What is Hess's Law and how is it applied in Chemical Energetics?

Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the pathway taken, provided the initial and final conditions are the same. In Chemical Energetics, particularly at the Cambridge International A Levels, Hess's Law is used to calculate enthalpy changes for reactions where direct measurement is difficult. By using known enthalpy changes of related reactions, students can apply Hess's Law to determine the enthalpy change of the target reaction.

Why is Hess's Law important in AS-Level Physical Chemistry?

Hess's Law is crucial in AS-Level Physical Chemistry because it allows students to calculate enthalpy changes indirectly. This is particularly useful for reactions that are impractical to measure directly in a laboratory setting. Understanding Hess's Law helps students grasp the concept of energy conservation in chemical reactions, a fundamental principle in thermodynamics and Chemical Energetics.

How can Hess's Law be used to calculate the enthalpy change of a reaction?

To use Hess's Law for calculating the enthalpy change of a reaction, students need to identify a series of reactions that add up to the overall reaction of interest. By summing the enthalpy changes of these individual reactions, which are typically provided or can be found in data tables, the enthalpy change for the overall reaction can be determined. This method relies on the principle that enthalpy is a state function, meaning its change is independent of the path taken.

What are some common mistakes students make when applying Hess's Law?

Common mistakes include not correctly balancing the chemical equations involved, neglecting to reverse the sign of the enthalpy change when a reaction is reversed, and failing to multiply the enthalpy change by the stoichiometric coefficients when necessary. It's important for students to carefully track these details to ensure accurate calculations when applying Hess's Law in Chemical Energetics.

Can you provide an example of a problem involving Hess's Law?

Certainly! Suppose you need to find the enthalpy change for the formation of CO2 from C and O2, but you only have the enthalpy changes for the combustion of C to CO and CO to CO2. Using Hess's Law, you can add these two reactions: C + 1/2 O2 → CO and CO + 1/2 O2 → CO2. By summing their enthalpy changes, you can determine the enthalpy change for the overall reaction C + O2 → CO2. This approach illustrates how Hess's Law facilitates the calculation of enthalpy changes for complex reactions.

Why is "Mixing up the formation and combustion formulae." a common mistake in Hess's Law?

Why it happens: Memorising formulae without the cycle. How to avoid it: Formation = products − reactants; combustion = reactants − products. Or draw the cycle and follow the arrows. Source: 9701 Examiner Reports 2022-2024

Why is "Forgetting to multiply ΔH by the number of moles." a common mistake in Hess's Law?

Why it happens: Ignoring coefficients. How to avoid it: Multiply each ΔHf/ΔHc by its coefficient (e.g. 3 × ΔHc(CO)).

Why is "Using a non-zero ΔHf for an element." a common mistake in Hess's Law?

Why it happens: Not knowing elements have ΔHf = 0. How to avoid it: ΔHf of an element in its standard state is zero.

Why is "Sign slips when subtracting a negative value." a common mistake in Hess's Law?

Why it happens: −(−890) handled as −890. How to avoid it: Subtracting a negative adds: −[(−890)] = +890. Write the brackets out before evaluating.

Chemical Energetics(AS-Level Physical Chemistry)Cambridge International A Level Chemistry 9701

Hess's Law

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Hess's Law — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

Hess's law and energy cycles: how to find an enthalpy change that cannot be measured directly, using formation data, combustion data and bond energies.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

5.2.1 — Apply Hess's law to construct simple energy cycles.
5.2.2 — Carry out calculations using energy cycles and relevant energy terms, including (a) determining enthalpy changes that cannot be found by direct experiment and (b) using bond energy data.

Hess's law and why it works

Enthalpy change depends only on start and end points, not the route between them.

Hess's law. The total enthalpy change of a reaction is independent of the route taken from reactants to products (provided the initial and final conditions are the same).

This works because enthalpy is a state function — it depends only on the current state, not how you got there. So you can replace a reaction you can't measure with a sequence of reactions you can, and the total ΔH is identical.

This is the key to finding enthalpy changes that cannot be measured directly, such as the enthalpy of formation of a compound that won't form cleanly from its elements.

ΔH is route-independent (state function).
Swap an unmeasurable reaction for a measurable alternative route.
Total ΔH is the same either way.

Cycles using enthalpy of formation

Arrows go UP from elements to reactants and products; ΔHr = ΣΔHf(products) − ΣΔHf(reactants).

When you are given enthalpy of formation data, build a cycle with the elements at the bottom. The formation arrows point up from the elements to the reactants and to the products.

$\Delta H_r = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$

Both formation arrows point up from the elements. Going reactants → elements → products: ΔHr = ΣΔHf(products) − ΣΔHf(reactants).

Worked. Find ΔHr for $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ , given ΔHf: CaCO₃ −1207, CaO −635, CO₂ −394 (kJ mol⁻¹): $\Delta H_r = [(-635) + (-394)] - [-1207] = -1029 + 1207 = +178\,\text{kJ mol}^{-1}$ (Endothermic — consistent with thermal decomposition needing heat.)

