Brown et al, Chemistry, 2nd ed (AUS), Ch. 12:

Transcription

1 Kinetics: Contents Brown et al, Chemistry, 2 nd ed (AUS), Ch. 12: Why kinetics? What is kinetics? Factors that Affect Reaction Rates Reaction Rates Concentration and Reaction Rate The Change of Concentration with Time Temperature and Reaction Rate Reaction Mechanisms Catalysis 1

2 Learning Outcomes After completing these sections, you should be able to: 1. Explain what is meant by the terms: rate of reaction and indicate what general factors affect the rates of reactions. 2. Write the rate of reaction in terms of any of the reactants or products given the stoichiometric equation. 3. Explain the terms rate expression, rate constant, reaction order and write the rate expression for zero order, first order and second order reactions. 4. Calculate initial reaction rates in terms of concentration of reactants in appropriate units. 5. Explain elementary steps in chemical reactions in terms of molecular collisions or molecular dissociation. 2

3 Learning Outcomes cont. 6. Distinguish between order and molecularity and elementary step and reaction mechanism. 7. Use initial rates to determine reaction order (zero, first and second order reactions). 8. Calculate reaction rates given a rate constant and reaction order (zero, first and second order reactions) and reactant concentrations. 9. Distinguish between zero, first and second order reactions on the basis of the concentration dependence of the reaction rate. 10. Calculate rate constants using integrated rate expressions. 11. Draw a typical energy profile for a reaction and describe how the temperature affects the proportion of molecules colliding with sufficient energy to react. 3

4 Learning Outcomes cont. 12. Explain the terms Arrhenius equation, activation energy and pre-exponential factor. 13. Obtain the activation energy and pre-exponential factor given the rate constant as a function of temperature, graphically or by calculation. 14. Calculate the rate constant at a particular temperature given the activation energy and the rate constant at some other temperature. 4

5 Why Kinetics? What is Kinetics? in Chemistry, we deal with reactions reactions mean change of components Thermodynamics: WHAT happens? Kinetics: HOW FAST does it happen? In addition, kinetics can tell us BY WHICH MECHANISM the reaction occurs, i.e. what happens on a molecular level 5

6 Reaction Rates Reaction rates span many orders of magnitude 6

7 Factors Affecting Reaction Rates Physical State of Reactants For a reaction to occur, the reactants need to be in contact. 7

8 Factors Affecting Reaction Rates cont. Concentration of Reactants A reaction usually proceeds faster for higher concentrations. Steel wool heated in air (20% O 2 ) Steel wool heated in oxygen (100% O 2 ) 8

9 Factors Affecting Reaction Rates cont. Temperature of Reactants A reaction usually proceeds faster at higher temperatures. Salmonella E. coli Listeria 9

10 Factors Affecting Reaction Rates cont. Presence of a Catalyst A catalyst is a substance that increases the reaction rate without being consumed during the reaction. 10

11 Factors Affecting Reaction Rates cont. On a Molecular Level: Reactants must have contact to react! Physical State of the Reactants: Reactants with higher surface area react more readily than those with low surface area. Gases react more readily than solids. Concentration of Reactants: Reactants collide more readily at higher concentrations. Temperature: The probability of collision is higher at higher temperature. Presence of a Catalyst: Catalysts affect the mechanism of reactions. 11

12 Progress of a Reaction 12

13 Reaction Rate Reaction rate: speed of a chemical reaction, i.e. change of concentration of reactants or products per time unit Units: M/s (molarity per second) Rates are always given as positive quantities = Average rate of appearance of B change in concentration of B change in time [B] at t 2 [B] at t = 1 = t 2 t 1 Δ [B] Δt 13

14 Reaction Rate For the disappearance of A the rate is accordingly: Average rate of disappearance of A = - change in concentration of A change in time [A] at t 2 [A] at t = - 1 = - t 2 t 1 Δ [A] Δt = Δ [B] Δt 14

15 LECTURE EXAMPLE 1(a): Average Rate of Reaction Each red sphere represents 0.01 mol A, each blue sphere represents 0.01 mol B. The vessel has a volume of 1.00 dm 3. At time t = 0, the vessel contains 1.00 mol A and o mol B. After 20 s, the vessel contains 0.54 mol A and 0.46 mol B. After 40 s, the vessel contains 0.30 mol A and 0.70 mol B. Calculate the average rate at which A disappears over the time interval from 20 s to 40 s. 15

