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Consider an ionic compound, MX3, composed of generic metal M and generic gaseous halogen X. - The enthalpy of formation of MX3 is ΔHi =−993 kJ/mol. - The enthalpy of sublimation of M is ΔH sub =159 kJ/mol. - The first, second, and third ionization energies of M are IE1 = 689 kJ/mol,IE2 = 1725 kJ/mol, and IE3 =2747 kJ/mol. - The electron affinity of X is ΔH EA =−321 kJ/mol. (Refer to the hint). - The bond energy of X2 is BE=197 kJ/mol. Determine the lattice energy of MX3
Consider an ionic compound, MX3, composed of generic metal M and generic gaseous halogen X. - The enthalpy of formation of MX3 is ΔHi =−993 kJ/mol. - The enthalpy of sublimation of M is ΔH sub =159 kJ/mol. - The first, second, and third ionization energies of M are IE1 = 689 kJ/mol,IE2 = 1725 kJ/mol, and IE3 =2747 kJ/mol. - The electron affinity of X is ΔH EA =−321 kJ/mol. (Refer to the hint). - The bond energy of X2 is BE=197 kJ/mol. Determine the lattice energy of MX3
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Final answer:The lattice energy of MX3 can be calculated by using the Born-Haber cycle.Explanation:Thelattice energyof MX3can be calculated using the Born-Haber cycle. The lattice energy is equal to the sum of the enthalpies of formation, sublimation, and ionization energies, minus the electron affinity and bond energy. In this case, the lattice energy would be:Lattice Energy = ΔHf + ΔHsub + 3IE1 + 3IE2 + 3IE3 - ΔHEA - 3BEPlugging in the given values, we get:Lattice Energy = -993 kJ/mol + 159 kJ/mol + 3(689 kJ/mol) + 3(1725 kJ/mol) + 3(2747 kJ/mol) - (-321 kJ/mol) - 3(197 kJ/mol)Learn more about Lattice Energy here:brainly.com/question/34679410#SPJ11...
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