Hess's law

Result

Hess's law

T-028Home CU-106Threads thermo

Statement

Because enthalpy \(H\) is a state function — its value depends only on a system's current state, never on the path taken to reach it — the total enthalpy change for a reaction is the same whether it proceeds in one direct step or via any number of intermediate steps. Consequently, thermochemical equations can be reversed (flipping the sign of \(\Delta H\)), scaled (scaling \(\Delta H\) by the same factor), and added together like ordinary algebraic equations, to compute the enthalpy change of a target reaction from the measured enthalpy changes of other, related reactions.

Why it matters

ionic-born-haber already applied exactly this principle, without deriving it explicitly, to extract lattice energy from a five-step thermodynamic cycle; this result establishes the underlying justification directly. Many reactions cannot be measured cleanly or safely in a calorimeter — a reaction might be too slow, might produce an uncontrolled mixture of products, or might be outright hazardous to run directly — and Hess's law is the standard tool for obtaining such a reaction's enthalpy anyway, purely by algebraic combination of other, more easily measured reactions.

Hypotheses

Enthalpy \(H=U+PV\) is a genuine state function.This follows because internal energy \(U\) is itself a state function (a direct consequence of the first law of thermodynamics, energy conservation: the total energy of an isolated system cannot depend on how it arrived at a given state) and because pressure \(P\) and volume \(V\) are themselves state variables; a quantity built entirely from state variables is itself necessarily a state function. All reactions being combined are compared under the same conditions (most commonly constant pressure and a fixed reference temperature, the standard-state convention).Even though \(H\) is path-independent for a fixed initial and final state, \(\Delta H\) itself generally does depend on the temperature and pressure at which a reaction is run; combining tabulated \(\Delta H\) values measured under inconsistent conditions would not give a physically meaningful result. This is why thermochemical data is conventionally tabulated under standard conditions (typically \(298\,\text{K}\), \(1\,\text{bar}\)).

Proof

\Delta H_{A\to D} = \Delta H_{A\to B} + \Delta H_{B\to C} + \Delta H_{C\to D}

Since \(H\) depends only on state (Hypotheses), the total enthalpy change from state \(A\) to state \(D\) equals the sum of the enthalpy changes along any chosen sequence of intermediate states, whether or not the reaction actually, physically proceeds through those specific intermediates. A

\Delta H_{\text{reverse}} = -\Delta H_{\text{forward}}

Consider the closed round trip \(A\to B\to A\): since \(H\) is a state function, returning to the identical starting state gives exactly zero net enthalpy change, \(\Delta H_{A\to B}+\Delta H_{B\to A}=0\); rearranging gives the reversal rule directly. A

\Delta H_{n\times\text{reaction}} = n\times\Delta H_{\text{reaction}}

Enthalpy is an extensive property (it scales with the amount of substance present), so running a reaction on \(n\) times the molar scale multiplies its enthalpy change by exactly \(n\). A

\text{Combine (reverse/scale/add) a set of measured reactions so all intermediate species cancel, leaving exactly the target reaction.}

Applying Steps 1–3 together, any target reaction's \(\Delta H\) can be obtained as the identical algebraic combination of measured \(\Delta H\) values, provided a valid combination of reversed and/or scaled given reactions sums to exactly the target equation. A

Result

\Delta H_{\text{reverse}}=-\Delta H_{\text{forward}}, \qquad \Delta H_{n\times}=n\,\Delta H, \qquad \Delta H_{\text{sum}}=\textstyle\sum\Delta H_i

Reading. Three simple algebraic rules — reversing flips the sign, scaling scales the value, and adding reactions adds their enthalpies — are all that is needed to combine any set of measured thermochemical equations into the enthalpy of a target reaction that need never be run directly.

Scope. Requires every combined reaction to be reported under consistent conditions (Hypotheses) and, critically, requires every cancelled intermediate species to match exactly in physical state (solid/liquid/gas/aqueous), not merely in chemical formula (Common errors).

Corollaries & converses

ionic-born-haber's five-step thermodynamic cycle is a direct application of this same result: an unmeasurable target quantity (there, lattice energy) is isolated by writing an alternative multi-step path between the same start and end states and demanding the two paths' total \(\Delta H\) agree.
This result is the direct theoretical basis for the enthalpy-of-formation shortcut (the next result in this unit), \(\Delta H_{\text{rxn}}=\sum\Delta H_f(\text{products})-\sum\Delta H_f(\text{reactants})\), which is itself just a particularly convenient, standardised application of Hess's law using formation reactions as the universal set of reference "intermediate" steps.
Converse: any closed cycle of reactions that returns exactly to its own starting state must have a net \(\Delta H\) of precisely zero (Step 2's underlying logic, generalised beyond a simple two-step reversal), a useful internal consistency check on any proposed thermochemical cycle.

