Q1(b). Formation enthalpy of lithium fluoride

Answer in brief

Taking the supplied 79 kJ/mol as the fluorine atomisation contribution for ½F₂ → F, ΔfH°(LiF) = −796 kJ mol⁻¹.

Key points

• The required reaction is Li(s) + ½F₂(g) → LiF(s).
• Atomise and ionise lithium: Li(s) → Li(g) costs 161 and Li(g) → Li⁺(g) + e⁻ costs 531 kJ/mol.
• Atomise fluorine for one mole of F atoms (79 kJ/mol) and attach an electron: F(g) + e⁻ → F⁻(g), −328 kJ/mol.
• Form the crystal from gaseous ions, releasing 1239 kJ/mol. Hess's law gives ΔfH° = 161 + 79 + 531 − 328 − 1239 = −796 kJ mol⁻¹.
• The negative result shows that lattice formation more than compensates for atomisation and ionisation.