If, `H_(2)(g)+Cl_(2)(g)rarr 2HCl(g) , Delta H^(@)=-44` Kcal
`2Na(s)+2HCl(g)rarr 2NaCl(s)+H_(2)(g), Delta H=-152` Kcal Then, `Na(s)+0.5Cl_(2)(g)rarr NaCl(s) , Delta H^(@) = ?`

Explanation

To find the enthalpy change for the reaction:

\[ \text{Na}(s) + 0.5 \text{Cl}_2(g) \rightarrow \text{NaCl}(s) \]

we will use the given reactions and their enthalpy changes.

### Step-by-Step Solution:

1. **Identify the Given Reactions:**
- Reaction 1:
\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g), \quad \Delta H = -44 \text{ kcal} \]
- Reaction 2:
\[ 2 \text{Na}(s) + 2 \text{HCl}(g) \rightarrow 2 \text{NaCl}(s) + \text{H}_2(g), \quad \Delta H = -152 \text{ kcal} \]

2. **Write the Enthalpy of Formation Concept:**
- The enthalpy change for the formation of a compound from its elements in their standard states is defined as the enthalpy of formation.
- For NaCl, the reaction can be represented as:
\[ \text{Na}(s) + 0.5 \text{Cl}_2(g) \rightarrow \text{NaCl}(s) \]

3. **Express the Enthalpy Change for the Reactions:**
- For Reaction 1, the enthalpy change can be expressed as:
\[ \Delta H_1 = 2 \times \Delta H_f(\text{HCl}) - \Delta H_f(\text{H}_2) - \Delta H_f(\text{Cl}_2) \]
Since the enthalpy of formation for elements in their standard state is zero:
\[ -44 = 2 \Delta H_f(\text{HCl}) \]
Thus,
\[ \Delta H_f(\text{HCl}) = -22 \text{ kcal} \]

4. **Using Reaction 2 to Find Enthalpy of Formation of NaCl:**
- For Reaction 2, we can write:
\[ \Delta H_2 = 2 \Delta H_f(\text{NaCl}) + \Delta H_f(\text{H}_2) - 2 \Delta H_f(\text{HCl}) - 2 \Delta H_f(\text{Na}) \]
Again, since the enthalpy of formation for elements in their standard state is zero:
\[ -152 = 2 \Delta H_f(\text{NaCl}) - 2(-22) \]
Simplifying this gives:
\[ -152 = 2 \Delta H_f(\text{NaCl}) + 44 \]
Rearranging:
\[ 2 \Delta H_f(\text{NaCl}) = -152 - 44 \]
\[ 2 \Delta H_f(\text{NaCl}) = -196 \]
Thus,
\[ \Delta H_f(\text{NaCl}) = -98 \text{ kcal} \]

5. **Conclusion:**
- Therefore, the enthalpy change for the reaction:
\[ \text{Na}(s) + 0.5 \text{Cl}_2(g) \rightarrow \text{NaCl}(s) \]
is:
\[ \Delta H = -98 \text{ kcal} \]

### Final Answer:
\[ \Delta H = -98 \text{ kcal} \]