2.1.5Unit 2. Energetics, Group Chemistry, Halogenoalkanes and Alcohols
Bond Enthalpies — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 14 of 53 Unit quiz
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 6Bond Enthalpy
The energy change in a reaction comes from breaking bonds in the reactants and making new bonds in the products.
Definition
- Definition The bond enthalpy is the energy needed to break one mole of a covalent bond in gaseous molecules.
Sign
- Sign Breaking bonds is endothermic, so bond enthalpies are positive; forming bonds is exothermic.
Average values
- Average values Tables give mean values taken over many compounds, because a bond has a slightly different strength in different molecules.
Strength
- Strength The larger the value, the stronger the bond, so C≡C is stronger than C=C, which is stronger than C–C.
Part 2 of 6Typical Values
Typical Values
C–H and C–C
- C–H and C–C 413 and 347 kJ mol⁻¹ respectively.
O–H and O=O
- O–H and O=O 464 and 498 kJ mol⁻¹, with the O=O value used for oxygen gas.
C=O in CO₂
- C=O in CO₂ 805 kJ mol⁻¹ is used for the double bond in carbon dioxide.
H–H, Cl–Cl and H–Cl
- H–H, Cl–Cl and H–Cl 436, 243 and 432 kJ mol⁻¹ respectively.
| Bond | Enthalpy / kJ mol⁻¹ |
|---|---|
| C–H | 413 |
| O=O | 498 |
| C=O (in CO₂) | 805 |
| O–H | 464 |
Part 3 of 6Calculation Method
1
Write a balanced equation and draw out all the bonds in each molecule so none are missed.
2
Add up the enthalpies of all bonds broken in the reactants.
3
Add up the enthalpies of all bonds formed in the products.
4
ΔH = total for bonds broken − total for bonds formed.
5
A negative answer means the new bonds are stronger overall, so the reaction is exothermic.
Part 4 of 6Limitations
Limitations
Average values
- Average values Real bond strengths vary between molecules, so the answer is only an estimate.
Gas phase only
- Gas phase only The method applies to gaseous species, so a liquid such as water must be treated as gas or corrected for vaporisation.
Comparison
- Comparison Results may differ by tens of kJ mol⁻¹ from values obtained using Hess cycles with formation data.
Usefulness
- Usefulness The method is quick and works for any reaction where bond data are available, even if formation data are not.
Part 5 of 6Worked Example
Question Use bond enthalpies to estimate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), with C–H = 413, O=O = 498, C=O = 805 and O–H = 464 kJ mol⁻¹.
- Bonds broken4 × 413 + 2 × 498 = 1652 + 996 = 2648 kJ.
- Bonds formed2 × 805 + 4 × 464 = 1610 + 1856 = 3466 kJ.
- ΔH = 2648 − 3466 = −818 kJ mol⁻¹.
Answer ΔH ≈ −818 kJ mol⁻¹
Part 6 of 6Exam Checklist & Tips
2.1.5 – Key Points for the Exam
- Bond enthalpy The energy needed to break one mole of a covalent bond in the gas phase.
- Average bond enthalpy The mean value of a bond enthalpy taken over a range of compounds.
- Bonds broken Bonds in the reactants, which require energy input.
- Bonds formed Bonds in the products, which release energy.
- Gaseous state The state required for bond enthalpy calculations.
Common Mistakes
- Calculating bonds formed minus bonds broken, which gives the wrong sign.
- Missing bonds in a molecule, such as the four C–H bonds in methane.
- Using liquid water in a bond enthalpy calculation without correction.
- Treating the value as exact rather than as an average.
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