Chapter 12, Chemical Kinetics

Transcription

1 Chapter 12, Chemical Kinetics This chapter is about: 1. numerical descriptions of how fast rxns. occur 2. the intermediates that form during a rxn (re. mechanism) 3. applying thermodynamics & the kinetic molecular theory to go from the descriptive learning to understanding. Our focus will be on gas and liquid phase (soln.) reactions. Why care? Your health! 1

2 I. Reaction Rates A. Defining rate: change per unit time B. Example: rate = concentration change time change 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) How could you measure the rate? 1. Measure [N 2 O 5 ], [NO 2 ], or [O 2 ] at various times. 2. Method: Pressure change, color change, etc. 2

3 C. Define rate as increase in product conc. per unit time or decrease in reactant conc./unit time. D. Let s try O 2 using data in Table 12.1, p rate O 2 formation = Δ O 2 = [O 2 ] at t 2! [O 2 ] at t 1 Δ t t 2! t 1 Your text uses the s interval, rate = 9 x 10!6 M/s. You should try a different time interval. (Pick one!) Re. Fig. 12.1, does your answer seem reasonable relative to the value at t = s? 3

4 E. Is rate of NO 2 formation equal to the rate of O 2 formation (or the rate of N 2 O 5 decomposition)? 1. Look at Fig (& Instructor animation: N 2 O 5 ) 2. The balanced equation gives us the mole ratios. a) Stoichiometry: 1 O 2, : 4 NO 2 :!2 N 2 O 5 b) Therefore equivalent rates (on M/s basis) are: Δ [O 2 ] = ¼ Δ [NO 2 ] =!½ Δ [N 2 O 5 ] Δ t Δ t Δ t As Δ t approaches zero, Δ[X]'Δ t becomes the slope. 4

5 So, if the O 2 formation rate was 0.80 M/s, the NO 2 formation rate should be, and the N 2 O 5 decomposition rate should be. (Remember, these rates are always positive numbers.) F. Does the rxn. rate change with time? Would you have predicted this? Logic? Try problem 12.2, p

6 II. Rate Laws & Rxn. Order (Demolition derby?) A. We are going to relate our numerical analysis of the rxn. rate to a mental picture (model) of what sort of collisions are occurring during the rxn. B. Consider the following general rxn.: a A + b B Products 1. I can write a rate law for this reaction in the form: Rate = k [A] m [B] n k: rate constant specific for this rxn., T, etc. 6

7 2. Why does this equation work? (Collision-theory) a) Why does a k value have to be included? b) Why [A] times [B]? (Boltzmann?) c) Why is [A] raised to the m power? (Aside: Is a sometimes equal to m [and b to n]?) C. Reaction order 1. The overall reaction order = the sum of all the exponents in the rate law. 2. The reaction order with respect to a specific reactant = the exponent for that reactant. Do prob. 12.3, p. 436 to illustrate. 7

8 D. If you have a balanced chemical rxn., you do not necessarily know the rate law. The rate law must be determined experimentally!!! How can we do that? 8

9 III. Experimental Determination of a Rate Law One method is to measure initial rate at different [reactant]. Then see what exponent (remember m and n?) best fits the data. Rxn.: 2 NO (g) + O 2 (g) ÿ 2 NO 2 (g) General Rate law: rate = k [NO] m [O 2 ] n Use Table 12.3 data, p. 436, to get values for variables. 9

10 A. To get m, compare exp. #1 & #2, where [O 2 ] is constant, but [NO] changes by a factor of 2 (0.015 M vs M). (Do math on board!) If m = 1, the rate should double, if m = 2, the rate should quadruple, if m = 3, the rate should go up by factor of 8, etc. Logic 2 1 = 2, 2 2 = 4, 2 3 = 8, etc. From the data, we see that the rate quadrupled: / = 4, therefore, m = 2. 10

11 B. Following same logic, we can determine value of n. What exps. should we compare in Table 12.3 to accomplish this? Exp. # and exp. #. Note: In this case we want [NO] to be constant. (Do math on board!) C. Now that we know m = 2 and n =, do we have enough information to determine k? Solve for k in the space below:(do math on board!) 11