16 LECTURE EXAMPLE 1(b): Average Rate of Reaction Each red sphere represents 0.01 mol A, each blue sphere represents 0.01 mol B. The vessel has a volume of 1.00 dm 3. At time t = 0, the vessel contains 1.00 mol A and o mol B. After 20 s, the vessel contains 0.54 mol A and 0.46 mol B. After 40 s, the vessel contains 0.30 mol A and 0.70 mol B. Calculate the average rate of appearance of B over the time interval from 0 s to 40 s. 16

17 Change of Reaction Rate with Time Reaction of butyl chloride (C 4 H 9 Cl) with water: 17

18 Change of Rate with Time Instantaneous rate = - Δ [C 4 H 9 Cl] Δt 18

19 LECTURE EXAMPLE 2(a): Instantaneous Rate of Reaction Calculate the instantaneous rate of disappearance of C 4 H 9 Cl at the time t = 0 (the initial rate). 19

20 LECTURE EXAMPLE 2(b): Instantaneous Rate of Reaction Calculate the instantaneous rate of disappearance of C 4 H 9 Cl at the time t = 300 s. 20

21 Reaction Rates and Stoichiometry For each reaction there is a relationship between the rate at which a product appears and the rate at which a reactant disappears given by the stoichiometry of the reaction: For a reaction aa + bb cc + dd Rate = - 1 Δ[A] a Δt = - 1 Δ[B] b Δt = 1 Δ[C] c Δt = 1 Δ[D] d Δt 21

22 LECTURE EXAMPLE 3(a): Reaction Rates and Stoichiometry (i) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O 3 (g) 3 O 2 (g)? (ii) If the rate at which O 2 appears, Δ[O 2 ]/Δt, is 6.0 x 10-5 M s -1 at a particular instant, at what rate is O 3 disappearing at this same time, - Δ[O 3 ]/Δt? 22

23 LECTURE EXAMPLE 3(b): Reaction Rates and Stoichiometry The decomposition of N 2 O 5 proceeds according to the following equation 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g). If the rate of decomposition of N 2 O 5 at a particular instant in a reaction vessel is 4.2 x 10-7 M s -1, what is the rate of appearance of (i) NO 2, (ii) O 2? 23

24 Concentration and Rate Reaction of Ammonium and Nitrite Ions in Water: The reaction rate depends on the concentration of reactants: Rate = k [NH 4+ ] [NO 2- ] 24

25 Rate Laws An equation that shows how the reaction rate depends on the concentration of reactants is called a rate law: aa + bb cc + dd Rate = k [A] m [B] n k is the rate constant and depends on temperature. The rate constant must be determined experimentally. 25

26 Reaction Order From the rate law we can determine the reaction order: Rate = k [A] m [B] n The rate is (m)th order in A and (n)th order in B. The overall reaction order is (m+n). 26

27 Units of Rate Constants The units of rate constants depend on the order of the reaction: Overall reaction order Units of k 0 Ms -1 1 s -1 2 M -1 s -1 3 M -2 s -1 The rate constant DOES NOT depend on concentration. It DOES depend on temperature and the presence of a catalyst. 27

28 LECTURE EXAMPLE 4(a): Effect of Concentration on the Rate Law Consider a reaction A + B C with reaction rate = k [A][B] 2. Each of the boxes represents a reaction mixture in which A is shown as red spheres and B as purple spheres. Rank these mixtures in order of increasing rate of reaction. 28

29 LECTURE EXAMPLE 4(b): Effect of Concentration on the Rate Law Consider a reaction A + B C. Assuming that rate = k [A][B], rank the mixtures in order of increasing rate of reaction. (Each of the boxes represents a reaction mixture in which A is shown as red spheres and B as purple spheres). 29

