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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
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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