TL;DR Enthalpy of neutralisation is the energy change when an acid reacts with a base to form one mole of water. It is always exothermic. The standard value for a strong acid reacting with a strong base is approximately −57.1 kJ/mol. You measure it with a polystyrene-cup calorimeter, applying Q=mcΔT and then scaling to per mole of water formed. Most marks are lost by using the wrong mass, forgetting to convert joules to kilojoules, or omitting the negative sign.
Quick practical map
Neutralisation releases heat: Temperature should rise.
Measure acid/base volumes and temperature change cleanly: Use the same starting-temperature method for both solutions.
Convert heat released into enthalpy per mole of water formed: Use Q=mcΔT
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Concrete example: If 25.0 cm³ of acid reacts with 25.0 cm³ of alkali and the temperature rises by 6.5 °C, use a total mass of about 50.0 g. Calculate Q, find the moles of water formed, then write ΔH with a negative sign because heat was released.
For marking priorities and examiner expectations, pair this walkthrough with the Paper 3 Practical Guide.
1 | What enthalpy of neutralisation is
Enthalpy of neutralisation is defined as the enthalpy change when an acid reacts with a base to produce one mole of water under standard conditions. The ionic equation that underlies every neutralisation of a strong acid with a strong base is:
H+(aq)+OH−(aq)→H2O(l)
Because the same bond (O--H in water) is formed in every case, the enthalpy change is essentially the same for all combinations of strong acid and strong base. The accepted standard value is approximately −57.1 kJ/mol.
The negative sign tells you the reaction is exothermic: energy is released to the surroundings, so the temperature of the solution rises. This temperature rise is what you measure in the experiment.
2 | Apparatus
Item
Purpose
Polystyrene cup (acts as the calorimeter)
Insulates the reaction mixture and minimises heat loss to the surroundings
Lid for the polystyrene cup
Reduces heat loss through evaporation and convection from the top
Measuring cylinder (25 cm³, two needed)
Measures 25 cm³ of acid and 25 cm³ of alkali separately
Thermometer (±0.5 °C or ±0.1 °C)
Measures the temperature of each solution before and after mixing
Glass stirring rod
Stirs the mixture gently to ensure uniform temperature
Dilute hydrochloric acid, HCl (1.0 mol/dm³)
The acid reactant
Sodium hydroxide solution, NaOH (1.0 mol/dm³)
The alkali reactant
Other valid acid--base pairs include sulfuric acid (HâSOâ) with potassium hydroxide (KOH). For strong acid + strong base combinations, the enthalpy of neutralisation per mole of water formed should be close to −57.1 kJ/mol regardless of the pair chosen.
3 | Step-by-step method
Rinse and dry the polystyrene cup. Using a clean measuring cylinder, measure exactly 25.0 cm³ of 1.0 mol/dm³ NaOH solution and pour it into the cup.
Record the initial temperature of the NaOH solution. Let it stand for about one minute so the reading stabilises, then note θNaOH.
Using a second clean measuring cylinder, measure exactly 25.0 cm³ of 1.0 mol/dm³ HCl. Record its initial temperature θHCl.
Calculate the average initial temperature: θi=2θNaOH+θHCl. Ideally both solutions should be at the same temperature; if they differ, the average is used.
Add the acid to the alkali quickly and place the lid on the cup immediately.
Stir gently with the stirring rod.
Record the highest temperature reached by the mixture, θf.
Calculate the temperature rise: ΔT=θf−θi.
4 | Key calculation - Q=mcΔT
The thermal energy released by the reaction is absorbed by the solution. Assuming the solution behaves like water:
Total volume of the mixture = 25 + 25 = 50 cm³
Density of the solution = 1.0 g/cm³, so mass m = 50 g
Specific heat capacityc = 4.18 J/(g · °C)
The energy absorbed by the solution is:
Q=mcΔT
To find the enthalpy of neutralisation per mole of water, you need the number of moles of water formed. From the equation:
HCl+NaOH→NaCl+H2O
the mole ratio is 1 : 1 : 1 : 1. Using 25 cm³ of 1.0 mol/dm³ of each reactant:
n=100025×1.0=0.025 mol
So 0.025 mol of water is formed. The enthalpy of neutralisation is:
ΔH=−nQ
The negative sign is included because the reaction is exothermic (the solution gained energy, so the reaction lost it).
