How to Calculate Formation Enthalpy from Reaction Data

Suppose you need to calculate the enthalpy change when carbon monoxide forms from graphite and oxygen. You have data for two reactions, but neither one is the reaction you’re looking for. One produces carbon dioxide from graphite, while the other converts carbon monoxide into carbon dioxide.

How can you use those measurements to find the enthalpy change for forming carbon monoxide?

You don’t need to measure the target reaction directly. You can manipulate the supplied thermochemical equations until they combine to give the reaction you want. The key is to keep track of both the chemical equations and their enthalpy changes.

Before we start, predict what happens to an enthalpy change when you reverse a chemical equation. Does its sign stay the same?

It changes sign. Reversing a reaction reverses the direction of energy transfer, so an exothermic reaction becomes endothermic when written backwards.

This is the foundation of Hess’s Law, and it is the main tool used to calculate formation enthalpies from supplied reaction data.

Syllabus scope (November 2026): Hess’s Law and the manipulation of thermochemical equations belong to the NSW Chemistry Stage 6 Syllabus (2017), which remains in use for students studying Chemistry in 2026 and 2027. The Chemistry 11-12 Syllabus (2025) begins implementation for Year 11 in 2028. Its specified energy-change calculations include calorimetry and \(\Delta H=-q/n\), but do not explicitly include Hess cycles or formation-enthalpy calculations. Students following the 2025 syllabus should treat the method in this article as extension material rather than a required syllabus dot point.

0101 What is an enthalpy of formation?

The standard enthalpy of formation, written as \(\Delta H_f^\circ\), is the enthalpy change when exactly one mole of a compound forms from its constituent elements in their standard states.

There are three important parts to this definition:

• Exactly one mole of the compound must be formed.
• The starting substances must be the elements that make up the compound.
• Each element must be in its standard state.

A standard state is the most stable physical form of a substance at the specified temperature and standard pressure. Standard enthalpy values are commonly reported at 298 K and 100 kPa.

For example, consider the formation of carbon monoxide:

\[ \ce{C(graphite) + 1/2O2(g) -> CO(g)} \]

This equation forms exactly one mole of carbon monoxide from carbon and oxygen in their standard states.

Its enthalpy change is the standard enthalpy of formation of carbon monoxide:

\[ \Delta H_f^\circ(\ce{CO})=-110.5\ \mathrm{kJ\,mol^{-1}} \]

The negative sign tells us that forming one mole of carbon monoxide releases energy.

Why do elements have zero formation enthalpy?

By convention, the standard enthalpy of formation of an element in its standard state is zero.

For example:

This makes calculations much easier.

However, be careful: the zero only applies to the element in its standard state.

For carbon, graphite is the standard state, not diamond. Diamond has a different enthalpy and cannot automatically be assigned zero.

Formation equations must produce one mole

Consider the formation of ammonia:

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

This reaction produces two moles of ammonia.

If the reaction enthalpy is \(-92.0\ \mathrm{kJ}\), the standard enthalpy of formation of ammonia is:

\[ \Delta H_f^\circ(\ce{NH3}) =\frac{-92.0}{2} \]

\[ \boxed{-46.0\ \mathrm{kJ\,mol^{-1}}} \]

The reaction enthalpy belongs to the complete equation as written. The formation enthalpy is expressed per mole of compound formed.

That distinction becomes important when you start manipulating equations.

0202 Why Hess’s Law allows us to calculate formation enthalpy

Imagine walking from the bottom of a hill to the top.

You could take a steep path, a winding path, or a path that goes around the hill. The total change in height between your starting point and finishing point remains the same.

Enthalpy works in a similar way.

The enthalpy change depends on the initial and final chemical states, not on the pathway taken between them.

This is Hess’s Law, which follows from the conservation of energy.

The analogy is useful, but it has a limit: enthalpy is not literally height. It is a thermodynamic property of a chemical system.

A chemical example

Consider these two reactions:

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

\[ \Delta H_1=-393.5\ \mathrm{kJ} \]

And:

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

\[ \Delta H_2=-283.0\ \mathrm{kJ} \]

Both reactions produce carbon dioxide, but they start with different substances.

