OCR CHEF0314 Topic 3 Reaction Kinetics PDF

(CHEF0314) TOPIC 3 REACTION KINETICS  3.0 Reaction kinetics 3.1 Rate expression, rate equations/rate law, order of reactions and rate constants. 3.2 Effect of concentration of reactant on rate 3.3 Effect of temperature on rate 3.4 Effect of presence of catal...

(CHEF0314) TOPIC 3 REACTION KINETICS  3.0 Reaction kinetics 3.1 Rate expression, rate equations/rate law, order of reactions and rate constants. 3.2 Effect of concentration of reactant on rate 3.3 Effect of temperature on rate 3.4 Effect of presence of catalyst on activation energy and rate constant 3.5 Reaction mechanism 16-2  3.0 Reaction kinetics Kinetics is the study of how fast chemical reactions occur. Speed of reaction is measured by the change in concentration with time. 16-3  Factors that affect Reaction Rates Particles must collide in order to react. (the basic tenet of collision theory)  Concentration of reactants  The higher the concentration of reactants, the greater the reaction rate.  A higher concentration of reactant particles allows a greater number of collisions. Physical states of reactants (same physical state or increase the surface area, increase the reaction rate) Temperature of reaction  The higher the temperature, the greater the reaction rate.  At higher temperatures particles have more energy and therefore collide more often and more effectively. Catalyst - Effect on activation energy and rate constant 16-4  3.1 Expressing The Reaction rate Reaction rates - The change in concentration of reactants or products over period of time. Rate = [ ]  t Change (): final - initial Units of rate: M/s OR mol/L.s OR Ms-1 OR mol L-1 s-1 Time can also be in minutes, hours, days or years For the general reaction A → B, there are 2 ways of measuring the rate: A is a reactant where the concentration is decreasing. change in concentration of A conc Af – conc Ai A] Rate = – =– =– change in time tf – ti t The negative sign is used to show the rate of disappearance of reactant. The rate is always a positive value. B is a product where the concentration is increasing. change in concentration of B conc Bf – conc Bi ] Rate = + =+ =+ change in time tf – ti t The positive sign is used to show the rate of appearance of product. 16-5  A graph of concentration versus time shows that as the reactants (purple A) disappear and the products (green B) appear 16-6 Figure 16.5 Three types of reaction rates for the reaction of O3 and C2H4. 16-7 16-8  Expressing Rate in Terms of Reactant and Product Concentrations In general, for the reaction aA + bB → cC + dD where a, b, c, and d are the coefficients for the balanced equation, the rate is expressed as: 1 [A] 1 [B] = 1 [C] = 1 [D] Rate = – =– a t b t c t d t 16-9  Example 2H2O2 (aq) → 2H2O (l) + O2 (g) 1 [H2O2] 1 [H2O] [O2 ] Rate = – =+ =+ 2 t 2 t t Rate of disappearance of [H2O2] is same as rate of appearance of [H2O] Rate of appearance of [O2] is one half of rate of disappearance of [H2O2 ] Rate of appearance of [H2O] is twice of rate of appearance of [O2 ] 16-10  Questions can be asked whether to write rate expression from a balanced equation OR write a balanced equation from the rate expression. 16-11  Figure 16.6A Plots of [reactant] and [product] vs. time. C2H4 + O3 → C2H4O + O2 [O2] increases just as fast as [C2H4] decreases. [C2H4] [O3] Rate = – = – t t [C2H4O] [O2] = = t t 16-12  Figure 16.6B Plots of [reactant] and [product] vs. time. H2 + I2 → 2HI [H2] [I2] [HI] Rate = – =– = 1 t t 2 t [HI] [H2] [I2] Rate = = –2 = –2 t t t Refer the graph: Rate of appearance of [HI] is twice of rate of disappearance of [H2 ] 16-13 Sample Problem 16.1 Hydrogen gas has a nonpolluting combustion product (water vapor). It is used as a fuel aboard the space shuttle and in earthbound cars with prototype engines: 2H2(g) + O2(g) → 2H2O(g) (a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time. [O2] 1 [H2] 1 [H2O] Rate = – =– = t 2 t 2 t (b) When [O2] is decreasing at 0.23 mol/L·s, at what rate is [H2O] increasing? 1 [H2O] [O2] =– = – (– 0.23 mol/L·s) 2 t t [H2O] = 2(0.23 mol/L·s) = 0.46 mol/L·s t 16-14 For a particular reaction, the reaction rate in terms of the change in concentration with time for each substance is [NH3] [O2] 1 [NO] 1 [H2O] Rate = – 1 =– 1 = + =+ 4 t 5 t 4 t 6 t Write a balanced equation for this gaseous reaction. When [NH3] is decreasing at 0.52 M/s, at what rate is [H2O] increasing? 