Page 46 - Basic physical chemistry for the atmospheric sciences

32 Basic physical chemistry

R T
*
-
d = -dp (2. 3 9)
G
p
Integrating Eq. (2.39) between the pressure limits p0 ( = I atm) and p
(in atm) 3
pR * T
G - =
dG
-dp
L .,..o L p p
o
uT
or,
- -:::;( p
G - GT = R * T In 0 = R * T lnp (2.40)
p
where G is the Gibbs free energy per mole at pressure p and tempera­
ture T and G/ the Gibbs free energy at I atm and temperature T.
(
For the general chemical reaction 1 . 5 )
!l.G = [gG(G) + hG(H) + . . . ] - [aG(A) + bG(B) + . . . ]
U s ing Eq. (2.40) in this last expression
!l.G = [gG\1(G) + hG¥(H) + . . . - aG¥{A) - bGi(B) - . . . ]
-
.
+ gR T * lnp0 + hR* T lnpH + . . - aR* T lnpA b R*T lnp8 - . . .
or,

(2.4 1 )

(
From Eqs. l . 9b) and (2.4 1 )
I
.:lG = !l.G/+ R*T n K P (2.42)
If the pressures P A ,P8, . . P0,PH . . . are those that exist when the re­
.
.
actants and products are in chemical equilibrium, then .:lG = 0, and
ilGY-= - R* T In Kp (2.43)

Equation (2.43) can be used to obtain the value of .:lG/, which is the
G
change in the i bbs free energy for a reaction at I atm and tempera­
ture T, from the equilibrium constant for the reactant at pressure p
and temperature T. For standard conditions 1 atm and 25.00°C), Eq.
(
(2.43) becomes
.:lG" = - R*(298 . 1 5 )ln Kp = - 2478.9lnK = - 5707.9log10K P (2.44)
P