Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ba. △ H°(kJ) Ba(s) → Ba(g) 194 Ba(g) → Ba⁺(g) + e⁻ 579 2 O(g) → O2(g) -484 O(g) + e⁻ → O⁻(g) -139 O⁻(g) + e⁻ → O2⁻(g) 865 Ba(s) + O2(g) → BaO(s) -636 Ba2⁺(g) + O2⁻(g) → BaO(s) -3,500
Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ba. △ H°(kJ) Ba(s) → Ba(g) 194 Ba(g) → Ba⁺(g) + e⁻ 579 2 O(g) → O2(g) -484 O(g) + e⁻ → O⁻(g) -139 O⁻(g) + e⁻ → O2⁻(g) 865 Ba(s) + O2(g) → BaO(s) -636 Ba2⁺(g) + O2⁻(g) → BaO(s) -3,500
Chemistry by OpenStax (2015-05-04)
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Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
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Chapter6: Electronic Structure And Periodic Properties Of Elements
Section: Chapter Questions
Problem 83E: Of the five elements Sn, Si, Sb, O, Te, which has the most endothermic reaction? (E represents an...
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Question 9.
Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ba.
△ H°(kJ)
Ba(s) → Ba(g) 194
Ba(g) → Ba⁺(g) + e⁻ 579
2 O(g) → O2(g) -484
O(g) + e⁻ → O⁻(g) -139
O⁻(g) + e⁻ → O2⁻(g) 865
Ba(s) + O2(g) → BaO(s) -636
Ba2⁺(g) + O2⁻(g) → BaO(s) -3,500
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