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Chapter 20 – Thermodynamics

Chapter 20 – Thermodynamics. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change 20.2 – Calculating the Change in Entropy of a Reaction 20.3 – Entropy, Free Energy, and Work 20.4 – Free Energy, Equilibrium, and Reaction Direction. stopcock opened. 0.5 atm. 0.5 atm.

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Chapter 20 – Thermodynamics

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  1. Chapter 20 – Thermodynamics • 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change • 20.2 – Calculating the Change in Entropy of a Reaction • 20.3 – Entropy, Free Energy, and Work • 20.4 – Free Energy, Equilibrium, and Reaction Direction

  2. stopcock opened 0.5 atm 0.5 atm 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change Spontaneous expansion of a gas Figure 20.2 stopcock closed 1 atm evacuated

  3. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change Figure 20.3 Expansion of a gas and the increase in number of microstates.

  4. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change 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 The second law of thermodynamics.

  5. Random motion in a crystal Figure 20.4 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change The third law of thermodynamics. A perfect crystal has zero entropy at a temperature of absolute zero. Ssystem = 0 at 0 K

  6. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change The increase in entropy from solid to liquid to gas. Figure 20.5

  7. MIX solution 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change Figure 20.6 The entropy change accompanying the dissolution of a salt. pure solid pure liquid

  8. Solution of ethanol and water Ethanol Water 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change Figure 20.7 The small increase in entropy when ethanol dissolves in water.

  9. O2 gas O2 gas in H2O 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change Figure 20.8 The large decrease in entropy when a gas dissolves in a liquid.

  10. N2O4 NO2 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change Entropy and vibrational motion. Figure 20.9 NO

  11. 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. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change 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

  12. 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. 20.2 – Calculating the Change in Entropy of a Reaction 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

  13. 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. 20.2 – Calculating the Change in Entropy of a Reaction 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.

  14. 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. 20.3 – Entropy, Free Energy, and Work 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

  15. 20.3 – Entropy, Free Energy, and Work 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

  16. D 4KClO3(s) 3KClO4(s) + KCl(s) PLAN: Use the DG summation equation. SOLUTION: DG0rxn= SmDG0products - SnDG0reactants 20.3 – Entropy, Free Energy, and Work 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

  17. 2SO2(g) + O2(g) 2SO3(g) 20.3 – Entropy, Free Energy, and Work 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.

  18. 20.3 – Entropy, Free Energy, and Work 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

  19. 20.3 – Entropy, Free Energy, and Work Figure B20.3 The coupling of a nonspontaneous reaction to the hydrolysis of ATP.

  20. 20.3 – Entropy, Free Energy, and Work The cycling of metabolic free enery through ATP Figure B20.4

  21. 20.3 – Entropy, Free Energy, and Work Why is ATP a high-energy molecule? Figure B20.5

  22. 20.4 – Free Energy, Equilibrium, and Reaction Direction 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

  23. 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 20.4 – Free Energy, Equilibrium, and Reaction Direction 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

  24. 2SO2(g) + O2(g) 2SO3(g) 20.4 – Free Energy, Equilibrium, and Reaction Direction 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.

  25. (-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) 20.4 – Free Energy, Equilibrium, and Reaction Direction 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 = 7x1024 = e57.2 At 973, the exponent is -DG0/RT = 1.50 = 4.5 = e1.50

  26. pSO32 (0.100)2 = (pSO2)2(pO2) (0.500)2(0.0100) 20.4 – Free Energy, Equilibrium, and Reaction Direction 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

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