Volume Changes (Equation of State)Slide 2Slide 3Slide 4Reference StatesHeat of ReactionEntropy of reactionJ. Willard GibbsG is a measure of driving forceFree Energy ExamplesSlide 11Phase RelationsSlide 13Slide 14Example – Diamond-graphiteClausius-Clapyron ExampleSlide 17Phase diagramGibbs Phase RuleVariance and fVolume Changes (Equation of State)VdPdGTPTVV1TPVV1Volume is related to energy changes:Mineral volume changes as a function of T: , coefficient of thermal expansionMineral volume changes as a function of P: , coefficient of isothermal expansionFor Minerals: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, aRw and bRK are constants)(2/1RKRwRKbVVTabVRTPHess’s LawKnown 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).CH4(g) + O2(g) --> CO2(g) + 2H2O(l) H = -890 kJIf the same reaction was carried out in two steps:CH4(g) + O2(g) --> CO2(g) + 2H2O(g) H = -802 kJ2H2O(g) --> 2H2O(l) H = -88 kJCH4(g) + O2(g) --> CO2(g) + 2H2O(l) H = -890 kJNet equationHess’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 H0R• H0R is positive  exothermic• H0R is negative  endothermic•Example: 2A + 3B  A2B3• H0R =H0f(A2B3)-[2H0f(A) + 3H0f(B)])()(000reactantsHnproductsHnHif iifiiiRHeat of ReactionEntropy of reaction•Just as was done with enthalpies:•Entropy of reaction S0R:•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)()(000reactantsSnproductsSnSiiiiiiR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•Tabulated values of G0R are in Appendix)reactants()(000iiiiiiRGnproductsGnG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 equilibriumFree Energy ExamplesG0R = H0R – TS0RH2O(l)=-63.32 kcal/mol (NIST value: http://webbook.nist.gov/chemistry/)•Fe2+ + ¼ O2 + H+  Fe3+ + ½ H2O=[-4120+(-63320*0.5)]-[-21870+(3954*0.25)] =[-67440]-[-19893]=-47547 cal/mol)reactants()(000iiiiiiRGnproductsGnG•Now, how does free energy change with T and P?•From G=H-TS: 21212)1(21)1(11222)(121,1,,TTPPTPTTPTPTPTPdPVdTTCTdTCTTSGGPPPhase 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:RRGVSTP 0Clausius-Clapeyron equation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 –)RRGVSTP 0PRTRRTVVSTCTSPPRSR change with T or P?V = Vº(1-P)21220000(21212PPPVSVdPSdPPSSSPPTPPPRRGVSTP 0Example – 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:graphite diamond(K-1)1.05E-05 7.50E-06(MPa-1)3.08E-05 2.27E-06Sº(J/mol K)5.74 2.38Vº(cm3/mol)5.2982 3.417Clausius-Clapyron ExamplePhase diagram•Need to represent how mineral reactions at equilibrium vary with P and TRRGVSTP 0PRRRTVVSTCTSPPR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 systemVariance 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 =