How to Explain a Cold Reaction Mixture in HSC Chemistry

Two reactants are mixed at 23.0°C. The mixture starts reacting, but the thermometer falls to 19.0°C. Where has the heat gone? Has the reaction released energy, or absorbed it?

If the temperature drop is caused by the reaction, the reacting chemical system has absorbed heat from its surroundings – mainly the water in the mixture. The water loses thermal energy and becomes colder. This is an endothermic process, with a positive enthalpy change, \(\Delta H > 0\).

The important distinction is that the reaction absorbs energy overall, even though its particles must also overcome an activation energy barrier to react.

There is one complication. A cold mixture does not automatically prove that a particular chemical reaction is endothermic. Dissolving a solid, mixing liquids at different initial temperatures, and evaporation can also affect the reading.

To explain the observation properly, you need to identify the system, track where energy moves, and decide what the measurement actually establishes.

01Identify the system and surroundings

Imagine two dissolved substances reacting in a cup of water.

For a simple energy calculation, we choose the reacting chemical species and their transformation into products as the system. The surrounding water, cup, temperature probe, and external air form the surroundings.

This boundary is a choice for our model, not a physical wall separating the reacting particles from the water.

If the chemical process is endothermic, energy is transferred from the surrounding solution into the chemical system. The solution cools because it loses thermal energy.

The sign convention

Heat transferred into the chosen system is positive. Heat transferred out is negative.

Here, \(q_{\mathrm{sol}}\) is the heat gained or lost by the surrounding solution, and \(q_{\mathrm{rxn}}\) is the heat absorbed or released by the reacting process. Both are measured in joules (J) or kilojoules (kJ).

Assuming negligible heat exchange with the air and cup, and no other significant energy transfers:

\[ q_{\mathrm{rxn}} \approx -q_{\mathrm{sol}} \]

The negative sign comes from conservation of energy. Energy lost by the solution is gained by the chemical process.

For a reaction at approximately constant pressure, the heat absorbed or released by the reaction gives its enthalpy change. When expressed per mole of reaction as written, this is \(\Delta H\), usually in \(\mathrm{kJ\,mol^{-1}}\).

An endothermic forward reaction has \(\Delta H>0\), meaning its products have higher enthalpy than its reactants under the stated conditions.

What if the whole mixture is called the system?

You can choose the entire cup and its contents as the system instead.

In a well-insulated cup, that larger system exchanges approximately no heat with the outside world. Yet its temperature can still fall. Thermal energy within the mixture is being converted into increased chemical enthalpy.

That is why writing “the system loses heat because its temperature falls” can be wrong. It depends on which system you have defined.

For the calculations in this lesson, we use the chemical process as the system and the bulk solution as its immediate surroundings.

02How an experiment can test the explanation

Consider a simple investigation involving two hypothetical aqueous reactants, R and S, which form a dissolved product without releasing gas.

The aim is to determine whether the chemical change causes the temperature to fall.

All numerical readings below are illustrative data, not measurements from a real experiment.

First design: an open beaker

Place the first reactant solution in a glass beaker, insert a temperature probe, and record its temperature. Add the second solution, stir, and measure the temperature as the reaction proceeds.

Suppose the temperature falls from 23.0°C to 19.0°C.

Before revealing the result, what would we predict for an exothermic reaction?

If the reaction released heat into the surrounding water faster than that heat escaped elsewhere, the water would warm. The observed cooling therefore supports an endothermic explanation.

But this first design has a weakness. The thermometer measures the combined effects of the reaction, mixing, and heat exchange with the environment.

If one reagent started colder than the other, the temperature could fall even without an endothermic reaction. If the mixture were open to air, evaporation could also contribute.

The first observation is evidence, but it is not enough to isolate the reaction’s enthalpy change.

Revised design: an insulated cup and a control

Use two nested insulated cups as a simple calorimeter. Fit a lid with a small hole for the temperature probe.

Measure both aqueous reactants separately, ensuring they start at 23.0°C. Pour them into the cup, replace the lid, stir gently, and record the temperature over time.

Next, perform a control experiment using comparable non-reacting solutions with the same initial temperatures, total mass, and approximately the same heat capacity.

The control tests how much the temperature changes from mixing and the apparatus alone.

What the temperature readings could show

The reacting mixture falls by 4.0°C during the first 60 seconds. The control remains close to its initial temperature.

