Why a Negative ΔH Does Not Mean a Fuel Burns Faster

Imagine a data sheet gives methane a standard enthalpy of combustion of \(-890.7\ \mathrm{kJ\,mol^{-1}}\), while liquid ethanol has a value of \(-1367.6\ \mathrm{kJ\,mol^{-1}}\).

It is tempting to predict that ethanol must burn faster because its enthalpy change is more negative. But that prediction is wrong. A negative \(\Delta H\) tells you how much energy is released by a specified amount of reaction, not how quickly the reaction occurs.

These are two different questions. Thermochemistry tells us about energy changes. Chemical kinetics tells us about reaction rates. Understanding the difference also helps you interpret thermochemical equations, calorimetry results, and fuel comparisons correctly.

01Energy released per mole is not energy released per second

Imagine two fuels.

• Fuel A releases 900 kJ for every mole that burns completely.
• Fuel B releases 500 kJ for every mole that burns completely.

Fuel A has the more negative enthalpy change. Does that mean it releases energy faster?

Not necessarily. We also need to know how quickly each fuel is being consumed.

Think about the difference between the capacity of a battery and its power output. Capacity tells you how much energy is available. Power tells you how quickly that energy is delivered.

The same distinction applies to combustion.

The relationship is:

\[ P = |\Delta_\mathrm{c}H|\times r \]

where:

• \(P\) is the rate of energy release, in kJ s\(^{-1}\)
• \(\Delta_\mathrm{c}H\) is the molar enthalpy of combustion, in kJ mol\(^{-1}\)
• \(r\) is the molar rate of fuel consumption, in mol s\(^{-1}\)

We use the absolute value of the enthalpy change because we are calculating the magnitude of energy released. The actual enthalpy change remains negative.

Consider these illustrative results:

Fuel A releases more energy per mole. However, Fuel B releases energy much faster because it is being consumed at a higher rate.

This is the key distinction: a more exothermic reaction does not necessarily have a higher reaction rate or greater power output.

02What a thermochemical equation actually tells you

To understand the distinction properly, you need to know what a thermochemical equation represents.

A thermochemical equation is a balanced chemical equation accompanied by its enthalpy change. It tells you the energy transferred when the reaction occurs in the quantities specified by the equation.

For example, complete combustion of methane is:

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

Its standard enthalpy of combustion is approximately:

\[ \Delta_\mathrm{c}H^\circ = -890.7\ \mathrm{kJ\,mol^{-1}} \]

This means that burning one mole of methane completely, under the specified standard-state conditions, releases approximately 890.7 kJ of energy.

The negative sign indicates an exothermic reaction. The products have lower enthalpy than the reactants, and energy is transferred to the surroundings.

Now compare this with ethanol:

\[ \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \]

Its standard enthalpy of combustion is approximately:

\[ \Delta_\mathrm{c}H^\circ = -1367.6\ \mathrm{kJ\,mol^{-1}} \]

These representative reference values are compiled from experimental thermochemical data by the NIST Chemistry WebBook for methane and ethanol.

Both equations represent complete combustion, with liquid water as a product. The standard reference temperature is 298.15 K and the standard pressure is 1 bar.

Ethanol releases more energy per mole than methane in these reactions. However, neither value tells us how long combustion takes.

There is another important detail here: the physical states matter. If water is produced as a gas instead of a liquid, the enthalpy change is different because condensation releases additional energy.

Always include state symbols when writing complete thermochemical equations.

Manipulating thermochemical equations

You also need to understand how the enthalpy value changes when you manipulate a thermochemical equation.

There are two essential rules.

Rule 1: Multiplying an equation multiplies its enthalpy change by the same factor.

For example:

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

\[ \Delta H=-890.7\ \mathrm{kJ} \]

If we double every coefficient:

\[ \ce{2CH4(g) + 4O2(g) -> 2CO2(g) + 4H2O(l)} \]

Then:

\[ \Delta H=2(-890.7) \]

\[ \Delta H=-1781.4\ \mathrm{kJ} \]

Twice as much methane burns, so twice as much energy is released.

Rule 2: Reversing an equation reverses the sign of its enthalpy change.

The reverse reaction is:

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

Its enthalpy change is:

\[ \Delta H=+890.7\ \mathrm{kJ} \]

The reverse reaction requires energy rather than releasing it.

