Activation Energy vs Enthalpy Change: HSC Chemistry
Understand how activation energy differs from enthalpy change, interpret reaction energy profiles, calculate forward and reverse activation energies, and explore how catalysts affect reaction rates without changing overall enthalpy.
A reaction can release a lot of energy and still be surprisingly slow to start. Hydrogen peroxide, for example, decomposes into water and oxygen, but the reaction happens much faster when a suitable catalyst is added.
Predict what the catalyst changes. Does it make the reaction release more energy, or does it make the reaction easier to get started?
It changes the second. The overall energy change remains the same, but the energy barrier becomes lower.
This is the difference between activation energy and enthalpy change. Activation energy describes the energy barrier a reaction must overcome. Enthalpy change describes the difference in energy between the reactants and products.
Both are measured in \(\mathrm{kJ\,mol^{-1}}\), but they describe different parts of a reaction.
01The difference in one sentence
Activation energy (\(E_a\)) is the energy required to reach the transition state from the reactants.
Enthalpy change (\(\Delta H\)) is the difference between the enthalpy of the products and the enthalpy of the reactants.
Their equations make this distinction clear:
\[ E_{a,\mathrm{forward}}=H^\ddagger-H_R \]
\[ \Delta H=H_P-H_R \]
Where:
- \(E_{a,\mathrm{forward}}\) is the forward activation energy.
- \(H^\ddagger\) is the enthalpy of the transition state.
- \(H_R\) is the enthalpy of the reactants.
- \(H_P\) is the enthalpy of the products.
- \(\Delta H\) is the reaction enthalpy change.
The key difference is the reference point.
Activation energy measures the climb from the reactants to the highest-energy point. Enthalpy change measures the difference between the starting and finishing points.
Comparing the two quantities
| Feature | Activation energy (\(E_a\)) | Enthalpy change (\(\Delta H\)) |
|---|---|---|
| What does it measure? | Energy barrier to the transition state | Overall energy difference between reactants and products |
| Which energy levels matter? | Reactants and transition state | Reactants and products |
| What does it tell us? | How difficult it is for a reaction to proceed | Whether the reaction absorbs or releases heat |
| Can a catalyst change it? | Yes | No, for the same overall reaction |
| Does it affect reaction rate? | Yes | Not directly |
| Units | \(\mathrm{kJ\,mol^{-1}}\) | \(\mathrm{kJ\,mol^{-1}}\) |
Notice that both quantities use the same units. This is a common source of confusion in HSC Chemistry.
Don’t assume that two quantities measuring energy must describe the same thing.
02Understanding both quantities on an energy profile
A reaction energy profile shows how the enthalpy of a chemical system changes as a reaction progresses.
The vertical axis represents enthalpy, usually in \(\mathrm{kJ\,mol^{-1}}\). The horizontal axis represents reaction progress, not time.
For a simple reaction with one energy barrier, the profile has three important features:
- The reactants, which establish the starting energy level.
- The transition state, which represents the highest-energy arrangement along the pathway.
- The products, which establish the final energy level.
Activation energy: the climb to the peak
Imagine the reactants starting at an enthalpy of \(40\ \mathrm{kJ\,mol^{-1}}\).
The transition state sits at \(105\ \mathrm{kJ\,mol^{-1}}\).
The forward activation energy is:
\[ E_{a,\mathrm{forward}}=105-40 \]
\[ \boxed{E_{a,\mathrm{forward}}=65\ \mathrm{kJ\,mol^{-1}}} \]
The reaction must overcome this energy barrier to proceed through the pathway.
The transition state is a temporary, high-energy arrangement of atoms. It is not a stable product that can be collected in a test tube.
Enthalpy change: the difference between the endpoints
Now suppose the products have an enthalpy of \(15\ \mathrm{kJ\,mol^{-1}}\).
The reaction enthalpy change is:
\[ \Delta H=H_P-H_R \]
\[ \Delta H=15-40 \]
\[ \boxed{\Delta H=-25\ \mathrm{kJ\,mol^{-1}}} \]
The negative sign tells us that the reaction is exothermic.
The products have \(25\ \mathrm{kJ\,mol^{-1}}\) less enthalpy than the reactants. The difference is released to the surroundings as heat under the usual constant-pressure conditions.
