Why Bond-Energy Calculations Are Estimates in HSC Chemistry

A bond-energy calculation might give you \(-184\ \mathrm{kJ\,mol^{-1}}\), but that doesn't mean an experiment would measure exactly \(-184\ \mathrm{kJ\,mol^{-1}}\). The arithmetic can be precise while the chemical model is approximate.

The reason is that tabulated bond enthalpies are generally **average values for bonds in gaseous molecules**. A particular C-H bond does not necessarily require exactly the same energy to break in methane, methanol, and ethane. Its surrounding molecular environment affects its strength, but a bond-energy table simplifies these differences into representative values.

There is a second issue: bond enthalpies describe gaseous molecules. If your reaction involves liquid water, solid carbon, or a liquid fuel, additional energy changes may be needed to account for those physical states.

This gives us an important distinction:

- Hess's Law is an exact thermochemical principle: the enthalpy change depends on the initial and final states.
- A calculation using average bond enthalpies is an estimate because its input values are averages.
- An experimentally measured reaction enthalpy refers to a particular reaction, with specified substances and physical states.

Understanding that distinction helps you decide when to trust a bond-energy calculation, when to apply a correction, and when an experimental value should take priority.

## 1. Why one type of bond can have different energies

Imagine breaking a C-H bond in methane:

\[
\ce{CH4(g) -> CH3(g) + H(g)}
\]

Now imagine breaking a C-H bond in methanol:

\[
\ce{CH3OH(g) -> CH3O(g) + H(g)}
\]

Both processes involve breaking a C-H bond, but the surrounding atoms are different. Methanol contains an oxygen atom, which changes the electronic environment around the carbon and affects the bond.

The energy required for each process is therefore not necessarily identical.

A bond dissociation enthalpy is the enthalpy change for breaking a particular bond in a particular gaseous molecule, with the other atoms remaining as a specified fragment. The exact value depends on that molecular environment.

[IMAGE: Side-by-side molecular representations of methane and methanol. Highlight one C-H bond in each molecule, showing that the carbon in methanol is also bonded to oxygen. Include the corresponding bond-breaking products as separate molecular fragments, making clear that the surrounding molecular structures differ.]

### Why do tables give one number, then?

For practical calculations, chemists use average bond enthalpies. These represent typical values for a particular bond type across a range of gaseous molecules.

For example, an illustrative data table might give:

| Bond | Average bond enthalpy |
|---|---:|
| C-H | \(413\ \mathrm{kJ\,mol^{-1}}\) |
| O-H | \(463\ \mathrm{kJ\,mol^{-1}}\) |
| C-C | \(347\ \mathrm{kJ\,mol^{-1}}\) |
| C=C | \(614\ \mathrm{kJ\,mol^{-1}}\) |
| O=O | \(498\ \mathrm{kJ\,mol^{-1}}\) |

These are representative values, not universal constants for every molecule containing those bonds.

When you use \(413\ \mathrm{kJ\,mol^{-1}}\) for a C-H bond, you are assuming that the average is a suitable approximation for the particular molecule in your reaction.

That assumption makes calculations manageable. It also introduces uncertainty.

02What a bond-energy calculation actually assumes

The bond-energy method imagines a reaction taking place in two stages:

1. Every bond in the gaseous reactants is broken, producing separated gaseous atoms.
2. Those atoms form the bonds found in the gaseous products.

The enthalpy change is estimated using:

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

Here:

• \(\Delta H_{\mathrm{rxn}}\) is the estimated reaction enthalpy, in \(\mathrm{kJ\\,mol^{-1}}\).
• \(D\) is the average bond enthalpy, in \(\mathrm{kJ\\,mol^{-1}}\).
• \(\sum D_{\mathrm{broken}}\) is the total energy required to break the reactant bonds.
• \(\sum D_{\mathrm{formed}}\) is the total energy released when the product bonds form.

The approximation sign is important. It tells you that the result comes from a simplified model.

There are two important limitations

Limitation 1: Average bond enthalpies replace molecule-specific values.

Consider methane, \(\ce{CH4}\). Breaking its first C-H bond produces \(\ce{CH3}\) and a hydrogen atom.

Breaking another C-H bond means breaking a bond in \(\ce{CH3}\), not in the original methane molecule. The electronic structure has changed, so the next bond dissociation enthalpy is different.

The same issue applies to many polyatomic molecules. A table’s average C-H value smooths over these differences.

