The oxidation of iron occurs as:
`4Fe(s) + 3 O_(2)(g) to 2Fe_(2)O_(3)(s)`
The enthalpy of formation of `Fe_(2)O_(3) is - 824.2 kJ mol^(-1)` and entropy change for the reaction is `-549 J K^(-1) mol^(-1)` at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous at 298 K?

Explanation

To determine why the oxidation of iron is spontaneous at 298 K despite a negative entropy change, we can analyze the thermodynamic principles involved, specifically the Gibbs free energy change (ΔG). The spontaneity of a reaction is determined by the sign of ΔG, which can be calculated using the following equation:

\[ \Delta G = \Delta H - T\Delta S \]

### Step 1: Identify the given data
- Enthalpy of formation of \( \text{Fe}_2\text{O}_3 \): \( \Delta H_f = -824.2 \, \text{kJ/mol} \)
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