Notice something important: the reaction has an activation energy of \(65\ \mathrm{kJ\,mol^{-1}}\), but its enthalpy change is only \(-25\ \mathrm{kJ\,mol^{-1}}\).
There is no contradiction. One quantity measures the barrier, while the other measures the overall change.
A useful analogy: walking over a hill
Imagine walking from one campsite to another across a hill.
- Your starting campsite represents the reactants.
- The highest point of the hill represents the transition state.
- Your destination represents the products.
The height you must climb from your starting point to reach the summit represents activation energy.
The difference in height between your starting and finishing campsites represents enthalpy change.
You could climb a very high hill and finish lower than where you started. That is exactly what happens in an exothermic reaction with a significant activation energy.
The analogy has a limit: molecules are not physically travelling over hills. The energy profile represents the changes in energy as bonds break, form, and rearrange.
Worked example: Calculate activation energy and enthalpy change
A reaction has the following energy levels:
- Reactants: \(55\ \mathrm{kJ\,mol^{-1}}\)
- Transition state: \(125\ \mathrm{kJ\,mol^{-1}}\)
- Products: \(20\ \mathrm{kJ\,mol^{-1}}\)
Calculate the forward activation energy and reaction enthalpy change. State whether the reaction is exothermic or endothermic.
Step 1
Use the difference between the transition state and reactants:
\[ E_{a,\mathrm{forward}}=H^\ddagger-H_R \]
\[ =125-55 \]
\[ =70\ \mathrm{kJ\,mol^{-1}} \]
Step 2
Use the difference between products and reactants:
\[ \Delta H=H_P-H_R \]
\[ =20-55 \]
\[ =-35\ \mathrm{kJ\,mol^{-1}} \]
Step 3
The negative enthalpy change means the reaction is exothermic.
The reaction releases \(35\ \mathrm{kJ}\) per mole of reaction as written, even though it must overcome a forward activation barrier of \(70\ \mathrm{kJ\,mol^{-1}}\).
The important lesson is that the size of the activation energy does not determine how much energy the reaction releases.
03How do experiments distinguish activation energy and enthalpy change?
Here is where the distinction becomes especially useful.
Imagine that a chemist wants to investigate both quantities for a chemical reaction.
Would measuring the temperature change of the reaction mixture tell them everything?
No. A temperature measurement can help determine the heat transferred, but it cannot reveal the height of the activation barrier.
The two quantities require different experimental evidence.
1. Measuring enthalpy change using calorimetry
Calorimetry measures the heat transferred during a chemical reaction.
A simple coffee-cup calorimeter consists of two nested insulated cups, a lid, and a thermometer or temperature probe.
A known quantity of reactants is mixed inside the cup. The thermometer measures the temperature before and after the reaction.
Suppose the temperature increases.
The solution has gained heat, meaning the reaction has transferred heat to its surroundings. Assuming negligible heat loss, the reaction is exothermic.
If the temperature decreases, the solution has lost heat to the reaction, indicating an endothermic process.
The heat absorbed or released by the solution can be calculated using:
\[ q=mc\Delta T \]
Where:
- \(q\) is the heat transferred, in joules.
- \(m\) is the mass of the solution, in grams.
- \(c\) is the specific heat capacity, in \(\mathrm{J\,g^{-1}\,^\circ C^{-1}}\).
- \(\Delta T\) is the temperature change, in degrees Celsius.
The heat gained by the solution is approximately equal in magnitude and opposite in sign to the heat transferred by the reaction:
\[ q_{\mathrm{reaction}}=-q_{\mathrm{solution}} \]
At constant pressure, the reaction enthalpy change per mole is:
\[ \Delta H=\frac{q_{\mathrm{reaction}}}{n} \]
Here, \(n\) is the amount of reaction in moles, on the reaction basis being used.
Calorimetry therefore allows us to estimate enthalpy change.
However, it does not tell us how high the transition state is. A reaction could release a large amount of heat while having either a high or low activation energy.
Worked example: Calculate enthalpy change using calorimetry
A reaction takes place in an insulated calorimeter containing \(80.0\ \mathrm{g}\) of solution.
The temperature increases from \(21.0^\circ\mathrm{C}\) to \(23.6^\circ\mathrm{C}\).
The reaction consumes \(0.0125\ \mathrm{mol}\) of reactants.
Assume the specific heat capacity of the solution is \(4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\), and all heat released by the reaction is absorbed by the solution.
