1 / 60

Chapter 16

nonspontaneous. spontaneous. Chapter 16. Thermodynamics: Entropy, Free Energy, and Equilibrium. In this chapter we will determine the direction of a chemical reaction and calculate equilibrium constant using thermodynamic values: Entropy and Enthalpy. Spontaneous Processes.

lam
Download Presentation

Chapter 16

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. nonspontaneous spontaneous Chapter 16 • Thermodynamics: Entropy,Free Energy,and Equilibrium In this chapter we will determine the direction of a chemical reaction and calculate equilibrium constant using thermodynamic values: Entropy and Enthalpy.

  2. Spontaneous Processes Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.

  3. Universe: System + Surroundings The system is what you observe; surroundings are everything else.

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

  5. Laws of Thermodynamics 1) Energy is neither created nor destroyed. 2) In any spontaneous process the total Entropy of a system and its surrounding always increases. We will prove later: ∆G = ∆Hsys – T∆Ssys ∆G < 0 Reaction is spontaneous in forward direction 3) The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0)

  6. Enthalpy, Entropy, and Spontaneous Processes State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state. Enthalpy Change (DH): The heat change in a reaction or process at constant pressure. Entropy (DS): The amount of Molecular randomness change in a reaction or process at constant pressure.

  7. DH° DH° DH° = +3.88 kJ = -890.3 kJ = +57.1 kJ CH4(g) + 2O2(g) N2O4(g) H2O(s) H2O(l) H2O(l) CO2(g) + 2H2O(l) 2NO2(g) H2O(g) DHfusion DHvap = +6.01 kJ = +40.7 kJ H2O NaCl(s) Na1+(aq) + Cl1-(aq) Enthalpy Change Exothermic: Endothermic:

  8. Entropy Change DS = Sfinal - Sinitial

  9. Entropy

  10. Entropy

  11. The sign of entropy change, DS, associated with the boiling of water is _______. • Positive • Negative • Zero

  12. Correct Answer: • Positive • Negative • Zero Vaporization of a liquid to a gas leads to a large increase in volume and hence entropy; DS must be positive.

  13. Entropy and Temperature 02

  14. Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0) Ssolid < Sliquid < Sgas

  15. 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 . The total number of gas molecules goes down, DS is negative.

  16. Standard Molar Entropies and Standard Entropies of Reaction Standard Molar Entropy (S°): The entropy of 1 mole of a pure substance at 1 atm pressure and a specified temperature. Calculated by using S = k ln W , W = Accessible microstates of translational, vibrational and rotational motions

  17. ∆G = ∆H – T∆S see page 301 of your book Review of Ch 8: We will show the proof of this formula later in this chapter • Calculating DH° for a reaction: • DH° = DH°f (Products) – DH°f (Reactants) • For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient. • aA + bB cC + dD • DH° = [cDH°f(C) + dDH°f(D)] – [aDH°f(A) + bDH°f(B)] See Appendix B, end of your book

  18. The standard entropy change 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) = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] = 427.2 – [395.2 + 205.0] = -173.0J/K•mol Entropy Changes in the System (DSsys) S0(CO) = 197.6 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol

  19. Calculating ∆S for a Reaction ∆So =  So (products) -  So (reactants) Consider 2 H2(g) + O2(g) ---> 2 H2O(liq) ∆So = 2 So (H2O) - [2 So (H2) + So (O2)] ∆So = 2 mol (69.9 J/K•mol) - [2 mol (130.6 J/K•mol) + 1 mol (205.0 J/K•mol)] ∆So = -326.4 J/K Note that there is a decrease in Entropy because 3 mol of gas give 2 mol of liquid.

  20. Calculate DS° for the equation below using the standard entropy data given: 2 NO(g) + O2(g) 2 NO2(g) DS° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205. 147 J/mol 76 J/mol +176 J/mol +147 J/mol

  21. Correct Answer: 147 J/K 76 J/K +176 J/K +147 J/K ( ) ( )     D  = D  D S S products S reactants DS° = 2(240)  [2(211) + 205] DS° = 480  [627] DS° =  147

  22. Thermite Reaction

  23. Spontaneous Processesand Entropy 07 • Consider the gas phase reaction of A2 molecules (red) with B atoms (blue). • (a) Write a balanced equation for the reaction. • (b) Predict the sign of ∆S for the reaction.

