Consider the following reactions. `DeltaH^(@)` values of the reactions hve been given as `-x, -y` and z kJ.
`Fe_(3)O_(4)(s) to 3Fe(s) + 2O_(2)(g), DeltaH^(@) = z kJ`
`2Fe(s) + O_(2)(g) to 2FeO(s), DeltaH^(@) =-x kJ`
`4Fe(s) + 3O_(2)(g) to 2Fe_(2)O_(3)(s), DeltaH^(@) =-y kJ`
In the given set of reaction, `-x//2` kJ refers to

Options

A. Heat of formation of FeO(s)
B. Standard heat of formation of FeO(s)
C. Half the heat of formation of FeO(s)
D. All of these are correct.

Explanation

To solve the problem, we need to analyze the given reactions and understand what the term `-x/2` refers to in the context of thermodynamics, specifically in relation to standard enthalpy changes.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Enthalpy Changes**:
- The first reaction is:
\[ \text{Fe}_3\text{O}_4(s) \rightarrow 3\text{Fe}(s) + 2\text{O}_2(g), \quad \Delta H^\circ = z \, \text{kJ} \]
- The second reaction is:
\[ 2\text{Fe}(s) + \text{O}_2(g) \rightarrow 2\text{FeO}(s), \quad \Delta H^\circ = -x \, \text{kJ} \]
- The third reaction is:
\[ 4\text{Fe}(s) + 3\text{O}_2(g) \rightarrow 2\text{Fe}_2\text{O}_3(s), \quad \Delta H^\circ = -y \, \text{kJ} \]

2. **Understanding Standard Enthalpy of Formation**:
- The standard enthalpy of formation (\(\Delta H_f^\circ\)) is defined as the change in enthalpy when one mole of a compound is formed from its elements in their standard states.
- In this context, the reaction \(2\text{Fe}(s) + \text{O}_2(g) \rightarrow 2\text{FeO}(s)\) indicates that the formation of 2 moles of FeO from its elements releases \(-x\) kJ. Therefore, the formation of 1 mole of FeO would release \(-x/2\) kJ.

3. **Conclusion**:
- The term \(-x/2\) kJ refers to the standard enthalpy of formation of 1 mole of FeO from its elements in their standard states. Thus, it can be denoted as:
\[ \Delta H_f^\circ \text{ of FeO} = -\frac{x}{2} \, \text{kJ} \]

### Final Answer:
The value of \(-\frac{x}{2}\) kJ refers to the standard enthalpy of formation of FeO(s) from its elements.