The Hess's law experiments are the thermal decomposition of NaHCO₃ and the hydration of MgSO₄ and CuSO₄. Free AQA notes and 15 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Hess's Law: Experimental Applications is part of Energetics in A Level AQA Chemistry.
7 min
Reading time
68
Lessons in this topic
A Level AQA
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
The Hess's law experiments are the thermal decomposition of NaHCO₃ and the hydration of MgSO₄ and CuSO₄. Free AQA notes and 15 questions answered.
Explain how Hess's Law allows enthalpy changes to be determined indirectly through experimental routes
Describe the experimental procedures for measuring enthalpy changes for the thermal decomposition of NaHCO3, hydration of MgSO4, and hydration of CuSO4
Calculate enthalpy changes using Hess's Law cycles constructed from experimental calorimetry data
Evaluate sources of error and uncertainty in calorimetric experiments and suggest improvements
Revise Right Mocks '26
AQA A-Level Chemistry mock draft, 26–31 Oct. Draft coverage: Chemistry 7405/1 · Paper 1. Reviewed for an original-practice sitting; check student availability before registering. Marking and results will follow later, after separate approval. Eligible new accounts receive a five-day Pro trial when signup email verification completes. Previous trials keep their original expiry. After expiry, new papers need Pro unless you have an explicitly confirmed free sitting offer for this event. That offer adds no Pro time or AI learning. Saved attempts and released results remain readable. No card or automatic charge.
Use this lesson to stay inside Energetics while connecting the detail back to the wider Physical chemistry topic.
Hess's Law: Thermodynamic principle stating that the total enthalpy change for a reaction is independent of the route taken, provided initial and final states are the same. Used when a direct enthalpy measurement is experimentally impractical.
Enthalpy of hydration: The enthalpy change when one mole of an anhydrous salt dissolves in water to form a hydrated salt under standard conditions. Used in Hess's Law cycles comparing anhydrous and hydrated salt dissolution.
Enthalpy of solution: The enthalpy change when one mole of a substance dissolves completely in water to form an infinitely dilute solution. Measured directly by calorimetry and used as a step in Hess's Law cycles.
Calorimetry: Experimental technique for measuring heat energy changes by monitoring temperature changes in a known mass of water. The relationship q = mcΔT is used to calculate the heat transferred.
Revision notes
Read the core explanations from Hess's Law: Experimental Applications before testing yourself from memory.
1. Why We Cannot Always Measure Enthalpy Directly
Some reactions refuse to cooperate in a calorimeter. The thermal decomposition of sodium hydrogencarbonate, for example, requires sustained high temperatures and produces a gas — making a clean, direct calorimetric measurement almost impossible to carry out accurately at A-Level. This is where Hess's Law becomes genuinely powerful.
Hess's Law tells us that enthalpy is a state function: the total enthalpy change depends only on the initial and final states, not on the route taken between them. So if you cannot measure a reaction directly, you design two alternative reactions that share the same start and end points, measure those instead, and use the cycle to calculate the value you need.
The three experimental applications you need to know for AQA are: the thermal decomposition of $\text{NaHCO}_3$, the hydration of $\text{MgSO}_4$, and the hydration of $\text{CuSO}_4$. Each one uses the same underlying logic — two measurable dissolution reactions linked by a Hess's Law cycle — but the chemistry of each system is worth understanding on its own terms.
2. The Core Principle: Building a Hess's Law Cycle
Picture the cycle as a triangle. The top of the triangle is the reaction you want ($\Delta H_1$). The two sides of the triangle are the reactions you can measure ($\Delta H_2$ and $\Delta H_3$). Because enthalpy is a state function, going around the triangle must give zero net change: $\Delta H_1 = \Delta H_2 - \Delta H_3$ (or $\Delta H_2 + \Delta H_3$, depending on how the arrows are drawn). Getting the signs right is where most marks are lost.
3. Thermal Decomposition of NaHCO3
The reaction we want to find the enthalpy change for is the thermal decomposition of sodium hydrogencarbonate:
Measuring this directly would require heating the solid to high temperatures and accounting for the escaping $\text{CO}_2$ gas — neither of which is straightforward in a school calorimeter. Instead, both $\text{NaHCO}_3$ and $\text{Na}_2\text{CO}_3$ are dissolved separately in dilute hydrochloric acid, and the two temperature changes are measured. Both reactions share the same final state (a solution of sodium chloride, water, and carbon dioxide), so Hess's Law connects them.
