Question

Fe2O3(s) 2Al(s)Al2O3(s) 2Fe(s) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.73 moles of Fe2O3(s) react at standard conditions.

Answer 1

Answer:

$\Delta S=54.3\frac{J}{K}$

Explanation:

Hello!

In this case, for the given reaction, we can write the equation to compute the entropy change as shown below:

$\Delta s=2\Delta S_{Fe}+\Delta S_{Al_2O_3}-2\Delta S_{Al}-\Delta S_{Fe_2O_3}$

Letting:

$\Delta s_{Fe}=27.3\frac{J}{mol*K}\\\\ \Delta s_{Fe_2O_3}=84.4\frac{J}{mol*K}\\\\\Delta s_{Al}=28.3\frac{J}{mol*K}\\\\\Delta s_{Al_2O_3}=51.00\frac{J}{mol*K}$

We obtain the entropy change per mole of Fe2O3(s):$\Delta s=2*27.3\frac{J}{mol*K}+84.4\frac{J}{mol*K}-2*28.3\frac{J}{mol*K}-51.0\frac{J}{mol*K} \\\\\Delta s=31.4\frac{J}{mol*K}$

Finally, the total entropy change when 1.73 moles of Fe2O3(s) react turns out:

$\Delta s=31.4\frac{J}{mol*K}*1.73mol\\\\\Delta S=54.3\frac{J}{K}$

Best regards!