AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/6/6-7)
Unit 6 · Topic 6.7
6.7 Bond Enthalpies
Breaking a bond always takes energy and forming a bond always releases energy. By adding up the average bond enthalpies of the bonds broken and subtracting those of the bonds formed, you can estimate ΔH for a reaction and explain why it is exothermic or endothermic.
Key terms
- bond enthalpy
- bonds broken
- bonds formed
- average bond energy
Bond enthalpy
A bond enthalpy is the energy needed to break one mole of a particular bond in gas-phase molecules. It is always positive, because pulling bonded atoms apart always takes energy. Forming that same bond releases the same amount of energy.
This connects to the potential energy curve in topic 2.2: the bond energy is the depth of the well. Stronger bonds have larger bond enthalpies. Between the same two atoms, triple bonds are stronger than double bonds, which are stronger than single bonds.
| Bond | Average bond enthalpy (kJ/mol) |
|---|---|
| H–H | 436 |
| C–H | 413 |
| C–C | 348 |
| C=C | 614 |
| O=O | 495 |
| C=O (in CO₂) | 799 |
| O–H | 463 |
| N≡N | 941 |
| N–H | 391 |
| Cl–Cl | 242 |
| H–Cl | 431 |
The method
ΔH ≈ Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed). Broken bonds are in the reactants; formed bonds are in the products.
- Draw a Lewis structure for each molecule so you can see every bond, including double and triple bonds.
- Count each type of bond in one molecule, then multiply by the coefficient.
- Add up the energy needed to break all reactant bonds.
- Add up the energy released by forming all product bonds.
- Subtract: broken minus formed.
What the sign means
If forming the product bonds releases more energy than it took to break the reactant bonds, the reaction is exothermic overall (ΔH < 0). If breaking takes more than forming gives back, it's endothermic (ΔH > 0).
Combustion reactions are strongly exothermic because the products, CO₂ and H₂O, contain very strong C=O and O–H bonds. Energy is not 'stored in' the bonds of the fuel and released when they break. The energy comes from forming stronger bonds than the ones you broke.
Why the answer is an estimate
A C–H bond in methane is not exactly as strong as a C–H bond in ethanol. Tables list averages over many molecules, so bond-enthalpy answers usually differ by a few percent from values found with enthalpies of formation (topic 6.8). Bond enthalpies also assume every species is a gas, so they ignore the energy of condensing liquids or solids.
A shortcut: you only need to count bonds that actually change. In C₂H₄ + H₂ → C₂H₆, the four original C–H bonds survive, so you only break C=C and H–H and form C–C and two new C–H bonds: (614 + 436) − (348 + 2 × 413) = −124 kJ/mol. Counting every bond gives the same answer.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
A simple gas-phase reaction
Use the bond enthalpies in the table to estimate ΔH for H₂(g) + Cl₂(g) → 2HCl(g).
Show the solutionHide the solution
- Step 1: Bonds broken: one H–H (436 kJ) and one Cl–Cl (242 kJ). Total = 678 kJ.
- Step 2: Bonds formed: two H–Cl, 2 × 431 = 862 kJ.
- Step 3: ΔH ≈ broken − formed = 678 − 862 = −184 kJ/mol. More energy is released forming the H–Cl bonds than is used breaking H–H and Cl–Cl, so the reaction is exothermic.
Answer: ΔH ≈ −184 kJ/mol
- Example 2
Combustion of methane
Estimate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) using the table.
Show the solutionHide the solution
- Step 1: Lewis structures: CH₄ has four C–H bonds; O₂ has one O=O; CO₂ (O=C=O) has two C=O; each H₂O has two O–H.
- Step 2: Broken: 4 C–H + 2 O=O = 4(413) + 2(495) = 1652 + 990 = 2642 kJ.
- Step 3: Formed: 2 C=O + 4 O–H (two waters × two bonds) = 2(799) + 4(463) = 1598 + 1852 = 3450 kJ.
- Step 4: ΔH ≈ 2642 − 3450 = −808 kJ/mol.
- Step 5: The measured value for this reaction (water as a gas) is about −802 kJ/mol, so the estimate is close but not exact, as expected for average bond enthalpies.
Answer: ΔH ≈ −808 kJ/mol
- Example 3
Trap: bonds per molecule times coefficients
Estimate ΔH for N₂(g) + 3H₂(g) → 2NH₃(g).
Show the solutionHide the solution
- Step 1: Broken: one N≡N and three H–H = 941 + 3(436) = 941 + 1308 = 2249 kJ. Use the triple-bond value, not a single N–N value.
- Step 2: Formed: each NH₃ has three N–H bonds, and there are two NH₃, so six N–H bonds: 6(391) = 2346 kJ. Counting only two N–H bonds (one per coefficient) is the trap.
- Step 3: ΔH ≈ 2249 − 2346 = −97 kJ/mol. (The value from enthalpies of formation is −92 kJ/mol.)
Answer: ΔH ≈ −97 kJ/mol
Common mistakes
- Subtracting in the wrong order. With bond enthalpies it's broken minus formed (reactants minus products), the opposite order from the enthalpy-of-formation formula.
- Counting molecules instead of bonds: two NH₃ molecules contain six N–H bonds.
- Using a single-bond value for a double or triple bond. Draw the Lewis structure first.
- Saying energy is released when bonds break. Breaking always costs energy; forming always releases it.
On the exam
- You will be given a table of bond enthalpies and asked to estimate ΔH, often with Lewis structures to interpret. Show which bonds are broken and formed, with counts.
- A common follow-up asks why the bond-enthalpy estimate differs from a measured or table-based value. Answer: bond enthalpies are averages over many different molecules.
Connected topics
Videos
Check yourself
4 questions on 6.7 Bond Enthalpies. Pick an answer to see if you got it, and why.
| Bond | Average bond enthalpy (kJ/mol) |
|---|---|
| C–H | 413 |
| Cl–Cl | 242 |
| C–Cl | 328 |
| H–Cl | 431 |
Bond enthalpies for the reaction CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)
Based on the bond enthalpies, what is the approximate ΔH for the reaction?
The measured ΔH for this reaction differs slightly from the value calculated from the table. Which of the following best explains the difference?
A reaction is exothermic. Which of the following must be true?
| Bond | Average bond enthalpy (kJ/mol) |
|---|---|
| C=C | 614 |
| C–C | 348 |
| C–H | 413 |
| H–H | 436 |
Average bond enthalpies. Reaction: H₂C=CH₂(g) + H₂(g) → H₃C–CH₃(g)
Use the bond enthalpies to estimate ΔH for the reaction.
0 of 4 answered