Physical Chemistry Third Edition

306 7 Chemical Equilibrium

In Chapter 6 we expressed the chemical potential of a general substanceiin terms

of its activityai:

μiμ◦i+RTln(ai) (7.1-11)

whereμ◦iis the chemical potential of substanceiin whatever standard state applies.

The rate of change of the Gibbs energy per mole of reaction is now expressed in terms

of activities:

(

∂G

∂ξ

)

T,P



∑c

i 1

viμ◦i+RT

∑c

i 1

viln(ai) (7.1-12)

If it is possible to maintain all of the reactants and products in their standard states,

all activities are equal to unity. The second sum in the right-hand side of Eq. (7.1-12)

vanishes and the change inGfor 1 mol of reaction is

∆G◦

∫ 1

0

[(

∂G

∂ξ

)

T,P

∣

∣

∣

∣

∣

standard state

]

dξ



∫ 1

0

(c

∑

i 1

viμ◦i

)

dξ

(c

∑

i 1

viμ◦i

)∫

1

0

dξ

so that for 1 mol of reaction

∆G◦

∑c

i 1

viμ◦i (standard-state reaction) (7.1-13)

The quantity∆G◦is thestandard-state Gibbs energy change for 1mol of reaction.It

applies to a reaction in which all reactants and products are in their standard states.

If the standard-state Gibbs energy change is negative, the forward reaction would be

spontaneous under standard conditions, and if it is positive, the reverse reaction would

be spontaneous under standard conditions.

TheGibbs energy change of formationis similar to the enthalpy change of formation

defined in Chapter 2. It is the Gibbs energy change to produce 1 mol of the substance in

question from the appropriate elements in their most stable forms.^1 If substance number

iand the elements are in their standard states, the Gibbs energy change of formation

of substanceiis denoted by∆fG◦(i). The Gibbs energy change for the standard-state

reaction can be calculated from standard-state Gibbs energy changes of formation in

the same way as∆H◦was calculated in Chapter 2:

∆G◦

∑c

i 1

vi∆fG◦(i) (7.1-14)

(^1) An exception to the practice of taking the most stable form of the element in defining the formation
reaction is in the case of phosphorus, for which the less reactive white phosphorus is taken instead of the
more stable but more reactive red phosphorus.