1 / 39

Chapter 20

Chapter 20. Thermodynamics:. Entropy, Free Energy and the Direction of Chemical Reactions. Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions. 20.1 The Second Law of Thermodynamics: Predicting Spontaneous Change.

chantale-irwin
Download Presentation

Chapter 20

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. Chapter 20 Thermodynamics: Entropy, Free Energy and the Direction of Chemical Reactions

  2. Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 20.1The Second Law of Thermodynamics: Predicting Spontaneous Change 20.2Calculating the Change in Entropy of a Reaction 20.3Entropy, Free Energy, and Work 20.4Free Energy, Equilibrium, and Reaction Direction

  3. Limitations of the First Law of Thermodynamics DE = q + w Euniverse= Esystem + Esurroundings DEsystem = -DEsurroundings DEsystem + DEsurroundings = 0 = DEuniverse The total energy-mass of the universe is constant. However, this does not tell us anything about the direction of change in the universe.

  4. water . Ba(OH)2 8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l) Figure20.1 A spontaneous endothermic chemical reaction. DH0rxn = +62.3 kJ

  5. more order less order solid liquid gas more order less order crystal + liquid ions in solution more order less order crystal + crystal gases + ions in solution The Concept of Entropy (S) Entropy refers to the state of order. A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process.

  6. stopcock opened 0.5 atm 0.5 atm Spontaneous expansion of a gas Figure 20.2 stopcock closed 1 atm evacuated

  7. Figure 20.3 Expansion of a gas and the increase in number of microstates.

  8. S = k ln W 1877 Ludwig Boltzman where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant), R/NA (R = universal gas constant, NA = Avogadro’s number. • A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy. • A system with many equivalent ways to arrange its components (larger W) has relatively more disorder and high entropy. DSuniverse = DSsystem + DSsurroundings > 0 This is the second law of thermodynamics.

  9. Random motion in a crystal Figure 20.4 The third law of thermodynamics. A perfect crystal has zero entropy at a temperature of absolute zero. Ssystem = 0 at 0 K

  10. Predicting Relative S0 Values of a System 1. Temperature changes S0 increases as the temperature rises. 2. Physical states and phase changes S0 increases as a more ordered phase changes to a less ordered phase. 3. Dissolution of a solid or liquid S0 of a dissolved solid or liquid is usually greater than the S0 of the pure solute. However, the extent depends upon the nature of the solute and solvent. 4. Dissolution of a gas A gas becomes more ordered when it dissolves in a liquid or solid. 5. Atomic size or molecular complexity In similar substances, increases in mass relate directly to entropy. In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy.

  11. The increase in entropy from solid to liquid to gas. Figure 20.5

  12. MIX solution Figure 20.6 The entropy change accompanying the dissolution of a salt. pure solid pure liquid

  13. Solution of ethanol and water Ethanol Water Figure 20.7 The small increase in entropy when ethanol dissolves in water.

  14. O2 gas O2 gas in H2O Figure 20.8 The large decrease in entropy when a gas dissolves in a liquid.

  15. N2O4 NO2 Entropy and vibrational motion. Figure 20.9 NO

  16. PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]: PLAN: In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature. Sample Problem 20.1: Predicting Relative Entropy Values (a) 1mol of SO2(g) or 1mol of SO3(g) (b) 1mol of CO2(s) or 1mol of CO2(g) (c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3) (d) 1mol of KBr(s) or 1mol of KBr(aq) (e) Seawater in midwinter at 20C or in midsummer at 230C (f) 1mol of CF4(g) or 1mol of CCl4(g) SOLUTION: (a) 1mol of SO3(g) - more atoms (d) 1mol of KBr(aq) - solution > solid (b) 1mol of CO2(g) - gas > solid (e) 230C - higher temperature (c) 3mol of O2(g) - larger #mols (f) CCl4 - larger mass

  17. Hsystem Ssurroundings = - T Entropy Changes in the System S0rxn - the entropy change that occurs when all reactants and products are in their standard states. S0rxn = S0products - S0reactants The change in entropy of the surroundings is directly related to an opposite change in the heat of the system and inversely related to the temperature at which the heat is transferred.

