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Use a Born-Haber Cycle to calculate the missing ∆H (bold reaction) of the following: Be(g) yields Be2+(g) + 2 e– ∆H= +2660 kJ/mole Be(s) yields Be(g) ∆H= +302 kJ/mole 1/2 I2(s) yields I(g) ∆H= +107 kJ/mole I(g) + e– yields I–(g) ∆H= –295 kJ/mole Be2+(g) + 2 I–(g) yields BeI2(s) ∆H= ? Be(s) + I2(s) yields BeI2(s) ∆H= –208 kJ/mole BeI2

Use a Born-Haber Cycle to calculate the missing ∆H (bold reaction) of the following: Be(g) yields Be2+(g) + 2 e– ∆H= +2660 kJ/mole Be(s) yields Be(g) ∆H= +302 kJ/mole 1/2 I2(s) yields I(g) ∆H= +107 kJ/mole I(g) + e– yields I–(g) ∆H= –295 kJ/mole Be2+(g) + 2 I–(g) yields BeI2(s) ∆H= ? Be(s) + I2(s) yields BeI2(s) ∆H= –208 kJ/mole BeI2

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 ∆H (bold reaction) of the following:

 

 

  1. Be(g) yields  Be2+(g) + 2 e–                                   ∆H= +2660 kJ/mole

             Be(s)  yields  Be(g)                                                ∆H= +302 kJ/mole

             1/2 I2(s)  yields  I(g)                                             ∆H= +107 kJ/mole 

           I(g) + e–   yields   I–(g)                                            ∆H= –295 kJ/mole

           Be2+(g) + 2 I–(g) yields   BeI2(s)                        ∆H= ?

                   Be(s) + I2(s) yields  BeI2(s)                      ∆H= –208 kJ/mole BeI2

1. Use a Born-Haber Cycle to calculate the missing AH (bold reaction) of the following: Вe(g) —> Ве"(g) + 2 e Be(s) → Be(g) 1/2 I2(s) → I(g) а. AH= +2660 kJ/mole AH= +302 kJ/mole AH= +107 kJ/mole I I(g) + e -> Г(g) Be*(g) + 2I (g) → Bel,(s) Be(s) + I2(s) → Bel>(s) AH= -295 kJ/mole AH= AH=-208 kJ/mole Bel,
Transcribed Image Text:1. Use a Born-Haber Cycle to calculate the missing AH (bold reaction) of the following: Вe(g) —> Ве"(g) + 2 e Be(s) → Be(g) 1/2 I2(s) → I(g) а. AH= +2660 kJ/mole AH= +302 kJ/mole AH= +107 kJ/mole I I(g) + e -> Г(g) Be*(g) + 2I (g) → Bel,(s) Be(s) + I2(s) → Bel>(s) AH= -295 kJ/mole AH= AH=-208 kJ/mole Bel,
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