Why Breaking Bonds Requires Energy in Exothermic Reactions

A candle releases heat when it burns, yet it won’t spontaneously burst into flames just because its fuel and oxygen are sitting together. Someone still needs to light the wick.

Here’s the apparent contradiction: if an exothermic reaction releases energy, why does it need energy supplied to get started?

The answer is that activation energy and enthalpy change describe two different things. Breaking existing bonds and rearranging atoms requires energy. Forming new bonds releases energy. An exothermic reaction releases more energy through bond formation than it absorbs through bond breaking, but it must still overcome the initial energy barrier.

Let’s separate those two energy changes and see how they fit together.

01Why breaking bonds requires energy

Imagine trying to pull two magnets apart when they’re stuck together. You have to work against the attraction holding them together.

Chemical bonds have a similar energy relationship. Bonded atoms occupy a more stable, lower-potential-energy arrangement than the same atoms separated from each other.

To separate bonded atoms, energy must be transferred into the chemical system.

The reverse process releases energy. When atoms form a bond, they move into a more stable arrangement, and energy is released.

This gives us the first important rule: breaking bonds never releases the energy needed to break them.

So why do some reactions release energy overall?

Because the energy released when product bonds form can be greater than the energy absorbed when reactant bonds break.

Consider a hypothetical reaction:

• Breaking the reactant bonds requires 650 kJ mol⁻¹.
• Forming the product bonds releases 900 kJ mol⁻¹.

The net energy change is:

\[ \Delta H=650-900 \]

\[ \boxed{\Delta H=-250\text{ kJ mol}^{-1}} \]

The reaction releases 250 kJ mol⁻¹ overall.

Notice that we haven’t said bond breaking releases energy. It absorbs 650 kJ mol⁻¹, but bond formation releases 900 kJ mol⁻¹.

The extra 250 kJ mol⁻¹ is transferred to the surroundings.

For a more detailed treatment of this energy accounting, see Bond Enthalpy and Conservation of Energy in HSC Chemistry.

02The energy barrier is different from the overall energy change

Here’s where the distinction becomes important.

Imagine riding a skateboard over a hill. Your starting platform is higher than the final platform, so the overall journey is downhill.

Does that mean you can reach the other side without first getting up the hill?

No. You still need to supply energy to get over the highest point.

A chemical reaction can behave similarly. The reactants may have more enthalpy than the products, but the reaction still needs to pass through a higher-energy arrangement before the products can form.

This is an energy profile diagram.

The vertical axis represents enthalpy, usually measured in kJ mol⁻¹. The horizontal axis represents reaction progress, showing how the chemical arrangement changes from reactants to products.

The horizontal axis is not time.

There are three important energy levels:

• Reactants: the starting substances.
• Transition state: the unstable, high-energy arrangement that must be reached during the reaction.
• Products: the substances formed.

The transition state is the highest-energy arrangement along this simple reaction pathway. Reaching it requires an input of energy.

The energy difference between the reactants and this peak is the forward activation energy.

Activation energy versus enthalpy change

These two quantities answer different questions.

Activation energy, \(E_a\):

How much energy must the reacting system gain to reach the transition state from its starting state?

Enthalpy change, \(\Delta H\):

How much does the enthalpy of the system change between the reactants and products?

For the forward reaction:

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

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

Here, \(H\) represents enthalpy.

Look at the illustrative diagram. The reactants begin at 100 kJ mol⁻¹, the peak is at 170 kJ mol⁻¹, and the products finish at 45 kJ mol⁻¹.

Therefore:

\[ E_{a,\mathrm{forward}}=170-100 \]

\[ E_{a,\mathrm{forward}}=70\text{ kJ mol}^{-1} \]

But:

\[ \Delta H=45-100 \]

\[ \Delta H=-55\text{ kJ mol}^{-1} \]

The activation energy is positive, while the enthalpy change is negative.

There is no contradiction.

The reaction absorbs energy to reach the transition state, then releases a greater amount of energy as the products form.

An exothermic reaction can have a positive activation energy because the energy needed to start the reaction is not the same as the net energy released by the reaction.

03What actually happens to the bonds?

It’s worth being a little more precise here.

A chemical reaction doesn’t necessarily break every reactant bond completely before forming any product bonds.

As reacting particles approach, existing bonds can stretch and weaken. At the transition state, bonds are being rearranged. Some original bonds may be partly broken while new bonds are partly formed.

The system passes through this unstable arrangement before reaching the more stable products.

The transition state is important because the reaction has to reach this arrangement before it can proceed along the pathway towards products.

