How to Find an Unknown Enthalpy by Cancelling Intermediate Species

Imagine you have a sample of white, anhydrous copper(II) sulfate and want to calculate the enthalpy change when it becomes blue copper(II) sulfate pentahydrate.

The reaction is:

\[ \ce{CuSO4(s) + 5H2O(l) -> CuSO4.5H2O(s)} \]

You could try to measure the heat released during hydration directly. However, there is a more useful approach: measure the enthalpy changes for dissolving each solid separately, then combine the two complete thermochemical equations.

Both dissolutions produce the same dissolved ions. When the equations are added in the correct directions, those ions cancel, leaving the hydration reaction and its unknown enthalpy change.

This is the central technique behind Hess’s law: find an unknown reaction enthalpy by combining known reaction enthalpies and cancelling intermediate species.

01Why can intermediate species be cancelled?

Consider two reactions:

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

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

If we add the equations, we obtain:

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

The substance B appears on both sides. It is produced in the first reaction and consumed in the second.

Cancelling it gives:

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

B is called an intermediate species because it exists during the reaction pathway but does not appear in the overall reaction.

The same principle works with chemical equations involving several substances, fractional coefficients, and different reaction pathways.

The connection to enthalpy

Enthalpy, represented by \(H\), is a state function.

This means that its value depends on the state of a chemical system, not the particular sequence of reactions used to reach that state.

For a reaction:

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

If you reach the same final state from the same initial state, the overall enthalpy change is the same, even if you take a different route.

Think about travelling from Sydney to Canberra. You might drive directly, or travel through another town. Your route changes, but your starting point and destination remain the same.

The analogy helps explain why intermediate steps do not change the overall enthalpy difference.

However, chemical systems have an extra complication: physical states and solution conditions matter. Liquid water and water vapour are not interchangeable, and aqueous solutions at different concentrations are not necessarily the same thermodynamic state.

This is why cancelling species requires more care than simply spotting matching chemical formulas.

02How Hess’s law developed from experiments

The idea has an experimental history.

In 1840, chemist Germain Henri Hess investigated the heat released during chemical reactions. One of the questions he explored was whether the total heat released when substances combined depended on the route taken.

His experiments involved calorimetry.

A calorimeter measures energy transfer by monitoring temperature changes. Hess used an apparatus containing a large quantity of water, glass vessels, and measured quantities of reactants.

He recorded the masses of the apparatus and reactants, along with the resulting temperature increases. He also accounted for the heat capacities of the different materials.

What was Hess trying to establish?

Consider a substance that can be converted into a final product either directly or through several intermediate reactions.

Hess’s prediction was that the total heat evolved should be the same for both pathways, provided the initial and final states were identical.

His observations supported this prediction. The total heat changes agreed, within the limitations of his measurements, even when the reaction pathways differed.

His 1840 paper explicitly concluded that the heat evolved for a chemical combination remained constant whether the combination occurred directly or through different steps.

There were also experimental limitations.

In one particularly difficult experiment involving sulfur trioxide and water, the heat generated was sufficient to shatter a glass tube. Hess recognised that estimating the amount of reactant from the recovered glass fragments was unreliable, so he used an independent chemical analysis to check the amount present.

His records also distinguish between heat capacities measured directly and those estimated approximately.

These details matter. The underlying principle was supported by experimental evidence, but the accuracy of individual measurements still depended on the apparatus and measurement methods.

Modern thermodynamics explains Hess’s law through the state-function nature of enthalpy.

You can read a transcription of Hess’s original 1840 thermochemical paper.

03Measuring the reactions before combining them

Let’s return to copper(II) sulfate.

We want to determine the enthalpy change for:

\[ \ce{CuSO4(s) + 5H2O(l) -> CuSO4.5H2O(s)} \]

Instead of trying to measure this reaction directly, we measure two dissolution reactions.

The first dissolves anhydrous copper(II) sulfate:

\[ \ce{CuSO4(s) -> Cu^2+(aq) + SO4^2-(aq)} \]

The second dissolves copper(II) sulfate pentahydrate:

\[ \ce{CuSO4.5H2O(s) -> Cu^2+(aq) + SO4^2-(aq) + 5H2O(l)} \]

Notice something important.

Both reactions produce the same aqueous ions. This gives us a common intermediate state through which we can connect the two reactions.