16-15  The Rate Law/Rate Equation For any general reaction occurring at a fixed temperature aA + bB → cC + dD Rate = k[A]m[B]n The term k is the rate constant, which is specific for a given reaction at a given temperature. Rate constant, k also depends on catalyst. The exponents m and n are reaction orders and are determined by experiment ONLY - The values of m and n are not necessarily related in any way to the coefficients a and b. 16-16  Reaction Orders A reaction has an individual order “with respect to” or “in” each reactant. For the simple reaction: A → products If the rate doubles when [A] doubles, the rate depends on [A]1 and the reaction is first order in ( or with respect to) A. If the rate quadruples when [A] doubles, the rate depends on [A]2 and the reaction is second order in [A]. -A reacting is nth order if doubling the concentration causes an 2n increase in rate. If the rate does not change when [A] doubles, the rate does not depend on [A], and the reaction is zero order with respect to A. 16-17  Plots of Reactant Concentration, [A], versus Time for First-, Second-, and Zero-Order Reactions Figure 16.7 © McGraw Hill  Figure 16.8 Plots of rate vs. reactant concentration, [A], for first-, second-, and zero-order reactions. 16-19  Individual and Overall Reaction Orders For the reaction 2NO(g) + 2H2(g) → N2(g) + 2H2O(g): The rate law is: Rate = k[NO]2[H2] The reaction is second order in NO and first order in H2. The overall reaction order is third order. Note that the reaction is first order with respect to or in H2 even though the coefficient for H2 in the balanced equation is 2. Reaction orders MUST be determined from experimental data and cannot be deduced from the balanced equation. 16-20  Sample Problem 16.2 Determining Reaction Orders from Rate Laws PROBLEM: For each of the following reactions, use the given rate law to determine the reaction order with respect to each reactant and the overall order. (a) 2NO(g) + O2(g) → 2NO2(g); Rate = k[NO]2[O2] (b) H2O2(aq) + 3I–(aq) + 2H+(aq) →I3–(aq) + 2H2O(l); Rate = k[H2O2][I–] SOLUTION: (a) The exponent of [NO] is 2 and the exponent of [O2] is 1, so the reaction is second order with respect to NO, first order with respect to O2 and third order overall. (b) The reaction is first order in H2O2, first order in I-, and second order overall. The reactant H+ does not appear in the rate law, so the reaction is zero order with respect to H+. 16-21  Using Initial Rates Data to Determine Rate Laws For the general reaction A + 2B → C + D, the rate law will have the form Rate = k[A]m[B]n To determine the values of m and n, we run a series of experiments in which one reactant concentration changes while the other is kept constant, and we measure the effect on the initial rate in each case. 16-22  Rate = k[A]m[B]n Table 16.1 Initial Rates for the Reaction between A and B Initial Rate Initial [A] Initial [B] Experiment (mol/L·s) (mol/L) (mol/L) 1 1.75x10-3 2.50x10-2 3.00x10-2 2 3.50x10-3 5.00x10-2 3.00x10-2 3 3.50x10-3 2.50x10-2 6.00x10-2 4 7.00x10-3 5.00x10-2 6.00x10-2 [B] is kept constant for experiments 1 and 2, while [A] is doubled. Then [A] is kept constant while [B] is doubled. - Determined quantitatively or qualitatively (depends on question asked.) 16-23  Finding m, the order with respect to A: We compare experiments 1 and 2, where [B] is kept constant but [A] doubles: Rate 2 k[A] m [B] n 2 2 = Rate 1 k[A] m 1 [B] n1 m n 3.50x10-3 mol/L·s k 5.00x10-2 mol/L 3.00x10-2 mol/L = 1.75x10-3mol/L·s -2 k 2.50x10 mol/L 3.00x10-2 mol/L 2.00 = (2.00)m m = 1, Since m = 1, so the reaction is first order in A. 16-24  Finding n, the order with respect to B: We compare experiments 3 and 1, where [A] is kept constant but [B] doubles: Rate 3 k[A] m [B] n 3 3 = Rate 1 k[A] m 1 [B] n1 m n 3.50x10-3 mol/L·s 2.50x10-2 mol/L =k 6.00x10-2 mol/L 1.75x10-3mol/L·s k 2.50x10 -2 mol/L 3.00x10-2 mol/L 2.00 = (2.00)n n=1 Since n = 1, so the reaction is first order in B. The rate law is given by: Rate = k[A][B] Overall order is second order 16-25  Determine