Question

The formation constants at 25°C for Fe(CN)64– and Fe(EDTA)2– are 1.00x1037 and 2.10x1014, respectively. Answer the...

The formation constants at 25°C for Fe(CN)64– and Fe(EDTA)2– are 1.00x1037 and 2.10x1014, respectively. Answer the questions below.

1)Calculate K under standard conditions for the reaction Fe(EDTA)2−(aq) + 6CN−(aq) -----> Fe(CN)4−6​(aq) + EDTA4−(aq)

2)Calculate ΔG° for the reaction. (kJ/mol)

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Answer #1

Fe^2+(aq) + 6CN^-(aq) ----> Fe(CN)64–(aq) Kf = 1.00*10^37

Fe^2+(aq) + EDTA^4-(aq) ---> Fe(EDTA)2–(aq) Kf = 2.10*10^14

form the given data,

Fe(EDTA)2–(aq)      -----> Fe^2+(aq) + EDTA^4-(aq) K1 = 1/(2.10*10^14) = 4.762*10^-15

Fe^2+(aq) + 6CN^-(aq) ----> Fe(CN)64–(aq)         Kf = 1.00*10^37
-------------------------------------------------------------------------

Fe(EDTA)2−(aq) + 6CN−(aq) -----> Fe(CN)4−6​(aq) + EDTA4−(aq) K2 =

-------------------------------------------------------------------------
    k2 = k1*kf

       = (4.762*10^-15)*(1.00*10^37)

       = 4.762*10^22

2) DG0 = - RTlnK2

       = -8.314*298.15ln(4.762*10^22)

      = -129.44 kj/mol

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