1/16/2013. Thermodynamics.

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1 Thermodynamics 1 1

2 Study of energy changes and flow of energy Answers several fundamental questions: Is it possible for reaction to occur? Will reaction occur spontaneously (without outside interference) at given T? Will reaction release or absorb heat? Tells us nothing about time frame of reaction Kinetics Thermodynamics Two major considerations must be balanced Enthalpy changes, H(heats of reaction) Heat exchange between system and surroundings Nature's trend to randomness or disorder 2 2

3 Examples of Spontaneous Reactions 3 3

4 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? 4 4

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? 5 5

6 Consider the Laws of Thermodynamics 1 st Law: The change in internal energy of a system U, equals q + w E = 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 6 6

7 Review of First Law of Thermodynamics Internal energy, E or U System's total energy Sum of KEand PEof all particles in system E = ( KE ) + ( PE ) system E or for chemical reaction E E+ energy into system system = E final E initial = E products E reactants E energy out of system system 7 7

8 Two Methods of Energy Exchange Between System and Surroundings Heat q = H Work w = P V U = E = q + w Conventions of heat and work q + Heat absorbed by system E system q Heat released by system E system w + Work done on system E system w Work done by system E system 8 8

9 U(Internal Energy) is a state function: Most often we are interested in the change: Internal Energy = Ε = U = U f - U i q = Energy that moves in and out of a system w = Force x distance = F x d = P V 9 9

10 First Law of Thermodynamics Energy can neither be created nor destroyed It can only be converted from one form to another KE PE Chemical Electrical Electrical Mechanical E and E are state functions Path independent E = - U = q + w E = ( KE ) + ( PE ) system system system 10 10

11 Work in Chemical Systems 1. Electrical 2. Pressure-volume or P V w = P V Where P = external pressure If P V only work in chemical system, then E = q + ( P V ) = q P V 11 11

12 Heat vs. Work 12 12

13 Showing work is P V w = - F x h = - F x V/A = -F/A x V = -P V 13 13

14 Heat at Constant Volume Reaction done at constant V V= 0 P V= 0, so E= q V Entire Echange due to heat absorbed or lost 14 14

15 Heat at Constant Pressure More common Reactions open to atmosphere Constant P Enthalpy H = E + PV Enthalpy change H = E + P V Substituting in first law for Egives H = (q P V) + P V = q P H = q P Heat of reaction at constant pressure 15 15

16 P V = -92 J q = 165 J U = q + w = (+165) + (-92) = +73 J 16 16

17 Heat of Reaction and Internal Energy Zn (s) + HCl (l) ZnCl 2 (aq) + H 2 (g) 17 17

18 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

19 q = - w = - Zn (s) + HCl (l) ZnCl 2 (aq) + H 2 (g)

20 w = -P V = -(1.01 x 10 5 pa)(24.5 l) = x 10 5 pa)(24.5 x 10-3 m 3 ) = x 10 3 J = kj q = -152 kj U = -152 kj + (-2.47 kj) = kj 20 20

21 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 = kj w U = P V = +(1.01 x 10 5 pa)(24.5 l)(2) = +(1.01 x 10 5 pa)(24.5 x 10-3 m 3 )(2) = kj = kj + (+4.95 kj) = kj 21 21

22 Enthalpy and Enthalpy Change Enthalpy is defined 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 H o f (reactants) H o f = Standard enthalpy change (25 o C) 22 22

23 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) = kj = for CO 2 (g) = kj = for NH 2 CONH 2 = kj = for H 2 O (l) = kj H o = [( ) (-2 x )] kj = Since H o has a negative sign, heat is evolved 23 23

24 Converting Between Eand H For Chemical Reactions H E Differ by H E = P V Only differ significantly when gases formed or consumed Assume gases are ideal nrt V = nrt V = P P Since P and T are constant RT V = n P 24 24

25 Converting Between Eand H For Chemical Reactions When reaction occurs Vcaused by nof gas Not all reactants and products are gases So redefine as n gas Where n gas = (n gas ) products (n gas ) reactants Substituting into H = E + P V gives RT H = E + P n gas or P H = E + n RT gas 25 25

26 Ex. 1. What is the difference between H and E for the following reaction at 25 C? 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) What is the % difference between H and E? Step 1: Calculate H using data (Table 7.2) Recall o H H o = o ( H f ) ( H o products H = (4 mol)(33.8 kj/mol) + (1 mol)(0.0 kj/mol) (2 mol)(11 kj/mol) o o f ) reactants = 4 Hf (NO2) + Hf (O2) 2 Hf (N2O5 ) H = 113 kj 26 o 26

