How to Compare Reaction Pathways with the Same Reactants and Products

Two reaction pathways can convert exactly the same reactants into exactly the same products, yet one pathway can be much faster than the other. How is that possible if the overall reaction is identical?

Imagine two pathways for a reaction, R → P. Both start at an enthalpy of 0 kJ mol⁻¹ and finish at −40 kJ mol⁻¹. However, pathway A reaches a maximum of +55 kJ mol⁻¹, while pathway B reaches +95 kJ mol⁻¹.

Before reading on, predict: which pathway releases more energy?

Neither. Both have the same reaction enthalpy, ΔH = −40 kJ mol⁻¹. Pathway A simply has a lower activation energy, meaning the reaction can proceed more easily through that route.

This is the central distinction when comparing reaction pathways: the overall enthalpy change depends on the reactants and products, while the activation energy depends on the pathway.

01Separate reaction enthalpy from activation energy

These two quantities both describe energy changes during a reaction, but they answer different questions.

Reaction enthalpy: how much energy changes overall?

The reaction enthalpy, ΔH, measures the difference in enthalpy between the products and reactants.

\[ \Delta H=H_{\mathrm{products}}-H_{\mathrm{reactants}} \]

Here:

• \(H_{\mathrm{products}}\) is the total enthalpy of the products.
• \(H_{\mathrm{reactants}}\) is the total enthalpy of the reactants.
• \(\Delta H\) is the enthalpy change, measured in kJ mol⁻¹.

The sign tells you whether the reaction is exothermic or endothermic.

For example, suppose a reaction starts at 80 kJ mol⁻¹ and finishes at 30 kJ mol⁻¹.

\[ \Delta H=30-80=-50\ \mathrm{kJ\,mol^{-1}} \]

The reaction is exothermic because the products have 50 kJ mol⁻¹ less enthalpy than the reactants.

The important point is that this calculation doesn’t involve the reaction pathway at all.

If you change the mechanism but keep the same reactants, products, physical states, and conditions, the reaction enthalpy stays the same.

Activation energy: how high is the barrier?

Activation energy, \(E_a\), is the energy difference between the starting substances and the transition state.

The transition state is the short-lived, highest-energy arrangement of atoms that must be reached as bonds break and form.

Think of two walking tracks that start at the same car park and finish at the same lookout. Both routes have the same overall change in elevation, but one has a much steeper hill.

The difference in starting and finishing heights represents reaction enthalpy. The height of the hill represents the activation barrier.

The analogy has a limit: molecules don’t physically travel along a road. A reaction pathway represents the changing arrangement of atoms and the associated energy as a reaction proceeds.

For a simple, single-step reaction, the forward activation energy is:

\[ E_{a,\mathrm{forward}}= H_{\mathrm{TS}}-H_{\mathrm{reactants}} \]

The reverse activation energy is:

\[ E_{a,\mathrm{reverse}}= H_{\mathrm{TS}}-H_{\mathrm{products}} \]

Here, \(H_{\mathrm{TS}}\) represents the enthalpy of the transition state.

Notice that the same transition state appears in both equations. This becomes important when we compare forward and reverse reactions.

02How to compare two pathways with the same endpoints

Let’s return to pathways A and B.

Suppose their energy levels are:

We can calculate the three important quantities.

Reaction enthalpy

For both pathways:

\[ \Delta H=-40-0=-40\ \mathrm{kJ\,mol^{-1}} \]

Forward activation energy

Pathway A:

\[ E_{a,\mathrm{forward}}=55-0=55 \]

Pathway B:

\[ E_{a,\mathrm{forward}}=95-0=95 \]

Reverse activation energy

Pathway A:

\[ E_{a,\mathrm{reverse}}=55-(-40)=95 \]

Pathway B:

\[ E_{a,\mathrm{reverse}}=95-(-40)=135 \]

The complete comparison is:

Pathway A has a lower activation energy in both directions.

