Chemical Kinetics Chapter 12 PDF

Topic 2: Chemical Kinetics Chapter 12 Instructor: Liliane Halab 1 Learning Objectives of Chapter 12: 12.1 Reaction Rates 12.2 Rate Laws: An Introduction 12.3 Determining the Form of the Rate Law 12.4 The Integrated Rate Law 12.5 Reaction Mechanism 12.6 A Mo...

Topic 2: Chemical Kinetics Chapter 12 Instructor: Liliane Halab 1 Learning Objectives of Chapter 12: 12.1 Reaction Rates 12.2 Rate Laws: An Introduction 12.3 Determining the Form of the Rate Law 12.4 The Integrated Rate Law 12.5 Reaction Mechanism 12.6 A Model for Chemical Kinetics 2  Kinetics versus Thermodynamics Thermodynamics (will see in chapter 17) tells us which direction a reaction will go (e.g. at standard condition, carbon is stable in graphite form instead of diamond). Kinetics tells us how quickly it will get there (e.g. diamond will not convert to graphite in our lifetime since this process is very slow under standard condition). Chemical Kinetics is the area in chemistry that is concerned with the speeds, or rates of reactions. Cdiamond (s) → Cgraphite (s) G°rxn = −3 kJ 3  Aims of Chemical Kinetics At the macroscopic level: To define rate of reaction, order of reaction and rate law To examine how the rates and orders are determined. At the molecular level: From experimental rate laws, to predict plausible reaction mechanisms, which is the sequence of bond-making and bond-breaking steps that occurs during the conversion of reactants to products. 4  Reaction Rates Reaction rate is the increase in molar concentration of product of a reaction per unit time or the decrease in molar concentration of reactant per unit time. The unit is usually mol/(L s) or M/s, however, can be any unit of concentration per unit of time. The average rate for a chemical reaction is the change in concentration of a substance over a period of time. The rate of reaction is always defined as a positive quantity. change in concentration average Rate = change in time Let’s consider the following rxn: A → B B at t2 − B at t1 ∆[B] Rate for formation of product B = = t2 − t1 ∆t A at t2 − A at t1 ∆[A] Rate for consumption of reactant A = − = − t2 − t1 ∆t 5  How to Determine Reaction Rates In order to experimentally determine reaction rates, we need to measure the concentrations of reactants and/or products over the course of a chemical reaction. We can design experiments that use physical properties to measure reaction rates. Physical measurements can be performed on a system while it is reacting. Examples: measurements of changes in pressure (for gases) or by measuring light absorbance of a solution (Beer’s law) when a color change is part of the reaction. 6 Expressing average reaction rates: The rate of this reaction can be found by measuring the concentration of O2 or NO2 at various times. Both concentrations increase with time. The rate could also be determined by measuring the concentration of N2O5, which would decrease over time. Δ[O2 ] Rate of formation of NO2 = Δ[NO 2 ] Rate of formation of O2 = Δt Δt Δ[N2 O 5 ] Rate of decomposition of N2O5 = − Δt We can relate these expressions by considering the reaction stoichiometry: Δ[NO 2 ] Δ[O2 ] Δ[N2 O 5 ] Δ[O2 ] = 4 − = 2 Δt Δt Δt Δt Δ[O2 ] 1 Δ[NO 2 ] 1 Δ[N2 O 5 ] Rate = = =− Δt 4 Δt 2 Δt Rate of ¼ (Rate of ½ (Rate of production = production = consumption of O2 of NO2) of N2O5) 7  Reaction Rates This can be generalized for the following : a A + bB → c C + d D 1 ∆[A] 1 ∆[B] 1 ∆[C] 1 ∆[D] Rate = − = − = = a ∆t b ∆t c ∆t d ∆t Example: If the average rate of consumption of N2O5 is 5.0  10−5 M/s , determine the: a) Rate of formation of O2: b) Rate of formation of NO2 : 8 Calculation of the average rate of formation of O2 during the decomposition of N2O5: Note that the rate is not constant but decreases with time. The average rate of formation of O2 decreases as the reaction proceeds. 