Question

(a) Use the Born–Haber cycle, with data from Appendices D and F, to calculate the lattice energy of LiF.

(b) Compare the result of part (a) with the Coulomb energy calculated by using an Li-F separation of $\mathbf{2}\mathbf{.014}\stackrel{\mathbf{°}}{\mathbf{\text{A}}}$in the LiF crystal, which has the rock-salt structure.

Short answer

1. ${\text{1046.3kJmol}}^{\text{-1}}$
2. ${\text{1205.5kJmol}}^{\text{-1}}$
   

Step-by-step solution

Subpart (a) The Born–Haber cycle to calculate the lattice energy of LiF.

A thermodynamic cycle that was created by Max Born and Fritz Haber called the Born–Haber cycle measures the Ionic lattice energies experimentally.

It is an application of Hess’s law which again is the first law of thermodynamics.

Neglecting the small differences present between $\Delta \text{U}$and$\Delta H$; the determination of the lattice energy of LiF, which$\Delta \text{U}$is for the reaction is shown below.

$\text{LiF(s)}\to {\text{Li}}^{+}\left(g\right)+{\text{F}}^{-}\left(g\right)\Delta \text{U=?}$

A series of steps with different enthalpy changes represents the reaction.

Here, in the first step, the ionic solid is transformed to its elements in their standard states as shown below.

$\text{LiF(s)}\to \text{Li}\left(s\right)+\frac{1}{2}{\text{F}}_2\left(g\right)$

$\begin{array}{c}\Delta {\text{U}}_1\approx \Delta \text{H=-}\Delta {\text{H}}_{\text{f}}^{\text{o}}\left(\text{LiF(s)}\right)\\ =-(-615.97{\text{kJmol}}^{-1})\\ =+615.97{\text{kJmol}}^{-1}\end{array}$

Now, in the second step, the elements are changed into their gas-phase atoms:

$$\begin{gathered} \begin{array}{l}\text{Li(s)}\to \text{Li}\left(g\right)\Delta \text{U}\approx \Delta \text{H=}\Delta {\text{H}}_{\text{f}}^{\text{o}}\left(\text{Li(g)}\right)=+159.37{\text{kJmol}}^{-1}\\ \frac{1}{2}{\text{F}}_2\left(g\right)\to \text{F}\left(g\right)\Delta \text{U}\approx \Delta \text{H=}\Delta {\text{H}}_{\text{f}}^{\text{o}}\left(\text{F(g)}\right)=+78.99{\text{kJmol}}^{-1}\end{array} \\ ---------------------------------- \\ \text{Li(s)+}\frac{1}{2}{\text{F}}_2\to \text{Li}\left(g\right)+\text{F}\left(g\right)\Delta {\text{U}}_2=+238.36\text{kJ} \end{gathered}$$

In the end, i.e., in the third step, electrons are shifted from the lithium atoms to the fluorine atoms to give ions:

$$\begin{gathered} \begin{array}{l}\text{Li(g)}\to {\text{Li}}^{+}\left(g\right)+e^{-}\Delta {\text{U=IE}}_1\text{(Li)=520.2}\\ \text{F}\left(g\right)+e^{-}\to {\text{F}}^{-}\left(g\right)\Delta \text{U=-EA(F)=-328.0}\end{array} \\ -------------------------------- \\ \text{Li(g)+F}\left(g\right)\to {\text{Li}}^{+}{\text{(g)+F}}^{-}\text{(g)}\Delta {\text{U}}_3\text{=192kJ} \end{gathered}$$

Here EA(F) is the electron affinity of Fluorine and${\text{IE}}_1\text{(Li)}$ is the first ionization energy of the Lithium atom.

Therefore, the total energy change is:

$\begin{array}{c}\Delta \text{U=}\Delta {\text{U}}_1+\Delta {\text{U}}_2+\Delta {\text{U}}_3\\ \text{=+}615.97{\text{kJmol}}^{-1}+238.36\text{kJ+192kJ}\\ =1046.33{\text{kJmol}}^{-1}\end{array}$

Subpart (b) The Coulomb energy is calculated using a Li-F given separation value.

Using the Madelung constant of 1.7476for the structure of rock salt.

$\text{latticeenergy=}\frac{{\text{N}}_{\text{A}}{e}^2\text{M}}{4\pi {\epsilon }_0{R}_0}$

Here, the Li-F separation is $2.014\stackrel{\text{o}}{\text{A}}=2.014\times {10}^{-10}\text{m}$.

Putting the values in the equation;

$\begin{array}{c}\text{latticeenergy=}\frac{(6.02\times {10}^{23}mo{l}^{-1}){(1.602\times {10}^{-19})}^2\left(1.7476\right)}{4\left(3.14\right)(8.854\times {10}^{-12}{\text{C}}^2{\text{J}}^{-1}{\text{m}}^{-1})(2.014\times {10}^{-10}\text{m})}\\ =12.055\times {10}^5{\text{Jmol}}^{-1}\\ =1205.5{\text{kJmol}}^{-1}\end{array}$