Hess's Law

Diamond does not burn easily at room temperature, so you cannot measure its combustion enthalpy in a calorimeter with any convenience. Yet chemists quote a precise value for it. They do so without ever igniting a diamond directly, by exploiting a deep property of enthalpy itself. Hess's law says that the enthalpy change for a reaction depends only on initial and final states, not on the path taken between them. This single idea lets chemists build inaccessible reaction enthalpies from accessible ones, purely by algebra. It underlies nearly every thermochemical calculation you will do, from formation data tables to bond-enthalpy estimates. This article develops Hess's law from the state-function nature of enthalpy. You will see the formal statement and its proof. Then you will work through two fully numeric examples. Finally you will see where the law breaks down in practice, and meet the chemist who first stated it.

Key takeaways

Enthalpy $H$ is a state function, so $\Delta H$ for a reaction depends only on initial and final states, not on the route.
Hess's law lets you add or subtract chemical equations (with their $\Delta H$ values) to get the $\Delta H$ of a target reaction.
Standard enthalpies of formation are the universal "building blocks" for applying Hess's law to any reaction.
Reverse a reaction and you flip the sign of $\Delta H$ ; multiply a reaction by $n$ and you multiply $\Delta H$ by $n$ .
Hess's law is how you obtain enthalpies for reactions that are slow, dangerous, or practically impossible to run directly.
Key idea: because enthalpy is a state function, any thermochemical cycle that starts and ends at the same states must sum to zero, so unknown $\Delta H$ values can be solved for algebraically.

The core insight

Because enthalpy depends only on the current state of a system and not on how it got there, any two paths connecting the same initial and final states must have the same total $\Delta H$ ; this permits reaction equations, together with their enthalpy changes, to be added, subtracted, and scaled like algebraic equations to obtain the enthalpy of a reaction that was never actually carried out.

The physical picture

Imagine standing on a mountain with an altimeter. Your change in altitude between the base and the summit is fixed, whether you take the steep direct trail or the long winding switchback path. Altitude is a state function: it depends only on where you are, not on the route you walked. Enthalpy behaves the same way for a chemical system held at constant pressure. This matters because many reactions are experimentally awkward to run in one step. The combustion of graphite to carbon dioxide is easy to measure calorimetrically; the combustion of diamond is much harder to control and measure directly with the same precision, and the direct conversion of graphite to diamond is not a reaction you can run in a calorimeter at all — it requires enormous pressure and geological timescales. But if you know the combustion enthalpy of graphite and the combustion enthalpy of diamond, you can combine them to get the enthalpy of converting graphite into diamond, a reaction that seemingly cannot be measured directly. The trick is always the same: draw a cycle. Pick a common set of "reference" species — often the elements in their standard states, as used to define standard enthalpies of formation — and route both the reactants and the products through that reference set. Because enthalpy does not care about the path, going "reactants → reference → products" must give the same total $\Delta H$ as going "reactants → products" directly.

Figure

Hess's law as a closed thermochemical cycle: two paths, one Δ H

The second useful picture is not a cycle diagram but an enthalpy-level diagram — a bar chart of relative enthalpies that makes it visually obvious why adding and reversing steps works.

Figure

Enthalpy level diagram: the same total drop, whatever the intermediate stops

The core relations

\Delta H_{\text{rxn}}^{\circ} = \sum_{\text{path}} \Delta H_i^{\circ} \quad \text{for any path connecting the same initial and final states}

where: the sum runs over any sequence of steps that starts at the reactants and ends at the products; the total is path-independent because $H$ is a state function.

\Delta H_{\text{rxn}}^{\circ} = \sum_{\text{products}} \nu_p \, \Delta H_f^{\circ}(p) - \sum_{\text{reactants}} \nu_r \, \Delta H_f^{\circ}(r)

where: $\nu_p, \nu_r$ are stoichiometric coefficients and $\Delta H_f^{\circ}$ is the standard enthalpy of formation of each species, the standard "reference-route" application of Hess's law.

\Delta H_{\text{reverse}} = -\Delta H_{\text{forward}}, \qquad \Delta H_{n \times \text{rxn}} = n \, \Delta H_{\text{rxn}}

where: reversing a reaction flips the sign of $\Delta H$ , and scaling all coefficients by $n$ scales $\Delta H$ by the same factor.

Derivation AWhy enthalpies of reaction add: from the state-function property to the formation-enthalpy formula

Goal. Show that $\Delta H$ for a target reaction can be obtained by adding the $\Delta H$ values of any set of reactions that sum, equation by equation, to the target — and derive the formation-enthalpy formula as a special case.

Step 1 — Enthalpy is a state function. For a system at constant pressure, $H = U + pV$ , and $H$ is defined entirely by the state variables (composition, $T$ , $p$ ) of the system, not by its history. Consequently, for any process taking the system from state A to state B,

\Delta H_{A \to B} = H(B) - H(A)

regardless of the intermediate states visited along the way. If a second route also goes from A to B via intermediate states $C_1, C_2, \ldots, C_n$ , then

\Delta H_{A \to B} = \big[H(C_1)-H(A)\big] + \big[H(C_2)-H(C_1)\big] + \cdots + \big[H(B)-H(C_n)\big]

because the intermediate $H(C_i)$ terms cancel telescopically, leaving $H(B) - H(A)$ again.