  24. Second Law Of Thermodynamic: In any spontaneous process the total Entropy of a system and its surrounding always increases. Entropy Changes in the Surroundings (DSsurr) Endothermic Process DSsurr < 0 Exothermic Process DSsurr > 0

  25. 1 T - DH Dssurr = T Entropy and the Second Law of Thermodynamics Dssurra ( - DH) Dssurra ( ) See example of tossing a rock into a calm waters vs. rough waters Page 661 of your book

  26. 2nd Law of Thermodynamics 01 • The total entropy increases in a spontaneous process and remains unchanged in an equilibrium process. • Spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0 • Equilibrium: ∆Stotal = ∆Ssys + ∆Ssur = 0 • The system is what you observe; surroundings are everything else.

  27. 2nd Law of Thermodynamics H -D sys S = D surr T Calculate change of entropy of surrounding in the following reaction: 2 H2(g) + O2(g) ---> 2 H2O(liq) ∆H ° = -571.7 kJ ∆Sosurroundings = +1917 J/K

  28. H H -D -D sys sys S = D sur T T -D H sys T -T (∆Stotal = ∆Ssys + ( ) ) < 0 • ∆Stotal = ∆Ssys + ( ) > 0 Spontaneous Reactions 01 • The 2nd law tells us a process will be spontaneous if ∆Stotal > 0 which requires a knowledge of ∆Ssurr.spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0

  29. -T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0 Gibbs Free Energy 02 • The expression –T∆Stotal is equated asGibbs free energy change (∆G)*, or simply free energy change: • ∆G = ∆Hsys – T∆Ssys • ∆G < 0 Reaction is spontaneous in forward direction.∆G = 0 Reaction is at equilibrium.∆G > 0 Reaction is spontaneous in reverse direction. –T∆Stotal = ΔG *Driving force of a reaction, Maximum work you can get from a system.

  30. Calculate ∆Go rxn for the following: C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g) Use enthalpies of formation to calculate ∆Horxn = -1256 kJ Use standard molar entropies to calculate ∆Sorxn ( see page 658, appendix A-10) ∆Sorxn = -97.4 J/K or -0.0974 kJ/K ∆Gorxn = -1256 kJ - (298 K)(-0.0974 kJ/K) = -1227 kJ Reaction is product-favored in spite of negative ∆Sorxn. Reaction is “enthalpy driven”

  31. Calculate DG° for the equation below using the thermodynamic data given: N2(g) + 3 H2(g) 2 NH3(g) DHf° (NH3) = 46 kJ; S° values (J/mol-K) are: NH3 = 192.5, N2 = 191.5, H2 = 130.6. +33 kJ +66 kJ 66 kJ 33 kJ

  32. Correct Answer: +33 kJ +66 kJ 66 kJ 33 kJ DH° = 92 kJ DS° = 198.3J/mol . K DG° = DH° TDS° DG° = (92)  (298)(0.1983) DG° = (92) + (59.2) = 32.9 KJ

  33. Gibbs Free Energy 03 Using ∆G = ∆H – T∆S, we can predict the sign of ∆G from the sign of ∆H and ∆S. 1) If both ∆H and ∆S are positive, ∆G will be negative only when the temperature value is large. Therefore, the reaction is spontaneous only at high temperature. 2) If ∆H is positive and ∆S is negative, ∆G will always be positive. Therefore, the reaction is not spontaneous

  34. Gibbs Free Energy: 04 ∆G = ∆H – T∆S 3) If ∆H is negative and ∆S is positive, ∆G will always be negative. Therefore, the reaction is spontaneous 4) If both ∆H and ∆S are negative, ∆G will be negative only when the temperature value is small. Therefore, the reaction is spontaneous only at Low temperatures.

