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Chemical kinetics

The document discusses chemical kinetics and reaction rates. It begins by defining chemical kinetics as the study of the rate at which a chemical reaction occurs. It notes that kinetics provides information about both reaction speed and reaction mechanism. The document then discusses several factors that affect reaction rates, including physical state of reactants, concentration of reactants, temperature, and presence of a catalyst. It explains that reaction rates can be determined by monitoring changes in concentration over time. The document provides examples of calculating average and instantaneous reaction rates. It also discusses how reaction rates relate to stoichiometry and explains rate laws.

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1 Chemical Kinetics Chapter 14 Chemical Kinetics Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
2 Chemical Kinetics Kinetics • Chemical Kinetics is the study of the rate at which a chemical process occurs. • Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
3 Chemical Kinetics Factors That Affect Reaction Rates • Physical State of the Reactants In order to react, molecules must come in contact with each other. If the reaction is happening between a solid and a liquid it will react only on the surface. The more homogeneous the mixture of reactants, the faster the molecules can react. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
4 Chemical Kinetics Factors That Affect Reaction Rates • Concentration of Reactants As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
5 Chemical Kinetics Factors That Affect Reaction Rates • Temperature At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
6 Chemical Kinetics Factors That Affect Reaction Rates • Presence of a Catalyst Catalysts speed up reactions by changing the mechanism of the reaction. Catalysts are not consumed during the course of the reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
7 Chemical Kinetics Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
8 Chemical Kinetics Reaction Rates In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
9 Chemical Kinetics The average rate of the reaction over each interval is the change in concentration divided by the change in time: • Average rate = ∆[C4H9Cl] ∆t [ ] [ ] [ ]t2 t1 2 1 X - X ∆ X Average rate = = t -t ∆t PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
10 Chemical Kinetics Reaction Rates C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) [ ] [ ] -4 0.0905 - 0.1000 0.0095 Average rate = = - 1.9 x 10 50-0 50 = − PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
11 Chemical Kinetics Reaction Rates • Note that the average rate decreases as the reaction proceeds. • This is because as the reaction goes forward, there are fewer collisions between reactant molecules. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
12 Chemical Kinetics • calculate the average rate of appearance of B over the time interval from 0 to 40 s. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
13 Chemical Kinetics Reaction Rates • A plot of concentration vs. time for this reaction yields a curve like this. • The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
14 Chemical Kinetics Reaction Rates • All reactions slow down over time. • Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
15 Chemical Kinetics Average vs. instantaneous rxn rate • Average rxn rate • Instantaneous rxn rate (tangent to curve) Usually in chemistry, we are interested in the instantaneous rxn rate at the very beginning (at t = 0) INITIAL REACTION RATE ∆t ∆ rate ]A[ elapsedtimetotal endandrxnofbeginningbetween ionconcentratchemicalofchange )(reactionofrate =       = ∆t ∆ rate ]A[ elapsedshort timethe timeofperiodshortaover ionconcentratchemicalofchange )(reactionofrate =       = PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
16 Chemical Kinetics Average vs. instantaneous rxn rateNaHCO3 + citric acid --> sodium citrate + H2O + CO2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.5 1 1.5 2 time (minutes) concentration(tablet/100mL) initial [A] = 1 tablet initial [A] = 2 tablets Average rate Instantaneous rate at t = 0 (initial rate) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
17 Chemical Kinetics Comparing concentrations of chemicals in a reaction A --> B 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.5 1 1.5 2 time (hours) concentration(mol/L) [A] [B] A --> 2 C 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.5 1 1.5 2 time (hours) concentration(mol/L) [A] [C] PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
18 Chemical Kinetics What we’re after in kinetics • We want to be able to understand and predict how much reactants remain (or products get formed) in a certain amount of time • Change with time is rate • Rate means calculus must be used, so we need to be careful which rate we are talking about • We will use instantaneous rate at the start of a reaction (initial rate) • Several factors affect the rate of a reaction Initial concentration Temperature Amount of reactant exposed (surface area) Presence of a catalyst • A theory (equations) that predicts kinetics must explain these PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
19 Chemical Kinetics Calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 (the initial rate). PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
20 Chemical Kinetics Determine the instantaneous rate of disappearance of C4H9Cl at t = 300 s. Answer: 1.1 × 10 –4 M/s PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
21 Chemical Kinetics Reaction Rates and Stoichiometry • In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. • Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) Rate = -∆[C4H9Cl] ∆t = ∆[C4H9OH] ∆t PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
22 Chemical Kinetics Reaction Rates and Stoichiometry • What if the ratio is not 1:1? 2 HI(g) → H2(g) + I2(g) in such a case , [ ] [ ] [ ]2 2 HI H I1 Rate = - = 2 t t t ∆ ∆ ∆ = ∆ ∆ ∆ PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
23 Chemical Kinetics Reaction Rates and Stoichiometry • To generalize, then, for the reaction aA + bB cC + dD Rate = − 1 a ∆[A] ∆t = − 1 b ∆[B] ∆t = 1 c ∆[C] ∆t 1 d ∆[D] ∆t = PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
24 Chemical Kinetics Key points about kinetics so far • Study the vocabulary • Reaction rate has weird units • Be clear whether you are talking about average reaction rate or instantaneous reaction rate • The equations which we will study all talk about instantaneous reaction rate, most easily measured at the beginning of a reaction (initial rate) • Relating rates at which products appear (+) and reactants disappear (-) has to do with inverse stoichiometric coefficients • Reaction rate depends on several factors: initial concentration, temperature, surface area of reactants exposed, presence of a catalyst PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
25 Chemical Kinetics (a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction (b) If the rate at which O2 appears, Δ[O2]/Δ t, is 6.0 × 10–5 M/s at a particular instant, at what rate is O3 disappearing at this same time, –Δ[O3] /Δ t? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
26 Chemical Kinetics Answers: (a) 8.4 × 10 –7 M/s, (b) 2.1 × 10 –7 M/s SAMPLE EXERCISE 14.3 continued The decomposition of N2O5 proceeds according to the following equation: If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 × 10–7 M/s, what is the rate of appearance of (a) NO2, (b) O2? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
27 Chemical Kinetics Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
28 Chemical Kinetics Concentration and Rate Comparing Experiments 1 and 2, when [NH4 +] doubles, the initial rate doubles. NH4 +(aq) + NO2 −(aq) N2(g) + 2 H2O(l) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
29 Chemical Kinetics Concentration and Rate Likewise, comparing Experiments 5 and 6, when [NO2 −] doubles, the initial rate doubles. NH4 +(aq) + NO2 −(aq) N2(g) + 2 H2O(l) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
30 Chemical Kinetics Concentration and Rate • This means Rate ∝ [NH4 +] Rate ∝ [NO2 −] Rate ∝ [NH+] [NO2 −] or Rate = k [NH4 +] [NO2 −] • This equation is called the rate law, and k is the rate constant. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
31 Chemical Kinetics • If you want your concepts to be clearer do the visualizing concepts part of the problems at the back of your chapter. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
