Question

Repeat the calculations of problem 41 for crystalline KBr, which has the rock-salt structure with a K-Br separation of$\mathbf{3}\mathbf{.298}\stackrel{\mathbf{°}}{\mathbf{\text{A}}}$ .

Short answer

a.$689{\text{kJmol}}^{-1}$

b.$736.2{\text{kJmol}}^{-1}$

Step-by-step solution

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

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 KBr, which is for the reaction is shown below.

$\text{KBr(s)}\to {\text{K}}^{+}\left(g\right)+{\text{Br}}^{-}\left(g\right)\Delta \text{U=?}$

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

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

$\text{KBr(s)}\to \text{K}\left(s\right)+\frac{1}{2}{\text{Br}}_2\left(g\right)$

$\begin{array}{c}\Delta {\text{U}}_1\approx \Delta \text{H=-}\Delta {\text{H}}_{\text{f}}^{\text{o}}\left(\text{KBr(s)}\right)\\ =-(-393.80{\text{kJmol}}^{-1})\\ =+393.80{\text{kJmol}}^{-1}\end{array}$

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

$$\begin{gathered} \begin{array}{l}\text{K(s)}\to K\left(g\right)\Delta \text{U}\approx \Delta \text{H=}\Delta {\text{H}}_{\text{f}}^{\text{o}}\left(\text{K(g)}\right)=+89.24{\text{kJmol}}^{-1}\\ \frac{1}{2}{\text{Br}}_2\left(g\right)\to \text{Br}\left(g\right)\Delta \text{U}\approx \Delta \text{H=}\Delta {\text{H}}_{\text{f}}^{\text{o}}\left(\text{Br(g)}\right)=+111.88{\text{kJmol}}^{-1}\end{array} \\ ------------------------------------ \\ \text{K(s)+}\frac{1}{2}{\text{Br}}_2\to \text{K}\left(g\right)+\text{Br}\left(g\right)\Delta {\text{U}}_2=+201.12\text{kJ} \end{gathered}$$

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

$$\begin{gathered} \begin{array}{l}\text{K(g)}\to {\text{K}}^{+}\left(g\right)+e^{-}\Delta {\text{U=IE}}_1\text{(K)=}418.8\\ \text{Br}\left(g\right)+e^{-}\to {\text{Br}}^{-}\left(g\right)\Delta \text{U=-EA(Br)=-}324.7\end{array} \\ -------------------------------- \\ \text{K(g)+Br}\left(g\right)\to {\text{K}}^{+}{\text{(g)+Br}}^{-}\text{(g)}\Delta {\text{U}}_3\text{=94.1kJ} \end{gathered}$$

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

Therefore, the total energy changeis:

$\begin{array}{c}\Delta \text{U=}\Delta {\text{U}}_1+\Delta {\text{U}}_2+\Delta {\text{U}}_3\\ {\text{=+393.80kJmol}}^{-1}+201.12\text{kJ+94.1kJ}\\ =689{\text{kJmol}}^{-1}\end{array}$

Subpart (b) The Coulomb energy is calculated using a KBr 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 $3.298\stackrel{\text{o}}{\text{A}}=3.298\times {10}^{-10}\text{m}$.

Putting the values in the equation;

$\begin{array}{c}\text{latticeenergy=}\frac{(6.02\times {10}^{23}mo{l}^{-1}){(1.602\times {10}^{-19})}^2\left(1.7476\right)}{4\left(3.14\right)(8.854\times {10}^{-12}{\text{C}}^2{\text{J}}^{-1}{\text{m}}^{-1})(3.298\times {10}^{-10}\text{m})}\\ =7.362\times {10}^5{\text{Jmol}}^{-1}\\ =736.2{\text{kJmol}}^{-1}\end{array}$