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AP* Chapter 17 Spontaneity, Entropy, and Free Energy Section 17.1 Spontaneous Processes and Entropy Thermodynamics vs. Kinetics Domain of Kinetics Rate of a reaction depends on the pathway from reactants to products. Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products. Copyright © Cengage Learning. All rights reserved 2 Section 17.1 Spontaneous Processes and Entropy Spontaneous Processes and Entropy Thermodynamics lets us predict the direction in which a process will occur but gives no information about the speed of the process. A spontaneous process is one that occurs without outside intervention. Copyright © Cengage Learning. All rights reserved 3 Section 17.1 Spontaneous Processes and Entropy CONCEPT CHECK! Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant. a) What should happen to the gas when you open the valve? Copyright © Cengage Learning. All rights reserved 4 Section 17.1 Spontaneous Processes and Entropy CONCEPT CHECK! Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant. b) Calculate ΔH, ΔE, q, and w for the process you described above. All are equal to zero. Copyright © Cengage Learning. All rights reserved 5 Section 17.1 Spontaneous Processes and Entropy CONCEPT CHECK! Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant. c) Given your answer to part b, what is the driving force for the process? Entropy Copyright © Cengage Learning. All rights reserved 6 Section 17.1 Spontaneous Processes and Entropy The Expansion of An Ideal Gas Into an Evacuated Bulb Copyright © Cengage Learning. All rights reserved 7 Section 17.1 Spontaneous Processes and Entropy Entropy The driving force for a spontaneous process is an increase in the entropy of the universe. A measure of molecular randomness or disorder. A measure of how many micro states can exist. Copyright © Cengage Learning. All rights reserved 8 Section 17.1 Spontaneous Processes and Entropy Entropy Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state. Nature spontaneously proceeds toward the states that have the highest probabilities of existing. Copyright © Cengage Learning. All rights reserved 9 Section 17.1 Spontaneous Processes and Entropy The Microstates That Give a Particular Arrangement (State) Section 17.1 Spontaneous Processes and Entropy Positional Entropy A gas expands into a vacuum to give a uniform distribution because the expanded state has the highest positional probability of states available to the system. Therefore: Ssolid < Sliquid << Sgas Section 17.1 Spontaneous Processes and Entropy Interactive Example 17.1 - Positional Entropy For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature a. Solid CO2 and gaseous CO2 b. N2 gas at 1 atm and N2 gas at 1.0×10–2 atm Section 17.1 Spontaneous Processes and Entropy CONCEPT CHECK! Predict the sign of ΔS for each of the following, and explain: + a) The evaporation of alcohol – b) The freezing of water – c) Compressing an ideal gas at constant temperature + d) Heating an ideal gas at constant pressure + e) Dissolving NaCl in water Copyright © Cengage Learning. All rights reserved 13 Section 17.2 Entropy and the Second Law of Thermodynamics Second Law of Thermodynamics In any spontaneous process there is always an increase in the entropy of the universe. The entropy of the universe is increasing. The total energy of the universe is constant, but the entropy is increasing. Suniverse = ΔSsystem + ΔSsurroundings Copyright © Cengage Learning. All rights reserved 14 Section 17.2 Entropy and the Second Law of Thermodynamics ΔSsurr ΔSsurr = +; entropy of the universe increases ΔSsurr = -; process is spontaneous in opposite direction ΔSsurr = 0; process has no tendency to occur Copyright © Cengage Learning. All rights reserved 15 Section 17.2 Entropy and the Second Law of Thermodynamics Entropy Changes in the Surroundings (ΔSsurr) ΔSsurr is determined by flow of energy as heat Exothermic process increases ΔSsurr Important driving force for spontaneity Endothermic process decreases ΔSsurr Impact of transfer of energy as heat to or from the surroundings is greater at lower temperatures Copyright © Cengage Learning. All rights reserved 