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Volume Changes (Equation of State)

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

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Volume Changes (Equation of State)

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

  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, aRw and bRK are constants

  3. Hess’s Law Known values of DH for reactions can be used to determine DH’s for other reactions. DH 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 DH of the reaction will be the same regardless of the details of the process (single vs multi- step).

  4. CH4(g) + O2(g) --> CO2(g) + 2H2O(l) DH = -890 kJ Net equation CH4(g) + O2(g) --> CO2(g) + 2H2O(l) DH = -890 kJ If the same reaction was carried out in two steps: CH4(g) + O2(g) --> CO2(g) + 2H2O(g) DH = -802 kJ 2H2O(g) --> 2H2O(l) DH = -88 kJ Hess’s law : if a reaction is carried out in a series of steps, DH 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 • Heat of reaction DH0R • DH0R is positive  exothermic • DH0R is negative  endothermic • Example: 2A + 3B  A2B3 • DH0R =H0f(A2B3)-[2H0f(A) + 3H0f(B)]

  7. Entropy of reaction • Just as was done with enthalpies: • Entropy of reaction S0R: • When DS0R is positive  entropy increases as a result of a change in state • When DS0R 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 DG0R = DH0R – TDS0R • Gibbs Free Energy (G) is a state variable, measured in KJ/mol • Tabulated values of DG0R are in Appendix

  9. G is a measure of driving force • DG0R = DH0R – TDS0R • When DG0R is negative  forward reaction has excess energy and will occur spontaneously • When DG0R is positive  there is not enough energy in the forward direction, and the BACKWARD reaction will occur • When DG0R is ZERO  reaction is AT equilibrium

  10. Free Energy Examples DG0R = DH0R – TDS0R H2O(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

  11. Now, how does free energy change with T and P? • From DG=DH-TDS:

  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 DGR = DSRdT + DVRdP, at equilibrium, DG=0, rearranging: Clausius-Clapeyron equation

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

  14. 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:

  15. Clausius-Clapyron Example

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

  17. 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

  18. 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

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