Question

for lihium the enthalpy of sublimation is + 161 kj mol-¹, and the first ionisation energy is +520 kj mol-¹ and the eletron affinity of fluorine is -328 kj mol-¹ the lattice energy of fluorine is - 1047 kj mol-¹, calculate the overall enthalpy change for the reaction Li(s) + ½F2(g)---->lif(s) ΔH⁰​

Answer 1

The reaction can be broken down into the following steps:

1. Li(s) → Li(g) (enthalpy of sublimation)

2. Li(g) → Li⁺(g) + e⁻ (first ionization energy)

3. ½F2(g) → F(g) (½ of electron affinity of fluorine)

4. Li⁺(g) + F(g) → LiF(s) (lattice energy of LiF)

The overall enthalpy change for the reaction is:

ΔH⁰ = enthalpy change for step 1 + enthalpy change for step 2 + enthalpy change for step 3 + enthalpy change for step 4

ΔH⁰ = (+161 kJ/mol) + (+520 kJ/mol) + (½ x (-328 kJ/mol)) + (-1047 kJ/mol)

ΔH⁰ = -694 kJ/mol

Therefore, the overall enthalpy change for the reaction Li(s) + ½F2(g) → LiF(s) is -694 kJ/mol.