How to Calculate Enthalpy Change from a Temperature Rise

A reaction mixture warms from 22.0 °C to 28.6 °C. You know the mass of the solution, its specific heat capacity, and the amount of reactant used. Can you calculate the reaction's enthalpy change?

Yes. First calculate the heat gained by the solution. Then use conservation of energy to work out the heat lost or gained by the reacting chemicals. Finally, divide by the amount of reaction that actually occurred.

The key calculation is:

\[
\boxed{\Delta H = \frac{-mc\Delta T}{n_{\text{reaction}}}}
\]

This simplified expression assumes that the solution absorbs all the heat released or supplied by the reaction, and that the calorimeter itself absorbs negligible heat. We'll build in the more complete version as we go.

## How to calculate enthalpy change from a temperature rise

A temperature rise tells you that the surroundings have gained thermal energy. If the temperature rises because of a chemical reaction, the reacting system has usually released that energy.

This gives us two different quantities:

- The heat change of the solution, which we calculate from its temperature change.
- The heat change of the reaction, which has the opposite sign.

### Step 1: Identify the system and surroundings

Imagine mixing hydrochloric acid and sodium hydroxide in an insulated cup.

The reacting chemicals are the **system**. The solution and calorimeter are the **surroundings** for the purpose of the heat measurement.

When the reaction releases heat, energy transfers from the system to the surroundings. The solution gets warmer.

[IMAGE: An annotated cutaway diagram of a coffee-cup calorimeter. Show two nested polystyrene cups, a loose-fitting insulating lid, a thermometer passing through the lid, a stirring rod, and aqueous reactants inside. Use arrows to show heat transferring from the reacting system to the surrounding solution during an exothermic reaction. Include a small inset distinguishing the reaction system from the solution and calorimeter surroundings.]

Think of it like a hot drink warming your hands through a mug. The drink loses thermal energy while the mug and your hands gain it. The mug is not creating energy; it is receiving energy from somewhere else.

The analogy helps picture the direction of energy transfer, but a chemical reaction is not simply a hot object cooling down. Energy is transferred as the reacting chemicals undergo a chemical change.

For an ideal insulated calorimeter:

\[
q_{\text{system}}+q_{\text{surroundings}}=0
\]

Therefore:

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

The negative sign is essential. It tells us that the reaction and solution have opposite heat changes.

### Step 2: Calculate the heat gained or lost by the solution

The heat change of a substance depends on three things:

- Its mass, \(m\), in grams.
- Its specific heat capacity, \(c\), in joules per gram per degree Celsius.
- Its temperature change, \(\Delta T\), in degrees Celsius.

Use:

\[
q=mc\Delta T
\]

where:

\[
\Delta T=T_{\text{final}}-T_{\text{initial}}
\]

For an aqueous solution, questions commonly ask you to assume:

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

This is the specific heat capacity of water, used as an approximation for the solution.

Notice that the temperature change is final temperature minus initial temperature. Do not reverse this order just because the reaction is exothermic.

[IMAGE: A comparison illustration of two coffee-cup calorimeter experiments. In the first, the solution thermometer rises from 22 °C to 28 °C, with heat flowing from the reaction into the solution. In the second, the thermometer falls from 22 °C to 16 °C, with heat flowing from the solution into an endothermic reaction. Clearly distinguish the opposite temperature changes and energy-transfer directions.]

If the temperature rises, \(\Delta T\) is positive, so \(q_{\text{solution}}\) is positive.

If the temperature falls, \(\Delta T\) is negative, so \(q_{\text{solution}}\) is negative.

The reaction's heat change has the opposite sign in each case.

| Observation | Heat change of solution | Heat change of reaction |
|---|---:|---:|
| Temperature rises | Positive | Negative |
| Temperature falls | Negative | Positive |

This is the sign convention you need to remember:

- Exothermic reaction: \(\Delta H<0\).
- Endothermic reaction: \(\Delta H>0\).

