How to Calculate Reverse Activation Energy Using Delta H

A reaction releases \(38\ \mathrm{kJ\,mol^{-1}}\) of energy, but its forward activation energy is \(52\ \mathrm{kJ\,mol^{-1}}\). If you reverse the reaction, does it need 14, 52, or 90 \(\mathrm{kJ\,mol^{-1}}\) to overcome the energy barrier?

It needs \(90\ \mathrm{kJ\,mol^{-1}}\). The forward reaction releases energy overall, so the products finish at a lower energy than the reactants. To travel backwards over the same barrier, they have further to climb.

The calculation is:

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

where:

• \(E_{a,\mathrm{reverse}}\) is the reverse activation energy.
• \(E_{a,\mathrm{forward}}\) is the forward activation energy.
• \(\Delta H\) is the enthalpy change of the forward reaction.

All three quantities are expressed in \(\mathrm{kJ\,mol^{-1}}\) on the same reaction basis.

For this exothermic reaction, \(\Delta H=-38\ \mathrm{kJ\,mol^{-1}}\):

\[ \begin{aligned} E_{a,\mathrm{reverse}} &=52-(-38)\\ &=90\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The essential move is subtracting the signed enthalpy change, not simply subtracting the size of the energy released.

01Why the reverse activation energy is different

Picture a walking track that goes up a hill and down the other side.

You start at an elevation of 100 metres, climb to 160 metres, and finish at 70 metres.

The uphill climb is 60 metres. Your overall change in elevation is −30 metres. But if you walk the route backwards, the climb from the finish to the summit is 90 metres.

An energy profile works in a similar way.

• The starting elevation represents the energy of the reactants.
• The hilltop represents the highest-energy arrangement along the reaction pathway.
• The finishing elevation represents the energy of the products.
• The difference between the start and finish represents the reaction enthalpy.

The analogy has a limit. Molecules aren’t physically walking along a hill. The horizontal axis of an energy profile represents reaction progress, not distance or time. The curve models energy changes as atoms rearrange and bonds break and form.

Three energy levels, two activation barriers

For a simple reaction with one energy barrier, we can identify three important enthalpy levels:

The transition state is a short-lived, high-energy arrangement of atoms along the reaction pathway.

The forward activation energy is the difference between the transition state and the reactants:

\[ E_{a,\mathrm{forward}}=H^\ddagger-H_R \]

The reverse activation energy is measured from the products to that same transition state:

\[ E_{a,\mathrm{reverse}}=H^\ddagger-H_P \]

Meanwhile, reaction enthalpy describes the overall difference between products and reactants:

\[ \Delta H=H_P-H_R \]

Notice that these measurements share the same energy levels.

Subtracting the forward reaction’s enthalpy change from its activation energy gives:

\[ \begin{aligned} E_{a,\mathrm{forward}}-\Delta H &=(H^\ddagger-H_R)\\ &\quad -(H_P-H_R)\\ &=H^\ddagger-H_P\\ &=E_{a,\mathrm{reverse}} \end{aligned} \]

So the relationship is:

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

The equation isn’t a separate rule you need to memorise without understanding. It follows from measuring two climbs to the same peak from different starting points.

The assumptions matter: we are using a single-barrier energy-profile model, with the same reaction pathway, physical states, and conditions in both directions.

02What an experiment can measure – and what it cannot

There’s an important measurement problem hiding behind this equation.

Suppose you want to investigate a hypothetical reversible colour-changing reaction:

\[ \ce{A(aq) <=> B(aq)} \]

Imagine that A and B have different colours, allowing their concentrations to be monitored. This is an illustrative experimental design, not a report of a performed experiment.

You want to find both the overall enthalpy change and the forward activation energy.

Could you get both from a thermometer?

First attempt: Measure the heat change

Place a known quantity of reacting solution inside an insulated calorimeter. A temperature probe passes through the lid into the liquid.

As the net reaction proceeds, record the temperature change and determine how much substance has converted.

If the solution warms because of the reaction, you would expect the reaction to be exothermic. If it cools, you would expect it to be endothermic, assuming negligible external heat exchange and no other significant thermal effects.

Using calorimetry, you can estimate \(\Delta H\) from the heat transferred and the amount that reacts.

But the thermometer cannot tell you how high the activation barrier is.

Two reaction pathways could release exactly the same amount of heat while having very different activation energies.

Revised design: Measure how the reaction rate changes with temperature

Now imagine placing a small sample inside a transparent cuvette in a temperature-controlled holder.

