1 / 14

Understanding Spontaneity and Thermodynamics in Chemical Reactions

This text explores the principles of spontaneity in chemical processes through thermodynamics. It discusses the significance of system and surroundings entropy (ΔSuniverse) and enthalpy (ΔH) in determining reaction spontaneity, including specific examples like the phase change of water (H₂O) and gas reactions between nitrogen dioxide (NO₂) and dinitrogen tetroxide (N₂O₄). Key equations are presented, including Gibbs Free Energy (ΔG) calculations, revealing insights into spontaneity via entropy contributions and temperature effects. Ideal for students and professionals in chemistry.

amaya-bradshaw
Download Presentation

Understanding Spontaneity and Thermodynamics in Chemical Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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 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 Suniv > 0 At low T 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. - + Suniverse= Ssystem+ Ssurroundings > 0 2NO2(g)  N2O4(g) S = Soproducts - Soreactants So (J mol-1 K-1) N2O4 NO2 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 At high T At low T

  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 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 What is sign of So Go= Ho- TSo - - +

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

More Related