How to Build a Hess Cycle from Combustion Data

When you are given the enthalpies of combustion of several substances, how can you use them to calculate the enthalpy change of a reaction that hasn’t been measured directly?

The key is to make every substance reach the same final combustion products. Once the reactants and products have been taken to the same destination, their energy changes can be compared using Hess’s Law.

This is called a Hess cycle using enthalpies of combustion. The central calculation is:

\[
\boxed{
\Delta H_{\text{reaction}}
=
\sum \Delta H_c(\text{reactants})
–
\sum \Delta H_c(\text{products})
}
\]

The sums must include the correct stoichiometric coefficients. Combustion enthalpies are signed values, so keep their negative signs throughout the calculation.

Let’s build the cycle from the chemical equations rather than trying to memorise a diagram.

01The problem: finding an enthalpy change indirectly

Consider the reaction between ethene and hydrogen:

\[
\ce{C2H4(g) + H2(g) -> C2H6(g)}
\]

This reaction converts ethene into ethane. Suppose you want to know its enthalpy change, but you don’t have a reliable direct measurement.

You do, however, have the following illustrative combustion data.

These numbers are illustrative, not experimental results from a particular investigation.

Predict which should release more energy: burning one mole of ethene and one mole of hydrogen separately, or burning one mole of ethane?

The separate combustion reactions release a total of \(1411+286=1697\) kJ. Ethane combustion releases 1560 kJ per mole. The difference between these energy changes tells us about the reaction that converts ethene and hydrogen into ethane.

That is the information we want.

02Build the cycle from complete combustion equations

The most important decision is choosing a common destination.

For substances containing carbon and hydrogen, complete combustion produces:

• Carbon dioxide, \(\ce{CO2(g)}\), from carbon.
• Liquid water, \(\ce{H2O(l)}\), from hydrogen, when the supplied combustion data specify liquid water.

Every substance in the cycle must reach the same final products, in the same physical states.

For our example, the common destination is:

\[
\ce{2CO2(g) + 3H2O(l)}
\]

Why these amounts?

The original reaction contains two carbon atoms and six hydrogen atoms on each side. Complete combustion therefore produces two carbon dioxide molecules and three water molecules, regardless of whether the starting material is ethene plus hydrogen or ethane.

Now write the three complete combustion equations:

Route 1: Burn the reactants separately

\[
\ce{C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l)}
\]

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

Adding these equations gives:

\[
\ce{C2H4(g) + H2(g) + 7/2O2(g)}
\]

\[
\ce{-> 2CO2(g) + 3H2O(l)}
\]

The combined enthalpy change is:

\[
\begin{aligned}
\Delta H_{\text{reactants, combustion}}
&=-1411+(-286)\\
&=-1697\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Route 2: Burn the product

\[
\ce{C2H6(g) + 7/2O2(g) -> 2CO2(g) + 3H2O(l)}
\]

Its enthalpy change is:

\[
\Delta H_{\text{products, combustion}}
=-1560\ \mathrm{kJ\,mol^{-1}}
\]

Both routes finish at exactly the same chemical state.

This is what makes the cycle work. Hess’s Law tells us that the total enthalpy change depends on the initial and final states, not on the route taken between them.

03Calculate the unknown enthalpy change

We can now compare the two routes.

Starting with ethene and hydrogen, there are two possible paths:

• React the ethene with hydrogen to form ethane, then combust the ethane.
• Combust the ethene and hydrogen separately.

The total enthalpy change must be identical.

Therefore:

\[
\Delta H_{\text{reaction}}
+
\Delta H_c(\ce{C2H6})
=
\Delta H_c(\ce{C2H4})
+
\Delta H_c(\ce{H2})
\]

Rearrange:

\[
\Delta H_{\text{reaction}}
=
\Delta H_c(\ce{C2H4})
+
\Delta H_c(\ce{H2})
–
\Delta H_c(\ce{C2H6})
\]

Substitute the supplied values:

\[
\begin{aligned}
\Delta H_{\text{reaction}}
&=-1411-286-(-1560)\\
&=-1697+1560\\
&=-137\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The reaction is exothermic. Forming one mole of ethane from ethene and hydrogen releases 137 kJ under the stated conditions.

