Lattice Energy Part III: Born-Haber Cycle

Video by Janet Gray Coonce, MS

Continuing to solve the problem discussed at the end of Part II…..

Problem:  Use the Born-Haber Cycle to calculate the lattice energy of MgF₂

The known enthalpies of each of the steps of the Born-Haber cycle are obtained from a table:

• Enthalpy of Sublimation for Mg =       ΔH_sub = + 146 kJ/mol
• 1st ionization energy for Mg =            IE₁ = +738 kJ/mol
• 2nd ionization energy for Mg =           IE₂ = +1451 kJ/mol
• Bond dissociation Energy for F₂ (g) =   BE = +159 kJ/mol
• Electron Affinity of F (g) =                    EA = –328 k/J/mol
• Enthalpy of formation of MgF₂ (s) =      ΔH^o_f = –1124 kJ/mol

Now, writing the component reactions of the Born-Haber Cycle:

Mg(s) –> Mg(g)                ΔH_sub = + 146 kJ/mol

Mg(g) –> Mg⁺(g)  + 1 e^–    IE₁ = +738 kJ/mol

Mg⁺(g) –> Mg⁺²(g) + 1 e^–   IE₂ = +1451 kJ/mol

F₂(g) –> 2F(g)                    BE = +159 kJ/mol

F(g) + 1 e^– –> F^–(g)             EA = –328 k/J/mol

Mg(s) + F₂(g) –> MgF₂(s)    ΔH^o_f = –1124 kJ/mol

Hess’s law states that the total enthalpy change during the complete course of a reaction is same whether the reaction is made in one step or in several steps.  Remember the lattice energy is energy released when ions in the gaseous form combine to form a solid.  For the case of MgF₂(s):

Mg²⁺(g) + 2F^-(g) –> MgF₂(s)  ΔH  = Lattice Energy

To calculate the total enthalpy change using Hess’s law we will need to find the enthalpy change for each component of the Born-Haber cycle and add them together.  Remember if we reverse the direction of the reaction, the sign of the change in enthalpy will change.  So to write the enthalpy reactions required in the formation of MgF₂(s), we need to write the reactions in the direction required and we need to adjust for the number of moles of each reactant used:

1. 2F^–(g)  – –> 2F(g) + 2 e^–       2EA = 2(+328 k/J/mol) = 656 kJ/mol
2. Mg⁺²(g) + 1 e^–  –> Mg+(g)   IE₂ = -1451 kJ/mol
3. 2F(g) –> F₂(g)                     BE = –159 kJ/mol
4. Mg⁺(g)  + 1 e^–  –> Mg(g)      IE₁ = +738 kJ/mol
5. Mg(g) –> Mg(s)                   ΔH_sub = – 146 kJ/mol
6. Mg(s) + F₂(g) –> MgF₂(s)    ΔH^o_f = –1124 kJ/mol

Now, according to Hess’s Law all we need to do is to add all the enthalpies associated with each step of the Born-Haber Cycle to calculate the lattice energy.

Therefore lattice energy of MgF₂ ΔH= -2862 kJ/mol

Transcription by James C. Gray, MD FACOG