Step 2 — Translate to chemical equations. A chemical equation with its $\Delta H^{\circ}$ is just a statement " $H(\text{products}) - H(\text{reactants}) = \Delta H^{\circ}$ " for that particular pair of states. If reaction (1) plus reaction (2) plus … plus reaction ( $n$ ), added as chemical equations with cancellation of identical species on opposite sides, reproduces the target equation exactly, then the corresponding states A (initial) and B (final) are connected by that composite path. By Step 1, the sum of the $\Delta H^{\circ}$ values of the component reactions must equal $\Delta H^{\circ}$ of the target reaction — the path-dependence cancels out because $H$ doesn't care about the route.

Step 3 — Reversing and scaling follow immediately. If reaction (1) forward is A → B with $\Delta H_1$ , then B → A is the reverse path, and by Step 1 applied to the closed loop A → B → A,

\Delta H_1 + \Delta H_{\text{reverse}} = \Delta H_{A \to A} = 0

since a closed loop returns the system to its starting state (Hess's law applied to a cycle: the total around any closed cycle is zero). Hence $\Delta H_{\text{reverse}} = -\Delta H_1$ . Scaling by $n$ simply repeats the same process $n$ times, so $\Delta H_{n\times} = n\Delta H_1$ .

Step 4 — Specialize to the formation-enthalpy route. Choose, as the universal intermediate state, "the elements in their standard reference states." Every reactant can (in principle) be broken down into its elements, absorbing $-\Delta H_f^{\circ}(\text{reactant})$ per mole (the reverse of formation), and every product can be built up from its elements, releasing $+\Delta H_f^{\circ}(\text{product})$ per mole. Chaining reactants → elements → products via Step 1's telescoping argument gives

\Delta H_{\text{rxn}}^{\circo} = \sum_p \nu_p \Delta H_f^{\circ}(p) - \sum_r \nu_r \Delta H_f^{\circ}(r)

Boxed result.

\boxed{\Delta H_{\text{rxn}}^{\circ} = \sum_{\text{any complete path}} \Delta H_i^{\circ} = \sum_p \nu_p \Delta H_f^{\circ}(p) - \sum_r \nu_r \Delta H_f^{\circ}(r)}

Worked examples

Example 1 — Foundational: graphite to diamond, using combustion data. Given, at 298 K and 1 bar:

\text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g), \quad \Delta H_1^{\circ} = -393.51\ \text{kJ mol}^{-1}

\text{C(diamond)} + \text{O}_2(g) \rightarrow \text{CO}_2(g), \quad \Delta H_2^{\circ} = -395.40\ \text{kJ mol}^{-1}

Find $\Delta H^{\circ}$ for C(graphite) → C(diamond). Step 1. Write the target: C(graphite) → C(diamond). Step 2. Take reaction 1 as is (graphite + O₂ → CO₂), and reverse reaction 2 (CO₂ → diamond + O₂, with $\Delta H = +395.40$ kJ mol⁻¹). Step 3. Add them:

\text{C(graphite)} + \text{O}_2 \rightarrow \text{CO}_2 \qquad \Delta H = -393.51\ \text{kJ mol}^{-1}

\text{CO}_2 \rightarrow \text{C(diamond)} + \text{O}_2 \qquad \Delta H = +395.40\ \text{kJ mol}^{-1}

O₂ and CO₂ cancel on opposite sides, leaving C(graphite) → C(diamond). Step 4. Sum the enthalpies:

\Delta H^{\circ} = -393.51 + 395.40 = +1.89\ \text{kJ mol}^{-1}

Sanity check. Diamond is known to be very slightly higher in enthalpy (and free energy) than graphite at standard conditions — consistent with graphite being the thermodynamically stable form at 1 bar, and consistent with the small positive value obtained (diamond is metastable but the enthalpy difference is tiny, only ~2 kJ mol⁻¹, which is why diamond persists indefinitely at room temperature despite being metastable).

Example 2 — Applied: enthalpy of combustion of gaseous ethanol from formation data, cross-checked via a second route. Data (standard enthalpies of formation, $\Delta H_f^{\circ}$ , kJ mol⁻¹, 298 K):

C₂H₅OH(g): $-235.1$
CO₂(g): $-393.51$
H₂O(g): $-241.82$
O₂(g): $0$ (element in its reference state)

Target reaction:

\text{C}_2\text{H}_5\text{OH}(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(g)

Step 1 — apply the formation-enthalpy formula.

\Delta H_{\text{rxn}}^{\circ} = \big[2(-393.51) + 3(-241.82)\big] - \big[(-235.1) + 3(0)\big]

Step 2 — evaluate the products term.

2(-393.51) + 3(-241.82) = -787.02 - 725.46 = -1512.48\ \text{kJ mol}^{-1}

Step 3 — evaluate the reactants term.

(-235.1) + 0 = -235.1\ \text{kJ mol}^{-1}

Step 4 — subtract.

