Enthalpy Change Examples: Dissolution, Neutralization, Displacement, and Combustion Revision Notes | A Level AQA Chemistry

1. Why Enthalpy Changes Matter — and How We Measure Them

Every chemical reaction either releases or absorbs energy. Whether you are dissolving a salt, burning a fuel, or mixing an acid with an alkali, the energy change is real, measurable, and examinable. The question is: how do we pin down that energy change precisely?

The answer is calorimetry — measuring the temperature change of a known mass of solution and working backwards to find the heat transferred. The core equation you will use throughout this lesson is $q = mc\Delta T$, where $q$ is heat transferred in joules, $m$ is the mass of solution in grams, $c$ is the specific heat capacity (4.18 J g⁻¹ °C⁻¹ for aqueous solutions), and $\Delta T$ is the temperature change in °C.

Notice the sign convention: if the temperature of the solution rises, the reaction is exothermic and $\Delta H$ is negative (energy released to surroundings). If the temperature falls, the reaction is endothermic and $\Delta H$ is positive (energy absorbed from surroundings). Getting this sign right is worth marks every time.

2. Enthalpy of Dissolution — KCl and Na₂CO₃ Compared

When an ionic solid dissolves in water, two competing energy processes occur simultaneously. First, the ionic lattice must be broken apart — this requires energy input (lattice enthalpy). Second, water molecules surround and stabilise the freed ions through ion–dipole interactions, releasing energy (enthalpy of hydration). The overall enthalpy of dissolution is the net result of these two processes.

Potassium chloride (KCl) dissolves endothermically: the lattice enthalpy of KCl is larger in magnitude than the combined hydration enthalpies of K⁺ and Cl⁻, so the solution cools. Sodium carbonate (Na₂CO₃) dissolves exothermically: the high charge density of the Na⁺ ion and the doubly charged CO₃²⁻ ion produce large hydration enthalpies that outweigh the lattice energy cost, so the solution warms.

In a typical dissolution experiment, a known mass of solid is added to a measured volume of distilled water in a polystyrene cup (which acts as a simple calorimeter). The temperature is recorded before and after dissolution, and $q = mc\Delta T$ is applied using the total mass of solution. The moles of solute dissolved give the per-mole enthalpy value.

Worked example: Calculating Enthalpy of Dissolution of KCl

2.98 g of KCl (M_r = 74.5 g mol⁻¹) is dissolved in 50.0 cm³ of water. The temperature falls from 22.4 °C to 20.1 °C. Calculate the enthalpy of dissolution of KCl in kJ mol⁻¹.

1. $\Delta T = 20.1 - 22.4 = -2.3\ °\text{C}$ — The temperature fell, so ΔT is negative, confirming an endothermic process.
2. $m = 50.0\ \text{g} + 2.98\ \text{g} = 52.98\ \text{g}$ — Use the total mass of the solution (water + dissolved solid), not just the water mass.
3. $q = mc\Delta T = 52.98 \times 4.18 \times (-2.3) = -509.4\ \text{J}$ — The negative sign confirms heat was absorbed by the solution from the surroundings — endothermic. Working: 52.98 × 4.18 = 221.5 J °C⁻¹; 221.5 × 2.3 = 509.4 J.
4. $n = \frac{2.98}{74.5} = 0.0400\ \text{mol}$ — Moles = mass ÷ molar mass.
5. $\Delta H = \frac{+509.4\ \text{J}}{0.0400\ \text{mol}} = +12735\ \text{J mol}^{-1} = +12.7\ \text{kJ mol}^{-1}$ — Because the solution absorbed heat (q is negative for the solution), the reaction itself is endothermic and ΔH is positive: ΔH = −q/n = +509.4/0.0400. Divide by 1000 to convert J to kJ.

Answer: $\Delta H_{\text{dissolution}}(\text{KCl}) = +12.7\ \text{kJ mol}^{-1}$

The dissolution examples establish the core calorimetry method. Now we apply the same logic to a reaction between two solutions — neutralisation — where the setup changes slightly but the underlying calculation remains the same.

3. Enthalpy of Neutralisation — NaOH and HCl

When sodium hydroxide solution reacts with hydrochloric acid, the net ionic equation is $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$. This is the reaction that releases energy — the formation of a stable O–H bond in water. The spectator ions Na⁺ and Cl⁻ do not contribute to the enthalpy change.

