Chapter 9, Problem 9.118QP

Calculate the lattice energy of potassium fluoride, KF, using the Born–Haber cycle. Use thermodynamic data from Appendix C to obtain the enthalpy changes for each step. (Note: You will obtain a slightly different answer if you use values given in Chapter 8 for the ionization energy and electron affinity, which are energy values at 0 K rather than the enthalpy changes at 298 K.)

Interpretation Introduction

Interpretation:

The lattice energy of Potassium fluoride has to be calculated via the Born Haber cycle.

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.

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

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

Answer to Problem 9.118QP

The lattice energy of Potassium fluoride is $\text{996}\text{kJ/mol}$

Explanation of Solution

The above thermo chemical equations can be obtained by applying Hess’s law.

The lattice energy (last row) is obtained by the sum of the enthalpy of sublimation, dissociation, ionization, electron affinity.

${\text{K}}_{\text{(s)}}$	$\to$	${\text{K}}_{\text{(g)}}$	${\text{ΔH}}_{\text{sublimation}}\text{=}\text{+89kJ/mol}$
$\frac{\text{1}}{\text{2}}{\text{F}}_{\text{2(g)}}$	$\to$	${\text{F}}_{\text{(g)}}^{}$	${\text{ΔH}}_{\text{dissociation}}\text{=}\text{+}\text{79}\text{.39}\text{kJ/mol}$
${\text{K}}_{\text{(g)}}$	$\to$	${\text{K}}_{\text{(g)}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}$	${\text{ΔH}}_{\text{ionization}}\text{=}\text{+}\text{514}\text{kJ/mol}$
${\text{F}}_{\text{(g)}}^{}\text{+}{\text{e}}^{\text{-}}$	$\to$	${\text{F}}_{\text{(g)}}^{\text{-}}$	${\text{ΔH}}_{\text{EA}}\text{=}\text{-}\text{255}\text{.1}\text{kJ/mol}$
${\text{K}}_{\text{(g)}}^{\text{+}}{\text{+F}}_{\text{(g)}}^{\text{-}}$	$\to$	${\text{KF}}_{\text{(s)}}$	${\text{ΔH}}_{\text{lattice}\text{energy}}\text{=}\text{-}\text{U}\text{kJ/mol}$
${\text{K}}_{\text{(s)}}$+ $\frac{\text{1}}{\text{2}}{\text{F}}_{\text{2(g)}}$	$\to$	${\text{KF}}_{\text{(s)}}$	${\text{ΔH}}_{\text{f}}^{\text{o}}\text{=}\text{-}\text{568}\text{.6}\text{kJ/mol}$

$\begin{array}{l}\text{-}\text{568}\text{.6}\text{kJ/mol}\text{=}\text{(}\text{89}\text{+}\text{79}\text{.39+514}\text{.0-255}\text{.1-U)}\text{kJ/mol}\\ \text{U}\text{(kJ/mol)}\text{=}\text{(}\text{89}\text{+}\text{79}\text{.39+514}\text{.0-255}\text{.1+568}\text{.6)}\text{kJ/mol}\\ \text{=}\text{995}\text{.9}\text{=}\text{996}\text{kJ/mol}\end{array}$

The lattice energy of Potassium fluoride is $\text{996}\text{kJ/mol}$

Conclusion

The lattice energy of Potassium fluoride is found to be $\text{996}\text{kJ/mol}$