The activation energy represents the energy barrier associated with reaching it.

This also explains why activation energy is not simply the energy required to break all the reactant bonds completely into separate atoms.

Bond enthalpy calculations and activation energy calculations are related to the same underlying energy changes, but they describe different things:

• Bond enthalpy measures the energy required to break a particular type of bond.
• Activation energy measures the energy difference between the starting state and the transition state along a reaction pathway.

Don’t treat them as interchangeable quantities.

04What happens when the reaction runs backwards?

Now imagine reversing the reaction.

The products become the reactants, and the reactants become the products.

The transition state remains the same for the same reaction pathway, but the starting energy level has changed.

This means the reverse reaction has a different activation energy.

For a simple one-step pathway:

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

Compare this with the forward activation energy:

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

Subtracting the two equations gives:

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

This relationship is particularly useful in HSC Chemistry questions.

It assumes that the forward and reverse reactions follow the same one-step pathway, with the same transition state.

Why the reverse activation energy is larger for an exothermic forward reaction

Return to our earlier example.

The forward reaction has:

• Reactant enthalpy: 100 kJ mol⁻¹
• Transition-state enthalpy: 170 kJ mol⁻¹
• Product enthalpy: 45 kJ mol⁻¹

We already calculated:

\[ E_{a,\mathrm{forward}}=70\text{ kJ mol}^{-1} \]

For the reverse reaction:

\[ E_{a,\mathrm{reverse}}=170-45 \]

\[ E_{a,\mathrm{reverse}}=125\text{ kJ mol}^{-1} \]

Why is the reverse barrier larger?

The products of the forward reaction are at a lower enthalpy than the original reactants. When the reaction runs backwards, it starts from this lower energy level.

It therefore has a greater energy gap to overcome before reaching the same transition state.

There are two useful conclusions:

• An exothermic forward reaction has a negative enthalpy change and a larger reverse activation energy.
• An endothermic forward reaction has a positive enthalpy change and a smaller reverse activation energy.

When you reverse a chemical equation, the sign of its enthalpy change also reverses.

\[ \Delta H_{\mathrm{reverse}} =-\Delta H_{\mathrm{forward}} \]

The activation energy doesn’t simply change sign. It must be recalculated from the new starting level.

For additional practice rearranging this relationship, see How to Calculate Reverse Activation Energy Using Delta H.

05A contrasting case: an endothermic reaction

Let’s check that this reasoning works in the opposite direction.

Suppose another reaction has these illustrative energy levels:

Calculate the forward enthalpy change:

\[ \Delta H=105-80 \]

\[ \Delta H=+25\text{ kJ mol}^{-1} \]

The reaction is endothermic because the products have higher enthalpy than the reactants.

Now calculate the forward activation energy:

\[ E_{a,\mathrm{forward}}=130-80 \]

\[ E_{a,\mathrm{forward}}=50\text{ kJ mol}^{-1} \]

And the reverse activation energy:

\[ E_{a,\mathrm{reverse}}=130-105 \]

\[ E_{a,\mathrm{reverse}}=25\text{ kJ mol}^{-1} \]

Notice something interesting.

The forward reaction is endothermic, but it still needs activation energy. The reverse reaction is exothermic, and its activation energy is smaller.

Compare the two types of reaction:

The key point is that both types of reaction can have an activation-energy barrier.

The sign of \(\Delta H\) tells you about the overall energy change, not whether a barrier exists.

06Why an exothermic reaction can still be slow

If a reaction releases energy, you might expect it to happen quickly.

That prediction is tempting, but it confuses thermodynamics with reaction kinetics.

Enthalpy change describes the overall energy difference between reactants and products. Activation energy helps determine how quickly the reaction proceeds.

For a reaction to occur through the usual collision-theory model, reacting particles need:

• Sufficient energy to overcome the activation-energy barrier.
• The correct orientation when they collide.

At a given temperature, not every particle has enough energy to overcome the barrier.

If the activation energy is large, only a relatively small fraction of collisions may be successful.

This is why an exothermic reaction can be slow at room temperature.

Combustion is a useful example. The overall reaction releases energy, but an initial input of energy is needed to start it. Once combustion begins, the energy released can heat nearby reactants and help further reactions occur.

The same principle explains why a piece of paper doesn’t spontaneously burst into flames at room temperature, even though burning it releases energy overall.

Heating and catalysts: two different effects

Heating and adding a catalyst can both increase reaction rate, but they work differently.

