Heat of neutralisation is the heat released when one mole of water forms from the neutralisation of an acid by an alkali, for example HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l); the ionic equation is H+(aq) + OH−(aq) → H2O(l). It is exothermic.
The heat of neutralisation is the heat released when one mole of water is formed from the reaction between an acid and an alkali. Neutralisation is always exothermic, so the temperature of the mixture rises. This reaction anchors part of the Form 5 Thermochemistry chapter, where the exam tests the equation and ionic equation, the definition, why the value changes with the strength of the acid, and the calculation from the temperature rise.
The balanced equation and ionic equation
For a strong acid and a strong alkali, hydrochloric acid and sodium hydroxide:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Because the strong acid, strong alkali and soluble salt are all fully ionised in solution, the only real change is hydrogen ions joining hydroxide ions to make water. The ionic equation captures this:
H+(aq) + OH−(aq) → H2O(l)
This is why the heat of neutralisation of any strong acid with any strong alkali is almost the same: the reaction being measured is identical in every case. The value released is about 57 kJ per mole of water formed for strong acid and strong alkali.
Conditions required
- Known volumes and concentrations of the acid and the alkali are used.
- They are mixed in an insulated container (a plastic or polystyrene cup) to reduce heat loss.
- The highest temperature reached is recorded with a thermometer.
No heating is applied, the reaction supplies its own heat, and the point is to measure the temperature rise it causes.
Observations
- The temperature of the mixture rises on mixing, confirming an exothermic reaction.
- There is no gas and, for a neutral salt, no precipitate; the change you record is the temperature increase.
How the value is calculated
The heat released is found from the temperature rise using Q = mcθ, where m is the mass of solution, c is its specific heat capacity (taken as 4.2 J per g per degree C, the value for water) and θ is the temperature rise. Dividing this heat by the number of moles of water formed gives the heat of neutralisation per mole. Getting the moles of water right, from the acid and alkali in the smaller amount, is the step that decides the mark.
Why strong and weak differ
The exam often asks why a weak acid, such as ethanoic acid, gives a lower heat of neutralisation than a strong acid. A weak acid is only partly ionised in water; when it is neutralised, some of the heat released must be used to ionise the rest of the acid molecules completely. That energy is taken away from the measured heat, so the value comes out smaller. A strong acid is already fully ionised, so no heat is lost this way and the full value is released.
Common mistakes to avoid
Define the quantity per one mole of water, not per mole of acid. Give the ionic equation H+(aq) + OH−(aq) → H2O(l), the reason the strong-strong value is constant. When calculating, use the mass of the solution (not the water alone) and the moles of the limiting reactant. Explain the lower weak-acid value by the energy used for ionisation, not by “weaker bonds”.
How it appears in the SPM exam
In Paper 2 (4541/2) you may write the equation and ionic equation, define heat of neutralisation, calculate it from a temperature rise with Q = mcθ, or explain the strong-versus-weak difference. In Paper 3 (4541/3), measuring the temperature rise when acid and alkali mix is a set practical. Keep H+(aq) + OH−(aq) → H2O(l) at the centre, and both the theory and the calculation stay consistent.
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