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Unformatted text preview: Chemical Kinetics Chemical Rates, Mechanisms, and Rates, Catalysis Catalysis Kinetics Chemistry is most useful when it allows us Chemistry to control things. to Thus we are interested in 2 factors... HOW FAST. (I feel a need for speed). HOW FAR. (Really, where is the equilibrium?) Chemical Kinetics 2 Kinetics We approach kinetics in 2 major branches; 1) The macroscopic scale where we 1) observe behavior. observe 2) The microscopic scale where we 2) attempt to explain behavior. attempt Chemical Kinetics 3 Some Experiments to Think About Lycopodium powder, in bulk vs powdered NH3 + HCl H2 + energy source 2 H2 + O2 2 H2O A Simple Clock 2 N2O5 4 NO2 + O2 What would plots of concentration vs time look like What for these examples? for Chemical Kinetics 4 The rate of a chemical reaction is affected by: is a. b. c. d. the concentration of the reactants. the temperature of the reaction. the presence of a catalyst. all of the above. Decomposition of Dinitrogen Pentoxide Chemical Kinetics 6 The Plot for the Forward Reaction Chemical Kinetics 7 The Plot for the Reverse The Reaction Reaction Chemical Kinetics 8 Consider These… How do we normally describe the motion of a How vehicle? vehicle? How could you describe the factor that limits How how fast your bath tub fills? These are common examples of the concept of rate rate The reaction rate would be The the change in … change Chemical Kinetics 9 Expressing Rates -x c1-x Chemical Kinetics -x c2-x +2x c3+2x 10 10 N2 + 3 H2 2 NH3 The OVERALL rate is expressed as The x | /∆ t|mol/l.s where x = ∆ [a species] mol/l.s What does that mean in terms of: What N2, H2, and NH3? and Chemical Kinetics 11 11 Example CK.1 Give the relative rates for Give disappearance of reactants and formation of products for the production of hexamethyleneproduction tetramine by the following reaction: 6CH2O(aq) + 4NH3(aq) (CH2)6N4(aq) + 6H2O (aq) Chemical Kinetics 12 12 N2 + 3 H2 2 NH3 If the rate of formation of NH3 is 0.024 M/s, what are the rates of disappearance of H2 & N2? are Chemical Kinetics 13 13 4 NH3 + 3 O2 2 N2 + 6 H2O NH If the rate of formation of N2 is 0.34 M/s, what are the rates of disappearance of NH3 & O2 disappearance and the rate of appearance of H2O? Chemical Kinetics 14 14 As a chemical reaction proceeds, the rate of the reaction tends to: reaction a. b. c. d. increase. decrease. remain constant. oscillate. How Do You Determine a Rate? 1. Remember that the Remember data that you will use to determine a rate will look something like one of these curves. curves. 2. Further, you will only Further, use the data from the _______ region of the plot. the 3. What possibilities are What there to get some sort of x/∆ t values? sChemical Kinetics ort 16 16 Average Rates A+BC+D Focus on the kinetic region Take slope of line from Take start to beginning of equilibrium equilibrium Or 2 points in between Note slope gets smaller, Note and approaches value at one point in time. one Chemical Kinetics 17 17 Instantaneous Rates Keep taking shorter Keep time intervals, time Eventually you Eventually reach the tangent. reach This is an This INSTANTANEOUS rate. rate. Example 1 Chemical Kinetics Example 2 18 18 Initial Rate Note the Note instantaneous rate as close to time 0 as possible. possible. This is the INITIAL This rate, and is the most commonly used value. used Chemical Kinetics 19 19 Reconsider Concentration • The clock reaction shows plainly that The concentration changes rates. concentration • The difficult question is always – HOW? • Let us reason some together... What kind of mathematical relationship gives rise to a curve? Chemical Kinetics 20 20 Let’s Get Mathematical In simplest terms Rate α[A] or Rate α [B] … Rate [A] But, with the curve to consider, we include Rate α[A]m , Rate α [B]n … Rate α And, of course, we can do the same for each And, species that alters the rate. species Combining, we get Rate α… α Combining, Chemical Kinetics 21 21 Proportionality? How does a mathematician make proportionality How easier to deal with? easier Add a proportionality constant! Our equation becomes… Rate = k[A]m[B]n… This is known as the RATE LAW or the RATE EQUATION Chemical Kinetics 22 22 Rate Law (Rate Equation) An expression of the concentration An dependence