1.6.7Physical Chemistry
Bond Enthalpies — AQA AS Chemistry Revision Notes & Topical Questions
5 min read 5 parts Lesson 39 of 102
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 5Bond Enthalpy Basics
Definition
- Definition Bond enthalpy is the energy required to break one mole of a particular covalent bond in gaseous molecules.
Mean value
- Mean value Because the same bond differs slightly between compounds, tables give a mean bond enthalpy averaged over many compounds.
Sign
- Sign Bond breaking is endothermic so bond enthalpies are positive, and bond making releases the same quantity of energy.
Gas phase
- Gas phase The definition applies only to gaseous species, so liquids and solids must be treated as gases when using these values.
Strength
- Strength A larger bond enthalpy means a stronger bond, as with N≡N which is much larger than N–N.
Part 2 of 5Calculation
Calculation
Formula
- Formula ΔH = Σ(bonds broken) − Σ(bonds formed), with each term the bond enthalpy multiplied by the number of bonds.
Broken
- Broken Add up the energy to break every bond in the reactants, and count every bond in each molecule separately.
Formed
- Formed Add up the energy released by the bonds formed in every product molecule.
Why approximate
- Why approximate Mean values, not actual values for the specific molecule, are used, and water or other products may not be gaseous in practice.
| Bond | Mean bond enthalpy / kJ mol⁻¹ |
|---|---|
| C–H | 413 |
| O=O | 498 |
| C=O | 805 |
| O–H | 464 |
| N≡N | 945 |
| H–H | 436 |
| N–H | 391 |
Part 3 of 5Steps
1
Write the balanced equation and draw each molecule so that every bond is visible.
2
List the bonds broken in the reactants and multiply each bond enthalpy by how many are broken.
3
List the bonds formed in the products and multiply likewise.
4
Subtract the total for bonds formed from the total for bonds broken to find ΔH.
5
Interpret the sign and compare with the experimentally measured value, explaining any difference by mean bond enthalpies.
Part 4 of 5Worked Example
Question
- Use mean bond enthalpies to find ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
- (C–H 413, O=O 498, C=O 805, O–H 464 kJ mol⁻¹)
- Bonds broken
- 4 × C–H = 4 × 413 = 1652;
- 2 × O=O = 2 × 498 = 996;
- total = 2648 kJ.
- 4 × C–H = 4 × 413 = 1652;
- Bonds formed
- 2 × C=O = 2 × 805 = 1610;
- 4 × O–H = 4 × 464 = 1856;
- total = 3466 kJ.
- 2 × C=O = 2 × 805 = 1610;
- ΔH = 2648 − 3466 = −818 kJ mol⁻¹.
Answer ΔH = −818 kJ mol⁻¹, an exothermic reaction.
Part 5 of 5Exam Checklist & Tips
1.6.7 – Key Points for the Exam
- Bond enthalpy The energy needed to break one mole of a covalent bond in gaseous molecules.
- Mean bond enthalpy The average value of a bond enthalpy across a range of different compounds.
- Gaseous state The condition required for the definition, in which no intermolecular forces remain.
- Σ The summation sign meaning add up all of the values.
Common Mistakes
- Subtracting in the wrong direction; the correct form is bonds broken minus bonds formed.
- Forgetting to count all the bonds, such as 4 C–H bonds in methane.
- Using bond enthalpies for reactions in which water is liquid without correcting to the gas phase.
- Treating mean bond enthalpies as exact values for the molecule concerned.
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