The standard enthalpy of neutralisation for a strong acid reacting with a strong base is approximately −57 kJ mol⁻¹. This value is consistent across different strong acid–strong base combinations because the net reaction is always the same: one mole of water forms from H⁺ and OH⁻ ions. Weak acid or weak base combinations give less exothermic values because energy is also required to fully ionise the weak acid or base.

Worked example: Calculating Enthalpy of Neutralisation of NaOH with HCl

25.0 cm³ of 2.00 mol dm⁻³ NaOH is mixed with 25.0 cm³ of 2.00 mol dm⁻³ HCl in a polystyrene cup. The temperature rises from 19.5 °C to 32.8 °C. Calculate the enthalpy of neutralisation in kJ mol⁻¹.

1. $\Delta T = 32.8 - 19.5 = 13.3\ °\text{C}$ — Temperature rose, so the reaction is exothermic.
2. $m = 25.0 + 25.0 = 50.0\ \text{g}$ — Assume density of both solutions is 1.00 g cm⁻³, so 50.0 cm³ total = 50.0 g. This is a standard AQA assumption.
3. $q = 50.0 \times 4.18 \times 13.3 = 2779.7\ \text{J}$ — Working: 50.0 × 4.18 = 209.0 J °C⁻¹; 209.0 × 13.3 = 2779.7 J. Positive q means the solution gained heat.
4. $n(\text{H}_2\text{O}) = 0.0250\ \text{dm}^3 \times 2.00\ \text{mol dm}^{-3} = 0.0500\ \text{mol}$ — Both reactants are in equal moles (0.0500 mol each), so 0.0500 mol of water forms. Neither is in excess.
5. $\Delta H = -\frac{2779.7\ \text{J}}{0.0500\ \text{mol}} = -55594\ \text{J mol}^{-1} = -55.6\ \text{kJ mol}^{-1}$ — The negative sign is applied because the reaction is exothermic — the system released heat to the solution. The experimental value is close to the theoretical −57 kJ mol⁻¹; the difference arises from heat losses to the surroundings.

Answer: $\Delta H_{\text{neutralisation}} = -55.6\ \text{kJ mol}^{-1}$

4. Enthalpy of Displacement — Zinc and Copper Sulphate

Displacement reactions occur when a more reactive metal displaces a less reactive metal from a solution of its ions. Zinc is above copper in the reactivity series, so zinc displaces copper from copper(II) sulphate solution. The overall equation is:

$\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$

The ionic equation strips away the spectator sulphate ions: $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq}) + \text{Cu(s)}$. Zinc is oxidised (loses electrons) and copper(II) ions are reduced (gain electrons). This is a redox reaction, and it is strongly exothermic — the solution temperature rises noticeably when zinc powder is added to copper sulphate solution.

The experimental setup uses a polystyrene cup containing a measured volume of copper sulphate solution. An excess of zinc powder is added to ensure the copper sulphate is the limiting reagent, so all Cu²⁺ ions react. The mass used in $q = mc\Delta T$ is the mass of the copper sulphate solution (not including the zinc powder, since the zinc is a solid and does not contribute significantly to the solution's heat capacity under standard AQA assumptions).

Worked example: Calculating Enthalpy of Displacement for Zn + CuSO₄

25.0 cm³ of 0.200 mol dm⁻³ CuSO₄(aq) is placed in a polystyrene cup. Excess zinc powder is added and the temperature rises from 18.2 °C to 27.6 °C. Calculate the enthalpy of displacement in kJ mol⁻¹.

1. $\Delta T = 27.6 - 18.2 = 9.4\ °\text{C}$ — Temperature rose, confirming an exothermic displacement reaction.
2. $m = 25.0\ \text{g}$ — Assume density of CuSO₄(aq) = 1.00 g cm⁻³, so 25.0 cm³ = 25.0 g. Under standard AQA assumptions, the mass of zinc powder added is not included in m.
3. $q = 25.0 \times 4.18 \times 9.4 = 982.3\ \text{J}$ — Working: 25.0 × 4.18 = 104.5 J °C⁻¹; 104.5 × 9.4 = 982.3 J.
4. $n(\text{CuSO}_4) = 0.0250\ \text{dm}^3 \times 0.200\ \text{mol dm}^{-3} = 0.00500\ \text{mol}$ — CuSO₄ is the limiting reagent because zinc is added in excess.
5. $\Delta H = -\frac{982.3\ \text{J}}{0.00500\ \text{mol}} = -196460\ \text{J mol}^{-1} = -196\ \text{kJ mol}^{-1}$ — Negative sign applied because the reaction is exothermic. Divide by 1000 to convert to kJ mol⁻¹.

