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Gibbs Free Energy/Enthalpy/Entropy Question

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Assume that ΔH° and ΔS° are independent of temperature. Calculate ΔG° for the following reaction at 533 K.

    2Cu(+)(aq) ----> Cu(s) + Cu(2+)(aq)

    Cu(2+)(aq) enthalpy (H) of formation = 64.77 kJ mol^-1; molar entropy (S) = -99.6 J K^-1 mol^-1
    Cu(s) enthalpy (H) of formation = 0 kJ/mol; molar entropy (S) = 33.15 J K^-1 mol^-1
    Cu(+)aq enthalpy (H) of formation = 71.67 kJ/mol; molar entropy (S) = 40.6 J K^-1 mol^-1

    2. Relevant equations

    ΔG = ΔH - TΔS

    3. The attempt at a solution

    ΔH = (64.77) + 0 - 2(71.67) = -78.57 kJ mol^-1
    ΔS = (33.15) + -99.6 -2(40.6) = -.14765 kJ K^-1 mol^-1

    ΔG = ΔH - TΔS
    = -78.57 - 533(-.14765)
    = .12745 = wrong

    I know I'm messing up somewhere because the correct answer is negative.
     
  2. jcsd
  3. Nov 13, 2012 #2
    To find the enthalpy of formation and entropy of formation its the summation of products minus the summation of the reactants, each multiplied by their respective coefficients. Double check your entropy calculation.
     
  4. Nov 13, 2012 #3
    I'm still getting -147.65 J K^-1 mol^-1, or -.14765 kJ K^-1 mol^-1.
     
  5. Nov 14, 2012 #4

    Borek

    User Avatar

    Staff: Mentor

    Significant digits perhaps?
     
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