The lattice enthalpy of KI will be, if the enthalpy of (I) `Delta` `H^-` (KI) = -78.0 kcal `mol^-1` (II) Ionisation energy of K to `K^+` is 4.0 eV (III) Dissociation energy of `I_2` is 28.0 kcal `mol^-1` (IV) Sublimation energy of `K` is 20.0 kcal `mol^-1` (V) electron gain enthalpy for` I` to `I^-` is -70.0 kcal `mol^-1` (VI) Sublimation energy of `I_2` is 14.0 kcal `mol^-1` (1 eV =23. 0 kcal `mol^-1)

Options

A. `+14.1 kJ mol^-1 `
B. `-14.1 kJ mol^-1 `
C. `-141 kJ mol^-1 `
D. `+141 kJ mol^-1 `

Answer

To calculate the lattice enthalpy of potassium iodide (KI), we can use the Born-Haber cycle, which relates the lattice enthalpy to various other thermodynamic quantities. The formula we will use is:

\[ \Delta H_{\text{reaction}} = \Delta H_{\text{sublimation}}(K) + \Delta H_{\text{ionization}}(K) + \Delta H_{\text{dissociation}}(I_2) + \Delta H_{\text{electron gain}}(I) + \Delta H_{\text{lattice}} \]

Where:
- \(\Delta H_{\text{reaction}}\) is the enthalpy of formation of KI, which is given as -78.0 kcal/mol.
- \(\Delta H_{\text{sublimation}}(K)\) is the sublimation energy of potassium, given as 20.0 kcal/mol.
- \(\Delta H_{\text{ionization}}(K)\) is the ionization energy of potassium, which is given as 4.0 eV. We need to convert this to kcal/mol using the conversion factor \(1 \text{ eV} = 23.0 \text{ kcal/mol}\):
\[ \Delta H_{\text{ionization}}(K) = 4.0 \text{ eV} \times 23.0 \text{ kcal/eV} = 92.0 \text{ kcal/mol} \]
- \(\Delta H_{\text{dissociation}}(I_2)\) is the dissociation energy of iodine, given as 28.0 kcal/mol. Since we are dealing with I2, we need to divide this by 2 to get the energy for one mole of iodine atoms:
\[ \Delta H_{\text{dissociation}}(I) = \frac{28.0 \text{ kcal/mol}}{2} = 14.0 \text{ kcal/mol} \]
- \(\Delta H_{\text{electron gain}}(I)\) is the electron gain enthalpy for iodine, given as -70.0 kcal/mol.

Now, we can substitute these values into the equation:

\[ -78.0 \text{ kcal/mol} = 20.0 \text{ kcal/mol} + 92.0 \text{ kcal/mol} + 14.0 \text{ kcal/mol} - 70.0 \text{ kcal/mol} + \Delta H_{\text{lattice}} \]

Now, let's simplify the right side:

\[ -78.0 \text{ kcal/mol} = 20.0 + 92.0 + 14.0 - 70.0 + \Delta H_{\text{lattice}} \]
\[ -78.0 \text{ kcal/mol} = 56.0 + \Delta H_{\text{lattice}} \]

Now, we can isolate \(\Delta H_{\text{lattice}}\):

\[ \Delta H_{\text{lattice}} = -78.0 \text{ kcal/mol} - 56.0 \text{ kcal/mol} \]
\[ \Delta H_{\text{lattice}} = -134.0 \text{ kcal/mol} \]

Thus, the lattice enthalpy of KI is \(-134.0 \text{ kcal/mol}\).