Heat of neutralisation

This page covers one Form 5 Thermochemistry content standard: the heat of neutralisation. It builds directly on exothermic and endothermic reactions and on energy level diagrams, and it is one of the most frequently examined calculations in the whole chapter. Once you can define the quantity precisely and work the calculation confidently, you can answer almost any neutralisation question the exam sets.

What the heat of neutralisation is

The heat of neutralisation is defined as the heat released when one mole of water is formed in the reaction between an acid and an alkali. Neutralisation is the reaction between hydrogen ions from an acid and hydroxide ions from an alkali, and the essential change is always the same:

H⁺(aq) + OH⁻(aq) → H₂O(l)

Because this reaction releases heat to the surroundings, neutralisation is an exothermic process. The temperature of the mixture rises, and the change in heat content, ΔH, is negative. The definition ties the heat to a fixed amount of product, one mole of water, so that different experiments can be compared on the same footing.

Notice the two things that must both appear in a correct definition: the phrase one mole of water and the words acid and alkali (or, more precisely, hydrogen ions and hydroxide ions). Leaving out “one mole of water” is the single most common way students lose the definition mark.

Why strong acid–strong base values are close together

When a strong acid reacts with a strong alkali, both are fully ionised in water. The only reaction that actually happens is H⁺ combining with OH⁻ to make water; the other ions (for example Na⁺ and Cl⁻) are spectator ions and take no part. Because the same essential reaction occurs every time, the heat of neutralisation for any strong acid with any strong alkali comes out at roughly the same value.

When a weak acid or a weak alkali is used, the value is less exothermic (a smaller amount of heat is released per mole of water). This is because a weak acid or weak alkali is only partially ionised, and some of the released energy is used up in ionising the remaining molecules before they can be neutralised. This comparison, strong versus weak, is a favourite explanation question, so learn the reason, not just the fact.

The calculation you must be able to do

Almost every neutralisation question asks you to calculate the heat of neutralisation from experimental data. The method has three fixed steps.

1. Heat released, Q = mcθ, where m is the mass of the solution in grams (take the density of the solution as 1 g cm⁻³, so the volume in cm³ equals the mass in grams), c is the specific heat capacity of the solution (use the value given, 4.2 J g⁻¹ °C⁻¹), and θ is the temperature change in °C.
2. Moles of water formed, found from the moles of acid or alkali that react.
3. Heat of neutralisation, ΔH = −Q ÷ (moles of water), converted to kJ mol⁻¹ and written with a negative sign because the reaction is exothermic.

Worked example

Question. 50 cm³ of 2.0 mol dm⁻³ hydrochloric acid is mixed with 50 cm³ of 2.0 mol dm⁻³ sodium hydroxide in a polystyrene cup. The temperature rises from 28.0 °C to 41.5 °C. Calculate the heat of neutralisation. (Specific heat capacity of solution = 4.2 J g⁻¹ °C⁻¹; density of solution = 1 g cm⁻³.)

Step 1, Mass of solution. Total volume = 50 + 50 = 100 cm³ Mass, m = 100 × 1 = 100 g

Step 2, Temperature change. θ = 41.5 − 28.0 = 13.5 °C

Step 3, Heat released. Q = mcθ = 100 × 4.2 × 13.5 = 5670 J = 5.67 kJ

Step 4, Moles of water formed. Moles of HCl = 0.050 dm³ × 2.0 mol dm⁻³ = 0.10 mol Moles of NaOH = 0.050 × 2.0 = 0.10 mol HCl + NaOH → NaCl + H₂O, so 0.10 mol of water is formed.

Step 5, Heat of neutralisation. ΔH = −Q ÷ moles of water = −5.67 kJ ÷ 0.10 mol = −56.7 kJ mol⁻¹

Answer. The heat of neutralisation is −56.7 kJ mol⁻¹. The negative sign shows the reaction is exothermic.

Practice question

Question. 100 cm³ of 1.0 mol dm⁻³ nitric acid is mixed with 100 cm³ of 1.0 mol dm⁻³ potassium hydroxide. The temperature rises from 29.0 °C to 35.8 °C. Calculate the heat of neutralisation. (c = 4.2 J g⁻¹ °C⁻¹; density = 1 g cm⁻³.)

Answer. Mass of solution = 100 + 100 = 200 g. Temperature change θ = 35.8 − 29.0 = 6.8 °C. Heat released Q = mcθ = 200 × 4.2 × 6.8 = 5712 J = 5.712 kJ. Moles of HNO₃ = 0.100 × 1.0 = 0.100 mol, and moles of KOH = 0.100 mol, so 0.100 mol of water is formed. ΔH = −5.712 kJ ÷ 0.100 mol = −57.1 kJ mol⁻¹. The value is negative, confirming an exothermic reaction, and it is close to the earlier strong acid–strong base result, exactly as expected.

Exam tip

Set your working out in the same fixed order every time, mass, temperature change, Q = mcθ, moles of water, then ΔH, so you never drop a step under pressure. Three details win or lose the marks. First, use the total volume of the mixed solution for m, not the volume of one reactant. Second, keep the negative sign on ΔH because neutralisation is exothermic; a positive answer is wrong and shows you have misread the energy direction. Third, convert joules to kilojoules before you write the final unit of kJ mol⁻¹. When a question compares a strong acid with a weak acid, remember the standard explanation: the weak acid is only partially ionised, so some energy is used to ionise it and less heat is released per mole of water. Quote your answer to a sensible number of decimal places and always attach the unit kJ mol⁻¹.

Where this fits

This is content standard 11.3 of the Thermochemistry chapter. It relies on the ideas of exothermic reactions and the sign of ΔH from the earlier standards, and it uses the same Q = mcθ method that also appears in heat of displacement, precipitation and combustion. Reinforce the definition and the three-step calculation with the chapter revision notes, and drill the arithmetic with the worked examples. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, our teachers make students narrate each step aloud, because a neutralisation calculation loses marks through slips, not through difficulty, and this quantity is tested throughout SPM Chemistry.

Quick recap

• Definition: heat released when one mole of water forms from an acid and an alkali.
• Neutralisation is exothermic, so ΔH is negative.
• Calculation: Q = mcθ, then ΔH = −Q ÷ moles of water, in kJ mol⁻¹.
• Strong acid + strong alkali gives a value that is roughly constant; a weak acid or weak alkali gives a less exothermic value because energy is used up ionising it.