How to Draw a Catalysed Pathway on an Energy Diagram
Learn how to draw and label a catalysed pathway on a chemistry energy profile diagram. Includes worked examples calculating forward and reverse activation energies, distinguishing activation barriers from reaction enthalpy, and analysing multistep pathways.
Two equal samples of hydrogen peroxide are left to decompose. One contains manganese dioxide, a catalyst, and the other does not. The oxygen syringe connected to the catalysed reaction fills much faster, yet both reactions eventually produce the same amount of oxygen.
How should this change the energy profile diagram?
The correct answer is to draw a second pathway with a lower activation energy, while keeping the reactant and product energy levels exactly where they were. A catalyst changes the route the reaction takes, not the overall energy difference between reactants and products.
Getting this right means understanding three separate quantities: forward activation energy, reverse activation energy, and reaction enthalpy. Let’s work through the experimental reasoning before drawing the diagram.
01What does the experiment actually show?
Consider the decomposition of hydrogen peroxide:
\[
\ce{2H2O2(aq) -> 2H2O(l) + O2(g)}
\]
Manganese dioxide, \(\ce{MnO2}\), can act as a catalyst for this reaction. It provides an alternative reaction pathway that allows oxygen to form more quickly.
Picture the apparatus
Imagine two identical conical flasks on a laboratory bench.
- Both contain the same volume and concentration of hydrogen peroxide.
- The first flask contains no catalyst.
- The second contains a measured amount of manganese dioxide.
- Each flask has a bung connected to a delivery tube.
- Each delivery tube leads into a gas syringe.
As the reaction proceeds, oxygen travels from the flask, through the delivery tube, and into the gas syringe. The gas pushes the syringe plunger outwards, allowing its volume to be measured over time.
What should we expect to observe?
The catalysed reaction should produce oxygen more quickly. However, if both samples contain the same initial amount of hydrogen peroxide and decompose completely, they should produce the same final amount of oxygen.
This gives us two important observations:
- The catalyst increases the reaction rate.
- The catalyst does not change the overall amount of product formed under these conditions.
What would the measurements look like?
The following is illustrative experimental data, not a record of an actual experiment.
| Time (s) | Without catalyst: oxygen volume (cm³) | With catalyst: oxygen volume (cm³) |
|---|---|---|
| 0 | 0 | 0 |
| 10 | 6 | 24 |
| 20 | 12 | 40 |
| 60 | 45 | 45 |
After 20 seconds, the catalysed reaction has produced 40 cm³ of oxygen, compared with just 12 cm³ in the uncatalysed reaction.
We can calculate the average oxygen production rate over the first 20 seconds:
\[
r_{\mathrm{avg}}=\frac{\Delta V}{\Delta t}
\]
Without the catalyst:
\[
r_{\mathrm{avg}}=\frac{12}{20}
=0.60\ \mathrm{cm^3\,s^{-1}}
\]
With the catalyst:
\[
r_{\mathrm{avg}}=\frac{40}{20}
=2.0\ \mathrm{cm^3\,s^{-1}}
\]
The catalysed reaction produces oxygen at an average rate approximately 3.3 times greater over this interval.
What can we conclude?
The experiment supports the idea that the catalyst makes the reaction faster.
However, notice an important limitation: measuring oxygen volume over time does not directly measure activation energy. It measures the rate of oxygen production.
The energy profile is a model that helps explain the observed change in reaction rate. It is not a direct measurement of the transition state.
There is also a useful lesson about experimental design. If we measured only the final oxygen volume, both reactions would appear identical. We would miss the difference in reaction rate entirely.
Recording gas volume at regular time intervals fixes this problem. We must also control temperature, hydrogen peroxide concentration, initial volume, and the apparatus used.
Now we can connect the experimental observation to the energy diagram.
02How to draw a catalysed pathway on an energy diagram
An energy profile diagram shows how the energy of a reacting system changes as a reaction progresses.
The horizontal axis represents reaction progress, sometimes called the reaction coordinate. It does not represent time or physical distance.
The vertical axis represents relative energy or enthalpy, usually measured in \(\mathrm{kJ\,mol^{-1}}\).