At the same temperature, and assuming comparable kinetic factors, pathway A would generally allow a faster reaction.

However, neither pathway is more exothermic. They release the same amount of energy overall.

The relationship between forward and reverse activation energies

There is a useful equation that makes these comparisons much quicker.

Starting with:

\[ E_{a,\mathrm{forward}}= H_{\mathrm{TS}}-H_{\mathrm{reactants}} \]

and:

\[ E_{a,\mathrm{reverse}}= H_{\mathrm{TS}}-H_{\mathrm{products}} \]

Subtracting the reaction enthalpy gives:

\[ \boxed{ E_{a,\mathrm{reverse}}= E_{a,\mathrm{forward}}-\Delta H } \]

This relationship is particularly useful in HSC-style calculations because you can calculate the reverse activation energy without needing to redraw the entire energy profile.

But pay attention to the sign of ΔH.

What changes between exothermic and endothermic reactions?

For an exothermic reaction, ΔH is negative.

Suppose:

\[ \Delta H=-35\ \mathrm{kJ\,mol^{-1}} \]

and:

\[ E_{a,\mathrm{forward}}=60\ \mathrm{kJ\,mol^{-1}} \]

Then:

\[ E_{a,\mathrm{reverse}}=60-(-35) \]

\[ =95\ \mathrm{kJ\,mol^{-1}} \]

The reverse activation energy is larger.

This makes sense because the products are at a lower enthalpy than the reactants. Starting from the products means climbing a greater energy difference to reach the same transition state.

For an endothermic reaction, the opposite happens.

Suppose:

\[ \Delta H=+35\ \mathrm{kJ\,mol^{-1}} \]

and the forward activation energy is 60 kJ mol⁻¹.

\[ E_{a,\mathrm{reverse}}=60-35 \]

\[ =25\ \mathrm{kJ\,mol^{-1}} \]

The reverse activation energy is smaller because the products are already at a higher enthalpy than the reactants.

Remember this relationship:

This relationship applies to the same single-step pathway. It is not a rule for comparing unrelated reactions with different endpoints.

03A reliable method for comparing reaction pathways

When an exam gives you two energy profiles, use this sequence.

Step 1: Check the reactants and products.

Make sure both pathways start and finish with the same substances in the same physical states.

This matters because gaseous and liquid substances have different enthalpies, even if they have the same chemical formula.

Step 2: Compare the starting and finishing energy levels.

Calculate:

\[ \Delta H=H_{\mathrm{products}}-H_{\mathrm{reactants}} \]

If the endpoints are identical, both pathways must have the same reaction enthalpy.

Step 3: Identify the transition-state peak for each pathway.

The peak represents the transition state.

Do not confuse the peak with the products or an intermediate.

Step 4: Calculate the activation energies.

Measure from the reactants to the peak for the forward reaction.

Measure from the products to the peak for the reverse reaction.

Step 5: Explain what the difference means.

A lower activation energy generally means a faster reaction at the same temperature, assuming comparable kinetic factors.

A lower activation energy does not mean a more exothermic reaction.

What if the pathway contains several steps?

Some reaction profiles contain several peaks and valleys.

Each peak represents a transition state, while each valley between peaks represents an intermediate.

For a multi-step pathway, calculate the activation energy of each elementary step from the preceding minimum to the next transition-state peak.

Don’t automatically measure every activation energy from the original reactants. That can give you the wrong value.

Also, identifying the largest elementary-step barrier is a useful starting point for considering the slow step, but an actual overall rate can depend on the mechanism and the concentrations of intermediates.

04What happens when a catalyst provides another pathway?

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

It does not change the enthalpy of the reactants or products.

Consider an exothermic reaction with:

• Reactant enthalpy: 0 kJ mol⁻¹
• Product enthalpy: −45 kJ mol⁻¹
• Uncatalysed transition-state enthalpy: +85 kJ mol⁻¹
• Catalysed transition-state enthalpy: +35 kJ mol⁻¹

The uncatalysed forward activation energy is:

\[ E_a=85-0=85\ \mathrm{kJ\,mol^{-1}} \]

The catalysed forward activation energy is:

\[ E_a=35-0=35\ \mathrm{kJ\,mol^{-1}} \]

The catalyst has lowered the forward activation barrier by 50 kJ mol⁻¹.

