Lecture Notes 9

1 kjmol the formation reaction can be carried out in

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Unformatted text preview: 2 ◦ ∆ H f = −594.1 kJ/mol The formation reaction can be carried out in five steps instead: (1) Convert solid lithium to lithium vapor (sublimation): ◦ ∆ H1 = 155.2 kJ/mol Li(s) −→ Li(g) (2) Dissociate 1 2 1 F2 (g) −→ F(g) 2 mole of F2 gas into gaseous F atoms: ◦ ∆ H2 = 75.3 kJ/mol (3) Ionize one mole of gaseous Li atoms: Li(g) −→ Li+ (g) + e− GChem I ◦ ∆ H3 = 520 kJ/mol 9.2 (4) Add one mole of electrons to one mole of gaseous F atoms: F(g) + e− −→ F− (g) ◦ ∆ H4 = −328 kJ/mol + (5) Combine one mole of gaseous Li and one mole of gaseous F− to form one mole of solid LiF: Li+ (g) + F− (g) −→ LiF(s) ◦ ∆ H5 =? kJ/mol ◦ Note: lattice energy = −∆ H5 Li(s) −→ Li(g) 1 F2 (g) −→ F(g) 2 Li(g) −→ Li+ (g) + e− F(g) + e− −→ F− (g) Li+ (g) + F− (g) −→ LiF(s) 1 Li(s) + F2 (g) −→ LiF(s) 2 ◦ ∆ H1 = 155.2 kJ/mol ◦ ∆ H2 = 75.3 kJ/mol ◦ ∆ H3 = 520 kJ/mol ◦ ∆ H4 = −328 kJ/mol ◦ ∆ H5 =? kJ/mol ◦ ∆ H f = −594.1 kJ/mol Hess’s law: ◦ ◦ ◦ ◦ ◦ ◦ ∆ H f = ∆ H1 + ∆ H2 + ∆ H3 + ∆ H4 + ∆ H5 GChem I 9.3 −594.1 kJ/mol = 155.2 kJ/mol + 75.3 kJ/mol + 520 kJ/mol ◦ + (−328 kJ/mol) + ∆ H5 =⇒ ◦ ∆ H5 = −1017 kJ/mol =⇒ lattice energy of LiF = 1017 kJ/mol Compoun...
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This document was uploaded on 03/18/2014 for the course CHEM 1303 at SMU.

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