Which of the reactions defined `Delta_f H^@`?

Options

A. `C_("diamond") + O_2(g) to CO_2(g)`
B. `1/2H_2(g) + 1/2 F_2(g) to HF(g)`
C. `N_2(g) + 3H_2(g) to 2NH_3(g)`
D. `CO(g) + 1/2O_(2)(g) to CO_(2)(g)`

Answer

To determine which of the reactions defines the standard enthalpy of formation (\( \Delta_f H^\circ \)), we need to understand the definition of formation enthalpy. The standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its pure elements in their standard states.

### Step-by-Step Solution:

1. **Understanding Formation Enthalpy**:
- The formation enthalpy (\( \Delta_f H^\circ \)) is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.

2. **Analyzing the Reactions**:
- We need to evaluate each reaction provided to see if it fits the criteria for formation enthalpy.

3. **First Reaction**:
- \( C_{(diamond)} + O_2 \rightarrow CO_2 \)
- Here, carbon is in the diamond form. The standard state for carbon is graphite, not diamond. Therefore, this reaction does not define the formation enthalpy.

4. **Second Reaction**:
- \( H_2 + F_2 \rightarrow 2HF \)
- In this reaction, one mole of HF is formed from its elements (hydrogen and fluorine) in their standard states. This fits the definition of formation enthalpy, but note that it produces 2 moles of HF, which is not what we want for formation enthalpy.

5. **Third Reaction**:
- \( CO + \frac{1}{2} O_2 \rightarrow C \)
- Here, CO is not an element; it is a compound formed from carbon and oxygen. Thus, this reaction does not define the formation enthalpy.

6. **Conclusion**:
- The only reaction that fits the criteria for defining the formation enthalpy is the second reaction, despite producing 2 moles of HF. The correct answer is option 2, as it involves the formation of a compound from its elements.

### Final Answer:
The reaction that defines \( \Delta_f H^\circ \) is the second reaction: \( H_2 + F_2 \rightarrow 2HF \).