Notice the subtraction of a negative number. This is a common source of errors. The product’s combustion enthalpy is negative, but the cycle requires us to subtract it.

The useful rule is:

\[
\boxed{
\Delta H_{\text{reaction}}
=
\text{combustion of reactants}
–
\text{combustion of products}
}
\]

Remember that this rule uses the complete, coefficient-weighted combustion enthalpies, not just the numbers copied from a data table.

04Why the combustion data are negative

Before manipulating the equations, it helps to understand what the supplied numbers actually represent.

Combustion is exothermic. The reacting chemicals transfer energy to their surroundings, so their enthalpy decreases.

For example:

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

\[
\Delta H_c=-286\ \mathrm{kJ\,mol^{-1}}
\]

The negative sign means that the chemical system loses 286 kJ for every mole of hydrogen combusted.

It does not mean that the surroundings lose energy. The surroundings gain it.

Where does a combustion enthalpy value come from?

A combustion value can originate from an experiment.

In a simple calorimetry investigation, a known amount of fuel is burned beneath a container of water. The water temperature increases. Its energy gain can be estimated using:

\[
q=mc\Delta T
\]

Here:

• \(q\) is the heat absorbed by the water, in joules.
• \(m\) is the mass of water, in grams.
• \(c\) is the specific heat capacity, in \(\mathrm{J\,g^{-1}\,^\circ C^{-1}}\).
• \(\Delta T\) is the temperature increase, in degrees Celsius.

The reaction releases approximately the same amount of energy that the measured surroundings absorb, but with the opposite sign.

The amount of fuel burned is then converted into moles, allowing an experimental molar enthalpy of combustion to be calculated.

However, a school calorimeter does not capture all the released energy. Heat can escape into the air, the apparatus can absorb energy, and incomplete combustion can occur. A measured result is therefore not automatically equal to an accepted standard enthalpy value.

For more precise combustion measurements, a bomb calorimeter uses a sample in a sealed, oxygen-filled vessel surrounded by water. An ignition wire starts the reaction, and the temperature rise of the surrounding water and calibrated apparatus is measured. The fixed-volume measurement requires appropriate treatment before it is reported as an enthalpy change.

The important distinction is that experimental measurements, tabulated values, and calculated Hess cycle results are not interchangeable.

A Hess cycle does not magically make inaccurate experimental data accurate. It uses the supplied enthalpy values consistently to calculate an unknown.

A useful guide to the calorimetry side of this process is Heat of Combustion: Calculations for HSC Chemistry. The important point here is what happens after those combustion values have been obtained: they become the alternative pathways in a Hess cycle.

What the original experiment can and cannot tell us

Imagine a student using a spirit burner to heat water.

The burner sits beneath a metal container holding a known mass of water. A thermometer measures the temperature rise. The student weighs the burner before and after burning to estimate how much fuel was consumed.

The prediction is straightforward: the energy released by combustion should equal the energy absorbed by the surroundings.

But the observation often falls short of that prediction. Some energy heats the air and apparatus, some fuel may evaporate without burning, and the fuel may undergo incomplete combustion.

A more controlled design uses a bomb calorimeter:

1. A measured fuel sample sits in a small crucible inside a strong sealed steel vessel.
2. The vessel is filled with excess oxygen.
3. An ignition wire connected to an external electrical supply starts combustion.
4. The sealed vessel sits inside a known quantity of water.
5. A thermometer measures the water’s temperature increase while a stirrer helps distribute the energy evenly.

The sealed oxygen environment encourages complete combustion. The surrounding water and calibrated apparatus provide a more controlled measurement of the released energy.

There are still limitations. A bomb calorimeter operates at constant volume, so its measured energy change requires appropriate treatment when reporting an enthalpy change. Heat losses and calibration also matter.

