Chapter 7, Problem 135IP

Using data from Chapter 2, calculate the change in energy expected for each of the following processes.

a. $\text{Na}\left(g\right)+\text{Cl}\left(g\right)\to {\text{Na}}^{\text{+}}\left(g\right)+{\text{Cl}}^{-}\left(g\right)$

b. $\text{Mg}\left(g\right)+\text{F}\left(g\right)\to {\text{Mg}}^{\text{+}}\left(g\right)+{\text{F}}^{-}\left(g\right)$

c. ${\text{Mg}}^{\text{+}}\left(g\right)+\text{F}\left(g\right)\to {\text{Mg}}^{\text{2+}}\left(g\right)+{\text{F}}^{-}\left(g\right)$.

d. $\text{Mg}\left(g\right)+2\text{F}\left(g\right)\to {\text{Mg}}^{\text{2+}}\left(g\right)+2{\text{F}}^{-}\left(g\right)$

Interpretation Introduction

Interpretation: The energy change of given reaction should be determined.

Concept introduction:

                 The energy change of ionic reactions is consider as the sum of ionization energy and electron affinity values of the reaction.

                                                      $\text{ΔΗ}\text{=}\text{IE+EA}\mathrm{.....}\text{(1)}$

                                              IE is of ionization energy

                                              EA electron affinity

                                              $\text{ΔΗ}$ enthalpy change

Answer to Problem 135IP

The enthalpy change (energy) of the given reactions are found to be $\text{146}\text{kJ}$

Explanation of Solution

To calculate the enthalpy change of the reaction.

The reaction is,

                   $\text{a)}{\text{Na}}_{\text{(g)}}{\text{+Cl}}_{\text{(g)}}\to {\text{Na}}^{\text{+}}{}_{\text{(g)}}{\text{+Cl}}^{\text{-}}{}_{\text{(g)}}$

                        ${\text{Na}}_{\text{(g)}}\to {\text{Na}}^{\text{+}}{\text{+ e}}^{\text{-}}$

The first ionization energy of sodium is $\text{495}\text{kJ}$

                           ${\text{Cl}}_{{}^{\text{(g)}}}{\text{+e}}^{\text{-}}\to {\text{Cl}}^{\text{-}}{}_{\text{(g)}}$

The electron affinity of chlorine is $\text{-348}\text{.7}\text{kJ}$

                                 $\begin{array}{l}\text{ΔΗ}\text{=}{\text{I}}_{\text{1}}\text{+EA}\\ \text{ΔΗ}\text{=}\text{495-348}\text{.7}\\ \text{=}\text{146}\text{kJ}\end{array}$

• The ionization energy of sodium and electron affinity of chlorine are add to give enthalpy change of the reaction.
• Enthalpy change of the reaction (a) is $\text{146}\text{kJ}$.

Interpretation Introduction

Interpretation: The energy change of given reaction should be determined.

Concept introduction:

                 The energy change of ionic reactions is consider as the sum of ionization energy and electron affinity values of the reaction.

                                                      $\text{ΔΗ}\text{=}\text{IE+EA}\mathrm{.....}\text{(1)}$

                                              IE is of ionization energy

                                              EA electron affinity

                                             $\text{ΔΗ}$ enthalpy change

Answer to Problem 135IP

The enthalpy change (energy) of the given reactions are found to be,

$407\text{kJ}$

Explanation of Solution

To calculate the enthalpy change of the reaction.