A light source shines through the cuvette towards a colourimeter detector. The detector records how much light is absorbed as the concentrations of A and B change.

The measurement sequence is:

1. Prepare the mixture at a controlled starting concentration and temperature.
2. Record absorbance at regular time intervals.
3. Use a suitable calibration to estimate concentration changes and reaction rate.
4. Repeat at several temperatures, keeping other conditions controlled.
5. Analyse the temperature dependence of the forward rate to estimate its activation energy.

For a reversible reaction, the forward and reverse contributions must be distinguished using an appropriate kinetic model.

A faster change in absorbance at a higher temperature would support a temperature-dependent reaction rate. It would not, by itself, provide a numerical activation energy.

The revised approach addresses the original measurement limitation: calorimetry measures an overall energy change, while temperature-dependent kinetic measurements help estimate an activation barrier.

The measurements answer different questions.

There are limitations. Heat loss affects calorimetry, while changing temperature can also change equilibrium composition and reaction mechanisms. An activation energy inferred from kinetic measurements is model-dependent.

For HSC-style single-barrier energy profiles, however, once the forward activation energy and reaction enthalpy are supplied, the reverse barrier follows directly.

For a refresher on why temperature and collision energy affect reaction rates, see Collision Theory and Activation Energy: HSC Chemistry Explained.

05The sign of delta H is your fastest error check

Compare the two worked examples.

Here, \(E_{a,f}\) and \(E_{a,r}\) are shortened symbols for forward and reverse activation energy. The symbol \(\lvert\Delta H\rvert\) means the positive magnitude of the enthalpy change.

This table assumes the same single-barrier pathway in both directions.

Three mistakes worth catching

Mistake 1: Using a positive enthalpy change for every calculation.

If the reaction releases 40 \(\mathrm{kJ\,mol^{-1}}\), then \(\Delta H=-40\ \mathrm{kJ\,mol^{-1}}\), not +40.

Always identify the sign before substitution.

Mistake 2: Treating activation energy as the heat absorbed or released.

Reaction enthalpy compares the two endpoints. Activation energy compares a starting point with a barrier.

An exothermic reaction can have a large forward activation energy. Releasing energy overall doesn’t mean the reaction can start without overcoming a barrier.

Mistake 3: Assuming the direction with the lower activation energy must always be faster.

For a given pathway, a lower barrier generally makes reaction events easier, all else being comparable.

But actual reaction rates also depend on temperature, concentrations, and kinetic factors such as molecular arrangement and collision frequency.

Likewise, activation energy alone doesn’t tell you which substances are favoured at equilibrium.

A quick decision process

When an exam gives you \(E_{a,\mathrm{forward}}\) and \(\Delta H\):

1. Check which chemical equation is defined as the forward reaction.
2. Write \(\Delta H\) with its correct positive or negative sign.
3. Calculate \(E_{a,\mathrm{reverse}}=E_{a,\mathrm{forward}}-\Delta H\).
4. Check the result against the relative reactant and product energy levels.
5. Confirm that the answer is non-negative and expressed in the correct units.

A negative reverse activation energy from the basic single-hump diagram is a warning that the data or assumptions are inconsistent. The crest cannot sit below the product level if it represents the barrier that the products must climb.

These checks are more reliable than trying to remember a different equation for every reaction type.

06Questions and solutions

All numerical reaction data in the following questions are illustrative. They are not measured properties of particular chemical substances.

Question 1

A hypothetical reversible reaction has a forward activation energy of \(46\ \mathrm{kJ\,mol^{-1}}\) and an enthalpy change of \(-18\ \mathrm{kJ\,mol^{-1}}\).

Calculate the reverse activation energy and identify whether the forward reaction is exothermic or endothermic.

Solution 1

The reverse activation energy is \(64\ \mathrm{kJ\,mol^{-1}}\), and the forward reaction is exothermic.

The negative enthalpy change shows that the products have lower enthalpy than the reactants.

Using the reverse activation energy equation:

\[ \begin{aligned} E_{a,\mathrm{reverse}} &=E_{a,\mathrm{forward}}-\Delta H\\ &=46-(-18)\\ &=64\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The reverse barrier is 18 greater than the forward barrier because the reverse reaction starts from the lower-energy products.

The common error would be calculating \(46-18=28\). That ignores the negative sign of the enthalpy change.