27 Step 2: Calculate n gas = (n gas ) products (n gas ) reactants n gas = ( ) mol = 3 mol Step 3: Calculate E using R = J/K mol Ex. 1. (cont.) T = 298 K E = 113 kj (3 mol)( J/K mol)(298 K)(1 kj/1000 J) E = 113 kj 7.43 kj = 106 kj 27 27

28 Ex. 1. finish Step 4: Calculate % difference 7. 43kJ % difference = 100% = 113kJ Bigger than most, but still small 6. 6% Note: Assumes that volumes of solids and liquids are negligible V solid V liquid << V gas 28 28

29 Is Assumption that V solid V liquid << V gas Justified? Consider CaCO 3 (s)+ 2H + (aq) Ca 2+ (aq)+ H 2 O(l) + CO 2 (g) 37.0 ml 2*18.0 ml 18 ml 18 ml 24.4 L Volumes assuming each coefficient equal number of moles So V= V prod V reac = L 24.4 L Yes, assumption is justified Note: If No gasesare present reduces to E H 29 29

30 Learning Check Consider the following reaction for picric acid: 8O 2 (g) + 2C 6 H 2 (NO 2 ) 3 OH(l) 3 N 2 (g) + 12CO 2 (g) + 6H 2 O(l) What type of reaction is it? Calculate ΔΗ, ΔΕ 8O 2 (g) + 2C 6 H 2 (NO 2 ) 3 OH(l) 3N 2 (g) + 12CO 2 (g) + 6H 2 O(l) ΔΗ f (kj/mol) ΔH 0 = 12mol( kj/mol) + 6mol( kJ/mol) + 6mol(0.00kJ/mol) 8mol(0.00kJ/mol) 2mol( kJ/mol) ΔH 0 = 13,898.9 kj ΔΕ = ΔH Δn gas RT = ΔH (15 8)mol*298K* kj/(mol K) Ε = 13,898.9 kj 29.0 kj = 13,927.9 kj 30 30

31 Your Turn! Given the following: 3H 2 (g) + N 2 (g) 2NH 3 (g) H o = kj mol -1 Determine E for the reaction. A kj mol -1 B kj mol -1 C kj mol -1 D kj mol -1 H = E + nrt E = H - nrt E = kj mol ( NRT) -(-2 mol)(8.314 J K -1 mol -1 )(298K)(1 kj/1000 J) E = kj mol

32 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) = kj = for CO 2 (g) = kj = for NH 2 CONH 2 = kj = for H 2 O (l) = kj H o = [( ) (-2 x )] kj = Since H o has a negative sign, heat is evolved 32 32

33 Spontaneity and Entropy Definition of Spontaneous Process: Physical or chemical process that occurs by itself. Why?? Give several spontaneous processes: Still cannot predict spontaneity

34 Enthalpy Changes and Spontaneity What are relationships among factors that influence spontaneity? Spontaneous Change Occurs by itself Without outside assistance until finished Ex. Water flowing over waterfall Melting of ice cubes in glass on warm day 34 34

35 Nonspontaneous Change Occurs only with outside assistance Never occurs by itself: Room gets straightened up Pile of bricks turns into a brick wall Decomposition of H 2 O by electrolysis Continues only as long as outside assistance occurs: Person does work to clean up room Bricklayer layers mortar and bricks Electric current passed through H 2 O 35 35

36 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

37 Reaction Rate and Spontaneity H indicates if reaction has tendency to occur Rate of reaction also plays role Some very rapid: Neurons firing in nerves in response to pain Detonation of stick of dynamite Some gradual: Erosion of stone Ice melting Iron rusting Some so slow, appear to be nonspontaneous: Gasoline and O 2 at RT Many biochemical processes 37

38 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 38 38

39 Direction of Spontaneous Change Many reactions which occur spontaneously are exothermic: Iron rusting Early thoughts were that Fuel burning exothermic reactions were spontaneous) Hand Eare negative Heat given off Energy leaving system Thus, H is one factor that influences spontaneity 39 39

40 Direction of Spontaneous Change Some endothermicreactions occur spontaneously: Ice melting Evaporation of water from lake Expansion of CO 2 gas into vacuum Hand Eare positive Heat absorbed Energy entering system Clearly other factors influence spontaneity 40 40

41 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? 41 41

42 There are how many Laws of thermodynamics? a. 1 b. 2 c. 3 d. 4 e

43 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 divided by the absolute temperature

44 Entropy and Molecular Disorder 44 44

45 Thermodynamic vs. Kinetics Thermodynamics tells us: Direction of reaction Is it possible for reaction to occur? Will reaction occur spontaneously at given T? Will reaction release or absorb heat? Kinetics tells us: Speed of reaction Pathway between reactants and products Energy Reactants Domain of Thermodynamics (initial and final states) Domain of Kinetics (reaction pathway) Reaction Progress Products 45 45