This explains why a supplied table of standard molar enthalpies of combustion is not necessarily a list of raw thermometer readings.

Keep these three levels separate:

• A temperature rise is an experimental observation.
• An experimental enthalpy is calculated from measured quantities and assumptions.
• A Hess cycle produces a derived reaction enthalpy from the supplied combustion enthalpies.

Average bond enthalpies are different again. They are model-based estimates, not direct measurements of the particular reaction.

05A reliable method for building a combustion Hess cycle

The easiest way to avoid sign errors is to build the chemical equations first and calculate the numbers afterwards.

Step 1: Write the target reaction

Start with the reaction whose enthalpy change you need.

For example:

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

Balance it before doing anything else. The coefficients determine how many moles of each substance participate.

Step 2: Write the complete combustion equation for each substance

For carbon-containing fuels, balance carbon first, hydrogen second, and oxygen last.

Check that every combustion equation is balanced.

For a hydrocarbon, the general pattern is:

\[
\ce{C_xH_y + (x+y/4)O2 -> xCO2 + (y/2)H2O}
\]

The equation assumes complete combustion and liquid water as the product. Change the physical-state symbols if the supplied data specify otherwise.

Step 3: Make the final products identical

This is the defining feature of a combustion Hess cycle.

Count the total carbon and hydrogen atoms on both routes. Both must end with the same quantities of \(\ce{CO2}\) and \(\ce{H2O}\).

Do not assume that two routes are equivalent just because both produce carbon dioxide and water. The quantities and physical states must also match.

Step 4: Apply the coefficients

If the target reaction uses two moles of a substance, its combustion enthalpy must also be multiplied by two.

For example, if:

\[
\Delta H_c(\ce{H2})=-286\ \mathrm{kJ\,mol^{-1}}
\]

then combusting two moles gives:

\[
2\Delta H_c(\ce{H2})
=2(-286)=-572\ \mathrm{kJ}
\]

This is one of the most common places students lose marks.

Step 5: Compare the two routes

The route through the reactants must have the same total enthalpy change as the route through the products.

Therefore:

\[
\boxed{
\Delta H_{\mathrm{reaction}}
=
\sum n\Delta H_c(\mathrm{reactants})
–
\sum n\Delta H_c(\mathrm{products})
}
\]

Here, \(n\) represents the stoichiometric coefficient of each substance in the target equation.

The subtraction is not an arbitrary shortcut. It comes directly from comparing two routes to the same final state.

Step 6: Check the result

Before accepting your answer, check:

• Does the final reaction match the target equation?
• Have all combustion equations been balanced?
• Have all coefficients been applied?
• Do both routes reach identical final products?
• Have you kept the negative signs?
• Is the final unit appropriate for the reaction as written?

A negative answer means the target reaction is exothermic. A positive answer means it is endothermic.

06Worked examples: applying the method

The first example introduces a useful special case: one of the reactants is already a combustion product.

07A contrasting case: when the water states do not match

Here’s a situation where the usual combustion formula can give a very convincing wrong answer.

Suppose a data table gives the combustion of propane to carbon dioxide and water vapour:

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

Its supplied enthalpy change is:

\[
\Delta H_c=-2044\ \mathrm{kJ\,mol^{-1}}
\]

But the other combustion values in your Hess cycle produce liquid water.

Can you use \(-2044\) directly?

No. The two routes do not reach the same final state.

The propane reaction produces four moles of water vapour. To reach the liquid-water endpoint, those four moles must condense.

Suppose the enthalpy of vaporisation of water is \(+44.0\ \mathrm{kJ\,mol^{-1}}\). Condensation has the opposite sign:

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

\[
\Delta H=-44.0\ \mathrm{kJ\,mol^{-1}}
\]

For four moles of water:

\[
\begin{aligned}
\Delta H_{\mathrm{condensation}}
&=4(-44.0)\\
&=-176.0\ \mathrm{kJ}
\end{aligned}
\]

Add this to the combustion enthalpy:

\[
\begin{aligned}
\Delta H_c(\text{liquid water})
&=-2044-176\\
&=-2220\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The corrected combustion enthalpy is \(-2220\ \mathrm{kJ\,mol^{-1}}\).

