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# Chapter18

## by Robert Craig,  lecturer, College of Staten Island at College of Staten Island on Jan 15, 2014

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## Chapter18Presentation Transcript

• Chapter 18 Thermodynamics and Equilibrium
• Contents and Concepts 1. First Law of Thermodynamics Spontaneous Processes and Entropy A spontaneous process is one that occurs by itself. As we will see, the entropy of the system increases in a spontaneous process. 2. Entropy and the Second Law of Thermodynamics 3. Standard Entropies and the Third Law of Thermodynamics Copyright © Cengage Learning. All rights reserved. 18 | 2
• Free−Energy Concept The quantity ∆H – T∆S can function as a criterion for the spontaneity of a reaction at constant temperature, T, and pressure, P. By defining a quantity called the free energy, G = H – TS, we find that ∆G equals the quantity ∆H – T∆S, so the free energy gives us a thermodynamic criterion of spontaneity. 4. Free Energy and Spontaneity 5. Interpretation of Free Energy Copyright © Cengage Learning. All rights reserved. 18 | 3
• Free Energy and Equilibrium Constants The total free energy of the substances in a reaction mixture decreases as the reaction proceeds. As we discuss, the standard free−energy change for a reaction is related to its equilibrium constant. 6. Relating ∆G° to the Equilibrium Constant 7. Change of Free Energy with Temperature Copyright © Cengage Learning. All rights reserved. 18 | 4
• Learning Objectives 1. First Law of Thermodynamics; Enthalpy a. Define internal energy, state function, work, and first law of thermodynamics. b. Explain why the work done by the system as a result of expansion or contraction during a chemical reaction is −P∆V. Copyright © Cengage Learning. All rights reserved. 18 | 5
• 1. First Law of Thermodynamics; Enthalpy (cont.) c. Relate the change of internal energy, ∆U, and heat of reaction, q. d. Define enthalpy, H. e. Show how heat of reaction at constant pressure, qp, equals the change of enthalpy, ∆H. Copyright © Cengage Learning. All rights reserved. 18 | 6
• Spontaneous Processes and Entropy 2. Entropy and the Second Law of Thermodynamics a. Define spontaneous process. b. Define entropy. c. Relate entropy to disorder in a molecular system (energy dispersal). d. State the second law of thermodynamics in terms of system plus surroundings. Copyright © Cengage Learning. All rights reserved. 18 | 7
• 2. Entropy and the Second Law of Thermodynamics (cont.) e. State the second law of thermodynamics in terms of the system only. f. Calculate the entropy change for a phase transition. g. Describe how ∆H − T∆S functions as a criterion of a spontaneous reaction. Copyright © Cengage Learning. All rights reserved. 18 | 8
• 3. Standard Entropies and the Third Law of Thermodynamics a. State the third law of thermodynamics. b. Define standard entropy (absolute entropy). c. State the situations in which the entropy usually increases. d. Predict the sign of the entropy change of a reaction. e. Express the standard change of entropy of a reaction in terms of standard entropies of products and reactants. f. Calculate ∆So for a reaction. Copyright © Cengage Learning. All rights reserved. 18 | 9
• Free−Energy Concept 4. Free Energy and Spontaneity a. Define free energy, G. b. Define the standard free−energy change. c. Calculate ∆Go from ∆Ho and ∆So. d. Define the standard free energy of formation, ∆Go. e. Calculate ∆Go from standard free energies of formation. f. State the rules for using ∆Go as a criterion for spontaneity. g. Interpret the sign of ∆Go. Copyright © Cengage Learning. All rights reserved. 18 | 10
• 5. Interpretation of Free Energy a. Relate the free−energy change to maximum useful work. b. Describe how the free energy changes during a chemical reaction. Copyright © Cengage Learning. All rights reserved. 18 | 11
• Free Energy and Equilibrium Constants 6. Relating ∆Go to the Equilibrium Constant a. Define the thermodynamic equilibrium constant, K. b. Write the expression for a thermodynamic equilibrium constant. c. Indicate how the free−energy change of a reaction and the reaction quotient are related. Copyright © Cengage Learning. All rights reserved. 18 | 12
