`2Fe+(3)/(2)O_(2)rarr Fe_(2)O_(3)" "xkJ//mol e`
`2Fe+O_(2) rarr 2FeO" " ykJ //mol e`
The `DeltaH` to form one mole of `Fe_(3)O_(4)` from `Fe` and `O_(2)` is `:`

Options

A. `x/2 + y`
B. `(2x + y)/2`
C. `(2x - y)/2`
D. `(y - 2x)/2`

Explanation

To solve the problem of finding the enthalpy change (ΔH) to form one mole of Fe₃O₄ from Fe and O₂, we will follow these steps:

### Step 1: Write the reactions and their enthalpy changes

We have two reactions given:

1. \( 2Fe + \frac{3}{2}O_2 \rightarrow Fe_2O_3 \) with ΔH = x kJ/mol
2. \( 2Fe + O_2 \rightarrow 2FeO \) with ΔH = y kJ/mol

### Step 2: Write the target reaction

We need to find the ΔH for the formation of one mole of Fe₃O₄ from Fe and O₂. The balanced equation for this reaction is:

\[ 6Fe + 4O_2 \rightarrow 2Fe_3O_4 \]

### Step 3: Relate Fe₃O₄ to the given reactions

Fe₃O₄ can be formed from Fe₂O₃ and FeO. The reaction can be written as:

\[ 2Fe_2O_3 + 2FeO \rightarrow 2Fe_3O_4 \]

### Step 4: Write the enthalpy change for the new reaction

We can express the ΔH for the reaction forming Fe₃O₄ in terms of the ΔH values of the previous reactions:

- The ΔH for forming 2 moles of Fe₃O₄ from 2 moles of Fe₂O₃ and 2 moles of FeO is:

\[ \Delta H_{Fe_3O_4} = 2 \Delta H_{Fe_2O_3} + 2 \Delta H_{FeO} \]

### Step 5: Substitute the known ΔH values

From the reactions, we know:

- For Fe₂O₃: ΔH = x (for 1 mole, it is x)
- For FeO: ΔH = y (for 2 moles, it is y)

Thus, the ΔH for the formation of 2 moles of Fe₃O₄ is:

\[ \Delta H_{Fe_3O_4} = 2(x) + 2(y) = 2x + y \]

### Step 6: Calculate ΔH for one mole of Fe₃O₄

Since the reaction we wrote gives us 2 moles of Fe₃O₄, to find the ΔH for 1 mole, we divide the total ΔH by 2:

\[ \Delta H_{1 \text{ mole of } Fe_3O_4} = \frac{2x + y}{2} \]

### Final Answer

Thus, the ΔH to form one mole of Fe₃O₄ from Fe and O₂ is:

\[ \Delta H = x + \frac{y}{2} \]