Hess's Law of Heat Summation

The overall enthalpy change of a reaction equals the sum of the enthalpy changes of the steps that compose it.

\Delta H^{\circ}_{\text{rxn}} = \sum \Delta H^{\circ}_{\text{steps}}

ΔH°rxn = Σ ΔH°(steps) — overall enthalpy = sum of step enthalpies

What it means

Hess's law is the practical payoff of enthalpy being a state function: because the enthalpy difference between start and finish does not depend on the route taken, you can compute an unmeasurable reaction's ΔH by walking a measurable detour. If you can arrange known steps so their equations add up to your target equation — reversing and rescaling steps as needed — the enthalpies add up the same way. When to use it: for reactions too slow, too fast, or too messy to measure directly (carbon burning to CO2 goes through CO), and for building formation enthalpies out of combustion data.

Variables

Symbol	Name	Unit	Meaning
$ΔH°rxn$	Reaction enthalpy	kJ/mol	The enthalpy change of the overall target reaction.
$ΔH°_step$	Step enthalpy	kJ/mol	The enthalpy change of each component step, signed and scaled to match how it is used.

Assumptions

All steps are added under consistent standard-state conditions (25°C, 1 atm).
Enthalpy is a state function — only initial and final states matter, never the pathway.
Steps that are reversed flip the sign of their ΔH; steps multiplied by a factor scale their ΔH by the same factor.

Worked example

Problem

Find ΔH° for C(s) + O2(g) → CO2(g) given the two formation steps above.

Given:

Step 1: C(s) + ½O2(g) → CO(g), ΔH° = −110.5 kJ/mol
Step 2: CO(g) + ½O2(g) → CO2(g), ΔH° = −283.0 kJ/mol
Steps add to the target: C(s) + O2(g) → CO2(g)

Check that the steps sum to the target: Step 1 + Step 2 gives C + ½O2 + CO + ½O2 → CO + CO2.
Cancel the CO that appears on both sides, and ½O2 + ½O2 = O2: net C + O2 → CO2. The steps match the target.
Add the enthalpies: ΔH° = (−110.5) + (−283.0) kJ/mol.
Sum: −110.5 − 283.0 = −393.5 kJ/mol, which rounds to −394 kJ/mol.

ΔH° = −394 kJ/mol for C + O2 → CO2

Carbon releases the same total heat whether it burns to CO2 in one step or via CO — the pathway is irrelevant, only the start and end matter.