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Chapter 19: Chemical Thermodynamics

Chapter 19: Chemical Thermodynamics. Chapter 19: Chemical Thermodynamics. Spontaneous processes. at 25 o C. H 2 O (s) H 2 O (l). Chapter 19: Chemical Thermodynamics. Spontaneous processes. at 25 o C. ‘dry ice’. at 25 o C.

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Chapter 19: Chemical Thermodynamics

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  1. Chapter 19: Chemical Thermodynamics

  2. Chapter 19: Chemical Thermodynamics Spontaneous processes at 25oC H2O (s) H2O (l)

  3. Chapter 19: Chemical Thermodynamics Spontaneous processes at 25oC ‘dry ice’ at 25oC CO2 (s) → CO2 (g)

  4. Chapter 19: Chemical Thermodynamics Spontaneous processes at 25oC Fe (s) + O2 (g) Fe2O3 (s) rust

  5. Chapter 19: Chemical Thermodynamics Spontaneous processes Burning Paper at 25oC C6H10O5 (s) + 6 O2 (g) 5 H2O (g) + 6 CO2 (g)

  6. Chapter 19: Chemical Thermodynamics Spontaneous processes …happen “on their own” although they sometimes require an initial push to get going …are “product favored” occur in a definite direction: towards the formation of product at 25oC H2O (s) → H2O (l) CO2 (s) → CO2 (g) 4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)

  7. Chapter 19: Chemical Thermodynamics The direction of a spontaneous processes may depend on temperature below 0oC H2O (s) H2O (l)

  8. Chapter 19: Chemical Thermodynamics • At a given temperature and pressure, processes are • spontaneous only in one direction • If a processes is spontaneous in one direction it is non-spontaneous in the other direction at 25oC H2O (l) → H2O (s) non-spontaneous CO2 (s) → CO2 (g) spontaneous non-spontaneous 2 Fe2O3 (s) → 4 Fe (s) + 3 O2 (g)

  9. Chapter 19: Chemical Thermodynamics

  10. Chapter 19: Chemical Thermodynamics Which reactions are spontaneous? • many spontaneous reactions are exothermic (DH < 0) … but not all! • Some reactions are endothermic(DH > 0) and still spontaneous NH4NO3 (s) → NH4+ (aq) + NO3- (aq) DH > 0

  11. Chapter 19: Chemical Thermodynamics Entropy can be thought of as a measure of disorder Ludwig Boltzmann (1844-1906) S = k log W W = Wahrscheinlichkeit (probability) k = 1.38 x 10-23 J / K

  12. Chapter 19: Chemical Thermodynamics The change in entropy for any process is: DS = Sfinal - Sinitial What is the sign of DS for the following processes at 25oC ? H2O (s) → H2O (l) DS > 0 (positive) DS < 0 (negative) CO2 (g) → CO2 (s)

  13. Chapter 19: Chemical Thermodynamics Second Law of Thermodynamics For any spontaneous process, the entropy of the universe increases DSouniverse = DSosystem + DSosurroundings> 0

  14. Chapter 19: Chemical Thermodynamics Of all phase states, gases have the highest entropy Ssolid < Sliquid < Sgas gas solid liquid In a gas, molecules are more randomly distributed

  15. Chapter 19: Chemical Thermodynamics Larger molecules/atoms generally have a larger entropy Ssmall < Smedium < Slarge Larger molecules have more internal motion

  16. Chapter 19: Chemical Thermodynamics Often, dissolving a solid or liquid will increase the entropy dissolves lower entropy more disordered arrangement higher entropy

  17. Chapter 19: Chemical Thermodynamics Dissolving a gas in a liquid decreases the entropy dissolves overall more disordered arrangement: higher entropy

  18. Chapter 19: Chemical Thermodynamics The entropy of a substance increases with temperature

  19. Chapter 19: Chemical Thermodynamics } size of molecules increases } Sgas > Sliquid } dissolving a gas in a liquid is accompanied by a lowering of the entropy } dissolving a liquid in another liquid is accompanied by an increase in entropy

  20. Chapter 19: Chemical Thermodynamics What is the sign of DS for the following reactions? FeCl2 (s) + H2 (g) → Fe (s) + 2 HCl (g) DS > 0 solid gas solid gas 1 mol 2 mol Ba(OH)2 (s) → BaO (s) + H2O (g) DS > 0 solid solid gas 2 SO2 (g) + O2 (g) → 2 SO3 (g) DS < 0 gas gas gas 2 mol 1 mol 2 mol DS < 0 Ag+ (aq) + Cl-(aq) → AgCl (s) in solution insoluble

  21. Chapter 19: Chemical Thermodynamics For each of the following pairs, which substance has a higher molar entropy at 25oC ? HCl (l) HCl (s) Li (s) Cs (s) C2H2 (g) C2H6 (g) Pb2+ (aq) Pb (s) O2 (g) O2 (aq) HCl (l) HBr (l) N2 (l) N2 (g) CH3OH (l) CH3OH (aq)

  22. Chapter 19: Chemical Thermodynamics

  23. Chapter 19: Chemical Thermodynamics If you know the standard molar entropies of reactants and products, you can calculate DS for a reaction: DSorxn = Σ n So(products) – Σ m So(reactants)

  24. substanceSo (J/K-mol) H2 130.6 C2H4 (g) 219.4 C2H6 (g) 229.5 So for elements are NOT zero Chapter 19: Chemical Thermodynamics What is DSo for the following reaction? C2H4 (g) + H2 (g) → C2H6 (g) Do you expect DS to be positive or negative? DSorxn = [229.5] – [219.4 + 130.6] = -120.5 J/K

