Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 12 The Solid State - Study Questions - Page 467b: 16


-691.6 kJ/mol

Work Step by Step

Dissociation enthalpy: $1/2\ Cl_2(g)\rightarrow Cl(g),\ \Delta H_{1A}=+121.3\ kJ/mol$ Enthalpy of formation, gaseous Rb: $Rb(s)\rightarrow Rb(g),\ \Delta H_{2A}=+80.9\ kJ/mol$ Electron attachment energy, Cl: $Cl(g)+e^-\rightarrow Cl^-(g),\ \Delta H_{1B}=-349.0\ kJ/mol$ Ionization energy, Rb: $Rb(g)\rightarrow Rb^+(g)+e^-,\ \Delta H_{2B}=+403.0\ kJ/mol$ Lattice energy: $Rb^+(g)+Cl^-(g)\rightarrow RbCl(s),\ \Delta H_l$ Enthalpy of formation, RbCl: $Rb(s)+1/2\ Cl_2(g)\rightarrow RbCl(s),\ \Delta H_{f}=-435.4\ kJ/mol$ Combining equations we get: $\Delta H_{1A}+\Delta H_{2A}+\Delta H_{1B}+\Delta H_{2B}+\Delta H_{l}=\Delta H_{f}$, so: $\Delta H_{l}=-691.6\ kJ/mol$
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