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Lattice Energy Part III: Born-Haber Cycle

December 15, 2012

Video by Janet Gray Coonce, MS

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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 =       ΔHsub = + 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) =      ΔHof = –1124 kJ/mol

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

Mg(s) –> Mg(g)                ΔHsub = + 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)    ΔHof = –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)                   ΔHsub = – 146 kJ/mol
  6. Mg(s) + F2(g) –> MgF2(s)    ΔHof = –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.

1. -1124
2. + 656
3. -1451
4. – 159
5 – 738
6 – 146
Total -2862

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

 

Transcription by James C. Gray, MD FACOG

From → Lattice Energy

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