Page 142 - Essentials of physical chemistry

104 Essentials of Physical Chemistry

FIGURE 6.1 Portrait of Josiah Willard Gibbs (Jr.) (1839–1903) from the Williams Haynes Portrait Collec-
tion, Chemical Heritage Foundation Collection. He earned his PhD from Yale (one of the first in the United
States) and spent his career there as a professor of mathematical physics.

ð 2 ð P 2 dP
a change in m as dm ¼ RT , so that we find m (T, P 2 ) m (T, P 1 ) ¼ RT ln P 2 . We can
2
1
1 P 1 P P 1
use this general formula to reference the chemical potential to standard conditions, such as 2988K
and 1 bar.

P
0
m(T, P) ¼ m þ RT ln or using G 0 298 values G(T, P) ¼ G 0 298 (1 bar) þ RT ln P. Many old
1
texts standardized on 1 atm but the new handbook values are relative to 1 bar (1 atm ¼ 1.01325 bar),
so there is not much difference in the numbers and the equation is the same.
Now let us consider a typical equilibrium for gases where we can use pressures according to
Dalton’s law.
aA þ bB ƒ cC þ dD
ƒ!
DG ¼ cm þ dm am bm
C D A B
0
0
0
0
DG ¼ cm þ dm am bm þ RT[c ln P C þ d ln P D a ln P A b ln P B ]
C D A B
c
but c ln P C ¼ ln P , etc.
C
c d c d
P P 0 P P
C D
C D
DG ¼ DG 0 þ RT ln ¼ DG þ RT ln K P where ¼ K P . Note DG ¼ 0at
298 a b 298 a b
P P P P
A B A B
equilibrium by the very definition of the meaning of equilibrium as a balance of decreasing energy
and increasing entropy. Therefore as a result of the equilibrium condition, we have
DG 0 298 DG 0 DG 0 298
0
DG and that leads to K P ¼ e RT ¼ exp .
298 ¼ RT ln K P and ln K P ¼
RT RT