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PowerPoint Lecture Presentation by J. David Robertson University of Missouri Chemical Kinetics Chapter 13 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chemical Kinetics a study of the rates of chemical reactions Basically….How fast does a chemical reaction occur? Rates of Chemical Reactions Rate of a chemical reaction refers to the change in concentration of a substance per unit of time Let’s consider the rate at which you give me 3 dollars…. your 3 dollars  my 3 dollars Let’s say that it took you 5 seconds to give it to me. reactants products your 3 dollars  my 3 dollars What is the rate of the reaction with respect to me [products]? Rate = change in concentration of money change in time Remember, change () is always [ final – initial] Positive because I am the product which gains the money Rate = + [3-0 dollars] [5-0 secs] Rate = 0.6 dollars/sec What is the rate with respect to you? Negative because you are the reactant and you are losing money Rate of reaction = - [0-3 dollars] [5-0 secs] Rate of reaction = 0.6 dollars/sec ***Therefore, you can determine the rate of reaction either by using the reactants or the products. It will give you the same rate of reaction**** Let’s consider the rates for chemical reaction NO(g) + ½ O2 (g)  NO2(g) Rate of the disappearance of NO: Rate = -[NO] t Rate of the disappearance of O2: Rate = -[O2] t Rate of the appearance of NO2: Rate = +[NO2] t A B time [A] rate = t [B] rate = t 13.1 If you know the rate of one species in a reaction you can determine the rate of any other species by the mole to mole ratio. Consider the previous reaction: NO(g) + ½ O2 (g)  NO2(g) The nitrogen dioxide increases at a rate of 0.0396 mol/L •s. What is the rate of oxygen decreasing? 0.0396 mol NO2 x L • sec ½ mol O2 = 1 mol NO2 0.0198 mol O2/L • s NO(g) + ½ O2 (g)  NO2(g) To determine the rate of reaction, you take the rate that you are given and divide by the stoichiometry in the balance reaction. The rate of disappearance of O2 is 0.0198 mol/L •s. What is the rate of reaction? 0.0198 mol O2 = ½ L•s 0.0396 mol/L• s Rate of the reaction Test Your Skill: For the reaction: 2NO(g) + Cl2(g)  2NOCl(g) The NOCl concentration increases at a rate of 0.030 mol/L•s under a particular set of conditions. Calculate the rate of disappearance of chlorine at this time and the rate of reaction. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g) slope of tangent [Br2] average rate = =t slope of tangent slope of tangent [Br2]final – [Br2]initial tfinal - tinitial instantaneous rate = rate for specific instance in time 13.1 rate  [Br2] rate = k [Br2] rate = rate constant k= [Br2] = 3.50 x 10-3 s-1 13.1 H2(g) + I2(g)  2HI(g) Concentration (mol/L) Rate of HI Formation 1 0.8 0.6 0.4 0.2 0 0 50 100 150 Time 200 250 Using the previous graph, answer these questions: (a) Write an expression for the rate of reaction in terms a changing concentration (b) Calculate the average rate of reaction between 20 and 60 s (c) Calculate the instantaneous rate of reaction after 40 s (d) Calculate the initial rate of reaction (e) Calculate the instantaneous rate of consumption of hydrogen 60 s after the start of the reaction H2(g) + I2(g)  2HI(g) Concentration (mol/L) Rate of HI Formation 1 0.8 0.6 0.4 0.2 0 0 50 100 150 Time 200 250 Reaction Conditions and Rates Several factors affect the speed of a reaction are: Concentration of reactants Temperature Catalyst Effect of Concentration on Reaction Rates The best way to describe how concentration of reactants affect rates is to use a rate law Consider the reaction: A+BC Rate constant Rate = k[A]x[B]y x and y are NOT necessarily the from the stiochiometry in the reaction The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A]x[B]y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall 13.2 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O2 (g) 2NO2 (g) N2O2 is detected during the reaction! Elementary step: NO + NO N 2 O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 13.5 Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. Elementary step: NO + NO N 2 O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 13.5 The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: Step 1: NO2 + NO2 NO + NO3 Step 2: NO3 + CO NO2 + CO2 What is the equation for the overall reaction? NO2+ CO NO + CO2 What is the intermediate? NO3 What can you say about the relative rates of steps 1 and 2? rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2 13.5 Pg 647A Pg 647B Mechanism I Step I-1 Step I-2 NO2  NO + O O + NO2  O2 + NO Overall reaction: 2NO2  2NO + O2 (rate determining) (fast) Mechanism II Step II-1 Step II-2 2NO2  NO3 + NO (rate determining) NO3  O2 + NO (fast) Overall reaction: 2NO2  2NO + O2 A rate law equation may be written directly from the balanced reaction if one of the following are noted: • rate-determining step • one step reaction • slow step If one of the above are mentioned, the rate law is written using just the reactants of that reaction. Mechanism I Step