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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.

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Chapter 20

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  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. First law of thermodynamics • Law of conservation of energy: Energy can be neither created nor destroyed. • ∆E= q + w (q=heat, w=work, E=internal energy) • E univ= E sys + E surr • Heat gained by system is lost by surroundings and vice-versa. • Total energy of Universe is constant ∆E sys + ∆E surr=0=∆E univ

  4. 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.

  5. Spontaneous • It occurs without outside intervention

  6. 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

  7. Enthalpy • H=E+PV, • E=internal energy, • P=pressure, • V=volume

  8. 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.

  9. 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.

  10. Entropy: • S, A measure of molecular randomness or disorder. • Thermodynamic function that describes number of arrangements that are available to a system existing in a given state. • Probability of occurrence of a particular arrangement(state) depends on the number of ways(microstates) in which it can be arranged. • S=k ln W, k= Boltzmann constant=R/Av number=1.38 x 10-23J/K. • Un its of S are J/K

  11. Entropy • More the microstates available, higher is the entropy. Ex: solid →liquid→gas • Entropy is a state function depends only on present state and not its path. • ∆ Ssys= Sfinal – Sinitial , ∆ Ssys > 0, as entropy increases, • ∆ Ssys < 0, as entropy decreases.

  12. Entropy • If there are n number of molecules, relative probability of finding all molecules in one place is: (1/2)n • ∆ Ssys= 5.76 J/K for 1 mol • ∆ Ssys=qrev/T , T=temp, q=heat absorbed, rev=reversible process

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

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

  15. Entropy and Second Law of Thermodynamics: • Second Law of Thermo: In any spontaneous process there is always an increase in the entropy of the universe. OR The entropy of the universe is increasing. • ∆Suniv= ∆Ssys + ∆Ssurr • If ∆Suniv= +ve, process is spontaneous in direction written. • ∆Suniv= -ve, process is spontaneous in the opposite direction. • ∆Suniv= 0, process has no tendency to occur, system is at equilibrium

  16. Third Law of Thermodynamics • the entropy of a perfect crystal at 0 K is zero

  17. 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

  18. 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.

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

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

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

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

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

  24. Answer the following: • 1. Will solid or gaseous CO2 have higher positional entropy? • 2. Will N2 gas at 1 atm or 1.0 x 10-2 atm have higher positional entropy? • 3. When solid sugar is dissolved in water to form a solution, what will be the sign of entropy change? • 4. When iodine vapors are condensed to form crystals, what will be the sign of entropy change?

  25. 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

  26. 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.

  27. 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

  28. Entropy changes in the surrounding: • Exothermic change: Heat is lost by the system and gained by the surroundings. • q sys< 0 , q surr >0, ∆ Ssurr > 0 • Endothermic Change: Heat gained by the system is lost by the surroundings. • q sys> 0 , q surr <0, ∆ Ssurr < 0 • The change is entropy of the surroundings is α opposite charge in the heat of a system 1/α temperature at which heat is transferred.

  29. Entropy changes in the surrounding • ∆Ssurr = -q sys/T • ∆Ssurr = -∆H sys/T • Sign of ∆Ssurr depends on the direction of heat flow. When an exothermic process occurs ∆Ssurr is + ve. Opposite is true for endothermic process. Nature tends to seek the lowest possible energy. • Magnitude of ∆Ssurr depends on the temperature. Transfer of heat at low temperature produces a greater percent change in the randomness of the surroundings.

  30. 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.

  31. Problem • P.3. Sb2 S3(s) + 3Fe(s)______ 2Sb (s) + 3FeS (s) ∆H=-125kJ • Sb4 O6(s) + 6C(s)______ 4Sb (s) + 6 CO (s) ∆H= 778 kJ • Calculate ∆Ssurr for both of these reactions at 25 C and 1 atm

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

  33. Components of DS0universe for spontaneous reactions Figure 20.10 exothermic endothermic system becomes more disordered exothermic system becomes more disordered ∆Ssurr must be smaller system becomes more ordered ∆Ssurr must be larger

  34. 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

  35. Free Energy • G= H-TS, H= enthalpy, T=temp in K, S= entropy. • Also called as Gibbs Free Energy. • Free energy change (∆G) is a measure of the spontaneity of a process and of the useful energy available from it. • ∆ G= ∆H-T∆S(constant temp) • ∆Suniv = -∆G/T • Process is spontaneous in the direction in which the free energy decreases. (constant T and P)

  36. Free Energy • ∆S univ =0, at equilibrium • ∆S univ > 0, for a spontaneous process. • ∆S univ < 0, for a nonspontaneous process. • ∆G=0, at equilibrium • ∆G > 0, for a nonspontaneous process. • ∆G< 0, for a spontaneous process.

  37. Standard free energy of formation • ∆Gf0 is defined as a change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states. • Standard free energy of formation of an element in its standard state is zero. • An equation coefficient x ∆Gf0 by the number. • Reversing a reaction changes ∆Gf0

  38. Problem • P.4. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO 2(g) + 4 H2O (g) • Calculate ∆G0 -163 0 -394 -229 ∆Gf0 kJ/mol

  39. 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.

  40. 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

  41. Problems P.6. Use ∆Gf0 values to calculate free energy change at 25 C for the following: 2NO (g) + O2(g) → 2NO2 (g) P.7. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO 2(g) + 4 H2O (g) Calculate ∆G0 -163 0 -394 -229 ∆Gf0 kJ/mol P.8. Is the reaction C2H4 (g) + H2O (l) _______ C2H5OH (l) spontaneous under standard conditions?

  42. 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

  43. ∆G and work a system can do: • ∆G < 0, for a spontaneous process., ∆G=-wmax, maximum useful work obtainable from the system. • ∆G > 0, for a nonspontaneous process., minimum work that must be done to the system to make the process take place. • In any real process work is performed irreversibly and some of the free energy is lost to the surroundings as heat. • You can never obtain maximum possible work. • You need more than minimum energy to start the process for a nonspontaneous type. • At equilibrium a reaction can no longer do any work.

  44. Effect of Temperature on Reaction spontaneity 1.Reaction is spontaneous at all temperatures: ∆H < 0, ∆S >0, -T ∆S= -ve, ∆G =-ve 2. Reaction is nonspontaneous at all temperatures: ∆H > 0, ∆S <0, -T ∆S= +ve,∆G =+ve 3. Reaction is spontaneous at higher temperatures: ∆H > 0, ∆S >0, -T ∆S favors spontaneity at high temp , ∆G =-ve and reaction is spontaneous. 4. Reaction is spontaneous at lower temperatures: ∆H < 0, ∆S <0, ∆H favors spontaneity at low temp , Reaction occurs if -T ∆S< ∆H

  45. - + - - + - + + + + - + 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

  46. The effect of temperature on reaction spontaneity. Figure 20.11 Reaction between Cu2O and C. Relatively const ∆H and steadily increasing T∆S

  47. 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

  48. 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

  49. Free Energy and Equilibrium: • G = G + RT ln(Q) • Q = reaction quotient from the law of mass action. • Equilibrium point occurs at the lowest value of free energy available to the reaction system. • G = RT ln(K)

  50. 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 .

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