If ∆G° = 18.1 kJ/mol, determine the value of ∆G assuming that a mixture contains 57.6 g of NiO, 182.3 g of Ni, 0.24 atm of NH₃, 8.54 atm of N₂ and 10.07 atm of H₂O.

Answer 1

Since you didn't provide a chemical reaction for which the ∆Gº value applies, I will assume one of the following:

(i) 3NiO(s) + 2NH₃(g) <==> 3Ni(s) + N₂(g) + 3H₂O(g)

or it could be ...

(ii) 3Ni(s) + N₂(g) + 3H₂O(g) <==> 3NiO(s) + 2NH₃(g)

Since ∆Gº has a positive value, this would be a non spontaneous reaction, and perhaps ∆S would decrease. That would make (ii) more likely than (i) but without knowing ∆H, we can't say for sure.

3Ni(s) + N₂(g) + 3H₂O(g) <==> 3NiO(s) + 2NH₃(g)

Q = (NH3)² / (N2)(H2O)³ = (0.24)² / (8.54)(10.07)³ = 6.61x10^-6

∆G = ∆Gº + RT ln Q = 18.1 + (0.008314)(298) * (-11.9)

∆G = 18.1 - 29.5

∆G = - 11.4 (spontaneous??)