46 Heat Transfer Between Hot and Cold Objects Consider system of two objects Initially one hot and one cold Hot = higher KE ave of molecules = faster Cold = lower KE ave of molecules = slower When they collide, what is most likely to occur? Faster objects bump into colder objects and transfer energy so Hot objects cool down and slow down Cold objects warm up and speed up The reverse doesn t occur 46

47 Result: Heat Transfer Between Hot and Cold Objects Heat flows spontaneously from hot to colder object Heat flows because of probable outcome of intermolecular collisions Spontaneous processes tend to proceed from states of low probability to states of higher probability Spontaneous processes tend to disperse energy 47

48 Entropy (symbol S) Thermodynamic quantity Describes number of equivalent ways that energy can be distributed Quantity that describes randomness of system Greater statistical probability of particular state means greater the entropy! Larger S, means more random and more probable 48 48

49 Entropy If Energy = money Entropy (S) describes number of different ways of counting it 49

50 Criterion for Spontaneity Clear order to things Things get rusty spontaneously Don't get shiny again Sugar dissolves in coffee Stir more it doesn't undissolve Ice liquid water at RT Opposite does NOT occur Fire burns wood, smoke goes up chimney Can't regenerate wood Common factor in all of these: Increase in randomness and disorder of system Something that brings about randomness more likely to occur than something that brings order 50

51 Entropy, S Measure of randomness and disorder Measure of chaos State function Independent of path S = Change in Entropy Also state function S = S final S initial For chemical reaction S = S products S reactants 51 51

52 Entropy S products > S reactants means S+ Entropy 's Probability of state 's Randomness 's Favors spontaneity S products < S reactants means S Entropy 's Probability of state 's Randomness 's Does not favor spontaneity Any reaction that occurs with in entropy tends to occur spontaneously 52 52

53 Effect of Volume on Entropy For gases, Entropy as Volume A. Gas separated from vacuum by partition B. Partition removed C. Gas expands to achieve more probable particle distribution More random, higher probability, more positive S 53 53

54 Effect of Temperature on Entropy As T, entropy A. T = 0 K, particles ( ) in equilibrium lattice positions and S relatively low B. T > 0 K, molecules vibrate, S C. T further, more violent vibrations occur and S higher than in B 54 54

55 Effect of Physical State on Entropy Crystalline solid very low S Liquid higher S, molecules can move freely More ways to distribute KE among them Gas highest S, particles randomly distributed throughout container Many, many ways to distribute KE 55 55

56 Entropy Affected by Number of Particles Adding particles to system number of ways energy can be distributed in system So all other things being equal Reaction that produces more particles will have positive S 56 56

57 Larger V, greater S Expansion of gas S + Higher T, greater S Summary Higher T, means more KE in particles, move more, so random distributions favored S solid < S liquid << S gas Solids more ordered than liquids, which are much more ordered than gases Reactions involving gases Simply calculate change in number of mole gas, n gas If n gas +, S + If n gas, S 57 57

58 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. 58

59 Did You Get It! Which represents an increase in entropy? A. water vapor condensing to liquid B. carbon dioxide subliming C. liquefying helium gas D. proteins forming from amino acids 59 59

60 Entropy Changes in Chemical Reactions Ex. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) n reactant = 4 n product = 2 n = 2 4 = 2 Predict S rxn < 0 Higher positional probability Lower positional probability 60

61 Entropy Changes in Chemical Reactions Reactions without gases Simply calculate number of mole molecules n = n products n reactants If n +, S + If n, S More molecules, means more disorder Usually the side with more molecules, has less complex molecules Smaller, fewer atoms per molecule 61

62 Ex. 2 Predict Sign of S for Following Reactions CaCO 3 (s) + 2H + (aq) Ca 2+ (aq) + H 2 O(l) + n gas = 1 mol 0 mol = 1 mol n gas +, S + 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) n gas = 4 mol + 1 mol 2 mol = 3 mol n gas +, S + OH (aq) + H + (aq) H 2 O (l) n gas = 0 mol n = 1 mol 2 mol = 1 mol n, S CO 2 (g) 62

63 Predict Sign of S in Following: Dry ice carbon dioxide gas CO 2 (s) CO 2 (g) positive Moisture condenses on a cool window H 2 O(g) H 2 O(l) negative AB A + B positive A drop of food coloring added to a glass of water disperses positive 2Al(s) + 3Br 2 (l) 2AlBr 3 (s) negative 63

64 Did You Get It? Which of the following has the most entropy at standard conditions? A. H 2 O(l) B. NaCl(aq) C. AlCl 3 (s) D. Can t tell from the information Which reaction would have a negative entropy? A. Ag + (aq) + Cl - (aq) AgCl(s) B. N 2 O 4 (g) 2NO 2 (g) C. C 8 H 18 (l ) + 25/2 O 2 (g) 8CO 2 (g) + 9H 2 O(g) D. CaCO 3 (s) CaO(s) + CO 2 (g) 64 64

