Lecture Notes 9

# 1 kjmol the formation reaction can be carried out in

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 03/18/2014 for the course CHEM 1303 at SMU.

Ask a homework question - tutors are online