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