We want to find:

\[ \ce{C(graphite) + 1/2O2(g) -> CO(g)} \]

The important observation is that we can reach the same final state, carbon dioxide, through two different pathways.

The total enthalpy change must be identical for both pathways.

Therefore:

\[ \Delta H_1=\Delta H_f^\circ(\ce{CO})+\Delta H_2 \]

Rearranging:

\[ \Delta H_f^\circ(\ce{CO}) =\Delta H_1-\Delta H_2 \]

\[ =-393.5-(-283.0) \]

\[ \boxed{-110.5\ \mathrm{kJ\,mol^{-1}}} \]

We have calculated the formation enthalpy without directly measuring the formation reaction.

This is the central idea behind calculating formation enthalpies from supplied reaction data.

0303 How reaction enthalpies are measured

Before manipulating experimental data, it helps to understand where the numbers come from.

A reaction enthalpy can be determined using calorimetry.

For a typical constant-pressure solution reaction, imagine an insulated polystyrene cup containing a known mass of solution. A lid holds a thermometer and stirring rod. Reactants are mixed inside the cup, and the temperature change is recorded.

The temperature change provides evidence of the energy transferred between the reaction and its surroundings.

The heat absorbed or released by the solution is calculated using:

\[ q=mc\Delta T \]

Where:

• \(q\) is heat energy transferred, in joules (J).
• \(m\) is the mass of the solution, in grams (g).
• \(c\) is the specific heat capacity, in \(\mathrm{J\,g^{-1}\,K^{-1}}\).
• \(\Delta T\) is the temperature change, in K.

For an exothermic reaction, the solution gains heat:

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

At constant pressure, the heat transferred by the reaction is its enthalpy change.

The molar enthalpy change can then be calculated by dividing the reaction heat by the amount of substance that reacted:

\[ \Delta H=\frac{q_{\mathrm{reaction}}}{n} \]

Here, \(n\) is the amount of reaction expressed in moles according to the balanced equation.

Combustion reactions require suitable sealed, calibrated combustion calorimeters rather than an open solution cup. Their measurements also require appropriate treatment of the calorimeter conditions.

In many examination questions, the experimental work has already been completed. You are supplied with the resulting reaction enthalpies and must use them to calculate an unknown value.

Measured values, calculated values, and model estimates

These are not interchangeable.

A Hess cycle gives you an inferred value for a reaction that may be difficult to measure directly. It does not mean that the target reaction itself was experimentally measured.

Likewise, a bond-enthalpy calculation is a model estimate. Average bond enthalpies do not represent the exact energy of every bond in every chemical environment.

Keep these distinctions clear when evaluating supplied data.

0404 The two calculation methods

There are two common ways to calculate formation enthalpy from supplied information.

Method 1: Manipulate complete thermochemical equations

Use this method when you are given balanced chemical equations with their enthalpy changes.

Remember these three rules:

For example, reversing:

\[ \ce{A -> B} \qquad \Delta H=-50\ \mathrm{kJ} \]

Gives:

\[ \ce{B -> A} \qquad \Delta H=+50\ \mathrm{kJ} \]

Doubling the original reaction gives:

\[ \ce{2A -> 2B} \qquad \Delta H=-100\ \mathrm{kJ} \]

The enthalpy change scales with the amount of reaction.

When adding equations, cancel substances appearing on both sides. The equation remaining must match your target reaction exactly.

Method 2: Use standard enthalpies of formation

If the question supplies formation enthalpies for individual substances, you can use:

\[ \Delta H^\circ_{\mathrm{rxn}} = \sum \nu\Delta H_f^\circ(\mathrm{products}) - \sum \nu\Delta H_f^\circ(\mathrm{reactants}) \]

Here:

• \(\Delta H^\circ_{\mathrm{rxn}}\) is the standard enthalpy change of the balanced reaction.
• \(\Delta H_f^\circ\) is the standard enthalpy of formation of each substance.
• \(\nu\) is the stoichiometric coefficient of that substance.
• The summation means you include every substance on the relevant side of the equation.