16 Section 17.3 The Effect of Temperature on Spontaneity CONCEPT CHECK! For the process A(l) A(s), which direction involves an increase in energy randomness? Positional randomness? Explain your answer. As temperature increases/decreases (answer for both), which takes precedence? Why? At what temperature is there a balance between energy randomness and positional randomness? Copyright © Cengage Learning. All rights reserved 17 Section 17.3 The Effect of Temperature on Spontaneity CONCEPT CHECK! Describe the following as spontaneous/non-spontaneous/cannot tell, and explain. A reaction that is: a) Exothermic and becomes more positionally random Spontaneous b) Exothermic and becomes less positionally random Cannot tell a) Endothermic and becomes more positionally random Cannot tell a) Endothermic and becomes less positionally random Not spontaneous Explain how temperature affects your answers. Section 17.3 The Effect of Temperature on Spontaneity ΔSsurr The sign of ΔSsurr depends on the direction of the heat flow. The magnitude of ΔSsurr depends on the temperature. $50 to a millionaire or a high school student Copyright © Cengage Learning. All rights reserved 19 Section 17.3 The Effect of Temperature on Spontaneity ΔSsurr Copyright © Cengage Learning. All rights reserved 20 Section 17.3 The Effect of Temperature on Spontaneity ΔSsurr Copyright © Cengage Learning. All rights reserved 21 Section 17.3 The Effect of Temperature on Spontaneity ΔSsurr Heat flow (constant P) = change in enthalpy = ΔH Ssurr H = T If the reaction is exothermic: ΔH has a negative sign ΔSsurr is positive since heat flows into the surroundings Copyright © Cengage Learning. All rights reserved 22 Section 17.3 The Effect of Temperature on Spontaneity Copyright © Cengage Learning. All rights reserved 23 Section 17.3 The Effect of Temperature on Spontaneity Interactive Example 17.4 - Determining ΔSsurr In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore For example, iron is used to reduce antimony in sulfide ores Sb2S3 (s) + 3Fe(s) 2Sb(s) + 3FeS(s) H = 125kJ Section 17.3 The Effect of Temperature on Spontaneity Interactive Example 17.4 - Determining ΔSsurr (Continued) Carbon is used as the reducing agent for oxide ores: Sb4O6 (s) + 6C(s) 4Sb(s) + 6CO(g ) ΔH = 778 kJ Calculate ΔSsurr for each of these reactions at 25°C and 1 atm Section 17.3 The Effect of Temperature on Spontaneity Interactive Example 17.4 - Solution ΔH ΔSsurr = , where T = 25 + 273 = 298 K T For the sulfide ore reaction, 125 kJ ΔSsurr = = 0.419 kJ/K = 419 J/K 298 K ΔSsurr is positive since this reaction is exothermic, and heat flow occurs to the surroundings, increasing the randomness of the surroundings Section 17.3 The Effect of Temperature on Spontaneity Interactive Example 17.4 - Solution (Continued) For the oxide ore reaction, 778 kJ ΔSsurr = = 2.61 kJ/K = 2.61 × 103 J/K 298 K In this case ΔSsurr is negative because heat flow occurs from the surroundings to the system Section 17.4 Free Energy Free Energy (G) Suniv G = (at constant T and P ) T A process (at constant T and P) is spontaneous in the direction in which the free energy decreases. Negative ΔG means positive ΔSuniv. Copyright © Cengage Learning. All rights reserved 28 Section 17.4 Free Energy Free Energy (G) ΔG = ΔH – TΔS (at constant T and P) All quantities refer to the system Copyright © Cengage Learning. All rights reserved 29 Section 17.4 Free Energy CONCEPT CHECK! A liquid is vaporized at its boiling point. Predict the signs of: – w + q + ΔH + ΔS – ΔSsurr 0 ΔG Explain your answers. Copyright © Cengage Learning. All rights reserved 30 Section 17.4 Free Energy EXERCISE! The value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C. Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm. ΔS = 132 J/K·mol ΔSsurr = -132 J/K·mol ΔG = 0 kJ/mol Copyright © Cengage Learning. All rights reserved 31 Section 17.4 Free Energy Spontaneous Reactions To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 32 Section 17.4 Free Energy Effect of ΔH and ΔS on Spontaneity Copyright © Cengage Learning. All rights reserved 33 Section 17.4 Free Energy Interactive Example 17.5 - Free Energy and Spontaneity At what temperatures is the following process spontaneous at 1 atm? Br2 (l ) Br2 (g ) ΔH o = 31.0 kJ/mol and ΔS o = 93.0 J/K mol What is the normal boiling point of liquid Br2? Section 17.4 Free Energy Interactive Example 17.5 - Solution The vaporization process will be spontaneous at all temperatures where ΔG°is negative Note that ΔS°favors the vaporization process because of the increase in positional entropy, and ΔH°favors the opposite process, which is exothermic These opposite tendencies will exactly balance at the boiling point of liquid Br2, since at this temperature liquid and gaseous Br2 are in equilibrium (ΔG°= 0) Section 17.4 Free Energy Interactive Example 17.5 - Solution (Continued 1) We can find this temperature by setting ΔG°= 0 in the following equation: ΔG= ΔH TΔS 0 ΔH TΔS ΔH =TΔS ΔH 3.10 ×104 J/mol T= = = 333K ΔS 93.0 J/K mol Section 17.4 Free Energy Interactive Example 17.5 - Solution (Continued 2) At temperatures above 333 K, TΔS°has a larger magnitude than ΔH°, and ΔG°is negative Above 333 K, the vaporization process is spontaneous The opposite process occurs spontaneously below this temperature At 333 K, liquid and gaseous Br2 coexist in equilibrium Section 17.4 Free Energy Interactive Example 17.5 - Solution (Continued 3) Summary of observations (the pressure is 1 atm in each case) T > 333 K The term ΔS°controls, and the increase in entropy when liquid Br2 is vaporized is dominant T < 333 K The process is spontaneous in the direction in which it is exothermic, and the term ΔH°controls Section 17.4 Free Energy Interactive Example 17.5 - Solution (Continued 4) T = 333 K The opposing driving forces are just balanced (ΔH°= 0), and the liquid and gaseous phases of bromine coexist This is the normal boiling point Section 17.5 Entropy Changes in Chemical Reactions Entropy Changes and Chemical Reactions Positional probability determines the changes that occur in a chemical system Fewer the molecules, fewer the possible configurations Ammonia synthesis : N (g) + 3H (g) 2NH (g) 2 Copyright © Cengage Learning. All rights reserved 2 40 3 Section 17.5 Entropy Changes in Chemical Reactions Figure 17.6 - Entropy of Water H2O molecule can vibrate and rotate in several ways Freedom of motion leads to a higher entropy for water Section 17.5 Entropy Changes in Chemical Reactions Entropy Changes in Reactions That Involve Gaseous Molecules Change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products If the number of product molecules is greater than the number of reactant molecules: Positional entropy increases ΔS is positive Copyright © Cengage Learning. All rights reserved 42 Section 17.5 Entropy Changes in Chemical Reactions Interactive Example 17.6 - Predicting the Sign of ΔS° Predict the sign of ΔS°for each of the following reactions a. Thermal decomposition of solid calcium carbonate CaCO3 (s) CaO(s) + CO2 (g ) b. Oxidation of SO2 in air 2SO2 (g ) + O2 (g ) 2SO3 (g ) Section 17.5 Entropy Changes in Chemical Reactions CONCEPT CHECK! Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant. Predict the signs of: ΔH ΔSsurr – + ΔS ΔSuniv 0 + Explain. Copyright © Cengage Learning. All rights reserved 44 Section 17.5 Entropy Changes in Chemical Reactions Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero. The entropy of a substance increases with temperature. Copyright © Cengage Learning. All rights reserved 45 Section 17.5 Entropy Changes in Chemical Reactions Standard Entropy Values (S°) Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure. ΔS°reaction = ΣnpS°products – ΣnrS°reactants Copyright © Cengage Learning. All rights reserved 46 Section 17.5 Entropy Changes in Chemical Reactions EXERCISE! Calculate ΔS° for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: S° (J/K·mol) Na(s) 51 H2O(l) 70 NaOH(aq) 50 H2(g) 131 ΔS°= –11 J/K Copyright © Cengage Learning. All rights reserved 47 Section 17.6 Free Energy and Chemical Reactions AP Learning Objectives, Margin Notes and References Learning Objectives LO 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of (either quantitatively or qualitatively) the signs of both Ho and So, and calculation or estimation of Go when needed. LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy. LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable. LO 