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Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Thermodynamics: Entropy, Free Energy and the Direction of Chemical Reactions Dr.ssa Rossana Galassi 320 4381420 [email protected] 20-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 1 The Second Law of Thermodynamics: Predicting Spontaneous Change 2 Calculating the Change in Entropy of a Reaction 3 Entropy, Free Energy, and Work 4 Free Energy, Equilibrium, and Reaction Direction 20-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. 20-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A spontaneous endothermic chemical reaction. water Ba(OH)2.8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l) DH0rxn = +62.3 kJ 20-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. more order solid more order crystal + liquid more order crystal + crystal 20-5 less order liquid gas less order ions in solution less order gases + ions in solution Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Spontaneous expansion of a gas stopcock closed 1 atm evacuated stopcock opened 0.5 atm 20-6 0.5 atm Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Expansion of a gas and the increase in number of microstates. 20-7 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1877 Ludwig Boltzman S = k ln W 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. 20-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Random motion in a crystal The third law of thermodynamics. A perfect crystal has zero entropy at a temperature of absolute zero. Ssystem = 0 at 0 K 20-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. 20-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The increase in entropy from solid to liquid to gas. 20-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The entropy change accompanying the dissolution of a salt. pure solid MIX pure liquid solution 20-12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The small increase in entropy when ethanol dissolves in water. Ethanol 20-13 Water Solution of ethanol and water Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The large decrease in entropy when a gas dissolves in a liquid. O2 gas O2 gas in H2O 20-14 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Entropy and vibrational motion. NO NO2 20-15 N2O4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.1: Predicting Relative Entropy Values 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)]: (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) PLAN: In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature. 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 20-16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Entropy Changes in the System DS0rxn - the entropy change that occurs when all reactants and products are in their standard states. DS0rxn = DS0products - DS0reactants 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. DSsurroundings = - 20-17 DHsystem T Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.2: PROBLEM: Calculating the Standard Entropy of Reaction, DS0rxn 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. 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 20-18 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.3: PROBLEM: Determining Reaction Spontaneity At 298K, the formation of ammonia has a negative DS0sys; N2(g) + 3H2(g) 2NH3(g) DS0sys = -197 J/K Calculate DS0rxn, and state whether the reaction occurs spontaneously at this temperature. PLAN: DS0universe must be > 0 in order for this reaction to be spontaneous, so DS0surroundings must be > 197 J/K. To find DS 0surr, first find DHsys; DHsys = DHrxn which can be calculated using DH0f values from tables. DS0universe = DS0surr + DS0sys. 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. 20-19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.4: PROBLEM: Calculating DG0 from Enthalpy and Entropy Values 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). +5 D 4KClO3(s) +7 -1 3KClO4(s) + KCl(s) Use DH0f and S0 values to calculate DG0sys (DG0rxn) at 250C for this reaction. PLAN: Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve. SOLUTION: DH0rxn = mDH0products - nDH0reactants DH0rxn = (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) (4 mol)(-397.7 kJ/mol) DH0rxn = -144 kJ 20-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.4: Calculating DG0 from Enthalpy and Entropy Values continued DS0rxn = mDS0products - nDS0reactants 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 20-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Lavoisier studying human respiration as a form of combustion. A whole-body calorimeter. 20-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Components of DS0universe for spontaneous reactions exothermic endothermic system becomes more disordered exothermic system becomes more disordered system becomes more ordered 20-23 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Gibbs Free Energy (G) DG, 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 DG < 0 for a spontaneous process DG > 0 for a nonspontaneous process DG = 0 for a process at equilibrium DG0rxn = mDG0products - nDG0reactants 20-24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.6: PROBLEM: Determining the Effect of Temperature on DG0 An important reaction in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g): 2SO2(g) + O2(g) 2SO3(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 25 0C, 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 DG 0 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. 20-25 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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 20-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. DG and the Work a System Can Do For a spontaneous process, DG is the maximum work obtainable from 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 20-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Reaction Spontaneity and the Signs of DH0, DS0, and DG0 DH0 DS0 -TDS0 DG0 - + - - Spontaneous at all T + - + + Nonspontaneous at all T + + - + or - Spontaneous at higher T; nonspontaneous at lower T - - + + or - Spontaneous at lower T; nonspontaneous at higher T 20-28 Description Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.5: PROBLEM: PLAN: Calculating DG0rxn from DG0f Values Use DG0f values to calculate DGrxn for the reaction in Sample Problem 20.4: D 4KClO3(s) 3KClO4(s) + KCl(s) Use the DG summation equation. SOLUTION: DG0rxn = mDG0products - nDG0reactants DG0rxn = (3mol)(-303.2kJ/mol) + (1mol)(-409.2kJ/mol) (4mol)(-296.3kJ/mol) DG0rxn = -134kJ 20-29 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The effect of temperature on reaction spontaneity. 20-30 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The coupling of a nonspontaneous reaction to the hydrolysis of ATP. 20-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The cycling of metabolic free enery through ATP 20-32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Why is ATP a high-energy molecule? 20-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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 20-34 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Relationship Between DG0 and K at 250C DG0(kJ) 100 3x10-18 50 2x10-9 10 2x10-2 1 7x10-1 0 1 1.5 -10 5x101 -50 6x108 -100 3x1017 -200 1x1035 Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent Forward reaction goes to completion; essentially no reverse reaction REVERSE REACTION 9x10-36 Significance FORWARD REACTION 200 -1 20-35 K Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.7: Calculating DG at Nonstandard Conditions PROBLEM: The oxidation of SO2, which we considered in Sample Problem 20.6 2SO2(g) + O2(g) 2SO3(g) 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 20-36 . Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.7: Calculating DG at Nonstandard Conditions continued (2 of 3) SOLUTION: (a) Calculating K at the two temperatures: DG0 (DG / RT) = -RTlnK so K e 0 (-141.6kJ/mol)(103J/kJ) At 298, the exponent is -DG0/RT = - = 57.2 (8.314J/mol*K)(298K) (DG / RT) Ke 0 = e57.2 = 7x1024 (-12.12kJ/mol)(103J/kJ) At 973, the exponent is -DG0/RT = 1.50 (8.314J/mol*K)(973K) (DG / RT) Ke 20-37 0 = e1.50 = 4.5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Calculating DG at Nonstandard Conditions Sample Problem 20.7: continued (3 of 3) pSO32 (b) The value of Q = (pSO2)2(pO2) (0.100)2 = = 4.00 (0.500)2(0.0100) 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 20-38 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The relation between free energy and the extent of reaction. DG0 < 0 K >1 20-39 DG0 > 0 K <1