Setting up the calorimetry experiment

A simple school experiment uses a polystyrene cup calorimeter.

The cup sits inside a beaker for support. A lid sits over the cup, with a thermometer or temperature probe passing through it. A stirrer allows the solution to be mixed evenly.

The procedure is:

1. Measure a known mass of water into the cup.
2. Record the initial temperature.
3. Add an accurately weighed sample of the solid.
4. Stir until the solid dissolves.
5. Record the highest or lowest temperature reached.
6. Repeat with the other solid.

The temperature change tells us whether the dissolution is exothermic or endothermic.

• Temperature increases: the solution gains heat, so the reaction releases heat.
• Temperature decreases: the solution loses heat, so the reaction absorbs heat.

The heat transferred to or from the solution is calculated using:

\[ q=mc\Delta T \]

where:

• \(q\) is the heat transferred to the solution, in joules (J).
• \(m\) is the mass of the solution, in grams (g).
• \(c\) is the specific heat capacity, in \(\mathrm{J\,g^{-1}\,K^{-1}}\).
• \(\Delta T\) is the temperature change, in kelvin (K).

For this example, assume the solution has the same specific heat capacity as water:

\[ c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}} \]

At approximately constant atmospheric pressure, the heat transferred to the reaction is the opposite of the heat transferred to the solution:

\[ q_{\text{reaction}}=-q_{\text{solution}} \]

If the solution warms up, the reaction has released energy. If the solution cools down, the reaction has absorbed energy.

To calculate the molar enthalpy change:

\[ \Delta H_{\text{reaction}} = \frac{q_{\text{reaction}}}{n} \]

where \(n\) is the amount of the limiting reactant in moles.

The second dissolution measurement

Now consider the hydrated salt.

For an illustrative second experiment, use:

• \(12.49\text{ g}\) of \(\ce{CuSO4.5H2O}\), equivalent to \(0.0500\text{ mol}\).
• \(95.5\text{ g}\) of water.
• An initial temperature of \(22.0^\circ\text{C}\).
• A final temperature of \(20.7^\circ\text{C}\).

The solution cools by \(1.3\text{ K}\), indicating an endothermic dissolution.

The hydrate contains five moles of water per mole of copper(II) sulfate. Therefore, \(0.0500\text{ mol}\) of hydrate releases:

\[ \begin{aligned} n(\ce{H2O})&=5(0.0500)\\ &=0.250\text{ mol} \end{aligned} \]

This corresponds to:

\[ m=0.250\times18.0=4.50\text{ g} \]

The \(4.50\text{ g}\) of water released from the crystal joins the original \(95.5\text{ g}\) of water.

The final solvent mass is therefore \(100.0\text{ g}\), matching the first experiment.

This adjustment helps ensure that both dissolution reactions reach the same final solution composition.

The approximate total solution mass is \(108.0\text{ g}\).

\[ \begin{aligned} q_{\text{solution}} &=108.0\times4.18\times(-1.3)\\ &=-587\text{ J}\\ &=-0.587\text{ kJ} \end{aligned} \]

Therefore:

\[ q_{\text{reaction}}=+0.587\text{ kJ} \]

And:

\[ \begin{aligned} \Delta H_{\text{sol}} &=\frac{+0.587}{0.0500}\\ &=+11.7\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The dissolution of the hydrated salt is endothermic.

Our illustrative experimental data are now:

These are illustrative values constructed for this worked scenario, not claimed measurements from an actual experiment.

04How to cancel the intermediate species and find the unknown enthalpy

We now have the two known enthalpy changes.

Our target reaction is:

\[ \ce{CuSO4(s) + 5H2O(l) -> CuSO4.5H2O(s)} \]

Write the two supplied reactions again:

Reaction 1:

\[ \ce{CuSO4(s) -> Cu^2+(aq) + SO4^2-(aq)} \]

\[ \Delta H_1=-66.8\ \mathrm{kJ\,mol^{-1}} \]

Reaction 2:

\[ \ce{CuSO4.5H2O(s) -> Cu^2+(aq) + SO4^2-(aq) + 5H2O(l)} \]

\[ \Delta H_2=+11.7\ \mathrm{kJ\,mol^{-1}} \]

The problem is that Reaction 2 has the hydrated salt as a reactant. Our target needs it as a product.

We therefore reverse Reaction 2.

