Lecture Presentation Chapter 14 Yonsei University 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). 2
14.1 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. Concentration of reactants. As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. 3 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. 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. 4
14.2 Reaction Rates 5 C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times. - average rate Average rate = [C 4H 9 Cl] t 6 Note that the average rate decreases as the reaction proceeds. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. 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. Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq)
Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. Rate = 7 [C 4H 9 Cl] t = [C 4H 9 OH] t Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI(g) H 2 (g) + I 2 (g) In such a case, Rate = 1 2 [HI] t = [I 2] t 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 8
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 O 3 (g) 3 O 2 (g)? (b) If the rate at which O 2 appears, [O 2 ]/ t, is 6.0 10 5 M/s at a particular instant, at what rate is O 3 disappearing at this same time, [O 3 ]/ t? Solution (a) (b) Practice Exercise If the rate of decomposition of N 2 O 5 in the reaction 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) at a particular instant is 4.2 10 7 M/s, what is the rate of appearance of (a) NO 2 and (b) O 2 at that instant? Answers: (a) 8.4 10 7 M/s, (b) 2.1 10 7 M/s 9 14.3 Concentration and Rate Law One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration. 10
Rate Law from Experiments NH 4+ (aq) + NO 2 (aq) N 2 (g) + 2 H 2 O(l) 11 If we compare Experiments 1 and 2, we see that when [NH 4+ ] doubles, the initial rate doubles. Rate Law from Experiments NH 4+ (aq) + NO 2 (aq) N 2 (g) + 2 H 2 O(l) 12 Likewise, when we compare Experiments 5 and 6, we see that when [NO 2 ] doubles, the initial rate doubles.
Rate Law This means Rate [NH 4+ ] Rate [NO 2 ] Therefore, Rate [NH 4+ ] [NO 2 ] which, when written as an equation, becomes Rate = k [NH 4+ ] [NO 2 ] This equation is called the rate law, and k is the rate constant. First-order in [NH 4+ ] and first-order in [NO 2 ] : This reaction is second-order overall. 13 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 2 < 1 < 3 Box 1: Rate = k(5)(5) 2 = 125k Box 2: Rate = k(7)(3) 2 = 63k Box 3: Rate = k(3)(7) 2 = 147 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 14
Reaction Orders: The Exponents in the Rate Law A rate law shows the relationship between the reaction rate and the concentrations of reactants. For a general reaction with rate law: Rate = k[reactant 1] m [reactant 2] n The exponents m and n are called reaction orders. The overall reaction order is the sum of the reaction orders. The overall order of reaction is m + n +. Note that reaction orders must be determined experimentally. They do not necessarily correspond to the stoichiometric coefficients in the balanced chemical equation! We commonly encounter reaction orders of 0, 1, or 2. Even fractional or negative values are possible. 15 Magnitudes and Units of Rate Constants In comparing reactions to evaluate which ones are relatively fast and which are relative slow, the rate constants are compared: A large value of k ( 10 9 ): the reaction is fast. A small value of k ( 10 ): the reaction is slow. Units of the rate constant depend on the overall reaction order. 16
17 Using Initial Rates to Determine Rate Laws To determine the rate law, we observe the effect of changing initial concentrations. If a reaction is zero order in a reactant, changing the initial concentration of that reactant will have no effect on rate (as long as some reactant is present). If a reaction is first order, doubling the concentration will cause the rate to double. If a reaction is second order, doubling the concentration will result in a 2 2 increase in rate. Similarly, tripling the concentration results in a 3 2 increase in rate. A reaction is n th order if doubling the concentration causes a 2 n increase in rate. Note that the rate, not the rate constant(k), depends on concentration. k is affected by temperature and by the presence of a catalyst. Sample Exercise 14.6 Determining a Rate Law from Initial Rate Data The initial rate of a reaction A + B C 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 rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M. Solution (a) Rate = k[a] 2 [B] 0 = k[a] 2 (b) (c) Rate = k[a] 2 = (4.0 10 3 M 1 s 1 )(0.050 M) 2 = 1.0 10 5 M/s 18
14.4 The Change of Concentration with Time Integrated Rate Laws Goal: Convert the rate law into a convenient equation that gives concentration as a function of time. 19 Integrated Rate Laws For a first-order reaction the rate doubles as the concentration of a reactant doubles. Therefore, we can write the differential rate law: A Rate k A t Integrating, we get the integrated rate law: ln A ln A kt t 0 Rearranging: An alternate form: ln ln A t kt ln A 0 A t kt A 0 A plot of ln[a] t versus t is a straight line with slope k and intercept ln[a] 0. Note that in this equation we use the natural logarithm, ln (log to the base e). 20