Order of Reaction Qualitatively For experiments 1 and 2: At constant [B], while doubling the [A], the rate is also doubled. Hence the order of reaction for reactant A is first order. For experiments 1 and 3: At constant [A], while doubling the [B], the rate is also doubled. Hence the order of reaction for reactant B is first order. The rate law is given by: Rate = k[A][B] 16-26  Table 16.2 Initial Rates for the Reaction between O2 and NO O2(g) + 2NO(g) → 2NO2(g) Rate = k[O2]m[NO]n Initial Reactant Concentrations (mol/L) Initial Rate Experiment (mol/L·s) [O2] [NO] 1 3.21x10-3 1.10x10-2 1.30x10-2 2 6.40x10-3 2.20x10-2 1.30x10-2 3 12.8x10-3 1.10x10-2 2.60x10-2 4 9.60x10-3 3.30x10-2 1.30x10-2 5 28.8x10-3 1.10x10-2 3.90x10-2 16-27  Finding m, the order with respect to O2: We compare experiments 1 and 2, where [NO] is kept constant but [O2] doubles: Rate 2 k[O2] m 2 [NO] n 2 = m n Rate 1 k[O2] 1 [NO] 1 m n 6.40x10-3 mol/L·s 2.20x10-2 mol/L =k 1.30x10-2 mol/L 3.21x10-3mol/L·s k 1.10x10-2 mol/L 1.30x10-2 mol/L 1.99 = (2.00)m or 2 = 2m m=1 Since m = 1, so the reaction is first order in O2 16-28  Sometimes the exponent is not easy to find by inspection. In those cases, we solve for m with an equation of the form a = bm: log a log 1.99 m= = = 0.993 log b log 2.00 This confirms that the reaction is first order with respect to O2. Reaction orders may be positive integers, zero, negative integers, or fractions. 16-29  Finding n, the order with respect to NO: We compare experiments 1 and 3, where [O2] is kept constant but [NO] doubles: Rate 3 k[O2] m 3 [NO] n 3 = m n Rate 1 k[O2] 1 [NO] 1 m n 12.8x10-3 mol/L·s 1.10x10-2 mol/L 2.60x10-2 mol/L =k 3.21x10-3mol/L·s -2 k 1.10x10 mol/L 1.30x10-2 mol/L 3.99 = (2.00)n or 4 = 2n, so n = 2. log a log 3.99 Alternatively: n= = = 2.00 log b log 2.00 Since n = 2, so the reaction is second order in NO The rate law is given by: Rate = k[O2][NO]2 Overall order is third order 16-30  Determine Order of Reaction Qualitatively For experiments 1 and 2: At constant [NO], while doubling the [O2], the rate is also doubled. Hence the order of reaction for reactant O2 is first order. For experiments 1 and 3: At constant [O2], while doubling the [NO], the rate increases by a factor of 4 or increases four times. Hence the order of reaction for reactant NO is second order. The rate law is given by: Rate = k[O2][NO]2 16-31  Sample Problem 16.3 Determining Reaction Orders from Rate Data PROBLEM: Many gaseous reactions occur in a car engine and exhaust systems. One of these is NO2(g) + CO(g) → NO(g) + CO2(g) rate = k[NO2]m[CO]n Use the following data to determine the individual and overall reaction orders: Initial Rate Initial [NO2] Initial [CO] Experiment (mol/L·s) (mol/L) (mol/L) 1 0.0050 0.10 0.10 2 0.080 0.40 0.10 3 0.0050 0.10 0.20 16-32  Sample Problem 16.3 PLAN: We need to solve the general rate law for m and for n and then add those orders to get the overall order. We proceed by taking the ratio of the rate laws for two experiments in which only the reactant in question changes concentration. SOLUTION: To calculate m, the order with respect to NO2, we compare experiments 1 and 2: rate 2 k [NO2]m2[CO]n2 = rate 1 k [NO2]m1 [CO]n1 m n 0.080 k 0.40 0.10 = 0.0050 k 0.10 0.10 16 = (4.0)m so m = 2 The reaction is second order in NO2. 16-33  Sample Problem 16.3 To calculate n, the order with respect to CO, we compare experiments 1 and 3: rate 3 k [NO2]m3[CO]n3 = rate 1 k [NO2]m1 [CO]n1 m n 0.0050 k 0.10 0.20 = 0.0050 k 0.10 0.10 1.0 = (2.0)n so n = 0 The reaction is zero order in CO. Rate = k[NO2]2[CO]0 or Rate = k[NO2]2 16-34  Table 16.3 Units of the Rate Constant k for Several Overall Reaction Orders Overall Reaction Units of k General formula: Order (t in seconds) L order-1 0 mol/L·s (or mol L-1s-1) mol Units of k = 1 1/s (or s-1) unit of t 2 L/mol·s (or L mol-1s-1) 3 L2/mol2·s (or L2 mol-2s-1) ** The value of k is easily determined from experimental rate data. The units of k depend on the overall reaction order. 