9  The instantaneous rate of the reaction The instantaneous rate is the rate at a particular time, it can be obtained by computing the slope of a line tangent to the curve at that specific time. As t gets smaller and approaches zero, the hypotenuse becomes a tangent line at that point. The slope of the tangent line equals the rate at that point. In calculus, the instantaneous rate is given by the derivative. change in y Instantaneous Rate = d[O2] = slope of the tangent line = change in x dt Calculate the instantaneous rate for the formation of O2 at t = 2400 s. Here are 2 points from the tangent line obtained from the graph: (x1, y1) = (1200, 0.0040) (x2, y2) = (3400, 0.0066) 10  Conditions that affect the rate of a reaction Variables that affect the rate of reaction: 1. The concentrations of the reactants: Often the rate of reaction increases when the concentration of a reactant is increased. In some reactions, however, the rate is unaffected by the concentration of a particular reactant, as long as it is present at some concentration (Rate Laws). 2. The temperature at which the reaction occurs: Usually reactions speed up when the temperature increases (Arrhenius Equation). 3. Catalyst is a substance that increases the rate of reaction without being consumed in the overall reaction (mechanism). 4. The surface area of the solid reactant or catalyst: If a reaction involves a solid with a gas or liquid, then the surface area of the solid affects the reaction rate. Because the reaction occurs at the surface of the solid, the rate increases with increasing surface area. 11  Rate Laws There are two forms of the Rate Law: 1. Differential Rate Law: “The Rate Law” Expresses how the reaction rate changes with concentration for each reactant We can derive it using the method of initial rates (experimentally) 2. Integrated Rate Law Expresses how the concentration of reactant varies with time We can derive it: By calculus: integrate the differential rate law (integration) By experiment: plot of the concentration versus time 12 The Rate Law: The effect of Concentration on Reaction Rate The rate of a reaction often depends on the concentration of one or more of the reactants. As long as the reverse reaction is negligibly slow, we can write a relationship between the rate of the reaction and the concentration of the reactants called the rate law (also called differential rate law). Consider the general rxn where the rate is determined based on the consumption of the reactant A : a A + bB → c C + d D ∆A Rate = − = k [A]m [B]n ∆t k is a proportionality constant called rate constant. It is specific to each reaction at a given temperature and is determined experimentally, k has units that vary depending on the overall order. The larger the value of k, the faster a reaction goes. n and m are the order of the reactants, it can be an integer (including 0) or a fraction, they are determined experimentally. We say that the reaction is mth order to respect to reactant A and nth order to respect to reactant B. The overall order of a reaction is the sum of the orders of the reactant species from the experimentally determined rate law, the rxn has an overall order of m+n. 13 The Rate Law: The effect of Concentration on Reaction Rate Consider the general rxn: a A + bB → c C + d D Rate = k [A]m [B]n If m = 0 and n = 0, the reaction is zero order and the rate is independent of the concentration of A and B ([A]0 = 1 and [B]0 = 1, so the rate is equal to k) If m = 1 and n = 0, the reaction is first order and the rate is directly proportional to [A] If m = 1 and n = 1, the reaction is second order and the rate is dependent on [A] and [B] If m = 2 and n = 0, the reaction is second order and the rate is proportional to the square of [A] And so on… The rate constant will have different units depending on the overall order of the reaction. Write units for k: 14  Example: Determining order of reaction Hydrogen peroxide oxidizes iodide ion in acidic solution: H2O2(aq) + 3I−(aq) + 2H+(aq) → I3−(aq) + 2H2O(l) The rate law for the reaction is Rate = k [H2O2]2 [I−] a. What is the order of reaction with respect to each reactant species? b. What is the overall order? c. What is the units for k? 15  Determining the Rate Law: Method of Initial Rates Method of initial rates is a method for determining