65 Both Entropy and Enthalpy Can Affect Reaction Spontaneity Sometimes they work together Building collapses PE H Stones disordered S + Sometimes work against each other Ice melting (ice/water mix) Endothermic H + nonspontaneous Disorder of molecules S + spontaneous 65

66 Which Prevails? Hard to tell depends on temperature! At 25 C, ice melts At 25 C, water freezes So three factors affect spontaneity: H S T 66 66

67 Second Law of Thermodynamics When a spontaneous event occurs, total entropy of universe s ( S total > 0) In a spontaneous process, S system can decrease as long as total entropy of universe increases S total = S system + S surroundings It can be shown that S surroundin gs = q surroundings T 67

68 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 divided by the absolute temperature

69 Does This Make Sense? Yes? As heat added Disorder or entropy, S q But how much T, S, depends on T at which it occurs Low T, larger S High T, smaller S S 1/T 69

70 What do you think? When ice melts in your hand (assume your hand is 30 o C), A. the entropy change of the system is less then the entropy change of the surroundings. B.the entropy change of the surroundings is less than the entropy change of the system. C.the entropy change of the system equals the entropy change of the surroundings. S sys = H/273K S surr = H/303K S sys > S surr 70

71 Law of Conservation of Energy Says q lost by system must be gained by surroundings q surroundings = q system If system at constant P, then q system = H So q surroundings = H system and S surroundings = q surroundings T = H T system 71

72 Thus Entropy for Entire Universe is S total = S system H system Multiplying both sides by Twe get T S total = T S system H system or T S total = ( H system T S system ) For reaction to be spontaneous T S total > 0 (+) So, ( H system T S system ) < 0 ( ) for reaction to be spontaneous T 72

73 Standard Entropy (Absolute Entropy): Entropy value for the standard state of a species. See Table B-13 p. B-17 and Table B-16 p. B-28 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

74 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 13.2 p

75 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 S o = Σn S o (products) - Σm S o (reactants) S o = [( ) - (2x )] J/K= -356 J/K See exercise 13.3 p 584 and problems 8-12 and

76 Third Law of Thermodynamics At absolute zero (0 K), Entropy of perfectly ordered, pure crystalline substance is zero S= 0 at T= 0 K Since S= 0 at T= 0 K Define absolute entropy of substance at higher temperatures Standard entropy, S Entropy of 1 mole of substance at 298 K (25 C) and 1 atm pressure S = S for warming substance from 0 K to 298 K (25 C) 76 76

77 Consequences of Third Law 1. All substances have positive entropies as they are more disordered than at 0 K Heating randomness S is biggest for gases most disordered 2. For elements in their standard states S 0 (but H f = 0) Units of S J/(mol K) Standard Entropy Change To calculate S for reaction, do Hess's Law type calculation Use S rather than entropies of formation S o = n S o (products) m S o (reactants) 77 77

78 Learning Check Calculate S 0 for the following: CO 2 (s) CO 2 (g) S 0 (J/mol K) S 0 = ( ) J/mol K S 0 = 26.1 J/mol K CaCO 3 (s) CO 2 (g) + CaO(s) S 0 (J/mol K) S 0 = ( ) J/mol K S 0 = 161 J/mol K 78 78

79 Ex. 3.Calculate S for reduction of aluminum oxide by hydrogen gas Al 2 O 3 (s) + 3 H 2 (g) 2 Al (s) + 3 H 2 O (g) Substance S (J/ K mol) Al (s) 28.3 Al 2 O 3 (s) H 2 (g) H 2 O (g) S o = S o o nps products = [2S o Al [ S ( s ) o Al O 2 + 3S 3 ( s ) o H O 2 n + 3S o r S reactants ( g ) o H ] 2 ( g ) ] 79 79

80 S o = 2 mol * 1 mol * 28.3 J mol K Ex J mol K J + 3 mol * mol K 130.6J + 3 mol * mol K S = 56.5 J/K J/K J/K J/K S = J/K 80 80

81 Gibbs Free Energy Would like one quantity that includes all three factors that affect spontaneity of a reaction Define new state function Gibbs Free Energy Maximum energy in reaction that is "free" or available to do useful work G H TS At constant P and T, changes in free energy G = H T S 81 81

82 Gibbs Free Energy G = H T S G < 0 Spontaneous process G = 0 At equilibrium G > 0 Nonspontaneous G state function Made up of T, H and S = state functions Has units of energy Extensive property G = G final G initial 82 82

