Calculate the lattice energy from the following data (given `1 eV = 23.0 kcal mol^(-1)`)
i. `Delta_(f) H^(ɵ) (KI) = -78.0 kcal mol^(-1)`
ii. `IE_(1)` of `K = 4.0 eV`
iii. `Delta_("diss")H^(ɵ)(I_(2)) = 28.0 kcal mol^(-1)`
iv. `Delta_("sub")H^(ɵ)(K) = 20.0 kcal mol^(-1)`
v. `EA` of `I = -70.0 kcal mol^(-1)`
vi. `Delta_("sub")H^(ɵ)` of `I_(2) = 14.0 kcal mol^(-1)`

Options

A. `+14.1 kcal mol^(-1)`
B. `-14.1 kcal mol^(-1)`
C. `-141 kcal mol^(-1)`
D. `+141 kcal mol^(-1)`

Explanation

To calculate the lattice energy from the provided data, we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We will break down the formation of KI into several steps and calculate the lattice energy accordingly.

### Step-by-Step Solution:

1. **Write the Formation Reaction:**
The formation of KI can be represented as:
\[ \text{I}_2(s) + 2\text{K}(s) \rightarrow 2\text{KI}(s) \]
The enthalpy of formation (\(\Delta_f H^\circ\)) for KI is given as:
\[ \Delta_f H^\circ(\text{KI}) = -78.0 \text{ kcal mol}^{-1} \]

2. **Sublimation of Potassium:**
To convert solid potassium to gaseous potassium, we use the sublimation enthalpy:
\[ \text{K}(s) \rightarrow \text{K}(g) \quad \Delta H_1 = 20.0 \text{ kcal mol}^{-1} \]
Since we have 2 moles of K, the total enthalpy change for this step is:
\[ \Delta H_1 = 2 \times 20.0 = 40.0 \text{ kcal mol}^{-1} \]

3. **Ionization Energy of Potassium:**
The ionization energy (IE) of potassium is given as \(4.0 \text{ eV}\). We convert this to kcal/mol:
\[ \Delta H_2 = 4.0 \text{ eV} \times 23.0 \text{ kcal mol}^{-1} = 92.0 \text{ kcal mol}^{-1} \]

4. **Sublimation of Iodine:**
The sublimation of iodine involves converting solid iodine to gaseous iodine. The enthalpy of sublimation for \(I_2\) is given as:
\[ \Delta H_3 = \frac{1}{2} \times 14.0 \text{ kcal mol}^{-1} = 7.0 \text{ kcal mol}^{-1} \]

5. **Dissociation of Iodine Molecule:**
To dissociate \(I_2\) into two iodine atoms, we use the dissociation enthalpy:
\[ \Delta H_4 = \frac{1}{2} \times 28.0 \text{ kcal mol}^{-1} = 14.0 \text{ kcal mol}^{-1} \]

6. **Electron Affinity of Iodine:**
The electron affinity (EA) of iodine is given as:
\[ \Delta H_5 = -70.0 \text{ kcal mol}^{-1} \]

7. **Combine All Steps:**
The overall reaction can be summed up as follows:
\[ \Delta_f H^\circ(\text{KI}) = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 + \Delta H_5 + \Delta H_{lattice} \]
Rearranging gives:
\[ \Delta H_{lattice} = \Delta_f H^\circ(\text{KI}) - (\Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 + \Delta H_5) \]

8. **Substituting the Values:**
Now we substitute the values:
\[ \Delta H_{lattice} = -78.0 - (40.0 + 92.0 + 7.0 + 14.0 - 70.0) \]
\[ = -78.0 - (40.0 + 92.0 + 7.0 + 14.0 - 70.0) \]
\[ = -78.0 - (40.0 + 92.0 + 7.0 + 14.0 - 70.0) = -78.0 - 83.0 = -161.0 \text{ kcal mol}^{-1} \]

### Final Calculation:
Thus, the lattice energy is:
\[ \Delta H_{lattice} = -161.0 \text{ kcal mol}^{-1} \]