Think of the reaction as travelling between two points across a mountain range. The reactants and products are at fixed energy levels, while the reaction must pass through an energy barrier.
A catalyst provides an alternative route with a lower barrier.
The mountain analogy is useful for picturing the barrier, but it has a limitation: molecules do not physically travel along a drawn curve. The diagram represents changes in the energy of the reacting system.
Step 1: Draw the reactant and product energy levels
Start by identifying the energy levels of the reactants and products.
These levels are determined by the overall reaction enthalpy, not by whether a catalyst is present.
For an exothermic reaction:
- Products have lower energy than reactants.
- The reaction releases energy.
- The enthalpy change is negative.
For an endothermic reaction:
- Products have higher energy than reactants.
- The reaction absorbs energy.
- The enthalpy change is positive.
The enthalpy change is calculated using:
\[
\Delta H=H_{\mathrm{products}}-H_{\mathrm{reactants}}
\]
Here, \(\Delta H\) is the reaction enthalpy change, measured in \(\mathrm{kJ\,mol^{-1}}\).
Step 2: Draw the uncatalysed pathway
Draw the original reaction pathway as a curve that rises from the reactant energy level, reaches a maximum, and then falls towards the product energy level.
The highest point represents the transition state.
The activation energy is the energy required to reach this transition state from the starting energy level.
For the forward reaction:
\[
E_{a,\mathrm{forward}}
=
E_{\mathrm{transition\ state}}
–
E_{\mathrm{reactants}}
\]
Do not confuse the activation energy with the overall energy difference between reactants and products.
They measure different things.
Step 3: Add the catalysed pathway
This is the most important part.
Draw a second pathway that:
- Starts at exactly the same reactant energy level.
- Ends at exactly the same product energy level.
- Has a lower activation barrier than the uncatalysed pathway.
For a simple single-step energy profile, this means drawing a second curve with a lower maximum.
Why must the two pathways have the same starting and finishing levels?
Because the reactants and products have not changed. The catalyst changes the mechanism by which the reaction occurs, but the overall chemical reaction remains the same.
A catalyst is regenerated during the overall reaction. It does not change the net chemical equation or the enthalpy difference between the initial and final states.
Step 4: Label the activation energies correctly
This is where many students lose marks.
Activation energy is always measured from the starting energy level to the transition state for the direction being considered.
There are therefore two different activation energies for a reversible reaction.
Forward activation energy
The energy difference between the reactants and the transition state.
Reverse activation energy
The energy difference between the products and the transition state.
For the same single-hump pathway:
\[
E_{a,\mathrm{reverse}}
=
E_{a,\mathrm{forward}}-\Delta H
\]
This relationship is particularly useful because it allows you to calculate the reverse activation energy even when it is not directly supplied.
Consider an exothermic reaction.
The products are lower in energy than the reactants. Therefore, the reverse reaction must climb a larger energy barrier to reach the same transition state.
This means:
\[
E_{a,\mathrm{reverse}}
>
E_{a,\mathrm{forward}}
\]
For an endothermic reaction, the opposite is true:
\[
E_{a,\mathrm{reverse}}
<
E_{a,\mathrm{forward}}
\]
Step 5: Label the enthalpy change separately
The enthalpy change, \(\Delta H\), is the vertical difference between the reactant and product energy levels.
It is not measured from either energy level to the transition state.
For a catalysed and uncatalysed pathway:
\[
\Delta H_{\mathrm{catalysed}}
=
\Delta H_{\mathrm{uncatalysed}}
\]
This is true because both pathways connect the same reactants to the same products.
A catalyst lowers the activation barrier, but it does not change the energy released or absorbed by the overall reaction.
A quick checklist for drawing the diagram
| Feature | What your diagram should show |
|---|---|
| Reactant level | Identical for both pathways |
| Product level | Identical for both pathways |
| Uncatalysed pathway | Higher activation barrier |
| Catalysed pathway | Lower activation barrier |
| Forward activation energy | Reactants to transition state |
| Reverse activation energy | Products to transition state |
| Enthalpy change | Reactants to products |
| Horizontal axis | Reaction progress |
| Vertical axis | Relative energy or enthalpy |
One final detail: a real catalytic mechanism can involve several reaction steps, producing multiple peaks and valleys. The simple lower-hump diagram is a useful model for a single-step comparison. For a multistep pathway, each activation barrier must be measured from its preceding intermediate, not automatically from the original reactants.