32 Chemical Kinetics Rate Law • An equation that shows how the rate depends on the concentration of the reactants is called the rate law • The exponents tell the order of the reaction with respect to each reactant. • This reaction is First-order in [NH4 +] First-order in [NO2 −] PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
33 Chemical Kinetics Rate Laws • The overall reaction order can be found by adding the exponents on the reactants in the rate law. • This reaction is second-order overall. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
34 Chemical Kinetics • For a general reaction: aA + bB cC + dD Rate law has the form Rate =k [A]m[B]n where k is the rate constant and is dependant on temperature. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
35 Chemical Kinetics If we know the rate law and the rate for a set of reaction conditions we can calculate the value of k. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
36 Chemical Kinetics To calculate the rate constant: Rate law Rate = k [A]m [B]n Rate = k[NH4] [NO2] For experiment 1: 5.4X10-7 M/s = k (0.0100M)(0.200M) k = 5.4X10-7 M/s k (0.0100M)(0.200M) = 2.7X10 -4 /Ms PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
37 Chemical Kinetics Exponents in the rate law The exponent in the rate law indicates how the rate is affected by the concentration of each reactant. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
38 Chemical Kinetics SAMPLE EXERCISE 14.4 Relating a Rate Law to the Effect of Concentration on Rate Consider a reaction for which rate = k[A][B]2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
39 Chemical Kinetics SAMPLE EXERCISE 14.4 Relating a Rate Law to the Effect of Concentration on Rate Consider a reaction for which rate = k[A][B]2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. Each box contains 10 spheres. The rate law indicates that in this case [B] has a greater influence on rate than [A] because B has a higher reaction order. Hence, the mixture with the highest concentration of B (most purple spheres) should react fastest. This analysis confirms the order 2 < 1 < 3. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
40 Chemical Kinetics Assuming that rate = k[A][B], rank the mixtures represented in this Sample Exercise in order of increasing rate. Answer: 2 = 3 < 1 PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
41 Chemical Kinetics Unit for k depends on the overall order of the reaction. For example for a second order reaction 2 Unit of Rate k = (Units of concentration) 2 2 M/s M = = = /Ms M M s PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
42 Chemical Kinetics (a)What are the overall reaction orders for the rate laws described in Equations 14.9 and 14.10? (b) What are the units of the rate constant for the rate law for Equation 14.9? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
43 Chemical Kinetics (a)What are the overall reaction orders for the rate laws described in Equations 14.9 and 14.10? (b) What are the units of the rate constant for the rate law for Equation 14.9? Answer: [ ]2 5 2 2 2 4 1/2 3 2 4 3 2 2N O (g) 4NO (g) + O (g) Rate = k N O CHCl (g) + Cl (g) CCl (g) + HCl(g) Rate = k[CHCl ] [Cl ] → → Answer: The rate of the reaction in Equation 14.9 is first order in N2O5 and first order overall. The reaction in Equation 14.10 is first order in CHCl3 and one-half order in Cl2. The overall reaction order is three halves. So Notice that the units of the rate constant change as the overall order of the reaction changes. [ ]2 4Rate = k N O PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
44 Chemical Kinetics 1.The reaction is first order in a reactant If the concentration of a particular component is doubled the rate doubles and triples when the concentration triples. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
45 Chemical Kinetics 2. If the reaction is second order with respect to a component: If the rate is second order with respect to a particular component, doubling its concentration increases the rate by a factor of 22 = 4 Tripling its concentration increases the rate by a factor of 32= 9 and so forth PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
46 Chemical Kinetics • 3. If a reaction is zero order with respect to a component: If a reaction is zero order in a particular reactant, changing its concentration will have no effect on the rate as long as long as the reactant is present. It is because any concentration raised to zero is one PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
47 Chemical Kinetics Order of reaction rate • Rate of reaction can depend on how much reactant is present in more than one way: Zero order: rate ∝ [A]0 First order: rate ∝ [A] Second order: rate ∝ [A]2 PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
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49 Chemical Kinetics [ ]2 5 2 2 2 4 1/2 3 2 4 3 2 2 2 2 2 2N O (g) 4NO (g) + O (g) Rate = k N O CHCl (g) + Cl (g) CCl (g) + HCl(g) Rate = k[CHCl ] [Cl ] H (g) + I (g) 2HI(g) Rate = k[H ][I ] → → → PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
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51 Chemical Kinetics • Although the exponents in the rate law are often the same as the coefficients in the balanced reaction, this is not necessarily the case. • The values of these exponents must be determined experimentally. • Most reaction orders are 0, 1 or 2.However we may encounter reaction orders in fraction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
52 Chemical Kinetics • The rate of the reaction depends on the concentration but the rate constant , k, does not. • k is dependent only on temperature. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
53 Chemical Kinetics 4 2 4 4 CH + 4 Cl CCl + 4HCl 1 [CH4] 1 [ ] Rate = - 1 4 1 Rate = - rate of disappearence of CH = rate of appearence of HCl 4 Rate x 4 = Rate of appearence of HCl HCl t t → ∆ ∆ = ∆ ∆ PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
54 Chemical Kinetics Change in Concentration with Time From the rate law we can calculate the rate of the reaction if we know the the rate constant and the reactant concentration Rate = k [A] Rate laws can also be converted into equations that tell us what the concentration of reactants and products are at any time during the reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
55 Chemical Kinetics Change of Concentration with Time for the First Order Reactions The first order reaction is one whose rate depends on the concentration of a single reactant raised to the first power reactant A product Rate = -Δ[A] = k [A] Δt This form of rate law, which expresses how rate depends upon concentration is called differential rate law. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
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58 Chemical Kinetics Integrated Rate Laws Using calculus to integrate the rate law for a first-order process we get: ln [A]t [A]0 = −kt Where [A]0 is the initial concentration of A. [A]t is the concentration of A at some time, t, during the course of the reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
59 Chemical Kinetics Integrated Rate Laws Manipulating this equation produces… ln [A]t [A]0 = −kt ln [A]t − ln [A]0 = − kt ln [A]t = − kt + ln [A]0 Where m is the slope of the line and b is th ey intersept of the line. Which is a general equation of a straight line. …which is in the form y = mx + b PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
60 Chemical Kinetics Integrated Rate Laws ln [A]t = − kt + ln [A]0 Where m is the slope of the line and b is the y intersept of the line. Which is a general equation of a straight line. …which is in the form y = mx + b PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
61 Chemical Kinetics First-Order Processes Therefore, if a reaction is first-order, a plot of ln [A]t vs. t will yield a straight line, and the slope of the line will be -k. ln [A]t = -kt + ln [A]0 PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
62 Chemical Kinetics First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH3NC CH3CN PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
63 Chemical Kinetics First-Order Processes This data was collected for this reaction at 198.9°C. CH3NC CH3CN PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
64 Chemical Kinetics • The experiments show this reaction to be first order so we can write the rate equation as In[CH3NC]t= - kt + ln [CH3NC]o PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
