`Fe_(2)O_(2)(s)+(3)/(2)C(s)to(3)/(2)CO_(2)(g)+2Fe(s)`
`DeltaH^(@)=+234.12KJ `
`C(s)+O_(2)(g)toCO_(2)(g) DeltaH^(@)=-393.5KJ`
Use these equations and `DeltaH^(@)` value to calculate `DeltaH^(@)` for this reaction :
`4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s)`

Options

A. `-1648.7kJ`
B. `-1255.3kJ`
C. `-1021.2kJ`
D. `-129.4 kJ`

Answer

To calculate the enthalpy change (ΔH°) for the reaction:

\[ 4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s) \]

we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step 1: Write down the given reactions and their ΔH° values.

1. **Reaction 1:**
\[ Fe_2O_3(s) + \frac{3}{2}C(s) \rightarrow \frac{3}{2}CO_2(g) + 2Fe(s) \]
\[ \Delta H° = +234.12 \, \text{kJ} \]

2. **Reaction 2:**
\[ C(s) + O_2(g) \rightarrow CO_2(g) \]
\[ \Delta H° = -393.5 \, \text{kJ} \]

### Step 2: Manipulate Reaction 1.

To find the desired reaction, we need to reverse Reaction 1 and multiply it by 2 to get \( 2Fe_2O_3 \) on the right side.

- **Reversed Reaction 1 (multiplied by 2):**
\[ 4Fe(s) + 3CO_2(g) \rightarrow 2Fe_2O_3(s) + 3C(s) \]
\[ \Delta H° = -2 \times 234.12 \, \text{kJ} = -468.24 \, \text{kJ} \]

### Step 3: Manipulate Reaction 2.

We need 3 moles of \( O_2 \) for our desired reaction, so we multiply Reaction 2 by 3.

- **Modified Reaction 2 (multiplied by 3):**
\[ 3C(s) + 3O_2(g) \rightarrow 3CO_2(g) \]
\[ \Delta H° = 3 \times (-393.5 \, \text{kJ}) = -1180.5 \, \text{kJ} \]

### Step 4: Add the manipulated reactions.

Now we add the two manipulated reactions together:

1. From Reversed Reaction 1:
\[ 4Fe(s) + 3CO_2(g) \rightarrow 2Fe_2O_3(s) + 3C(s) \]

2. From Modified Reaction 2:
\[ 3C(s) + 3O_2(g) \rightarrow 3CO_2(g) \]

Combining these gives:
\[ 4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s) \]

### Step 5: Calculate the total ΔH°.

Now, we can find the total ΔH° for the desired reaction:
\[ \Delta H° = (-468.24 \, \text{kJ}) + (-1180.5 \, \text{kJ}) = -1648.74 \, \text{kJ} \]

### Final Answer:
\[ \Delta H° = -1648.74 \, \text{kJ} \]

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