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Solutions

  • Page ID
    11066
  • Using the tables on the web, calculate DSo and DGo for the following reactions using:
    DSo = SSoproducts - SSoreactants and

    DGorxn=SDGoproducts - SDGoreactants

    1. CH4 (g) + 2 O2 (g) ---------> CO2 (g) + 2 H2O (l)

    a) DSo = SSoproducts - SSoreactants

    Soproducts Soreactants
    (1 mol CO2)(213.7 J/Kmol) = 213.7 J/K
    (1 mol CH4)(186.1 J/Kmol) = 186.1 J/K
    (2 mol H2O)(69.940 J/Kmol) = 139.88 J/K
    (2 mol O2)(205 J/Kmol) = 410 J/K
    SSoproducts= 353.58 J/K
    SSoreactants = 596.1 J/K
    DSo= 353.58 J/K - 596.1 J/K = -242.52 J/K

    b) DGorxn=SDGoproducts - SDGoreactants

    DGoproducts DGoreactants
    (1 mol CO2)(-394.4 kJ/mol) = -394.4 kJ
    (1 mol CH4)(-50.81 kJ/mol) = -50.81 kJ
    (2 mol H2O)(-237.192 J/Kmol) = -474.384 kJ
    (2 mol O2)(0 kJ/mol) = 0 kJ
    SDGoproducts = -868.784 kJ
    SDGoreactants= -50.81 kJ
    DGorxn= -868.784 kJ - (-50.81 kJ) = -817.974 kJ
    spontaneous reaction

    2. 2 MgO (s) --------> 2 Mg (s) + O2 (g)

    a) DSo = SSoproducts - SSoreactants

    Soproducts Soreactants
    (2 mol Mg)(32.69 J/Kmol) = 65.38J/K
    (2 mol MgO)(26.9 J/Kmol) = 53.8 J/K
    (1 mol O2)(205 J/Kmol) = 205 J/K
    SSoproducts = 270.38 J/K
    SSoreactants = 53.8 J/K
    DSo = 270.38 J/K - 53.8 J/K = 216.58 J/K

    b) DGorxn=SDGoproducts - SDGoreactants

    DGoproducts DGoreactants
    (2 mol Mg)( 0 kJ/mol) = 0 kJ
    (2 mol MgO)(-569.0 kJ/mol) = -1138.0 kJ
    (1 mol O2)( 0 kJ/mol) = 0 kJ
    SDGoproducts = 0 kJ
    SDGoreactants = -1138.0 kJ
    DGorxn= 0 kJ - (-1138.0 kJ) = +1138.0 kJ
    nonspontaneous reaction