03Worked example 1: Calculate both activation energies for an exothermic reaction
A reaction has the following illustrative energy levels:
- Reactants: \(0\ \mathrm{kJ\,mol^{-1}}\)
- Products: \(-60\ \mathrm{kJ\,mol^{-1}}\)
- Uncatalysed transition state: \(+100\ \mathrm{kJ\,mol^{-1}}\)
- Catalysed transition state: \(+55\ \mathrm{kJ\,mol^{-1}}\)
Calculate the enthalpy change and the forward and reverse activation energies for both pathways.
Step 1
Use the energy difference between products and reactants:
\[
\begin{aligned}
\Delta H
&=H_{\mathrm{products}}-H_{\mathrm{reactants}}\\
&=-60-0\\
&=-60\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]
The reaction is exothermic because its enthalpy change is negative.
Step 2
For the uncatalysed pathway:
\[
E_{a,\mathrm{forward}}=100-0
\]
\[
=100\ \mathrm{kJ\,mol^{-1}}
\]
For the catalysed pathway:
\[
E_{a,\mathrm{forward}}=55-0
\]
\[
=55\ \mathrm{kJ\,mol^{-1}}
\]
The catalyst has reduced the forward activation energy by \(45\ \mathrm{kJ\,mol^{-1}}\).
Step 3
Remember that the reverse reaction starts at the product energy level, which is \(-60\ \mathrm{kJ\,mol^{-1}}\).
For the uncatalysed pathway:
\[
E_{a,\mathrm{reverse}}=100-(-60)
\]
\[
=160\ \mathrm{kJ\,mol^{-1}}
\]
For the catalysed pathway:
\[
E_{a,\mathrm{reverse}}=55-(-60)
\]
\[
=115\ \mathrm{kJ\,mol^{-1}}
\]
Step 4
| Quantity | Uncatalysed | Catalysed |
|---|---|---|
| Forward activation energy | 100 kJ mol⁻¹ | 55 kJ mol⁻¹ |
| Reverse activation energy | 160 kJ mol⁻¹ | 115 kJ mol⁻¹ |
| Enthalpy change | -60 kJ mol⁻¹ | -60 kJ mol⁻¹ |
The catalyst reduces both activation energies by \(45\ \mathrm{kJ\,mol^{-1}}\), while the enthalpy change remains unchanged.
Notice that the reverse activation energy is larger than the forward activation energy because the reaction is exothermic.
04Worked example 2: Find the reverse activation energy for an endothermic reaction
An endothermic reaction has a forward activation energy of \(92\ \mathrm{kJ\,mol^{-1}}\) without a catalyst.
A catalyst provides an alternative pathway with a forward activation energy of \(58\ \mathrm{kJ\,mol^{-1}}\).
The reaction enthalpy is \(+35\ \mathrm{kJ\,mol^{-1}}\).
Calculate the reverse activation energy for both pathways. Explain why the reverse activation energy is smaller.
Step 1
We know:
\[
E_{a,\mathrm{reverse}}
=
E_{a,\mathrm{forward}}-\Delta H
\]
The enthalpy change is positive because the reaction is endothermic.
Step 2
\[
E_{a,\mathrm{reverse}}
=92-35
\]
\[
=57\ \mathrm{kJ\,mol^{-1}}
\]
Step 3
\[
E_{a,\mathrm{reverse}}
=58-35
\]
\[
=23\ \mathrm{kJ\,mol^{-1}}
\]
Step 4
| Quantity | Uncatalysed | Catalysed |
|---|---|---|
| Forward activation energy | 92 kJ mol⁻¹ | 58 kJ mol⁻¹ |
| Reverse activation energy | 57 kJ mol⁻¹ | 23 kJ mol⁻¹ |
The reverse activation energy is smaller because the products of an endothermic reaction are at a higher energy level than the reactants.
