Skip to notes
wise otter.Chemistry
Contents
8873 / 2027
Chemical Energetics: Thermochemistry

Lesson 2 of 3

Measuring enthalpy

Turn a temperature rise into a molar result.

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

  • I can calculate experimental enthalpy changes.Syllabus

    Syllabus 8873, 6(d). Calculate experimental enthalpy changes.

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 exampleNeutralisation 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.

Try it first. Then show the worked stepsHide the worked steps
  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 exampleA 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.

Try it first. Then show the worked stepsHide the worked steps
  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.

Try it step by step

Work it out in steps: Heat gained by the water, q = mc delta T, then Moles of fuel burned, then Enthalpy of combustion.

  1. Problem 1. Burning 0.320 g of methanol (M = 32.0 g mol-1) raises the temperature of 150 g of water by 11.5 K. Use c = 4.18 J g-1 K-1. Find the enthalpy of combustion in kJ mol-1 to 3 significant figures.
  2. Problem 2. Burning 0.580 g of propanone (M = 58.0 g mol-1) raises the temperature of 100 g of water by 20.0 K. Use c = 4.18 J g-1 K-1. Find the enthalpy of combustion in kJ mol-1 to 3 significant figures.
  3. Problem 3. Burning 0.230 g of ethanol (M = 46.0 g mol-1) raises the temperature of 100 g of water by 12.0 K. Use c = 4.18 J g-1 K-1. Give each answer to 3 significant figures, with the sign of delta H.
Show the answer

Problem 1.

Heat gained by the water, q = mc delta T: q = 150 x 4.18 x 11.5 = 7211 J = 7.21 kJ, gained by the water.

Moles of fuel burned: n(methanol) = 0.320 / 32.0 = 0.0100 mol.

Enthalpy of combustion: delta H = -7.21 / 0.0100 = -721 kJ mol-1. The sign is negative because burning gave heat to the water.

Problem 2.

Heat gained by the water, q = mc delta T: q = 100 x 4.18 x 20.0 = 8360 J = 8.36 kJ.

Moles of fuel burned: n(propanone) = 0.580 / 58.0 = 0.0100 mol.

Enthalpy of combustion: delta H = -8.36 / 0.0100 = -836 kJ mol-1. Heat lost to the air makes this much smaller in size than the data-book value.

Problem 3.

Heat gained by the water, q = mc delta T: q = 100 x 4.18 x 12.0 = 5016 J, which is 5.02 x 103 J to 3 significant figures.

Moles of fuel burned: n(ethanol) = 0.230 / 46.0 = 0.00500 mol.

Enthalpy of combustion: delta H = -5.016 / 0.00500 = -1003 kJ mol-1, which is -1.00 x 103 kJ mol-1 to 3 significant figures.

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 understanding

In the neutralisation example, 50 g of acid was mixed with 50 g of alkali. What happens if only 50 g is used for m in Q = mc(delta T)?

  1. Nothing changes, because the temperature rise is still the same.
  2. The heat calculated, and the size of delta H, come out half as big.
  3. The heat calculated comes out twice as big, as each solution counts once.
Show the answer

The heat calculated, and the size of delta H, come out half as big.

Both solutions were mixed and warmed together, so the mass heated is about 100 g. Using 50 g halves Q and so halves the molar enthalpy change.