Chapter 9, Problem 9.25QP

Use the Born-Haber cycle outlined in Section 9.3 for LiF to calculate the lattice energy of NaCl. [The heat of sublimation of Na is 108 kJ/mol and ΔH°_f(NaCl ) = −411 kJ/mol. Energy needed to dissociate $\frac{1}{2}$ mole of Cl₂ into Cl atoms = 121.4 kJ.]

Interpretation Introduction

Interpretation:

Using the Born-Haber cycle for $LiF$ lattice energy of $NaCl$ 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.

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 9.25QP

Lattice energy of $NaCl$ is $787kJ/mol.$

Explanation of Solution

Given data:

$\begin{array}{l}\text{heat of sublimation of Na}\text{= 108 kJ/mol}\\ {\text{ΔH}}_{\text{f}}^{\text{°}}\left(NaCl\right)=-411kJ/mol\\ energyneededtodissociate\\ halfmoleofC{l}_2=121.4kJ\end{array}$

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

$N{a}_{(s)}\stackrel{}{\to }N{a}_{(g)}\Delta H_1^{°}=108kJ/mol$

The second step of Born-Haber cycle involves dissociation of gaseous $C{l}_2$ into gaseous $Cl$ atoms.

$\frac{1}{2}C{l}_{2(g)}\stackrel{}{\to }C{l}_{(g)}\Delta H_2^{°}=121.4kJ/mol$

The third step of Born-Haber cycle is ionization of gaseous $Na$ into gaseous $N{a}^{+}$ ions.

${\text{Na}}_{\text{(g)}}\stackrel{}{\to }{\text{Na}}_{\text{(g)}}^{\text{+}}{\text{ + e}}^{\text{-}}{\text{ΔH}}_{\text{3}}^{\text{°}}\text{ = 495}\text{.9 kJ/mol}$

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

${\text{Cl}}_{\text{(g)}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Cl}}_{\text{(g)}}^{-}{\text{ΔH}}_4^{\text{°}}\text{ = -349 kJ/mol}$

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.

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

${\text{ΔH}}_5^{\text{°}}$ is the enthalpy change for the above reaction and it is equivalent to the lattice energy of $NaCl.$  It is calculated by Hess’s law as follows,

$\begin{array}{l}{\text{Na}}_{\text{(s)}}\stackrel{}{\to }\cancel{{\text{Na}}_{\text{(g)}}}{\text{ΔH}}_{\text{1}}^{\text{°}}\text{ = 108 kJ/mol}\\ \cancel{\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2}}{}_{\text{(g)}}}\stackrel{}{\to }\cancel{{\text{Cl}}_{\text{(g)}}}{\text{ΔH}}_{\text{2}}^{\text{°}}\text{ = 121}\text{.4 kJ/mol}\\ \cancel{{\text{Na}}_{\text{(g)}}}\stackrel{}{\to }{\text{ Na}}_{\text{(g)}}^{\text{+}}\text{ + }\cancel{{\text{e}}^{\text{-}}}{\text{ΔH}}_{\text{3}}^{\text{°}}\text{ = 495}\text{.9 kJ/mol}\\ \cancel{{\text{Cl}}_{\text{(g)}}}\text{ + }{\cancel{\text{e}}}^{\text{-}}\stackrel{}{\to }\cancel{{\text{Cl}}_{\text{(g)}}^{\text{-}}}{\text{ΔH}}_{\text{4}}^{\text{°}}\text{ = -349 kJ/mol}\\ \cancel{{\text{Na}}^{\text{+}}{}_{\text{(g)}}}\text{ + }\cancel{{\text{Cl}}_{\text{(g)}}^{\text{-}}}\stackrel{}{\to }{\text{ NaCl}}_{\text{(s)}}{\text{ΔH}}_{\text{5}}^{\text{°}}\text{ = ?}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ N{a}_{(s)}\text{+}\frac{1}{2}{\text{Cl}}_{2(g)}\stackrel{}{\to }{\text{ NaCl}}_{(s)}{\text{ΔH}}_{overall}^{°}\text{ = -411 kJ/mol}\end{array}$

$\begin{array}{l}latticeenergyofNaCl,\Delta H_5^{°}=\Delta H_{overall}^{°}-\Delta H_1^{°}-\Delta H_2^{°}-\Delta H_3^{°}-\Delta H_4^{°}\\ =\left(-411\right)-108-121.4-495.9-\left(-349\right)\\ =-787kJ/mol\end{array}$

$\text{787 kJ/mol}$ of energy is released (negative sign) when one mole of $NaCl$ is formed.  It also means $\text{787 kJ/mol}$ of energy is required to break bond between $N{a}^{+}$ and $C{l}^{-}$ ion in one mole of $NaCl.$