Question 2

A hypothetical reaction is written:

\[ \ce{J <=> K} \]

The forward conversion of J into K is endothermic, with:

\[ \Delta H=+36\ \mathrm{kJ\,mol^{-1}} \]

An energy-profile analysis indicates that the activation energy for K changing back into J is \(57\ \mathrm{kJ\,mol^{-1}}\).

Take the enthalpy of K to be zero.

(a) Calculate the relative enthalpy of J.

(b) Calculate the relative enthalpy of the transition state.

(c) Determine the forward activation energy for J changing into K.

(d) A student argues that K must always convert into J faster than J converts into K because the reverse activation energy is smaller. Evaluate the claim.

Solution 2

The enthalpy of J is \(-36\ \mathrm{kJ\,mol^{-1}}\), the transition state is at \(+57\ \mathrm{kJ\,mol^{-1}}\), and the forward activation energy is \(93\ \mathrm{kJ\,mol^{-1}}\). The smaller reverse barrier does not, by itself, guarantee a faster observed reverse reaction.

(a) Relative enthalpy of J

The reaction is endothermic, so K is 36 higher than J.

Using:

\[ \Delta H=H_K-H_J \]

and \(H_K=0\):

\[ \begin{aligned} 36&=0-H_J\\ H_J&=-36\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

J must be below K on the energy profile.

(b) Transition-state enthalpy

The reverse activation energy is measured from K to the transition state:

\[ E_{a,\mathrm{reverse}}=H^\ddagger-H_K \]

Therefore:

\[ \begin{aligned} 57&=H^\ddagger-0\\ H^\ddagger&=57\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

(c) Forward activation energy

Starting from J at −36, the reaction must reach the transition state at +57.

\[ \begin{aligned} E_{a,\mathrm{forward}} &=H^\ddagger-H_J\\ &=57-(-36)\\ &=93\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

This agrees with the original equation:

\[ \begin{aligned} E_{a,\mathrm{reverse}} &=93-36\\ &=57\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

(d) Evaluate the rate claim

The reverse direction has the lower activation barrier. With comparable kinetic factors, that would favour faster reverse reaction events at the same temperature.

However, the activation energies alone do not determine the actual rates.

The concentrations of J and K, their molecular behaviour, and other kinetic factors also matter.

The tempting mistake is treating activation energy as though it were a complete rate equation.

Question 3

A hypothetical reversible reaction is investigated using two methods.

In the forward reaction, one mole of U is converted into V.

An insulated calorimeter shows that the surrounding solution warms as the reaction proceeds. The measured heat transfer has a magnitude of \(24\ \mathrm{kJ}\) per mole of U converted.

However, the data summary lists:

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

A separate investigation of the same single-step reaction pathway reports:

Assume negligible heat loss, no external heating, correctly calibrated instruments, and no significant heat effects other than the reaction.

(a) Explain why the calorimetry summary has a sign error.

(b) Use the activation energies to calculate the forward enthalpy change independently.

(c) State the corrected enthalpy change and explain why the two independent methods now agree.

(d) Explain why the calorimeter alone could not have supplied either activation energy.

Solution 3

The correct forward enthalpy change is \(-24\ \mathrm{kJ\,mol^{-1}}\). Both the calorimetry observation and the two activation energies support an exothermic reaction.

(a) Calorimetry sign

The surrounding solution warms, meaning it gains heat from the reacting system under the stated assumptions.

The reaction therefore releases heat.

With the usual sign convention:

\[ q_{\mathrm{reaction}} =-q_{\mathrm{surroundings}} \]

Here, \(q\) represents heat transferred.

For one mole converted, the reaction loses 24 kJ while the surroundings gain approximately 24 kJ.

The summary’s positive sign is incorrect.

(b) Calculate enthalpy change from the barriers

Rearrange the relationship:

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

to obtain:

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

Substitute:

\[ \begin{aligned} \Delta H &=48-72\\ &=-24\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

(c) Reconcile the measurements

The corrected value is:

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

Both methods now agree.

The calorimeter indicates that the reaction releases heat. The energy-profile data indicate that the products are 24 lower than the reactants, because the reverse barrier exceeds the forward barrier by 24.

If the printed positive enthalpy change had been correct, it would have contradicted both observations.

(d) Why calorimetry cannot measure activation energy directly

Calorimetry measures heat transferred during the overall conversion. It does not reveal the energy of the transition state.

To estimate activation energy experimentally, suitable rate measurements at multiple temperatures and an appropriate kinetic analysis would be needed.

