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Chapter 6 endothermic reactions with G°negative and products favored at room temperature,
Reaction Equilibrium in Ideal Gas because of the T S° term.
Mixtures
For very high temperatures, the factor T makes the second term on the right side
of (6.28) the dominant one, and we have the following rough relation:
K P ¢G° T ¢S° high T (6.30)

At high temperatures, a reaction with a positive S° has a negative G°, and products
are favored.
Consider the breaking of a chemical bond, for example, N (g) ∆ 2N(g). Since a
2
bond is broken, the reaction is highly endothermic ( H° W 0). Therefore at reason-
ably low temperatures, G° is highly positive, and N is not significantly dissociated
2
at low temperatures (including room temperature). For N (g) ∆ 2N(g), the number of
2
moles of gases increases, so we expect this reaction to have a positive S° (Sec. 5.10).
K P
(Appendix data give S° 115 J mol 1 K 1 for this reaction.) Thus for high
298
temperatures, we expect from (6.30) that G° for N ∆ 2N will be negative, favoring
2
dissociation to atoms.
Figure 6.4 plots K° versus T for N (g) ∆ 2N(g). At 1 bar, significant dissociation
P 2
occurs only above 3500 K. Calculation of the composition of nitrogen gas above 6000

K must also take into account the ionization of N and N to N e and N e ,
2
2
Figure 6.4 respectively. Calculation of the high-T composition of air must take into account the
dissociation of O and N , the formation of NO, and the ionization of the molecules
K° versus T for N (g) ∆ 2N(g). 2 2
P
2
The vertical scale is logarithmic. and atoms present. Figure 6.5 plots the composition of dry air at 1 bar versus T.

Thermodynamic data for gaseous ions and for e (g) can be found in the NIST-JANAF
tables (Sec. 5.9).
6.3 TEMPERATURE DEPENDENCE OF THE
EQUILIBRIUM CONSTANT
The ideal-gas equilibrium constant K° is a function of temperature only. Let us derive
P
its temperature dependence. Equation (6.14) gives ln K° G°/RT. Differentiation
P
with respect to T gives
d ln K° ¢G° 1 d1¢G°2
P
(6.31)
dT RT 2 RT dT
Use of G° n G° [Eq. (6.9)] gives
i
m,i
i
d d dG° m,i
¢G° a n G° a n i (6.32)
m,i
i
dT dT i i dT
Figure 6.5 From dG S dT V dP, we have (
G /
T) S for a pure substance. Hence
m m m m P m
Mole-fraction equilibrium dG° >dT S° m,i (6.33)
m,i
composition of dry air versus T at
1 bar pressure. CO and other The degree superscript indicates the pressure of pure ideal gas i is fixed at the standard
2
minor components are omitted. value 1 bar. Hence G° depends only on T, and the partial derivative becomes an
m,i
The vertical scale is logarithmic. ordinary derivative. Using (6.33) in (6.32), we have
The mole fraction of Ar (x
Ar
n /n ) decreases above 3000 K d ¢G°
Ar
tot
i
m,i
because the dissociation of O 2 dT a n S° ¢S° (6.34)
increases the total number of i
moles present. Above 6000 K, where S° is the reaction’s standard entropy change, Eq. (5.36). Hence (6.31)

formation of O and N becomes becomes
significant. At 15000 K, only d ln K° ¢G° ¢S° ¢G° T ¢S°
charged species are present in P (6.35)
significant amounts. dT RT 2 RT RT 2