For the oxidation of iron:
$4 Fe_{(s)} + 3 O_{2(g)} \rightarrow 2 Fe_2O_{3(s)}$
The entropy change is $-549.4 \ J \ K^{-1} \ mol^{-1}$ at $298 \ K$. Despite the negative entropy change of this reaction,why is the reaction spontaneous?
($\Delta_r H^\Theta$ for this reaction is $-1648 \times 10^3 \ J \ mol^{-1}$)

(N/A) The spontaneity of a reaction is determined by the total entropy change,$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$.
To calculate $\Delta S_{surr}$,we consider the heat released to the surroundings,which is equal to $-\Delta_r H^\Theta$. At constant pressure and temperature $T$,the entropy change of the surroundings is given by:
$\Delta S_{surr} = -\frac{\Delta_r H^\Theta}{T}$
Substituting the given values:
$\Delta S_{surr} = -\frac{(-1648 \times 10^3 \ J \ mol^{-1})}{298 \ K} = 5530 \ J \ K^{-1} \ mol^{-1}$
Now,the total entropy change for the reaction is:
$\Delta_r S_{total} = \Delta S_{sys} + \Delta S_{surr} = -549.4 \ J \ K^{-1} \ mol^{-1} + 5530 \ J \ K^{-1} \ mol^{-1} = 4980.6 \ J \ K^{-1} \ mol^{-1}$
Since $\Delta_r S_{total} > 0$,the reaction is spontaneous.