Modern Engineering Thermodynamics

594 CHAPTER 15: Chemical Thermodynamics
With this proposition, he was able to show that hydrogen, oxygen, nitrogen, and the like exist as diatomic
molecules in nature. Modern experiments have determined that the actual number of molecules in any substance
whose mass in kilograms is equal to its (relative) molecular mass M is 6.022 × 10 26 . Therefore, the mass
of one molecule of this substance is
M
kilograms:
26
6:022 × 10
The acceptance of Avogadro’s law soon led to the development of the mole (sometimes abbreviated mol, both
being a contraction of the German word molekul) as a convenient chemical mass unit. Originally a “mole” was
defined as a mass in grams that was equal to the molecular mass M of a substance (i.e., originally a “mole” of
carbon-12 contained 12 grams of carbon-12). Today, we need to recognize that this “mole” is really a gram-mole,
or gmole, becausethe“mole” unit is used in so many different units systems today. Now, it must carry a prefix
showing the units system being used, such as gmole, kgmole, orlbmole. But remember that these “moles” are not
equal to each other, since 1 kgmole = 1000 gmole = 2.2046 lbmole.
Since equal “moles” of different substances contain the same number of molecules, the stoichiometric
coefficients, which initially represented only individual molecules, can also be used to represent the number
of moles of each element present. Therefore, Eq. (15.1) can be written in the equivalent form
or
or
2 gmole of A + 1 gmole of B = 1 gmole of C
2 kgmole of A + 1 kgmole of B = 1 kgmole of C
2 lbmole of A + 1 lbmole of B = 1 lbmole of C
and Eq. (15.2) can be interpreted as a reaction between 1 kgmole of H 2 and 0.5 kgmole of O 2 producing
1 kgmole of H 2 O or 1 lbmole of H 2 and 0.5 lbmole of O 2 producing 1 lbmole of H 2 O and so forth.
Most combustion processes occur in air, not pure oxygen. The composition of air used to determine its thermodynamic
properties on a molar or volume basis is 78.09% nitrogen, 20.95% oxygen, 0.93% argon, and 0.03%
carbon dioxide and trace elements. For convenience, we round this off to 79.0% N 2 and 21.0% O 2 .Indoing
this, we are essentially dividing air into two components: pure oxygen and a mixture of noncombustibles
(N 2 , Ar, CO 2 ). The noncombustible group, called atmospheric nitrogen, has a mole fraction composition of
and
n N2
78:09
χ N2
=
=
n N2 + n Ar + n CO2 78:09 + 0:93 + 0:03 = 78:09 = 0:9879 = 98:79%
79:05
n
χ Ar =
Ar
= 0:93 = 0:0118 = 1:18%
n N2 + n Ar + n CO2 79:05
χ CO2 =
n CO2
= 0:03 = 0:00038 = 0:038%
n N2 + n Ar + n CO2 79:05
where n N2 , n Ar , and n CO2 are the number of moles of nitrogen, argon, and carbon dioxide present in the mixture.
Then, from Eq. (12.11), the equivalent molecular mass of this mixture is
ðMÞ atmospheric
nitrogen
= ∑ χ i M i = ð0:9879Þð28:016Þ + ð0:0118Þð39:944Þ + ð0:00038Þð44:01Þ
= 28:16 kg/kgmole = 28:16 lbm/lbmole
From this point on, we refer to the atmospheric nitrogen mixture as simply nitrogen, and we assume that air has
a molar composition of 21.0% oxygen and 79.0% nitrogen, where the molecular mass of oxygen is still 32.00,
but the molecular mass of nitrogen is now 28.16 instead of 28.016. The equivalent molecular mass of air is still
28.97, since the argon and carbon dioxide are now merely grouped with the nitrogen.
For this air composition, each mole of oxygen is accompanied by 79:0/21:0 = 3:76 moles of nitrogen. Thus, if
the hydrogen combustion reaction described by Eq. (15.2) were carried out in air instead of pure oxygen, it
would be written as
H 2 + 0:5½O 2 + 3:76ðN 2 ÞŠ ! H 2 O + 1:88ðN 2 Þ
In this equation, the nitrogen is assumed to be inert and therefore passes through the reaction unchanged.