First-Order Processes CH 3 NC CH 3 CN: at 198.9 C. 21 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. Sample Exercise 14.7 Using the Integrated First-Order Rate Law The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr 1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 10 7 g/cm 3. Assume that the average temperature of the lake is 12 ºC. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to decrease to 3.0 10 7 g/cm 3? Solution (a) ln[insecticide] t = 1 yr = (1.45 yr 1 )(1.00 yr) + ln(5.0 10 7 ) ln[insecticide] t = 1 yr = 1.45 + ( 14.51) = 15.96 [insecticide] t = 1 yr = e 15.96 = 1.2 10 7 g/cm 3 (b) ln(3.0 10 7 ) = (1.45 yr 1 )(t) + ln(5.0 10 7 ) t = [ln(3.0 10 7 ) ln(5.0 10 7 )]/1.45 yr 1 = ( 15.02 + 14.51)/1.45 yr 1 = 0.35 yr 22
Integrated Rate Laws For a Second-Order Reaction A second-order reaction is one whose rate depends on the reactant concentration to the second power or on the concentration of two reactants, each raised to the first power. For a second-order reaction with just one reactant, we write the differential rate law: A Rate t k A 2 Integrating, we get the integrated form of the rate law: t 0 A plot of 1/[A] t versus t is a straight line with slope k and intercept 1/[A] 0. 1 A kt 1 A 23 Note that a second-order process can have a rate constant expression of the form: Rate = k[a][b] That is, the reaction is second order overall, but has first-order dependence on A and B. Second-Order Processes The decomposition of NO 2 at 300 C 1 NO 2 (g) NO(g) + 2 O 2 (g) Time (s) [NO 2 ], M 1/[NO 2 ] 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 24
Zero-Order Reactions A zero-order reaction is one whose rate is independent of the reactant concentration. A Rate t k The integrated rate law for a zero-order reaction is: A kt t A 25 Half-life of a Reaction 26
Half-Life Half-life, t 1/2, is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0. 27 Half-Life For a first-order process, this becomes 0.5 [A] ln 0 [A] = kt 1/2 0 ln 0.5 = kt 1/2 28 0.693 = kt 1/2 0.693 k = t 1/2 Note: For a first-order process, then, the half-life does not depend on [A] 0.
29 Half-Life For a second-order process, 1 0.5 [A] 0 = kt 1/2 + 2 [A] 0 = kt 1/2 + 2 1 1 = [A] 0 [A] 0 = kt 1/2 1 k[a] 0 = t 1/2 1 [A] 0 1 [A] 0 14.5 Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature-dependent. We see an approximate doubling of the rate of the reaction with each 10 C increase in temperature. 30
31 The Collision Model Rates of reactions are affected by concentration and temp. An explanation is provided by the collision model, based on kinetic-molecular theory. In order for molecules to react, they must collide. The greater the number of collisions, the faster the rate. The more molecules present, the greater the probability of collision and the faster the rate. Thus, reaction rate should increase with an increase in the concentration of reactant molecules. The higher the temperature, the more energy available to the molecules and the more frequently the molecules collide. Thus, reaction rate should increase with an increase in temperature. However, not all collisions lead to products. In fact, only a small fraction of collisions lead to products. In order for a reaction to occur, the reactant molecules must collide in the correct orientation and with enough energy to form products. The Collision Model The Orientation Factor Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. Cl + NOCl NO + Cl 2 32
Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy, E a. In order to form products, bonds must be broken in the reactants. Bond breakage requires energy. Molecules moving too slowly, with too little kinetic energy, don t react when they collide. 33 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(e a ) barrier. 34
Sample Exercise 14.10 Relating Energy Profiles to Activation Energies and Speeds of Reaction Consider a series of reactions having these energy profiles: Rank the reactions from slowest to fastest assuming that they have nearly the same value for the frequency factor A. Solution The lower the activation energy, the faster the reaction. The value of E does not affect the rate. Hence, the order from slowest reaction to fastest is 2 < 3 < 1. Practice Exercise Rank the reverse reactions from slowest to fastest. Answer: 2 < 1 < 3 because, if you approach the barrier from the right, the E a values are 40 kj/mol for reverse reaction 2, 25 kj/mol for reverse reaction 1, and 15 kj/mol for reverse reaction 3. 35 Maxwell Boltzmann Distributions 36 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. As the temperature increases, the curve flattens and broadens. Thus, at higher temperatures, a larger population of molecules has higher energy.
