Chapter 8, Problem 20QP

Calculate the lattice energy of ${\text{CaCl}}_{\text{2}}$. Use data from Figures 7.8 and 7.10 and Appendix 2. (The second ionization energy of Ca, $I{E}_{\text{2}}\text{, is 1145 kJ/mol}\text{.}$)

Interpretation Introduction

Interpretation:

The lattice energy of ${\text{CaCl}}_2$ is to be calculated.

Concept introduction:

Lattice energy is defined as the amount of energy needed to convert a mole of a solid ionic compound to its respective gaseous ions.

The lattice energy of ${\text{CaCl}}_2$ is calculated by the expression as follows:

$\left\{\Delta H_f{}^{\circ }\left[\text{Ca}\left(g\right)\right]+2\Delta H_f{}^{\circ }\left[\text{Cl}\left(g\right)\right]+{\text{IE}}_{\text{1}}\left(\text{Ca}\right)+{\text{IE}}_{\text{2}}\left(\text{Ca}\right)+\left|\Delta H_f{}^{\circ }\left[{\text{CaCl}}_2\left(s\right)\right]\right|\right\}-2\text{EA}\left(\text{Cl}\right)$

Here, $\Delta H_f{}^{\circ }\left[\text{Ca}\left(g\right)\right]$ is the standard enthalpy of the formation of Ca, $\Delta H_f{}^{\circ }\left[\text{Cl}\left(g\right)\right]$ is the standard enthalpy of the formation of Cl, ${\text{IE}}_{\text{1}}\left(\text{Ca}\right)$ is the ionization enthalpy of Ca, ${\text{IE}}_2\left(\text{Ca}\right)$ is the second ionization enthalpy of Ca, $\text{EA}\left(\text{Cl}\right)$ is the electron affinity of Cl, and $\Delta H_f{}^{\circ }\left[{\text{CaCl}}_2\left(s\right)\right]$ is the standard enthalpy of the formation of ${\text{CaCl}}_2$.

Answer to Problem 20QP

Solution: $2255\text{ kJ/mol}$.

Explanation of Solution

Given information: The value of $\Delta H_f{}^{\circ }\left[\text{Ca}\left(g\right)\right]$ from Appendix $2$ is $\text{179}\text{.3 kJ/mol}$.

The value of $\Delta H_f{}^{\circ }\left[\text{Cl}\left(g\right)\right]$ from Appendix $2$ is $\text{121}\text{.7 kJ/mol}$.

The value of ${\text{IE}}_{\text{1}}\left(\text{Ca}\right)$ from table $7.8$ is $\text{590 kJ/mol}$.

The value of ${\text{IE}}_{\text{2}}\left(\text{Ca}\right)$ is $\text{1145 kJ/mol}$.

The value of $\Delta H_f{}^{\circ }\left[{\text{CaCl}}_2\left(s\right)\right]$ from Appendix $2$ is $-794.96\text{ kJ/mol}$.

The value of $\text{EA}\left(\text{Cl}\right)$ from table $7.10$ is $\text{349 kJ/mol}$.

The lattice energy of ${\text{CaCl}}_2$ is calculated by the expression as follows:

$\text{Lattice energy}=\left\{\Delta H_f{}^{\circ }\left[\text{Ca}\left(g\right)\right]+2\Delta H_f{}^{\circ }\left[\text{Cl}\left(g\right)\right]+{\text{IE}}_{\text{1}}\left(\text{Ca}\right)+{\text{IE}}_{\text{2}}\left(\text{Ca}\right)+\left|\Delta H_f{}^{\circ }\left[{\text{CaCl}}_2\left(s\right)\right]\right|\right\}-2\text{EA}\left(\text{Cl}\right)$

Substitute the value $\text{179}\text{.3 kJ/mol}$ for $\Delta H_f{}^{\circ }\left[\text{Ca}\left(g\right)\right]$, $\text{121}\text{.7 kJ/mol}$ for $\Delta H_f{}^{\circ }\left[\text{Cl}\left(g\right)\right]$, $\text{590 kJ/mol}$ for ${\text{IE}}_{\text{1}}\left(\text{Ca}\right)$, $\text{1145 kJ/mol}$ for ${\text{IE}}_{\text{2}}\left(\text{Ca}\right)$, $-794.96\text{ kJ/mol}$ for $\Delta H_f{}^{\circ }\left[{\text{CaCl}}_2\left(s\right)\right]$, and $\text{349 kJ/mol}$ for $\text{EA}\left(\text{Cl}\right)$ in the above expression.

$\begin{array}{c}\text{Lattice energy}=\left(\text{179}\text{.3 kJ/mol}+2\times 121.7\text{ kJ/mol}+590\text{ kJ/mol}+1145\text{ kJ/mo+794}\text{.96}\text{kJ/mol}\right)-\left(\text{2}\times \text{349kJ/mol}\right)\\ =\left(\text{ 2952}\text{.66 kJ/mol}-698\text{ kJ/mol}\right)\\ =2254.66\text{ kJ/mol}\\ \cong 2255\text{ kJ/mol}\end{array}$.

Conclusion

The lattice energy of themolecule ${\text{CaCl}}_2$ is $2255\text{ kJ/mol}$.