Limitation 2: The method describes gaseous molecules.

A bond enthalpy table does not automatically account for the energy needed to vaporise a liquid or separate a solid structure.

For example, the equation

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

describes vaporisation, not the breaking of O-H covalent bonds. It requires energy because the water molecules must overcome intermolecular attractions.

If your reaction starts with liquid water but your bond-energy calculation uses gaseous water molecules, you need to include this additional enthalpy change.

03How experimental enthalpy values are obtained

A bond-energy calculation predicts an enthalpy change using a molecular model. An experiment approaches the problem differently: it measures the heat transferred during a reaction.

One common method is calorimetry.

Picture two measured solutions, initially kept separate. They are poured into an insulated polystyrene cup nested inside a beaker. A lid holds a thermometer or temperature probe, while a stirrer helps distribute heat evenly.

The initial temperature is recorded. The solutions are mixed, and the temperature is monitored as the reaction proceeds.

What should happen if the reaction is exothermic?

The solution temperature should increase. The chemical reaction releases heat, which is transferred to the solution and the calorimeter.

We use:

\[ q_{\mathrm{solution}}=mc\Delta T \]

where:

• \(q_{\mathrm{solution}}\) is the heat gained or lost by the solution, in joules.
• \(m\) is the solution mass, in grams.
• \(c\) is the specific heat capacity, in \(\mathrm{J\\,g^{-1}\\,K^{-1}}\).
• \(\Delta T\) is the final temperature minus the initial temperature, in kelvin.

Assuming negligible heat loss and negligible heat absorbed by the cup:

\[ q_{\mathrm{reaction}}=-q_{\mathrm{solution}} \]

The negative sign is essential. If the solution gains heat, the reaction loses it.

The measured heat change can then be divided by the amount of limiting reactant to obtain the molar enthalpy change.

Does an experimental value have to be exact?

No. The experiment has its own limitations.

Some heat may escape into the surroundings. The cup itself may absorb energy. The temperature probe may have limited resolution, and the solutions may not have exactly the assumed density or specific heat capacity.

The original simple setup can be improved by:

• Using a lid and additional insulation to reduce heat loss.
• Calibrating the calorimeter to account for the heat absorbed by the apparatus.
• Stirring consistently to produce an even temperature.
• Repeating measurements and comparing results.

A measured value therefore has experimental uncertainty. However, it refers to the actual substances and physical states involved, rather than relying on average bond strengths.

04Manipulating complete thermochemical equations

A thermochemical equation contains three pieces of information:

1. The balanced chemical equation.
2. The physical state of every reactant and product.
3. The enthalpy change, \(\Delta H\).

For example:

\[ \ce{CO(g) + \frac{1}{2}O2(g) -> CO2(g)} \]

\[ \Delta H^\circ=-282.98\ \mathrm{kJ\\,mol^{-1}} \]

The enthalpy change belongs to the equation as written. Change the equation, and you must change its enthalpy value accordingly.

There are three operations you need to recognise.

Reversing an equation

Start with:

\[ \ce{CO(g) + \frac{1}{2}O2(g) -> CO2(g)} \]

\[ \Delta H^\circ=-282.98\ \mathrm{kJ\\,mol^{-1}} \]

Reverse the reaction:

\[ \ce{CO2(g) -> CO(g) + \frac{1}{2}O2(g)} \]

The new enthalpy change is:

\[ \Delta H^\circ=+282.98\ \mathrm{kJ\\,mol^{-1}} \]

The sign changes because the reaction now requires energy rather than releasing it.

Multiplying an equation

Suppose you want twice as much carbon dioxide to form.

Multiply every coefficient by two:

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

The enthalpy change also doubles:

\[ \Delta H^\circ=2(-282.98) \]

\[ \Delta H^\circ=-565.96\ \mathrm{kJ\\,mol^{-1}} \]

You cannot double the chemical equation while leaving its enthalpy unchanged. Twice the reaction means twice the energy transfer.

Adding equations

This is where Hess’s Law becomes useful.

Suppose you have:

\[ \ce{A -> B} \qquad \Delta H_1 \]

and

\[ \ce{B -> C} \qquad \Delta H_2 \]

Adding the equations gives:

\[ \ce{A -> C} \]

The intermediate substance B appears on both sides and cancels.

