Hess's Law — Enthalpy by Indirect Routes

By Aniket Bhardwaj · 16 September 2026 · Calculator/Formula Guide

Some enthalpy changes simply cannot be measured. You cannot burn carbon to carbon monoxide and nothing else — some CO₂ always forms. You cannot put graphite and hydrogen in a calorimeter and get methane. Hess's law is what rescues these cases: it lets you reach an enthalpy change you cannot measure by adding up several that you can.

The law, and why it is true

Hess's law: the enthalpy change of a reaction is the same whether it happens in one step or in several, provided the initial and final states are the same.

This is not a separate law of nature. Enthalpy is a state function — its value depends only on the state of the system, not on how the system got there — so the change ΔH between two states cannot depend on the path. That in turn follows from the first law of thermodynamics: if a path dependence existed, you could go around a cycle and come back with free energy.

The three manipulation rules

Nearly every Hess's law question is solved by rearranging given equations until they add up to the target. Three rules govern that:

What you do to the equation	What happens to ΔH
Reverse it	Change the sign
Multiply it by a number	Multiply ΔH by the same number
Add two equations	Add their ΔH values

A practical method: look at the target equation, decide where each species must end up, and arrange the given equations so that everything not in the target cancels between the two sides.

Two shortcut formulas — and the reversal that catches people out

From formation enthalpies: Δ_rH = Σ Δ_fH(products) − Σ Δ_fH(reactants)

From combustion enthalpies: Δ_rH = Σ Δ_cH(reactants) − Σ Δ_cH(products)

The two orders are opposite, and that is not a misprint. Formation enthalpies point towards the compounds, so products come first. Combustion enthalpies point away from them towards a common set of oxides, so reactants come first. Both are just Hess's law with the cancelling already done. Every term must be multiplied by its stoichiometric coefficient.

Remember also that Δ_fH of an element in its standard state is zero — but only in its standard state. Δ_fH(C, graphite) = 0 while Δ_fH(C, diamond) = +1.9 kJ mol⁻¹, and Δ_fH(O₂, g) = 0 while Δ_fH(O₃, g) is strongly positive.

Worked example 1 — the classic methane problem

Find Δ_fH of methane, C(s) + 2H₂(g) → CH₄(g), from:

(1) C(s) + O₂(g) → CO₂(g)    ΔH₁ = −393.5 kJ mol⁻¹
(2) H₂(g) + ½O₂(g) → H₂O(l)    ΔH₂ = −285.8 kJ mol⁻¹
(3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)    ΔH₃ = −890.3 kJ mol⁻¹

Step 1 — place the carbon. C(s) must be a reactant in the target, and it already is in (1). Keep equation (1) as it is.

Step 2 — place the hydrogen. The target needs 2H₂ as a reactant. Equation (2) has one H₂ as a reactant, so use 2 × (2), which also doubles ΔH₂.

Step 3 — place the methane. CH₄ must be a product, but in (3) it is a reactant. Reverse (3), which changes the sign of ΔH₃.

Step 4 — add.
Δ_fH = ΔH₁ + 2ΔH₂ − ΔH₃
= (−393.5) + 2(−285.8) − (−890.3)
= −393.5 − 571.6 + 890.3
= −965.1 + 890.3 = −74.8 kJ mol⁻¹

Check the cancellation. On the left we have C + 2H₂ + 3O₂ + CO₂ + 2H₂O; on the right CO₂ + 2H₂O + 3O₂ + CH₄. Everything except C, 2H₂ and CH₄ appears on both sides and cancels, leaving exactly the target equation ✔

The tabulated value of Δ_fH(CH₄, g) is −74.8 kJ mol⁻¹.

Worked example 2 — carbon monoxide, the one you cannot measure

Find ΔH for C(s) + ½O₂(g) → CO(g), given:

(1) C(s) + O₂(g) → CO₂(g) ΔH₁ = −393.5 kJ mol⁻¹
(2) CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = −283.0 kJ mol⁻¹

CO must end up as a product, so reverse (2) and add it to (1):

ΔH = ΔH₁ − ΔH₂ = (−393.5) − (−283.0) = −393.5 + 283.0 = −110.5 kJ mol⁻¹

This is the value listed in data tables as Δ_fH(CO, g), and it was obtained in exactly this way — the direct experiment is impossible because burning carbon in limited oxygen always gives a mixture.