Heating the reaction

Increasing temperature gives particles a greater average kinetic energy and increases the proportion of particles with enough energy to overcome the activation barrier.

In the standard HSC energy-profile model, heating does not lower the activation energy itself.

Adding a catalyst

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

It lowers the energy required to reach the relevant transition state.

For a simple one-step reaction, the catalyst lowers both forward and reverse activation energies by the same amount because the reactant and product energy levels remain unchanged.

It does not change the overall enthalpy change.

This is an important distinction in exam questions:

• Heating increases the proportion of particles able to overcome the existing barrier.
• A catalyst lowers the barrier by providing another pathway.
• Neither changes the overall enthalpy change in the standard model.

09Questions and solutions

Question 1

A hypothetical reaction has the following energy levels:

Calculate:

a. The forward enthalpy change.

b. The forward activation energy.

c. The reverse activation energy.

d. Classify the forward and reverse reactions.

Solution 1

The forward reaction has an enthalpy change of \(-35\text{ kJ mol}^{-1}\), a forward activation energy of \(55\text{ kJ mol}^{-1}\), and a reverse activation energy of \(90\text{ kJ mol}^{-1}\).

Part a: Forward enthalpy change

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

\[ \Delta H=25-60 \]

\[ \boxed{\Delta H=-35\text{ kJ mol}^{-1}} \]

The forward reaction is exothermic.

Part b: Forward activation energy

\[ E_{a,\mathrm{forward}} =115-60 \]

\[ \boxed{E_{a,\mathrm{forward}}=55\text{ kJ mol}^{-1}} \]

Part c: Reverse activation energy

\[ E_{a,\mathrm{reverse}} =115-25 \]

\[ \boxed{E_{a,\mathrm{reverse}}=90\text{ kJ mol}^{-1}} \]

Part d: Classify both directions

The forward reaction is exothermic because its enthalpy change is negative.

The reverse reaction is endothermic because its enthalpy change is positive:

\[ \Delta H_{\mathrm{reverse}}=+35\text{ kJ mol}^{-1} \]

The reverse activation energy is larger because the reverse reaction begins at the lower product enthalpy level and must reach the same transition state.

Question 2

For a reversible, one-step reaction:

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

the forward enthalpy change is \(-42\text{ kJ mol}^{-1}\), and the reverse activation energy is \(108\text{ kJ mol}^{-1}\).

Calculate:

a. The forward activation energy.

b. The reverse enthalpy change.

c. Explain why the reverse activation energy is greater than the forward activation energy.

Solution 2

The forward activation energy is \(66\text{ kJ mol}^{-1}\), and the reverse enthalpy change is \(+42\text{ kJ mol}^{-1}\).

Part a: Forward activation energy

Start with:

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

Substitute:

\[ E_{a,\mathrm{forward}}-108=-42 \]

Therefore:

\[ E_{a,\mathrm{forward}}=108-42 \]

\[ \boxed{E_{a,\mathrm{forward}} =66\text{ kJ mol}^{-1}} \]

Part b: Reverse enthalpy change

Reversing the chemical equation reverses the sign of the enthalpy change.

\[ \Delta H_{\mathrm{reverse}} =-\Delta H_{\mathrm{forward}} \]

\[ \boxed{\Delta H_{\mathrm{reverse}} =+42\text{ kJ mol}^{-1}} \]

Part c: Explain the difference

The forward reaction is exothermic, so B has a lower enthalpy than A.

The reverse reaction begins at B and must reach the same transition state as the forward reaction.

Because B is lower in enthalpy, the reverse reaction has a larger energy gap to overcome.

A tempting mistake is to reverse the sign of the activation energy when reversing the equation. Activation energy is not an overall energy change. It is a positive energy barrier measured from the starting level to the transition state.

Question 3

Two reactions occur at the same temperature.

Their illustrative energy data are:

Assume both reactions have similar collision frequencies and orientation requirements.

a. Which reaction is more exothermic?

b. Which reaction is expected to proceed faster?

c. Explain why the reaction releasing more energy is not necessarily the faster reaction.

Solution 3

Reaction C → D is more exothermic, but reaction E → F is expected to proceed faster under the stated assumptions.

Part a: Compare the enthalpy changes

Reaction C → D:

\[ \Delta H=-130\text{ kJ mol}^{-1} \]

Reaction E → F:

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

Reaction C → D has the more negative enthalpy change.

It therefore releases more energy per mole of reaction.

Part b: Compare the activation energies

Reaction C → D:

\[ E_a=150\text{ kJ mol}^{-1} \]

Reaction E → F:

\[ E_a=45\text{ kJ mol}^{-1} \]

Reaction E → F has a much lower activation-energy barrier.