05What can and cannot be cancelled?

The most important rule is:

You can only cancel the same chemical species in the same physical state and, for solutions, the same thermodynamic state.

This means you need to check three things.

The common mistake: Adding enthalpies without reversing the reaction

In our copper sulfate example, a student might calculate:

\[ -66.8+11.7=-55.1 \]

This is incorrect.

Why?

The second reaction describes dissolving hydrated copper sulfate. But our target reaction requires forming hydrated copper sulfate.

The second equation must be reversed, so its enthalpy changes from \(+11.7\) to \(-11.7\).

The correct calculation is:

\[ -66.8+(-11.7)=-78.5 \]

Always manipulate the chemical equation first. Then make the corresponding adjustment to its enthalpy.

Another trap: Liquid water is not water vapour

Consider these two reactions:

Reaction 1:

\[ \ce{H2(g) + \frac{1}{2}O2(g) -> H2O(l)} \]

\[ \Delta H_1=-285.8\ \mathrm{kJ\,mol^{-1}} \]

Reaction 2:

\[ \ce{H2O(l) -> H2O(g)} \]

\[ \Delta H_2=+44.0\ \mathrm{kJ\,mol^{-1}} \]

Add the equations.

The liquid water cancels because it is produced in Reaction 1 and consumed in Reaction 2.

However, water vapour remains as a product:

\[ \ce{H2(g) + \frac{1}{2}O2(g) -> H2O(g)} \]

The overall enthalpy change is:

\[ \begin{aligned} \Delta H&=-285.8+44.0\\ &=-241.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The enthalpy change for producing water vapour is different from the enthalpy change for producing liquid water.

If you had cancelled liquid water against water vapour, you would have lost the enthalpy of vaporisation and obtained the wrong result.

The physical state symbols are part of the chemical equation. Never ignore them when constructing a Hess cycle.

06Measured enthalpies versus model estimates

There is another important distinction when working with unknown enthalpies.

Not every enthalpy value comes from the same type of evidence.

1. Experimentally measured enthalpies

In calorimetry, the temperature change is measured directly.

You then calculate the heat transferred using:

\[ q=mc\Delta T \]

The molar enthalpy change is calculated from the heat transferred and the number of moles reacted.

These values are experimental results. They depend on the quality of the apparatus and the assumptions used in the calculation.

For example, a simple polystyrene cup calorimeter may lose energy to the surrounding air. If an exothermic reaction transfers some heat to the surroundings rather than the measured solution, the observed temperature increase will be smaller.

The calculated enthalpy change will then be less negative than the actual value.

Other limitations include:

• Neglecting the heat capacity of the cup and thermometer.
• Assuming the solution has the same specific heat capacity as water.
• Incomplete dissolution of a solid.
• Delays in recording the maximum or minimum temperature.
• Differences in final solution concentrations.

2. Hess’s law calculations using measured data

A Hess’s law calculation can produce an enthalpy value for a reaction that is difficult to measure directly.

The result is an indirect experimental determination if the supplied reaction enthalpies come from experiments.

It is not a completely independent measurement. Any uncertainty in the original measurements carries into the final answer.

For example, our calculated hydration enthalpy of \(-78.5\ \mathrm{kJ\,mol^{-1}}\) depends on the accuracy of both dissolution enthalpies.

The mathematical manipulation can be perfectly correct while the final experimental value still contains uncertainty.

3. Model estimates using bond enthalpies

Bond enthalpy calculations work differently.

They estimate the energy change by comparing the energy required to break bonds with the energy released when new bonds form.

\[ \Delta H_{\text{estimate}} \approx \sum E_{\text{bonds broken}} - \sum E_{\text{bonds formed}} \]

For example, consider:

\[ \ce{H2(g) + Cl2(g) -> 2HCl(g)} \]

Suppose the supplied average bond enthalpies are:

Energy required to break the reactant bonds:

\[ E_{\text{broken}}=436+243=679 \]

Energy released when forming the product bonds:

\[ E_{\text{formed}}=2(432)=864 \]

Therefore:

\[ \begin{aligned} \Delta H_{\text{estimate}} &=679-864\\ &=-185\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

This is an estimate because average bond enthalpies are used rather than exact measurements of the individual molecular environments.

The distinction is important:

Hess’s law does not automatically make an estimate exact. The quality of the final result depends on the quality of the values supplied.