Neither manipulation tells us anything about reaction speed. They change the quantities or direction of the reaction, not the time taken for it to occur.

03How chemists measure energy released during combustion

If a negative enthalpy change doesn’t tell us how fast a fuel burns, how do we actually determine its enthalpy change?

We need an experiment that measures energy transfer.

A common school-level approach uses a spirit burner to heat water.

Consider the apparatus:

• A spirit burner containing ethanol sits beneath a metal calorimeter containing a known mass of water.
• A thermometer extends into the water to measure its temperature.
• A stirrer distributes heat throughout the water.
• The burner is weighed before and after combustion to estimate the mass of fuel consumed.

The flame heats the bottom of the calorimeter. Energy transfers into the metal and water, causing the water temperature to rise.

What should the experiment show?

Suppose we burn a known quantity of ethanol.

If the experiment captures the released energy, we should be able to calculate the heat transferred to the water and use the amount of ethanol burned to estimate the molar enthalpy of combustion.

The water’s temperature change is measured, not predicted.

We calculate the heat absorbed by the water using:

\[ q=mc\Delta T \]

where:

• \(q\) is the heat absorbed, in joules
• \(m\) is the mass of water, in grams
• \(c\) is the specific heat capacity of water, \(4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\)
• \(\Delta T\) is the temperature change, in °C

Because combustion is exothermic, the heat released by the fuel has the opposite sign to the heat absorbed by the water.

\[ q_\mathrm{reaction}\approx -q_\mathrm{water} \]

We can then divide by the number of moles of ethanol burned to estimate its molar enthalpy change.

Notice what this experiment measures: a temperature change and a quantity of fuel consumed.

Unless we also record the time taken, we have not measured the rate of combustion.

How a revised calorimeter improves the measurement

The open spirit-burner arrangement has limitations. Some energy escapes into the air, and the fuel may not undergo complete combustion.

A more controlled instrument is a bomb calorimeter.

In this arrangement:

• A small fuel sample sits in a crucible inside a strong, sealed steel vessel called the bomb.
• The bomb contains excess oxygen to support complete combustion.
• An ignition wire is connected to the sample.
• The bomb is surrounded by a known quantity of water.
• A thermometer and stirrer measure and distribute the temperature change.
• An insulated outer jacket reduces energy exchange with the surroundings.

When combustion is initiated, the released energy heats the bomb and surrounding water. The temperature rise is measured using a calibrated calorimeter.

This revised design improves the measurement by providing excess oxygen and more controlled heat capture.

However, it still does not automatically tell us how quickly the fuel burns. The usual calculation uses the total temperature rise to determine the energy released.

A time-resolved temperature or heat-flow measurement would be needed to investigate the rate.

Precision note – extension: A bomb calorimeter operates at constant volume, so its direct heat measurement corresponds to the internal energy change, \(\Delta U\), rather than directly to the enthalpy change, \(\Delta H\). Converting between them requires a correction:

\[ \Delta H=\Delta U+\Delta n_\mathrm{g}RT \]

Here, \(\Delta n_\mathrm{g}\) is the change in the amount of gaseous substances, \(R\) is the gas constant, and \(T\) is the absolute temperature. This correction is useful for understanding precise calorimetric measurements but is not needed for the central distinction between enthalpy and reaction rate.

04Bond enthalpies give model estimates, not measured values

There is another important distinction when working with thermochemical equations.

You might be given average bond enthalpies and asked to estimate the enthalpy change of a combustion reaction.

This method is different from calorimetry.

The bond enthalpy method uses a model of chemical bonds. It estimates the energy required to break bonds and the energy released when new bonds form.

The relationship is:

\[ \Delta H\approx \sum E_\mathrm{broken} - \sum E_\mathrm{formed} \]

Breaking bonds requires energy. Forming bonds releases energy.

Consider methane combustion producing water vapour:

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

The bonds broken are:

• Four C-H bonds in methane.
• Two O=O bonds in oxygen.

The bonds formed are:

• Two C=O bonds in carbon dioxide.
• Four O-H bonds in water.