In the Hess's Law cycle, the route through the measurable dissolution reactions must equal the route through the target decomposition reaction because enthalpy depends only on the initial and final states. For the NaHCO3 cycle, the correct relationship is ΔH1 = 2ΔH2 - ΔH3, where ΔH2 is the enthalpy change for NaHCO3 reacting with acid and ΔH3 is the enthalpy change for Na2CO3 reacting with acid. The factor of 2 appears because the target decomposition equation contains 2 mol of NaHCO3.
4. Hydration of MgSO4 and CuSO4
The hydration reactions for magnesium sulphate and copper(II) sulphate follow the same Hess's Law logic, but here the two measurable steps are the enthalpies of solution of the anhydrous salt and the hydrated salt — both dissolved in water rather than acid.
For magnesium sulphate, the target reaction is:
The Hess's Law cycle links three states: anhydrous $\text{MgSO}_4\text{(s)}$, hydrated $\text{MgSO}_4 \cdot 7\text{H}_2\text{O(s)}$, and aqueous solution $\text{MgSO}_4\text{(aq)}$. Dissolving the anhydrous salt directly gives $\Delta H_{\text{sol, anhydrous}}$; dissolving the hydrated salt gives $\Delta H_{\text{sol, hydrated}}$. Because both routes end at the same aqueous solution, the cycle gives:
The same cycle applies to copper(II) sulphate, where the hydrated form is $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$:
Experimentally, the procedure is the same for both salts: dissolve a known mass of the anhydrous salt in a measured volume of water in a polystyrene cup, record the temperature change, then repeat with the hydrated salt. The key difference is that dissolving anhydrous $\text{CuSO}_4$ is strongly exothermic (temperature rises sharply), while dissolving $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ is endothermic or only mildly exothermic. This contrast makes the copper sulphate experiment a particularly clear demonstration of the Hess's Law principle.
5. Calorimetry: The Experimental Method
All three experiments rely on the same calorimetric calculation. You measure the temperature change of the water (or solution), use the mass and specific heat capacity of water to find the heat transferred, then convert to a molar enthalpy value using the number of moles of solid used.
Once you have $q$ in joules, divide by 1000 to convert to kilojoules, then divide by the number of moles of solid dissolved to get $\Delta H$ in kJ mol$^{-1}$. The sign convention is critical: if the temperature of the solution rises, the reaction is exothermic and $\Delta H$ is negative. If the temperature falls, the reaction is endothermic and $\Delta H$ is positive.
Worked example: Calculating Enthalpy of Solution from Calorimetry Data
2.40 g of anhydrous MgSO4 is dissolved in 50.0 cm3 of water. The temperature rises from 19.5 °C to 29.3 °C. Calculate the enthalpy of solution of MgSO4 in kJ mol-1. (Mr of MgSO4 = 120.4 g mol-1)
$\Delta T = 29.3 - 19.5 = 9.8$ °C — Final temperature minus initial temperature. A positive value confirms the solution warmed up — the dissolution is exothermic.
$q = 50.0 \times 4.18 \times 9.8 = 2048.2$ J — We approximate the mass of solution as 50.0 g for this A-level calorimetry method and use c = 4.18 J g-1 °C-1. The arithmetic gives q = 50.0 × 4.18 × 9.8 = 2048.2 J.
$q = \frac{2048.2}{1000} = 2.0482$ kJ — Divide by 1000 to convert from J to kJ. Keep extra significant figures at this stage to avoid rounding errors.
$n = \frac{2.40}{120.4} = 0.01993$ mol — Moles = mass ÷ molar mass. Use the exact mass weighed out, not a rounded value.
$\Delta H_{\text{sol}} = -\frac{2.0482}{0.01993} = -102.8$ kJ mol$^{-1}$ — Divide heat released by moles. The negative sign is applied because the temperature rose — the reaction is exothermic, so the enthalpy change is negative. Round to 3 significant figures: -103 kJ mol-1.
Experimental values often differ from published data because some heat is lost to the surroundings and the mass/specific heat capacity of the whole solution are approximated. — The sign and order of magnitude are reasonable, but the exact literature value depends on data source and experimental conditions, so the comparison should be treated as an approximate check rather than a fixed mark-scheme fact.
6. Applying the Hess's Law Cycle: Full Worked Example
Now that you can calculate individual enthalpies of solution, the next step is combining them in the Hess's Law cycle to find the enthalpy of hydration. This is where the two experimental measurements come together.
