What you'll learn
Energetics is the study of the energy changes that accompany chemical reactions. Every reaction involves breaking bonds (which absorbs energy) and making bonds (which releases energy), and the balance between the two determines whether a reaction gives out heat — like the combustion of natural gas in a Trinidadian power station — or takes it in. This guide covers exothermic and endothermic reactions and energy profile diagrams, standard enthalpy changes and their definitions, measuring enthalpy changes by calorimetry, Hess's law and energy cycles, bond energies, and lattice energy with Born–Haber cycles, including the factors that affect lattice energy and the enthalpies of solution and hydration.
CAPE questions combine definitions, calculations and explanations. You must learn the precise definitions, set out calculations clearly with signs and units, and explain trends in lattice energy using ionic charge and radius.
Key terms and definitions
Enthalpy change, ΔH — The heat energy change at constant pressure, in kJ mol⁻¹.
Exothermic — Releases energy to the surroundings; ΔH is negative.
Endothermic — Absorbs energy from the surroundings; ΔH is positive.
Standard conditions — 298 K (25 °C), 1 atm (101 kPa), solutions at 1 mol dm⁻³, substances in their standard states.
Standard enthalpy change of formation, ΔH°f — The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.
Standard enthalpy change of combustion, ΔH°c — The enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions.
Standard enthalpy change of neutralisation, ΔH°neut — The enthalpy change when an acid and alkali react to form one mole of water under standard conditions.
Standard enthalpy change of atomisation, ΔH°at — The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state.
Bond energy — The average energy needed to break one mole of a given bond in gaseous molecules.
Hess's law — The total enthalpy change for a reaction is independent of the route taken, provided initial and final conditions are the same.
Lattice energy, ΔH°latt — The enthalpy change when one mole of an ionic compound is formed from its gaseous ions (an exothermic process).
Electron affinity — The enthalpy change when one mole of electrons is added to one mole of gaseous atoms to form gaseous 1− ions.
Enthalpy of hydration — The enthalpy change when one mole of gaseous ions dissolves in water to form an infinitely dilute solution.
Enthalpy of solution — The enthalpy change when one mole of a solute dissolves in a solvent to form an infinitely dilute solution.
Core concepts
Exothermic and endothermic reactions
In an exothermic reaction, more energy is released by forming bonds in the products than is absorbed breaking bonds in the reactants, so energy is given to the surroundings and the temperature rises. Combustion, respiration and neutralisation are exothermic.
In an endothermic reaction, more energy is absorbed than released, so the surroundings cool. Thermal decomposition of calcium carbonate, photosynthesis and dissolving ammonium nitrate are endothermic.
Energy profile diagrams show the enthalpy of reactants and products, the activation energy (Eₐ) — the minimum energy needed for reaction — and ΔH. In exothermic reactions, products are lower than reactants; in endothermic reactions, they are higher.
Measuring enthalpy changes: calorimetry
The heat change is calculated from the temperature change of a known mass of water or solution:
q = m c ΔT
where q is heat energy (J), m is mass of the solution (g), c is specific heat capacity (4.18 J g⁻¹ K⁻¹ for water) and ΔT is the temperature change (K).
Then ΔH = −q ÷ n, where n is the moles of the limiting reactant (the negative sign is used because a temperature rise means the reaction is exothermic).
Sources of error include heat loss to the surroundings, incomplete combustion, heat absorbed by the container, and evaporation. Using an insulated polystyrene cup with a lid, and extrapolating a temperature–time graph back to the moment of mixing, improves accuracy.
Hess's law
Because enthalpy depends only on the start and end states, an enthalpy change that cannot be measured directly can be calculated using a cycle of other reactions.
Using enthalpies of formation:
ΔH°r = ΣΔH°f(products) − ΣΔH°f(reactants)
Using enthalpies of combustion:
ΔH°r = ΣΔH°c(reactants) − ΣΔH°c(products)
The ΔH°f of an element in its standard state is zero.
Bond energies
ΔH = Σ(bond energies of bonds broken) − Σ(bond energies of bonds made)
Bond energies are average values from many compounds, so calculations using them give approximate answers that may differ from values measured experimentally. They also apply only to substances in the gaseous state.
