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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

Respuesta :

kendrich

Answer:

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

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