Why Exothermic Reactions Produce Lower-Energy Products

Mix hydrochloric acid and sodium hydroxide solution in an insulated cup. The temperature rises from \(22.0^\circ\text{C}\) to \(28.3^\circ\text{C}\). The products are now hotter than the reactants, so how can they have lower chemical energy?

Predict what happened to the energy during the reaction. Did the chemicals gain energy because their temperature increased, or did they release energy?

The reaction released energy. The products of an exothermic reaction have lower chemical energy because the new arrangement of atoms and their interactions is energetically lower than the original arrangement. The difference is transferred to the surroundings as heat.

There’s a subtle point here: the products have lower enthalpy when compared with the reactants at the same temperature and pressure. Their increased temperature after reacting is a separate effect. Chemical energy can decrease while the thermal energy of the mixture increases.

01Why lower chemical energy means energy has been released

Think about a ball rolling from a high shelf to a lower shelf. When it loses height, its gravitational potential energy decreases. That energy doesn’t vanish. It can become motion, sound, or heat.

Chemical reactions involve a similar energy change, except the relevant energy comes from interactions between particles rather than their height above the ground.

Atoms are held together by electrical attractions involving nuclei and electrons. Breaking chemical bonds requires energy, while forming bonds releases energy.

During an exothermic reaction, the energy released as new bonds and other stabilising interactions form is greater than the energy required to disrupt the original interactions.

The products therefore occupy a lower-energy arrangement.

The shelf analogy has a limit. Molecules aren’t physically falling downwards, and chemical energy doesn’t depend on their position in a laboratory. It depends on their microscopic arrangement and interactions.

A molecular example: neutralisation

Consider the reaction:

\[ \ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)} \]

Hydrochloric acid supplies hydrogen ions, while sodium hydroxide supplies hydroxide ions.

The essential change is often written as:

\[ \ce{H+(aq) + OH-(aq) -> H2O(l)} \]

This is a useful shorthand. In aqueous solution, hydrogen ions are hydrated, so a more explicit representation is:

\[ \ce{H3O+(aq) + OH-(aq) -> 2H2O(l)} \]

As the particles rearrange and water forms, the overall chemical system moves to a lower-enthalpy state.

The excess energy is transferred into the surrounding solution and laboratory apparatus.

The sodium and chloride ions remain dissolved. They are spectator ions, not the particles responsible for the main neutralisation change.

This doesn’t mean every individual product bond must be stronger than every reactant bond. In aqueous reactions, interactions with water also contribute to the energy change.

The comparison concerns the total enthalpy of the chemical system, not one isolated bond.

The precise statement: reaction enthalpy

For reactions at constant pressure, the useful energy quantity is enthalpy, represented by \(H\).

The reaction enthalpy change is:

\[ \Delta H=H_P-H_R \]

where:

• \(\Delta H\) is the enthalpy change of the reaction.
• \(H_P\) is the total enthalpy of the products.
• \(H_R\) is the total enthalpy of the reactants.

Both sides must refer to the same amount of reaction, temperature, and pressure, with the physical states specified.

For an exothermic reaction:

\[ \Delta H<0 \]

This means \(H_P<H_R\).

For example, if a reaction has an enthalpy change of \(-45\ \text{kJ mol}^{-1}\), the products have 45 kJ less enthalpy per mole of reaction as written.

The negative sign describes the chemical system’s energy change. It doesn’t mean the surroundings lose energy.

We usually report molar reaction enthalpy in \(\text{kJ mol}^{-1}\). An experiment involving a particular amount of substance involves a corresponding total energy change in kJ or J.

One more limitation: enthalpy differences can be measured, but absolute enthalpy values depend on the chosen reference. An energy diagram showing reactants at zero is choosing a convenient reference, not claiming that the reactants contain no energy.

02How an experiment reveals the energy difference

If the products have lower enthalpy, how could we establish that experimentally without seeing the energy inside individual molecules?

We can measure the thermal energy transferred during the reaction.

The method is called calorimetry.

