Consider the halogenation of ethene, where X is a generic

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

Problem 15SAQ 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 15SAQ

Consider the halogenation of ethene, where X is a generic halogen:

Use bond energies to determine which halogen produces the most exothermic halogenation reaction with ethene. The C—F, C —Br, and C —I bond energies are 552 kJ/mol, 280 kJ/mol, and 209 kJ/mol, respectively. Find all other necessary bond energies in Table 9.3 .

a) fluorine

b) chlorine

c) bromine

d) iodine

Step-by-Step Solution:

Solution 15 SAQ Let’s demonstrate the correct answer corresponds to the letter a) Step 1 of 5 Consider the halogenation of ethene is as follows: The halogenation of an alkene consists in breaking the C=C bond and forming two C-X bond. In order to break a bond we have to spend energy in form of heat, so the reaction is endothermic; when a bonds is formed it releases a certain amount of energy in form of heat (so the a bond formation is an exothermic process). So we can calculate the overall enthalpy change (H )rxnH’s bond broken) - (H’s bond formed). NOTE 1: since the halogenations consists in forming two C-X bond, we have two consider two times the Hentalpy value for C-X formation NOTE 2: (H’s bond formed) has to be considered as a positive value before of changing its sign because of the minus sign before the bracket Step 2 of 5 For fluorine: We can find that the H value for C-F bond formation is – 552 KJ. But, as just said, we have two multiply for 2, so H= 1104 KJ For C=C bond break. The enthalpy value is +611 KJ. H rxn= +611 KJ – 1104 KJ= - 493 KJ Step 3 of 5 For chlorine: We can find that the H value for C-Cl bond formation is – 339 KJ. But, as just said, we have two multiply for 2, so H= -678 KJ For C=C bond break. H =+611 KJ. H rxn= +611 KJ – 678 KJ= - 67 KJ

Step 4 of 5

Chapter 9, Problem 15SAQ is Solved
Step 5 of 5

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

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