Chapter 9, Problem 9.118SP

Interpretation Introduction

Interpretation:

Using the Born-Haber cycle for $MgO$ lattice energy of $O^{-}$ has to be calculated.

Concept Introduction:

Born-Haber cycle is based on Hess’s law to calculate the lattice enthalpy of ionic compounds and deals with energy changes in formation of ionic compounds.

The energy released when gaseous state ions of unlike charges that are infinitely farther apart combine to form a stable ionic solid is called Lattice energy.  Conversely, the energy required to break the electrostatic force of attraction between the ions of unlike charges in the ionic solid and revert them to gaseous state is also termed as Lattice energy of an ionic solid.

Electron affinity of an atom refers to the energy released when one electron is added to neutral atom in gaseous state.

Hess’s law is applied to calculate the enthalpy changes in a reaction.  According to Hess’s law – “The overall enthalpy change of a reaction is equal to the sum of the enthalpy changes involving in each and every individual steps in the reaction.”  Thus if a reaction involves ‘n’ steps then enthalpy change $\Delta H°$ of the reaction is,

  $\Delta H°=\Delta {\text{H}}_1^{°}+\Delta {\text{H}}_2^{°}+\Delta {\text{H}}_3^{°}\mathrm{....}+\Delta {\text{H}}_n^{°}$

Answer to Problem 9.118SP

Electron affinity of $O^{-}$ is calculated as $-844kJ/mol.$

Explanation of Solution

Given data:

    $\text{heat of sublimation of Mg}\text{= 148 kJ/mol}$

The first step of Born-Haber cycle involves sublimation of solid $Mg$ into gaseous $Mg$

    $M{g}_{(s)}\stackrel{}{\to }M{g}_{(g)}\Delta H_1^{°}=148kJ/mol$

The second step of Born-Haber cycle involves dissociation of gaseous $O_2$ into gaseous $O$ atoms.

    $\frac{1}{2}{O}_{2(g)}\stackrel{}{\to }{O}_{(g)}\Delta H_2^{°}=\frac{1}{2}\left(498.7\right)kJ/mol$

The third step of Born-Haber cycle is ionization of gaseous $Mg$ into gaseous $M{g}^{2+}$ ions.

    $\begin{array}{l}{\text{Mg}}_{\text{(g)}}\stackrel{}{\to }{\text{Mg}}_{\text{(g)}}^{\text{+}}{\text{ + e}}^{\text{-}}{\text{ΔH}}_{\text{3}}^{\text{°'}}\text{ = 738}\text{.1 kJ/mol}\\ {\text{Mg}}_{\text{(g)}}^{\text{+}}\stackrel{}{\to }{\text{Mg}}_{\text{(g)}}^{\text{2+}}{\text{ + e}}^{\text{-}}{\text{ΔH}}_{\text{3}}^{\text{°''}}\text{ = 1450 kJ/mol}\end{array}$

The fourth step of Born-Haber cycle is ionization of gaseous $O$ into gaseous $O^{-}$ ions.

    $\begin{array}{l}{\text{O}}_{\text{(g)}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{O}}_{\text{(g)}}^{-}{\text{ΔH}}_4^{\text{°'}}\text{ = -141 kJ/mol}\\ {\text{O}}_{\text{(g)}}^{-}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{O}}_{\text{(g)}}^{2-}{\text{ΔH}}_4^{\text{°''}}\text{ = ? kJ/mol}\end{array}$

The fifth and final step of Born-Haber cycle is formation of solid $NaCl$ as a result of binding gaseous $N{a}^{+}$ and $C{l}^{-}$ ions together by electrostatic force of attraction.

    ${\text{Mg}}^{2+}{}_{\text{(g)}}{\text{+O}}_{\text{(g)}}^{2-}\stackrel{}{\to }{\text{ MgO}}_{(s)}\Delta H_5^{°}=-3890kJ/mol$

${\text{ΔH}}_4^{\text{°''}}$ corresponding to electron affinity of $O^{-}$ ion is calculated by Hess’s law as follows,

    $\begin{array}{l}M{g}_{(s)}\stackrel{}{\to }M{g}_{(g)}\Delta H_1^{°}=148kJ/mol\\ \frac{1}{2}{O}_{2(g)}\stackrel{}{\to }{O}_{(g)}\Delta H_2^{°}=\frac{1}{2}\left(498.7\right)kJ/mol\\ {\text{Mg}}_{\text{(g)}}\stackrel{}{\to }{\text{Mg}}_{\text{(g)}}^{\text{+}}{\text{ + e}}^{\text{-}}{\text{ΔH}}_{\text{3}}^{\text{°'}}\text{= 738}\text{.1 kJ/mol}\\ {\text{Mg}}_{\text{(g)}}^{\text{+}}\stackrel{}{\to }{\text{Mg}}_{\text{(g)}}^{\text{2+}}{\text{ + e}}^{\text{-}}{\text{ΔH}}_{\text{3}}^{\text{°''}}\text{= 1450 kJ/mol}\\ {\text{O}}_{\text{(g)}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{O}}_{\text{(g)}}^{-}{\text{ΔH}}_4^{\text{°'}}\text{= -141 kJ/mol}\\ {\text{O}}_{\text{(g)}}^{-}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{O}}_{\text{(g)}}^{2-}{\text{ΔH}}_4^{\text{°''}}\text{= ? kJ/mol}\\ {\text{Mg}}^{2+}{}_{\text{(g)}}{\text{+O}}_{\text{(g)}}^{2-}\stackrel{}{\to }{\text{ MgO}}_{(s)}\Delta H_5^{°}=-3890kJ/mol\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ M{g}_{(s)}+\frac{1}{2}{O}_{2(g)}\stackrel{}{\to }Mg{O}_{(s)}\Delta H_{overall}^{°}=-601.8kJ/mol\end{array}$

$\begin{array}{l}electronaffinityof{O}^{-} ion,\Delta H_4^{°\text{'}\text{'}}=\Delta H_{overall}^{°}-\Delta H_1^{°}-\Delta H_2^{°}-\Delta H_3^{°}-\Delta H_4^{°}-\Delta H_5^{°}\\ =\left(-601.8-148-249.4-738.1-1450+141+3890\right)\\ =844kJ/mol\end{array}$

$\text{844 kJ/mol}$ of energy is released (negative sign) when an electron is added to O atom in gaseous state.

Hence electron affinity of $O^{-}$ is $-844kJ/mol.$