The enthalpy of formation of `Fe_(2)O_(3)(s)` is `-824.2 kJ "mol"^(-1)`.Calculate the enthalpy change for the reaction :
`4Fe(s) + 3O_(2)(g) to 2Fe_(2)O_(3)(s)`

Explanation

To calculate the enthalpy change for the reaction:

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

we will follow these steps:

### Step 1: Understand the Enthalpy of Formation
The enthalpy of formation (\( \Delta H_f \)) of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states. For \( Fe_2O_3(s) \), the enthalpy of formation is given as \( -824.2 \, \text{kJ/mol} \).

### Step 2: Write the Reaction for Formation
The formation reaction for \( Fe_2O_3(s) \) from its elements is:

\[ 2Fe(s) + \frac{3}{2}O_2(g) \rightarrow Fe_2O_3(s) \]

### Step 3: Calculate the Enthalpy Change for the Given Reaction
The given reaction produces 2 moles of \( Fe_2O_3(s) \). Therefore, we need to consider the enthalpy change for forming 2 moles of \( Fe_2O_3(s) \):

1. Since the enthalpy of formation for 1 mole of \( Fe_2O_3(s) \) is \( -824.2 \, \text{kJ} \), for 2 moles it will be:

\[ \Delta H = 2 \times (-824.2 \, \text{kJ}) = -1648.4 \, \text{kJ} \]

### Step 4: Consider the Enthalpy of Reactants
The enthalpy of formation for the elements in their standard states (solid iron and gaseous oxygen) is defined as zero:

- \( \Delta H_f \) for \( Fe(s) = 0 \)
- \( \Delta H_f \) for \( O_2(g) = 0 \)

### Step 5: Calculate the Overall Enthalpy Change
Using the formula for enthalpy change:

\[ \Delta H_{reaction} = \Delta H_{products} - \Delta H_{reactants} \]

Substituting the values:

\[ \Delta H_{reaction} = (-1648.4 \, \text{kJ}) - (0 + 0) = -1648.4 \, \text{kJ} \]

### Final Answer
Thus, the enthalpy change for the reaction \( 4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s) \) is:

\[ \Delta H = -1648.4 \, \text{kJ} \]

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