how do you solve this problem?? In a gaseous KCl molecule, the internuclear dist

Question

how do you solve this problem?? In a gaseous KCl molecule, the internuclear distance is2.67X10-10 m. Using data from Appendix F( is aboutphysical properties of elements) and neglecting the smallshort-range repulsion between the ion cores ofK+ and Cl-, estimate thedissociation energy of gaseous KCl into K and Cl atoms ( inkilojoules per mole)
i'll provide some information from appendix F regarding K andCl Potassium Ionization energy = 418.8 Electron affinity = 48.384 Electronegativity= 0.82
Chlorine Ionization energy = 1251.1 Electron affinity = 328.0 Electronegativity= 3.98
don't know if that's the information you need to solve theproblem Thanks In a gaseous KCl molecule, the internuclear distance is2.67X10-10 m. Using data from Appendix F( is aboutphysical properties of elements) and neglecting the smallshort-range repulsion between the ion cores ofK+ and Cl-, estimate thedissociation energy of gaseous KCl into K and Cl atoms ( inkilojoules per mole)
i'll provide some information from appendix F regarding K andCl Potassium Ionization energy = 418.8 Electron affinity = 48.384 Electronegativity= 0.82
Chlorine Ionization energy = 1251.1 Electron affinity = 328.0 Electronegativity= 3.98
don't know if that's the information you need to solve theproblem Thanks Ionization energy = 1251.1 Electron affinity = 328.0 Electronegativity= 3.98
don't know if that's the information you need to solve theproblem Thanks

Explanation / Answer

Given that -   KCl (s) -------> K (s) + 1/2Cl2 (g) We know that the lattice energy of the KCl is    KCl (s) ---------> K+(g) + Cl-(g)                      H1 (lattice) = +788 kJ/mol step 2 K+ (g) + 1 e  ---------> K(s)                  H2 = -H (ionization energy) = - 418.8kJ/mol step 3 Cl- (g) ----------> 1/2 Cl2 (g) + 1e        H3 = - H(electron affinity) = - 328.0 kJ/mol Therefore for the overall reaction H =  H1+H2+H3                                                           = 788 - 418.8 - 328                                                           = 41.2 kJ

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