Question

For oxidation of iron $4Fe\left(s\right)+3O_2\left(g\right)\to 2F{e}_2{O}_3\left(s\right)$ Entropy change is $-549.4J{K}^{-1}mo{l}^{-1}\text{at}298K$. Inspite of negative entropy change of this reaction, why is the reaction spontaneous? $({\Delta }_f{H}^{\Theta }\text{for this reaction is}-1648\times {10}^{3}Jmo{l}^{-1})$

Answer 1

Total entropy changes

One decides the spontaneity of a reaction by considering
$\Delta S_{\text{total}}(\Delta S_{\text{sys}}+\Delta S_{\text{surr}})$. For calculating $\Delta S_{\text{surr}}$, we have to consider the heat absorbed by the surroundings which is equal to $-{\Delta }_r{H}^{\ominus }$.

At temperature $T$, entropy change of the surroundings is

$$\begin{array}{cc}{\Delta }_{S_{\text{surr}}}& =-\frac{{\Delta }_r{H}^{\ominus }}{T}\left(atconstantpressure\right)\\ & =-\frac{(1648\times {10}^{3}Jmo{l}^{-1})}{298K}\\ & =5530J{K}^{-1}mo{l}^{-1}\end{array}$$
Thus, total entropy changes for this reaction:
$$\begin{array}{cc}\Delta S_{\text{total}}& =(\Delta S_{\text{sys}}+\Delta S_{\text{surr}}\\ {\Delta }_r{S}_{\text{total}}& =5530J{K}^{-1}mo{l}^{-1}+(-549.4J{K}^{-1}mo{l}^{-1})\\ {\Delta }_r{S}_{\text{total}}& =4980.6J{K}^{-1}mo{l}^{-1}\end{array}$$
This shows that the above reaction is $\text{spontaneous}$.

Final answer:
The total entropy of the reaction is $\text{positive}$,
so the reaction is $\text{spontaneous}$.