Elements at the bottom; ΔHf arrows point up.
ΔHr = ΣΔHf(products) − ΣΔHf(reactants).
ΔHf of an element = 0.

Cycles using enthalpy of combustion

Arrows go DOWN to the combustion products; ΔHr = ΣΔHc(reactants) − ΣΔHc(products).

When given enthalpy of combustion data, build a cycle with the combustion products at the bottom (CO₂ and H₂O). The combustion arrows point down.

$\Delta H_r = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$

Note this is the opposite arrangement to the formation cycle (reactants − products), because the arrows now point down rather than up.

Worked. Find ΔHf of ethanol from combustion data — a classic 'can't measure directly' problem. For $2\text{C} + 3\text{H}_2 + \tfrac{1}{2}\text{O}_2 \rightarrow \text{C}_2\text{H}_5\text{OH}$ , using ΔHc: C −394, H₂ −286, ethanol −1367 (kJ mol⁻¹): $\Delta H_f = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$ $= [2(-394) + 3(-286)] - [-1367] = (-788 - 858) + 1367 = -279\,\text{kJ mol}^{-1}$

Tip for direction. A reliable method is to follow the arrows of the cycle: travelling with an arrow adds its ΔH, travelling against an arrow subtracts it.

Combustion products (CO₂, H₂O) at the bottom; arrows down.
ΔHr = ΣΔHc(reactants) − ΣΔHc(products).
Opposite arrangement to the formation cycle.

How it’s examined

Hess cycles are a guaranteed multi-mark Paper 2 question — usually finding a ΔHf or ΔHr from formation or combustion data. Examiner reports flag: using the wrong formula direction (products − reactants vs reactants − products), missing stoichiometric multipliers, and arithmetic sign slips. Drawing the cycle prevents most of these.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 5.2; 9701 Examiner Reports 2022-2024; 9701/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-20.

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Step-by-step worked examples — Hess's Law

Step-by-step solutions to past-paper-style questions on hess's law, written exactly the way a tutor would explain them at the board.

1Stating and using Hess's law (2 marks)
Getting started• concept
Question
State Hess's law and explain why it lets the enthalpy of formation of methane be found indirectly.
Step-by-step solution
1. Step 1
  Hess's law: the total enthalpy change is independent of the route taken (provided start and end conditions are the same).
2. Step 2
  Application: methane cannot be made cleanly from C and H₂ directly, so a measurable route (via combustion) gives the same ΔH.
Answer
ΔH is route-independent, so a measurable alternative route gives the unmeasurable ΔHf.
Examiner tip
'Independent of route' is the key phrase — it is why energy cycles work.
2ΔHr from formation data (2 marks)
Getting started• formation
Question
For $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ , ΔHf(NH₃) = −46 kJ mol⁻¹. Calculate ΔHr.
Step-by-step solution
1. Step 1
  Products − reactants; elements (N₂, H₂) have ΔHf = 0.
  $\Delta H_r = [2(-46)] - [0 + 0]$
2. Step 2
  Evaluate.
  $\Delta H_r = -92\,\text{kJ mol}^{-1}$
Answer
$\Delta H_r = -92\,\text{kJ mol}^{-1}$ .
Examiner tip
Multiply ΔHf by the coefficient (2 × NH₃) and remember elements contribute zero.
3ΔHr from formation data, full cycle (3 marks)
Building confidence• formation
Question
Find ΔHr for $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ . ΔHf: Fe₂O₃ −824, CO −111, CO₂ −394 (kJ mol⁻¹).
Step-by-step solution
1. Step 1
  ΔHr = ΣΔHf(products) − ΣΔHf(reactants). Fe is an element (ΔHf = 0).
  $= [2(0) + 3(-394)] - [(-824) + 3(-111)]$
2. Step 2
  Evaluate.
  $= (-1182) - (-824 - 333) = -1182 + 1157$
3. Step 3
  Result.
  $\Delta H_r = -25\,\text{kJ mol}^{-1}$
Answer
$\Delta H_r = -25\,\text{kJ mol}^{-1}$ .
Examiner tip
Use ΔHf(Fe) = 0 and multiply CO and CO₂ by 3. Products − reactants for formation cycles.
4ΔHf from combustion data (4 marks)
Building confidence• combustion
Question
Find ΔHf of methane for $\text{C} + 2\text{H}_2 \rightarrow \text{CH}_4$ . ΔHc: C −394, H₂ −286, CH₄ −890 (kJ mol⁻¹).
Step-by-step solution
1. Step 1
  ΔHf = ΣΔHc(reactants) − ΣΔHc(products).
  $= [(-394) + 2(-286)] - [(-890)]$
2. Step 2
  Evaluate.
  $= (-394 - 572) + 890 = -966 + 890$
3. Step 3
  Result.
  $\Delta H_f = -76\,\text{kJ mol}^{-1}$
Answer
$\Delta H_f(\text{CH}_4) = -76\,\text{kJ mol}^{-1}$ .
Examiner tip
For combustion cycles use reactants − products (the arrows point down). Multiply H₂ by 2.
5ΔHc from formation data (4 marks)
Stretch• formation, combustion
Question
Find ΔHc of propane: $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$ . ΔHf: C₃H₈ −104, CO₂ −394, H₂O(l) −286 (kJ mol⁻¹).
Step-by-step solution
1. Step 1
  ΔHc = ΣΔHf(products) − ΣΔHf(reactants); O₂ is an element (0).
  $= [3(-394) + 4(-286)] - [(-104) + 5(0)]$
2. Step 2
  Evaluate the brackets.
  $= (-1182 - 1144) - (-104) = -2326 + 104$
3. Step 3
  Result.
  $\Delta H_c = -2222\,\text{kJ mol}^{-1}$
Answer
$\Delta H_c(\text{C}_3\text{H}_8) = -2222\,\text{kJ mol}^{-1}$ .
Examiner tip
Apply the coefficients (3 CO₂, 4 H₂O) carefully — a missed multiplier is the usual lost mark.
6ΔHr of hydrogenation from combustion data (4 marks)
Stretch• combustion
Question
Find ΔHr for $\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6$ . ΔHc: C₂H₄ −1411, H₂ −286, C₂H₆ −1560 (kJ mol⁻¹).
Step-by-step solution
1. Step 1
  ΔHr = ΣΔHc(reactants) − ΣΔHc(products).
  $= [(-1411) + (-286)] - [(-1560)]$
2. Step 2
  Evaluate.
  $= -1697 + 1560$
3. Step 3
  Result.
  $\Delta H_r = -137\,\text{kJ mol}^{-1}$
Answer
$\Delta H_r = -137\,\text{kJ mol}^{-1}$ .
Examiner tip
Combustion data → reactants − products. The double negative (−(−1560)) is where signs go wrong.