Suppose the following average bond enthalpies are provided:

The energy required to break the bonds is:

\[ \begin{aligned} E_\mathrm{broken} &=4(413)+2(498)\\ &=2648\ \mathrm{kJ} \end{aligned} \]

The energy released when the new bonds form is:

\[ \begin{aligned} E_\mathrm{formed} &=2(805)+4(463)\\ &=3462\ \mathrm{kJ} \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta H&\approx2648-3462\\ &\approx-814\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The model predicts an exothermic reaction, with an estimated enthalpy change of approximately \(-814\ \mathrm{kJ\,mol^{-1}}\).

But why doesn’t this exactly match the standard methane combustion value of \(-890.7\ \mathrm{kJ\,mol^{-1}}\)?

There are two important reasons.

First, average bond enthalpies are approximations. A particular bond can have a slightly different enthalpy depending on its molecular environment.

Second, the two equations do not have the same product state. The bond-enthalpy calculation produces water vapour, whereas the standard combustion equation above produces liquid water.

Condensing the product water releases additional energy, making the standard combustion enthalpy more negative.

This is why you must compare complete thermochemical equations, including their state symbols, rather than comparing numerical values in isolation.

Model estimate versus experimental value

Most importantly, none of the first three methods automatically gives you the combustion rate.

05Why an exothermic fuel can still burn slowly

We now know that enthalpy change and reaction rate describe different properties.

But what actually controls the speed of combustion?

The answer involves activation energy and collision theory.

Activation energy is the barrier to reaction

Even when a reaction releases energy overall, the reactants usually need an initial input of energy before the reaction can proceed.

This minimum energy requirement is called the activation energy, \(E_\mathrm{a}\).

Think of a skateboarder travelling down a ramp.

The total drop in height tells you how much gravitational potential energy is converted during the descent. But it doesn’t tell you how high the initial ramp is.

A steep initial ramp could require considerable effort to climb before the descent begins.

In a reaction energy profile:

• The vertical difference between reactants and products represents \(\Delta H\).
• The energy difference between reactants and the peak represents the activation energy.
• The peak represents the transition state, a temporary high-energy arrangement of atoms.

A reaction can therefore have a very negative \(\Delta H\) but still have a high activation energy.

Notice that both pathways have the same overall enthalpy change.

However, one pathway has a lower activation energy.

At the same temperature, the lower-barrier pathway generally allows a greater proportion of reacting particles to overcome the energy barrier, increasing the reaction rate, assuming other relevant factors are comparable.

This is why a reaction’s enthalpy change cannot tell you its activation energy.

What does a catalyst change?

A catalyst provides an alternative reaction pathway with a lower activation energy.

It increases the reaction rate without changing the overall enthalpy change.

Why?

Because the reactants and products remain the same. Their enthalpies have not changed. Only the pathway between them is different.

This gives us a useful comparison:

A catalyst can therefore make a fuel react more quickly without making its combustion more exothermic.

Other factors affecting combustion rate

Activation energy is not the only factor influencing combustion.

The rate also depends on conditions that affect the frequency of successful collisions.

For example:

• Temperature: Increasing temperature increases the proportion of particles with sufficient energy to overcome the activation energy.
• Oxygen availability: A greater oxygen concentration can increase the frequency of collisions between fuel and oxygen particles.
• Surface area: A finely divided solid fuel exposes more surface for reaction than the same mass in one large piece.
• Mixing: Better mixing can improve contact between fuel and oxygen.
• Physical state: A fuel that has vaporised can mix with oxygen differently from the same fuel in liquid form.

These factors can change how rapidly fuel is consumed without necessarily changing the enthalpy change per mole for the same complete combustion reaction.

There is one important qualification.

If oxygen becomes insufficient, combustion may become incomplete. The products can include carbon monoxide or carbon rather than only carbon dioxide.

The overall reaction has now changed, so its enthalpy change may also be different. You cannot automatically apply the standard enthalpy of complete combustion to an incomplete reaction.

For a broader comparison of fuel types, energy content, and biofuels, see Organic Fuels and Biofuels: HSC Chemistry Comparison Guide.

06A quick decision rule for HSC questions

When you see a negative enthalpy change in a question, identify exactly what the question is asking before choosing an equation.

Keep these quantities separate. A negative enthalpy change establishes that a reaction is exothermic. It does not establish that the reaction is fast.

07Questions and solutions

The following questions build from basic energy calculations towards situations where enthalpy, reaction rate, and experimental measurements interact.

