Chapter 18 Thermodynamics and Equilibrium Contents and Concepts

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 18 | 2

Free-Energy Concept The quantity DH – TDS 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 DG equals the quantity DH – TDS, so the free energy gives us a thermodynamic criterion of spontaneity. 4. Free Energy and Spontaneity 5. Interpretation of Free Energy 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 DG° to the Equilibrium Constant 7. Change of Free Energy with Temperature 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 -PDV. 18 | 5

1. First Law of Thermodynamics; Enthalpy (cont) c. Relate the change of internal energy, DU, and heat of reaction, q. d. Define enthalpy, H. e. Show heat of reaction at constant pressure, qp, equals the change of enthalpy, DH. 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. 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 DH - TDS functions as a criterion of a spontaneous reaction. 18 | 8

3. Standard Entropies and the Third Law of Thermodynamics a. State 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 DSo for a reaction. 18 | 9

Free-Energy Concept 4. Free Energy and Spontaneity a. Define free energy, G. b. Define the standard free-energy change. c. Calculate DGo from DHo and DSo. d. Define the standard free energy of formation, DGo. e. Calculate DGo from standard free energies of formation. f. State the rules for using DGo as a criterion for spontaneity g. Interpret the sign of DGo. 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. 18 | 11

Free Energy and Equilibrium Constants 6. Relating DGo to the Equilibrium Constant a. Define 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. 18 | 12

6. Relating DGo to the Equilibrium Constant (cont) d. Relate the standard free-energy change to 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). 18 | 13

7. Change of Free Energy with Temperature a. Describe how DGo at a given temperature (DGo. T) is approximately related to DHo and DSo 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 DHo and DSo. c. Calculate DGo and K at various temperatures. 18 | 14

Thermodynamics is the study of heat and other forms of energy involved in chemical or physical processes. 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: DU = Uf – Ui 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. DU = q + w 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. 18 | 18

Sign Convention for w Recall from Chapter 6 that w = –PDV. When the system expands, DV 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, DV is negative, so w is positive. The surroundings do work on the system, which increases the internal energy of the system. 18 | 19

Here the system expands and evolves heat from A to B. Zn 2+(aq) + 2 Cl-(aq) + H 2(g) DV is positive, so work is negative. 18 | 20

At constant pressure: q. P = DH The first law of thermodynamics can now be expressed as follows: DU = DH – PDV 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. 18 | 22

A ball will roll downhill A ball will not roll uphill spontaneously. It will spontaneously. It eventually reach requires work. equilibrium at the bottom. 18 | 23

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. 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. 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. 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. 18 | 27

In each of the preceding examples, energy has been dispersed (that is, spread out). 18 | 28

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. 18 | 29

Entropy is measured in units of J/K. Entropy change is calculated as follows: DS = Sf – Si 18 | 30

The iodine has spread out, so its entropy has increased. 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. 18 | 32

For a spontaneous process at a constant temperature, we can state the second law of thermodynamics in terms of only the system: DS = entropy created + For a spontaneous process: DS > 18 | 33

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. 18 | 34

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. 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. 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. 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. 18 | 38

When ice melts, the molecules become more disordered, again dispersing energy more widely. When one molecule decomposes to give two, as in N 2 O 4(g) 2 NO 2(g) more disorder is created because the two molecules produced can move independently of each other. Energy is more dispersed. 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. 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. DS = This concept can be applied to melting using DHfus for q and to vaporization using DHvap for q. 18 | 41

? Acetone, CH 3 COCH 3, is a volatile liquid solvent; it is used in nail polish, for example. The standard enthalpy of formation, DHf°, of the liquid at 25°C is – 247. 6 k. J/mol; the same quantity for the vapor is – 216. 6 k. J/mol. What is the entropy change when 1. 00 mol liquid acetone vaporizes at 25°C? 18 | 42

DHf° n n. DHf° CH 3 COCH 3(l) CH 3 COCH 3(g) – 247. 6 k. J/mol – 216. 6 k. J/mol 1 mol – 247. 6 k. J – 216. 6 k. J DH° = 31. 0 k. J DS = 104 J/K 18 | 43

Criterion for a Spontaneous Reaction The criterion is that the entropy of the system and its surroundings must increase. 18 | 44

Third Law of Thermodynamics A substance that is perfectly crystalline at zero Kelvin (0 K) has an entropy of zero. 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. 18 | 46

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Standard Entropy of Bromine, Br 2, at Various Temperatures 18 | 49

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. 18 | 50

? The opening of Chapter 6, on thermochemistry, describes the endothermic reaction of solid barium hydroxide octahydrate and solid ammonium nitrate: Ba(OH)2 8 H 2 O(s) + 2 NH 4 NO 3(s) 2 NH 3(g) + 10 H 2 O(l) + Ba(NO 3)2(aq) Predict the sign of DS° for this reaction. 3 moles of reactants produces 13 moles of products. Solid reactants produce gaseous, liquid, and aqueous products. DS° is positive. 18 | 51

