The sublimation energy of `I_(2)` (solid) is 57.3 KJ/mole and enthalpy of fusion is 15.5 KJ/mole. The enthalpy of vapourisation of `I_(2)` is

Options

A. 41.8 KJ/mole
B. `-41.8 KJ//"mole"`
C. `72.8 KJ//"mole"`
D. `-72.8 KJ//"mole"`

Explanation

To find the enthalpy of vaporization of iodine (I₂), we can use the given sublimation energy and enthalpy of fusion. Here’s a step-by-step solution:

### Step 1: Understand the Processes
1. **Sublimation**: This is the process where solid iodine (I₂(s)) turns directly into gaseous iodine (I₂(g)). The enthalpy change for this process is given as:
\[ \Delta H_{\text{sublimation}} = 57.3 \, \text{kJ/mol} \]
This can be represented as:
\[ I_2(s) \rightarrow I_2(g) \quad \Delta H = 57.3 \, \text{kJ/mol} \]

2. **Fusion**: This is the process where solid iodine (I₂(s)) turns into liquid iodine (I₂(l)). The enthalpy change for this process is given as:
\[ \Delta H_{\text{fusion}} = 15.5 \, \text{kJ/mol} \]
This can be represented as:
\[ I_2(s) \rightarrow I_2(l) \quad \Delta H = 15.5 \, \text{kJ/mol} \]

### Step 2: Write the Vaporization Reaction
3. **Vaporization**: This is the process where liquid iodine (I₂(l)) turns into gaseous iodine (I₂(g)). We need to find the enthalpy change for this process, represented as:
\[ I_2(l) \rightarrow I_2(g) \quad \Delta H = \Delta H_{\text{vaporization}} \]

### Step 3: Relate the Processes
4. To find the enthalpy of vaporization, we can relate the three processes:
- Starting from solid iodine, we can first convert it to liquid (fusion), and then from liquid to gas (vaporization).
- The sublimation process can be thought of as going directly from solid to gas.

### Step 4: Set Up the Equation
5. The relationship can be expressed as:
\[ \Delta H_{\text{sublimation}} = \Delta H_{\text{fusion}} + \Delta H_{\text{vaporization}} \]
Rearranging this gives:
\[ \Delta H_{\text{vaporization}} = \Delta H_{\text{sublimation}} - \Delta H_{\text{fusion}} \]

### Step 5: Substitute the Values
6. Now, substitute the known values into the equation:
\[ \Delta H_{\text{vaporization}} = 57.3 \, \text{kJ/mol} - 15.5 \, \text{kJ/mol} \]

### Step 6: Calculate the Result
7. Perform the calculation:
\[ \Delta H_{\text{vaporization}} = 41.8 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of vaporization of iodine (I₂) is:
\[ \Delta H_{\text{vaporization}} = 41.8 \, \text{kJ/mol} \]