Answer:
[tex]\Delta S=54.3\frac{J}{K}[/tex]
Explanation:
Hello!
In this case, for the given reaction, we can write the equation to compute the entropy change as shown below:
[tex]\Delta s=2\Delta S_{Fe}+\Delta S_{Al_2O_3}-2\Delta S_{Al}-\Delta S_{Fe_2O_3}[/tex]
Letting:
[tex]\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}[/tex]
We obtain the entropy change per mole of Fe2O3(s):[tex]\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}[/tex]
Finally, the total entropy change when 1.73 moles of Fe2O3(s) react turns out:
[tex]\Delta s=31.4\frac{J}{mol*K}*1.73mol\\\\\Delta S=54.3\frac{J}{K}[/tex]
Best regards!