83 Criteria for Spontaneity? At constant P and T, process spontaneous only if it is accompanied by in free energy of system H S Spontaneous? + G = ( ) [T(+)] = Always,regardless of T + G = (+) [T( )] = + Never,regardless of T + + G = (+) [T(+)] =? Depends; spontaneous at hight, G G = ( ) [T( )] =? Depends; spontaneous at lowt, G 83 83

84 Summary When Hand Shave same sign, Tdetermines whether spontaneous or non-spontaneous Temperature-controlled reactions are spontaneous at one temperature and not at another 84 84

85 Free Energy Concept G = H T S 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 = kj S o = -365 J/K = kj/k H o T S o = ( kj) (298 K) x (-0.365kJ/K = kj H o T S o is a negativity quantity, from which we can conclude that the reaction is spontaneous under standard conditions

86 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 86 86

87 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 87 87

88 Standard Free Energy Changes Standard Free Energy Change, G Gmeasured at 25 C (298 K) and 1 atm Two ways to calculate, depending on what data is available Method 1. G = H ( K) S Ex. 4. Calculate G for reduction of aluminum oxide by hydrogen gas Al 2 O 3 (s) + 3 H 2 (g) 2 Al(s) + 3H 2 O(g) 88 88

89 Ex. 4. Method 1 Al 2 O 3 (s) + 3 H 2 (g) 2 Al (s) + 3 H 2 O (g) Step 1: Calculate H for reaction using H o = H Substance Al (s) 0.0 Al 2 O 3 (s) H 2 (g) 0.0 H 2 O (g) o n p H = [2 H [ H o f products o f Al( s ) o f Al 2 O 3 ( s ) H f + 3 H o (kj/mol) o f H n r o f H O( g ) H 2 H ] ( g ) ] o f o f H reactants 89 89

90 Ex. 4. Method 1 Step 1( H ) H o kj kj = 2 mol* + 3 mol* mol mol kj kj 1 mol* + 3 mol* mol mol H = 0.0 kj kj 0.00 kj ( kj) H = kj 90 90

91 Ex. 4 Method 1 Step 2: Calculate S see Ex. 4 S = J/K Step 3: Calculate G = H ( K) S G = kj (298 K)(179.9 J/K)(1 kj/1000 J) G = kj 53.6 kj = kj G = + not spontaneous 91 91

92 Method 2 Use Standard Free Energies of Formation o G f Energy to form 1 mole of substance from its elements in their standard states at 1 atm and 25 C G o = n p G o f products n r G o f reactants 92 92

93 Ex. 4. Method 2 Calculate G for reduction of aluminum oxide by hydrogen gas. Al 2 O 3 (s)+ 3 H 2 (g) 2 Al (s)+ 3 H 2 O (g) Substance G f Al (s) 0.0 Al 2 O 3 (s) H 2 (g) 0.0 H 2 O (g) o (kj/mol) G o = [2 G [ G o f Al o f Al( s ) 2 O 3 ( s ) + 3 G + 3 G o f H O( g ) 2 o f H 2 ( g ) ] ] 93 93

94 Ex. 4. Method 2 G o kj kj = 2 mol* + 3 mol* mol mol kj kj 1 mol* + 3 mol* mol mol G = 0.0 kj kj 0.00 kj ( kj) G = kj Both methods same within experimental error 94 94

95 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

96 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) = kj/mol KCl (s) = kj/mol O 2 (g) = 0 G fo are as follows: KClO 3 (s) = kj/mol KCl (s) = kj/mol O 2 (g) =

97 2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g) H f o 2 x ( ) 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

98 Spontaneous Reactions Produce Useful Work Fuels burned in engines to power cars or heavy machinery Chemical reactions in batteries Start cars Run cellular phones, laptop computers, mp3 players Energy not harnessed if reaction run in an open dish All energy lost as heat to surroundings Engineers seek to capture energy to do work Maximize efficiency with which chemical energy is converted to work Minimize amount of energy transformed to unproductive heat 98 98

99 Thermodynamically Reversible Process that can bereversed and is always very close to equilibrium Change in quantities is infinitesimally small Example - expansion of gas Done reversibly, it does most work on surroundings 99 99

100 G = Maximum Possible Work Gis maximum amount of energy produced during a reaction that can theoretically be harnessed as work Amount of work if reaction done under reversible conditions Energy that need not be lost to surroundings as heat Energy that is free or available to do work

101 Ex. 5 CalculateΔG for reaction below at 1 atm and 25 C, given ΔH = 246.1kJ/mol, ΔS = /(mol K). H 2 C 2 O 4 (s) + ½ O 2 (g) 2CO 2 (g) + H 2 O(l) G 25 = H T S o kj J 1J G = (298K ) mol K mol 1000kJ ΔG =( )kJ/mol ΔG = 358.5kJ/mol