This supports the conclusion that the chemical process is responsible for most of the cooling.

Notice that the reacting mixture begins warming again after 60 seconds.

That does not mean the chemical reaction has suddenly become exothermic. Once the mixture is colder than its surroundings, heat can enter through the cup walls and lid.

The recorded minimum is influenced by both the reaction and heat transfer from the room.

The revised design reduces these effects but does not eliminate every source of error. The cup and probe absorb heat, insulation is imperfect, and a different reaction might involve dissolving solids or escaping gases.

For an accurate reaction enthalpy, those additional processes may need separate measurements or corrections.

03Calculate the heat absorbed by a cold reaction mixture

For a dilute aqueous solution, we can estimate its heat change using:

\[ q_{\mathrm{sol}}=mc\Delta T \]

where:

• \(q_{\mathrm{sol}}\) is the heat change of the solution, in J.
• \(m\) is the solution mass, in g.
• \(c\) is the specific heat capacity, in \(\mathrm{J\,g^{-1}\,^\circ C^{-1}}\).
• \(\Delta T\) is the final temperature minus the initial temperature, in °C.

For dilute aqueous solutions, we often approximate \(c\) as \(4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\), the specific heat capacity of water.

This is an approximation, not an exact value for every solution.

04Why an endothermic reaction still needs activation energy

Suppose our reaction cools the solution, so its products have higher enthalpy than its reactants.

Does that mean the activation energy equals the heat absorbed?

No. They describe different energy differences.

The reaction enthalpy, \(\Delta H\), compares the starting and finishing enthalpies:

\[ \Delta H=H_P-H_R \]

Here \(H_P\) and \(H_R\) are the product and reactant enthalpies, respectively.

Activation energy, \(E_a\), describes the energy barrier associated with the reaction pathway. On a simple energy profile, it is the difference between the reactant level and the highest-energy point along the pathway.

For the forward reaction:

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

For the reverse reaction along the same single-peak pathway:

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

\(H^\ddagger\) represents the energy level at the pathway’s peak, while \(E_{a,f}\) and \(E_{a,r}\) are the forward and reverse activation energies.

The important consequences are:

• An endothermic forward reaction has products at higher enthalpy than its reactants.
• Its reverse reaction is exothermic.
• Both directions may have activation energy barriers.
• For the same single-peak pathway, the forward barrier is larger than the reverse barrier when the forward reaction is endothermic.

Activation energy helps explain how readily reactions occur and how their rates respond to conditions. Reaction enthalpy describes the overall energy change between specified reactants and products.

05A contrasting case: cold to touch, but exothermic

Imagine two reactant solutions that have been stored in a refrigerator.

Both are initially at 11.0°C. The room is at 22.0°C.

After mixing them in an insulated cup, the solution reaches 15.0°C.

The mixture would still feel cold to your hand. But did the reaction absorb or release heat?

It released heat, assuming the temperature rise was caused by the chemical process.

The relevant comparison is not between the final mixture and room temperature. It is between the actual mixture and the temperature it would have reached without the reaction.

Here:

\[ \Delta T=15.0-11.0=+4.0^\circ\mathrm C \]

The surrounding solution gained heat, so the chemical process released it. The reaction is exothermic, with \(\Delta H<0\).

What if the reactants begin at different temperatures?

Suppose one solution begins at 30°C and another at 20°C.

Even without a reaction, mixing them can change their individual temperatures as they exchange heat.

If the two solutions have the same specific heat capacity and no heat escapes, their expected mixing temperature is:

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

Here \(m_1\) and \(m_2\) are the masses of the two solutions, while \(T_1\) and \(T_2\) are their initial temperatures.

A measured temperature below this expected mixing temperature supports an endothermic contribution, not necessarily a temperature below both initial temperatures.

This is why a well-matched non-reacting control is so useful.

06A reliable method for explaining a cold reaction mixture

In an HSC Chemistry response, build the explanation in this order:

1. Describe the observation. State the measured temperature change relative to the appropriate starting or control temperature.
2. Define the system. Identify the reacting chemical species as the system and the surrounding solution as the heat source or sink.
3. Explain the transfer. If the solution cools because of the reaction, thermal energy has moved from the surroundings into the reacting system.
4. State the enthalpy sign. The chemical process is endothermic, so \(\Delta H>0\).
5. Check other effects. Ask whether different starting temperatures, dissolution, evaporation, incomplete reaction, or heat exchange could affect the inference.

