Concise Physical Chemistry

c08 JWBS043-Rogers September 13, 2010 11:25 Printer Name: Yet to Come

DEGENERACY AND EQUILIBRIUM 109

A(g) B(g)

· · · εB

· · · · · εA

FIGURE 8.1 A two-level equilibrium. There are 5 molecules in A and 3 molecules in B.

Conversely, knowing the energy separation, one can calculate the equilibrium

constant. If the energy of B islowerthan the energy of A by 2. 10 × 10 −^21 J, level B

will contain more molecules at equilibrium than will level A andKeqwill be larger

than 1.0 (Fig. 8.2).

Keq=

nB

nA

=e−(εB−εA)/kBT=

5

3

= 1. 67

In these simple models, the equilibrium constants are inverses and(εB−εA)of the

first equilibrium is equal but opposite in sign to(εB−εA)of the second reaction. This

analysis gives a description of the relationship between energy and the equilibrium

constant that is useful as far as it goes, but it has left out an important consideration:

the influence of entropy.

8.2 DEGENERACY AND EQUILIBRIUM

If the upper level is split into two levels, each of which has the same energy, its

capacity to accommodate molecules of kind B is doubled (Fig. 8.3). The equilibrium

constantKeq=

6

5

has been made greater than 1.0. The reaction now favors products

rather than reactants even though the energy change is uphill. The second law has

come into play; the product state is more disordered. There are more places for the

molecule to go. It is like having two stacks of paper on your desk rather than one.

There are now two places where that important piece of paper you are looking for may

be, whereas before there was only one. Increased disorder means increased entropy.

· · ·

· · · · ·

A B

FIGURE 8.2 A two-level equilibrium.