\Delta H_{\text{rxn}}^{\circ} = -1512.48 - (-235.1) = -1277.38\ \text{kJ mol}^{-1}

Cross-check by an independent Hess's-law route. Build the reaction from elements explicitly. Formation of ethanol is

2\text{C(s)} + 3\text{H}_2(g) + \tfrac12\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(g), \quad \Delta H_f^{\circ} = -235.1\ \text{kJ mol}^{-1}

Reverse this (decompose ethanol into elements, $+235.1$ kJ mol⁻¹). Then form 2 mol CO₂ from 2 mol C + 2 mol O₂ ( $2 \times -393.51 = -787.02$ kJ mol⁻¹), and form 3 mol H₂O(g) from 3 mol H₂ + 1.5 mol O₂ ( $3 \times -241.82 = -725.46$ kJ mol⁻¹). Adding:

\Delta H = 235.1 - 787.02 - 725.46 = -1277.38\ \text{kJ mol}^{-1}

Both routes agree exactly, as Hess's law guarantees. This value (about $-1277$ kJ mol⁻¹ for gaseous ethanol; the commonly tabulated liquid-ethanol combustion value near $-1367$ kJ mol⁻¹ differs because it includes the extra enthalpy released condensing water and ethanol vapor) is close to typical literature figures for gas-phase ethanol combustion, confirming the arithmetic.

Limiting cases & common traps

Situation	Correct handling
Reaction run in reverse	Flip the sign of $\Delta H$ ; magnitude unchanged
Reaction coefficients doubled	Double $\Delta H$ (extensive property)
Elements in their standard reference state	$\Delta H_f^{\circ} = 0$ by definition (e.g., O₂(g), graphite, not diamond)
Physical state of a species changes (g vs l)	Must match the phase in the given data; add phase-change enthalpy if needed
Combining data measured at different temperatures	Strictly requires heat-capacity (Kirchhoff) correction; ordinary Hess's-law sums assume common $T$ , usually 298 K

The chemist behind it

Germain Henri Hess was a Swiss-born chemist who spent most of his career in St. Petersburg, working at a time — the 1830s and 1840s — when calorimetry was primitive and many reactions of interest simply could not be measured directly. Hess performed careful series of dissolution and neutralization experiments and noticed that the total heat evolved was the same however he broke a process into steps. He published this generalization in 1840, well before the first law of thermodynamics was formally established by Joule, Helmholtz, and others in the following decade. Hess's law is thus often called the first quantitative statement of energy conservation applied to chemical reactions, even though it predates a rigorous theoretical foundation for why it must be true.

Origins · 1 minGermain Henri Hess and how this was discovered

Germain Henri Hess

1802–1850 · Swiss-Russian chemist who established, from calorimetric measurements alone, that reaction heat is path-independent — a principle later explained by the first law of thermodynamics

Ask the tutor: How do I extend this same cycle-building logic to compute a lattice enthalpy using a Born-Haber cycle?

Does Hess's law only work for enthalpy, or for other state functions too?

Hess's law is really a statement about any state function, not enthalpy specifically. Internal energy, entropy, Gibbs energy, and Helmholtz energy are all state functions, so additive cycle arguments work identically for each: you can build $\Delta S^{\circ}$ or $\Delta G^{\circ}$ of a reaction from tabulated formation values exactly as shown here for $\Delta H^{\circ}$ . Enthalpy gets the spotlight in introductory thermochemistry because it is the quantity measured directly by constant-pressure calorimetry, the everyday laboratory technique.

Why does reversing a reaction flip the sign of ΔH?

Because a reaction and its reverse, run consecutively, return the system to its exact starting state, tracing a closed loop. Since enthalpy is a state function, the total change around any closed loop must be zero, so the forward and reverse enthalpy changes must be exact negatives of each other. If reversing did not flip the sign, you could complete the round trip and net-gain or net-lose energy for free, violating the first law.

Can Hess's law be applied at any temperature, or only 298 K?

Hess's law itself is temperature-independent as a logical principle: at any fixed temperature and pressure, $\Delta H$ for a reaction is path-independent. The complication is practical: most tabulated $\Delta H_f^{\circ}$ values are referenced to 298.15 K. If your component reactions are measured at different temperatures, you must first bring them to a common temperature using heat capacities (the Kirchhoff equation) before adding them, or the cycle will not close correctly.

How does Hess's law relate to bond enthalpies and the Born-Haber cycle?

Both are direct applications of the same state-function logic, just with a different choice of intermediate reference state. Bond-enthalpy calculations use "atoms in the gas phase" as the intermediate, breaking reactant bonds and forming product bonds. The Born-Haber cycle uses a chain of ionization, atomization, and lattice-formation steps to obtain lattice enthalpies that cannot be measured directly. Both are Hess's law cycles built around a convenient common reference point.

What data do I actually need to apply Hess's law to a new reaction?

At minimum, you need a set of reactions whose sum, added and subtracted appropriately, reproduces your target equation exactly, along with each reaction's $\Delta H^{\circ}$ . In practice, standard enthalpies of formation for every species in the target reaction are the most convenient and complete data set, since any reaction can be built from the "elements" reference route without hunting for a custom combination of side reactions.