Thermodynamics Entropy, Free Energy and the Direction of Chemical Reactions 20-1 Thermodynamics Entropy,GIBBS Free Energy, and the Direction of Chemical Reactions 20.1 The Second Law of Thermodynamics: ENTROPY 20.2 OBJECTIVES:1. Understanding Entropy in our life 2.Predicting spontaneous reaction by its increase or decrease 3. Calculating the change in entropy of a reaction by the amount of heat 20.3 Free energy, equilibrium and reaction direction (we will consider later in the semester) 20-2 The Larger Question A B What factors must be identified in order to predict whether the reaction will proceed spontaneously in the direction written (as opposed to in the opposite direction)? Enthalpy and entropy changes are combined to yield a new term, the Gibbs free energy. The sign of the latter allows predictions of reaction spontaneity. These are thermodynamic, not kinetic, considerations! 20-3 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, these energy changes do not explain the direction of change in the universe! 20-4 A spontaneous endothermic chemical reaction Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. water Ba(OH)2.8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l) DHorxn = + 62.3 kJ Figure 20.1 20-5 The sign of the enthalpy change is insufficient to predict reaction spontaneity! The Missing Factor: Entropy (S) Entropy refers to the state of order/disorder of a system. A change in order is a change in the number of ways of arranging the particles. It is a key factor in determining the direction of a spontaneous process. more order solid more order crystal + liquid more order crystal + crystal 20-6 less order liquid gas less order ions in solution less order gases + ions in solution The number of ways to arrange a deck of playing cards Figure 20.2 20-7 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Spontaneous expansion of a gas This process occurs spontaneously without a change in the total internal energy of the system! stopcock closed evacuated 1 atm This is an entropydriven process! stopcock opened Figure 20.3 20-8 0.5 atm Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 0.5 atm ENTROPY, SI UNIT IS J/K. 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. The Second Law of Thermodynamics DSuniv = DSsys + DSsurr > 0 Entropy is a state function,means it depends only on the start and endpoint. 20-9 All processes occur spontaneously in the direction that increases the entropy of the universe (system + surroundings). 20-10 Random motion in a crystal (increased arrangements, greater S) The Third Law of Thermodynamics A perfect crystal has zero entropy at a temperature of absolute zero. Ssystem = 0 at 0 K Figure 20.4 20-11 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Third Law leads to absolute values for the entropies of substances! Standard Molar Entropies (So) Standard states: 1 atm for gases; 1 M for solutions; pure substance in its most stable form for solids and liquids. units: J/mol/K at 25 oC This value equals the entropy increase in a substance upon raising its temperature from 0 K to the specified temperature. 20-12 Predicting relative So values of a system 1. Temperature changes So increases as temperature rises. 2. Physical states and phase changes So increases as a more ordered phase changes to a less ordered phase. 3. Dissolution of a solid or liquid So of a dissolved solid or liquid is usually greater than So of the pure solute. However, the extent depends on 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. 