Lattice EnergyLattice EnergyBy: Shelby Toler, Courtney Matson, By: Shelby Toler, Courtney Matson, Andrew DiLosa, and Jordan FikeAndrew DiLosa, and Jordan Fike

Definition of Lattice Definition of Lattice EnergyEnergy

The energy required to separate one The energy required to separate one mole of a solid ionic compound into its mole of a solid ionic compound into its gaseous ionsgaseous ions

This means that no lattice energy could This means that no lattice energy could be measured if one of the ions can not be be measured if one of the ions can not be made gaseousmade gaseous

Periodic TrendPeriodic Trend

As the mass of the ionic compound As the mass of the ionic compound decreases and the ion charges increase, decreases and the ion charges increase, the lattice energy increasesthe lattice energy increases

Potential Magnitude of Potential Magnitude of Lattice EnergyLattice Energy

EEelel= (kQ= (kQ11QQ22)/d)/d

QQ1 1 and Qand Q22 are the charges of the ions are the charges of the ions

k is a constant; 8.99 x 10k is a constant; 8.99 x 1099 J-m/C J-m/C22

d is the distance between the centersd is the distance between the centers Hint: To find d, add the bonding radii of the Hint: To find d, add the bonding radii of the

ions in the compound. Use table on page ions in the compound. Use table on page 231 (old book) and page 266 (new book)231 (old book) and page 266 (new book)

Example 1:Finding Example 1:Finding Potential Lattice EnergyPotential Lattice Energy

Find the potential lattice energy of MgO.Find the potential lattice energy of MgO. Eel= (kQ1Q2)/dEel= (kQ1Q2)/d

d= Mg (1.30) + O (0.73) = 2.03d= Mg (1.30) + O (0.73) = 2.03 k= constant: 8.99 x 109 J-m/C2 k= constant: 8.99 x 109 J-m/C2 Q1= charge of Mg = 2Q1= charge of Mg = 2 Q2= charge of O = 2Q2= charge of O = 2

Eel= ((8.99 x 109 J-m/C2 )(2)(2))/(2.03)Eel= ((8.99 x 109 J-m/C2 )(2)(2))/(2.03)

Example 2: Finding Example 2: Finding Lattice EnergyLattice Energy

By using the data from Appendix C and By using the data from Appendix C and the other reference materials given, the other reference materials given, calculate the lattice energy of NaCl. calculate the lattice energy of NaCl.

NaCl(s) NaCl(s) Na(g)+Cl(g) Na(g)+Cl(g) ∆∆Hf°(NaCl)= ∆Hf°(Na (g))+ Hf°(NaCl)= ∆Hf°(Na (g))+

∆Hf°(Cl (g))+I1(Na)+EA(Cl)- ∆Hlattice∆Hf°(Cl (g))+I1(Na)+EA(Cl)- ∆Hlattice

∆∆Hf°(-411)= ∆Hf°(108(g))+ Hf°(-411)= ∆Hf°(108(g))+ ∆Hf°(122(g))+I1(496)+EA(-349)- ∆Hlattice∆Hf°(122(g))+I1(496)+EA(-349)- ∆Hlattice

∆∆Hf°(-411) = 377 - ∆HlatticeHf°(-411) = 377 - ∆Hlattice ∆∆Hlattice = 788 kJ/molHlattice = 788 kJ/mol

Example 3: Finding Example 3: Finding Lattice EnergyLattice Energy

By using the data from Appendix C and By using the data from Appendix C and the other reference materials given, the other reference materials given, calculate the lattice energy of CaFcalculate the lattice energy of CaF22. The . The

II22 value of Ca is 1145 kJ/mol. value of Ca is 1145 kJ/mol.

CaF2(s) CaF2(s) Ca(g) + 2F(g) Ca(g) + 2F(g) ∆∆Hf°(CaF2)= ∆Hf°(Ca (g))+ Hf°(CaF2)= ∆Hf°(Ca (g))+

2∆Hf°(F (g))+I1(Ca)+I2(Ca)+2EA(F)- 2∆Hf°(F (g))+I1(Ca)+I2(Ca)+2EA(F)- ∆Hlattice∆Hlattice

∆∆Hf°(-1219.6)= ∆Hf°(179.3)+ Hf°(-1219.6)= ∆Hf°(179.3)+ 2∆Hf°(80)+I1(590)+I2(1145)+2EA(-328)- 2∆Hf°(80)+I1(590)+I2(1145)+2EA(-328)- ∆Hlattice∆Hlattice

∆∆Hf°(-1219.6) = Hf°(-1219.6) = 1418.3- 1418.3- ∆Hlattice∆Hlattice ∆∆Hlattice = 2637.9 kJ/molHlattice = 2637.9 kJ/mol