Measuring Enthalpy Changes

Enthalpy changes can be measured experimentally using a technique called calorimetry. At A-Level, this typically involves measuring the temperature change of water (or a solution) when a reaction takes place, and then using that temperature change to calculate the energy transferred. This lesson covers the key equation, worked calculations, sources of error, and the graphical extrapolation technique you need for your exams.

The Principle of Calorimetry

The core idea is straightforward: if a reaction occurs in aqueous solution (or heats a known mass of water), the energy released or absorbed by the reaction is transferred to or from the water. By measuring how much the water temperature changes, we can calculate how much energy was transferred.

The key equation is:

q = mcΔT

where:

q = energy transferred (in joules, J)
m = mass of the water or solution being heated (in grams, g)
c = specific heat capacity of water = 4.18 J g⁻¹ K⁻¹
ΔT = temperature change (in K or °C — the magnitude is the same for both)

Common error: Students sometimes use the mass of the solid added rather than the mass of the solution. In solution-based experiments, m is always the mass of the total solution (or the mass of water if you are heating water with a flame).

Worked Example 1: Combustion of Ethanol

A student burns 0.46 g of ethanol (C₂H₅OH, Mᵣ = 46.0) in a spirit burner beneath a copper calorimeter containing 150 g of water. The water temperature rises from 21.0 °C to 33.5 °C.

Step 1: Calculate the energy transferred to the water.

q = mcΔT q = 150 × 4.18 × (33.5 − 21.0) q = 150 × 4.18 × 12.5 q = 7837.5 J = 7.84 kJ

Step 2: Calculate the moles of ethanol burned.

moles = mass / Mᵣ = 0.46 / 46.0 = 0.010 mol

Step 3: Calculate the enthalpy of combustion per mole.

ΔH°c = −q / moles = −7.84 / 0.010 = −784 kJ mol⁻¹

The negative sign is included because combustion is exothermic.

Step 4: Compare with the data book value.

The accepted value for ΔH°c of ethanol is −1367 kJ mol⁻¹. The experimental value (−784 kJ mol⁻¹) is significantly less exothermic. This discrepancy is due to heat losses and other experimental errors (see below).

Worked Example 2: Neutralisation in Solution

A student adds 50.0 cm³ of 1.00 mol dm⁻³ HCl(aq) to 50.0 cm³ of 1.00 mol dm⁻³ NaOH(aq) in a polystyrene cup. The temperature rises from 22.0 °C to 28.8 °C.

Step 1: Calculate the energy transferred.

Total volume of solution = 50.0 + 50.0 = 100.0 cm³

We assume the density of the solution is 1.00 g cm⁻³, so mass = 100.0 g.

q = mcΔT = 100.0 × 4.18 × 6.8 = 2842 J = 2.84 kJ

Step 2: Calculate the moles of water formed.

Moles of HCl = (50.0/1000) × 1.00 = 0.0500 mol Moles of NaOH = (50.0/1000) × 1.00 = 0.0500 mol

The reaction produces one mole of water per mole of acid: moles of H₂O = 0.0500 mol.

Step 3: Calculate the enthalpy of neutralisation.

ΔH°neut = −q / moles = −2.84 / 0.0500 = −56.8 kJ mol⁻¹

This is close to the accepted value of −57.1 kJ mol⁻¹ for strong acid–strong base neutralisation.

Worked Example 3: Dissolving Ammonium Nitrate

A student dissolves 4.00 g of NH₄NO₃ (Mᵣ = 80.0) in 50.0 cm³ of water in a polystyrene cup. The temperature drops from 21.5 °C to 16.3 °C.

Step 1: Calculate the energy transferred.

We assume the mass of solution ≈ 50.0 g (the mass of solute is small compared to the water).

q = mcΔT = 50.0 × 4.18 × (21.5 − 16.3) = 50.0 × 4.18 × 5.2 = 1086.8 J = 1.087 kJ

Step 2: Calculate moles of NH₄NO₃.

moles = 4.00 / 80.0 = 0.0500 mol

Step 3: Calculate ΔH°sol.

The temperature decreased, so the reaction is endothermic. ΔH is positive.

ΔH°sol = +q / moles = +1.087 / 0.0500 = +21.7 kJ mol⁻¹

The accepted value is +25.7 kJ mol⁻¹. The experimental value is less positive, likely because the solution absorbed some heat from the surroundings during the experiment.

Sign check: Temperature drops → endothermic → ΔH is positive. Temperature rises → exothermic → ΔH is negative. The sign of ΔH is always opposite to the direction of the temperature change in the surroundings.

Sources of Error

Calorimetry experiments at A-Level are imprecise. The most important sources of error include:

Source of error	Effect on result	Applies to
Heat lost to surroundings	Less temperature change measured; calculated ΔH is less exothermic than true value	Combustion, neutralisation
Incomplete combustion	Less energy released per mole; calculated ΔH is less exothermic	Combustion
Heat absorbed by the calorimeter (container)	Less temperature change in water; ΔH underestimated	All experiments
Evaporation of volatile liquids	Some fuel lost without heating water	Combustion
Assuming c = 4.18 J g⁻¹ K⁻¹ for solutions	Solutions have slightly different specific heat capacities	Neutralisation, dissolving
Assuming density = 1.00 g cm⁻³	Slight inaccuracy in mass	Neutralisation, dissolving
Heat absorbed from surroundings (endothermic)	Less temperature drop measured; calculated ΔH is less positive	Dissolving endothermic salts

Improving Experimental Accuracy

Several practical modifications can reduce experimental error:

Insulation: Use a polystyrene cup with a lid, or a vacuum flask, to minimise heat loss to the surroundings.
Stirring: Stir the solution continuously to ensure even heat distribution.
Lid and thermometer through lid: Reduces evaporation and convection losses.
Shielding the flame (combustion experiments): A draught screen around the spirit burner reduces heat loss to air currents.
Use a bomb calorimeter (the most accurate method for combustion): The reaction takes place in a sealed, insulated vessel surrounded by a known mass of water, so virtually all heat is captured.

