A-Level Chemistry May/June 2025 Q4(a)(iv): The lattice energy, ∆Hlatt, of calcium fluoride, CaF2, is –2602 kJ mol−1. Calculate the…

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  The relationship between enthalpy of solution, lattice energy, and hydration enthalpies is given by the energy cycle:

  $ΔH_{sol} = ΣΔH_{hyd}(ions) - ΔH_{latt}$

  For $CaF_2$, this becomes:

  $ΔH_{sol}(CaF_2) = [ΔH_{hyd}(Ca^{2+}) + 2 × ΔH_{hyd}(F^{-})] - ΔH_{latt}(CaF_2)$

  Substituting the values provided in the question paper:

  $ΔH_{hyd}(Ca^{2+}) = -1650$ kJ mol⁻¹ $ΔH_{hyd}(F^{-}) = -506$ kJ mol⁻¹ $ΔH_{latt}(CaF_2) = -2602$ kJ mol⁻¹

  $ΔH_{sol} = [(-1650) + 2 × (-506)] - (-2602)$ $ΔH_{sol} = -1650 - 1012 + 2602$ $ΔH_{sol} = -2662 + 2602$ $ΔH_{sol} = -60$ kJ mol⁻¹

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  How the marks are awarded

  • M1 — This mark is for the correct working, showing the sum of the hydration enthalpies for one Ca²⁺ ion and two F⁻ ions, and correctly subtracting the given lattice energy value. This is demonstrated in the line: [-1650 - (2 × 506)] + 2602.
  • M2 — This mark is for the correct final answer of '-60', including the negative sign.

  Common mistakes

  • Forgetting to multiply the enthalpy of hydration of the fluoride ion (F⁻) by two, which is required by the stoichiometry of CaF₂.
  • Using an incorrect sign for the lattice energy. The calculation requires the lattice dissociation enthalpy (endothermic, +2602), which is the negative of the given lattice formation enthalpy (exothermic, -2602).
  • Incorrectly summing the hydration enthalpies, for example by getting the signs mixed up during the calculation.
  • Making a simple arithmetic error when calculating the final answer from the correct working.

  Examiner tip: Always construct a simple Hess's Law energy cycle to visualise the relationship between lattice energy, hydration enthalpies, and the enthalpy of solution, ensuring all signs and stoichiometric ratios are correctly applied.

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