This mini-lesson covers AQA 3.1.4 Energetics: enthalpy change ΔH and standard conditions, calorimetry (q = mcΔT), Hess’s law cycles using enthalpies of formation and combustion, and mean bond enthalpy calculations — plus why the two methods disagree.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Everything here is A-level standard — the maths is done properly, not skipped. Press Start when you're ready.
Energetics · enthalpy
Enthalpy change and standard conditions
Enthalpy change ΔH is the heat energy change measured at constant pressure.
Exothermic: heat is released to the surroundings, the products are at lower enthalpy, so ΔH is negative (combustion, neutralisation, most oxidations).
Endothermic: heat is absorbed from the surroundings, so ΔH is positive (thermal decomposition, photosynthesis, most dissolving of ammonium salts).
Standard conditions (⦵): a pressure of 100 kPa, a stated temperature (usually 298 K), and all solutions at a concentration of 1.00 mol dm⁻³, with every substance in its standard state.
Bonds:breaking bonds is always endothermic; making bonds is always exothermic. A reaction is exothermic overall when the bonds made are stronger (release more energy) than the bonds broken.
Energetics · calorimetry
Measuring ΔH: q = mcΔT
q = m c ΔTq = heat in J · m = mass of the WATER in g · c = 4.18 J g⁻¹ K⁻¹ · ΔT = temperature change in K
Then divide by the moles of the substance that reacted, and flip the sign (a temperature rise means the reaction gave out heat, so ΔH is negative):
ΔH = − q / nthe answer is per mole, so the units are kJ mol⁻¹ — remember to divide q by 1000
Worked example
100 g of water rises by 20.0 K when 0.0100 mol of a fuel burns.
q = 100 × 4.18 × 20.0 = 8360 J = 8.36 kJ
ΔH = −8.36 ÷ 0.0100 = −836 kJ mol⁻¹
Why experimental values are always less exothermic than the data book:heat loss to the surroundings, incomplete combustion (soot on the beaker), evaporation of the fuel, and heat absorbed by the apparatus rather than the water.
Calculate
Your turn
1100 g of water rises in temperature by 20.0 K. Calculate the heat energy transferred, in kJ. (c = 4.18 J g⁻¹ K⁻¹)
2That heat came from burning 0.0100 mol of a fuel. Calculate the enthalpy of combustion, in kJ mol⁻¹. (Include the sign.)
kJ mol⁻¹
Hint: ΔH = −q ÷ n = −8.36 ÷ 0.0100. The temperature rose, so ΔH must be negative.
Quick check
Quick check
?A student’s experimental enthalpy of combustion for ethanol is −980 kJ mol⁻¹; the data book value is −1367 kJ mol⁻¹. Which explanation is best?
Energetics · Hess’s law
Hess’s law and enthalpy cycles
Hess’s law: the enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. It follows directly from conservation of energy — and it lets us find ΔH for reactions we cannot measure directly.
ΔH = Σ ΔHf(products) − Σ ΔHf(reactants)formation data: arrows point UP from the elements, so go up-and-over — products minus reactants
ΔH = Σ ΔHc(reactants) − Σ ΔHc(products)combustion data: arrows point DOWN to the combustion products — reactants minus products
Two rules that stop sign errors: the standard enthalpy of formation of an element in its standard state is zero (O₂, C(graphite), N₂ …). And multiply each ΔH by its balancing number in the equation before you add.
Calculate
Your turn
3For CH₄ + 2O₂ → CO₂ + 2H₂O(l), use ΔHf: CH₄ = −74.8, CO₂ = −393.5, H₂O(l) = −285.8, O₂ = 0 (all kJ mol⁻¹). Calculate ΔH for the reaction.
ΔH = 678 − 862 = −184 kJ mol⁻¹ (exothermic — the bonds made are stronger)
Why bond-enthalpy answers disagree with Hess-cycle answers: mean bond enthalpies are averages taken across many different molecules, so they are not exact for the specific compound in question. They also apply to gaseous species only, so any enthalpy of vaporisation is ignored.
Calculate
Your turn
4For H₂ + Cl₂ → 2HCl, mean bond enthalpies (kJ mol⁻¹) are H–H = 436, Cl–Cl = 242, H–Cl = 431. Calculate ΔH for the reaction.
Tap a term on the left, then its definition on the right.
Enthalpy term
Definition (per mole, standard conditions)
Quick check
Quick check
?Why is a ΔH value calculated from mean bond enthalpies only approximate?
Quick check
Quick check
?A reaction has ΔH = +178 kJ mol⁻¹. Which statement is correct?
Energetics · exam traps
Getting the cycle and the signs right
Formation data: ΔH = Σ ΔHf(products) − Σ ΔHf(reactants). Combustion data: ΔH = Σ ΔHc(reactants) − Σ ΔHc(products). The two are the other way round — because the arrows point in opposite directions.
Multiply by the balancing numbers BEFORE you add. Forgetting the 2 in front of H₂O is the single commonest error in the whole topic.
ΔHf of an element in its standard state = 0. So the O₂ term always vanishes.
q = mcΔT uses the mass of the WATER, not the mass of the fuel. And divide q by 1000 to get kJ before you divide by moles.
The definition trap: a standard enthalpy of formation forms ONE mole of the compound from its elements. So for ethanol it is 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) — the half is compulsory, because the equation must produce exactly one mole of ethanol.
Quick check
Quick check
?Which equation correctly represents the standard enthalpy of formation of ethanol?
Calculate
Your turn
5Use enthalpies of combustion to find ΔHf of ethanol. ΔHc: C(s) = −393.5, H₂(g) = −285.8, C₂H₅OH(l) = −1367.0 kJ mol⁻¹. For 2C + 3H₂ + ½O₂ → C₂H₅OH, calculate ΔHf.