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74 CHAPTER 4 Thermochemistry

The prime in the integral indicates a “dummy variable” that is otherwise identical to
the temperature. This notation is needed because T appears in the upper limit of the inte-
gral. In Equation (4.18), H° 298.15 K is the absolute enthalpy at 1 bar and 298.15 K.
However, because there are no unique values for absolute enthalpies, it is useful to com-
bine similar equations for all reactants and products with the appropriate stoichiometic
coefficients to obtain the following equation for the reaction enthalpy at temperature T:
T
¢H° R,T =¢H° R,298.15 K + ¢C (T¿)dT¿ (4.19)
P
3
298.15 K
where

¢C (T¿) = a i P,i (4.20)
n C (T¿)
P
i
Recall that in our notation, ¢H° R or ¢H° f without an explicit temperature value implies
that T = 298.15 K. In Equation (4.20), the sum is over all reactants and products,
including both elements and compounds. A calculation of ¢H° R at an elevated tempera-
ture is shown in Example Problem 4.2.

EXAMPLE PROBLEM 4.2
Calculate ¢H° R, 1450 K for the reaction 1>2 H (g) + 1>2 Cl (g) ¡ HCl(g) and
2
2
1 bar pressure given that ¢H°(HCl,g) =-92.3 kJ mol -1 at 298.15 K and that
f
T -7 T 2 -1 -1
-3
C P,m (H ,g) = a29.064 - 0.8363 * 10 + 20.111 * 10 2 bJ K mol
2
K K
T -7 T 2 -1 -1
-3
C P,m (Cl ,g) = a31.695 + 10.143 * 10 - 40.373 * 10 bJ K mol
2
K K 2
T -7 T 2 -1 -1
-3
C P,m (HCl,g) = a28.165 + 1.809 * 10 + 15.464 * 10 bJ K mol
K K 2
2
over this temperature range. The ratios T>K and T >K 2 appear in these equations in
order to have the right units for the heat capacity.
Solution
1450
¢H° R,1450K =¢H° R,298.15K + ¢C (T)dT
P
3
298.15
T -7 T 2
-3
¢C (T) = c28.165 + 1.809 * 10 + 15.464 * 10
P
K K 2
1 -3 T -7 T 2
- a29.064 - 0.8363 * 10 + 20.111 * 10 b
2 K K 2
1 -3 T -7 T 2 -1 -1
- a31.695 + 10.143 * 10 - 40.373 * 10 bd J K mol
2 K K 2
T -7 T 2 -1 -1
-3
= a-2.215 - 2.844 * 10 + 25.595 * 10 b J K mol
K K 2
¢H° R,1450K =-92.3kJmol -1

1450
2
T -7 T T -1
+ a-2.215 - 2.844 * 10 + 25.595 * 10 b * d Jmol
-3
K K 2 K
3
298.15
=-92.3kJmol -1 - 2.836kJmol -1 =-95.1kJmol -1