Physical Chemistry Third Edition

3.5 The Third Law of Thermodynamics and Absolute Entropies 143

Solution

Using values from Table A.8,

∆S

◦

(2)(213.64JK−^1 mol−^1 )+(−2)(197.564JK−^1 mol−^1 )

+(−1)(205.029JK−^1 mol−^1 )

− 172 .88JK−^1 mol−^1

Exercise 3.19

a.Explain in molecular terms why∆Sof the reaction in Example 3.16 is negative.

b.Calculate∆H◦for this reaction.

c.Without doing any calculation make a prediction about the entropy change of the surroundings

for this reaction.

d.Assuming that the surroundings remain at equilibrium at 298.15 K, calculate the entropy

change of the surroundings and of the universe for 1 mol of the reaction in Example 3.16.

Entropy Changes of Chemical Reactions at Various

Temperatures

If you desire an entropy change for a chemical reaction at some temperature not found

in a table, you can calculate it by using the fact that entropy is a state function. The

procedure is analogous to that used with the enthalpy in Chapter 2. Assume that the

entropy change at temperatureT 1 is known and its value at temperatureT 2 is desired.

The entropy change at temperatureT 2 is equal to the entropy change to bring the

reactants from temperatureT 2 to temperatureT 1 plus the entropy change of the reaction

at temperatureT 1 plus the entropy change to bring the products from temperatureT 1 to

temperatureT 2. If there are no phase changes betweenT 2 andT 1 , the result is analogous

to Eq. (2.7-19),

∆S(T 2 )∆S(T 1 )+

∫T 2

T 1

∆CP

T

dT (3.5-8)

where∆CPis defined in Eq. (2.7-20). If∆CPis temperature-independent,

∆S(T 2 )∆S(T 1 )+∆CPln

(

T 2

T 1

)

(∆CPconstant) (3.5-9)

EXAMPLE3.17

Calculate∆S◦for the reaction of the previous example at 373.15 K. Assume that the heat

capacities are temperature-independent.

Solution

Using values from the appendix for 298.15 K,

∆CP2(37.129JK−^1 mol−^1 )−2(29.142JK−^1 mol−^1 )− 29 .376JK−^1 mol−^1

− 13 .402JK−^1 mol−^1