A concise explanation for the opening scenario would be:

“The reaction mixture cooled relative to its initial temperature because the net chemical process absorbed heat from the surrounding solution. The solution therefore lost thermal energy. Assuming the temperature change is attributable to the reaction and heat exchange with the environment is negligible, the reaction is endothermic and has a positive enthalpy change.”

You would need a separate argument to explain how rapidly the cooling occurs.

A catalyst can lower activation energy and change the reaction’s rate without changing the enthalpy difference between the same reactants and products. That distinction becomes important when interpreting temperature-versus-time data.

07Questions and solutions

The first question tests the basic sign convention. The later questions require you to separate thermal effects, activation energies, and competing explanations.

Question 1

Two aqueous reactants are mixed in an insulated cup. Both initially have a temperature of 23.0°C. The final solution mass is 80.0 g, and the temperature decreases to 20.0°C.

Assume the specific heat capacity is \(4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\), the reaction occurs at approximately constant pressure, and heat transfer to the cup and room is negligible.

Calculate the heat absorbed or released by the chemical process and identify whether it is endothermic or exothermic.

Solution 1

The reaction is endothermic and absorbs approximately 1.00 kJ of heat from the surrounding solution.

The temperature change is:

\[ \begin{aligned} \Delta T&=20.0-23.0\\ &=-3.0^\circ\mathrm C \end{aligned} \]

Using the solution mass and specific heat capacity:

\[ \begin{aligned} q_{\mathrm{sol}}&=mc\Delta T\\ &=(80.0)(4.18)(-3.0)\\ &=-1003.2\ \mathrm J \end{aligned} \]

The solution loses 1003.2 J of thermal energy.

Therefore:

\[ \begin{aligned} q_{\mathrm{rxn}}&\approx-q_{\mathrm{sol}}\\ &=+1003.2\ \mathrm J\\ &\approx+1.00\ \mathrm{kJ} \end{aligned} \]

The positive heat change of the reacting system means it absorbs energy. The negative value calculated using \(mc\Delta T\) belongs to the surrounding solution.

The common mistake is to see the negative temperature change and assign a negative enthalpy change to the reaction. That reverses the energy transfer.

Question 2

A student investigates a reaction using two solutions with different initial temperatures:

• Solution X: 40.0 g at 30.0°C.
• Solution Y: 60.0 g at 20.0°C.

The solutions have the same specific heat capacity, \(4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\).

A control trial using comparable non-reacting liquids reaches 24.0°C after mixing.

In the reacting trial, the final temperature is 21.0°C. Chemical analysis confirms that 0.0150 mol of reaction as written has occurred.

Assume both trials have the same total heat capacity, negligible external heat exchange, and no other significant thermal effects.

Calculate the reaction’s molar enthalpy change. Explain why comparing 21.0°C directly with 20.0°C would produce the wrong conclusion.

Solution 2

The reaction is endothermic, with an estimated molar enthalpy change of \(+83.6\ \mathrm{kJ\,mol^{-1}}\).

The reacting mixture reaches 21.0°C, but the non-reacting control reaches 24.0°C. That is the correct reference temperature.

Step 1: Establish the temperature change attributable to the reaction.

\[ \begin{aligned} \Delta T_{\mathrm{rxn}}&=21.0-24.0\\ &=-3.0^\circ\mathrm C \end{aligned} \]

The solution is 3.0°C colder than expected from mixing alone.

Step 2: Calculate the solution’s heat change attributable to reaction.

The combined mass is \(40.0+60.0=100.0\) g.

\[ \begin{aligned} q_{\mathrm{sol}}&=mc\Delta T_{\mathrm{rxn}}\\ &=(100.0)(4.18)(-3.0)\\ &=-1254\ \mathrm J\\ &=-1.254\ \mathrm{kJ} \end{aligned} \]

Step 3: Find the reaction enthalpy.

The reacting system absorbs approximately 1.254 kJ.

\[ \begin{aligned} \Delta H&\approx\frac{1.254}{0.0150}\\ &=+83.6\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The reaction absorbs approximately 83.6 kJ per mole of reaction as written.

Step 4: Explain the misleading comparison.

A student might notice that 21.0°C is warmer than the initially cooler solution at 20.0°C and conclude that the reaction is exothermic.