30 LECTURE EXAMPLE 5(a): Reaction Orders and Units of Reaction Constants (i) What are the overall reaction orders for the following reactions: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) rate = k[n 2 O 5 ] CHCl 3 (g) + Cl 2 (g) CCl 4 (g) + HCl(g) rate = k[chcl 4 ][Cl 2 ] 1/2 (ii) What are the units of the rate constant for the rate law for the reaction 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) rate = k[n 2 O 5 ] 30

31 LECTURE EXAMPLE 5(b): Reaction Orders and Units of Reaction Constants (i) What is the reaction order of the reactant H 2 in the following equation: H 2 (g) + I 2 (g) 2 HI(g) rate = k[h 2 ][I 2 ] (ii) What are the units of the rate constant? 31

32 LECTURE EXAMPLE 6(a): Using Initial Rates to Determine Rate Laws The initial rate of a reaction A + B C was measured for several different starting concentrations of A and B: Using these data, determine (a) the rate law for the reaction, (b) the magnitude of the rate constant, (c) the rate of the reaction, when [A] = M and [B] = M. 32

33 LECTURE EXAMPLE 6(b): Using Initial Rates to Determine Rate Laws The following data were measured for the reaction of nitric oxide with hydrogen: 2 NO(g) + 2 H 2 (g) N 2 (g) + 2 H 2 O(g) (a) Determine the rate law for the reaction. (b) Calculate the rate constant. (c) Calculate the rate, when [NO] = M and [H 2 ] = M. 33

34 Change of Concentration over Time Zero-Order Reactions The rate is independent of the concentration of reactants: Rate = - Δ[A] Δt = k [A] 0 = k Integrated rate law: [A] t = - k t + [A] 0 34

35 Change of Concentration over Time First-Order Reactions The rate depends on the concentration of one reactant raised to the first power: Rate = - Δ[A] Δt = k [A] Integrated rate law: ln [A] t ln [A] 0 = - k t OR ln [A] t [A] 0 = - k t 35

36 LECTURE EXAMPLE 7(a): Integrated First-Order Rate Law The decomposition of a certain insecticide in water follows firstorder kinetics with a rate constant of 1.45 yr -1 at 12 C. A quantity of this insecticide is washed into a lake on 1 June, leading to a concentration of 5.0 x 10-7 g cm -3. Assume that the average temperature of the lake is 12 C. (a) What is the concentration of the insecticide on 1 June of the following year? (b) How long will it take for the concentration of the insecticide to drop to 3.0 x 10-7 g cm -3? 36

37 LECTURE EXAMPLE 7(b): Integrated First-Order Rate Law The decomposition of dimethyl ether, (CH 3 ) 2 O, at 510 C is a first-order process with a rate constant of 6.8 x 10-4 s -1 : (CH 3 ) 2 O(g) CH 4 (g) + H 2 (g) + CO(g). If the initial pressure of (CH 3 ) 2 O is 18 kpa, what is its partial pressure after 1420 s? 37

38 Change of Concentration over Time Graphic representation: First-Order Reactions ln [A] t = - k t + ln [A] 0 has the form of the equation for a straight line: y = m x + b Therefore, in a plot ln [A] t against t, we get a straight line if the reaction is first order. The slope is k and the y-intercept is ln [A] 0. 38

39 Change of Concentration over Time Second-Order Reactions The rate depends on the reactant concentration raised to the second power or on the concentrations of two different reactants, each raised to the first power. Rate = - Δ[A] Δt = k [A] 2 Integrated rate law: 1 1 = k t + [A] t [A] 0 A plot of 1/[A] t against t is a straight line the reaction is second order. The slope is k. The intercept is 1/[A] t. 39

40 LECTURE EXAMPLE 8(a): Integrated Rate Law The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300 C: NO 2 (g) NO(g) + ½ O 2 (g) Is the reaction first or second order in NO 2? 40

41 LECTURE EXAMPLE 8(b): Integrated Rate Law Consider the decomposition of NO 2 : NO 2 (g) NO(g) + ½ O 2 (g) The reaction is second order in NO 2 with k = M -1 s -1. If the initial concentration of NO 2 in a closed vessel is M, what is the remaining concentration after h? Is the reaction first or second order in NO 2? 41

