Bond Enthalpy (Bond Energy)

The Bond Enthalpy is the energy required to break a chemical bond. It is usually expressed in units of kJ mol^-1, measured at 298 K. The exact bond enthalpy of a particular chemical bond depends upon the molecular environment in which the bond exists. Therefore, bond enthalpy values given in chemical data books are averaged values.

ΔH = ∑ ΔH_{(bonds broken)} - ∑ ΔH_{(bonds formed)}

This basically means that you add up all the energies of the broken bonds; add up all the energies of the bonds that are reformed and subtract one from the other.

A specific example can be made from our old familiar combustion of methane reaction. We calculated the enthalpy change during this transformation before from traditional thermochemcial methods. We can do this again by using the average bond enthalpies of C-H, C=O, {O=O}, and O-H bonds.

EXAMPLE #1

FindH for the following reaction given the following bond energies:

Bond Bond Energy
(kJ/mol)

H-H 436

O=O 499

O-H 463

We have to figure out which bonds are broken and which bonds are formed.

2 H-H bonds are broken.

1 O=O bond is broken

2 O-H bonds are formed per water molecule, and there are 2 water molecules formed, therefore 4 O-H bonds are formed

Kent --------->

Calculator

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EXAMPLE #2

The complete combustion of propane can be represented by the following equation:

or we could redraw it to represent the bonds present:

Bond Type

Average bond enthalpy kJ/mol

C-H +413

C-C +347

O=O +498

C=O +805

H-O +464

Highlight below to reveal the answer

Total endothermic change
for bond breaking:
8xC-H

2xC-C

5xO=O

= (8 x +413) + 2(+347) + 5(+498)

Total exothermic change
for bond making:
6 x C=O
8 x O-H
= (6 x -805) + (8 x -464)

Bond Enthalpies
in kJ/mol
C-H = 413
C-C = 347
O-H = 464
O=O = 498
C=O = 805

DH= [(8x413)+(2x347)+(5x498)] - [(6x805)+(8x464)]
= - 2054 kJ mol^-1

EXAMPLE #3

The method involves breaking chemical bonds in the reactant molecules (an endothermic process) and forming new bonds in the products (an exothermic process).

Highlight below to reveal the answer

Total endothermic change
for bond breaking:
3 x C-H
1 x C-O
1 x O-H
1½ x O=O
= (3 x +413) + +336 + +464 + (1½ x +498)
= +2786 kJ
Total exothermic change
for bond making:
2 x C=O
4 x O-H
= (2 x -805) + (4 x -464)
= -3466 kJ
Bond Enthalpies
in kJ/mol
C-H = 413
C-O = 336
O-H = 464
O=O = 498
C=O = 805

\ DH = + 2786 + - 3466 = - 680 kJ/mol

Bond Type	Average bond enthalpy kJ/mol
C-H	+413
C-C	+347
O=O	+498
C=O	+805
H-O	+464

Total endothermic change for bond breaking: 3 x C-H 1 x C-O 1 x O-H 1½ x O=O = (3 x +413) + +336 + +464 + (1½ x +498) = +2786 kJ	Total exothermic change for bond making: 2 x C=O 4 x O-H = (2 x -805) + (4 x -464) = -3466 kJ	Bond Enthalpies in kJ/mol C-H = 413 C-O = 336 O-H = 464 O=O = 498 C=O = 805
\ DH = + 2786 + - 3466 = - 680 kJ/mol