How to Decide Which Bonds Are Broken and Formed in a Reaction

A reaction equation tells you which substances react and which products form. It doesn’t immediately tell you which bonds have to be broken, which new bonds appear, or how many of each are involved.

Consider the addition of hydrogen chloride to propene:

\[ \ce{CH3-CH=CH2 + HCl -> CH3-CH(Cl)-CH3} \]

The carbon-carbon double bond becomes a single bond. Hydrogen and chlorine end up attached to different carbon atoms.

Before looking at the energy changes, predict which bonds disappear and which bonds appear. Is the double bond the only bond that changes?

No. The H-Cl bond also disappears, while a new C-H bond and a new C-Cl bond form. The carbon-carbon double bond is replaced by a single C-C bond.

This is the central skill: compare the bonds present in the reactants with the bonds present in the products, using their structural formulas and the coefficients in the balanced equation.

Once you can do that reliably, calculating reaction enthalpy from bond enthalpies becomes much easier.

01How to identify bonds broken and formed

The most reliable method is to build a bond inventory.

A bond inventory is simply a count of every type of bond on each side of a balanced chemical equation.

Follow this process:

1. Write the balanced chemical equation.
2. Draw or inspect the structural formula of every reactant and product.
3. Count each type of bond in one molecule of each substance.
4. Multiply each count by the substance’s stoichiometric coefficient.
5. Compare the total bond counts on the reactant and product sides.
6. Identify the bonds that disappear and the bonds that appear.

Let’s apply this to the propene reaction.

Count the bonds in the reactants

Propene has the structural formula:

\[ \ce{CH3-CH=CH2} \]

It contains:

• One C-C single bond.
• One C=C double bond.
• Six C-H bonds.

Hydrogen chloride, \(\ce{HCl}\), contains one H-Cl bond.

The reactants therefore contain the following bond inventory:

Count the bonds in the products

The product, 2-chloropropane, has the structural formula:

\[ \ce{CH3-CH(Cl)-CH3} \]

It contains:

• Two C-C single bonds.
• Seven C-H bonds.
• One C-Cl bond.

Its bond inventory is:

Compare the two inventories

Now compare the totals.

Therefore:

Bonds broken:

• One C=C double bond.
• One H-Cl bond.

Bonds formed:

• One C-C single bond.
• One C-H bond.
• One C-Cl bond.

Notice that the number of carbon atoms and hydrogen atoms is conserved. However, their bonding arrangements change.

This is why counting atoms alone is not enough. You need to know how those atoms are connected.

02Three bond-counting rules that prevent mistakes

Before calculating bond energies, get these details right.

Rule 1: A double bond is not two single bonds

A C=C double bond is counted as one C=C bond.

A C≡C triple bond is counted as one C≡C bond.

Each bond type has its own bond enthalpy.

For example:

\[ D(\ce{C-C})\neq D(\ce{C=C}) \]

and

\[ D(\ce{C=C})\neq 2D(\ce{C-C}) \]

Here, \(D\) represents bond enthalpy.

A double bond contains two shared electron pairs, but its bond enthalpy is not simply twice that of a single bond. The bonding interactions are different.

Rule 2: Multiply every bond count by the coefficient

Consider the balanced equation:

\[ \ce{2H2(g) + O2(g) -> 2H2O(g)} \]

One hydrogen molecule contains one H-H bond.

One oxygen molecule contains one O=O bond.

One water molecule contains two O-H bonds.

But the equation produces two water molecules.

Therefore:

Bonds broken:

• Two H-H bonds.
• One O=O bond.

Bonds formed:

• Four O-H bonds.

The coefficient of 2 multiplies every bond in the water molecule, not just one selected bond.

Rule 3: The molecular formula may not tell you enough

Consider two substances with the molecular formula \(\ce{C4H10}\):

• Butane: \(\ce{CH3-CH2-CH2-CH3}\)
• 2-methylpropane: \(\ce{CH3-CH(CH3)-CH3}\)

They have the same number of carbon and hydrogen atoms, but their structures are different.