The order is always:

Products minus reactants.

If you are finding the formation enthalpy of an unknown product, substitute the known values and rearrange the equation.

For example, for:

\[ \ce{A + B -> 2X + C} \]

You would write:

\[ \begin{aligned} \Delta H^\circ_{\mathrm{rxn}} ={}&[2\Delta H_f^\circ(X)+\Delta H_f^\circ(C)]\\ &-[\Delta H_f^\circ(A)+\Delta H_f^\circ(B)] \end{aligned} \]

Then solve for \(\Delta H_f^\circ(X)\).

Remember that elements in their standard states contribute zero.

The coefficients matter. If a substance appears twice in the balanced equation, its formation enthalpy must also be multiplied by two.

The same coefficient bookkeeping is used when connecting reacting quantities, as explained in How to Calculate CO₂ Volume from a Carbonate Reaction. The additional step here is that enthalpy changes must scale with the complete thermochemical equation.

0505 Worked examples

0606 The common misconception: bond enthalpy is not formation enthalpy

There is another type of energy data you may encounter in the 2017 Chemistry syllabus: average bond enthalpies.

These are useful, but they answer a slightly different question.

Bond enthalpies estimate the energy required to break chemical bonds and the energy released when new bonds form.

The calculation is:

\[ \Delta H_{\mathrm{reaction}} \approx \sum E(\text{bonds broken}) - \sum E(\text{bonds formed}) \]

The approximation symbol matters.

Average bond enthalpies are based on average values for bonds in gaseous molecules. The energy of a particular bond can vary depending on its chemical environment.

Consequently, a bond-enthalpy calculation produces an estimate, not necessarily the exact standard enthalpy change of a reaction.

For example, calculating the energy change for producing gaseous water using average bond enthalpies does not automatically give the standard formation enthalpy of liquid water. You would also need to account for the difference between gaseous and liquid water.

Keep this decision table in mind:

Do not mix these methods just because they all involve energy.

0707 Questions and solutions

The following questions progress from straightforward formation calculations to more demanding Hess cycles, phase changes, and competing standard-state choices.

All numerical datasets identified as illustrative are supplied specifically for these questions.

Question 1

Easy exam style

The following thermochemical equation is supplied:

\[ \ce{2Al(s) + 3Cl2(g) -> 2AlCl3(s)} \]

The standard enthalpy change is \(-1408\ \mathrm{kJ}\) for the reaction as written.

Calculate the standard enthalpy of formation of solid aluminium chloride, \(\ce{AlCl3(s)}\).

Solution 1

The standard enthalpy of formation of aluminium chloride is \(-704\ \mathrm{kJ\,mol^{-1}}\).

The supplied reaction produces two moles of aluminium chloride, whereas the formation enthalpy is defined per mole.

Step 1: Write the formation equation.

\[ \ce{Al(s) + 3/2Cl2(g) -> AlCl3(s)} \]

Step 2: Identify the number of moles produced.

The supplied equation produces two moles of \(\ce{AlCl3}\).

Therefore, its reaction enthalpy must be divided by two.

Step 3: Calculate the formation enthalpy.

\[ \begin{aligned} \Delta H_f^\circ(\ce{AlCl3}) &=\frac{-1408}{2}\\ &=-704\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The negative result indicates that forming aluminium chloride from aluminium and chlorine releases energy.

A tempting mistake is to report \(-1408\ \mathrm{kJ\,mol^{-1}}\). That is the enthalpy change for the complete equation, which forms two moles of product, not one.

Question 2

Tough exam style

Three illustrative thermochemical equations are supplied at 298 K and 100 kPa:

Reaction A

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

\[ \Delta H_A=-393.5\ \mathrm{kJ} \]

Reaction B

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

\[ \Delta H_B=-285.8\ \mathrm{kJ} \]

Reaction C

\[ \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \]

\[ \Delta H_C=-890.3\ \mathrm{kJ} \]

(a) Use Hess’s Law to calculate the standard enthalpy of formation of methane, \(\ce{CH4(g)}\).

(b) Write the complete thermochemical equation representing the formation of one mole of methane.