First design: mixing chemicals in an open beaker

Imagine placing hydrochloric acid in an ordinary glass beaker, measuring its temperature, and then adding sodium hydroxide solution.

A thermometer sits in the liquid, and the mixture is stirred.

If the reaction is exothermic, what should happen?

Provided the solutions begin at the same temperature and there are no significant competing thermal effects, the mixture should warm as the reaction releases energy.

A temperature rise would support the prediction.

But an open glass beaker has several problems:

• Some energy heats the glass rather than the solution.
• Some energy escapes into the surrounding air.
• If the solutions begin at different temperatures, simple mixing can change the temperature even without a reaction.
• A slow thermometer may miss the highest temperature.

The observed rise is useful evidence, but this design cannot capture all the released energy accurately.

Revised design: an insulated calorimeter

A better design uses a polystyrene cup with a lid.

The two solutions are measured separately, and their initial temperatures are recorded. They are then combined inside the cup.

A temperature probe passes through the lid and extends into the solution without touching the bottom. The liquid is stirred gently, and the highest temperature is recorded.

The energy transfer follows this sequence:

1. The acid and alkali react inside the solution.
2. The chemical rearrangement releases energy.
3. The liquid gains thermal energy.
4. Its temperature rises.
5. The probe measures that change.

The insulation and lid reduce energy losses. A control using non-reacting liquids can help identify temperature changes caused by mixing rather than chemical reaction.

The revised apparatus doesn’t eliminate every error. The cup and probe still absorb energy, and some heat can escape. However, it provides a better estimate of the reaction enthalpy.

03Why lower-energy products don’t mean the reaction has no energy barrier

Imagine a ball starting on a platform, rolling over a small hump, and then dropping to a lower platform.

Its final position is lower than its starting position, but it still had to get over the hump.

A chemical reaction can behave similarly.

The reactants must pass through a higher-energy arrangement before the products form. This temporary arrangement is called the transition state.

The energy barrier is associated with the activation energy, \(E_a\).

For a simple reaction pathway:

• The forward activation energy is measured from the reactants to the transition state.
• The overall enthalpy change is measured from the reactants to the products.
• The reverse activation energy is measured from the products to the transition state.

These are three different energy differences.

A common mistake is to think that because the products have lower energy, the reaction should happen immediately.

But even a strongly exothermic reaction can proceed slowly if its activation energy is high.

Reacting particles must have suitable energy and orientation to reach the transition state. At any particular temperature, only some collisions will meet these conditions.

The overall energy drop tells us how much enthalpy changes. The activation barrier helps determine how readily the reaction proceeds.

04What happens when an exothermic reaction runs backwards?

Suppose a reaction converts A into B and releases \(40\ \text{kJ mol}^{-1}\):

\[ \ce{A -> B} \qquad \Delta H=-40\ \text{kJ mol}^{-1} \]

The products B have lower enthalpy than A.

Now reverse the reaction:

\[ \ce{B -> A} \]

To restore the original higher-enthalpy chemical arrangement, the reverse transformation must absorb the same amount of energy under the same conditions.

Therefore:

\[ \Delta H_{\text{reverse}} =+40\ \text{kJ mol}^{-1} \]

The reverse process is endothermic.

This is not the same as saying that every molecule needs exactly \(40\ \text{kJ mol}^{-1}\) to react. That value is the overall enthalpy change, not the activation energy.

A reverse reaction can still have a substantial activation barrier above its starting products.

The forward and reverse barriers share the same peak

For a simple, single-barrier energy profile:

\[ E_{a,r}=E_{a,f}-\Delta H_f \]

Here, \(E_{a,r}\) and \(E_{a,f}\) are the reverse and forward activation energies, and \(\Delta H_f\) is the signed enthalpy change of the forward reaction.

For an exothermic forward reaction, \(\Delta H_f\) is negative, so the reverse activation energy is larger than the forward activation energy along the same pathway.

This doesn’t make the reverse reaction impossible. It means the energy requirements are different.

