This chemistry grades 11-12 lesson on enthalpy changes notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on enthalpy changes and check yourself with the quiz.
Quick answer: An enthalpy change ΔH is the heat transferred at constant pressure; it is negative for exothermic reactions. Calorimetry gives it from q = mcΔT and ΔH = −q/n. Hess’s law says ΔH is route independent, so it can be found from formation or combustion data or estimated from bond enthalpies.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R1.1-R1.2
Enthalpy changes notes: the key ideas
What is an enthalpy change?
The enthalpy change ΔH is the heat energy transferred between a reacting system and its surroundings at constant pressure. It is negative for an exothermic reaction (heat released) and positive for an endothermic reaction (heat absorbed). Standard values refer to 100 kPa and a stated temperature, usually 298 K.
| Enthalpy change | Meaning (per mole) | Typical sign |
|---|---|---|
| Formation, ΔHf | One mole of a compound formed from its elements in their standard states | Usually negative |
| Combustion, ΔHc | One mole of a substance burned completely in oxygen | Always negative |
| Neutralisation | One mole of water formed from an acid and an alkali | Negative |
How do you find ΔH from a temperature change?
In a calorimeter the heat absorbed by the water is q, and the reaction transferred the same amount in the opposite direction. Divide by the amount of the limiting reactant to get ΔH per mole.
q = m × c × ΔT and ΔH = −q / n
q = heat absorbed by the solution (J), m = mass of solution (g), c = 4.18 J/(g K), ΔT = temperature change (K), n = moles of limiting reactant (mol), ΔH in J/mol (divide by 1000 for kJ/mol).
Hannah in Boston mixes 40.0 cm³ of 1.50 mol/dm³ nitric acid with 40.0 cm³ of 1.50 mol/dm³ potassium hydroxide. The temperature rises by 10.0 K.
Mass = 80.0 g, so q = 80.0 × 4.18 × 10.0 = 3344 J. Moles = 0.0400 × 1.50 = 0.0600 mol. ΔH = −3344 / 0.0600 = −55 733 J/mol = −55.7 kJ/mol.
What does Hess’s law say?
Hess’s law: the total enthalpy change of a reaction is the same whichever route is taken, provided the initial and final conditions are the same. So ΔH of a reaction equals the sum of ΔH values of any route linking the same reactants and products.
For combustion data the working is ΔH = Σ ΔHc(reactants) − Σ ΔHc(products). For formation data it is ΔH = Σ ΔHf(products) − Σ ΔHf(reactants).
Calculate ΔHf of ethane from ΔHc: carbon −393.5, hydrogen −285.8, ethane −1560 kJ/mol.
Route through the elements burning: 2(−393.5) + 3(−285.8) = −787.0 − 857.4 = −1644.4 kJ. ΔHf = −1644.4 − (−1560) = −84.4 kJ/mol.
How do bond enthalpies estimate ΔH?
Mean bond enthalpy is the average energy needed to break one mole of a given bond in gaseous molecules. Breaking bonds absorbs energy and making bonds releases it, so ΔH = Σ(bonds broken) − Σ(bonds formed).
| Bond | Mean enthalpy (kJ/mol) | Bond | Mean enthalpy (kJ/mol) |
|---|---|---|---|
| H–H | 436 | C=O (in CO₂) | 805 |
| O=O | 498 | O–H | 464 |
| C–H | 413 | C–C | 347 |
| C=C | 612 | H–Cl | 432 |
Estimate ΔH for 2H₂(g) + O₂(g) → 2H₂O(g).
Broken: 2 × 436 + 498 = 1370 kJ. Formed: 4 × 464 = 1856 kJ. ΔH = 1370 − 1856 = −486 kJ (for the equation as written).
Do not use the mass of the solid in q = m c ΔT: the heated mass is the solution or water. Also do not forget the minus sign in ΔH = −q / n: a temperature rise means an exothermic reaction and a negative ΔH.
Draw the Hess cycle before you calculate. Label each arrow with its ΔH, then follow the two routes. Arrows against the arrow direction change sign.
Bond enthalpy results are approximate because the values are means taken from many compounds and refer to the gas phase, so they differ a little from calorimetric data.
- q = m c ΔT with c = 4.18 J/(g K); ΔH = −q / n in J/mol.
- Hess’s law: ΔH depends only on the initial and final states, not on the route.
- ΔH = Σ ΔHf(products) − Σ ΔHf(reactants) and ΔH = Σ ΔHc(reactants) − Σ ΔHc(products).
- Bond enthalpies: ΔH = Σ(bonds broken) − Σ(bonds formed), a gas-phase estimate.
- Exothermic reactions have ΔH below zero; endothermic reactions have ΔH above zero.
Heat changes, energy cycles and bond enthalpies: frequently asked questions
What is the difference between enthalpy of formation and combustion?
Formation is the enthalpy change when one mole of a compound forms from its elements in their standard states. Combustion is the change when one mole of a substance burns completely in oxygen.
Why is ΔH negative for an exothermic reaction?
The products have less enthalpy than the reactants because heat is released to the surroundings. ΔH is the enthalpy of products minus reactants, so it is below zero.
How do you use Hess’s law with an energy cycle?
Draw the cycle with the unknown enthalpy change on one arrow and known values on the others. Add the values along one route and compare it with the other route, changing the sign of any arrow followed backwards.
Why are bond enthalpy results only approximate?
Mean bond enthalpies are averages over many different molecules and apply to gases. The real bond energy in a particular molecule differs a little, so the result differs from calorimetric values.
Why is heat loss a problem in calorimetry?
Some heat escapes to the surroundings, so the measured temperature change is too small. The calculated ΔH is then less exothermic than the true value. Lids, insulation and graph extrapolation reduce the error.