The tempting approach is to use the supplied number because the chemical formula of the product is still \(\ce{H2O}\). But physical state is part of the thermochemical equation. Condensation releases additional energy, so ignoring it changes the result.

Decision rule: If the final products of every combustion equation are identical in quantity and physical state, use the simple combustion formula. If they are not, add the necessary phase-change steps before comparing the routes.

08Manipulating thermochemical equations without losing the sign

Sometimes a question asks you to construct the cycle by adding and reversing complete equations rather than using the shortcut formula.

The same three rules always apply.

For example, hydrogen combustion is:

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

\[
\Delta H=-286\ \mathrm{kJ}
\]

If you reverse it:

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

then:

\[
\Delta H=+286\ \mathrm{kJ}
\]

If you double the original reaction:

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

then:

\[
\Delta H=2(-286)=-572\ \mathrm{kJ}
\]

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

Two valid ways to calculate the answer

You can use either of these approaches:

Method A: Use the combustion shortcut.

\[
\Delta H_{\mathrm{reaction}}
=
\Delta H_{c,\mathrm{reactants}}
–
\Delta H_{c,\mathrm{products}}
\]

Method B: Reverse the product combustion equations and add them to the reactant combustion equations.

Both approaches give the same result.

Be careful not to mix them. The subtraction of product combustion enthalpies in Method A already accounts for reversing those reactions. You must not reverse their signs and then subtract them again.

09Questions and solutions

The following questions progress from a straightforward combustion cycle to problems involving unchanged reactants, competing estimates, experimental limitations, and mismatched product states.

Question 1

Methane can be formed from solid graphite and hydrogen:

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

The following illustrative standard molar enthalpies of combustion are supplied:

All combustion reactions produce \(\ce{CO2(g)}\) and \(\ce{H2O(l)}\).

Calculate the enthalpy change for the formation of one mole of methane.

Solution 1

The enthalpy change is \(-74.8\ \mathrm{kJ\,mol^{-1}}\). Forming one mole of methane from graphite and hydrogen is exothermic.

Step 1: Write the combustion equations.

For graphite:

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

For hydrogen:

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

For methane:

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

All three routes reach the same final products.

Step 2: Calculate the total combustion enthalpy of the reactants.

Two moles of hydrogen are required:

\[
\begin{aligned}
\Delta H_{\mathrm{reactants}}
&=-393.5+2(-285.8)\\
&=-393.5-571.6\\
&=-965.1\ \mathrm{kJ}
\end{aligned}
\]

Step 3: Calculate the product combustion enthalpy.

One mole of methane is produced:

\[
\Delta H_{\mathrm{products}}
=-890.3\ \mathrm{kJ}
\]

Step 4: Apply the Hess relationship.

\[
\begin{aligned}
\Delta H_{\mathrm{reaction}}
&=-965.1-(-890.3)\\
&=-965.1+890.3\\
&=-74.8\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The negative sign confirms that methane formation releases energy.

The main trap is forgetting that two moles of hydrogen react. If you only use \(-285.8\) once, the reactant combustion pathway will not contain enough hydrogen to reach the final products.

Question 2

Propene reacts with liquid water to produce propan-2-ol:

\[
\ce{C3H6(g) + H2O(l) -> C3H7OH(l)}
\]

The following illustrative combustion data are provided:

Both combustion reactions produce carbon dioxide and liquid water.

a. Calculate the enthalpy change for the hydration of propene.

b. Explain why the reactant water does not need its own combustion enthalpy in this cycle.

c. A student incorrectly adds an extra \(-286\ \mathrm{kJ\,mol^{-1}}\) to the reactant combustion total, arguing that water contains hydrogen. Calculate the student’s incorrect result and explain the error.

