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Chapter 19: Chemical Thermodynamics

Chapter 19: Chemical Thermodynamics. Spontaneous processes…. …happen without outside help. …are “product favored”. Chapter 19: Chemical Thermodynamics. Spontaneous processes. at 25 o C. H 2 O (s) → H 2 O (l). Chapter 19: Chemical Thermodynamics.

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Chapter 19: Chemical Thermodynamics

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  1. Chapter 19: Chemical Thermodynamics Spontaneous processes… …happen without outside help …are “product favored”

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

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

  4. Chapter 19: Chemical Thermodynamics Spontaneous processes at 25oC 4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s) rust

  5. Chapter 19: Chemical Thermodynamics Spontaneous processes …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)

  6. Chapter 19: Chemical Thermodynamics The direction of a spontaneous processes may depend on temperature at -10oC H2O (l) → H2O (s)

  7. 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) CO2 (s) → CO2 (g) 2 Fe2O3 (s)→ 4 Fe (s) + 3 O2 (g)

  8. 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

  9. Chapter 19: Chemical Thermodynamics Which reactions are spontaneous? Reactions proceed towards a more probable state In general, the more probable state is associated with more disorder

  10. 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

  11. 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) CO2 (g) → CO2 (s)

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

  13. Chapter 19: Chemical Thermodynamics Of all phase states, gases have the highest entropy Ssolid < Sliquid < Sgas gas solid liquid

  14. Chapter 19: Chemical Thermodynamics Larger Molecules generally have a larger entropy Ssmall < Smedium < Slarge

  15. Chapter 19: Chemical Thermodynamics Often, dissolving a solid or liquid will increase the entropy dissolves

  16. Chapter 19: Chemical Thermodynamics Dissolving a gas in a liquid decreases the entropy dissolves

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

  18. Chapter 19: Chemical Thermodynamics S = k ln W • W is a measure for how many different ways there are • of arranging a molecule or an ensemble of molecules (the system) • W reflects the number of microstates of a system • The larger the possiblenumber of microstates the higher the entropy

  19. Chapter 19: Chemical Thermodynamics Of all phase states, gases have the highest entropy Ssolid < Sgas gas solid

  20. Chapter 19: Chemical Thermodynamics Of all phase states, gases have the highest entropy Ssolid < Sgas gas solid

  21. Chapter 19: Chemical Thermodynamics Larger Molecules generally have a larger entropy Ssmall < Slarge

  22. Chapter 19: Chemical Thermodynamics Often, dissolving a solid or liquid will increase the entropy

  23. Chapter 19: Chemical Thermodynamics Often, dissolving a solid or liquid will increase the entropy

  24. Chapter 19: Chemical Thermodynamics Dissolving a gas in a liquid decreases the entropy

  25. Chapter 19: Chemical Thermodynamics } } } }

  26. Chapter 19: Chemical Thermodynamics What is the sign of DS for the following reactions? FeCl2 (s) + H2 (g) → Fe (s) + 2 HCl (g) Ba(OH)2 (s) → BaO (s) + H2O (g) 2 SO2 (g) + O2 (g) → 2 SO3 (g) Ag+ (aq) + Cl-(aq) → AgCl (s)

  27. 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)

  28. 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)

  29. 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 =

  30. 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

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

  32. 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? What is the sign of DGo? What is the sign of DHo? What is the sign of DSo?

  33. 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

  34. Chapter 19: Chemical Thermodynamics DGo = DHo - TDSo DGo DHoDSo + - - + - - + +

  35. Chapter 19: Chemical Thermodynamics DGo = DHo - TDSo at 25oC (298K): H2O (s) → H2O (l) spontaneous: DGo < 0 > 0 DHo > 0 DSo TDSo > 0 > DHo => DGo < 0 at 298K: but: if T becomes very small: TDSo > 0 < DHo => DGo> 0

  36. 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 ? 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 CO2 (g) -393.5 -394.4 213.6

  37. 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)

  38. 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 DGo = - 280.2 kJ

  39. 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 = -280.1 kJ

  40. 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. At what temperature does the reaction become spontaneous in the opposite direction? DG = DH - TDS DG < 0 at 298 K DH < 0 DS < 0 DG > 0 at high T The reaction “switches” direction if DH = TDS, i.e. if DG = 0

  41. Chapter 19: Chemical Thermodynamics 0 = DH - TDS TDS = DH T = -571.7 kJ T = -326.3 J/K x 1 kJ / 1000J T = 1752.1 K At temperatures greater than 1752.1 K liquid water will spontaneously decompose into H2 and O2 !!

  42. Chapter 19: Chemical Thermodynamics At the normal melting point, the Gibbs free energies of the solid and liquid phase of any substance are equal: at 0oC DGo = 0 H2O (s) → H2O (l) At the normal boiling point, the Gibbs free energies of the liquid and gas phase of any substance are equal: at 100oC DGo = 0 H2O (l) → H2O (g)

  43. At a phase change: DGo = 0 T Chapter 19: Chemical Thermodynamics DGo = DHo - TDSo if we know DHofus (or DHovap) , we can calculate DSo at the melting (or boiling point): 0 = DHo - TDSo DHo = TDSo DSo = DHo

  44. units! Chapter 19: Chemical Thermodynamics The entropy of melting for H2O DSofus = DHofus = 6.02 kJ/mol = 22.0 J/mol-K T 273.15 K melting temp The entropy of vaporization for H2O DSovap = DHovap = 40.7 kJ/mol = 109 J/mol-K T 373.15 K boiling temp

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