• 6. Relating ∆Go to the Equilibrium Constant (cont.) d. Relate the standard free−energy change to the thermodynamic equilibrium constant. e. Calculate K from the standard free−energy change (molecular equation). f. Calculate K from the standard free−energy change (net ionic equation). Copyright © Cengage Learning. All rights reserved. 18 | 13
• 7. Change of Free Energy with Temperature a. Describe how ∆Go at a given temperature (∆GoT) is approximately related to ∆Ho and ∆So at that temperature. b. Describe how the spontaneity or nonspontaneity of a reaction is related to each of the four possible combinations of signs of ∆Ho and ∆So. c. Calculate ∆Go and K at various temperatures. Copyright © Cengage Learning. All rights reserved. 18 | 14
• Thermodynamics is the study of heat and other forms of energy involved in chemical or physical processes. Copyright © Cengage Learning. All rights reserved. 18 | 15
• First Law of Thermodynamics The first law of thermodynamics is essentially the law of conservation of energy applied to a thermodynamic system. Recall that the internal energy, U, is the sum of the kinetic and potential energies of the particles making up the system: ∆U = Uf – Ui Copyright © Cengage Learning. All rights reserved. 18 | 16
• Exchanges of energy between the system and its surroundings are of two types: heat, q, and work, w. Putting this in an equation gives us the first law of thermodynamics. ∆U = q + w Copyright © Cengage Learning. All rights reserved. 18 | 17
• Sign Convention for q When heat is evolved by the system, q is negative. This decreases the internal energy of the system. When heat is absorbed by the system, q is positive. This increases the internal energy of the system. Copyright © Cengage Learning. All rights reserved. 18 | 18
• Sign Convention for w Recall from Chapter 6 that w = –P∆V. When the system expands, ∆V is positive, so w is negative. The system does work on the surroundings, which decreases the internal energy of the system. When the system contracts, ∆V is negative, so w is positive. The surroundings do work on the system, which increases the internal energy of the system. Copyright © Cengage Learning. All rights reserved. 18 | 19
• Here the system expands and evolves heat from A to B. Zn2+(aq) + 2Cl−(aq) + H2(g) ∆V is positive, so work is negative. Copyright © Cengage Learning. All rights reserved. 18 | 20
• At constant pressure: qP = ∆H The first law of thermodynamics can now be expressed as follows: ∆U = ∆H – P∆V Copyright © Cengage Learning. All rights reserved. 18 | 21
• To understand why a chemical reaction goes in a particular direction, we need to study spontaneous processes. A spontaneous process is a physical or chemical change that occurs by itself. It does not require any outside force, and it continues until equilibrium is reached. Copyright © Cengage Learning. All rights reserved. 18 | 22
• The first law of thermodynamics cannot help us to determine whether a reaction is spontaneous as written. We need a new quantity—entropy. Entropy, S, is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy. Copyright © Cengage Learning. All rights reserved. 18 | 24
• Examining some spontaneous processes will clarify this definition. First, heat energy from a cup of hot coffee spontaneously flows to its surroundings—the table top, the air around the cup, or your hand holding the cup. The entropy of the system (the cup of hot coffee) and its surroundings has increased. Copyright © Cengage Learning. All rights reserved. 18 | 25
• The rock rolling down the hill is a bit more complicated. As it rolls down, the rock’s potential energy is converted to kinetic energy. As it collides with other rocks on the way down, it transfers energy to them. The entropy of the system (the rock) and its surroundings has increased. Copyright © Cengage Learning. All rights reserved. 18 | 26
• Now consider a gas in a flask connected to an equal−sized flask that is evacuated. When the stopcock is open, the gas will flow into the evacuated flask. The kinetic energy has spread out, and the entropy of the system has increased. Copyright © Cengage Learning. All rights reserved. 18 | 27
