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Entropy, Free Energy, and Equilibrium

Entropy, Free Energy, and Equilibrium. Chapter 16. nonspontaneous. spontaneous. Spontaneous Physical and Chemical Processes. A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm., water freezes below 0 0 C and ice melts above 0 0 C

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Entropy, Free Energy, and Equilibrium

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  1. Entropy, Free Energy, and Equilibrium Chapter 16

  2. nonspontaneous spontaneous Spontaneous Physical and Chemical Processes • A waterfall runs downhill • A lump of sugar dissolves in a cup of coffee • At 1 atm., water freezes below 00C and ice melts above 00C • Heat flows from a hotter object to a colder object • The expansion of a gas in an evacuated bulb • Iron exposed to oxygen and water forms rust 18.2

  3. spontaneous nonspontaneous 18.2

  4. CH4(g) + 2O2(g) CO2(g) + 2H2O (l)DH0 = -890.4 kJ H+(aq) + OH-(aq) H2O (l)DH0 = -56.2 kJ H2O (s) H2O (l)DH0 = 6.01 kJ H2O NH4NO3(s) NH4+(aq) + NO3-(aq)DH0 = 25 kJ Does a decrease in enthalpy mean a reaction proceeds spontaneously? Spontaneous reactions Spontaneous reactions could be either exothermic or endothermic. Therefore, to predict the spontaneity of a reaction, another thermodynamic entity must be considered, i.e., entropy. 18.2

  5. S order disorder S H2O (s) H2O (l) Entropy (S) is a measure of the randomness or disorder of a system. DS = Sf - Si If the change from initial to final results in an increase in randomness Sf > Si DS > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state Ssolid < Sliquid << Sgas DS > 0 18.2

  6. Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, enthalpy, pressure, volume, temperature , entropy Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 6.7

  7. At higher temperatures, • molecular motion increases • randomness increases. • entropy increases.

  8. Microstates and Entropy • 2 possible arrangements • 50 % chance of finding the left empty

  9. 4 possible arrangements • 25% chance of finding the left empty • 50 % chance of them being evenly dispersed

  10. 4 possible arrangements • 8% chance of finding the left empty • 50 % chance of them being evenly dispersed

  11. Processes that lead to an increase in entropy (DS > 0) 18.2

  12. How does the entropy of a system change for each of the following processes? (a) Condensing water vapor Dec (b) Forming sucrose crystals from a supersaturated solution Dec (c) Heating hydrogen gas from 600C to 800C Inc (d) Subliming dry ice Inc 18.2

  13. How does the entropy of a system change for each of the following processes? Inc (a) A solid melts Dec (b) A liquid freezes Inc (c) Sugar dissolves in water (d) A vapor condenses to a liquid Dec 18.2

  14. First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 18.3

  15. The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aS0(A) bS0(B) - [ + ] cS0(C) dS0(D) [ + ] = aA + bB cC + dD - S mS0(reactants) S nS0(products) = DS0 DS0 DS0 DS0 rxn rxn rxn rxn What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2(g) 2CO2(g) S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] Entropy Changes in the System (DSsys) 18.3

  16. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2(g) 2ZnO (s) Entropy Changes in the System (DSsys) When gases are produced (or consumed) • If a reaction produces more gas molecules than it consumes, DS0 > 0. • If the total number of gas molecules diminishes, DS0 < 0. • If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number. The total number of gas molecules goes down, and a gas is converted to a solid, DS is negative. 18.3

  17. a) 3 O2(g) 2 O3(g) b) 2 NaHCO3(s) Na2CO3(s) + H2O (l) + CO2(g) c) N2(g) + O2(g) 2 NO (g) d) I2(s)2 I (g) e) 2 Hg (l) + O2(g)2 HgO (s) Predict the sign of the entropy change for the following reactions? - + - + - 18.3

  18. - DHsys DSsurr= T Entropy Changes in the Surroundings (DSsurr) Endothermic Process DSsurr < 0 Exothermic Process DSsurr > 0 Dssurris proportional to - DHsys 18.3

  19. Is the following reaction spontaneous at 25°C? N2(g) + 3H2(g) 2NH3(g); DH° = -92.6 kJ Spontaneous process: DSuniv = DSsys + DSsurr > 0 - DHsys - (-92600 J) DSsurr= = = 311 J/K T 298 K DSsys = -199 J/K = [2 x S0(NH3) ] – [S0(N2) + 3 x S0(H2)] DHsys = -92.6 kJ DSuniv = DSsys + DSsurr = -199 J/K + 311 J/K = 112 J/K > 0 Process is spontaneous, but may be very slow.

