Lesson 3 · 25 min

Hess’s law and enthalpies of formation

Enthalpy only cares where you start and finish. Snap measurable reactions together like building blocks to get ones nobody can measure; one table of standard enthalpies of formation covers thousands of reactions.

In this lesson

Explain Hess’s law from the fact that enthalpy is a state function
Reverse, scale and add known reactions to obtain the ΔH of a target reaction
Define the standard state and the standard enthalpy of formation, and know which substances have ΔfH° = 0
Compute reaction enthalpies with ΔrH° = ΣνΔfH°(products) − ΣνΔfH°(reactants)

What is the enthalpy of formation of carbon monoxide? Measuring it looks easy: burn carbon in a limited supply of oxygen. Unfortunately, once carbon catches fire some of it always burns all the way to $\ce{CO2}$ ; you get a mixture of CO and CO₂, and the heat tells you nothing about either reaction. In 1840 Germain Hess, working in St Petersburg, found a way out: the heat of a reaction depends only on where it starts and where it ends. Burning carbon completely to $\ce{CO2}$ and burning carbon monoxide to $\ce{CO2}$ can both be measured cleanly — “subtract” one from the other and out comes the reaction nobody can measure.

Predict first

Given $\ce{C(s) + O2(g)} \longrightarrow \ce{CO2(g)}$ , ΔH = −393.5 kJ, and $\ce{CO(g) + 1/2O2(g)} \longrightarrow \ce{CO2(g)}$ , ΔH = −283.0 kJ, what is ΔH for $\ce{C(s) + 1/2O2(g)} \longrightarrow \ce{CO(g)}$ ?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

−676.5 kJ −110.5 kJ +110.5 kJ −283.0 kJ

Enthalpy only remembers the start and the finish

Hess’s law: the total enthalpy change of a reaction is the same whether it happens in one step or several. It is not an extra law of nature but a direct consequence of enthalpy being a state function — the height gained from valley to summit does not depend on the path. So thermochemical equations can be handled like algebra:

reverse a reaction and ΔH changes sign;
multiply an equation by any number (fractions included) and ΔH is multiplied too;
add equations and their ΔH values add; species appearing on both sides cancel.

Assemble the target from measurable reactions and you can find the enthalpy change of any reaction — even one that never happens in practice.

Enthalpy is a state function · data from the formation table

Hess’s law cycle builder

Problem

Target reaction

\ce{C}\text{(s, graphite)} + 2\,\ce{H2(g)} \longrightarrow \ce{CH4(g)}

ΔH° = ?

Carbon and hydrogen do not combine directly to methane, but all three combustions are clean.

Known reactions (pick a multiplier; negative = reversed)

1 $\ce{C}\text{(s, graphite)} + \ce{O2(g)} \longrightarrow \ce{CO2(g)}$ ΔH° = −393.5 kJ

Multiplier
2 $\ce{H2(g)} + \tfrac{1}{2}\,\ce{O2(g)} \longrightarrow \ce{H2O(l)}$ ΔH° = −285.8 kJ

Multiplier
3 $\ce{CH4(g)} + 2\,\ce{O2(g)} \longrightarrow \ce{CO2(g)} + 2\,\ce{H2O(l)}$ ΔH° = −890.6 kJ

Multiplier

Sum of the chosen steps

No steps chosen yet.

Each problem offers a few known reactions; choose a multiplier for each until the sum matches the target. Graphite never turns into diamond at ordinary pressure, yet you can compute +1.9 kJ for it; ethyne has a positive enthalpy of formation, which is why an oxy-acetylene flame tops 3000 °C.

Standard enthalpies of formation: one table for every reaction

So that laboratories can compare their numbers, chemists agree on a standard state: pressure $p^\circ = 1$ bar; pure substances in their most stable form at the temperature of interest; solutes at 1 mol/kg. Data are usually listed at 298.15 K, and the superscript ° means “standard”.

The standard enthalpy of formation $\Delta_\mathrm{f}H^\circ$ is the enthalpy change for making one mole of a substance from its elements in their standard states. For example $\Delta_\mathrm{f}H^\circ(\ce{H2O}, \mathrm{l}) = -285.83$ kJ/mol belongs to $\ce{H2(g) + 1/2O2(g)} \longrightarrow \ce{H2O(l)}$ . By definition the reference elements themselves have $\Delta_\mathrm{f}H^\circ = 0$ : $\ce{O2(g)}$ , $\ce{N2(g)}$ , graphite, $\ce{Br2(l)}$ , $\ce{Hg(l)}$ and so on.