  25. DGo = Gibbs free energy… Chapter 19: Chemical Thermodynamics Which reactions are spontaneous? We need to know the magnitudes of both DS and DH ! DGo = DHo - TDSo … is a measure of the amount of “useful work” a system can perform

  26. Chapter 19: Chemical Thermodynamics DGo = DHo - TDSo A reaction is spontaneous if DGo is negative J. Willard Gibbs (1839 – 1903)

  27. Chapter 19: Chemical Thermodynamics The reaction of sodium metal with water: 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g) Is the reaction spontaneous? Yes What is the sign of DGo? DGo = negative What is the sign of DHo? DHo = negative (exothermic!) What is the sign of DSo? DSo = positive

  28. Chapter 19: Chemical Thermodynamics DGo< 0 => reaction is spontaneous (“product favored”) “exergonic” DGo> 0 => reaction is non-spontaneous “endergonic” DGo= 0 => reaction is at equilibrium

  29. Chapter 19: Chemical Thermodynamics DGo = DHo - TDSo DGo DHoDSo + - + - + - - - sign depends on T ! low T => DGo is negative high T => DGo is positve + + sign depends on T ! low T => DGo is positive high T => DGo is negative

  30. Chapter 19: Chemical Thermodynamics e. I have no idea how to even start thinking about this Spontaneity: DG must be negative - think about the signs of DH and DS endothermic DH > 0

  31. Chapter 19: Chemical Thermodynamics

  32. Chapter 19: Chemical Thermodynamics e. I have no idea how to even start thinking about this Spontaneity: DG must be negative - think about the signs of DH and DS endothermic DH > 0 number of gas molecules doubles DS > 0

  33. Chapter 19: Chemical Thermodynamics DGo = DHo - TDSo DGo DHoDSo + - + - + - - - sign depends on T ! low T => DGo is negative high T => DGo is positve + + sign depends on T ! low T => DGo is positive high T => DGo is negative

  34. Chapter 19: Chemical Thermodynamics e. I have no idea how to even start thinking about this Spontaneity: DG must be negative - think about the signs of DH and DS endothermic DH < 0 number gas molecules doubles DS > 0

  35. Chapter 19: Chemical Thermodynamics Consider the following reaction: 2 H2 (g) + O2 (g) → 2 H2O (l) DS = -326.3 J/K, DH = -571.7 kJ , and DG = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous? If the decomposition of water is spontaneous, then the formation of water is non-spontaneous: DG > 0 2 H2 (g) + O2 (g) → 2 H2O (l) i.e. is there a temperature at which DG becomes > 0 (non-spontaneous)?

  36. DH < 0 DS < 0 Chapter 19: Chemical Thermodynamics Consider the following reaction: 2 H2 (g) + O2 (g) → 2 H2O (l) DS = -326.3 J/K, DH = -571.7 kJ , and DG = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous? DG = DH - TDS

  37. Chapter 19: Chemical Thermodynamics DGo = DHo - TDSo DGo DHoDSo + - + - + - - - sign depends on T ! low T => DGo is negative high T => DGo is positve + + sign depends on T ! low T => DGo is positive high T => DGo is negative

  38. DH < 0 DS < 0 Chapter 19: Chemical Thermodynamics Consider the following reaction: 2 H2 (g) + O2 (g) → 2 H2O (l) DS = -326.3 J/K, DH = -571.7 kJ , and DG = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous? DG = DH - TDS The formation of water becomes non-spontaneous at high temperatures  The decomposition of water becomes spontaneous at high temperatures!

  39. substanceDHfo (kJ/mol)DGfo(kJ/mol)So (J/K-mol) O2 0 0 205.0 C (diamond, s) 1.88 2.84 2.43 C (graphite, s) 0 0 5.69 CO (g) -110.5 -137.2 197.9 Chapter 19: Chemical Thermodynamics A diamond left behind in a burning house reacts according to 2 C (s) + O2 (g) → 2 CO (g) What is the value of DGo for the reaction at 298K ?

  40. Chapter 19: Chemical Thermodynamics There are two possible ways to calculate DGo: DGo = DHo - TDSo I) calculate DGo from DHo and So : DGo = Σ n DGfo (products) – Σ m DGfo (reactants) II)

  41. Chapter 19: Chemical Thermodynamics 2 C (s) + O2 (g) → 2 CO (g) DGo = DHo - TDSo I) calculate DGo from DHo and DSo : substanceDHfo (kJ/mol)DGfo(kJ/mol)So (J/K-mol) O2 0 0 205.0 C (diamond, s) 1.88 2.84 2.43 C (graphite, s) 0 0 5.69 CO (g) -110.5 -137.2 197.9 DHo =[2 x (-110.5] – [2 x 1.88 + 0] = -224.8 kJ DSo = [2 x 197.9] – [2 x 2.43 + 205.0] = 185.9 J/K DGo = -224.8 kJ - = - 280.2 kJ 298K x 185.9 J/K

  42. Chapter 19: Chemical Thermodynamics 2 C (s) + O2 (g) → 2 CO (g) DGo = Σ n DGfo (products) – Σ m DGfo (reactants) II) substanceDHfo (kJ/mol)DGfo(kJ/mol)So (J/K-mol) O2 0 0 205.0 C (diamond, s) 1.88 2.84 2.43 C (graphite, s) 0 0 5.69 CO (g) -110.5 -137.2 197.9 DGo = [2 x (-137.2)] – [2 x 2.84 + 0] = -280.1 kJ

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