20-13 The increase in entropy from solid to liquid to gas Large changes in S occur at phase transitions! Figure 20.5 20-14 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 salt becomes more disordered (dissociation): S increases water becomes more ordered (ion-dipole interactions): S decreases Figure 20.6 20-15 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The small increase in entropy when ethanol dissolves in water ethanol water Freedom of movement remains essentially unchanged; S increase is due solely to random mixing effects. Figure 20.7 20-16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. solution of ethanol and water The large decrease in entropy when a gas dissolves in a liquid O2 gas O2 dissolved Figure 20.8 20-17 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Entropy and vibrational motion more motions, greater S NO NO2 N 2O 4 Internal molecular motions influence entropy! Figure 20.9 20-18 20-19 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) 1 mol of SO2(g) or 1 mol of SO3(g) (b) 1 mol of CO2(s) or 1 mol of CO2(g) (c) 3 mol of oxygen gas (O2) or 2 mol of ozone gas (O3) (d) 1 mol of KBr(s) or 1 mol of KBr(aq) (e) seawater in mid-winter at 2 oC or in mid-summer at 23 oC (f) 1 mol of CF4(g) or 1 mol of CCl4(g) PLAN: In general less ordered systems have higher entropy than ordered systems and entropy increases with increasing temperature. SOLUTION: (a) 1 mol of SO3(g) - more atoms (d) 1 mol of KBr(aq) - solution > solid (b) 1 mol of CO2(g) - gas > solid (e) 23 oC - higher temperature (c) 3 mol of O2(g) - larger # mols (f) CCl4 - larger mass 20-20 Standard Entropies of Reaction, DSorxn By analogy to calculating standard heats of reaction: DHorxn DHorxn = S mDHof (products) - S nDHof (reactants) From Chapter 6 DSorxn = S mSo (products) - S nSo (reactants) As before, m and n are the appropriate coefficients in the balanced chemical equation. 20-21 Sample Problem 20.2 PROBLEM: Calculating the standard entropy of reaction, DSorxn Calculate DSorxn for the combustion of 1 mol of propane at 25 oC. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) PLAN: Use summation equations. Entropy is expected to decrease because the reaction goes from 6 moles of gas to 3 moles of gas. SOLUTION: Find standard entropy values in an appropriate data table. DSorxn = [(3 mol)(So CO2) + (4 mol)(So H2O)] - [(1 mol)(So C3H8) + (5 mol)(So O2)] DSorxn = [(3 mol)(213.7 J/mol.K) + (4 mol)(69.9 J/mol.K)] - [(1 mol)(269.9 J/mol.K) + (5 mol)(205.0 J/mol.K)] DSorxn = - 374 J/K 20-22 We can’t forget entropy changes in the surroundings! For spontaneous reactions in which a decrease in the entropy of the system occurs: must be outweighed by a concomitant increase in the entropy of the surroundings (DSuniv > 0) For an exothermic process: qsys < 0, qsurr > 0, DSsurr > 0 For an endothermic process: qsys > 0, qsurr < 0, DSsurr < 0 The initial temperature of the surroundings affects the magnitude of DSsurr. DSsurr a -qsys 20-23 and DSsurr a 1/T Combining these ideas...... DSsurr = -qsys/T For a process of constant pressure, where qp = DH: DSsurr = -DHsys/T Implication: DSsurr can be calculated by measuring DHsys and the temperature at which the change occurs. 20-24 Sample Problem 20.3 PROBLEM: Determining reaction spontaneity At 298 K, the formation of ammonia has a negative DSosys. N2(g) + 3H2(g) 2NH3(g) DSosys = -197 J/K Calculate DSouniv and state whether the reaction occurs spontaneously at this temperature. PLAN: DSouniv must be > 0 in order for the reaction to be spontaneous, so DSosurr must be > +197 J/K. To find DS osurr, first find DHsys. DHosys = DHorxn which can be calculated using DHof values from data tables. Then apply DSouniv = DSosurr + DSosys. SOLUTION: DHorxn = [(2 mol)(DHof NH3)] - [(1 mol)(DHof N2) + (3 mol)(DHof H2)] DHorxn = -91.8 kJ DSosurr = -DHosys/T = -(-91.8 x 103J/298 K) = +308 J/K (>+197 J/K) DSouniv = DSosurr + DSosys = 308 J/K + (-197 J/K) = +111 J/K DSouniv > 0: the reaction is spontaneous! 20-25 A whole-body calorimeter Living systems do not violate the Second Law DSuniv = DSsys + DSsurr > 0 Figure B20.2 20-26 What happens when equilibrium is reached? DSouniv = DSosurr + DSosys = 0 Examples: phase changes (fusion, vaporization) When a system reaches equilibrium, neither the forward nor the reverse reaction is spontaneous; neither proceeds further because there is no driving force. 