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# Lecture 8 Summary

Lecture 8 Summary. Spontaneous, exothermic. Spontaneous, exothermic. Spontaneous, endothermic. Lecture 9: Gibbs Free Energy G. Reading: Zumdahl 10.7, 10.9 Outline Defining the Gibbs Free Energy ( D G) Calculating D G (several ways to) Pictorial Representation of D G.

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## Lecture 8 Summary

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1. Lecture 8 Summary Spontaneous, exothermic Spontaneous, exothermic Spontaneous, endothermic

2. Lecture 9: Gibbs Free Energy G Reading: Zumdahl 10.7, 10.9 Outline Defining the Gibbs Free Energy (DG) Calculating DG (several ways to) Pictorial Representation of DG

3. Defining DG Recall, the second law of thermodynamics: DSuniv = DStotal = DSsys + DSsurr Also recall: DSsurr = -DHsys/T Substituting,  DStotal = DSsys + DSsurr -DHsys/T Multipling all by (-T) gives -TDStotal = -TDSsys+DHsys

4. We then define: DG = -TDStotal Substituting DG = -TDSsys+DHsys Giving finally  DG = DH - TDS DG = The Gibbs Free Energy @const P

5. DG and Spontaneous Processes Three possibilities: If DSuniv > 0…..process is spontaneous If DSuniv < 0…..process is spontaneous in opposite direction. If DSuniv = 0….equilibrium Recall from the second law the conditions of spontaneity: In our derivation of DG, we divided by -T; therefore, the direction of the inequality relative to entropy is now reversed.

6. Three possibilities in terms of DS: If DSuniv > 0…..process is spontaneous. If DSuniv < 0…..process is spontaneous in opposite direction. If DSuniv = 0…. system is in equilibrium. Three possibilities in terms of DG: If DG < 0…..process is spontaneous. If DG > 0…..process is spontaneous in opposite direction. If DG = 0….system is in equilibrium

7. Spontaneous Processes: temperature dependence Note that DG is composed of both DH and DS terms  DG = DH - TDS A reaction is spontaneous if DG < 0. So, If DH < 0 and DS > 0….spontaneous at all T If DH > 0 and DS < 0….not spontaneous at any T If DH < 0 and DS < 0…. becomes spontaneous at low T If DH > 0 and DS > 0….becomes spontaneous at high T

8. Example: At what T is the following reaction spontaneous? Br2(l) Br2(g) where DH° = 30.91 kJ/mol, DS° = 93.2 J/mol.K DG° = DH° - TDS°

9. Try 298 K just to see result at standard conditions DG° = DH° - TDS° DG° = 30.91 kJ/mol - (298K)(93.2 J/mol.K) DG° = (30.91 - 27.78) kJ/mol = 3.13 kJ/mol > 0 Not spontaneous at 298 K

10. At what T then does the process become spontaneous? DG° = DH° - TDS° = 0 T = DH°/DS° T = (30.91 kJ/mol) /(93.2 J/mol.K) T = 331.65 K Just like our previous calculation

11. Calculating DG° In our previous example, we needed to determine DH°rxn and DS°rxn separately to determine DG°rxn But ∆ G is a state function; therefore, we can use known DG° to determine DG°rxn using:

12. Standard DG of Formation: DGf° Like DHf° and S°, the standard Gibbs free energy of formation DGf° is defined as the “change in free energy that accompanies the formation of 1 mole of that substance for its constituent elements with all reactants and products in their standard state.” As for DHf° , DGf° = 0 for an element in its standard state: Example: DGf° (O2(g)) = 0

13. Example • Determine the DG°rxn for the following: C2H4(g) + H2O(l) C2H5OH(l) • Tabulated DG°f from Appendix 4: DG°f(C2H5OH(l)) = -175 kJ/mol DG°f(C2H4(g)) = 68 kJ/mol DG°f(H2O (l)) = -237 kJ/mol

14. Using these values: C2H4(g) + H2O(l) C2H5OH(l) DG°rxn = DG°f(C2H5OH(l)) - DG°f(C2H4(g)) -DG°f(H2O (l)) DG°rxn = -175 kJ - 68 kJ -(-237 kJ) DG°rxn = -6 kJ < 0 ; therefore, spontaneous

15. More DG° Calculations • Similar to DH°, one can use the DG° for various reactions to determine DG° for the reaction of interest (a “Hess’ Law” for DG°) • Example: C(s, diamond) + O2(g) CO2(g) DG° = -397 kJ C(s, graphite) + O2(g) CO2(g) DG° = -394 kJ

16. C(s, diamond) + O2(g) CO2(g) DG° = -397 kJ C(s, graphite) + O2(g) CO2(g) DG° = -394 kJ CO2(g) C(s, graphite) + O2(g) DG° = +394 kJ C(s, diamond) C(s, graphite) DG° = -3 kJ DG°rxn < 0…..rxn is spontaneous

17. DG°rxn ≠ Reaction Rate • Although DG°rxn can be used to predict if a reaction will be spontaneous as written, it does not tell us how fast a reaction will proceed. • Example: C(s, diamond) + O2(g) CO2(g) DG°rxn = -397 kJ <<0 But diamonds are forever…. DG°rxn ≠ rate

18. Example Problem • Is the following reaction spontaneous under standard conditions?

19. Example Problem Solution • Calulating DH°rxn • Calulating DS°rxn

20. Example Problem Solution • Calulating DG°rxn DG°rxn < 0 ;therefore, reaction is spontaneous.

21. Example Problem Continued For what temperatures will this reaction be spontaneous? Answer: For T in which DGrxn < 0. Spontaneous as long as T < 3446 K.

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