Solution 2

The hydration enthalpy is \(-52\ \mathrm{kJ\,mol^{-1}}\). Water is already a fully oxidised product, so it passes through the cycle unchanged. The student’s incorrect result is \(-338\ \mathrm{kJ\,mol^{-1}}\), because they have counted water as though it were hydrogen gas undergoing combustion.

Step 1: Write the combustion equation for propene.

\[
\ce{C3H6(g) + 9/2O2(g) -> 3CO2(g) + 3H2O(l)}
\]

There are three carbon atoms and six hydrogen atoms, giving three carbon dioxide molecules and three water molecules.

However, the original reaction also contains one mole of water.

That water remains in the final products, giving:

\[
\ce{3CO2(g) + 4H2O(l)}
\]

Step 2: Write the combustion equation for propan-2-ol.

\[
\ce{C3H7OH(l) + 9/2O2(g) -> 3CO2(g) + 4H2O(l)}
\]

Both routes now reach the same final state.

Step 3: Calculate the unknown enthalpy.

\[
\begin{aligned}
\Delta H_{\mathrm{reaction}}
&=\Delta H_c(\text{propene})\\
&\quad-\Delta H_c(\text{propan-2-ol})\\
&=-2058-(-2006)\\
&=-52\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The reaction is exothermic.

Step 4: Explain why water has no combustion term.

Water already contains hydrogen in its fully oxidised form. It cannot undergo further combustion to release additional energy.

It simply remains present as the cycle proceeds.

This is different from hydrogen gas, which can react with oxygen to form water and release energy.

Step 5: Identify the student’s incorrect calculation.

The student has incorrectly treated water as though it were hydrogen gas.

Their calculation becomes:

\[
\begin{aligned}
\Delta H_{\mathrm{incorrect}}
&=(-2058-286)-(-2006)\\
&=-2344+2006\\
&=-338\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

This is incorrect because no additional hydrogen combustion occurs. The water is already a final combustion product, not another fuel.

Question 3

A chemist wants to calculate the enthalpy change for forming ethene from its elements:

\[
\ce{2C(s, graphite) + 2H2(g) -> C2H4(g)}
\]

The following illustrative standard molar combustion enthalpies are available:

All combustion reactions produce \(\ce{CO2(g)}\) and \(\ce{H2O(l)}\).

a. Calculate the enthalpy change for the formation of one mole of ethene.

b. A student calculates \(-52.4\ \mathrm{kJ\,mol^{-1}}\). Identify the likely error.

c. Determine the enthalpy change for the reverse reaction, in which ethene decomposes into graphite and hydrogen.

Solution 3

The enthalpy change for forming ethene is \(+52.4\ \mathrm{kJ\,mol^{-1}}\). The reaction is endothermic. The student has most likely subtracted the two combustion pathways in the wrong order. The reverse reaction has an enthalpy change of \(-52.4\ \mathrm{kJ\,mol^{-1}}\).

Step 1: Calculate the total combustion enthalpy of the reactants.

There are two moles of graphite and two moles of hydrogen.

\[
\begin{aligned}
\Delta H_{\mathrm{reactants}}
&=2(-393.5)+2(-285.8)\\
&=-787.0-571.6\\
&=-1358.6\ \mathrm{kJ}
\end{aligned}
\]

Step 2: Identify the combustion enthalpy of the product.

One mole of ethene undergoes complete combustion:

\[
\ce{C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l)}
\]

Therefore:

\[
\Delta H_{\mathrm{products}}=-1411\ \mathrm{kJ}
\]

Step 3: Calculate the target enthalpy.

\[
\begin{aligned}
\Delta H_{\mathrm{reaction}}
&=-1358.6-(-1411)\\
&=-1358.6+1411\\
&=+52.4\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The positive value means energy must be absorbed to form ethene from graphite and hydrogen.

Step 4: Explain the student’s mistake.

The student has likely calculated:

\[
-1411-(-1358.6)
=-52.4\ \mathrm{kJ\,mol^{-1}}
\]

This subtracts the reactant combustion total from the product combustion total.

