Calculate the lattice enthalpy of AgCl (s) using the following thermodynamic data. Note that the data given has been perturbed, so looking up the answer is probably not a good idea. CI - CI (g) Enthalpy of dissociation = 223 kJ/mol Ag (g) Enthalpy of formation = 265 kJ/mol CI (g) Electron attachment enthalpy = -369 kJ/mol Ag (g) Enthalpy of ionization = 711 kJ/mol Enthalpy of formation = -147 kJ/mol AgCl (s) kJ/mol Do you expect the magnitude (absolute value) of this to be larger or smaller than the lattice enthalpy of AgBr (s)?
Calculate the lattice enthalpy of AgCl (s) using the following thermodynamic data. Note that the data given has been perturbed, so looking up the answer is probably not a good idea. CI - CI (g) Enthalpy of dissociation = 223 kJ/mol Ag (g) Enthalpy of formation = 265 kJ/mol CI (g) Electron attachment enthalpy = -369 kJ/mol Ag (g) Enthalpy of ionization = 711 kJ/mol Enthalpy of formation = -147 kJ/mol AgCl (s) kJ/mol Do you expect the magnitude (absolute value) of this to be larger or smaller than the lattice enthalpy of AgBr (s)?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:Calculate the lattice enthalpy of AgCl (s) using the following thermodynamic data. Note that the data given has been
perturbed, so looking up the answer is probably not a good idea.
CI - CI (g) Enthalpy of dissociation = 223 kJ/mol
Ag (g)
265 kJ/mol
Cl (g)
Enthalpy of formation
Electron attachment enthalpy
Enthalpy of ionization
Enthalpy of formation = -147 kJ/mol
Ag (g)
711 kJ/mol
AgCl (s)
=
=
=
-369 kJ/mol
kJ/mol
Do you expect the magnitude (absolute value) of this to be larger or smaller than the lattice enthalpy of AgBr (s)?
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