Mini Chemistry
Part of Mini EducationSwitch subject

Bond Enthalpy Calculations

Learn and apply Bond Enthalpy Calculations in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
On this page

Bond Enthalpy Calculations: Orientation

Bond enthalpy questions test one skill: can you translate a reaction into “bonds broken” and “bonds formed”? Once the bond counting is correct, the arithmetic is usually straightforward.

Keep Enthalpy Changes and Energy Profiles and the Energetics and Thermodynamics hub in view, because most questions mix definitions with cycle reasoning.

Definitions (Must Know)

A. Bond enthalpy

A bond enthalpy is the enthalpy change when 1 mol of a particular bond is broken in gaseous molecules (units: kJ mol⁻¹).

B. Mean (average) bond enthalpy

Most tables give mean bond enthalpies: average values taken over many different compounds. Therefore calculations using them give estimates, not exact values.

Detailed Explanations

A. Workflow (always works)

  1. Write the balanced equation and identify the reactants/products structures (what bonds exist).
  2. Count bonds broken (reactant side).
  3. Count bonds formed (product side).
  4. Substitute into: Δ H ≈ ∑ E(broken) - ∑ E(formed)
  5. State whether the reaction is exothermic/endothermic from the sign of Δ H.

Mini example (structure-count idea):

  • In H₂ + Cl₂ → 2HCl, you break 1 H–H and 1 Cl–Cl, and you form 2 H–Cl.

B. Why the value is not exact

Because “mean bond enthalpy” values are averages across many compounds, the actual bond enthalpy depends on the molecule it is in.

Therefore your calculated Δ H may not match a data-booklet value exactly, even if your method is correct.

Worked Examples

Modelled example 1

Estimate enthalpy from changed bonds

Core

Problem

Estimate Δ H for H₂(g) + Cl₂(g) → 2HCl(g) using H–H 436, Cl–Cl 242, and H–Cl 431 kJ mol⁻¹.
Study the worked solution

  1. Total bonds broken

    Method

    Add one H–H and one Cl–Cl bond enthalpy.

    Reason

    Breaking reactant bonds requires energy.

    Working

    E_broken = 436 + 242 = 678 kJ mol⁻¹.

  2. Total bonds formed

    Method

    Multiply the H–Cl value by two.

    Reason

    The balanced equation forms two H–Cl bonds.

    Working

    E_formed = 2(431) = 862 kJ mol⁻¹.

  3. Subtract formed from broken

    Method

    Use Δ H ≈ E_broken-E_formed.

    Reason

    Bond formation releases the energy represented by the second total.

    Working

    Δ H ≈ 678-862 = -184 kJ mol⁻¹.

Guided practice 2

Track bond changes in hydrogenation

About 7 min

Problem

Estimate Δ H for C₂H₄(g) + H₂(g) → C₂H₆(g) using C=C 612, H–H 436, C–C 348, and C–H 412 kJ mol⁻¹.

Try this before viewing the solution

Hints

Hint 1: changed bonds
The C=C and H–H disappear; a C–C and two additional C–H bonds appear.
Hint 2: signed relationship
Calculate [612 + 436]-[348 + 2(412)].
View solution step by step

  1. Count broken bonds

    Method

    Include one C=C and one H–H.

    Reason

    These reactant bonds are replaced during hydrogenation.

    Working

    E_broken = 612 + 436 = 1048 kJ mol⁻¹.

  2. Count formed bonds

    Method

    Include one C–C and two new C–H bonds.

    Reason

    The original four C–H bonds are unchanged and cancel from a full count.

    Working

    E_formed = 348 + 2(412) = 1172 kJ mol⁻¹.

  3. Estimate delta H

    Method

    Subtract the formed total from the broken total.

    Reason

    This combines endothermic bond breaking with exothermic bond formation.

    Working

    Δ H ≈ 1048-1172 = -124 kJ mol⁻¹.

Common misconception 3

Cancel unchanged bonds consistently

Find and correct the mistake

Learner method

For CH₄ + Cl₂ → CH₃Cl + HCl, a learner counts all four reactant C–H bonds as broken but no product C–H bonds as formed. Diagnose the count.