Calculate the molar enthalpy change.
Step 1
\[ \Delta T=T_{\mathrm{final}}-T_{\mathrm{initial}} \]
\[ =23.6-21.0 \]
\[ =2.6^\circ\mathrm{C} \]
Step 2
\[ q_{\mathrm{solution}}=mc\Delta T \]
\[ =80.0\times4.18\times2.6 \]
\[ =869.4\ \mathrm{J} \]
Convert to kilojoules:
\[ q_{\mathrm{solution}}=0.8694\ \mathrm{kJ} \]
Step 3
The solution gained heat, so the reaction lost heat.
\[ q_{\mathrm{reaction}}=-0.8694\ \mathrm{kJ} \]
Step 4
\[ \Delta H=\frac{q_{\mathrm{reaction}}}{n} \]
\[ =\frac{-0.8694}{0.0125} \]
\[ \boxed{\Delta H=-69.6\ \mathrm{kJ\,mol^{-1}}} \]
The reaction is exothermic. It releases approximately \(69.6\ \mathrm{kJ}\) per mole of reaction as written.
Notice what we have not calculated: activation energy.
The temperature change tells us about heat transfer, not the energy required to reach the transition state.
Real calorimeters also have limitations. Heat loss to the surroundings, the heat capacity of the apparatus, and incomplete reaction can affect the result.
2. Investigating activation energy using reaction rates
To investigate activation energy, chemists need to examine how quickly a reaction occurs.
One possible experimental system is the decomposition of hydrogen peroxide:
\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]
Manganese dioxide can act as a catalyst for this reaction.
An illustrative investigation could use a conical flask containing hydrogen peroxide, connected through a delivery tube to a gas syringe.
The flask sits in a temperature-controlled water bath. When the catalyst is added, oxygen gas forms and pushes the gas syringe’s piston outwards.
The volume of oxygen is recorded at regular time intervals.
The investigation can then be repeated at different temperatures.
To make the comparison fair, the chemist keeps the following conditions constant:
- Initial hydrogen peroxide concentration.
- Volume of hydrogen peroxide.
- Mass and surface area of manganese dioxide.
- Apparatus and measurement conditions.
Only the temperature is deliberately changed.
The gas syringe measures oxygen production over time. The gradient of a volume-versus-time graph gives an indication of the reaction rate.
For example, consider these illustrative measurements:
| Temperature | Oxygen produced in first 10 seconds | Average rate over first 10 seconds |
|---|---|---|
| \(20^\circ\mathrm{C}\) | 6.0 mL | 0.60 mL/s |
| \(30^\circ\mathrm{C}\) | 11.0 mL | 1.10 mL/s |
| \(40^\circ\mathrm{C}\) | 19.0 mL | 1.90 mL/s |
The reaction becomes faster as temperature increases.
Why?
At a higher temperature, a greater proportion of particles have enough kinetic energy to overcome the activation energy barrier. Heating increases the number of successful collisions per unit time.
Importantly, heating has not necessarily lowered the activation energy itself. It has changed the distribution of particle energies.
A numerical activation energy requires more detailed kinetic analysis using measurements at multiple temperatures. A single rate measurement cannot establish its value.
What does each experiment actually tell us?
| Experimental measurement | What we can determine | What we cannot determine from it alone |
|---|---|---|
| Temperature change in a calorimeter | Heat transferred and reaction enthalpy | Activation energy |
| Gas volume produced over time | Reaction rate | Reaction enthalpy |
| Reaction rates at several temperatures | Temperature dependence of reaction rate and an estimate of activation energy using a kinetic model | Reaction enthalpy |
| Energy profile with known energy levels | Activation energy and enthalpy change | Actual reaction rate without further information |
This is the experimental distinction worth remembering:
Calorimetry measures the overall heat change. Kinetic measurements investigate the energy barrier.
04What happens to activation energy in the reverse reaction?
The distinction between activation energy and enthalpy change becomes even more important when considering a reversible reaction.
The forward and reverse reactions travel between the same reactants and products, but they start at opposite energy levels.
Both directions must reach the same transition state when following the same single-step reaction pathway.