of the reaction rate. dependence For simple reactions it is the product of a For simple temperature dependent rate constant, k, rate and the reactant concentrations each raised to an exponent – the reactant order. order Reactant orders MUST be determined Reactant experimentally. experimentally. Chemical Kinetics 23 23 Reactant Orders The reactant orders are often, but NOT The ALWAYS, equal to the coefficients in the balanced equation. balanced H + I 2 HI Rate = k[H2][I2] 2 2 2 HI H 2 + I 2 HI 2 N2O5 4 NO2 + O2 Rate = k[N2O5] Chemical Kinetics Rate = k[HI]2 24 24 Reactant Orders Essentially, data are gathered to set up 2 equations for each reactant. (i.e., at least 3 experiments) Then we use the rate law equation at each condition. Chemical Kinetics 25 25 In Abstract Rate1 = k[A]1m and Rate2 = k[A]2m Since k is the same in each Rate1/[A]1m = Rate2 / [A]2m = k Rate1/ Rate2 = [A]1m / [A]2m Or Rate1/ Rate2 = ([A]1 / [A]2)m Chemical Kinetics 26 26 Rate Constant The rate constant, k, is found experimentally. The The rate constant is affected by temperature. The rate constant is usually different for the forward and reverse reactions. Chemical Kinetics 27 27 Finding Orders Here’’s a set from an other text book s Chemical Kinetics 28 28 A+BC+D Rate = k[A][B] Rate The overall order of this reaction The is: is: a. b. c. d. first. second. third. fourth. If tripling the concentration of reactant A multiplies the initial rate of the reaction by nine, the reaction is _______ order in A. reaction a. b. c. d. d. zeroth first second second third Finding the Rate Constant This is an easy task now we have This the orders. the Pick any set of conditions in the data table and solve the RATE LAW for k. k = rate /([A]m[B]n[C]p. (msd) Chemical Kinetics 31 31 A Simple Case First The rate of decomposition of 0.10M N2O5 at 298K is 0.22 M/min. What is the first order rate constant at 298 K? order 2 N2O5 4 NO2 + O2 1 -(of )(∆ [N2O5]/∆ t) k[N2O5] Rate = (rate /2 decomposition) = (first order) k= 1 -(1 )(∆ [N x / ∆ (0.22/2M/min2O51]/2)t) x 1/[N2OM /0.10 5] Chemical Kinetics More = 1.1 min-1 32 32 Continuing… What is the rate of formation of NO2 at 298 K when [N2O5] = 0.042 M? at 2 N2O5 4 NO2 + O2 Rate = 1.1kmin-1 x [N2O5]M 0.042 = 0.046 M/min ∆ [NO2]/∆ t = 4 x 0.046 M/min = n reaction rate Chemical Kinetics 0.18 M/min 33 33 Concentration vs Time Borrowing from calculus, we can do some Borrowing nifty things. nifty Consider a 1st order reaction, for example. Rate = -∆ [R]/∆ t. (the expression) But rate = k[R]. Hence -∆ [R]t/∆ t = k[R]0. Gather terms to get ∆ [R]t/[R]0 = -k∆ t Chemical Kinetics 34 34 This is a USEFUL EQUATION Note that ln([R]t/[R]o) = ln[R]t – ln[R]o The integrated equation is for a straight line. Specifically: ln[R]t = ln[R]0 - kt A plot of ln[R] vs time gives a line with a slope = -k and proves the reaction 1st order. A useful graphical method thus derived. Chemical Kinetics 35 35 Other orders Without any math… Zero-order reactions -> [R]o – [R]t = kt 2nd order -> (1/[R]t) – (1/[R]o) = kt We’ll only use these occasionally, but again they are straight line plots which give order and k Chemical Kinetics 36 36 Back to 1 Order st Let’’s look at the integrated rate law at a s special time. special Every reaction must reach a point where the [R]t is exactly half the original ([R]0). If this becomes a new “original [R]0”, a new half way point will be reached eventually. Every one of these half way points takes the same time. Chemical Kinetics 37 37 (Continuing) This time is called the half-life for the This reaction, represented as t½ reaction, In a first order reaction this happens when ([R]t/[R]o) = 1/2 So ln(1/2) = -kt½ = ln1 – ln2 Thus ln2 = kt½ = 0.693 Time-related problems can then be dealt with in numbers of half-lives. Chemical Kinetics 38 38 The time needed for half of the reactant to be consumed is called the __________ of the reaction. reaction. a. b. c. d. midpoint equivalence point half-rate half-life The half-life of a first-order reaction is equal to _________, reaction where k is the rate constant. where a. 0.693/k 0.693/ b. 0.693k 0.693 c. k/2 d. 2k Why Do Reactions Happen? Let’s review a couple of things briefly. 1. bonds are dynamic things. 