What about the reverse reaction?

Uncatalysed:

\[ E_{a,\mathrm{reverse}}=85-(-45)=130 \]

Catalysed:

\[ E_{a,\mathrm{reverse}}=35-(-45)=80 \]

The reverse activation energy has also decreased by 50 kJ mol⁻¹.

This leads to an important distinction:

• A catalyst increases the rates of both the forward and reverse reactions.
• A catalyst does not change the reaction enthalpy.
• A catalyst does not change the equilibrium position or equilibrium constant.
• A catalyst allows equilibrium to be reached more quickly.

A common misconception is that lowering the forward activation energy makes the products more stable.

It doesn’t. The catalyst changes the route taken, not the energy difference between the starting and finishing substances.

05Worked examples

06Questions and solutions

The following questions move from straightforward energy-profile calculations to more challenging comparisons involving physical states, catalysts, and consistency between forward and reverse activation energies.

Question 1

A reaction converts substance A into substance B through a single-step pathway.

The enthalpy levels are:

• Reactants: 0 kJ mol⁻¹
• Products: −24 kJ mol⁻¹
• Transition state: +61 kJ mol⁻¹

Calculate:

(a) The reaction enthalpy.

(b) The forward activation energy.

(c) The reverse activation energy.

(d) State whether the reaction is exothermic or endothermic.

Solution 1

The answers are ΔH = −24 kJ mol⁻¹, forward activation energy = 61 kJ mol⁻¹, and reverse activation energy = 85 kJ mol⁻¹. The reaction is exothermic.

(a) Reaction enthalpy

Use:

\[ \Delta H=H_P-H_R \]

Substitute:

\[ \Delta H=-24-0 \]

\[ \boxed{\Delta H=-24\ \mathrm{kJ\,mol^{-1}}} \]

The negative value indicates an exothermic reaction.

(b) Forward activation energy

Measure from the reactants to the transition state.

\[ E_{a,\mathrm{forward}}=61-0 \]

\[ \boxed{61\ \mathrm{kJ\,mol^{-1}}} \]

(c) Reverse activation energy

Measure from the products to the transition state.

\[ E_{a,\mathrm{reverse}}=61-(-24) \]

\[ \boxed{85\ \mathrm{kJ\,mol^{-1}}} \]

(d) Reaction type

The reaction is exothermic because its enthalpy change is negative.

A common mistake is to use 61 kJ mol⁻¹ for both activation energies. That value only measures the forward barrier. The reverse reaction starts from the lower product enthalpy, so its barrier is larger.

Question 2

An endothermic reaction converts C into D. Its reaction enthalpy is +32 kJ mol⁻¹.

Two alternative single-step pathways are available:

• Pathway P has a forward activation energy of 74 kJ mol⁻¹.
• Pathway Q has a forward activation energy of 103 kJ mol⁻¹.

(a) Calculate the reverse activation energy for each pathway.

(b) Identify which pathway is expected to be faster in the forward direction and which is expected to be faster in the reverse direction, assuming comparable kinetic factors.

(c) Explain why the reverse activation energy is smaller than the forward activation energy in both pathways.

Solution 2

Pathway P has a reverse activation energy of 42 kJ mol⁻¹, while pathway Q has a reverse activation energy of 71 kJ mol⁻¹. Pathway P is expected to be faster in both directions.

(a) Reverse activation energies

Use:

\[ E_{a,\mathrm{reverse}}= E_{a,\mathrm{forward}}-\Delta H \]

For pathway P:

\[ E_{a,\mathrm{reverse}}=74-32 \]

\[ \boxed{42\ \mathrm{kJ\,mol^{-1}}} \]

For pathway Q:

\[ E_{a,\mathrm{reverse}}=103-32 \]

\[ \boxed{71\ \mathrm{kJ\,mol^{-1}}} \]

(b) Compare the pathways

All values are in kJ mol⁻¹.