16-35  Initial Reactant Concentrations (mol/L) Initial Rate Experiment (mol/L·s) [O2] [NO] 1 3.21x10-3 1.10x10-2 1.30x10-2 2 6.40x10-3 2.20x10-2 1.30x10-2 3 12.8x10-3 1.10x10-2 2.60x10-2 4 9.60x10-3 3.30x10-2 1.30x10-2 5 28.8x10-3 1.10x10-2 3.90x10-2 Rate = k[O2][NO]2 16-36 16-37 16-38  Figure 16.9 Information sequence to determine the kinetic parameters of a reaction. Series of plots of concentration vs. time Determine slope of tangent at t0 for each plot. Initial rates Compare initial rates when [A] changes and [B] is held constant (and vice versa). Reaction orders Substitute initial rates, orders, and concentrations into rate = k[A]m[B]n, and solve for k. Rate constant (k) and actual rate law 16-39  Integrated Rate Laws An integrated rate law includes time as a variable. First-order rate equation: rate = – [A] = k [A] [A]0 ln = kt t [A]t Second-order rate equation: rate = – [A] = k [A]2 1 1 – = kt t [A]t [A]0 Zero-order rate equation: rate = – [A] = k [A]0 [A]0 – [A]t = kt t 16-40 16-41  Figure 16.11A Graphical method for finding the reaction order from the integrated rate law. First-order reaction Rate = k [A] Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. integrated rate law ln [A]0 = kt [A]t straight-line form ln[A]t = –kt + ln[A]0 A plot of ln [A] vs. time gives a straight line for a first-order reaction. 16-42  Figure 16.11B Graphical method for finding the reaction order from the integrated rate law. Second-order reaction Rate = k [A]2 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. integrated rate law 1 1 – = kt [A]t [A]0 straight-line form 1 1 = kt + [A]t [A]0 1 A plot of vs. time gives a straight line for a second-order reaction. [A] 16-43  Figure 16.11C Graphical method for finding the reaction order from the integrated rate law. Zero-order reaction Rate = k [A]0 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. integrated rate law [A]0 - [A]t = kt straight-line form [A]t = –kt + [A]0 A plot of [A] vs. time gives a straight line for a zero-order reaction. 16-44  Graphical determination of the reaction order for the Figure 16.12 decomposition of N2O5. The concentration data is used to construct three different plots. Since the plot of ln [N2O5] vs. time gives a straight line, the reaction is first order. 16-45 Sample Problem 16.5 Determining the Reactant Concentration after a Given Time At 1000oC, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87 s-1, to two molecules of ethylene (C2H4). (a) If the initial C4H8 concentration is 2.00 M, what is the concentration after 0.010 s? [C4H8 ]0 2.00 mol/L ln = kt ln = (87 s-1)(0.010 s) = 0.87 [C4H8 ]t [C4H8 ]t 2.00 mol/L = e0.87 = 2.4 [C4H8 ]t [C4H8 ]t = 2.00 mol/L = 0.83 mol/L 2.4 (b) How long will it take for 70.0% of the C4H8 to decompose? Means [C4H8 ]t is 30% remains (out of 2.00 M) = 0.600 M 2.00 mol/L ln = (87 s-1)(t) 0.600 mol/L t = 0.014 s 16-46  Identify carefully whether the % given in the question meant for % of compound to decompose OR % of compound left (current) Example: (a) How long does it take for 75.0% of the compound to decompose? (b) How long does it take for [reactant] to reach or to decrease to 12.5% of its original value? % that compound has decomposed = [initial] - [current ] X 100 [initial ] 16-47  Reaction Half-life The half-life (t½ ) for a reaction is the time taken for the concentration of a reactant to drop to half its initial value. For a first-order reaction, t1/2 does not depend on the starting concentration. (initial [ ]) ln 2 0.693 t ½= = k k The half-life for a first-order reaction is a constant. 16-48  Figure 16.10 A plot of [N2O5] vs. time for three reaction half-lives. for a first-order process ln 2 0.693 t1/2 = = k k 16-49  Sample Problem 16.7 Determining the Half-Life of a First-Order Reaction Cyclopropane is the smallest cyclic hydrocarbon. Because its 60° bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000ºC via the following first-order reaction: The rate constant is 9.2 s–1. (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? 