The correct order is:

\[
\text{Reactant combustion}
–
\text{Product combustion}
\]

The chemical equations must determine the subtraction order, not whichever numerical arrangement looks easiest.

Step 5: Calculate the reverse reaction.

Reversing a thermochemical equation changes the sign of its enthalpy change:

\[
\ce{C2H4(g) -> 2C(s, graphite) + 2H2(g)}
\]

Therefore:

\[
\Delta H=-52.4\ \mathrm{kJ\,mol^{-1}}
\]

Decomposing ethene into its elements releases the same amount of energy that its formation absorbs.

Question 4

A chemist investigates the hydrogenation of ethyne:

\[
\ce{C2H2(g) + 2H2(g) -> C2H6(g)}
\]

The following illustrative combustion enthalpies are supplied:

All combustion reactions produce carbon dioxide and liquid water.

The chemist also estimates the reaction enthalpy using the following average bond enthalpies:

a. Calculate the reaction enthalpy using the combustion data.

b. Calculate an estimated reaction enthalpy using the average bond enthalpies.

c. Explain why the two answers do not have to be identical.

d. If a question specifically asks for the enthalpy change calculated from the combustion data, which value should be reported? Explain why averaging the two answers is not appropriate.

Solution 4

The combustion data give \(-312\ \mathrm{kJ\,mol^{-1}}\), while the average bond enthalpies give an estimate of \(-288\ \mathrm{kJ\,mol^{-1}}\). If the question asks for the combustion-data result, report \(-312\ \mathrm{kJ\,mol^{-1}}\). Do not average the two methods.

Step 1: Calculate the enthalpy change using combustion data.

The combined combustion of the reactants is:

\[
\begin{aligned}
\Delta H_{\mathrm{reactants}}
&=-1300+2(-286)\\
&=-1300-572\\
&=-1872\ \mathrm{kJ}
\end{aligned}
\]

The combustion enthalpy of one mole of ethane is:

\[
\Delta H_{\mathrm{products}}=-1560\ \mathrm{kJ}
\]

Therefore:

\[
\begin{aligned}
\Delta H_{\mathrm{reaction}}
&=-1872-(-1560)\\
&=-312\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Step 2: Identify the bonds broken and formed.

For the bond enthalpy method:

\[
\Delta H_{\mathrm{estimate}}
=
\text{energy to break bonds}
–
\text{energy released forming bonds}
\]

The net changes are:

• Break one C≡C bond.
• Break two H-H bonds.
• Form one C-C bond.
• Form four additional C-H bonds.

The two original C-H bonds are present on both sides, so they cancel from the net bond changes.

Step 3: Calculate the estimated enthalpy change.

\[
\begin{aligned}
\Delta H_{\mathrm{estimate}}
&=(839+2(436))\\
&\quad-(347+4(413))\\
&=1711-1999\\
&=-288\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Step 4: Explain the difference.

The combustion method uses supplied molar enthalpies for the specific substances involved.

The bond enthalpy method uses average values for particular bond types. These averages represent a simplified model and are not exact measurements of the individual bonds in every molecule.

Consequently, the two calculations can produce different answers without either method being incorrectly applied.

Step 5: Decide which result to report.

If the question explicitly asks for a calculation using combustion data, the answer is:

\[
\boxed{-312\ \mathrm{kJ\,mol^{-1}}}
\]

The \(-288\ \mathrm{kJ\,mol^{-1}}\) value is a separate model-based estimate.

Averaging them to obtain \(-300\ \mathrm{kJ\,mol^{-1}}\) would have no chemical justification. The two methods use different types of input data and answer slightly different estimation questions.

Question 5

A chemist investigates the hydrogenation of propene:

\[
\ce{C3H6(g) + H2(g) -> C3H8(g)}
\]

Three combustion enthalpies are supplied:

The enthalpy of vaporisation of water is \(+44.0\ \mathrm{kJ\,mol^{-1}}\).

