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Chapter 5 EXAMPLE 5.5 Change in H° with temperature
Standard Thermodynamic
Functions of Reaction
Use Appendix data and the approximation that C° is independent of T to esti-
P
mate H° for the reaction
1200
2CO1g2 O 1g2 S 2CO 1g2
2
2
Equation (5.19) gives
298
¢H° 1200 ¢H° 1200 K ¢C° dT (5.21)
P
298 K
Appendix H° and C° data give
f 298 P
¢H° >1kJ>mol2 21 393.5092 21 110.5252 0 565.968
298
¢C° >1J>mol-K2 2137.112 2129.1162 29.355 13.37
P,298
With the approximation T 2 C° dT C° T 2 dT, Eq. (5.21) becomes
T 1 P P,T 1 T 1
¢H° 1200 565968 J>mol 1 13.37 J>mol-K211200 K 298.15 K2

578.03 kJ>mol

Exercise
Use Appendix data and neglect the temperature dependence of C° to estimate
P
H° for O (g) → 2O(g). (Answer: 508.50 kJ/mol.)
1000 2

EXAMPLE 5.6 Change in H° with T

The C° ’s of the gases O , CO, and CO in the range 298 to 1500 K can each
2
2
P,m
be represented by Eq. (5.20) with these coefficients:
4
3
2
a/(J/mol-K) b/(J/mol-K ) c/(J/mol-K ) d/(J/mol-K )
O (g) 25.67 0.01330 3.764 10 6 7.310 10 11
2
CO(g) 28.74 0.00179 1.046 10 5 4.288 10 9
CO (g) 21.64 0.06358 4.057 10 5 9.700 10 9
2
Use these data and Appendix data to find for the reaction 2CO(g) O (g) →
2
2CO (g) an expression for H° in the range 298 K to 1500 K and calculate
T
2
H° 1200 . Was the approximation made in Example 5.5 justified?
We use Eq. (5.21). We have
¢C° 2C° P,m,CO 2 2C° P,m,CO C° P,m,O 2
P
Substitution of the series (5.20) for each C° P,m gives
2
3
¢C° ¢a T ¢b T ¢c T ¢d
P
where a 2a 2a a with similar equations for b, c, and d.
CO 2 CO O 2
Substitution of C° into (5.19) and integration gives
P
1 2 2 1 3 3 1 4 4
¢H° ¢H° ¢a1T T 2 ¢b1T T 2 ¢c1T T 2 ¢d1T T 2
4
1
2
2
2
2
1
2
3
1
1
T 2 T 1
Substitution of values in the table gives
¢a>1J>mol-K2 2121.642 2128.742 25.67 39.87