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Review

= - H sys T. = q r T. Review. S system. + S surroundings. = S universe. > 0. for spontaneous processes. S surroundings. energetic disorder. S system. positional disorder. = - H T. -. S system + S surroundings = S universe. > 0.

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Review

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

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

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

  4. - + 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

  5. + Suniverse= Ssystem+ Ssurroundings > 0  N2O4(g) 2NO2(g) NO2 - brown, toxic gas N2O4- colorless gas Ssurr = -H /T H = HofN2O4 - 2 HofNO2 = (9.66) - (33.5) -58 kJ mol-1 2 = Ssurr 0 favors spontaneity >

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

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

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

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

  10. G= H - TS calculate Go for:  CO2 (g) + 2H2O (l) CH4(g) + 2O2 (g) Gorxn = [Gof CO2 (g)+ ] 2 (Gof H2O (l)) - [ ] Gof CH4 (g) + 2(Gof O2 (g)) = - 819 kJ a) > b) < Sorxn 0 Gorxn =Horxn - TSorxn = [-892 kJ] - [ ] (298K) (-242 J/K) = -819 kJ

  11. 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 - - +

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

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

  14. Napoleon - 1812 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

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