Modern Engineering Thermodynamics

598 CHAPTER 15: Chemical Thermodynamics
EXAMPLE 15.1
Determine the stoichiometric reaction equation for methane (CH 4 ) burned in
a. 100.% theoretical air.
b. 150.% excess air.
c. 20.0% deficit air.
Solution
We could solve this problem for methane, CH 4 = C n H m by setting n = 1andm = 4inEqs.(15.3a–c).However,since
methane is a simple compound, it is more enlightening to carry out the individual atomic balances to obtain the correct
reaction equations.
a. The general combustion equation for 1 mole of methane in 100% theoretical air is
The element 3 balances are
Carbon (C) balance: 1 = b
Hydrogen (H 2 ) balance: 2 = c
Oxygen (O 2 ) balance: a = b + c/2 = 1 + 2/2 = 2
Nitrogen (N 2 ) balance: að3:76Þ = d = 2ð3:76Þ = 7:52
CH 4 + a½O 2 + 3:76ðN 2 ÞŠ ! bðCO 2 Þ + cðH 2 OÞ + dðN 2 Þ
The resulting stoichiometric equation for 100.% theoretical air is
CH 4 + 2½O 2 + 3:76ðN 2 ÞŠ ! CO 2 + 2ðH 2 OÞ + 7:52ðN 2 Þ
b. The 150.% excess air corresponds to 250.% theoretical air. The reaction equation now has O 2 in the products and
consequently has the form
CH 4 + 2:5ð2Þ½O 2 + 3:76ðN 2 ÞŠ ! aðCO 2 Þ + bðH 2 OÞ + cðO 2 Þ + dðN 2 Þ
and again element balances can be used to find that a = 1, b = 2, c = 3, and d = 2.5(2)(3.76) = 18.8; so that, for 150.%
excess air, we have
CH 4 + 5½O 2 + 3:76ðN 2 ÞŠ ! CO 2 + 2ðH 2 OÞ + 3ðO 2 Þ + 18:8ðN 2 Þ
c. The 20.0% deficit air corresponds to 80.0% theoretical air. Again, assuming all the hydrogen reacts to water, the reaction
now has CO in the products and has the form
CH 4 + 0:8ð2Þ½O 2 + 3:76ðN 2 ÞŠ ! aðCO 2 Þ + bðCOÞ + cðH 2 OÞ + dðN 2 Þ
and again the element balances can be used to yield the coefficients a = 0:2, b = 0:8, c = 2:0, and d = 6:016: Note that
these results correspond to the same coefficients one would obtain using Eq. (15.3c). The final reaction equation for
20.0% deficit air is
CH 4 + 1:6½O 2 + 3:76ðN 2 ÞŠ ! 0:2ðCO 2 Þ + 0:8ðCOÞ + 2ðH 2 OÞ + 6:016ðN 2 Þ
Exercises
1. Determine the number of kgmoles of water produced in the reaction of Example 15.1 per kgmole of methane burned in
200.% excess air. Answer: 2 kgmoles per kgmole of CH 4 .
2. Determine the molar and mass A/F ratios for parts a, b, and c in Example 15.1. Answer:
a. (A/F) molar = 9.52 lbmole air/lbmole CH 4 = 9.52 kgmole air/kgmole CH 4
(A/F) mass = 17.24 lbm air/lbm CH 4 = 17.24 kg air/kg CH 4
b. (A/F) molar = 23.8 lbmole air/lbmole CH 4 = 23.8 kg air/kg CH 4
(A/F) mass = 43.1 lbm air/lbm CH 4 = 43.1 lbm air/lbm CH 4
c. (A/F) molar = 7.616 lbmole air/lbmole CH 4 = 7.616 kg air/kg CH 4
(A/F) mass = 13.79 lbm air/lbm CH 4 = 13.79 lbm air/lbm CH 4
3. Rework Example 15.1 for combustion in 200% theoretical air. Answer: CH 4 + 4[O 2 + 3.76(N 2 )] → CO 2 + 2(H 2 O) +
2(O 2 ) + 15.04(N 2 ).
3 In modern chemistry, the term element refers to the stable form of a substance composed of only one kind of atom. The chemically stable forms of
carbon, hydrogen, oxygen, and nitrogen in these reactions are C, H 2 ,O 2 , and N 2 , rather than C, H, O, and N. So, even though H 2 ,O 2 , and N 2 are
really diatomic molecules, they are considered to be the proper forms for these elements in common chemical reactions.