The reverse reaction therefore starts closer to the transition state.
The catalyst reduces both barriers by \(34\ \mathrm{kJ\,mol^{-1}}\), without changing the enthalpy change.
05Common mistakes to avoid
| Common mistake | Why it is incorrect |
|---|---|
| Lowering the product energy when adding a catalyst | This changes the enthalpy change of the reaction. |
| Keeping the activation energy unchanged | A catalyst provides an alternative pathway with a lower activation barrier. |
| Measuring reverse activation energy from the reactants | The reverse reaction starts at the products. |
| Treating activation energy as the enthalpy change | Activation energy measures a barrier; enthalpy change measures the difference between the endpoints. |
| Assuming a catalyst only speeds up the forward reaction | A catalyst increases the rates of both forward and reverse reactions. |
| Drawing reaction progress against time | Reaction progress is not elapsed time. |
| Assuming every catalytic pathway has exactly one peak | Real catalytic mechanisms can involve several elementary steps and intermediates. |
For a more complete explanation of reading reactant levels, transition states, and reaction enthalpy, see Energy Profile Diagrams for HSC Chemistry: How to Read Them.
06Questions and solutions
Question 1
An exothermic reaction has the following energy levels:
- Reactants: \(0\ \mathrm{kJ\,mol^{-1}}\)
- Products: \(-30\ \mathrm{kJ\,mol^{-1}}\)
- Uncatalysed transition state: \(+70\ \mathrm{kJ\,mol^{-1}}\)
- Catalysed transition state: \(+45\ \mathrm{kJ\,mol^{-1}}\)
Calculate:
a) The enthalpy change of the reaction.
b) The forward activation energy for both pathways.
c) The reverse activation energy for both pathways.
Solution 1
The enthalpy change is \(-30\ \mathrm{kJ\,mol^{-1}}\), and the catalyst reduces both activation energies by \(25\ \mathrm{kJ\,mol^{-1}}\).
Part a: Enthalpy change
\[
\Delta H=-30-0
\]
\[
\boxed{\Delta H=-30\ \mathrm{kJ\,mol^{-1}}}
\]
Part b: Forward activation energies
Uncatalysed:
\[
E_{a,\mathrm{forward}}=70-0
\]
\[
=70\ \mathrm{kJ\,mol^{-1}}
\]
Catalysed:
\[
E_{a,\mathrm{forward}}=45-0
\]
\[
=45\ \mathrm{kJ\,mol^{-1}}
\]
Part c: Reverse activation energies
The reverse reaction starts at the product energy level of \(-30\ \mathrm{kJ\,mol^{-1}}\).
Uncatalysed:
\[
E_{a,\mathrm{reverse}}=70-(-30)
\]
\[
=100\ \mathrm{kJ\,mol^{-1}}
\]
Catalysed:
\[
E_{a,\mathrm{reverse}}=45-(-30)
\]
\[
=75\ \mathrm{kJ\,mol^{-1}}
\]
The reverse activation energies are larger because the reaction is exothermic.
The catalyst reduces both barriers by the same amount in this single-hump model. The enthalpy change remains \(-30\ \mathrm{kJ\,mol^{-1}}\).
Question 2
An endothermic reversible reaction has an enthalpy change of \(+26\ \mathrm{kJ\,mol^{-1}}\).
Its forward activation energy is:
- \(91\ \mathrm{kJ\,mol^{-1}}\) without a catalyst.
- \(57\ \mathrm{kJ\,mol^{-1}}\) with a catalyst.
Calculate the reverse activation energies for both pathways.
Explain which direction has the smaller activation energy and why. Also explain whether adding the catalyst changes the equilibrium position.
Solution 2
The reverse activation energies are \(65\ \mathrm{kJ\,mol^{-1}}\) without the catalyst and \(31\ \mathrm{kJ\,mol^{-1}}\) with the catalyst. The reverse direction has the smaller activation energy because the reaction is endothermic.