12 Should we do Worked Key Concept Ex (p. 439) in class? Comment on Key Concept problems!!! IV. Integrated Rate Law: 1st-order Reaction A. For rxn, A ÿ products, that is 1 st order overall: rate =!ΔA / Δt = k [A] (m = 1) Recall from your calculus:!δa / Δt = k [A] is ~same as:!da / dt = k [A] which we can rearrange to get: da / [A] =!k dt which we can integrate to get: ln [A] /[A] =!k t t 0 12

13 This is the integrated 1 st order rate equation. If you prefer base 10: log [A] t /[A] 0 =!k t /2.303 B. Because [A] 0 means [A] at time = 0, it is a constant, we can simplify to: log [A] t =!k t / log [A] 0 1. This should remind you of the equation for a straight line: y = m x + b (See Fig. 12.6, p. 441, next page.) Not the same m as the exponent in the general rate law. 13

14 2. Can you do a linear regression on your calculator to solve for m & b? (You do Probs & 12.7.) V. Half-life of a First-order Reaction A. A number of interesting rxns follow 1 st order decay kinetics. Example: decay of 3 H (tritium, t = 12.3 yr), re. SRS (Savanna River Site). ½ B. Because of the nature of this type of rxn. the half-life is a constant. (What is half-life?) (1 st order CD animation) 14

15 C. Rxn. half-life: time required for amount of reactant to decrease to ½ of its original value. (Fig. 12.7, p. 445) D. Mathematically: log [A] t /[A] 0 =!k t /2.303 log (½) =!k t ½ / /k! =!k t ½ /2.303 So, t ½ = (!0.3010!2.303)/ k = 0.693/ k 15

16 E. This means that: t ½ % 1'k Surprising? (Do Prob.12.9, p 446, Key Concept Prob 12.10) VI. Radioactive Decay Rates A. Applications: 1. Medicine, particularly internal exposure. 2. Archeology, etc. 14 C dating 3. Geology: How old are rocks, age of the earth? B. Try an application with 14 C dating: prob , p

17 VII. Second-order Reactions Many reactions do not fit 1 st order kinetics. Of these, many do fit 2 nd order kinetics. A. For rxn that is 2 nd order in A: ΔA / Δt = k [A] 2 Example: 2 NO 2 (g) ÿ 2 NO (g) + O 2 (g) B. The integrated rate law is: 1/[A] t = k t + 1/ [A] 0 C. Does this look like y = m x + b? Look at Figure 12.8 & the figure on p. 452 in Ex

18 Worked Example 12.8 (Graphical determination of rate law) Left-hand graph should give a straight line if rxn is 1 st order. Right-hand graph should give a straight line if rxn is 2 nd order. What is the order of this rxn with respect to NO 2? A copy of Table 12.4 will be provided w/exam I. 1. Fig. 12.8: plot of [A] vs. t for a 2 nd order rxn. 2. Which of the figures on p. 452 is a better fit to a straight line? What does this tell you about the kinetics of NO 2 decomposition? D. Try prob outside of class. 18

19 VIII. Zeroth- Order Reactions (See Fig. 12.9, p. 453) Rare. Text example notes Pt catalyzed decomposition of NH 3. Collisions appear to be unimportant. This is zeroth order (with respect to?) under the conditions given: 1. Relatively large amount of NH 3 2. Relatively small Pt surface See CD animation for Surface Reaction-hydrogenation Can you imagine conditions where [NH 3 ] & collisions would become important re. rate? 19

20 3. Enzymes can sometimes be viewed as surface catalysts. Assays to determine the amount of a given enzyme (ex.: the enzyme PAH for PKU diagnosis) are often set up to be zeroeth order with respect to the reactants. Would the slope change if we changed [A] 0? 20