5.16 The student can use Le Châtelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product. LO 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction. LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions. Section 17.6 Free Energy and Chemical Reactions Standard Free Energy Change (ΔG°) The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. ΔG° = ΔH° – TΔS° ΔG°reaction = ΣnpG°products – ΣnrG°reactants Copyright © Cengage Learning. All rights reserved 49 Section 17.6 Free Energy and Chemical Reactions CONCEPT CHECK! A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: ΔH° ΔS° – – ΔG° – Explain. Copyright © Cengage Learning. All rights reserved 50 Section 17.6 Free Energy and Chemical Reactions CONCEPT CHECK! Consider the following system at equilibrium at 25°C. PCl3(g) + Cl2(g) PCl5(g) ΔG° = −92.50 kJ What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain. The ratio will decrease. Copyright © Cengage Learning. All rights reserved 51 Section 17.6 Free Energy and Chemical Reactions Interactive Example 17.11 - Calculating ΔG° Methanol is a high-octane fuel used in highperformance racing engines Calculate ΔG °for the following reaction: 2CH3OH(g ) + 3O2 (g ) 2CO2 (g ) + 4H2O(g ) The following energies of formation are provided: Section 17.6 Free Energy and Chemical Reactions Interactive Example 17.11 - Solution Use the following equation: ΔG = npΔGf (products) npΔGf (reactants) = 2ΔGf CO2 g + 4ΔGf H2O g 3ΔGf O2 g 2ΔGf CH3OH g = 2 mol 394 kJ/mol + 4 mol 229 kJ/mol 3 0 2 mol 163 kJ/mol G = 1378 kJ Section 17.6 Free Energy and Chemical Reactions Interactive Example 17.11 - Solution (Continued) The large magnitude and the negative sign of ΔG°indicate that this reaction is very favourable thermodynamically Section 17.7 The Dependence of Free Energy on Pressure AP Learning Objectives, Margin Notes and References Learning Objectives LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions. Additional AP References LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”) Section 17.7 The Dependence of Free Energy on Pressure Free Energy and Pressure System under constant P and T proceeds spontaneously in the direction that lowers its free energy Free energy of a reaction system changes as the reaction proceeds Dependent on the pressure of a gas or on the concentration of species in solution Equilibrium - Point where free energy value is at its lowest Copyright © Cengage Learning. All rights reserved 56 Section 17.7 The Dependence of Free Energy on Pressure Free Energy and Pressure (Continued 1) For ideal gases: Enthalpy is not pressure-dependent Entropy depends on pressure due to its dependence on volume At a given temperature for 1 mole of ideal gas: Slarge volume > Ssmall volume Or, Slow pressure > Shigh pressure Copyright © Cengage Learning. All rights reserved 57 Section 17.7 The Dependence of Free Energy on Pressure Free Energy and Pressure (Continued 2) G = G + RTln P G° - Free energy of a gas at 1 atm G - Free energy of the gas at a pressure of P atm R - Universal gas constant T - Temperature in Kelvin Copyright © Cengage Learning. All rights reserved 58 Section 17.7 The Dependence of Free Energy on Pressure Free Energy and Pressure (Continued 3) G = G+ RTln Q Q - Reaction quotient T - Temperature in Kelvin R - Universal gas constant (8.3145 J/K·mol) ΔG°- Free energy change at 1 atm ΔG - Free energy change at specified pressures Copyright © Cengage Learning. All rights reserved 59 Section 17.7 The Dependence of Free Energy on Pressure Interactive Example 17.13 - Calculating ΔG° One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases CO g + 2H2 g CH3OH l Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol Section 17.7 The Dependence of Free Energy on Pressure Interactive Example 17.13 - Solution To calculate ΔG for this process, use the following equation: G = G+ RTln Q ₒ First compute ΔG from standard free energies of formation ΔGfo CH3OHl 166 kJ ΔGfo H2 g 0 ΔGfoCO g 137 kJ ΔGo = 166 kJ 137 kJ 0 = 29 kJ = 2.9×104 J Section 17.7 The Dependence of Free Energy on Pressure Interactive Example 17.13 - Solution (Continued 1) One might call