How to calculate heat of neutralisation

Heat of neutralisation is one of the friendliest calculations in the thermochemistry chapter because the definition is fixed and the numbers are tidy. What separates a full-mark answer from a near miss is understanding what you divide by, and being able to explain why a strong acid and a weak acid give different values. This guide covers the definition, the method, a worked example, and the theory examiners like to test.

The definition, get this exactly right

Heat of neutralisation is the heat released when one mole of water is formed from the neutralisation between an acid and an alkali. Two words in that sentence do the heavy lifting: one mole of water. You divide the heat released not by moles of acid in general, but by the moles of water produced.

Neutralisation is always exothermic, so the temperature rises and ΔH is negative. The essential reaction, in ionic form, is simply hydrogen ions joining hydroxide ions:

H⁺ + OH⁻ → H₂O

You can revise the reaction itself on our heat of neutralisation page.

The two formulas

Heat released: Q = mcθ, where m is the mass of the mixture in grams (assume density 1 g cm⁻³, so cm³ ≈ g), c is 4.2 J g⁻¹ °C⁻¹, and θ is the temperature rise in °C. The step-by-step for this part alone is on our heat change of reaction page.

Per mole of water: ΔH = −Q ÷ n, where n is the moles of water formed, and ΔH is quoted in kJ mol⁻¹ with a negative sign.

The method in four steps

1. Mass, m. Add the two volumes and treat the total as grams. Acid + alkali of 25 cm³ + 25 cm³ gives m = 50 g.
2. Temperature change, θ. Highest temperature minus starting temperature.
3. Heat released, Q = mcθ, then divide by 1000 for kilojoules.
4. Moles of water, n, then ΔH = −Q ÷ n. For a monoprotic acid and a monobasic alkali reacting completely, moles of water = moles of acid = moles of alkali.

Worked example

25 cm³ of 2.0 mol dm⁻³ nitric acid is mixed with 25 cm³ of 2.0 mol dm⁻³ potassium hydroxide. The temperature rises from 27.5 °C to 41.0 °C. Calculate the heat of neutralisation.

Step 1, mass. Total volume = 25 + 25 = 50 cm³, so m = 50 g.

Step 2, temperature change. θ = 41.0 − 27.5 = 13.5 °C.

Step 3, heat released.

Q = mcθ = 50 × 4.2 × 13.5 = 2835 J = 2.835 kJ

Step 4, moles of water and ΔH. Moles of HNO₃ = 2.0 × (25 ÷ 1000) = 0.05 mol. KOH is also 0.05 mol, they react 1 : 1, so 0.05 mol of water forms.

ΔH = −2.835 ÷ 0.05 = −56.7 kJ mol⁻¹

The value comes out close to −57 kJ mol⁻¹, which is a useful sanity check for a strong acid and a strong alkali.

Why strong acid + strong alkali is always about −57 kJ mol⁻¹

Here is the theory examiners reward. A strong acid and a strong alkali are fully ionised in water. So whichever strong acid and strong alkali you pick, the only reaction actually happening is H⁺ + OH⁻ → H₂O. Because it is always the same reaction, the heat released per mole of water is always roughly the same value, about −57 kJ mol⁻¹. That is why hydrochloric, nitric or sulfuric acid with sodium or potassium hydroxide all land near the same figure.

Why a weak acid gives a smaller value

If you replace the strong acid with a weak acid such as ethanoic acid, the heat of neutralisation is smaller in magnitude (less negative, for example around −55 kJ mol⁻¹ or lower). A weak acid is only partially ionised. Some of the energy released is used up to ionise the remaining acid molecules completely before their H⁺ ions can react with OH⁻. Because part of the energy is diverted to ionisation, less heat is available to raise the temperature, so ΔH is smaller. Being able to say that sentence is often worth a whole mark.

Common mistakes to avoid

• Dividing by moles of acid when the acid is in excess. Use the moles of water formed, which is set by the limiting reactant.
• Forgetting a diprotic acid gives two moles of water per mole of acid. For sulfuric acid neutralised completely, base the moles of water on the balanced equation.
• Using the mass of solid or just one volume for m. m is the mass of the whole mixture.
• Dropping the negative sign. Neutralisation is exothermic, so ΔH is negative.

Make the method automatic

The order never changes: mass, temperature rise, Q = mcθ, moles of water, then ΔH with a negative sign. Add the strong-versus-weak explanation and you can answer almost any neutralisation question in the thermochemistry chapter. If your child can do the arithmetic but stumbles on the explanation marks, that is exactly what our online one-to-one teachers drill, in English, from RM50 an hour, with a paid one-hour trial. See how it works.