5 | Worked example
Given: 25.0 cm³ of 1.0 mol/dm³ HCl is mixed with 25.0 cm³ of 1.0 mol/dm³ NaOH. The temperature rises from 22.0 °C to 28.5 °C.
Step 1 -- Temperature change:
ΔT=28.5−22.0=6.5Aˆ°C
Step 2 -- Heat absorbed by solution:
Q=mcΔT=50×4.18×6.5=1358.5 J
Step 3 -- Convert to kJ:
Q=10001358.5=1.3585 kJ
Step 4 -- Moles of water formed:
n=100025×1.0=0.025 mol
Step 5 -- Enthalpy of neutralisation:
ΔH=−0.0251.3585=−54.3 kJ/mol
The experimental value of −54.3 kJ/mol is slightly less exothermic than the accepted −57.1 kJ/mol. The difference is mainly due to heat loss to the surroundings during the experiment.
6 | Temperature correction graph (extrapolation method)
In a more precise version of the experiment, you record the temperature of the solution at regular intervals (e.g. every 30 seconds) before, during, and after mixing. You then plot temperature against time and use extrapolation to correct for heat loss.
How to draw the extrapolation
Before mixing: Record the temperature of the alkali every 30 seconds for about 3 minutes. Plot these points. They should form a roughly horizontal line (or a very slight slope).
Mix the acid in at a noted time (e.g. at t=3.0 min). Continue recording the temperature every 30 seconds for another 5 minutes.
After mixing: The temperature rises rapidly, then gradually falls as heat is lost. Plot all these points.
Draw two best-fit lines: one through the pre-mixing points (extended forward) and one through the cooling points after the peak (extended backward). Both lines are extrapolated to the mixing time.
The corrected ΔT is the vertical distance between the two extrapolated lines at the mixing time. This value is always larger than the raw ΔT because it accounts for the heat that was already escaping while you were mixing and recording.
Use this corrected ΔT in your Q=mcΔT calculation for a more accurate result.
7 | Sources of error
For a per-experiment error bank that compares calorimetry alongside titration, QA, rate of reaction, electrolysis, and salt prep, see the O-Level Chemistry sources-of-error reference.
Error
Effect on ΔH
How to reduce it
Heat loss to the surroundings
Measured ΔT is too low, so ΔH is less exothermic than the true value
Use a lid, insulate the cup with cotton wool, or use the extrapolation method
Incomplete or slow mixing
Temperature is not uniform; recorded peak may be lower than the true maximum
Stir gently but promptly; add acid quickly
Thermometer lag
The thermometer reads the peak temperature after the actual peak has passed
Use a digital temperature sensor with a fast response time
Heat absorbed by the calorimeter
Some energy heats the cup and thermometer rather than the solution
Polystyrene absorbs very little energy (low thermal mass), which minimises this error
Imprecise volume measurement
Incorrect mass used in calculation
Use a clean, properly read measuring cylinder; read at the bottom of the meniscus
Solutions not at the same initial temperature
Average initial temperature may not reflect the true starting energy of the system
Allow both solutions to equilibrate to room temperature before starting
8 | Why polystyrene and not glass?
Polystyrene has a much lower thermal conductivity than glass. A polystyrene cup transfers very little heat to the bench or the air, so more of the reaction energy stays in the solution and is detected as a temperature rise.
Glass beakers, by contrast, conduct heat relatively well and have a higher thermal mass, meaning they absorb a noticeable fraction of the reaction energy themselves. Using a glass beaker would produce a smaller ΔT and an experimental ΔH further from the accepted value.
A polystyrene cup is lightweight, disposable, and inexpensive, making it the standard choice for calorimetry experiments at O-Level.
