Answered: Explain how the thermodynamics of ionic…

Explain how the thermodynamics of ionic crystal formation can be used to determine electron affinities

General Chemistry - Standalone book (MindTap Course List)

11th Edition

ISBN:9781305580343

Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell

Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell

Chapter9: Ionic And Covalent Bonding

Section: Chapter Questions

Problem 9.7QP

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Explain how the thermodynamics of ionic crystal formation can be used to determine electron affinities

Science that deals with the amount of energy transferred from one equilibrium state to another equilibrium state.

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Look in Appendix D and compare the electron configurations shown there with the fusion enthalpies for the metals shown in Table 9.7. Is there any correlation between these configurations and this property? Does strength of attraction among metal atoms correlate with number of valence electrons? Explain your answers.
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Use Born-Mayer equation to calculate the lattice energy for PbS (it crystallizes in theNaCl structure). Then, use the Born–Haber cycle to obtain the value of lattice energy for PbS.You will need the following data following data : ΔH Pb(g) = 196 kJ/mol; ΔHf PbS = –98kJ/mol; electron affinities for S(g)→S- (g) is -201 kJ/mol; ) S- (g) →S2-(g) is 640kJ/mol. Ionizationenergies for Pb are listed in Resource section 2, p.903. Remember that enthalpies of formationare calculated beginning with the elements in their standard states (S8 for sulfur). Diatomicsulfur, S2, is formed from S8 (ΔHf: S2 (g) = 535 kJ/mol. Can you just do the Born-Haber part?
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The standard formation enthalpy of KCl is -571 kJmol-1. Plot the Born-Haber cycle and calculate the electron affinity of chlorine with the help of the data below. ΔH°a(K) = 89 kJmol-1 BE(Cl-Cl) = 244 kJmol-1 IE(K) = 425 kJmol-1ΔHL(KCl) = 719 kJmol-1
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