this the value of ΔG°for one round of the reaction or for 1 mole of the reaction Thus, the ΔG°value might better be written as –2.9×104 J/mol of reaction, or –2.9×104 J/mol rxn Use this value to calculate the value of ΔG Section 17.7 The Dependence of Free Energy on Pressure Interactive Example 17.13 - Solution (Continued 2) ΔG°= – 2.9×104 J/mol rxn R = 8.3145 J/K·mol T = 273 + 25 = 298 K 1 1 2 Q= = = 2.2×10 2 2 PCO PH2 5.0 3.0 Note that the pure liquid methanol is not included in the calculation of Q Section 17.7 The Dependence of Free Energy on Pressure Interactive Example 17.13 - Solution (Continued 3) G = G+ RTln Q = 2.9×104 J/mol rxn + 8.3145 J/K mol rxn 298K ln 2.2×102 = 2.9×104 J/mol rxn 9.4×103 J/mol rxn 3.8×104 J/mol rxn = 38 kJ/mol rxn Section 17.7 The Dependence of Free Energy on Pressure Interactive Example 17.13 - Solution (Continued 4) Note that ΔG is significantly more negative than ΔG°, implying that the reaction is more spontaneous at reactant pressures greater than 1 atm This result can be expected from Le Châtelier’s principle Section 17.7 The Dependence of Free Energy on Pressure CONCEPT CHECK! Sketch graphs of: 1. G vs. P 2. H vs. P 3. ln(K) vs. 1/T (for both endothermic and exothermic cases) Copyright © Cengage Learning. All rights reserved 66 Section 17.7 The Dependence of Free Energy on Pressure The Meaning of ΔG for a Chemical Reaction A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion. Copyright © Cengage Learning. All rights reserved 67 Section 17.8 Free Energy and Equilibrium AP Learning Objectives, Margin Notes and References Learning Objectives LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions. LO 6.25 The student is able to express the equilibrium constant in terms of Go and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process. Additional AP References LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 6.25 (see Appendix 7.11, “Non-Spontaneous Reactions”) Section 17.8 Free Energy and Equilibrium The equilibrium point occurs at the lowest value of free energy available to the reaction system. ΔG = 0 = ΔG° + RT ln(K) ΔG° = –RT ln(K) Copyright © Cengage Learning. All rights reserved 69 Section 17.8 Free Energy and Equilibrium Change in Free Energy to Reach Equilibrium Copyright © Cengage Learning. All rights reserved 70 Section 17.8 Free Energy and Equilibrium Copyright © Cengage Learning. All rights reserved 71 Section 17.8 Free Energy and Equilibrium Interactive Example 17.15 - Free Energy and Equilibrium II Section 17.8 Free Energy and Equilibrium Interactive Example 17.15 - Solution Calculate ΔG° from ΔG° = ΔH°– TΔS° ΔH = 2ΔH f(Fe2O3 (s )) 3ΔH f(O2 (g )) 4ΔH f(Fe(s )) = 2 mol 826 kJ/mol 0 0 = 1652 kJ = 1.652×106 J ΔS = 2S Fe2O3 3S O2 4S Fe = 2 mol 90 J/K mol 3 mol 205 J/K mol 4 mol 27 J/K mol 543 J/K T = 273 + 25 = 298 K Section 17.8 Free Energy and Equilibrium Interactive Example 17.15 - Solution (Continued 1) Therefore, ΔG° = ΔH°– TΔS° = (– 1.652×106 J) – (298 K) (– 543 J/K) = – 1.490 ×106 J ΔG= RT ln K = 1.490 ×106 J = 8.3145 J/K mol 298K ln K 1.490×106 ln K = = 601 3 2.48×10 Section 17.8 Free Energy and Equilibrium Interactive Example 17.15 - Solution (Continued 2) Therefore, K = e601 This is a very large equilibrium constant The rusting of iron is clearly very favourable from a thermodynamic point of view Section 17.9 Free Energy and Work Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy. wmax = ΔG Copyright © Cengage Learning. All rights reserved 76 Section 17.9 Free Energy and Work Reversible and Irreversible Processes Reversible process: Universe is exactly the same as it was before a cyclic process Irreversible process: Universe is different after a cyclic process All real processes are irreversible Characteristics of a real cyclic process Work is changed to heat Entropy of the universe increases Copyright © Cengage Learning. All rights reserved 77 Section 17.9 Free Energy and Work Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy. All real processes are irreversible. First law: You can’t win, you can only break even. Second law: You can’t break even. As we use energy, we degrade its usefulness. Copyright © Cengage Learning. All rights reserved 78