Also, ordinary average covalent bond enthalpies are not a suitable substitute for the lattice and solvation energy contributions involved in dissolving an ionic solid such as copper(II) sulfate.

For further practice with different types of supplied data, see How to Build a Hess Cycle from Combustion Data and How to Calculate Formation Enthalpy from Reaction Data.

07A reliable method for finding an unknown enthalpy

When you encounter a Hess’s law question, use the following sequence.

Remember the three mathematical operations:

A useful final check is to balance the overall equation again.

If the final equation contains an extra intermediate, an incorrect coefficient, or a missing physical state, the cancellation is not finished.

Syllabus note: Hess’s law is explicitly included in the NSW Chemistry Stage 6 Syllabus (2017), which covers stepped reaction enthalpies and energy cycles. NESA schedules the Chemistry 11-12 Syllabus (2025) for implementation from 2028, with the first HSC examination under that syllabus in 2029. This article follows the 2017 syllabus edition. Check the applicable course edition for your cohort. See NESA’s 2017 Chemistry syllabus and 2025 syllabus implementation timeline.

08Questions and solutions

The following questions progress from a straightforward equation reversal to more demanding problems involving multiple reaction steps, physical states, and experimental limitations.

Question 1

Sulfur dioxide can be formed from solid sulfur and oxygen.

The following standard enthalpy changes are supplied:

Reaction 1:

\[ \ce{S(s) + \frac{3}{2}O2(g) -> SO3(g)} \]

\[ \Delta H_1^\circ=-395.7\ \mathrm{kJ\,mol^{-1}} \]

Reaction 2:

\[ \ce{SO2(g) + \frac{1}{2}O2(g) -> SO3(g)} \]

\[ \Delta H_2^\circ=-98.9\ \mathrm{kJ\,mol^{-1}} \]

Calculate the standard enthalpy change for:

\[ \ce{S(s) + O2(g) -> SO2(g)} \]

Solution 1

The standard enthalpy change is \(-296.8\ \mathrm{kJ\,mol^{-1}}\).

Reaction 2 must be reversed because sulfur dioxide needs to be a product, not a reactant.

Step 1: Reverse Reaction 2.

\[ \ce{SO3(g) -> SO2(g) + \frac{1}{2}O2(g)} \]

The enthalpy sign changes:

\[ \Delta H_2^\circ=+98.9\ \mathrm{kJ\,mol^{-1}} \]

Step 2: Add the reversed reaction to Reaction 1.

\[ \begin{aligned} \ce{S(s) + \frac{3}{2}O2(g)} &\ce{-> SO3(g)}\\ \ce{SO3(g)} &\ce{-> SO2(g) + \frac{1}{2}O2(g)} \end{aligned} \]

The sulfur trioxide cancels.

Oxygen also appears on both sides. Cancelling half a mole leaves one mole of oxygen on the reactant side.

The overall equation becomes:

\[ \ce{S(s) + O2(g) -> SO2(g)} \]

Step 3: Add the enthalpy changes.

\[ \begin{aligned} \Delta H^\circ &=-395.7+98.9\\ &=-296.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The negative result indicates that forming sulfur dioxide from solid sulfur and oxygen is exothermic.

Question 2

Iron forms two different oxides, iron(II) oxide and iron(III) oxide.

The following standard enthalpy changes are supplied:

Reaction 1:

\[ \ce{2Fe(s) + \frac{3}{2}O2(g) -> Fe2O3(s)} \]

\[ \Delta H_1^\circ=-824.2\ \mathrm{kJ\,mol^{-1}} \]

Reaction 2:

\[ \ce{2FeO(s) + \frac{1}{2}O2(g) -> Fe2O3(s)} \]

\[ \Delta H_2^\circ=-280.2\ \mathrm{kJ\,mol^{-1}} \]

Calculate the standard enthalpy change for forming one mole of iron(II) oxide:

\[ \ce{Fe(s) + \frac{1}{2}O2(g) -> FeO(s)} \]

Explain why both supplied reactions must be adjusted before they can be combined.

Solution 2

The standard enthalpy change for forming iron(II) oxide is:

\[ \boxed{-272.0\ \mathrm{kJ\,mol^{-1}}} \]

Both supplied equations contain two moles of iron atoms or iron-containing formula units. Our target equation produces one mole of \(\ce{FeO}\), so we must halve both equations.

