Calculate the formation constant for the formation of [Cu(NH3)2(H2O)4]2+ from [Cu(H2O)6]2+, given that ΔG∘ is −44.7kJ⋅mol−1 at 298 K.

Answer 1

∆Gº = -RT lnK

∆Gº = -44.7 kJ/mol

R = 8.314 J/mol-K = 0.008314 kJ/mol-K (to keep units consistent with those of ∆Gº)

T = 298K

K = ?

-44.7 = -(0.008314)(298) lnK

-44.7 = -2.48 lnK

lnK = 18.0

K = 6.6x10⁷