With such a table, any reaction can be imagined to travel “through the elements”: take the reactants apart into elements, then build the products from them.

\Delta_\mathrm{r}H^\circ = \sum \nu\,\Delta_\mathrm{f}H^\circ(\text{products}) - \sum \nu\,\Delta_\mathrm{f}H^\circ(\text{reactants})

Taking reactants apart costs −ΣνΔfH°(reactants); building products from the elements gives +ΣνΔfH°(products); the two steps add up to this formula.

$\nu$: stoichiometric coefficient in the balanced equation
$\Delta_\mathrm{f}H^\circ$: standard enthalpy of formation / (kJ/mol), see the atlas data table

Enthalpy is a state function · data from the formation table

Hess’s law cycle builder

Reaction

$\ce{CH4(g)} + 2\,\ce{O2(g)} \longrightarrow \ce{CO2(g)} + 2\,\ce{H2O(l)}$

Species	ν	ΔfH° / (kJ/mol)	ν·ΔfH° / kJ
$\ce{CH4(g)}$	−1	−74.60	74.6
$\ce{O2(g)}$	−2	0.00	0.0
$\ce{CO2(g)}$	+1	−393.51	−393.5
$\ce{H2O(l)}$	+2	−285.83	−571.7

Take apart to elements: −Σν·ΔfH°(reactants) = 74.6 kJ
Build from elements: +Σν·ΔfH°(products) = −965.2 kJ

Reactants on the left (enthalpy relative to the elements), elements in the middle (0), products on the right. The blue arrow takes the reactants apart, the orange one builds the products, and the dashed purple arrow is the direct route. Try thermite and photosynthesis: one falls 850 kJ, the other has to climb 2800 kJ.

Selected standard enthalpies of formation (298.15 K)
Substance	ΔfH° / (kJ/mol)	Substance	ΔfH° / (kJ/mol)
$\ce{H2O(l)}$	−285.83	$\ce{CO2(g)}$	−393.51
$\ce{H2O(g)}$	−241.83	$\ce{CO(g)}$	−110.53
$\ce{CH4(g)}$	−74.6	$\ce{C2H2(g)}$	+227.4
$\ce{NH3(g)}$	−45.94	$\ce{NO(g)}$	+91.3
$\ce{NaCl(s)}$	−411.15	$\ce{CaCO3(s)}$	−1207.6
$\ce{Fe2O3(s)}$	−824.2	$\ce{Al2O3(s)}$	−1675.7
Diamond	+1.895	$\ce{O3(g)}$	+142.7

Source: CODATA Key Values (1989); NBS Tables (1982); CRC Handbook, 97th ed. — full table in the atlas: /atlas/data/thermodynamic-data/.

Worked example

How much heat does thermite release?

Thermite is a mixture of aluminium powder and iron oxide: $\ce{2Al(s) + Fe2O3(s)} \longrightarrow \ce{Al2O3(s) + 2Fe(s)}$ . Use enthalpies of formation to find ΔrH° and the heat released per gram of mixture.

Concept check

Which substance has a standard enthalpy of formation of zero?

\ce{O3(g)}

\ce{Br2(l)}

diamond

\ce{H(g)}

Try the calculation

Quicklime heats up with water (“slaking”): $\ce{CaO(s) + H2O(l)} \longrightarrow \ce{Ca(OH)2(s)}$ . With ΔfH° values CaO −634.92, H₂O(l) −285.83, Ca(OH)₂ −986.09 kJ/mol, find ΔrH° (kJ).

Key takeaways

Hess’s law: the total enthalpy change is path-independent because enthalpy is a state function.
Thermochemical equations may be reversed (ΔH changes sign), scaled (ΔH scales) and added (ΔH adds).
Standard enthalpy of formation: the enthalpy change for making 1 mol from reference-state elements; those elements have ΔfH° = 0.
ΔrH° = ΣνΔfH°(products) − ΣνΔfH°(reactants): one table covers thousands of reactions.

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Data and references

G. H. Hess, “Recherches thermochimiques”, Bull. Sci. Acad. Imp. Sci. St.-Pétersbourg 8, 257 (1840).
J. D. Cox, D. D. Wagman, V. A. Medvedev, CODATA Key Values for Thermodynamics, Hemisphere (1989).
D. D. Wagman et al., “The NBS tables of chemical thermodynamic properties”, J. Phys. Chem. Ref. Data 11, Suppl. 2 (1982).
W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016), “Standard thermodynamic properties of chemical substances”.
P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 2C (thermochemistry).