42 Half-Life Half-life is the amount of time until the concentration of a reactant has reached [A] 1/2 =1/2 [A] 0 : For a zero-order reaction: 1/2 [A] 0 = - k t 1/2 + [A] 0 t 1/2 = [A] 0 2 k The half-life is directly proportional to the initial concentration. For a first-order reaction: ln 1/2 [A] 0 = - k t 1/2 [A] 0 t 1/2 = - ln 1/2 k = k The half-life is independent of the initial concentration. 42

43 Half-Life cont. First-order rearrangement of methyl isonitrile: In a first-order reaction, the concentration of the reactant decreases by a factor of ½ in each of a series of regularly spaced time intervals t 1/2. 43

44 Half-Life cont. For a second-order reaction: 1 1 = k t 1/2 + t 1/2[A] 0 [A] 1/2 = 0 1 k [A] 0 The half-life inversely proportional to the initial concentration. 44

45 LECTURE EXAMPLE 9(a): Half-Life The reaction of C 4 H 9 Cl with water is a first-order reaction. (a) From the graph, estimate the half-life for this reaction. (b) Use the half-life to calculate the rate constant. 45

46 LECTURE EXAMPLE 9(b): Half-Life 46

47 Temperature and Rate The rates of most chemical reactions increase as temperature increases. 47

48 The Collision Model: Orientation Factor Cl + NOCl NO + Cl 2 Effective collision: reactants are suitably oriented Ineffective collision: reactants are not suitably oriented 48

49 Activation Energy In order for a reaction to occur, the colliding molecules need to have a minimum kinetic energy (Activation Energy, E a ). This energy is necessary to crack, stretch or bend bonds and react. If the kinetic energy of the colliding molecules is two low, the two molecules just bounce off each other when the electron clouds repel each other. 49

50 Activation Energy cont. (Activated complex) The reaction rate depends on E a, not on ΔE. 50

51 Effect of Temperature on the Kinetic Energy of Molecules f = exp (- E a / RT) 51

52 Arrhenius Equation The reaction rate depends on The fraction of molecules with an energy > E a, The number of collisions per second, The fraction of collisions that have the appropriate orientation. Arrhenius Equation: k = A exp (- E a / RT ) A is a constant. The rate decreases as E a increases. k = reaction rate A = frequency factor E a = activation energy T = temperature R = gas constant 52

53 LECTURE EXAMPLE 10(a): Energy Profile, Activation Energies, Speed of Reaction Consider a series of reactions with the following energy profiles: Assuming that all three reactions have nearly the same frequency factors, rank the reactions from slowest to fastest. 53

54 LECTURE EXAMPLE 10(b): Energy Profile, Activation Energies, Speed of Reaction Imagine that these reactions are reversed: Rank the reverse reactions from slowest to fastest. 54

55 Determination of the Activation Energy (Graphical) k = A exp (- E a / RT ) ln k = - E a / RT + ln A A plot of ln k against 1/T gives a straight line with -E a /R as the slope and ln A as the intercept. 55

56 Determination of the Activation Energy (Non-Graphical) ln k 1 = - E a / RT 1 + ln A and ln k 2 = - E a / RT 2 + ln A ln A = ln k 1 + E a / RT 1 and ln A = ln k 2 + E a / RT 2 ln k 1 + E a / RT 1 = ln k 2 + E a / RT 2 ln k 1 - ln k 2 = + E a / RT E a / RT 1 ln k 1 / k 2 = + E a / R (1/T 2 1/T 1 ) 56

57 LECTURE EXAMPLE 11(a): Activation Energy The table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures: (a) Calculate the activation energy for the reaction. (b) What is the value of the rate constant at K? 57

58 LECTURE EXAMPLE 11(b): Activation Energy The following table shows the rate constants for the rearrangement of methyl isonitrile at different temperatures: Calculate the rate constant for the reaction at 280ºC. 58

59 Reaction Mechanisms Chemical reactions describe starting materials and end products. They don t give information HOW the reaction happens. The reaction mechanism describes the process including which bonds are broken and formed, which atoms rearrange themselves in a molecule and the intermediate products. 59

60 Elementary Reactions Simple, one-step reactions: Collisions between methyl isonitrile (CH3NC) provide energy for rearrangement of atoms. NO(g) + O 3 (g) NO 2 (g) + O 2 (g) Single collisions between NO and ozone provides energy for the reaction. 60