A reaction that releases 24 \(\mathrm{kJ\,mol^{-1}}\) could still have many different activation energies.

The crucial distinction is between energy released overall and energy required to reach the reaction barrier.

Question 4

Consider the hypothetical reversible reaction:

\[ \ce{R <=> P} \]

The forward reaction is endothermic:

\[ \Delta H=+44\ \mathrm{kJ\,mol^{-1}} \]

Two reaction pathways are available:

Both pathways have the same reactants, products, and overall enthalpy change. Each can be represented by a single main activation barrier.

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

(b) Determine how much the catalyst lowers the activation energy in each direction.

(c) A mixture of R and P is initially at chemical equilibrium at constant temperature. Explain what happens to the forward and reverse reaction rates when a suitable catalyst is introduced, and whether the equilibrium composition changes.

(d) A student argues that the catalyst favours formation of P because it lowers the forward activation energy by 30. Explain the flaw.

Solution 4

The reverse activation energies are \(54\ \mathrm{kJ\,mol^{-1}}\) without the catalyst and \(24\ \mathrm{kJ\,mol^{-1}}\) with the catalyst. The catalyst lowers both barriers by 30, but it does not change the equilibrium composition.

(a) Calculate the reverse barriers

For the uncatalysed pathway:

\[ \begin{aligned} E_{a,\mathrm{reverse}} &=98-44\\ &=54\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

For the catalysed pathway:

\[ \begin{aligned} E_{a,\mathrm{reverse}} &=68-44\\ &=24\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

(b) Compare the reductions

The forward activation energy decreases by:

\[ 98-68=30\ \mathrm{kJ\,mol^{-1}} \]

The reverse activation energy decreases by:

\[ 54-24=30\ \mathrm{kJ\,mol^{-1}} \]

Both directions gain a lower-barrier pathway.

This happens because the catalyst changes the pathway without changing the reactant or product enthalpy levels.

(c) Effect at equilibrium

The catalyst increases the rates of both forward and reverse reactions by making the alternative pathway available.

At equilibrium, the forward and reverse rates are equal. After introducing the catalyst, they remain equal at the equilibrium composition, although both can be faster.

Consequently, there is no net shift in equilibrium composition.

We cannot calculate the exact rate increase from the activation energy values alone because other kinetic information is missing.

(d) Why the catalyst does not favour P

The student has considered only the forward barrier.

The catalyst lowers the reverse barrier as well. It does not change the enthalpy difference between R and P or the equilibrium position at the same temperature.

The tempting inference is that making the forward reaction easier must lead to more products at equilibrium. But equilibrium depends on the balance between the two directions, not the forward barrier considered in isolation.

The energy profile tells us about the available reaction pathway. It does not imply that the catalyst changes the final thermodynamic balance.

Question 5

Extension challenge: What happens when there is more than one energy barrier?

A hypothetical reversible reaction proceeds through two elementary steps:

\[ \ce{A <=> I <=> B} \]

Here, I is a genuine intermediate that can exist in an energy well between two transition states.

The following relative enthalpies describe the pathway at the same temperature and conditions.

The pathway follows this order:

A → first transition state → I → second transition state → B.

A student makes the following argument:

“The largest activation barrier for a forward elementary step must be the activation energy of the overall forward reaction. Therefore, I can apply \(E_{a,\mathrm{reverse}}=E_{a,\mathrm{forward}}-\Delta H\) to find the largest reverse elementary-step barrier.”

(a) Calculate the activation energies of both elementary steps in the forward direction.

(b) Calculate the activation energies of both elementary steps in the reverse direction.

(c) Calculate the overall enthalpy change from A to B.

(d) Test the student’s argument numerically. Explain precisely why the shortcut fails when applied to the largest elementary-step barriers.

(e) The highest transition state is at +45. Its energy is 45 above A but 65 above B. Explain why these two numbers satisfy the usual forward-reverse relationship, even though neither is the largest elementary-step barrier in its direction.

(f) Can the experimentally observed activation energy for the overall A-to-B conversion be determined uniquely from these five energy levels alone? Explain.

Assume the intermediate can relax into its stated energy well between the two steps, and no alternative pathway is available.

Solution 5

The largest forward elementary-step barrier is \(70\ \mathrm{kJ\,mol^{-1}}\), while the largest reverse elementary-step barrier is \(95\ \mathrm{kJ\,mol^{-1}}\). The student’s shortcut is invalid because those barriers belong to different steps and start from different energy levels.

