5) The equilibrium separation of the K and CI ions in KCl is about 0.267 nm. (a) Calculate the potential energy of attraction of the ions, assuming them to be point charges at this separation. (b) The ionization energy of potassium is 4.34 eV and the electron affinity of chlorin 3.61 eV. Find the dissociation energy for KCI, neglecting any energy of repulsion. (c) The measured dissociation energy is 4.43 eV. What is the energy due to repulsion of the ions at the equilibrium separation?
5) The equilibrium separation of the K and CI ions in KCl is about 0.267 nm. (a) Calculate the potential energy of attraction of the ions, assuming them to be point charges at this separation. (b) The ionization energy of potassium is 4.34 eV and the electron affinity of chlorin 3.61 eV. Find the dissociation energy for KCI, neglecting any energy of repulsion. (c) The measured dissociation energy is 4.43 eV. What is the energy due to repulsion of the ions at the equilibrium separation?
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Transcribed Image Text:5) The equilibrium separation of the K and CI ions in KCl is about 0.267 nm.
(a) Calculate the potential energy of attraction of the ions, assuming them to be point
charges at this separation.
(b) The ionization energy of potassium is 4.34 eV and the electron affinity of chlorin 3.61
eV. Find the dissociation energy for KCl, neglecting any energy of repulsion.
(c) The measured dissociation energy is 4.43 eV. What is the energy due to repulsion of
the ions at the equilibrium separation?
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