For which one of the following equation is `DeltaH_(reaction)^(@)` equal to `DeltaH_(f)^(@)` for the product ?

Options

A. `N_(2)(g)+O_(3)(g)rarrN_(2)O_(3)(g)`
B. `CH_(4)(g)+2Cl_(2)(g)rarrCH_(2)Cl_(2)(l)+2HCl(g)`
C. `Xe(g)+2F_(2)(g)rarrXeF_(4)(g)`
D. `2CO(g)+O_(2)(g)rarr2CO_(2)(g)`

Explanation

To determine which equation has the reaction enthalpy change (ΔH_reaction^(@)) equal to the standard enthalpy of formation (ΔH_f^(@)) for the product, we need to analyze each option based on the definition of standard enthalpy of formation.

### Step-by-Step Solution:

1. **Understanding Standard Enthalpy of Formation:**
The standard enthalpy of formation (ΔH_f^(@)) of a compound is defined as the change in enthalpy when one mole of the compound is formed from its elements in their standard states. Therefore, for ΔH_reaction^(@) to equal ΔH_f^(@) for a product, the reaction must represent the formation of that product from its elements in their standard states.

2. **Analyzing Each Option:**
- **Option 1:**
\[ \text{N}_2 + \text{O}_3 \rightarrow \text{N}_2\text{O}_3 \]
- Here, N2 is in its elemental state (ΔH_f^(@) = 0), but O3 is not in its standard elemental state. Thus, this option is incorrect.

- **Option 2:**
\[ \text{C} + \text{H}_2 + \text{Cl}_2 \rightarrow \text{CH}_2\text{Cl}_2 \]
- The reactants are not all in their elemental forms. C is in its elemental state, but H2 and Cl2 are not. Therefore, this option is also incorrect.

- **Option 3:**
\[ \text{Xe} + \text{F}_2 \rightarrow \text{XeF}_4 \]
- Both xenon (Xe) and fluorine (F2) are in their elemental states. Thus, the ΔH_reaction^(@) for this equation is equal to the ΔH_f^(@) for XeF4. This option is correct.

- **Option 4:**
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \]
- Here, C is in its elemental state, but O2 is not in its standard state as it is diatomic. Thus, this option is incorrect.

3. **Conclusion:**
The correct option is **Option 3**, where ΔH_reaction^(@) is equal to ΔH_f^(@) for XeF4.

### Final Answer:
**Option 3: Xe + F2 → XeF4**