AP Chemistry · Lesson 3

Hess's law and bond energies

Know Know it

Enthalpy is a state function: ΔH depends only on where a reaction starts and ends. So by Hess's law, if equations add up to a target equation, their ΔH values add up to the target's ΔH. Reverse an equation and its ΔH changes sign; multiply it and its ΔH is multiplied too.

A standard enthalpy of formation, ΔH°f, is ΔH for making 1 mol of a compound from its elements in their standard states. For an element in its standard state, such as O₂(g), it is zero. Then ΔHrxn∘=∑nΔHf∘(products)−∑nΔHf∘(reactants)\Delta H^\circ_{\text{rxn}} = \sum n\Delta H^\circ_f(\text{products}) - \sum n\Delta H^\circ_f(\text{reactants}).

Breaking a bond always absorbs energy, and forming one always releases energy. With average bond energies, ΔH≈∑(bonds broken)−∑(bonds formed)\Delta H \approx \sum(\text{bonds broken}) - \sum(\text{bonds formed}). A reaction is exothermic when the bonds formed release more energy than the bonds broken absorb.

Worked example Easy

Use these two reactions. (1) C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ. (2) C(s) + ½O₂(g) → CO(g), ΔH₂ = −110.5 kJ.

What is ΔH for the reaction 2CO(g) + O₂(g) → 2CO₂(g)?

  1. −283.0 kJ
  2. −504.0 kJ
  3. −1008 kJ
  4. −566.0 kJAnswer

How to solve it

  1. The target has CO on the left, so reverse (2) and double it: 2CO → 2C + O₂, ΔH = +221.0 kJ.
  2. The target has 2CO₂ on the right, so double (1): 2C + 2O₂ → 2CO₂, ΔH = −787.0 kJ.
  3. Add: 2C and one O₂ cancel, leaving 2CO + O₂ → 2CO₂, ΔH = +221.0 − 787.0 = −566.0 kJ.

Why each choice is right or wrong

  • A. Incorrect. −283.0 kJ is ΔH for 1 mol of CO burning; the target equation has 2CO, so both reactions must be doubled.
  • B. Incorrect. −504.0 kJ adds the two reactions as given; reaction (2) must be reversed, which changes the sign of ΔH₂.
  • C. Incorrect. −1008 kJ doubles both reactions but forgets to reverse reaction (2), so it adds −110.5 kJ instead of +110.5 kJ twice.
  • D. Correct. Double reaction (1) and double the reverse of (2): 2(−393.5) + 2(+110.5) = −566.0 kJ.

Trap Bond energies run the other way

With formation enthalpies, ΔH is products minus reactants. With bond energies, ΔH is bonds broken (in the reactants) minus bonds formed (in the products). Mixing up the two gives the right size with the wrong sign.

Explain Explain it

Task verbs: calculate, explain

When you use formation values, multiply each by its coefficient and leave out elements in their standard states, which are zero. When you use bond energies, draw or list every bond in each molecule first, and count how many of each break and form.

Worked example Medium

Calcium carbonate decomposes when strongly heated: CaCO₃(s) → CaO(s) + CO₂(g).

Standard enthalpies of formation at 25 °C

SubstanceΔH°f (kJ/mol)
CaCO₃(s)−1206.9
CaO(s)−635.1
CO₂(g)−393.5

What is ΔH° for the decomposition?

  1. +178.3 kJ/molAnswer
  2. −178.3 kJ/mol
  3. −2235.5 kJ/mol
  4. −1028.6 kJ/mol

How to solve it

  1. Products: −635.1 + (−393.5) = −1028.6 kJ.
  2. Reactants: −1206.9 kJ.
  3. ΔH° = −1028.6 − (−1206.9) = +178.3 kJ/mol rxn.

Why each choice is right or wrong

  • A. Correct. Products minus reactants: (−635.1 − 393.5) − (−1206.9) = +178.3 kJ/mol, so the decomposition is endothermic.
  • B. Incorrect. −178.3 kJ/mol takes reactants minus products, the reverse of the formation rule, which flips the sign.
  • C. Incorrect. −2235.5 kJ/mol adds all three formation values instead of subtracting the reactant's value.
  • D. Incorrect. −1028.6 kJ/mol is the sum for the products only; the reactant's value, −1206.9, still has to be subtracted.

Explain Your turn

For the reaction H₂(g) + Cl₂(g) → 2HCl(g), use these average bond energies: H–H 436 kJ/mol, Cl–Cl 242 kJ/mol, H–Cl 431 kJ/mol. Calculate ΔH for the reaction, and explain in terms of bond breaking and bond forming why the reaction is exothermic.

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A full answer does this

  • Calculates ΔH = −184 kJ/mol rxn
  • Explains that forming the bonds releases more energy than breaking the bonds absorbs

Model answer

ΔH ≈ bonds broken − bonds formed = (436 + 242) − 2(431) = 678 − 862 = −184 kJ/mol rxn. Breaking one H–H bond and one Cl–Cl bond absorbs 678 kJ, but forming two H–Cl bonds releases 862 kJ. More energy is released by forming the new bonds than is absorbed by breaking the old ones, so the reaction releases heat overall and is exothermic.

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