an increase in entropy of a system. G = H –TS or  G 0 R =  H 0 R – T  S 0 R Gibbs Free Energy (G) is a state variable, measured in KJ/mol Tabulated values of  G 0 R are in Appendix

9. G is a measure of driving force  G 0 R =  H 0 R – T  S 0 R When  G 0 R is negative  forward reaction has excess energy and will occur spontaneously When  G 0 R is positive  there is not enough energy in the forward direction, and the BACKWARD reaction will occur When  G 0 R is ZERO  reaction is AT equilibrium

10. Free Energy Examples  G 0 R =  H 0 R – T  S 0 R H 2 O (l) =-63.32 kcal/mol (NIST value: http://webbook.nist.gov/chemistry/) Fe 2+ + ¼ O 2 + H +  Fe 3+ + ½ H 2 O =[-4120+(-63320*0.5)]-[-21870+(3954*0.25)] =[-67440]-[-19893]=-47547 cal/mol

11. Now, how does free energy change with T and P? From  G=  H-T  S:

12. Phase Relations Rule: At equilibrium, reactants and products have the same Gibbs Energy –For 2+ things at equilibrium, can investigate the P-T relationships  different minerals change with T-P differently… For  G R =  S R dT +  V R dP, at equilibrium,  G  rearranging: Clausius-Clapeyron equation

13.  V for solids stays nearly constant as P, T change,  V for liquids and gases DOES NOT Solid-solid reactions linear  S and V nearly constant,  S/  V constant  + slope in diagram For metamorphic reactions involving liquids or gases, volume changes are significant,  V terms large and a function of T and P (and often complex functions) – slope is not linear and can change sign (change slope + to –)  S R change with T or P? V = Vº(1-  P)

15. Example – Diamond-graphite To get C from graphite to diamond at 25ºC requires 1600 MPa of pressure, let’s calculate what P it requires at 1000ºC: graphitediamond  (K -1 ) 1.05E-057.50E-06  (MPa -1 ) 3.08E-052.27E-06 Sº (J/mol K) 5.742.38 Vº (cm3/mol) 5.29823.417

16. Clausius-Clapyron Example

18. Phase diagram Need to represent how mineral reactions at equilibrium vary with P and T

19. Gibbs Phase Rule The number of variables which are required to describe the state of a system: p+f=c+2 f=c-p+2 –Where p=# of phases, c= # of components, f= degrees of freedom –The degrees of freedom correspond to the number of intensive variables that can be changed without changing the number of phases in the system

20. Variance and f f=c-p+2 Consider a one component (unary) diagram If considering presence of 1 phase (the liquid, solid, OR gas) it is divariant 2 phases = univariant 3 phases = invariant