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CHEM 160 General Chemistry II Lecture Presentation Chemical Thermodynamics

CHEM 160 General Chemistry II Lecture Presentation Chemical Thermodynamics. Chapter 19. Thermodynamics. Thermodynamics study of energy and its transformations heat and energy flow Helps determine natural direction of reactions

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CHEM 160 General Chemistry II Lecture Presentation Chemical Thermodynamics

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  1. CHEM 160 General Chemistry IILecture PresentationChemical Thermodynamics Chapter 19 Chapter 19

  2. Thermodynamics • Thermodynamics • study of energy and its transformations • heat and energy flow • Helps determine natural direction of reactions • Allows us to predict if a chemical process will occur under a given set of conditions • Organized around three fundamental laws of nature • 1st law of thermodynamics • 2nd law of thermodynamics • 3rd law of thermodynamics Chapter 19

  3. First Law of Thermodynamics • 1st Law: • Energy can be neither created nor destroyed • Energy of the universe is constant • Important concepts from thermochemistry • Enthalpy • Hess’s law • Purpose of 1st Law • Energy bookkeeping • How much energy? • Exothermic or endothernic? • What type of energy? Chapter 19

  4. Spontaneous vs Nonspontaneous • Spontaneous process • Occurs without outside assistance in the form of energy (occurs naturally) • Product-favored at equilibrium • May be fast or slow • May be influenced by temperature • Nonspontaneous process • Does not occur without outside assistance • Reactant-favored at equilibrium • All processes which are spontaneous in one direction cannot be spontaneous in the reverse direction • Spontaneous processes have a definite direction • Spontaneous processes are irreversible Chapter 19

  5. Spontaneous vs. Nonspontaneous • Spontaneous Processes • Gases expand into larger volumes • H2O(s) melts above 0C • H2O(l) freezes below 0C • NH4NO3 dissolves spontaneously in H2O • Steel (iron) rusts in presence of O2 and H2O • Wood burns to form CO2 and H2O • CH4 gas burns to form CO2 and H2O • Reverse processes are not spontaneous! • nonspontaneous Chapter 19

  6. Spontaneous vs. Nonspontaneous • Spontaneous Processes • H2O(s) melts above 0C (endothermic) • NH4NO3 dissolves spontaneously in H2O (endothermic) • Steel (iron) rusts in presence of O2 and H2O (exothermic) • Wood burns to form CO2 and H2O (exothermic) • CH4 gas burns to form CO2 and H2O (exothermic) • Heat change alone is not enough to predict spontaneity because energy is conserved • Many, but not all, spontaneous processes are exothermic Chapter 19

  7. Factors That Favor Spontaneity • Two thermodynamic properties of a system are considered when determining spontaneity: • Enthalpy, H • Many, but not all, spontaneous processes tend to be exothermic as already noted • Entropy, S (J/K) • Measure of the disorder of a system • Many, but not all, spontaneous processes tend to increase disorder of the system Chapter 19

  8. Entropy • Entropy, S (J/K) • describes # of ways the particles in a system can be arranged in a given state • More arrangements = greater entropy • S = heat change/T = q/T • S = Sfinal - Sinitial •  S > 0 represents increased randomness or disorder Chapter 19

  9. Patterns of Entropy Change • For the same or similar substances: Ssolid < Sliquid < Sgas solid vapor Chapter 19 liquid

  10. Patterns of Entropy Change Particles farther apart, occupy larger volume of space; even more positions available to particles Rigidly held particles; few positions available to particles Particles free to flow; more positions available for particles solid vapor Chapter 19 liquid

  11. Patterns of Entropy Change least ordered less ordered most ordered solid vapor Chapter 19 liquid

  12. Patterns of Entropy Change • Solution formation usually leads to increased entropy particles more disordered solvent solute solution Chapter 19

  13. Patterns of Entropy Change • Chemical Reactions • If more gas molecules produced than consumed, S increases. (Srxn > 0) • If only solids, ions and/or liquids involved, S increases if total # particles increases. Chapter 19

  14. fewer arrangements possible so lower entropy more arrangements possible so higher entropy Chapter 19

  15. Patterns of Entropy Change • Increasing temperature increases entropy System at T1 (S1) System at T2 (T2 > T1) (S2 > S1) Chapter 19

  16. Patterns of Entropy Change more energetic molecular motions less energetic molecular motions System at T1 (S1) System at T2 (T2 > T1) (S2 > S1) Chapter 19

  17. Third Law of Thermodynamics • If increasing T increases S, then the opposite should be true also. • Is it possible to decrease T to the point that S is zero? • At what T does S = 0? • If entropy is zero, what does that mean? Chapter 19

  18. Third Law of Thermodynamics • 3rd Law of Thermodynamics • Entropy of a perfect crystalline substance at 0 K is zero • No entropy = highest order possible • Why? • Purpose of 3rd Law • Allows S to be measured for substances • S = 0 at 0 K • S = heat change/temperature = q/T • S = standard molar entropy Chapter 19

  19. 50 40 Standard entropy,S° (J/K) 30 20 10 0 50 100 150 200 250 300 Chapter 19 Temperature (K)

  20. 50 Gas Liquid 40 Standard entropy,S° (J/K) 30 Solid 20 10 0 50 100 150 200 250 300 Chapter 19 Temperature (K)

  21. Standard Molar Entropies of Selected Substances at 298 K Chapter 19

  22. Entropy versus Probability • Systems tend to move spontaneously towards increased entropy. Why? • Entropy is related to probability • Disordered states are more probable than ordered states • S = k(lnW) • k (Boltzman’s constant) = 1.38 x 10-23 J/K • W = # possible arrangements in system Chapter 19

