1 / 16

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.

thina
Télécharger la présentation

CHAPTER

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related