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A Born–Haber cycle links standard formation enthalpy to atomisation, ionisation, electron affinity and lattice enthalpy. Lattice formation joins gaseous ions into one mole of solid and is exothermic. Lattice dissociation separates the solid into gaseous ions and is endothermic. State which convention a value uses. Remember coefficients: making Mg2+ requires the first and second ionisation energies, while making chloride ions requires atomised chlorine and its electron affinity. Higher ionic charge and smaller separation usually give stronger lattice attractions.
Video by Eliot Rintoul. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.
Viewing focus: Construct Born–Haber cycles with a stated sign convention
After watching, attempt these questions before opening the answers. You can also use the written lesson below.
After watching 1: What is the sign of lattice dissociation enthalpy?
Positive. Separating an ionic lattice requires energy.
After watching 2: What can a negative ΔG establish?
Thermodynamic feasibility under stated conditions. Kinetics remains a separate question.
What you'll be able to do
A Born–Haber cycle links standard formation enthalpy to atomisation, ionisation, electron affinity and lattice enthalpy. Lattice formation joins gaseous ions into one mole of solid and is exothermic. Lattice dissociation separates the solid into gaseous ions and is endothermic. State which convention a value uses. Remember coefficients: making Mg2+ requires the first and second ionisation energies, while making chloride ions requires atomised chlorine and its electron affinity. Higher ionic charge and smaller separation usually give stronger lattice attractions.
Entropy measures the dispersal of energy among available arrangements. More gas particles usually means a positive system entropy change, but calculate from tabulated standard entropies when supplied: products minus reactants, including coefficients. Unlike formation enthalpies, standard entropies of elements are not generally zero. Dissolving a salt includes both lattice dissociation and hydration; a large exothermic hydration contribution may offset the energy needed to separate the lattice. Solubility reflects free energy rather than enthalpy alone.
At constant temperature and pressure, . Use kelvin and matching energy units: entropy is often given in J K⁻¹ mol⁻¹ while enthalpy is in kJ mol⁻¹. A negative value indicates thermodynamic feasibility under the stated conditions, but a high activation barrier may make the reaction extremely slow. If both ΔH and ΔS are positive, higher temperature favours feasibility. Setting ΔG to zero gives an approximate crossover temperature if their temperature dependence can be neglected.
Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.
Common mistakes to avoid
Key takeaways
Question 1: Formation from elements is −400 kJ mol⁻¹ and all non-lattice steps sum to +600. Find lattice formation enthalpy.
−1000 kJ mol⁻¹; dissociation would be +1000.
Check the assessed scope and practical or mathematical requirements.
An external explanation or interactive activity to use alongside the lesson. It may include extension material.
Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.
Test yourself
Answer a few questions, then review anything you missed.
Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.