Fe3O4 (s) + 4H2 (g) 3Fe (s) + 4H2O (g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.79 moles of Fe3O4 (s) react at standard conditions. S°surroundings = J/K

dS = 1.79*169.504

j/k = 303.41 j/k

Explanation:

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