65 Chemical Kinetics • For gas we can use pressure as a unit of concentration as from the ideal gas law equation the pressure is directly proportional to the number of moles per unit volume PV = nRT n P = RT V At constant temperature n P α V PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
66 Chemical Kinetics First-Order Processes • When ln P is plotted as a function of time, a straight line results. • Therefore, The process is first-order. k is the negative slope: 5.1 × 10-5 s−1. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
67 Chemical Kinetics • How to get the natural log value on your calculator LN will give the natural log value. Pressing LN will show In ( key in the value) and press enter PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
68 Chemical Kinetics The initial rate of a reaction was measured for several different starting concentrations of A and B, and the results are as follows: 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] = 0.050 M and [B] = 0.100 M. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
69 Chemical Kinetics The initial rate of a reaction was measured for several different starting concentrations of A and B, and the results are as follows: 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] = 0.050 M and [B] = 0.100 M a. (b) Using the rate law and the data from experiment 1, we have (c) Using the rate law from part (a) and the rate constant from part (b), we have PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
70 Chemical Kinetics PRACTICE EXERCISE The following data were measured for the reaction of nitric oxide with hydrogen: (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
71 Chemical Kinetics PRACTICE EXERCISE The following data were measured for the reaction of nitric oxide with hydrogen: Answers: (a) rate = k[NO]2[H2]; (b) k =1.2 M–2s–1; (c) rate = 4.5 × 10–4 M/s (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
72 Chemical Kinetics PRACTICE EXERCISE The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 × 10–4s–1: If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
73 Chemical Kinetics PRACTICE EXERCISE The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 × 10–4s–1: If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s? Answer: 51 torr PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
74 Chemical Kinetics Half-Life • Half-life is defined as the time required for one-half of a reactant to react. • Because [A] at t1/2 is one-half of the original [A], [A]t = 1/2 [A]0. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
75 Chemical Kinetics Half-Life For a first-order process, this becomes 0.5 [A]0 [A]0 ln = −kt1/2 ln 0.5 = −kt1/2 −0.693 = −kt1/2 = t1/2 0.693 kNOTE: For a first-order process, the half-life does not depend on [A]0. ln [A]t [A]0 = −kt PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
76 Chemical Kinetics • The half life of a first order reaction remains the same throughout the reaction. • The concentration of the reactant keeps becoming half of the previous value after every half life. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
77 Chemical Kinetics • In the first order reaction, the concentration of reactant decreases by ½ in each of a series of regularly spaced time intervals, namely t1/2 • The concept of a first order reaction is widely used in describing radioactive decay, a first order process. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
78 Chemical Kinetics SAMPLE EXERCISE 14.7 Using the Integrated First-Order Rate Law • The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1 at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 × 10–7 g/cm3. Assume that the average temperature of the lake is 12°C. • (a) What is the concentration of the insecticide on June 1 of the following year? • (b) How long will it take for the concentration of the insecticide to drop to 3.0 × 10–7 g/cm3? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
79 Chemical Kinetics SAMPLE EXERCISE 14.7 Using the Integrated First-Order Rate Law • The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1 at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 × 10–7 g/cm3. Assume that the average temperature of the lake is 12°C. • (a) What is the concentration of the insecticide on June 1 of the following year? • (b) How long will it take for the concentration of the insecticide to drop to 3.0 × 10–7 g/cm3? ln [A]t = -kt + ln [A]0 Solve: (a) Substituting the known quantities into Equation 14.13, we have Note that the concentration units for [A]t and [A]0 must be the same. (b) Again substituting into Equation 14.13, with [insecticide]t = 3.0 × 10–7 g/cm3, gives Solving for t gives PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
80 Chemical Kinetics Temperature and Rate • Generally, as temperature increases, so does the reaction rate. • This is because k is temperature dependent. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
81 Chemical Kinetics The Collision Model • In a chemical reaction, bonds are broken and new bonds are formed. • Molecules can only react if they collide with each other. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
82 Chemical Kinetics The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
83 Chemical Kinetics Activation Energy • In other words, there is a minimum amount of energy required for reaction: the activation energy, Ea. • Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
84 Chemical Kinetics H3C N H3C C N H3C C NC- • This is a reaction where methyl isonitrile is rearranging to acetonitrile • This reaction passes through an intermediate state in which the N C is sitting sideways. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
85 Chemical Kinetics Reaction Coordinate Diagrams It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
86 Chemical Kinetics Reaction Coordinate Diagrams • It shows the energy of the reactants and products (and, therefore, ∆E). • The high point on the diagram is the transition state. • The species present at the transition state is called the activated complex. • The energy gap between the reactants and the activated complex is the activation energy barrier. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
87 Chemical Kinetics • Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. • At any temperature there is a wide distribution of kinetic energies. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
88 Chemical Kinetics Maxwell–Boltzmann Distributions • As the temperature increases, the curve flattens and broadens. • Thus at higher temperatures, a larger population of molecules has higher energy. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
89 Chemical Kinetics Maxwell–Boltzmann Distributions • If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. • As a result, the reaction rate increases. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
90 Chemical Kinetics Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression where R is the gas constant (8.314J/mol-K) and T is the Kelvin temperature. Ea is the minimum energy needed for reaction. f = e−Ea/RT PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
91 Chemical Kinetics • Arrhenious noted that the increase in reaction rate is not linear with increase in temperature in most reactions. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
92 Chemical Kinetics • Arrhenius observed that most reaction rates depend on the following three factors A. Fraction of molecules possessing Ea or greater B. number of collisions per second C. fraction of molecules having the correct orientation PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
93 Chemical Kinetics Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k (the rate constant) and Ea: k = A e−Ea/RT where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
94 Chemical Kinetics Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes ln k = -Ea ( ) + ln A 1 RT y = mx + b Therefore, if k is determined experimentally at several temperatures, and lnk vs 1/T is plotted Ea can be calculated from the slope of the plot. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
95 Chemical Kinetics • If we know the constant k at two or more temperatures At temperatures T1 and T2 the rate constant is k1 and k , ln k1 = -Ea + lnA and ln k2 = -Ea + lnA RT1 RT2 lnk1-lnk2 = [ -Ea/RT1 +lnA] - [ -Ea/RT2+lnA] lnk1/k2 = Ea (1/T1-1/T2) R PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
