Contents and Concepts

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 1 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 2 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 3 1

Learning Objectives 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. 4 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, q p, equals the change of enthalpy, H. 5 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. 6 2

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. 7 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 S o for a reaction. 8 Free-Energy Concept 4. Free Energy and Spontaneity a. Define free energy, G. b. Define the standard free-energy change. c. Calculate G o from H o and S o. d. Define the standard free energy of formation, G o. e. Calculate G o from standard free energies of formation. f. State the rules for using G o as a criterion for spontaneity g. Interpret the sign of G o. 9 3

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. 10 Free Energy and Equilibrium Constants 6. Relating G o 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. 11 6. Relating G o to the Equilibrium Constant (cont) d. Relate the standard free-energy change to the thermodynamic equilibrium constant. e. Calculate K from the standard freeenergy change (molecular equation). f. Calculate K from the standard freeenergy change (net ionic equation). 12 4

7. Change of Free Energy with Temperature a. Describe how G o at a given temperature ( G o T) is approximately related to H o and S o 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 H o and S o. c. Calculate G o and K at various temperatures. 13 Thermodynamics and Equilibrium Thermodynamics: Define The study of the relationship between heat and other forms of energy involved in a chemical or physical process How do we use this knowledge in chemistry? 14 Consider this Reaction 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) Concerning this reaction: 1. Does this reaction naturally occur as written? 2. Will the reaction mixture contain sufficient amount of product at equilibrium? 15 5

We can answer these questions with heat measurements only!!! 1. We can predict the natural direction. 2. We can determine the composition of the mixture at equilibrium. How? 16 Consider the Laws of Thermodynamics 1 st Law: The change in internal energy of a system U, equals q + w U = q + w You can t win 2 nd Law: The total entropy of a system and its surroundings increases for a spontaneous process. You cant break even. 3 rd Law: A substance that is perfectly crystalline at 0 K has an entropy of zero. You can t quit the game 17 1st Law: The change in internal energy of a system U, equals q + w U = q + w U = Internal energy = Sum of kinetic energy and potential energy of the particle making up the system. Kinetic Energy = Energy of motion of electrons, protons and molecules. Potential Energy = Energy from chemical bonding of atoms and from attraction between molecules. 18 6

U is a state function: Most often we are interested in the change: U = U f - U i q = Energy that moves in and out of a system w = Force x distance = F x d = P V 19 Heat vs. Work 20 Showing work is P V w = - F x h = - F x V/A = -F/A x V = -P V 21 7

P V = -92 J q = 165 J U = q + w = (+165) + (-92) = +73 J 22 Heat of Reaction and Internal Energy Zn (s) + HCl (l) ZnCl 2 (aq) + H 2 (g) 23 Here the system expands and evolves heat from A to B. Zn 2+ (aq) + 2Cl - (aq) + H 2 (g) V is positive, so work is negative. 24 8

q = - w = - 25 w = -P V = -(1.01 x 10 5 pa)(24.5 l) = -1.01 x 10 5 pa)(24.5 x 10-3 m 3 ) = -2.47 x 10 3 J = -2.47 kj q = -152 kj U = -152 kj + (-2.47 kj) = -154.9 kj 26 Diagram and explain the change in internal energy for the following reaction. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) q = -890.2 kj w U = P V = +(1.01 x 10 5 pa)(24.5 l)(2) = +(1.01 x 10 5 pa)(2.45 x 10-3 m 3 )(2) = +4.95 kj = -890.2 kj + (+4.95 kj) = 885.2 kj Do exercise 18.1 See problems 18.31& 32 27 9

Enthalpy and Enthalpy Change Chapter 6 defined Enthalpy as q p Enthalpy = H = U + PV All are state functions. H = H f - H i H = (U f + PV f ) (U i + PV i ) = (U f U i ) + P(V f V i ) U = q p - P V H = (q p - P V) + P V = q p H o f = Σn H o f (products) - Σm Ho f (reactants) H o f = Standard enthalpy change (25 o C) 28 H o f = Σn H o f (products) - Σm H o f (reactants) Calculate H o f for the reaction in slide 15 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) H o f = for NH 3 (g) = -45.9 kj = for CO 2 (g) = -393.5 kj = for NH 2 CONH 2 = -319.2 kj = for H 2 O (l) = -285.8 kj H o = [(-319.2 285.8) (-2 x 45.9 393.50)] kj = -119.7 Since H o has a negative sign, heat is evolved Do Exercise 18.2 See Problems 18.33 and 18.34 29 Spontaneity and Entropy Definition of Spontaneous Process: Physical or chemical process that occurs by itself. Why?? Give several spontaneous processes: Still cannot predict spontaneity.. 30 10