Question 1

A hypothetical fuel has a molar enthalpy of combustion of \(-750\ \mathrm{kJ\,mol^{-1}}\). It is consumed at a constant rate of \(0.020\ \mathrm{mol\,s^{-1}}\).

Calculate the rate at which energy is released during combustion.

Solution 1

The energy release rate is 15 kJ s\(^{-1}\), or 15 kW.

Use the relationship:

\[ P=|\Delta_\mathrm{c}H|\times r \]

Substitute the values:

\[ \begin{aligned} P&=750\times0.020\\ &=15\ \mathrm{kJ\,s^{-1}} \end{aligned} \]

Since 1 kJ s\(^{-1}\) equals 1 kW:

\[ \boxed{P=15\ \mathrm{kW}} \]

The fuel releases 750 kJ for each mole consumed, but the rate at which it delivers that energy depends on how quickly the fuel reacts.

A more negative enthalpy change alone would not have been enough to calculate the energy release rate.

Question 2

The standard enthalpies of combustion of methane and liquid ethanol are:

• Methane: \(-890.7\ \mathrm{kJ\,mol^{-1}}\)
• Ethanol: \(-1367.6\ \mathrm{kJ\,mol^{-1}}\)

Their molar masses are:

• Methane: \(16.04\ \mathrm{g\,mol^{-1}}\)
• Ethanol: \(46.07\ \mathrm{g\,mol^{-1}}\)

Calculate the energy released per gram of each fuel during complete combustion.

Determine which fuel releases more energy per gram. Explain why this comparison still tells us nothing about which fuel burns faster.

Solution 2

Methane releases more energy per gram, even though ethanol has the more negative molar enthalpy of combustion. Neither value establishes the combustion rate.

Start with methane.

Step 1: Calculate the energy released per gram.

One mole of methane has a mass of 16.04 g and releases 890.7 kJ.

\[ \frac{890.7}{16.04} =55.53\ \mathrm{kJ\,g^{-1}} \]

Expressed as an enthalpy change per gram:

\[ \Delta H=-55.53\ \mathrm{kJ\,g^{-1}} \]

Step 2: Calculate the energy released per gram of ethanol.

\[ \frac{1367.6}{46.07} =29.68\ \mathrm{kJ\,g^{-1}} \]

Therefore:

\[ \Delta H=-29.68\ \mathrm{kJ\,g^{-1}} \]

Step 3: Compare the results.

Methane releases more energy per gram.

However, ethanol releases more energy per mole because one mole of ethanol contains substantially more mass than one mole of methane.

This exposes an important distinction: comparing fuels per mole and comparing fuels per gram are different calculations.

Neither comparison tells us how quickly the fuel reacts. We would need information about the molar or mass consumption rate to determine that.

Question 3

A student estimates the enthalpy change for methane combustion using average bond enthalpies.

The reaction produces gaseous water:

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

The following average bond enthalpies are provided:

Calculate the estimated enthalpy change per mole of methane combusted.

A reference value for methane’s standard enthalpy of combustion, with liquid water as a product, is \(-890.7\ \mathrm{kJ\,mol^{-1}}\).

Explain why the estimated value should not be expected to match this reference value exactly.

Solution 3

The bond-enthalpy model estimates an enthalpy change of approximately \(-814\ \mathrm{kJ\,mol^{-1}}\). It differs from the reference value because the calculation uses average bond enthalpies and produces gaseous rather than liquid water.

Step 1: Identify the bonds broken.

The reactants contain:

• Four C-H bonds.
• Two O=O bonds.

Therefore:

\[ \begin{aligned} E_\mathrm{broken} &=4(413)+2(498)\\ &=2648\ \mathrm{kJ} \end{aligned} \]

Step 2: Identify the bonds formed.

The products contain:

• Two C=O bonds in carbon dioxide.
• Four O-H bonds in water.

Therefore:

\[ \begin{aligned} E_\mathrm{formed} &=2(805)+4(463)\\ &=3462\ \mathrm{kJ} \end{aligned} \]

Step 3: Calculate the estimated enthalpy change.

\[ \begin{aligned} \Delta H &\approx E_\mathrm{broken}-E_\mathrm{formed}\\ &=2648-3462\\ &=-814\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Step 4: Explain the difference from the reference value.

There are two reasons.

First, average bond enthalpies are approximate values. Actual bond enthalpies depend on the molecular environment.

