1. Systems/processes want to lower their overall energy
Free Energy
Systems/processes want to lower their overall energy

2. DG = DH – TDS Signs for free energy - DG + DG
Entropy
(J/K)
Free energy
(kJ)
Enthalpy
(kJ)
Temp
(K)
Spon
nonspon

3. G = H – TS Cases result +S, -H +S, +H -S, -H -S, +H
Spon at all temps
Spon at high temps
Spon at low temps
Nonspon at all temps
(reverse is spon)

4. Math with free energy
At what temps will the following process be spon at 1 atm? Br2 (l)  Br2(g) DHo = 31.0 kJ/mol DSo = 93.0 J/(K mol) DGo = DHo – TDSo DHo = TDSo T = DHo/DSo = When T > 333K, G will be negative Which means Br2(l) boils at 333 K.
Go = 0
31.0 kJ/mol
0.093 kJ/mol K
= 333 K

5. Sorxn = ∑Soproducts - ∑Soreactants
2 NiS(s) + 3 O2(g)  2 SO2(g) + NiO(s) Predict the sign of S: - goes from 3 gasses to 2 gasses Get So numbers from appendix A21-A23 Sorxn =[2mol(248J/K mol)) + 38] – [2(53) + 3(205)] Sorxn = J/K nonspon

6. Finding free energy (2 ways)
First: use enthalpy and entropy values Horxn = ∑Hoproducts - ∑Horeactants Sorxn = ∑Soproducts - ∑Soreactants G = H – TS Or use free energy values Gorxn = ∑Goproducts - ∑Goreactants