96 Chemical Kinetics •lnk1=Ea(1/T2-1/T1) •R PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
97 Chemical Kinetics PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
98 Chemical Kinetics Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
99 Chemical Kinetics Reaction Mechanisms • Reactions may occur all at once or through several discrete steps. • Each of these processes is known as an elementary reaction or elementary process. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
100 Chemical Kinetics Rate laws for elementary reactions • The rate law of an elementary reaction is directly based on its molecularity or concentration. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
101 Chemical Kinetics Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
102 Chemical Kinetics • In a multistep reaction each step will have its own rate constant and activation energy. • The slowest step determines the overall rate. Example Relay race PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
103 Chemical Kinetics Multistep Mechanisms • In a multistep process, one of the steps will be slower than all others. • The overall reaction cannot occur faster than this slowest, rate-determining step. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
104 Chemical Kinetics • A balanced chemical equation often occurs in a series of elementary reactions • A balanced chemical reaction itself could have a multi step mechanism PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
105 Chemical Kinetics Slow Initial Step • The rate law for this reaction is found experimentally to be Rate = k [NO2]2 • CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. • This suggests the reaction occurs in two steps. NO2 (g) + CO (g) → NO (g) + CO2 (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
106 Chemical Kinetics Slow Initial Step • A proposed mechanism for this reaction is Step 1: NO2 + NO2 → NO3 + NO (slow) Step 2: NO3 + CO → NO2 + CO2 (fast) • The NO3 intermediate is consumed in the second step. • As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. • Since step 1 is the slow rate determining step the rate law is going to be Rate = [NO2] 2 NO2 (g) + CO (g) → NO (g) + CO2 (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
107 Chemical Kinetics Practice exercise Ozone reacts with NO2 to give dinitrogen pentoxide and oxygen: O3 (g) + 2NO2 (g) N2O5(g) + O2 (g) The reaction occurs in the following two steps: O3(g) + NO2 (g) NO3 (g) + O2(g) NO3(g) + NO2(g) N2O5 (g) The experimental rate law is Rate = [ O3][NO2] What can you sat about the relative rate of the two steps of the mechanism? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
108 Chemical Kinetics • How many elementary reactions are there in the mechanism? • How many intermediates? • Which step is rate limiting? • Which is the fastest step? • Exo or endothermic? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
109 Chemical Kinetics Problems • How many elementary reactions are there in the mechanism? • How many intermediates? • Which step is rate limiting? • Which is the fastest step? • Exo or endothermic? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
110 Chemical Kinetics Fast Initial Step • Experimentally the rate law for this reaction is found to be Rate = k [NO]2 [Br2] • Because ter molecular processes are rare, this rate law suggests a two-step mechanism. 2 NO (g) + Br2 (g) → 2 NOBr (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
111 Chemical Kinetics Fast Initial Step • A proposed mechanism is Step 2: NOBr2 + NO → 2 NOBr (slow) Step 1 includes the forward and reverse reactions. The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k2 [NOBr2] [NO] But how can we find [NOBr2]? Step 1: NO + Br2 NOBr2 (fast) 2 NO (g) + Br2 (g) → 2 NOBr (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
112 Chemical Kinetics Fast Initial Step • NOBr2 can react two ways: With NO to form NOBr By decomposition to reform NO and Br2 • The reactants and products of the first step are in equilibrium with each other. • Therefore, Ratef = Rater Step 1: NO + Br2 NOBr2 (fast) Step 2: NOBr2 + NO → 2 NOBr (slow) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
113 Chemical Kinetics Fast Initial Step • Because Ratef = Rater , k1 [NO] [Br2] = k−1 [NOBr2] Pay attention to the fact that the reverse k is k-1 • Solving for [NOBr2] gives us k1 k−1 [NO] [Br2] = [NOBr2] PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
114 Chemical Kinetics Fast Initial Step From before Rate = k2 [NOBr2] [NO] From the previous equation: Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives k2k1 k−1 Rate = [NO] [Br2] [NO] = k [NO]2 [Br2] k1 k−1 [NO] [Br2] = [NOBr2] PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
115 Chemical Kinetics • In general when ever a fast step precedes a slow step we can solve the problem of the concentration of the intermediate step by assuming that an equilibrium is established in the fast step. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
116 Chemical Kinetics • Write down the rate law for the overall reaction. SAMPLE EXERCISE 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the experimentally observed one: PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
117 Chemical Kinetics PRACTICE EXERCISE The first step of a mechanism involving the reaction of bromine is What is the expression relating the concentration of Br (g) to that of Br2(g)? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
118 Chemical Kinetics Catalysts Arrhenius equation • K = A e –E/RT A= frequency factor and E= Energy of activation • Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. • Catalysts change the mechanism by which the process occurs. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
119 Chemical Kinetics • Homogeneous catalysts 2H2O2(aq) 2H2O (l) + O2 (g) NaBr in an acidic solution acts as a catalyst for this reaction 2Br -(aq) + H2O2(aq) + 2 H+ (aq) Br2(aq) + 2H2O(l) Br2(aq) + H2O2(aq) 2Br –(aq) + 2H+(aq) + O2 (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
120 Chemical Kinetics PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
121 Chemical Kinetics PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
122 Chemical Kinetics Heterogeneous Catalyst One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. The reactants are adsorbed on the surface of the catalyst. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
123 Chemical Kinetics Enzymes • Enzymes are catalysts in biological systems. • The substrate fits into the active site of the enzyme much like a key fits into a lock. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com

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Chemical kinetics

  • 1.
    1 Chemical Kinetics Chapter 14 Chemical Kinetics Chemistry,The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 2.
    2 Chemical Kinetics Kinetics • Chemical Kineticsis the study of the rate at which a chemical process occurs. • Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 3.
    3 Chemical Kinetics Factors That AffectReaction Rates • Physical State of the Reactants In order to react, molecules must come in contact with each other. If the reaction is happening between a solid and a liquid it will react only on the surface. The more homogeneous the mixture of reactants, the faster the molecules can react. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 4.
    4 Chemical Kinetics Factors That AffectReaction Rates • Concentration of Reactants As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 5.
    5 Chemical Kinetics Factors That AffectReaction Rates • Temperature At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 6.
    6 Chemical Kinetics Factors That AffectReaction Rates • Presence of a Catalyst Catalysts speed up reactions by changing the mechanism of the reaction. Catalysts are not consumed during the course of the reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 7.
    7 Chemical Kinetics Reaction Rates Rates ofreactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 8.
    8 Chemical Kinetics Reaction Rates In thisreaction, the concentration of butyl chloride, C4H9Cl, was measured at various times. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 9.
    9 Chemical Kinetics The average rateof the reaction over each interval is the change in concentration divided by the change in time: • Average rate = ∆[C4H9Cl] ∆t [ ] [ ] [ ]t2 t1 2 1 X - X ∆ X Average rate = = t -t ∆t PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 10.
