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nor isothermal, some heat flows to the surroundings and the temperature of the system Section 5.4
rises by an amount that is less than T under adiabatic conditions. Determination of Standard Enthalpies
of Formation and Reaction

EXAMPLE 5.4 Calculation of H° and U°from H°
c
f
f
The standard enthalpy of combustion H° of C H (g) to CO (g) and H O(l)
2
c
2
6
298
2
is 1559.8 kJ/mol. Use this H°and Appendix data on CO (g) and H O(l) to
c
2
2
find H° and U° of C H (g).
f
298
f
2
6
298
Combustion means burning in oxygen. The combustion reaction for one
mole of ethane is
C H 1g2 7 2 O 1g2 S 2CO 1g2 3H O1l2
6
2
2
2
2
The relation H° n H°[Eq. (5.6)] gives for this combustion
i
i
f
i
¢ H° 2 ¢ H°1CO , g2 3 ¢ H°1H O, l2 ¢ H°1C H , g2 7 2 ¢ H°1O , g2

2
f
f
2
2
c
2
f
f
6
Substitution of the values of H° of CO (g) and H O(l) and H°gives at 298 K
c
f
2
2
1559.8 kJ>mol 21 393.51 kJ>mol2 31 285.83 kJ>mol2
¢ H°1C H , g2 0
f
6
2
¢ H°1C H , g2 84.7 kJ>mol
6
f
2
Note that this example essentially repeats the preceding Hess’s law calculation.
Reactions (2) and (3) above are the formation reactions of CO (g) and H O(l).
2
2
To find U° from H° ,we must write the formation reaction for C H ,
298
2
f
298
6
f
which is 2C(graphite) 3H (g) → C H (g). This reaction has n /mol 1 3
6
g
2
2
2, and Eq. (5.10) gives
U° 84.7 kJ mol ( 2)(0.008314 kJ mol-K)(298.1 K) 79.7 kJ mol
f
298
A common student error is to find n from the combustion reaction instead of
g
from the formation reaction.
Exercise
4
H° 298 of crystalline buckminsterfullerene, C (cr), is 2.589 10 kJ/mol
c
60
[H. P. Diogo et al., J. Chem. Soc. Faraday Trans., 89, 3541 (1993)]. With the aid
3
of Appendix data, find H° of C (cr). (Answer: 2.28 10 kJ/mol.)
60
298
f
Calculation of H H re
id
The standard state of a gas is the hypothetical ideal gas at 1 bar. To find H°ofa gaseous
f
compound or a compound formed from gaseous elements, we must calculate the differ-
ence between the standard-state ideal-gas enthalpy and the enthalpy of the real gas
(steps 1 and 6 in the first part of Sec. 5.4). Let H (T, P°) be the enthalpy of a (real)
re
gaseous substance at T and P°, and let H (T, P°) be the enthalpy of the corresponding
id
fictitious ideal gas at T and P°, where P° 1 bar. H (T, P°) is the enthalpy of a hypo-
id
thetical gas in which each molecule has the same structure (bond distances and angles
and conformation) as in the real gas but in which there are no forces between the mole-
cules. To find H H ,we use the following hypothetical isothermal process at T:
id
re
1a2 1b2 1c2
Real gas at P° S real gas at 0 bar S ideal gas at 0 bar S ideal gas at P° (5.13)