experimentally the form of the rate law for a reaction, the orders of each reactant and the rate constant k. The initial rate of a reaction is the instantaneous rate determined just after the reaction begins, before the initial concentration of reactants have changed significantly, essentially t = 0 (where the reverse reaction is negligible). The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants. 16 Determining the Rate Law: Method of Initial Rates Determine the orders of both reactants, the overall order and the value for the rate constant for the following reaction using the data from the table: NH4+ (aq) + NO2‒ (aq) → N2 (g) + 2 H2O (l) Experiment Initial [NH4+] Initial [NO2‒] Initial rate 1 0.100 M 0.0050 M 1.35 x 10-7 M/s 2 0.100 M 0.010 M 2.70 x 10-7 M/s 3 0.200 M 0.010 M 5.40 x 10-7 M/s 1) Determine the orders for each reactant: Rate = k [NH4+]m [NO2−]n For experiment 1: 1.35 x 10-7 M/s = k (0.100 M)m (0.0050 M)n For experiment 2: 2.70 x 10-7 M/s = k (0.100 M)m (0.010 M)n For experiment 3: 5.40 x 10-7 M/s = k (0.200 M)m (0.010 M)n To determine the value of m, we choose the 2 To determine the value of n, we choose the 2 experiments where only [NH4+] changes experiments where only [NO2−] changes Rate 2 = 2.70 x 10-7 M/s = k (0.100 M)m (0.010 M)n Rate 2 = 2.70 x 10-7 M/s = k (0.100 M)m (0.010 M)n Rate 3 5.40 x 10-7 M/s k (0.200 M)m (0.010 M)n Rate 1 1.35 x 10-7 M/s k (0.100 M)m (0.0050 M)n 17 Determining the Rate Law: Method of Initial Rates Determine the orders of both reactants, the overall order and the value for the rate constant for the following reaction using the data from the table: NH4+ (aq) + NO2‒ (aq) → N2 (g) + 2 H2O (l) Experiment Initial [NH4+] Initial [NO2‒] Initial rate Rate constant 1 0.100 M 0.0050 M 1.35 x 10-7 M/s 2.70 x 10-4 M-1·s-1 2 0.100 M 0.010 M 2.70 x 10-7 M/s 2.70 x 10-4 M-1·s-1 3 0.200 M 0.010 M 5.40 x 10-7 M/s 2.70 x 10-4 M-1·s-1 2) Write the rate law: 3) Determine the rate constant : 18 Exercise: The initial rate of the reaction between bromate ions and bromide ions in acidic solution was measured for various initial concentrations of the reactants. BrO3− (aq) + 5 Br− (aq) + 6 H+ (aq) → 3 Br2 (l) + 3 H2O (l) Experiment [BrO3−]0 [Br−]0 [H+]0 Initial rate for the formation of Br2 1 0.10 M 0.10 M 0.10 M 8.0  10−4 M/s 2 0.20 M 0.10 M 0.10 M 1.6  10−3 M/s 3 0.20 M 0.20 M 0.10 M 3.2  10−3 M/s 4 0.10 M 0.10 M 0.20 M 3.2  10−3 M/s a) Determine the rate Law (equation) for this reaction. b) Determine the rate constant for this reaction at this temperature. 19 Integrated Rate Laws: Concentration-Time Relationship A rate law (differential rate law) tells us how the rate of a reaction depends on reactant concentration at a particular moment. The integrated rate law of a reaction is a mathematical relationship showing how a reactant concentration changes over a period of time. The integrated rate law depends on the order of the reaction. How to derive the integrated rate law : By calculus: integrate the differential rate law (integration) By experimental: plot of the concentration versus time 20  The Integrated Rate Law: Zero Order Reaction A → products Rate = − d[A] = k [A]0 = k dt Integrated Rate Law: [A]t = −k t + [A]0 equation of straight line: y = m x + b [A]t = concentration of A at time t k = rate constant t = time [A]0 = initial concentration of A at t = 0 In a zero order reaction: The rate of the reaction is equal to a constant, k, and remains constant throughout the reaction. The units for k are the same as the units of the rate of reaction. The concentration-time plot is a straight line with a negative slope. The slope is equal to −k. 21  The Integrated Rate Law: First Order Reaction A → products Rate = − d[A] = k [A] dt [A]t Integrated Rate Law: ln[A]t = −k t + ln[A]0 or ln = −k t [A]0 equation of straight line: y = mx +b In a first order reaction: The rate of the reaction is equal to the product of k and [A]. [A] The units for k are (time)−1 such as s−1 The concentration-time plot is a straight line with a negative slope. The slope is equal to −k. 22  The Integrated Rate Law: Second Order