Chemistry
Chemistry
3rd Edition
ISBN: 9780073402734
Author: Julia Burdge
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 8, Problem 86AP
Interpretation Introduction

Interpretation:

The energy change for the given reaction is to be calculated.

Concept introduction:

Ionization energy is the energy required when an electron is removed from a gaseous atom.

Electron affinity is the energy released when an electron is added to a gaseous atom.

Expert Solution & Answer
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Answer to Problem 86AP

Solution:

(a) 225 kJ/mol

(b) 168 kJ/mol

(c) 70 kJ/mol

(a)

Explanation of Solution

Given information: The reaction is as follows:

Li(g)+I(g)Li +(g)+I(g)

The given reaction is as follows:

Li(g)+I(g)Li +(g)+I(g)

In this reaction, lithium loses an electron whereas iodine gains an electron.

With reference tofigure 7.8,

The ionization energy of lithium is 520 kJ/mol

Li(g)Li +(g)+e                                      ΔH°=520 kJ/mol…… (1)

With reference tofigure 7.10,

The electron affinity of iodine is 295 kJ/mol

I(g)+eI(g)                                          ΔH°=295 kJ/mol…… (2)

Add equations(1) and (2) as follows:

Li(g)Li +(g)+e                                      ΔH°=520 kJ/molI(g)+eI-(g)                                                 ΔH°=295 kJ/molLi(g)+I(g)¯Li +(g)+I-(g)¯                                  ΔH°=225 kJ/mol¯

Therefore, the energy change for the reactionis 225 kJ/mol.

(b)

Given information: The reaction is as follows:

Na(g)+F(g)Na +(g)+F(g)

The given reaction is as follows:

Na(g)+F(g)Na +(g)+F(g)

In this reaction, sodium loses an electron whereas fluorine gains an electron.

With referencetofigure 7.8,

The ionization energy of sodium is 496 kJ/mol

Na(g)Na +(g)+e                                   ΔH°=496 kJ/mol…… (3)

With referencetofigure 7.10,

The electron affinity of fluorine is 328 kJ/mol

F(g)+eF(g)                                        ΔH°=328 kJ/mol…… (4)

Add equation (3) and (4) as follows:

Na(g)Na +(g)+e                                    ΔH°=496 kJ/molF(g)+eF(g)                                                ΔH°=328 kJ/molNa(g)+F(g)¯Na +(g)+F(g) ¯                             ΔH°=168 kJ/mol¯

Therefore, the energy change for the reaction is 168 kJ/mol.

(c)

Given information: The reaction is as follows:

K(g)+Cl(g)K+(g)+Cl(g)

The given reaction is as follows:

K(g)+Cl(g)K+(g)+Cl(g)

In the above reaction, potassium loses an electron whereas chlorine gains an electron.

With referencetofigure 7.8,

The ionization energy of lithium is 419 kJ/mol

K(g)K+(g)+e                                                  ΔH°=419 kJ/mol…… (5)

With referencetofigure 7.10,

The electron affinity of iodine is 349 kJ/mol

Cl(g)+eCl(g)                                                 ΔH°=349 kJ/mol…… (6)

Add equations(5) and (6) as follows:

K(g)K+(g)+e                                        ΔH°=419 kJ/molCl(g)+eCl(g)                                               ΔH°=349 kJ/molK(g)+Cl(g)¯K+(g)+Cl(g)¯                                ΔH°=70 kJ/mol¯

Therefore, the energy change for the reaction is 70 kJ/mol.

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Chapter 8 Solutions

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