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

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

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)

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

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 = - 187 J/K nonspon

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