2. Lecture 8 Summary

3. Lecture 9: Gibbs Free Energy Reading: Zumdahl 10.7, 10.9 Outline –Defining the Gibbs Free Energy (  G) –Calculating  G –Pictorial Representation of  G

4. Defining  G Recall, the second law of thermodynamics:  S univ =  S total =  S system +  S surr Also recall:  S surr = -  H sys /T Then  S total =  S system +  S surr -  H sys /T  S total = -T  S system +  H sys

5. Defining  G (cont.) We then define:  G = -T  S total Then:  Or  G =  H - T  S  G = -T  S sys +  H sys  G = The Gibbs Free Energy w/ P const

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

7.  G and Spontaneous Processes (cont.) Three possibilities: –If  S univ > 0…..process is spontaneous –If  S univ < 0…..process is spontaneous in opposite direction. –If  S univ = 0….equilibrium In terms of  G: –If  G < 0…..process is spontaneous –If  G > 0…..process is spontaneous in opposite direction. –If  G = 0….equilibrium

8.  G and Spontaneous Processes (cont.) Note that  G is composite of both  H and  S   G =  H - T  S If  H 0….spontaneous at all T A reaction is spontaneous if  G < 0. Such that  If  H > 0 and  S < 0….not spontaneous at all T If  H < 0 and  S < 0….spontaneous at low T If  H > 0 and  S > 0….spontaneous at high T

9. Example At what T is the following reaction spontaneous? Br 2 (l) Br 2 (g) where  H° = 30.91 kJ/mol,  S° = 93.2 J/mol.K Ans:  G° =  H° - T  S°

10. Example (cont.) Try 298 K just to see:  G° =  H° - T  S°  G° =  kJ/mol - (298K)  J/mol.K   G° =  kJ/mol = 3.13 kJ/mol > 0 Not spontaneous at 298 K

11. Example (cont.) At what T then?  G° =  H° - T  S°  S   = (  kJ/mol)  J/mol.K  = 0 Just like our previous calculation  = 331.65 K

12. Calculating  G° In our previous example, we needed to determine  H° rxn and  S° rxn to determine  G° rxn Now,  G is a state function; therefore, we can use known  G° to determine  G° rxn using:

13. Standard  G of Formation:  G f ° Like  H f ° and S°,  G f ° 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  H f °,  G f ° = 0 for an element in its standard state: Example:  G f ° (O 2 (g)) = 0

14. Example Determine the  G° rxn for the following: C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) Tabulated  G° f from Appendix 4:  G° f (C 2 H 5 OH(l)) = -175 kJ/mol  G° f (C 2 H 4 (g)) = 68 kJ/mol  G° f (H 2 O (l)) = -237 kJ/mol

15. Example (cont.) Using these values: C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l)  G° rxn  G° f (C 2 H 5 OH(l)) -  G° f (C 2 H 4 (g))  G° f (H 2 O (l))  G° rxn  -175 kJ - 68 kJ -(-237 kJ)  G° rxn  -6 kJ < 0 ; therefore, spontaneous

16. More  G° Calculations Similar to  H°, one can use the  G° for various reactions to determine  G° for the reaction of interest (a “Hess’ Law” for  G°) Example: C(s, diamond) + O 2 (g) CO 2 (g)  G° = -397 kJ C(s, graphite) + O 2 (g) CO 2 (g)  G° = -394 kJ

17. More  G° Calculations (cont.) C(s, diamond) + O 2 (g) CO 2 (g)  G° = -397 kJ C(s, graphite) + O 2 (g) CO 2 (g)  G° = -394 kJ CO 2 (g) C(s, graphite) + O 2 (g)  G° = +394 kJ C(s, diamond) C(s, graphite)  G° = -3 kJ  G° rxn < 0…..rxn is spontaneous

18.  G° rxn ≠ Reaction Rate Although  G° 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) + O 2 (g) CO 2 (g)  G° rxn = -397 kJ But diamonds are forever…. <<0  G° rxn ≠ rate

19. Example Problem Is the following reaction spontaneous under standard conditions?  H° f (kJ/mol) S° (J/mol.K) KClO 3 (s)-397.7143.1 KClO 4 (s)-432.8151.0 KCl (s)-436.782.6

20. Example Problem Solution Calulating  H° rxn Calulating  S° rxn

21. Example Problem Solution Calulating  G° rxn  G°rxn < 0 ;therefore, reaction is spontaneous.

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