## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 12 The Solid State - 12-3 Bonding in Ionic Compounds: Lattice Energy - Review & Check for Section 12-3 - Page 452: 3

#### Answer

$\Delta H_{l}=-702.0\ kJ/mol$

#### Work Step by Step

Dissociation enthalpy: $1/2\ I_2(g)\rightarrow I(g),\ \Delta H_{1A}=+106.838\ kJ/mol$ Enthalpy of formation, gaseous sodium: $Na(s)\rightarrow Na(g),\ \Delta H_{2A}=+107.3\ kJ/mol$ Electron attachment energy, iodine: $I(g)+e^-\rightarrow I^-(g),\ \Delta H_{1B}=-295.16\ kJ/mol$ Ionization energy, sodium: $Na(g)\rightarrow Na^+(g)+e^-,\ \Delta H_{2B}=+496\ kJ/mol$ Lattice energy: $Na^+(g)+I^-(g)\rightarrow NaI(s),\ \Delta H_l$ Enthalpy of formation, sodium iodide: $Na(s)+1/2\ I_2(g)\rightarrow NaI(s),\ \Delta H_{f}=-287\ kJ/mol$ Combining equations we get: $\Delta H_{1A}+\Delta H_{2A}+\Delta H_{1B}+\Delta H_{2B}+\Delta H_{l}=\Delta H_{f}$, so: $\Delta H_{l}=-702.0\ kJ/mol$

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