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Published byDonald Day Modified over 9 years ago
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Bond Enthalpy Chem 12
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The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. H 2 (g) H (g) + H 0 = 436.4 kJ Cl 2 (g) Cl (g) + H 0 = 242.7 kJ HCl (g) H (g) +Cl (g) H 0 = 431.9 kJ O 2 (g) O (g) + H 0 = 498.7 kJ OO N 2 (g) N (g) + H 0 = 941.4 kJ N N Bond Energy Bond Energies Single bond < Double bond < Triple bond 9.10
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Average bond energy in polyatomic molecules H 2 O (g) H (g) +OH (g) H 0 = 502 kJ OH (g) H (g) +O (g) H 0 = 427 kJ Average OH bond energy = 502 + 427 2 = 464 kJ 9.10
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Bond Energies (BE) and Enthalpy changes in reactions H 0 = total energy input – total energy released = BE(reactants) – BE(products) Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. 9.10
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H 2 (g) + Cl 2 (g) 2HCl (g)2H 2 (g) + O 2 (g) 2H 2 O (g)
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Use bond energies to calculate the enthalpy change for: H 2 (g) + F 2 (g) 2HF (g) H 0 = BE(reactants) – BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) HH1436.4 FF 1156.9 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) HF2568.21136.4 H 0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ 9.10
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