Chapter 6, Problem 6.47P

(a)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 6.47P

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{0}\text{.32aJ}\text{.}$

Explanation of Solution

The given equation is,

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

The first electron affinity of lithium is written as,

  ${\text{Li}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Li}}^{\text{+}}{}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}{\text{I}}_{\text{1}}\text{=0}\text{.86aJ}$

The first ionization energy of bromine is written as,

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

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

  $\begin{array}{l}{\text{E}}_{\text{rxtn}}\text{=0}\text{.86aJ+}\left(\text{-0}\text{.540}\right)\text{aJ}\\ {\text{E}}_{\text{rxtn}}\text{=0}\text{.32aJ}\end{array}$

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{0}\text{.32aJ}\text{.}$

(b)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 6.47P

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{-0}\text{.090aJ}\text{.}$

Explanation of Solution

The given equation is,

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

The first electron affinity of lithium is written as,

  ${\text{I}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{I}}_{\left(\text{g}\right)}{\text{E}}_{\text{a}}\text{=0}\text{.490aJ}$

The first ionization energy of chlorine is written as,

  ${\text{Cl}}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{Cl}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a1}}\text{=-0}\text{.580aJ}$

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

  $\begin{array}{l}{\text{E}}_{\text{rxtn}}\text{=0}\text{.490aJ+}\left(\text{-0}\text{.580}\right)\text{aJ}\\ {\text{E}}_{\text{rxtn}}\text{=-0}\text{.090aJ}\end{array}$

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{-0}\text{.090aJ}\text{.}$

(c)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 6.47P

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{2}\text{.308aJ}\text{.}$

Explanation of Solution

The given equation is,

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

The first electron affinity of sodium is written as,

  ${\text{Na}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Na}}^{\text{+}}{}_{\left(\text{g}\right)}{\text{+2e}}^{\text{-}}{\text{I}}_{\text{1}}\text{=0}\text{.83aJ}$

Since, there are two moles of sodium ${\text{I}}_{\text{1}}\text{=0}\text{.83aJ}\times 2\text{=1}\text{.66aJ}$

The first ionization energy of sulphur is written as,

  ${\text{S}}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{S}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a1}}\text{=-0}\text{.332aJ}$

The second ionization energy of sulphur is written as,

  ${\text{S}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{S}}^{\text{2-}}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a2}}\text{=0}\text{.980aJ}$

The total energy for sulphur is,

  $\begin{array}{l}{\text{E=E}}_{\text{a2}}{\text{+E}}_{\text{a1}}\\ \text{E=}\left(\text{-0}\text{.332+0}\text{.980}\right)\text{aJ}\\ \text{E=0}\text{.648aJ}\end{array}$

The total energy for sulphur is $\text{0}\text{.648aJ}\text{.}$

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

  $\begin{array}{l}{\text{E}}_{\text{rxtn}}\text{=1}\text{.66aJ+}\left(\text{0}\text{.648}\right)\text{aJ}\\ {\text{E}}_{\text{rxtn}}\text{=2}\text{.308aJ}\end{array}$

The ${\text{E}}_{\text{rxtn}}$ of the reaction is $\text{2}\text{.308aJ}\text{.}$