16-50  Sample Problem 16.7 SOLUTION: 0.693 0.693 (a) t½ = = = 0.075 s k 9.2 s–1 (b) For the initial concentration to drop to one-quarter of its value requires 2 half-lives to pass. Time = 2(0.075 s) = 0.15 s 16-51  Half-life Equations For a first-order reaction, t1/2 does not depend on the initial concentration. For a second-order reaction, t1/2 is inversely proportional to the initial concentration: 1 t1/2 = (second order process; rate = k[A]2) k[A]0 For a zero-order reaction, t1/2 is directly proportional to the initial concentration: [A]0 t1/2 = (zero order process; rate = k) 2k 16-52  Table 16.4 An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions Zero Order First Order Second Order Rate law rate = k rate = k[A] rate = k[A]2 Units for k mol/L·s 1/s L/mol·s Half-life [A]0 ln 2 1 2k k k[A]0 Integrated rate law [A]t = -kt + [A]0 ln[A]t = -kt + ln[A]0 1 1 = kt + in straight-line form [A]t [A]0 1 Plot for straight line [A]t vs. t ln[A]t vs. t vs. t [A]t Slope, y intercept –k, [A]0 –k, ln[A]0 k, 1 [A]0 16-53  Collision Theory and Concentration The basic principle of collision theory is that particles must collide in order to react. An increase in the concentration of a reactant leads to a larger number of collisions, hence increasing reaction rate. The number of collisions depends on the product of the numbers of reactant particles, not their sum. Concentrations are multiplied in the rate law, not added. 16-54  Figure 16.13 The number of possible collisions is the product, not the sum, of reactant concentrations. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. add another 6 collisions 4 collisions add another 9 collisions 16-55  Effective collisions – molecules of reactants must collide with each other through the correct orientation and with sufficient (enough) energy where energy of the reactants must be more or equal than activation energy to lead to product. Activation energy – the minimum energy required to initiate a chemical reaction (where from reactants possible to get products) When the concentration of reactants increases, the number of molecules increases. Thus the frequency of collision increases where the frequency of effective collision also increases and cause an increase in the rate of reaction. 16-56  Figure 16.15 The importance of molecular orientation to an effective collision. NO(g) + NO3(g) → 2NO2(g) There is only one relative orientation of these two molecules that leads to an effective collision. It is important to note that effective collision only result when two reactants molecules having energy equal to or more than Ea collide at the correct orientation 16-57  Activation Energy In order to be effective, collisions between particles must exceed a certain energy threshold. When particles collide effectively, they reach an activated state (or transition state). The energy difference between the reactants and the activated state is the activation energy (Ea) for the reaction. The lower the activation energy, the faster the reaction. Smaller Ea larger f larger k increased rate 16-58  Figure 16.15 Energy-level diagram for a reaction. Collisions must occur with This particular reaction is reversible sufficient energy to reach an and is exothermic in the forward activated state. direction. 16-59  Transition State Theory An effective collision between particles leads to the formation of a transition state or activated complex. The transition state is an unstable species that contains partial bonds. It is a transitional species partway between reactants and products. Transition states cannot be isolated. The transition state exists at the point of maximum potential energy. The energy required to form the transition state is the activation energy. 16-60  Depicting the Reaction Between BrCH3 and OH− Figure 16.17 © McGraw Hill Figure 16.16 The transition state of the reaction between BrCH3 and OH-. BrCH3 + OH– → Br– + CH3OH Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The transition state contains partial bonds (dashed) between C and Br and between C and O. 16-62  -Ea/RT Based on Arrhenius equation; k = Ae k = rate constant, A =frequency factor of effective collision, Ea = activation energy, R = gas constant and T = absolute temperature (a) The higher the temperature, the larger the value of rate constant and leads to the higher reaction rate. Higher T larger k increased rate (b) Catalyst lowers an activation energy, the value of rate constant becomes larger and leads to the higher reaction rate. Even though there are no T and catalyst in rate law, but the value of rate constant, k depends on these two factors. So when the value of k is large, it will cause increasing in reaction rate. 16-63  Temperature and Collision Energy An increase in temperature causes an increase in the kinetic energy of the reactant particles. This leads to a larger fraction of collisions that have sufficient (enough) energy to exceed activation energy, Ea. So rate of reaction increases when temperature increases. Explanation with the aid of Maxwell-Boltzmann Distribution Curve diagram. (Figure 16.20) 16-64  Figure 16.20 The effect of temperature on the distribution of collision energies. 