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 (c_water = 4.18 J g⁻¹ K⁻¹) → ΔH = −q / n

Hess's law

Hess's law says the enthalpy change of a reaction is independent of the route taken. That lets you find a ΔH you cannot measure directly by building a cycle from data you can.

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 bonds is endothermic (+), making them is exothermic (−). Because the values are means over many compounds, 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 the combustion products: ΔH°f = Σ ΔH°c(elements) − ΔH°c(compound).
Substituting gives [−394 + 2(−286)] − (−890).
That works out to −966 + 890 = −76 kJ mol⁻¹.

3.1.4.2Calorimetry checklist

In q = m c ΔT, m is the mass of the solution (or water) being heated, c = 4.18 J g⁻¹ K⁻¹, and ΔT comes from a cooling-corrected graph. Divide q by moles of the limiting reagent, then convert J → kJ; the sign follows the temperature change (a rise means negative ΔH).

Evaluation points examiners pay for: heat loss to the surroundings; incomplete combustion (fuels); assuming the solution has water's density and specific heat capacity; and 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

Bonds broken: 4 C–H + 2 O=O = 4(413) + 2(498) = 2648 kJ.
Bonds made: 2 C=O + 4 O–H = 2(805) + 4(464) = 3466 kJ.
So ΔH ≈ 2648 − 3466 = −818 kJ mol⁻¹ (data-book −890: mean values are averages over many compounds).

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.

The two Hess cycles: via formation data (arrows up from elements) and via combustion data (arrows down to combustion products) — note the direction flip in each rule.Diagram: ChemLab original

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, and attach the sign by logic: if the temperature rose, the reaction is exothermic, so ΔH is negative.

The main error sources are heat loss to the 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.

The two profiles mark schemes expect: exothermic (ΔH negative, products below) and endothermic (ΔH positive) — label reactants, products, Ea and ΔH.Diagram: ChemLab original

Extended🎓 Beyond the standard course

ΔH vs ΔU: enthalpy is heat at constant pressure, while internal energy is heat at constant volume. They differ by the expansion work pΔV = Δn(gas)·RT — small but real when the number of gas moles changes.

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. And 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.

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, then build the cycle. Any step you traverse AGAINST its arrow flips sign; any step scaled ×n scales its ΔH ×n.

With formation data, ΔrH = ΣΔfH(products) − ΣΔfH(reactants); with 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, because halving "to make it neat" halves your answer.

Where energetics meets life

Self-heating cans run on CaO + H₂O → Ca(OH)₂, which is exothermic, while sports cold packs dissolve NH₄NO₃, which is endothermic (entropy-driven — a Year-2 callback). Reusable hand-warmers crystallise supersaturated sodium ethanoate, releasing enthalpy as the solid orders itself.

Fuel comparisons weigh energy per gram against energy per mole and CO₂ per kJ. Hydrogen wins per gram (143 kJ g⁻¹) but loses on storage density; methane beats coal on CO₂ per kJ.

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.

Energetics II: Combustion Calorimetry and Formation Cycles

This section pushes past the basics: a full spirit-burner combustion, an enthalpy-of-solution calorimetry, formation via a Hess cycle, and the exact vs mean bond-enthalpy distinction.

Four enthalpies, one glance:

ΔH_c^⦵ combustion — 1 mol substance burned completely in excess O₂. Always negative.

ΔH_f^⦵ formation — 1 mol compound from elements in standard states. Can be + or −; element ΔH_f = 0.

ΔH_r^⦵ reaction — per mole as written in the equation.

ΔH_neut^⦵ neutralisation — 1 mol H₂O formed from H⁺(aq) + OH⁻(aq).

Worked: ΔH_c of ethanol from a spirit burner. Burning ethanol (M = 46.0) heats 150 g water from 19.0 °C to 41.5 °C. Burner mass falls from 218.60 g to 217.42 g.

Step 1 — heat gained by water. q = mcΔT = 150 × 4.18 × (41.5 − 19.0) = 150 × 4.18 × 22.5 = 14 108 J = 14.108 kJ.

Step 2 — mass of fuel burnt. 218.60 − 217.42 = 1.18 g.

Step 3 — moles of ethanol. n = 1.18 ÷ 46.0 = 0.02565 mol.

Step 4 — enthalpy per mole. ΔH_c = −(14.108 ÷ 0.02565) = −550 kJ mol⁻¹ (3 s.f.). The sign is negative because the water gained the energy the fuel released.

The data-book value is −1367 kJ mol⁻¹ — the experiment is far less exothermic. Standard mark points: heat loss to the air, beaker and unburnt vapour; incomplete combustion (soot on the base = carbon, not CO₂, so less energy released); evaporation of ethanol without burning; and non-standard conditions (water is not at 298 K, product water is liquid but conditions differ). Never write "the equipment was faulty".

Worked: enthalpy of solution. Dissolving 4.00 g NH₄NO₃ (M = 80.0) in 50.0 g water drops the temperature from 21.0 °C to 14.6 °C. q = 50.0 × 4.18 × (21.0 − 14.6) = 1338 J. n = 4.00 ÷ 80.0 = 0.0500 mol. ΔH_sol = +1338 ÷ 0.0500 = +26.8 kJ mol⁻¹ (endothermic — temperature fell, so ΔH is positive).

Worked: ΔH_f of methane by Hess cycle. Route via combustion: C(s) + 2H₂(g) → CH₄(g). Using ΔH_c: C = −394, H₂ = −286, CH₄ = −890 kJ mol⁻¹.

ΔH_f = Σ ΔH_c(reactants) − Σ ΔH_c(products)

= [(−394) + 2(−286)] − [(−890)] = (−966) − (−890) = −76 kJ mol⁻¹. Watch the double sign flip on the product term.

Exact vs mean bond enthalpy. Breaking the four C–H bonds in CH₄ one at a time gives four different bond dissociation enthalpies (the molecule changes each time). The mean bond enthalpy is their average, +413 kJ mol⁻¹, quoted for a bond averaged across many molecules. That is why bond-enthalpy answers only ever approximate ΔH — the values are not exact for one specific compound.

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