Consider the following equations:
\[ E_{a,\mathrm{forward}}=H^\ddagger-H_R \]
\[ E_{a,\mathrm{reverse}}=H^\ddagger-H_P \]
And remember:
\[ \Delta H=H_P-H_R \]
Combining these relationships gives:
\[ \boxed{ E_{a,\mathrm{reverse}} = E_{a,\mathrm{forward}}-\Delta H } \]
The enthalpy change must be substituted with its sign.
Exothermic forward reaction
For an exothermic forward reaction, \(\Delta H\) is negative.
The products have lower enthalpy than the reactants, so the reverse reaction has a greater activation energy.
For example, suppose:
\[ E_{a,\mathrm{forward}}=50 \]
\[ \Delta H=-30 \]
Then:
\[ E_{a,\mathrm{reverse}}=50-(-30) \]
\[ =80\ \mathrm{kJ\,mol^{-1}} \]
The reverse reaction has a larger energy barrier because it starts from a lower energy level.
Endothermic forward reaction
For an endothermic forward reaction, \(\Delta H\) is positive.
The products have higher enthalpy than the reactants, so the reverse reaction has a smaller activation energy.
For example:
\[ E_{a,\mathrm{forward}}=80 \]
\[ \Delta H=+30 \]
Then:
\[ E_{a,\mathrm{reverse}}=80-30 \]
\[ =50\ \mathrm{kJ\,mol^{-1}} \]
The reverse reaction starts closer to the transition state.
| Forward reaction | Sign of \(\Delta H\) | Relationship between activation energies |
|---|---|---|
| Exothermic | Negative | Reverse activation energy is greater |
| Endothermic | Positive | Reverse activation energy is smaller |
| Thermoneutral | Zero | Forward and reverse activation energies are equal |
These relationships assume the same single-step pathway and transition state in both directions.
For more practice with this particular calculation, see How to Calculate Reverse Activation Energy Using Delta H.
05Common misconceptions to avoid
A few mistakes appear repeatedly when students interpret reaction energy profiles.
| Incorrect idea | Correct understanding |
|---|---|
| An exothermic reaction must have a low activation energy. | A reaction can release substantial energy but still have a high activation barrier. |
| Activation energy is the amount of energy released by a reaction. | Activation energy is the barrier to the transition state. |
| A catalyst makes the reaction more exothermic. | A catalyst changes the reaction pathway and can lower activation energy, but does not change the overall enthalpy change. |
| A reaction with a larger negative \(\Delta H\) must be faster. | Enthalpy change alone does not determine reaction rate. |
| Heating lowers activation energy. | Heating increases the proportion of particles with sufficient energy to overcome the barrier. |
| Forward and reverse activation energies are always equal. | They are equal only when the reactants and products have the same enthalpy, assuming the same pathway. |
06Questions and solutions
The following questions move from reading an energy profile to combining calorimetry, reaction rates, and activation energy.
Question 1
An energy profile for a single-step reaction contains the following information:
- Enthalpy of reactants: \(48\ \mathrm{kJ\,mol^{-1}}\)
- Enthalpy of transition state: \(108\ \mathrm{kJ\,mol^{-1}}\)
- Enthalpy of products: \(18\ \mathrm{kJ\,mol^{-1}}\)
Calculate:
a) The forward activation energy.
b) The reaction enthalpy change.
c) Whether the reaction is exothermic or endothermic.
Solution 1
The forward activation energy is \(60\ \mathrm{kJ\,mol^{-1}}\), and the reaction enthalpy change is \(-30\ \mathrm{kJ\,mol^{-1}}\). The reaction is exothermic.
Part a: Forward activation energy
Use the energy difference between the transition state and reactants:
\[ E_{a,\mathrm{forward}}=H^\ddagger-H_R \]
\[ =108-48 \]
\[ \boxed{60\ \mathrm{kJ\,mol^{-1}}} \]
Part b: Enthalpy change
Use the energy difference between products and reactants:
\[ \Delta H=H_P-H_R \]
\[ =18-48 \]
\[ \boxed{-30\ \mathrm{kJ\,mol^{-1}}} \]
Part c: Reaction classification
The enthalpy change is negative.
Therefore, the reaction is exothermic because the products have lower enthalpy than the reactants.
The important distinction is that the activation energy describes the climb to the peak, whereas enthalpy change describes the difference between the endpoints.
Question 2
A student investigates the decomposition of hydrogen peroxide using manganese dioxide as a catalyst.
