Question

Consider the reaction: Fe3O4(s) + 4H2(g)—>3Fe(s) + 4H2O(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.79 moles of Fe3O4(s) react at standard conditions. ASºsystem = J/K

Answer 1

Fe3O4(s) + 4H2(g) --> 3Fe (s)+ 4H2O(g)
  

dS(Fe3O4) =146.4 j/k
dS(H2) =130.684

dS(Fe) =27.78

dS(H2O) =188.825

dSrxn = dS[product]-dS[reactants]
= 3*dS(Fe)+ 4*dS(H2O)-[1*dS(Fe3O4)+ 4dS(H2)]
= [3*27.78 +4*188.825-146.4 -4*130.684] j/k = 169.504 j/k
This is the dS for 1mole Fe3O4
for 1.79 mols Fe3O4

dS= 1.79*169.504 j/k = 303.41 j/k
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Thank you !!