    10 Chemical Kinetics Reaction Rates C4H9Cl(aq) +H2O(l) → C4H9OH(aq) + HCl(aq) [ ] [ ] -4 0.0905 - 0.1000 0.0095 Average rate = = - 1.9 x 10 50-0 50 = − PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 11.
    11 Chemical Kinetics Reaction Rates • Notethat the average rate decreases as the reaction proceeds. • This is because as the reaction goes forward, there are fewer collisions between reactant molecules. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 12.
    12 Chemical Kinetics • calculate theaverage rate of appearance of B over the time interval from 0 to 40 s. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 13.
    13 Chemical Kinetics Reaction Rates • Aplot of concentration vs. time for this reaction yields a curve like this. • The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 14.
    14 Chemical Kinetics Reaction Rates • Allreactions slow down over time. • Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 15.
    15 Chemical Kinetics Average vs. instantaneousrxn rate • Average rxn rate • Instantaneous rxn rate (tangent to curve) Usually in chemistry, we are interested in the instantaneous rxn rate at the very beginning (at t = 0) INITIAL REACTION RATE ∆t ∆ rate ]A[ elapsedtimetotal endandrxnofbeginningbetween ionconcentratchemicalofchange )(reactionofrate =       = ∆t ∆ rate ]A[ elapsedshort timethe timeofperiodshortaover ionconcentratchemicalofchange )(reactionofrate =       = PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 16.
    16 Chemical Kinetics Average vs. instantaneousrxn rateNaHCO3 + citric acid --> sodium citrate + H2O + CO2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.5 1 1.5 2 time (minutes) concentration(tablet/100mL) initial [A] = 1 tablet initial [A] = 2 tablets Average rate Instantaneous rate at t = 0 (initial rate) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 17.
    17 Chemical Kinetics Comparing concentrations of chemicalsin a reaction A --> B 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.5 1 1.5 2 time (hours) concentration(mol/L) [A] [B] A --> 2 C 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.5 1 1.5 2 time (hours) concentration(mol/L) [A] [C] PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 18.
    18 Chemical Kinetics What we’re afterin kinetics • We want to be able to understand and predict how much reactants remain (or products get formed) in a certain amount of time • Change with time is rate • Rate means calculus must be used, so we need to be careful which rate we are talking about • We will use instantaneous rate at the start of a reaction (initial rate) • Several factors affect the rate of a reaction Initial concentration Temperature Amount of reactant exposed (surface area) Presence of a catalyst • A theory (equations) that predicts kinetics must explain these PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 19.
    19 Chemical Kinetics Calculate the instantaneousrate of disappearance of C4H9Cl at t = 0 (the initial rate). PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 20.
    20 Chemical Kinetics Determine the instantaneousrate of disappearance of C4H9Cl at t = 300 s. Answer: 1.1 × 10 –4 M/s PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 21.
    21 Chemical Kinetics Reaction Rates andStoichiometry • In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. • Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH. C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) Rate = -∆[C4H9Cl] ∆t = ∆[C4H9OH] ∆t PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 22.
    22 Chemical Kinetics Reaction Rates andStoichiometry • What if the ratio is not 1:1? 2 HI(g) → H2(g) + I2(g) in such a case , [ ] [ ] [ ]2 2 HI H I1 Rate = - = 2 t t t ∆ ∆ ∆ = ∆ ∆ ∆ PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 23.
    23 Chemical Kinetics Reaction Rates andStoichiometry • To generalize, then, for the reaction aA + bB cC + dD Rate = − 1 a ∆[A] ∆t = − 1 b ∆[B] ∆t = 1 c ∆[C] ∆t 1 d ∆[D] ∆t = PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 24.
    24 Chemical Kinetics Key points aboutkinetics so far • Study the vocabulary • Reaction rate has weird units • Be clear whether you are talking about average reaction rate or instantaneous reaction rate • The equations which we will study all talk about instantaneous reaction rate, most easily measured at the beginning of a reaction (initial rate) • Relating rates at which products appear (+) and reactants disappear (-) has to do with inverse stoichiometric coefficients • Reaction rate depends on several factors: initial concentration, temperature, surface area of reactants exposed, presence of a catalyst PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 25.
    25 Chemical Kinetics (a) How isthe rate at which ozone disappears related to the rate at which oxygen appears in the reaction (b) If the rate at which O2 appears, Δ[O2]/Δ t, is 6.0 × 10–5 M/s at a particular instant, at what rate is O3 disappearing at this same time, –Δ[O3] /Δ t? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 26.
    26 Chemical Kinetics Answers: (a) 8.4× 10 –7 M/s, (b) 2.1 × 10 –7 M/s SAMPLE EXERCISE 14.3 continued The decomposition of N2O5 proceeds according to the following equation: If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 × 10–7 M/s, what is the rate of appearance of (a) NO2, (b) O2? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 27.
    27 Chemical Kinetics Concentration and Rate Onecan gain information about the rate of a reaction by seeing how the rate changes with changes in concentration. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 28.
    28 Chemical Kinetics Concentration and Rate ComparingExperiments 1 and 2, when [NH4 +] doubles, the initial rate doubles. NH4 +(aq) + NO2 −(aq) N2(g) + 2 H2O(l) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 29.
    29 Chemical Kinetics Concentration and Rate Likewise,comparing Experiments 5 and 6, when [NO2 −] doubles, the initial rate doubles. NH4 +(aq) + NO2 −(aq) N2(g) + 2 H2O(l) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 30.
    30 Chemical Kinetics Concentration and Rate •This means Rate ∝ [NH4 +] Rate ∝ [NO2 −] Rate ∝ [NH+] [NO2 −] or Rate = k [NH4 +] [NO2 −] • This equation is called the rate law, and k is the rate constant. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 31.
    31 Chemical Kinetics • If youwant your concepts to be clearer do the visualizing concepts part of the problems at the back of your chapter. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 32.
    32 Chemical Kinetics Rate Law • Anequation that shows how the rate depends on the concentration of the reactants is called the rate law • The exponents tell the order of the reaction with respect to each reactant. • This reaction is First-order in [NH4 +] First-order in [NO2 −] PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 33.
    33 Chemical Kinetics Rate Laws • Theoverall reaction order can be found by adding the exponents on the reactants in the rate law. • This reaction is second-order overall. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 34.
    34 Chemical Kinetics • For ageneral reaction: aA + bB cC + dD Rate law has the form Rate =k [A]m[B]n where k is the rate constant and is dependant on temperature. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 35.
    35 Chemical Kinetics If we knowthe rate law and the rate for a set of reaction conditions we can calculate the value of k. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 36.
    36 Chemical Kinetics To calculate therate constant: Rate law Rate = k [A]m [B]n Rate = k[NH4] [NO2] For experiment 1: 5.4X10-7 M/s = k (0.0100M)(0.200M) k = 5.4X10-7 M/s k (0.0100M)(0.200M) = 2.7X10 -4 /Ms PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 37.
