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Lecture 8 Summary

Lecture 8 Summary. Lecture 9: Gibbs Free Energy. Reading: Zumdahl 10.7, 10.9 Outline Defining the Gibbs Free Energy ( D G) Calculating D G Pictorial Representation of D G. Defining D G. Recall, the second law of thermodynamics: D S univ = D S total = D S system + D S surr.

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Lecture 8 Summary

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  1. Lecture 8 Summary

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

  3. Defining DG • Recall, the second law of thermodynamics: DSuniv = DStotal = DSsystem + DSsurr • Also recall: DSsurr = -DHsys/T • Then, DStotal = DSsystem + DSsurr -DHsys/T -TDStotal = -TDSsystem+DHsys

  4. Defining DG (cont.) • We then define: DG = -TDStotal • Then: DG = -TDSsys+DHsys • Or: DG = DH - TDS DG = The Gibbs Free Energy w/ Pconst

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

  6. DG and Spontaneous Processes (cont.) • Three possibilities: • If DSuniv > 0…..process is spontaneous • If DSuniv < 0…..process is spontaneous in opposite direction. • If DSuniv = 0….equilibrium • In terms of DG: • If DG < 0…..process is spontaneous • If DG > 0…..process is spontaneous in opposite direction. • If DG = 0….equilibrium

  7. DG and Spontaneous Processes (cont.) • Note that DG is composite of both DH and DS DG = DH - TDS • A reaction is spontaneous if DG < 0. Such that: If DH < 0 and DS > 0….spontaneous at all T If DH > 0 and DS < 0….not spontaneous at all T If DH < 0 and DS < 0….spontaneous at low T If DH > 0 and DS > 0….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 • Ans: DG° = DH° - TDS°

  9. Example (cont.) • Try 298 K just to see: 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. Example (cont.) • At what T then? 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 to determine DG°rxn • Now, DG 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°, 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.” • Like 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. Example (cont.) • 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. More DG° Calculations (cont.) 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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