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CHAPTER

CHAPTER. 19. Gibbs Free Energy , ∆ G. Under standard conditions — ∆ G o sys = ∆ H o sys - T∆S o sys. free energy = total energy change for system - energy change in disordering the system. ∆G o = ∆H o - T ∆S o.

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CHAPTER

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  1. CHAPTER 19

  2. Gibbs Free Energy, ∆G Under standard conditions — ∆Gosys = ∆Hosys - T∆Sosys free energy = total energy change for system - energy change in disordering the system

  3. ∆Go = ∆Ho- T∆So If reaction is exothermic (negative ∆ Ho) (energy dispersed) and entropy increases (positive ∆So) (matter dispersed) then∆Gomust beNEGATIVE reaction is spontaneous (and product-favored).

  4. ∆Go = ∆Ho- T∆So If reaction is endothermic (positive ∆Ho) and entropy decreases (negative ∆So) then ∆Gomust be POSITIVE reaction is NOT spontaneous (and is reactant-favored).

  5. ENTHALPY and ENTROPY

  6. 1st method:calculating ∆Go use Gibb’s equation ∆Go = ∆Ho - T∆So Determine ∆Horxn and ∆Sorxn

  7. Combustion of acetylene C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g) Use enthalpies of formation to calculate ∆Horxn= -1238 kJ Use standard molar entropies to calculate ∆Sorxn= -97.4 J/K or -0.0974 kJ/K ∆Gorxn= -1238 kJ - (298 K)(-0.0974 J/K) = -1209 Kj Reaction is product-favored in spite of negative ∆Sorxn. Reaction is “enthalpy driven”

  8. NH4NO3(s) + heat ---> NH4NO3(aq) Is the dissolution of ammonium nitrate product-favored? If so, is it enthalpy- or entropy-driven?

  9. NH4NO3(s) + heat ---> NH4NO3(aq) From tables of thermodynamic data we find ∆Horxn = +25.7 kJ ∆Sorxn = +108.7 J/K or +0.1087 kJ/K ∆Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) = -6.7 kJ Reaction is product-favored in spite of positive ∆Horxn. Reaction is “entropy driven”

  10. 2nd method of calculating ∆Go Use tabulated values of ∆Gfo, free energies of formation. ∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)

  11. Free Energies of Formation Note that ∆G˚f for an element = 0

  12. C(graphite) + O2(g) --> CO2(g) free energy of a standard state element is 0. ∆Gorxn =∆Gfo(CO2) - [∆Gfo(graph) + ∆Gfo(O2)] ∆Gorxn = -394.4 kJ - [ 0 + 0] ∆Gorxn = -394.4 kJ Reaction is product-favored as expected. ∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)

  13. More thermo? You betcha!

  14. Keqand Thermodynamics ∆Gorxn is the change in free energy when pure reactants convert COMPLETELY to pure products. Product-favored systems have Keq > 1. Therefore, both ∆G˚rxn and Keq are related to reaction favorability.

  15. Thermodynamics and Keq Keq is related to reaction favorability and so to ∆Gorxn. ∆Gorxn = - RT lnK where R = 8.31 J/K•mol The larger the value of K the more negative the value of ∆Gorxn

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