Entropy and the 2 nd Law of Thermodynamics 2 nd Law: The total entropy of a system and its surroundings increases for a spontaneous process. Entropy = S = A thermodynamic quantity that is a measure of how dispersed the energy is among the different possible ways that a system can contain energy. Consider: 1. A hot cup of coffee on the table 2. Rock rolling down the side of a hill. 3. Gas expanding. 4. Stretching a rubber band. 31 Flask connected to an evacuated flask by a valve or stopcock S = S f - S i H 2 O (s) H 2 O (l) S = (63 41) J/K = 22 J/K 32 Concept Check: You have a sample of solid iodine at room temperature. Later you notice that the iodine has sublimed. What can you say about the entropy change of the iodine? 33 11

2 nd Law: The total entropy of a system and its surroundings increases for a spontaneous process. Process occurs naturally as a result of energy dispersal In the system. S = entropy created + q/t S > q/t For a spontaneous process at a given temperature, the change in entropy of the system is greater than the heat 34 divided by the absolute temperature. Entropy and Molecular Disorder 35 Entropy Change for a Phase Transition S > q/t (at equilibrium) What processes can occur under phase change at equilibrium? 36 12

Was thought that in order to be spontaneous A reaction had to be exothermic. H < 0 For water undergoing fusion. S = H fus /T = 6.0 x 10 3 J/273 K S = 22J/K The heat of vaporization, H vap, of carbon tetrachloride, CCl 4, at 25 o C is 39.4 kj/mol. CCl 4 (l) CCl 4 (g) H vap = 39.4 kj/mol If 1 mol of liquid carbon tetrachloride has an entropy of 216 J/K at 25 o C, what is the entropy of 1 mol of vapor 37 at equilibrium with the liquid at this temperature? Solution: S = H vap /T = (39.4 x 10 3 J/mol)/ 298 K = 132 j/(mol K) Entropy of Vapor = (216 + 132) J/(mol K) = 348 J/(mol K) Solve exercise 18.3 on page 740 of text. See problems 18.35 and 36 38 Criterion for a Spontaneous Reaction 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) Is this reaction spontaneous? That is, does it go left to right as written? S > q/t q p = H at constant pressure If we know S and H we can make prediction. S > q p /T = H/T H/T - S < 0 H - T S < 0 (spontaneous rxn, constant T and P) 39 13

Standard Entropies and the Third Law of Thermodynamics To determine the entropy of a substance you first measure the heat absorbed by the substance by warming it at various temperatures (heat capacity at different temperatures) Determination of entropy is based on the 3 rd Law. 3 rd Law: A substance that is perfectly crystalline at 0 K has an entropy of zero. See page 742 - Read to understand how slowly heating a substance from near zero K. 40 Standard entropy of methyl chloride, CHCl 3, at various Temperatures (approximate schematic graph) 41 Entropy Change for a Reaction S o may be positive for reactions with the following: 1. The reaction is one in which a molecule is broken into two or more smaller molecules. 2. The reaction is one in which there is an increase in moles of gas. 3. The process is one in which a solid changes to a liquid or a liquid changes to a gas. 42 14

Standard Entropy (Absolute Entropy): Entropy value for the standard state of a species. See Table 18.1 page 743 Entropy values of substances must be positive. S o must be >0 but H o can be plus or minus (Why?) How about ionic species? S o for H 3 O + is set at zero. 43 Predict The Entropy sign for the following reactions: a. C 6 H 12 O 11 (s) 2CO 2 (g)+ C 2 H 5 OH (l) b. 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) c. CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) d. Stretching a rubber band Exercise 18.4 Page 744 44 S o Calculating S o for a reaction = Σn S o (products) - Σm S o (reactants) Calculate the entropy change for the following reaction At 25 o C. 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) S o 2 x 193 214 174 70 S o = Σn S o (products) - Σm S o (reactants) S o = [(174 + 70) - (2x 193 + 214)] J/K= -356 J/K See exercise 18.5 then do problems 18.41 and 42 45 15

