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Gibbs Free Energy: G (See pages 13-15; 309- 318 Horton) Determine the change in free energy of a...
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Transcript of Gibbs Free Energy: G (See pages 13-15; 309- 318 Horton) Determine the change in free energy of a...
Gibbs Free Energy: G(See pages 13-15; 309- 318 Horton)
Determine the change in free energy of a reaction
G
G = H –TS H = heat of systemS = entropy of system
G = amount of energy available to do work
G = Gproducts - Greactants
G = H –T/S
A + B C + D
G = (GC + ) - (GA + GB) GD
G < 0 exergonic, rx tends to be spontaneousG > 0 endergonic, rs requires input of energyG = 0 at equilibrium
Units: Joules or kiloJoulesJoule = amount of energy required to apply 1 newton of force over 1m
Go’
Standard free energy change:
Reactants and products present at Concentration of 1M and pH = 7.0
A B
Keq = [B] / [A]
Reaction may be: exergonic and A BGo’ < 0
Reaction may be: endergonic and B AGo’ > 0
Reaction may be at equilibrium Go’ = 0
• Go’ is independent of pathway
A B Cor
A B E F G C
Both pathways have the same Go’
Calculation of G: change in Gibbs free energy
Go’ = -RTlnKeq
R = 8.315 J/mol/oKT = 298oK (25oC)
= 2.48lnKeq (kJ/mol)
1.
i. A B Go’ = +16.7 KJ/mol
Go’ = -RTlnKeq
16.7 = -2.48lnKeq
lnKeq = -(16.7/2.48) = -6.73
Keq = 1.19 x 10-3 = [B] / [A]
What is Keq for this reaction?
If Go’ = +22.4 kJ/mol
22.4 = -2.48lnKeq
lnKeq = -(22.4/2.48) = -9.03
Keq = 1.19 x 10-4
• small changes inGo’ produce large changes in Keq
Increase in Go’ from 16.7 to 22.4, a 35%
increase, results in a 10-fold change in Keq.
•An unfavorable reaction may be made to proceed by coupling it to a favorable reaction
e.g. A B Go’ = +15kJ/mol
B C Go’ = -20kJ/mol
Net rx: A C NetGo’ = -5kJ/mol
Coupling of unfavorable reaction to a favorable one.
i. A B Go’ = +16.7 kJ/mol
ii. ATP ADP + Pi Go’ = -30kJ/mol
A + ATP B + ADP + Pi Go’ = -13.8kJ/mol
Go’ = -RTlnKeq
-13.8 = -2.48lnKeq
Keq = 2.6 x 102
Keq = [A]
[B] [ADP][Pi]
[ATP]X
Assume that [ATP]
[ADP][Pi]= 500
and: [A]
[B]= Keq x
[ATP]
[ADP][Pi]
= 2.6 x 102 x 500
= 1.32 x 105
Keq in the presence of ATP hydrolysis:
= 1.32 x 105
Keq in the absence of ATP hydrolysis:
= 1.19 x 10-3
An increase of 108-fold
G = Go’ + RTln[B]
[A]2.
[B]
[A]= Q, the mass action ratio
Actual free energy change
If [A] = 2 x 10-4 M , and [B] = 3 x 10-6 M
Then G = Go’ + RTln[B]
[A]
= + RTln 3 x 10-6 M
2 x 10-4 M
= -2.86kJ/mol actual conditions
BUT
e.g. A B
from Go’ = -RTlnKeq
Go’ = 7.55kJ/mol standard conditions
Keq = 0.0475
Control of metabolic flux
•Reactions that operate near equilibrium are readily reversible - rate and direction ofreaction effectively controlled by concentrationsof substrate and products
• Reactions that operate far from equilibriumare metabolically irreversible – rate can only be altered by changing enzyme activity
e.g. Phosphofructokinase
F-6-P + ATP F-1,6-bisP + ADP
Keq = 300
But under intracellular conditions Q = 0.03
Insuffcient enzyme activity to equilibrate reactionand enzyme operates near Vmax at all times
Can only increase rate of product formationby increasing enzyme activity
This a potential control point
In metabolic pathways intermediates are not allowed to “pile up”
All reactions in a sequence proceed at the same rate and concentration of intermediatesis 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 isfaster than the slowest enzyme downstream
A B C D E1 2 3 4