A reaction can also be reversible without containing equal amounts of reactants and products. At equilibrium, the forward and reverse reaction rates are equal, even though the activation energies may differ.

05The important contrast: a higher temperature isn’t proof of higher chemical energy

Return to the opening experiment.

The reactants initially had a temperature of \(22.0^\circ\text{C}\). After neutralisation, the mixture reached \(28.3^\circ\text{C}\).

There are two changes happening:

1. The chemical arrangement changes to a lower-enthalpy arrangement.
2. The thermal energy of the mixture increases, raising its temperature.

Those changes aren’t contradictory. They are connected by energy conservation.

Here’s a useful way to separate the evidence.

That final row matters when the starting solutions have different temperatures.

Imagine mixing a hot solution with a cooler solution. The cooler liquid can warm simply because energy transfers from the hotter liquid.

Even if a chemical change absorbs energy overall, the final mixture could still be warmer than the initially cooler solution.

To identify the reaction’s enthalpy change, you must account for what the temperature would have been from mixing alone.

There is another limitation to the phrase “products have lower energy”.

A sample of hot products may have greater total enthalpy than a much colder sample of reactants, depending on their temperatures and heat capacities.

The proper comparison for reaction enthalpy holds temperature and pressure constant.

In a well-insulated reaction vessel, some of the decrease in chemical enthalpy can remain in the material as increased thermal energy. Once the products cool back to the original temperature, that energy can be transferred to the surroundings.

This distinction becomes especially important when interpreting calorimetry experiments, temperature profiles, and reaction energy diagrams.

06Questions and solutions

Question 1

A hypothetical reaction has the following relative enthalpy values:

Calculate:

a) The reaction enthalpy.

b) The forward activation energy.

c) The reverse activation energy.

d) Whether the forward reaction is exothermic or endothermic.

Solution 1

The forward reaction is exothermic, with \(\Delta H=-40\ \text{kJ mol}^{-1}\), a forward activation energy of \(75\ \text{kJ mol}^{-1}\), and a reverse activation energy of \(115\ \text{kJ mol}^{-1}\).

a) Reaction enthalpy

\[ \begin{aligned} \Delta H&=H_P-H_R\\ &=-25-(+15)\\ &=-40\ \text{kJ mol}^{-1} \end{aligned} \]

The negative result means the products have lower enthalpy.

b) Forward activation energy

\[ \begin{aligned} E_{a,f}&=90-15\\ &=75\ \text{kJ mol}^{-1} \end{aligned} \]

c) Reverse activation energy

\[ \begin{aligned} E_{a,r}&=90-(-25)\\ &=115\ \text{kJ mol}^{-1} \end{aligned} \]

d) Classification

The reaction is exothermic because its products lie below its reactants.

The tempting error is to treat the forward activation energy as the reaction enthalpy. The activation energy measures the climb to the transition state, while \(\Delta H\) measures the difference between the starting and finishing levels.

Question 2

Two solutions react in an insulated calorimeter, but they begin at different temperatures.

The first solution has a mass of \(40.0\ \text{g}\) and a temperature of \(18.0^\circ\text{C}\).

The second has a mass of \(60.0\ \text{g}\) and a temperature of \(26.0^\circ\text{C}\).

After mixing and reacting, the final temperature is \(29.0^\circ\text{C}\).

The reaction produces \(0.0400\ \text{mol}\) of water.

Assume:

• Both solutions have a specific heat capacity of \(4.00\ \text{J g}^{-1}\text{K}^{-1}\).
• The reaction is the only chemical change.
• Heat losses and heat absorbed by the calorimeter are negligible.
• Both solutions would mix without an additional heat of mixing.
• Their heat capacities remain approximately constant.

a) Calculate the temperature expected if no reaction occurred.

b) Calculate the molar reaction enthalpy.

c) Explain why using \(18.0^\circ\text{C}\) as the initial temperature for the entire mixture would produce an incorrect result.

