As an example, consider how the Gibbs energy can be calculated
from the equilibrium constant for the reaction catalyzed by the enzyme
phosphoglucomutase:
glucose 1 −phosphate ↔glucose 6 −phosphate
Biochemical analysis of the reaction at 25°C and pH equal to 7 shows
that the final equilibrium mixture will contain 1 mM glucose 1-phosphate
for every 19 mM glucose 6-phosphate. The equilibrium constant is given
by the ratio of the relative concentrations:
(3.21)
Using eqn 3.20, the change in the Gibbs energy can be calculated:
ΔG=−RTlnK=−(8.315 J/(mol K))(298 K)(ln 19)
=−7,296 J mol−^1 (3.22)
Thus, the reaction proceeds spontaneously, producing a large decrease in
the Gibbs energy, as expected for a large positive value for the equilibrium
constant.
Research direction: drug design II
In Chapter 2, calorimetry was described as a technique that could accurately
measure the heat change in response to drug binding. The binding affinity
can now be related to the Gibbs energy according to eqn 3.20 and con-
sequently also related to the changes in enthalpy and entropy associated
with the binding:
K=e−ΔG/kT=e−(ΔH−TΔS)/kT (3.23)
K
[]
[
=
Glucose 6 – phosphate
Glucose 1 – phosphate]]
==
19
1
19
mM
mM
56 PARTI THERMODYNAMICS AND KINETICS
Table 3.1
Relationships among the equilibrium constant, K, Gibbs energy change,
ΔG, and direction of a chemical reaction*.
K ΔΔG Direction
>1.0 Negative Proceeds forward
1.0 Zero At equilibrium
<1.0 Positive Proceeds in reverse
*Normally the change in the Gibbs energy is the standard value, as discussed in
Chapter 6.
