Concise Physical Chemistry

c06 JWBS043-Rogers September 13, 2010 11:25 Printer Name: Yet to Come

88 THE GIBBS FREE ENERGY

4

2 2

2 3

1

1 bar 1 bar

(, ) (, )

(, ) (, )

G

AgpBgp

GG

G

AgpBgp

Δ

→

ΔΔ

Δ

→

↑↑

FIGURE 6.1 A reaction diagram for
G 4.

Integrating to find
G 2 , we obtain

G 2 =

∫Gp

2

G1bar

dG=RT

∫Gp

2

G1bar

dp

p

=RTln

p 2

1

where the pressure at the lower limit of integration is 1 bar. A similar equation

describes
G 3. Combination of
G 1 ,
G 2 , and
G 3 with attention to the sign

differences gives
G 4 , the desired change in the Gibbs function for reaction at the

new pressurep 2. One way of getting the signs straight is to set up the diagram as a

cyclic process, and remember that the sum of changes in any thermodynamic function

must be zero around a cycle.

6.5.1 The Equilibrium Constant as a Quotient of Quotients

In writing equilibrium constant expressions such asKeq=pB/pAfor a reaction

A → B in the gas phase, it should be remembered that the pressures are quotients

relative to a pressure in the standard statep/p 0 , hence they are unitless and so isKeq.

If the gas is not ideal, the fugacityfis used in place of the pressure or in the case

of liquids and solutions, the activityamay be used. Fugacities or activities are also

ratiosf/f 0 ora/a 0 relative to a standard state soKeqis still unitless, as it must be for

some of the mathematical manipulations to come. Changes toKeqexpressed in terms

of fugacities or activities bring about complication in the mathematical expression

and in the experimental determination of the quantities involved; but once again, the

principles are the same as in the ideal case.

6.6 THE TEMPERATURE DEPENDENCE OF THE FREE ENERGY

From the usual expression for the derivative of a quotient,

d

(u

v

)

dx

=

v

du

dx

−u

dv

dx

v^2

,

and stipulating constant pressure, we get

d

(

G

T

)

dT

=

T

(

dG

dT

)

−G

(

dT

dT

)

T^2

=

−TS−G

T^2