`Delta_(f)H^(@)` of `Fe_(2)O_(3)` is -825 kJ. What is the enthalpy change for the reaction?
`4Fe_(2)O_(3) to 8Fe + 6O_(2)`

Explanation

To find the enthalpy change for the reaction \( 4 \text{Fe}_2\text{O}_3 \rightarrow 8 \text{Fe} + 6 \text{O}_2 \), we will follow these steps:

### Step 1: Understand the Enthalpy of Formation
The given enthalpy of formation (\( \Delta_f H^\circ \)) for \( \text{Fe}_2\text{O}_3 \) is -825 kJ. This value indicates that when 1 mole of \( \text{Fe}_2\text{O}_3 \) is formed from its elements in their standard states, 825 kJ of heat is released (exothermic reaction).

### Step 2: Write the Formation Reaction
The formation of \( \text{Fe}_2\text{O}_3 \) can be represented as:
\[
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