1. Example Problem 9.49 CO2(g) 2 NH3(g)  + N2H4CO(g) + H2O(g)  + + 2 H
Use bond energies to determine the H for the following reaction:
CO2(g)
2 NH3(g)  +
N2H4CO(g) +
H2O(g) H N C O 2 H N O H  + + O C

2. Tabulated Bond Energies

3. Example Problem 9.49 CO2(g) 2 NH3(g)  + N2H4CO(g) + H2O(g)  + + 2 H
Use bond energies to determine the H for the following reaction:
CO2(g)
2 NH3(g)  +
N2H4CO(g) +
H2O(g) H N C O 2 H N O H  + + O C
Bonds Broken
Bonds Made
2 C=O Bonds
@ 799 kJ/mol
4 N-H Bonds
@ 391 kJ/mol
6 N-H Bonds
@ 391 kJ/mol
2 C-N Bonds
@ 305 kJ/mol
1 C=O Bond
@ 745 kJ/mol
2 O-H Bonds
@ 467 kJ/mol

4. Example Problem 9.49  + + 2 3944 kJ/mol 3853 kJ/mol H N C O H N O H O
Bonds Broken
Bonds Made
2 C=O Bonds
@ 799 kJ/mol
4 N-H Bonds
@ 391 kJ/mol
6 N-H Bonds
@ 391 kJ/mol
2 C-N Bonds
@ 305 kJ/mol
1 C=O Bond
@ 745 kJ/mol
2 O-H Bonds
@ 467 kJ/mol
Total Energy (endothermic)
Total Energy (exothermic)
(2  799) + (6  391) =
(4  391) + (2  305) + (1  745) + (2  467) =
3944 kJ/mol
3853 kJ/mol

5. Example Problem 9.49 3944 kJ/mol 3853 kJ/mol
Total Energy (endothermic)
Total Energy (exothermic)
(2  799) + (6  391) =
(4  391) + (2  305) + (1  745) + (2  467) =
3944 kJ/mol
3853 kJ/mol
H = B.E.reactants B.Eproducts
H = kJ/mol kJ/mol
H = +91 kJ/mol
(rxn is endothermic)