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# Bond Enthalpy

Bond enthalpy, also known as bond dissociation energy or, simply, ‘bond energy’, refers to the amount of energy you will need to break up the bonds in one mole of a covalent substance into separate atoms.

Bond enthalpy (E) is the amount of energy required to break one mole of a specific covalent bond in the gas phase.

If you are asked for the definition of bond enthalpy in your exams, you must include the part about the substance being in the gas phase. Additionally, you can only do bond enthalpy calculations on substances in the gas phase.

We show the specific covalent bond being broken by putting it in brackets after the symbol E. For example, you write the bond enthalpy of one mole of diatomic hydrogen (${H}_{2}$) as E (H-H).

A diatomic molecule is simply one that has two atoms in it like H2 or O2 or HCl.

• Over the course of this article, we will define bond enthalpy.
• Discover mean bond energies.
• Learn how to use mean bond enthalpies to work out ΔH of a reaction.
• Learn how to use enthalpy of vapourisation in bond enthalpy calculations.
• Uncover the relationship between bond enthalpy and trends in the enthalpies of combustion of a homologous series.

## What is meant by bond enthalpy?

What happens if the molecule we are dealing with has more than one bond to break? As an example, methane (${\mathrm{CH}}_{4}$) has four C-H bonds. All four hydrogens in methane are bonded to carbon with a single bond. You might expect the bond enthalpy for all four bonds to be the same. In reality, each time we break one of those bonds we change the environment of the bonds left over. The strength of a covalent bond is affected by the other atoms in the molecule. This means the same type of bond can have different bond energies in different environments. The O-H bond in water, for example, has a different bond energy to the O-H bond in methanol. Since bond energies are affected by the environment, we use the mean bond enthalpy.

Mean bond energy (also called average bond energy) is the amount of energy needed to break a covalent bond into gaseous atoms averaged over different molecules.

Average bond enthalpies are always positive (endothermic) as breaking bonds always requires energy.

Essentially, an average is taken from the bond enthalpies of the same type of bonds in different environments. The values of bond enthalpy you see in a data book may vary slightly because they are average values. As a result, calculations using bond enthalpies will only be approximate.

## How to find ∆H of a reaction using bond enthalpies

We can use mean bond enthalpy figures to calculate the enthalpy change of a reaction when it is not possible to do so experimentally. We can apply Hess' Law by using the following equation:

${\mathbf{H}}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\sum }\mathbf{}\mathbf{Bond}\mathbf{}\mathbf{enthalpies}\mathbf{}\mathbf{broken}\mathbf{}\mathbf{in}\mathbf{}\mathbf{reactants}\mathbf{}\mathbf{-}\mathbf{}\mathbf{\sum }\mathbf{}\mathbf{bond}\mathbf{}\mathbf{enthalpies}\mathbf{}\mathbf{formed}\mathbf{}\mathbf{in}\mathbf{}\mathbf{products}$

Fig. 1 - Using bond enthalpies to find ∆H

Calculating ΔH of a reaction using bond enthalpies will not be as accurate as using enthalpy of formation/combustion data, because bond enthalpy values are usually the mean bond energy - an average over a range of different molecules.

Now let's practice bond enthalpy calculations with some examples!

Remember you can only use bond enthalpies as long as all the substances are in the gas phase.

Calculate ∆H for the reaction between carbon monoxide and steam in the manufacture of hydrogen. The bond enthalpies are listed below.

${\mathrm{CO}}_{\left(\mathrm{g}\right)}+{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{g}\right)}\to {\mathrm{H}}_{{2}_{\left(\mathrm{g}\right)}}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)}$

 Bond Type Bond Enthalpy (kJ${\mathrm{mol}}^{-1}$) C-O (carbon monoxide) +1077 C=O (carbon dioxide) +805 O-H +464 H-H +436

We will use a Hess cycle in this example. Let’s begin by drawing a Hess cycle for the reaction.