Maxwell Boltzmann Distributions This fraction of molecules can be found through the expression f = e E a/rt 37 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. Arrhenius Equation Arrhenius discovered that most reaction-rate data obeyed an equation based on three factors: The number of collisions per unit time. The fraction of collisions that occur with the correct orientation. The fraction of the colliding molecules that have an energy equal to or greater than Ea. Svante Arrhenius developed a mathematical relationship between k and E a : 38 k=ae -E a/rt where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. Both A and Ea are specific to a given reaction.
Determining the Activation Energy ln k = ( ) + ln A or k=ae -E a/rt k ln k 2 E a R 1 T E a 1 R T2 T1 1 1 39 Therefore, if k is determined experimentally at several temperatures, E a 1 can be calculated from the slope of a plot of ln k vs.. T 40 14.6 Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. The balanced chemical equation provides information about substances present at the beginning and end of the reaction. The reaction mechanism is the process by which the reaction occurs. Mechanisms provide a picture of which bonds are broken and formed during the course of a reaction. Each of these processes is known as an elementary reaction or elementary process.
Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. 41 Multistep Mechanisms 42 A multistep mechanism consists of a sequence of elementary steps. The elementary steps must add to give the balanced chemical equation. Some multistep mechanisms will include intermediates. These are species that appear in an elementary step but are neither a reactant nor product. Intermediates are formed in one elementary step and consumed in another. They are not found in the balanced equation for the overall reaction. Intermediates are NOT the same as transition states.
Rate Laws of Elementary Reactions The rate laws of the elementary steps determine the overall rate law of the reaction. The rate law of an elementary step is determined by its molecularly. Unimolecular processes are first order. Bimolecular processes are second order. Termolecular processes are third order. 43 The Rate-Determining Step for a 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. 44
Slow Initial Step NO 2 (g) + CO(g) NO(g) + CO 2 (g) The rate law for this reaction is found experimentally to be Rate = k[no 2 ] 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. 45 Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2 NO 3 + NO (slow) Step 2: NO 3 + CO NO 2 + CO 2 (fast) The NO 3 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. 46
47 Fast Initial Step 2 NO(g) + Br 2 (g) 2 NOBr(g) The rate law for this reaction is found to be Rate = k[no] 2 [Br 2 ] Because termolecular processes are rare, this rate law suggests a two-step mechanism. A proposed mechanism is Step 1: NO + Br 2 NOBr 2 (fast) Step 2: NOBr 2 + NO 2 NOBr (slow) Step 1 includes the forward and reverse reactions. Fast Initial Step Step 1: NO + Br 2 NOBr 2 (fast) Step 2: NOBr 2 + NO 2 NOBr (slow) rate rate 2 Rate = k 2 [NOBr 2 ] [NO] 48 From step 1 (equilibrium) rate f = rate r k 1 [NO] [Br 2 ] = k 1 [NOBr 2 ] k 1 k 1 [NO] [Br 2 ] = [NOBr 2 ] k 2 k 1 Rate = k [NO] [Br 2 ] [NO] = k[no] 2 [Br 2 ] 1
Sample Exercise 14.14 Determining the Rate Law for a Multistep Mechanism The decomposition of nitrous oxide, N 2 O, is believed to occur by a two-step mechanism: N 2 O(g) N 2 (g) + O(g) (slow) N 2 O(g) + O(g) N 2 (g) + O 2 (g) (fast) (a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction. Solution (a) Adding the two elementary reactions gives 2 N 2 O(g) + O(g) 2 N 2 (g) + 2 O 2 (g) + O(g) Omitting the intermediate, O(g), which occurs on both sides of the equation, gives the overall reaction: 2 N 2 O(g) 2 N 2 (g) + O 2 (g) (b) The rate law for the overall reaction is just the rate law for the slow, rate-determining elementary reaction. Because that slow step is a unimolecular elementary reaction, the rate law is first order: Rate = k[n 2 O] 49 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: Solution The second step is rate determining, so the overall rate is Rate = k 2 [N 2 O 2 ][Br 2 ] We solve for the concentration of the intermediate N 2 O 2 by assuming that an equilibrium is established in step 1; thus, the rates of the forward and reverse reactions in step 1 are equal: k 1 [NO] 2 = k -1 [N 2 O 2 ] 50