The reaction is:
\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]
The student repeats the experiment at two temperatures, keeping the initial hydrogen peroxide concentration and catalyst amount constant.
The following illustrative data are obtained. Assume the gas volumes have been corrected to the same temperature and pressure.
| Time (s) | Oxygen volume at \(25^\circ\mathrm{C}\) (mL) | Oxygen volume at \(35^\circ\mathrm{C}\) (mL) |
|---|---|---|
| 0 | 0 | 0 |
| 10 | 7 | 13 |
| 20 | 12 | 22 |
| 30 | 16 | 29 |
a) Calculate the average rate of oxygen production during the first 10 seconds at each temperature.
b) Explain why the reaction proceeds faster at the higher temperature.
c) Explain why these measurements do not directly determine the enthalpy change or numerical activation energy.
Solution 2
The average oxygen production rates are \(0.70\ \mathrm{mL\,s^{-1}}\) at \(25^\circ\mathrm{C}\) and \(1.30\ \mathrm{mL\,s^{-1}}\) at \(35^\circ\mathrm{C}\). The higher temperature increases the reaction rate because a greater proportion of particles can overcome the activation energy barrier.
Part a: Calculate the average rates
Use:
\[ \text{Average rate}=\frac{\Delta V}{\Delta t} \]
At \(25^\circ\mathrm{C}\):
\[ \text{Rate}=\frac{7-0}{10-0} \]
\[ =0.70\ \mathrm{mL\,s^{-1}} \]
At \(35^\circ\mathrm{C}\):
\[ \text{Rate}=\frac{13-0}{10-0} \]
\[ =1.30\ \mathrm{mL\,s^{-1}} \]
The reaction produces oxygen approximately \(1.86\) times as quickly at the higher temperature during this interval.
Part b: Explain the effect of temperature
Increasing temperature increases the average kinetic energy of the particles.
More importantly, a greater proportion of particles now have sufficient energy to overcome the activation energy barrier.
This increases the number of successful collisions per second, resulting in a faster reaction.
The activation energy itself has not necessarily changed.
Part c: Identify the experimental limitations
The gas syringe measures the volume of oxygen produced over time. It therefore provides evidence about reaction rate.
It does not directly measure the heat released or absorbed by the reaction, so it cannot independently establish \(\Delta H\).
The measurements also do not provide a numerical activation energy by themselves. This requires more extensive kinetic measurements across several temperatures and an appropriate kinetic analysis.
The tempting mistake is to treat a faster reaction as proof of a smaller activation energy. Temperature has changed here, not the reaction pathway.
Question 3
Two different chemical reactions, A and B, are investigated at the same temperature.
Their energy and experimental conditions are summarised below.
| Quantity | Reaction A | Reaction B |
|---|---|---|
| Enthalpy change | \(-52\ \mathrm{kJ\,mol^{-1}}\) | \(-52\ \mathrm{kJ\,mol^{-1}}\) |
| Forward activation energy | \(88\ \mathrm{kJ\,mol^{-1}}\) | \(43\ \mathrm{kJ\,mol^{-1}}\) |
Assume that the reactions have comparable concentrations, collision frequencies, and orientation factors.
a) Which reaction would you expect to proceed faster? Explain.
b) Which reaction releases more energy per mole?
c) A student argues that Reaction B must be approximately twice as fast because its activation energy is roughly half that of Reaction A.
Evaluate this claim.
Solution 3
Reaction B is expected to proceed faster, but both reactions release the same amount of energy per mole. The student’s claim about the reaction being approximately twice as fast is not justified.
Part a: Compare reaction rates
Reaction B has a lower activation energy:
\[ E_{a,B}=43\ \mathrm{kJ\,mol^{-1}} \]
Compared with:
\[ E_{a,A}=88\ \mathrm{kJ\,mol^{-1}} \]
At the same temperature, a greater proportion of particles in Reaction B can overcome its activation energy barrier.
Given the comparable collision and orientation factors, Reaction B is expected to have a faster reaction rate.
Part b: Compare energy released
Both reactions have:
\[ \Delta H=-52\ \mathrm{kJ\,mol^{-1}} \]
Therefore, both release \(52\ \mathrm{kJ}\) per mole of reaction as written.
The activation energy does not determine the overall heat released.
Part c: Evaluate the student’s claim
The student’s conclusion is incorrect.