(a) Forward elementary-step activation energies

The first step is A → I, passing through the first transition state at +45.

\[ \begin{aligned} E_{a,\mathrm{A\to I}} &=45-0\\ &=45\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The second step begins at intermediate I, whose enthalpy is −50.

It must reach the second transition state at +20:

\[ \begin{aligned} E_{a,\mathrm{I\to B}} &=20-(-50)\\ &=70\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The larger forward elementary-step barrier is 70.

Notice that this barrier is larger than the first transition state’s height above A. The intermediate begins much lower than A, so it faces a substantial climb despite the second peak being lower on the overall energy profile.

(b) Reverse elementary-step activation energies

In reverse, the reaction travels B → I → A.

For B → I, the system starts at −20 and climbs to the second transition state at +20:

\[ \begin{aligned} E_{a,\mathrm{B\to I}} &=20-(-20)\\ &=40\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

For I → A, it starts at −50 and climbs to the first transition state at +45:

\[ \begin{aligned} E_{a,\mathrm{I\to A}} &=45-(-50)\\ &=95\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The larger reverse elementary-step barrier is 95.

The highest barrier faced by an individual elementary reverse step occurs after the system has entered the low-energy intermediate I.

(c) Overall reaction enthalpy

The enthalpy change depends only on A and B:

\[ \begin{aligned} \Delta H &=H_B-H_A\\ &=-20-0\\ &=-20\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The overall forward reaction is exothermic.

(d) Test the student’s shortcut

The student uses the largest forward elementary-step barrier, 70, as though it were the activation energy for a single overall step.

That gives:

\[ \begin{aligned} E_{a,\mathrm{reverse}} &=70-(-20)\\ &=90\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

But the largest actual reverse elementary-step barrier is 95.

Why the difference?

The forward barrier of 70 belongs to I → B and is measured from I at −50 to the second transition state at +20.

Its matching reverse barrier is B → I, measured from B at −20 to that same second transition state.

The enthalpy change of this individual step is:

\[ \begin{aligned} \Delta H_{\mathrm{I\to B}} &=-20-(-50)\\ &=+30\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Apply the formula to the matching step:

\[ \begin{aligned} E_{a,\mathrm{B\to I}} &=70-30\\ &=40\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

This is correct.

The student instead subtracted the overall A-to-B enthalpy change from the barrier of the individual I-to-B step.

Those values refer to different starting and finishing states, so they cannot be combined in that way.

(e) Why the two endpoint-to-crest differences still work

The highest transition state is +45.

Measured from A at 0, its height is:

\[ 45-0=45\ \mathrm{kJ\,mol^{-1}} \]

Measured from B at −20, its height is:

\[ 45-(-20)=65\ \mathrm{kJ\,mol^{-1}} \]

These two differences obey:

\[ \begin{aligned} 65&=45-(-20)\\ &=45-\Delta H \end{aligned} \]

The energy-level subtraction still works perfectly when both differences refer to the same transition-state peak.

But 65 is not the activation barrier for the elementary I → A step.

After travelling from B into intermediate I, the system is at −50, not −20.

It must then climb from −50 to +45, a rise of 95.

The fair but subtle distinction is between:

• The height of a selected transition-state peak relative to the overall reactant or product level.
• The activation barrier for an individual elementary step, measured from that step’s own starting state.

These are not automatically the same quantity in a multistep mechanism.

(f) Can the overall experimental activation energy be found?

Not uniquely from the supplied energy levels alone.

An experimental activation energy for an overall multistep reaction depends on how the elementary reactions combine to produce the measured rate.

The populations of intermediates, their formation and consumption rates, and other kinetic factors can affect which part of the mechanism controls the observed temperature dependence.

The energy diagram establishes the individual barrier heights. It does not provide a complete kinetic model.

This is why the single-barrier equation must not be applied blindly to a multistep reaction using whichever barrier looks largest.

The original equation remains useful. What changes is the care required in defining the forward and reverse processes being compared.

For ordinary HSC energy-profile questions, the practical rule is simple: identify the forward enthalpy change, preserve its sign, and subtract it from the forward activation energy to obtain the reverse barrier for the same single-step pathway.

The next useful skill is recognising what happens when the pathway itself changes. A catalyst can lower both activation barriers without changing the overall enthalpy difference, while alternative mechanisms can introduce different transition states and intermediates. That is the starting point for comparing reaction pathways with the same reactants and products.