  35. Gibbs Free Energy 04

  36. Gibbs Free Energy 06 • What are the signs (+, –, or 0) of ∆H, ∆S, and ∆G for the following spontaneous reaction of A atoms (red) and B atoms (blue)?

  37. -T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0 Gibbs Free Energy 02 • The expression –T∆Stotal is equated asGibbs free energy change (∆G)*, or simply free energy change: • ∆G = ∆Hsys – T∆Ssys • ∆G < 0 Reaction is spontaneous in forward direction.∆G = 0 Reaction is at equilibrium.∆G > 0 Reaction is spontaneous in reverse direction. –T∆Stotal = ΔG *Driving force of a reaction, Maximum work you can get from a system.

  38. N2(g) + 3H2(g) 2NH3(g) -198.7 J 1 kJ K 1000 J Standard Free-Energy Changes for Reactions Calculate the standard free-energy change at 25 °C for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes: DH° = -92.2 kJ DS° = -198.7 J/K DG° = DH° - TDS° -92.2 kJ - 298 K = x x = -33.0 kJ

  39. Gibbs Free Energy 05 • Iron metal can be produced by reducing iron(III) oxide with hydrogen: Fe2O3(s) + 3 H2(g)  2 Fe(s) + 3 H2O(g) ∆H° = +98.8 kJ; ∆S° = +141.5 J/K • Is this reaction spontaneous at 25°C? • At what temperature will the reaction become spontaneous?

  40. Decomposition of CaCO3 has a H° = 178.3 kJ/mol and S° = 159 J/mol K. At what temperature does this become spontaneous? 121°C 395°C 848°C 1121°C

  41. Correct Answer: 121°C 395°C 848°C 1121°C T = H°/S° T = 178.3 kJ/mol/0.159 kJ/mol K T = 1121 K T (°C) = 1121 – 273 = 848

  42. aA + bB cC + dD Standard Free Energies of Formation DG° = DG°f (products) - DG°f (reactants) DG° = [cDG°f (C) + dDG°f (D)] - [aDG°f (A) + bDG°f (B)] Products Reactants

  43. Standard free energy of formation (DG0f) is the free-energy change that occurs when1 mole of the compound is formed from its elements in their standard ( 1 atm) states.

  44. 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 DG0 DG0 rxn rxn DG0 of any element in its stable form is zero. f dDG0 (D) nDG0 (products) bDG0 (B) cDG0 (C) aDG0 (A) f f f f f --

  45. Calculate DG° for the equation below using the Gibbs free energy data given: 2 SO2(g) + O2(g) 2 SO3(g) DGf° values (kJ): SO2(g) = 300.4, SO3(g) = 370.4 +70 kJ +140 kJ 140 kJ 70 kJ

  46. Correct Answer: +70 kJ +140 kJ 140 kJ 70 kJ ( ) ( )   ° °  ° =   G G products G reactants  f f DG° = (2  370.4)  [(2  300.4) + 0] DG° = 740.8  [600.8] = 140

  47. Gibbs Free Energy 07

  48. Calculation of Nonstandard ∆G • The sign of ∆G° tells the direction of spontaneous reaction when both reactants and products are present at standard state conditions. • Under nonstandard( condition where pressure is not 1atm or concentrations of solutions are not 1M) conditions, ∆G˚ becomes ∆G.∆G = ∆G˚ + RT lnQ • The reaction quotient is obtained in the same way as an equilibrium expression.

  49. 2C(s) + 2H2(g) C2H4(g) P P H2 C2H4 2 ln = -11.51 0.10 1002 Free Energy Changes and the Reaction Mixture DG = DG° + RT ln Q Calculate DG for the formation of ethylene (C2H4) from carbon and hydrogen at 25 °C when the partial pressures are 100 atm H2 and 0.10 atm C2H4. Qp = Calculate ln Qp:

  50. 1 kJ 1000 J J K mol Free Energy Changes and the Reaction Mixture Calculate DG: DG = DG° + RT ln Q 8.314 = 68.1 kJ/mol + (298 K)(-11.51) ( see page A-13 of your book, page 667) DG = 39.6 kJ/mol

More Related