3.5 The Third Law of Thermodynamics and Absolute Entropies
143
Solution
Using values from Table A.8,
∆S
◦ (2)(213.64 J K −1 mol −1 ) + (−2)(197.564 J K −1 mol −1 )
+ (−1)(205.029 J K −1 mol −1 )
−172.88 J K −1 mol −1
Exercise 3.19
a. Explain in molecular terms why ∆S of 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 temperature T 1 is known and its value at temperature T 2 is desired.
The entropy change at temperature T 2 is equal to the entropy change to bring the
reactants from temperature T 2 to temperature T 1 plus the entropy change of the reaction
at temperature T 1 plus the entropy change to bring the products from temperature T 1 to
temperature T 2 . If there are no phase changes between T 2 and T 1 , the result is analogous
to Eq. (2.7-19),
∆S(T 2 ) ∆S(T 1 ) +
T 2
T 1
∆C P
T
dT
(3.5-8)
where ∆C P is defined in Eq. (2.7-20). If ∆C P is temperature-independent,
∆S (T 2 ) ∆S (T 1 ) + ∆C P ln
T 2
T 1
(∆C P constant)
(3.5-9)
E X A M P L E 3.17
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,
∆C P 2(37.129 J K −1 mol −1 ) − 2(29.142 J K −1 mol −1 ) − 29.376 J K −1 mol −1
−13.402 J K −1 mol −1
143
Solution
Using values from Table A.8,
∆S
◦ (2)(213.64 J K −1 mol −1 ) + (−2)(197.564 J K −1 mol −1 )
+ (−1)(205.029 J K −1 mol −1 )
−172.88 J K −1 mol −1
Exercise 3.19
a. Explain in molecular terms why ∆S of 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 temperature T 1 is known and its value at temperature T 2 is desired.
The entropy change at temperature T 2 is equal to the entropy change to bring the
reactants from temperature T 2 to temperature T 1 plus the entropy change of the reaction
at temperature T 1 plus the entropy change to bring the products from temperature T 1 to
temperature T 2 . If there are no phase changes between T 2 and T 1 , the result is analogous
to Eq. (2.7-19),
∆S(T 2 ) ∆S(T 1 ) +
T 2
T 1
∆C P
T
dT
(3.5-8)
where ∆C P is defined in Eq. (2.7-20). If ∆C P is temperature-independent,
∆S (T 2 ) ∆S (T 1 ) + ∆C P ln
T 2
T 1
(∆C P constant)
(3.5-9)
E X A M P L E 3.17
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,
∆C P 2(37.129 J K −1 mol −1 ) − 2(29.142 J K −1 mol −1 ) − 29.376 J K −1 mol −1
−13.402 J K −1 mol −1