The overall enthalpy change is:

\[ \Delta H_{\mathrm{overall}}=\Delta H_1+\Delta H_2 \]

One detail is easy to overlook: you can only cancel a substance when it is in the same physical state on both sides.

For example, \(\ce{H2O(l)}\) and \(\ce{H2O(g)}\) cannot simply cancel. They represent different energetic states.

05Comparing a bond-energy estimate with a reference value

Let’s bring the two approaches together.

Consider the oxidation of carbon monoxide:

\[ \ce{CO(g) + \frac{1}{2}O2(g) -> CO2(g)} \]

We can estimate its enthalpy using average bond enthalpies. We can also obtain a reference value using standard enthalpies of formation.

What happens if you combine measured and estimated values?

Suppose a Hess cycle contains two steps:

• One step has a measured enthalpy change of \(-150\ \mathrm{kJ\\,mol^{-1}}\).
• Another step has an estimated bond-energy enthalpy change of \(-80\ \mathrm{kJ\\,mol^{-1}}\).

The overall result is:

\[ \Delta H_{\mathrm{overall}}\approx-150-80 \]

\[ \Delta H_{\mathrm{overall}}\approx-230\ \mathrm{kJ\\,mol^{-1}} \]

The result remains an estimate.

Adding an exact thermochemical relationship to an approximate bond-energy calculation does not remove the approximation. The final answer inherits the limitations of the estimated step.

06Common mistakes when interpreting bond-energy results

For a more detailed explanation of the energy involved in breaking and forming bonds, read Bond Enthalpy and Conservation of Energy in HSC Chemistry.

For further practice constructing energy pathways and estimating reaction enthalpies, see Hess’s Law with Bond Energies: Estimate Reaction Enthalpy.

06Questions and solutions

Question 1

The synthesis of ammonia is represented by:

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

The enthalpy change for the reaction as written is:

\[ \Delta H=-92.4\ \mathrm{kJ\\,mol^{-1}} \]

(a) Write a thermochemical equation for the formation of one mole of ammonia.

(b) Write the thermochemical equation for the reverse of the original reaction, including its enthalpy change.

Solution 1

(a) The enthalpy change is \(-46.2\ \mathrm{kJ\\,mol^{-1}}\) for the formation of one mole of ammonia.

(b) The reverse reaction has an enthalpy change of \(+92.4\ \mathrm{kJ\\,mol^{-1}}\).

Part (a)

The original equation produces two moles of ammonia.

To produce one mole, divide every coefficient by two:

\[ \ce{\frac{1}{2}N2(g) + \frac{3}{2}H2(g) -> NH3(g)} \]

The enthalpy change must also be halved:

\[ \Delta H=\frac{-92.4}{2} \]

\[ \boxed{\Delta H=-46.2\ \mathrm{kJ\\,mol^{-1}}} \]

Part (b)

Reverse the original equation:

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

Changing the direction of the reaction changes the sign of the enthalpy:

\[ \boxed{\Delta H=+92.4\ \mathrm{kJ\\,mol^{-1}}} \]

The common mistake is to change the coefficients without changing the enthalpy, or to reverse the equation while keeping its original negative sign.

Remember that the equation and its enthalpy change must always be manipulated together.

Question 2

Ethanal can be hydrogenated to form ethanol:

\[ \ce{CH3CHO(g) + H2(g) -> CH3CH2OH(g)} \]

Use the following illustrative average bond enthalpies:

(a) Identify the bonds that are broken and formed.

(b) Estimate the reaction enthalpy.

(c) Explain why your answer should not be reported as an exact experimental value.

Solution 2

The estimated reaction enthalpy is \(-53\ \mathrm{kJ\\,mol^{-1}}\). The reaction is exothermic.

Part (a)

Ethanal contains:

• Four C-H bonds.
• One C-C bond.
• One C=O bond.

Hydrogen contributes one H-H bond.

Ethanol contains:

• Five C-H bonds.
• One C-C bond.
• One C-O bond.
• One O-H bond.

The C-C bond and four C-H bond contributions cancel from the accounting.

The remaining changes are:

Bonds broken:

• One C=O bond.
• One H-H bond.

Bonds formed:

• One C-O bond.
• One C-H bond.
• One O-H bond.