Pathway P has the lower activation energy in both directions.

At the same temperature, and assuming comparable kinetic factors, pathway P is expected to be faster.

(c) Explain the relationship

The reaction is endothermic, meaning the products have a higher enthalpy than the reactants.

The reverse reaction starts from these higher-energy products. Therefore, it has a shorter energy difference to overcome before reaching the same transition state.

The reverse activation energy is consequently smaller.

The tempting mistake is to assume that reversing a reaction always produces a larger activation energy. That is only true for an exothermic forward reaction.

Question 3

A chemist investigates two possible ways of converting gaseous substance A into substance B.

The chemical formula of B is identical in both cases, but the physical states are different.

The measured enthalpy levels are:

Pathway L produces liquid B, while pathway G produces gaseous B.

The enthalpy of vaporisation of B is +30 kJ mol⁻¹.

(a) Are these genuinely two pathways with identical thermodynamic endpoints? Explain.

(b) Calculate the reaction enthalpy, forward activation energy, and reverse activation energy for each pathway.

(c) If the gaseous B subsequently condenses into liquid B, calculate the enthalpy change for the complete conversion from A(g) to B(l).

(d) Explain why the two pathways cannot automatically be compared as alternative routes to the same final product state.

Solution 3

No. The two pathways do not have identical thermodynamic endpoints because liquid B and gaseous B have different enthalpies. Pathway L has ΔH = −48 kJ mol⁻¹, while pathway G has ΔH = −18 kJ mol⁻¹.

(a) Compare the endpoints

Although both products have the same chemical formula, their physical states differ.

Gaseous B has a higher enthalpy than liquid B because energy is absorbed during vaporisation.

Therefore, the two pathways do not finish at the same thermodynamic state.

This is an important qualification to the statement that identical reactants and products give identical reaction enthalpies. The physical states must also be identical.

(b) Calculate the energy quantities

For pathway L:

Reaction enthalpy:

\[ \Delta H_L=-48-0 \]

\[ \boxed{\Delta H_L=-48\ \mathrm{kJ\,mol^{-1}}} \]

Forward activation energy:

\[ E_{a,\mathrm{forward}}=64-0 \]

\[ \boxed{64\ \mathrm{kJ\,mol^{-1}}} \]

Reverse activation energy:

\[ E_{a,\mathrm{reverse}}=64-(-48) \]

\[ \boxed{112\ \mathrm{kJ\,mol^{-1}}} \]

For pathway G:

Reaction enthalpy:

\[ \Delta H_G=-18-0 \]

\[ \boxed{\Delta H_G=-18\ \mathrm{kJ\,mol^{-1}}} \]

Forward activation energy:

\[ E_{a,\mathrm{forward}}=82-0 \]

\[ \boxed{82\ \mathrm{kJ\,mol^{-1}}} \]

Reverse activation energy:

\[ E_{a,\mathrm{reverse}}=82-(-18) \]

\[ \boxed{100\ \mathrm{kJ\,mol^{-1}}} \]

The results are:

Notice something interesting: pathway L has the lower forward activation energy, but pathway G has the lower reverse activation energy.

This does not contradict the relationship between forward and reverse activation energies. The pathways have different product states and therefore different reaction enthalpies.

(c) Calculate the enthalpy change after condensation

The enthalpy change for condensation is the negative of the enthalpy of vaporisation.

\[ \Delta H_{\mathrm{condensation}}=-30\ \mathrm{kJ\,mol^{-1}} \]

The overall conversion is:

\[ \Delta H_{\mathrm{overall}}= \Delta H_G+\Delta H_{\mathrm{condensation}} \]

\[ =-18+(-30) \]

\[ \boxed{-48\ \mathrm{kJ\,mol^{-1}}} \]

The overall enthalpy change now matches pathway L because both routes finish with liquid B.

