23. Chemical energetics

Enthalpy of atomisation is the enthalpy change when one mole of gaseous atoms forms from an element in its standard state. Lattice energy here is the enthalpy change when gaseous ions form one mole of an ionic solid lattice.

State the direction and physical states before using a value in a cycle. Reversing a step changes its sign; lattice energy of dissociation is the opposite direction to lattice formation.

Na(s) → Na(g) contributes atomisation, while Na⁺(g) + Cl⁻(g) → NaCl(s) is lattice formation. Both are different from sublimation or hydration.

Do not call any solid-to-gas process atomisation, and do not mix lattice formation and separation conventions in one calculation.

First electron affinity is the enthalpy change for X(g) + e⁻ → X⁻(g). It is usually exothermic when the incoming electron is attracted to the nucleus, but the numerical sign depends on the convention used.

Across a period, nuclear attraction generally increases; down a group, distance and shielding increase. Filled or half-filled subshell effects can create small deviations, so explain trends rather than drawing a perfect line.

Halogens have strongly favourable first electron affinities because gaining one electron completes the p subshell. Group 16 values differ from Group 17 because electron pairing and nuclear attraction compete.

Electron affinity is not ionisation energy in reverse, and the second electron affinity is a different process involving an anion.

A Born–Haber cycle applies Hess’s law to an ionic solid. Typical steps are atomisation, ionisation energies, electron affinities, bond dissociation for a non-metal, and lattice formation, linked to the standard enthalpy of formation.

Write every species and state on the energy path. For +2 cations include two successive ionisation energies; for −2 anions include the second electron affinity with its correct sign.

For MgO(s), the cycle includes Mg atomisation, IE₁ + IE₂, oxygen atomisation, EA₁ + EA₂ and lattice formation, summing to ΔHf°.

A cycle is not a list of numbers: omitting a gaseous atom or using the wrong electron-affinity direction invalidates the result.

Use Hess’s law: the algebraic sum of the enthalpy changes around the cycle equals the standard enthalpy of formation. Rearrange only after deciding whether each step forms or separates a species.

Keep a sign table and isolate the unknown lattice energy or electron affinity. Check units and compare the magnitude with the expected ionic attraction.

If all steps except lattice formation are known, ΔHlatt = ΔHf° − (atomisation + ionisation + electron-affinity terms). A negative formation lattice energy is expected in the formation direction.

Do not change signs because a value “looks too large”, and do not use lattice dissociation data as formation data without reversal.

Lattice energy reflects electrostatic attraction between oppositely charged ions. Greater ionic charge strengthens attraction, while a smaller internuclear distance also increases the magnitude.

Compare like structures and state what is held constant. Charge effects are often larger than radius effects, but both act through Coulombic attraction and lattice packing.

MgO has a more negative lattice-formation enthalpy than NaCl because Mg²⁺/O²⁻ charges are larger, despite the ions not being identical in size.

Do not claim that radius alone determines lattice energy or compare compounds with different structures without qualification.

Hydration enthalpy is the enthalpy change when one mole of gaseous ions becomes aqueous. Solution enthalpy is the enthalpy change when one mole of an ionic solid dissolves to form aqueous ions.

Hydration involves ion–water attractions; solution combines lattice separation and hydration. State the ion charge, state symbols and direction before comparing values.

Dissolving NaCl(s) gives Na⁺(aq) and Cl⁻(aq); the solution enthalpy equals lattice dissociation plus the two hydration enthalpies.

Hydration is not the same as dissolving a molecular solute, and solution enthalpy is not automatically exothermic.

A solution cycle connects the direct dissolution route with an indirect route: lattice dissociation of the solid followed by hydration of the gaseous ions. Hess’s law makes the two routes equivalent.

Choose lattice formation or dissociation consistently. The indirect route is ΔHsol = ΔHlatt(dissociation) + ΣΔHhyd, with signs determined by direction.

For MgCl₂, the cycle contains one Mg²⁺ hydration and two Cl⁻ hydration steps. Omitting the coefficient on chloride gives a wrong energy balance.

Do not use one hydration value for the whole salt or mix lattice formation and dissociation arrows.

Write the known enthalpy values on a labelled cycle, choose a sign convention, then use Hess’s law to isolate the unknown. Include every ion with its stoichiometric coefficient.

Check units and direction after rearrangement. A quick magnitude check should match the expected strength of ion–water or lattice attraction.

If ΔHsol, lattice dissociation and one hydration term are known, the missing hydration term is the residual after subtracting the other route contributions.