(c) Explain why Reaction C must be reversed when constructing the required Hess cycle.

Solution 2

The standard enthalpy of formation of methane is \(-74.8\ \mathrm{kJ\,mol^{-1}}\).

The important step is to combine the combustion reactions of carbon and hydrogen with the reversed combustion reaction of methane.

Step 1: Write the target formation equation.

\[ \ce{C(graphite) + 2H2(g) -> CH4(g)} \]

We need carbon and hydrogen on the reactant side and methane on the product side.

Step 2: Keep Reaction A unchanged.

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

\[ \Delta H_A=-393.5\ \mathrm{kJ} \]

Step 3: Double Reaction B.

We need two moles of hydrogen to form the two moles of water that appear in Reaction C.

Therefore:

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

\[ \Delta H_B=2(-285.8)=-571.6\ \mathrm{kJ} \]

Step 4: Reverse Reaction C.

Reaction C consumes methane, but our target equation produces it.

Therefore:

\[ \ce{CO2(g) + 2H2O(l) -> CH4(g) + 2O2(g)} \]

The enthalpy changes sign:

\[ \Delta H_C=+890.3\ \mathrm{kJ} \]

Step 5: Add all three equations.

The combined enthalpy change is:

\[ \begin{aligned} \Delta H &=-393.5-571.6+890.3\\ &=-965.1+890.3\\ &=-74.8\ \mathrm{kJ} \end{aligned} \]

When the equations are added, carbon dioxide, water, and oxygen cancel.

The remaining equation is:

\[ \ce{C(graphite) + 2H2(g) -> CH4(g)} \]

This produces exactly one mole of methane.

Therefore:

\[ \boxed{\Delta H_f^\circ(\ce{CH4(g)}) =-74.8\ \mathrm{kJ\,mol^{-1}}} \]

The complete formation equation is:

\[ \ce{C(graphite) + 2H2(g) -> CH4(g)} \]

Why reversing Reaction C matters: Simply adding the three original equations would leave methane on the reactant side. Reversing Reaction C puts methane on the product side and changes its enthalpy contribution from negative to positive.

The cancellation of the other substances is what makes the target formation equation possible.

Question 3

Tough exam style

A chemist wants to determine the standard enthalpy of formation of sulfur dioxide gas, \(\ce{SO2(g)}\), using supplied reaction data.

The following illustrative thermochemical equations are available:

Reaction A

\[ \ce{2S(s) + 3O2(g) -> 2SO3(g)} \]

\[ \Delta H_A=-791.4\ \mathrm{kJ} \]

Reaction B

\[ \ce{2SO2(g) + O2(g) -> 2SO3(g)} \]

\[ \Delta H_B=-197.8\ \mathrm{kJ} \]

(a) Calculate the standard enthalpy of formation of \(\ce{SO2(g)}\).

(b) Explain why using the enthalpy changes as supplied, without reversing either reaction, cannot produce the required formation equation.

Solution 3

The standard enthalpy of formation of sulfur dioxide is \(-296.8\ \mathrm{kJ\,mol^{-1}}\).

We need to reverse Reaction B so that sulfur dioxide appears as a product rather than a reactant.

Step 1: Write the target formation equation.

\[ \ce{S(s) + O2(g) -> SO2(g)} \]

This equation produces one mole of sulfur dioxide.

Step 2: Keep Reaction A unchanged.

\[ \ce{2S(s) + 3O2(g) -> 2SO3(g)} \]

\[ \Delta H_A=-791.4\ \mathrm{kJ} \]

Step 3: Reverse Reaction B.

\[ \ce{2SO3(g) -> 2SO2(g) + O2(g)} \]

Therefore:

\[ \Delta H_B=+197.8\ \mathrm{kJ} \]

Step 4: Add the equations.

\[ \begin{aligned} \ce{2S(s) + 3O2(g)} &\ce{-> 2SO3(g)}\\ \ce{2SO3(g)} &\ce{-> 2SO2(g) + O2(g)} \end{aligned} \]

Cancel the two moles of \(\ce{SO3}\).

