Review Ssystem + Ssurroundings = Suniverse > 0 for spontaneous processes Ssurroundings = - Hsys T energetic disorder Ssystem = qr T positional disorder

- Ssystem + Ssurroundings = Suniverse > 0 for spontaneous processes at 298 K H2O(s)  H2O(l) spontaneous Ssurroundings = - H T H a) > b) < Ssurr < 0 Ssurr does not contribute to spontaneity

- + Ssystem + Ssurroundings = Suniverse for spontaneous processes at 298 K H2O(s)  H2O(l) spontaneous solid  liquid increasing disorder S > 0 S positional disorder S does contribute to spontaneity

- + Ssystem + Ssurroundings = Suniverse H2O(s)  H2O(l) S - H / T solid  liquid endothermic positive negative At high T a) spontaneous b) non-spontaneous Suniv > 0 At low T non-spontaneous Suniv < 0

Suniverse= Ssystem+ Ssurroundings > 0 2NO2 (g)  N2O4 (g) NO2 - brown, toxic gas N2O4 - colorless gas Ssurr = - H / T H = Hof N2O4 - 2 Hof NO2 = (9.66) - 2 (33.5) -58 kJ mol-1 = Ssurr 0 > favors spontaneity

Suniverse= Ssystem+ Ssurroundings > 0 - + Suniverse= Ssystem+ Ssurroundings > 0 2NO2 (g)  N2O4 (g) S = Soproducts - Soreactants So (J mol-1 K-1) N2O4 NO2 a) > b) < So N2O4 So NO2 304.18 239.95 S = [304.18 - (239.95)] 2 = -175.7 J/mol K 2 mol gas  1 mol gas non-spontaneous

Suniverse= Ssystem+ Ssurroundings > 0 - + Suniverse= Ssystem+ Ssurroundings > 0 2NO2 (g)  N2O4 (g) S -H / T -175.7 J mol-1 K-1 +58 kJ mol-1 T(K) At high T a) spontaneous b) non-spontaneous At low T spontaneous

G = H - TS Free Energy G Ssystem+ Ssurroundings = Suniverse > 0 - H T > 0 - T S = Suniverse H - TS = - TSuniverse < 0 G  H - TS G = H - TS

elements in standard states G = H - TS G < 0 spontaneous reaction G > 0 non-spontaneous reaction G = 0 equilibrium G = wmax maximum useful work G is an extensive State function Gof = 0 elements in standard states Gorxn = Gof products - Gof reactants

G = H - TS calculate Go for: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) Gorxn = [Gof CO2 (g)+ ] 2 (Gof H2O (l)) - [ ] Gof CH4 (g) + 2(Gof O2 (g)) = - 819 kJ a) > b) < Sorxn Gorxn = Horxn - TSorxn = [-892 kJ] - [ ] (298K) (-242 J/K) = -819 kJ

Rubber band Thermodynamics State 1 = relaxed State 2 = stretched go from State 1 to State 2 What is sign of Go + - What is sign of Ho a) + b) - What is sign of So Go = Ho - TSo + - -

Go = Ho - TSo Ho So Go + - always positive - + always negative + - always positive - + always negative negative a) high T b) low T + + positive low T a) high T b) low T - - negative positive high T

Equilibrium Go = Ho - TS Go = 0 phase changes chemical reactions TSo = 0 Ho Ho = So T Hof prod - Hof react T = Ho = T ___________________________ Soprod - Soreact So 2NO2  N2O4 -58 kJ = 331 K -175.7 J/K

Napoleon - 1812 Snwhite Sngrey tin buttons ΔHof (kJ/mol) So (J/mol K) white tin 0.0 51.55 grey tin -2.1 44.14 ΔHo = -2.1 - 0.0 = -2.1 kJ Snwhite Sngrey  ΔSo = 44.14 - 51.55 = -7.4 J/mol K T = -2100 J = 283 K = 10oC -7.4 J/K ∆G298 = .105 kJ ∆G233 = -.376 kJ