Choose a valid count

A consistent method is

View solution step by step

  1. Compare both structures

    Method

    Identify three C–H bonds in CH₃Cl as well as four in CH₄.

    Reason

    A full inventory must count bonds on both sides of the balanced equation.

    Working

    Three C–H contributions occur in both totals.

  2. Cancel the unchanged contributions

    Method

    Reduce the calculation to the net bond changes.

    Reason

    Equal C–H terms subtract to zero.

    Working

    Break one C–H and one Cl–Cl; form one C–Cl and one H–Cl.

Examiner practice 4

Estimate methane chlorination enthalpy

4 marks

Problem

Estimate Δ H for CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g) using C–H 413, Cl–Cl 242, C–Cl 338, and H–Cl 431 kJ mol⁻¹. State whether the reaction is exothermic or endothermic. [4 marks]

Try this before viewing the solution

View solution step by step

  1. List bonds broken

    1 mark

    Method

    Count one changed C–H and one Cl–Cl.

    Reason

    The three unchanged C–H bonds cancel.

    Working

    E_broken = 413 + 242 = 655 kJ mol⁻¹.

  2. List bonds formed

    1 mark

    Method

    Count one C–Cl and one H–Cl.

    Reason

    These are the two new bonds in the products.

    Working

    E_formed = 338 + 431 = 769 kJ mol⁻¹.

  3. Calculate the estimate

    1 mark

    Method

    Use broken minus formed.

    Reason

    Formation contributions release energy.

    Working

    Δ H ≈ 655-769 = -114 kJ mol⁻¹.

  4. Interpret the sign

    1 mark

    Method

    Classify the reaction from the negative value.

    Reason

    A negative enthalpy change means net energy release.

    Working

    The reaction is exothermic.

Challenge 5

Recover an unknown bond enthalpy

Minimal support

Problem

For H₂(g) + Cl₂(g) → 2HCl(g), Δ H ≈ -184 kJ mol⁻¹. Given H–H 436 and H–Cl 431 kJ mol⁻¹, estimate the Cl–Cl bond enthalpy.

Try this before viewing the solution

Hints

Hint 1: retain the unknown
Write the broken-bond total as 436 + E(Cl–Cl).
Hint 2: form two bonds
Set -184 = [436 + E]-2(431) and rearrange.
View solution step by step

  1. Write the estimate equation

    Method

    Place the unknown in the broken-bond total.

    Reason

    The Cl–Cl bond is broken, while two known H–Cl bonds are formed.

    Working

    -184 = [436 + E(Cl–Cl)]-862.

  2. Rearrange

    Method

    Add 862 and subtract 436.

    Reason

    This isolates the energy required to break one mole of Cl–Cl bonds.

    Working

    E(Cl–Cl) = -184 + 862-436 = 242 kJ mol⁻¹.

Mind Stretchers

Mind stretcher 1Extension

For: H₂(g) + I₂(g) → 2HI(g) the reaction enthalpy is estimated as Δ H ≈ -12.0 kJ mol⁻¹.

Data (kJ mol⁻¹): H–H 436, I–I 151. Estimate the H–I bond enthalpy.

Show Hint

Count every bond on each side. Breaking is positive; forming releases energy and is subtracted.

Show Answer

Mark scheme:

  • Bonds broken: 436 + 151 = 587
  • Bonds formed: 2E(H–I)
  • Δ H ≈ 587 - 2E
  • -12.0 = 587 - 2E ⇒ 2E = 599 ⇒ E = 300 kJ mol⁻¹ (3 s.f.)

Mind stretcher 2: Explaining disagreement with calorimetryExtension

Question. Average bond enthalpies give Δ H = -184 kJ mol⁻¹ for a reaction, while calorimetry gives -201 kJ mol⁻¹. Explain why both values may be defensible.

Show Hint

One method uses gas-phase averages; the other measures a particular experiment with particular physical states.

Show Answer

Average bond enthalpies are mean gas-phase values across many molecular environments, so their result is an estimate. Calorimetry measures the stated reaction and includes its actual physical states, although it may contain heat-loss and heat-capacity errors.