Fails without

Drop the state-function property of enthalpy (Hypotheses): if \(\Delta H\) genuinely depended on the specific path or mechanism a reaction followed (the way, for example, work \(w\) alone generally does in a non-reversible process), then combining measured step-wise \(\Delta H\) values algebraically would carry no guarantee of matching the true, direct reaction's \(\Delta H\) at all — Hess's law's entire validity rests specifically on \(H\), unlike \(q\) or \(w\) individually, being path-independent.
Cancel a species between two combined equations despite a phase mismatch (e.g. cancelling \(\text{H}_2\text{O}(l)\) from one equation against \(\text{H}_2\text{O}(g)\) from another): these are not the same thermodynamic state, and cancelling across the mismatch silently omits the phase-change enthalpy (e.g. water's enthalpy of vaporisation, \(\approx+44.0\,\text{kJ/mol}\) at \(298\,\text{K}\)) that should have been explicitly included, introducing a real, non-negligible error into the final result (Problems gives a worked case).

Common errors

Reversing a reaction's equation (swapping reactants and products) without also flipping the sign of its \(\Delta H\), or vice versa — both must change together, per Step 2.
Scaling a reaction's stoichiometric coefficients without scaling its \(\Delta H\) by the identical factor (Step 3).
Cancelling a species between two combined equations without checking that its physical state matches exactly on both sides, not merely its chemical formula (Fails without, second bullet).
Assuming Hess's law requires identifying the reaction's true physical or kinetic mechanism; it does not — any algebraically valid combination of reactions works equally well, whether or not it reflects how the reaction actually proceeds step-by-step in reality (Discussion).

Discussion

Germain Henri Hess, a Swiss-born chemist working in Russia, published this law in 1840, based on his own careful calorimetric measurements — notably before the first law of thermodynamics (energy conservation) was formally established by Joule, Helmholtz, and Mayer through the 1840s. Hess's law was therefore an early, specifically chemical empirical discovery of what would soon be recognised as a direct consequence of the more general physical principle of energy conservation, another instance of the empirical-regularity-before-full-theoretical-explanation pattern recurring throughout this network's content (ideal-gas-law, grahams-law-effusion, chirality-optical-activity, and the bonding unit's VSEPR and molecular orbital theory results all share this same historical arc).

A genuinely important and sometimes underappreciated feature of Hess's law is its complete independence from reaction mechanism: the combined "intermediate" reactions used in a Hess's law calculation need not correspond in any way to the actual, physical sequence of elementary steps a reaction follows kinetically. Thermodynamics (governed by state functions like enthalpy, and hence subject to Hess's law) and kinetics (governed by the actual reaction mechanism and rate) are, in this specific sense, entirely separate questions — a reaction's enthalpy change can be computed via Hess's law using a purely hypothetical combination of steps that never physically occur, while its actual rate and mechanism remain governed by an entirely different, unrelated set of considerations.

Common misconception: that a Hess's law calculation must be built from reactions representing the true physical mechanism of the target reaction. As the Discussion establishes, this is not required at all — any mathematically valid combination of measured reactions summing to the target equation gives the correct \(\Delta H\), entirely independent of mechanism.

Worked examples

\text{Target: C(graphite)} + \tfrac12\text{O}_2(g)\to\text{CO}(g),\ \Delta H=?

Direct combustion of carbon to pure \(\text{CO}\) is experimentally difficult, since it tends to over-oxidise partially to \(\text{CO}_2\) as well, contaminating any direct measurement; Hess's law provides the standard practical route around this difficulty. Given: \(\text{C(graphite)}+\text{O}_2(g)\to\text{CO}_2(g)\), \(\Delta H_1=-393.5\,\text{kJ/mol}\) (easily, cleanly measured); \(\text{CO}(g)+\tfrac12\text{O}_2(g)\to\text{CO}_2(g)\), \(\Delta H_2=-283.0\,\text{kJ/mol}\) (also easily, cleanly measured). A

\Delta H_{\text{target}} = \Delta H_1 - \Delta H_2 = -393.5 - (-283.0) = -110.5\,\text{kJ/mol}

