Thermodynamics
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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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Free energy
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


Thermodynamics

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?


D g d h t d s
DG = DH – TDS


Thermodynamics

Is this reaction possible?

CH4 (g)  C (diamond) + 2H2 (g)

Calculate DHorxn, DSorxn, and DGorxn

DG = DH-TDS


Thermodynamics

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


Thermodynamics

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!


Thermodynamics

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


Biological pathways
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?


Coupled reactions
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


Thermodynamics

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


Diamonds are not forever

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