I-1 Step I-2 NO2  NO + O O + NO2  O2 + NO Overall reaction: 2NO2  2NO + O2 Rate = k[NO2] (rate determining) (fast) Mechanism II Step II-1 Step II-2 2NO2  NO3 + NO (rate determining) NO3  O2 + NO (fast) Overall reaction: 2NO2  2NO + O2 Rate = k[NO2]2 If a slow step is not noted, experimental data must be used to calculate the rate law. 2NO(g) + Cl2(g)  2NOCl Exp 1 2 3 4 [NO] 0.01 0.01 0.02 0.02 [Cl2] 0.01 0.02 0.02 0.02 Initial Rate of Rxn 2.5 5.1 5.2 10.3 The Sum of the Exponents in a Rate Law Gives the Overall Order of the Reaction. We will be mostly dealing with overall orders of: Zero Order First Order Second Order F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Rate doubles x=1 Quadruple [ClO2] with [F2] constant rate = k [F2][ClO2] Rate quadruples y=1 13.2 Rate Laws • Rate laws are always determined experimentally unless you have been told a rate-limiting or slow step. • Reaction order is always defined in terms of reactant (not product) concentrations. • The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2][ClO2] 1 13.2 Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Experiment [S2O82-] [I-] Initial Rate (M/s) 1 0.08 0.034 2.2 x 10-4 2 0.08 0.017 1.1 x 10-4 3 0.16 0.017 2.2 x 10-4 rate = k [S2O82-]x[I-]y y=1 x=1 rate = k [S2O82-][I-] Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) 2.2 x 10-4 M/s rate k= = = 0.08/M•s 2- [S2O8 ][I ] (0.08 M)(0.034 M) 13.2 First-Order Reactions A k= product [A] rate = t rate M/s = = 1/s or s-1 M [A] [A] = [A]0exp(-kt) rate = k [A] [A] = k [A] t [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 ln[A] = ln[A]0 - kt 13.3 Decomposition of N2O5 13.3 The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? [A]0 = 0.88 M ln[A] = ln[A]0 - kt [A] = 0.14 M kt = ln[A]0 – ln[A] ln[A]0 – ln[A] = t= k ln [A]0 [A] k ln = 0.88 M 0.14 M 2.8 x 10 s -2 -1 = 66 s 13.3 First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 ln t½ = [A]0 [A]0/2 k ln2 0.693 = = k k What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? 0.693 t½ = ln2 = = 1200 s = 20 minutes -4 -1 5.7 x 10 s k How do you know decomposition is first order? units of k (s-1) 13.3 First-order reaction A product # of half-lives 1 [A] = [A]0/n 2 4 3 8 4 16 2 13.3 13.3 Second-Order Reactions A product [A] rate = t rate M/s = k= 2 = 1/M•s 2 M [A] 1 1 = + kt [A] [A]0 rate = k [A]2 [A] = k [A]2 t [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 t½ = 1 k[A]0 13.3 Zero-Order Reactions A product [A] rate = t rate = M/s k= 0 [A] [A] = [A]0 - kt rate = k [A]0 = k [A] =k t [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 t½ = [A]0 2k 13.3 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Order 0 Concentration-Time Equation Rate Law [A] = [A]0 - kt rate = k 1 rate = k [A] ln[A] = ln[A]0 - kt 2 rate = k [A] 1 1 = + kt [A] [A]0 2 Half-Life t½ = [A]0 2k t½ = ln2 k 1 t½ = k[A]0 13.3 How temperature and Catalysts Affect Reaction Rates Consider Arrhenius’ Equation k= A e-Ea/RT Frequency of collisions Temperature Activation energy Gas constant 8.314 J/mol • K The A term in the Arrenhius Equation In order for molecules to react, they must collide. Collision Theory 1. The reaction molecules must collide with one another. 2. The reaction molecules must collide with sufficient energy. 3. The molecules must collide in an orientation that can lead to rearrangement of the atoms. Effect of Concentration on the frequency of collisions Therefore, increase the number of collisions, increase the rate Effect of Temperature k= A e-Ea/RT The larger the temperature, the smaller the negative term. The smaller negative term, the larger the value once you have applied ex. Therefore, increase temperature, increase reaction rate. Activation Energy, Ea Ea A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A • exp( -Ea/RT ) Ea uncatalyzed k catalyzed ratecatalyzed > rateuncatalyzed Ea‘ < Ea 13.6 k= A e-Ea/RT Take the ln of both sides. ln k = ln A – Ea (1/T) R y b m x If we were going to graph this, what would we graph? ln k Slope = -Ea/R 1/T If we have two different k values at different temperatures, you can combine the two reactions to give: ln k2 = -Ea 1 k1 R T2 1 T1 Effect of a Catalyst Homogeneous catalyst: a catalyst is present in the same phase as the reacting substances Heterogeneous catalyst: a catalyst is present in a different phase as the reaction substances In heterogeneous catalysis, the reactants and the catalysts are in different phases. • Haber synthesis of ammonia • Ostwald process for the production of nitric acid • Catalytic converters In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. • Acid catalysis • Base catalysis 13.6 Haber Process N2 (g) + 3H2 (g) Fe/Al2O3/K2O catalyst 2NH3 (g) 13.6 Ostwald Process 4NH3 (g) + 5O2 (g) Pt catalyst 2NO (g) + O2 (g) 2NO2 (g) + H2O (l) 4NO (g) + 6H2O (g) 2NO2 (g) HNO2 (aq) + HNO3 (aq) Pt-Rh catalysts used in Ostwald process Hot Pt wire over NH3 solution 13.6 Catalytic Converters CO + Unburned Hydrocarbons + O2 2NO + 2NO2 catalytic converter catalytic converter CO2 + H2O 2N2 + 3O2 13.6 Enzyme Catalysis 13.6