Modern Engineering Thermodynamics

15.6 Heat of Formation 605
and
H 2 O ! H 2 + ð1/2ÞðO 2 Þ + 285:8 MJ/kgmole H 2 O
Then, the combustion equation for the compound C n H m can be written as
C n H m + aðO 2 Þ!nðC + O 2 + 393:5Þ + ðm/2ÞðH 2 + ð1/2ÞO 2 + 285:8Þ+ HHV CnH m
! nðCÞ + ðm/2ÞðH 2 Þ+ ðn + m/4ÞðO 2 Þ
+
nð393:5Þ+ ðm/2Þð285:8Þ+ HHV Cn H m
Now, an oxygen balance on the original compound combustion equation gives a = n + m/4, so the O 2 terms in
the previous equation cancel, and again using Hess’s law to rearrange this equation, we get
nðCÞ+ ðm/2ÞðH 2 Þ ! C n H m −
nð393:5Þ+ ðm/2Þð285:8Þ+ HHV CnH m
(15.4)
! C n H m + q
f
C ° n H m
where the HHV is in MJ/kgmole of compound. Consequently, the heat of formation in MJ/kgmole of the hydrocarbon
fuel C n H m at the standard reference state is
q
f
° = −½nð393:5Þ + ðm/2Þð285:5Þ + HHV CnH m
Š in MJ/kgmole (15.5)
C nH m
The use of this relation is illustrated in the following example.
EXAMPLE 15.6
To prevent the Universe from collapsing in a deadly hypergeometric spiral, you must quickly determine the heat of formation
of methane gas CH 4 (g) at the standard reference state. Normally, you would react carbon and hydrogen gas in your
laboratory to form methane and measure its heat of formation during the reaction. Unfortunately, there is no known reaction
by which you can form methane by reacting solid carbon with hydrogen gas. How will you save the Universe?
Solution
Even though we do not know how to form CH 4 (g) from a direct reaction of solid carbon and hydrogen gas, we can use
Eqs. (15.4) and (15.5) with Tables 15.2 and 15.3 to calculate the heat of formation of CH 4 (g) at the standard reference
state. Tables 15.2 and 15.3 give the higher heating value (HHV) of CH 4 as −890.4 MJ/kgmole. Then Eq. (15.5) gives
ðq ° f Þ CH4
= − ½nð393:5Þ + ðm/2Þð285:5Þ + HHV CnHm Š
= − ½1ð393:5Þ + ð4/2Þð285:5Þ + HHV CH4 Š
= − ½393:5 + 2ð285:8Þ + ð−890:4ÞŠ = −74:7 MJ/kgmole of CH 4
and then Eq. (15.4) becomes
CðsÞ + 2½H 2 ðgÞŠ ! CH 4 ðgÞ − 74:7 MJ/kgmole of CH 4
In this example, we denote the physical state of the substances in parentheses as solid (s), liquid (l), or gas (g). Note that
the negative sign on the HHV of methane indicates that the combustion of methane is an exothermic (heat-producing)
reaction.
Exercises
16. The methane in Example 15.6 is replaced by acetylene gas. You must now determine the heat of formation of acetylene
gas, C 2 H 2 (g), at the standard reference state to save the Universe. Answer: ðq
f ° Þ acetylene
=+227 MJ=kgmole.
17. Oops, it is not methane or acetylene gas in Example 15.6, it is ethylene gas. So now, to save the Universe, you must
determine the heat of formation of ethylene gas, C 2 H 4 (g), at the standard reference state. Answer:
ðq
f ° Þ ethylene = −52:4MJ=kgmole.
18. Well, I bet you are surprised to find out that, at the last minute, it was ethane gas that was actually used to generate the
deadly hypergeometric collapse of the Universe. To save all life in the Universe, you must now determine the heat of
formation of ethane gas, C 2 H 6 (g), at the standard reference state. Answer: ðq ° f Þ ethane = −84:5MJ=kgmole:
Notice that, in this example, we do not take into account the heat of formation of H 2 from atomic hydrogen H
or O 2 from atomic oxygen O. This is because the elements used in the formation of a compound must be in
their stable molecular forms at the standard reference state. In the case of methane, the elements are solid carbon
(graphite), C, and diatomic hydrogen gas, H 2 .