A-level chemistry · Chapter 4
Energetics
Enthalpy changes, calorimetry, Hess's law, bond enthalpies
Standard enthalpy of combustion ΔH°c — enthalpy change when one mole of a substance burns completely in oxygen, all species in standard states, 100 kPa and a stated temperature (298 K).
Standard enthalpy of formation ΔH°f — enthalpy change when one mole of a compound forms from its elements, all in standard states.
q = m × c × ΔT (cwater = 4.18 J g⁻¹ K⁻¹) → ΔH = −q / n
Hess's law
- The enthalpy change is independent of route.
- From formation data: ΔH = Σ ΔH°f(products) − Σ ΔH°f(reactants).
- From combustion data: ΔH = Σ ΔH°c(reactants) − Σ ΔH°c(products). Note the direction flips.
Bond enthalpies
- ΔH ≈ Σ(bonds broken) − Σ(bonds made). Breaking is endothermic (+), making is exothermic (−).
- Values are means over many compounds, so answers differ from Hess-law values — a favourite exam point.
See the energy profile simulation to connect Ea and ΔH visually.
3.1.4.4Hess cycles with combustion data
Worked example — ΔH°f of methane. ΔH°c: C(s) −394, H₂(g) −286, CH₄(g) −890 (kJ mol⁻¹).
- Route via combustion products: ΔH°f = Σ ΔH°c(elements) − ΔH°c(compound)
- = [−394 + 2(−286)] − (−890)
- = −966 + 890 = −76 kJ mol⁻¹
3.1.4.2Calorimetry checklist
- q = m c ΔT: m is the mass of solution (or water) heated, c = 4.18 J g⁻¹ K⁻¹, ΔT from a cooling-corrected graph.
- Divide by moles of the limiting reagent, then convert J → kJ; sign follows the temperature change (rise → negative ΔH).
- Evaluation points examiners pay for: heat loss to surroundings; incomplete combustion (fuels); assuming the solution has water's density and specific heat capacity; evaporation.
Exam tip. Definitions of ΔH°c and ΔH°f must include "one mole", "standard states" and "100 kPa with a stated temperature". Write the equation your definition describes — e.g. for ΔH°f of ethanol: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l). Fractions are fine; scaling to whole numbers is not (it must form ONE mole).
3.1.4.5Mean bond enthalpies (AQA-convention values, kJ mol⁻¹)
| Bond | kJ mol⁻¹ | Bond | kJ mol⁻¹ | Bond | kJ mol⁻¹ |
|---|---|---|---|---|---|
| C–H | 413 | C–C | 347 | C=C | 612 |
| O–H | 464 | O=O | 498 | C=O (in CO₂) | 805 |
| H–H | 436 | N≡N | 945 | C–Cl | 346 |
Worked example — combustion of methane by bond enthalpies. CH₄ + 2O₂ → CO₂ + 2H₂O
- Broken: 4 C–H + 2 O=O = 4(413) + 2(498) = 2648 kJ
- Made: 2 C=O + 4 O–H = 2(805) + 4(464) = 3466 kJ
- ΔH ≈ 2648 − 3466 = −818 kJ mol⁻¹ (data-book −890: mean values are averages over many compounds).
Extended🎓 Beyond the standard course
- ΔH vs ΔU: enthalpy is heat at constant pressure; internal energy at constant volume. They differ by the expansion work pΔV = Δn(gas)·RT — small but real when gas moles change.
- Standard states matter: carbon's is graphite, not diamond — diamond sits 1.9 kJ mol⁻¹ above it, so ΔH°f(diamond) ≠ 0.
- Hess's law is the first law of thermodynamics in disguise: if enthalpy depended on route you could run a cycle and create energy.
- Real calorimetry calibrates the whole apparatus (its "heat capacity" in J K⁻¹) with an electrical heater or standard reaction, sidestepping the water-only assumption schools make.
Deep dive📚 The rest of the chapter, in full
Definitions that must be word-perfect
- Standard enthalpy of combustion, ΔcH°: the enthalpy change when one mole of a substance burns completely in oxygen, all species in standard states at 100 kPa and a stated temperature.
- Standard enthalpy of formation, ΔfH°: the enthalpy change when one mole of a compound forms from its elements in their standard states at 100 kPa. ΔfH° of any element in its standard state is zero.
- Standard enthalpy of neutralisation: the enthalpy change when one mole of water is formed in the reaction of an acid with a base under standard conditions (≈ −57 kJ mol⁻¹ for strong acid + strong base).
Hess's law: the two cycle recipes
Using formation data: ΔrH = ΣΔfH(products) − ΣΔfH(reactants) — arrows go UP from elements to both sides. Using combustion data: ΔrH = ΣΔcH(reactants) − ΣΔcH(products) — arrows go DOWN to the combustion products. If you can only remember one thing: reverse an arrow, reverse its sign; multiply an equation, multiply its ΔH.
Worked example (formation route). For CH₄ + 2O₂ → CO₂ + 2H₂O(l): ΔrH = [−394 + 2(−286)] − [−74.8 + 0] = −966 + 74.8 = −891 kJ mol⁻¹ — matching the data-book combustion value, as it must.
