Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 8, Problem 63E
Consider the following energy changes:
| ∆E(kJ/mol) | |
| Mg(g)→ Mg+(g) + e− | 735 |
| Mg+ (g) → Mg2+(g) + e− | 1445 |
| O(g) + e− → + O− (g) | −141 |
| O−(g) + e− → O2− (g) | 878 |
Magnesium oxide exists as Mg2+O2− and not as Mg+O−. Explain.
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Use the data in the table below to calculate the lattice energy of KCI.
K(s) → K(g)
K(g) → K+ (g) + e¯
Cl₂(g) → Cl(g)
Cl(g) + e Cl¯ (g)
→
K(s) + ½Cl₂(g) → KCl(s)
155.2
90.0 kJ/mol
418.8 kJ/mol
122 kJ/mol
-349 kJ/mol
-437 kJ/mol
Draw the energy diagram of the reaction: Li(s) + ½F2 → LiF(s).
Enthalpy (kJ/mol)
Li(s) → Li(g) +155.2
Li(g) → Li+(g) + e +520
½F2 →F(g) +75.3
F(g) + e → F-(g) -3
1. Below is a list of enthalpy changes for the Born-Haber cycle for the formation of solid LiF from Li(s) and F(g). Use these data to determine the lattice energy for the formation LiF(s). Li(s) → Li(g) ΔH1 = +162 kJ/mol Li(g) → Li+(g) + e- ΔH2 = +520.2 kJ/molF2(g) → 2F(g) ΔH3 = 154 kJ/mol F(g) + e- → F-(g) ΔH4 = -328 kJ/molLi(s) + 1/2F2(g) → LiF(s) ΔHf = -612 kJ/mol
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a. 1371 kJ/mol
b. -1371 kJ/mol
c. 1043 kJ/mol
d. -1043 kJ/mol
Chapter 8 Solutions
Chemistry
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