Part (b)

Calculate the energy required to break the bonds:

\[ \begin{aligned} \sum D_{\mathrm{broken}} &=745+436\\\ &=1181\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Calculate the energy released when the new bonds form:

\[ \begin{aligned} \sum D_{\mathrm{formed}} &=358+413+463\\\ &=1234\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta H &=1181-1234\\\ &=-53\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

\[ \boxed{\Delta H\approx-53\ \mathrm{kJ\\,mol^{-1}}} \]

The negative result indicates that bond formation releases more energy than bond breaking absorbs.

Part (c)

The answer is an estimate because the supplied bond enthalpies are averages.

The C=O bond in ethanal has a particular molecular environment. The average C=O value is not guaranteed to equal its exact dissociation enthalpy.

The same applies to the new bonds in ethanol. The calculation gives a useful prediction, but an experiment or appropriate reference thermochemical data would be needed to establish the reaction enthalpy more accurately.

Question 3

A chemist has obtained the following illustrative experimental thermochemical data:

Equation A

\[ \ce{N2(g) + O2(g) -> 2NO(g)} \]

\[ \Delta H_A=+180.5\ \mathrm{kJ\\,mol^{-1}} \]

Equation B

\[ \ce{N2(g) + 2O2(g) -> 2NO2(g)} \]

\[ \Delta H_B=+66.3\ \mathrm{kJ\\,mol^{-1}} \]

Determine the enthalpy change for:

\[ \ce{2NO(g) + O2(g) -> 2NO2(g)} \]

Show how the complete thermochemical equations must be manipulated.

Solution 3

The required enthalpy change is \(-114.2\ \mathrm{kJ\\,mol^{-1}}\).

The key is to reverse Equation A before adding it to Equation B.

Step 1: Reverse Equation A.

The original equation is:

\[ \ce{N2(g) + O2(g) -> 2NO(g)} \]

The required equation needs NO on the reactant side, so reverse it:

\[ \ce{2NO(g) -> N2(g) + O2(g)} \]

Its enthalpy changes sign:

\[ \Delta H_A=-180.5\ \mathrm{kJ\\,mol^{-1}} \]

Step 2: Keep Equation B unchanged.

\[ \ce{N2(g) + 2O2(g) -> 2NO2(g)} \]

\[ \Delta H_B=+66.3\ \mathrm{kJ\\,mol^{-1}} \]

Step 3: Add the equations.

\[ \begin{aligned} \ce{2NO(g)}&\ce{-> N2(g) + O2(g)}\\\ \ce{N2(g) + 2O2(g)}&\ce{-> 2NO2(g)} \end{aligned} \]

The nitrogen molecules cancel, as does one mole of oxygen gas.

The resulting equation is:

\[ \ce{2NO(g) + O2(g) -> 2NO2(g)} \]

Step 4: Add the enthalpy changes.

\[ \begin{aligned} \Delta H &=-180.5+66.3\\\ &=-114.2\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{\Delta H=-114.2\ \mathrm{kJ\\,mol^{-1}}} \]

The tempting mistake is to add the two original enthalpies without reversing Equation A. That would give the wrong result because the original equations do not combine to produce the required reaction.

Always manipulate the chemical equations first. The enthalpy values follow those operations.

Question 4

A student mixes \(40.0\ \mathrm{mL}\) of \(1.50\ \mathrm{mol\\,L^{-1}}\) hydrochloric acid with \(60.0\ \mathrm{mL}\) of \(0.800\ \mathrm{mol\\,L^{-1}}\) sodium hydroxide in an insulated calorimeter.

The temperature increases from \(21.5^\circ\mathrm{C}\) to \(27.8^\circ\mathrm{C}\).

Assume:

• The solution density is \(1.00\ \mathrm{g\\,mL^{-1}}\).
• The specific heat capacity is \(4.18\ \mathrm{J\\,g^{-1}\\,K^{-1}}\).
• Heat absorbed by the calorimeter and heat lost to the surroundings are negligible.

(a) Determine the limiting reactant.

(b) Calculate the heat released by the reaction.

(c) Calculate the molar enthalpy change of neutralisation.

(d) Write the complete thermochemical equation.

Solution 4

The limiting reactant is NaOH, and the molar enthalpy change is approximately \(-54.9\ \mathrm{kJ\\,mol^{-1}}\).

Part (a)

Calculate the amount of each reactant.

For hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=cV\\\ &=1.50\times0.0400\\\ &=0.0600\ \mathrm{mol} \end{aligned} \]

For sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=0.800\times0.0600\\\ &=0.0480\ \mathrm{mol} \end{aligned} \]

The balanced equation is:

\[ \ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)} \]

The mole ratio is 1:1.