A common mistake is to see a temperature rise and write a positive enthalpy change. The solution has gained heat, but the reaction has lost it. Keep those two signs separate.

### Step 3: Work out how much reaction occurred

Calculating the heat released is only half the job. Enthalpy change is usually reported per mole of reaction, so you must calculate the amount of reaction that actually occurred.

Start with the balanced chemical equation. Then calculate the number of moles of each reactant:

\[
n=cV
\]

where \(n\) is the amount in moles, \(c\) is concentration in \(\mathrm{mol\,L^{-1}}\), and \(V\) is volume in litres.

If a question gives mass instead, use:

\[
n=\frac{m}{M}
\]

where \(m\) is mass in grams and \(M\) is molar mass in \(\mathrm{g\,mol^{-1}}\).

Next, identify the limiting reactant. This is the reactant that runs out first and determines how far the reaction can proceed.

There is one extra detail worth catching: the amount of reaction is not always equal to the number of moles of the limiting reactant.

For a balanced equation:

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

two moles of hydrogen correspond to one mole of reaction as written.

Therefore, if hydrogen is the limiting reactant:

\[
n_{\text{reaction}}=
\frac{n(\ce{H2})}{2}
\]

In general:

\[
n_{\text{reaction}}=
\frac{n_{\text{limiting reactant}}}
{\text{its stoichiometric coefficient}}
\]

This gives the amount of reaction represented by the balanced equation, rather than simply counting the moles of one chemical.

### Step 4: Calculate the molar enthalpy change

Once you know the reaction's heat change and the amount of reaction, calculate:

\[
\boxed{
\Delta H=
\frac{q_{\text{reaction}}}
{n_{\text{reaction}}}
}
\]

The usual unit is:

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

If your heat calculation gives joules, convert to kilojoules before dividing:

\[
1000\ \mathrm{J}=1\ \mathrm{kJ}
\]

For an exothermic reaction, your final answer should be negative. For an endothermic reaction, it should be positive.

Step 5: Check the calorimetry assumptions

The simple calculation assumes that the solution is the only part of the surroundings absorbing heat. Real calorimeters are not quite that simple.

Before substituting, check the following:

If the question provides a calorimeter heat capacity, \(C_{\text{cal}}\), include it:

\[ q_{\text{surroundings}}= mc\Delta T+C_{\text{cal}}\Delta T \]

Here, \(C_{\text{cal}}\) is measured in \(\mathrm{J\\,^{\circ}C^{-1}}\).

The reaction heat is then:

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

You should not ignore a supplied calorimeter constant. The cup absorbs some energy as well as the solution, so leaving it out underestimates the magnitude of the reaction’s heat change.

A few other details matter:

• Use the combined mass of the solution after mixing, not just the mass of one reactant solution.
• If a solid dissolves and remains in the mixture, include its mass when calculating the solution mass.
• A temperature difference has the same numerical value in degrees Celsius and kelvin.
• A coffee-cup calorimeter operates approximately at constant atmospheric pressure. Under these conditions, the heat transferred at constant pressure corresponds to the enthalpy change of the reaction.

These assumptions explain why experimental enthalpy values are often estimates rather than perfectly accurate theoretical values.

01Worked examples

02What if the temperature decreases?

The same method works for an endothermic reaction or dissolution process. You simply need to keep the signs consistent.

Imagine dissolving 5.00 g of ammonium chloride in 100.0 g of water. The temperature falls from \(22.0\\,^{\circ}\mathrm{C}\) to \(18.4\\,^{\circ}\mathrm{C}\).

Assume the resulting solution has a specific heat capacity of \(4.18\ \mathrm{J\\,g^{-1}\\,^{\circ}C^{-1}}\), and ignore heat absorbed by the calorimeter.

