Chapter 8, Problem 8.88QP

(a)

Interpretation Introduction

Interpretation:

The change in energy of the reaction $M{g}_{\left(g\right)}+\text{2}{\text{F}}_{\text{(g)}}\underset{}{\overset{}{\to }}{\text{Mg}}^{\text{2+}}{}_{\text{(g)}}+\text{2}{\text{F}}^{\text{-}}{}_{\text{(g)}}$ has to be determined.

Concept Introduction:

First ionization energy: The ionization energy is the minimum energy required to remove the electron from an isolated atom which is in the gaseous state results to give gaseous ion with one positive charge.

Second ionization energy: The minimum energy needed to remove an electron from a unipositive gaseous ion to form a dipositive ion in the ground state is known as second ionization energy.

Electron Affinity: It is energy change taking place when one electron is added to an atom which is neutral in its gaseous state forming an anion.

Answer to Problem 8.88QP

The change in energy of the given reaction is $1532\text{kJ/mol}$

Explanation of Solution

The reaction is given below:

$M{g}_{\left(g\right)}+\text{2}{\text{F}}_{\text{(g)}}\underset{}{\overset{}{\to }}{\text{Mg}}^{\text{2+}}{}_{\text{(g)}}+\text{2}{\text{F}}^{-}{}_{\text{(g)}}$

Here, two electrons are removed from $Mg$ and one electron is accepted by two $F$ atoms

Hence the change in energy for the given reaction is the sum of first and second ionization energy of $Mg$ and the electron affinity of $F$

Therefore,

The change in energy can be given as:

$\begin{array}{l}FirstionizationofMg,\\ {\text{Mg}}_{\text{(g)}}\underset{}{\to }{\text{Mg}}^{\text{+}}{}_{\text{(g)}}+\stackrel{-}{e}\text{ΔH}\text{=}\text{738}\text{.1kJ/mol}\\ \\ \text{Second}\text{i}onizationofMg,\\ {\text{Mg}}^{+}{}_{\text{(g)}}\underset{}{\to }{\text{Mg}}^{\text{2+}}{}_{\text{(g)}}+\stackrel{-}{e}\text{ΔH}\text{=}\text{1450}\text{kJ/mol}\\ \\ ElectronaffinityofF,\\ 2F_{\text{(g)}}+2\stackrel{-}{e}\underset{}{\to }2F^{-}{}_{\left(g\right)}\text{ΔH}\text{=}2\left(-\text{328}\text{kJ/mol}\right)\\ \underset{\_}{}\\ \\ M{g}_{\left(g\right)}+\text{2}{\text{F}}_{\text{(g)}}\underset{}{\overset{}{\to }}{\text{Mg}}^{\text{2+}}{}_{\text{(g)}}+\text{2}{\text{F}}^{-}{}_{\text{(g)}}\text{ΔH}\text{=}1532\text{kJ/mol}\end{array}$

The change in energy of the given reaction is $1532\text{kJ/mol}$

(b)

Interpretation Introduction

Interpretation: The change in energy of the reaction ${\text{2Al}}_{\left(g\right)}+\text{3}{\text{O}}_{\text{(g)}}\underset{}{\overset{}{\to }}\text{2}{\text{Al}}^{\text{3+}}{}_{\text{(g)}}+\text{3}{\text{O}}^{2-}{}_{\text{(g)}}$ has to be determined.

Concept Introduction:

First ionization energy: The ionization energy is the minimum energy required to remove the electron from an isolated atom which is in the gaseous state results to give gaseous ion with one positive charge.

Second ionization energy: The minimum energy needed to remove an electron from a unipositive gaseous ion to form a dipositive ion in the ground state is known as second ionization energy.

Third ionization energy: The minimum energy needed to remove an electron from a dipositive gaseous ion to form a tripositive ion in the ground state is known as third ionization energy.

First electron Affinity: It is energy change taking place when one electron is added to an atom which is neutral in its gaseous state forming an anion with one negative charge.

Second electron Affinity: It is energy change taking place when one electron is added to an anion having a negative charge in its gaseous state forming an anion with two negative charges.

Answer to Problem 8.88QP

The change in energy of the given reaction is $\text{12,405 kJ/mol}$

Explanation of Solution

The reaction is given below:

${\text{2Al}}_{\left(g\right)}+\text{3}{\text{O}}_{\text{(g)}}\underset{}{\overset{}{\to }}\text{2}{\text{Al}}^{\text{3+}}{}_{\text{(g)}}+\text{3}{\text{O}}^{2-}{}_{\text{(g)}}$

Here, three electrons are removed from $Al$ and two electrons is accepted by $O$

Hence the change in energy for the given reaction is the sum of first, second and ionization energy of $Al$ and the first and second electron affinity of $O$

The first, second and ionization energy of $Al$ can be given as:

$\begin{array}{l}FirstionizationofAl,\\ 2\left(A{l}_{\text{(g)}}\underset{}{\to }A{l}^{\text{+}}{}_{\text{(g)}}+\stackrel{-}{e}\right)\text{ΔH}\text{=}2\left(\text{577}\text{.9kJ/mol}\right)\\ \\ \text{Second}\text{i}onizationofAl,\\ 2\left(A{l}^{+}{}_{\text{(g)}}\underset{}{\to }A{l}^{\text{2+}}{}_{\text{(g)}}+\stackrel{-}{e}\right)\text{ΔH}\text{=}2\left(\text{1820}\text{kJ/mol}\right)\\ \\ \text{Third}\text{i}onizationofAl,\\ 2\left(A{l}^{2+}{}_{\text{(g)}}\underset{}{\to }A{l}^{\text{3+}}{}_{\text{(g)}}+\stackrel{-}{e}\right)\text{ΔH}\text{=}2\left(\text{2750}\text{kJ/mol}\right)\end{array}$

The first and second electron affinity of $O$ can be given as:

$\begin{array}{l}FirstelectronaffinityofO,\\ 3\left(O_{\text{(g)}}+\stackrel{-}{e}\underset{}{\to }{O}^{-}{}_{\left(g\right)}\right)\text{ΔH}\text{=}3\left(-141\text{kJ/mol}\right)\\ \\ \text{Second}electronaffinityofO,\\ 3\left(O^{-}{}_{\text{(g)}}+\stackrel{-}{e}\underset{}{\to }{O}^{2-}{}_{\left(g\right)}\right)\text{ΔH}\text{=}3\left(844\text{kJ/mol}\right)\end{array}$

The change in energy of the given reaction can be calculated as:

$\begin{array}{c}Thechangeinenergy=\left(\begin{array}{c}2\left(\text{577}\text{.9kJ/mol}\right)+2\left(\text{1820}\text{kJ/mol}\right)+2\left(\text{2750}\text{kJ/mol}\right)+\\ 3\left(-141\text{kJ/mol}\right)+3\left(844\text{kJ/mol}\right)\end{array}\right)\\ \\ =\text{12,405 kJ/mol}\end{array}$

The change in energy of the given reaction is $\text{12,405 kJ/mol}$