'Standard gibbs free energy' diaporamas de présentation

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Chapter 9: Chemical Equilibrium

Chapter 9: Chemical Equilibrium. The forward and reverse reaction are both taking place at the same rate. Production and Decomposition of Ammonia. Forward Reaction: N 2 (g) + 3H 2 (g) ? 2NH 3 (g). Reverse Reaction: 2NH 3 (g) ? N 2 (g) + 3H 2 (g).

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??????? Electroanalytical Chemsitry and Its Biological Applications ?????? ?? 2014 ??

??????? Electroanalytical Chemsitry and Its Biological Applications ?????? ?? 2014 ??. Questions:. How an electrolytic reaction is influenced by electrode potential? How is the electrode potential is established at equilibrium?. Electrode Potential/Kinetics ( ???? / ????? ).

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??????? Electroanalytical Chemsitry and Its Biological Applications ?????? ?? 2014 ??

??????? Electroanalytical Chemsitry and Its Biological Applications ?????? ?? 2014 ??. Questions:. How an electrolytic reaction is influenced by electrode potential? How is the electrode potential is established at equilibrium?. Electrode Potential/Kinetics ( ???? / ????? ).

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Special Note: Endothermic Reactions

Special Note: Endothermic Reactions. As long as the total entropy is slightly positive, endothermic reactions will go forward Spontaneity is determined by the increase in the entropy of the system, NOT a decrease in the ENERGY (heat) of the system!. ?. Equilibrium.

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Chapter 19

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Review

Review. Standard Gibbs Free Energy. ? G o =. ? H o. - T ? S o. spontaneous. ?G o < 0. ?G o = - 30.5 kJ/mol. non-spontaneous. ?G o > 0. ?G o = + 16.7 kJ/mol. ?G o = 0. equilibrium. ?G o = - 13.8 kJ/mol. ?. glucose-6-phosphate. + H 2 O. + phosphate. glucose. + H 2 O ?.

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Review

Review. Standard Gibbs Free Energy. ? G o =. ? H o. - T ? S o. spontaneous. ?G o < 0. ?G o = - 30.5 kJ/mol. non-spontaneous. ?G o > 0. ?G o = + 16.7 kJ/mol. ?G o = 0. equilibrium. ?G o = - 13.8 kJ/mol. ?. glucose-6-phosphate. + H 2 O. + phosphate. glucose. + H 2 O ?.

230 views • 11 slides

Review

Review. Standard Gibbs Free Energy. ? G o =. ? H o. - T ? S o. spontaneous. ?G o < 0. ?G o = - 30.5 kJ/mol. non-spontaneous. ?G o > 0. ?G o = + 16.7 kJ/mol. ?G o = 0. equilibrium. ?G o = - 13.8 kJ/mol. ?. glucose-6-phosphate. + H 2 O. + phosphate. glucose. + H 2 O ?.

118 views • 11 slides