21 IX. Reaction Mechanisms (major shift in approach) Time to see if there is more to the analysis that we have been doing than meets the eye. A rxn mechanism describes the specific changes that occur during a chemical rxn. This stuff is massively useful in drug development! A. Terms 1. A single step in mechanism is called an elementary reaction or an elementary step. 2. The molecularity of a rxn refers to the # of separate molecules/atoms on reactant side of elementary rxn. a) unimolecular = one b) bimolecular = two c) termolecular = three (These are uncommon.) 21

22 B. Let s look at a mechanism for: NO 2 (g) + CO (g) ÿ NO (g) + CO 2 (g) 1. Experiments have suggested a two step process: 1 st step: NO 2 (g) + NO 2 (g) ÿ NO(g) + NO 3 (g) 2 nd step: NO 3 (g) + CO(g) ÿ NO 2 (g) + CO 2 (g) See Fig , p. 454 (Think about molecules colliding and/or breaking apart!) 22

23 The molecularity of the 1 st step is. The molecularity of the 2 nd step is. 2.While a number of the components in the balanced molecular equation are unstable, one component of elementary rxns. is very unstable. That is NO 3. A component like NO 3 that exists transiently (made & then used up) in the mechanism is a reaction intermediate. 3. Notice that the sum of the individual elementary steps must add up to the overall, balanced molecular equation: 23

24 1 st step: NO 2 (g) + NO 2 (g) ÿ NO(g) + NO 3 (g) 2 nd step: NO 3 (g) + CO(g) ÿ NO 2 (g) + CO 2 (g) NO 2(g) +NO 2(g) + NO 3(g) + CO (g) ÿ NO (g) + NO 3(g) + NO 2(g) + CO 2(g) Net: NO 2 (g) + CO(g) ÿno(g) + CO 2 (g) The terms that occur on both reactant and products sides of this summed equation are shown in bold. When these terms are dropped, you must regenerate the original, balanced overall rxn. for the mechanism to have a chance of being correct. Do Prob , p a) Equation for overall rxn? b) Molecularity of each step? 24

25 X. Rate Laws for Elementary Reactions (simple) A. 1st order example, decay of O 3 : O 3 ÿ O 2 + O If this is 1 st order in O 3, what is the rate law? B. Second order example, rxn. of CH 3 Br w/ OH! : See S N 2 Molecular Movies animation. Also on 4 th floor computers. CH 3 Br + OH! ÿ CH 3 OH + Br! If this is 1 st order in both CH 3 Br & OH! what is the rate law? 25

26 Important: When the overall rxn. occurs in a single step, then the rate law is determined by the molecular equation. Because this rxn. has a single step mechanism: CH 3 Br (aq) + OH! (aq) ÿ CH 3 OH (aq) + Br! (aq) rate =!Δ [CH 3 Br] / Δt = k [CH 3 Br] [OH! ] Do prob , p

27 XI. Rate Laws for Overall Reactions Back to the crossword puzzle analogy. The balanced molecular equation for a rxn. only gives information about stoichiometry, not necessarily about the mechanism. However, the rate law for the rate limiting step in a mechanism does define the rate law for the overall rxn. A. Introductory comments: 1. Find an every day analogy of a multistep process with a clear rate limiting step. The rate limiting step is the slowest step. (Shopping?) Ever worked on or seen an assembly line? When a rxn. occurs with more than one elementary step, usually one of the elementary steps is much slower than the others. This is the rate limiting step. 27

28 B. Multi step Reactions with an Initial Slow Step 1. Let s go back and look at the rxn. of NO 2 with CO: Balanced rxn.: NO 2 (g) + CO (g) ÿ NO (g) + CO 2 (g) Proposed mechanism: k 1 1 st step: NO 2 (g) + NO 2 (g) ÿ NO(g) + NO 3 (g) k 2 2 nd step: NO 3 (g) + CO(g) ÿ NO 2 (g) + CO 2 (g) The 1 st step is the slow step in this sequence. By slow, I mean slow relative to the 2 nd step. That is, k 1 << k 2. 28

