1. 1 Gibbs Free Energy, G 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 ∆S univ = ∆S surr + ∆S sys

2. 2 ∆G o = ∆H o - T∆S o Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is exothermic (negative ∆ H o ) (energy dispersed) exothermic (negative ∆ H o ) (energy dispersed) and entropy increases (positive ∆S o ) (matter dispersed)and entropy increases (positive ∆S o ) (matter dispersed) then ∆G o must be NEGATIVEthen ∆G o must be NEGATIVE reaction is spontaneous (and product- favored).

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

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

5. 5 Gibbs Free Energy, G ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o Two methods of calculating ∆G o a)Determine ∆H o rxn and ∆S o rxn and use GIbbs equation. b)Use tabulated values of free energies of formation, ∆G f o. ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

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

7. 7 Calculating ∆G o rxn Combustion of acetylene C 2 H 2 (g) + 5/2 O 2 (g) --> 2 CO 2 (g) + H 2 O(g) Use enthalpies of formation to calculate ∆H o rxn = -1238 kJ ∆H o rxn = -1238 kJ Use standard molar entropies to calculate ∆S o rxn = -97.4 J/K or -0.0974 kJ/K ∆S o rxn = -97.4 J/K or -0.0974 kJ/K ∆G o rxn = -1238 kJ - (298 K)(-0.0974 J/K) = -1209 kJ = -1209 kJ Reaction is product-favored in spite of negative ∆S o rxn. Reaction is “enthalpy driven”

8. 8 Calculating ∆G o rxn Is the dissolution of ammonium nitrate product- favored? If so, is it enthalpy- or entropy-driven? NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

9. 9 Calculating ∆G o rxn From tables of thermodynamic data we find ∆H o rxn = +25.7 kJ ∆H o rxn = +25.7 kJ ∆S o rxn = +108.7 J/K or +0.1087 kJ/K ∆G o rxn = +25.7 kJ - (298 K)(+0.1087 J/K) = -6.7 kJ = -6.7 kJ Reaction is product-favored in spite of negative ∆H o rxn. Reaction is “entropy driven” NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

10. 10 Gibbs Free Energy, G ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o Two methods of calculating ∆G o a)Determine ∆H o rxn and ∆S o rxn and use GIbbs equation. b)Use tabulated values of free energies of formation, ∆G f o. ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

11. 11 Calculating ∆G o rxn Combustion of carbon C(graphite) + O 2 (g) --> CO 2 (g) ∆G o rxn = ∆G f o (CO 2 ) - [∆G f o (graph) + ∆G f o (O 2 )] ∆G o rxn = -394.4 kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0. ∆G o rxn = -394.4 kJ Reaction is product-favored as expected. ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

12. 12 Free Energy and Temperature 2 Fe 2 O 3 (s) + 3 C(s) ---> 4 Fe(s) + 3 CO 2 (g) ∆H o rxn = +467.9 kJ ∆S o rxn = +560.3 J/K ∆G o rxn = +300.8 kJ Reaction is reactant-favored at 298 K At what T does ∆G o rxn just change from being (+) to being (-)? When ∆G o rxn = 0 = ∆H o rxn - T∆S o rxn

13. 13 More thermo? You betcha!

14. 14  FACT: ∆G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products.  FACT: Product-favored systems have K eq > 1.  Therefore, both ∆G˚ rxn and K eq are related to reaction favorability. Thermodynamics and K eq

15. 15 K eq is related to reaction favorability and so to ∆G o rxn. The larger the value of K the more negative the value of ∆G o rxn ∆G o rxn = - RT lnK ∆G o rxn = - RT lnK where R = 8.31 J/Kmol Thermodynamics and K eq

16. 16 Calculate K for the reaction N 2 O 4 --->2 NO 2 ∆G o rxn = +4.8 kJ ∆G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K ∆G o rxn = - RT lnK Thermodynamics and K eq K = 0.14 When ∆G o rxn > 0, then K 0, then K < 1

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

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

19. 19 Product favored reactionProduct favored reaction –∆G o and K > 1–∆G o and K > 1 In this case ∆G rxn is < ∆G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion.In this case ∆G rxn is < ∆G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion. ∆G, ∆G˚, and K eq

20. 20 Product-favored reaction. 2 NO 2 ---> N 2 O 4 ∆G o rxn = – 4.8 kJ Here ∆G rxn is less than ∆G o rxn, so the state with both reactants and products present is more stable than complete conversion. ∆G, ∆G˚, and K eq

21. 21 Reactant-favored reaction. N 2 O 4 --->2 NO 2 ∆G o rxn = +4.8 kJ Here ∆G o rxn is greater than ∆G rxn, so the state with both reactants and products present is more stable than complete conversion. ∆G, ∆G˚, and K eq

22. 22  K eq is related to reaction favorability.  When ∆G o rxn < 0, reaction moves energetically “downhill”  ∆G o rxn is the change in free energy when reactants convert COMPLETELY to products. Thermodynamics and K eq