Solution 2

The reaction is exothermic, with an estimated molar enthalpy of \(-62.0\ \text{kJ mol}^{-1}\). The correct no-reaction mixing temperature is \(22.8^\circ\text{C}\), not \(18.0^\circ\text{C}\).

a) Temperature without reaction

Since the solutions have the same specific heat capacity, use the mass-weighted average:

\[ T_{\text{mix}} =\frac{m_1T_1+m_2T_2}{m_1+m_2} \]

Here, \(m_1\) and \(m_2\) are the individual masses, while \(T_1\) and \(T_2\) are their starting temperatures.

\[ \begin{aligned} T_{\text{mix}} &=\frac{40.0(18.0)+60.0(26.0)} {100.0}\\ &=22.8^\circ\text{C} \end{aligned} \]

Without reaction, the hotter solution would warm the cooler one until both reached \(22.8^\circ\text{C}\).

b) Reaction enthalpy

The temperature increase caused by the reaction is:

\[ \begin{aligned} \Delta T&=29.0-22.8\\ &=6.2^\circ\text{C} \end{aligned} \]

The total mass is \(100.0\ \text{g}\).

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=100.0(4.00)(6.2)\\ &=2480\ \text{J}\\ &=2.48\ \text{kJ} \end{aligned} \]

The chemical change therefore released approximately \(2.48\ \text{kJ}\).

\[ \begin{aligned} \Delta H&=\frac{-2.48}{0.0400}\\ &=-62.0\ \text{kJ mol}^{-1} \end{aligned} \]

The products have lower enthalpy than the reactants when compared at the same temperature and pressure.

c) Why the initial temperature matters

Using \(18.0^\circ\text{C}\) would give an apparent rise of \(11.0^\circ\text{C}\).

That would incorrectly attribute energy transferred from the initially warmer solution to the chemical reaction.

The governing constraint is conservation of energy across both starting solutions. Their different starting temperatures must be accounted for before isolating the reaction’s thermal contribution.

A final temperature higher than either starting temperature is strong evidence of additional heating in this idealised experiment, but the correct enthalpy calculation still requires the mixing baseline.

Question 3

A sealed, rigid container contains gases A and B at equilibrium.

They participate in a hypothetical reversible reaction:

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

The forward reaction has:

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

and:

\[ E_{a,f}=65\ \text{kJ mol}^{-1} \]

The temperature is rapidly increased from \(300\ \text{K}\) to \(330\ \text{K}\).

Assume a simple reversible pathway, no competing reactions, and that the gases remain gaseous.

a) Calculate the reverse activation energy.

b) Explain what happens to the forward and reverse reaction rates immediately after heating, before the concentrations have significantly changed.

c) Predict the direction of the subsequent equilibrium shift.

d) A student argues: “Because B is lower in energy than A, increasing temperature must produce more B.” Explain why this reasoning fails.

e) Would adding a catalyst at the new temperature restore the original equilibrium composition?

Solution 3

Heating initially increases both reaction rates, but the equilibrium shifts towards A, the reactant in the exothermic forward direction. Adding a catalyst does not restore the original equilibrium composition.

a) Reverse activation energy

\[ \begin{aligned} E_{a,r}&=E_{a,f}-\Delta H_f\\ &=65-(-35)\\ &=100\ \text{kJ mol}^{-1} \end{aligned} \]

The reverse barrier is greater because B starts at a lower enthalpy.

b) Immediate effect of heating

Increasing temperature raises the average kinetic energy of the gas particles and changes the distribution of their collision energies.

More collisions can overcome the activation barriers, so both reaction rates increase.

In a simple Arrhenius model with approximately constant pre-exponential factors, the reverse reaction has the greater activation energy and therefore experiences the greater proportional increase in rate.

Immediately before heating, the system was at equilibrium, so its forward and reverse rates were equal.

Immediately after heating, the reverse rate exceeds the forward rate.

c) Equilibrium shift

The reverse reaction is endothermic because:

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

Increasing temperature favours the endothermic direction.

The system shifts towards A, decreasing the equilibrium proportion of B.

