Hess's Law & Enthalpy of Formation

You’ll be able to

Apply Hess's law by adding, reversing, and scaling thermochemical equations
Compute ΔH°rxn from standard enthalpies of formation
Estimate ΔH from bond enthalpies (bonds broken minus bonds formed)

Enthalpy is a state function

Enthalpy depends only on the state of a system, not the path taken to get there — like altitude, which depends on where you stand, not the trail you hiked. This means the total ΔH for a reaction is the same whether it happens in one step or ten. That single idea powers every technique in this lesson.

Hess's law: build a reaction from known steps

Hess's law says that if a reaction can be written as the sum of several steps, its ΔH is the sum of the steps' ΔH values. Two moves let you assemble any target: reverse an equation (flip the sign of its ΔH) and scale an equation (multiply its ΔH by the same factor). Line up the steps so unwanted species cancel, then add.

Enthalpy from formation enthalpies

ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants)

Each ΔH°f is multiplied by its coefficient. The ΔH°f of any element in its standard state is exactly 0.

Standard enthalpy of formation

The standard enthalpy of formation ΔH°f is the enthalpy change to make 1 mole of a compound from its elements in their standard states. Because forming an element from itself changes nothing, ΔH°f of a pure element (O₂, N₂, C(graphite), Fe…) is defined as 0. Tabulated ΔH°f values are really a shortcut version of Hess's law: every compound is referenced to its elements.

Worked example

Calculate ΔH°rxn for the combustion of methane: CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l). Use ΔH°f = −74.8 (CH₄), −393.5 (CO₂), −285.8 (H₂O, l), 0 (O₂), all in kJ·mol⁻¹.

1.Apply ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants), weighting each by its coefficient.
2.Products: ΔH°f(CO₂) + 2·ΔH°f(H₂O) = (−393.5) + 2(−285.8) = −393.5 − 571.6 = −965.1 kJ.
3.Reactants: ΔH°f(CH₄) + 2·ΔH°f(O₂) = (−74.8) + 2(0) = −74.8 kJ.
4.Subtract: ΔH°rxn = −965.1 − (−74.8) = −965.1 + 74.8 = −890.3 kJ·mol⁻¹.

Answer: ΔH°rxn = −890.3 kJ·mol⁻¹ (strongly exothermic, as expected for combustion)

Watch out

Products minus reactants — never the other way around. A very common error is flipping the subtraction, which reverses the sign of the whole answer. Also remember to multiply each ΔH°f by its coefficient before summing.

Checkpoint

Given C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ, and CO(g) + ½ O₂(g) → CO₂(g), ΔH = −283.0 kJ, find ΔH for C(s) + ½ O₂(g) → CO(g).

Bond enthalpies: an estimate from breaking and making

Breaking a bond costs energy; forming a bond releases energy. So the enthalpy of a gas-phase reaction is roughly ΔH ≈ Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed). It is only an estimate because tabulated bond enthalpies are averages across many molecules, but it captures the right sign and magnitude.

Checkpoint

Estimate ΔH for H₂(g) + Cl₂(g) → 2 HCl(g) using bond enthalpies: H−H = 436, Cl−Cl = 243, H−Cl = 431 kJ·mol⁻¹.

Checkpoint

What is the standard enthalpy of formation, ΔH°f, of O₂(g)?

On the exam

Three routes, one answer: Hess's law with given steps, the ΔH°f formula with a table, and bond enthalpies for gas-phase estimates. If a problem hands you ΔH°f values, reach for Σproducts − Σreactants first — it is the fastest and the most exam-common.

Answer the 3 checkpoints as you read.