Lattice Energy
By: Shelby Toler, Courtney Matson,
Andrew DiLosa, and Jordan Fike
Definition of Lattice
Energy
 The energy required to separate one
mole of a solid ionic compound into its
gaseous ions
 This means that no lattice energy could
be measured if one of the ions can not be
made gaseous
Periodic Trend
 As the mass of the ionic compound
decreases and the ion charges increase,
the lattice energy increases
Potential Magnitude of
Lattice Energy





Eel= (kQ1Q2)/d
Q1 and Q2 are the charges of the ions
k is a constant; 8.99 x 109 J-m/C2
d is the distance between the centers
Hint: To find d, add the bonding radii of
the ions in the compound. Use table on
page 231 (old book) and page 266 (new
book)
Example 1:Finding
Potential Lattice Energy
 Find the potential lattice energy of MgO.
 Eel= (kQ1Q2)/d




d= Mg (1.30) + O (0.73) = 2.03
k= constant: 8.99 x 109 J-m/C2
Q1= charge of Mg = 2
Q2= charge of O = 2
 Eel= ((8.99 x 109 J-m/C2 )(2)(2))/(2.03)
Example 2: Finding Lattice
Energy
 By using the data from Appendix C and
the other reference materials given,
calculate the lattice energy of NaCl.
 NaCl(s)  Na(g)+Cl(g)
 ∆Hf°(NaCl)= ∆Hf°(Na (g))+
∆Hf°(Cl (g))+I1(Na)+EA(Cl)- ∆Hlattice
 ∆Hf°(-411)= ∆Hf°(108(g))+
∆Hf°(122(g))+I1(496)+EA(-349)- ∆Hlattice
 ∆Hf°(-411) = 377 - ∆Hlattice
 ∆Hlattice = 788 kJ/mol
Example 3: Finding Lattice
Energy
 By using the data from Appendix C and
the other reference materials given,
calculate the lattice energy of CaF2. The
I2 value of Ca is 1145 kJ/mol.
 CaF2(s)  Ca(g) + 2F(g)
 ∆Hf°(CaF2)= ∆Hf°(Ca (g))+
2∆Hf°(F (g))+I1(Ca)+I2(Ca)+2EA(F)∆Hlattice
 ∆Hf°(-1219.6)= ∆Hf°(179.3)+
2∆Hf°(80)+I1(590)+I2(1145)+2EA(-328)∆Hlattice
 ∆Hf°(-1219.6) = 1418.3- ∆Hlattice
 ∆Hlattice = 2637.9 kJ/mol