Calculating bond enthalpies

In a nutshell

During a reaction, bonds are broken and formed and this can be measured as bond enthalpy. The total enthalpy change of a reaction can be calculated from the mean bond enthalpies of the individual bonds in the reactants and products.

Equations

Below is the equation that will be used to calculate total enthalpy change.

$Total\ \varDelta H=Bond\ enthalpy\ (reactants)-Bond \ enthlalpy\ (products)$

Bond breaking and making

Reactant bonds are broken and product bonds are formed during a reaction. Energy is absorbed when reactant bonds break so this process is endothermic and $\varDelta H$ is positive. Covalent and ionic bonding both have attractive forces holding the nuclei/ions together, so energy is needed to break this attraction. Energy is released when product bonds form so the process is exothermic and $\varDelta H$ is negative.

For an overall reaction, $\varDelta H$ will be negative if the energy released by the formation of new bonds is greater than the one absorbed to break bonds in the reactants. The opposite will happen if $\varDelta H$ is positive.

Mean bond enthalpies

Definitions

Bond enthalpy is the amount of energy needed to break one mole of a bond in the gaseous state. Bond enthalpy is measured in units of $kJ \, mol^{-1}$ .

Mean bond enthalpy is the average amount of energy required to break one mole of a bond in the gaseous state for various molecules. Mean bond enthalpies are an average therefore individual bond enthalpies of the same type of bond can vary.

Note: Bond enthalpies will always have a positive value as energy is absorbed (endothermic) to break bonds.

Example

$Bond\ enthalpy(CO_2)=+805\, kJ\ mol^{-1}$

$Bond\ enthalpy(CO)=+1077 \, kJ \, mol^{-1}$

More enthalpy is needed to break a $C-O$ bond for $CO$ than for $CO_2$ . This is because $CO_2$ consists of double bonds whereas $CO$ consists of triple bonds which are stronger and therefore require more enthalpy to break.

Mean bond enthalpies take the average of these bond enthalpies resulting in a value of:

$Mean\ bond\ enthalpy (C-O)=\underline{941 \, kJ \,mol^{-1}}$

Calculating enthalpy changes

Total enthalpy changes for reactions can be calculated if you know the mean bond enthalpies for the reactants and products.

$Total\ \varDelta H=Bond\ enthalpy\ (reactants)-Bond \ enthlalpy\ (products)$

PRocedure

1.	Find the sum of the mean bond enthalpy of reactants
2.	Find the sum of the mean bond enthalpy of products
3.	Subtract the mean bond enthalpy of products from reactants

Example

Determine the total $\varDelta H$ of the following reaction using mean bond enthalpies provided.

$2H_2+O_2\rightarrow 2H_2O$

Bond	Mean bond enthalpy
$H-H$	$432 \,kJ \, mol^{-1}$
$O=O$	$494 \,kJ \,mol^{-1}$
$O-H$	$459 \, kJ \,mol^{-1}$

Using the mean bond enthalpies, input the different values for the reactant and product bonds into this formula:

$Total\ \varDelta H=Bond\ enthalpy\ (reactants)-Bond \ enthlalpy\ (products)$

There are four $O-H$ bonds in total as each $H_2O$ has two $O-H$ bonds:

$Total\ \varDelta H=[(2\times H-H)+(O=O)]-[4\times O-H]$

$Total\ \varDelta H=[(2\times432)+494]-[4\times456]$

$Total\ \varDelta H=1358-1824$

$Total\ \varDelta H=-466 \,kJ \,mol^{-1}$

The total enthalpy change for the formation of water is $\underline{-466\ kJ\ mol^{-1}}$ .

Tip: If you know if a reaction is exothermic or endothermic, you can use this to check your answer. This is an exothermic reaction so $\varDelta H$ will be negative.

Calculating mean bond enthalpies

Mean bond enthalpies can be calculated if you know the total enthalpy change of a reaction and the mean bond enthalpies of all the other bonds involved.

Example

Determine the mean bond enthalpy for the bonds in methane using the data provided.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

$\varDelta H=-802 \,kJ \, mol^{-1}$

BOND	Mean bond enthalpy
$O=O$	$494 \,kJ \,mol^{-1}$
$C=O$	$799 \, kJ \,mol^{-1}$
$O-H$	$459 \, kJ \,mol^{-1}$

Using the mean bond enthalpies and enthalpy change, input the different values for the bonds of the reactants and products into this formula:

$Total\ \varDelta H=Bond\ enthalpy\ (reactants)-Bond \ enthlalpy\ (products)$

$\varDelta H=[(4\times C-H)+(2\times O-O)]-[(2\times C-O)+(4\times O-H)]$

$-802=[(4\times C-H)+(2\times 494)]-[(2\times 799)+(4\times 459)]$

$-802=[(4\times C-H)+988]-3434$

$-802=(4\times C-H)-2466$

Rearrange to find the bond enthalpy of the four bonds:

$4\times C-H=1664 \,kJ \,mol^{-1}$

Divide by four to find the mean bond enthalpy of a single bond:

$C-H=416 \,kJ \,mol^{-1}$

The mean bond enthalpy for a single $C-H$ bond in methane is $\underline{+416\ kJ\ mol^{-1}}$ .

Note: Always check that the calculated mean bond enthalpy is positive.