A reaction with approximately half the activation energy does not necessarily proceed at twice the rate.
The relationship between activation energy and reaction rate is not a simple linear relationship. It depends on the fraction of particles able to overcome the barrier, as well as other kinetic factors.
The correct conclusion is that Reaction B is expected to be faster under the stated assumptions, but the exact rate ratio cannot be determined from the activation energies alone.
Question 4
An endothermic reaction follows a single-step reaction pathway.
Its enthalpy change is:
\[ \Delta H=+34\ \mathrm{kJ\,mol^{-1}} \]
Two students propose different forward activation energies:
- Student A: \(28\ \mathrm{kJ\,mol^{-1}}\)
- Student B: \(58\ \mathrm{kJ\,mol^{-1}}\)
a) Determine which proposed activation energy is physically consistent with the reaction profile.
b) Calculate the reverse activation energy for the physically consistent proposal.
c) Explain why the other proposal cannot describe a valid single-step energy profile.
Solution 4
Student B’s proposed activation energy of \(58\ \mathrm{kJ\,mol^{-1}}\) is physically consistent. Its reverse activation energy is \(24\ \mathrm{kJ\,mol^{-1}}\).
Part a: Identify the valid proposal
For an endothermic reaction, the products have higher enthalpy than the reactants.
The products are already \(34\ \mathrm{kJ\,mol^{-1}}\) above the reactants.
The transition state must be above the products in the single-step energy profile.
Therefore, the forward activation energy must be greater than \(34\ \mathrm{kJ\,mol^{-1}}\).
Student A’s value of \(28\ \mathrm{kJ\,mol^{-1}}\) is impossible under these assumptions.
Student B’s value of \(58\ \mathrm{kJ\,mol^{-1}}\) is consistent.
Part b: Calculate the reverse activation energy
Use:
\[ E_{a,\mathrm{reverse}} = E_{a,\mathrm{forward}}-\Delta H \]
Substitute the signed values:
\[ =58-34 \]
\[ \boxed{24\ \mathrm{kJ\,mol^{-1}}} \]
Part c: Explain why Student A is incorrect
If the forward activation energy were \(28\ \mathrm{kJ\,mol^{-1}}\), the reverse activation energy would be:
\[ E_{a,\mathrm{reverse}}=28-34 \]
\[ =-6\ \mathrm{kJ\,mol^{-1}} \]
A negative activation barrier is inconsistent with the proposed single-step profile.
The problem is that the transition state would sit below the products, even though the reaction must reach that transition state before forming the products.
This is a useful consistency check: for an endothermic single-step reaction, the forward activation energy must exceed the positive enthalpy change.
Question 5
A chemist investigates the effect of a catalyst on an exothermic reaction:
\[ \ce{X -> Y} \]
Two identical, well-insulated calorimeters are used. Both initially contain \(0.0200\ \mathrm{mol}\) of reactant X.
One reaction proceeds without a catalyst. The other uses a catalyst that provides an alternative pathway.
After 30 seconds, the following illustrative measurements are recorded:
| Measurement | Without catalyst | With catalyst |
|---|---|---|
| Amount of X reacted | \(0.0050\ \mathrm{mol}\) | \(0.0150\ \mathrm{mol}\) |
| Heat gained by solution | \(0.350\ \mathrm{kJ}\) | \(1.050\ \mathrm{kJ}\) |
| Solution temperature rise | \(0.70^\circ\mathrm{C}\) | \(2.10^\circ\mathrm{C}\) |
Assume:
- Both experiments occur at constant pressure.
- Both reactions produce the same product Y.
- Heat losses are negligible.
- The heat gained by the solution equals the heat released by the reaction in magnitude.
- Both reactions eventually consume all the available X.
A separate energy-profile analysis gives:
| Energy quantity | Without catalyst | With catalyst |
|---|---|---|
| Forward activation energy | \(92\ \mathrm{kJ\,mol^{-1}}\) | \(57\ \mathrm{kJ\,mol^{-1}}\) |
| Enthalpy change | \(-70\ \mathrm{kJ\,mol^{-1}}\) | \(-70\ \mathrm{kJ\,mol^{-1}}\) |
A student argues:
“The catalyst makes the reaction three times more exothermic because the solution temperature rises three times as much. It also releases more total energy.”
Evaluate this argument.