If you only look at the molecular formula, you miss the arrangement of the bonds.

Always use a displayed or structural formula when the connectivity matters.

03What does it actually mean to break and form bonds?

There is an important distinction between bond counting and describing a reaction mechanism.

When calculating reaction enthalpy using average bond enthalpies, we use an imaginary two-stage process:

1. Break all the bonds in the gaseous reactants, separating the atoms.
2. Rearrange those atoms and form all the bonds in the gaseous products.

This is an energy-accounting model. It is not necessarily how the actual reaction happens.

For example, a reaction might involve several intermediate steps, collisions, and temporary molecular arrangements. Our bond-energy calculation does not attempt to describe those events.

It only compares the energy required to separate the reactant atoms with the energy released when the product bonds form.

Can you cancel bonds that appear on both sides?

Yes, provided you understand what the cancellation means.

Suppose a bond inventory contains six C-H bonds in the reactants and seven C-H bonds in the products.

You can cancel six matching C-H contributions, leaving one additional C-H bond formed.

This is a mathematical simplification.

It does not prove that six particular C-H bonds physically remain untouched during the reaction.

There is also a limitation: when using average bond enthalpies, cancellation assumes that the same average value is used for the bond type on both sides. If a question supplies different bond enthalpies for different molecular environments, keep those contributions separate.

04Using the bond inventory to calculate reaction enthalpy

Once you have identified the bonds, you can estimate the reaction enthalpy.

Bond enthalpy is the energy required to break one mole of a particular type of bond in gaseous molecules.

It is usually represented by \(D\) and measured in \(\mathrm{kJ\,mol^{-1}}\).

Breaking a bond requires energy, so its contribution is positive.

Forming a bond releases energy, so its contribution is negative.

This gives the bond-enthalpy equation:

\[ \boxed{ \Delta H_{\mathrm{rxn}} \approx \sum D_{\mathrm{broken}} - \sum D_{\mathrm{formed}} } \]

where:

• \(\Delta H_{\mathrm{rxn}}\) is the estimated reaction enthalpy.
• \(D_{\mathrm{broken}}\) represents the bond enthalpies of bonds broken.
• \(D_{\mathrm{formed}}\) represents the bond enthalpies of bonds formed.
• The summations include every bond, multiplied by the number of moles of that bond involved.

The sign convention is important:

If the result is negative, the reaction is exothermic.

If the result is positive, the reaction is endothermic.

05Complete thermochemical equations: a different kind of calculation

So far, we have used average bond enthalpies to estimate reaction enthalpy.

But sometimes a question gives you complete thermochemical equations with their enthalpy changes already supplied.

In that situation, you should use Hess’s Law rather than trying to count bonds.

A complete thermochemical equation contains:

• A balanced chemical equation.
• Physical states for every substance.
• An enthalpy change, \(\Delta H\), for the reaction as written.

For example:

\[ \ce{H2(g) + \frac{1}{2}O2(g) -> H2O(g)} \]

\[ \Delta H^\circ=-241.8\ \mathrm{kJ\,mol^{-1}} \]

The physical state matters. Water vapour and liquid water are different thermodynamic states, even though both contain the same H-O-H bonds.

Three rules for manipulating thermochemical equations

When adding equations, cancel identical substances appearing on opposite sides.

The substances must be in the same physical state to cancel directly.

For example, \(\ce{H2O(g)}\) and \(\ce{H2O(l)}\) cannot simply be cancelled against each other.

06Why bond-energy estimates differ from measured values

Here is a distinction worth remembering for exam questions.

A bond-energy calculation and a measured reaction enthalpy are not necessarily the same thing.

Average bond enthalpies produce a model estimate.

They use average values for particular bond types, generally measured across different gaseous molecular environments.

A C-H bond in one molecule may not have exactly the same bond enthalpy as a C-H bond in another molecule.

The average value simplifies this variation.

Experimental reaction enthalpies describe the actual reaction.

They can be determined experimentally using calorimetry or obtained from reliable thermochemical data.