The resulting equation is:

\[ \ce{2S(s) + 2O2(g) -> 2SO2(g)} \]

Step 5: Add the enthalpy changes.

\[ \begin{aligned} \Delta H &=-791.4+197.8\\ &=-593.6\ \mathrm{kJ} \end{aligned} \]

This reaction forms two moles of sulfur dioxide.

Divide by two:

\[ \Delta H_f^\circ(\ce{SO2}) =\frac{-593.6}{2} \]

\[ \boxed{-296.8\ \mathrm{kJ\,mol^{-1}}} \]

The original reactions cannot simply be added because they both produce sulfur trioxide. Sulfur dioxide must be produced in the final equation, so Reaction B must be reversed.

The other important detail is dividing by two. The combined equation forms two moles of sulfur dioxide, not one.

Question 4

Tough exam style

A chemistry student is given two illustrative combustion enthalpies for propane at 298 K and 100 kPa.

Reaction A: Liquid water produced

\[ \ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)} \]

\[ \Delta H_A=-2219.0\ \mathrm{kJ} \]

Reaction B: Water vapour produced

\[ \ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)} \]

\[ \Delta H_B=-2043.0\ \mathrm{kJ} \]

The following standard formation enthalpies are also supplied:

(a) Calculate the standard enthalpy of formation of propane using Reaction A.

(b) Repeat the calculation using Reaction B.

(c) Calculate the standard enthalpy change for the vaporisation of one mole of water.

(d) Explain why the two combustion enthalpies are different, even though the same fuel is combusted in both reactions.

Solution 4

Both combustion datasets give the same standard enthalpy of formation for propane:

\[ \boxed{-104.7\ \mathrm{kJ\,mol^{-1}}} \]

The standard enthalpy of vaporisation of water is:

\[ \boxed{+44.0\ \mathrm{kJ\,mol^{-1}}} \]

The difference between the two combustion enthalpies comes from the physical state of the water produced.

Part (a): Calculate using liquid water.

Step 1: Write the formation enthalpy equation.

Let:

\[ \Delta H_f^\circ(\ce{C3H8})=x \]

Apply products minus reactants:

\[ \begin{aligned} \Delta H_A ={}&[3\Delta H_f^\circ(\ce{CO2})\\ &+4\Delta H_f^\circ(\ce{H2O(l)})]\\ &-[x+5\Delta H_f^\circ(\ce{O2})] \end{aligned} \]

Step 2: Substitute the values.

\[ \begin{aligned} -2219.0 ={}&[3(-393.5)+4(-285.8)]\\ &-[x+5(0)] \end{aligned} \]

Step 3: Calculate the products.

\[ 3(-393.5)=-1180.5 \]

\[ 4(-285.8)=-1143.2 \]

Therefore:

\[ -2219.0=-2323.7-x \]

Step 4: Rearrange.

\[ \begin{aligned} 104.7&=-x\\ x&=-104.7 \end{aligned} \]

So:

\[ \boxed{\Delta H_f^\circ(\ce{C3H8}) =-104.7\ \mathrm{kJ\,mol^{-1}}} \]

Part (b): Calculate using water vapour.

Now use the formation enthalpy of gaseous water rather than liquid water.

\[ \begin{aligned} -2043.0 ={}&[3(-393.5)+4(-241.8)]\\ &-[x+5(0)] \end{aligned} \]

Calculate the products:

\[ 3(-393.5)=-1180.5 \]

\[ 4(-241.8)=-967.2 \]

Therefore:

\[ -2043.0=-2147.7-x \]

Rearranging:

\[ 104.7=-x \]

\[ \boxed{x=-104.7\ \mathrm{kJ\,mol^{-1}}} \]

Both methods give the same formation enthalpy.

Part (c): Calculate the enthalpy of vaporisation.

Vaporisation is the change from liquid water to water vapour:

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

Apply products minus reactants:

\[ \begin{aligned} \Delta H_{\mathrm{vap}} &=\Delta H_f^\circ(\ce{H2O(g)})\\ &\quad-\Delta H_f^\circ(\ce{H2O(l)})\\ &=-241.8-(-285.8)\\ &=+44.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Vaporisation is endothermic because energy is required to separate the water molecules into the gas phase.

