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Use the Born–Haber cycle and data from Appendix IIB and

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 48E Chapter 9

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 48E

Use the Born–Haber cycle and data from Appendix IIB and Table 9.3 to calculate the lattice energy of CaO. (ΔHsub for calcium is 178 kJ/mol; IE1 and IE2 for calcium are 590 kJ/mol and 1145 kJ/mol, respectively; EA1 and EA2 for O are –141 kJ/mol and 744 kJ/mol, respectively.)

Step-by-Step Solution:

Solution 48E Step 1 of 7 First, we have to consider the overall reaction and its H. In this case Ca + ½ O CaO H= - 639 kJ (s) 2 (g) After that we have to remember that Born-Haber the reaction above is divided in five sub-reactions Step 2 of 7 First reaction: Sublimation of Metal (in this case sublimation of Calcium) Ca (s) (g) H 1- 177.8 kJ Step 3 of 7 Second Reaction: Atomization of the other atom concurring in final molecule’s formation (in this case: atomization of Oxygen) ½ O 2(g) O H =2 bond energy of O = 249 kJ 2 Step 4 of 7 Third reaction: Ionization of gaseous metal In this case ionization of Ca happens in two steps because Ca can lose two electrons and only one at time + Ca (g) Ca (g) H 3+590 kJ Ca (g) Ca (g)+ H 4+1145.4 kJ

Step 5 of 7

Chapter 9, Problem 48E is Solved
Step 6 of 7

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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Use the Born–Haber cycle and data from Appendix IIB and

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