1 / 12

Gibbs Free Energy: G

D G = D H –T/ D S. D G = amount of energy available to do work. D G = G products - G reactants. A + B C + D. D G = (G C + ) - (G A + G B ). G D. Gibbs Free Energy: G. (See pages 13-15; 309- 318 Horton). G = H –TS H = heat of system

javen
Download Presentation

Gibbs Free Energy: G

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. DG = DH –T/DS DG = amount of energy available to do work DG = Gproducts - Greactants A + B C + D DG = (GC + ) - (GA + GB) GD Gibbs Free Energy: G (See pages 13-15; 309- 318 Horton) G = H –TS H = heat of system S = entropy of system Determine the change in free energy of a reaction DG DG < 0 exergonic, rx tends to be spontaneous DG > 0 endergonic, rs requires input of energy DG = 0 at equilibrium Units: Joules or kiloJoules Joule = amount of energy required to apply 1 newton of force over 1m

  2. A B Reaction may be: exergonic and A B DGo’ < 0 Keq = [B] / [A] Reaction may be: endergonic and B A DGo’ > 0 Reaction may be at equilibrium DGo’ = 0 Standard free energy change: DGo’ Reactants and products present at Concentration of 1M and pH = 7.0

  3. A B C or A B E F G C Both pathways have the same DGo’ • DGo’ is independent of pathway

  4. 1. DGo’ = -RTlnKeq R = 8.315 J/mol/oK T = 298oK (25oC) = 2.48lnKeq (kJ/mol) • A B DGo’ =+16.7 KJ/mol Calculation ofDG: change in Gibbs free energy What is Keq for this reaction? DGo’ = -RTlnKeq 16.7 = -2.48lnKeq lnKeq = -(16.7/2.48) = -6.73 Keq = 1.19 x 10-3 = [B] / [A]

  5. DGo’produce large changes • in Keq • small changes in If DGo’ =+22.4 kJ/mol 22.4 = -2.48lnKeq lnKeq = -(22.4/2.48) = -9.03 Keq = 1.19 x 10-4 Increase in DGo’ from 16.7 to 22.4, a 35% increase, results in a 10-fold change in Keq.

  6. e.g. A B DGo’ = +15kJ/mol B C DGo’ = -20kJ/mol Net DGo’ = -5kJ/mol Net rx: A C • An unfavorable reaction may be made to • proceed by coupling it to a favorable reaction

  7. A B DGo’ =+16.7 kJ/mol • ATP ADP + Pi DGo’ =-30kJ/mol A + ATP B + ADP + Pi DGo’ =-13.8kJ/mol Keq = X [B] [B] [ADP][Pi] [ATP] [ATP] [A] [A] [ATP] Assume that = 500 [ADP][Pi] [ADP][Pi] and: = Keq x Coupling of unfavorable reaction to a favorable one. DGo’ = -RTlnKeq -13.8 = -2.48lnKeq Keq = 2.6 x 102 = 2.6 x 102 x 500 = 1.32 x 105

  8. Keq in the absence of ATP hydrolysis: = 1.19 x 10-3 Keq in the presence of ATP hydrolysis: = 1.32 x 105 An increase of 108-fold

  9. [B] DG = DGo’ + RTln 2. [A] Keq = 0.0475 e.g. A B from DGo’ = -RTlnKeq DGo’ = 7.55kJ/mol standard conditions [B] = Q, the mass action ratio [A] [B] Then DG = DGo’ + RTln [A] 3 x 10-6 M = 7.55 + RTln 2 x 10-4 M = -2.86kJ/mol actual conditions Actual free energy change BUT If [A] = 2 x 10-4 M , and [B] = 3 x 10-6 M

  10. Control of metabolic flux • Reactions that operate near equilibrium are • readily reversible - rate and direction of • reaction effectively controlled by concentrations • of substrate and products • Reactions that operate far from equilibrium • are metabolically irreversible – rate can only • be altered by changing enzyme activity

  11. F-1,6-bisP + ADP F-6-P + ATP This a potential control point e.g. Phosphofructokinase Keq = 300 But under intracellular conditions Q = 0.03 Insuffcient enzyme activity to equilibrate reaction and enzyme operates near Vmax at all times Can only increase rate of product formation by increasing enzyme activity

  12. A B C D E 3 4 2 1 In metabolic pathways intermediates are not allowed to “pile up” All reactions in a sequence proceed at the same rate and concentration of intermediates is constant – the steady state condition This is achieved by having several points of control First enzyme of a pathway does not feed substrate into the pathway at a rate that is faster than the slowest enzyme downstream

More Related