Caesium chlorides is formed according to the following equation `Cs(s)+0.5Cl_(2)(g)rarrCaCl(s)`. The enthalpy of sublimation of Cs, enthalpy of dissocitation of chlorine, ionization energy of Cs and electron affinity of chlorine are 81.2,243.0,375.7 and -348.3 kJ`mol^(-1)`. The energy change involved in the formation of CsCl is -388.6 kJ `mol^(-)`. Calculate the lattice energy of CsCl.

Options

A. 618.7 kJ `mol^(-1)`
B. 1237.4 kJ `mol^(-1)`
C. `1237.4 kJ mol^(-1)`
D. `-532.7 KJ mol^(-1)`

Explanation

To calculate the lattice energy of CsCl, we will use Hess's law and the given thermodynamic data. The formation of CsCl can be broken down into several steps, and we can express the enthalpy change for the formation of CsCl in terms of the individual steps.

### Step-by-Step Solution:

1. **Write the Formation Reaction:**
The formation of cesium chloride (CsCl) from its elements can be represented as:
\[ Cs(s) + 0.5Cl_2(g) \rightarrow CsCl(s) \]

2. **Identify the Given Data:**
- Enthalpy of sublimation of Cs (\( \Delta H_{sublimation} \)): 81.2 kJ/mol
- Enthalpy of dissociation of chlorine (\( \Delta H_{dissociation} \)): 243.0 kJ/mol (for 1 mole of Cl2, hence for 0.5 mole: \( \frac{243.0}{2} = 121.5 \) kJ/mol)
- Ionization energy of Cs (\( \Delta H_{ionization} \)): 375.7 kJ/mol
- Electron affinity of Cl (\( \Delta H_{electron\ affinity} \)): -348.3 kJ/mol
- Enthalpy change for the formation of CsCl (\( \Delta H_{formation} \)): -388.6 kJ/mol

3. **Set Up the Enthalpy Change Equation:**
According to Hess's law, the total enthalpy change for the formation of CsCl can be expressed as:
\[ \Delta H_{formation} = \Delta H_{sublimation} + \Delta H_{ionization} + \Delta H_{dissociation} + \Delta H_{electron\ affinity} + \Delta H_{lattice} \]

4. **Substitute the Known Values:**
Plugging in the values we have:
\[ -388.6 = 81.2 + 375.7 + 121.5 - 348.3 + \Delta H_{lattice} \]

5. **Calculate the Lattice Energy:**
Simplifying the right side:
\[ -388.6 = 81.2 + 375.7 + 121.5 - 348.3 + \Delta H_{lattice} \]
\[ -388.6 = 81.2 + 375.7 + 121.5 - 348.3 \]
\[ -388.6 = 229.1 + \Delta H_{lattice} \]
Now, isolate \( \Delta H_{lattice} \):
\[ \Delta H_{lattice} = -388.6 - 229.1 \]
\[ \Delta H_{lattice} = -617.7 \text{ kJ/mol} \]

6. **Final Result:**
The lattice energy of CsCl is:
\[ \Delta H_{lattice} = 617.7 \text{ kJ/mol (as a positive value)} \]