    37 Chemical Kinetics Exponents in therate law The exponent in the rate law indicates how the rate is affected by the concentration of each reactant. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 38.
    38 Chemical Kinetics SAMPLE EXERCISE 14.4Relating a Rate Law to the Effect of Concentration on Rate Consider a reaction for which rate = k[A][B]2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 39.
    39 Chemical Kinetics SAMPLE EXERCISE 14.4Relating a Rate Law to the Effect of Concentration on Rate Consider a reaction for which rate = k[A][B]2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. Each box contains 10 spheres. The rate law indicates that in this case [B] has a greater influence on rate than [A] because B has a higher reaction order. Hence, the mixture with the highest concentration of B (most purple spheres) should react fastest. This analysis confirms the order 2 < 1 < 3. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 40.
    40 Chemical Kinetics Assuming that rate= k[A][B], rank the mixtures represented in this Sample Exercise in order of increasing rate. Answer: 2 = 3 < 1 PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 41.
    41 Chemical Kinetics Unit for kdepends on the overall order of the reaction. For example for a second order reaction 2 Unit of Rate k = (Units of concentration) 2 2 M/s M = = = /Ms M M s PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 42.
    42 Chemical Kinetics (a)What are theoverall reaction orders for the rate laws described in Equations 14.9 and 14.10? (b) What are the units of the rate constant for the rate law for Equation 14.9? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 43.
    43 Chemical Kinetics (a)What are theoverall reaction orders for the rate laws described in Equations 14.9 and 14.10? (b) What are the units of the rate constant for the rate law for Equation 14.9? Answer: [ ]2 5 2 2 2 4 1/2 3 2 4 3 2 2N O (g) 4NO (g) + O (g) Rate = k N O CHCl (g) + Cl (g) CCl (g) + HCl(g) Rate = k[CHCl ] [Cl ] → → Answer: The rate of the reaction in Equation 14.9 is first order in N2O5 and first order overall. The reaction in Equation 14.10 is first order in CHCl3 and one-half order in Cl2. The overall reaction order is three halves. So Notice that the units of the rate constant change as the overall order of the reaction changes. [ ]2 4Rate = k N O PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 44.
    44 Chemical Kinetics 1.The reaction isfirst order in a reactant If the concentration of a particular component is doubled the rate doubles and triples when the concentration triples. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 45.
    45 Chemical Kinetics 2. If thereaction is second order with respect to a component: If the rate is second order with respect to a particular component, doubling its concentration increases the rate by a factor of 22 = 4 Tripling its concentration increases the rate by a factor of 32= 9 and so forth PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 46.
    46 Chemical Kinetics • 3. Ifa reaction is zero order with respect to a component: If a reaction is zero order in a particular reactant, changing its concentration will have no effect on the rate as long as long as the reactant is present. It is because any concentration raised to zero is one PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 47.
    47 Chemical Kinetics Order of reactionrate • Rate of reaction can depend on how much reactant is present in more than one way: Zero order: rate ∝ [A]0 First order: rate ∝ [A] Second order: rate ∝ [A]2 PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 48.
    48 Chemical Kinetics PDF Created withdeskPDF PDF Writer - Trial :: http://www.docudesk.com
  • 49.
    49 Chemical Kinetics [ ]2 52 2 2 4 1/2 3 2 4 3 2 2 2 2 2 2N O (g) 4NO (g) + O (g) Rate = k N O CHCl (g) + Cl (g) CCl (g) + HCl(g) Rate = k[CHCl ] [Cl ] H (g) + I (g) 2HI(g) Rate = k[H ][I ] → → → PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 50.
    50 Chemical Kinetics PDF Created withdeskPDF PDF Writer - Trial :: http://www.docudesk.com
  • 51.
    51 Chemical Kinetics • Although theexponents in the rate law are often the same as the coefficients in the balanced reaction, this is not necessarily the case. • The values of these exponents must be determined experimentally. • Most reaction orders are 0, 1 or 2.However we may encounter reaction orders in fraction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 52.
    52 Chemical Kinetics • The rateof the reaction depends on the concentration but the rate constant , k, does not. • k is dependent only on temperature. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 53.
    53 Chemical Kinetics 4 2 4 4 CH+ 4 Cl CCl + 4HCl 1 [CH4] 1 [ ] Rate = - 1 4 1 Rate = - rate of disappearence of CH = rate of appearence of HCl 4 Rate x 4 = Rate of appearence of HCl HCl t t → ∆ ∆ = ∆ ∆ PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 54.
    54 Chemical Kinetics Change in Concentrationwith Time From the rate law we can calculate the rate of the reaction if we know the the rate constant and the reactant concentration Rate = k [A] Rate laws can also be converted into equations that tell us what the concentration of reactants and products are at any time during the reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 55.
    55 Chemical Kinetics Change of Concentrationwith Time for the First Order Reactions The first order reaction is one whose rate depends on the concentration of a single reactant raised to the first power reactant A product Rate = -Δ[A] = k [A] Δt This form of rate law, which expresses how rate depends upon concentration is called differential rate law. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 56.
    56 Chemical Kinetics PDF Created withdeskPDF PDF Writer - Trial :: http://www.docudesk.com
  • 57.
    57 Chemical Kinetics PDF Created withdeskPDF PDF Writer - Trial :: http://www.docudesk.com
  • 58.
    58 Chemical Kinetics Integrated Rate Laws Usingcalculus to integrate the rate law for a first-order process we get: ln [A]t [A]0 = −kt Where [A]0 is the initial concentration of A. [A]t is the concentration of A at some time, t, during the course of the reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 59.
    59 Chemical Kinetics Integrated Rate Laws Manipulatingthis equation produces… ln [A]t [A]0 = −kt ln [A]t − ln [A]0 = − kt ln [A]t = − kt + ln [A]0 Where m is the slope of the line and b is th ey intersept of the line. Which is a general equation of a straight line. …which is in the form y = mx + b PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 60.
    60 Chemical Kinetics Integrated Rate Laws ln[A]t = − kt + ln [A]0 Where m is the slope of the line and b is the y intersept of the line. Which is a general equation of a straight line. …which is in the form y = mx + b PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 61.
    61 Chemical Kinetics First-Order Processes Therefore, ifa reaction is first-order, a plot of ln [A]t vs. t will yield a straight line, and the slope of the line will be -k. ln [A]t = -kt + ln [A]0 PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 62.
    62 Chemical Kinetics First-Order Processes Consider theprocess in which methyl isonitrile is converted to acetonitrile. CH3NC CH3CN PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 63.
    63 Chemical Kinetics First-Order Processes This datawas collected for this reaction at 198.9°C. CH3NC CH3CN PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 64.
