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
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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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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