Reversing the second reaction (\(\text{CO}_2\to\text{CO}+\tfrac12\text{O}_2\), flipping its sign to \(+283.0\)) and adding it to the first cancels \(\text{CO}_2\) exactly and leaves \(\tfrac12\text{O}_2\) net on the reactant side, reproducing the target equation exactly; this is, in fact, the real, standard method by which \(\Delta H_f(\text{CO})\) is determined in practice, matching the well-established literature value of \(-110.5\,\text{kJ/mol}\) exactly. A

\Delta H_f(\text{CO}) = -110.5\,\text{kJ/mol}\ (\text{matches literature exactly})

Reading. A quantity that resists direct calorimetric measurement is obtained exactly and reliably by combining two other, cleanly measurable reactions according to the three algebraic rules of the Result.

Scope. The identical technique applies to any target reaction expressible as an algebraic combination of reactions with independently known \(\Delta H\) values.

Problems

Given: \(2\text{C(graphite)}+\text{H}_2(g)\to\text{C}_2\text{H}_2(g)\), \(\Delta H_A=+227.0\,\text{kJ/mol}\); \(\text{C(graphite)}+\text{O}_2(g)\to\text{CO}_2(g)\), \(\Delta H_B=-393.5\,\text{kJ/mol}\); \(\text{H}_2(g)+\tfrac12\text{O}_2(g)\to\text{H}_2\text{O}(l)\), \(\Delta H_C=-285.8\,\text{kJ/mol}\). Find \(\Delta H\) for the combustion of acetylene, \(\text{C}_2\text{H}_2(g)+\tfrac52\text{O}_2(g)\to2\text{CO}_2(g)+\text{H}_2\text{O}(l)\).
Solution
Reverse reaction A (\(\Delta H=-227.0\)), double reaction B (\(\Delta H=2\times-393.5=-787.0\)), keep reaction C as given (\(\Delta H=-285.8\)); summing cancels \(2\text{C}\) and \(\text{H}_2\) exactly, leaving the target equation with total oxygen \(2+\tfrac12=\tfrac52\), matching. \(\Delta H_{\text{target}}=-227.0-787.0-285.8=-1299.8\,\text{kJ/mol}\).
A student combines two given reactions, one producing \(\text{H}_2\text{O}(g)\) and another consuming \(\text{H}_2\text{O}(l)\), and cancels the water directly between them without adjustment. Explain the error and state what additional term must be included to correct it.
Solution
\(\text{H}_2\text{O}(g)\) and \(\text{H}_2\text{O}(l)\) are different physical states with different enthalpy content, not interchangeable species for cancellation purposes (Fails without, second bullet). The missing term is water's enthalpy of vaporisation (or condensation, with appropriate sign), \(\text{H}_2\text{O}(l)\to\text{H}_2\text{O}(g)\), \(\Delta H_{\text{vap}}\approx+44.0\,\text{kJ/mol}\) at \(298\,\text{K}\); this phase-change reaction must be explicitly added (with the correct sign, depending on which direction the mismatch runs) to the combination before the two water terms can be validly cancelled.
Confirm, using only the state-function property of enthalpy, that a two-step cycle consisting of melting a solid and then re-freezing it back to the identical starting solid state must have a net \(\Delta H\) of exactly zero.
Solution
Melting (\(\Delta H_{\text{fus}}\)) followed by re-freezing back to the same starting state is a closed round trip, \(A\to B\to A\); by Step 2's reasoning (itself a direct consequence of \(H\) depending only on state, Hypotheses), any closed round trip has \(\Delta H_{\text{net}}=\Delta H_{A\to B}+\Delta H_{B\to A}=0\) exactly — equivalently, freezing's enthalpy change is exactly the negative of melting's (\(\Delta H_{\text{freeze}}=-\Delta H_{\text{fus}}\)), so the two exactly cancel.
Explain why a Hess's law calculation for a given target reaction remains valid even if the specific combination of intermediate reactions used does not correspond to the reaction's actual physical mechanism.
Solution
Hess's law depends entirely on enthalpy being a state function (Hypotheses): \(\Delta H\) for a transformation from a given initial state to a given final state is fixed by those two states alone, regardless of the path taken between them. Since any algebraically valid combination of reactions connects the identical overall initial and final states as the target reaction, the state-function property guarantees the combined \(\Delta H\) must equal the target reaction's true \(\Delta H\), independent of whether that particular combination of steps is what actually happens mechanistically — thermodynamics (state functions, governed by Hess's law) and kinetics (actual reaction pathway and rate) are separate questions, as the Discussion explains.