    64 Chemical Kinetics • The experimentsshow this reaction to be first order so we can write the rate equation as In[CH3NC]t= - kt + ln [CH3NC]o PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 65.
    65 Chemical Kinetics • For gaswe can use pressure as a unit of concentration as from the ideal gas law equation the pressure is directly proportional to the number of moles per unit volume PV = nRT n P = RT V At constant temperature n P α V PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 66.
    66 Chemical Kinetics First-Order Processes • Whenln P is plotted as a function of time, a straight line results. • Therefore, The process is first-order. k is the negative slope: 5.1 × 10-5 s−1. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 67.
    67 Chemical Kinetics • How toget the natural log value on your calculator LN will give the natural log value. Pressing LN will show In ( key in the value) and press enter PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 68.
    68 Chemical Kinetics The initial rateof a reaction was measured for several different starting concentrations of A and B, and the results are as follows: 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] = 0.050 M and [B] = 0.100 M. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 69.
    69 Chemical Kinetics The initial rateof a reaction was measured for several different starting concentrations of A and B, and the results are as follows: 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] = 0.050 M and [B] = 0.100 M a. (b) Using the rate law and the data from experiment 1, we have (c) Using the rate law from part (a) and the rate constant from part (b), we have PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 70.
    70 Chemical Kinetics PRACTICE EXERCISE The followingdata were measured for the reaction of nitric oxide with hydrogen: (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 71.
    71 Chemical Kinetics PRACTICE EXERCISE The followingdata were measured for the reaction of nitric oxide with hydrogen: Answers: (a) rate = k[NO]2[H2]; (b) k =1.2 M–2s–1; (c) rate = 4.5 × 10–4 M/s (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 72.
    72 Chemical Kinetics PRACTICE EXERCISE The decompositionof dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 × 10–4s–1: If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 73.
    73 Chemical Kinetics PRACTICE EXERCISE The decompositionof dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 × 10–4s–1: If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s? Answer: 51 torr PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 74.
    74 Chemical Kinetics Half-Life • Half-life isdefined as the time required for one-half of a reactant to react. • Because [A] at t1/2 is one-half of the original [A], [A]t = 1/2 [A]0. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 75.
    75 Chemical Kinetics Half-Life For a first-orderprocess, this becomes 0.5 [A]0 [A]0 ln = −kt1/2 ln 0.5 = −kt1/2 −0.693 = −kt1/2 = t1/2 0.693 kNOTE: For a first-order process, the half-life does not depend on [A]0. ln [A]t [A]0 = −kt PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 76.
    76 Chemical Kinetics • The halflife of a first order reaction remains the same throughout the reaction. • The concentration of the reactant keeps becoming half of the previous value after every half life. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 77.
    77 Chemical Kinetics • In thefirst order reaction, the concentration of reactant decreases by ½ in each of a series of regularly spaced time intervals, namely t1/2 • The concept of a first order reaction is widely used in describing radioactive decay, a first order process. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 78.
    78 Chemical Kinetics SAMPLE EXERCISE 14.7Using the Integrated First-Order Rate Law • The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1 at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 × 10–7 g/cm3. Assume that the average temperature of the lake is 12°C. • (a) What is the concentration of the insecticide on June 1 of the following year? • (b) How long will it take for the concentration of the insecticide to drop to 3.0 × 10–7 g/cm3? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 79.
    79 Chemical Kinetics SAMPLE EXERCISE 14.7Using the Integrated First-Order Rate Law • The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1 at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 × 10–7 g/cm3. Assume that the average temperature of the lake is 12°C. • (a) What is the concentration of the insecticide on June 1 of the following year? • (b) How long will it take for the concentration of the insecticide to drop to 3.0 × 10–7 g/cm3? ln [A]t = -kt + ln [A]0 Solve: (a) Substituting the known quantities into Equation 14.13, we have Note that the concentration units for [A]t and [A]0 must be the same. (b) Again substituting into Equation 14.13, with [insecticide]t = 3.0 × 10–7 g/cm3, gives Solving for t gives PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 80.
    80 Chemical Kinetics Temperature and Rate •Generally, as temperature increases, so does the reaction rate. • This is because k is temperature dependent. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 81.
    81 Chemical Kinetics The Collision Model •In a chemical reaction, bonds are broken and new bonds are formed. • Molecules can only react if they collide with each other. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 82.
    82 Chemical Kinetics The Collision Model Furthermore,molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 83.
    83 Chemical Kinetics Activation Energy • Inother words, there is a minimum amount of energy required for reaction: the activation energy, Ea. • Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 84.
    84 Chemical Kinetics H3C N H3C C N H3CC NC- • This is a reaction where methyl isonitrile is rearranging to acetonitrile • This reaction passes through an intermediate state in which the N C is sitting sideways. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 85.
    85 Chemical Kinetics Reaction Coordinate Diagrams Itis helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 86.
    86 Chemical Kinetics Reaction Coordinate Diagrams •It shows the energy of the reactants and products (and, therefore, ∆E). • The high point on the diagram is the transition state. • The species present at the transition state is called the activated complex. • The energy gap between the reactants and the activated complex is the activation energy barrier. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 87.
    87 Chemical Kinetics • Temperature is definedas a measure of the average kinetic energy of the molecules in a sample. • At any temperature there is a wide distribution of kinetic energies. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 88.
    88 Chemical Kinetics Maxwell–Boltzmann Distributions • Asthe temperature increases, the curve flattens and broadens. • Thus at higher temperatures, a larger population of molecules has higher energy. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 89.
    89 Chemical Kinetics Maxwell–Boltzmann Distributions • Ifthe dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. • As a result, the reaction rate increases. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 90.
    90 Chemical Kinetics Maxwell–Boltzmann Distributions This fractionof molecules can be found through the expression where R is the gas constant (8.314J/mol-K) and T is the Kelvin temperature. Ea is the minimum energy needed for reaction. f = e−Ea/RT PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 91.
    91 Chemical Kinetics • Arrhenious notedthat the increase in reaction rate is not linear with increase in temperature in most reactions. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 92.
    92 Chemical Kinetics • Arrhenius observedthat most reaction rates depend on the following three factors A. Fraction of molecules possessing Ea or greater B. number of collisions per second C. fraction of molecules having the correct orientation PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 93.
    93 Chemical Kinetics Arrhenius Equation Svante Arrheniusdeveloped a mathematical relationship between k (the rate constant) and Ea: k = A e−Ea/RT where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 94.
    94 Chemical Kinetics Arrhenius Equation Taking thenatural logarithm of both sides, the equation becomes ln k = -Ea ( ) + ln A 1 RT y = mx + b Therefore, if k is determined experimentally at several temperatures, and lnk vs 1/T is plotted Ea can be calculated from the slope of the plot. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 95.
