Enthalpy Change Examples: Dissolution, Neutralization, Displacement, and Combustion Revision Notes | A Level AQA Chemistry
The standard enthalpy of neutralisation for a strong acid and strong base is about −57 kJ mol⁻¹. Free A Level AQA Chemistry notes and 15 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Enthalpy Change Examples: Dissolution, Neutralization, Displacement, and Combustion is part of Energetics in A Level AQA Chemistry.
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What this lesson covers
The standard enthalpy of neutralisation for a strong acid and strong base is about −57 kJ mol⁻¹. Free A Level AQA Chemistry notes and 15 questions answered.
Calculate enthalpy changes for dissolution, neutralisation, displacement, and combustion reactions using calorimetric data
Explain the energy changes occurring at a molecular level for each reaction type
Compare the experimental methods used to measure enthalpy changes across different reaction types
Evaluate sources of error and suggest improvements in calorimetric experiments
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Use this lesson to stay inside Energetics while connecting the detail back to the wider Physical chemistry topic.
Enthalpy change (ΔH): The heat energy exchanged with the surroundings at constant pressure during a chemical reaction. Measured in kJ mol⁻¹; negative values indicate exothermic reactions, positive values indicate endothermic reactions.
Calorimetry: An experimental technique that measures heat energy transferred during a reaction by monitoring temperature changes in a known mass of solution, using q = mcΔT.
Enthalpy of dissolution: The enthalpy change when one mole of a solute dissolves completely in sufficient solvent to form a solution of infinite dilution, under standard conditions.
Enthalpy of combustion: The enthalpy change when one mole of a substance burns completely in excess oxygen under standard conditions (298 K, 100 kPa), producing specified products.
Revision notes
Read the core explanations from Enthalpy Change Examples: Dissolution, Neutralization, Displacement, and Combustion before testing yourself from memory.
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.
$\Delta T = 20.1 - 22.4 = -2.3\ °\text{C}$ — The temperature fell, so ΔT is negative, confirming an endothermic process.
$m = 50.0\ \text{g} + 2.98\ \text{g} = 52.98\ \text{g}$ — Use the total mass of the solution (water + dissolved solid), not just the water mass.
$q = mc\Delta T = 52.98 \times 4.18 \times (-2.3) = -509.4\ \text{J}$ — 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.
$\Delta H = \frac{+509.4\ \text{J}}{0.0400\ \text{mol}} = +12735\ \text{J mol}^{-1} = +12.7\ \text{kJ mol}^{-1}$ — 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.
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.
$\Delta T = 32.8 - 19.5 = 13.3\ °\text{C}$ — Temperature rose, so the reaction is exothermic.
$m = 25.0 + 25.0 = 50.0\ \text{g}$ — 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.
$n(\text{H}_2\text{O}) = 0.0250\ \text{dm}^3 \times 2.00\ \text{mol dm}^{-3} = 0.0500\ \text{mol}$ — Both reactants are in equal moles (0.0500 mol each), so 0.0500 mol of water forms. Neither is in excess.
$\Delta H = -\frac{2779.7\ \text{J}}{0.0500\ \text{mol}} = -55594\ \text{J mol}^{-1} = -55.6\ \text{kJ mol}^{-1}$ — 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.
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:
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.
$\Delta T = 27.6 - 18.2 = 9.4\ °\text{C}$ — Temperature rose, confirming an exothermic displacement reaction.
$m = 25.0\ \text{g}$ — 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.
$n(\text{CuSO}_4) = 0.0250\ \text{dm}^3 \times 0.200\ \text{mol dm}^{-3} = 0.00500\ \text{mol}$ — CuSO₄ is the limiting reagent because zinc is added in excess.
$\Delta H = -\frac{982.3\ \text{J}}{0.00500\ \text{mol}} = -196460\ \text{J mol}^{-1} = -196\ \text{kJ mol}^{-1}$ — Negative sign applied because the reaction is exothermic. Divide by 1000 to convert to kJ mol⁻¹.
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:
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.
$\Delta T = 38.5 - 18.5 = 20.0\ °\text{C}$ — A rise of 20.0 °C — the combustion is strongly exothermic.
$q = 100.0 \times 4.18 \times 20.0 = 8360\ \text{J}$ — 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.
$\Delta H = -\frac{8360\ \text{J}}{0.0200\ \text{mol}} = -418000\ \text{J mol}^{-1} = -418\ \text{kJ mol}^{-1}$ — 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.
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.
Reaction type
Example
Mass used in q = mcΔT
Limiting reagent / moles used
Typical sign of ΔH
Dissolution
KCl in water
Total mass of solution (solvent + solute)
Moles of solute dissolved
KCl: positive; Na₂CO₃: negative
Neutralisation
NaOH + HCl
Total mass of combined solutions
Moles of water formed (from limiting reagent)
Negative (exothermic)
Displacement
Zn + CuSO₄(aq)
Mass of solution only (not the solid metal)
Moles of solution-phase limiting reagent (CuSO₄)
Negative (exothermic)
Combustion
Ethanol burned
Mass of water in calorimeter
Moles of fuel burned (from mass loss)
Negative (exothermic)
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.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
What is the ionic equation for the reaction of zinc with copper sulphate?