  18. PROBLEM: Calculate DS0rxn for the combustion of 1mol of propane at 250C. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) PLAN: Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6 mols of gas to 3 mols of gas. SOLUTION: Find standard entropy values in the Appendix or other table. Sample Problem 20.2: Calculating the Standard Entropy of Reaction, DS0rxn DS = [(3 mol)(S0 CO2) + (4 mol)(S0 H2O)] - [(1 mol)(S0 C3H8) + (5 mol)(S0 O2)] DS = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) + (5 mol)(205.0J/mol*K)] DS = - 374 J/K

  19. PROBLEM: At 298K, the formation of ammonia has a negative DS0sys; N2(g) + 3H2(g) 2NH3(g) DS0sys = -197 J/K PLAN: DS0universe must be > 0 in order for this reaction to be spontaneous, so DS0surroundings must be > 197 J/K. To find DS0surr, first find DHsys; DHsys = DHrxn which can be calculated using DH0f values from tables. DS0universe = DS0surr + DS0sys. Sample Problem 20.3: Determining Reaction Spontaneity Calculate DS0rxn, and state whether the reaction occurs spontaneously at this temperature. SOLUTION: DH0rx = [(2 mol)(DH0fNH3)] - [(1 mol)(DH0fN2) + (3 mol)(DH0fH2)] DH0rx = -91.8 kJ DS0surr = -DH0sys/T = -(-91.8x103J/298K) = 308 J/K DS0universe = DS0surr + DS0sys = 308 J/K + (-197 J/K) = 111 J/K DS0universe > 0 so the reaction is spontaneous.

  20. PROBLEM: Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated. Note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation). D 4KClO3(s) 3KClO4(s) + KCl(s) PLAN: Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve. Sample Problem 20.4: Calculating DG0 from Enthalpy and Entropy Values +5 +7 -1 Use DH0f and S0 values to calculate DG0sys (DG0rxn) at 250C for this reaction. SOLUTION: DH0rxn= SmDH0products - SnDH0reactants DH0rxn= (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) - (4 mol)(-397.7 kJ/mol) DH0rxn= -144 kJ

  21. Sample Problem 20.4: Calculating DG0 from Enthalpy and Entropy Values continued DS0rxn= SmDS0products - SnDS0reactants DS0rxn= (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) - (4 mol)(143.1 J/mol*K) DS0rxn= -36.8 J/K DG0rxn= DH0rxn - T DS0rxn DG0rxn= -144 kJ - (298K)(-36.8 J/K)(kJ/103 J) DG0rxn= -133 kJ

  22. Figure B20.1 Lavoisier studying human respiration as a form of combustion. Figure B20.2 A whole-body calorimeter.

  23. Components of DS0universe for spontaneous reactions Figure 20.10 exothermic endothermic system becomes more disordered exothermic system becomes more disordered system becomes more ordered

  24. Gibbs Free Energy (G) G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it. DG0system = DH0system - TDS0system G < 0 for a spontaneous process G > 0 for a nonspontaneous process G = 0 for a process at equilibrium DG0rxn = S mDG0products - SnDG0reactants

  25. 2SO2(g) + O2(g) 2SO3(g) Sample Problem 20.6: Determining the Effect of Temperature on DG0 PROBLEM: An important reaction in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g): At 298K, DG0 = -141.6kJ; DH0 = -198.4kJ; and DS0 = -187.9J/K (a) Use the data to decide if this reaction is spontaneous at 250C, and predict how DG0 will change with increasing T. (b) Assuming DH0 and DS0 are constant with increasing T, is the reaction spontaneous at 900.0C? PLAN: The sign of DG0 tells us whether the reaction is spontaneous and the signs of DH0 and DS0 will be indicative of the T effect. Use the Gibbs free energy equation for part (b). SOLUTION: (a) The reaction is spontaneous at 250C because DG0 is (-). Since DH0 is (-) but DS0 is also (-), DG0 will become less spontaneous as the temperature increases.