At the same temperature, and assuming similar collision and orientation factors, a greater proportion of its collisions can overcome the barrier.

Therefore, E → F is expected to proceed faster.

Part c: Explain the difference

The mistake is assuming that a greater energy release automatically means a faster reaction.

Enthalpy change describes the energy difference between the reactants and products. Activation energy describes the barrier that must be overcome for the reaction to proceed.

These are independent features of the energy profile.

Reaction C → D releases more energy overall, but it has a much larger activation-energy barrier.

Reaction E → F releases less energy, but its lower barrier makes successful collisions more likely at the same temperature.

The amount of energy released does not directly determine the reaction rate.

Question 4

A reaction has the following energy levels:

Calculate:

a. The enthalpy change.

b. The uncatalysed forward and reverse activation energies.

c. The catalysed forward and reverse activation energies.

d. Evaluate the statement:

“The catalyst makes the reaction less exothermic because it lowers the energy of the reaction pathway.”

Solution 4

The reaction has an enthalpy change of \(-50\text{ kJ mol}^{-1}\). The uncatalysed activation energies are 60 kJ mol⁻¹ forward and 110 kJ mol⁻¹ reverse. The catalysed activation energies are 25 kJ mol⁻¹ forward and 75 kJ mol⁻¹ reverse.

Part a: Enthalpy change

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

\[ \Delta H=-10-40 \]

\[ \boxed{\Delta H=-50\text{ kJ mol}^{-1}} \]

The reaction is exothermic.

Part b: Uncatalysed activation energies

Forward:

\[ E_{a,\mathrm{forward}}=100-40 \]

\[ \boxed{60\text{ kJ mol}^{-1}} \]

Reverse:

\[ E_{a,\mathrm{reverse}}=100-(-10) \]

\[ \boxed{110\text{ kJ mol}^{-1}} \]

Part c: Catalysed activation energies

Forward:

\[ E_{a,\mathrm{forward,catalysed}}=65-40 \]

\[ \boxed{25\text{ kJ mol}^{-1}} \]

Reverse:

\[ E_{a,\mathrm{reverse,catalysed}}=65-(-10) \]

\[ \boxed{75\text{ kJ mol}^{-1}} \]

Both activation energies have decreased by:

\[ 60-25=35\text{ kJ mol}^{-1} \]

and

\[ 110-75=35\text{ kJ mol}^{-1} \]

Part d: Evaluate the statement

The statement is incorrect.

The catalyst lowers the transition-state energy by providing an alternative pathway, but the reactant and product enthalpies remain unchanged.

Therefore:

\[ \Delta H=-10-40 \]

\[ \boxed{\Delta H=-50\text{ kJ mol}^{-1}} \]

The reaction releases exactly the same enthalpy change per mole along either pathway under the stated assumptions.

The tempting mistake is to treat a lower transition-state peak as a smaller overall energy release.

But the catalyst changes the barrier between the endpoints, not the difference between the endpoints themselves.

Question 5

A reaction proceeds through two elementary steps:

\[ \ce{R -> I -> P} \]

The energy profile has the following illustrative enthalpy levels:

Assume the same pathway operates in reverse, so the reverse reaction passes through TS₂ first, followed by TS₁.

A proposed change in reaction conditions stabilises the intermediate I by 30 kJ mol⁻¹. The enthalpies of R, P, TS₁, and TS₂ remain unchanged.

Answer the following:

a. Calculate the overall enthalpy change for R → P.

b. Calculate the activation energy of each forward elementary step before the change.

c. Calculate the activation energy of each reverse elementary step before the change.

d. After stabilising I, calculate all four elementary activation energies again.

e. Identify the largest local activation-energy barrier in each direction before and after the change.

f. Explain why making the intermediate more stable can actually make the second forward step harder, even though the overall reaction enthalpy remains unchanged.

Solution 5

The overall enthalpy change is \(-30\text{ kJ mol}^{-1}\). Before stabilising the intermediate, the forward barriers are 60 and 50 kJ mol⁻¹, while the reverse barriers are 100 and 40 kJ mol⁻¹. After stabilisation, the forward barriers become 60 and 80 kJ mol⁻¹, and the reverse barriers become 100 and 70 kJ mol⁻¹.

The important detail is that activation energy is measured from the starting level of each individual step, not from the original reactant level every time.

