Chapter 6, Problem 6.69P

Interpretation Introduction

Interpretation:

The reason why the energy evolved in the process decrease in the order has to be given.

Concept Introduction:

Coulomb’s Law of attraction:

Coulomb’s law states that the energy of interaction of two ions is directly proportional to the product their electric charges and is inversely proportional to the distance between them.  The formula is,

  ${\text{E}}_{\text{Coulomb}}\text{=k}\cdot \frac{{\text{Q}}_{\text{1}}{\text{Q}}_{\text{2}}}{\text{d}}$

${\text{Q}}_{\text{1}}{\text{,Q}}_{\text{2}}\text{=}$ Charges of the two ions

$\text{d=}$ Distance between the centers of the two ions

$\text{k=}$ Proportionality constant

The value of proportionality constant depends on the units of charges and distance.

Explanation of Solution

The given reaction is,

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

Sodium chloride:

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

This equation can be simplified into two reactions of first electron affinity of sodium and first ionization of chlorine.

  $\begin{array}{l}{\text{Na}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Na}}^{\text{+}}{}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}{\text{I}}_{\text{1}}\text{=0}\text{.824aJ}\\ {\text{Cl}}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{Cl}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a1}}\text{=-0}\text{.580aJ}\end{array}$

Sodium forms monopositive ion $\left({\text{Na}}^{\text{+}}\right)$ and its radius is 102pm.

Chlorine forms mononegative ion $\left({\text{Cl}}^{\text{-}}\right)$ and its radius is 181pm.

The distance d is sum of the distance between chlorine and sodium.  The distance d is calculated as,

  $\begin{array}{l}\text{d=102pm+181pm}\\ \text{d=283pm}\end{array}$

The value of energy is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{Coulomb}}\text{=}\left(\text{231aJ}\text{.pm}\right)\text{×}\frac{\left(\text{+1}\right)\left(\text{-1}\right)}{\text{283pm}}\\ {\text{E}}_{{}_{\text{Coulomb}}}\text{=-0}\text{.816aJ}\end{array}$

The energy of reaction is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{rxtn}}{\text{=I}}_{\text{1}}{\text{+E}}_{\text{a1}}{\text{+E}}_{\text{coulomb}}\\ {\text{E}}_{\text{rxtn}}\text{=}\left(\text{0}\text{.824aJ}\right)\text{+}\left(\text{-0}\text{.580aJ}\right)\text{+}\left(\text{-0}\text{.816aJ}\right)\\ {\text{E}}_{\text{rxtn}}\text{=-0}\text{.572aJ}\end{array}$

Sodium bromide:

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

This equation can be simplified into two reactions of first electron affinity of sodium and first ionization of bromine.

  $\begin{array}{l}{\text{Na}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Na}}^{\text{+}}{}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}{\text{I}}_{\text{1}}\text{=0}\text{.824aJ}\\ {\text{Br}}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{Br}}^{\text{-}}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a1}}\text{=-0}\text{.540aJ}\end{array}$

Sodium forms monopositive ion $\left({\text{Na}}^{\text{+}}\right)$ and its radius is 102pm.

Bromine forms mononegative ion $\left({\text{Br}}^{\text{-}}\right)$ and its radius is 196pm.

The distance d is sum of the distance between bromine and sodium.  The distance d is calculated as,

  $\begin{array}{l}\text{d=102pm+196pm}\\ \text{d=298pm}\end{array}$

The value of energy is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{Coulomb}}\text{=}\left(\text{231aJ}\text{.pm}\right)\text{×}\frac{\left(\text{+1}\right)\left(\text{-1}\right)}{\text{298pm}}\\ {\text{E}}_{{}_{\text{Coulomb}}}\text{=-0}\text{.775aJ}\end{array}$

The energy of reaction is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{rxtn}}{\text{=I}}_{\text{1}}{\text{+E}}_{\text{a1}}{\text{+E}}_{\text{coulomb}}\\ {\text{E}}_{\text{rxtn}}\text{=}\left(\text{0}\text{.824aJ}\right)\text{+}\left(\text{-0}\text{.540aJ}\right)\text{+}\left(\text{-0}\text{.775aJ}\right)\\ {\text{E}}_{\text{rxtn}}\text{=-0}\text{.491aJ}\end{array}$

Sodium iodide:

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

This equation can be simplified into two reactions of first electron affinity of sodium and first ionization of iodine.

  $\begin{array}{l}{\text{Na}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{Na}}^{\text{+}}{}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}{\text{I}}_{\text{1}}\text{=0}\text{.824aJ}\\ {\text{I}}_{\left(\text{g}\right)}{\text{+1e}}^{\text{-}}\stackrel{}{\to }{\text{I}}^{-}{}_{\left(\text{g}\right)}{\text{E}}_{\text{a1}}\text{=-0}\text{.490aJ}\end{array}$

Sodium forms monopositive ion $\left({\text{Na}}^{\text{+}}\right)$ and its radius is 102pm.

Iodine forms mononegative ion $\left({\text{I}}^{\text{-}}\right)$ and its radius is 220pm.

The distance d is sum of the distance between iodine and sodium.  The distance d is calculated as,

  $\begin{array}{l}\text{d=102pm+220pm}\\ \text{d=321pm}\end{array}$

The value of energy is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{Coulomb}}\text{=}\left(\text{231aJ}\text{.pm}\right)\text{×}\frac{\left(\text{+1}\right)\left(\text{-1}\right)}{\text{321pm}}\\ {\text{E}}_{{}_{\text{Coulomb}}}\text{=-0}\text{.720aJ}\end{array}$

The energy of reaction is calculated as,

  $\begin{array}{l}{\text{E}}_{\text{rxtn}}{\text{=I}}_{\text{1}}{\text{+E}}_{\text{a1}}{\text{+E}}_{\text{coulomb}}\\ {\text{E}}_{\text{rxtn}}\text{=}\left(\text{0}\text{.824aJ}\right)\text{+}\left(\text{-0}\text{.490aJ}\right)\text{+}\left(\text{-0}\text{.720aJ}\right)\\ {\text{E}}_{\text{rxtn}}\text{=-0}\text{.386aJ}\end{array}$

It can be seen that ${\text{E}}_{\text{rxn}}\left(\text{NaCl}\right){\text{<E}}_{\text{rxn}}\left(\text{NaBr}\right){\text{<E}}_{\text{rxn}}\left(\text{NaI}\right)$ and hence, the energy evolved in the process decreases.