Chemistry
9th Edition
ISBN: 9781133611097
Author: Steven S. Zumdahl
Publisher: Cengage Learning
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Chapter 8, Problem 61E
Interpretation Introduction
Interpretation: The energy of sublimation is to be calculated from the given data of lithium iodide reaction.
Concept introduction: The sublimation is the process in which change of phase occur from solid to gas. The energy change occur during this process is called sublimation energy
To determine: The value of energy of sublimation for the given reaction.
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Calculate the second ionization energy of the metal M (AHion2 in kJ/mol) using the
following data:
Lattice enthalpy of MO(s), AH = -2278 kJ/mol
Bond dissociation enthalpy of O2(g) = +498 kJ/mol
First electron affinity of O = -141 kJ/mol
Second electron affinity of O = +744 kJ/mol
Enthalpy of sublimation of M = + 125 kJ/mol
First ionization energy of M = + 309 kJ/mol
Standard enthalpy of formation of MO(s), AH = -341 kJ/mol
Use the Born-Haber cycle to calculate the lattice energy of KF. [The heat of sublimation of K is 91.6 kJ·mol−1 and
ΔfH(KF) = −567.3 kJ·mol−1.
Bond enthalpy for F2 is
158.8 kJ·mol−1.
Other data may be found in the Ionization Energies Table and the Electron Affinities Table.]
Calculate the lattice energy for LiBr(s) given the following:
sublimation energy for Li(s)
+166 kJ/mol
ΔHf for Br(g)
+97 kJ/mol
first ionization energy of Li(g)
+520. kJ/mol
electron affinity of Br(g)
–325 kJ/mol
enthalpy of formation of LiBr(s)
–351 kJ/mol
Chapter 8 Solutions
Chemistry
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