The total solution mass is:

\[ m=100.0+5.00=105.0\ \mathrm{g} \]

The temperature change is:

\[ \begin{aligned} \Delta T&=18.4-22.0\\\ &=-3.6\\,^{\circ}\mathrm{C} \end{aligned} \]

Calculate the heat change of the solution:

\[ \begin{aligned} q_{\text{solution}} &=105.0\times4.18\times(-3.6)\\\ &=-1580\ \mathrm{J}\\\ &=-1.58\ \mathrm{kJ} \end{aligned} \]

The solution has lost heat, so the dissolving process must have gained it:

\[ q_{\text{process}}=+1.58\ \mathrm{kJ} \]

The molar mass of ammonium chloride is approximately \(53.5\ \mathrm{g\\,mol^{-1}}\).

\[ \begin{aligned} n(\ce{NH4Cl}) &=\frac{5.00}{53.5}\\\ &=0.0935\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta H_{\text{solution}} &=\frac{+1.58}{0.0935}\\\ &=+16.9\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

The positive enthalpy change means the dissolution process is endothermic.

The important distinction is that a falling thermometer does not mean the reaction has a negative enthalpy change. It means the surroundings have lost heat, which the endothermic process has absorbed.

03Questions and solutions

Question 1

A student mixes 25.0 mL of \(1.00\ \mathrm{mol\\,L^{-1}}\) hydrochloric acid with 25.0 mL of \(1.00\ \mathrm{mol\\,L^{-1}}\) sodium hydroxide.

The initial temperature is \(20.5\\,^{\circ}\mathrm{C}\), and the final temperature is \(27.3\\,^{\circ}\mathrm{C}\).

The balanced equation is:

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

Calculate the molar enthalpy change.

Assume the combined solution has a mass of 50.0 g, a specific heat capacity of \(4.18\ \mathrm{J\\,g^{-1}\\,^{\circ}C^{-1}}\), and that heat absorbed by the calorimeter is negligible.

Solution 1

The molar enthalpy change is approximately \(-56.8\ \mathrm{kJ\\,mol^{-1}}\). The temperature rises, so the solution gains heat and the reaction releases it.

Step 1: Calculate the temperature change.

\[ \Delta T=27.3-20.5=+6.8\\,^{\circ}\mathrm{C} \]

Step 2: Calculate the heat gained by the solution.

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\\ &=50.0\times4.18\times6.8\\\ &=1421.2\ \mathrm{J}\\\ &=+1.42\ \mathrm{kJ} \end{aligned} \]

Step 3: Calculate the heat released by the reaction.

\[ q_{\text{reaction}}=-1.42\ \mathrm{kJ} \]

Step 4: Calculate the amount of reaction.

\[ \begin{aligned} n(\ce{HCl})&=1.00\times0.0250\\\ &=0.0250\ \mathrm{mol} \end{aligned} \]

The sodium hydroxide also has 0.0250 mol.

The balanced equation has a 1:1 ratio, so:

\[ n_{\text{reaction}}=0.0250\ \mathrm{mol} \]

Step 5: Calculate the molar enthalpy change.

\[ \begin{aligned} \Delta H &=\frac{-1.42}{0.0250}\\\ &=-56.8\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

The negative result confirms that the reaction is exothermic. The solution’s temperature rise is evidence of energy being transferred from the reacting system.

Question 2

Sulfuric acid reacts with sodium hydroxide according to:

\[ \ce{H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)} \]

A student mixes:

• 40.0 mL of \(0.500\ \mathrm{mol\\,L^{-1}}\) sulfuric acid.
• 30.0 mL of \(0.500\ \mathrm{mol\\,L^{-1}}\) sodium hydroxide.

The initial temperature is \(23.0\\,^{\circ}\mathrm{C}\), and the final temperature is \(25.8\\,^{\circ}\mathrm{C}\).

Assume a combined solution mass of 70.0 g, a specific heat capacity of \(4.18\ \mathrm{J\\,g^{-1}\\,^{\circ}C^{-1}}\), and negligible heat absorption by the calorimeter.

Calculate the molar enthalpy change. Identify the limiting reactant and explain why simply dividing by the moles of sulfuric acid would give the wrong result.