2. molecules are always in motion. Chemical Kinetics 41 41 Why Do Reactions Happen? (2) Simple Collision Collision Effectiveness Bimolecular Collision & Ea Introduction to Mechanism Chemical Kinetics 42 42 Conditions for Reaction? Collisions MUST occur. 1. Collisions MUST have sufficient energy. 2. Collisions MUST be in correct orientation. 3. Collisions won’t necessarily succeed. Chemical Kinetics 43 43 A Microscopic View of Microscopic Reactions Reactions Activation energy barriers Nature of the reactants Concentrations Concentrations Molecular orientations Temperature and the Arrhenius equation Effect of catalysts Catalysts in industry and the environment Chemical Kinetics 44 44 Let’s Just Play for a Minute ORA Chemical Kinetics 45 45 Factors Influencing Collisions We’ve seen concentration affects rate… But why? What effect would temperature have? Why? Remember the Boltzmann distribution? Let’s link 2 plots! Chemical Kinetics 46 46 Energy Relationships Chemical Kinetics 47 47 The minimum energy a collision between molecules must have to produce the products is called the __________ energy. the a. b. c. d. initiation internal external activation Arrhenius We met Arrhenius in 105 in connection We with acids and bases. with But he also got into the act in kinetics. His most interesting addition was an equation which summarizes the factors we’ve been talking about. It is k = Ae-Ea/RT. Chemical Kinetics 49 49 k = Ae -Ea/RT A iis called a pre-exponential factor and is s made up of 2 parts; made One part to account for the number of collisions The second to account for the fraction of collisions which are effective The exponential term e-Ea/RT is always less than one and is thought of as the fraction with sufficient energy to react. Chemical Kinetics 50 50 Utility of Arrhenius’ Equation Let’s remove the exponential term. Take the natural log of the equation. Yields ln k = ln A – Ea/RT. Find k at several temperatures; A plot of ln k vs.1/T should be a straight line with intercept ln A and slope related to Ea. Chemical Kinetics 51 51 Reaction Mechanisms Introduction Introduction Molecularity of elementary steps Molecularity Rate equations for elementary steps Molecularity and reaction order Mechanism and rate equations Chemical Kinetics 52 52 Reaction Mechanism How does the reaction actually occur How on the atomic level? on usually a series of steps called usually "elementary reactions“ "elementary Stoichiometry of these elementary Stoichiometry reactions IS related to the powers in the IS rate law. rate Chemical Kinetics 53 53 Let’s Just Play for a Minute ORA Chemical Kinetics 54 54 Elementary Reaction Bimolecular Collision of 2 atoms or molecules Termolecular Collision of 3 atoms or molecules Unimolecular Reaction of 1 atom or molecule Reaction mechanism: Sequence of Elementary Reaction Steps Chemical Kinetics Each step has its own rate 55 55 Rate and Mechanism one approach: assume one elementary reaction much assume slower than the rest slower this is the "bottleneck" that determines the this overall rate overall Called Rate Limiting Reaction Or Rate Determining Step Chemical Kinetics 56 56 Simplest Case 2 NO2 + F2 2 NO2F NO2 + F2 = NO2F + F (slow) NO2 + F = NO2F (fast) Since first step is slow: -(1/1) ∆ [F2]/∆ T = k[NO2][F2] -(1/1) Chemical Kinetics 57 57 iif any other step is rate f determining, determining, more complicated! Chemical Kinetics 58 58 e.g., H2(g) + Br2(g) = 2HBr(g) the rate law is very complex The elementary reactions are Br2 = 2 Br fast Br + H2 = HBr + H slow H + Br2 = HBr + Br fast H + HBr = H2 + Br HBr fast Chemical Kinetics 59 59 2NO(g) +2H2(g) = N2(g) + 2H2O(g) Sequence of Elementary Reaction Steps NO + NO = N2O2 N 2O 2 + H 2 = N 2O + H 2O N2O + H2 = N2 + H2O each has its own equilibrium each has forward rate constant, eg k1 each has reverse rate constant, eg k-1 Chemical Kinetics 60 60 Return to Catalysts Now we know a lot about reactions, let’’s s reconsider catalysts. reconsider How does it work? provides a new "pathway" or mechanism for the reaction. with lower activation energy. Chemical Kinetics 61 61 Catalysis Movie Chemical Kinetics 62 62 Catalyzed Pathway I Chemical Kinetics 63 63 Catalyzed Pathway II Chemical Kinetics 64 64 Catalyzed Pathway III End of Show Chemical Kinetics 65 65 A catalyst increases the rate of a reaction by: reaction a. causing the molecules to move more causing rapidly. rapidly. b. causing the molecules