From the information provided in the question, the second ionization energy of the metal is 578 kJ/mol.
From the question, we have the following information;
Lattice enthalpy of MO(s) = -2383 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
First ionization energy of M = + 267 kJ/mol
Heat of sublimation of M = + 130 kJ/mol
Standard enthalpy of formation of MO(s) = -307 kJ/mol
Using Hess law of constant heat summation;
ΔHf = ΔHs + BE + ∑IE + ∑EA + U
ΔHs = Heat of sublimation of metal
ΔHf = Heat of formation MO
BE = Bond energy of O2
∑EA = sum of electron affinities of Oxygen
∑IE = Sum of the ionization energies of M
U = Lattice energy of MO
Let the second ionization energy be x
Substituting values;
(-307) = 130 + 498 + (267 + x) + 603 + (-2383)
(-307) = -885 + x
-x = -885 + 307
-x = -578
x = 578 kJ/mol
The second ionization energy of the metal is 578 kJ/mol.
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