They apply to the specified reactants and products, including their physical states.

This gives us an important distinction:

There is another important limitation.

Average bond enthalpies describe the breaking and forming of covalent bonds in gaseous molecules. They do not automatically include energy changes associated with vaporisation, condensation, dissolving, or separating ions in a crystal lattice.

For example, changing liquid water into water vapour does not require breaking the covalent O-H bonds inside each water molecule. It requires overcoming intermolecular attractions between water molecules.

If a bond-energy estimate and a supplied experimental value differ, do not automatically assume that your calculation is wrong.

First check:

1. Did you count the correct bonds?
2. Did you apply every stoichiometric coefficient?
3. Did you use the correct bond enthalpy for each bond type?
4. Are the physical states consistent?
5. Is the difference reasonably explained by using average rather than molecule-specific bond enthalpies?

For a more detailed explanation of why these estimates are approximate, see Hess’s Law with Bond Energies: Estimate Reaction Enthalpy.

07Questions and solutions

The following questions progress from straightforward bond counting to situations where the bond-energy model and experimental evidence appear to disagree.

Question 1

Consider the synthesis of ammonia:

\[ \ce{N2(g) + 3H2(g) -> 2NH3(g)} \]

The following illustrative average bond enthalpies are supplied:

a. Identify all the bonds broken and formed.

b. Calculate the estimated reaction enthalpy.

c. State whether the reaction is exothermic or endothermic.

Solution 1

The reaction is exothermic, with an estimated enthalpy change of \(-93\ \mathrm{kJ}\) for the balanced equation as written.

Step 1: Count the bonds broken.

One nitrogen molecule contains one N≡N triple bond.

Three hydrogen molecules contain three H-H bonds.

Therefore:

\[ \begin{aligned} E_{\mathrm{broken}} &=945+3(436)\\ &=2253\ \mathrm{kJ} \end{aligned} \]

Step 2: Count the bonds formed.

Each ammonia molecule, \(\ce{NH3}\), contains three N-H bonds.

The equation produces two ammonia molecules, so:

\[ 2\times3=6 \]

Six N-H bonds are formed.

Their total bond enthalpy is:

\[ E_{\mathrm{formed}}=6(391) \]

\[ E_{\mathrm{formed}}=2346\ \mathrm{kJ} \]

Step 3: Calculate the reaction enthalpy.

\[ \begin{aligned} \Delta H_{\mathrm{rxn}} &\approx2253-2346\\ &=-93\ \mathrm{kJ} \end{aligned} \]

The negative sign indicates an exothermic reaction.

Forming six N-H bonds releases more energy than is required to break the N≡N and three H-H bonds.

The common mistake is to count only three N-H bonds because one ammonia molecule contains three. The coefficient of 2 means two ammonia molecules are produced.

Question 2

Carbon monoxide reacts with oxygen to form carbon dioxide:

\[ \ce{2CO(g) + O2(g) -> 2CO2(g)} \]

Use the following illustrative average bond enthalpies:

a. Identify the bonds broken and formed.

b. Estimate the reaction enthalpy.

c. An experimental thermochemical value for this reaction is supplied as \(-566\ \mathrm{kJ}\). Calculate the difference between the estimate and the experimental value.

d. Explain why the two values are not identical.

Solution 2

The estimated reaction enthalpy is \(-578\ \mathrm{kJ}\), which is \(12\ \mathrm{kJ}\) more exothermic than the supplied experimental value.

Step 1: Identify the bonds broken.

Each CO molecule contains one C≡O bond.

There are two CO molecules, so two C≡O bonds are broken.

One O2 molecule contains one O=O bond.

Therefore:

\[ \begin{aligned} E_{\mathrm{broken}} &=2(1072)+498\\ &=2642\ \mathrm{kJ} \end{aligned} \]

Step 2: Identify the bonds formed.

Each CO2 molecule contains two C=O double bonds.

