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

`\Delta S=54.3(J)/(K)`

Step-by-step 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(J)/(mol*K)\\\\ \Delta s_(Fe_2O_3)=84.4(J)/(mol*K)\\\\\Delta s_(Al)=28.3(J)/(mol*K)\\\\\Delta s_(Al_2O_3)=51.00(J)/(mol*K)`

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

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

`\Delta s=31.4(J)/(mol*K)*1.73mol\\\\\Delta S=54.3(J)/(K)`

Best regards!