102 You Try! Calculate G o for the following reaction, H 2 O 2 (l ) H 2 O(l ) + O 2 (g) given H o = kj mol -1 and S o = J K -1 mol -1. A kj mol -1 B kj mol -1 C kj mol -1 D. 3.7 x 10 5 kj mol -1 G o = kj mol K ( kj K -1 mol -1 ) G o = kj mol

103 System at Equilibrium Neither spontaneous nor nonspontaneous In state of dynamic equilibrium G products = G reactants G= 0 Consider freezing of water at 0 o C H 2 O(l) H 2 O(s) System remains at equilibrium as long as no heat added or removed Both phases can exist together indefinitely Below 0 o C, G < 0freezing spontaneous Above0 o C, G> 0 freezing nonspontaneous Define equilibrium

104 Ex. 6 CalculateT bp for reaction below at 1 atm and 25 C, given ΔH = 31.0kJ/mol, ΔS = 92.9 J/(mol K) Br 2 (l) Br 2 (g) o H 31.0kJ / mol T bp = 334K o S kJ /( mol K ) For T > 334 K, ΔG < 0 and reaction is spontaneous (ΔS dominates) For T < 334 K, ΔG > 0 and reaction is nonspontaneous (ΔH dominates) For T = 334 K, ΔG = 0 and T = normal boiling point

105 Free energy change during reaction

106 Free energy change during reaction

107 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

108 Equilibrium Expression aa + bb cc + dd Kc = [C] c [D] d [A] a [B] b Solids and liquids considered unity (1)

109 Write Kp and Kc Consider again N 2 (g) + 3H 2 (g) 2 NH 3 (g) K c = [NH 3 ] 2 [N 2 ][H 2 ]

110 In terms of partial pressures, N 2 (g) + 3H 2 (g) 2 NH 3 (g) K p = [P NH3 ] 2 [P N2 ][P H2 ]

111 K c and K p are related. PV = nrt P = [n/v]rt K p =K c (RT) n n = (number of moles of product gas) (number of moles of reactant gas) For this reaction, n = -2. N 2 (g) + 3H 2 (g) 2 NH 3 (g)

112 No Work Done at Equilibrium G = 0 No free energy available to do work Consider fully charged battery Initially All reactants, no products G large and negative Lots of energy available to do work As battery discharges Reactants converted to products G less negative Less energy available to do work At Equilibrium G = G products G reactants = 0 No further work can be done Dead battery

113 Phase Change = Equilibrium H 2 O(l) H 2 O(g) G= 0 = H T S Only one temperature possible for phase change at equilibrium Solid-liquid equilibrium Melting/freezing temperature (point) Liquid-vapor equilibrium Boiling temperature (point) S = H T = Thus H = T S and or S H T

114

115 G and Position of Equilibrium When G > 0 (positive) Position of equilibrium lies close to reactants Little reaction occurs by the time equilibrium is reached Reaction appears nonspontaneous When G < 0 (negative) Position of equilibrium lies close to products Mainly products exist by the time equilibrium is reached Reaction appears spontaneous

116 G and Position of Equilibrium When G = 0 Position of equilibrium lies ~ halfway between products and reactants Significant amount of both reactants and products present at time equilibrium is reached Reaction appears spontaneous, whether start with reactants or products Can Use G to Determine Reaction Outcome G large and positive No observable reaction occurs G large and negative Reaction goes to completion

117 Learning Check Ex. 7 Given that ΔH = 97.6 kj/mol, ΔS = 122 J/(mol K), at 1atm and 298K, will the following reaction occur spontaneously? MgO(s)+ 2HCl(g) H 2 O(l)+ MgCl 2 (s) G = H T S = 97.6kJ/mol 298K( kj/mol K) G = 97.6kJ/mol +36.4kJ/mol = 61.2 kj/mol

118 Effect of Temperature on G Reactions often run at T s other that 298 K Position of equilibrium can change as G depends on T G = H T S For T s near 298 K, expect only very small changes in H and S For reaction at T,we can write: G o T = H o T T S o T o G T H o o 298 T S

119 Ex. 8 Determining Effect of T on Spontaneity Calculate G at 25 C and 500 C for the Haber process N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Assume that H and S do not change with T Solving Strategy Step 1.Using data from Tables, calculate H and S for the reaction at 25 C H = kj S = J/K

120 Ex. 8 Determining Effect of T on Spontaneity Step 2.Calculate G for the reaction at 25 C using H and S N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H = kj 1kJ S = J/K 1000J G = H T S G = kj (298 K)( J/K) G = kj kj = 33.3 kj So the reaction is spontaneous at 25 C

