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Gibbs Free Energy, G

Gibbs Free Energy, G. ∆S univ = ∆S surr + ∆S sys. Multiply through by -T -T∆S univ = ∆H sys - T∆S sys -T∆S univ = change in Gibbs free energy for the system = ∆G system Under standard conditions — ∆G o sys = ∆H o sys - T∆S o sys. ∆G o = ∆H o - T∆S o.

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Gibbs Free Energy, G

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  1. Gibbs Free Energy, G ∆Suniv = ∆Ssurr + ∆Ssys Multiply through by -T -T∆Suniv = ∆Hsys - T∆Ssys -T∆Suniv = change in Gibbs free energy for the system = ∆Gsystem Under standard conditions — ∆Gosys = ∆Hosys - T∆Sosys

  2. ∆Go = ∆Ho - T∆So Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is • exothermic (negative ∆ Ho) (energy dispersed) • and entropy increases (positive ∆So) (matter dispersed) • then ∆Gomust be NEGATIVE • reaction is spontaneous (and product-favored).

  3. ∆Go = ∆Ho - T∆So Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is • endothermic (positive ∆Ho) • and entropy decreases (negative ∆So) • then ∆Go must be POSITIVE • reaction is not spontaneous (and is reactant-favored).

  4. Gibbs Free Energy, G ∆Go = ∆Ho - T∆So ∆Ho ∆So ∆Go Reaction exo(–) increase(+) – Prod-favored endo(+) decrease(-) + React-favored exo(–) decrease(-) ? T dependent endo(+) increase(+) ? T dependent

  5. Gibbs Free Energy, G ∆Go = ∆Ho - T∆So Two methods of calculating ∆Go a) Determine ∆Horxn and ∆Sorxn and use GIbbs equation. b) Use tabulated values of free energies of formation, ∆Gfo. ∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)

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

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

  9. Calculating ∆Gorxn 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 J/K) = -6.7 kJ Reaction is product-favored in spite of negative ∆Horxn. Reaction is “entropy driven”

  10. Gibbs Free Energy, G ∆Go = ∆Ho - T∆So Two methods of calculating ∆Go a) Determine ∆Horxn and ∆Sorxn and use GIbbs equation. b) Use tabulated values of free energies of formation, ∆Gfo. ∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)

  11. Calculating ∆Gorxn ∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants) Combustion of carbon C(graphite) + O2(g) --> CO2(g) ∆Gorxn = ∆Gfo(CO2) - [∆Gfo(graph) + ∆Gfo(O2)] ∆Gorxn = -394.4 kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0. ∆Gorxn = -394.4 kJ Reaction is product-favored as expected.

  12. Free Energy and Temperature 2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g) ∆Horxn = +467.9 kJ∆Sorxn = +560.3 J/K ∆Gorxn = +300.8 kJ Reaction is reactant-favored at 298 K At what T does ∆Gorxn just change from being (+) to being (-)? When ∆Gorxn = 0 = ∆Horxn - T∆Sorxn

  13. More thermo? You betcha!

  14. Thermodynamics and Keq • FACT: ∆Gorxn is the change in free energy when pure reactants convert COMPLETELY to pure products. • FACT: 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. The larger the value of K the more negative the value of ∆Gorxn ∆Gorxn = - RT lnK where R = 8.31 J/K•mol

  16. Thermodynamics and Keq ∆Gorxn = - RT lnK Calculate K for the reaction N2O4 --->2 NO2 ∆Gorxn = +4.8 kJ ∆Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K K = 0.14 When ∆Gorxn > 0, then K < 1

  17. ∆G is change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient When Q < K or Q > K, reaction is spontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K ∆G, ∆G˚, and Keq

  18. ∆G, ∆G˚, and Keq But systems can reach equilibrium when reactants have NOT converted completely to products. In this case ∆Grxn is < ∆Gorxn , so state with both reactants and products present is MORE STABLE than complete conversion. Figure 19.10

  19. ∆G, ∆G˚, and Keq • Product favored reaction • –∆Go and K > 1 • In this case ∆Grxn is < ∆Gorxn , so state with both reactants and products present is MORE STABLE than complete conversion.

  20. ∆G, ∆G˚, and Keq Product-favored reaction. 2 NO2 ---> N2O4 ∆Gorxn = – 4.8 kJ Here ∆Grxn is less than ∆Gorxn , so the state with both reactants and products present is more stable than complete conversion.

  21. ∆G, ∆G˚, and Keq Reactant-favored reaction. N2O4 --->2 NO2 ∆Gorxn = +4.8 kJ Here ∆Gorxn is greater than ∆Grxn , so the state with both reactants and products present is more stable than complete conversion.

  22. Thermodynamics and Keq • Keq is related to reaction favorability. • When ∆Gorxn < 0, reaction moves energetically “downhill” • ∆Gorxn is the change in free energy when reactants convert COMPLETELY to products.

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