Lesson 4: Energetics

Every chemical reaction involves energy changes — the blast of a firework, the warmth of a hand warmer, the chill of an instant cold pack. Energetics quantifies these changes as enthalpy, and A-level takes you from simple calorimetry to Born–Haber cycles, entropy and Gibbs free energy. This lesson covers standard enthalpy definitions, measuring enthalpy changes by calorimetry, Hess's law, bond enthalpies, Born–Haber cycles for ionic compounds, and the entropy and free-energy ideas that predict feasibility. Energetics questions reward careful definitions, correct signs and methodical cycles — get those right and the marks follow.

Enthalpy changes and standard definitions

Enthalpy change, ΔH, is the heat energy change at constant pressure, in kJ mol⁻¹. Exothermic reactions release heat (ΔH negative) — combustion, neutralisation; endothermic reactions absorb heat (ΔH positive) — thermal decomposition. Enthalpy profile diagrams show reactants, products and activation energy: in an exothermic profile the products sit lower than the reactants, and the activation-energy hump is the minimum energy needed for reaction.

Standard conditions are 100 kPa, a stated temperature (usually 298 K) and 1 mol dm⁻³ concentrations. Key definitions, each specifying one mole and standard conditions: standard enthalpy of combustion — the enthalpy change when 1 mole of a substance burns completely in oxygen under standard conditions; standard enthalpy of formation — the enthalpy change when 1 mole of a compound forms from its elements under standard conditions; standard enthalpy of neutralisation — the energy released when 1 mole of water forms in the neutralisation of an acid by an alkali.

The enthalpy of neutralisation for a strong acid–strong alkali pair is always about −57 kJ mol⁻¹ because the only reaction occurring is H⁺(aq) + OH⁻(aq) → H₂O(l). Weak acids give less exothermic values because energy is absorbed in dissociating the acid: ethanoic acid gives about −55 kJ mol⁻¹ for exactly this reason.

Calorimetry — measuring enthalpy changes

Simple calorimetry measures ΔH using q = mcΔT: heat transferred = mass heated × specific heat capacity × temperature change. For a spirit burner heating water: burn a known mass of fuel, measure the temperature rise, calculate q, then convert to per mole of fuel burned. Worked example: 0.50 g of ethanol (Mᵣ 46.0) heats 100 g of water by 22 °C.

q = 100 × 4.18 × 22 = 9196 J
n(ethanol) = 0.50 / 46.0 = 0.0109 mol
ΔH = −9196 / 0.0109 = −846 kJ mol⁻¹   (negative: exothermic)

Experimental values are always less exothermic than data-book values because of heat loss to the apparatus and surroundings, incomplete combustion and non-standard conditions. Improvements: insulate with a lid, use a draught shield, stir, and use a digital thermometer. For neutralisation, use a polystyrene cup as the calorimeter and take the maximum temperature from a cooling curve extrapolated back to the mixing time.

Required Practical 2 measures an enthalpy change calorimetrically. Key skills: record temperature every 30 seconds, plot temperature against time, extrapolate to find the true maximum, and calculate percentage uncertainty from the thermometer (±0.5 °C per reading, so ±1.0 °C across a temperature change). Always show the q = mcΔT step, the moles step and the per-mole step separately for method marks.

Hess's law

Hess's law states that the total enthalpy change of a reaction is independent of the route taken — enthalpy is a state function depending only on the initial and final states. This allows ΔH values to be calculated for reactions that cannot be measured directly, such as the formation of carbon monoxide (which burns further to CO₂). Construct a cycle with the target reaction as one side and measurable reactions completing the triangle.

Using combustion data for the formation of methane, C + 2H₂ → CH₄: burn C, 2H₂ and CH₄ separately, and Hess gives ΔHբ = ΣΔH_c(reactants) − ΣΔH_c(products). Using formation data the formula is ΔHᵣ = ΣΔHբ(products) − ΣΔHբ(reactants). Memorise both formulas — and multiply every ΔH by its stoichiometric coefficient.

Cycle technique: label every arrow with its ΔH and direction, apply signs consistently, and check the final sign makes chemical sense (formation of a stable compound is usually exothermic). The most common error is forgetting to multiply ΔH values by the moles in the balanced equation. Practise drawing cycles from scratch rather than memorising examples — the exam always uses unfamiliar reactions.

Bond enthalpies and Born–Haber cycles

Mean bond enthalpy is the average energy needed to break one mole of a particular covalent bond in the gaseous state, averaged over many compounds. Reaction enthalpies can be estimated as Σ(bond enthalpies broken) − Σ(bond enthalpies formed): breaking bonds is endothermic (+), forming bonds is exothermic (−). For H₂ + Cl₂ → 2HCl: (436 + 242) − 2(431) = −184 kJ mol⁻¹. Values differ from data-book ΔH because mean bond enthalpies are averages, not exact for the specific molecules.

Born–Haber cycles analyse the formation of ionic compounds through gaseous atoms and ions. For NaCl the steps are: atomisation of sodium, Na(s) → Na(g) (ΔH_at); first ionisation, Na(g) → Na⁺(g) + e⁻; atomisation of ½Cl₂; electron affinity, Cl(g) + e⁻ → Cl⁻(g) (exothermic); then the ions come together, releasing the lattice enthalpy — defined as the enthalpy change when 1 mole of solid ionic compound forms from its gaseous ions, exothermic by this definition. The cycle lets you calculate one unknown, often the lattice enthalpy, from the others. For MgCl₂ remember two ionisation energies and two electron affinities.

