Chapter 15, Problem 15.18P

Interpretation Introduction

Interpretation:

The lattice enthalpy of $CaO$ 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,

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

Explanation of Solution

The overall chemical equation for the formation of $CaO$ is represented as,

$Ca\left(s\right)+O_2\left(g\right)\stackrel{}{\to }CaO\left(s\right)$

Using the given standard data, the calculations are done,

The reaction involves more than one step and they are given below. The corresponding enthalpy changes to the each step also given.

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}C{a}_{\text{(s)}}\stackrel{}{\to }\cancel{{\text{Ca}}_{\text{(g)}}}{\text{ΔH}}_{\text{1}}^{\text{°}}\text{ = +178 kJ/mol}\\ \cancel{{\text{Ca}}_{\text{(g)}}}\stackrel{}{\to }\cancel{C{a}^{2+}{}_{\text{(g)}}}\text{+ }\cancel{2e^{\text{-}}}{\text{ΔH}}_{\text{2}}^{\text{°}}\text{ =}\text{+1735 kJ/mol}\\ \frac{1}{2}{O}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }\cancel{O_{\left(g\right)}}{\text{ΔH}}_3^{°}\text{ = +249 kJ/mol}\\ \cancel{{\text{O}}_{\left(g\right)}}+\cancel{e^{-}}\stackrel{}{\to }\cancel{O_{(g)}^{-}}{\text{ΔH}}_4^{°}\text{ = }\left(\text{-141 kJ/mol}\right)\\ \cancel{O_{(g)}^{-}}+\cancel{e^{-}}\stackrel{}{\to }\cancel{{\text{O}}_{\left(g\right)}^{{}^{2-}}}{\text{ΔH}}_5^{°}\text{ = }\left(\text{+844 kJ/mol}\right)\\ \cancel{C{a}^{2+}{}_{\text{(g)}}}+\cancel{{\text{O}}_{\left(g\right)}^{{}^{2-}}}\stackrel{}{\to }CaO\left(s\right){\text{ΔH}}_6^{°}\text{ =}\text{?}\end{array}}\\ {\text{Ca}}_{\text{(s)}}\text{ + }\frac{1}{2}{O}_{2\left(l\right)}\stackrel{}{\to }{\text{ CaO}}_{\left(s\right)}{\text{Δ}}_f{\text{H}}^{\circ }\text{ = -635}\text{kJ/mol}\end{array}}$

The enthalpy formation of $CaO$ is shown above, from that the lattice enthalpy $\left({\text{Δ}}_6{\text{H}}^{\circ }{}_{\text{LE}}\right)$ obtained as follows,

$\begin{array}{l}{\text{Δ}}_f{\text{H}}^{\circ }={\text{ΔH}}_{\text{1}}^{\text{°}}+{\text{ΔH}}_{\text{2}}^{\text{°}}+{\text{ΔH}}_3^{°}+{\text{ΔH}}_4^{°}+{\text{ΔH}}_5^{°}+{\text{ΔH}}_6^{°}\\ \\ \text{-}635\text{kJ/mol}=\left(\text{+ 178 kJ/mol}\right)+\left(\text{+1735 kJ/mol}\right)+\left(\text{+249 kJ/mol}\right)+\left(\text{-141 kJ/mol}\right)+\left(\text{ +844 kJ/mol}\right)+{\text{ΔH}}_6^{°}\\ \\ {\text{ΔH}}_6^{°}=-3500kJ/mol\end{array}$

Hence, the lattice enthalpy of $CaO$ is $-3500kJ/mol$