20-27 Components of DSouniv for spontaneous reactions exothermic endothermic system becomes more disordered exothermic system becomes more disordered system becomes more ordered Figure 20.10 20-28 But measuring DSsys and DSsurr is inconvenient to determine reaction spontaneity More convenient to use parameters that apply only to the system! The Gibbs free energy (G) accomplishes this goal, where G = H - TS or DG = DH - TDS G combines the system’s enthalpy and entropy; thus G is a state function. Where does the equation come from? 20-29 Deriving the Gibbs free energy equation DSuniv = DSsys + DSsurr DSsurr = -DHsys/T DSuniv = DSsys - DHsys/T -TDSuniv = DHsys - TDSsys DGsys = DHsys - TDSsys DGsys = -TDSuniv DSuniv > 0 or DGsys < 0 for a spontaneous process DSuniv < 0 or DGsys > 0 for a nonspontaneous process DSuniv = 0 or DGsys = 0 for a process at equilibrium 20-30 Some Key Concepts If a process is nonspontaneous in one direction (DG > 0), then it is spontaneous in the opposite direction (DG < 0). The magnitude and sign of DG are unrelated to the rate (speed) of the reaction. 20-31 Standard Free Energy Changes A reference state; similar to DHo and So All components are in their standards states. DGosys = DHosys - TDSosys 20-32 Sample Problem 20.4 PROBLEM: Calculating DGo from enthalpy and entropy values Potassium chlorate, one of the common oxidizing agents in explosives, fireworks and match heads, undergoes a solid-state redox reaction when heated. In this reaction, the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (a disproportionation reaction). +5 +7 -1 D 4KClO3(s) 3KClO4(s) + KCl(s) Use DHof and So values to calculate DGosys (DGorxn) at 25 oC for this reaction. PLAN: Obtain appropriate thermodynamic data from a data table; insert them into the Gibbs free energy equation and solve. SOLUTION: DHorxn = S mDHof (roducts) - S nDHof (reactants) DHorxn = [(3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol)] [(4 mol)(-397.7 kJ/mol)] DHorxn = -144 kJ 20-33 Sample Problem 20.4 (continued) DSorxn = S mSoproducts - S nSoreactants DSorxn = [(3 mol)(151 J/mol.K) + (1 mol)(82.6 J/mol.K)] [(4 mol)(143.1 J/mol.K)] DSorxn = -36.8 J/K DGorxn = DHorxn - TDSorxn DGorxn = -144 kJ - (298 K)(-36.8 J/K)(kJ/103J) DG0rxn = -133 kJ The reaction is spontaneous. 20-34 Another way to calculate DGorxn From standard free energies of formation, DGof DGof = the free energy change that occurs when 1 mol of compound is made from its elements, with all components in their standard states. Thus, DGorxn = S mDGof (products) - S nDGof (reactants) DGof values have properties similar to DHof values. 20-35 Sample Problem 20.5 PROBLEM: Calculating DGorxn from DGof values Use DGof values to calculate DGorxn for the following reaction: 4KClO3(s) PLAN: D 3KClO4(s) + KCl(s) Use the DGorxn summation equation. SOLUTION: DGorxn = S mDGof (products) - S nDGof (reactants) DGorxn = [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] [(4 mol)(-296.3 kJ/mol)] DGorxn = -134 kJ 20-36 DG and Work 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 minimum work that must be done to the system to make the process take place. An example of a gas doing work 20-37 More Key Concepts A reversible process: one that can be changed in either direction by an infinitesimal change in a variable. The maximum work from a spontaneous process is obtained only if the work is carried out reversibly. In most real processes, work is performed irreversibly; thus, maximum work is not obtained; some free energy is lost to the surroundings as heat and is thus unavailable to do work. A reaction at equilibrium cannot do work. 20-38 Effect of Temperature on Reaction Spontaneity DGosys = DHosys - TDSosys The sign of DGosys is T-independent when DHosys and DSosys have opposite signs. The sign of DGosys is T-dependent when DHosys and DSosys have the same signs. 