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Chapter 4 EXAMPLE 4.8 Condition for reaction equilibrium
Material Equilibrium
Write out the equilibrium condition (4.98) for

(a) 2C H 15O S 12CO 6H O
2
2
6
6
2
(b) aA bB S cC dD
Since reactants have negative stoichiometric numbers, the equilibrium con-
dition for (a) is n m 2m 15m 12m 6m 0 or
i i i C 6 H 6 O 2 CO 2 H 2 O
2m 15m 12m 6m
C 6 H 6 O 2 CO 2 H 2 O
which has the same form as the chemical reaction. For the general reaction (b),
the equilibrium condition (4.98) is
am bm cm dm D
C
A
B
Exercise
Write the equilibrium condition for 2H O → 2H O. (Answer: 2m m
2 2 2 H 2 O 2
2m .)
H 2 O

The equilibrium condition n m 0 looks abstract, but it simply says that at
i i i
reaction equilibrium, the chemical potentials of the products “balance” those of the
reactants.
If the reacting system is held at constant T and P, the Gibbs energy G is minimized
at equilibrium. Note from the Gibbs equation (4.78) for dG that the sum m dn in
i i i
(4.96) equals dG at constant T and P; dG m dn . Use of dn n dj [Eq. (4.97)]
T,P i i i i i
gives dG n m dj:
T,P i i i
dG
a n m const. T, P (4.99)
dj i i i
At equilibrium, dG/dj 0, and G is minimized. The m ’s in (4.99) are the chemical po-
i
tentials of the substances in the reaction mixture, and they depend on the composition
of the mixture (the n ’s). Hence the chemical potentials vary during the reaction. This
i
variation continues until G (which depends on the m ’s and the n ’s at constant T and P)
i i
is minimized (Fig. 4.2) and (4.98) is satisfied. Figure 4.7 sketches G versus j for a re-
G
action run at constant T and P. For constant T and V, G is replaced by A in the preced-
ing discussion.
Const.
T and P The quantity n m in the equilibrium condition (4.98) is often written as G
r
i
i
i
(where r stands for reaction) or as G, so with this notation (4.98) becomes G 0,
r
where G n m . However, n m is not the actual change in G in the reacting
r i i i i i i
system and the in G really means (
/
j) . (See Sec. 11.9 for further discussion.)
r r T,P
Note the resemblance of the reaction-equilibrium condition (4.98) to the phase-
equilibrium condition (4.88). If we regard the movement of substance A from phase b
i
b
d
d
b
to phase d to be the chemical reaction A → A , then n 1 for A and n 1 for A .
i i i i
b
d
Equation (4.98) gives m m 0, which is the same as (4.88).
i i
j
j eq
4.9 ENTROPY AND LIFE
Figure 4.7
The second law of thermodynamics is the law of increase in entropy. Increasing en-
Gibbs energy versus extent of tropy means increasing disorder. Living organisms maintain a high degree of internal
reaction in a system held at order. Hence one might ask whether life processes violate the second law.
constant T and P.