61 Molecularity Elementary reactions can be classified according to the number of molecules that participate: Molecularity = number of molecules that participate as reactants in an elementary reaction. Unimolecular: a single molecule is involved in the reaction Bimolecular: collision of two reactant molecules Termolecular: simultaneous collision of three molecules 61

62 Multistep Mechanisms Overall reaction: NO 2 (g) + CO(g) NO(g) + CO 2 (g) Elementary reaction steps: NO 2 (g) + NO 2 (g) NO 3 (g) + CO(g) NO 3 (g) + NO(g) NO 2 (g) + CO 2 (g) Elementary reaction steps must add up to the equation for the overall reaction! NO 3 is an intermediate, i.e. it is not a reactant and not a product in this reaction. 62

63 Multistep Mechanisms cont. Multistep mechanisms involve one or more intermediates. 63

64 LECTURE EXAMPLE 12(a): Molecularity and Intermediates It has been proposed that the conversion of ozone into O 2 proceeds by a two-step mechanism: O 3 (g) O 2 (g) + O(g) O 3 (g) + O(g) 2 O 2 (g) (i) Describe the molecularity of each elementary reaction in this mechanism. (ii) Write the equation for the overall reaction. (iii) Identify the intermediates. 64

65 LECTURE EXAMPLE 12(b): Molecularity and Intermediates The following mechanism has been proposed for the reaction of NO with H 2 to form N 2 O and H 2 O: NO(g) + NO(g) N 2 O 2 (g) N 2 O 2 (g) + H 2 (g) 2N 2 O(g) + H 2 O(g) (i) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (ii) Write the rate law for each elementary reaction in the mechanism. (iii) Identify any intermediates in the mechanism. (iv) The observed rate law is rate = k [NO] 2 [H 2 ]. If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions? 65

66 Rate Laws for Elementary Reactions Rate laws and relative speeds for elementary reaction steps determine the overall rate law for the reaction. If we know a reaction is an elementary reaction, we know the rate law: Unimolecular: A products Rate = k [A] As the concentration of A increases, so does the rate of decomposition. The reaction is first order. Bimolecular: A + B products Rate = k [A] [B] As the concentration of A and B increases, so does the number of collisions between A and B so does the rate. The reaction is first order in [A] and [B], second order overall 66

67 Elementary Reactions and Rate Laws 67

68 LECTURE EXAMPLE 13(a): Predicting the Rate Law If the following reaction occurs in a single elementary reaction, predict the rate law: H 2 (g) + Br 2 (g) 2 HBr(g) 68

69 LECTURE EXAMPLE 13(b): Predicting the Rate Law Consider the following reaction: 2 NO(g) + Br 2 (g) 2 NOBr(g) (i) Write the rate law for the reaction, assuming it involves a single elementary reaction. (ii) Is a single-step mechanism likely for this reaction? 69

70 Rate-Determining Steps Most chemical reactions occur by mechanisms that involve 2 or more elementary reactions. The overall rate of a reaction can not exceed the rate of the slowest elementary step of its mechanism. The rate-determining step governs the rate law for the overall reaction. 70

71 Mechanisms with a Slow Initial Step If the first step in a reaction is rate-determining, the rate law of the overall reaction equals the rate law of the first step. 71

72 Mechanisms with a Slow Initial Step cont. Step 1: Step 2: Overall: NO 2 (g) + NO 2 (g) NO 3 (g) + NO(g) slow NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) fast NO 2 (g) + CO(g) NO(g) + CO 2 (g) Step 1: rate = k 1 [NO 2 ] 2 Step 2: rate = k 2 [NO 3 ] [CO] Overall: since k 2 >> k 1, the rate of the overall reaction equals the rate of the first step (which agrees with the experiment). 72

73 LECTURE EXAMPLE 14(a): Rate Law for Multistep Mechanisms The decomposition of nitrous oxide, N 2 O, is believed to occur by a two-step mechanism: N 2 O(g) N 2 (g) + O(g) N 2 O(g) +O (g) N 2 (g) + O 2 (g) slow fast (a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction. 73