Graphical Extrapolation (Cooling Correction)

In some experiments, the temperature change is not instantaneous. Heat is lost to the surroundings while the reaction is still occurring, so the maximum temperature recorded is lower than the true maximum. We correct for this using graphical extrapolation.

Method:

Record the temperature at regular intervals before adding the reagent.
Add the reagent and continue recording temperature at regular intervals.
Plot temperature (y-axis) against time (x-axis).
Draw the best-fit line for the temperature readings before the reaction.
Draw the best-fit line for the temperature readings after the maximum (the cooling curve).
Extrapolate both lines back to the time of mixing.
The corrected ΔT is the vertical distance between the two extrapolated lines at the time of mixing.

This technique gives a more accurate ΔT by compensating for the heat lost during the experiment.

Worked Example 4: Using Graphical Extrapolation

A student's cooling-corrected graph shows the extrapolated temperature at the time of mixing is 35.2 °C. The initial temperature was 22.0 °C. The corrected ΔT = 35.2 − 22.0 = 13.2 °C.

If the student had used the raw maximum temperature of 33.8 °C, the ΔT would be only 11.8 °C. This underestimates the energy transfer by:

(13.2 − 11.8) / 13.2 × 100 = 10.6%

Graphical extrapolation therefore significantly improves the accuracy of the result.

Real-World Applications

Hand warmers use exothermic reactions. Reusable hand warmers often contain a supersaturated sodium ethanoate solution. Clicking a metal disc triggers crystallisation: NaCH₃COO(aq) → NaCH₃COO(s), releasing heat. Disposable hand warmers use the oxidation of iron powder: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).
Instant cold packs contain a bag of water and a sachet of ammonium nitrate. When the inner sachet is broken, NH₄NO₃ dissolves endothermically, cooling the pack for treating sports injuries.
Calorimetry in food science: The energy content of food (in kJ or kcal) is determined by burning the food in a bomb calorimeter and measuring the temperature rise of the surrounding water.

Summary Checklist

q = mcΔT is the core equation. Get the mass, specific heat capacity, and temperature change correct before anything else.
Always convert q to kJ before dividing by moles.
Always include the sign: negative for exothermic, positive for endothermic.
Be able to explain why experimental values differ from accepted values.
Be able to describe and apply the graphical extrapolation technique.

A-Level Deep Dive: Measuring Enthalpy Changes

Spec mapping

Edexcel 9CH0 specification Topic 8.1 requires use of the relationship q = mcΔT (or Q = mcΔT) to determine enthalpy changes from calorimetric data, including reactions in solution, combustion of liquid fuels, and dissolution (refer to the official specification document for exact wording). The associated Core Practical 1 (CP1) is the experimental anchor: measuring the enthalpy change of a chosen reaction. Calorimetry questions appear on Paper 1 (quantitative physical chemistry) and Paper 2 (combustion of organic liquids); error analysis appears on Paper 3 (synoptic, with strong AO3 weighting). Candidates must be fluent with the cooling-curve extrapolation technique to correct for heat losses, and must be able to state and justify limitations of bench-top calorimetry. The data booklet provides the specific heat capacity of water (4.18 J g⁻¹ K⁻¹) which is to be used in q = mcΔT unless otherwise specified.

Worked example with full mark scheme

Question (8 marks):

A student burned 0.92 g of ethanol (M = 46.0 g mol⁻¹) under a copper calorimeter containing 100.0 g of water. The water temperature rose from 22.4 °C to 47.6 °C. The specific heat capacity of water is 4.18 J g⁻¹ K⁻¹.

(a) Calculate the heat energy gained by the water. (2) (b) Calculate the experimental enthalpy of combustion of ethanol, in kJ mol⁻¹. (3) (c) The data-book value is −1367 kJ mol⁻¹. Calculate the percentage error and identify three reasons why the experimental value differs. (3)

Solution with mark scheme:

(a) ΔT = 47.6 − 22.4 = 25.2 K. q = mcΔT = 100.0 × 4.18 × 25.2.

M1 — correct substitution of m, c, ΔT into the equation. A1 — q = 10 533.6 J ≈ 10 534 J (or 10.53 kJ).

(b) Moles of ethanol burned: n = 0.92 / 46.0 = 0.0200 mol.

M1 — calculation of moles using n = m/M.

ΔH = −q / n = −10 534 / 0.0200 = −526 700 J mol⁻¹ ≈ −527 kJ mol⁻¹.

M1 — division of energy by moles. A1 — sign included (negative for combustion) and units kJ mol⁻¹.

A1 — correct percentage error to appropriate sig figs.

B1 — reason 1: heat lost to surroundings (calorimeter, air, draughts) means q underestimates the true energy released.

B1 — reason 2: incomplete combustion (yellow flame, soot on calorimeter) means less than 0.92 g actually combusted to CO₂ and H₂O.

(Other accepted reasons: evaporation of ethanol before combustion; non-standard conditions; specific heat capacity of the calorimeter not accounted for; use of c for water only.)

Total: 8 marks.

Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks):