3.1.8 Thermodynamics

Lattice formation enthalpy is the enthalpy change when one mole of ionic solid forms from its separate gaseous ions. Lattice dissociation enthalpy is the exact reverse: one mole of solid is separated into gaseous ions.

The two values have equal magnitude and opposite sign. Decide which definition your data use before you put a lattice term into a cycle; the word “formation” or “dissociation” controls the sign.

In a Born–Haber cycle, build the ionic solid from elements by an alternative route through gaseous atoms and gaseous ions. Atomisation, ionisation enthalpies, bond enthalpy and electron affinities each represent a specific chemical step.

Keep stoichiometry visible. If you need two gaseous atoms, two ions or two electron-affinity steps, multiply the corresponding enthalpy contribution before closing the Hess cycle.

For Born–Haber definitions, the species and physical states are part of the chemistry. Ionisation enthalpy starts with gaseous atoms or ions; electron affinity involves gaseous atoms/ions; lattice formation uses separate gaseous ions and produces one mole of solid.

I’d learn each definition with its equation. If the equation and words disagree, the equation usually exposes the missing state or wrong direction immediately.

To construct a Born–Haber cycle, write the standard formation reaction for the ionic solid, then build an alternative route to the same solid through gaseous atoms and gaseous ions. Hess’s law connects the two routes.

Put each supplied enthalpy on the step it actually describes and apply any stoichiometric multiplier. Once we’ve made every species and state explicit, the missing lattice term is just the remaining part of the energy balance.

A Born–Haber cycle can solve for any missing term, not just lattice enthalpy. Write the complete energy balance first, with signs and coefficients attached to the defined processes, then rearrange algebraically.

Do not change an enthalpy sign merely because you move a term across the equals sign; distinguish the chemical direction of the process from the later algebraic rearrangement.

Compare a lattice enthalpy from an experimental cycle with the perfect-ionic-model value: look at the size of the disagreement. A large discrepancy can be evidence that polarisation has introduced appreciable covalent character.

If the experimental lattice formation enthalpy is more exothermic than the purely ionic model predicts, extra covalent character can account for stronger attraction than the simple spherical-ion model allows.

When you build a solution cycle, separate it into two jobs: lattice dissociation takes the ionic solid to gaseous ions, then hydration enthalpies take those gaseous ions to aqueous ions. The overall route gives ΔHₛₒₗ.

If the formula contains more than one of an ion, multiply its hydration enthalpy accordingly. A cycle for MX₂, for example, contains one hydration term for M²⁺ but two for X⁻.

For the definition, I’d learn: enthalpy of hydration is the enthalpy change when one mole of gaseous ions becomes one mole of aqueous ions under standard conditions.

Hydration is usually exothermic because ion–dipole attractions form between the ion and water. The definition is per mole of the specified ion, so keep its charge and state explicit.

For a quantitative solution cycle, keep one lattice sign convention throughout, multiply hydration values by the number of each ion produced and then use Hess’s law to connect solid → gaseous ions → aqueous ions.

A quick sign check helps: lattice dissociation is endothermic, whereas hydration enthalpies are normally exothermic. If your calculation has reversed those physical processes, inspect the cycle before blaming the arithmetic.

ΔH alone cannot tell you whether a process is thermodynamically feasible. The Gibbs equation combines enthalpy with the entropy contribution: ΔG = ΔH − TΔS.

An endothermic process can still have ΔG < 0 if TΔS is sufficiently positive; an exothermic process can become unfavourable if the entropy term works strongly against it.

Use entropy as a measure of how dispersed the matter and energy are. Entropy often increases when gases form, when the number of gas particles increases or when particles become freer to move.

Justify a sign using the actual species and states. “More disorder” is too vague if you can instead say, for example, that two moles of gas become three moles of gas.

Use ΔG = ΔH − TΔS with T in kelvin and consistent energy units. If ΔH is in kJ mol⁻¹ and ΔS in J K⁻¹ mol⁻¹, convert one before calculating TΔS.

Write the unit conversion explicitly. A factor of 1000 in the entropy term can give a plausible-looking but incorrect sign for ΔG, so write the conversion explicitly.

Under the stated conditions, ΔG < 0 means the forward process is thermodynamically feasible and ΔG = 0 corresponds to equilibrium. A positive ΔG means the forward direction is not thermodynamically favoured.

Keep feasibility separate from rate. A reaction can have ΔG < 0 and still be extremely slow because it has a large activation-energy barrier.

Calculate reaction entropy with ΔS° = ΣS°(products) − ΣS°(reactants), including every stoichiometric coefficient. Absolute entropy values are positive, but the reaction change can have either sign.

Before calculating, multiply each tabulated S° by the number of moles in the balanced equation. The final unit is normally J K⁻¹ mol⁻¹ for the reaction as written.

Temperature changes the size of the TΔS term, so the signs of ΔH and ΔS tell you how ΔG will move as T changes. If ΔS is positive, increasing T makes −TΔS more negative; if ΔS is negative, it becomes more positive.

You can often predict whether high or low temperature favours feasibility before doing any arithmetic. Then use the calculation to locate the boundary if one exists.

At the temperature where the sign of ΔG changes, set ΔG = 0. You can then rearrange to T = ΔH/ΔS, provided you have put ΔH and ΔS into consistent units.

The calculated T is the boundary. Use the signs of ΔH and ΔS to decide whether temperatures above or below it give ΔG < 0; do not assume “higher temperature helps” without checking the entropy sign.