# Bond Enthalpies

The enthalpy of a reaction can be estimated by the bonds that are broken and bonds that are formed in a chemical reaction.
A covalent bond is a state of low energy between two atoms. When in a covalent bond, atoms need energy to move toward or away from each other. If atoms that form a bond receive sufficient energy, the bond may be broken. Bond enthalpy, or bond energy, is the energy needed to break a bond in 1 mole of a gaseous substance. For example, the energy needed to remove one hydrogen atom from a methane molecule is 438 kJ/mol:
${\rm{CH}}_{4}(g)\rightarrow{\rm{CH}}_{3}(g)+{\rm{H}}(g)\;\;\;\;\;\Delta H=438\;{\rm{kJ}}/\rm{mol}$
Note that both the CH3 and hydrogen atoms in this reaction are highly unstable. They will react with other substances quickly. Bond enthalpies in different compounds are not exactly the same, but they are similar. For example, it takes approximately 1,660 kJ/mol of energy to remove four hydrogen atoms from a methane molecule:
${\rm{CH}}_{4}(g)\rightarrow{\rm{C}}(g)+4{\rm{H}}(g)\;\;\;\;\;\Delta H=1660\;{\rm{kJ}}/\rm{mol}$
To determine the energy needed to remove each hydrogen atom, take 1660 kJ/mol divided by four for the number of hydrogen atoms, which yields 415 kJ/mol. This is less than the energy needed to remove the first hydrogen atom, which was 438 kJ/mol. However, the difference is not large. Scientists measure the strength of the covalent bonds between two atoms in a variety of compounds and calculate average bond strengths. The average ${\rm{C}{-}\rm{H}}$ bond strength, for example, is 413 kJ/mol.

### Average Bond Enthalpies of Common Bonds

Bond Average Bond Enthalpy (kJ/mol)
${\rm{C}{-}\rm{H}}$ 413
${\rm{C}{-}\rm{C}}$ 348
${\rm{C}{-}\rm{O}}$ 358
${\rm{C}{-}\rm{F}}$ 485
${\rm{C}{-}\rm{Cl}}$ 328
${\rm{O}{-}\rm{H}}$ 463
${\rm{O}{-}\rm{O}}$ 146
${\rm{F}{-}\rm{F}}$ 155
${\rm{Cl}{-}\rm{Cl}}$ 242
${\rm{H}{-}\rm{H}}$ 436
${\rm{H}{-}\rm{F}}$ 567
${\rm{H}{-}\rm{Cl}}$ 431
${\rm{Si}{-}\rm{H}}$ 323
${\rm{Si}{-}\rm{Si}}$ 226
${\rm{C}{=}\rm{C}}$ 614
${\rm{C}{\equiv}\rm{C}}$ 839
${\rm{C}{=}\rm{O}}$ 799
${\rm{N}{\equiv}\rm{N}}$ 941
${\rm{O}{=}\rm{O}}$ 495

The average bond enthalpy for a molecular bond can vary depending on the molecule in which the atoms are found.

A chemical reaction involves breaking and forming new bonds. A simplified model of a reaction would first involve the breaking of bonds, which requires energy, represented by positive values. Then the formation of new bonds releases energy, represented by negative values. Using this model and Hess's law, average bond strengths could be used to estimate the reaction enthalpy for a chemical reaction. The $\Delta H$ in this case would be:
$\Delta H=\text{Sum of enthalpies for broken bonds}-\text{Sum of enthalpies for bonds formed}$
Consider the combustion of methane:
${\rm{CH}}_{4}(g)+2{\rm{O}}_{2}(g)\rightarrow{\rm{CO}}_{2}(g)+2{\rm{H}}_{2}{\rm{O}}(g)$

### Bond Enthalpies of ${\rm{C}{-}\rm{H}}$ and ${\rm{O}{=}\rm{O}}$ for Methane Combustion

Bond Type Number of Bonds Bond Enthalpy kJ/mol
${\rm{C}{-}\rm{H}}$ 4 +413
${\rm{O}{=}\rm{O}}$ 2 +495

In the combustion reaction of methane, the ${\rm{C}{-}\rm{H}}$ and ${\rm{O}{=}\rm{O}}$ bonds are broken, so the bond enthalpies are positive.

### Bond Enthalpies of ${\rm{C}{=}\rm{O}}$ and ${\rm{O}{-}\rm{H}}$ for Methane Combustion

Bond Type Number of Bonds Bond Enthalpy kJ/mol
${\rm{C}{=}\rm{O}}$ 2 −799
${\rm{O}{-}\rm{H}}$ 4 −463

In the combustion reaction of methane, the ${\rm{C}{=}\rm{O}}$ and ${\rm{O}{-}\rm{H}}$ bonds are formed, so the bond enthalpies are negative.

In order to calculate the change in enthalpy for the combustion of methane, the sum of enthalpies for bonds formed is subtracted from the sum of enthalpies for broken bonds.
\begin{aligned}\Delta H&=\text{Sum of enthalpies for broken bonds}-\text{Sum of enthalpies for bonds formed}\\&=\left[(4)\Delta {H_{\rm{C-H\;bond}}}+(2)\Delta {H_{\rm{O=O\;bond}}}\right]-\left[(2)\Delta {H_{\rm{C=O\;bond}}}+(4)\Delta {H_{\rm {O-H\;bond}}}\right]\\&=\left[(4)(413\;{\rm{kJ}}/{\rm{mol}})+(2)(495\;{\rm{kJ}}/{\rm{mol}})\right]-\left[(2)(799\;{\rm{kJ}}/{\rm{mol}})+(4)(463\;{\rm{kJ}}/{\rm{mol}})\right]\\&=-808\;{\rm{kJ}}/{\rm{mol}}\end{aligned}
Note that a simplified version of this calculation would be to add all the bond enthalpies. For example:
\begin{aligned}\Delta H&=\left[(4)(+413\;{\rm{kJ}}/{\rm{mol}})\right]+\left[(2)(+495\;{\rm{kJ}}/{\rm{mol}})\right]+\left[(2)(-799\;{\rm{kJ}}/{\rm{mol}})\right]+\left[(4)(-463\;{\rm{kJ}}/{\rm{mol}})\right]\\&=-808\;{\rm{kJ}}/{\rm{mol}}\end{aligned}
Experimental measurements find $\Delta H$ for combustion of methane as –802.3 kJ/mol. There is a difference of about 6 kJ/mol. There will always be a difference between the real values and estimates based on bond enthalpies because bond enthalpies use average values.