• Entropy is a state function. It depends on variables, such as temperature and pressure, that determine the state of the substance. Entropy is an extensive property. It depends on the amount of substance present. Copyright © Cengage Learning. All rights reserved. 18 | 29
• Entropy is measured in units of J/K. Entropy change is calculated as follows: ∆S = Sf – Si Copyright © Cengage Learning. All rights reserved. 18 | 30
• Concept Check 18.1 You have a sample of 1.0 mg of solid iodine at room temperature. Later, you notice that the iodine has sublimed (passed into the vapor state). What can you say about the change of entropy of the iodine? The iodine has spread out, so its entropy has increased. Copyright © Cengage Learning. All rights reserved. 18 | 31
• Second Law of Thermodynamics The total entropy of a system and its surroundings always increases for a spontaneous process. Note: Entropy is a measure of energy dispersal, not a measure of energy itself. Copyright © Cengage Learning. All rights reserved. 18 | 32
• For a spontaneous process at a constant temperature, we can state the second law of thermodynamics in terms of only the system: q ∆S = entropy created + T For a spontaneous process: q ∆S > T Copyright © Cengage Learning. All rights reserved. 18 | 33
• A C B A pendulum is put in motion, with all of its molecules moving in the same direction, as shown in Figures A and B. As it moves, the pendulum collides with air molecules. When it comes to rest in Figure C, the pendulum has dispersed its energy. This is a spontaneous process. Copyright © Cengage Learning. All rights reserved. 18 | 34
• D E F Now consider the reverse process, which is not spontaneous. While this process does not violate the first law of thermodynamics, it does violate the second law because the dispersed energy becomes more concentrated as the molecules move together. Copyright © Cengage Learning. All rights reserved. 18 | 35
• Entropy and Molecular Disorder Entropy is essentially related to energy dispersal. The entropy of a molecular system may be concentrated in a few energy states and later dispersed among many more energy states. The energy of such a system increases. Copyright © Cengage Learning. All rights reserved. 18 | 36
• In the case of the cup of hot coffee, as heat moves from the hot coffee, molecular motion becomes more disordered. In becoming more disordered, the energy is more dispersed. Copyright © Cengage Learning. All rights reserved. 18 | 37
• Likewise, when the gas moves from one container into the evacuated container, molecules become more disordered because they are dispersed over a larger volume. The energy of the system is dispersed over a larger volume. Copyright © Cengage Learning. All rights reserved. 18 | 38
• When ice melts, the molecules become more disordered, again dispersing energy more widely. When one molecule decomposes to give two, as in N2O4(g) → 2NO2(g) more disorder is created because the two molecules produced can move independently of each other. Energy is more dispersed. Copyright © Cengage Learning. All rights reserved. 18 | 39
• In each of these cases, molecular disorder increases, as does entropy. Note: This understanding of entropy as disorder applies only to molecular situations in which increasing disorder increases the dispersion of energy. It cannot be applied to situations that are not molecular—such as a desk. Copyright © Cengage Learning. All rights reserved. 18 | 40
• Entropy Change for a Phase Transition In a phase transition process that occurs very close to equilibrium, heat is very slowly absorbed or evolved. Under these conditions, no significant new entropy is created. q ∆S = T This concept can be applied to melting using ∆Hfus for q and to vaporization using ∆Hvap for q. Copyright © Cengage Learning. All rights reserved. 18 | 41
• ? Acetone, CH3COCH3, is a volatile liquid solvent; it is used in nail polish, for example. The standard enthalpy of formation, ∆Hf°, of the liquid at 25°C is –247.6 kJ/mol; the same quantity for the vapor is –216.6 kJ/mol. What is the entropy change when 1.00 mol liquid acetone vaporizes at 25°C? Copyright © Cengage Learning. All rights reserved. 18 | 42
