Bond Energy Calculations (Higher Tier)

This lesson covers the quantitative calculation of energy changes using bond energy data. Bond energy calculations are a core Higher Tier skill for AQA GCSE Combined Science Trilogy (8464). You will learn the method, work through examples step by step, and practise interpreting your results.

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What Is Bond Energy?

Bond energy is the amount of energy needed to break one mole of a particular covalent bond in a gaseous molecule. It is also the energy released when one mole of that bond forms.

Bond energy is measured in kilojoules per mole (kJ/mol).

Bond	Bond Energy (kJ/mol)
H–H	436
O=O	498
C–H	413
C=O	799
O–H	463
N≡N	945
N–H	391
Cl–Cl	242
H–Cl	432
Br–Br	193
H–Br	366
C–C	347
C=C	614
C–O	358

Exam Tip: You do not need to memorise bond energies — they are always given in the question. But you DO need to know how to use them.

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The Key Formula

$\Delta H = \sum \text{(energy to break bonds in reactants)} - \sum \text{(energy released forming bonds in products)}$ΔH=∑(energy to break bonds in reactants)−∑(energy released forming bonds in products)

Or more simply:

$\boxed{\Delta H = \text{Bonds Broken} - \text{Bonds Made}}$ΔH=Bonds Broken−Bonds Made​

graph LR
    A["Step 1: Draw structural \nformulae of reactants \nand products"] --> B["Step 2: Count and list \nall bonds BROKEN \nin reactants"]
    B --> C["Step 3: Count and list \nall bonds MADE \nin products"]
    C --> D["Step 4: Calculate total \nenergy IN \n(bonds broken)"]
    D --> E["Step 5: Calculate total \nenergy OUT \n(bonds made)"]
    E --> F["Step 6: ΔH = \nIN − OUT"]

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Worked Example 1: Combustion of Hydrogen

Reaction: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$2H2​+O2​→2H2​O

Step 1: Bonds broken in the reactants

Bond	Number	Bond Energy (kJ/mol)	Total (kJ)
H–H	2	436	872
O=O	1	498	498
Total energy to break bonds			1370

Step 2: Bonds formed in the products

Bond	Number	Bond Energy (kJ/mol)	Total (kJ)
O–H	4	463	1852
Total energy to form bonds			1852

Step 3: Calculate the overall energy change

$\Delta H = 1370 - 1852 = -482 \text{ kJ/mol}$ΔH=1370−1852=−482 kJ/mol

The answer is negative, so the reaction is exothermic. This makes sense because combustion always releases energy.

graph TD
    A["Bonds Broken (Reactants)"] --> B["2 × H–H = 872 kJ"]
    A --> C["1 × O=O = 498 kJ"]
    B --> D["Total IN = 1370 kJ"]
    C --> D
    E["Bonds Formed (Products)"] --> F["4 × O–H = 1852 kJ"]
    F --> G["Total OUT = 1852 kJ"]
    D --> H["ΔH = 1370 − 1852 = −482 kJ"]
    G --> H
    H --> I["EXOTHERMIC \n(negative value)"]
    style I fill:#ff9999,stroke:#cc0000

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Worked Example 2: Reaction of Nitrogen and Hydrogen (Haber Process)

Reaction: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$N2​+3H2​→2NH3​

Step 1: Bonds broken

Bond	Number	Bond Energy (kJ/mol)	Total (kJ)
N≡N (triple bond)	1	945	945
H–H	3	436	1308
Total energy to break bonds			2253

Step 2: Bonds formed

Bond	Number	Bond Energy (kJ/mol)	Total (kJ)
N–H	6	391	2346
Total energy to form bonds			2346

Step 3: Calculate

$\Delta H = 2253 - 2346 = -93 \text{ kJ/mol}$ΔH=2253−2346=−93 kJ/mol

The Haber process is exothermic.

Exam Tip: When counting bonds in NH₃, remember each molecule has 3 N–H bonds. Since the equation makes 2NH₃, there are $2 \times 3 = 6$2×3=6 N–H bonds formed. Always use the balanced equation to count bonds correctly.

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Worked Example 3: Decomposition of Water (Endothermic)

Reaction: $2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$2H2​O→2H2​+O2​

Bonds broken: 4 × O–H = $4 \times 463 = 1852$4×463=1852 kJ

Bonds formed: 2 × H–H = $2 \times 436 = 872$2×436=872 kJ + 1 × O=O = $498$498 kJ → Total = $1370$1370 kJ

$\Delta H = 1852 - 1370 = +482 \text{ kJ/mol}$ΔH=1852−1370=+482 kJ/mol

The reaction is endothermic (positive $\Delta H$ΔH). This is the reverse of Worked Example 1 — the energy change is equal in magnitude but opposite in sign.

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Worked Example 4: Combustion of Methane

Reaction: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$CH4​+2O2​→CO2​+2H2​O

Bond energies: C–H = 413, O=O = 498, C=O = 799, O–H = 463

Bonds broken:

• 4 × C–H = $4 \times 413 = 1652$4×413=1652 kJ
• 2 × O=O = $2 \times 498 = 996$2×498=996 kJ
• Total = 2648 kJ

Bonds formed:

• 2 × C=O = $2 \times 799 = 1598$2×799=1598 kJ
• 4 × O–H = $4 \times 463 = 1852$4×463=1852 kJ
• Total = 3450 kJ

$\Delta H = 2648 - 3450 = -802 \text{ kJ/mol}$ΔH=2648−3450=−802 kJ/mol

The reaction is exothermic — consistent with combustion.

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Why Bond Energy Calculations Are Approximate

Bond energy values used in calculations are averages. The actual energy of a specific bond varies depending on the molecule it is in.

Factor	Explanation
Average bond energies	Values are averaged across many different compounds
Molecular environment	Surrounding atoms and bonds affect the exact bond energy
State of matter	Bond energy data assumes all substances are gases; liquids and solids may differ

This means calculated $\Delta H$ΔH values are approximate, not exact.

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Interpreting the Sign of the Answer

Sign of $\Delta H$ΔH	Meaning	Type of Reaction
Negative (−)	More energy released forming bonds than needed to break bonds	Exothermic
Positive (+)	More energy needed to break bonds than released forming bonds	Endothermic
Zero	Energy in = energy out	No overall energy change

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Common Mistakes

Mistake	How to Avoid It
Not balancing the equation first	Always balance the equation, then count bonds
Counting bonds incorrectly	Draw structural formulae and count every single bond
Subtracting the wrong way round	Always: BROKEN − MADE. If reversed, the sign will be wrong
Forgetting double/triple bonds	O=O is a double bond; N≡N is a triple bond — use the correct energy
Wrong units	Bond energies are kJ/mol; your answer is also in kJ/mol
Not showing working	Show every step — method marks are available even if the final answer is wrong

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Practice Problem

Calculate $\Delta H$ΔH for: $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$H2​+Cl2​→2HCl

Bond energies: H–H = 436, Cl–Cl = 242, H–Cl = 432

Bonds broken: H–H = 436 + Cl–Cl = 242 → Total = 678 kJ

Bonds formed: 2 × H–Cl = $2 \times 432 = 864$2×432=864 kJ → Total = 864 kJ

$\Delta H = 678 - 864 = -186 \text{ kJ/mol}$ΔH=678−864=−186 kJ/mol

The reaction is exothermic.

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Summary