Standard enthalpy changes, Hess's law and the Born-Haber cycle

Chemistry · Thermochemistry. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

Test this topic free All Chemistry topics

The idea

A standard enthalpy change (ΔH°) is measured at 25 °C (298 K) and 1 atm, with every substance in its standard state. Each named enthalpy change refers to one mole of a stated substance, so learn exactly which substance the one mole is in each definition.

Hess's law says the total enthalpy change of a reaction is the same whether it happens in one step or in several. This works because enthalpy is a state function. It lets us calculate enthalpy changes that cannot be measured directly.

The Born-Haber cycle applies Hess's law to an ionic compound. It breaks the formation of the solid from its elements into steps, so that the lattice energy can be found.

Rules to remember

Enthalpy of formation, ΔH°_f: one mole of a compound is formed from its elements in their standard states. It is zero for an element in its standard state, e.g. O₂(g), C(graphite).
Enthalpy of combustion, ΔH°_c: one mole of a substance is burned completely in excess oxygen. Always negative.
Enthalpy of atomisation, ΔH°_at: one mole of gaseous atoms is formed from the element. Always positive.
Enthalpy of neutralisation: one mole of H⁺ from an acid reacts with one mole of OH⁻ from a base to form one mole of water. About −57.4 kJ mol⁻¹ for any strong acid with any strong base.
Enthalpy of solution: one mole of a substance dissolves in so much solvent that further dilution causes no heat change.
ΔH° of reaction = sum of ΔH°_f of products − sum of ΔH°_f of reactants. Multiply each value by its coefficient in the equation.
Born-Haber cycle for NaCl: ΔH°_f = atomisation of Na + ionisation energy of Na + atomisation of ½Cl₂ + electron affinity of Cl + lattice energy.

Common mistake

Students subtract the wrong way round or forget the coefficients. It is products minus reactants, each ΔH°_f multiplied by its number of moles. In a Born-Haber cycle keep the electron affinity negative.

Worked example

ΔH°_f of C₂H₄(g) = +52.3 kJ mol⁻¹ and of C₂H₆(g) = −84.7 kJ mol⁻¹. Calculate ΔH° for C₂H₄(g) + H₂(g) → C₂H₆(g).

H₂(g) is an element in its standard state, so its ΔH°_f = 0.
ΔH° = ΔH°_f(products) − ΔH°_f(reactants) = (−84.7) − (+52.3 + 0).
ΔH° = −84.7 − 52.3 = −137.0 kJ mol⁻¹.

Answer: ΔH° = −137.0 kJ mol⁻¹ (exothermic).

Ask about this

Type what you are stuck on. The helper finds the explanation, worked examples and the notes to read, free and instant.

Read the full notes

Chapter 7: Thermochemistry

Keep going

← Energy, the first law and enthalpy Law of mass action, Kc and Kp →Take a Chemistry test