H2 Chemistry (9476): Bond Enthalpy Calculations (A Level)
Key idea: Estimate reaction enthalpy using average bond enthalpies (ΔH ≈ Σ bonds broken − Σ bonds formed), with sign conventions and exam pitfalls.
Before you start: Hess’ Law and Cycles
By the end, you can
- Bond Enthalpy Calculations
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.
1. 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.
2. Key Ideas (What Earns Marks)
- Use: Δ H ≈ sum E(bonds broken) - sum E(bonds formed)
- Breaking bonds is endothermic (+); forming bonds is exothermic (−).
- Multiply each bond enthalpy by the number of that bond broken/formed.
- Your final Δ H is an estimate because bond enthalpies are averages (and usually refer to gaseous species).
- List bonds broken (reactants) and bonds formed (products).
- Multiply each bond enthalpy by how many times it appears.
- Do “broken − formed”, then interpret the sign.
Data table
| Contribution | Energy |
|---|---|
| Bonds broken | 678 |
| Bonds formed | -862 |
| Net ΔH | -184 |
3. Detailed Explanations
A. Workflow (always works)
- Write the balanced equation and identify the reactants/products structures (what bonds exist).
- Count bonds broken (reactant side).
- Count bonds formed (product side).
- Substitute into: Δ H ≈ sum E(broken) - sum E(formed)
- 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.
4. Common Mistakes
- Reversing the formula.
- Forgetting to multiply by the number of each bond type.
- Treating the result as exact (it’s an estimate).
- Forgetting the values are usually for gaseous species and are mean values (so small differences from data-booklet values are normal).
When you can explain this confidently, use the Energetics Thermodynamics quiz and the Exam Skills hub to pressure-test exam wording.
5. Exam Tips
- Use Δ H ≈ sum E(bonds broken) - sum E(bonds formed) and show the bond counts.
- Always write “estimate” or “approximate” in your conclusion if asked to comment on accuracy.
- Don’t forget to multiply bond enthalpies by the number of each bond in the equation.
6. Worked Examples
Example 1Core
Use the bond enthalpies below to estimate Δ H for: H₂(g) + Cl₂(g) → 2HCl(g)
Data (kJ mol⁻¹): H–H 436, Cl–Cl 242, H–Cl 431.
Show Answer
Mark scheme:
- Bonds broken: 1(H–H) + 1(Cl–Cl) = 436 + 242 = 678
- Bonds formed: 2(H–Cl) = 2(431) = 862
- Δ H ≈ 678 - 862 = -184 kJ mol⁻¹
Example 2Core
Use bond enthalpies to estimate Δ H for hydrogenation of ethene: C₂H₄(g) + H₂(g) → C₂H₆(g)
Data (kJ mol⁻¹): C=C 612, H–H 436, C–C 348, C–H 412.
Show Answer
Mark scheme:
- Bonds broken: 1(C=C) + 1(H–H) = 612 + 436 = 1048
- Bonds formed: 1(C–C) + 2(C–H) = 348 + 2(412) = 1172
- Δ H ≈ 1048 - 1172 = -124 kJ mol⁻¹
Example 3Core
Estimate Δ H for: CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)
Data (kJ mol⁻¹): C–H 413, Cl–Cl 242, C–Cl 338, H–Cl 431.
Show Answer
Mark scheme:
- Bonds broken: 1(C–H) + 1(Cl–Cl) = 413 + 242 = 655
- Bonds formed: 1(C–Cl) + 1(H–Cl) = 338 + 431 = 769
- Δ H ≈ 655 - 769 = -114 kJ mol⁻¹
7. 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 Rightarrow 2E = 599 Rightarrow 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.
8. Practice, Quiz and Next Step
Fixed task
Average bond enthalpies give $\Delta H=-184\ \text{kJ mol}^{-1}$ for a reaction, while calorimetry gives $-201\ \text{kJ mol}^{-1}$. Explain why both values may be defensible.
Expected answer and marking feedback
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.
Repair before continuing
Reversing the formula.
Open course practice Next registered lesson: Hess Law And Cycles A Level