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//mol`
B. `-41.8 kJ//mol`
C. `72.8 kJ//mol`
D. `-72.8 kJ//mol`

Explanation

To find the enthalpy of vaporization of iodine (I₂), we can 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.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- **Enthalpy of Sublimation (ΔH_sub)**: The energy required to convert one mole of solid directly into gas. For I₂, this is given as 57.3 kJ/mol.
- **Enthalpy of Fusion (ΔH_fus)**: The energy required to convert one mole of solid into liquid. For I₂, this is given as 15.5 kJ/mol.
- **Enthalpy of Vaporization (ΔH_vap)**: The energy required to convert one mole of liquid into gas. This is what we need to find.

2. **Write the Hess's Law Equation**:
According to Hess's law, the enthalpy of sublimation can be expressed as the sum of the enthalpy of fusion and the enthalpy of vaporization:
\[ ΔH_{sub} = ΔH_{fus} + ΔH_{vap} \]

3. **Substitute the Known Values**:
We can rearrange the equation to solve for the enthalpy of vaporization:
\[ ΔH_{vap} = ΔH_{sub} - ΔH_{fus} \]
Substituting the known values:
\[ ΔH_{vap} = 57.3 \, \text{kJ/mol} - 15.5 \, \text{kJ/mol} \]

4. **Calculate the Enthalpy of Vaporization**:
\[ ΔH_{vap} = 57.3 - 15.5 = 41.8 \, \text{kJ/mol} \]

5. **Conclusion**:
The enthalpy of vaporization of iodine (I₂) is **41.8 kJ/mol**.