9 | Strong acid vs weak acid - why the values differ
When a strong acid (e.g. HCl) neutralises a strong base (e.g. NaOH), both are fully ionised in solution. All the energy released comes from the formation of water:
H+(aq)+OH−(aq)→H2O(l)ΔH≈−57.1 kJ/mol
When a weak acid (e.g. ethanoic acid, CHâCOOH) is used instead, the acid is only partially ionised. Some of the energy released by forming water must first be used to ionise the undissociated acid molecules:
This endothermic ionisation step partially offsets the exothermic neutralisation. The net result is a measured ∣ΔH∣ that is less than 57.1 kJ/mol. For example, neutralising ethanoic acid with NaOH gives roughly −55 kJ/mol.
The same reasoning applies to weak bases such as ammonia (NHâ). If both the acid and the base are weak, the deviation from −57.1 kJ/mol is even larger.
10 | Common exam mistakes
Mistake 1 -- Forgetting to convert joules to kilojoules
The Q=mcΔT calculation gives the answer in joules. Enthalpy of neutralisation is conventionally reported in kJ/mol. Divide by 1000 before dividing by the number of moles.
Mistake 2 -- Using the wrong mass
The mass in Q=mcΔT is the total mass of the final solution, not the mass of one reactant alone. If you mix 25 cm³ of acid with 25 cm³ of alkali, the total volume is 50 cm³, so the mass is 50 g (assuming density = 1.0 g/cm³).
Mistake 3 -- Omitting or reversing the sign
Enthalpy of neutralisation is exothermic, so ΔH must carry a negative sign. Writing +54.3 kJ/mol instead of −54.3 kJ/mol loses the mark. The sign convention is: negative for exothermic, positive for endothermic.
Mistake 4 -- Not ensuring complete reaction
If one reactant is in excess, some of it remains unreacted and does not contribute to the temperature rise but does contribute to the mass of the solution. This dilutes the temperature change and reduces the apparent ΔH. Use equal moles of acid and alkali (or a known excess and adjust the calculation accordingly).
Mistake 5 -- Misidentifying the moles of water formed
For diprotic acids like HâSOâ, one mole of acid produces two moles of water:
H2SO4+2NaOH→Na2SO4+2H2O
If you use 25 cm³ of 1.0 mol/dm³ HâSOâ with excess NaOH, you form 0.050 mol of water, not 0.025 mol. Always check the balanced equation to find the correct number of moles of water before calculating ΔH per mole.
Mistake 6 -- Confusing specific heat capacity units
At O-Level the specific heat capacity of the solution is taken as 4.18 J/(g · °C), with mass in grams. Do not accidentally use kg in this context without adjusting the value to 4180 J/(kg · °C).
11 | Frequently asked questions
Why is enthalpy of neutralisation always exothermic?
Neutralisation forms water from hydrogen ions and hydroxide ions. The O--H bonds in water are strong, so forming them releases a large amount of energy. This energy release exceeds any energy input needed, making the overall process exothermic.
Can I use a glass beaker instead of a polystyrene cup?
You can, but a glass beaker has higher thermal conductivity and thermal mass. More heat escapes to the surroundings and is absorbed by the beaker itself, giving a smaller measured ΔT and a less accurate result. Polystyrene is strongly preferred.
What if my experimental value is much lower than 57.1 kJ/mol?
The most likely causes are significant heat loss (no lid, slow mixing, long measurement time) or a systematic error in volume measurement. Using the temperature correction (extrapolation) method and ensuring both solutions start at the same temperature will improve accuracy.
Does the concentration of the acid or base affect the enthalpy of neutralisation?
The enthalpy of neutralisation per mole of water formed should be the same regardless of concentration, provided both reactants are strong and fully ionised. However, using very dilute solutions produces a small temperature change that is harder to measure accurately, increasing percentage error.
How does this experiment link to the burette skills tested in Paper 3?
Although this calorimetry experiment typically uses measuring cylinders, the principles of accurate volume measurement and meniscus reading are the same skills tested in titration practicals. For a refresher on burette technique, see the Burette Skills guide.
What would happen if I used excess acid or excess base?
The unreacted excess adds to the total mass of the solution without contributing to the energy release. This means the measured temperature rise is smaller than it would be with stoichiometric amounts, and the calculated ∣ΔH∣ appears lower. Always use equal moles for the most accurate result.
