Chapter 14 Chemical Kinetics Lecture Presentation Chapter 14 Chemical Kinetics John D. Bookstaver St. Charles Community College Cottleville, MO © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Kinetics In kinetics we study 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). © 2012 Pearson Education, Inc.

Factors That Affect Reaction Rates Physical state of the reactants. In order to react, molecules must come in contact with each other. The more homogeneous the mixture of reactants, the faster the molecules can react. © 2012 Pearson Education, Inc.

Factors That Affect Reaction Rates Concentration of reactants. As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. © 2012 Pearson Education, Inc.

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. © 2012 Pearson Education, Inc.

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. © 2012 Pearson Education, Inc.

A. The effect of increasing the partial pressures of the reactive components of a gaseous mixture depends on which side of the chemical equation has the most gas molecules. B. Increasing the partial pressures of the reactive components of a gaseous mixture has no effect on the rate of reaction if each reactant pressure is increased by the same amount. C Increasing the partial pressures of the reactive components of a gaseous mixture increases the rate of reaction. D. Increasing the partial pressures of the reactive components of a gaseous mixture decreases the rate of reaction. Answer: C

If A converts completely to B, what type of molecules will the container hold? A. Mixture of A and B molecules B. A molecules only C. B molecules only D. AB molecules Answer: C

© 2012 Pearson Education, Inc. Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. © 2012 Pearson Education, Inc.

Sample Exercise 14.1 Calculating an Average Rate of Reaction From the data in Figure 14.3, calculate the average rate at which A disappears over the time interval from 20 s to 40 s. Solution Analyze We are given the concentration of A at 20 s (0.54 M) and at 40 s (0.30 M) and asked to calculate the average rate of reaction over this time interval. Plan The average rate is given by the change in concentration, [A], divided by the change in time, t. Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity. Solve Practice Exercise Use the data in Figure 14.3 to calculate the average rate of appearance of B over the time interval from 0 s to 40 s. Answer: 1.8  102 M/s

© 2012 Pearson Education, Inc. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average rate = [C4H9Cl] t © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) 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. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) A plot of [C4H9Cl] versus 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. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) 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. Rate = [C4H9Cl] t = [C4H9OH] © 2012 Pearson Education, Inc.

Sample Exercise 14.2 Calculating an Instantaneous Rate of Reaction Using Figure 14.4, calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 s (the initial rate). Solution Analyze We are asked to determine an instantaneous rate from a graph of reactant concentration versus time. Plan To obtain the instantaneous rate at t = 0s, we must determine the slope of the curve at t = 0. The tangent is drawn on the graph as the hypotenuse of the tan triangle. The slope of this straight line equals the change in the vertical axis divided by the corresponding change in the horizontal axis (that is, change in molarity over change in time). Solve The tangent line falls from [C4H9Cl] = 0.100 M to 0.060 M in the time change from 0 s to 210 s. Thus, the initial rate is

Sample Exercise 14.2 Calculating an Instantaneous Rate of Reaction Continued Practice Exercise Using Figure 14.4, determine the instantaneous rate of disappearance of C4H9Cl at t = 300 s. Answer: 1.1  104 M/s

An instantaneous rate is measured at a particular point in time of a reaction, whereas average rate is measured over a time period; Yes. An instantaneous rate is measured at a particular point in time of a reaction, whereas average rate is measured over a time period; No. An instantaneous rate is measured over a short time period, whereas average rate is measured over a large time period; Yes. An instantaneous rate is measured over a short time period, whereas average rate is measured over a large time period; No. Answer: A

Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI(g)  H2(g) + I2(g) In such a case, Rate =  1 2 [HI] t = [I2] © 2012 Pearson Education, Inc.

Reaction Rates and Stoichiometry To generalize, then, for the reaction aA + bB cC + dD Rate =  1 a [A] t =  b [B] = c [C] d [D] © 2012 Pearson Education, Inc.

Sample Exercise 14.3 Relating Rates at Which Products Appear and Reactants Disappear (a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O3(g)  3 O2(g)? (b) If the rate at which O2 appears, [O2]/t, is 6.0  10–5 M/s at a particular instant, at what rate is O3 disappearing at this same time,  [O3]/t? Solution Analyze We are given a balanced chemical equation and asked to relate the rate of appearance of the product to the rate of disappearance of the reactant. Solve (a) Using the coefficients in the balanced equation and the relationship given by Equation 14.4, we have: (b) Solving the equation from part (a) for the rate at which O3 disappears, [O3]/t, we have: Check We can apply a stoichiometric factor to convert the O2 formation rate to the O3 disappearance rate: Plan We can use the coefficients in the chemical equation as shown in Equation 14.4 to express the relative rates of reactions.

