Question

Calculate the second ionization energy of the metal M (ΔHion2° in kJ/mol) using the following data:...

Calculate the second ionization energy of the metal M (ΔHion2° in kJ/mol) using the following data: Lattice enthalpy of MO(s), ΔHl° = -2297 kJ/mol Bond dissociation enthalpy of O2(g) = +498 kJ/mol First electron affinity of O = -141 kJ/mol Second electron affinity of O = +744 kJ/mol Enthalpy of sublimation of M = + 102 kJ/mol First ionization energy of M = + 340 kJ/mol Standard enthalpy of formation of MO(s), ΔHf° = -336 kJ/mol Refer to the textbook for definitions of ionization energy and electron affinity. Do not use scientific notation for your answer. Do not enter units.

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Answer #1
  1. Ionization Energy: It is the energy required to remove an electron from a neutral gaseous atom or an ion.
  2. Electron Affinity: It is the energy released when an electron is added to an isolated neutral gaseous atom or an ion.
  3. Dissociation energy: The energy required to dissociate a compound is called as dissociation energy. Dissociation of a compound is always an endothermic process and requires an input of energy.
  4. Sublimation energy: The energy required to change the phase from solid to gas, by passing the liquid phase is called as sublimation energy.
  5. Heat of formation: The energy change during the formation of a compound from its elements is known as heat of formation.Heat of formation ( Δ Hf°) = Heat of atomization( Δ Ha°)+ Dissociation energy( Δ Hd°)+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy

∆Hf° = ∆ Hs°+ ∆ HIE°+ ∆ Ha°+ ∆ Hea°+U
Or Lattice enthalpy

-336=+102+(340+x)+498+(-141+744)+(-2297)

(x+340)= -336-102-498-603+2297

           = 438-1101+2297

           = -663+2297

           x= 1634-340

            x= 1294 KJ/mol

second ionization energy of the metal M (ΔHion2° in kJ/mol) is 1294 KJ/mol .

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