For oxidation of iron,4Fe(s)+3O_2(g) 2Fe_2O_3(s) entropy change i… - Vidyadip

Question

For oxidation of iron,$4\text{Fe(s)}+3\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ 2\text{Fe}_2\text{O}_3(\text{s})$
entropy change is $-549.4 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ at 298 K . Inspite of negative entropy change of this reaction, why is the reaction spontaneous? $\left(\Delta_{\mathrm{r}} \mathrm{H}^{\circ}\right.$ for this reaction is $\left.-1648 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}\right)$

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Answer

$4\text{Fe(s)}+3\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ 2\text{Fe}_2\text{O}_3(\text{s})$$\Delta\text{S}_\text{surr}=-\frac{\Delta_\text{r}\text{H}^\circ}{\text{T}}=\frac{-(-1648\times10^3\text{ J mol}^{-1})}{289\text{K}}$
$\Delta\text{S}_\text{surr}=5530\text{J K}^{-1}\text{mol}^{-1}$
$\Delta\text{S}_\text{sys}=-549.4\text{J K}^{-1}\text{mol}^{-1}$
$\Delta_\text{r}\text{S}_\text{total}=\Delta\text{S}_\text{sys}+\Delta\text{S}_\text{sys}$
$\Rightarrow\Delta_\text{r}\text{S}_\text{total}=5530-549.4$
$=4980.6\text{J K}^{-1}\text{mol}^{-1}$
Since $\Delta_\text{r}\text{S}_\text{total}$ is +ve, therefore, reaction is spontaneous.
$\because\Delta\text{G}=-\text{T}\Delta\text{S}_\text{total}$
$\because\Delta\text{G}=-\text{ve}$
When $\text{S}_\text{total}$ is positive.

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