41
dAG/dT AG/T = -AH/T (1-43)
Equation (1-43) may be expressed as
T (AG/T) = -AH/T (1-44)
since, upon differentiation, it would become
m T(dAG/dT)-AG dAG/dT
1 2
T T
Equation (1-45) may be written
and, in that
AG = -RT In K (1-46)
AG
T
(1-45)
in terms of standard states,
it can be stated as
d In K/dT = AH/RT2
which is equivalent to
d log K
d (1/T)
-AH
2.303R
+ Constant
(1-47)
(1-48)
Assuming that AH is temperature independent, equation (1-48)
can be integrated to yield equation (1-31). The above
relationship may be applied to reactions in solution whose
equilibrium association constants are based on the activities
of the species involved [see equations (1-23) through (1-30)].
Plots of log K vs. 1/T should yield straight lines having
slopes equal to -AH/2.303R. If the relationship between
log K and 1/T is not linear then the enthalpy of binding is