At the new equilibrium, the forward and reverse rates become equal again.

d) Why lower product energy doesn’t settle the prediction

The student is treating lower enthalpy as though it guarantees that increasing temperature always favours those products.

But equilibrium depends on the competing forward and reverse processes under the new conditions.

The forward reaction releases heat, so increasing temperature favours the reverse, heat-absorbing direction.

A lower-enthalpy product isn’t automatically the species favoured by every temperature change.

e) Effect of a catalyst

A catalyst lowers the activation barriers of both directions through an alternative pathway.

It increases the rates at which equilibrium is approached, but it does not change the equilibrium composition at a fixed temperature.

The original composition cannot be restored simply by adding a catalyst.

The key misconception is to confuse reaction speed with the position of equilibrium. Equal forward and reverse rates at equilibrium do not imply equal activation energies.

Question 4

An idealised reaction occurs in two separate stages inside an insulated, constant-pressure calorimeter:

\[ \ce{A -> B -> C} \]

Initially, \(0.0100\ \text{mol}\) of A is present.

The first stage converts all A into B. The second stage then converts all B into C.

The temperature readings are:

The combined effective heat capacity of the calorimeter and its contents is \(100\ \text{J K}^{-1}\), assumed constant.

No heat enters or leaves the insulated apparatus. There are no phase changes or side reactions.

a) Determine the molar enthalpy change for each stage.

b) Determine the overall enthalpy change from A to C.

c) Rank A, B, and C from highest to lowest enthalpy at the same temperature.

d) A student claims that C must have higher enthalpy than A because the second stage cools the mixture. Assess the claim.

Solution 4

The overall reaction from A to C is exothermic by \(20\ \text{kJ mol}^{-1}\), even though the second stage is endothermic. The enthalpy order is A highest, C intermediate, and B lowest.

a) Enthalpy change of the first stage

The temperature increases by:

\[ \Delta T_1=27.0-20.0=7.0\ \text{K} \]

The thermal energy gained by the apparatus and its contents is:

\[ \begin{aligned} q_1&=C_{\text{total}}\Delta T_1\\ &=100(7.0)\\ &=700\ \text{J} \end{aligned} \]

Here, \(C_{\text{total}}\) is the combined effective heat capacity.

The chemical change released this energy:

\[ \begin{aligned} \Delta H_1&=\frac{-700}{0.0100}\\ &=-70000\ \text{J mol}^{-1}\\ &=-70\ \text{kJ mol}^{-1} \end{aligned} \]

The conversion from A to B is exothermic.

For the second stage:

\[ \Delta T_2=22.0-27.0=-5.0\ \text{K} \]

\[ \begin{aligned} q_2&=100(-5.0)\\ &=-500\ \text{J} \end{aligned} \]

The thermal energy of the mixture decreased. The chemical change absorbed the corresponding energy.

\[ \begin{aligned} \Delta H_2&=\frac{+500}{0.0100}\\ &=+50\ \text{kJ mol}^{-1} \end{aligned} \]

The conversion from B to C is endothermic.

b) Overall reaction enthalpy

The overall enthalpy change is the sum of the two stage changes:

\[ \begin{aligned} \Delta H_{\text{overall}} &=-70+50\\ &=-20\ \text{kJ mol}^{-1} \end{aligned} \]

A to C is exothermic overall.

This is consistent with the final temperature being \(2.0^\circ\text{C}\) higher than the starting temperature.

c) Ranking the chemical states

Set the enthalpy of A to zero as a reference.

After the first stage, B is \(70\ \text{kJ mol}^{-1}\) lower.

After the second stage, C is \(50\ \text{kJ mol}^{-1}\) higher than B.

Therefore, the relative enthalpies are:

So:

\[ H_A>H_C>H_B \]

These levels compare the substances at the same temperature and pressure.

d) Assessing the claim

The student’s claim is incorrect.

The second stage absorbs energy, raising the chemical enthalpy from B to C. However, B had already fallen much further below A during the first stage.

C can therefore have higher enthalpy than B while still having lower enthalpy than A.