Trends: lattice enthalpy becomes more exothermic with smaller, more highly charged ions — hence MgO (−3795 kJ mol⁻¹) dwarfs NaCl (−787 kJ mol⁻¹). Comparing theoretical (perfect ionic model) and experimental lattice enthalpies reveals covalent character: a large difference, as in AgI, indicates polarisation of the anion by the cation (Fajans' rules, Lesson 3).

Entropy and Gibbs free energy

Entropy, S, measures the dispersal of energy and disorder in a system (J K⁻¹ mol⁻¹). Entropy increases going solid → liquid → gas, with more moles of gas, and with more complex molecules. Calculate ΔS = ΣS(products) − ΣS(reactants). For CaCO₃(s) → CaO(s) + CO₂(g), ΔS is positive because a gas is produced from a solid.

A reaction is feasible when the Gibbs free energy change is negative: ΔG = ΔH − TΔS. Four cases: ΔH negative and ΔS positive → feasible at all temperatures (e.g. combustion); ΔH positive and ΔS negative → never feasible; ΔH negative and ΔS negative → feasible at low temperatures; ΔH positive and ΔS positive → feasible at high temperatures. The decomposition of calcium carbonate becomes feasible above T = ΔH/ΔS ≈ 1110 K, matching lime-kiln temperatures.

Crucially, ΔG predicts feasibility, not rate. A reaction with negative ΔG may be immeasurably slow if its activation energy is huge — diamond → graphite has negative ΔG but an astronomical barrier, so diamonds persist. And feasible does not mean complete: equilibrium position (Lesson 6) determines the final mixture. Examiners love testing these distinctions.

Worked examples — Hess and ΔG calculations

Worked example 1: combustion enthalpy from formation data. CH₄ + 2O₂ → CO₂ + 2H₂O, using ΔHբ values (kJ mol⁻¹): CH₄ −75, CO₂ −394, H₂O −286 (elements in their standard states contribute zero). ΔH = ΣΔHբ(products) − ΣΔHբ(reactants) = [−394 + 2(−286)] − [−75] = −966 + 75 = −891 kJ mol⁻¹, matching the data-book combustion value for methane. Note how the ×2 coefficients on H₂O and O₂ must be carried through — the single most common slip in this calculation.

Worked example 2: feasibility temperature. For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. The reaction becomes feasible when ΔG = 0, at T = ΔH/ΔS — but first convert units: ΔS = 0.161 kJ K⁻¹ mol⁻¹. T = 178 / 0.161 = 1106 K ≈ 1110 K. Lime kilns operate above this temperature, exactly as the calculation predicts. Forget the J → kJ conversion and you would report 1.11 K — an instant sanity check.

Going deeper — lattice enthalpy comparisons and enthalpy of solution

Comparing experimental lattice enthalpies from Born–Haber cycles with theoretical values from the perfect ionic model tests the bonding model itself. For NaCl and MgO the two agree within about 1–2%, confirming nearly perfect ionic bonding. For AgCl and AgI the experimental value is far more exothermic than the ionic model predicts — evidence of additional covalent character from polarisation of the large, soft anion by the small Ag⁺ cation (Fajans' rules). Examiners ask "what does the difference suggest?" — the answer is always about covalency increasing with polarisation.

The same enthalpies explain solubility through the enthalpy of solution: ΔH_sol = −ΔH_lattice(formation) + ΣΔH_hydration, informally "break the lattice, hydrate the ions". For MgSO₄ the very exothermic hydration of the small Mg²⁺ outweighs the lattice cost, so it dissolves exothermically; for BaSO₄ the large-ion lattice is too stable and hydration too weak, so it stays solid (Lesson 8). Entropy matters too — dissolving always mixes particles, so a slightly endothermic salt such as NH₄NO₃ still dissolves, cooling its surroundings (the cold-pack reaction), which connects ΔH, ΔS and ΔG in one everyday example.

Common mistakes

• Hess-cycle sign errors. With combustion data the formula is reactants − products; with formation data it is products − reactants. Drawing the cycle and following the arrows prevents the swap.
• Forgetting coefficients in ΣΔHբ sums. Every ΔH must be multiplied by its stoichiometric number — the two waters in methane combustion carry 2 × −286, not −286.
• Using ΔS in J K⁻¹ mol⁻¹ inside ΔG = ΔH − TΔS with ΔH in kJ. Convert ΔS to kJ (÷ 1000) first, or the feasibility temperature is wrong by three orders of magnitude.
• Quoting lattice enthalpy for the dissociation direction when the definition used is formation (gaseous ions → solid, exothermic). Check which direction the question defines before signing your answer.

Exam Experience

Q1. Use the ΔHբ values (kJ mol⁻¹) — C₂H₄: +52; CO₂: −394; H₂O(l): −286 — to calculate the enthalpy change for C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l). [2] Q2. A student burns 0.30 g of propan-1-ol (Mᵣ 60.0) and uses the energy to heat 150 g of water; the temperature rises by 9.5 °C. (a) Calculate the enthalpy of combustion obtained from these results. [3] (b) The data-book value is −2021 kJ mol⁻¹. Give two reasons why the experimental value is smaller in magnitude. [2]

Mark scheme. Q1: ΔH = ΣΔHբ(products) − ΣΔHբ(reactants) = [2(−394) + 2(−286)] − [+52] [1] = −1360 − 52 = −1412 kJ mol⁻¹ [1]. Q2(a): q = 150 × 4.18 × 9.5 = 5957 J [1]; n = 0.30/60.0 = 5.00 × 10⁻³ mol [1]; ΔH = −5957/5.00 × 10⁻³ = −1.19 × 10⁶ J mol⁻¹ = −1190 kJ mol⁻¹ [1]. (b) Any two: heat lost to the surroundings/apparatus; incomplete combustion of the fuel (soot/CO formed); some fuel evaporates without burning; conditions not standard [2].