This AP Chemistry lesson on thermochemistry notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on thermochemistry and check yourself with the quiz.
Quick answer: Thermochemistry measures heat changes in reactions. At constant pressure the heat is the enthalpy change ΔH: negative for exothermic, positive for endothermic. Calorimetry uses q = mcΔT, then divides by moles. Hess’s law adds step enthalpies, and ΔH can also come from formation enthalpies or bond enthalpies.
Level: AP Chemistry · Curriculum: AP Chem Unit 6
Thermochemistry notes: the key ideas
What is an enthalpy change?
At constant pressure, the heat exchanged with the surroundings in a reaction is the enthalpy change ΔH. An exothermic reaction releases heat (ΔH < 0) and the surroundings warm up; an endothermic reaction absorbs heat (ΔH > 0) and they cool down.
Enthalpy change ΔH = H(products) − H(reactants), in kJ/mol of reaction as written. The standard enthalpy of formation ΔH°f is the enthalpy change when 1 mol of a compound forms from its elements in their standard states; for an element in its standard state it is 0.
How does calorimetry measure heat?
A calorimeter isolates the reaction so that its heat goes into a known mass of water. We use q = mcΔT, then divide by the amount in moles to get ΔH. The heat gained by the solution is equal in size and opposite in sign to the heat released by the reaction.
q = mcΔT, with q in J, m in g, c in J/(g °C) (4.18 for dilute solutions) and ΔT = Tfinal − Tinitial in °C. Then ΔH = −q(solution)/n, with n in mol, giving J/mol.
Wanjiru in Nairobi mixes 100.0 mL of 1.00 M HCl with 100.0 mL of 1.00 M NaOH, both at 22.0 °C. The mixture reaches 28.6 °C. Take the density as 1.00 g/mL and c = 4.18 J/(g °C).
Mass = 200.0 g; ΔT = 6.6 °C; q = 200.0 × 4.18 × 6.6 = 5.52 × 10³ J.
Water formed: n = 0.1000 L × 1.00 mol/L = 0.100 mol.
ΔH = −5.52 kJ/0.100 mol = −55 kJ/mol.
How does Hess’s law work?
The enthalpy change of a reaction is the same whether it takes place in one step or several, because enthalpy is a state function. Reversing an equation changes the sign of ΔH; multiplying an equation by n multiplies ΔH by n.
Find ΔH for C(s) + ½O₂(g) → CO(g) given: (1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ; (2) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ.
Keep (1). Reverse (2): CO₂ → CO + ½O₂, ΔH = +283.0 kJ. Add: C + O₂ → CO + ½O₂, and cancel ½O₂.
ΔH = −393.5 + 283.0 = −110.5 kJ.
How do you use enthalpies of formation and bond enthalpies?
With formation data: ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient. With bond enthalpies: ΔH ≈ Σ(bonds broken) − Σ(bonds formed). Breaking bonds costs energy; forming bonds releases it. Bond-enthalpy answers are estimates because the values are averages for gases.
| Bond | Enthalpy (kJ/mol) | Bond | Enthalpy (kJ/mol) |
|---|---|---|---|
| C–H | 413 | O=O | 498 |
| C=O (in CO₂) | 799 | O–H | 463 |
| H–H | 436 | C–C | 347 |
Estimate ΔH for CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g).
Broken: 4 C–H + 2 O=O = 4(413) + 2(498) = 1652 + 996 = 2648 kJ.
Formed: 2 C=O + 4 O–H = 2(799) + 4(463) = 1598 + 1852 = 3450 kJ.
ΔH ≈ 2648 − 3450 = −802 kJ.
For C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l), use ΔH°f (kJ/mol): C₃H₈ −103.8, CO₂ −393.5, H₂O(l) −285.8, O₂ 0.
ΔH° = [3(−393.5) + 4(−285.8)] − [−103.8 + 0] = (−1180.5 − 1143.2) + 103.8 = −2219.9 kJ.
Students often swap the order and use reactants minus products with formation data, or forget that bond enthalpies work the other way round: broken minus formed. Also, do not drop the sign of ΔT or of q: the water gains heat, so the reaction has the opposite sign.
Write the sign rule on the page margin: formation data are products − reactants, bond data are broken − formed. Then check the sign with your common sense: if the beaker got warmer, ΔH must be negative.
Which enthalpy words should you know?
Standard conditions mean 1 bar and a stated temperature, usually 298 K. Combustion enthalpy is for burning 1 mol in oxygen. Neutralisation enthalpy is for forming 1 mol of water from an acid and a base, about −57 kJ/mol for strong acids and strong bases.
- q = mcΔT; ΔH(reaction) = −q(solution)/n.
- Hess’s law: reverse an equation, change the sign; multiply, multiply ΔH.
- ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants); ΔH°f of an element is 0.
- Bond enthalpies: ΔH ≈ Σ(broken) − Σ(formed), an estimate for gases.
- Exothermic: ΔH < 0, surroundings warm; endothermic: ΔH > 0, surroundings cool.
Heat, enthalpy and Hess's law: frequently asked questions
What is the difference between heat and temperature?
Temperature measures the average kinetic energy of particles, while heat is energy transferred because of a temperature difference. A large bath at 40 °C can carry more heat than a spark at 1000 °C.
Why is ΔH negative for an exothermic reaction?
Because the system loses enthalpy: the products have less enthalpy than the reactants and the difference goes to the surroundings as heat.
Why does Hess’s law work?
Enthalpy is a state function, so ΔH depends only on the initial and final states, not on the route taken between them.
Why are bond-enthalpy answers only estimates?
The tabulated values are averages over many different molecules and refer to gaseous species, so the result differs a little from values measured by calorimetry.
Why is the standard enthalpy of formation of O₂(g) zero?
It is an element in its standard state, so forming it from itself involves no change. Only compounds and non-standard forms have non-zero values.