Step 1: Divide Reaction 1 by two.

\[ \ce{Fe(s) + \frac{3}{4}O2(g) -> \frac{1}{2}Fe2O3(s)} \]

Its enthalpy change also halves:

\[ \Delta H_1^\circ=\frac{-824.2}{2}=-412.1 \]

Step 2: Halve and reverse Reaction 2.

First, halving gives:

\[ \ce{FeO(s) + \frac{1}{4}O2(g) -> \frac{1}{2}Fe2O3(s)} \]

The enthalpy change becomes:

\[ -280.2\div2=-140.1 \]

However, we need iron(II) oxide as a product. Reverse the equation:

\[ \ce{\frac{1}{2}Fe2O3(s) -> FeO(s) + \frac{1}{4}O2(g)} \]

The enthalpy change is now:

\[ \Delta H_2^\circ=+140.1 \]

Step 3: Add the two adjusted equations.

\[ \begin{aligned} \ce{Fe(s)+\frac{3}{4}O2(g)} &\ce{-> \frac{1}{2}Fe2O3(s)}\\ \ce{\frac{1}{2}Fe2O3(s)} &\ce{-> FeO(s)+\frac{1}{4}O2(g)} \end{aligned} \]

The iron(III) oxide cancels.

Cancelling one-quarter mole of oxygen from both sides leaves:

\[ \ce{Fe(s)+\frac{1}{2}O2(g)->FeO(s)} \]

Step 4: Calculate the enthalpy change.

\[ \begin{aligned} \Delta H^\circ &=-412.1+140.1\\ &=-272.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The result is exothermic.

The important detail is that halving a reaction also halves its enthalpy change. Reversing it then changes the sign. Both adjustments are required to obtain the correct overall reaction.

Question 3

Ethene can react with water to form ethanol:

\[ \ce{C2H4(g) + H2O(l) -> C2H5OH(l)} \]

Directly determining the enthalpy change for this reaction is difficult.

Instead, a chemist supplies the following standard combustion enthalpies:

Reaction 1: Ethene combustion

\[ \ce{C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l)} \]

\[ \Delta H_1^\circ=-1411\ \mathrm{kJ\,mol^{-1}} \]

Reaction 2: Ethanol combustion

\[ \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \]

\[ \Delta H_2^\circ=-1367\ \mathrm{kJ\,mol^{-1}} \]

Calculate the standard enthalpy change for the formation of ethanol from ethene and water.

Explain which substances cancel when the equations are combined.

Solution 3

The standard enthalpy change for converting ethene into ethanol is:

\[ \boxed{-44\ \mathrm{kJ\,mol^{-1}}} \]

The key is to reverse the ethanol combustion equation so that ethanol becomes a product.

Step 1: Keep the ethene combustion reaction unchanged.

\[ \ce{C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l)} \]

\[ \Delta H_1^\circ=-1411 \]

Step 2: Reverse the ethanol combustion reaction.

\[ \ce{2CO2(g) + 3H2O(l) -> C2H5OH(l) + 3O2(g)} \]

The enthalpy change becomes:

\[ \Delta H_2^\circ=+1367 \]

Step 3: Add the equations and cancel intermediate species.

The following substances cancel:

• Two moles of \(\ce{CO2(g)}\).
• Three moles of \(\ce{O2(g)}\).
• Two moles of \(\ce{H2O(l)}\).

One mole of liquid water remains on the reactant side.

The resulting equation is:

\[ \ce{C2H4(g) + H2O(l) -> C2H5OH(l)} \]

This matches the target.

Step 4: Add the enthalpy changes.

\[ \begin{aligned} \Delta H^\circ &=-1411+1367\\ &=-44\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The negative result means that the hydration of ethene to form liquid ethanol is exothermic under the stated standard conditions.

Notice that water did not cancel completely. Three moles appeared on one side of the reversed reaction, while two appeared on the other side of the original reaction. Subtracting the coefficients leaves one mole of water as a reactant.

Question 4

A student is given the following reactions:

Reaction 1:

\[ \ce{H2(g) + \frac{1}{2}O2(g) -> H2O(l)} \]

\[ \Delta H_1^\circ=-285.8\ \mathrm{kJ\,mol^{-1}} \]

Reaction 2:

\[ \ce{H2O(l) -> H2O(g)} \]

\[ \Delta H_2^\circ=+44.0\ \mathrm{kJ\,mol^{-1}} \]

The student claims that liquid water and water vapour can be cancelled completely because they have the same chemical formula.

