Chapter 6, Problem 6.48P

(a)

Interpretation Introduction

Interpretation:

The ${\text{E}}_{\text{rxtn}}$ of the reaction has to be written.

    ${\text{Na}}_{\left(\text{g}\right)}{\text{+H}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Na}}^{\text{+}}{}_{\left(\text{g}\right)}{\text{+H}}^{\text{-}}{}_{\left(\text{g}\right)}$

Answer to Problem 6.48P

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{0}\text{.71aJ}\text{.}$

Explanation of Solution

The given equation is,

  ${\text{Na}}_{\left(\text{g}\right)}{\text{+H}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Na}}^{\text{+}}{}_{\left(\text{g}\right)}{\text{+H}}^{\text{-}}{}_{\left(\text{g}\right)}$

The first electron affinity of sodium is written as,

  ${\text{Na}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Na}}^{\text{+}}{}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}{\text{I}}_{\text{1}}\text{=0}\text{.83aJ}$

The first ionization energy of hydrogen is written as,

  ${\text{H}}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{H}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a1}}\text{=-0}\text{.12aJ}$

The ${\text{E}}_{\text{rxtn}}$ of the reaction is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{rxtn}}\text{=0}\text{.83aJ+}\left(\text{-0}\text{.12}\right)\text{aJ}\\ {\text{E}}_{\text{rxtn}}\text{=0}\text{.71aJ}\end{array}$

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{0}\text{.71aJ}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The ${\text{E}}_{\text{rxtn}}$ of the reaction has to be written.

    ${\text{Mg}}_{\left(\text{g}\right)}{\text{+O}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Mg}}^{\text{2+}}{}_{\left(\text{g}\right)}{\text{+O}}^{\text{2-}}{}_{\left(\text{g}\right)}$

Answer to Problem 6.48P

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{4}\text{.71aJ}\text{.}$

Explanation of Solution

The given equation is,

  ${\text{Mg}}_{\left(\text{g}\right)}{\text{+O}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Mg}}^{\text{2+}}{}_{\left(\text{g}\right)}{\text{+O}}^{\text{2-}}{}_{\left(\text{g}\right)}$

The first electron affinity of magnesium is written as,

  ${\text{Mg}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Mg}}_{\text{(g)}}{}^{\text{+}}{\text{+2e}}^{\text{-}}{\text{I}}_{\text{1}}\text{=1}\text{.23aJ}$

The second electron affinity of magnesium is written as,

  ${\text{Mg}}^{+}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Mg}}_{\text{(g)}}{}^{\text{2+}}{\text{+2e}}^{\text{-}}{\text{I}}_2\text{=2}\text{.41aJ}$

The total electron affinity of magnesium is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{1}}{\text{=I}}_{\text{1}}{\text{+I}}_{\text{2}}\\ {\text{E}}_{\text{1}}\text{=}\left(\text{1}\text{.23+2}\text{.41}\right)\text{aJ}\\ {\text{E}}_{\text{1}}\text{=3}\text{.64aJ}\end{array}$

The total electron affinity is $\text{3}\text{.64aJ}\text{.}$

The first ionization energy of oxygen is written as,

  ${\text{O}}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{O}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a1}}\text{=-0}\text{.234aJ}$

The second ionization energy of oxygen is written as,

  ${\text{O}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{O}}^{\text{2-}}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a1}}\text{=1}\text{.30aJ}$

The total ionization energy is calculated as,

  $\begin{array}{l}{\text{E=E}}_{\text{a2}}{\text{+E}}_{\text{a1}}\\ \text{E=}\left(\text{-0}\text{.234+1}\text{.30}\right)\text{aJ}\\ \text{E=1}\text{.07aJ}\end{array}$

The total ionization energy of oxygen is $\text{1}\text{.07aJ}\text{.}$

The ${\text{E}}_{\text{rxtn}}$ of the reaction is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{rxtn}}\text{=3}\text{.64aJ+1}\text{.07aJ}\\ {\text{E}}_{\text{rxtn}}\text{=4}\text{.71aJ}\end{array}$

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{4}\text{.71aJ}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The ${\text{E}}_{\text{rxtn}}$ of the reaction has to be written.

    ${\text{Mg}}_{\left(\text{g}\right)}{\text{+2Br}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Mg}}^{\text{2+}}{}_{\left(\text{g}\right)}{\text{+2Br}}^{\text{-}}{}_{\left(\text{g}\right)}$

Answer to Problem 6.48P

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{2}\text{.56 aJ}\text{.}$

Explanation of Solution

The given equation is,

  ${\text{Mg}}_{\left(\text{g}\right)}{\text{+2Br}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Mg}}^{\text{2+}}{}_{\left(\text{g}\right)}{\text{+2Br}}^{\text{-}}{}_{\left(\text{g}\right)}$

The first electron affinity of magnesium is written as,

  ${\text{Mg}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Mg}}_{\text{(g)}}{}^{\text{+}}{\text{+2e}}^{\text{-}}{\text{I}}_{\text{1}}\text{=1}\text{.23aJ}$

The second electron affinity of magnesium is written as,

  ${\text{Mg}}^{+}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Mg}}_{\text{(g)}}{}^{\text{2+}}{\text{+2e}}^{\text{-}}{\text{I}}_2\text{=2}\text{.41aJ}$

The total electron affinity of magnesium is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{1}}{\text{=I}}_{\text{1}}{\text{+I}}_{\text{2}}\\ {\text{E}}_{\text{1}}\text{=}\left(\text{1}\text{.23+2}\text{.41}\right)\text{aJ}\\ {\text{E}}_{\text{1}}\text{=3}\text{.64aJ}\end{array}$

The first ionization energy of bromine is written as,

  ${\text{Br}}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{Br}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a1}}\text{=-0}\text{.540aJ}$

Since there are two bromines present, ${\text{E}}_{\text{a1}}\text{=}\left(\text{2}\right)\left(\text{-0}\text{.540aJ}\right)\text{=-1}\text{.08aJ}$

The ${\text{E}}_{\text{rxtn}}$ of the reaction is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{rxtn}}\text{=3}\text{.64aJ+}\left(\text{-1}\text{.08}\right)\text{aJ}\\ {\text{E}}_{\text{rxtn}}\text{=2}\text{.56aJ}\end{array}$

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{2}\text{.56 aJ}\text{.}$