Fig. 2 - Bond enthalpy calculation

Now let’s break the covalent bonds in each molecule into single atoms using their given bond enthalpies. Remember:

• There are two O-H bonds in ${\mathrm{H}}_{2}\mathrm{O}$,
• One C-O bond in CO,
• Two C-O bonds in ${\mathrm{CO}}_{2}$,
• And one H-H bond in ${\mathrm{H}}_{2}$.

Fig. 3 - Bond enthalpy calculation

You can now use Hess’ Law to find an equation for the two routes.

$∆{\mathrm{H}}_{\mathrm{r}}=\mathrm{\Sigma }\mathrm{bond}\mathrm{enthalpies}\mathrm{broken}\mathrm{in}\mathrm{reactants}-\mathrm{\Sigma }\mathrm{bond}\mathrm{enthalpies}\mathrm{formed}\mathrm{in}\mathrm{products}$

∆H = [ 2(464) +1077 ] - [ 2(805) + 436 ]

∆H = -41 kJ ${\mathrm{mol}}^{-1}$

In the next example, we won't use a Hess cycle - you simply count the number of bond enthalpies broken in the reactants and the number of bond enthalpies formed in the products. Let’s have a look!

Some exams might specifically ask you to calculate ∆H using the following method.

Calculate the enthalpy of combustion for ethylene shown below, using the given bond enthalpies.

$2{\mathrm{C}}_{2}{\mathrm{H}}_{2\left(\mathrm{g}\right)}+5{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{g}\right)}+4{\mathrm{CO}}_{2\left(\mathrm{g}\right)}$

 Bond Type Bond Enthalpy (kJ${\mathrm{mol}}^{-1}$) C-H +414 C=C +839 O=O +498 O-H +463 C=O +804

Enthalpy of combustion is the change in enthalpy when one mole of a substance reacts in excess oxygen to make water and carbon dioxide.

You must begin by rewriting the equation so that we have one mole of ethylene.

$2{\mathrm{C}}_{2}{\mathrm{H}}_{2}+5{\mathrm{O}}_{2}\to 2{\mathrm{H}}_{2}\mathrm{O}+4{\mathrm{CO}}_{2}$

${{\mathbf{C}}}_{{\mathbf{2}}}{{\mathbf{H}}}_{{\mathbf{2}}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{2}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{O}}}_{{\mathbf{2}}}{\mathbf{}}{\mathbf{\to }}{\mathbf{}}{{\mathbf{H}}}_{{\mathbf{2}}}{\mathbf{O}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{2}}{{\mathbf{CO}}}_{{\mathbf{2}}}$

Count the number of bonds being broken and the number of bonds being formed:

 Bonds Broken Bonds Formed 2 x (C-H) = 2(414) 2 x (O-H) = 2(463) 1 x (C=C) = 839 4 x (C=O) = 4(804) ${2}\frac{1}{2}$x (O=O) = $2\frac{1}{2}$(498) Total 2912 4142

Fill the values in the equation below

$∆{\mathrm{H}}_{\mathrm{r}}=\mathrm{\Sigma }\mathrm{bond}\mathrm{enthalpies}\mathrm{broken}\mathrm{in}\mathrm{reactants}-\mathrm{\Sigma }\mathrm{bond}\mathrm{enthalpies}\mathrm{formed}\mathrm{in}\mathrm{products}$

$∆{\mathrm{H}}_{\mathrm{r}}$ = 2912 - 4142

$∆{\mathrm{H}}_{\mathrm{r}}$ = -1230 ${\mathrm{kJmol}}^{-1}$

That's it! You have calculated the enthalpy change of reaction! You can see why this method might be easier than using a Hess cycle.

Perhaps you are curious about how you would calculate ∆H of a reaction if some of the reactants are in the liquid phase. You will need to change the liquid to a gas using what we call the enthalpy change of vaporisation.

Enthalpy of vaporisation ($∆{\mathrm{H}}_{\mathrm{vap}}$) is simply the enthalpy change when one mole of a liquid turns to gas at its boiling point.

To see how this works, let's do a calculation where one of the products is a liquid.

The combustion of methane is shown below.

