2. THERMODYNAMICS OF LIVING SYSTEMS
39
From the practical point of view, Eq. 50, AF° = — RT In K, represents one of the most useful relationships developed in this section since
it permits calculating the equilibrium constant for a reaction if AF° is
known or AF° if the value of K can be determined. It must be remembered that AF° is the standard free energy change, i.e., the change
in free energy in a reaction in which the reactants in their standard
states (unit activity for solutions, one atmosphere for gases, etc.) react
to form products in their standard states. For illustration we will take
an example used by West (5). The following reaction, one of the steps
in glycolysis, has an equilibrium constant at 38° of approximately 19.
Glucose-1-phosphate ;=± Glucose-6-phosphate
Thus
AF° = - RT In K = -ET
7 (2.303) log 19 = -1800cal.
If the reactants and products are not at their equilibrium activities, as
is usually the case in living systems, Eq. 51 can be used for calculating
the actual free energy change in a reaction. For a reversible reaction
A -f B^±C -\- D, Eq. 51 may be put in the simplified form
AF = AF° + RT In gjgj
(58)
where, in dilute solutions, [C], [D], etc. represent the concentrations
of the products and reactants.
6. The Rehtion Between AF° and Standard Oxidation Potentials
There is one more biologically important result that will be developed, the relation between free energy changes and standard oxidation potentials.
Consider the general reaction
aA + bB + . . . ^± IL + mM + . . .
The free energy of a given mixture may be expressed in terms of
chemical potentials by the following equations
^reactants = &VA + bßß + . . .
(59)
/^products
=
*ML +
m ^M + · · ·
where we have assumed T, P, and the composition to be constant. The
free energy change of the reaction is
AFT,P
— -^products ~~ -^reactants
= (W + ηΐμ Μ +...) — (αμ Α + bß B + . . .)
(60)
39
From the practical point of view, Eq. 50, AF° = — RT In K, represents one of the most useful relationships developed in this section since
it permits calculating the equilibrium constant for a reaction if AF° is
known or AF° if the value of K can be determined. It must be remembered that AF° is the standard free energy change, i.e., the change
in free energy in a reaction in which the reactants in their standard
states (unit activity for solutions, one atmosphere for gases, etc.) react
to form products in their standard states. For illustration we will take
an example used by West (5). The following reaction, one of the steps
in glycolysis, has an equilibrium constant at 38° of approximately 19.
Glucose-1-phosphate ;=± Glucose-6-phosphate
Thus
AF° = - RT In K = -ET
7 (2.303) log 19 = -1800cal.
If the reactants and products are not at their equilibrium activities, as
is usually the case in living systems, Eq. 51 can be used for calculating
the actual free energy change in a reaction. For a reversible reaction
A -f B^±C -\- D, Eq. 51 may be put in the simplified form
AF = AF° + RT In gjgj
(58)
where, in dilute solutions, [C], [D], etc. represent the concentrations
of the products and reactants.
6. The Rehtion Between AF° and Standard Oxidation Potentials
There is one more biologically important result that will be developed, the relation between free energy changes and standard oxidation potentials.
Consider the general reaction
aA + bB + . . . ^± IL + mM + . . .
The free energy of a given mixture may be expressed in terms of
chemical potentials by the following equations
^reactants = &VA + bßß + . . .
(59)
/^products
=
*ML +
m ^M + · · ·
where we have assumed T, P, and the composition to be constant. The
free energy change of the reaction is
AFT,P
— -^products ~~ -^reactants
= (W + ηΐμ Μ +...) — (αμ Α + bß B + . . .)
(60)