(d) Explain the comparison

The two routes can only be treated as alternative pathways to the same reaction if they finish in the same physical state.

Once gaseous B condenses, the final product state matches pathway L, and the overall enthalpy change becomes identical.

However, the activation energy of the complete gas-forming-and-condensation route cannot be calculated from the supplied data alone. We would also need information about the energy barrier associated with the condensation step.

This is why checking physical states before comparing energy profiles is so important.

Question 4

A reversible exothermic reaction has a reaction enthalpy of −42 kJ mol⁻¹.

An uncatalysed single-step pathway has a forward activation energy of 88 kJ mol⁻¹.

A catalyst provides an alternative single-step pathway with a forward activation energy of 38 kJ mol⁻¹.

Assume both pathways have the same reactants and products.

(a) Calculate the reverse activation energy for the uncatalysed reaction.

(b) Calculate the reverse activation energy for the catalysed reaction.

(c) Calculate how much the catalyst reduces each activation energy.

(d) Explain whether adding the catalyst changes the equilibrium position of the reaction.

Solution 4

The uncatalysed reverse activation energy is 130 kJ mol⁻¹, while the catalysed reverse activation energy is 80 kJ mol⁻¹. The catalyst reduces both forward and reverse activation energies by 50 kJ mol⁻¹ without changing the equilibrium position.

(a) Uncatalysed reverse activation energy

Use:

\[ E_{a,\mathrm{reverse}}= E_{a,\mathrm{forward}}-\Delta H \]

Substitute:

\[ =88-(-42) \]

\[ \boxed{130\ \mathrm{kJ\,mol^{-1}}} \]

(b) Catalysed reverse activation energy

\[ E_{a,\mathrm{reverse}}=38-(-42) \]

\[ \boxed{80\ \mathrm{kJ\,mol^{-1}}} \]

(c) Calculate the reductions

Forward reduction:

\[ 88-38=50\ \mathrm{kJ\,mol^{-1}} \]

Reverse reduction:

\[ 130-80=50\ \mathrm{kJ\,mol^{-1}} \]

Both activation energies decrease by the same amount.

This happens because the catalyst provides a different transition state while the starting and finishing enthalpies remain unchanged.

(d) Effect on equilibrium

The catalyst does not change the equilibrium position.

It increases the rates of both the forward and reverse reactions, allowing equilibrium to be reached more quickly.

At equilibrium, the forward and reverse reaction rates remain equal.

The important distinction is that a catalyst changes reaction kinetics, not the overall thermodynamics of the reaction.

Question 5

A reversible single-step reaction, A ⇌ B, has a reaction enthalpy of −36 kJ mol⁻¹.

Two proposed pathways have the following reported activation energies:

Both pathways are claimed to connect exactly the same reactants and products, in the same physical states, through one transition state each.

A student argues that pathway Y is a valid catalyst-assisted route because its forward activation energy decreases by 23 kJ mol⁻¹, while its reverse activation energy decreases by only 11 kJ mol⁻¹.

(a) Determine whether the reported activation energies for pathway X are consistent with the reaction enthalpy.

(b) Determine whether the reported activation energies for pathway Y are consistent with the reaction enthalpy.

(c) If the forward activation energy of pathway Y is correct, calculate the reverse activation energy it must have.

(d) Alternatively, if both reported activation energies for pathway Y are correct, what reaction enthalpy do they imply?

(e) Explain why the student’s argument about a catalyst lowering the forward and reverse barriers by different amounts is problematic. Discuss whether pathway Y could change the equilibrium position if it were genuinely a catalyst-assisted pathway with the same endpoints.

Solution 5

Pathway X is consistent with the stated reaction enthalpy, but pathway Y is not. If pathway Y has a forward activation energy of 31 kJ mol⁻¹ and the reaction enthalpy is −36 kJ mol⁻¹, its reverse activation energy must be 67 kJ mol⁻¹, not 79 kJ mol⁻¹.