29 Rate of the overall process can t be faster than rate of step 1. Can we write a rate law for the 1 st elementary rxn.? Rate =!Δ [NO 2 ] / Δt = k 1 [NO 2 ] 2 The experimentally determined rate law for this rxn. does turn out to be: Rate =!Δ [NO 2 ] / Δt = k [NO 2 ] 2 Therefore, our mechanism predicted the correct rate law. There is something of a mutually reinforcing relationship between the mechanism and the rate law: The rate law helps to confirm the mechanism and the mechanism gives us insight into the rate law. 29

30 What is the order of the NO 2 -CO rxn.: Overall? With respect to NO 2? With respect to CO? Does this seem logical? In other words: does it bother you that the rate law for the rxn.: NO 2(g) + CO (g) ÿ NO (g) + CO 2(g) has no [CO] term in it? Try Prob , p

31 C. Multi step Reactions with Initial Fast Step New! What if step 1 in the mechanism isn t the rate limiting step? 1. Example: 2 NO (g) + 2 H 2(g) ÿ N 2(g) + 2 H 2 O (g) 2. Proposed mechanism: k 1 Step 1: 2 NO W 2 N 2 O 2 fast, reversible k!1 k 2 Step 2: N 2 O 2 + H 2 ÿ N 2 O + H 2 O slow, rate limiting k 3 Step 3: N 2 O + H 2 ÿ N 2 + H 2 O fast Do these steps sum to give the overall balanced equation? 31

32 3. Based on our previous logic, the rate law should be: rate = k 2 [N 2 O 2 ] [H 2 ] This isn t practical or useful. a) Not practical because we can t measure [N 2 O 2 ] easily. b) Not useful because we can t control [N 2 O 2 ] easily. 4. Is there a way to cope? a) Because the 1 st rxn is fast, it is essentially at equilibrium. (By definition: at equilibrium, rate forward = rate reverse ) b) Therefore rate f = k 1 [NO] 2 and rate r = k!1 [N 2 O 2 ] c) Since rate f = rate r then k 1 [NO] 2 = k!1 [N 2 O 2 ] d) Rearrange k 1 [NO] 2 = k!1 [N 2 O 2 ] to solve for [N 2 O 2 ]: 32

33 k 1 [N 2 O 2 ] = )) [NO] 2 & substitute this back into: k!1 rate = k 2 [N 2 O 2 ] [H 2 ] to get: k 1 rate = )) k 2 [NO] 2 [H 2 ] k!1 Practical and useful! We usually combine all of the constant terms into a new constant k. 33

34 XII. Rxn Rates & Temp: Arrhenius Equation This equation will help us relate the rate constant, k, to other chemical ideas (collision theory, steric hindrance, etc.). It helps to unify our thoughts about kinetics. A. Reaction rates tend to increase with increasing temperature. Why? 1. Consider the rxn.: O(g) + HCl(g) ÿ OH(g) + Cl(g) In this rxn. the H!Cl bond must be broken. Think about overlap of electron clouds. ² O H!Cl ÿ OþHþCl ÿ O!HCl 34

35 2. Do electrons normally repel each other? (Y or N?) 3. The intermediates in a rxn. are usually very unstable & high in potential energy. (See Fig 12.15) Important terms: a) activation energy (E a ) b) transition state (synonym: activated complex) Try to relate the process of going from reactants to the transition state to some human activity. (Mountain climbing?) 35

36 B. Some of the thoughts behind this part of kinetics started with analysis of collision rates in the gas phase. 1. Air, at normal T & P: each gas molecule has about 10 9 collisions/sec. Imagine a reaction occurring with each collision. While this does seem consistent with the rates of chemical rxns. that cause explosions, it clearly is not consistent with the rates of the majority of chemical rxns. Therefore, for most rxns., only a small percentage of the collisions lead to a reaction occurring. These productive collisions occur with appropriate energy and orientation. 36

