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.

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Answer 1

Question

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

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 $−295 kJ/mol$

$I(g)+e−→I−(g) ΔH°=−295 kJ/mol$ …… (2)

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

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

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 $−328 kJ/mol$

$F(g)+e−→F−(g) ΔH°=−328 kJ/mol$ …… (4)

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

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

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 $−349 kJ/mol$

$Cl(g)+e−→Cl−(g) ΔH°=−349 kJ/mol$ …… (6)

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

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

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

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Answer 2

Question

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

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 $−295 kJ/mol$

$I(g)+e−→I−(g) ΔH°=−295 kJ/mol$ …… (2)

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

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

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 $−328 kJ/mol$

$F(g)+e−→F−(g) ΔH°=−328 kJ/mol$ …… (4)

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

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

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 $−349 kJ/mol$

$Cl(g)+e−→Cl−(g) ΔH°=−349 kJ/mol$ …… (6)

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

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

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

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Answer 3

Question

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

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 $−295 kJ/mol$

$I(g)+e−→I−(g) ΔH°=−295 kJ/mol$ …… (2)

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

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

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 $−328 kJ/mol$

$F(g)+e−→F−(g) ΔH°=−328 kJ/mol$ …… (4)

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

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

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 $−349 kJ/mol$

$Cl(g)+e−→Cl−(g) ΔH°=−349 kJ/mol$ …… (6)

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

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

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

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Answer 4

Question

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

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 $−295 kJ/mol$

$I(g)+e−→I−(g) ΔH°=−295 kJ/mol$ …… (2)

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

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

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 $−328 kJ/mol$

$F(g)+e−→F−(g) ΔH°=−328 kJ/mol$ …… (4)

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

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

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 $−349 kJ/mol$

$Cl(g)+e−→Cl−(g) ΔH°=−349 kJ/mol$ …… (6)

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

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

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

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Answer 5

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

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 $−295 kJ/mol$

$I(g)+e−→I−(g) ΔH°=−295 kJ/mol$ …… (2)

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

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

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 $−328 kJ/mol$

$F(g)+e−→F−(g) ΔH°=−328 kJ/mol$ …… (4)

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

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

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 $−349 kJ/mol$

$Cl(g)+e−→Cl−(g) ΔH°=−349 kJ/mol$ …… (6)

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

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

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

Answer 6

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

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 $−295 kJ/mol$

$I(g)+e−→I−(g) ΔH°=−295 kJ/mol$ …… (2)

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

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

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 $−328 kJ/mol$

$F(g)+e−→F−(g) ΔH°=−328 kJ/mol$ …… (4)

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

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

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 $−349 kJ/mol$

$Cl(g)+e−→Cl−(g) ΔH°=−349 kJ/mol$ …… (6)

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

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

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

Answer 7

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.

Answer 8

Expert Solution & Answer

Answer to Problem 86AP

Solution:

(a) $225 kJ/mol$

(b) $168 kJ/mol$

(c) $70 kJ/mol$

(a)

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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 $−295 kJ/mol$

$I(g)+e−→I−(g) ΔH°=−295 kJ/mol$ …… (2)

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

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

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 $−328 kJ/mol$

$F(g)+e−→F−(g) ΔH°=−328 kJ/mol$ …… (4)

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

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

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 $−349 kJ/mol$

$Cl(g)+e−→Cl−(g) ΔH°=−349 kJ/mol$ …… (6)

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

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

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

Answer 12

Ch. 8.1 - Practice ProblemATTEMPT Write Lewis dot symbols...Ch. 8.1 - Practice Problem BUILD Indicate the charge on...Ch. 8.1 - Practice ProblemCONCEPTUALIZE For each of the...Ch. 8.1 - 8.1.1 Using only a periodic table, determine the...Ch. 8.1 - 8.1.2 Using only a periodic table, determine the...Ch. 8.1 - To which group does element X belong if its Lewis...Ch. 8.1 - Prob. 4CP Ch. 8.2 - Prob. 1PPA Ch. 8.2 - Practice ProblemBUILD Arrange the compounds NaF,...Ch. 8.2 - Practice ProblemCONCEPTUALIZE Common ions of four...

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.

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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.

Answer to Problem 86AP

Explanation of Solution

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Chapter 8 Solutions

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.

Videos

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.

Answer to Problem 86AP

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions

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.