• CH3COCH3(l) → CH3COCH3(g) ∆Hf° n n∆Hf° –247.6 kJ/mol –216.6 kJ/mol 1 mol –247.6 kJ 1 mol –216.6 kJ ∆H° = 31.0 kJ ∆H ΔS = T 31.0 × 10 3 J ΔS = 298 K ∆S = 104 J/K Copyright © Cengage Learning. All rights reserved. 18 | 43
• Criterion for a Spontaneous Reaction The criterion is that the entropy of the system and its surroundings must increase. q ΔS > T ΔH ΔS > T TΔS > ΔH 0 > Δ H - TΔ S or Δ H - TΔ S < 0 Copyright © Cengage Learning. All rights reserved. 18 | 44
• Third Law of Thermodynamics A substance that is perfectly crystalline at zero Kelvin (0 K) has an entropy of zero. Copyright © Cengage Learning. All rights reserved. 18 | 45
• The standard entropy of a substance—its absolute entropy, S°—is the entropy value for the standard state of the species. The standard state is indicated with the superscript degree sign. For a pure substance, its standard state is 1 atm pressure. For a substance in solution, its standard state is a 1 M solution. Copyright © Cengage Learning. All rights reserved. 18 | 46
• Entropy Change for a Reaction Entropy usually increases in three situations: 1. A reaction in which a molecule is broken into two or more smaller molecules. 2. A reaction in which there is an increase in the number of moles of a gas. 3. A process in which a solid changes to a liquid or gas or a liquid changes to a gas. Copyright © Cengage Learning. All rights reserved. 18 | 50
• The opening of Chapter 6, on thermo−chemistry, describes the endothermic reaction of solid barium hydroxide octahydrate and solid ammonium nitrate: Ba(OH)2  8H2O(s) + 2NH4NO3(s) → 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq) Predict the sign of ∆S° for this reaction. ? 3 moles of reactants produces 13 moles of products. Solid reactants produce gaseous, liquid, and aqueous products. ∆S° is positive. Copyright © Cengage Learning. All rights reserved. 18 | 51
• To compute ∆S° where n = moles: ΔS =  ∑ nS  (products) − Copyright © Cengage Learning. All rights reserved. ∑ nS  (reactants) 18 | 52
• When wine is exposed to air in the presence of certain bacteria, the ethyl alcohol is oxidized to acetic acid, giving vinegar. Calculate the entropy change at 25°C for the following similar reaction: CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l) ? The standard entropies, S°, of the substances in J/(K  mol) at 25°C are CH3CH2OH(l),161; O2(g), 205; CH3COOH(l), 160; H2O(l), 69.9. Copyright © Cengage Learning. All rights reserved. 18 | 53
• CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l) S° 161 J/(mol  K) n mol 1 nS° J/K 161 366 205 160 69.9 1 205 1 160 1 69.9 229.9 ∆S = 229.9 J/K – 366 J/K ∆S = –136 J/K Copyright © Cengage Learning. All rights reserved. 18 | 54
• Free Energy and Spontaneity Physicist J. Willard Gibbs introduced the concept of free energy, G. Free energy is a thermodynamic quantity defined as follows: G = H – TS Copyright © Cengage Learning. All rights reserved. 18 | 55
• As a reaction proceeds, G changes: ∆G = ∆H – T∆S Standard free energy change: ∆G° = ∆H° – T∆S° Copyright © Cengage Learning. All rights reserved. 18 | 56
• ? Using standard enthalpies of formation, ∆Hf° and the value of ∆S° from the previous problem, calculate ∆G° for the oxidation of ethyl alcohol to acetic acid. CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l) Copyright © Cengage Learning. All rights reserved. 18 | 57
• CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l) ∆Hf° kJ/mol –277.6 n mol n∆Hf° kJ 0 1 1 –277.6 0 –277.6 kJ –487.0 –285.8 1 –487.0 1 –285.8 –772.8 kJ ∆H° = –495.2 kJ ∆S° = –136 J/K T = 298 K Copyright © Cengage Learning. All rights reserved. 18 | 58
• ∆H° = –495.2 kJ ∆S° = –136 J/K = –0.136 kJ/K T = 298 K ∆G° = ∆H° – T∆S° ∆G° = –495.2 kJ – (298 K)(–0.136 kJ/K) ∆G° = –495.2 kJ + 40.5 kJ ∆G° = –454.7 kJ The reaction is spontaneous. Copyright © Cengage Learning. All rights reserved. 18 | 59
• Standard Free Energies of Formation, ∆Gf° The standard free energy of formation is the free−energy change that occurs when 1 mol of substance is formed from its elements in their standard states at 1 atm and at a specified temperature, usually 25°C. The corresponding reaction for the standard free energy of formation is the same as that for standard enthalpy of formation, ∆Hf°. Copyright © Cengage Learning. All rights reserved. 18 | 60