${\mathrm{CH}}_{4\left(\mathrm{g}\right)}+2{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)}$

Calculate the enthalpy of combustion using the bond dissociation energies in the table.

 Bond Type Bond Enthalpy C-H +413 O=O +498 C=O (carbon dioxide) +805 O-H +464

One of the products, ${\mathrm{H}}_{2}\mathrm{O}$, is a liquid. We have to change it to a gas before we can use bond enthalpies to calculate ∆H. The enthalpy of vaporisation of water is +41 kJ${\mathrm{mol}}^{-1}$.

 Bonds Broken (kJ${\mathrm{mol}}^{-1}$) Bonds Formed (kJ${\mathrm{mol}}^{-1}$) 4 x (C-H) = 4(413) 4 x (O-H) = 4(464) + 2(41) 2 x (O=O) = 2(498) 2 x (C-O) = 2(805) Total 2648 3548

Use the equation:

$∆{\mathrm{H}}_{\mathrm{r}}$ = ∑bond enthalpies broken in reactants - ∑bond enthalpies formed in products

∆H = 2648 - 3548

∆H = -900 kJ${\mathrm{mol}}^{-1}$

Before we round up this lesson, here’s one last interesting thing related to bond enthalpy. We can observe a trend in the enthalpies of combustion in a 'homologous series'.

A homologous series is a family of organic compounds. Members of a homologous series share similar chemical properties and a general formula. For example, alcohols contain an -OH group in their molecules and the suffix ‘-ol’.

Have a look at the table below. It shows the number of carbon atoms, the number of hydrogen atoms and enthalpy of combustion of members of the alcohol homologous series. Can you see a pattern?

Fig. 4 - Trends in the combustion enthalpies of a homologous series

Notice there is a steady increase in the enthalpy of combustion as:
• The number of carbon atoms in the molecule increases.
• The number of hydrogen atoms in the molecule increases.

This is due to the number of C bonds and H bonds being broken in the combustion process. Each successive alcohol in the homologous series has an extra$-{\mathrm{CH}}_{2}$ bond. Each extra $-{\mathrm{CH}}_{2}$ increases the enthalpy of combustion for this homologous series by approximately 650kJmol-1.

This is actually really handy if you want to calculate the enthalpies of combustion for a homologous series because you can use a graph to predict the values! The calculated values from the graph are, in a sense, ‘better’ than the experimental values obtained from calorimetry. The experimental values end up being a lot smaller than the calculated ones due to factors such as heat loss and incomplete combustion.

Fig. 5 - Combustions enthalpy of a homologous series, calculated and experimental values

## Bond Enthalpy - Key takeaways

• Bond enthalpy (E) is the amount of energy required to break one mole of a specific covalent bond in the gas phase.
• Bond enthalpies are affected by their environment; the same type of bond can have different bond energies in different environments.
• Enthalpy values use the mean bond energy which is an average over different molecules.
• We can use the mean bond energy to calculate the ΔH of a reaction by using the formula: ΔH = Σ bond energies broken - Σ bond energies made.
• You can only use bond enthalpies to calculate ∆H when all substances are in the gas phase.
• There is a steady increase in the enthalpies of combustion in a homologous series due to the number of C bonds and H bonds being broken in the combustion process.
• We can graph this trend to calculate the enthalpies of combustion of a homologous series without needing calorimetry.
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What is bond enthalpy?

Bond enthalpy (E) is the amount of energy required to break one mole of a specific covalent bond in the gas phase. We show the specific covalent bond being broken by putting it in brackets after the symbol E. For example, you write the bond enthalpy of one mole of diatomic hydrogen (H2) as E (H-H).

How do you calculate average bond enthalpy?

Chemists find bond enthalpies by measuring the energy required to break one mole of a specific covalent molecule into single gaseous atoms. Bond enthalpy is calculated as an average over different molecules known as mean bond enthalpy. This is because the same type of bond can have different bond enthalpies in different environments.

Why do bond enthalpies have positive values?

Average bond enthalpies are always positive (endothermic), as breaking bonds always requires energy from the environment.

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