Solution 2

The molar enthalpy change is approximately \(-109\ \mathrm{kJ\\,mol^{-1}}\), and sodium hydroxide is the limiting reactant. The important detail is that two moles of sodium hydroxide are required for every mole of reaction.

Step 1: Calculate the temperature change.

\[ \Delta T=25.8-23.0=+2.8\\,^{\circ}\mathrm{C} \]

Step 2: Calculate the heat gained by the solution.

\[ \begin{aligned} q_{\text{solution}} &=70.0\times4.18\times2.8\\\ &=819.28\ \mathrm{J}\\\ &=+0.819\ \mathrm{kJ} \end{aligned} \]

Therefore:

\[ q_{\text{reaction}}=-0.819\ \mathrm{kJ} \]

Step 3: Calculate the moles of each reactant.

Sulfuric acid:

\[ \begin{aligned} n(\ce{H2SO4}) &=0.500\times0.0400\\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

Sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=0.500\times0.0300\\\ &=0.0150\ \mathrm{mol} \end{aligned} \]

Step 4: Identify the limiting reactant and calculate the amount of reaction.

The balanced equation requires two moles of sodium hydroxide for every mole of sulfuric acid.

The amount of reaction supported by the sulfuric acid is:

\[ \frac{0.0200}{1}=0.0200\ \mathrm{mol} \]

The amount supported by the sodium hydroxide is:

\[ \frac{0.0150}{2}=0.00750\ \mathrm{mol} \]

Sodium hydroxide is therefore limiting.

The actual amount of reaction is:

\[ n_{\text{reaction}}=0.00750\ \mathrm{mol} \]

Step 5: Calculate the molar enthalpy change.

\[ \begin{aligned} \Delta H &=\frac{-0.819}{0.00750}\\\ &=-109.2\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

The final answer is approximately \(-109\ \mathrm{kJ\\,mol^{-1}}\).

The tempting mistake is to divide by 0.0200 mol because that is the amount of sulfuric acid present. But not all of that acid can react. The sodium hydroxide runs out first, and its coefficient of two must be included when calculating the amount of reaction.

Question 3

A neutralisation reaction is performed in a calorimeter with a heat capacity of \(60.0\ \mathrm{J\\,^{\circ}C^{-1}}\).

The student mixes:

• 75.0 mL of \(0.800\ \mathrm{mol\\,L^{-1}}\) hydrochloric acid.
• 50.0 mL of \(1.00\ \mathrm{mol\\,L^{-1}}\) sodium hydroxide.

The initial temperature is \(19.5\\,^{\circ}\mathrm{C}\), and the final temperature is \(24.0\\,^{\circ}\mathrm{C}\).

The combined solution has a mass of 125 g and a specific heat capacity of \(4.18\ \mathrm{J\\,g^{-1}\\,^{\circ}C^{-1}}\).

Calculate the molar enthalpy change, including the heat absorbed by the calorimeter.

Explain why ignoring the calorimeter’s heat capacity would produce an inaccurate result.

Solution 3

The molar enthalpy change is approximately \(-52.4\ \mathrm{kJ\\,mol^{-1}}\). The calorimeter must be included because it absorbs additional energy from the reaction.

Step 1: Calculate the temperature change.

\[ \Delta T=24.0-19.5=+4.5\\,^{\circ}\mathrm{C} \]

Step 2: Calculate the heat absorbed by the solution.

\[ \begin{aligned} q_{\text{solution}} &=125\times4.18\times4.5\\\ &=2351.25\ \mathrm{J} \end{aligned} \]

Step 3: Calculate the heat absorbed by the calorimeter.

Use:

\[ q_{\text{cal}}=C_{\text{cal}}\Delta T \]

Substitute:

\[ \begin{aligned} q_{\text{cal}} &=60.0\times4.5\\\ &=270\ \mathrm{J} \end{aligned} \]

Step 4: Calculate the total heat gained by the surroundings.