  23. Isothermal Gas Expansion Consider why gases tend to isothermally expand into larger volumes. Gas Container = two bulbed flask Ordered State Gas Molecules Chapter 19

  24. Isothermal Gas Expansion Gas Container Ordered State S = k(ln 1) = (1.38 x 10-23 J/K)(0) = 0 J/K Chapter 19

  25. Chapter 19

  26. Chapter 19

  27. Chapter 19

  28. Chapter 19

  29. Chapter 19

  30. Disordered States Chapter 19

  31. Disordered States Chapter 19

  32. Disordered States More probable that the gas molecules will disperse between two halves than remain on one side Chapter 19

  33. Disordered States Driving force for expansion is entropy (probability); gas molecules have a tendency to spread out Chapter 19

  34. Disordered States S = k(ln 7) = (1.38 x 10-23 J/K)(1.95) = 2.7 x 10-23 J/K Chapter 19

  35. Total Arrangements Stotal = k(ln 23) = k(ln 8) = (1.38 x 10-23 J/K)(1.79) = 2.9 x 10-23 J/K Chapter 19

  36. 2nd Law of Thermodynamics • 2nd Law of Thermodynamics • The entropy of the universe increases in any spontaneous process (Suniv > 0) • Increased disorder in the universe is the driving force for spontaneity Chapter 19

  37. 2nd Law of Thermodynamics • 2nd Law of Thermodynamics • Suniv = Ssysyem + Ssurroundings • Suniv > 0, process is spontaneous • Suniv = 0, process at equilibrium • Suniv < 0, process is nonspontaneous • Process is spontaneous in reverse direction Chapter 19

  38. 2nd Law of Thermodynamics • Spontaneous Processes • H2O(s) melts above 0C (endothermic) • Steel (iron) rusts in presence of O2 and H2O (exothermic) • 4Fe(s) + 3O2(g)  2Fe2O3(s) • CH4 gas burns to form CO2 and H2O (exothermic) • CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) • Each process increases Suniverse. Chapter 19

  39. 2nd Law of Thermodynamics • To determine Suniv for a process, both Ssystem and Ssurroundings need to be known: • Ssysyem • related to matter dispersal in system • Ssurroundings • determined by heat exchange between system and surroundings and T at which it occurs • Sign of Ssurroundings depends on whether process in system is endothermic or exothermic • Why? • Magnitude of Ssurroundings depends on T • Ssurroundings = -Hsystem/T Chapter 19

  40. Entropy Changes in a System For any reaction: S°rxn = nS°(products) - mS°(reactants) Where n and m are stoichiometric coefficients Chapter 19

  41. Example 1(1 on Example Problem Handout) Using standard molar entropies, calculate S°rxn for the following reaction at 25°C: 2SO2(g) + O2(g) --> 2SO3(g) S° = 248.1 205.1 256.6 (J · K-1mol-1) (Ans.: -187.9 J/K) Chapter 19

  42. Gibbs Free Energy, G 2nd law: Suniv = Ssys + Ssurr = Ssys - Hsys/T Rearrange (multiply by -T) -TSuniv = -TSsys + Hsys = Hsys - TSsys -TSuniv = G (-TSuniv = G) G = Gibbs free energy (J or kJ) G = H - TS and G = Hsys - TSsys G° = H°sys - TS°sy (if at standard state) Chapter 19

  43. Gibbs Free Energy • Summary of Conditions for Spontaneity • G < 0 • G > 0 • G = 0 Chapter 19

  44. Gibbs Free Energy • Summary of Conditions for Spontaneity • G < 0 • reaction is spontaneous in the forward direction • (Suniv > 0) • G > 0 • G = 0 Chapter 19

  45. Gibbs Free Energy • Summary of Conditions for Spontaneity • G < 0 • reaction is spontaneous in the forward direction • (Suniv > 0) • G > 0 • reaction is nonspontaneous in the forward direction • (Suniv < 0) • G = 0 Chapter 19

  46. Gibbs Free Energy • Summary of Conditions for Spontaneity • G < 0 • reaction is spontaneous in the forward direction • (Suniv > 0) • G > 0 • reaction is nonspontaneous in the forward direction • (Suniv < 0) • G = 0 • system is at equilibrium • (Suniv = 0) Chapter 19

  47. Example 2(2 in Example Problem Handout) For a particular reaction, Hrxn = 53 kJ and Srxn = 115 J/K. Is this process spontaneous a) at 25°C, and b) at 250°C? (c) At what temperature does Grxn = 0? (ans.: a)G = 18.7 kJ, nonspontaneous; b) –7.1 kJ, spontaneous; c) 460.9 K or 188C) Chapter 19

  48. Gibbs Free Energy and Temperature • G = H - TS • S > 0 H < 0 • Spontaneous at all T • S < 0 H > 0 • Not spontaneous at any T • S > 0 H > 0 • Spontaneous at high T; nonspontaneous at low T • S < 0 H < 0 • Spontaneous at low T; nonspontaneous at high T Chapter 19

  49. Gibbs Free Energy and Temperature • Spontaneous Processes • H2O(s) melts above 0C (endothermic) • H > 0, S > 0 • Steel (iron) rusts in presence of O2 and H2O at 25 C (exothermic) • 4Fe(s) + 3O2(g)  2Fe2O3(s) • H < 0, S < 0 • G < 0 for each process • T determines sign of G: G = H -TS Chapter 19

  50. Calculating Free Energy Changes For any reaction: G°rxn = Gf°(products) - mGf°(reactants) Where n and m are stoichiometric coefficients and Gf° = standard free energy of formation Chapter 19

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