Answer: $\Delta H_{\text{displacement}} = -196\ \text{kJ mol}^{-1}$

5. Enthalpy of Combustion — Burning Alcohols

Combustion is the most energetically significant reaction type in this lesson. When an alcohol burns completely in excess oxygen, carbon dioxide and water are the only products. The general equation for a primary alcohol $\text{C}_n\text{H}_{2n+1}\text{OH}$ burning completely is:

$\text{C}_n\text{H}_{2n+1}\text{OH}(\text{l}) + \frac{3n+1}{2}\text{O}_2\text{(g)} \rightarrow n\text{CO}_2\text{(g)} + (n+1)\text{H}_2\text{O}(\text{l})$

For example, ethanol burns as: $\text{C}_2\text{H}_5\text{OH}(\text{l}) + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O}(\text{l})$. As the carbon chain length increases from methanol to butanol, more C–H and C–C bonds are broken and more C=O and O–H bonds are formed, releasing progressively more energy per mole.

The experimental method uses a spirit burner containing the alcohol, placed beneath a copper calorimeter holding a known mass of water. The alcohol is burned for a measured time, and the temperature rise of the water is recorded. The mass of alcohol burned is found by weighing the spirit burner before and after combustion.

Worked example: Calculating Enthalpy of Combustion of Ethanol

A spirit burner containing ethanol (M_r = 46.0 g mol⁻¹) is used to heat 100.0 g of water. The water temperature rises from 18.5 °C to 38.5 °C. The mass of the spirit burner decreases by 0.920 g. Calculate the enthalpy of combustion of ethanol in kJ mol⁻¹.

1. $\Delta T = 38.5 - 18.5 = 20.0\ °\text{C}$ — A rise of 20.0 °C — the combustion is strongly exothermic.
2. $q = 100.0 \times 4.18 \times 20.0 = 8360\ \text{J}$ — Working: 100.0 × 4.18 = 418.0 J °C⁻¹; 418.0 × 20.0 = 8360 J. Only the water mass is used here, as the copper calorimeter's heat capacity is typically ignored under standard AQA assumptions unless stated.
3. $n = \frac{0.920}{46.0} = 0.0200\ \text{mol}$ — Moles = mass burned ÷ molar mass.
4. $\Delta H = -\frac{8360\ \text{J}}{0.0200\ \text{mol}} = -418000\ \text{J mol}^{-1} = -418\ \text{kJ mol}^{-1}$ — Negative because combustion is exothermic. The data book value for ethanol is −1367 kJ mol⁻¹, so this experimental value is significantly lower — a large systematic error due to heat losses.

Answer: $\Delta H_{\text{combustion}}(\text{ethanol}) = -418\ \text{kJ mol}^{-1}$ (experimental)

The trend is clear: each additional CH₂ unit in the carbon chain adds approximately 650 kJ mol⁻¹ to the magnitude of the enthalpy of combustion. This is because each extra CH₂ group contains two C–H bonds and one C–C bond, all of which are oxidised to CO₂ and H₂O during complete combustion, releasing additional energy.

6. Comparing the Four Reaction Types

Now that all four reaction types have been covered, it is worth stepping back to compare their experimental setups and the key decisions you must make in each calculation. The underlying method — $q = mc\Delta T$ followed by conversion to kJ mol⁻¹ — is the same throughout, but the choice of mass, the identification of the limiting reagent, and the sign convention each require careful thought.

7. Exam Technique and Common Pitfalls

Enthalpy calculation questions are among the most reliably mark-winning questions on AQA papers — if you follow the method precisely. The most common errors are sign mistakes, using the wrong mass, and forgetting to convert joules to kilojoules. Each of these costs marks that are entirely avoidable.