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Thermodynamics

Thermodynamics. Part 2. Free Energy. D G - Gibbs free energy - maximum possible useful work at constant T and P D G = D H – T D S Our predictive tool for reactions D G > 0 non-spontaneous D G = 0 at equilibrium D G < 0 spontaneous For a spontaneous process:

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Thermodynamics

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  1. Thermodynamics Part 2

  2. Free Energy • DG - Gibbs free energy - maximum possible useful work at constant T and P • DG = DH – TDS • Our predictive tool for reactions DG > 0 non-spontaneous DG = 0 at equilibrium DG < 0 spontaneous • For a spontaneous process: maximize entropy minimize energy

  3. Standard molar free energy of formation - DGof DGof = 0 for free elements in standard state Calculate free energy change for a reaction CaCO3 (s)  CaO (s) + CO2 (g) DGof : -1129 kJ/mole -604 kJ/mole -394 kJ/mole Is this a spontaneous reaction?

  4. DG = DH – TDS

  5. Is this reaction possible? CH4 (g)  C (diamond) + 2H2 (g) Calculate DHorxn, DSorxn, and DGorxn DG = DH-TDS

  6. Can the reaction become spontaneous? DG = 0 = DH - TDS Teq = DH/DS (estimate) Can it rain diamonds on Neptune and Uranus? Neptune/Uranus blue color – CH4 10-50GPa 2000-3000K

  7. Hess’ Law C (graphite) + O2 (g)  CO2 (g) DH1 C (graphite) + ½ O2 (g)  CO (g) DH2 Can't run this reaction, you get CO and CO2 CO (g) + ½ O2 (g)  CO2 (g) DH3 Reactions are additive!

  8. C (graphite) + O2 (g)  CO2 (g) DH1 CO (g) + ½ O2 (g)  CO2 (g) DH3 C (graphite) + O2 (g)  CO2 (g) DH1 CO2 (g)  CO (g) + ½ O2 (g) -DH3 C (graphite) + ½ O2 (g)  CO (g) DH2 = DH1-DH3

  9. Biological pathways Important in the metabolism of glucose Glucose + 6O2 CO2 + H2O First step in glycolytic pathway: Glucose + HPO4-2 + H+ [glucose-6-phosphate]- + H2O DGo’ = 13.8 kJ Is this reaction spontaneous?

  10. Coupled reactions Glucose + HPO4-2 + H+ [glucose-6-phosphate]- + H2O DGo’ = 13.8 kJ ATP-4 + H2O  ADP-3 + HPO4-2 + H+ DGo’ = -30.5 kJ Glucose + ATP-4 [glucose-6-phosphate]- + ADP-3 DGo’ = 13.8 + -30.5 = -16.7 kJ

  11. Free energy and the equilibrium constant DGo = -2.303RTlog K T - kelvin temperature R - gas constant – 8.314 J/K-mole • DGo > 0 K < 1 reactant favored • DGo = 0 K = 1 • DGo < 0 K > 1 product favored

  12. log K = DGorxn x 1000 -2.303 x 8.314 J/K-mole x 298 K Diamonds are NOT forever! Cdiamond Cgraphite DGof : 3 kJ/mole 0 DGorxn = 0 – 3 = -3 kJ log K = 0.53 K = 100.53 = 3.4 very very very very slow kinetics

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