To compute DS° where n = moles: 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: CH 3 CH 2 OH(l) + O 2(g) CH 3 COOH(l) + H 2 O(l) The standard entropies, S°, of the substances in J/(K mol) at 25°C are CH 3 CH 2 OH(l), 161; O 2(g), 205; CH 3 COOH(l), 160; H 2 O(l), 69. 9. 18 | 53

CH 3 CH 2 OH(l) + O 2(g) CH 3 COOH(l) + H 2 O(l) S° J/(mol K) 161 205 n mol 1 1 n. S° J/K 161 205 160 1 160 69. 9 1 69. 9 DS = 229. 9 J/K – 366 J/K DS = – 136 J/K 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 18 | 55

As a reaction proceeds, G changes: DG = DH – TDS Standard free energy change: DG° = DH° – TDS° 18 | 56

? Using standard enthalpies of formation, DHf° and the value of DS° from the previous problem, calculate DG° for the oxidation of ethyl alcohol to acetic acid. CH 3 CH 2 OH(l) + O 2(g) CH 3 COOH(l) + H 2 O(l) 18 | 57

CH 3 CH 2 OH(l) + O 2(g) CH 3 COOH(l) + H 2 O(l) DHf° k. J/mol – 277. 6 n mol 1 n. DHf° k. J – 277. 6 0 1 0 – 487. 0 1 – 487. 0 – 277. 6 k. J – 285. 8 1 – 285. 8 – 772. 8 k. J DH° = – 495. 2 k. J DS° = – 136 J/K T = 298 K 18 | 58

DH° = – 495. 2 k. J DS° = – 136 J/K = – 0. 136 k. J/K T = 298 K DG° = DH° – TDS° DG° = – 495. 2 k. J – (298 K)(– 0. 136 k. J/K) DG° = – 495. 2 k. J + 40. 5 k. J DG° = – 454. 7 k. J The reaction is spontaneous. 18 | 59

Standard Free Energies of Formation, DGf° The standard free energy of formation is the freeenergy 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, DHf°. 18 | 60

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To find the standard free energy change for a reaction where n = moles: 18 | 63

? Calculate the free-energy change, DG°, for the oxidation of ethyl alcohol to acetic acid using standard free energies of formation. CH 3 CH 2 OH(l) + O 2(g) CH 3 COOH(l) + H 2 O(l) 18 | 64

CH 3 CH 2 OH(l) + O 2(g) CH 3 COOH(l) + H 2 O(l) DGf°, k. J/mol n, mol n. DGf°, k. J – 174. 8 1 – 174. 8 0 1 0 – 392. 5 1 – 392. 5 – 174. 8 k. J – 237. 2 1 – 237. 2 – 629. 7 k. J DG° = – 629. 7 – (– 174. 8) DG° = – 454. 9 k. J 18 | 65

DG° as a Criterion for Spontaneity The spontaneity of a reaction can now be determined by the sign of DG° < – 10 k. J: spontaneous DG° > +10 k. J: nonspontaneous DG° = very small or zero (< +10 k. J and – 10 k. J): at equilibrium > 18 | 66

; DG° = 173. 2 k. J For the reaction as written, DG° = 173. 20 k. J. For 1 mol NO(g), DGf° = 86. 60 k. J/mol. The reaction is nonspontaneous. 18 | 67

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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 = DG 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. 18 | 71

? Write the expression for thermodynamic equilibrium constant for these reactions: a. N 2 O 4(g) 2 NO 2(g) b. Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g) a. b. 18 | 72

Standard free energy change is related to thermodynamic equilibrium constant, K, at equilibrium: DG = DG° + RT ln Q At equilibrium: DG = 0 and Q = K DG° = –RT ln K 18 | 73

? Calculate the value of thermodynamic equilibrium constant at 25°C for the reaction N 2 O 4(g) 2 NO 2(g) The standard free energy of formation at 25°C is 51. 30 k. J/mol for NO 2 and 97. 82 k. J/mol for N 2 O 4(g). DG° = 2 mol(51. 30 k. J/mol) – 1 mol(97. 82 k. J/mol) DG° = 102. 60 k. J – 97. 82 k. J DG° = 4. 78 k. J 18 | 74

DG = –RT ln K 18 | 75

? Sodium carbonate, Na 2 CO 3, can be prepared by heating sodium hydrogen carbonate, Na. HCO 3: 2 Na. HCO 3(s) Na 2 CO 3(s) + H 2 O(g) + CO 2(g) Estimate the temperature at which the reaction proceeds spontaneously at 1 atm. See Appendix C for data. 18 | 76

2 Na. HCO 3(s) Na 2 CO 3(s) + H 2 O(g) + CO(g) DHf°, k. J/mol – 947. 7 – 1130. 8 – 241. 8 – 393. 5 n, mol 2 1 1 1 n. DHf°, k. J – 1895. 4 – 1130. 8 – 241. 8 – 393. 5 – 1895. 4 k. J DH° = 129. 3 k. J Sf°, J/mol K 102 n, mol 2 n. Sf°, J/K 204 J/K – 1766. 1 139 188. 7 1 188. 7 213. 7 1 213. 7 541. 4 18 | 77

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