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

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).
  1. Route via the combustion products: ΔH°f = Σ ΔH°c(elements) − ΔH°c(compound).
  2. Substituting gives [−394 + 2(−286)] − (−890).
  3. 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)

BondkJ mol⁻¹BondkJ mol⁻¹BondkJ mol⁻¹
C–H413C–C347C=C612
O–H464O=O498C=O (in CO₂)805
H–H436N≡N945C–Cl346
Worked example — combustion of methane by bond enthalpies. CH₄ + 2O₂ → CO₂ + 2H₂O
  1. Bonds broken: 4 C–H + 2 O=O = 4(413) + 2(498) = 2648 kJ.
  2. Bonds made: 2 C=O + 4 O–H = 2(805) + 4(464) = 3466 kJ.
  3. So ΔH ≈ 2648 − 3466 = −818 kJ mol⁻¹ (data-book −890: mean values are averages over many compounds).
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