NaOH has fewer moles, so it is the limiting reactant.

Part (b)

Calculate the temperature change:

\[ \Delta T=27.8-21.5=6.3\ \mathrm{K} \]

The combined solution volume is:

\[ 40.0+60.0=100.0\ \mathrm{mL} \]

Therefore, its mass is:

\[ m=100.0\times1.00=100.0\ \mathrm{g} \]

Calculate the heat gained by the solution:

\[ \begin{aligned} q_{\mathrm{solution}} &=mc\Delta T\\\ &=100.0\times4.18\times6.3\\\ &=2633.4\ \mathrm{J}\\\ &=2.6334\ \mathrm{kJ} \end{aligned} \]

The reaction releases this heat:

\[ \boxed{q_{\mathrm{reaction}}=-2.6334\ \mathrm{kJ}} \]

Part (c)

The reaction consumes \(0.0480\ \mathrm{mol}\) of NaOH.

Because the mole ratio is 1:1, this is also the amount of reaction that has occurred.

\[ \begin{aligned} \Delta H &=\frac{-2.6334}{0.0480}\\\ &=-54.8625\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

To three significant figures:

\[ \boxed{\Delta H=-54.9\ \mathrm{kJ\\,mol^{-1}}} \]

Part (d)

The complete thermochemical equation is:

\[ \ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)} \]

\[ \boxed{\Delta H\approx-54.9\ \mathrm{kJ\\,mol^{-1}}} \]

The negative sign indicates an exothermic reaction.

The important trap is dividing by the initial amount of hydrochloric acid. Not all of it reacts because NaOH is limiting.

Also, this is an experimentally determined value under the stated assumptions. It is not calculated by adding average covalent bond enthalpies.

Question 5

A student is investigating the decomposition of liquid water into hydrogen and oxygen.

The target reaction is:

\[ \ce{3H2O(l) -> 3H2(g) + \frac{3}{2}O2(g)} \]

The following illustrative rounded data are supplied at \(298.15\ \mathrm{K}\) and \(1\ \mathrm{bar}\):

Thermochemical data

\[ \Delta_fH^\circ(\ce{H2O(g)}) =-241.8\ \mathrm{kJ\\,mol^{-1}} \]

\[ \Delta H^\circ_{\mathrm{vap}}(\ce{H2O}) =+44.0\ \mathrm{kJ\\,mol^{-1}} \]

Average bond enthalpies

A student makes the following prediction:

Each water molecule contains two O-H bonds. Breaking those bonds and forming H-H and O=O bonds requires approximately 241 kJ per mole of water. Therefore, decomposing three moles of liquid water requires approximately 723 kJ.

Evaluate this prediction.

Calculate:

(a) The reference enthalpy change for the decomposition of three moles of liquid water using the supplied thermochemical data.

(b) The bond-energy estimate for the corresponding gaseous reaction.

(c) The corrected bond-energy estimate for the liquid-water reaction.

(d) Explain why the student’s original prediction is incomplete, and why the corrected estimate still differs slightly from the reference result.

Solution 5

The student’s original prediction is incomplete because the average bond enthalpies describe gaseous water molecules, not liquid water.

The reference enthalpy change is \(+857.4\ \mathrm{kJ}\) for the reaction as written. The corrected bond-energy estimate is \(+855\ \mathrm{kJ}\).

The remaining difference is \(2.4\ \mathrm{kJ}\), which comes from the approximation in the average bond enthalpies.

Part (a): Calculate the reference enthalpy change.

Start by converting liquid water into gaseous water:

\[ \ce{3H2O(l) -> 3H2O(g)} \]

The enthalpy change is:

\[ \begin{aligned} \Delta H_1 &=3(44.0)\\\ &=+132.0\ \mathrm{kJ} \end{aligned} \]

Next, reverse the formation reaction for gaseous water:

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

Its enthalpy change is \(-241.8\ \mathrm{kJ\\,mol^{-1}}\).

