HookThe railway weld that carries its own furnace
Somewhere on Britain's railways tonight, a maintenance gang is fusing two rail ends into one continuous ribbon of steel with no furnace and no power supply — just a clay crucible clamped over the gap, filled with a grey powder of iron(III) oxide and aluminium. One igniter later, the mixture is running at roughly 2,500 °C and white-hot liquid iron is pouring into the mould around the rails: \(\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow 2\text{Fe} + \text{Al}_2\text{O}_3\), ΔH ≈ −852 kJ mol⁻¹. Hans Goldschmidt patented this aluminothermic process in 1895, and rail crews still perform tens of thousands of thermite welds a year. There is no external heat source because the reaction is the heat source: the energy was sitting in the chemicals all along.
Energetics is the accounting system for that energy. Section 3.1.4 asks exactly four things of you: define enthalpy change and pin it to standard conditions; measure it with calorimetry and \(q = mc\Delta T\); calculate it with a Hess's law cycle when direct measurement is impossible; and estimate it from mean bond enthalpies when there is no data at all — then explain, honestly, why the estimate misses. Every energetics mark on Paper 1 is one of those four moves, and the sign convention is the thread running through all of them.
ModelEnthalpy change — the sign convention that runs the topic
The enthalpy change, ΔH, is the heat energy change of a reaction measured at constant pressure. Exothermic reactions transfer energy to the surroundings, so the chemicals finish with less stored energy than they started with and ΔH is negative — combustion, neutralisation, the thermite weld. Endothermic reactions absorb energy from the surroundings and ΔH is positive — thermal decomposition of calcium carbonate is the standard example. The sign belongs to the chemicals, not to you: a bigger temperature rise means a more negative ΔH, and examiners watch that direction like hawks.
Because ΔH varies with conditions, comparisons only mean something under standard conditions: a pressure of 100 kPa and a stated temperature, almost always 298 K, with every substance in its standard state — the physical state it holds under those conditions. Two definitions then have to be word-perfect. The standard enthalpy of combustion is the enthalpy change when one mole of a substance burns completely in oxygen, all reactants and products in their standard states under standard conditions. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements, all in their standard states under standard conditions — which makes the standard enthalpy of formation of any element in its standard state exactly zero, a fact Hess cycles quietly depend on.
Every word in those definitions is load-bearing: one mole, complete combustion, standard states. Drop one and the definition mark goes with it.
DataCalorimetry — q = mcΔT, then divide by the moles
You cannot see enthalpy, but you can see a thermometer. Calorimetry routes the reaction's energy into something with a known heat capacity — usually water or a dilute aqueous solution — and measures the temperature change. The machine has three steps, and AQA marks each one separately. First, \(q = mc\Delta T\): m is the mass in grams of the substance being heated, c is its specific heat capacity (4.18 J g⁻¹ K⁻¹ for water), ΔT the temperature change in kelvin. Second, convert q from joules to kilojoules and work out the moles of the substance the question defines the enthalpy per mole of. Third, \(\Delta H = -q/n\) for a temperature rise — negative because energy left the chemicals — quoted in kJ mol⁻¹ to three significant figures.
The classic trap sits in step one: when a spirit burner heats a beaker of water, m is the mass of the water, not the mass of fuel burned. The fuel mass enters at step two, as moles. Mixing the two is the single most common way to lose all three marks at once.
Burning 0.46 g of ethanol (\(M_r = 46.0\), so \(n = 0.010\ \text{mol}\)) beneath a beaker holding 200 g of water raises the water temperature by 10.5 K. Energy transferred: \(q = 200 \times 4.18 \times 10.5 = 8{,}778\ \text{J} = 8.78\ \text{kJ}\). Enthalpy of combustion: \(\Delta H = -8.78 \div 0.010 = -878\ \text{kJ mol}^{-1}\). The data-book value is −1,367 kJ mol⁻¹, so the experiment has captured barely two-thirds of the energy. That gap is not careless technique — it is heat escaping to the air and the beaker, incomplete combustion (the yellow, sooty flame), and ethanol evaporating unburned from the wick. AQA regularly asks you to name these and to state the direction of the error: every one of them makes the measured value less exothermic than the true one.
ModelHess's law — enthalpy doesn't care about the route
Some enthalpy changes cannot be measured directly. You cannot form methane by pointing carbon and hydrogen at each other — the reaction \(\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}\) does not proceed cleanly in any calorimeter on Earth. Hess's law rescues the situation: the total enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same. Enthalpy behaves like altitude on a mountain — the height gained between two points does not depend on the path you walked.
So you build a cycle through substances whose enthalpy changes you can measure. Two patterns cover almost every exam question. With combustion data, both sides of the target equation burn down to the same CO₂ and H₂O, and \(\Delta H = \Sigma\Delta_c H(\text{reactants}) - \Sigma\Delta_c H(\text{products})\). With formation data, both sides are built up from the same elements, and \(\Delta H = \Sigma\Delta_f H(\text{products}) - \Sigma\Delta_f H(\text{reactants})\). If you prefer drawing the triangle, one rule does all the work: follow the arrows, and when you travel against an arrow, reverse the sign of its ΔH.
Find the standard enthalpy of formation of methane from combustion data: \(\Delta_c H\) of C(s) = −393.5, of H₂(g) = −285.8, of CH₄(g) = −890.3, all in kJ mol⁻¹. The cycle runs from the elements down to the combustion products by two routes — either directly through burning C and H₂, or via methane first. Applying the combustion rule to \(\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}\): \(\Delta_f H = [(-393.5) + 2(-285.8)] - (-890.3) = -965.1 + 890.3 = -74.8\ \text{kJ mol}^{-1}\). Watch the two places candidates bleed marks: forgetting to double the H₂ value for the 2 mol in the equation, and mangling the double negative when subtracting −890.3.