Step 1: Calculate the uncatalysed reverse activation energy.
\[
E_{a,\mathrm{reverse}}=91-26
\]
\[
=65\ \mathrm{kJ\,mol^{-1}}
\]
Step 2: Calculate the catalysed reverse activation energy.
\[
E_{a,\mathrm{reverse}}=57-26
\]
\[
=31\ \mathrm{kJ\,mol^{-1}}
\]
Step 3: Explain the difference between forward and reverse barriers.
Because the reaction is endothermic, the products sit at a higher energy level than the reactants.
The reverse reaction therefore starts at a higher energy level and has a smaller barrier to overcome.
Step 4: Consider the equilibrium position.
Adding a catalyst does not change the equilibrium position or equilibrium constant at a fixed temperature.
It increases the rates of both forward and reverse reactions, allowing equilibrium to be reached more quickly.
The tempting mistake is to assume that lowering the forward activation energy makes the forward reaction thermodynamically more favourable. It does not. Reaction rate and equilibrium position are different concepts.
Question 3
A student investigates the decomposition of hydrogen peroxide using two identical experimental setups.
Both contain the same initial amount of hydrogen peroxide and are maintained at the same temperature.
The first setup contains no catalyst. The second contains manganese dioxide.
After 10 seconds, the following oxygen volumes are recorded:
| Measurement | Without catalyst | With catalyst |
|---|---|---|
| Oxygen volume after 10 s | 4.0 cm³ | 16.0 cm³ |
| Final oxygen volume | 64.0 cm³ | 64.0 cm³ |
a) Calculate the average oxygen production rate over the first 10 seconds for each experiment.
b) Determine how many times greater the catalysed average oxygen production rate is.
c) Explain why both experiments produce the same final oxygen volume.
d) A student concludes that the catalysed reaction releases more energy because it produces oxygen more quickly. Evaluate this conclusion.
Solution 3
The catalysed reaction has an average oxygen production rate four times greater than the uncatalysed reaction, but the student is incorrect about the energy released.
Part a: Calculate the average rates
Use:
\[
r_{\mathrm{avg}}=\frac{\Delta V}{\Delta t}
\]
Without the catalyst:
\[
r_{\mathrm{avg}}=\frac{4.0}{10}
\]
\[
=0.40\ \mathrm{cm^3\,s^{-1}}
\]
With the catalyst:
\[
r_{\mathrm{avg}}=\frac{16.0}{10}
\]
\[
=1.60\ \mathrm{cm^3\,s^{-1}}
\]
Part b: Compare the rates
\[
\text{Rate ratio}=\frac{1.60}{0.40}
\]
\[
\boxed{4}
\]
The catalysed reaction produces oxygen at four times the average rate over the measured interval.
Part c: Explain the final oxygen volumes
Both experiments begin with the same amount of hydrogen peroxide.
Assuming complete decomposition, no competing reactions, and no oxygen loss, both produce the same final amount of oxygen.
The catalyst changes how quickly the reaction occurs, not the stoichiometric amount of product.
Part d: Evaluate the student’s conclusion
The conclusion is incorrect.
Producing oxygen more quickly indicates an increased reaction rate, not a greater enthalpy change.
The catalyst provides an alternative reaction pathway with a lower activation barrier. It does not change the energy difference between the reactants and products.
The equal final oxygen volumes are consistent with both reactions undergoing the same overall chemical change.
The experiment demonstrates a kinetic effect, not an increase in the energy released per mole of reaction.
Question 4
An exothermic reaction has an enthalpy change of \(-36\ \mathrm{kJ\,mol^{-1}}\).
The reverse activation energy for its catalysed pathway is \(64\ \mathrm{kJ\,mol^{-1}}\).
A catalyst reduces the reverse activation energy by \(20\ \mathrm{kJ\,mol^{-1}}\) compared with the uncatalysed reaction.
Calculate:
a) The reverse activation energy without the catalyst.
b) The forward activation energy with the catalyst.
c) The forward activation energy without the catalyst.
Explain how the results relate to the shape of the energy profile.