121 Ex. 8 Determining Effect of T on Spontaneity Step 3.Calculate G for the reaction at 500 C using H and S. T = 500 C = 773 K 1kJ H = kj 1000J S = J/K G = H T S G = kj (773 K)( J/K) G = kj+ 153 kj = 61 kj So the reaction is NOT spontaneous at 500 C

122 Ex. 8 Does this answer make sense? G = H T S H = kj S = J/K Since both H and S are negative At low T G will be negative and spontaneous At high T T S will become a bigger positive number and G will become more positive and thus eventually, at high enough T, will become nonspontaneous

123 Effect of Change in Pressure or Concentration on G Gat nonstandard conditions is related to G at standard conditions by an expression that includes reaction quotient Q o This important expression allows for any concentration or pressure Recall: G = G Q + RT lnq y [ products] = [ reactants] x

124 Ex. 9 Calculating G at Nonstandard conditions Calculate Gat 298 K for the Haber process N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G = 33.3 kj For a reaction mixture that consists of 1.0 atm N 2, 3.0 atm H 2 and 0.5 atm NH 3 Step 1 Calculate Q Q = P 2 PNH 3 N PH 2 2 = (0.50) 2 3 (1.0)(3.0) 3 =

125 Ex. 9 Calculating G at Nonstandard conditions Step 2Calculate G = G + RT lnq G = 33.3 kj/mol + (8.314J/K mol)*(1kj/1000j)* (298K)*ln( ) = 33.3 kj/mol + (2.479 kj/mol)*ln( ) = 33.3 kj + ( 11.6 kj/mol) = 44.9 kj/mol At standard conditions all gases (N 2, H 2 and NH 3 ) are at 1 atm of pressure Gbecomes more negative when we go to 1.0 atm N 2, 3.0 atm H 2 and 0.5 atm NH 3 Indicates larger driving force to form NH 3 P reactants > P products

126 How K is related to G Use relation G = G + RT lnq to derive relationship between K and G At Equilibrium G = 0 and Q = K So 0= G + RT lnk G = RTlnK Taking antilog (e x ) of both sides gives K= e G /RT

127 At Equilibrium G = RTlnK and K= e G /RT Provides connection between G and K Can estimate K s at various T s if know G Can get G in know K s Relationship between K and G K eq G Reaction > 1 Spontaneous Favored Energy released < 1 + non-spontaneous Unfavorable Energy needed = 1 0 At Equilibrium

128 Ex. 10 Calculating G from K K sp for AgCl(s) at 25 C is Determine G for the process Ag + (aq) + Cl (aq) AgCl (s) Reverse of K sp equation, so K = 1 K sp 1 = = G = RT lnk = (8.3145J/K mol)(298k)* ln( )*(1kJ/1000J) G = 56 kj/mol Negative G indicates precipitation will occur

129 Ex. 11 Calculating K from G Calculate K at 25 C for the Haber process N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G = 33.3 kj/mol = 33,300 J/mol o G RT K = e Step 1 Solve for exponent G o = RT J K Step 2Take e x to obtain K / ( 33,300J / mol ) ( / mol )(298K ) = 13.4 K = e G o / RT 13.4 Large Kindicates NH 3 favored at RT = e =

130 You Try! Calculate the equilibrium constant for the decomposition of hydrogen peroxide at 298 K given G o = kj mol. A. 8.5 x B. 1.0 x C. 3.4 x D x o G ( 234,300 J / mol ) = = RT ( J / K mol )(298 K ) K = e = 1.17 x

131 Ex. 12 Calculating K from G, First Calculate G Calculate the equilibrium constant at 25 C for the decarboxylation of liquid pyruvic acid to form gaseous acetaldehyde and CO 2 O O H 3 C C C OH H 3 C C H + CO 2 O

132 First Calculate G from G f Compound G f, kj/mol CH 3 COH CH 3 COCOOH CO G o G o o f o o 3 f f COCOOH = G ( CH COH ) + G ( CO2 ) G ( CH 3 = ( ) ( ) G o = kJ 132 ) 132

133 Next Calculate Equilibrium Constant K = e G o RT RT G o = 64.28kJ 1000J (8.314J / K)(298K) kj = ( ) K = e = e K =

134 Temperature Dependence of K G = RT lnk = H T S Rearranging gives lnk = H R 1 T Equation for line o S R Slope = H /RT Intercept = S /R Also way to determine Kif you know H and S + o

135 Ex. 12 Calculate K given H and S Calculate Kat 500 C for Haber process N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Given H = kj and S = J/K Assume that H and S do not change with T o o H S lnk = + RT R 92,380J 198.4J / K lnk = + (8.314J / K )(773K ) (8.314J / K ) lnk = = 9.49 K = e 9.49 =

136 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 = and K p = 1.1 x atm What do these values tell you about CaCO 3? What happens when the reaction is carried out at a higher temperature?