74 LECTURE EXAMPLE 14(b): Rate Law for Multistep Mechanisms Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen: O 3 (g) + 2 NO 2 (g) N 2 O 5 (g) + O 2 (g) The reaction is believed to occur in two steps: O 3 (g) + NO 2 (g) NO 3 (g) + O 2 (g) NO 3 (g) + NO 2 (g) N 2 O 5 (g) The experimental rate law is rate = k [O 3 ] [NO 2 ]. What can you say about the relative rates of the two steps of the mechanism? 74

75 Mechanisms with a Fast Initial Step If step 2 is the slow, rate-determining step, the rate of the overall reaction is governed by it. 75

76 Mechanisms with a Fast Initial Step cont. 2 NO(g) + Br 2 (g) 2 NOBr(g) Step 1: NO(g) + Br 2 (g) NOBr 2 (g) fast Step 2: NOBr 2 (g) + NO(g) 2 NOBr(g) slow Since step 2 is rate-determining: overall rate = k [NOBr 2 ] [NO] But: the concentration of the intermediate is not known! 76

77 Mechanisms with a Fast Initial Step cont. Assumptions: intermediate is unstable compound and doesn t accumulate intermediate is in dynamic equilibrium with starting materials Therefore: K 1 [NO] [Br 2 ] = k -1 [NOBr 2 ] [NOBr 2 ] = k 1 /k -1 [NO] [Br 2 ] Therefore: overall rate = k 2 k 1 /k -1 [NO] [Br 2 ] [NO] overall rate = k [NO] 2 [Br 2 ] Result is consistent with the experimentally determined rate law. The experimental rate constant k is k 1 k 2 /k

78 LECTURE EXAMPLE 15(a): Rate Law for a Multistep Mechanism The gas phase reaction of nitric oxide with Bromine follows this overall reaction: 2 NO(g) + Br 2 (g) 2 NOBr(g) Show that the following mechanism produces a rate law consistent with the experimentally observed one, which is rate = k [NO] 2 [Br]. Step 1: NO(g) + NO(g) N 2 O 2 (g) fast, equilibrium Step 2: N 2 O 2 (g) + Br 2 (g) 2 NOBr(g) slow 78

79 LECTURE EXAMPLE 15(b): Rate Law for a Multistep Mechanism The first step of a mechanism involving the reaction of bromine is: Br 2 (g) 2 Br(g) (fast, equilibrium) What is the expression relating the concentration of Br(g) to that of Br 2 (g)? 79

80 Catalysis Catalyst: A substance that changes the speed of a chemical reaction without undergoing permanent chemical change itself. Homogeneous Catalysis: The catalyst that is present in the same phase as the reacting molecules. Heterogeneous Catalysis: The catalyst exists in a different phase from the reactant molecules. 80

81 Homogeneous Catalysis Without a catalyst: With a catalyst: 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g) very slow Step 1: Step 2: Overall: 2 Br (aq) + 2 H 2 O 2 (aq) + 2 H + Br 2 (aq) + 2 H 2 O(l) Br 2 (aq) + H 2 O 2 (aq) 2 Br (aq) + 2H + (aq) + O 2 (g) 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g) Bromide ion is the catalyst because it speeds up the reaction without undergoing a net change itself. Br 2 is an intermediate, first formed, then consumed. 81

82 Homogeneous Catalysis H 2 O 2 (aq) + NaBr(aq) Overall reaction: 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) 82

83 Energy Profiles with and without Catalyst A catalyst lowers the overall activation energy for a chemical reaction and/or increases the frequency factor by providing a different reaction mechanism. 83

84 Heterogeneous Catalysis Without a catalyst: C 2 H 4 (g) + H 2 C 2 H 6 (g) ΔH 0 = kj/mol exothermic, but very slow With a catalyst, i.e. in the presence of a metal powder: adsorption of reactants to the metal surface less energy is needed to break bonds, i.e. the activation energy is lowered formation of new bonds releases energy new species desorbs from the metal surface 84

85 Heterogeneous Catalysis Adsorption of C 2 H 4 and H 2 Breaking of H-H and C-C bonds Diffusion of H Desorption of C 2 H 6 85