Page 200 - Physical Chemistry

lev38627_ch06.qxd 3/3/08 10:07 AM Page 181

181
Except for reactions with n 0, the equilibrium constant K depends on P as well Section 6.2
x
as on T and so is not as useful as K°. Ideal-Gas Reaction Equilibrium
P
Introduction of K° and K is simply a convenience, and any ideal-gas equilibrium
c x
problem can be solved using only K°. Since the standard state is defined as having
P
1 bar pressure, G° is directly related to K° by G° RT ln K° [Eq. (6.14)] but is
P P
only indirectly related to K° and K through (6.25) and (6.27).
c x
Qualitative Discussion of Chemical Equilibrium
The following discussion applies in a general way to all kinds of reaction equilibria,
not just ideal-gas reactions.
The standard equilibrium constant K° is the product and quotient of positive
P
numbers and must therefore be positive: 0 K° q. If K° is very large (K° W 1),
P P P
its numerator must be much greater than its denominator, and this means that the equi-
librium pressures of the products are usually greater than those of the reactants.
Conversely, if K° is very small (K° V 1), its denominator is large compared with its
P P
numerator and the reactant equilibrium pressures are usually larger than the product
equilibrium pressures. A moderate value of K° usually means substantial equilibrium
P
pressures of both products and reactants. (The word “usually” has been used because
it is not the pressures that appear in the equilibrium constant but the pressures raised
to the stoichiometric coefficients.) A large value of the equilibrium constant favors
products; a small value favors reactants.
We have K° 1/e G°/RT [Eq. (6.15)]. If G° W 0, then e G°/RT is very large and
P K P
K° is very small. If G° V 0, then K° e G°/RT is very large. If G° 0, then K°
P P P
1. A large positive value of G°favors reactants; a large negative G°favors prod-
ucts. More precisely, it is G°/RT, and not G°, that determines K°. If G° 12RT,
P
5
12
6
then K° e 12 6 10 . If G° 12RT, then K° e 2 10 . If G°
P P
50RT, then K° 2 10 22 . Because of the exponential relation between K° and G°,
P P
unless G° is in the approximate range 12RT G° 12RT, the equilibrium
constant will be very large or very small. At 300 K, RT 2.5 kJ/mol and 12RT
30 kJ/mol, so unless G° 30 kJ/mol, the equilibrium amounts of products or of
300
reactants will be very small. The Appendix data show G° values are typically a
f 298
couple of hundred kJ/mol, so for the majority of reactions, G° will not lie in the
range 12RT to 12RT and K° will be very large or very small. Figure 6.3 plots K°
P P
versus G° at two temperatures using a logarithmic scale for K°. A small change in
P
G° produces a large change in K° e G°/RT . For example, at 300 K, a decrease of
P
only 10 kJ/mol in G° increases K° by a factor of 55.
P
Since G° H° TS°, we have for an isothermal process K P
¢G° ¢H° T ¢S° const. T (6.28)
so G° is determined by H°, S°, and T. If T is low, the factor T in (6.28) is small
and the first term on the right side of (6.28) is dominant. The fact that S° goes to zero
as T goes to zero (the third law) adds to the dominance of H° over T S° at low tem-
peratures. Thus in the limit T → 0, G° approaches H°. For low temperatures, we Figure 6.3
have the following rough relation:
Variation of K° with G°for two
P
¢G° ¢H° low T (6.29) temperatures. The vertical scale is
logarithmic.
For an exothermic reaction, H° is negative, and hence from (6.29) G° is nega-
tive at low temperatures. Thus at low T, products of an exothermic reaction are favored
over reactants. (Recall from Sec. 4.3 that a negative H increases the entropy of the
surroundings.) For the majority of reactions, the values of H° and T S° are such that
at room temperature (and below) the first term on the right side of (6.28) dominates.
Thus, for most exothermic reactions, products are favored at room temperature.
However, H° alone does not determine the equilibrium constant, and there are many