However, heat would have transferred from the 30.0°C solution to the 20.0°C solution even without a reaction. Because the masses and heat capacities differ, the expected no-reaction temperature is 24.0°C, as the control confirms.

The reaction has caused a decrease relative to that expected value. It is therefore endothermic.

The initial temperatures must be accounted for before interpreting the chemical heat change.

Question 3

Two identical cups contain the same amount of an endothermic reaction mixture. Each begins at 23.0°C in a room also at 23.0°C.

Cup A contains a catalyst. Cup B does not.

Both reactions eventually convert the same amount of reactants into the same products, with the same physical states. Assume the catalyst produces no net chemical change and its addition has negligible thermal effect.

The following temperature minima are observed:

The cups are insulated but not perfectly.

A student claims that the catalyst makes the reaction more endothermic because Cup A reaches a lower temperature.

Evaluate this claim. Explain how both temperature measurements could be consistent with the same reaction enthalpy, and describe a way to test that explanation experimentally.

Solution 3

The student’s claim is incorrect. The deeper temperature drop in Cup A does not show that the catalyst changes the reaction enthalpy.

A catalyst provides an alternative reaction pathway with a lower activation energy. It can increase the rate at which the reaction proceeds, but it does not change the enthalpy difference between the same initial reactants and final products.

The two cups can show different minimum temperatures because their reactions happen over different time intervals.

In Cup A, the endothermic reaction absorbs energy rapidly. There is little time for heat from the room to pass through the cup walls before the reaction has largely finished. A relatively deep temperature drop is possible.

In Cup B, the same endothermic process occurs more slowly. As the solution cools below room temperature, it starts gaining heat from the air through the cup walls. That incoming heat partly offsets the energy absorbed by the reaction.

Cup B can therefore have a smaller measured temperature drop even though its reaction has the same overall enthalpy change.

The key is that a temperature probe records the net thermal effect at a particular time, not the total reaction enthalpy directly.

To test the explanation, repeat both trials using better-insulated calorimeters with calibrated heat capacities. Record the full temperature-versus-time curves and estimate or correct for heat transfer through the apparatus.

If both reactions proceed to the same extent and the appropriate corrections are made, their calculated molar enthalpy changes should agree within experimental uncertainty.

A tempting mistake is to assume that a larger temperature drop always means more energy has been absorbed by the chemical reaction. That is only a reliable comparison when the amounts reacted, heat capacities, and heat-transfer conditions are appropriately controlled.

Question 4

A reversible reaction is represented by:

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

Its single-peak energy profile has these relative energy levels:

Two otherwise identical insulated calorimeters are prepared at 22.0°C. Each contains 80.0 g of solution with a specific heat capacity of \(4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\).

• In Cup A, chemical analysis shows a net conversion of 0.0100 mol of R into P.
• In Cup B, chemical analysis shows a net conversion of 0.0150 mol of P into R.

Assume the given enthalpy changes remain approximately constant over the temperature range, the reactions occur at constant pressure, and other heat transfers are negligible.

Calculate the expected final temperature in each cup, and determine the activation energy for each direction.

A student also argues that the reverse reaction must always dominate because it has the smaller activation energy. Explain why this conclusion does not follow.

Solution 4

Cup A cools to approximately 20.8°C, Cup B warms to approximately 23.8°C, and the forward and reverse activation energies are 100 and 60 kJ/mol, respectively.

Both thermal outcomes are possible for the same reversible reaction because the cups undergo different net reaction directions.

Step 1: Determine the forward enthalpy change.

\[ \begin{aligned} \Delta H_f&=H_P-H_R\\ &=40-0\\ &=+40\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The forward conversion from R to P is endothermic.

The reverse conversion from P to R is therefore exothermic:

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

Step 2: Calculate the final temperature in Cup A.

Cup A undergoes 0.0100 mol of net forward reaction.

The chemical process absorbs:

\[ \begin{aligned} q_{\mathrm{rxn,A}}&=(0.0100)(40)\\ &=+0.400\ \mathrm{kJ}\\ &=+400\ \mathrm J \end{aligned} \]

The solution loses 400 J, so \(q_{\mathrm{sol,A}}=-400\ \mathrm J\).

Using \(q_{\mathrm{sol}}=mc\Delta T\):

\[ \begin{aligned} \Delta T_A&=\frac{-400}{(80.0)(4.18)}\\ &=-1.20^\circ\mathrm C \end{aligned} \]

Therefore:

\[ T_{f,A}=22.0-1.20=20.8^\circ\mathrm C \]

Cup A cools because the net reaction is endothermic.