Lattice energy and Born–Haber cycles
Lattice energy cannot be measured directly, so it is calculated using a Born–Haber cycle, an application of Hess's law. For sodium chloride, the steps are:
- Atomisation of sodium: Na(s) → Na(g)
- First ionisation energy of sodium: Na(g) → Na⁺(g) + e⁻
- Atomisation of chlorine: ½Cl₂(g) → Cl(g)
- Electron affinity of chlorine: Cl(g) + e⁻ → Cl⁻(g)
- Lattice energy: Na⁺(g) + Cl⁻(g) → NaCl(s)
The enthalpy of formation of NaCl equals the sum of these steps, so:
ΔH°latt = ΔH°f − (ΔH°at(Na) + IE₁(Na) + ΔH°at(Cl) + EA(Cl))
Factors affecting lattice energy
Lattice energy becomes more exothermic (larger in magnitude) when:
- ionic charges are higher — MgO has a much more exothermic lattice energy than NaCl;
- ionic radii are smaller — the ions can approach more closely, so attraction is stronger. LiF is more exothermic than KF.
Higher lattice energies mean higher melting points and lower solubility in some cases.
Enthalpies of solution and hydration
When an ionic solid dissolves, the lattice must be broken (endothermic, the reverse of lattice energy) and the ions are hydrated by water molecules (exothermic).
ΔH°sol = −ΔH°latt + ΣΔH°hyd
Small, highly charged ions have more exothermic hydration enthalpies because they attract water molecules more strongly. Whether a salt dissolves depends on the balance between these two quantities (and on entropy).
Worked examples
Example 1: Calorimetry
50.0 cm³ of 1.00 mol dm⁻³ HCl was mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rose by 6.8 °C. Calculate ΔH of neutralisation. (Assume density 1.00 g cm⁻³, c = 4.18 J g⁻¹ K⁻¹.)
Mass of solution = 100 g. q = 100 × 4.18 × 6.8 = 2842 J. Moles of water formed = 0.0500 mol. ΔH = −2842 ÷ 0.0500 = −56 840 J mol⁻¹ = −56.8 kJ mol⁻¹.
Example 2: Hess's law using formation enthalpies
Calculate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). (ΔH°f: CH₄ = −75, CO₂ = −394, H₂O(l) = −286 kJ mol⁻¹)
ΔH = [−394 + 2(−286)] − [−75 + 0] = (−394 − 572) + 75 = −891 kJ mol⁻¹.
Example 3: Bond energies
Estimate ΔH for H₂ + Cl₂ → 2HCl. (Bond energies: H–H 436, Cl–Cl 242, H–Cl 431 kJ mol⁻¹)
Bonds broken = 436 + 242 = 678. Bonds made = 2 × 431 = 862. ΔH = 678 − 862 = −184 kJ mol⁻¹.
Example 4: Born–Haber cycle
Calculate the lattice energy of NaCl. (kJ mol⁻¹: ΔH°f(NaCl) = −411; ΔH°at(Na) = +107; IE₁(Na) = +496; ΔH°at(Cl) = +122; EA(Cl) = −349)
ΔH°latt = −411 − (107 + 496 + 122 − 349) = −411 − 376 = −787 kJ mol⁻¹.
Example 5: Enthalpy of combustion from a spirit burner
0.92 g of ethanol was burned under a copper can containing 200 g of water. The temperature rose by 25.0 °C. Calculate the enthalpy of combustion of ethanol and compare it with the data-book value of −1367 kJ mol⁻¹. (Mᵣ C₂H₅OH = 46)
q = 200 × 4.18 × 25.0 = 20 900 J = 20.9 kJ. Moles of ethanol = 0.92 ÷ 46 = 0.0200 mol. ΔH = −20.9 ÷ 0.0200 = −1045 kJ mol⁻¹.
The experimental value is much less exothermic than the data-book value, mainly because of heat loss to the surroundings (the air, the can and the clamp), incomplete combustion (soot on the can) and evaporation of ethanol from the wick. Improvements include using a draught shield, insulating the can, reducing the distance between flame and can, and using a bomb calorimeter.
Example 6: Hess's law using combustion data
Calculate the enthalpy change for C₂H₄(g) + H₂(g) → C₂H₆(g). (ΔH°c / kJ mol⁻¹: C₂H₄ = −1411, H₂ = −286, C₂H₆ = −1560)
ΔH = ΣΔH°c(reactants) − ΣΔH°c(products) = (−1411 + −286) − (−1560) = −1697 + 1560 = −137 kJ mol⁻¹.