  26. Sample Problem 20.6: Determining the Effect of Temperature on DG0 continued (b) DG0rxn= DH0rxn - T DS0rxn DG0rxn= -198.4kJ - (1173K)(-187.9J/mol*K)(kJ/103J) DG0rxn= 22.0 kJ; the reaction will be nonspontaneous at 900.0C

  27. DG and the Work a System Can Do For a spontaneous process, DG is the maximum work obtainablefrom the system as the process takes place: DG = workmax For a nonspontaneous process, DG is the maximum work that must be done to the system as the process takes place: DG = workmax An example

  28. - + - - + - + + + + - + or - - - + + or - Table 20.1 Reaction Spontaneity and the Signs of DH0, DS0, and DG0 DH0 DS0 -TDS0 DG0 Description Spontaneous at all T Nonspontaneous at all T Spontaneous at higher T; nonspontaneous at lower T Spontaneous at lower T; nonspontaneous at higher T

  29. D 4KClO3(s) 3KClO4(s) + KCl(s) PLAN: Use the DG summation equation. SOLUTION: DG0rxn= SmDG0products - SnDG0reactants Sample Problem 20.5: Calculating DG0rxn from DG0f Values PROBLEM: Use DG0f values to calculate DGrxn for the reaction in Sample Problem 20.4: DG0rxn= (3mol)(-303.2kJ/mol) + (1mol)(-409.2kJ/mol) - (4mol)(-296.3kJ/mol) DG0rxn= -134kJ

  30. The effect of temperature on reaction spontaneity. Figure 20.11

  31. Figure B20.3 The coupling of a nonspontaneous reaction to the hydrolysis of ATP.

  32. The cycling of metabolic free enery through ATP Figure B20.4

  33. Why is ATP a high-energy molecule? Figure B20.5

  34. Free Energy, Equilibrium and Reaction Direction • If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (DG < 0) • If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (DG > 0) • If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (DG = 0) DG = RT ln Q/K = RT lnQ - RT lnK Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln Q = 0 so DG0 = - RT lnK

  35. Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent FORWARD REACTION REVERSE REACTION Forward reaction goes to completion; essentially no reverse reaction Table 20.2 The Relationship Between DG0 and K at 250C DG0(kJ) K Significance 200 9x10-36 100 3x10-18 50 2x10-9 10 2x10-2 1 7x10-1 0 1 -1 1.5 -10 5x101 -50 6x108 -100 3x1017 -200 1x1035

  36. 2SO2(g) + O2(g) 2SO3(g) Sample Problem 20.7: Calculating DG at Nonstandard Conditions PROBLEM: The oxidation of SO2, which we considered in Sample Problem 20.6 is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature. (a) Calculate K at 298K and at 973K. (DG0298 = -141.6kJ/mol of reaction as written using DH0 and DS0 values at 973K. DG0973 = -12.12kJ/mol of reaction as written.) (b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO2, 0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate DG for the system in part (b) at each temperature. PLAN: Use the equations and conditions found on slide .

  37. (-12.12kJ/mol)(103J/kJ) (-141.6kJ/mol)(103J/kJ) At 298, the exponent is -DG0/RT = - (8.314J/mol*K)(298K) (8.314J/mol*K)(973K) Sample Problem 20.7: Calculating DG at Nonstandard Conditions continued (2 of 3) SOLUTION: (a) Calculating K at the two temperatures: DG0 = -RTlnK so = 57.2 = e57.2 = 7x1024 At 973, the exponent is -DG0/RT = 1.50 = e1.50 = 4.5

  38. pSO32 (0.100)2 = (pSO2)2(pO2) (0.500)2(0.0100) Sample Problem 20.7: Calculating DG at Nonstandard Conditions continued (3 of 3) (b) The value of Q = = 4.00 Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight. (c) The nonstandard DG is calculated using DG = DG0 + RTlnQ DG298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(ln4.00) DG298 = -138.2kJ/mol DG973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(ln4.00) DG298 = -0.9kJ/mol

  39. Figure 20.12 The relation between free energy and the extent of reaction. DG0 < 0 K >1 DG0 > 0 K <1

More Related