The chemist also has the following average bond enthalpies:

In a separate calorimetry experiment, 0.0100 mol of propene undergoes complete combustion. The measured water and apparatus absorb 18.5 kJ.

Assume all enthalpy data refer to the same temperature, the supplied combustion reactions are complete, and the calorimetry sample burns completely.

Answer the following:

a. Adjust the propane combustion enthalpy so that its products include liquid water rather than water vapour.

b. Calculate the enthalpy change for the hydrogenation of propene using the corrected combustion data.

c. Calculate the enthalpy change for the reverse reaction, the dehydrogenation of propane.

d. Estimate the hydrogenation enthalpy using average bond enthalpies. Explain why this value may differ from your answer to part b.

e. Calculate the percentage of the combustion energy absorbed by the calorimeter. Explain why the measured 18.5 kJ should not replace the supplied standard combustion enthalpy in the Hess cycle.

Solution 5

The corrected propane combustion enthalpy is \(-2220\ \mathrm{kJ\,mol^{-1}}\). The hydrogenation enthalpy is \(-124\ \mathrm{kJ\,mol^{-1}}\), and the reverse reaction has an enthalpy change of \(+124\ \mathrm{kJ\,mol^{-1}}\). The bond enthalpy estimate is \(-123\ \mathrm{kJ\,mol^{-1}}\), while the calorimeter captures approximately 89.9% of the energy released.

This question combines four ideas: phase changes, Hess cycles, bond enthalpy estimates, and experimental heat transfer.

Step 1: Correct the propane combustion enthalpy.

The supplied propane combustion reaction produces four moles of water vapour.

To match the other combustion equations, all four moles must condense.

The condensation enthalpy is the negative of the vaporisation enthalpy:

\[
\Delta H_{\mathrm{condensation}}
=-44.0\ \mathrm{kJ\,mol^{-1}}
\]

For four moles:

\[
\begin{aligned}
\Delta H_{\mathrm{condensation}}
&=4(-44.0)\\
&=-176.0\ \mathrm{kJ}
\end{aligned}
\]

Add this to the propane combustion enthalpy:

\[
\begin{aligned}
\Delta H_c(\text{propane, liquid water})
&=-2044-176\\
&=-2220\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The corrected value is now consistent with the other combustion data.

Step 2: Calculate the hydrogenation enthalpy.

The total combustion enthalpy of the reactants is:

\[
\begin{aligned}
\Delta H_{\mathrm{reactants}}
&=-2058-286\\
&=-2344\ \mathrm{kJ}
\end{aligned}
\]

The corrected combustion enthalpy of the product is:

\[
\Delta H_{\mathrm{products}}=-2220\ \mathrm{kJ}
\]

Apply Hess’s Law:

\[
\begin{aligned}
\Delta H_{\mathrm{hydrogenation}}
&=-2344-(-2220)\\
&=-124\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Hydrogenation releases 124 kJ per mole of propene reacted.

Step 3: Calculate the reverse reaction.

The reverse reaction is:

\[
\ce{C3H8(g) -> C3H6(g) + H2(g)}
\]

Reversing a thermochemical equation changes the sign:

\[
\boxed{
\Delta H_{\mathrm{dehydrogenation}}
=+124\ \mathrm{kJ\,mol^{-1}}
}
\]

Dehydrogenation is endothermic. It requires 124 kJ per mole of propane decomposed.

Step 4: Calculate the bond enthalpy estimate.

Hydrogenation involves:

• Breaking one C=C bond.
• Breaking one H-H bond.
• Forming one C-C bond.
• Forming two additional C-H bonds.

Therefore:

\[
\begin{aligned}
\Delta H_{\mathrm{estimate}}
&=(614+436)\\
&\quad-(347+2(413))\\
&=1050-1173\\
&=-123\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

This is close to the combustion-data result of \(-124\ \mathrm{kJ\,mol^{-1}}\).

