1. CHM 102 Sinex Thermodynamics Part 2

2. CHM 102 Sinex Free Energy  G - Gibbs free energy - maximum possible useful work at constant T and P  G =  H – T  S Our predictive tool for reactions  G > 0non-spontaneous  G = 0 at equilibrium  G < 0 spontaneous For a spontaneous process: maximize entropy minimize energy

3. CHM 102 Sinex Standard molar free energy of formation -  G o f  G o f = 0 for free elements in standard state CaCO 3 (s)  CaO (s) + CO 2 (g) Calculate free energy change for a reaction  G o f : -1129 kJ/mole-604 kJ/mole-394 kJ/mole Is this a spontaneous reaction?

4. CHM 102 Sinex  G =  H – T  S CASERESULT- the sign of  G  S > 0  H < 0 Spontaneous at ALL temperatures  S > 0  H > 0 Spontaneous at high temperatures  S < 0  H < 0 Spontaneous at low temperatures  S < 0  H > 0 Non-spontaneous at ANY temperature

5. CHM 102 Sinex Is this reaction possible? CH 4 (g)  C (diamond) + 2H 2 (g) @298K CH 4 diamondH2H2  H o f (kJ/mole) -7520 S o (J/K-mole) 1862131  G o f ( kJ/mole) -5130 Calculate  H o rxn,  S o rxn, and  G o rxn  G =  H-T  S

6. CHM 102 Sinex Can it rain diamonds on Neptune and Uranus? Neptune/Uranus blue color – CH 4 10-50GPa 2000-3000K Can the reaction become spontaneous?  G = 0 =  H - T  S T eq =  H/  S(estimate)

7. CHM 102 Sinex Hess’ Law C (graphite) + O 2 (g)  CO 2 (g)  H 1 C (graphite) + ½ O 2 (g)  CO (g)  H 2 CO (g) + ½ O 2 (g)  CO 2 (g)  H 3 Can't run this reaction, you get CO and CO 2 Reactions are additive!

8. CHM 102 Sinex C (graphite) + O 2 (g)  CO 2 (g)  H 1 CO (g) + ½ O 2 (g)  CO 2 (g)  H 3 C (graphite) + O 2 (g)  CO 2 (g)  H 1 CO 2 (g)  CO (g) + ½ O 2 (g) -  H 3 C (graphite) + ½ O 2 (g)  CO (g)  H 2 =  H 1 -  H 3

9. CHM 102 Sinex Biological pathways Glucose + HPO 4 -2 + H +  [glucose-6-phosphate] - + H 2 O Important in the metabolism of glucose  G o’ = 13.8 kJ Glucose + 6O 2  CO 2 + H 2 O First step in glycolytic pathway: Is this reaction spontaneous?

10. CHM 102 Sinex Coupled reactions Glucose + HPO 4 -2 + H +  [glucose-6-phosphate] - + H 2 O  G o’ = 13.8 kJ ATP -4 + H 2 O  ADP -3 + HPO 4 -2 + H +  G o’ = -30.5 kJ Glucose + ATP -4  [glucose-6-phosphate] - + ADP -3  G o’ = 13.8 + -30.5 = -16.7 kJ

11. CHM 102 Sinex Free energy and the equilibrium constant  G o = -2.303RTlog K T - kelvin temperature R - gas constant – 8.314 J/K-mole  G o > 0K < 1 reactant favored  G o = 0K = 1  G o 1 product favored

12. CHM 102 Sinex Diamonds are NOT forever! C diamond  C graphite  G o f : 3 kJ/mole0  G o rxn = 0 – 3 = -3 kJ log K =  G o rxn x 1000 -2.303 x 8.314 J/K-mole x 298 K log K = 0.53 K = 10 0.53 = 3.4 very very very very slow kinetics