Reaction A → products Rate = − d[A] = k [A]2 dt 1 1 Integrated Rate Law: =kt+ [A]t [A]0 equation of straight line: y = mx +b In a second order reaction: [A] The rate of the reaction is equal to the product of k and [A]2. The units for k are M−1 s−1 The concentration-time plot is a straight line with a positive slope. The slope is equal to k. 23 Experimentally determining the Integrated Rate Law Plots of concentration and time can also be used to determine the order of reaction. To experimentally determine the integrated rate law for a reaction, concentrations are measured at various values of time as the reaction proceeds. Then, the data is plotted based on which integrated rate law gives a straight line. For zero-order reactions, [A] versus t is linear. For first-order reactions, ln[A] versus t is linear. For second-order reactions, 1/[A] versus t is linear. 24  The Half-life of a Reaction, t1/2 The half-life of a reaction is the time required for a reactant to reach half its initial concentration. It is designated by the symbol t1/2 It is a convenient way to describe the rate at which a reactant is consumed in a chemical reaction: the longer the half-life, the slower the reaction. Successive half-lives can also be determined. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Half-life is used in chemistry (radioactive decay) and medicine (drug stability and elimination) to predict the stability of substances. 25  The Half-Life of the Different Orders of Reactions The half-life will vary between different type of reactions with different orders. To obtain the equation for the half-life for each reaction order: Replace t in the integrated rate Laws by t½ and [A]t by ½ [A]0 To determine a half-life, t½, we need to know: The order of the reaction or enough information to determine it. The rate constant, k, for the reaction or enough information to determine it. In some cases, we need to know the initial concentration, [A]0 Substitute this information into the equation for the half-life of a reaction with the correct order and solve for t½. 26  The Half-Life of Zero Order Reaction Half-life depends on the initial concentration of the reactant. Each successive t1/2 is half the preceding one. Half-life gets shorter as the reaction progresses and the concentration of reactants decrease.  A 0 t1 = 2 2k Derive the equation: Integrated Rate Law: [A]t = −k t + [A]0 27  The Half-Life of First Order Reaction Half-life does not depend on the concentration of reactants. Each successive t1/2 is equal to the preceding one. 0.693 t1 = 2 k Derive the equation: [A]t Integrated Rate Law: ln = −k t [A]0 28  The Half-Life of Second Order Reaction Half-life depends on the initial concentration of the reactant. Each successive t1/2 is double the preceding one. Half-life gets longer as the reaction progresses, and the concentration of reactants decrease. 1 t1 = 2 k  A 0 Derive the equation: 1 1 Integrated Rate Law: =kt+ [A]t [A]0 29 Summary of the Rate Laws 30  Radioactive decay Radioactive decay is the disintegration of an unstable atom to a more stable one (different nucleus) with emission of radiation. The decay of radioactive elements occurs by first order kinetics. The half-life of a radioisotope is the time required for half of the amount of unstable material to degrade into a more stable material. Example: Americium is used in smoke detectors and in medicine. One isotope of americium, 241Am, has a rate constant, k, for radioactive decay of 0.0016 year-1. In contrast radioactive iodine-125, used for studies of thyroid functioning, has a rate constant for decay of 0.011 day-1 a) What are the half-lives of these isotopes? And which one decays faster? b) If you are given a dose of iodine-125 containing 1.6 x 1015 atoms, how many atoms remain after 7 days? 31 Example: If the initial concentration of a particular drug in the patient’s body decreases from 5.00 × 10‒4 M to 2.50 × 10‒4 M in 2.00 hours and from 2.50 × 10‒4 M to 1.25 × 10‒4 M in 1.00 hour, what is the rate constant for the elimination process of this drug? How many hours will it take for 