Part a: Overall enthalpy change

The overall reaction begins at R and finishes at P.

\[ \Delta H=H_P-H_R \]

\[ \Delta H=-30-0 \]

\[ \boxed{\Delta H=-30\text{ kJ mol}^{-1}} \]

The overall reaction is exothermic.

Part b: Forward activation energies before stabilisation

The first forward step is:

\[ \ce{R -> I} \]

Its activation energy is measured from R to TS₁:

\[ E_{a,1}=60-0 \]

\[ \boxed{E_{a,1}=60\text{ kJ mol}^{-1}} \]

The second forward step is:

\[ \ce{I -> P} \]

Its activation energy is measured from I to TS₂:

\[ E_{a,2}=70-20 \]

\[ \boxed{E_{a,2}=50\text{ kJ mol}^{-1}} \]

The first forward step has the larger local activation-energy barrier.

Notice that TS₂ is the highest point on the entire profile, but it does not have the largest forward activation-energy barrier.

Why?

Because the second step begins at the intermediate, not at the original reactants.

Part c: Reverse activation energies before stabilisation

The reverse reaction begins at P.

Its first step is:

\[ \ce{P -> I} \]

This passes through TS₂.

\[ E_{a,\mathrm{reverse},1} =70-(-30) \]

\[ \boxed{E_{a,\mathrm{reverse},1} =100\text{ kJ mol}^{-1}} \]

The second reverse step is:

\[ \ce{I -> R} \]

This passes through TS₁.

\[ E_{a,\mathrm{reverse},2} =60-20 \]

\[ \boxed{E_{a,\mathrm{reverse},2} =40\text{ kJ mol}^{-1}} \]

The first reverse step has the larger local activation-energy barrier.

Part d: Activation energies after stabilising I

Stabilising I by 30 kJ mol⁻¹ changes its enthalpy:

\[ H_I=20-30 \]

\[ H_I=-10\text{ kJ mol}^{-1} \]

All other energy levels remain unchanged.

For the forward reaction:

First step:

\[ E_{a,1}=60-0 \]

\[ \boxed{E_{a,1}=60\text{ kJ mol}^{-1}} \]

Second step:

\[ E_{a,2}=70-(-10) \]

\[ \boxed{E_{a,2}=80\text{ kJ mol}^{-1}} \]

For the reverse reaction:

First step:

\[ E_{a,\mathrm{reverse},1} =70-(-30) \]

\[ \boxed{E_{a,\mathrm{reverse},1} =100\text{ kJ mol}^{-1}} \]

Second step:

\[ E_{a,\mathrm{reverse},2} =60-(-10) \]

\[ \boxed{E_{a,\mathrm{reverse},2} =70\text{ kJ mol}^{-1}} \]

Part e: Compare the largest local barriers

Before stabilisation:

• Forward: first step, 60 kJ mol⁻¹.
• Reverse: first step, 100 kJ mol⁻¹.

After stabilisation:

• Forward: second step, 80 kJ mol⁻¹.
• Reverse: first step, 100 kJ mol⁻¹.

The largest forward barrier has shifted from the first step to the second step.

The largest reverse barrier remains unchanged.

These comparisons identify the largest local energy barriers, not exact overall reaction rates. Calculating rates for a multistep mechanism would require additional kinetic information.

Part f: Explain the effect of stabilising the intermediate

This is the central idea.

A more stable intermediate has a lower enthalpy.

That sounds as though the reaction should become easier, but consider what happens next.

The second forward step must now begin from an even lower energy level. Its transition-state energy has not changed.

Therefore, the energy gap between the intermediate and TS₂ becomes larger.

Before stabilisation:

\[ E_{a,2}=70-20=50 \]

After stabilisation:

\[ E_{a,2}=70-(-10)=80 \]

The second forward step has become harder, even though the intermediate is more stable.

The reverse second step also has a larger barrier because it starts from the same newly stabilised intermediate.

However, the first reverse step starts at P, and its transition state remains at TS₂. Neither of those energy levels changed, so its activation energy remains 100 kJ mol⁻¹.

Finally, the overall enthalpy change remains:

\[ \Delta H=-30-0 \]

\[ \boxed{\Delta H=-30\text{ kJ mol}^{-1}} \]

The initial and final states have not changed. Only the intermediate has moved.

This is the subtle distinction: changing an intermediate can alter the activation energies of individual steps without changing the overall enthalpy change.

The next time you interpret an energy profile, don’t automatically measure every activation energy from the original reactants. Identify the starting point of the particular reaction step first. That decision becomes essential when comparing multistep pathways, catalysts, and reverse reactions.