Answer: 138 J/K
Explanation:
1) ΔS reaction = ΔS products - ΔS reactants
2) ΔS products = 2ΔS Fe(s) + 3 ΔS H₂O (g)
3) ΔS reactants = ΔS Fe₂O₃ (s) + 3 ΔS H₂ (g)
4) Now you have to search the standard entropy values for each product and each reactant in a table.
I found these values at 25° and 1 bar.
ΔS H₂O = 188.8 J/Kₓmol
ΔS Fe(s) = 27.3 J/Kmol
ΔS Fe₂O₃ (s) = 87.4 J/Kmol
ΔS H₂ (g) = 130.7 J/Kmol
5) Replace those values into above equations:
ΔS products = 2ΔS Fe(s) + 3ΔS H₂O (g) =
2 mol × 27.3 J/K×mol + 3 mol ×188.8 J/K×mol = 621.0 J/K
ΔS reactants = ΔS Fe₂O₃ (s) + 3 ΔS H₂ (g) =
1mol × 87.4 J/K×mol + 3×130.7 J/K×mol = 479.5 J/K
ΔS reaction = 621.0 J/K - 479.5 J/K = 141.5 J/K
Taking into account the differences in the values from different sources (specially due to the temperature), you can consider that the value 141.5 J/K is pretty much close to 138 J/K, and take that answer.
