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Enthalpy experiments: analysis, errors and improvements

Edexcel IAL Chemistry · WCH13 Practical Skills in Chemistry I

Calorimetry questions appear on almost every Unit 3 paper. The method and Core Practical 2 are in the Unit 2 calorimetry and Hess cycle lessons; here you practise the analysis and evaluation that examiners test.

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Taking the readings

Put the solution in a polystyrene cup (insulation) standing in a glass beaker (for stability, not insulation).
Record the temperature every minute for a few minutes before mixing, to get a steady starting value and let the apparatus reach the same temperature.
Add the second reactant at a set time (say 3 minutes). Do not record at that moment: you are adding and stirring.
Keep recording every 30 s or minute afterwards, stirring.
Plot temperature against time and extrapolate the cooling line back to the time of mixing to find ΔT.

The calculation

1. q = m c ΔT: m is the mass of solution (take 1.00 g cm^-3, so 50.0 cm³ is 50.0 g), not including any solid; c = 4.18 J g^-1 °C^-1; ΔT in °C (a change in °C equals a change in K, so do not convert).

2. Find the moles of the limiting reactant (check which is in excess).

3. ΔH = -q / n, converted to kJ mol^-1. Exothermic means negative: include the sign.

4. For neutralisation, give ΔH per mole of water: H2SO4 + 2NaOH makes 2 mol of water per mole of acid.

Assumptions, errors and improvements

Assumptions: the solution has the density and specific heat capacity of water; all the heat goes into the solution; the cup absorbs negligible heat; the reaction is complete.

Main error: heat loss to the surroundings, which makes ΔT too small and ΔH less negative.

Specific improvements: a lid; extra insulation; readings every 30 s or a data logger; a thermometer with higher resolution (0.1 °C) if the change is small; plotting and extrapolating.

With solid salts, the solid may not dissolve fully, or a hydrated salt may not fully form, so the result is unreliable.

Core Practical 2 and Hess's law

Some enthalpy changes cannot be measured directly (such as the decomposition of potassium hydrogencarbonate). Measure two reactions with an acid that can be measured, then combine them with a Hess cycle. The uncertainties of both experiments add together in the final value.

Worked example

50.0 cm³ of 1.00 mol dm^-3 zinc sulfate reacts with excess magnesium powder. ΔT = 35.8 °C. Calculate ΔH for Mg + Zn2+ -> Mg2+ + Zn.

q = 50.0 x 4.18 x 35.8 = 7482.2 J
n(Zn2+) = 0.0500 x 1.00 = 0.0500 mol (Mg in excess)
ΔH = -7482.2 / 0.0500 = -149644 J mol^-1

Answer: -150 kJ mol^-1

Key idea: q = mcΔT with the mass of solution only. Divide by moles of the limiting reactant; give the sign. Per mole of water for neutralisation. Main error is heat loss; improve with a lid, insulation, frequent readings and extrapolation.

Check yourself

25.0 cm³ of 1.00 mol dm^-3 HCl is mixed with 25.0 cm³ of 1.00 mol dm^-3 NaOH. The temperature rises by 6.70 °C. Calculate the enthalpy change of neutralisation (c = 4.18 J g^-1 °C^-1; density 1.00 g cm^-3).
Show answer
-56.0 kJ mol^-1. q = 50.0 x 4.18 x 6.70 = 1400.3 J; n(H2O) = 0.0250 mol; ΔH = -1400.3 / 0.0250 = -56.0 kJ mol^-1.
25.0 cm³ of 1.00 mol dm^-3 H2SO4 is mixed with 50.0 cm³ of 1.00 mol dm^-3 NaOH. ΔT = 8.80 °C. Calculate the enthalpy change per mole of water formed.
Show answer
-55.2 kJ mol^-1. q = 75.0 x 4.18 x 8.80 = 2758.8 J; H2SO4 + 2NaOH -> Na2SO4 + 2H2O, so n(H2O) = 0.0500 mol; ΔH = -55.2 kJ mol^-1.
25.0 cm³ of 0.500 mol dm^-3 copper(II) sulfate reacts with excess zinc powder. ΔT = 26.0 °C. Calculate ΔH for Zn + Cu2+ -> Zn2+ + Cu.
Show answer
-217 kJ mol^-1. q = 25.0 x 4.18 x 26.0 = 2717 J; n(Cu2+) = 0.0125 mol; ΔH = -217 kJ mol^-1.
Why is the polystyrene cup placed inside a glass beaker?
- To stop the cup tipping over
- To insulate the cup
- To act as a heat sink
- To measure the volume
Show answer
To stop the cup tipping over. The June 2024 report: the beaker gives stability, not insulation.
The reactant is added at exactly 3 minutes. Why is no temperature recorded at 3 minutes?
- The temperature does not change at 3 minutes
- The thermometer must be removed to add the reactant
- The student is adding the reactant and stirring at that moment
- Readings are only taken every 2 minutes
Show answer
The student is adding the reactant and stirring at that moment. The thermometer is being used to stir; the value is found by extrapolation instead.
What is the main reason an experimental enthalpy of neutralisation is less exothermic than the data-book value?
- Heat is lost to the surroundings
- The reaction is incomplete
- Too much stirring
- The acid was too concentrated
Show answer
Heat is lost to the surroundings. Heat loss makes ΔT too small.
In one run the thermometer shows no temperature change, although a small change was expected. What is the best explanation?
- The reaction is too slow
- The thermometer is broken
- The thermometer's resolution is too low to show such a small change
- The reaction is endothermic
Show answer
The thermometer's resolution is too low to show such a small change. If the change is smaller than one division, it cannot be seen.

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Last updated 8 November 2026. Spotted a mistake? Tell us.