Volume Changes Equation of State For Minerals Volume is related to energy changes dG V dP T Mineral volume changes as a function of T coefficient of thermal expansion 1 V V T P Mineral volume changes as a function of P coefficient of isothermal expansion 1 V V P T 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 a Rw RT P 1 2 V bRK T V V bRK V bar denotes a molar quatity aRw and bRK are constants 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 multistep CH4 g O2 g CO2 g 2H2O l H 890 kJ If the same reaction was carried out in two steps CH4 g O2 g CO2 g 2H2O g H 802 kJ H 88 kJ 2H2O g 2H2O l CH4 g O2 g CO2 g 2H2O 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 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 Heat of Reaction Heat of reaction H0R H R0 ni H 0fi products i 0 n H i fi reactants i H0R is positive exothermic H0R is negative endothermic Example 2A 3B A2B3 H0R H0f A2B3 2H0f A 3H0f B Entropy of reaction Just as was done with enthalpies Entropy of reaction S0R S ni S products 0 R 0 i i n S i 0 i reactants i When S0R is positive entropy increases as a result of a change in state When S0R is negative entropy decreases as a result of a change in state 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 G0R H0R T S0R Gibbs Free Energy G is a state variable measured in KJ mol G ni G products 0 R 0 i i nG i 0 i reactants i Tabulated values of G0R are in Appendix G is a measure of driving force G0R H0R T S0R When G0R is negative forward reaction has excess energy and will occur spontaneously When G0R is positive there is not enough energy in the forward direction and the BACKWARD reaction will occur When G0R is ZERO reaction is AT equilibrium Free Energy Examples G0R H0R T S0R GR0 ni Gi0 products i nG i 0 i i H2O l 63 32 kcal mol NIST value http webbook nist gov chemistry Fe2 O2 H Fe3 H2O 4120 63320 0 5 21870 3954 0 25 67440 19893 47547 cal mol reactants Now how does free energy change with T and P From G H T S T2 T2 T1 T1 C P P1 GP2 T2 GP1 T1 S P1 T 1 T2 T1 C P P1 dT T 2 T P2 dT VT2 dP P1 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 GR SRdT VRdP at equilibrium G rearranging S R P T G 0 VR Clausius Clapeyron equation S R P T G 0 VR SR change with T or P C P S R T T P S V R R VR T T P V V 1 P P1 P1 S S P S 0 dP S 0 VdP P T P2 P2 S 0 V 0 P P22 P12 2 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 P T G 0 VR Example Diamond graphite To get C from graphite to diamond at 25 C requires 1600 MPa of pressure let s calculate what P it requires at 1000 C graphite diamond K 1 MPa 1 S J mol K 1 05E 05 7 50E 06 3 08E 05 2 27E 06 5 74 2 38 V cm3 mol 5 2982 3 417 Clausius Clapyron Example Phase diagram Need to represent how mineral reactions at equilibrium vary with P and T S R VR V T P R S R P T G 0 VR C P S R T T P 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 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