Bond Breaking and Bond Forming

Key idea: Explain required bond breaking and making, with numerical bond-energy calculations clearly marked as out-of-syllabus enrichment.

  • About 7 minutes
  • Reviewed Jul 26, 2026

Before you start: Represent exothermic reactions using energy profile diagramsRepresent endothermic reactions using energy profile diagrams

By the end, you can

  • Explain overall enthalpy changes using bond-breaking and bond-making energies
Core content with out-of-syllabus enrichment
This resource begins with required K324 / 6092 content. Sections marked “out-of-syllabus enrichment” do not contribute to course completion or mastery evidence.

The required idea is the energy balance between bonds broken and bonds formed: breaking bonds absorbs energy, while forming bonds releases energy. The numerical Δ H = Ebᵣₑₐₖ - Efₒᵣₘ method later on this page is out-of-syllabus enrichment.

K324 / 6092 scope

The qualitative explanation—breaking bonds takes in energy, making bonds releases energy, and their balance determines the overall enthalpy change—is required. The numerical bond-energy method and calculation examples are explicitly out of syllabus for K324 / 6092.

1. Definition

A. Breaking vs Forming Bonds

Breaking bonds is endothermic (energy is absorbed). Forming bonds is exothermic (energy is released).

B. Bond Energy

Bond energy is the energy required to break one mole of a specified covalent bond in gaseous molecules (units: kJ mol⁻¹).

2. Key Ideas

  • Energy changes in reactions come from bond breaking and bond forming.
  • Breaking bonds absorbs energy: endothermic.
  • Forming bonds releases energy: exothermic.
  • A reaction is exothermic when bond making releases more energy than bond breaking absorbs.
  • A reaction is endothermic when bond breaking absorbs more energy than bond making releases.
  • Out-of-syllabus enrichment: quantify that comparison with Δ H = Ebᵣₑₐₖ - Efₒᵣₘ. Bond energies are supplied; do not memorise random tables.
Core explanation

Compare both energy transfers: energy is absorbed to break reactant bonds and released when product bonds form. State which transfer is greater to explain the overall reaction.

3. Detailed Explanations

Quick recall
  • Breaking bonds absorbs energy (endothermic step); forming bonds releases energy (exothermic step).
  • If bond making releases more energy than bond breaking absorbs, the overall reaction is exothermic.
  • If bond breaking absorbs more energy than bond making releases, the overall reaction is endothermic.
  • Out-of-syllabus calculations use Δ Hᵣₑₐcₜᵢₒₙ = Ebᵣₑₐₖ - Efₒᵣₘ after balancing the equation and counting bonds.

A. What “Bond Energy” Really Means

If a bond energy is 436kJ mol⁻¹, that means:

  • 436 kJ of energy is absorbed to break 1 mol of that bond.
  • Forming the same bond releases the same amount of energy (but in the opposite direction).
Prerequisite

Recall covalent bonding basics in Covalent Bonds. If you cannot count bonds correctly, you cannot do these questions.

B. Keep the Representations Separate

LevelWhat to say
MacroscopicThe surroundings warm for an exothermic reaction or cool for an endothermic reaction, if heat exchange is controlled.
Particle/bond modelEnergy is absorbed to break reactant bonds and released when product bonds form.
Symbolic calculationΔ H = sum E(bonds broken) - sum E(bonds formed).

Do not describe bond breaking itself as releasing energy because “atoms become free”. Breaking an attraction always requires energy; the overall reaction can still be exothermic when forming product bonds releases more energy.

C. Out-of-Syllabus Enrichment: Calculating Δ H Using Bond Energies

  1. Write a balanced equation.
  2. Count bonds in reactants (bonds broken).
  3. Count bonds in products (bonds formed).
  4. Calculate:
Δ Hᵣₑₐcₜᵢₒₙ = Ebᵣₑₐₖ - Efₒᵣₘ
  1. State the reaction type:
  • Δ H < 0 → exothermic
  • Δ H > 0 → endothermic
Why answers can be “off”

Bond energies are average values (they vary slightly between different molecules). Your calculated Δ H is usually an estimate.

4. Common Mistakes

  • Saying that breaking bonds releases energy because separated atoms are “free”. Breaking an attraction requires energy.
  • Mentioning only bond breaking or only bond making. The overall energy change depends on the balance of both transfers.
  • Calling every bond-breaking step endothermic without distinguishing it from the overall reaction.

For the optional numerical method:

  • Subtracting the wrong way round (writing Efₒᵣₘ - Ebᵣₑₐₖ).
  • Not balancing the equation first, then counting the wrong number of bonds.
  • Counting “atoms” instead of “bonds” (e.g., H₂ has one H-H bond).
  • Forgetting to multiply by the number of bonds (e.g., 2 × H-Cl).
  • Missing units: bond energies and the final molar enthalpy change are in kJ mol⁻¹. The value refers to one mole of reaction as written.

5. Exam Tips

Core qualitative answer

Write both sides of the comparison: “Energy is absorbed to break bonds in the reactants, while energy is released when bonds form in the products. More energy is released than absorbed, so the reaction is exothermic.” Reverse the final comparison for an endothermic reaction.

