The lattice enthalpy of KI will be, if the enthalpy of
(I) `Delta_f` `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 KI, we will use the given data and apply Hess's law. The lattice enthalpy can be derived from the enthalpy of formation and the various energy changes involved in forming KI from its elements in their standard states.

### Step-by-Step Solution:

1. **Identify Given Data**:
- Enthalpy of formation of KI, \( \Delta_f H^- \) (KI) = -78.0 kcal/mol
- Ionization energy of K to \( K^+ \) = 4.0 eV
- Dissociation energy of \( I_2 \) = 28.0 kcal/mol
- Sublimation energy of K = 20.0 kcal/mol
- Electron gain enthalpy for I to \( I^- \) = -70.0 kcal/mol
- Sublimation energy of \( I_2 \) = 14.0 kcal/mol (for half of \( I_2 \))

2. **Convert Ionization Energy to kcal/mol**:
\[ \text{Ionization energy (K)} = 4.0 \, \text{eV} \times 23.0 \, \text{kcal/eV} = 92.0 \, \text{kcal/mol} \]

3. **Calculate the Energy Changes**:
- **Sublimation of K**: +20.0 kcal/mol
- **Ionization of K to \( K^+ \)**: +92.0 kcal/mol
- **Sublimation of \( I_2 \)** (for half of \( I_2 \)):
\[ \text{Sublimation energy for } \frac{1}{2} I_2 = \frac{14.0 \, \text{kcal/mol}}{2} = 7.0 \, \text{kcal/mol} \]
- **Dissociation of \( I_2 \)** (for half of \( I_2 \)):
\[ \text{Dissociation energy for } \frac{1}{2} I_2 = \frac{28.0 \, \text{kcal/mol}}{2} = 14.0 \, \text{kcal/mol} \]
- **Electron gain enthalpy for I to \( I^- \)**: -70.0 kcal/mol

4. **Set Up the Equation Using Hess's Law**:
The formation of KI can be represented as:
\[ \Delta_f H^- \text{(KI)} = \text{Sublimation of K} + \text{Ionization of K} + \text{Sublimation of } \frac{1}{2} I_2 + \text{Dissociation of } \frac{1}{2} I_2 + \text{Electron gain enthalpy} + \text{Lattice Energy} \]
Rearranging gives:
\[ \text{Lattice Energy} = \Delta_f H^- \text{(KI)} - (\text{Sublimation of K} + \text{Ionization of K} + \text{Sublimation of } \frac{1}{2} I_2 + \text{Dissociation of } \frac{1}{2} I_2 + \text{Electron gain enthalpy}) \]

5. **Substituting Values**:
\[ \text{Lattice Energy} = -78.0 \, \text{kcal/mol} - (20.0 + 92.0 + 7.0 + 14.0 - 70.0) \]
\[ = -78.0 \, \text{kcal/mol} - (20.0 + 92.0 + 7.0 + 14.0 - 70.0) \]
\[ = -78.0 \, \text{kcal/mol} - 63.0 \, \text{kcal/mol} \]
\[ = -141.0 \, \text{kcal/mol} \]

6. **Final Calculation**:
Therefore, the lattice enthalpy of KI is:
\[ \text{Lattice Energy} = -141.0 \, \text{kcal/mol} \]

### Summary:
The lattice enthalpy of KI is approximately -141.0 kcal/mol.