(a) Check pathway X

For a single-step reversible pathway:

\[ E_{a,\mathrm{reverse}}= E_{a,\mathrm{forward}}-\Delta H \]

Substitute the values:

\[ E_{a,\mathrm{reverse}}=54-(-36) \]

\[ =90\ \mathrm{kJ\,mol^{-1}} \]

This matches the reported reverse activation energy.

Therefore, pathway X is internally consistent.

(b) Check pathway Y

Apply the same relationship:

\[ E_{a,\mathrm{reverse}}=31-(-36) \]

\[ =67\ \mathrm{kJ\,mol^{-1}} \]

However, the reported reverse activation energy is 79 kJ mol⁻¹.

The reported value is 12 kJ mol⁻¹ higher than expected.

Therefore, pathway Y’s reported data cannot all be correct if it connects the same endpoints through a single transition state.

(c) Calculate the correct reverse activation energy

If the forward activation energy of 31 kJ mol⁻¹ and reaction enthalpy of −36 kJ mol⁻¹ are both correct:

\[ E_{a,\mathrm{reverse}}= 31-(-36) \]

\[ \boxed{67\ \mathrm{kJ\,mol^{-1}}} \]

(d) Find the enthalpy change implied by the reported barriers

Rearrange the relationship:

\[ E_{a,\mathrm{reverse}}= E_{a,\mathrm{forward}}-\Delta H \]

Therefore:

\[ \Delta H= E_{a,\mathrm{forward}}- E_{a,\mathrm{reverse}} \]

Substitute the reported pathway Y values:

\[ \Delta H=31-79 \]

\[ \boxed{\Delta H=-48\ \mathrm{kJ\,mol^{-1}}} \]

The reported activation energies actually imply an enthalpy change of −48 kJ mol⁻¹.

This is different from the stated reaction enthalpy of −36 kJ mol⁻¹.

The data alone cannot tell us which individual measurement is incorrect. Either the reported reverse activation energy, the forward activation energy, or the stated endpoint enthalpy needs to be checked.

(e) Explain the catalyst argument

The student’s argument is incorrect for the stated single-step comparison.

The forward and reverse reactions pass through the same transition state. Lowering that transition state’s enthalpy by 23 kJ mol⁻¹ reduces both activation energies by 23 kJ mol⁻¹ when the reactants and products remain unchanged.

The reverse reaction is not associated with an entirely separate transition state simply because it proceeds in the opposite direction.

If pathway Y genuinely provides a catalyst-assisted route between the same endpoints, its forward and reverse activation energies must remain consistent with the same reaction enthalpy.

It also cannot change the equilibrium position. A catalyst increases both reaction rates without changing the relative enthalpies of the reactants and products.

This question tests an important skill: don’t just calculate activation energies. Check whether the numbers in a proposed energy profile can all be true at the same time.

07Where to take this next

Once you can distinguish reaction enthalpy from activation energy, you can apply the same reasoning to other reaction questions.

The next useful decision is identifying what the question actually wants you to compare:

• If it asks which chemical transformation occurs, focus on the reactants, products, functional groups, reagents, and conditions. See How to Read Organic Reaction Pathways in HSC Chemistry.
• If it asks about combustion products or balancing combustion equations, focus on the available oxygen and the resulting products. See Combustion Reactions: Products and Equations for HSC Chemistry.

For energy-profile questions, keep one distinction firmly in mind: ΔH tells you the overall energy change, while activation energy tells you the barrier along a particular pathway.

Syllabus context: As at November 2026, NSW students are still following the Chemistry Stage 6 (2017) syllabus. The Chemistry 11-12 Syllabus (2025) begins implementation for Year 11 in 2028 and Year 12 later in 2028. This explanation uses the current 2017 syllabus context rather than assuming the new syllabus is already in effect. See NESA’s syllabus transition information.