Reversing it gives:

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

\[ \Delta H_2=+241.8\ \mathrm{kJ\\,mol^{-1}} \]

For three moles:

\[ \begin{aligned} \Delta H_2 &=3(241.8)\\\ &=+725.4\ \mathrm{kJ} \end{aligned} \]

Add the two stages:

\[ \begin{aligned} \Delta H_{\mathrm{reference}} &=132.0+725.4\\\ &=+857.4\ \mathrm{kJ} \end{aligned} \]

Therefore:

\[ \boxed{\Delta H_{\mathrm{reference}}=+857.4\ \mathrm{kJ}} \]

The positive sign indicates that energy must be absorbed to decompose the liquid water.

Part (b): Calculate the gaseous bond-energy estimate.

First consider one mole of gaseous water:

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

Breaking the two O-H bonds requires:

\[ \begin{aligned} \sum D_{\mathrm{broken}} &=2(463)\\\ &=926\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Forming one H-H bond and half an O=O bond releases:

\[ \begin{aligned} \sum D_{\mathrm{formed}} &=436+\frac{1}{2}(498)\\\ &=436+249\\\ &=685\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta H_{\mathrm{gas,estimate}} &=926-685\\\ &=+241\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

For three moles:

\[ \begin{aligned} \Delta H_{\mathrm{gas,estimate}} &=3(241)\\\ &=+723\ \mathrm{kJ} \end{aligned} \]

This is where the student’s original answer comes from.

The calculation is reasonable for gaseous water. It does not yet describe the specified liquid-water reaction.

Part (c): Correct the estimate for liquid water.

The bond-energy calculation starts with gaseous water, but our actual reactant is liquid water.

We must include the energy required to vaporise three moles:

\[ \Delta H_{\mathrm{vap}}=3(44.0) \]

\[ \Delta H_{\mathrm{vap}}=+132.0\ \mathrm{kJ} \]

Add this to the bond-energy estimate:

\[ \begin{aligned} \Delta H_{\mathrm{corrected}} &=723+132.0\\\ &=+855\ \mathrm{kJ} \end{aligned} \]

Therefore:

\[ \boxed{\Delta H_{\mathrm{corrected}}\approx+855\ \mathrm{kJ}} \]

Part (d): Explain the remaining difference.

Compare the two results:

The difference is:

\[ 857.4-855=2.4\ \mathrm{kJ} \]

The student’s original answer missed the vaporisation energy, which accounts for \(132.0\ \mathrm{kJ}\) of the difference.

After that correction, the remaining \(2.4\ \mathrm{kJ}\) difference comes from using average rather than molecule-specific bond enthalpies.

There are two separate effects here:

1. The physical state of the reactant changes the enthalpy of the reaction.
2. The average bond-energy model introduces an additional approximation.

This is the important conceptual point. The bond-energy calculation was not necessarily badly performed. It was applied to an incomplete representation of the actual reaction.

NIST’s evaluated formation enthalpies for gaseous and liquid water are approximately \(-241.826\) and \(-285.830\ \mathrm{kJ\\,mol^{-1}}\), respectively. Their difference confirms the approximately \(44.0\ \mathrm{kJ\\,mol^{-1}}\) vaporisation correction used here. 【turn253745search3】

07What to remember when answering HSC questions

When a question involves bond energies and thermochemical equations, make three decisions before calculating.

First: Identify the type of data.

• Average bond enthalpies mean you are estimating.
• Calorimetry results are experimentally determined values, subject to experimental uncertainty.
• Standard enthalpies of formation and other reference thermochemical values can be combined using Hess’s Law.

Second: Check the physical states.

Average bond enthalpies apply to gaseous molecules. If the equation contains liquids or solids, identify any additional physical changes required.

Third: Manipulate the entire thermochemical equation.

Reverse the equation and change the sign. Multiply every coefficient and multiply the enthalpy by the same factor. Add equations and cancel only identical substances in identical states.

The distinction between a model estimate and an experimentally determined enthalpy is the foundation for making sensible decisions in more complicated Hess’s Law calculations.

This resource follows the Chemistry Stage 6 Syllabus (2017) context for enthalpy and Hess’s Law. NESA schedules implementation of the Chemistry 11-12 Syllabus (2025) from 2028, with Year 12 continuing under the 2017 syllabus until Term 4 2028 and the first HSC examination under the new syllabus in 2029. Students should check the syllabus applying to their cohort rather than assume the two editions have identical requirements. 【turn811815view4】

The next useful step is learning to calculate reaction enthalpies from standard enthalpies of formation. That method uses complete thermochemical equations and experimentally evaluated data, rather than relying on average bond strengths.