01
Enthalpy and its sign
The heat content of a system at constant pressure is its enthalpy, H. We never measure H itself, only the change, ΔH, during a reaction. Exothermic reactions release heat, so ΔH is negative; endothermic reactions absorb it, so ΔH is positive — the deeper, quantitative version of the energetics you met earlier.
Enthalpy changes are quoted per mole under standard conditions (298 K, 1 atm), written ΔH°.
02
The standard enthalpy changes
A few named enthalpy changes come up constantly. Enthalpy of formation (ΔHf°): forming one mole of a compound from its elements. Enthalpy of combustion (ΔHc°): burning one mole completely in oxygen. Enthalpy of reaction: for the reaction as written. Elements in their standard state have ΔHf° = 0 by definition.
03
Hess's law
Enthalpy is a state function — it depends only on the start and end states, not the route between them. That is Hess's law: the total enthalpy change is the same whether a reaction happens in one step or several.
It is enormously useful because it lets you calculate a ΔH you cannot measure directly, by adding known steps in an energy cycle. Reverse a step and you flip the sign of its ΔH; multiply a step and you multiply its ΔH.
ΔH_reaction = Σ ΔHf°(products) − Σ ΔHf°(reactants)a direct consequence of Hess's law
04
Measuring it: calorimetry
You find an enthalpy change experimentally with a calorimeter: carry out the reaction and measure the temperature change of a known mass of water (or solution). The heat transferred is q = mcΔT — mass × specific heat capacity × temperature change — and dividing by the moles reacted gives ΔH per mole.
Worked example
Burning 0.010 mol of a fuel warms 100 g of water by 12 °C. Find the enthalpy of combustion. (c = 4.18 J g⁻¹ °C⁻¹)
- Heat gained by water: q = mcΔT = 100 × 4.18 × 12 = 5016 J ≈ 5.0 kJ.The water absorbs the heat the reaction released.
- That heat came from 0.010 mol of fuel, so per mole: 5.0 / 0.010 = 500 kJ.Scale the measured heat up to one mole.
- Combustion releases heat, so ΔHc = −500 kJ mol⁻¹.Exothermic → negative sign.
ΔHc ≈ −500 kJ mol⁻¹ (exothermic). Real experiments lose some heat to the surroundings, so measured values are usually a little low.
05
Bond enthalpies
Every bond stores energy: breaking a bond absorbs energy (endothermic) and making one releases energy (exothermic). The average bond enthalpy is the mean energy to break one mole of a given bond, averaged over many molecules. You can estimate a reaction's enthalpy change by totalling the bonds broken in the reactants and subtracting the bonds made in the products.
Because the values are averages, bond-enthalpy calculations give an approximation, not the exact figure a calorimeter or Hess cycle would.
ΔH = Σ(bonds broken) − Σ(bonds made)positive terms for breaking, negative overall if more energy is released making bonds
06
Lattice energy and Born–Haber cycles (A Level)
For an ionic compound, the lattice energy is the energy released when gaseous ions come together to form one mole of the solid lattice — a measure of how strongly the ions are held. It cannot be measured directly, so a Born–Haber cycle uses Hess's law to find it from steps that can be measured (atomisation, ionisation, electron affinity and formation).
Lattice energy is larger (more exothermic) for ions with higher charges and smaller radii, because the electrostatic attraction is stronger — which is why magnesium oxide melts far higher than sodium chloride.
Practice questions
6 questions · 23 marks · full working on every one
Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.
Short questions
4 · 10 marksTwo marks each, in the style of the short-question section of the paper. Answer in two or three lines.
SQ1[2 marks]
Define the standard enthalpy change of reaction, ΔH.
Model answer
The heat energy change for a reaction measured at standard conditions (298 K, 1 atm), with substances in their standard states.
Examiner tip. ΔH is negative for exothermic, positive for endothermic reactions.
SQ2[3 marks]
State Hess's law and explain why it is useful.
Model answer
The total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. It lets us calculate enthalpy changes that cannot be measured directly.
Examiner tip. Hess's law follows from conservation of energy — energy change depends only on start and end points.
SQ3[2 marks]
Write the expression used to calculate heat energy released when a solution's temperature changes (q = mcΔT), defining each term.
Model answer
q = mcΔT, where q = heat energy, m = mass of solution, c = specific heat capacity, ΔT = temperature change.
Examiner tip. Use the mass of the SOLUTION (usually water) heated, not the mass of the fuel.
SQ4[3 marks]
For the reaction H₂ + Cl₂ → 2HCl, use the bond enthalpies H–H = 436, Cl–Cl = 242, H–Cl = 431 kJ/mol to calculate ΔH.
Model answer
Bonds broken = 436 + 242 = 678 kJ. Bonds made = 2 × 431 = 862 kJ. ΔH = 678 − 862 = −184 kJ/mol (exothermic).
Examiner tip. Sum the bonds broken, subtract the bonds made; a negative answer is exothermic.
Long questions
2 · 13 marksTheory and numerical together, as they appear in the long-question section.
LQ1[7 marks]
Burning 0.50 g of ethanol raises the temperature of 100 g of water by 30 °C. (c of water = 4.2 J g⁻¹ °C⁻¹; Mr ethanol = 46)
- Calculate the heat energy released to the water. [2]
- Calculate the number of moles of ethanol burned. [1]
- Calculate the enthalpy change of combustion of ethanol in kJ/mol. [3]
- Suggest why the experimental value is smaller in magnitude than the data-book value. [1]
Mark scheme
- q = mcΔT = 100 × 4.2 × 30[1]
- = 12600 J = 12.6 kJ[1]
- Moles ethanol = 0.50 ÷ 46 = 0.0109 mol[1]
- ΔH = energy ÷ moles = 12.6 ÷ 0.0109[1]
- = 1156 kJ/mol (magnitude)[1]
- ΔH is negative (exothermic): ≈ −1160 kJ/mol[1]
- Heat is lost to the surroundings / incomplete combustion[1]
q = 12.6 kJ; 0.0109 mol; ΔH ≈ −1160 kJ/mol; lower magnitude due to heat loss / incomplete combustion.
Examiner tip. Combustion is exothermic so ΔH is negative; heat lost to surroundings always lowers the measured value.
LQ2[6 marks]
Magnesium oxide and sodium chloride are both ionic solids.
- Define the term lattice energy. [2]
- Explain why the lattice energy of magnesium oxide is much more exothermic than that of sodium chloride. [4]
Mark scheme
- The energy released when gaseous ions[1]
- combine to form one mole of solid ionic lattice[1]
- MgO contains Mg²⁺ and O²⁻ ions with 2+ and 2− charges[1]
- NaCl contains only 1+ and 1− ions[1]
- the ions in MgO are also smaller[1]
- higher charges and smaller radii give stronger electrostatic attraction, so a much larger (more exothermic) lattice energy[1]
Lattice energy = energy released forming a mole of lattice from gaseous ions; MgO's higher charges and smaller ions give much stronger attraction than NaCl.
Examiner tip. Lattice energy rises with ionic charge and falls with ionic radius.