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Li(g) → Li (g) +e Li(s) → Li(g) 1/2 O2(g) → 0(g) O(g)+ 2 e→ O²(g) 2 Li"(g) + O*(g) → Li,O(s) c. AH=+540 kJ/mole AH=+162 kJ/mole ΔΗ- AH=+598 kJ/mole AH=-2800 kJ/mole 2 Li(s) + 1/2 O2(g) →Li¿O AH=-554 kJ/mole Li,O

Li(g) → Li (g) +e Li(s) → Li(g) 1/2 O2(g) → 0(g) O(g)+ 2 e→ O²(g) 2 Li"(g) + O*(g) → Li,O(s) c. AH=+540 kJ/mole AH=+162 kJ/mole ΔΗ- AH=+598 kJ/mole AH=-2800 kJ/mole 2 Li(s) + 1/2 O2(g) →Li¿O AH=-554 kJ/mole Li,O

Chemistry
Chemistry
10th Edition
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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1. Use a Born-Haber Cycle to calculate the missing AH (bold reaction) of the following: 2+ Вe(g) > Ве"(g) + 2 е Be(s)→ Be(g) 1/2 I2(s) →H I(g) + e → I(g) Be“(g) + 2 I (g) → Bel(s) Be(s) + I2(s) → Bel2(s) AH=+2660 kJ/mole AH=+302 kJ/mole AH=+107 kmole I а. AH=-298 kJ/mole 2+ ΔΗ AH=-208 kJ/mole Bel2 Mn(g) → Mn“(g)+2 e Mn(s) → Mn(g) S(s) → S(g) S(g) + 2 eA s* Mn + S(g) → MnS(s) Mn(s) + S(s) –→ MnS(s) b. AH=+2230 kJ/mole AN= AH=264 kJ/mole AH= +246 kJ/mole AH=-3176 k/mole AH=-212 kJ/mole (g) 2+ Li(g) → Li"(g)+e Li(s) → Li(g) 1/2 02(g) → O(g) O(g) + 2 e→ 0²(g) 2 Li"(g) + O (g) → Li,O(s) с. AH= +540 kJ/mole AH= +162 kJ/mole ΔΗ- AH=+598 kJ/mole AH= -2800 kJ/mole 2 Li(s) + 1/2 O2(g) → Li,O AH=-554 kJ/mole Li,O
Transcribed Image Text:1. Use a Born-Haber Cycle to calculate the missing AH (bold reaction) of the following: 2+ Вe(g) > Ве"(g) + 2 е Be(s)→ Be(g) 1/2 I2(s) →H I(g) + e → I(g) Be“(g) + 2 I (g) → Bel(s) Be(s) + I2(s) → Bel2(s) AH=+2660 kJ/mole AH=+302 kJ/mole AH=+107 kmole I а. AH=-298 kJ/mole 2+ ΔΗ AH=-208 kJ/mole Bel2 Mn(g) → Mn“(g)+2 e Mn(s) → Mn(g) S(s) → S(g) S(g) + 2 eA s* Mn + S(g) → MnS(s) Mn(s) + S(s) –→ MnS(s) b. AH=+2230 kJ/mole AN= AH=264 kJ/mole AH= +246 kJ/mole AH=-3176 k/mole AH=-212 kJ/mole (g) 2+ Li(g) → Li"(g)+e Li(s) → Li(g) 1/2 02(g) → O(g) O(g) + 2 e→ 0²(g) 2 Li"(g) + O (g) → Li,O(s) с. AH= +540 kJ/mole AH= +162 kJ/mole ΔΗ- AH=+598 kJ/mole AH= -2800 kJ/mole 2 Li(s) + 1/2 O2(g) → Li,O AH=-554 kJ/mole Li,O
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