Why is the enthalpy of sublimation equal to the sum of enthalpy of fusion and enthalpy of vaporisation ?

Explanation

To understand why the enthalpy of sublimation is equal to the sum of the enthalpy of fusion and the enthalpy of vaporization, we can break down the processes involved step by step.

### Step-by-Step Solution:

1. **Definition of Sublimation**:
- Sublimation is the process where a solid changes directly into a gas without passing through the liquid phase. For example, dry ice (solid CO₂) sublimates to carbon dioxide gas.

2. **Enthalpy Changes**:
- The enthalpy change associated with sublimation is denoted as \( \Delta H_s \).
- The enthalpy change for the melting of a solid into a liquid is called the enthalpy of fusion, denoted as \( \Delta H_f \).
- The enthalpy change for the vaporization of a liquid into a gas is called the enthalpy of vaporization, denoted as \( \Delta H_v \).

3. **Process Breakdown**:
- To understand the relationship, we can visualize the process in two steps:
- First, the solid is converted to a liquid (melting), which requires energy equal to \( \Delta H_f \).
- Second, the liquid is converted to a gas (vaporization), which requires energy equal to \( \Delta H_v \).

4. **Applying Hess's Law**:
- According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction, regardless of the pathway taken.
- Therefore, the total enthalpy change for the sublimation process can be expressed as:
\[ \Delta H_s = \Delta H_f + \Delta H_v \]

5. **Conclusion**:
- Thus, the enthalpy of sublimation (\( \Delta H_s \)) is equal to the sum of the enthalpy of fusion (\( \Delta H_f \)) and the enthalpy of vaporization (\( \Delta H_v \)):
\[ \Delta H_s = \Delta H_f + \Delta H_v \]