Evaluate this claim.

Calculate the standard enthalpy change for:

\[ \ce{H2(g) + \frac{1}{2}O2(g) -> H2O(g)} \]

Explain why the student’s proposed cancellation is incorrect.

Solution 4

The student’s claim is incorrect. The overall enthalpy change is:

\[ \boxed{-241.8\ \mathrm{kJ\,mol^{-1}}} \]

Liquid water and water vapour have different physical states and different enthalpies.

Step 1: Add the two supplied equations.

\[ \begin{aligned} \ce{H2(g)+\frac{1}{2}O2(g)} &\ce{-> H2O(l)}\\ \ce{H2O(l)} &\ce{-> H2O(g)} \end{aligned} \]

Step 2: Cancel the liquid water.

Liquid water is produced in Reaction 1 and consumed in Reaction 2.

It cancels, but water vapour remains as a product.

The overall reaction is:

\[ \ce{H2(g)+\frac{1}{2}O2(g)->H2O(g)} \]

Step 3: Add the enthalpy changes.

\[ \begin{aligned} \Delta H^\circ &=-285.8+44.0\\ &=-241.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The positive \(44.0\ \mathrm{kJ\,mol^{-1}}\) represents the energy required to convert liquid water into water vapour.

If the student cancelled both forms of water completely, this phase-change energy would be lost.

The lesson is that matching chemical formulas are not sufficient. Physical states must also match before species can be cancelled.

Question 5

Two laboratory groups are investigating the enthalpy change for hydrating anhydrous copper(II) sulfate:

\[ \ce{CuSO4(s) + 5H2O(l) -> CuSO4.5H2O(s)} \]

Both groups dissolve \(0.0500\text{ mol}\) of each copper sulfate solid.

However, they use different experimental procedures.

Group A uses:

• \(7.98\text{ g}\) of anhydrous \(\ce{CuSO4}\).
• \(100.0\text{ g}\) of water.
• A measured temperature increase of \(7.40\text{ K}\).

Group B initially uses:

• \(12.49\text{ g}\) of \(\ce{CuSO4.5H2O}\).
• \(100.0\text{ g}\) of water.

One student argues that the two dissolution equations can be combined because they both produce the same quantities of \(\ce{Cu^2+(aq)}\) and \(\ce{SO4^2-(aq)}\).

Another student points out that the pentahydrate releases water of crystallisation into the solution, changing its final concentration.

Group B therefore repeats its experiment using \(95.5\text{ g}\) of water instead.

The adjusted experiment gives a temperature decrease of \(1.30\text{ K}\).

For the matched experiments, assume:

• Both final solutions contain \(0.0500\text{ mol}\) of each aqueous ion in \(100.0\text{ g}\) of solvent water.
• The total solution mass is approximately \(108.0\text{ g}\) in each experiment.
• The specific heat capacity is \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\).
• Heat absorbed by the cup and thermometer is negligible.
• There is no heat exchange with the surroundings.

Answer the following:

(a) Explain why Group B must reduce its initial water mass.

(b) Calculate the experimental molar enthalpy of dissolution for each solid.

(c) Use Hess’s law to calculate the enthalpy change for the hydration reaction.

(d) Evaluate the first student’s argument. Why is matching the chemical formulas of the aqueous ions not necessarily enough to justify cancellation?

Solution 5

The hydration enthalpy is:

\[ \boxed{\Delta H_{\text{hydration}}=-78.5\ \mathrm{kJ\,mol^{-1}}} \]

However, the second student is also correct: the final solution conditions must match for the aqueous ions to cancel exactly.

(a) Group B must account for water of crystallisation.

Each mole of copper(II) sulfate pentahydrate contains five moles of water.

For \(0.0500\text{ mol}\) of hydrate:

\[ \begin{aligned} n(\ce{H2O})&=5(0.0500)\\ &=0.250\text{ mol} \end{aligned} \]

The mass of this water is:

\[ \begin{aligned} m&=nM\\ &=0.250\times18.0\\ &=4.50\text{ g} \end{aligned} \]

The hydrate contributes \(4.50\text{ g}\) of water to the solution.

