Question

How do I calculate the cohesive energy of sodium chloride (NaCl) given the ionic radii of Na+ and Cl−, assuming the Born exponent n is 9 and a Madelung constant of 1.8?

Answer 1

To calculate the cohesive energy of sodium chloride (NaCl), we can use the Born-Lande equation, which provides a way to determine the lattice energy of an ionic solid. The expression is given by:

\[
U = -\frac{N_A \cdot M \cdot z_+^2 \cdot z_-^2 \cdot e^2}{4 \pi \epsilon_0 \cdot r_0} \cdot \left(1 - \frac{1}{n}\right)
\]

Where:

- \( U \) is the lattice energy (or cohesive energy) in joules per mole.
- \( N_A \) is Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\)).
- \( M \) is the Madelung constant (for NaCl, \(M = 1.8\)).
- \( z_+ \) and \( z_- \) are the charges of the cation and anion, respectively (for NaCl, \( z_+ = +1\) and \( z_- = -1\)).
- \( e \) is the elementary charge (\(1.602 \times 10^{-19} \, \text{C}\)).
- \( \epsilon_0 \) is the permittivity of free space (\(8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)\)).
- \( r_0 \) is the distance between the ions in meters, which can be approximated as the sum of the ionic radii of Na+ and Cl−.
- \( n \) is the Born exponent (given as 9).

### Step 1: Calculate \( r_0 \)
Let's assume the ionic radii are provided. For typical ionic radii, we have:

- \( r_{\text{Na}^+} \approx 1.02 \, \text{Å} = 1.02 \times 10^{-10} \, \text{m} \)
- \( r_{\text{Cl}^-} \approx 1.81 \, \text{Å} = 1.81 \times 10^{-10} \, \text{m} \)

Adding these together gives:

\[
r_0 = r_{\text{Na}^+} + r_{\text{Cl}^-} = (1.02 + 1.81) \times 10^{-10} \, \text{m} = 2.83 \times 10^{-10} \, \text{m}
\]

### Step 2: Substitute values into the Born-Lande equation

Now substituting the values into the Born-Lande equation:

\[
U = -\frac{(6.022 \times 10^{23}) \cdot (1.8) \cdot (1^2) \cdot (1^2) \cdot (1.602 \times 10^{-19})^2}{4 \pi (8.854 \times 10^{-12}) (2.83 \times 10^{-10})} \cdot \left(1 - \frac{1}{9}\right)
\]

### Step 3: Calculate

Let's calculate each component:
- First, compute the elementary charge squared:
\[
(1.602 \times 10^{-19})^2 \approx 2.566 \times 10^{-38} \text{ C}^2
\]

- Now, compute the denominator:
\[
4 \pi (8.854 \times 10^{-12}) \cdot (2.83 \times 10^{-10}) \approx 3.147 \times 10^{-21} \text{ C}^2/\text{N} \cdot \text{m}^2
\]

- Now substituting everything:
\[
U \approx -\frac{(6.022 \times 10^{23}) \cdot (1.8) \cdot (2.566 \times 10^{-38})}{3.147 \times 10^{-21}} \cdot \left( \frac{8}{9} \right)
\]

Calculating:
\[
U \approx -\frac{(1.557 \times 10^6)}{3.147 \times 10^{-21}} \cdot \left(\frac{8}{9}\right)
\]

Calculating the lattice energy, we find:
\[
U \approx -4.96 \times 10^7 \text{ J/mol} \cdot \left(\frac{8}{9}\right) \approx -4.43 \times 10^7 \text{ J/mol}
\]

### Conclusion:
The cohesive energy of sodium chloride (NaCl) is approximately \(-4.43 \times 10^7 \, \text{J/mol}\) (or in other units, approximately \(-443 \, \text{kJ/mol}\)).