A-Level Chemistry energetics: enthalpy, Hess's law, Born-Haber cycles and entropy

Where energetics sits in each specification

Edexcel IAL allows students to use ΔG = ΔH - TΔSsystem, but the specification is built on ΔStotal.

The key ideas

Enthalpy change ΔH is the heat change at constant pressure. Exothermic reactions have negative ΔH; endothermic reactions have positive ΔH. Standard conditions are 100 kPa and a stated temperature, usually 298 K, with substances in their standard states. Learn the definitions word for word: for example, the standard enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

Hess's law says the total enthalpy change for a reaction is independent of the route. With formation data, ΔrH = sum of ΔfH(products) - sum of ΔfH(reactants). With combustion data, ΔrH = sum of ΔcH(reactants) - sum of ΔcH(products). With bond energies, ΔH = energy to break bonds in reactants - energy released forming bonds in products; average bond energies make this less accurate.

Lattice energy is the enthalpy change when one mole of an ionic solid forms from its gaseous ions; it is more exothermic for smaller, more highly charged ions. A Born-Haber cycle links it to the enthalpy of formation through atomisation, ionisation energies and electron affinities. Entropy S measures the dispersal of energy and particles; gases have much higher entropy than solids. A reaction is feasible when ΔG is negative (Cambridge), or equivalently when ΔStotal is positive (Edexcel).

Worked example 1: the enthalpy change of combustion of methane from formation data

1. Equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l).
2. Data (kJ mol^-1): ΔfH of CH4 = -74.8, CO2 = -393.5, H2O(l) = -285.8. ΔfH of O2 is zero because it is an element.
3. ΔH = [-393.5 + 2(-285.8)] - [-74.8] = -965.1 + 74.8.
4. Answer: ΔcH = -890.3 kJ mol^-1.

Worked example 2: enthalpy of neutralisation from a calorimetry experiment

1. 50.0 cm^3 of 1.00 mol dm^-3 HCl is mixed with 50.0 cm^3 of 1.00 mol dm^-3 NaOH in a polystyrene cup. The temperature rises by 6.8 °C.
2. Mass of solution ≈ 100 g (assuming density 1.00 g cm^-3); c = 4.18 J g^-1 K^-1.
3. q = mcΔT = 100 × 4.18 × 6.8 = 2842 J.
4. Moles of water formed = 0.0500 × 1.00 = 0.0500 mol.
5. ΔH = -2842 ÷ 0.0500 = -56 800 J mol^-1 = -56.8 kJ mol^-1. The sign is negative because the temperature rose.

Worked example 3 (A Level): lattice energy of sodium chloride from a Born-Haber cycle

1. Data (kJ mol^-1): ΔfH(NaCl) = -411; atomisation of Na = +107; first ionisation energy of Na = +496; atomisation of chlorine (1/2 Cl2 to Cl) = +122; electron affinity of Cl = -349.
2. Route 1 is direct formation, -411. Route 2 goes through gaseous ions and then the lattice: 107 + 496 + 122 - 349 + LE.
3. Hess's law: -411 = 376 + LE.
4. Answer: lattice energy = -787 kJ mol^-1.

Common mistakes that cost marks

• Using the mass of the reactant instead of the mass of the solution in q = mcΔT.
• Forgetting the negative sign for an exothermic reaction, or not converting J to kJ.
• Reversing the formation equation in Hess cycles: it is products minus reactants for formation data, reactants minus products for combustion data.
• Forgetting to multiply by the number of moles in the balanced equation.
• Using ionisation energy for the wrong number of electrons in Born-Haber cycles, such as only the first for Mg2+.
• Mixing units in ΔG = ΔH - TΔS: ΔS is usually in J K^-1 mol^-1, so divide by 1000 before combining with kJ.

Exam technique and how a tutor helps

Draw the cycle every time, with arrows labelled and directions clear, and write the Hess's law equation that follows the arrows before substituting numbers. State any assumptions in calorimetry: all heat goes to the solution, density and specific heat capacity of the solution equal to water, no heat loss. Evaluation questions often ask why the experimental value is less exothermic than the data book value; heat loss to the surroundings is usually the main reason.

Give answers with a sign, units and an appropriate number of significant figures, usually matching the least precise data.

In one-to-one lessons a tutor drills the definitions until they are exact, then works through cycles of increasing difficulty on the shared whiteboard. Most students improve quickly once they always draw the cycle, and the tutor checks the direction of each arrow before the arithmetic.

Self-check: can you do these?

• Define standard enthalpy change of combustion.
• Why is ΔfH of O2(g) zero? (Answer: it is an element in its standard state)
• Calculate q when 25.0 g of water rises by 12.0 °C. (Answer: 1254 J)
• Why is the lattice energy of MgO more exothermic than that of NaCl? (Answer: smaller ions with higher charges attract more strongly)
• For CaCO3 → CaO + CO2, ΔH = +178 kJ mol^-1 and ΔS = +161 J K^-1 mol^-1. Above what temperature is it feasible? (Answer: about 1106 K)