Step 3: Calculate the final temperature in Cup B.

Cup B undergoes 0.0150 mol of net reverse reaction.

\[ \begin{aligned} q_{\mathrm{rxn,B}}&=(0.0150)(-40)\\ &=-0.600\ \mathrm{kJ}\\ &=-600\ \mathrm J \end{aligned} \]

The surrounding solution gains 600 J.

\[ \begin{aligned} \Delta T_B&=\frac{+600}{(80.0)(4.18)}\\ &=+1.79^\circ\mathrm C \end{aligned} \]

Therefore:

\[ T_{f,B}=22.0+1.79=23.8^\circ\mathrm C \]

Cup B warms because the net reaction is exothermic.

Step 4: Calculate both activation energies.

For the forward reaction:

\[ \begin{aligned} E_{a,f}&=100-0\\ &=100\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

For the reverse reaction:

\[ \begin{aligned} E_{a,r}&=100-40\\ &=60\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Step 5: Evaluate the student’s claim.

The reverse activation energy is smaller, but that alone does not tell us which direction dominates.

A reversible reaction has both forward and reverse processes occurring. Their rates depend on concentrations and other kinetic conditions, not just the relative heights of the energy barriers.

Cup A and Cup B have different compositions, and chemical analysis tells us their actual net directions.

A lower reverse barrier does not force every mixture to undergo net reverse reaction. Nor does an endothermic forward reaction need to have an exothermic temperature response.

The crucial distinction is between the net direction in which the chemicals are changing and the activation barriers for the possible pathways.

Question 5

Advanced challenge: this question uses multistep enthalpy accounting associated with Hess’s law in the NSW Chemistry Stage 6 Syllabus (2017). It extends the usual calorimetry question by separating two energy changes hidden inside one temperature reading.

A hypothetical solid A is added to an aqueous solution containing B.

Two processes occur in sequence:

First, A dissolves:

\[ \ce{A(s) -> A(aq)} \]

Then dissolved A reacts with B:

\[ \ce{A(aq) + B(aq) -> C(aq)} \]

A student observes that the complete mixture becomes colder and concludes that the second, chemical reaction must be endothermic.

To investigate, two calorimetry trials are performed. Both begin at 25.0°C.

In both trials, the total solution mass used for the heat calculation is 100.0 g, with \(c=4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\).

Assume the dissolution enthalpy of A is identical in both trials, there are no other significant mixing or thermal effects, the cup and room exchange negligible heat with the solution, and the changes occur at approximately constant pressure.

The aqueous chemical reaction has a forward activation energy of 70.0 kJ/mol along a single-peak pathway.

Determine:

a. The molar enthalpy change of the complete forward process.

b. The molar enthalpy change of the aqueous chemical reaction alone.

c. The reverse activation energy of the aqueous chemical reaction.

d. Whether the solution would warm or cool if the entire sequence were reversed to completion in an otherwise identical insulated setup starting at 25.0°C. Explain how this can differ from the thermal effect of reversing the aqueous chemical reaction alone.

For part (d), assume both reverse steps can occur completely, with no additional heat transfer or external work, and neglect the temperature dependence of enthalpy and heat capacity.

Solution 5

The complete forward process is endothermic, at \(+20.9\ \mathrm{kJ\,mol^{-1}}\), but the aqueous chemical reaction alone is exothermic, at \(-41.8\ \mathrm{kJ\,mol^{-1}}\). The reverse aqueous reaction has an activation energy of \(111.8\ \mathrm{kJ\,mol^{-1}}\), and reversing the complete process would warm the solution by about 1.0°C under the stated assumptions.

The surprising result comes from separating dissolution from the chemical reaction. The thermometer measures their combined thermal effect, not the enthalpy change of the aqueous reaction by itself.

Part (a): Complete forward process

The reacting trial begins at 25.0°C and finishes at 24.0°C.

\[ \Delta T_{\mathrm{total}}=-1.0^\circ\mathrm C \]

The heat lost by the solution is:

\[ \begin{aligned} q_{\mathrm{sol}}&=mc\Delta T\\ &=(100.0)(4.18)(-1.0)\\ &=-418\ \mathrm J \end{aligned} \]

The combined chemical and dissolution processes absorb 418 J, or 0.418 kJ.