to collide more causing frequently. frequently. c. lowering the energy of activation. d. all of the above. Biological Catalysts Biological Chemical Kinetics 67 Model Enzyme Model Chemical Kinetics 68 Chemical Kinetics 69 69 THE STEPS of IRON TRANSPORTATION TRANSPORTATION the first step that should be noted is that the amino acids in the transferrin are structured already ready to bind with iron transferrin once iron and transferrin join they flow through the blood stream once in search of a transferrin receptor (TFR) on the surface of a cell in transferrin and the transferrin receptor bind and create a pocket transferrin that holds the iron that this is then drawn into the cell the cell acidifies the inside of the pocket and transferrin lets the the iron go iron the receptor and the already used transferrin are let back out of the the cell cell with the blood stream's pH the TFR releases the transferrin and with they start all over they Chemical Kinetics 70 Transferrin and Receptor Transferrin Chemical Kinetics 71 An Indicator of Sensitivity An Chemical Kinetics 72 Sample Rate Determination Tangent 2 = -0.0006 0.132659 0.108703 0.084747 0.060792 0.036836 [ N 2O 5] / M Decomposition of N2O5 1.000 0.900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 0.000 600 560 520 480 440 400 360 320 280 240 200 160 0.3 0.086 0 -0.00296 time (s) -0.00296 0.799098 0 360 500 Tangent 1 = 120 0.8 0.442 0 0.68061 0.56211 0.44362 0.32513 0.20664 80 0 120 270 2N2O5 4NO2 + O2 40 Concentration 1.000 0.993208 0.762 1 0.580 0.442 0.336 0.256 0.195 0.148 0.113 0.086 0.066 0.065 0.050 0.038 0.029 0.022 0.017 0.013 0.010 0.007 0.004 0.003 0.002 0.002 0.001 0.001 0 Time 0 40 80 120 160 200 240 280 320 360 400 440 480 520 560 600 640 680 720 800 840 880 920 960 1000 -0.0006 0.300351 Chemical Kinetics Exit 73 73 Instantaneous Rate X Chemical Kinetics 74 74 Instantaneous Rates X Chemical Kinetics 75 75 Instantaneous Rates X Chemical Kinetics 76 76 More Examples X Chemical Kinetics 77 77 More Rates X Chemical Kinetics 78 78 More Rates X Chemical Kinetics 79 79 Half-lives Chemical Kinetics 80 80 Half-Lives (continued) At time t Multiplier OR [N2O5]t [N2O5]0/1 X (1/2 ) [N2O5]t [N2O5]0 /2 /2 X (1/21) [N2O5]t [N2O5]0 /4 /4 X (1/22) [N2O5]t [N2O5]0 /8 /8 X (1/23) Amt. Left = Chemical Kinetics 0 Init. Amt. X (1/2)0 X (1/2)1 Calc. In X (0.5)0 X (0.5)1 X (0.5) X (1/2)2 X (0.5)3 X (1/2)3 X (0.5)#half lives 2 81 81 Example Iodine-131 is a radionuclide that decays in Iodine-131 a first-order process with a half life of 8.0 days. It is a β-emitter, which makes it safe for -emitter, medical use to develop images of the thyroid. thyroid. We can use our little equation to check the We effects of time on doses. effects Chemical Kinetics 82 82 Example a We inject a patient with 1.0000 mg of I131. How much will be left after 32 days? 32 days = ? Half lives? = 4 half lives half Amount left = 1.0000 mg x (0.5)4 = 0.0625 mg 0.0625 Chemical Kinetics 83 83 Example b We use a simple in-house generator to make 25.0 We mg of I-131 to meet expected diagnostic needs. How much would be left after 40. Days? How 40 days = ___ half lives? =5 Amount left = 25.0 mg x (0.5)5 = 25.0 mg x 0.0313 = 0.78 mg 25.0 Chemical Kinetics 84 84 Example c If we find a sample of 23.5 mg of I-131, how If much would we expect to have left after 5.0 days? days? 5.0 days = (5.0/8.0) half lives 5.0 = 0.625 half lives 0.625 Amount left = 23.5 mg x (0.5)0.625 = 23.5 mg x 0.648 23.5 = 15.2 mg 15.2 Chemical Kinetics 85 85 Ea calculations T(K) 338 1.50 x 10-3 318 Chemical Kinetics 4.87 x 10-3 328 Data for the temperature Data dependence of the reaction N2O5(g) 2NO2(g) + ½O2(g). (g) (g) ½O k(s-1) 4.98 x 10-4 308 1.35 x 10-4 298 3.46 x 10-5 273 7.87 x 10-7 86 86 One Important Catalysis Chemical Kinetics 87 87 Arrhenius Equation Activity Chemical Kinetics 88 88 Surface Reactions Chemical Kinetics 89 89 Equilibrium Defined Chemical Kinetics 90 90 Half Life First Order Process Chemical Kinetics 91 91 Second Order Chemical Kinetics 92 92 Reaction Rates Chemical Kinetics 93 93 Transition State Theory Chemical Kinetics 94 94 Reaction Rates Chemical Kinetics 95 95 ...
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This document was uploaded on 11/04/2011 for the course PWS 150 at BYU.

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