Question detail

Calculate the lattice enthalpy of NaCl given the following Born-Haber cycle data: Ionization energy of Na = 496 kJ/mol, Electron affinity of Cl = -349 kJ/mol, Sublimation energy of Na = 108 kJ/mol, Bond dissociation energy of Cl2 = 243 kJ/mol, and the enthalpy of formation of NaCl = -411 kJ/mol.

Try the question, check the answer, then read the explanation to understand the curriculum point.

At a glance

MCQ

Type

practice

Style

Topic

Thermodynamics (A-level only)

Question

  1. A. −787 kJ/mol
  2. B. −788 kJ/mol
  3. C. −786 kJ/mol
  4. D. −800 kJ/mol

Answer

−787 kJ/mol

Explanation

The correct option is −787 kJ/mol. −787 kJ/mol is the best answer because it directly supports the AQA A-Level Chemistry objective to calculate lattice enthalpy from Born-Haber data. This reasoning is anchored to Born-Haber cycles (A-level only) in Thermodynamics (A-level only), and it separates lattice enthalpy from similar A-Level ideas rather than relying on a vague recall statement. Other options are weaker if they use the wrong evidence, calculation, mechanism, observation, unit, or conclusion for this subtopic.

Common mistake

Incorrect Lattice Enthalpy Calculation

Students often forget to include the sign of the lattice enthalpy when calculating it from Born-Haber data, leading to incorrect conclusions about the stability of the ionic compound.

To fix this, remember that lattice enthalpy is exothermic and should be treated as a negative value in calculations. Use the formula: Lattice Enthalpy = ΔHf (formation enthalpy) - (ΔHsub + ΔHionization + ΔHelectron + ΔHbond). Substitute the values correctly, ensuring to apply the negative sign for lattice enthalpy. For example, if ΔHf = -400 kJ/mol, ΔHsub = +100 kJ/mol, ΔHionization = +200 kJ/mol, ΔHelectron = -100 kJ/mol, and ΔHbond = +50 kJ/mol, the calculation would be: Lattice Enthalpy = -400 - (100 + 200 - 100 + 50) = -400 - 250 = -650 kJ/mol. Thus, the lattice enthalpy is -650 kJ/mol.

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