• To find the standard free energy change for a reaction where n = moles: ΔG =  ∑ nΔG (products) − ∑ nΔG (reactants)  Copyright © Cengage Learning. All rights reserved.  18 | 63
• ? Calculate the free−energy change, ∆G°, for the oxidation of ethyl alcohol to acetic acid using standard free energies of formation. CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l) Copyright © Cengage Learning. All rights reserved. 18 | 64
• CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l) ∆Gf°, kJ/mol 237.2 n, mol n∆Gf°, kJ 237.2 –174.8 0 1 1 –174.8 0 –174.8 kJ –392.5 – 1 1 –392.5 – –629.7 kJ ∆G° = –629.7 – (–174.8) ∆G° = –454.9 kJ Copyright © Cengage Learning. All rights reserved. 18 | 65
• ∆G° as a Criterion for Spontaneity The spontaneity of a reaction can now be determined by the sign of ∆G°. ∆G° < –10 kJ: spontaneous ∆G° > +10 kJ: nonspontaneous ∆G° = very small or zero (< +10 kJ and > –10 kJ): at equilibrium Copyright © Cengage Learning. All rights reserved. 18 | 66
• Concept Check 18.2 Consider the reaction of nitrogen, N2, and oxygen, O2, to form nitrogen monoxide, NO: N2(g) + O2(g) → 2NO(g) From the standard free energy of formation of NO, what can you say about thisreaction? For the reaction as written, ∆G° = 173.20 kJ. For 1 mol NO(g), ∆Gf° = 86.60 kJ/mol. The reaction is nonspontaneous. Copyright © Cengage Learning. All rights reserved. 18 | 67
• Interpreting Free Energy Theoretically, spontaneous reactions can be used to perform useful work. In fact, we use reactions such as the combustion of gasoline to move a vehicle. We can also use spontaneous reactions to provide the energy needed for a nonspontaneous reaction. The maximum useful work is wmax = ∆G Copyright © Cengage Learning. All rights reserved. 18 | 70
• The thermodynamic equilibrium constant is the equilibrium constant in which the concentrations of gases are expressed as partial pressures in atmospheres and the concentrations of solutes in solutions are expressed in molarities. If only gases are present, K = Kp. If only solutes in liquid solution are present, K = Kc. Copyright © Cengage Learning. All rights reserved. 18 | 71
• ? Write the expression for the thermodynamic equilibrium constant for these reactions: a. N2O4(g)  2NO2(g) b. Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g) a. b. Copyright © Cengage Learning. All rights reserved. K= K= 2 PNO2 PN2O 4 [Zn 2+ ] PH 2 [H + ] 2 18 | 72
• Standard free energy change is related to the thermodynamic equilibrium constant, K, at equilibrium: ∆G = ∆G° + RT ln Q At equilibrium: ∆G = 0 and Q = K ∆G° = –RT ln K Copyright © Cengage Learning. All rights reserved. 18 | 73
• ? Calculate the value of the thermodynamic equilibrium constant at 25°C for the reaction N2O4(g)  2NO2(g) The standard free energy of formation at 25°C is 51.30 kJ/mol for NO2(g) and 97.82 kJ/mol for N2O4(g). ∆G° = 2 mol(51.30 kJ/mol) – 1 mol(97.82 kJ/mol) ∆G° = 102.60 kJ – 97.82 kJ ∆G° = 4.78 kJ Copyright © Cengage Learning. All rights reserved. 18 | 74
• ∆G = –RT ln K − ∆G ln K = RT J   3 −  4.78 × 10 ÷ mol   ln K = J 8.315 (298 K) mol K ln K = −1.929 K = 0.145 Copyright © Cengage Learning. All rights reserved. 18 | 75
• ? Sodium carbonate, Na2CO3, can be prepared by heating sodium hydrogen carbonate, NaHCO3: 2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) Estimate the temperature at which the reaction proceeds spontaneously at 1 atm. See Appendix C for data. Copyright © Cengage Learning. All rights reserved. 18 | 76
• 2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO(g) ∆Hf°, kJ/mol –947.7 –1130.8 –241.8 –393.5 n, mol 2 1 1 1 n∆Hf°, kJ –1895.4 –1130.8 –241.8 –393.5 –1895.4 kJ –1766.1 kJ ∆H° = 129.3 kJ Sf°, J/mol  K 102 n, mol 2 nSf°, J/K 204 204 J/K 139 1 139 188.7 1 188.7 541.4 J/K ∆S° = 337.4 J/K Copyright © Cengage Learning. All rights reserved. 213.7 1 213.7 18 | 77
• ΔH ° T = ΔS ° 129.3 × 10 3 J T= J 337.4 K T = 383 K T = 110°C Copyright © Cengage Learning. All rights reserved. 18 | 78