If enthalpy of fusion and enthalpy of vaporisation of sodium metal are 2.6 and `98.2 kJ "mol"^(-1)` respectively, what is the enthalpy of sublimation of sodium? (Keep your answer in three digits and KJ/mol)

Explanation

To find the enthalpy of sublimation of sodium, 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 of the reaction.

### Step-by-Step Solution:

1. **Understand the Processes**:
- Enthalpy of fusion (ΔH_fus) is the energy required to convert a solid into a liquid.
- Enthalpy of vaporization (ΔH_vap) is the energy required to convert a liquid into a gas.
- Enthalpy of sublimation (ΔH_sub) is the energy required to convert a solid directly into a gas.

2. **Write the Relationship**:
The enthalpy of sublimation can be expressed as the sum of the enthalpy of fusion and the enthalpy of vaporization:
\[ \Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap} \]

3. **Substitute the Given Values**:
We are given:
- ΔH_fus = 2.6 kJ/mol
- ΔH_vap = 98.2 kJ/mol

Now, substitute these values into the equation:
\[ \Delta H_{sub} = 2.6 \, \text{kJ/mol} + 98.2 \, \text{kJ/mol} \]

4. **Perform the Calculation**:
\[ \Delta H_{sub} = 2.6 + 98.2 = 100.8 \, \text{kJ/mol} \]

5. **Final Answer**:
The enthalpy of sublimation of sodium is:
\[ \Delta H_{sub} = 100.8 \, \text{kJ/mol} \]