Since 0.0200 mol of A undergoes the complete sequence:

\[ \begin{aligned} \Delta H_{\mathrm{total}} &=\frac{+0.418}{0.0200}\\ &=+20.9\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The complete forward process is endothermic.

That conclusion is correct, but it does not determine the sign of each individual step.

Part (b): Aqueous chemical reaction alone

The control experiment isolates dissolution.

Its temperature change is:

\[ \Delta T_{\mathrm{diss}}=-3.0^\circ\mathrm C \]

Therefore:

\[ \begin{aligned} q_{\mathrm{sol,diss}} &=(100.0)(4.18)(-3.0)\\ &=-1254\ \mathrm J \end{aligned} \]

Dissolution absorbs 1.254 kJ.

Its molar enthalpy change is:

\[ \begin{aligned} \Delta H_{\mathrm{diss}} &=\frac{+1.254}{0.0200}\\ &=+62.7\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The complete process consists of dissolution followed by aqueous reaction.

By conservation of energy, the enthalpy changes of the two steps add:

\[ \Delta H_{\mathrm{total}} = \Delta H_{\mathrm{diss}} + \Delta H_{\mathrm{chem}} \]

Here \(\Delta H_{\mathrm{chem}}\) is the enthalpy change of the aqueous chemical reaction alone.

Rearranging:

\[ \begin{aligned} \Delta H_{\mathrm{chem}} &=20.9-62.7\\ &=-41.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The aqueous reaction is exothermic.

It releases 41.8 kJ per mole, but the preceding dissolution absorbs 62.7 kJ per mole. Overall, more energy is absorbed than released, so the mixture cools.

The tempting assumption was that the entire observed temperature change came from the aqueous reaction. The control experiment shows why that assumption fails.

Part (c): Reverse activation energy

The forward aqueous reaction is exothermic, with:

\[ \Delta H_{\mathrm{chem}}=-41.8\ \mathrm{kJ\,mol^{-1}} \]

For the same single-peak reaction pathway:

\[ E_{a,r}=E_{a,f}-\Delta H_{\mathrm{chem}} \]

Substituting:

\[ \begin{aligned} E_{a,r}&=70.0-(-41.8)\\ &=111.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The reverse activation energy is larger than the forward activation energy.

This makes sense because the forward aqueous reaction releases energy, leaving C at a lower enthalpy than A(aq) and B(aq).

Reversing it requires climbing from that lower product level to the pathway’s peak.

The measured cooling of the complete mixture does not change this barrier relationship for the individual aqueous reaction.

Part (d): Reverse the entire process

For the forward sequence, the two enthalpy changes are:

Reversing each individual step changes the sign of its enthalpy change.

The reverse aqueous reaction, in which C forms A(aq) and B(aq), absorbs 41.8 kJ/mol.

However, the reverse dissolution step, in which dissolved A returns to solid A, releases 62.7 kJ/mol under the stated assumptions.

The complete reverse process therefore has:

\[ \begin{aligned} \Delta H_{\mathrm{reverse}} &=+41.8-62.7\\ &=-20.9\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

It is exothermic overall, even though the reverse aqueous chemical reaction is endothermic.

For 0.0200 mol of complete reverse process, the heat released is:

\[ \begin{aligned} q_{\mathrm{reverse}} &=(0.0200)(-20.9)\\ &=-0.418\ \mathrm{kJ} \end{aligned} \]

The surrounding solution therefore gains 418 J.

\[ \begin{aligned} \Delta T&=\frac{+418}{(100.0)(4.18)}\\ &=+1.0^\circ\mathrm C \end{aligned} \]

Starting at 25.0°C, the final solution temperature would be approximately 26.0°C.

There is one final limitation. Knowing a reaction’s enthalpy and activation energy does not establish whether its reverse will occur spontaneously, or how quickly. That requires information about equilibrium, the reaction conditions, and kinetics. The predicted warming is conditional on the specified complete reverse conversion.

The broader lesson is that a temperature drop describes the net thermal effect of everything occurring inside the mixture. It does not automatically identify the enthalpy change of every individual process.

Once you’ve identified which process produces the temperature change, the next decision is whether you’re explaining how much energy is transferred or how quickly the reaction occurs. For the second question, use reaction-rate reasoning. To connect rate changes to the correct energy representation, see how to draw a catalysed pathway on an energy diagram.