    95 Chemical Kinetics • If weknow the constant k at two or more temperatures At temperatures T1 and T2 the rate constant is k1 and k , ln k1 = -Ea + lnA and ln k2 = -Ea + lnA RT1 RT2 lnk1-lnk2 = [ -Ea/RT1 +lnA] - [ -Ea/RT2+lnA] lnk1/k2 = Ea (1/T1-1/T2) R PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 96.
    96 Chemical Kinetics •lnk1=Ea(1/T2-1/T1) •R PDF Created withdeskPDF PDF Writer - Trial :: http://www.docudesk.com
  • 97.
    97 Chemical Kinetics PDF Created withdeskPDF PDF Writer - Trial :: http://www.docudesk.com
  • 98.
    98 Chemical Kinetics Reaction Mechanisms The sequenceof events that describes the actual process by which reactants become products is called the reaction mechanism. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 99.
    99 Chemical Kinetics Reaction Mechanisms • Reactionsmay occur all at once or through several discrete steps. • Each of these processes is known as an elementary reaction or elementary process. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 100.
    100 Chemical Kinetics Rate laws forelementary reactions • The rate law of an elementary reaction is directly based on its molecularity or concentration. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 101.
    101 Chemical Kinetics Reaction Mechanisms The molecularityof a process tells how many molecules are involved in the process. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 102.
    102 Chemical Kinetics • In amultistep reaction each step will have its own rate constant and activation energy. • The slowest step determines the overall rate. Example Relay race PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 103.
    103 Chemical Kinetics Multistep Mechanisms • Ina multistep process, one of the steps will be slower than all others. • The overall reaction cannot occur faster than this slowest, rate-determining step. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
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    104 Chemical Kinetics • A balancedchemical equation often occurs in a series of elementary reactions • A balanced chemical reaction itself could have a multi step mechanism PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 105.
    105 Chemical Kinetics Slow Initial Step •The rate law for this reaction is found experimentally to be Rate = k [NO2]2 • CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. • This suggests the reaction occurs in two steps. NO2 (g) + CO (g) → NO (g) + CO2 (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 106.
    106 Chemical Kinetics Slow Initial Step •A proposed mechanism for this reaction is Step 1: NO2 + NO2 → NO3 + NO (slow) Step 2: NO3 + CO → NO2 + CO2 (fast) • The NO3 intermediate is consumed in the second step. • As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. • Since step 1 is the slow rate determining step the rate law is going to be Rate = [NO2] 2 NO2 (g) + CO (g) → NO (g) + CO2 (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 107.
    107 Chemical Kinetics Practice exercise Ozone reactswith NO2 to give dinitrogen pentoxide and oxygen: O3 (g) + 2NO2 (g) N2O5(g) + O2 (g) The reaction occurs in the following two steps: O3(g) + NO2 (g) NO3 (g) + O2(g) NO3(g) + NO2(g) N2O5 (g) The experimental rate law is Rate = [ O3][NO2] What can you sat about the relative rate of the two steps of the mechanism? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 108.
    108 Chemical Kinetics • How manyelementary reactions are there in the mechanism? • How many intermediates? • Which step is rate limiting? • Which is the fastest step? • Exo or endothermic? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 109.
    109 Chemical Kinetics Problems • How manyelementary reactions are there in the mechanism? • How many intermediates? • Which step is rate limiting? • Which is the fastest step? • Exo or endothermic? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 110.
    110 Chemical Kinetics Fast Initial Step •Experimentally the rate law for this reaction is found to be Rate = k [NO]2 [Br2] • Because ter molecular processes are rare, this rate law suggests a two-step mechanism. 2 NO (g) + Br2 (g) → 2 NOBr (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 111.
    111 Chemical Kinetics Fast Initial Step •A proposed mechanism is Step 2: NOBr2 + NO → 2 NOBr (slow) Step 1 includes the forward and reverse reactions. The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k2 [NOBr2] [NO] But how can we find [NOBr2]? Step 1: NO + Br2 NOBr2 (fast) 2 NO (g) + Br2 (g) → 2 NOBr (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 112.
    112 Chemical Kinetics Fast Initial Step •NOBr2 can react two ways: With NO to form NOBr By decomposition to reform NO and Br2 • The reactants and products of the first step are in equilibrium with each other. • Therefore, Ratef = Rater Step 1: NO + Br2 NOBr2 (fast) Step 2: NOBr2 + NO → 2 NOBr (slow) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 113.
    113 Chemical Kinetics Fast Initial Step •Because Ratef = Rater , k1 [NO] [Br2] = k−1 [NOBr2] Pay attention to the fact that the reverse k is k-1 • Solving for [NOBr2] gives us k1 k−1 [NO] [Br2] = [NOBr2] PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 114.
    114 Chemical Kinetics Fast Initial Step Frombefore Rate = k2 [NOBr2] [NO] From the previous equation: Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives k2k1 k−1 Rate = [NO] [Br2] [NO] = k [NO]2 [Br2] k1 k−1 [NO] [Br2] = [NOBr2] PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 115.
    115 Chemical Kinetics • In generalwhen ever a fast step precedes a slow step we can solve the problem of the concentration of the intermediate step by assuming that an equilibrium is established in the fast step. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 116.
    116 Chemical Kinetics • Write downthe rate law for the overall reaction. SAMPLE EXERCISE 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the experimentally observed one: PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 117.
    117 Chemical Kinetics PRACTICE EXERCISE The firststep of a mechanism involving the reaction of bromine is What is the expression relating the concentration of Br (g) to that of Br2(g)? PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 118.
    118 Chemical Kinetics Catalysts Arrhenius equation • K= A e –E/RT A= frequency factor and E= Energy of activation • Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. • Catalysts change the mechanism by which the process occurs. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 119.
    119 Chemical Kinetics • Homogeneous catalysts 2H2O2(aq)2H2O (l) + O2 (g) NaBr in an acidic solution acts as a catalyst for this reaction 2Br -(aq) + H2O2(aq) + 2 H+ (aq) Br2(aq) + 2H2O(l) Br2(aq) + H2O2(aq) 2Br –(aq) + 2H+(aq) + O2 (g) PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
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    120 Chemical Kinetics PDF Created withdeskPDF PDF Writer - Trial :: http://www.docudesk.com
  • 121.
    121 Chemical Kinetics PDF Created withdeskPDF PDF Writer - Trial :: http://www.docudesk.com
  • 122.
    122 Chemical Kinetics Heterogeneous Catalyst One waya catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. The reactants are adsorbed on the surface of the catalyst. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com
  • 123.
    123 Chemical Kinetics Enzymes • Enzymes are catalystsin biological systems. • The substrate fits into the active site of the enzyme much like a key fits into a lock. PDFCreatedwithdeskPDFPDFWriter-Trial::http://www.docudesk.com