                  ${\text{Mg}}_{\text{(g)}}{\text{+F}}_{\text{(g)}}\to {\text{Mg}}^{\text{+}}{}_{\text{(g)}}{\text{+F}}^{\text{-}}{}_{\text{(g)}}$

                                  ${\text{Mg}}_{\text{(g)}}\to {\text{Mg}}^{\text{+}}{\text{+ e}}^{\text{-}}$

The first ionization energy of magnesium is $\text{735 kJ}$

                                  ${\text{F}}_{{}^{\text{(g)}}}{\text{+e}}^{\text{-}}\to {\text{F}}^{\text{-}}{}_{\text{(g)}}$

The electron affinity of fluorine is $\text{-327}\text{.8}\text{kJ}$

                                 $\begin{array}{l}\text{ΔΗ}\text{=}{\text{I}}_{\text{1}}\text{+EA}\\ \text{ΔΗ}\text{=}735\text{-327}\text{.8}\\ \text{=}407\text{kJ}\end{array}$

• The ionization energy of magnesium and electron affinity of fluorine are add to give enthalpy change of the reaction (b).
• Enthalpy change of the reaction (b) is $407\text{kJ}$.

Interpretation Introduction

Interpretation: The energy change of given reaction should be determined.

Concept introduction:

                 The energy change of ionic reactions is consider as the sum of ionization energy and electron affinity values of the reaction.

                                                      $\text{ΔΗ}\text{=}\text{IE+EA}\mathrm{.....}\text{(1)}$

                                              IE is of ionization energy

                                              EA electron affinity

                                             $\text{ΔΗ}$ enthalpy change

Answer to Problem 135IP

The enthalpy change (energy) of the given reactions are found to be,

$1117\text{kJ}$

Explanation of Solution

To calculate the enthalpy change of the reaction.

The reaction is,

                  $\text{b)}{\text{Mg}}^{+}{}_{\text{(g)}}{\text{+F}}_{\text{(g)}}\to {\text{Mg}}^{\text{2+}}{}_{\text{(g)}}{\text{+F}}^{\text{-}}{}_{\text{(g)}}$

                                  ${\text{Mg}}_{\text{(g)}}\to {\text{Mg}}^{\text{+}}{\text{+ e}}^{\text{-}}$

The second ionization energy of magnesium is $\text{1445 kJ}$

                                  ${\text{F}}_{{}^{\text{(g)}}}{\text{+e}}^{\text{-}}\to {\text{F}}^{\text{-}}{}_{\text{(g)}}$

The electron affinity of fluorine is $\text{-327}\text{.8}\text{kJ}$

                                   $\begin{array}{l}\text{ΔΗ}\text{=}{\text{I}}_2\text{+EA}\\ \text{ΔΗ}\text{=}1445\text{-327}\text{.8}\\ \text{=}1117\text{kJ}\end{array}$

• The second ionization energy of magnesium and electron affinity of fluorine are add to give enthalpy change of the reaction (c).
• Enthalpy change of the reaction (c) is $1117\text{kJ}$.

Interpretation Introduction

Interpretation: The energy change of given reaction should be determined.

Concept introduction:

                 The energy change of ionic reactions is consider as the sum of ionization energy and electron affinity values of the reaction.

                                                       $\text{ΔΗ}\text{=}\text{IE+EA}\mathrm{.....}\text{(1)}$

                                              IE is of ionization energy

                                              EA electron affinity

                                             $\text{ΔΗ}$ enthalpy change

Answer to Problem 135IP

The enthalpy change (energy) of the given reactions are found to be,

$\text{1524}\text{kJ}$

Explanation of Solution

To calculate the enthalpy change of the reaction.

                      ${\text{d)Mg}}_{\text{(g)}}{\text{+2F}}_{\text{(g)}}\to {\text{Mg}}^{\text{2+}}{}_{\text{+}}{\text{2F}}^{\text{-}}{}_{\text{(g)}}$

From step 3and 4,

               $\begin{array}{l}\text{ΔΗ}\text{=}{\text{ΔΗ}}_{\text{step3}}{\text{+ΔΗ}}_{\text{step4}}\\ \text{ΔΗ}\text{=}\text{407+1117}\\ \text{=}\text{1524}\text{kJ}\end{array}$

• The enthalpy change of magnesium in step 3and step 4 are add to give enthalpy change of the reaction (d).
• Enthalpy change of the reaction (d) is $\text{1524}\text{kJ}$.