Physical Chemistry , 1st ed.

2. The second step is to evaluate the change in entropy at the reference tem-
   perature, 298 K. We will label this S 2. This was, in fact, calculated in
   Example 3.7. It is

S 2 326.7 

K

J



3. The third step is to evaluate the change in entropy as we bring the prod-
   ucts from the reference temperature to the specified reaction temperature
   (that is, from 298 K to 372 K). This entropy change is labeled S 3. According
   to equation 3.18:

S 3 (2 mol)75.3 

mo

J

lK

ln 

3

2

7

9

2

8

K

K



S 3 33.4 

K

J

The overall entropy change is the sum of the three individual entropy values:

rxnS19.3 326.7 33.4 

K

J



rxnS312.6 

K

J



Although the change in entropy is similar to that at 25°C, it is slightly differ-

ent. This is an example of a relatively minor change in conditions. If the

temperatures were hundreds of degrees different from the reference temper-

ature, large changes in Swould be seen. If this were the case, temperature-

dependent functions would have to be used for the heat capacities, since their

being constant is also an approximation.

Example 3.9

What is the entropy change of the reaction

2H 2 (g) O 2 (g) →2 H 2 O ()

at 25°C and 300 atm? Assume molar quantities based on the balanced chem-

ical reaction. Assume also that a pressure change does not affect the entropy

of the liquid water product (that is,S 3 0).

Solution

This example is similar to Example 3.8, except that the pressure is non-

standard. Since S 3 is approximated as zero, we need only evaluate the S’s

of the first two steps:

1. The change in entropy as the pressure of the reactants goes from 300 atm
   to the standard pressure of 1 atm is
   S 1 

(2 mol)8.314 

mo

J

lK

ln 

3

1

00

at

a

m

tm

(1 mol)8.314 

mo

J

lK

ln

3

1

00

at

a

m

tm



where the first term is for the hydrogen and the second term is for the oxy-

gen. Solving:

S 1 142.3 

K

J



84 CHAPTER 3 The Second and Third Laws of Thermodynamics