Both the solution and calorimeter absorb energy:

\[ \begin{aligned} q_{\text{surroundings}} &=2351.25+270\\\ &=2621.25\ \mathrm{J}\\\ &=+2.62\ \mathrm{kJ} \end{aligned} \]

Therefore:

\[ q_{\text{reaction}}=-2.62\ \mathrm{kJ} \]

Step 5: Calculate the amount of reaction.

The balanced equation is:

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

For hydrochloric acid:

\[ n(\ce{HCl})=0.800\times0.0750 =0.0600\ \mathrm{mol} \]

For sodium hydroxide:

\[ n(\ce{NaOH})=1.00\times0.0500 =0.0500\ \mathrm{mol} \]

The mole ratio is 1:1, so sodium hydroxide is limiting.

\[ n_{\text{reaction}}=0.0500\ \mathrm{mol} \]

Step 6: Calculate the molar enthalpy change.

\[ \begin{aligned} \Delta H &=\frac{-2.62}{0.0500}\\\ &=-52.4\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

The reaction is exothermic, with a molar enthalpy change of approximately \(-52.4\ \mathrm{kJ\\,mol^{-1}}\).

Why does the calorimeter matter?

If you ignore it, you calculate only the energy absorbed by the solution:

\[ q_{\text{reaction}}=-2.35\ \mathrm{kJ} \]

This would give:

\[ \Delta H=-47.0\ \mathrm{kJ\\,mol^{-1}} \]

The magnitude is too small because some energy was transferred into the calorimeter as well. Whenever the question supplies a calorimeter constant, include it.

Question 4

Magnesium reacts with hydrochloric acid in an insulated calorimeter.

A student adds 0.0486 g of magnesium ribbon to 50.0 mL of \(0.100\ \mathrm{mol\\,L^{-1}}\) hydrochloric acid.

The initial temperature is \(21.0\\,^{\circ}\mathrm{C}\), and the final temperature is \(24.8\\,^{\circ}\mathrm{C}\).

The reaction is:

\[ \ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \]

The following assumptions and measurements apply:

• Mass of solution: 50.0 g.
• Specific heat capacity of solution: \(4.18\ \mathrm{J\\,g^{-1}\\,^{\circ}C^{-1}}\).
• Calorimeter heat capacity: \(35.0\ \mathrm{J\\,^{\circ}C^{-1}}\).
• Hydrogen gas escapes and carries negligible heat.
• The reaction goes to completion.

Calculate the molar enthalpy change. Identify the limiting reactant and explain why using the initial moles of hydrochloric acid as the amount of reaction would be incorrect.

Solution 4

The molar enthalpy change is approximately \(-464\ \mathrm{kJ\\,mol^{-1}}\), and magnesium is the limiting reactant. The two-to-one mole ratio means the amount of reaction must be calculated using the magnesium, not simply the number of moles of hydrochloric acid.

Step 1: Calculate the temperature change.

\[ \Delta T=24.8-21.0=+3.8\\,^{\circ}\mathrm{C} \]

Step 2: Calculate the heat absorbed by the solution.

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\\ &=50.0\times4.18\times3.8\\\ &=794.2\ \mathrm{J} \end{aligned} \]

Step 3: Calculate the heat absorbed by the calorimeter.

\[ \begin{aligned} q_{\text{cal}} &=C_{\text{cal}}\Delta T\\\ &=35.0\times3.8\\\ &=133\ \mathrm{J} \end{aligned} \]

Step 4: Calculate the total heat released by the reaction.

\[ \begin{aligned} q_{\text{surroundings}} &=794.2+133\\\ &=927.2\ \mathrm{J} \end{aligned} \]

Therefore:

\[ q_{\text{reaction}}=-0.9272\ \mathrm{kJ} \]

Step 5: Calculate the moles of each reactant.

For hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=0.100\times0.0500\\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

For magnesium, use its molar mass of approximately \(24.3\ \mathrm{g\\,mol^{-1}}\):

\[ \begin{aligned} n(\ce{Mg}) &=\frac{0.0486}{24.3}\\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

Step 6: Identify the limiting reactant.

