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5.2 Hess’s Law

5.2 Hess’s Law — A Level Chemistry
Learning outcomes

Candidates should be able to:

  1. apply Hess’s Law to construct simple energy cycles
  2. carry out calculations using cycles and relevant energy terms, including:
    • determining enthalpy changes that cannot be found by direct experiment
    • use of bond energy data
Before you start

You can’t set a match to a mountain path and measure how much height you “burn” — but you could measure the height at the start and the height at the end, however you got there. Enthalpy works the same way. Keep that in mind as you read Hess’s Law below.

1 Hess’s Law: The Principle

The law of conservation of energy states that energy cannot be created or destroyed — this is also known as the first law of thermodynamics. It applies just as much to chemical reactions as to anything else: the total energy of the chemicals and their surroundings must stay constant.

In 1840, the chemist Germain Hess applied this idea specifically to enthalpy changes, giving us:

In other words: it doesn’t matter whether reactants turn into products in one step, or via several intermediate steps — the overall enthalpy change is exactly the same either way, because enthalpy only depends on the initial and final states, not the path taken.

Reactants Products Intermediate ΔH (direct route) ΔH1 ΔH2
Whichever path is taken from reactants to products, the total enthalpy change is the same — so ΔH = ΔH1 + ΔH2. This means an enthalpy change that’s difficult (or impossible) to measure directly can instead be calculated by adding up two or more enthalpy changes that can be measured.

Why Hess’s Law matters: an enthalpy change you cannot measure directly

Some standard enthalpy changes simply cannot be found by direct experiment. A classic example is the standard enthalpy change of formation of carbon monoxide, ΔHf⦵(CO):

C(s) + ½O2(g) → CO(g)

This can’t be measured directly, because burning carbon in a limited supply of oxygen always produces a mixture of CO and CO2 — never pure CO. However, two related enthalpy changes can be measured directly by burning each substance in excess oxygen:

Worked example

Calculate ΔHf⦵(CO) from the following data:

Data C(s) + O2(g) → CO2(g)   ΔH1 = −393.5 kJ mol−1
CO(g) + ½O2(g) → CO2(g)   ΔH2 = −283.0 kJ mol−1
Step 1 · set up the cycle Both routes from C(s) + O2(g) end at CO2(g): the direct route (ΔH1) and the indirect route via CO(g) (ΔHf(CO) then ΔH2). By Hess’s Law: ΔH1 = ΔHf(CO) + ΔH2
Step 2 · rearrange and solve ΔHf(CO) = ΔH1 − ΔH2 = −393.5 − (−283.0) = −110.5 kJ mol−1
2 Enthalpy Change from Enthalpy Changes of Formation

Every compound can be related back to elements in their standard states via its standard enthalpy change of formation, ΔHf⦵. Because every reactant and every product can be “built” from elements, elements make a convenient common reference point for a Hess cycle — one that works for any reaction, as long as you have formation data for every substance involved.

Reactants CH₄(g) + 2O₂(g) Products CO₂(g) + 2H₂O(l) Elements ΔHr (unknown) ΣΔHf(reactants) ΣΔHf(products)
Both arrows point up from elements, because ΔHf is defined as elements → compound. Going the “long way round”: reactants → elements (reverse of ΔHf(reactants)) → products (ΔHf(products)).
Enthalpy change of reaction from formation data ΔHr = ΣΔHf(products) − ΣΔHf(reactants)
Worked example

Calculate the enthalpy change of reaction for the complete combustion of methane:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Data (ΔHf⦵ / kJ mol−1) CH4(g) = −74.8  •  O2(g) = 0  •  CO2(g) = −393.5  •  H2O(l) = −285.8
Step 1 · ΣΔHf(products) [−393.5] + 2[−285.8] = −393.5 − 571.6 = −965.1 kJ mol−1
Step 2 · ΣΔHf(reactants) [−74.8] + 2[0] = −74.8 kJ mol−1
Step 3 · ΔHr = products − reactants ΔHr = −965.1 − (−74.8) = −890.3 kJ mol−1

This matches the experimental value used for methane combustion in Topic 5.1 — the same reaction, calculated by a completely different route.

3 Enthalpy Change from Enthalpy Changes of Combustion

For many organic reactions — especially converting one hydrocarbon or alcohol into another — formation data may not be available, but standard enthalpy changes of combustion, ΔHc⦵, usually are, since they’re easy to measure directly by burning a substance in a bomb calorimeter. Combustion cycles use the combustion products (typically CO2 and H2O) as the common reference point instead of elements.

Reactants C₂H₄(g) + H₂(g) Products C₂H₆(g) Combustion products ΔHr (unknown) ΣΔHc(reactants) ΣΔHc(products)
Both arrows point down to the combustion products, because ΔHc is defined as substance → combustion products. Going the “long way round”: reactants → combustion products (ΔHc(reactants)) → products (reverse of ΔHc(products)).
Enthalpy change of reaction from combustion data ΔHr = ΣΔHc(reactants) − ΣΔHc(products)
Worked example

Calculate the enthalpy change for the hydrogenation of ethene to ethane:

C2H4(g) + H2(g) → C2H6(g)

Data (ΔHc⦵ / kJ mol−1) C2H4(g) = −1411  •  H2(g) = −286  •  C2H6(g) = −1560
Step 1 · ΣΔHc(reactants) (−1411) + (−286) = −1697 kJ mol−1
Step 2 · ΣΔHc(products) −1560 kJ mol−1 (just one product)
Step 3 · ΔHr = reactants − products ΔHr = −1697 − (−1560) = −137 kJ mol−1
4 Bond Energy Cycles & Solving Any Hess Cycle
Connecting back to Topic 5.1

When you used bond energies to calculate ΔH by breaking bonds in the reactants (endothermic) and forming new bonds in the products (exothermic), you were already using a Hess cycle — just with a set of gaseous atoms as the intermediate step, instead of elements or combustion products.

