For oxidation of iron,
$4 F e (s) + {3 O}_{2} (g) \to 2 {F e}_{2} O_{3} (s)$
entropy change is - 549.4 ${J K}^{- 1} {m o l}^{- 1}$ at 298 K. In spite of negative entropy change of this reaction, why is the reaction spontaneous?
( $△, H^{⊝}$ for this reaction is $- 1648 \times 10^{3} J {m o l}^{- 1}$ )

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Solution

For a spontaneous reaction, total entropy must be greater than zero i.e. positive
$4 F e (s) + 3 O_{2} (g) ⟶ 2 F e_{2} O_{3} (s)$
$△ H_{f} = - 1648$ $K J m o l^{- 1}$
reaction is exothermic and surroundings absorbed $1648 K J$ of heat
$△ S_{s u r r} = \frac{△ H_{f}}{T} = \frac{1648}{298} = 5.53$ $K J m o l^{- 1} k^{- 1}$
$= 5530$ $J m o l^{- 1} k^{- 1}$
$△ S_{t o t a l} = △ S_{s y s t e m} + △ S_{s u r r o u n d i n g}$
$= - 549.4 + 5530$
$= 4980.6$ $J k^{- 1} m o l^{- 1}$

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For oxidation of iron, 4Fe(s)+3O2(g)→2Fe2O3(s)entropy change is - 549.4 JK−1mol−1 at 298 K. In spite of negative entropy change of this reaction, why is the reaction spontaneous?(△,H⊝ for this reaction is −1648×103Jmol−1)