Solution 4
The uncatalysed reverse activation energy is \(84\ \mathrm{kJ\,mol^{-1}}\), the catalysed forward activation energy is \(28\ \mathrm{kJ\,mol^{-1}}\), and the uncatalysed forward activation energy is \(48\ \mathrm{kJ\,mol^{-1}}\).
Part a: Uncatalysed reverse activation energy
The catalyst reduces the reverse activation energy by \(20\ \mathrm{kJ\,mol^{-1}}\).
Therefore:
\[
E_{a,\mathrm{reverse,uncat}}
=64+20
\]
\[
=84\ \mathrm{kJ\,mol^{-1}}
\]
Part b: Catalysed forward activation energy
Use:
\[
E_{a,\mathrm{reverse}}
=
E_{a,\mathrm{forward}}-\Delta H
\]
Rearrange:
\[
E_{a,\mathrm{forward}}
=
E_{a,\mathrm{reverse}}+\Delta H
\]
Substitute:
\[
E_{a,\mathrm{forward}}
=64+(-36)
\]
\[
=28\ \mathrm{kJ\,mol^{-1}}
\]
Part c: Uncatalysed forward activation energy
\[
E_{a,\mathrm{forward}}
=84+(-36)
\]
\[
=48\ \mathrm{kJ\,mol^{-1}}
\]
Interpretation
The results can be summarised as follows:
| Quantity | Uncatalysed | Catalysed |
|---|---|---|
| Forward activation energy | 48 kJ mol⁻¹ | 28 kJ mol⁻¹ |
| Reverse activation energy | 84 kJ mol⁻¹ | 64 kJ mol⁻¹ |
The reverse activation energy is \(36\ \mathrm{kJ\,mol^{-1}}\) greater than the forward activation energy in both cases.
This is consistent with an exothermic reaction, where the products sit below the reactants.
Notice the useful check: the catalyst reduces both barriers by \(20\ \mathrm{kJ\,mol^{-1}}\), but the enthalpy change remains \(-36\ \mathrm{kJ\,mol^{-1}}\).
Question 5
A reaction proceeds through a two-step catalysed pathway involving an intermediate.
The following illustrative relative energy levels are measured in \(\mathrm{kJ\,mol^{-1}}\):
| Species or transition state | Relative energy |
|---|---|
| Reactants, R | 0 |
| First transition state, TS1 | +35 |
| Intermediate, I | -45 |
| Second transition state, TS2 | +10 |
| Products, P | -25 |
For comparison, the uncatalysed reaction follows a single-step pathway with a transition state at \(+85\ \mathrm{kJ\,mol^{-1}}\).
Assume that the frequency factors for the elementary steps are comparable and that the step with the largest local activation energy controls the observed rate.
Answer the following:
a) Calculate the overall enthalpy change.
b) Calculate the activation energies of both elementary steps in the forward direction.
c) Calculate the activation energies of both elementary steps in the reverse direction.
d) Identify the rate-determining step in each direction under the stated assumption.
e) A student argues that the forward activation energy of the catalysed reaction must be \(35\ \mathrm{kJ\,mol^{-1}}\) because TS1 is the highest point on the diagram. Explain why this reasoning is incorrect.
f) Compare the largest local activation barriers of the catalysed pathway with the forward and reverse activation energies of the uncatalysed pathway. Does the catalyst speed up both directions?
g) Does the catalyst change the equilibrium position? Explain.
Solution 5
The overall enthalpy change is \(-25\ \mathrm{kJ\,mol^{-1}}\). The largest forward local activation barrier is \(55\ \mathrm{kJ\,mol^{-1}}\), while the largest reverse local activation barrier is \(80\ \mathrm{kJ\,mol^{-1}}\). The catalyst provides a faster pathway in both directions under the stated assumptions, without changing the equilibrium position.
Part a: Calculate the enthalpy change
The reactants are at 0, and the products are at \(-25\ \mathrm{kJ\,mol^{-1}}\).
\[
\Delta H=H_P-H_R
\]
\[
=-25-0
\]
\[
\boxed{\Delta H=-25\ \mathrm{kJ\,mol^{-1}}}
\]
The overall reaction is exothermic.
Part b: Calculate the forward activation energies
There are two elementary steps.
