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30 December, 02:11

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

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  1. 30 December, 05:02
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    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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