Do not average the hydration values or change a sign because the numerical answer is negative; the arrow direction determines the sign.

Hydration enthalpy measures attraction between an ion and polar water molecules. Higher charge density gives stronger attraction and a more negative hydration enthalpy.

Compare charge first, then radius among ions of the same charge. Smaller ions place charge closer to water dipoles, but hydration also depends on the ion’s coordination environment.

Mg²⁺ has a more negative hydration enthalpy than Na⁺ because of its higher charge density; Na⁺ is more strongly hydrated than K⁺ because it is smaller.

Do not claim that every larger ion has a more negative value; for a fixed charge, the trend is usually toward less exothermic hydration.

Entropy, S, describes the number of possible arrangements of particles and their energy. More accessible microstates correspond to greater entropy.

It is a system property, not simply “disorder”. Particle number, phase, temperature and mixing all affect the number of accessible arrangements.

A gas has much higher entropy than the same substance as a solid because particles can occupy many more positions and energy distributions.

Entropy is not a synonym for messiness and a reaction can increase one part of a system while decreasing another.

Entropy increases on melting, boiling, dissolving and heating, and decreases for the reverse changes. A reaction forming more gaseous molecules usually has positive ΔS; forming fewer has negative ΔS.

Use the strongest structural change first: gas count often dominates, while phase and temperature changes provide additional evidence. The sign prediction is qualitative unless data are supplied.

H₂O(s) → H₂O(l) gives ΔS>0; 2SO₂(g)+O₂(g) → 2SO₃(g) gives a negative gas-count contribution because three gas molecules become two.

A positive entropy change does not by itself prove spontaneity; Gibbs free energy also depends on enthalpy and temperature.

For a reaction, ΔS° = ΣS°(products) − ΣS°(reactants). Multiply each standard molar entropy by its stoichiometric coefficient before summing.

Keep units consistent, usually J K⁻¹ mol⁻¹, and interpret the sign after the subtraction. A positive result means the system has more accessible arrangements overall.

For A + 2B → C, ΔS° = S°(C) − [S°(A)+2S°(B)]. The coefficients belong inside the brackets, not only in the balanced equation.

Do not reverse the subtraction or use ΔS° = ΔSsurr + ΔSsys when the syllabus only asks for the tabulated-state calculation.

The standard Gibbs equation is ΔG° = ΔH° − TΔS°, where temperature is in kelvin and ΔH° and TΔS° use the same energy units.

Enthalpy favours exothermic change, while the entropy term becomes more influential as temperature increases. The equation combines these competing contributions for a specified standard state.

An endothermic reaction can still have negative ΔG° at high T if its positive ΔS° makes TΔS° larger than ΔH°.

Do not insert Celsius or combine kJ enthalpy with J entropy without conversion.

Substitute ΔH° and TΔS° into ΔG° = ΔH° − TΔS°. Convert entropy from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ when enthalpy is in kJ mol⁻¹.

Write the units beside each term and keep the sign of ΔS. Round only after the final calculation, then state what the sign means.

If ΔH° = 40 kJ mol⁻¹, ΔS° = 100 J K⁻¹ mol⁻¹ and T = 298 K, TΔS° = 29.8 kJ mol⁻¹ and ΔG° = +10.2 kJ mol⁻¹.

A negative entropy value makes −TΔS positive; do not drop that double sign.

If ΔG° < 0, the forward process is thermodynamically feasible under the specified standard conditions; if ΔG° > 0, the reverse direction is favoured. ΔG° = 0 indicates equilibrium.

Feasible does not mean fast, safe or complete. Kinetics, activation energy, heat transfer and actual concentrations can determine what is observed.

A combustion reaction may have a strongly negative ΔG° but still need ignition because its activation energy is high.

Do not equate negative ΔG° with an instantaneous reaction or 100% yield.

Because ΔG° = ΔH° − TΔS°, raising T makes a positive ΔS° more favourable and a negative ΔS° less favourable. The crossover occurs when ΔG° changes sign.

For ΔH°>0 and ΔS°>0, high temperature can make a process feasible; for ΔH°<0 and ΔS°<0, low temperature is favoured. The other sign combinations are usually temperature-independent in the simple model.

If ΔH°=+50 kJ mol⁻¹ and ΔS°=+150 J K⁻¹ mol⁻¹, feasibility begins above T≈333 K because ΔH°/ΔS° sets the threshold.

Do not say “higher temperature always makes reactions spontaneous”; inspect both signs.