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Energetics

Average bond enthalpies and their limitations

  • 3 things to know
  • Unit AS 2

Overview

Recall questions on the definition of average bond enthalpy and why it is positive, calculating an enthalpy change from bonds broken and made, worked examples on hydrogen chloride, ammonia, methane and ethene, working back to a single bond enthalpy, and why the method is less reliable than Hess's Law.

What you need to know

Answer each one before you reveal it.

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Key term

Why is a bond enthalpy always a positive quantity?

Key term

Why is the word average used in average bond enthalpy?

Numbers

The enthalpy change for CH₄(g) → C(g) + 4H(g) is +1652 kJ mol⁻¹. Calculate the average C–H bond enthalpy in methane.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Define the term average bond enthalpy.

    Model answer

    The average bond enthalpy is the mean energy required to break one mole of a given type of covalent bond in gaseous molecules, averaged over a range of different compounds.

  2. 2

    Why is a bond enthalpy always a positive quantity?

    Model answer

    Breaking a bond needs energy to overcome the attraction holding the atoms together, so the process is always endothermic.

  3. 3

    State how an enthalpy change of reaction is calculated from average bond enthalpies.

    Model answer

    The enthalpy change equals the total energy needed to break all the bonds in the reactants minus the total energy released when all the bonds in the products are made.

  4. 4

    Why is the word average used in average bond enthalpy?

    Model answer

    The same type of bond has a slightly different strength in different molecules, so a mean value taken over many compounds is quoted.

  5. 5

    Using bond enthalpies of 436 kJ mol⁻¹ for H–H, 243 kJ mol⁻¹ for Cl–Cl and 432 kJ mol⁻¹ for H–Cl, calculate ΔH for H₂(g) + Cl₂(g) → 2HCl(g).

    Model answer

    Breaking the bonds needs 436 + 243, which is 679 kJ mol⁻¹, and making the bonds releases 2 × 432, which is 864 kJ mol⁻¹, so ΔH is −185 kJ mol⁻¹.

  6. 6

    Using bond enthalpies of 945 kJ mol⁻¹ for N≡N, 436 kJ mol⁻¹ for H–H and 391 kJ mol⁻¹ for N–H, calculate ΔH for N₂(g) + 3H₂(g) → 2NH₃(g).

    Model answer

    Breaking the bonds needs 945 + 3 × 436, which is 2253 kJ mol⁻¹, and making six N–H bonds releases 2346 kJ mol⁻¹, so ΔH is −93 kJ mol⁻¹.

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