The trap is to treat the sign of the last stage as the sign of the overall process.

Energy differences must be tracked from the original reactants all the way to the final products.

Question 5

Three identical calorimeters are used to investigate the same hypothetical reaction:

\[ \ce{A -> B} \]

At \(20.0^\circ\text{C}\), its molar reaction enthalpy is:

\[ \Delta H=-48.0\ \text{kJ mol}^{-1} \]

Each calorimeter initially contains \(1.00\ \text{mol}\) of A at \(20.0^\circ\text{C}\).

The combined effective heat capacity of each calorimeter and its contents is \(2.00\ \text{kJ K}^{-1}\).

Assume that this heat capacity remains constant, the reaction enthalpy changes negligibly over the temperature range, there are no phase changes or competing reactions, and the reaction can proceed as specified.

The three experiments differ:

• Cup I: Perfectly insulated. All the A converts to B, with no energy transferred outside the cup.
• Cup II: The reaction also goes to completion, but the cup is maintained at \(20.0^\circ\text{C}\) by a temperature-controlled water bath that removes energy.
• Cup III: Perfectly insulated. After two minutes, its temperature is \(35.0^\circ\text{C}\), but some A remains unreacted.

Two students offer competing explanations.

Student X: “The products in Cup I must have higher enthalpy than the original reactants because they are hotter. Therefore, the reaction cannot really be exothermic.”

Student Y: “Because \(\Delta H=-48.0\ \text{kJ mol}^{-1}\), the products in Cup I must have exactly 48.0 kJ less total enthalpy than the original reactants, even at their higher temperature.”

Answer the following:

a) Calculate the final temperature in Cup I.

b) Calculate the energy transferred to the water bath in Cup II.

c) Determine the amount of A converted into B in Cup III after two minutes.

d) Evaluate both students’ explanations. In your response, distinguish total enthalpy at different temperatures from reaction enthalpy at the same temperature.

e) Could a catalyst change the final temperature of Cup I after complete conversion under the stated assumptions? Explain.

Solution 5

The reaction is exothermic in all three experiments. Cup I reaches \(44.0^\circ\text{C}\), Cup II transfers \(48.0\ \text{kJ}\) to the bath, and Cup III has converted \(0.625\ \text{mol}\) of A. Neither student’s explanation correctly distinguishes chemical enthalpy from the total enthalpy of the warmer system.

The central issue is that reaction enthalpy compares different chemical states at the same temperature, while these experiments compare systems at different temperatures and with different amounts of thermal energy transferred externally.

a) Final temperature in Cup I

Complete conversion of \(1.00\ \text{mol}\) of A releases:

\[ q_{\text{released}} =1.00(48.0) =48.0\ \text{kJ} \]

Because Cup I is perfectly insulated, this energy remains inside the apparatus, increasing its temperature.

Using:

\[ q=C_{\text{total}}\Delta T \]

we obtain:

\[ \begin{aligned} 48.0&=2.00\Delta T\\ \Delta T&=24.0\ \text{K} \end{aligned} \]

Therefore:

\[ T_f=20.0+24.0=44.0^\circ\text{C} \]

Cup I ends at \(44.0^\circ\text{C}\).

Notice that no energy has escaped to the external surroundings, even though the chemical change is exothermic.

The chemical enthalpy decrease has become increased thermal energy within the calorimeter.

b) Energy transferred in Cup II

Cup II remains at \(20.0^\circ\text{C}\).

Since its temperature does not increase, the chemical energy released cannot accumulate as a temperature increase in the cup.

The water bath must remove it.

\[ q_{\text{bath}}=+48.0\ \text{kJ} \]

The bath gains \(48.0\ \text{kJ}\).

Both Cups I and II undergo exactly the same chemical transformation.

Both have the same reaction enthalpy.

The difference is where the released energy ends up.

