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Chemical Energetics: Thermochemistry

Topic 2 of 3

Measuring enthalpy

Turn a temperature rise into a molar result.

A-Level 8873, revised syllabus (2026-2027)

Convert a temperature change into heat per mole

First find the surroundings heat, then attach the opposite sign to the reaction.

For a solution approximated as water, q = mcΔT, where m is the mass of the heated solution, c its specific heat capacity and ΔT its temperature change. If c is in J g-1 K-1, use grams and a temperature difference in K or degrees C. The reaction heat is approximately -q when heat exchange with other surroundings and the container is neglected.

Worked example

Neutralisation calorimetry

Mix 50.0 cm3 of 1.00 mol dm-3 HCl and 50.0 cm3 of 1.00 mol dm-3 NaOH. The temperature rises 6.70 K. Assume density 1.00 g cm-3 and c = 4.18 J g-1 K-1.

  1. Total solution mass = (50.0 + 50.0) × 1.00 = 100.0 g. Use both solutions, not only the acid.
  2. q(solution) = 100.0 × 4.18 × 6.70 = 2800.6 J = 2.8006 kJ.
  3. Each reactant supplies 0.0500 mol, forming 0.0500 mol water.
  4. Enthalpy per mole water = -2.8006/0.0500 = -56.012 kJ mol-1.
Answer

The measured enthalpy of neutralisation is -56.0 kJ mol-1, to three significant figures, under the stated approximation.

Worked example

A simple fuel-heating experiment

Burning 0.460 g of a fuel with molar mass 46.0 g mol-1 raises 200 g water by 12.0 K. Estimate its molar combustion enthalpy with c = 4.18 J g-1 K-1.

  1. q(water) = 200 × 4.18 × 12.0 = 10032 J = 10.032 kJ.
  2. Fuel burned = 0.460/46.0 = 0.0100 mol.
  3. Estimated enthalpy = -10.032/0.0100 = -1003.2 kJ mol-1.
Answer

About -1.00 × 103 kJ mol-1. This simple apparatus measurement need not equal a standard data-book value.

Heat lost to the room, heat absorbed by the container and incomplete combustion commonly make the measured temperature rise too small, giving an exothermic value that is less negative than the desired value. Insulation and a lid reduce heat loss; stirring reduces temperature gradients. In a temperature-time experiment, extrapolating the cooling trend back towards the mixing time can estimate the temperature rise before heat loss, provided the method is justified by the data.

Check your understandingIn the neutralisation example, what happens if only 50 g is used for m?Think it through, then reveal the answer
The calculated heat and magnitude of the molar enthalpy are halved. Both mixed solutions were heated, so the relevant mass is about 100 g.