Which of the following defines `DeltaH_(f)^@` ?

Options

A. `CO(g) + 1/O_(2)(g) rarr CO_(2)(g)`
B. `1/2 H_(2)(g) + 1/2 F_(2)(g) rarr HF (g)`
C. `N_(2)(g) + 2H_(2)(g) rarr 2NH_(3)(g)`
D. `C_("diamond") + O_(2)(g) rarr CO_(2)(g)`

Explanation

To determine which of the given reactions defines the standard enthalpy of formation (ΔH_f^0), we need to understand the criteria for ΔH_f^0:

1. **Definition of ΔH_f^0**: The standard enthalpy of formation (ΔH_f^0) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states at a specified temperature (usually 25°C or 298 K) and pressure (1 atm).

2. **Criteria for the Reaction**:
- The reaction must produce one mole of the compound.
- The reactants must be in their standard states.

Now, let's analyze the four reactions provided in the question:

### Step-by-Step Solution:

1. **Identify the Reactions**:
- Reaction A: CO + O2 → CO2
- Reaction B: 1/2 H2 + 1/2 F2 → HF
- Reaction C: N2 + 3H2 → 2NH3
- Reaction D: C (diamond) + O2 → CO2

2. **Analyze Each Reaction**:
- **Reaction A**: CO + O2 → CO2
- This reaction produces CO2, but it does not start from the elements in their standard states (C is not in graphite form). Therefore, it does not define ΔH_f^0.

- **Reaction B**: 1/2 H2 + 1/2 F2 → HF
- This reaction produces 1 mole of HF from its elements (H2 and F2) in their standard states. This reaction satisfies both criteria for ΔH_f^0. Thus, this reaction defines ΔH_f^0.

- **Reaction C**: N2 + 3H2 → 2NH3
- This reaction produces 2 moles of NH3, which does not meet the requirement of producing only 1 mole of the compound. Therefore, it does not define ΔH_f^0.

- **Reaction D**: C (diamond) + O2 → CO2
- In this reaction, carbon is in the diamond form, which is not its standard state (the standard state of carbon is graphite). Therefore, this reaction does not define ΔH_f^0.

3. **Conclusion**:
- The only reaction that defines ΔH_f^0 is **Reaction B: 1/2 H2 + 1/2 F2 → HF**.

### Final Answer:
The reaction that defines ΔH_f^0 is **Reaction B: 1/2 H2 + 1/2 F2 → HF**.