Chapter 10, Problem 10.82QP

Interpretation Introduction

Interpretation:

The lattice energy of $\text{LiCl}$ has to be identified.

Concept Introduction:

Lattice energy:

The amount of energy that is necessary for the conversion of one mole of ionic solid to its constituent ions in gaseous phase is called Lattice energy.

Hess’s law:

The enthalpy change for given set of reactants to the given set of products is the same, whether the process takes place in single or sequence of steps. This is called as Hess’s law.

Enthalpy is generally calculated from the standard enthalpy of formation.

${\text{ΔH°}}_{\text{reaction}}\text{=}\sum {\text{n}}_{\text{p}}{\text{ΔH}}^{\text{°}}{}_{\text{f}}\text{(products)-}\sum {\text{n}}_{\text{p}}{\text{ΔH}}^{\text{°}}\text{(reactants)}$

With the thermodynamic values from Hess’s law, the lattice energy of ionic compound can be determined.

To calculate: The lattice energy of $\text{LiCl}$

Answer to Problem 10.82QP

The lattice energy of $\text{LiCl}$ is calculated to be $\text{860}\text{kJ}{\text{mol}}^{\text{-1}}$ .

Explanation of Solution

${\text{ΔH°}}_{\text{f}}\left[{\text{Li}}_{\left(\text{g}\right)}\right]\text{=159}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}$

$\begin{array}{l}{\text{ΔH°}}_{\text{f}}\left[{\text{Cl}}_{\left(\text{g}\right)}\right]\text{=121}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{IE}}_{\left(\text{1}\right)}\text{(Li)=520}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{ΔH°}}_{\text{f}}\left[{\text{LiCl}}_{\left(\text{s}\right)}\right]\text{=-408}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ \text{EA(Cl)=349}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Lattice energy = ${\text{ΔH°}}_{\text{f}}\left[{\text{Li}}_{\left(\text{g}\right)}\right]{\text{+ΔH°}}_{\text{f}}\left[{\text{Cl}}_{\left(\text{g}\right)}\right]{\text{+IE}}_{\text{1}}\text{(Li)+}|{\text{ΔH°}}_{\text{f}}\left[{\text{LiCl}}_{\left(\text{s}\right)}\right]|\text{-EA(Cl)}$

$\begin{array}{l}{\text{Li}}_{\left(\text{s}\right)}\to {\text{Li}}_{\left(\text{g}\right)}{\text{ΔH°}}_{\text{f}}\left[{\text{Li}}_{\left(\text{g}\right)}\right]\text{=159}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ \frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}\to {\text{Cl}}_{\left(\text{g}\right)}{\text{ΔH°}}_{\text{f}}\left[{\text{Cl}}_{\left(\text{g}\right)}\right]\text{=121}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{Li}}_{\left(\text{g}\right)}\to {\text{Li}}^{\text{+}}{}_{\left(\text{g}\right)}{\text{+e}}^{\text{-}}{\text{IE}}_{\text{1}}\text{(Li)=520}\text{kJ}{\text{mol}}^{\text{-1}}\\ \end{array}$

$\begin{array}{l}{\text{Li}}_{\left(\text{s}\right)}\text{+}\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}\to {\text{LiCl}}_{\left(\text{s}\right)}{\text{ΔH°}}_{\text{f}}\left[{\text{LiCl}}_{\left(\text{s}\right)}\right]\text{=-408}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ |{\text{ΔH°}}_{\text{f}}\left[{\text{LiCl}}_{\left(\text{s}\right)}\right]|\text{=}\text{408}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{Cl}}_{\left(\text{g}\right)}{\text{+e}}^{\text{-}}\to {\text{Cl}}^{\text{-}}{}_{\left(\text{g}\right)}\text{EA(Cl)=349}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Lattice energy = ${\text{ΔH°}}_{\text{f}}\left[{\text{Li}}_{\left(\text{g}\right)}\right]{\text{+ΔH°}}_{\text{f}}\left[{\text{Cl}}_{\left(\text{g}\right)}\right]{\text{+IE}}_{\text{1}}\text{(Li)+}|{\text{ΔH°}}_{\text{f}}\left[{\text{LiCl}}_{\left(\text{s}\right)}\right]|\text{-EA(Cl)}$

= $\text{(159}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(121}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(520}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(408}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}\text{)-(349}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}$

= $\text{860}\text{kJ}{\text{mol}}^{\text{-1}}$

Lattice energy of $\text{LiCl}$ = $\text{860}\text{kJ}{\text{mol}}^{\text{-1}}$

Conclusion

The lattice energy of $\text{LiCl}$ was calculated using the standard thermodynamic data using Hess’s law.  The lattice energy of $\text{LiCl}$ was found to be $\text{860}\text{kJ}{\text{mol}}^{\text{-1}}$.