The first step takes the reaction from R to I through TS1.
\[
E_{a,1,\mathrm{forward}}=35-0
\]
\[
\boxed{35\ \mathrm{kJ\,mol^{-1}}}
\]
The second step takes the reaction from I to P through TS2.
This time, the starting point is the intermediate at \(-45\), not the original reactants at zero.
\[
E_{a,2,\mathrm{forward}}=10-(-45)
\]
\[
\boxed{55\ \mathrm{kJ\,mol^{-1}}}
\]
The second elementary step has the larger forward activation energy.
Part c: Calculate the reverse activation energies
The reverse reaction proceeds from P to I, then from I to R.
For the first reverse step, P to I through TS2:
\[
E_{a,1,\mathrm{reverse}}=10-(-25)
\]
\[
\boxed{35\ \mathrm{kJ\,mol^{-1}}}
\]
For the second reverse step, I to R through TS1:
\[
E_{a,2,\mathrm{reverse}}=35-(-45)
\]
\[
\boxed{80\ \mathrm{kJ\,mol^{-1}}}
\]
The second reverse step has the larger activation energy.
Part d: Identify the rate-determining steps
Under the assumption that the largest local activation barrier controls the observed rate:
- Forward direction: the second step, I to P, is rate-determining, with an activation energy of \(55\ \mathrm{kJ\,mol^{-1}}\).
- Reverse direction: the second step, I to R, is rate-determining, with an activation energy of \(80\ \mathrm{kJ\,mol^{-1}}\).
Notice that the rate-determining steps are different in the two directions.
Part e: Evaluate the student’s reasoning
The student’s conclusion is incorrect because the height of a transition state must be measured relative to the energy of the species immediately before that step.
TS1 is at \(+35\), while TS2 is at \(+10\). Therefore, TS1 is the highest point when measured against the same arbitrary zero reference.
However, the second forward step begins at the intermediate energy of \(-45\).
Its activation energy is:
\[
10-(-45)=55\ \mathrm{kJ\,mol^{-1}}
\]
This is greater than the first step’s activation energy of \(35\ \mathrm{kJ\,mol^{-1}}\).
The important distinction is that the highest absolute transition-state energy is not necessarily associated with the largest local activation barrier.
Always measure each barrier from its preceding reactant or intermediate.
Part f: Compare the catalysed and uncatalysed pathways
For the uncatalysed reaction, the forward activation energy is:
\[
E_{a,\mathrm{forward}}=85-0
\]
\[
=85\ \mathrm{kJ\,mol^{-1}}
\]
The reverse activation energy is:
\[
E_{a,\mathrm{reverse}}=85-(-25)
\]
\[
=110\ \mathrm{kJ\,mol^{-1}}
\]
Compare the largest local barriers:
| Direction | Catalysed pathway | Uncatalysed pathway |
|---|---|---|
| Forward | 55 kJ mol⁻¹ | 85 kJ mol⁻¹ |
| Reverse | 80 kJ mol⁻¹ | 110 kJ mol⁻¹ |
The catalysed pathway has a smaller largest local barrier in both directions.
Under the assumptions in the question, this supports a faster forward and reverse reaction.
The important point is that the catalyst does not need to make every individual step identical in speed. It needs to provide an overall alternative mechanism with more favourable kinetic barriers.
In real multistep mechanisms, the exact rate law also depends on intermediate concentrations and the kinetics of individual steps. An energy profile alone does not establish an exact reaction rate.
Part g: Does the catalyst change the equilibrium position?
No. The catalyst does not change the equilibrium position or equilibrium constant at a fixed temperature.
It provides an alternative pathway that increases the rates of both forward and reverse reactions.
The reaction therefore reaches equilibrium more quickly, but the equilibrium composition remains unchanged.
This is the final distinction to remember: activation energy controls reaction kinetics, while the overall energy difference between reactants and products remains a separate property of the reaction.
The most useful next step is to connect energy profiles with reaction-rate graphs and equilibrium. An energy diagram explains why a catalyst increases reaction rates; experimental measurements show that change, while equilibrium reasoning explains why the final equilibrium composition stays the same.