Two CO2 molecules therefore contain four C=O bonds.

\[ \begin{aligned} E_{\mathrm{formed}} &=4(805)\\ &=3220\ \mathrm{kJ} \end{aligned} \]

Step 3: Calculate the estimated enthalpy change.

\[ \begin{aligned} \Delta H_{\mathrm{estimate}} &\approx2642-3220\\ &=-578\ \mathrm{kJ} \end{aligned} \]

Step 4: Compare with the experimental value.

The experimental value is:

\[ \Delta H_{\mathrm{experimental}}=-566\ \mathrm{kJ} \]

The difference is:

\[ \begin{aligned} \text{Difference} &=|-578-(-566)|\\ &=12\ \mathrm{kJ} \end{aligned} \]

The bond-energy estimate is \(12\ \mathrm{kJ}\) more negative than the experimental value.

Step 5: Explain the discrepancy.

The experimental value describes the actual enthalpy change for the specified reaction.

The estimated value uses average bond enthalpies. These averages do not capture every difference in the bonding environments of the individual molecules.

The discrepancy does not violate conservation of energy. It demonstrates the limitation of the model.

Question 3

Two complete thermochemical equations are supplied:

\[ \ce{2C(s,graphite) + 2O2(g) -> 2CO2(g)} \]

\[ \Delta H_1^\circ=-787.0\ \mathrm{kJ} \]

and

\[ \ce{2CO(g) + O2(g) -> 2CO2(g)} \]

\[ \Delta H_2^\circ=-566.0\ \mathrm{kJ} \]

Use Hess’s Law to calculate the enthalpy change for:

\[ \ce{2C(s,graphite) + O2(g) -> 2CO(g)} \]

Then determine the enthalpy change per mole of CO produced.

Solution 3

The enthalpy change is \(-221.0\ \mathrm{kJ}\) for the equation as written, equivalent to \(-110.5\ \mathrm{kJ\,mol^{-1}}\) of CO produced.

Step 1: Identify the required equation.

We want carbon and oxygen to form carbon monoxide.

The first supplied equation forms carbon dioxide, so we need to remove carbon dioxide from the products.

Step 2: Reverse the second equation.

The second equation is:

\[ \ce{2CO + O2 -> 2CO2} \]

Reverse it:

\[ \ce{2CO2 -> 2CO + O2} \]

Its enthalpy change becomes:

\[ \Delta H_2^\circ=+566.0\ \mathrm{kJ} \]

Step 3: Add the equations.

\[ \ce{2C + 2O2 -> 2CO2} \]

\[ \ce{2CO2 -> 2CO + O2} \]

The carbon dioxide cancels because it appears on opposite sides.

The net equation is:

\[ \ce{2C + O2 -> 2CO} \]

Step 4: Add the enthalpy changes.

\[ \begin{aligned} \Delta H^\circ &=-787.0+566.0\\ &=-221.0\ \mathrm{kJ} \end{aligned} \]

Step 5: Convert to an enthalpy change per mole of CO.

The balanced equation produces two moles of CO.

Therefore:

\[ \begin{aligned} \Delta H^\circ_{\text{per mol CO}} &=\frac{-221.0}{2}\\ &=-110.5\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The negative value means the formation of CO from graphite and oxygen is exothermic.

The important detail is that reversing the second equation also reverses its enthalpy sign. Keeping the original negative sign would produce the wrong result.

Question 4

Two thermochemical equations are supplied:

\[ \ce{2H2(g) + O2(g) -> 2H2O(g)} \]

\[ \Delta H_1^\circ=-483.6\ \mathrm{kJ} \]

and

\[ \ce{2H2(g) + O2(g) -> 2H2O(l)} \]

\[ \Delta H_2^\circ=-571.6\ \mathrm{kJ} \]

a. Use Hess’s Law to calculate the enthalpy change for:

\[ \ce{2H2O(l) -> 2H2O(g)} \]

b. Calculate the enthalpy of vaporisation per mole of water.

c. Explain why this energy change cannot be found simply by counting the O-H covalent bonds broken and formed.

Solution 4

The enthalpy change is \(+88.0\ \mathrm{kJ}\) for two moles of water, giving an enthalpy of vaporisation of \(+44.0\ \mathrm{kJ\,mol^{-1}}\).