Reactants Products Gaseous atoms ΔHr (unknown) ΣE(bonds broken) ΣE(bonds made)
The same Hess’s Law logic gives ΔHr = ΣE(bonds broken) − ΣE(bonds made) — exactly the bond energy equation from Topic 5.1.

A general method for any Hess cycle question

Whatever data you’re given — formation, combustion, or bond energies — the same method solves every Hess cycle problem:

Write out the equation for the reaction you want ΔH for, and draw it as the top edge of a triangle.

Look at the data you’ve been given and decide which type of cycle it fits: formation data → elements at the bottom; combustion data → combustion products at the bottom; bond energies → gaseous atoms at the bottom.

Draw in the two remaining arrows in the correct direction for that data (formation arrows point away from elements; combustion arrows point towards combustion products).

Apply Hess’s Law: the enthalpy change going one way around the cycle equals the enthalpy change going the other way.

Rearrange to find the unknown ΔH, remembering to reverse the sign of any arrow you travel against.

Exam warning

Reversing the direction you travel along an arrow reverses the sign of that enthalpy change. This is the single most common source of errors in Hess cycle questions — always check whether you’re travelling with or against each arrow before adding or subtracting it.

Practice questions

State Hess’s Law.

Write your answer, then reveal the mark scheme.

Show mark scheme

The enthalpy change accompanying a chemical change is independent of the route by which the change occurs (1), provided the initial and final conditions (states) are the same (1).

Use the enthalpy changes of formation below to calculate ΔH for the complete combustion of ethanol, C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l). (ΔHf: C2H5OH(l) = −277.6, CO2(g) = −393.5, H2O(l) = −285.8, O2(g) = 0, all in kJ mol−1)

Show your working, then reveal the mark scheme.

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ΣΔHf(products) = 2(−393.5) + 3(−285.8) = −787.0 − 857.4 = −1644.4 kJ mol−1 (1)

ΣΔHf(reactants) = −277.6 + 0 = −277.6 kJ mol−1 (1)

ΔHr = −1644.4 − (−277.6) = −1366.8 kJ mol−1 (1)

Use the enthalpy changes of combustion below to calculate ΔH for the hydrogenation of propene, C3H6(g) + H2(g) → C3H8(g). (ΔHc: C3H6(g) = −2058, H2(g) = −286, C3H8(g) = −2220, all in kJ mol−1)

Show your working, then reveal the mark scheme.

Show mark scheme

ΣΔHc(reactants) = −2058 + (−286) = −2344 kJ mol−1 (1)

ΣΔHc(products) = −2220 kJ mol−1 (1)

ΔHr = −2344 − (−2220) = −124 kJ mol−1 (1)

Sulfur dioxide can be oxidised to sulfur trioxide: SO2(g) + ½O2(g) → SO3(g). Given ΔHf[SO2(g)] = −296.8 kJ mol−1 and ΔHf[SO3(g)] = −395.7 kJ mol−1, calculate ΔHr.

Show your working, then reveal the mark scheme.

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ΔHr = ΣΔHf(products) − ΣΔHf(reactants) (1)

ΔHr = (−395.7) − (−296.8) = −98.9 kJ mol−1 (1). (O2(g) is an element in its standard state, so contributes zero.) (1)

Explain, in terms of the law of conservation of energy, why Hess’s Law must be true.

Write your explanation, then reveal the mark scheme.

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Energy cannot be created or destroyed (1). If two different routes between the same initial and final states gave different overall enthalpy changes, energy could be created or destroyed simply by choosing a different route — which would violate the law of conservation of energy (1). So the enthalpy change must be the same regardless of route (1).

Exam Tips & Quick Summary

Do remember

  • Hess’s Law: ΔH depends only on initial and final states, not the route taken.
  • Formation cycles: elements sit at the bottom, with both arrows pointing up (elements → compound).
  • Combustion cycles: combustion products sit at the bottom, with both arrows pointing down (substance → combustion products).
  • Bond energy cycles: gaseous atoms sit at the bottom — one arrow down (breaking, endothermic), one arrow up (making, exothermic).
  • Reversing an arrow reverses the sign of that ΔH.

Do not confuse

  • Formation formula: products − reactants. Combustion formula: reactants − products — the order flips.
  • A Hess cycle is not “extra working” — it’s the only way to find some enthalpy changes at all (e.g. ΔHf of CO).
  • Elements in their standard states always have ΔHf = 0, but this is not true of ΔHc — elements still burn and release energy.
  • Don’t forget to multiply each ΔH by the correct stoichiometric number of moles from the balanced equation.

Recap in 20 seconds

Hess’s Law says enthalpy change depends only on the start and end points, not the path — a direct result of energy conservation. This lets you calculate enthalpy changes that are hard or impossible to measure directly, by building a triangular cycle through a common reference point: elements (formation data, arrows up), combustion products (combustion data, arrows down), or gaseous atoms (bond energies, one arrow each way). Whichever cycle you draw, the enthalpy change going one way round always equals the enthalpy change going the other way.

Hess’s Law Energy cycle Conservation of energy ΔHr from ΔHf ΔHr from ΔHc Bond energy cycles Enthalpy change of formation of CO

📥 Download PDF 5.2 Hess’s law – Notes

📥 Download PDF 5.2 Hess’s Law – MCQs

📥 Download PDF 5.2 Hess’s Law – Structured Questions

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