enthalpy of formation of all the stable elements (carbon atom C; molecular
hydrogen H 2 ; molecular nitrogen N 2 ; molecular oxygen O 2 ) at standard temperature
and pressure to be zero. Step 2 is an experimental program of determining the
enthalpy of formation of species that is directly related to the stable elements by
simple reactions. Take the example
H 2 þ 1 = 2 O 2 ! H 2 O þ Q to be removed
The Q to be removed is found experimentally to be 141,780 kJ/kg or (2.016)
(141780) = 285,830 kJ/kmol for the case of liquid water in the product mixture.
The energy balance implies that
H H 2 þ H1
2 O 2 ¼ H H 2 O l
ð Þ þ 285; 830
Since the enthalpy of formation of both hydrogen, H H 2 , and oxygen, H1
2 O 2 , are zero,
the enthalpy of formation of one kmol of water (liquid)—designated now by the
notation h
0
f
H 2 o
—satisfies the enthalpy balance
0 ¼ h
0
f
H 2 O l
ð Þ
þ 285; 830
Therefore,
h
0
f
H 2 O l
ð Þ
¼ À285; 830 kJ=kmol
The corresponding value for water in vapor phase in the product is
h
0
f
H 2 O g
ð Þ
¼ À241; 820 kJ=kmol
Consider another elementary reaction
C þ O 2 ! CO 2 þ Q to be removed
The Q to be removed is found experimentally in this case to be 393,520 kJ/kmol.
Therefore,
h
0
f
CO 2
¼ À393; 520 kJ=kmol
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9 Applications to Special States of Thermodynamic Equilibrium …