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Learn9701Paper 1 · Multiple ChoiceChemical energeticsEnthalpy changes and Hess's law
Chemical energetics · Subtopic 1 of 1

Enthalpy Changes and Hess's Law

Enthalpy change ΔH\Delta H is heat transferred at constant pressure. Standard conditions: 100 kPa, 298 K, 1 mol dm⁻³. Key definitions: formation (ΔfH⊖\Delta_fH^\ominus: 1 mol of compound from its elements in standard states), combustion (ΔcH⊖\Delta_cH^\ominus: 1 mol burned completely in oxygen), neutralisation (forming 1 mol of water). Calorimetry: q=mcΔTq=mc\Delta T, ΔH=−qn\Delta H=-\frac{q}{n}. Hess's law: the enthalpy change is independent of route, so ΔrH=∑ΔfH(products)−∑ΔfH(reactants)\Delta_rH=\sum\Delta_fH(\text{products})-\sum\Delta_fH(\text{reactants}), or ∑ΔcH(reactants)−∑ΔcH(products)\sum\Delta_cH(\text{reactants})-\sum\Delta_cH(\text{products}). Mean bond enthalpies give approximate values.

Cambridge 9701 Paper 1Cambridge 9701 Paper 2Edexcel YCH11 WCH12AQA 7405 Paper 1≈ 50 minRevision 1 · updated 2026-10-30
Section 1 of 14

Learning objectives

  • Define standard enthalpy changes of reaction, formation, combustion and neutralisation.
  • Calculate ΔH\Delta H from calorimetry.
  • Construct Hess's law cycles using formation and combustion data.
  • Use mean bond enthalpies and explain their limitations.
Pause and think

You can't measure the enthalpy change for C(s) + 2H₂(g) → CH₄(g) directly. How can you still find it?

Section 2 of 14

Core idea

Start hereMeasuring the unmeasurable

Nobody can directly measure the enthalpy change for turning graphite and hydrogen into methane, but chemists know it to the nearest kilojoule. Hess's law makes it possible.

The big idea

Enthalpy change doesn't depend on route. Measurable reactions (combustion, formation, neutralisation) combine in cycles to give any enthalpy change.

1. Hess cycles

Formation cycle

Both routes start from the elements.

Worked exampleDecomposing calcium carbonate

CaCO3→CaO+CO2\mathrm{CaCO_3\rightarrow CaO+CO_2}. ΔfH\Delta_fH: CaCO₃ −1207, CaO −635, CO₂ −394 kJ mol⁻¹.

Goal
ΔrH\Delta_rH.
Strategy
Products − reactants.
  1. ΔrH=(−635−394)−(−1207)\Delta_rH=(-635-394)-(-1207)
    Formation data: P − R.
  2. =+178 kJ mol−1=+178\ \text{kJ mol}^{-1}
    Endothermic.
Answer+178+178 kJ mol⁻¹.
✓ CheckLimestone must be heated strongly ✓.

2. Calorimetry

ΔH=−mcΔTn\Delta H=-\frac{mc\Delta T}{n}
where
m
mass of water or solution heated (g)
c
4.18 J g⁻¹ K⁻¹
n
moles of limiting reactant
Common misconceptionSigns and units qq comes out in J; divide by 1000 for kJ, divide by moles, and add the minus sign if the temperature rose.
Pause and think

Why is the enthalpy of neutralisation of a weak acid less exothermic than that of a strong acid?

Summary

  • Standard definitions.
  • q=mcΔTq=mc\Delta T.
  • Hess: P − R (formation), R − P (combustion).
Key words
Enthalpy change
Heat energy transferred at constant pressure.
Hess's law
Total enthalpy change is independent of the route.
Mean bond enthalpy
Average energy to break one mole of a bond in gaseous molecules.
HistoryGermain Hess. Germain Hess, a Swiss-born Russian chemist, published his law in 1840, before energy conservation was established as a general principle.
Fun factRocket fuel. Space engineers use Hess's law to compare fuels: liquid hydrogen releases about 286 kJ per mole of hydrogen burned, giving one of the highest energies per kilogram of any fuel.
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