For which of the following reaction `DeltaH_("reaction")^(0)` of product.

Options

A. `2CO(g) + O_(2)(g) rarr 2CO_(2)`
B. `N_(2)(g) + O_(3)(g) rarr N_(2)O_(3)(g)`
C. `CH_(4)(g) + 2Cl_(2)(g) rarr CH_(2)Cl_(2)(l) + 2HCl(g)`
D. `Xe(g) + 2F_(2)(g) rarr XeF_(4)(g)`

Explanation

To determine for which reaction the standard enthalpy change of the reaction (ΔH₀_reaction) is equal to the standard enthalpy of the products, we need to analyze the standard states of the reactants and products involved in each option.

### Step-by-step Solution:

1. **Understanding Standard Enthalpy of Formation (ΔH₀_f)**:
- The standard enthalpy of formation (ΔH₀_f) of a substance is defined as the change in enthalpy when one mole of a compound is formed from its elements in their standard states.
- For elements in their standard states, ΔH₀_f = 0.

2. **Analyzing Each Option**:
- **Option A**: 2 CO(g) → CO₂(g)
- CO (carbon monoxide) is not in its standard state (which is C(s) + 1/2 O₂(g)). Therefore, ΔH₀_f for CO is not zero. Thus, ΔH₀_reaction ≠ ΔH₀_product.

- **Option B**: N₂(g) + O₃(g) → NO(g)
- N₂ is in its standard state (ΔH₀_f = 0), but O₃ (ozone) is not in its standard state (which is O₂(g)). Therefore, ΔH₀_f for O₃ is not zero. Thus, ΔH₀_reaction ≠ ΔH₀_product.

- **Option C**: CH₄(g) → C(s) + 2 H₂(g)
- CH₄ is not in its standard state (which is C(s) + 2 H₂(g)). Therefore, ΔH₀_f for CH₄ is not zero. Thus, ΔH₀_reaction ≠ ΔH₀_product.

- **Option D**: Xe(g) + F₂(g) → XeF₂(g)
- Both Xe (xenon) and F₂ (fluorine) are in their standard states (ΔH₀_f = 0). Therefore, the ΔH₀_reaction will equal the ΔH₀_product since both reactants have ΔH₀_f = 0.

3. **Conclusion**:
- The only reaction where ΔH₀_reaction equals the standard enthalpy of the products is **Option D**: Xe(g) + F₂(g) → XeF₂(g).

### Final Answer:
**Option D** is the correct answer.