90.0% of the initial concentration of the drug to be eliminated from the body? 32  Reaction Mechanisms and Elementary Steps Reaction mechanism can have one or more steps of bond-making and bond- breaking steps that occurs during the conversion of reactants to products. Each step in a reaction mechanism is called an elementary step. An intermediate is a species produced during a reaction that does not appear in the net equation because it reacts in the subsequent elementary step in the mechanism. Example: H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g) rate = k[H2][ICl] (rate law determined from experiment) The following reaction mechanism is proposed: (1) H2 (g) + ICl (g) → HI (g) + HCl (g) (2) HI (g) + ICl (g) → I2 (g) + HCl (g) 33  Rate Laws for Elementary Steps Although we cannot deduce the rate law for an overall chemical reaction from the balanced equation, we can deduce the rate law for an elementary step by the stoichiometric coefficients in the equation for that elementary step. Elementary steps are characterized by their molecularity, the number of reactant particles involved in the step. Unimolecular – reaction involving one molecule in the reactant; first order. Bimolecular – reaction involving the collision of two species; second order. Termolecular (rare step) – reaction involving the collision of three species; third order. 34  Reaction Mechanism and Rate-Determining Step Multistep reactions often have one elementary step that is much slower than the others, this step is called the rate-determining step. The rate-determining step (slow step) in a reaction mechanism limits the overall rate of the reaction. A rxn is only as fast as its slowest step. Therefore, it is the slow step that is used to write the rate law for the overall reaction based on the molecularity of this elementary step. Requirement for a valid reaction mechanism (one that is plausible): 1) Elementary steps in the mechanism must add to the overall balanced equation 2) Mechanism (rate law for the rate-determining step) must agree with experimentally determined rate law. 35  Mechanism with Slow First Elementary Step Example: Is the proposed mechanism plausible for the given reaction? H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g) rate = k[H2][ICl] (experimentally determined rate law) The following mechanism is proposed where k1 < k2: k1 (1) H2 (g) + ICl (g) → HI (g) + HCl (g) slow step (rate-determining step) k2 (2) HI (g) + ICl (g) → I2 (g) + HCl (g) 36  Collision Model Collision model is a model used to explain the observed temperature dependence of reaction rates. For a reaction to take place, the reacting molecules must collide with each other. Once molecules collide, they may react together or they may not, depending on two factors: whether the reacting molecules collide in the proper orientation for new bonds to form and whether the collision has enough energy to break the bonds in the reactants and form the products The collision model considers that such energy is provided by kinetic energies of molecules upon their collisions. 37  Collision Model and Kinetic Energy The collision frequency is typically 1030 collisions per second for molecules in gas phase, however only 1 of 1010 collisions leads to a reaction. The rate of reaction is limited by the orientation of molecules at the time of their collision and their kinetic energy. The collision must involve enough energy to produce the reaction. Which collisions can lead to the reaction: 2 BrNO (g) → 2 NO (g) + Br2 38  Activation Energy For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy equal or exceeds the activation energy (Ea). Activation energy (Ea) is the energy that must be overcome to produce a chemical reaction. To get from the reactant to the product, the molecules must go through a high- energy state called the transition state or activated complex. The transition state or activated complex is a state with partially broken and partially formed bonds. Energy diagram for reaction with one step mechanism 2 BrNO (g) → 2 NO (g) + Br2 39 Energy Diagram for Reaction with a Two Step Mechanism NO2(g) + CO(g) → NO(g) + CO2(g) Rateexperimental = k[NO2]2 Proposed mechanism: 1. NO2(g) + NO2(g) → NO3(g) + NO(g) Slow 2. NO3(g) + CO(g) → NO2(g) + CO2(g) Fast 40  Catalysts A catalyst is a substance that is added to a reaction to lower the activation energy and increase the rate of the reaction without being consumed itself. A catalyst is not consumed in a reaction. Rather, it is present in the beginning, is used in one step, and is produced again in a subsequent step. As a result, the catalyst does not appear in the overall reaction equation. 