# For oxidation of iron, 4Fe(s) + 3O2(g) → 2F2O3(s) Entropy change is−549.4 JK^-1 at 298 K.

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For oxidation of iron,

4Fe(s) + 3O2(g) → 2F2O3(s)

Entropy change is−549.4 JK-1 at 298 K. Inspire of negative entropy of this reaction, why is the reaction spontaneous?

(∆rHΘ for this reaction is -1648 x 103 J mol-1)

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One decides the spontaneity of a reaction by considering

= ∆Stotal = ∆Ssys + ∆Ssurr

For calculating ∆Ssurr, we have to consider the heat absorbed by the surroundings which is equal to −∆rHΘ

At temperature T, entropy change of the surroundings is:

∆Ssurr = - $\frac{\Delta_rH^\ominus}{T}$ (at constant pressure)

$\frac{-(-1648\times10^3\,J\,mol^{-1})}{298\,K}$

= 5530 JK-1 mol-1

So, total entropy change for this reaction

ΔrStotal = 5530JK−1mol−1 + (−549.4 JK−1 mol−1)

= 4980.6 J K−1 mol−1

This Show that the above reaction is spontaneous.