The balanced equation requires:

\[ 1\ \mathrm{mol\ Mg}:2\ \mathrm{mol\ HCl} \]

The amount of reaction supported by the hydrochloric acid is:

\[ \frac{0.00500}{2}=0.00250\ \mathrm{mol} \]

The amount supported by magnesium is:

\[ \frac{0.00200}{1}=0.00200\ \mathrm{mol} \]

Magnesium is limiting, so:

\[ n_{\text{reaction}}=0.00200\ \mathrm{mol} \]

Step 7: Calculate the molar enthalpy change.

\[ \begin{aligned} \Delta H &=\frac{-0.9272}{0.00200}\\\ &=-463.6\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Therefore:

\[ \Delta H\approx-464\ \mathrm{kJ\\,mol^{-1}} \]

The reaction is exothermic.

The important trap is that 0.00500 mol of hydrochloric acid does not mean 0.00500 mol of reaction. Each mole of magnesium requires two moles of hydrochloric acid. Magnesium runs out first, leaving some acid unreacted.

Question 5

Extension: Combining enthalpy changes (Hess’s law connection, Chemistry Stage 6 Syllabus 2017).

A student wants to calculate the enthalpy change of neutralisation between hydrochloric acid and sodium hydroxide.

Instead of using an aqueous sodium hydroxide solution, the student adds 4.00 g of solid sodium hydroxide directly to 100.0 mL of \(1.00\ \mathrm{mol\\,L^{-1}}\) hydrochloric acid.

The initial temperature is \(20.0\\,^{\circ}\mathrm{C}\), and the final temperature is \(42.0\\,^{\circ}\mathrm{C}\).

The following data are provided:

Assume all the sodium hydroxide dissolves and reacts completely, there are no other significant heat effects, and the supplied dissolution enthalpy applies to these conditions.

The enthalpy change for dissolving solid sodium hydroxide in water is:

\[ \ce{NaOH(s) -> Na+(aq) + OH-(aq)} \]

\[ \Delta H_{\text{dissolution}} =-44.5\ \mathrm{kJ\\,mol^{-1}} \]

Answer the following:

(a) Calculate the total heat released by the observed process, including heat absorbed by the calorimeter.

(b) Calculate the molar enthalpy change for the overall process involving solid sodium hydroxide and hydrochloric acid.

(c) Determine the actual molar enthalpy change of neutralisation.

(d) Explain why a student who simply divides the total reaction heat by the moles of sodium hydroxide would obtain the wrong enthalpy change for neutralisation.

Solution 5

The enthalpy change of neutralisation is approximately \(-56.6\ \mathrm{kJ\\,mol^{-1}}\), not \(-101\ \mathrm{kJ\\,mol^{-1}}\). The measured temperature rise includes two energy-releasing processes: dissolving solid sodium hydroxide and neutralising the acid.

This is the crucial distinction. The experiment measures the combined enthalpy change, but the question asks for only one part of it.

Step 1: Calculate the temperature change.

\[ \begin{aligned} \Delta T&=42.0-20.0\\\ &=+22.0\\,^{\circ}\mathrm{C} \end{aligned} \]

Step 2: Calculate the heat absorbed by the solution.

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\\ &=104.0\times4.18\times22.0\\\ &=9543.8\ \mathrm{J} \end{aligned} \]

Step 3: Calculate the heat absorbed by the calorimeter.

\[ \begin{aligned} q_{\text{cal}} &=C_{\text{cal}}\Delta T\\\ &=25.0\times22.0\\\ &=550\ \mathrm{J} \end{aligned} \]

Step 4: Calculate the total heat released.

Add the heat absorbed by both parts of the surroundings:

\[ \begin{aligned} q_{\text{surroundings}} &=9543.8+550\\\ &=10093.8\ \mathrm{J}\\\ &=+10.09\ \mathrm{kJ} \end{aligned} \]

Therefore:

\[ q_{\text{total reaction}}=-10.09\ \mathrm{kJ} \]

Step 5: Calculate the amount of reaction.

