Question 10.11: Using data from Figures 4.8 and 4.10 and Appendix 2, calcula...
Using data from Figures 4.8 and 4.10 and Appendix 2, calculate the lattice energy of cesium chloride (\text{CsCl}).
Strategy Using Figure 10.14 as a guide, combine the pertinent thermodynamic data and use Hess’s law to calculate the lattice energy.
Setup From Figure 4.8, IE_1(\text{Cs}) = 376 \text{kJ/mol}. From Figure 4.10, EA_1(\text{Cl}) = 349.0 \text{kJ/mol}. From Appendix 2, ∆H^\circ_\text{f} [\text{Cs(g)}] = 76.50 \text{kJ/mol}, ∆H^\circ_\text{f} [\text{Cl(g)}] = 121.7 \text{kJ/mol}, ∆H^\circ_\text{f} [\text{CsCl}(s)] = –422.8 \text{kJ/mol}. Because we are interested in magnitudes only, we can use the absolute values of the thermodynamic data. And, because only the standard heat of formation of \text{CsCl}(s) is a negative number, it is the only one for which the sign changes.
\left\{∆H^\circ_\text{f}[\text{Cs(g)}] + ∆H^\circ_\text{f}[\text{Cl(g)}] + IE_1(\text{Cs}) + \left|∆H^\circ_\text{f}\text{[CsCl}(s)]\right|\right\} – EA_1(\text{Cl}) = \text{lattice energy}
= (76.50 \text{kJ/mol} + 121.7 \text{kJ/mol} + 376 \text{kJ/mol} + 422.8 \text{kJ/mol}) – 349.0 \text{kJ/mol}
= 648 \text{kJ/mol}