Modern Engineering Thermodynamics

626 CHAPTER 15: Chemical Thermodynamics
was done earlier with their h f ° values. Therefore, we define the molar specific entropy of formation, s f °,ofa
compound as
ðs f °Þ compound
= s° compound −
∑
elements
Then, the molar specific Gibbs function of formation of a compound, g f °, is given by
n i /n compound
si ° (15.28)
ðg f °Þ compound
= ðh f °Þ compound
− T°ðs f °Þ compound
(15.29)
where T° is either 298 K or 537 R depending on whether SI or English units are being used.
Table 15.7 lists values of s° and g f ° for the same substances found in Table 15.1.
Note that, because the elements of the reaction are not considered to be compounds themselves (even though
they may be diatomic molecules), they cannot have an entropy of formation. Therefore, for all these elements,
we can set h f °=s f °=g f °=0:
EXAMPLE 15.13
As in Example 15.6, the Universe is about to come to an end, except this time it can be saved if you can determine the
molar specific entropy of formation and the molar specific Gibbs function of formation for methane gas CH 4 (g). Would
you please save the Universe again?
Solution
In Example 15.6, we discovered that there is no known reaction by which we can form methane gas by reacting solid carbon
with hydrogen gas. However, in Example 15.6, we determined the specific molar heat of formation of CH 4 using the
(hypothetical) reaction
CðsÞ + 2ðH 2
ðgÞÞ ! CH 4
Consequently, we can also determine the specific molar entropy of formation and the specific molar Gibbs function of formation
using the same reaction. Equation (15.28) gives
ðs f °Þ CH4
= s CH4 ° −
n C
n CH4
s C °+
n H2
n CH4
s H2 °
and values for s , s CH4 C , and s ° are found in Table 15.7. Then, the specific molar entropy of formation of methane is
H2
ðs°Þ f CH4 = 186:256 5:740 + 2ð130:684Þ = 80:852 kJ/kgmole . K
Equation (15.29) and Table 15.1 can be used to find the specific molar Gibbs function of formation of methane as
ðg f °Þ CH4
= ðh f °Þ CH4 − T°ðs f °Þ CH4
= −74:873 MJ/kgmole − ð298:15 KÞð−80:852 kJ/kgmole . KÞð1 MJ/1000 kJÞ
= −50:782 MJ/kgmole
And you have saved the Universe again. Thanks.
Exercises
37. Using the technique of Example 15.13, determine the molar specific entropy of formation of carbon dioxide, CO 2 .
Answer: ðs f °Þ CO2
= 2.92 kJ/(kgmole ·K).
38. Using the results of Exercise 37 and the technique of Example 15.13, determine the molar specific Gibbs function of
formation of carbon dioxide, CO 2 . Answer: ðg f °Þ CO2 = −394.4 MJ/(kgmole·K).
39. Following the technique of Example 15.13, determine the molar specific entropy of formation and the molar specific
Gibbs function of formation of water vapor, H 2 O(g). Answer: ðs f °Þ H2O = −44.400 kJ/(kgmole · K) and ðg f °Þ H2O =
−228.600 MJ/(kgmole·K).
15.12 CHEMICAL EQUILIBRIUM AND DISSOCIATION
In 1877, the Dutch chemist Jacobus Hendricus van’t Hoff (1852–1911) developed the basic principles of chemical
equilibrium from the fundamental laws of thermodynamics. For this, among other things, he won the first
Nobel Prize for Chemistry in 1901.