Free Energy Concept H o T S o Can Serve as a criteria for Spontaneity 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) H o = -119.7 kj S o = -365 J/K = -0.365 kj/k H o T S o = (-119.7 kj) (298 K) x (-0.365kJ/K = -13.6 kj H o T S o is a negativity quantity, from which we can conclude that the reaction is spontaneous under standard conditions. 46 Free Energy and Spontaneity Free Energy: Thermodynamic quantity defined by the equation G = H - TS G = H - T S If you can show that G for a reaction at a given temperature and pressure is negative, you can predict that the reaction will be spontaneous 47 Standard Free Energy Changes Standard Conditions: 1 atm pressure 1 atm partial pressure 1 M concentration Temperature of 25 o C or 298 K Standard free energy is free-energy change that takes place when reactants in their standard states are converted to products in their standard states. G o = H o - T S o 48 16

Calculate G o from H o and S o What is the standard free energy change G o, for the following reaction at 25 o C? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Use values of H fo and S o from tables 6.1 and 18.1 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H f o : 0 0 2 x (-45.9) kj S o : 191.6 3 x 130.6 2 x 192.7 J/K 49 G o = H o - T S o H o = Σn H o (products) - Σm H o (reactants) = [2 x (-45.9) 0] = -91.8 kj S o = ΣnH o (products) - ΣmH o (reactants) = [2 x 192.7 (191.6 + 3 x 130.60] J/K = -198.0 J/K G o = H o - T S o G o = 91.8 kj - (298 K)(-0.1980 kj/k) = -32.8 kj See exercise 18.6 and do problems 18.45 and 18.4650 Standard Free Energies of Formation G o f = The free energy change that occurs when 1 mol of substance is formed from its elements in their most stable states at 1 atm and at a specific temperature (normally 25 o C ) G o = Σn G fo (products) - Σm G fo (reactants) ½ N 2 (g) + 3/2 H 2 (g) NH 3 (g) G o for 2 mols of NH 3 = -32.8 kj G o for 1 mol of NH 3 = -32.8 kj/2 mol = -16.4 kj 51 17

Calculation of G o from Standard Free Energies of Formation Calculate G o for the following reaction: C 2 H 5 OH (l) + 3 O 2 (g) 2 CO 2 (g) + 3 H 2 O (g) G fo = -174.9 0 2(-394.4) 3(-228.6) G o = Σn G fo (products) - Σm G fo (reactants) G o = [2(-394.4) + 3(-228.6) (- 174.9)] = -1299.7kJ Calculate G o for the following reaction: (Exercise 18.7) CaCO 3 (s) CaO (s) + CO 2 (g) G o = 130.9 kj Do Exercise 18.7 See problems 18.49 and 18.50 52 G o as a Criterion for Spontaneity G o = H o - T S o 1. When G o is a large negative number (more negative than about 10 kj), the reaction is spontaneous as written, and reactants transform almost entirely into products when equilibrium is reached. 2. When G o is a large positive number (more positive than about + 10 kj), the reaction is not spontaneous as written, and reactants transform almost entirely into products when equilibrium is reached. 3. When G o has a small positive or negative value(less than about 10 kj), the reaction mixture gives an equilibrium mixture with significant amounts of both reactants and products.. 53 Interpreting the Sign of G o Calculate H o and G o for the following reaction 2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g) Interpret the signs of H o and G o H fo are as follows: KClO 3 (s) = -397.7 kj/mol KCl (s) = -436.7 kj/mol O 2 (g) = 0 G fo are as follows: KClO 3 (s) = -296.3 kj/mol KCl (s) = -408.8 kj/mol O 2 (g) = 0 54 18

2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g) H f o 2 x (-397.70) 2 x (-436.7) 0 kj G f o 2 x (-296.3) 2 x (-408.8) 0 kj Then: H o = [2 x (-436.7) 2 x (-397.7)] kj = -78 kj G o = [2 x (-408.8) 2 x (-296.3)] kj = -225 kj The reaction is exothermic, liberating 78 kj of heat. The large negative value of G o indicates that the equilibrium is mostly KCl and O 2. Do exercise 18.8 See problems 18.53 and 54 55 G fo = 86.60 kj/mol 56 Interpretation of Free Energy Theoretically, spontaneous reactions can be used to obtain useful work. Combustion of gasoline Battery generating electricity Biochemical reaction in muscle tissue Often reaction are not carried out in in a way that does useful work. Reactants simply poured together.. 57 19