41  Effect of Temperature If the minimum energy required for a reaction to occur (activation energy) is indicated by the black dotted lines in the graph, then the fraction of molecules possessing this energy will be greater at T2 than at T1. 42  Effect of Temperature on Reaction Rate The temperature dependence of the reaction rate is contained in the rate constant, k, which varies with the temperature. An increase in temperature results in an increase in k which results in an increase of the rate of reaction. Arrhenius equation illustrates the relationship between the rate constant and the temperature. All rate constants show an exponential increase with absolute temperature. Arrhenius Equation: Ea  1    + ln ( A ) − Ea / RT k = Ae or ln(k ) = − R T A = frequency factor is a constant that represents the collision frequency and the fraction of collisions with effective orientation for a given reaction (same units as k) Ea = activation energy (J/mol) for a given reaction R = gas constant (8.314 J/K·mol) T = temperature (in K) 43  Determining the Activation Energy From Arrhenius Equation One procedure for finding Ea for a reaction is to measure the rate constant at different temperatures and plot ln k versus 1/T. The plot of ln k versus 1/T will give a straight line with a slope of −Ea/R. Ea  1  ln(k ) = −   + ln ( A ) R T equation of straight line: y = m x +b Calculate the activation energy for the reaction using the following information: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) 44  Temperature Effect on Rate Constant Arrhenius Equation can relate rate constants at 2 different temperatures. Ea can be calculated from the values of k at two different temperatures using the equation below. It also allows finding k for any T if Ea and k for one specific T is known. 𝑘1 Ea 1 1 T in Kelvin ln = − R = 8.314 J/K-mol 𝑘2 R T2 T1 Ea in J/mol Derive the equation: Ea 1 Ea 1 ln 𝑘1 = − + ln 𝐴 ln 𝑘2 = − + ln 𝐴 R T1 R T2 ln k1 − ln k2 = − Ea 1 E + ln 𝐴 − − a 1 + ln 𝐴 R T1 R T2 Ea 1 1 ln k1 − ln k2 = − R T 2 T1 𝑘1 Ea 1 1 ln = − 𝑘2 R T2 T1 45  Temperature Effect on Rate Constant Example: A first order reaction has rate constants of 4.6 x 10-2 s-1 and 8.1 x 10-2 s-1 at 0°C and 20°C respectively. What is the value of the activation energy? 46  Suggested Readings-Problems Topic 2 Read Chapter 12 (sections 12.1-12.6) from Zumdahl (same for all editions) Work on suggested problems: from Zumdahl, 11th edition: Reaction rates: 31, 35-36 Initial rates Method: 39, 43, 106 Integrated Rate Law: 51, 53, 57, 59 Half-lives: 61, 63, 65, 67, 105 Reaction mechanisms: 75, 77-78 Activation energy: 83, 138 temperature: 88-89, 91, 117 Or from Zumdahl, 10th edition: Or from Zumdahl, 9th edition: Reaction rates: 25, 27-28 Reaction rates: 23, 25-26 Initial rates Method: 31, 35, 88 Initial rates Method: 29, 33, 84 Integrated Rate Law: 41, 43, 47, 49 Integrated Rate Law: 39, 41, 45, 47 Half-lives: 51, 53, 55, 57, 87 Half-lives: 49, 51, 53, 55, also 55 from 10th edition Reaction mechanisms: 61, 63-64 Reaction mechanisms: 59, 61-62 Activation energy: 67, 116 Activation energy: 63, 106 temperature: 72-73, 75, 95 temperature: 68-69, 71, 89 Answers to the even numbers from the 11th edition: 36: 2A + 3B → 4C ; 78: rate = rate of step1 (slow step) = k[NO2]2 ; NO2 + CO → NO + CO2 ; 88: a) 91.5 kJ/mol b) 3.54 × 1014 s−1 c) 3.30 × 10−2 s−1 106: Rate = k[H2SeO3][H+]2[I−]3 ; k = 5.19 × 105 L5/mol5 s ; 138: e. the second step has the largest activation energy and will be the rate-determining step (the slow step).