Part (d): Explain the difference between the combustion enthalpies.

The combustion reaction producing liquid water is more exothermic.

The reason is that liquid water has a lower enthalpy than water vapour. Additional energy is released when the water formed during combustion is in the liquid state rather than remaining as vapour.

The difference between the two reaction enthalpies is:

\[ \begin{aligned} \Delta H_B-\Delta H_A &=-2043.0-(-2219.0)\\ &=+176.0\ \mathrm{kJ} \end{aligned} \]

Four moles of water are produced, and each mole requires 44.0 kJ to vaporise.

\[ 4(44.0)=176.0\ \mathrm{kJ} \]

The values agree exactly.

The tempting mistake is to assume that combustion enthalpy depends only on the fuel. It also depends on the physical states of the products.

Question 5

Extension challenge

A student is calculating the standard enthalpy of formation of carbon monoxide at 298 K and 100 kPa.

They receive the following illustrative reaction data:

Reaction A

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

\[ \Delta H_A=-393.5\ \mathrm{kJ} \]

Reaction B

\[ \ce{C(diamond) + O2(g) -> CO2(g)} \]

\[ \Delta H_B=-395.4\ \mathrm{kJ} \]

Reaction C

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

\[ \Delta H_C=-283.0\ \mathrm{kJ} \]

Two students produce different answers.

• Student A uses Reaction A and Reaction C to calculate the formation enthalpy of carbon monoxide.
• Student B uses Reaction B and Reaction C because the diamond combustion reaction releases more energy.

Student B argues that their result must be more accurate because they used a larger measured energy change.

Resolve the disagreement.

Your response must:

(a) Calculate the standard enthalpy of formation of carbon monoxide.

(b) Calculate the enthalpy change for the conversion of diamond into graphite:

\[ \ce{C(diamond) -> C(graphite)} \]

(c) Calculate the enthalpy change for forming carbon monoxide directly from diamond and oxygen:

\[ \ce{C(diamond) + 1/2O2(g) -> CO(g)} \]

(d) Demonstrate numerically that both pathways from diamond to carbon dioxide give the same total enthalpy change.

(e) Explain why Student B’s answer is not the standard enthalpy of formation of carbon monoxide, even if their Hess calculation is mathematically correct.

Solution 5

Student A has the correct standard enthalpy of formation:

\[ \boxed{\Delta H_f^\circ(\ce{CO}) =-110.5\ \mathrm{kJ\,mol^{-1}}} \]

Student B has calculated a different, valid enthalpy change: the formation of carbon monoxide from diamond rather than graphite.

The distinction is the standard state of carbon.

Part (a): Calculate the standard formation enthalpy of carbon monoxide.

Step 1: Identify the standard state.

At 298 K and 100 kPa, graphite is the standard state of carbon.

Therefore, the target formation reaction is:

\[ \ce{C(graphite) + 1/2O2(g) -> CO(g)} \]

Step 2: Reverse Reaction C.

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

\[ \Delta H_C=+283.0\ \mathrm{kJ} \]

Step 3: Add Reaction A and reversed Reaction C.

\[ \Delta H=-393.5+283.0 \]

\[ \boxed{-110.5\ \mathrm{kJ\,mol^{-1}}} \]

This is the standard enthalpy of formation because the reaction begins with graphite, the standard state of carbon.

Part (b): Calculate the conversion of diamond into graphite.

We want:

\[ \ce{C(diamond) -> C(graphite)} \]

Notice that both Reaction A and Reaction B produce the same substance, carbon dioxide.

Reverse Reaction A:

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

\[ \Delta H=+393.5\ \mathrm{kJ} \]

Now add it to Reaction B:

\[ \ce{C(diamond) + O2(g) -> CO2(g)} \]

\[ \Delta H=-395.4\ \mathrm{kJ} \]

Adding the enthalpy changes:

\[ \begin{aligned} \Delta H &=-395.4+393.5\\ &=-1.9\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The resulting equation is:

\[ \ce{C(diamond) -> C(graphite)} \]

Therefore:

\[ \boxed{\Delta H=-1.9\ \mathrm{kJ\,mol^{-1}}} \]

Diamond has a slightly higher enthalpy than graphite under these conditions. Converting diamond to graphite releases a small amount of energy.