Sample Exercise 14.3 Relating Rates at Which Products Appear and Reactants Disappear Continued Practice Exercise If the rate of decomposition of N2O5 in the reaction 2 N2O5(g)  4 NO2(g) + O2(g) at a particular instant is 4.2  107 M/s, what is the rate of appearance of (a) NO2 and (b) O2 at that instant? Answers: (a) 8.4  107 M/s, (b) 2.1  107 M/s

Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration. © 2012 Pearson Education, Inc.

Concentration and Rate NH4+(aq) + NO2(aq) N2(g) + 2 H2O(l) If we compare Experiments 1 and 2, we see that when [NH4+] doubles, the initial rate doubles. © 2012 Pearson Education, Inc.

Concentration and Rate NH4+(aq) + NO2(aq) N2(g) + 2 H2O(l) Likewise, when we compare Experiments 5 and 6, we see that when [NO2] doubles, the initial rate doubles. © 2012 Pearson Education, Inc.

Concentration and Rate This means Rate  [NH4+] Rate  [NO2] Rate  [NH4+] [NO2] which, when written as an equation, becomes Rate = k [NH4+] [NO2] This equation is called the rate law, and k is the rate constant. Therefore, © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Rate Laws A rate law shows the relationship between the reaction rate and the concentrations of reactants. The exponents tell the order of the reaction with respect to each reactant. Since the rate law is Rate = k[NH4+] [NO2] the reaction is First-order in [NH4+] and First-order in [NO2] © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Rate Laws Rate = k[NH4+] [NO2] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall. Add problem(s) after this slide. © 2012 Pearson Education, Inc.

Yes, only if the order of each concentration is one in the rate law. Yes, the rate is directly proportional to the rate constant in a rate law. Yes, only if the order of each concentration is one in the rate law. No, as the order of each concentration changes in a rate law, the units of the rate constant change, but those of the rate are always concentration/time. No, as the order of each concentration changes in a rate law, there are changes in the units of both. Answer: C

A. 2nd order in NO, 1st order in H2, 2nd order overall B. 2nd order in NO, 1st order in H2, 3rd order overall C. 1st order in NO, 1st order in H2, 2nd order overall D. 1st order in NO, 2nd order in H2, 3rd order overall Answer: B

(b) A. Yes B. No Answer: B

Sample Exercise 14.4 Relating a Rate Law to the Effect of Concentration on Rate Consider a reaction A + B  C 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. Solution Analyze We are given three boxes containing different numbers of spheres representing mixtures containing different reactant concentrations. We are asked to use the given rate law and the compositions of the boxes to rank the mixtures in order of increasing reaction rates. Plan Because all three boxes have the same volume, we can put the number of spheres of each kind into the rate law and calculate the rate for each box.

Sample Exercise 14.4 Relating a Rate Law to the Effect of Concentration on Rate Continued Solve Box 1 contains 5 red spheres and 5 purple spheres, giving the following rate: Box 1: Rate = k(5)(5)2 = 125k Box 2 contains 7 red spheres and 3 purple spheres: Box 2: Rate = k(7)(3)2 = 63k Box 3 contains 3 red spheres and 7 purple spheres: Box 3: Rate = k(3)(7)2 = 147k The slowest rate is 63k (box 2), and the highest is 147k (box 3). Thus, the rates vary in the order 2 < 1 < 3. Check 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. Practice Exercise Assuming that rate = k[A][B], rank the mixtures represented in this Sample Exercise in order of increasing rate. Answer: 2 = 3 < 1

to compare reactions to evaluate which ones are relatively fast and which ones are relatively slow, the quantity of interest is the rate constant. large value of k (~109 or higher) means a fast reaction and a small value of k (10 or lower) means a slow reaction.