20-39 Table 20.1 Reaction Spontaneity and the Signs of DHo, DSo and DGo DHo DSo -TDSo DGo - + - - 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 20-40 description Sample Problem 20.6 PROBLEM: Determining the effect of temperature on DGo 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 298 K, DGo = -141.6 kJ, DHo = -198.4 kJ, and DSo = -187.9 J/K. (a) Use these data to decide if this reaction is spontaneous at 25 oC, and predict how DGo will change with increasing T. (b) Assuming DHo and DSo are constant with increasing T, is the reaction spontaneous at 900. oC? PLAN: The sign of DG o tells us whether the reaction is spontaneous and the signs of DHo and DSo will be indicative of the T effect. Use the Gibbs free energy equation for part (b). SOLUTION: (a) The reaction is spontaneous at 25 oC because DGo is (-). Since DHo is (-) and DSo is (-), DGo will become less negative and the reaction less spontaneous as temperature increases. 20-41 Sample Problem 20.6 (continued) (b) DGorxn = DHorxn - TDSorxn DGorxn = -198.4 kJ - [(1173 K)(-187.9 J/mol.K)(kJ/103J)] DGorxn = +22.0 kJ; the reaction will be nonspontaneous at 900.oC 20-42 Determining the Temperature at which a Reaction Becomes Spontaneous (only when DH and DS have the same sign) DGo = DHo - TDSo = 0 (solve for this condition) DHo = TDSo T = DHo/DSo 20-43 The effect of temperature on reaction spontaneity Cu2O(s) + C(s) 2Cu(s) + CO(g) DHo = +58.1 kJ DSo = +165 J/K DHo: relatively insensitive to T DSo: increases with increasing T Figure 20.11 20-44 The cycling of metabolic free energy through ATP Figure B20.4 20-45 The coupling of a nonspontaneous reaction to the hydrolysis of ATP D-glucose Figure B20.3 20-46 + ATP D-glucose 6P + ADP (catalyzed by the enzyme, hexokinase) Why is ATP a high-energy molecule? The DGo of ATP hydrolysis is -31 kJ/mol; considerably more (-) under in vivo conditions! Figure B20.5 20-47 End of Assigned Material 20-48 Free energy, equilibrium and reaction direction DG = RT ln Q/K = RT ln Q - RT ln K 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) Under standard conditions (1 M concentrations, 1 atm for gases), Q = 1 and ln Q = 0. Therefore: DGo = - RT ln K 20-49 Table 20.2 The relationship between DGo and K at 25 oC DGo (kJ) 100 3 x 10-18 50 2 x 10-9 10 2 x 10-2 1 7 x 10-1 0 1 Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to the same extent 1.5 -10 5 x 101 -50 6 x 108 -100 3 x 1017 -200 1 x 1035 Forward reaction goes to completion; essentially no reverse reaction REVERSE REACTION 9 x 10-36 significance FORWARD REACTION 200 -1 20-50 K Sample Problem 20.7 PROBLEM: Calculating DG at non-standard conditions The oxidation of SO2, 2SO2(g) + O2(g) 2SO3(g) is too slow at 298 K 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 298 K and at 973 K. (DGo298 = -141.6 kJ/mol for the reaction as written using DHo and DSo values. At 973 K, DGo973 = 12.12 kJ/mol for the reaction as written.) (b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500 atm of SO2, 0.0100 atm of O2, and 0.100 atm of SO3 and kept at 25 oC and at 700. oC. 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 previous slides. 20-51 Sample Problem 20.7 (continued) SOLUTION: (a) Calculating K at the two temperatures: DGo (DG / RT) = -RTln K so K e At 298 K, the exponent is -DGo/RT (DG / RT) Ke 0 (DG / RT) 20-52 0 (-141.6 kJ/mol)(103 J/kJ) =- (8.314 J/mol.K)(298 K) = 57.2 = e57.2 = 7 x 1024 At 973 K, the exponent is -DGo/RT = - Ke 0 = e1.50 = 4.5 (-12.12 kJ/mol)(103 J/kJ) (8.314 J/mol.K)(973 K) = 1.50 Sample Problem 20.7 (continued) (b) The value of Q = pSO32 (pSO2)2(pO2) = (0.100)2 (0.500)2(0.0100) = 4.00 Since Q is < K at both temperatures the reaction will shift right; for 298 K there will be a dramatic shift while at 973 K the shift will be slight. (c) The non-standard DG is calculated using DG = DGo + RTlnQ DG298 = -141.6 kJ/mol + (8.314 J/mol.K)(kJ/103J)(298 K)(ln 4.00) DG298 = -138.2 kJ/mol DG973 = -12.12 kJ/mol + (8.314 J/mol.K)(kJ/103J)(973 K)(ln 4.00) DG298 = -0.9 kJ/mol 20-53 The relation between free energy and the extent of reaction DGo < 0 K >1 DGo > 0 K <1 Figure 20.12 20-54 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