Part (c): Calculate the formation of carbon monoxide from diamond.

Now use Reaction B rather than Reaction A.

Reverse Reaction C:

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

\[ \Delta H=+283.0\ \mathrm{kJ} \]

Add it to Reaction B:

\[ \ce{C(diamond) + O2(g) -> CO2(g)} \]

\[ \Delta H=-395.4\ \mathrm{kJ} \]

Therefore:

\[ \begin{aligned} \Delta H &=-395.4+283.0\\ &=-112.4\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The resulting reaction is:

\[ \ce{C(diamond) + 1/2O2(g) -> CO(g)} \]

So:

\[ \boxed{\Delta H=-112.4\ \mathrm{kJ\,mol^{-1}}} \]

Student B’s calculation is mathematically correct for the reaction they actually calculated.

However, it is not the standard enthalpy of formation because diamond is not carbon’s standard state.

Part (d): Demonstrate that both pathways give the same result.

We can now compare two pathways from diamond to carbon dioxide.

The direct pathway is Reaction B:

\[ \ce{C(diamond) + O2(g) -> CO2(g)} \]

\[ \Delta H=-395.4\ \mathrm{kJ} \]

The alternative pathway goes through graphite and carbon monoxide:

\[ \ce{C(diamond) -> C(graphite)} \]

\[ \Delta H=-1.9\ \mathrm{kJ} \]

Then:

\[ \ce{C(graphite) + 1/2O2(g) -> CO(g)} \]

\[ \Delta H=-110.5\ \mathrm{kJ} \]

Finally:

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

\[ \Delta H=-283.0\ \mathrm{kJ} \]

Add the three enthalpy changes:

\[ \begin{aligned} \Delta H_{\mathrm{total}} &=-1.9-110.5-283.0\\ &=-395.4\ \mathrm{kJ} \end{aligned} \]

This matches the direct combustion value:

\[ \boxed{-395.4\ \mathrm{kJ}} \]

Part (e): Explain the disagreement.

The two students have not calculated the same reaction.

Student A calculated the formation of carbon monoxide from graphite, which is carbon’s standard state.

Student B calculated the formation of carbon monoxide from diamond, which is a different starting state with a slightly higher enthalpy.

Both calculations are consistent with Hess’s Law.

The important distinction is that standard formation enthalpy has a specific definition. You cannot replace an element’s standard state with another allotrope simply because its reaction data are available or its measured energy change is larger.

This is the deeper lesson: Hess’s Law allows different pathways to be compared, but it does not allow you to change the starting or final chemical states without accounting for the enthalpy difference.

0808 A final checklist for formation enthalpy calculations

When you receive supplied reaction data, work through these checks before submitting your answer.

• [ ] Have I written the exact formation equation I need?
• [ ] Does the formation equation produce exactly one mole of compound?
• [ ] Are all starting elements in their standard states?
• [ ] Have I included the physical states of every substance?
• [ ] If I reversed an equation, did I reverse the sign of its enthalpy?
• [ ] If I multiplied an equation, did I multiply its enthalpy by the same factor?
• [ ] When adding equations, did I cancel all substances appearing on both sides?
• [ ] If I used formation enthalpy data, did I calculate products minus reactants?
• [ ] Did I multiply every formation enthalpy by its balanced equation coefficient?
• [ ] Have I distinguished an experimental or Hess-derived value from an approximate bond-enthalpy estimate?
• [ ] Does my final answer have the correct sign and units?

The most useful habit is to write the target chemical equation before touching your calculator. It tells you which substances must appear, which equations need reversing, and whether you need to divide your final enthalpy change.

Once you can manipulate complete thermochemical equations confidently, the next useful skill is recognising when average bond enthalpies are being used instead. Both approaches involve energy conservation, but one uses reaction enthalpy data to construct an exact Hess cycle, while the other estimates energy changes using a simplified model of chemical bonds.