14.7 Catalysts A catalyst is a substance that changes the rate of a chemical reaction without itself undergoing a permanent chemical change in the process. Catalysts change the mechanism by which the process occurs. 51 Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. catalysts can operate by increasing the number of effective collisions: increase k by increasing A or decreasing E a. When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction. 52
Homogeneous Catalysis 53 A homogeneous catalyst is one that is present in the same phase as the reacting molecules. For example, hydrogen peroxide decomposes very slowly in the absence of a catalyst: 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) In the presence of bromide ion, the decomposition occurs rapidly in acidic solution: 2Br (aq) + H 2 O 2 (aq) + 2H + (aq) Br 2 (aq) + 2H 2 O(l) Br 2 (aq) + H 2 O 2 (aq) 2Br (aq) + 2H + (aq) + O 2 (g) Br is a catalyst because it is regenerated at the end of the reaction. The net reaction is still: 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) Heterogeneous Catalysis 54 A heterogeneous catalyst exists in a different phase than the reactants. a solid catalyst in contact with gaseous reactants and gaseous products or with reactants in a liquid. Many industrial catalysts are heterogeneous. The first step is adsorption Adsorption occurs due to the high reactivity of atoms or ions on the surface of the solid. cf) Absorption Molecules are adsorbed onto the catalyst surface. The number of active sites on a given amount of catalyst depends on several factors such as: the nature of the catalyst. how the catalyst was prepared. how the catalyst was treated prior to use.
One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. Catalysts C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) H =137 kj/mol 55 Catalysts CH( g) H( g) CH( g) Pt 2 2 2 3 3 56
Heterogeneous Catalysis 57 C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) H = 137 kj/mol The reaction is slow in the absence of a catalyst. In the presence of a finely divided metal catalyst (Ni, Pt, or Pd) the reaction occurs quickly at room temperature. First, the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface. The H H bond breaks and the H atoms migrate about the metal surface. When an H atom collides with an ethylene molecule on the surface, the C C (π) bond breaks and a C H bond forms. An ethyl group, C 2 H 5, is weakly bonded to the metal surface with a metal-carbon bond. When C 2 H 6 forms, it desorbs from the surface. When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds. The activation energy for the reaction is lowered : Reaction rate is increased. Enzymes Enzymes are biological catalysts. Most enzymes are large protein molecules : 10 4 to 10 6 amu. Enzymes are capable of catalyzing very specific reactions. For example, catalase is an enzyme found in blood and liver cells. It catalyzes the decomposition of hydrogen peroxide: 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) This reaction is important in removing peroxide, a potentially harmful oxidizing agent. 58
Enzymes 59 Enzymes are catalysts in biological systems. The substrate fits into the active site of the enzyme much like a key fits into a lock. lock-and-key model 60 Enzyme Inhibitors If a molecule binds so tightly to an enzyme that substrate molecules cannot displace it, then the active site is blocked and the catalyst is inhibited. Such molecules are called enzyme inhibitors. Many poisons act by binding to the active site, blocking the binding of substrates. Some poisons bind to other locations on the enzyme. Binding ultimately causes a change in the enzyme that interferes with enzyme activity. Enzymes are extremely efficient catalysts. The number of individual catalytic events occurring at an active site per unit time is called the turnover number. Large turnover numbers correspond to very low Ea values. For enzymes, turnover numbers are very large (typically 10 1077 per 33 second).
Nitrogen Fixation and Nitrogenase Nitrogen gas cannot be used in the soil for plants or animals. The fixed nitrogen (NH 3, NO 2, and NO 3 ) is consumed by plants and then eaten by animals. 61 Problems 10, 26, 34, 51, 64, 80, 86, 118 62