Concise Physical Chemistry

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

PRESSURE DEPENDENCE OF THE CHEMICAL POTENTIAL 87

6.4 THE FREE ENERGY OF REACTION

We have procedures that give us the enthalpies of formation of the participants in

any reaction at 1.0 atm pressure, and we can calculate the absolute entropies of the

participants. Therefore we can construct a table of Gibbs free energies of formation

for many compounds. The free energy change of reaction is found in the usual way:

Gr=

∑

fG(products)−

∑

fG(reactants)

The Gibbs free energies of formation of all elements are defined as zero. This def-

inition is possible because no element can be formed from any other element by an

ordinary chemical reaction.

An insight into the difference between chemical potential and enthalpy can be

found in the stepwise hydrogenation of acetylene, first to ethene and then to ethane:

HC CH(g)+H 2 (g)→H 2 C CH 2 (g)

HC CH 2 (g)+H 2 (g)→H 3 C CH 3 (g)

The enthalpies of these two reactions are, respectively,−174 and−137 kJ mol−^1

and the Gibbs free energies of reaction are−141 and−101 kJ mol−^1 , respectively.

The numbers themselves are quite different, but thedifferencebetween them is

comparable: 33 kJ mol−^1 in the first case and 36 kJ mol−^1 in the second. This is

because each reaction involves “tying up” two moles of gas and releasing only one.

Each reaction involves more or less the same reduction of disorder, hence there is

less energy free to seek a minimum than one might expect considering the enthalpy

change alone. The chemical potential well is less deep than it would be without the

opposing entropy factor.

6.5 PRESSURE DEPENDENCE OF THE CHEMICAL POTENTIAL

The previous calculations were carried out for reactions at 1 atm pressure. All re-

actions are not carried out at 1 atm pressure, so we need a method of finding the

change in chemical potential at any pressure. One can construct a reaction diagram

(Fig. 6.1). If we can find
G 2 and
G 3 for a change in pressure from 1 bar (or

1 atm) to a new pressurep 2 for all the reaction components, the change in Gibbs

chemical potential for the reaction
G 4 can be found at any selected pressurep 2 .The

problem is already solved, however, because we have the identity (∂G/∂p)T=V

from Section 6.3. Explicitly stipulating constant temperature, we can go from partials

to total derivatives for an ideal gas:

dG=Vdp=

RT

p

dp