Out-of-syllabus numerical method

Two-line calculation method
  1. Ebᵣₑₐₖ (reactants) 2) Efₒᵣₘ (products) then Δ H = Ebᵣₑₐₖ - Efₒᵣₘ. Finish by writing “Δ H is negative/positive so the reaction is exothermic/endothermic”.
  • If the question says “using the bond energies below”, use those values even if you “remember” a different table.
  • Always show at least one line that links the sign of Δ H to exothermic/endothermic (that is often a mark).
  • Count bonds from the displayed formula or structure rather than guessing from the molecular formula.

6. Worked Examples

Example 1: Calculate Δ H (Formation of Hydrogen Fluoride)Optional

Use bond energies (in kJ mol⁻¹): H-H = 436, F-F = 158, H-F = 568.

Calculate Δ H for:

H₂(g) + F₂(g) → 2HF(g)
Show Answer

Bonds broken (reactants): 1 × H-H and 1 × F-F

Ebᵣₑₐₖ = 436 + 158; ; = 594kJ

Bonds formed (products): 2 × H-F

Efₒᵣₘ = 2(568); ; = 1136kJ
Δ Hᵣₑₐcₜᵢₒₙ = Ebᵣₑₐₖ - Efₒᵣₘ; ; = 594 - 1136; ; = -542kJ mol⁻¹

Δ H is negative, so the reaction is exothermic.

Example 2: Calculate Δ H (Decomposition of Ammonia)Optional

Use bond energies (in kJ mol⁻¹): N-H = 391, N#N = 945, H-H = 436.

Calculate Δ H for:

2NH₃(g) → N₂(g) + 3H₂(g)
Show Answer

Bonds broken (reactants): 6 × N-H

Ebᵣₑₐₖ = 6(391); ; = 2346kJ

Bonds formed (products): 1 × N#N and 3 × H-H

Efₒᵣₘ = 945 + 3(436); ; = 2253kJ
Δ Hᵣₑₐcₜᵢₒₙ = 2346 - 2253; ; = + 93kJ mol⁻¹

Δ H is positive, so the reaction is endothermic.

Example 3: Decide Exothermic or Endothermic (Formation of Hydrogen Chloride)Optional

Use bond energies (in kJ mol⁻¹): H-H = 436, Cl-Cl = 243, H-Cl = 432.

Determine whether this reaction is exothermic or endothermic:

H₂(g) + Cl₂(g) → 2HCl(g)
Show Answer

Bonds broken: 1 × H-H and 1 × Cl-Cl

Ebᵣₑₐₖ = 436 + 243 = 679kJ

Bonds formed: 2 × H-Cl

Efₒᵣₘ = 2(432) = 864kJ
Δ Hᵣₑₐcₜᵢₒₙ = 679 - 864 = -185kJ mol⁻¹

Δ H is negative, so the reaction is exothermic.

Example 4: Error Analysis (Fix the Method)Optional

A student writes: “Δ H = Efₒᵣₘ - Ebᵣₑₐₖ”. Identify the mistake and write the correct method.

Show Answer

They reversed the subtraction. The mark-scheme method is:

Δ Hᵣₑₐcₜᵢₒₙ = Ebᵣₑₐₖ - Efₒᵣₘ

You must also finish with the reaction type: negative = exothermic, positive = endothermic.

7. Mind Stretchers

Mind stretcher 1: “Why is my answer different from the data book?”Extension

You calculate Δ H using average bond energies and get a value that is different from the standard enthalpy change in a data book. Give one valid reason.

Show Answer

Bond energies are average values that depend on the molecule. Standard enthalpy changes are measured under standard conditions for specific substances, so the bond-energy method gives an estimate.

Mind stretcher 2: Missing State Symbols TrapExtension

A student writes H₂ + Cl₂ → 2HCl with no state symbols and no “as written” wording. What should they do to reduce mark loss?

Show Answer

Write the equation clearly and consistently with state symbols if provided/expected (e.g., H₂(g) + Cl₂(g) → 2HCl(g)). State that Δ H is for the reaction as written (per 1 mol of reaction).

8. Quiz

Quiz Time!

K324 and 6092 core practice check the required qualitative bond-breaking and bond-making explanation. The numerical questions on the lesson quiz are out-of-syllabus enrichment and do not contribute to course mastery.

G3 Pure / O-Level Core Practice  Out-of-Syllabus Bond-Energy Quiz

Recommended next step

Continue with objective-selected practice

Practise the shared G3 Pure / O-Level Chemistry objectives for K324 / 6092.

Mapped to G3 Pure / O-Level Chemistry (K324 / 6092) across 1 learning objective.

Created and internally reviewed by MiniEducation TeamSyllabus K324 / 6092Credibility details

Created and maintained by MiniEducation Team. Internal editorial team for Mini Chemistry and the Mini Education family.

Lessons are written against syllabus outcomes, exam-safe wording, and recurring mark-scheme pitfalls. Editorial policy · Review policy · Corrections policy

  • Years active: 2010-present
  • Syllabus scope: Secondary G1 Science | Secondary G2 Science (Chemistry) | Secondary G3 Science (Chemistry) | G3 Pure / GCE O Level Chemistry (6092) | GCE A Level H1 Chemistry (8873) | GCE A Level H2 Chemistry (9476)
  • Reviewed by: MiniEducation Team
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  • Syllabus: K324 / 6092
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