Second, the estimated reaction produces water vapour, whereas the standard combustion reaction produces liquid water.

Condensation of water releases additional energy, making the standard enthalpy of combustion more negative.

The student has therefore calculated an approximate model value for a different product state. It is not a direct experimental measurement of the standard enthalpy of combustion.

Question 4

A hypothetical fuel, F, can undergo complete or incomplete combustion depending on the oxygen available.

The following illustrative data are provided:

Assume all the fuel consumed follows the stated combustion pathway.

(a) Calculate the average molar rate of fuel consumption in each condition.

(b) Calculate the average rate of energy release in each condition.

(c) A student claims that condition B must have a slower combustion rate because its enthalpy change is less negative. Evaluate this claim.

Solution 4

Condition B has the faster combustion rate and the higher average energy release rate, despite having the less negative enthalpy change per mole.

(a) Calculate the average molar rate.

Use:

\[ r=\frac{\text{moles consumed}}{\text{time}} \]

For condition A:

\[ r_A=\frac{0.0100}{10.0} \]

\[ \boxed{r_A=0.00100\ \mathrm{mol\,s^{-1}}} \]

For condition B:

\[ r_B=\frac{0.0100}{5.0} \]

\[ \boxed{r_B=0.00200\ \mathrm{mol\,s^{-1}}} \]

Condition B consumes fuel at twice the molar rate.

(b) Calculate the average energy release rate.

Use:

\[ P=|\Delta H|\times r \]

For condition A:

\[ P_A=800\times0.00100 \]

\[ \boxed{P_A=0.800\ \mathrm{kJ\,s^{-1}}} \]

For condition B:

\[ P_B=500\times0.00200 \]

\[ \boxed{P_B=1.00\ \mathrm{kJ\,s^{-1}}} \]

Condition B releases energy at a higher average rate.

(c) Evaluate the student’s claim.

The claim is incorrect.

The student has confused energy released per mole with the rate at which fuel is consumed.

Condition B releases less energy per mole because it follows a different combustion pathway and produces different products. However, it consumes twice as many moles of fuel per second.

This demonstrates why comparing enthalpy changes alone is not sufficient to determine reaction rates.

It also highlights an important experimental consideration: incomplete combustion changes the overall reaction, so the enthalpy of complete combustion cannot automatically be applied.

Question 5

Two identical constant-pressure calorimeter experiments investigate the complete combustion of methane.

Each experiment begins with 0.0200 mol of methane and excess oxygen at 298 K.

Both reactions produce carbon dioxide gas and liquid water. The standard enthalpy of combustion is:

\[ \Delta_\mathrm{c}H^\circ =-890.7\ \mathrm{kJ\,mol^{-1}} \]

A catalyst is added to the first experiment.

The observations are:

The calibrated heat capacity of each calorimeter, including its water and apparatus, is \(10.0\ \mathrm{kJ\,^\circ C^{-1}}\).

Neither calorimeter is perfectly insulated.

Assume:

• Both experiments undergo complete combustion.
• The same quantity of methane reacts in each experiment.
• The catalyst is not consumed and does not change the products.
• All energy not recorded by the calorimeter is transferred to the surroundings.
• Temperature corrections to the supplied standard enthalpy are negligible.

A student concludes:

“The catalyst makes methane combustion more exothermic because Experiment A has a greater temperature increase. Therefore, its enthalpy change must be more negative.”

Evaluate this conclusion.

In your answer:

(a) Calculate the total enthalpy change for each experiment.

(b) Calculate the average rate of energy release in each experiment.

(c) Calculate the energy captured by each calorimeter and the apparent molar enthalpy change calculated from those measurements.

(d) Explain why the experimental results appear to show different enthalpy changes, even though the actual reaction is the same.

Solution 5

The student’s conclusion is incorrect. Both experiments have the same true enthalpy change of \(-17.8\ \mathrm{kJ}\). The catalyst increases the reaction rate, while the difference in measured temperature rises results from different amounts of heat escaping to the surroundings.

This question requires us to separate three things:

• The actual enthalpy change of the reaction.
• The rate at which energy is released.
• The amount of energy captured by the calorimeter.

(a) Calculate the total enthalpy change.

Each experiment burns 0.0200 mol of methane.