The ionic equation for zinc with copper sulphate is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s); the sulphate ions are spectator ions. Zinc is oxidised by losing electrons and copper(II) ions are reduced by gaining electrons, so this displacement is a redox reaction and is strongly exothermic.
Why is dissolving potassium chloride endothermic?
Potassium chloride dissolves endothermically because the lattice enthalpy of KCl is larger in magnitude than the combined hydration enthalpies of K⁺ and Cl⁻ ions, so the solution cools and ΔH is positive. In the worked example, dissolving 2.98 g of KCl gives ΔH = +12.7 kJ mol⁻¹.
What is the enthalpy of neutralisation of a strong acid and a strong base?
The standard enthalpy of neutralisation for a strong acid reacting with a strong base is approximately −57 kJ mol⁻¹. It is consistent across combinations because the net reaction is always H⁺(aq) + OH⁻(aq) → H₂O(l). Weak acids or bases give less exothermic values because energy is needed to fully ionise them.
How do you calculate the enthalpy change for zinc and copper sulphate?
For zinc and copper(II) sulphate, add excess zinc powder to a measured volume of CuSO₄ solution, use only the solution mass in q = mcΔT, and divide by the moles of CuSO₄, the limiting reagent. For 25.0 cm³ of 0.200 mol dm⁻³ CuSO₄ rising by 9.4 °C, ΔH = −196 kJ mol⁻¹.
Why does the enthalpy of combustion of alcohols increase with chain length?
The enthalpy of combustion of alcohols becomes more negative as chain length increases because each extra CH₂ unit contains two C–H bonds and one C–C bond that are oxidised to CO₂ and H₂O. Each additional CH₂ unit adds approximately 650 kJ mol⁻¹ to the magnitude of the enthalpy of combustion.
Which mass do you use in q = mcΔT?
The mass in q = mcΔT depends on the reaction: the total mass of solution for dissolution and neutralisation, the mass of solution only (not the solid metal) for displacement, and the mass of water in the calorimeter for combustion. Using the wrong mass is one of the most common errors.
What are the enthalpies of dissolution and neutralisation?
The enthalpy of dissolution is the net result of breaking the ionic lattice, which needs energy, and hydrating the freed ions, which releases energy. The enthalpy of neutralisation comes from H⁺ + OH⁻ → H₂O; for a strong acid and strong base it is approximately −57 kJ mol⁻¹.
Why is dissolving sodium carbonate exothermic?
Dissolving sodium carbonate is exothermic because the high charge density of the Na⁺ ion and the doubly charged CO₃²⁻ ion produce large hydration enthalpies. These outweigh the lattice energy cost of breaking the solid apart, so the overall ΔH is negative and the solution warms.
What happens when an ionic solid dissolves in water?
When an ionic solid dissolves in water, two energy processes happen at the same time. The ionic lattice is broken apart, which requires energy (lattice enthalpy), and water molecules surround and stabilise the freed ions through ion–dipole interactions, which releases energy (enthalpy of hydration). The overall enthalpy of dissolution is the net result.
Why is the enthalpy of neutralisation of a weak acid less exothermic?
The enthalpy of neutralisation of a weak acid or weak base is less exothermic because energy is also required to fully ionise the weak acid or base. For strong acid–strong base pairs the value is about −57 kJ mol⁻¹, because the net reaction is always one mole of water forming from H⁺ and OH⁻.
Why is zinc added in excess in the displacement experiment?
Zinc powder is added in excess in the displacement experiment so that copper sulphate is the limiting reagent and all the Cu²⁺ ions react. The moles of CuSO₄ are then used to calculate ΔH; for example, 25.0 cm³ of 0.200 mol dm⁻³ CuSO₄ contains 0.00500 mol.
How do you calculate the enthalpy of dissolution of potassium chloride?
The enthalpy of dissolution of potassium chloride is about +12.7 kJ mol⁻¹. Dissolving 2.98 g of KCl in 50.0 cm³ of water drops the temperature by 2.3 °C, so q = 52.98 × 4.18 × 2.3 = 509.4 J absorbed; 2.98 ÷ 74.5 = 0.0400 mol, and 509.4 ÷ 0.0400 = +12735 J mol⁻¹.
What is the general equation for the complete combustion of an alcohol?
The general equation for the complete combustion of a primary alcohol is CₙH₂ₙ₊₁OH + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O. When an alcohol burns completely in excess oxygen, carbon dioxide and water are the only products.
How do you calculate the enthalpy of neutralisation of NaOH and HCl?
The enthalpy of neutralisation of NaOH and HCl is found from the temperature rise of the mixed solutions. Mixing 25.0 cm³ of each at 2.00 mol dm⁻³ raises the temperature by 13.3 °C, so q = 50.0 × 4.18 × 13.3 = 2779.7 J; 0.0500 mol of water forms, giving ΔH = −55.6 kJ mol⁻¹.
What are the most common mistakes in enthalpy calculations?
The most common errors in enthalpy change calculations are sign mistakes, using the wrong mass in q = mcΔT, and forgetting to convert joules to kilojoules. A temperature rise means ΔH is negative, and q must be divided by 1000 before dividing by moles to give kJ mol⁻¹.