3.1.4.4 - Bond Enthalpies

What Mean Bond Enthalpy Means

Breaking a covalent bond always requires energy, so bond breaking is endothermic. Forming a covalent bond releases energy, so bond formation is exothermic. Bond enthalpy data give us a way to measure those energy changes.

There are three ideas packed into that definition. First, the value is for one mole of bonds, not one mole of molecules. Second, everything is in the gaseous state, because the data refer to bonds in gaseous molecules rather than liquids or solids. Third, it is a mean value. A C-H bond in methane is not exactly the same strength as a C-H bond in chloroethane, so the data book gives an average taken across many compounds.

That is why bond enthalpy values are useful but not exact for a specific reaction. They tell you the typical strength of a bond type. They do not tell you the precise bond enthalpy for that exact bond in that exact molecule.

One detail worth keeping in mind is that bond enthalpy values in a table are always positive. That is because the table tells you the energy needed to break bonds. The negative sign for an exothermic reaction appears only after you compare energy in with energy out.

The Core Calculation Method

For a gaseous reaction, the approximate enthalpy change comes from a simple energy balance. Imagine all the relevant bonds in the reactants being broken to give separate gaseous atoms. Then imagine those atoms joining to form the bonds in the products.

The first sum is positive because bond breaking needs energy input. The second sum also uses positive table values, but we subtract it because bond formation releases energy.

The diagram below shows this bond enthalpy method as an energy path. Energy is absorbed to break the reactant bonds into separate gaseous atoms, and energy is released when those atoms form the bonds in the products.

[DIAGRAM: asset_name: 1.4.4 - Bond Enthalpies - Diagram 1; asset_slug: 1.4.4 - Bond Enthalpies - Diagram 1; recommended_method: retained_png; description: Breaking H–H and Cl–Cl bonds requires energy; forming two H–Cl bonds releases energy. The panels are a process schematic, not an enthalpy axis.]

A reliable way to organise the calculation is:

1. Write the balanced equation with state symbols.
2. Identify the bonds broken in the reactants.
3. Identify the bonds formed in the products.
4. Add the bond enthalpies for each list.
5. Use broken − formed, then check whether the sign makes chemical sense.

Worked example:

For $\ce{H2(g)} + \ce{Cl2(g)} \rightarrow 2\ce{HCl(g)}$, use $\ce{H}-\ce{H} = 436\,\mathrm{kJ}\,\mathrm{mol^{-1}}$, $\ce{Cl}-\ce{Cl} = 243\,\mathrm{kJ}\,\mathrm{mol^{-1}}$, and $\ce{H}-\ce{Cl} = 432\,\mathrm{kJ}\,\mathrm{mol^{-1}}$.

Energy in to break bonds:

Energy out when bonds form:

So:

The reaction is exothermic because more energy is released when two H-Cl bonds form than is needed to break the H-H and Cl-Cl bonds.

This method is limited to reactions in the gaseous phase for a good reason. If a reactant or product is a liquid or solid, changes of state would also affect the enthalpy change, and simple bond enthalpy data would not tell the whole story.

Making Larger Calculations Efficient

In bigger molecules, counting every single bond can be slow. A quicker approach is to focus on the bonds that actually change. Any bond present on both sides of the equation cancels out, because you would count it once as broken and once as formed.

Consider the reaction:

Most of the molecule stays the same. You do not need to count the C-C bond or the five unchanged C-H bonds. Only these bonds change:

• broken: one C-H and one Cl-Cl
• formed: one C-Cl and one H-Cl

Using $\ce{C}-\ce{H} = 413$, $\ce{Cl}-\ce{Cl} = 243$, $\ce{C}-\ce{Cl} = 346$, and $\ce{H}-\ce{Cl} = 432$ in $\mathrm{kJ}\,\mathrm{mol^{-1}}$:

Energy in:

$413 + 243 = 656$

Energy out:

$346 + 432 = 778$

So:

This is not a different method. It is the same calculation, just simplified before the arithmetic. That shortcut is especially helpful in substitution reactions, where only a small number of bonds actually change.

Finding an unknown bond enthalpy

The same equation can be rearranged when a bond enthalpy is unknown. For $\ce{H2(g) + Br2(g) -> 2HBr(g)}$, suppose $\Delta H=-72\ \mathrm{kJ\,mol^{-1}}$, $E(\ce{H-H})=436$ and $E(\ce{Br-Br})=193\ \mathrm{kJ\,mol^{-1}}$. Let $x=E(\ce{H-Br})$:

The coefficient 2 counts the two new H–Br bonds. Solve for the energy of one mole of bonds, not the total energy released by all bonds formed.

A very common error is to reverse the subtraction and do formed − broken. Another is to miss the bond multiplicity: each $\ce{NH3}$ molecule contains three N-H bonds, so $\ce{2NH3}$ contains six, not three.

Why mean bond data give an estimate

The bond-enthalpy method itself applies Hess's law, using separate gaseous atoms as an intermediate state. The difference is the data: mean bond values are approximate, whereas measured formation or combustion enthalpies refer to the particular substances and physical states.

Mean bond data introduce two important limitations. First, the bond enthalpies are average values. The real O-H, C-H, or C-Cl bonds inside a particular molecule may be slightly stronger or weaker than the mean table value. Second, bond enthalpy data refer to gaseous species. If you compare them with values based on substances in their standard states, state effects can create extra differences unless everything is gaseous.

So when a Hess's law value and a mean bond enthalpy value do not match, that does not mean the method has failed. It means the bond enthalpy method is an estimate built from average bond strengths, while Hess's law can use data for the exact substances and states involved in the reaction.

For instance, an estimate for gaseous water omits the heat released when water condenses. To compare it with a combustion enthalpy forming liquid water, subtract the enthalpy of vaporisation for each mole of water formed. Both calculations obey Hess's law; their initial and final states must match.