If Group B initially adds \(100.0\text{ g}\) of water, its final solvent mass becomes:

\[ 100.0+4.50=104.50\text{ g} \]

This differs from Group A’s final solvent mass of \(100.0\text{ g}\).

Reducing the initially added water to \(95.5\text{ g}\) corrects this difference:

\[ 95.5+4.50=100.0\text{ g} \]

Both experiments now produce solutions with matching final solvent masses and ion concentrations under the stated assumptions.

(b) Calculate the dissolution enthalpies.

For Group A:

\[ \Delta T=+7.40\text{ K} \]

The heat absorbed by the solution is:

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=108.0\times4.18\times7.40\\ &=3341\text{ J}\\ &=3.34\text{ kJ} \end{aligned} \]

The reaction releases this energy:

\[ q_{\text{reaction}}=-3.34\text{ kJ} \]

There are \(0.0500\text{ mol}\) of anhydrous copper sulfate.

Therefore:

\[ \begin{aligned} \Delta H_1 &=\frac{-3.34}{0.0500}\\ &=-66.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

For Group B:

\[ \Delta T=-1.30\text{ K} \]

The heat absorbed by the solution is:

\[ \begin{aligned} q_{\text{solution}} &=108.0\times4.18\times(-1.30)\\ &=-587\text{ J}\\ &=-0.587\text{ kJ} \end{aligned} \]

The reaction absorbs this energy:

\[ q_{\text{reaction}}=+0.587\text{ kJ} \]

Therefore:

\[ \begin{aligned} \Delta H_2 &=\frac{+0.587}{0.0500}\\ &=+11.7\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The results are:

\[ \Delta H_1=-66.8\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta H_2=+11.7\ \mathrm{kJ\,mol^{-1}} \]

(c) Apply Hess’s law.

The target reaction is:

\[ \ce{CuSO4(s)+5H2O(l)->CuSO4.5H2O(s)} \]

The first dissolution reaction is already in the correct direction:

\[ \ce{CuSO4(s)->Cu^2+(aq)+SO4^2-(aq)} \]

The second must be reversed:

\[ \ce{Cu^2+(aq)+SO4^2-(aq)+5H2O(l)->CuSO4.5H2O(s)} \]

Its enthalpy change becomes:

\[ \Delta H_2=-11.7\ \mathrm{kJ\,mol^{-1}} \]

Adding the reactions cancels both aqueous ions.

The remaining equation is exactly the target reaction.

Therefore:

\[ \begin{aligned} \Delta H_{\text{hydration}} &=\Delta H_1-\Delta H_{\text{sol,hydrate}}\\ &=-66.8-11.7\\ &=-78.5\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

(d) Evaluate the first student’s argument.

The first student is only partly correct.

The two equations contain the same aqueous ions, but the ions must also exist in matching final solution states for exact cancellation.

At finite concentrations, the enthalpy of dissolution can depend on the concentration of the resulting solution because the interactions between dissolved ions and surrounding water change.

If the final solution compositions differ, the two aqueous states are not identical. Simply cancelling the ion symbols would ignore the additional enthalpy associated with the difference in solution conditions.

The second student correctly identifies this limitation and adjusts the experiment to account for the water released by the hydrate.

There is one further distinction: if the supplied data are standard enthalpies of solution defined at the same standard state, the standard-state convention already establishes comparable final states. The concentration-matching issue is particularly important when interpreting finite-concentration school experiments.

This is why a good Hess’s law calculation needs more than correct algebra. You must also check that the chemical states represented by the equations genuinely match.

09Where this method leads next

Once you can manipulate complete thermochemical equations, you can calculate enthalpy changes for reactions that are difficult to measure directly.

Combustion data are particularly useful because many substances have well-established enthalpies of combustion. By combining these reactions, you can determine unknown reaction enthalpies, including formation enthalpies and energy changes in biological processes such as photosynthesis.

The next useful step is learning how standard enthalpies of formation simplify these calculations. Instead of constructing every intermediate reaction individually, you can use the formation enthalpies of the reactants and products:

\[ \Delta H^\circ_{\text{reaction}} = \sum n\Delta H_f^\circ(\text{products}) - \sum n\Delta H_f^\circ(\text{reactants}) \]

This is another application of the same principle.

The reaction pathway can change. The initial and final chemical states determine the overall enthalpy change.