Bond enthalpy calculations — and their built-in error
ΔrH ≈ Σ(bonds broken) − Σ(bonds made). Breaking is endothermic (+), making is exothermic (−). Values are mean bond enthalpies averaged over many compounds and defined for the gaseous state — two reasons answers differ from Hess-law values: real bonds in this molecule are not average bonds, and any liquid species adds vaporisation terms the calculation ignores.
Worked example. H₂ + Cl₂ → 2HCl. Break H–H (436) + Cl–Cl (243) = +679; make 2 × H–Cl (432) = −864; ΔrH ≈ −185 kJ mol⁻¹ (data-book: −184.6).
Calorimetry pitfalls (they examine these, not the formula)
- q = mcΔT uses the mass of the liquid being heated, not the fuel or the solid added; c = 4.18 J g⁻¹ K⁻¹ is water's value.
- Convert J → kJ before dividing by moles; attach the sign by logic (temperature rose → exothermic → negative).
- Main error sources: heat loss to surroundings and apparatus, incomplete combustion, evaporation from the wick, and the assumption that the solution's specific heat capacity equals pure water's. Fixes: draught shields, lids, closer flames — or the graphical cooling correction from RP2.
Enthalpy level diagrams. Exothermic: products drawn below reactants, ΔH arrow pointing down, activation energy from reactants up to the peak. Endothermic: products above. Label ALL three features — the mark scheme checks each.
Mastery vault🏛 Every remaining spec point, banked
Bond enthalpy data bank (learn the ballpark, quote from the paper)
| Bond | H–H | C–H | C–C | C=C | C≡C | O–H | O=O | C=O (in CO₂) | Cl–Cl | H–Cl | N≡N |
|---|---|---|---|---|---|---|---|---|---|---|---|
| kJ mol⁻¹ | 436 | 412 | 347 | 612 | 837 | 463 | 496 | 805 | 243 | 432 | 945 |
Patterns worth knowing: N≡N's enormous value explains nitrogen's inertness (and why the Haber process needs a catalyst and heat); C=C is less than twice C–C, which is why addition across the π-bond is exothermic and alkenes are reactive.
Full combustion by bonds. CH₄ + 2O₂ → CO₂ + 2H₂O(g). Break: 4 C–H (1648) + 2 O=O (992) = +2640. Make: 2 C=O (1610) + 4 O–H (1852) = −3462. ΔH ≈ −822 kJ mol⁻¹. The data-book value (−891 for liquid water) differs because gaseous water is assumed and mean values were used — quote BOTH reasons when asked.
Flame calorimetry, end to end
Spirit-burner run. Burning 0.46 g of ethanol (Mᵣ 46) raises 100 g of water by 21.5 °C. q = 100 × 4.18 × 21.5 = 8987 J = 8.99 kJ. n = 0.010 mol → ΔcH ≈ −899 kJ mol⁻¹? No — −8.99 ÷ 0.010 = −899 kJ mol⁻¹ vs the true −1367: about 34% of the heat vanished into the surroundings, the copper, evaporation and incomplete combustion (sooty flame). List improvements: draught shield, lid, stirring, shorter flame-to-can distance, oxygen-rich flame.
Enthalpy of neutralisation experiments
Mix measured volumes of acid and alkali in a polystyrene cup, take the temperature rise, and use the TOTAL solution mass in q = mcΔT. Strong–strong pairs all give ≈ −57 kJ per mole of water — the reaction is just H⁺ + OH⁻ → H₂O. Weak acids give slightly less exothermic values because energy is spent dissociating the acid first — a favourite explain-the-difference question.
Hess with mixed data — the sign discipline
- Write the target equation. Build the cycle. Any step you traverse AGAINST its arrow flips sign; any step scaled ×n scales ΔH ×n.
- Formation data: ΔrH = ΣΔfH(products) − ΣΔfH(reactants). Combustion data: ΔrH = ΣΔcH(reactants) − ΣΔcH(products). If you mix data types, draw the full cycle rather than trusting either formula.
- Watch for elements (ΔfH = 0) and for equations already containing fractional coefficients — keep the fractions; halving "to make it neat" halves your answer.
Where energetics meets life
- Self-heating cans: CaO + H₂O → Ca(OH)₂ (exothermic). Sports cold packs: dissolving NH₄NO₃ (endothermic — entropy-driven, a Year-2 callback).
- Fuel comparisons: energy per gram vs per mole, CO₂ per kJ — hydrogen wins per gram (143 kJ g⁻¹) but loses on storage density; methane beats coal on CO₂ per kJ.
- Hand-warmers reusing supersaturated sodium ethanoate crystallisation — enthalpy released on ordering.
Three-term answer template for "why does the calculated value differ from the data book": (1) mean bond enthalpies are averages over many compounds; (2) data apply to gaseous species — state changes are ignored; (3) experimental: heat losses / incomplete combustion. Deploy whichever two the context supports.
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