137 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?

138 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 )

139 a. From Data Table you have the following: CaCO 3 (s) CaO (s) + CO 2 (g) H fo : kj S o : J/K H o = [( ) ( )] = kj S o = [( ) (92.9)] = J/K G To = H o T S o = 178.3kJ (1273 K)(0.1590kJ/K) = kj G o is negative reaction is spontaneous

140 b. Substitute the values of G o at 1273 K, which equals x 10 3 J, into the equation relating ln K and G o. ln K = G o -RT = x x 1273 = K = K p = e = 9.76 K p = P CO2 = 9.76 atm

141 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 = kj kj/k T = 1121 K = 848 o C

142 Bond Energy Amount of energy needed to break chemical bond into electrically neutral fragments Useful to know Within reaction Bonds of reactants broken New bonds formed as products appear Bond breaking 1 st step in most reactions One of the factors that determines reaction rate Ex. N 2 very unreactive due to strong N N bond

143 Bond Energies Can be determined spectroscopically for simple diatomic molecules H 2, O 2, Cl 2 More complex molecules, calculate using thermochemical data and Hess s Law Use H formation enthalpy of formation Need to define new term Enthalpy of atomization or atomization energy, H atom Energy required to rupture chemical bonds of 1 mole of gaseous molecules to give gaseous atoms

144 Determining Bond Energies Ex. CH 4 (g) C(g) + 4 H(g) H atom = energy needed to break all bonds in molecule H atom /4 = average bond C H dissociation energy in methane D = bond dissociation energy Average bond energy to required to break all bonds in molecule How do we calculate this? Use H f for forming gaseous atoms from elements in their standard states Hess s Law

145 Determining Bond Energies Path 1: bottom Formation of CH 4 from its elements = H f Path 2: top 3 step path Step 1 break H H bonds Step 2 break C C bonds Step 3: form 4 C H bonds 1.2H 2 (g) 4H(g) H 1 = 4 H f (H,g) 2.C(s) C(g) H 2 = H f (C,g) 3.4H(g) + C(g) CH 4 (g) 2H 2 (g) + C(s) CH 4 (g) H 3 = H atom H = H f (CH 4,g)

146 Calculating H atom and Bond Energy H f (CH 4,g) = 4 H f (H,g) + H f (C,g) H atom Rearranging gives H atom = 4 H f (H,g) + H f (C,g) H f (CH 4,g) Look these up in Table 18.3, 6.2 or appendix C H atom = 4(217.9kJ/mol) kJ/mol ( 74.8kJ/mol) H atom = kj/mol of CH kj/mol bond energy = 4 = kj/mol of C H bonds

147 Table 19.4 Some Bond Energies

148 Using Bond Energies to Estimate H f Calculate H f for CH 3 OH(g) (bottom reaction) Use 4 step path Step 1 break 1C C bonds Step 2 break 2H H bonds Step 3: break 1O O bond Step 4: form 3 C H, 1 O H, & 1O C bonds

149 Using Bond Energies H f (CH 3 OH,g) = H f (C,g) + 4 H f (H,g) + H f (O,g) H atom (CH 3 OH,g) H f (C,g) + 4 H f (H,g) + H f (O,g) = { (4*217.9) }kJ = kj H atom (CH 3 OH,g) = 3D C H + D C O + D O H = (3*412) = 2059 kj H f (CH 3 OH,g) = kj 2059 kj = 222 kj Experimentally find H f (CH 3 OH,g) = 201 kj/mol So bond energies give estimate within 10% of actual

150 Chapter 13/19 Chemical Thermodynamics 1. Spontaneous Chemical and Physical Processes 2. Entropy and Disorder 3. Entropy and the Second Law of Thermodynamics 4. Standard-State Entropies of Reaction 5. The Third Law of Thermodynamics 6. Calculating Entropy Changes for Chemical Reactions 7. Gibbs Free Energy 8. The Effect of Temperature on the Free Energy of a Reaction 9. Beware of Oversimplification 10. Stand-State Free Energies of Reaction 11. Equilibria Expressed in Partial Pressures 12. Interpreting Stand-State Free Energy of Reaction Data 13. Relationship Between Free Energy and Equilibrium Constants 14. Temperature Dependence of Equilibrium Constants 15. Gibbs Free Energies of Formation and Absolute Entropies 16. Calculate H with bond energies

151 Entropy Change for a Phase Transition S > q/t (at equilibrium) What processes can occur under phase change at equilibrium? Solid to liquid Liquid to gas Solid to gas

152 Solution: S = H vap /T = (39.4 x 10 3 J/mol)/ 298 K = 132 j/(mol K) Entropy of Vapor = ( ) J/(mol K) = 348 J/(mol K)

153