In your response:
a) Calculate the enthalpy change per mole of reaction for both experiments.
b) Explain why the temperature rises are different.
c) Calculate the total heat released when each reaction reaches completion.
d) Explain what the catalyst has changed and what it has not changed.
Solution 5
The student’s argument is incorrect. Both reactions have the same enthalpy change of \(-70\ \mathrm{kJ\,mol^{-1}}\), and both release \(1.40\ \mathrm{kJ}\) when they reach completion. The catalyst lowers the activation energy and allows the reaction to proceed faster.
Part a: Calculate the enthalpy change for both reactions
Start with the relationship:
\[ \Delta H=\frac{q_{\mathrm{reaction}}}{n} \]
Remember that the reaction releases heat, so:
\[ q_{\mathrm{reaction}}=-q_{\mathrm{solution}} \]
Without the catalyst:
\[ \Delta H=\frac{-0.350}{0.0050} \]
\[ \boxed{\Delta H=-70.0\ \mathrm{kJ\,mol^{-1}}} \]
With the catalyst:
\[ \Delta H=\frac{-1.050}{0.0150} \]
\[ \boxed{\Delta H=-70.0\ \mathrm{kJ\,mol^{-1}}} \]
Both reactions have exactly the same molar enthalpy change.
Part b: Explain the different temperature rises
The catalysed reaction has reacted:
\[ 0.0150\ \mathrm{mol} \]
The uncatalysed reaction has reacted:
\[ 0.0050\ \mathrm{mol} \]
Therefore, the catalysed reaction has converted three times as much reactant after 30 seconds.
Since both reactions release the same amount of heat per mole, the catalysed reaction has released three times as much total heat at this particular time.
This explains the temperature measurements:
\[ \frac{2.10}{0.70}=3 \]
The threefold temperature rise does not mean the catalyst has changed the molar enthalpy change.
It means more reaction has occurred in the same period.
This is the crucial distinction: total heat released depends on how much reaction has occurred, while molar enthalpy change describes the heat released per mole of reaction.
Part c: Calculate the total heat released at completion
Both calorimeters initially contain:
\[ n=0.0200\ \mathrm{mol} \]
Both reactions eventually consume all the reactant.
Using:
\[ q_{\mathrm{reaction}}=n\Delta H \]
\[ =0.0200\times(-70.0) \]
\[ \boxed{q_{\mathrm{reaction}}=-1.40\ \mathrm{kJ}} \]
Each reaction releases \(1.40\ \mathrm{kJ}\) when complete.
The catalyst changes how quickly this energy is released, not the total amount released for the same extent of reaction.
Part d: Explain the effect of the catalyst
Compare the activation energies:
Without catalyst:
\[ E_a=92\ \mathrm{kJ\,mol^{-1}} \]
With catalyst:
\[ E_a=57\ \mathrm{kJ\,mol^{-1}} \]
The reduction is:
\[ 92-57=35\ \mathrm{kJ\,mol^{-1}} \]
The catalyst provides an alternative reaction pathway with a lower activation energy.
This makes it easier for particles to overcome the energy barrier, increasing the reaction rate.
However, the reactants and products remain the same. Their enthalpy levels have not changed.
Consequently, the reaction enthalpy remains:
\[ \Delta H=-70\ \mathrm{kJ\,mol^{-1}} \]
The student’s mistake was comparing the total heat released after a fixed time rather than comparing heat released per mole of reaction.
This is a particularly important distinction in experimental questions. A faster reaction can release more heat over a short interval without being more exothermic.
07Where to go next
Once you can distinguish the activation barrier from the overall enthalpy change, the next step is understanding why changing temperature or adding a catalyst affects reaction rates.
That is where collision theory becomes important. It explains why a lower activation energy allows more collisions to become successful, without changing the overall energy difference between reactants and products.
Continue with Collision Theory and Activation Energy: HSC Chemistry Explained to connect energy profiles with particle behaviour and reaction rates.
Syllabus note: As at November 2026, the Chemistry Stage 6 Syllabus (2017) remains in use. The Chemistry 11-12 Syllabus (2025) is scheduled for implementation from 2028, with its first HSC examination in 2029. This explanation should not be interpreted as a claim that every technique or extension discussed is a prescribed dot point in the 2025 syllabus. See NESA’s syllabus implementation information.