The molar mass of sodium hydroxide is \(40.0\ \mathrm{g\\,mol^{-1}}\).

\[ \begin{aligned} n(\ce{NaOH}) &=\frac{4.00}{40.0}\\\ &=0.100\ \mathrm{mol} \end{aligned} \]

For hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=1.00\times0.1000\\\ &=0.100\ \mathrm{mol} \end{aligned} \]

The reaction has a 1:1 ratio, so both reactants are present in exactly the required proportions.

The amount of reaction is:

\[ n_{\text{reaction}}=0.100\ \mathrm{mol} \]

Step 6: Calculate the overall molar enthalpy change.

\[ \begin{aligned} \Delta H_{\text{total}} &=\frac{-10.09}{0.100}\\\ &=-100.9\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

This is the enthalpy change for the overall process involving solid sodium hydroxide and hydrochloric acid.

It is not yet the enthalpy change of neutralisation alone.

Step 7: Separate the dissolution and neutralisation processes.

The experiment involves two processes.

First, solid sodium hydroxide dissolves:

\[ \ce{NaOH(s) -> Na+(aq) + OH-(aq)} \]

Its enthalpy change is:

\[ \Delta H_{\text{dissolution}}=-44.5\ \mathrm{kJ\\,mol^{-1}} \]

Second, hydroxide ions react with hydrogen ions:

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

This is the neutralisation process we actually want to calculate.

Because the overall enthalpy change is the sum of the two individual changes:

\[ \Delta H_{\text{total}} = \Delta H_{\text{dissolution}} + \Delta H_{\text{neutralisation}} \]

Rearrange:

\[ \Delta H_{\text{neutralisation}} = \Delta H_{\text{total}} - \Delta H_{\text{dissolution}} \]

Substitute:

\[ \begin{aligned} \Delta H_{\text{neutralisation}} &=-100.9-(-44.5)\\\ &=-56.4\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Therefore, the neutralisation enthalpy is approximately:

\[ \Delta H_{\text{neutralisation}} =-56.4\ \mathrm{kJ\\,mol^{-1}} \]

Step 8: Explain the student’s mistake.

The student has treated all the heat released as though it came from neutralisation.

But the solid sodium hydroxide must first dissolve into aqueous ions. That process also releases heat.

The thermometer measures the combined effect of both processes. It cannot distinguish where the energy came from.

The student can calculate the overall enthalpy change directly from the temperature rise, but must use the supplied dissolution enthalpy to isolate the neutralisation enthalpy.

This is an application of enthalpy additivity, the principle behind Hess’s law.

The important lesson is that a temperature measurement tells you the total heat transferred across your chosen system boundary. It does not automatically tell you the enthalpy change of one particular step within a more complicated process.

04A quick method to remember

When calculating enthalpy change from a temperature rise, follow this sequence:

1. Write the balanced chemical equation.
2. Identify the reacting system and its surroundings.
3. Calculate the temperature change using final minus initial temperature.
4. Calculate the heat absorbed by the solution using \(q=mc\Delta T\).
5. Include the calorimeter heat capacity if it is provided.
6. Reverse the sign to obtain the reaction’s heat change.
7. Calculate the amount of reaction using the limiting reactant and balanced equation.
8. Divide reaction heat by the amount of reaction.

Keep the following distinctions clear:

A temperature rise usually indicates an exothermic reaction, but the temperature change alone does not determine the molar enthalpy. You also need the heat capacity of the surroundings and the amount of reaction that occurred.

Once you can calculate experimental enthalpy changes reliably, the next useful step is understanding what the sign and magnitude tell you about chemical energy. The tutorgum guide to entropy and enthalpy changes explains how enthalpy relates to energy transfer, and why it is different from entropy.

For students studying the Chemistry Stage 6 Syllabus (2017), the energy accounting used here also provides a foundation for more involved Hess’s law calculations.