37 2. The fraction of collisions with enough energy (see Fig ) to reach E a is: f = e!ea/rt (Units for T??) At T= 298K, a rxn. with E a = 75 kj/mol has f = 7 x 10!14. With a collision rate of 10 9 /s, how long (ave)would you have to wait to get a collision with sufficient energy? What would f be for this reaction at 373 K? Can you see why T has such a large effect on rxn rate? This takes care of part of the energy consideration mentioned before. What about orientation? 37

38 C. Back to our previous example w/gas phase rxn.: O(g) + HCl(g) ÿ OH(g) + Cl(g) 1. What if O collides with the Cl side of the HCl molecule, instead of the H side? ² H!Cl O ÿ No H to O bonding can occur. 2. Define a steric factor, p: fraction of collisions w/ appropriate orientation for the rxn. to occur. If two atoms of a diatomic reactant are about the same size, p (Electrostatics?) 38

39 3. For the example above with HCl reacting w/ O, would p be < 0.5 or > 0.5? Review molecular modeling: 3 representations of HCl cylinder ball and stick All to same scale. Align atom centers. space filling (geometrically more accurate than the 2 above) 39

40 º» a) Recall the O (red) has to hit the H (white) b) Is the surface area of the H greater than the surface area of the Cl (green)? 40

41 D. We can now try to set up an equation predicting rxn. rates w/ the collision theory described here: 1. collision rate = Z [A] [B] (This was 10 9 above.) 2. rxn rate = p f collision rate = p f Z [A][B] Remind you of our general rate law? rate = k [A][B] Clearly, p f Z = k. Therefore: k = p Z e!ea'rt We now can relate structure, velocity, energy, etc. to understanding & predicting rxn rates. Thanks Arrhenius! p Z is sometimes called A, the frequency factor. 41

42 XIII. Using the Arrhenius equation A. Rearrange k = A e!ea/rt to: ln k =! E a /RT + lna Can you see y = mx + b? By measuring k a range of different T values we can get a good estimate of E a. (Do prob on your own.) B. Why would anyone want to do this? 42

43 Knowledge is power! 1. If you understand a rxn., you can modify it to change the rate, improve efficiency, and so on. 2. Enzyme inhibitors based on transition-state analogs. By Baldwin et al., 1995 Green & blue represent the A & B subunits of HIV protease. The red molecule (KNI-272) is a transition state analog that is an effective inhibitor of the enzyme. From PDBSum accession code 1hpx. 43

44 2-dimensional map of inhibitor-enzyme interactions: (From PDBSum code 1hpx) Covalent bonds of the inhibitor are shown in purple. Everything else is part of the enzyme. (Ex.: Val 82(A) is the valine residue at position 82 of the A subunit.) Green dotted lines are ionic interactions or Hydrogen Bonds. Red eyelashes represent Hydrophobic Interactions. 44

45 3-D view of inhibitor-enzyme interactions: (From PDBSum code 1hpx) The thicker tube structure represents the inhibitor (KNI-272). The thinner tubes represent those parts of the enzyme that interact directly with the inhibitor. Standard CPK color-coding for elements. grey: carbon blue: nitrogen red: oxygen yellow: sulfur What element appears to be strangely absent? 45

46 The active site of the enzyme is a different environment than bulk solvent. Hence, different energy pathway. 3. Catalytic antibodies = made-to-order catalysts. XIV. Catalysis (PKU & PAH? HIV protease?) A. Catalysts work by decreasing activation energy. (See Fig ) B. Enzymes (and catalysts in general) do this by creating a different environment (different from what?), which allows for a different (lower energy) reaction pathway. 46

47 For some enzymes, the environment and pathway is very well understood. See HIV protease. C. Do catalysts speed up the rate of the reverse rxn. too? Do Key Concept Prob , p rxn: 2 A + C 2 ÿ 2 AC A = red B = blue C = yellow Relatives rates: a) What is the order of the rxn in A, B, C? b) Write the rate law. c) Write a mechanism that agrees with the rate law. d) Identify all catalysts & intermediates in mechanism. 47

48 XIII. Homogeneous & Heterogeneous Catalysis (Read on your own.) The Interlude on pp gives you a nice, brief introduction into an important part of biochemistry. 48