The molar enthalpy of combustion is:

\[ \Delta_\mathrm{c}H^\circ =-890.7\ \mathrm{kJ\,mol^{-1}} \]

Therefore:

\[ \begin{aligned} \Delta H &=n\Delta_\mathrm{c}H^\circ\\ &=0.0200(-890.7)\\ &=-17.814\ \mathrm{kJ} \end{aligned} \]

Thus:

\[ \boxed{\Delta H=-17.8\ \mathrm{kJ}} \]

This result applies to both experiments.

The same amount of methane undergoes the same complete combustion reaction, producing the same products in the same physical states.

The catalyst changes the reaction pathway, not the initial and final chemical states. Therefore, it does not change the overall enthalpy change.

(b) Calculate the average rate of energy release.

The total energy released is 17.814 kJ in both experiments.

For Experiment A:

\[ P_A=\frac{17.814}{4.0} \]

\[ \boxed{P_A=4.45\ \mathrm{kJ\,s^{-1}}} \]

For Experiment B:

\[ P_B=\frac{17.814}{40.0} \]

\[ \boxed{P_B=0.445\ \mathrm{kJ\,s^{-1}}} \]

Experiment A releases energy at ten times the average rate of Experiment B.

This is consistent with the catalyst lowering the activation energy and increasing the reaction rate.

Notice that both experiments release the same total energy. The difference is how quickly that energy is released.

(c) Calculate the energy captured and apparent molar enthalpy.

The calorimeter heat capacity is:

\[ C_\mathrm{cal}=10.0\ \mathrm{kJ\,^\circ C^{-1}} \]

Use:

\[ q_\mathrm{cal}=C_\mathrm{cal}\Delta T \]

For Experiment A:

\[ \begin{aligned} q_A&=10.0\times1.60\\ &=16.0\ \mathrm{kJ} \end{aligned} \]

For Experiment B:

\[ \begin{aligned} q_B&=10.0\times1.45\\ &=14.5\ \mathrm{kJ} \end{aligned} \]

The calorimeter captures 16.0 kJ in Experiment A and 14.5 kJ in Experiment B.

Now calculate the apparent molar enthalpy for each experiment.

For Experiment A:

\[ \begin{aligned} \Delta H_{\mathrm{app},A} &=\frac{-16.0}{0.0200}\\ &=-800\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

For Experiment B:

\[ \begin{aligned} \Delta H_{\mathrm{app},B} &=\frac{-14.5}{0.0200}\\ &=-725\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The apparent experimental results are:

Experiment A appears more exothermic.

However, these are not the actual enthalpy changes of the combustion reactions. They represent the energy captured by the calorimeters.

(d) Explain the difference.

The key is that neither calorimeter captures all the energy released.

For Experiment A:

\[ 17.814-16.0=1.814\ \mathrm{kJ} \]

Approximately 1.81 kJ is transferred to the surroundings rather than captured by the calorimeter.

For Experiment B:

\[ 17.814-14.5=3.314\ \mathrm{kJ} \]

Approximately 3.31 kJ is transferred to the surroundings.

The slower reaction in Experiment B releases energy over a longer period. This allows more energy to escape to the surroundings before the calorimeter records the final temperature rise.

Experiment A releases energy more rapidly, so a greater proportion is captured by its calorimeter.

This creates the misleading impression that the catalyst has made the reaction more exothermic.

But the true enthalpy change is identical in both experiments.

If the calorimeters were perfectly insulated and had the same heat capacity, both would record the same final temperature increase, even though Experiment A would reach that temperature much sooner.

This is the important conceptual distinction: a change in reaction rate can affect an experimental calorimetry result without changing the actual enthalpy change of the reaction.

08What to study next

When analysing fuel combustion, keep three questions separate:

1. How much energy does one mole of fuel release? Use the molar enthalpy of combustion.
2. How quickly is the fuel reacting? Investigate the reaction rate and activation energy.
3. How quickly is useful energy being delivered? Consider both the molar enthalpy and the rate of fuel consumption.

The next useful step is to become confident interpreting reaction energy profiles and explaining how temperature, catalysts, and collision frequency affect reaction rates.

For HSC Chemistry students studying the 2026-27 courses, this explanation is framed around the Chemistry Stage 6 (2017) syllabus. NESA lists implementation of the Chemistry 11-12 Syllabus (2025) from 2028. See the official NSW syllabus information.