Interpretation of Free Energy In principle, if a reaction is carried out to obtain the maximum useful work, no entropy is produced. It can be shown that maximum useful work, w max, for a spontaneous reaction is G, The free energy change is the maximum energy available, or free to do useful work. The sign of w (work) is defined so that a negative value means work (energy is subtracted from the system) obtained from the system). You can obtain work from a reaction if its G is negative. 58 Free energy change during reaction 59 Free energy change during reaction 60 20

Free Energy and Equilibrium Constant Very important relation is the relation between free energy and the equilibrium constant. Thermodynamic Equilibrium Constant- the equilibrium constant in which the concentration of gases are expressed in partial pressures in atmospheres, whereas the concentration of solutes in liquid are expressed in molarities. K = Kc for reactions involving only liquid solutions K = Kp for reactions involving only gases 61 Write Kp and Kc 62 Writing the Expression for a Thermodynamic Equilibrium Constant a. 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) K = [NH 2 CONH 2 ] P 2 NH3 P CO2 b. AgCl (s) Ag + (aq) + Cl - (aq) K = [Ag + ][Cl - ] Do Exercise 18.9 and work problems 19.59 and 19.60 63 21

Relating G o to the Equilibrium Constant G = G o + RT ln Q G 0 can be calculated from thermodynamic data If you want G for a none standard state use the above equation. At G = 0 the reaction is at equilibrium 0 = G o + RT ln K And Q = K the equilibrium constant G o = -RT ln K 64 Problem: Find the equilibrium constant for the following reaction: 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) G o for the reaction is -13.6 kj using G o = H o - T S o 0 = G o + RT ln K ln K = G o = -RT -13.6 kj -8.31 J/(K mol) x 298 K K = e 5.49 = 2.42 x 10 2 Do Exercise 18.10 and 11 See Problems 19.61-66 65 Understanding G o as a criterion for Spontaneity G o = -RT ln K When K is <1, ln is neg, G o is positive When K is >1, ln is pos, G o is negative 66 22

a. No change b. Increases 67 Change of Free Energy with Temperature Measuring G o and K at different temperatures is difficult G To = H o T S o (approximation for G To ) Spontaneity and Temperature Change 68 69 23

Increase in temperature increase in NO 2 70 Consider the decomposition of dinitrogen tetroxide,n 2 O 4, to nitrogen dioxide, NO 2 : N 2 O 4 (g) 2 NO 2 (g) How would you expect the spontaneity of the reaction to behave with temperature change? Increase in temperature increase in NO 2 71 Calculation of G o at Various Temperatures Consider the following reaction: CaCO 3 (s) CaO (s) + CO 2 (g) At 25 o C G o = +130.9 and K p = 1.1 x 10-23 atm What do these values tell you about CaCO 3? What happens when the reaction is carried out at a higher temperature? 72 24

Calculating G o and K at Various Temperatures a. What is G o at 1000 o C for the calcium carbonate reaction? CaCO 3 (s) CaO (s) + CO 2 (g) Is this reaction spontaneous at 1000 o C and 1 atm? b. What is the value of K p at 1000 o C for this reaction? What is the partial pressure of CO 2? 73 Strategy for solution a. Calculate H o and S o at 25 o C using standard enthalpies of formation and standard entropies. Then substitute into the equation for G fo. b. Use the G fo value to find K (=K p ), as in example 18.8. 74 a. From Table 6.2 and 18.1 you have the following: CaCO 3 (s) CaO (s) + CO 2 (g) H fo : -1206.9-635.1-393.5 kj S o : 92.9 38.2 213.7 J/K H o = [(-635.1 393.5) (-1206.9)] = 178.3 kj S o = [(38.2 + 213.7) (92.9)] = 159.0 J/K G To = H o T S o = 178.3kJ (1273 K)(0.1590kJ/K) = -24.1 kj G o is negative reaction is spontaneous 75 25

b. Substitute the values of G o at 1273 K, which equals -24.1 x 10 3 J, into the equation relating ln K and G o. ln K = G o -RT = -24.1 x 10 3-8.31 x 1273 = 2.278 K = K p = e 2.278 = 9.76 K p = P CO2 = 9.76 atm Do exercise 18.12 See problems 18.67 and 18.68 76 Where does the reaction change from spontaneous to non-spontaneous? Solve for T; G o = 0 = H o - T S o T = H o S o = 178.3 kj 0.1590 kj/k T = 1121 K = 848 o C Do exercise 18.13 See problems 18.69 and 18.70 77 That s it for Chapter 19 Quiz 2. Explain how you are able to determine whether a reaction is spontaneous. 78 26