Physical Chemistry , 1st ed.

Example 3.5

The absolute entropy of Fe (s) at 25.0°C and standard pressure is 27.28 J/molK.

Approximately how many possible combinations of states are available to a

collection of 25 Fe atoms under those conditions? Does the answer suggest

why the system is being limited to only 25 atoms?

Solution

Using Boltzmann’s equation for entropy:

27.28 mo

J

lK

1.381  10 ^23 

K

J

(ln )

Solving, we find

ln 

82.01



4.12  1035

which is an incredible number of possible states for just 25 atoms! However,

the sample is at a relatively high temperature, 298 K. We will see in later chap-

ters how this implies a huge kinetic energy for such a small system.

Example 3.6

Rationalize the following order of absolute molar entropies at 298 K:

S[N 2 O 5 (s)] S[NO (g)] S[N 2 O 4 (g)]

Solution

If we apply the idea that entropy is related to the number of states accessible

to the system, then we can argue immediately that a system of a solid phase

should have fewer states accessible to it. Therefore, it should have the lowest

entropy of the three materials given. Of the remaining two, both materials are

gases. However, one gas is composed of diatomic molecules while the other

is composed of molecules with six atoms. It can be argued that the diatomic

molecule will have fewer states available to it than will a hexatomic molecule,

so S[NO (g)] will probably be less than S[N 2 O 4 (g)]. You can verify this order

by consulting a table of experimental entropies for compounds (like the one

in Appendix 2).

3.7 Entropies of Chemical Reactions

We have already used the idea of combining the changes in entropy of various
individual steps to determine the change in entropy of the combination of those
steps. We can use such ideas to determine the changes in entropy that occur
with chemical reactions. The situation is only slightly different, because we can
determine the absoluteentropies of the chemical reactants and products. Figure
3.8 illustrates the concept for a process where Sis negative, that is, entropy is
going down. As such, we do not need to rely on formation reactions but can
state that the change in entropy of a chemical reaction equals the combined en-
tropies of the products minus the combined entropies of the reactants. Thus,

rxnS

0

products

S

0

reactants

S (3.27)

25 atoms



6.022  1023 atoms/mol

3.7 Entropies of Chemical Reactions 81