Key Formulae — Hess's Law

The formulae you need to memorise for hess's law on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Hess (formation data)
$\Delta H_r = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
When to use
When given enthalpies of formation (elements at the bottom of the cycle).
Hess (combustion data)
$\Delta H_r = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$
When to use
When given enthalpies of combustion (combustion products at the bottom of the cycle).
Hess's law (route independence)
$\Delta H_{\text{direct}} = \sum \Delta H_{\text{indirect steps}}$
When to use
The underlying principle: the direct ΔH equals the sum of any indirect route's steps.

Key Definitions and Keywords — Hess's Law

Definitions to memorise and the exact keywords mark schemes credit for hess's law answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Hess's law
Examiner keyword
The total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same.
Standard enthalpy of reaction (ΔHr⦵)
Examiner keyword
The enthalpy change when the molar quantities in the equation react under standard conditions.
Energy (Hess) cycle
Examiner keyword
A diagram of alternative routes between the same reactants and products, used to calculate an unknown ΔH.
ΔHf of an element
Examiner keyword
The standard enthalpy of formation of any element in its standard state is zero (it forms from itself).

Common Mistakes and Misconceptions — Hess's Law

The traps other students keep falling into on hess's law questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Mixing up the formation and combustion formulae.
9701 Examiner Reports 2022-2024
Why it happens
Memorising formulae without the cycle.
How to avoid it
Formation = products − reactants; combustion = reactants − products. Or draw the cycle and follow the arrows.
✕Forgetting to multiply ΔH by the number of moles.
Why it happens
Ignoring coefficients.
How to avoid it
Multiply each ΔHf/ΔHc by its coefficient (e.g. 3 × ΔHc(CO)).
✕Using a non-zero ΔHf for an element.
Why it happens
Not knowing elements have ΔHf = 0.
How to avoid it
ΔHf of an element in its standard state is zero.
✕Sign slips when subtracting a negative value.
Why it happens
−(−890) handled as −890.
How to avoid it
Subtracting a negative adds: −[(−890)] = +890. Write the brackets out before evaluating.

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Hess's Law

Short Study Notes in the form of Slides

Short Study Notes — Hess's Law

Detailed Notes

What you’ll learn

How it’s examined

1Stating and using Hess's law (2 marks)

2ΔHr from formation data (2 marks)

3ΔHr from formation data, full cycle (3 marks)

4ΔHf from combustion data (4 marks)

5ΔHc from formation data (4 marks)

6ΔHr of hydrogenation from combustion data (4 marks)

Hess (formation data)

Hess (combustion data)

Hess's law (route independence)

Hess's law

Standard enthalpy of reaction (ΔHr⦵)

Energy (Hess) cycle

ΔHf of an element

✕Mixing up the formation and combustion formulae.

✕Forgetting to multiply ΔH by the number of moles.

✕Using a non-zero ΔHf for an element.

✕Sign slips when subtracting a negative value.

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Hess's Law — frequently asked questions