Step 1: Reverse the liquid-water formation equation.

The supplied equation is:

\[ \ce{2H2 + O2 -> 2H2O(l)} \]

Reverse it:

\[ \ce{2H2O(l) -> 2H2 + O2} \]

The enthalpy changes sign:

\[ \Delta H=+571.6\ \mathrm{kJ} \]

Step 2: Keep the gaseous-water formation equation unchanged.

\[ \ce{2H2 + O2 -> 2H2O(g)} \]

\[ \Delta H=-483.6\ \mathrm{kJ} \]

Step 3: Add the equations.

\[ \ce{2H2O(l) -> 2H2 + O2} \]

\[ \ce{2H2 + O2 -> 2H2O(g)} \]

Hydrogen and oxygen cancel, leaving:

\[ \ce{2H2O(l) -> 2H2O(g)} \]

The total enthalpy change is:

\[ \begin{aligned} \Delta H &=571.6-483.6\\ &=+88.0\ \mathrm{kJ} \end{aligned} \]

Step 4: Calculate the enthalpy change per mole.

The equation vaporises two moles of water.

\[ \begin{aligned} \Delta H_{\mathrm{vap}} &=\frac{88.0}{2}\\ &=+44.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Step 5: Explain why bond counting is insufficient.

The tempting approach is to count the two O-H bonds in each water molecule and conclude that the same covalent bonds exist in both the liquid and gas.

That observation is correct, but it misses the energy change.

During vaporisation, the O-H covalent bonds remain intact. Instead, energy is absorbed to overcome intermolecular attractions between neighbouring water molecules.

Average covalent bond enthalpies do not account for this phase change.

This is why physical states must be included in complete thermochemical equations.

Question 5

Two structural isomers of butane are shown below:

• Straight-chain butane: \(\ce{CH3-CH2-CH2-CH3}\)
• Branched 2-methylpropane: \(\ce{CH3-CH(CH3)-CH3}\)

Both substances have the molecular formula \(\ce{C4H10}\).

A student counts the bonds in each structure and finds:

The student concludes:

Both molecules have exactly the same number of each bond type. Therefore, the enthalpy change for converting butane into 2-methylpropane must be zero.

However, the following illustrative experimental combustion data are supplied. All substances are gaseous, and both combustion reactions produce carbon dioxide gas and water vapour.

Combustion of butane:

\[ \ce{C4H10(g) + \frac{13}{2}O2(g) -> 4CO2(g) + 5H2O(g)} \]

\[ \Delta H_1=-2658\ \mathrm{kJ} \]

Combustion of 2-methylpropane:

\[ \ce{C4H10(g) + \frac{13}{2}O2(g) -> 4CO2(g) + 5H2O(g)} \]

\[ \Delta H_2=-2649\ \mathrm{kJ} \]

Answer the following:

a. Use the bond-energy model to explain why the student predicts an enthalpy change of approximately zero.

b. Use Hess’s Law and the supplied experimental data to calculate the actual enthalpy change for:

\[ \ce{CH3CH2CH2CH3(g) -> CH3CH(CH3)CH3(g)} \]

c. Explain why the student’s conclusion is incorrect, even though their bond counts are correct.

d. Does the identical bond inventory prove that no bonds need to change during the actual molecular rearrangement? Explain.

Solution 5

The experimental enthalpy change is \(-9\ \mathrm{kJ\,mol^{-1}}\). The reaction is exothermic, and the identical bond counts do not prove that the two molecules have identical enthalpies.

Step 1: Evaluate the bond-energy prediction.

Both molecules contain:

• Three C-C bonds.
• Ten C-H bonds.

If we use the same average bond enthalpy for every C-C bond and the same average bond enthalpy for every C-H bond, the total bond-energy contributions cancel.

The model therefore predicts:

\[ \begin{aligned} \Delta H_{\mathrm{estimate}} &\approx [3D(\ce{C-C})+10D(\ce{C-H})]\\ &\quad- [3D(\ce{C-C})+10D(\ce{C-H})]\\ &\approx0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Under this simplified average-bond model, the prediction is reasonable.

