Bond Enthalpies
A concise revision guide to average bond enthalpies: what average bond enthalpy means, why reactions are exothermic or endothermic, calculating ΔH from average bond enthalpies, and why the answers differ from Hess’s law values, for OCR A Level Chemistry.
- 3.2.1(f)(i)
- 3.2.1(f)(ii)-a
- 3.2.1(f)(ii)-b
- 3.2.1(f)(iii)
What these spec points say
- 3.2.1(f)(i) explanation of the term average bond enthalpy (as the breaking of 1 mol of bonds in gaseous molecules)
- 3.2.1(f)(ii)-a explanation of exothermic reactions in terms of enthalpy changes associated with the breaking and making of chemical bonds
- 3.2.1(f)(ii)-b explanation of endothermic reactions in terms of enthalpy changes associated with the breaking and making of chemical bonds
- 3.2.1(f)(iii) use of average bond enthalpies to calculate enthalpy changes and related quantities (see also 2.2.2 f)
Bond Enthalpies: Breaking In, Making Out
Choose a step to watch hydrogen and chlorine react: energy goes in to break the old bonds, comes out as the new bonds form, and the difference is the enthalpy change.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
Bonds broken and made when methane burns
Break every bond in CH4 and 2O2, form the bonds in CO2 and 2H2O, and add up the bond enthalpies to find the enthalpy change of combustion.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
GCSE Recap: Energy In and Energy Out
Before you start, check the GCSE rule this page puts numbers on.
What Bond Enthalpy Means
The bond enthalpy is the energy needed to break one mole of a given covalent bond in molecules in the gaseous state. Breaking a bond always needs energy, so average bond enthalpies are always positive; making the same bond releases the same amount of energy.
For a diatomic molecule such as H₂ the value is exact.
For bonds that appear in many compounds, such as C–H, the value quoted is an average bond enthalpy: the average energy needed to break that type of bond, taken over a range of compounds.
Average bond enthalpies for common bonds: multiple bonds are stronger than single bonds between the same atoms, and N≡N is the strongest shown.
Definition: Average bond enthalpy is the energy needed to break one mole of a given type of bond in gaseous molecules, averaged over a range of compounds.
Quick Check: Exact or Mean?
Drag the words into place to decide which of two quoted values is exact.
Why Reactions Are Exothermic or Endothermic
Every reaction breaks some bonds and makes others. Bond breaking is endothermic and bond making is exothermic. The overall enthalpy change depends on which is larger:
- If more energy is released making the bonds in the products than is absorbed breaking the bonds in the reactants, the reaction is exothermic.
- If more energy is absorbed breaking bonds than is released making them, the reaction is endothermic.
Key idea: ΔH = energy absorbed breaking bonds − energy released making bonds. A negative answer means the products are held by stronger bonds overall.
Quick Check: Reading the Sign
In each round, choose the one statement that is accurate.
Calculating ΔH from Average Bond Enthalpies
Key equation: ΔH = Σ(average bond enthalpies of bonds broken) − Σ(average bond enthalpies of bonds made)
Worked example. Calculate the enthalpy change for the combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), using C–H 413, O=O 498, C=O 805 and O–H 464 kJ mol⁻¹.
| Bonds | Number | Energy / kJ |
|---|---|---|
| Broken: C–H | 4 | 4 × 413 = 1652 |
| Broken: O=O | 2 | 2 × 498 = 996 |
| Total broken | +2648 | |
| Made: C=O | 2 | 2 × 805 = 1610 |
| Made: O–H | 4 | 4 × 464 = 1856 |
| Total made | 3466 |
ΔH = 2648 − 3466 = −818 kJ mol⁻¹
Counting every bond always works. Where a bond is unchanged on both sides you may leave it out and count only the bonds that break and form.
The unchanged bonds cancel, so the two routes give the same answer, and the shorter one saves a great deal of time.
Exam tip: Data books differ slightly in the values they quote, so always use the values printed in the question.
Exam focus: Draw out the displayed formulae first and count every bond. Most lost marks come from missing a bond, not from the arithmetic.
Quick Check: Calculate an Enthalpy Change
Work the calculation out on paper with the data in the question, then choose your answer.
Why Bond Enthalpy Values Differ from Hess Values
The data book enthalpy change of combustion of methane is −890 kJ mol⁻¹, but the bond enthalpy calculation gives −818 kJ mol⁻¹. There are two reasons:
- Mean values are used. The C–H value is an average over many compounds, not the exact value for methane, and the same is true of C=O and O–H.
- Everything is assumed to be gaseous. Average bond enthalpies apply to gaseous molecules, but standard enthalpy changes of combustion form liquid water. Condensing the water releases extra energy that the calculation leaves out. Put plainly, this calculation gives the value for H₂O(g), while the data book enthalpy change of combustion is quoted for H₂O(l).
Exam sentence: Values calculated from average bond enthalpies are less accurate than those from Hess’s law because average bond enthalpies are averages over many compounds and apply only to substances in the gaseous state.
The combustion of methane from data book values and from bond enthalpies, and the two reasons the answers differ.
Quick Check: Explain the Difference
Write a short explanation, then compare it with the mark points and the model answer.
Finding an Average Bond Enthalpy from ΔH
The same equation can be rearranged to find one unknown bond value when ΔH and the others are known.
Worked example. H₂(g) + Cl₂(g) → 2HCl(g), ΔH = −184 kJ mol⁻¹. Using H–H 436 and Cl–Cl 243 kJ mol⁻¹, find the H–Cl value.
- ΔH = (436 + 243) − 2E(H–Cl)
- −184 = 679 − 2E(H–Cl)
- 2E(H–Cl) = 863, so E(H–Cl) = 432 kJ mol⁻¹ (3 significant figures)
With the unknown on the left. H₂(g) + I₂(g) → 2HI(g), ΔH = −11 kJ mol⁻¹, using H–H 436 and H–I 299 kJ mol⁻¹. Bonds broken are 436 + E(I–I); bonds made are 2 × 299 = 598. So −11 = 436 + E(I–I) − 598, giving E(I–I) = 151 kJ mol⁻¹.
Remember: Two moles of HCl form, so two H–Cl bonds are made. Divide by the number of bonds of the unknown type at the end.
Quick Check: Find the Unknown Bond
Rearrange the key equation to find the missing value, then choose your answer.
Common Exam Points
Define average bond enthalpy
The energy needed to break one mole of a given type of bond in gaseous molecules, averaged over a range of compounds.
Calculate ΔH from average bond enthalpies
Bonds broken minus bonds made; count every bond from the displayed formulae.
Explain why the answer differs from the data book value
Average bond enthalpies are averages, and the calculation assumes all species are gases.
Do not say
“Bond making needs energy”; “average bond enthalpies are exact for every molecule”.
FAQs
Use these quick answers to check the average bond enthalpies ideas and calculations.
Why are average bond enthalpies always positive?
They are defined for breaking bonds, and breaking a bond always needs energy.
How do I calculate ΔH from bond values?
Add up the average bond enthalpies of all the bonds broken, add up those of all the bonds made, and subtract: bonds broken minus bonds made.
Why is my answer different from the data book value?
The average bond enthalpies are averages over many compounds, and the method assumes every substance is a gas, whereas standard values may involve liquids.
What makes a reaction exothermic in terms of bonds?
More energy is released making the bonds in the products than is absorbed breaking the bonds in the reactants.
Are any bond values exact?
Yes, for diatomic molecules such as H₂ or Cl₂, where there is only one bond in one environment. Bonds found in many compounds, such as C–H, have average values.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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