However, the two methods are not identical. The combustion cycle uses supplied molar combustion enthalpies, while the bond method uses average bond enthalpies.

The agreement is useful as a consistency check, but it does not make the bond enthalpy estimate an experimental measurement.

Step 5: Calculate the theoretical energy released in the calorimetry experiment.

The amount of propene burned is:

\[
n=0.0100\ \mathrm{mol}
\]

The standard molar enthalpy of combustion is:

\[
\Delta H_c=-2058\ \mathrm{kJ\,mol^{-1}}
\]

The magnitude of the expected energy release is:

\[
\begin{aligned}
q_{\mathrm{released}}
&=n|\Delta H_c|\\
&=(0.0100)(2058)\\
&=20.58\ \mathrm{kJ}
\end{aligned}
\]

Step 6: Calculate the percentage captured.

The calorimeter absorbs 18.5 kJ.

\[
\begin{aligned}
\text{Percentage captured}
&=\frac{18.5}{20.58}\times100\\
&=89.9\%
\end{aligned}
\]

Therefore, approximately 89.9% of the released energy is absorbed by the measured surroundings.

The difference is:

\[
\begin{aligned}
q_{\mathrm{unmeasured}}
&=20.58-18.5\\
&=2.08\ \mathrm{kJ}
\end{aligned}
\]

Because the question specifies complete combustion, this missing energy was still released chemically. It was simply not absorbed by the measured water and apparatus.

Step 7: Explain why the measured value cannot replace the combustion enthalpy.

The calorimeter records the energy absorbed by its surroundings, not necessarily the entire energy released by combustion.

The 18.5 kJ value relates to this particular sample and experimental setup. It is not the molar enthalpy of complete combustion.

The supplied value of \(-2058\ \mathrm{kJ\,mol^{-1}}\) describes the complete combustion reaction under the specified conditions. That is the appropriate value for the Hess cycle.

This is the important distinction:

• The calorimeter measures an energy transfer.
• The combustion enthalpy describes the chemical reaction.
• The Hess cycle uses the chemical reaction enthalpies to calculate another reaction enthalpy.
• The bond enthalpy method provides a separate approximate model.

Combining these ideas incorrectly can produce a numerically convincing answer that does not represent the reaction being investigated.

09Final checks before submitting a Hess cycle calculation

Use this short checklist when you finish an exam question.

Three mistakes deserve particular attention:

1. Subtracting in the wrong order. Always compare the combustion of the reactants with the combustion of the products.
2. Forgetting the coefficients. Two moles of hydrogen require twice the combustion enthalpy of one mole.
3. Ignoring physical states. Water vapour and liquid water have different enthalpies, so the combustion pathways must be adjusted before comparison.

10Where to go next

Combustion Hess cycles are one application of a broader idea: the enthalpy change of a reaction depends on its initial and final states, not on the pathway taken.

Once you can construct these cycles confidently, the next useful step is learning to combine more general thermochemical equations, including reactions that must be reversed or multiplied before they can be added together.

For that broader method, see Hess’s Law Energy Cycles: HSC Chemistry Calculations Explained.

You can also connect the results to bond enthalpies. The key distinction is that combustion data provide reaction-specific enthalpy values, while average bond enthalpies provide approximate model estimates.

Syllabus note: This article addresses Enthalpy and Hess’s Law in the NSW Chemistry Stage 6 Syllabus (2017), where this content appears in Year 11. The Chemistry 11-12 Syllabus (2025) begins implementation for Year 11 in 2028, with the first HSC examination under the new syllabus scheduled for 2029. Students should check the requirements for their syllabus edition rather than assuming that this material is assessable under the 2025 syllabus. 【turn249674search43】【turn275873view0】

Extension – bomb calorimetry: A bomb calorimeter operates at constant volume. Converting its measured internal-energy change into an enthalpy change requires an additional thermodynamic correction. This is a measurement detail beyond the combustion-cycle calculations developed here; when an HSC question supplies molar enthalpies of combustion directly, use those values without attempting that conversion.