Worked example: Calculating Enthalpy of Hydration of CuSO4 Using Hess's Law
A student dissolves 1.60 g of anhydrous CuSO4 in 50.0 cm3 of water and records a temperature rise of 4.20 °C. They then dissolve 2.50 g of CuSO4·5H2O in 50.0 cm3 of water and record a temperature fall of 0.60 °C. Calculate the enthalpy of hydration of CuSO4. (Mr: CuSO4 = 159.6 g mol-1; CuSO4·5H2O = 249.6 g mol-1)
$\Delta H_{\text{hyd}} = \Delta H_{\text{sol, anhydrous}} - \Delta H_{\text{sol, hydrated}}$
$\Delta H_{\text{hyd}} = -87.6 - (+12.5) = -100.1$ kJ mol$^{-1}$ — Subtracting a positive value makes the result more negative. This makes physical sense: forming the hydrated salt from the anhydrous salt releases energy (exothermic), so a negative value is expected.
$\Delta H_{\text{hyd}}(\text{CuSO}_4) = -100$ kJ mol$^{-1}$ (to 3 s.f.) — Always include the sign, the value, and the units. The negative sign confirms the hydration is exothermic, consistent with the strong temperature rise observed when anhydrous CuSO4 is added to water.
7. Evaluating the Experiments: Errors and Improvements
AQA questions frequently ask you to evaluate these experiments — not just describe them. That means identifying specific sources of error, explaining how they affect the result, and suggesting concrete improvements.
The most significant source of error in all three experiments is heat loss to the surroundings. The polystyrene cup is a reasonable insulator, but it is not perfect: heat escapes through the walls, the lid, and the thermometer. This means the measured temperature change is smaller than the true value, so the calculated $|\Delta H|$ is lower than the true value — the experiment underestimates the magnitude of the enthalpy change.
A second source of error specific to the hydration experiments is the reactivity of anhydrous salts with atmospheric moisture. Anhydrous $\text{CuSO}_4$ and $\text{MgSO}_4$ absorb water from the air rapidly, so the solid may already be partially hydrated before it is weighed. This reduces the effective amount of anhydrous salt reacting, making the temperature rise smaller than expected and again underestimating $|\Delta H_{\text{sol, anhydrous}}|$.
Use a lid on the polystyrene cup and surround it with additional insulation to reduce heat loss to the surroundings.
Weigh the anhydrous salt quickly and keep the container sealed until the moment of addition to minimise moisture absorption.
Use a temperature probe with data-logging to capture the maximum (or minimum) temperature more precisely, avoiding the delay inherent in reading a mercury thermometer.
Repeat the experiment at least three times and calculate a mean temperature change to reduce the effect of random errors.
Use a larger mass of solid (and correspondingly larger volume of water) to produce a bigger temperature change, reducing the percentage uncertainty in $\Delta T$.
For the NaHCO3 experiment, an additional consideration is that $\text{CO}_2$ gas escapes during the reaction with HCl. The loss of gas carries some thermal energy with it, contributing to a slightly lower measured temperature change. This is an inherent limitation of the indirect method rather than an avoidable experimental error.
8. Exam Technique: Scoring Full Marks
Hess's Law questions on AQA papers tend to test three things: constructing the cycle correctly, applying the correct sign convention, and evaluating the experimental method. Each of these has specific mark-winning habits worth building now.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
What Hess's law experiments are needed for A Level Chemistry?
The Hess's law experiments for A Level Chemistry are the thermal decomposition of NaHCO₃, the hydration of MgSO₄ and the hydration of CuSO₄. Each uses the same logic: two measurable dissolution reactions, linked by a Hess's law cycle, give the enthalpy change for a reaction that cannot easily be measured directly.
How do you find the enthalpy change for the thermal decomposition of sodium hydrogencarbonate?
The enthalpy change for the thermal decomposition of NaHCO3 is found indirectly: NaHCO3 and Na2CO3 are each reacted separately with dilute hydrochloric acid and the temperature changes are measured. Both reactions share the same final state, so Hess's law gives ΔH1 = 2ΔH2 − ΔH3, the factor of 2 coming from 2 mol of NaHCO3.
How do you calculate the enthalpy of hydration of copper sulphate?