Worked example 3 — straight from a table of formation enthalpies

Find Δ_rH for the complete combustion of ethane:
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)

Using Δ_fH values: C₂H₆(g) = −84.7, CO₂(g) = −393.5, H₂O(l) = −285.8, and O₂(g) = 0 because it is an element in its standard state.

Products: 4(−393.5) + 6(−285.8) = −1574.0 + (−1714.8) = −3288.8 kJ
Reactants: 2(−84.7) + 7(0) = −169.4 kJ

Δ_rH = (−3288.8) − (−169.4) = −3288.8 + 169.4 = −3119.4 kJ

That is for two moles of ethane, as the equation is written. Per mole: −3119.4 ÷ 2 = −1559.7 kJ mol⁻¹, which agrees with the tabulated enthalpy of combustion of ethane, close to −1560 kJ mol⁻¹ ✔

Worked example 4 — the same answer by the combustion route

To prove the two shortcut formulas really are one law, redo example 1 using combustion data. The target is again C(s) + 2H₂(g) → CH₄(g).

Δ_cH(C, s) = −393.5, Δ_cH(H₂, g) = −285.8, Δ_cH(CH₄, g) = −890.3 kJ mol⁻¹ — the same three numbers, read as combustion enthalpies this time.

Δ_rH = Σ Δ_cH(reactants) − Σ Δ_cH(products)
= [(−393.5) + 2(−285.8)] − [(−890.3)]
= (−965.1) + 890.3 = −74.8 kJ mol⁻¹ ✔

Identical to example 1, as it must be. If your two routes ever disagree, the fault is almost always a sign or a coefficient, not the data.

The liquid-versus-gas water question

More Hess's law answers are marked wrong for this than for anything else. Water appears in a large fraction of these cycles, and Δ_fH(H₂O, l) = −285.8 kJ mol⁻¹ while Δ_fH(H₂O, g) = −241.8 kJ mol⁻¹. The 44.0 kJ mol⁻¹ gap is the enthalpy of vaporisation. Read the state symbols in the question before you pick a value, and if a question mixes the two, insert the vaporisation step explicitly as one more equation in the cycle.

Common mistakes that cost marks

Multiplying the equation but not the ΔH. If you use 2 × (2), you must use 2 × ΔH₂. This is the single most frequent slip.
Forgetting the sign change on reversal. Reversing an equation flips the sign, always.
Using the formation-enthalpy order with combustion data. Formation: products minus reactants. Combustion: reactants minus products. Getting them the wrong way round reverses the sign of the entire answer.
Mixing H₂O(l) and H₂O(g) — a 44.0 kJ mol⁻¹ error per mole of water.
Ignoring stoichiometric coefficients in the shortcut formulas. The 4 and the 6 in example 3 are not optional.
Assuming every element has Δ_fH = 0. Only the standard allotrope does. Diamond, ozone and monatomic gaseous elements all have non-zero values.
Not checking the cancellation. Write out both sides of the summed equation and confirm that only the target species survive. It takes twenty seconds and catches nearly every error above.

Where this appears in exams

Level	Typical question
CBSE/ICSE Class 11	Δ_fH of a compound from three combustion equations; state Hess's law
JEE/NEET	Multi-step cycles, resonance energy of benzene from hydrogenation data
IIT-JAM / CUET-PG	Born–Haber cycles, lattice and hydration enthalpies, solution enthalpies
GATE / CSIR-NET	Thermochemical cycles combined with bond and atomisation enthalpies

Hess's law is addition, subtraction and careful signs, so there is no dedicated tool for it in the suite — and saying so is more useful than pointing you at the wrong screen. Open the suite for the Scientific Calculator while you work the cycle through.

Open the ABC Chemistry Calculator Suite →

Once the cycle has given you Δ_rH, the Gibbs Free Energy calculator turns it into ΔG = ΔH − TΔS and tells you whether the reaction is spontaneous.

Thermochemistry rewards method over memory. ABC Chemistry teaches Class 11–12 chemistry at its Gurugram centre and in online batches across India, with home tuition available in Delhi-NCR — abcchemistry.in.