Example 7: Enthalpy of solution
The lattice energy of NaCl is −787 kJ mol⁻¹. The hydration enthalpies of Na⁺ and Cl⁻ are −406 and −364 kJ mol⁻¹. Calculate the enthalpy of solution of NaCl.
ΔH°sol = −ΔH°latt + ΣΔH°hyd = +787 + (−406 − 364) = 787 − 770 = +17 kJ mol⁻¹.
The small positive value explains why dissolving salt in water causes a slight drop in temperature. Sodium chloride is still very soluble, because dissolving increases the disorder (entropy) of the system.
Example 8: Comparing lattice energies
Explain why the lattice energy of magnesium chloride is much more exothermic than that of sodium chloride.
Mg²⁺ has a double charge and a smaller ionic radius than Na⁺. Both factors increase the electrostatic attraction between the cation and chloride ions, so more energy is released when the lattice forms.
Energy cycle checklist
- Draw arrows in the direction of each process, labelled with ΔH values.
- When going against an arrow, reverse the sign of its ΔH.
- Remember that ΔH°f of elements in their standard states is zero.
- Include state symbols; the enthalpy change for H₂O(g) is different from H₂O(l).
- Check that the units are kJ mol⁻¹ and that signs make chemical sense (lattice formation and hydration are exothermic; atomisation and ionisation are endothermic).
Common mistakes and how to avoid them
- Wrong sign for ΔH — exothermic is negative.
- Using the mass of the solid instead of the solution in q = mcΔT.
- Forgetting to divide by moles to get ΔH per mole.
- Incomplete definitions — include "one mole", "standard conditions" and "standard states".
- Getting the Hess's law subtraction backwards — products minus reactants for formation; reactants minus products for combustion.
- Forgetting that atomisation of chlorine involves ½Cl₂.
Self-test questions
- Use ΔH°f values (kJ mol⁻¹: CO₂ = −394, H₂O(l) = −286, C₃H₈ = −104) to calculate ΔH°c of propane. Answer: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. ΔH = [3(−394) + 4(−286)] − (−104) = −1182 − 1144 + 104 = −2222 kJ mol⁻¹.
- Define standard enthalpy change of atomisation. Answer: The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions.
- Why are enthalpy changes calculated from average bond energies only approximate? Answer: Average bond energies are means taken from many different compounds, and the actual strength of a bond depends on its molecular environment; they also assume all species are gaseous.
- Explain why the lattice energy of LiF is more exothermic than that of KF. Answer: Li⁺ is smaller than K⁺, so the ions in LiF can approach more closely and the electrostatic attraction is stronger.
How this links to your Internal Assessment
Calorimetry is a classic CAPE SBA. You might measure the enthalpy of neutralisation of different acids, the enthalpy of combustion of a series of alcohols, or use Hess's law to find the enthalpy change for a reaction that cannot be measured directly (such as converting anhydrous to hydrated copper(II) sulfate). Strong reports include a clear method with controlled variables, temperature–time graphs extrapolated to find the true ΔT, calculations with units and signs, comparison with data-book values, and a detailed evaluation of heat losses and how to reduce them.
Exam technique for Energetics and enthalpy changes
- Learn definitions word for word; marks are lost for missing "one mole" or "standard conditions".
- In calculations, show q = mcΔT, the moles, and the division, with units and signs.
- Draw Hess cycles clearly with arrows and labels.
- For Born–Haber cycles, label each step and keep track of signs carefully.
- Explain lattice energy trends using charge and ionic radius.
Quick revision summary
- Exothermic: ΔH negative, products lower; endothermic: ΔH positive, products higher.
- Standard conditions: 298 K, 1 atm, 1 mol dm⁻³.
- Calorimetry: q = mcΔT; ΔH = −q ÷ n; main error is heat loss.
- Hess's law: route-independent; formation (products − reactants), combustion (reactants − products).
- Bond energies: broken − made; average values, approximate results.
- Born–Haber cycles calculate lattice energy; more exothermic with higher charges and smaller ions.
- ΔH°sol = −ΔH°latt + ΣΔH°hyd.