Page 166 - Physical Chemistry

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Neglecting the volumes of the solid and liquid substances A and E, we have V° Section 5.4
cV° dV° bV° . The standard state of a gas is an ideal gas, so V° RT/P° Determination of Standard Enthalpies
m,C m,D m,B m of Formation and Reaction
for each of the gases C, D, and B. Therefore V° (c d b)RT/P°. The quantity
c d b is the total number of moles of product gases minus the total number of
moles of reactant gases. Thus, c d b is the change in the number of moles of gas
for the reaction. We write c d b n /mol, where n stands for moles of gas.
g g
Since c d b is a dimensionless number, we divided n by the unit “mole” to
g
make it dimensionless. We thus have V° ( n /mol)RT/P°, and (5.9) becomes
g
¢H° ¢U° ¢n RT>mol (5.10)
T
T
g
For example, the reaction C H (g) 5O (g) → 3CO (g) 4H O(l) has n /mol
3 8 2 2 2 g
3 1 5 3 and (5.10) gives H° U° 3RT. At 300 K, H° U°
T T
7.48 kJ/mol for this reaction, which is small but not negligible.
EXAMPLE 5.3 Calculation of U°from H°
f f

For CO(NH ) (s), H° 333.51 kJ/mol. Find U° of CO(NH ) (s).
2 2 f 298 f 298 2 2
The formation reaction is
C1graphite2 1 2 O 1g2 N 1g2 2H 1g2 S CO1NH 2 1s2
2
2
2
2 2
1
7
and has n /mol 0 2 1 . Equation (5.10) gives
2
2
g
7 3
¢ U° 333.51 kJ>mol 1 218.314 10 kJ>mol-K21298.15 K2
2
f
298
324.83 kJ>mol
Exercise
For CF ClCF Cl(g), H° 890.4 kJ/mol. Find U° of CF ClCF Cl(g).
2
2
298
f
2
2
298
f
(Answer: 885.4 kJ/mol.)
Exercise
In Example 5.2, U° of glucose was found to be 2801 kJ/mol. Find H° 298
298
c
c
of glucose. (Answer: 2801 kJ/mol.)
For reactions not involving gases, n is zero, and H° is essentially the same
g
as U° to within experimental error. For reactions involving gases, the difference
between H° and U°, though certainly not negligible, is usually not great. The
quantity RT in (5.10) equals 2.5 kJ/mol at 300 K and 8.3 kJ/mol at 1000 K, and
n /mol is usually a small integer. These RT values are small compared with typi-
g
cal H°values, which are hundreds of kJ/mol (see the H°values in the
f
Appendix). In qualitative reasoning, chemists often don’t bother to distinguish be-
tween H° and U°.
Hess’s Law
Suppose we want the standard enthalpy of formation H° of ethane gas at 25°C.
f 298
This is H° for 2C(graphite) 3H (g) → C H (g). Unfortunately, we cannot react
298 2 2 6
graphite with hydrogen and expect to get ethane, so the heat of formation of ethane
cannot be measured directly. This is true for most compounds. Instead, we determine