For oxidation of iron,4Fe(s) + 3O2(g) → 2F2O3(s)Entropy change is−

1.	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)
Answer» One decides the spontaneity of a reaction by considering = ∆S_total = ∆S_sys+ ∆S_surr For calculating ∆S_surr, we have to consider the heat absorbed by the surroundings which is equal to −∆_rH^Θ. At temperature T, entropy change of the surroundings is: ∆S_surr = - \(\frac{\Delta_rH^\ominus}{T}\) (at constant pressure) = \(\frac{-(-1648\times10^3\,J\,mol^{-1})}{298\,K}\) = 5530 JK^-1 mol^-1 So, total entropy change for this reaction Δ_rS_total = 5530JK⁻¹mol⁻¹ + (−549.4 JK⁻¹ mol⁻¹) = 4980.6 J K⁻¹ mol⁻¹ This Show that the above reaction is spontaneous.

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)

Answer»

One decides the spontaneity of a reaction by considering

= ∆S_total = ∆S_sys+ ∆S_surr

For calculating ∆S_surr, we have to consider the heat absorbed by the surroundings which is equal to −∆_rH^Θ.

At temperature T, entropy change of the surroundings is:

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

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

= 5530 JK^-1 mol^-1

So, total entropy change for this reaction

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

= 4980.6 J K⁻¹ mol⁻¹

This Show that the above reaction is spontaneous.

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