  20. Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. 18.3

  21. DSuniv = DSsys - > 0 - DHsys T Gibbs Free Energy Spontaneous process: DSuniv = DSsys + DSsurr > 0 -TDSuniv = DHsys - TDSsys < 0 For a constant-temperature process: Gibbs free energy (G) DG = DHsys -TDSsys DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium. 18.4

  22. The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD - [ + ] [ + ] = - mDG0 (reactants) S S = f Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. DG0 DG0 rxn rxn f DG0 of any element in its stable form is zero. f dDG0 (D) nDG0 (products) aDG0 (A) bDG0 (B) cDG0 (C) f f f f f 18.4

  23. - mDG0 (reactants) S S = f 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) DG0 DG0 DG0 - [ ] [ + ] = rxn rxn rxn [12(–394.4) + 6(–237.2)] – [2(124.5)] = -6405 kJ = Is the reaction spontaneous at 25 0C? 12DG0 (CO2) 2DG0 (C6H6) f f 6DG0 (H2O) f nDG0 (products) f What is the standard free-energy change for the following reaction at 25 0C? DG0 = -6405 kJ < 0 spontaneous 18.4

  24. a) CaCO3(s) CaO (s) + CO2(g) b) 2 NaHCO3(s) Na2CO3(s) + H2O (l) + CO2(g) c) 2HgO (s) 2Hg (l) + O2(g) d) 2C2H4(g) + 6O2(g) 4CO2(g) + 4H2O (l) Calculate the standard free-energy change for the following reactions at 25 0C and determine whether the reactions are spontaneous. 18.4

  25. Calculate the standard free-energy change for the following reactions at 25 0C and determine whether the reactions are spontaneous. • [(-603.5 + -394.4) ] – [( -1128.8) ] = b) [-1044.4+ -237.2+ -394.4] - [2 NaHCO3(s) ] = c) [2 (0) + 0] - [2 (-58.49)] = d) [4 (-394.4) + 4 (237.2) ] - [2C2H4(g) + 0) ] = 18.4

  26. DG = DH - TDS 18.4

  27. CaCO3(s) CaO (s) + CO2(g) Temperature and Spontaneity of Chemical Reactions DH0 = 177.8 kJ DS0 = 160.5 J/K DG0 = DH0 – TDS0 At 25 0C, DG0 = 130.0 kJ nonspontaneous Tat which DG0 = 0 is when the reaction starts to become spontaneous DG0 = 0 at 835 0C 18.4

  28. MgCO3(s) MgO (s) + CO2(g) To what temperature must MgCO3 must be heated to decompose it? Is this higher or lower than the temperature to decompose CaCO3? DG0 = DH0 – TDS0 = 0 T = DH0 DS0 18.4

  29. H2O (l) H2O (g) 40.79 kJ = 373 K DHvap DSvap = T Gibbs Free Energy and Phase Transitions At phase transitions, the system is at equilibrium (DG0 = 0) DG0 = 0 = DH0 – TDS0 = 109 J/K 18.5

  30. Ar (l) Ar (g) The molar heat of vaporization of argon is 6.3 kJ/mol, and argon’s boiling point is -186°C. Calculate the entropy change of vaporization. DH vap DSvap = T DSvap = 6300 J 87 K = 72.4 J/K 18.5

  31. Gibbs Free Energy and Chemical Equilibrium DG = DG0 + RT lnQ R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium Q = K DG = 0 0 = DG0 + RT lnK DG0 = -RT lnK 18.5

  32. DG0 < 0 DG0 > 0 18.5

  33. DG0 = -RT lnK 18.5

  34. Using thermodynamic data, calculate the equilibrium constants for the following reactions at 25°C. a) CaCO3(s) CaO (s) + CO2(g) b) 2 NaHCO3(s) Na2CO3(s) + H2O (l) + CO2(g) c) 2HgO (s) 2Hg (l) + O2(g) d) 2C2H4(g) + 6O2(g) 4CO2(g) + 4H2O (l) DG0 = -RT lnK 18.5

  35. BaF2(s) Ba+2(aq) + 2 F-(aq) Calculate DG0 for the following process at 25C. The Ksp of CaCO3 is 8.7 x 10-9. CaCO3(s) Ca+2(aq) + CO3-2(aq) Calculate DG0 for the following process at 25°C. The Ksp of BaF2 is 1.7 x 10-6. 18.5

  36. Calculate the Ksp at 25°C for the following reaction, using standard free energy of formation (DG0f ) values. CaCO3(s) Ca+2(aq) + CO3-2(aq) 18.5

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