The rate of the reaction is very slow. There is no reaction. The rate constant can not be calculated. All of the above Answer: B

The units of the rate constant depend on the overall reaction order of the rate law. For Ex. A reaction that is 2nd order over all -

Sample Exercise 14.5 Determining Reaction Orders and Units for Rate Constants (a) What are the overall reaction orders for the reactions described in Equations 14.9 and 14.10? (b) What are the units of the rate constant for the rate law in Equation 14.9? Solution Analyze We are given two rate laws and asked to express (a) the overall reaction order for each and (b) the units for the rate constant for the first reaction. Plan The overall reaction order is the sum of the exponents in the rate law. The units for the rate constant, k, are found by using the normal units for rate (M/s) and concentration (M) in the rate law and applying algebra to solve for k. Solve (a) 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. (b) For the rate law for Equation 14.9, we have Units of rate = (units of rate constant)(units of concentration) so Notice that the units of the rate constant change as the overall order of the reaction changes. 40

Sample Exercise 14.5 Determining Reaction Orders and Units for Rate Constants Continued Practice Exercise (a) What is the reaction order of the reactant H2 in Equation 14.11? (b) What are the units of the rate constant for Equation 14.11? Answers: (a) 1, (b) M–1 s–1 41

© 2012 Pearson Education, Inc. Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us ln [A]t [A]0 = kt where [A]0 is the initial concentration of A, and [A]t is the concentration of A at some time, t, during the course of the reaction. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. 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 which is in the form y = mx + b © 2012 Pearson Education, Inc.

First-Order Processes ln [A]t = kt + ln [A]0 Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be k. © 2012 Pearson Education, Inc.

First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH3NC CH3CN © 2012 Pearson Education, Inc.

First-Order Processes CH3NC CH3CN This data were collected for this reaction at 198.9 C. © 2012 Pearson Education, Inc.

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 of the slope: 5.1  105 s1. © 2012 Pearson Education, Inc.

Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1 [A]t = kt + [A]0 also in the form y = mx + b © 2012 Pearson Education, Inc.

Second-Order Processes 1 [A]t = kt + [A]0 So if a process is second-order in A, a plot of vs. t yields a straight line, and the slope of that line is k. 1 [A] © 2012 Pearson Education, Inc.

Second-Order Processes The decomposition of NO2 at 300 °C is described by the equation NO2(g) NO(g) + O2(g) 1 2 and yields data comparable to this table: Time (s) [NO2], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 © 2012 Pearson Education, Inc.

Second-Order Processes Plotting ln [NO2] vs. t yields the graph at the right. The plot is not a straight line, so the process is not first-order in [A]. Time (s) [NO2], M ln [NO2] 0.0 0.01000 4.610 50.0 0.00787 4.845 100.0 0.00649 5.038 200.0 0.00481 5.337 300.0 0.00380 5.573 © 2012 Pearson Education, Inc.

Second-Order Processes 1 [NO2] Graphing ln vs. t, however, gives this plot Fig. 14.9(b). Because this is a straight line, the process is second-order in [A]. Time (s) [NO2], M 1/[NO2] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 154 200.0 0.00481 208 300.0 0.00380 263 Add problem after this slide. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. 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 = 0.5 [A]0. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. 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 k Note: For a first-order process, then, the half-life does not depend on [A]0. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Half-Life For a second-order process, 1 0.5 [A]0 = kt1/2 + [A]0 2 [A]0 = kt1/2 + 1 Add problem after this slide. 2  1 [A]0 = kt1/2 1 = = t1/2 1 k[A]0 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature-dependent. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. 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. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. 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. © 2012 Pearson Education, Inc.

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. © 2012 Pearson Education, Inc.

Reaction Coordinate Diagrams The diagram 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. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions 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. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus, at higher temperatures, a larger population of molecules has higher energy. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation-energy barrier. As a result, the reaction rate increases. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression where R is the gas constant and T is the Kelvin temperature. −Ea/RT f = e © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and Ea: k = Ae where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. −Ea/RT © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes ln k =  ( ) + ln A Ea R 1 T Add problem(s) after this slide. y = mx + b Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. . 1 T © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. 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. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. 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. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Slow Initial Step NO2(g) + CO(g)  NO(g) + CO2(g) 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 that the reaction occurs in two steps. © 2012 Pearson Education, Inc.

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. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Fast Initial Step 2 NO(g) + Br2(g)  2 NOBr(g) The rate law for this reaction is found to be Rate = k[NO]2 [Br2] Because termolecular processes are rare, this rate law suggests a two-step mechanism. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Fast Initial Step A proposed mechanism is Step 1: NO + Br2 NOBr2 (fast) Step 2: NOBr2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Fast Initial Step 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]? © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. 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 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Fast Initial Step Because Ratef = Rater, k1[NO] [Br2] = k1[NOBr2] Solving for [NOBr2], gives us k1 k1 [NO] [Br2] = [NOBr2] © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Fast Initial Step 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] © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Catalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. 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. © 2012 Pearson Education, Inc.