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Lattice Energy Part II: Born-Haber Cycle

December 13, 2012

Video by Janet Coonce

 

 

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Be sure to watch video’s Lattice Energy Part I and Part II before reviewing the transcription.  Part II begins with a continuation of the discussion on EA:  Electron Affinity.

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Chlorine is happier (more stable) as a charged gaseous ion (g) than a gaseous free radical (g) with an unshared electron.  Chlorine needs an electron to complete its octet.  This increased stability is confirmed by the change in enthalpy.  The electron affinity for chlorine can be found on a table of known electron affinities and is expressed as the delta H:

Δ H = – 349 kJ/mol

Since heat is released, the reaction is exothermic and the Δ H is a negative number.  That means that the chloride anion, Cl- (g) is more stable than the chlorine free radical, Cl (g).  A free radical is an atom with at least one lone electron in its outer shell.  Remember that Cl normally exists as a diatomic molecule, Cl2 gas.  We discussed the energy required to break the Cl—Cl bond in the discussion of dissociation energy.

The next term you will need to understand to calculate lattice energy using the Born-Haber Cycle is Lattice Energy itself.

 

5.  Lattice Energy is the enthalpy change (Δ H) that accompanies the formation of 1 mole of an ionic solid from its separated gaseous ions.

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In the case of reacting gaseous sodium cation and gaseous chloride anion to form a lattice of sodium chloride solid, the reaction is expressed symbolically as:

Na+ (g) + Cl- (g) ——> NaCl(s)

By looking at a table of known values we determine the change in enthalpy or lattice energy to be negative 787 kJ/mol.  The number is negative and energy is lost as heat.  The reaction is exothermic.  Symbolically the lattice energy is expressed as the change in enthalpy associated with forming the solid lattice:

Δ H = – 787 kJ/mol  Negative means heat was lost (exothermic) and the reaction resulted in a lower energy state

To convert the NaCl solid back to the gaseous ion state, enough heat would have to be added to break the lattice bond.  To indicate that heat is added, the lattice energy would be expressed as a positive 787 kJ/mol:

NaCl(s)  —->  Na+ (g) + Cl- (g)

Δ H = + 787 kJ/mol     Positive  means heat was used (endothermic) to move to a higher energy state

In your text book if lattice energy is defined as the amount of energy required to heat a solid to the gaseous ions, the lattice energy will be a positive number.  If it is defined as the energy released when the gaseous ions combine to form the solid lattice, then the lattice energy will be a negative number.  The absolute value is the same.  The sign reflects the direction of the reaction depending on whether energy (heat) was added or removed.

The final term you will need to know to calculate lattice energy using the Born-Haber Cycle is the standard enthalpy of formation.

6.  Standard Enthalpy of Formation:

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This is another value that is available in a table of known values.  The symbol to indicate the standard enthalpy of formation is:

ΔHof

The superscript ‘’o’’ indicates a ‘’standard’’ given temperature and 1 atomosphere pressure.  The subscript ‘’f’’ indicates that this is the formation of 1 mole of a substance from the reference form of an element in its standard state.

Sodium is a metal in group I of the periodic table.  In its standard state it is a solid.  Chlorine in its standard state is a diatomic molecule.  Since we want to form only 1 mole of NaCl we only need 1/2 mole of Cl2 (g) which explains the use of a fraction to balance the equation.  We look at a table of known values and find that the standard enthalpy of formation of NaCl is expressed as:

ΔHof  = –411 kJ/mol

 

So how do we use the Born-Haber Cycle to calculate the lattice energy of an ionic compound?  Let’s work a typical problem you might see in a textbook.

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

Given:

  • 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)                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

 

This transcription will continue in Lattice Energy Part III

 

 

Transcription by James C. Gray MD FACOG

From → Lattice Energy

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