ModelMean bond enthalpies — a good estimate, honestly labelled
A mean bond enthalpy is the energy needed to break one mole of a given bond, averaged over that bond in many different compounds, with everything in the gas phase. The averaging is the point: a C–H bond in methane is not identical to a C–H bond in ethanol, so the tabulated 412 kJ mol⁻¹ is a compromise across them all.
Bond breaking always costs energy (endothermic); bond making always releases it (exothermic). A reaction's overall enthalpy change is the difference between the bill and the refund: \(\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds made})\). If the new bonds are stronger than the old ones, the refund wins and the reaction is exothermic. That single sentence — energy in to break, energy out to make, and the balance decides the sign — is the correct answer to a whole family of explain-questions.
Because the values are averages, and because they describe gaseous species only, a bond-enthalpy answer is an estimate. When AQA asks why it differs from the data-book value, those are the two creditworthy reasons: mean values are not exact for the specific compounds in this reaction, and the real measurement may involve liquids — liquid water in a combustion, say — while the bond calculation assumes gases.
Estimate ΔH for methane combustion, \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O(g)}\), using mean bond enthalpies C–H 412, O=O 496, C=O 743, O–H 463 (kJ mol⁻¹). Bonds broken: \(4(412) + 2(496) = 1{,}648 + 992 = 2{,}640\ \text{kJ mol}^{-1}\). Bonds made: \(2(743) + 4(463) = 1{,}486 + 1{,}852 = 3{,}338\ \text{kJ mol}^{-1}\). So \(\Delta H = 2{,}640 - 3{,}338 = -698\ \text{kJ mol}^{-1}\). The calorimetric value is −890 kJ mol⁻¹ — a 20% miss, and exactly the discrepancy the mark scheme wants explained: mean values averaged over many compounds, and gaseous water in the calculation versus liquid water in the standard measurement.
CaseRequired practical 2 — the cooling correction is the experiment
RP2 measures an enthalpy change with apparatus honest enough to admit its own flaw. The usual version is the displacement reaction \(\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}\) in a polystyrene cup. Method: pipette 50.0 cm³ of 0.500 mol dm⁻³ copper(II) sulfate into the cup, record the temperature every minute for four minutes, add an excess of zinc powder at minute four, stir, and keep recording to minute fifteen. The excess zinc guarantees the copper sulfate is the limiting reagent, so the moles in your final division are the moles you actually pipetted.
The cup leaks heat the whole time, so the peak thermometer reading understates the true rise. The fix is graphical: plot temperature against time, draw a best-fit line through the cooling section, and extrapolate it back to the minute of mixing. The gap between that extrapolated temperature and the initial steady temperature is your corrected ΔT — an estimate of the rise you would have seen if the reaction had been instantaneous and lossless. The calculation then assumes the solution has water's specific heat capacity and a density of 1.00 g cm⁻³, and ignores the heat soaked up by the zinc, the cup and the thermometer — each assumption worth naming as an error source, and each biasing the result towards less exothermic.
The same technique powers the Hess variant of RP2: measure the enthalpy changes of potassium hydrogencarbonate and potassium carbonate each reacting with hydrochloric acid, then combine the two in a cycle to find the enthalpy of thermal decomposition of KHCO₃ — a reaction you could never wire to a thermometer directly, because it needs sustained heating. Calorimetry supplies the measurable legs; Hess's law supplies the unmeasurable answer.
In the zinc–copper sulfate run above, the extrapolated temperature rise is 22.3 K. Energy released: \(q = 50.0 \times 4.18 \times 22.3 = 4{,}661\ \text{J} = 4.66\ \text{kJ}\). Moles of CuSO₄: \(0.0500\ \text{dm}^3 \times 0.500\ \text{mol dm}^{-3} = 0.0250\ \text{mol}\). So \(\Delta H = -4.66 \div 0.0250 = -186\ \text{kJ mol}^{-1}\), against an accepted value near −217 kJ mol⁻¹ — about 86% recovered, respectable for a coffee-cup calorimeter. Note the mass used is the 50.0 g of solution, not the solution plus zinc.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Definitions are AO1 marks that must be word-perfect: one mole, complete combustion, elements in their standard states, standard conditions of 100 kPa and a stated temperature. AQA mark schemes list these as separate marking points, so a definition that gestures at the idea without the qualifiers scores partial credit at best.
Calculations are marked as a ladder — typically M1 for \(q = mc\Delta T\) with the correct mass, M2 for moles, M3 for the final value with sign and units — and error carried forward is live: a wrong q done right afterwards still earns M2 and M3. So never abandon a calculation, always show the substitution line, and give three significant figures with kJ mol⁻¹ and an explicit sign. A missing minus sign on an exothermic answer costs the final mark every single session.
The AO3 questions cluster around two patterns. 'Suggest why the experimental value differs from the data-book value' wants heat loss, incomplete combustion or evaporation — plus the direction of the bias (less exothermic). 'Why does the bond-enthalpy value differ' wants mean values and the gaseous-state assumption. And for RP2, be ready to justify the method itself: why excess zinc (so the limiting moles are known), why extrapolate (to correct for heat lost during the reaction), and what is still uncorrected (heat absorbed by the cup, thermometer and zinc). Practical-technique marks are about 15% of the whole qualification, and RP2 is one of AQA's favourite vehicles for them.