Cup I stores it as increased thermal energy within the cup. Cup II transfers it to the external water bath.

c) Reaction extent in Cup III

Cup III warms from \(20.0^\circ\text{C}\) to \(35.0^\circ\text{C}\).

\[ \Delta T=35.0-20.0=15.0\ \text{K} \]

The increase in thermal energy is:

\[ \begin{aligned} q&=C_{\text{total}}\Delta T\\ &=2.00(15.0)\\ &=30.0\ \text{kJ} \end{aligned} \]

Each mole of A converted releases \(48.0\ \text{kJ}\).

The amount converted is:

\[ \begin{aligned} n_{\text{converted}} &=\frac{30.0}{48.0}\\ &=0.625\ \text{mol} \end{aligned} \]

The amount of A remaining is:

\[ \begin{aligned} n_{\text{remaining}} &=1.00-0.625\\ &=0.375\ \text{mol} \end{aligned} \]

Cup III’s smaller temperature rise doesn’t mean the reaction has a different molar enthalpy.

It means less A has reacted by the measurement time.

This conclusion depends on the assumption that no other process contributes to the temperature change.

d) Evaluating the competing explanations

Student X is incorrect because a warmer final mixture doesn’t imply that the chemical products have higher reaction enthalpy.

At the same temperature:

\[ H_B-H_A=-48.0\ \text{kJ} \]

for the one-mole conversion.

The products have lower enthalpy at equal temperature and pressure.

However, the products in Cup I are no longer at the starting temperature.

Student Y also overlooks the change in thermal energy.

Take the total enthalpy of the original cup and its contents at \(20.0^\circ\text{C}\) as zero.

At the same temperature, completely converting A to B lowers the total enthalpy by \(48.0\ \text{kJ}\).

But warming the resulting cup and contents from \(20.0^\circ\text{C}\) to \(44.0^\circ\text{C}\) increases their sensible enthalpy by:

\[ \begin{aligned} \Delta H_{\text{thermal}} &=2.00(24.0)\\ &=+48.0\ \text{kJ} \end{aligned} \]

The two changes cancel:

\[ \begin{aligned} \Delta H_{\text{total}} &=-48.0+48.0\\ &=0\ \text{kJ} \end{aligned} \]

Under these assumptions, the total enthalpy of the final hot system in Cup I equals that of the original cool system.

This is precisely what energy conservation requires for the insulated, constant-pressure apparatus.

There is no contradiction.

The products have lower enthalpy than the reactants when compared at \(20.0^\circ\text{C}\), but the increased temperature adds enough thermal enthalpy to account for the difference.

If the completed Cup I were subsequently cooled from \(44.0^\circ\text{C}\) back to \(20.0^\circ\text{C}\), it would release \(48.0\ \text{kJ}\) to the surroundings.

At that point, the reactants and products would be compared at the same temperature, and the negative enthalpy change would appear directly in the energy transferred to the surroundings.

The important correction is that reaction enthalpy is not the same comparison as the total enthalpy of two samples at different temperatures.

e) Effect of a catalyst

A catalyst would not change Cup I’s \(44.0^\circ\text{C}\) final temperature if the same amount of A converted completely to the same B under the stated assumptions.

It could make the reaction reach completion sooner by providing a pathway with lower activation energy.

However, the enthalpy change would remain:

\[ \Delta H=-48.0\ \text{kJ mol}^{-1} \]

The total energy released by complete conversion would still be \(48.0\ \text{kJ}\).

The final temperature would therefore remain \(44.0^\circ\text{C}\).

A catalyst could affect the temperature measured after a fixed short time by changing how much has reacted, as illustrated by Cup III. It cannot change the energy difference between identical starting and finishing chemical states.

This final comparison connects three separate ideas:

• Reaction enthalpy determines the energy difference between the specified chemical states at the same conditions.
• Activation energy influences how readily the transformation occurs.
• Calorimetry measures where the transferred energy goes and how much chemical change has taken place, provided the experimental assumptions are valid.

When you next interpret an energy profile, keep those three questions separate. If you’re given a forward activation energy and reaction enthalpy, the next useful calculation is the barrier for the reverse process. That relationship is developed further in How to Calculate Reverse Activation Energy Using Delta H.