The enthalpy of hydration of CuSO4 is found by measuring the enthalpy of solution of anhydrous CuSO4 and of CuSO4·5H2O separately, then subtracting: enthalpy of hydration = enthalpy of solution of the anhydrous salt − enthalpy of solution of the hydrated salt. In the worked example, −87.6 − (+12.5) gives about −100 kJ mol⁻¹.
What is the main source of error in a calorimetry experiment?
The most significant source of error in calorimetry is heat loss to the surroundings through the walls, the lid and the thermometer of the polystyrene cup. This makes the measured temperature change smaller than the true value, so the experiment underestimates the magnitude of the enthalpy change.
How do you measure the enthalpy of solution of magnesium sulphate?
To measure the enthalpy of solution of magnesium sulphate, dissolve a known mass of the anhydrous salt in a measured volume of water in a polystyrene cup, record the temperature change, calculate q = mcΔT, then divide by the moles dissolved. Dissolving 2.40 g in 50.0 cm³ with a 9.8 °C rise gives −103 kJ mol⁻¹.
How can you improve the accuracy of a calorimetry experiment?
Calorimetry experiments can be improved by using a lid and extra insulation on the polystyrene cup, weighing anhydrous salts quickly and keeping them sealed, using a temperature probe with data-logging, repeating at least three times for a mean, and using a larger mass of solid to reduce the percentage uncertainty in ΔT.
Why can't the thermal decomposition of NaHCO₃ be measured directly?
The thermal decomposition of NaHCO₃ cannot easily be measured directly because it needs sustained high temperatures and produces CO₂ gas, which escapes. Instead, NaHCO₃ and Na₂CO₃ are each dissolved in dilute hydrochloric acid, the temperature changes are measured, and a Hess's law cycle connects the two reactions.
Why is there a factor of 2 in the NaHCO₃ Hess's law cycle?
The NaHCO₃ Hess's law cycle has a factor of 2 because the target decomposition equation contains 2 mol of NaHCO₃. The relationship is ΔH1 = 2ΔH2 − ΔH3, where ΔH2 is for NaHCO₃ reacting with acid and ΔH3 is for Na₂CO₃ reacting with acid.
How do you calculate enthalpy of solution from a temperature change?
Enthalpy of solution is calculated from a temperature change using q = mcΔT, with the mass of water and c = 4.18 J g⁻¹ °C⁻¹. The value of q in joules is divided by 1000 to give kJ, then divided by the moles of solid dissolved to give ΔH in kJ mol⁻¹.
Is ΔH positive or negative if the temperature rises in calorimetry?
A temperature rise in calorimetry means the reaction is exothermic, so ΔH is negative. If the temperature of the solution falls, the reaction is endothermic and ΔH is positive. Applying this sign convention correctly is critical in calculating enthalpies of solution and hydration.
Why does anhydrous copper sulphate get hot when dissolved but the hydrated salt does not?
Dissolving anhydrous CuSO₄ is strongly exothermic, so the temperature rises sharply, while dissolving CuSO₄·5H₂O is endothermic or only mildly exothermic. The anhydrous salt releases extra energy because forming the hydrated salt from the anhydrous salt is exothermic, with an enthalpy of hydration of about −100 kJ mol⁻¹.
What is the Hess's law equation for enthalpy of hydration?
The Hess's law equation for enthalpy of hydration is ΔH(hyd) = ΔH(sol, anhydrous) − ΔH(sol, hydrated). The anhydrous and hydrated salts are each dissolved in water and both routes end at the same aqueous solution, so the difference between their enthalpies of solution gives the hydration enthalpy.
Why does moisture absorption cause an error when using anhydrous salts?
Anhydrous CuSO₄ and MgSO₄ absorb water from the air rapidly, so the solid may already be partially hydrated before it is weighed. This reduces the amount of anhydrous salt reacting, making the temperature rise smaller and underestimating the enthalpy of solution. Weighing quickly and keeping the container sealed reduces this.
Why does escaping CO₂ affect the NaHCO₃ experiment?
Escaping CO₂ affects the NaHCO₃ experiment because the gas carries some thermal energy with it as it leaves during the reaction with HCl, giving a slightly lower measured temperature change. This is an inherent limitation of the indirect method rather than an avoidable experimental error.
How can the percentage uncertainty in a temperature change be reduced?
The percentage uncertainty in a temperature change is reduced by using a larger mass of solid with a correspondingly larger volume of water, which produces a bigger temperature change. Using a temperature probe with data-logging and repeating the experiment at least three times to find a mean also improves the result.