a) Since this reaction is not at equilibrium and the question asks for the free energy change, the equation needed is:

DG = DG^o + RTlnQ

Information given:

R= 0.008314 kJmol^-1K^-1

T = 298 K (room temperature is assumed unless otherwise given)

K_eq = 10^-7

Need to calculate Q & DG^o:

Q = [NH₃]⁶[Ni(H₂O)₆]/[Ni(NH₃)₆]

Q = (1)⁶(1)/(1) = 1.0

DG^o = -RTlnK_eq

= - (0.008314 kJmol^-1K^-1)(298 K)ln10^-7

= 40 kJ/mol

Now substitute into free energy equation:

DG = 40 kJ/mol + (0.008314kJmol^-1K^-1)(298 K)ln1

DG = 40 kJ/mol + 0

DG = 40 kJ/mol

b) At equilibrium Q = K_eq so the equation from part a becomes:

DG = DG^o + RTlnK_eq

DG = -RTlnK_eq + RTlnK_eq

DG = 0

Which is what you would predict. See multiple-choice question # 4 for confirmation.