But it is not an exact result.

Step 2: Reverse the combustion equation for 2-methylpropane.

The supplied equation is:

\[ \ce{C4H10(g) + \frac{13}{2}O2(g) -> 4CO2(g) + 5H2O(g)} \]

with:

\[ \Delta H_2=-2649\ \mathrm{kJ} \]

Reverse it:

\[ \ce{4CO2(g) + 5H2O(g) -> C4H10(g) + \frac{13}{2}O2(g)} \]

The enthalpy change becomes:

\[ \Delta H_2=+2649\ \mathrm{kJ} \]

Step 3: Add the combustion of butane to the reversed combustion of 2-methylpropane.

The first equation is:

\[ \ce{C4H10(g) + \frac{13}{2}O2(g) -> 4CO2(g) + 5H2O(g)} \]

with:

\[ \Delta H_1=-2658\ \mathrm{kJ} \]

Add the two equations.

Carbon dioxide and water cancel because they appear on opposite sides in identical physical states.

Oxygen also cancels.

The remaining equation is:

\[ \ce{CH3CH2CH2CH3(g) -> CH3CH(CH3)CH3(g)} \]

Step 4: Calculate the enthalpy change.

\[ \begin{aligned} \Delta H_{\mathrm{rxn}} &=\Delta H_1+\Delta H_2\\ &=-2658+2649\\ &=-9\ \mathrm{kJ} \end{aligned} \]

Therefore:

\[ \boxed{\Delta H_{\mathrm{rxn}}=-9\ \mathrm{kJ\,mol^{-1}}} \]

The reaction is exothermic. Under the supplied illustrative data, the branched isomer has a lower enthalpy than straight-chain butane by \(9\ \mathrm{kJ\,mol^{-1}}\).

Step 5: Identify the limitation of the bond-energy model.

The student’s bond counts are correct.

The mistake is assuming that every C-C bond has exactly the same bond enthalpy and every C-H bond has exactly the same bond enthalpy, regardless of its molecular environment.

In reality, the bonding environments differ between straight-chain and branched molecules.

Average bond enthalpies simplify these differences into representative values. They cannot distinguish every structural arrangement.

Consequently, two molecules can have identical average bond inventories but different actual enthalpies.

Step 6: Distinguish bond counting from the actual rearrangement.

The two isomers have different atomic connectivity.

Changing one structure into the other requires a rearrangement of the bonding network. The fact that both structures contain three C-C bonds and ten C-H bonds does not mean that the same individual connections remain unchanged.

The bond-energy calculation is not a reaction mechanism. It does not describe the sequence of molecular events or identify the precise bonds involved at every stage.

The experimental thermochemical equations, on the other hand, establish the net enthalpy change between the specified initial and final states.

This is the crucial distinction: identical average bond inventories can give an estimated enthalpy change of zero without proving that the actual reaction enthalpy is zero.

08What to do next in a bond-energy question

When you encounter a reaction and need to decide which bonds are broken and formed, use this sequence:

If a question supplies complete thermochemical equations instead of average bond enthalpies, switch methods. Reverse, multiply, and add the equations using Hess’s Law, keeping the physical states consistent.

For NSW Chemistry students, there is also a syllabus distinction worth knowing. The Chemistry Stage 6 Syllabus (2017) remains in use until the new Chemistry 11-12 Syllabus (2025) begins implementation in 2028. The new syllabus retains the study of total bond-energy changes, conservation of energy, and bond breaking and formation. However, do not assume that every detailed average-bond-enthalpy or Hess’s Law calculation from the 2017 course is automatically required in the 2025 course. Treat the more detailed calculation methods here as extension for 2025-syllabus students unless they are included in your school’s teaching program.

Once you can distinguish bond inventories, actual reaction mechanisms, and complete thermochemical equations, you can tackle a much wider range of energetics problems without confusing an approximate model with an experimental result.