16-65  Catalysis: Speeding up a Reaction A catalyst is a substance that increases the reaction rate without itself being consumed in the reaction. How does a catalyst increase a reaction rate? A catalyst operates by lowering the activation energy, provides an alternative reaction pathway where the number of effective collisions increases, thus the rate constant, k also increases and leads to an increase in reaction rate. 16-66  Using the energy profile for Catalyst 16-67  Particles which do not have enough energy to react 16-68  There are two types of catalysts: (a) Homogeneous – catalyst that is present in the same phase as the reacting molecules. (b) Heterogeneous – catalyst that is present in the different phase than the reactants. 16-69  Reaction Mechanisms The mechanism of a reaction is the sequence of single reaction steps that make up the overall equation. The individual steps of the reaction mechanism are called elementary steps because each one describes a single molecular event. Each elementary step is characterized by its molecularity, the number of particles involved in the reaction. The rate law for an elementary step can be deduced from the reaction stoichiometry – reaction order equals molecularity for an elementary step only. Reaction intermediate - a substance that is formed first in one elementary step and consumed (used up) in another elementary step. It does not appear in the overall reaction. 16-70  Table 16.6 Rate Laws for General Elementary Steps Elementary Step Molecularity Rate Law A product Unimolecular Rate = k[A] 2A product Bimolecular Rate = k[A]2 A+B product Bimolecular Rate = k[A][B] 2A + B product Termolecular Rate = k[A]2[B] 16-71  Sample Problem 16.11 Determining Molecularity and Rate Laws for Elementary Steps PROBLEM: The following elementary steps are proposed for a reaction mechanism: (1) Cl2(g) 2Cl(g) (2) Cl(g) + CHCl3(g) → HCl(g) + CCl3(g) (3) Cl(g) + CCl3(g) → CCl4(g) (a) Write the overall balanced equation. (b) Determine the molecularity of each step. (c) Write the rate law for each step. PLAN: We find the overall equation from the sum of the elementary steps. The molecularity of each step equals the total number of reactant particles. We write the rate law for each step using the molecularities as reaction orders. 16-72  Sample Problem 16.11 SOLUTION: (a) Writing the overall balanced equation: (1) Cl2(g) 2Cl(g) (2) Cl(g) + CHCl3(g) → HCl(g) + CCl3(g) (3) Cl(g) + CCl3(g) → CCl4(g) Cl2(g) + Cl(g) + CHCl3(g) + Cl(g) + CCl3(g) → 2Cl(g) + HCl(g) + CCl3(g) + CCl4(g) Cl2(g) + CHCl3(g) → HCl(g) + CCl4(g) (b) Determining the molecularity of each step: The first step has one reactant, Cl2, so it is unimolecular. The second and third steps have two reactants (Cl and CHCl3; Cl(g) + CCl3), so they are bimolecular. (c) Writing rate laws for the elementary steps: (1) Rate1 = k1[Cl2] (2) Rate2 = k2[Cl][CHCl3] (3) Rate3 = k3[Cl][CCl3] 16-73  The Rate-Determining Step of a Reaction Mechanism Most reactions occur by mechanism with more than one elementary step (multi-steps). Every elementary step in the mechanism has its own rate. (different rates) However there is usually one step much slower than others. The slowest step in a reaction is called the rate- determining or rate-limiting step. (the step with the highest Ea) This slow step limits the overall reaction rate where it governs the rate law for the overall reaction. The rate law for the rate-determining step becomes the rate law for the overall reaction. 16-74  Consider the reaction between nitrogen dioxide and carbon monoxide The reaction NO2(g) + CO(g) → NO(g) + CO2(g) has been proposed to occur by a two-step mechanism: (1) NO2(g) + NO2(g) → NO3(g) + NO(g) [slow; rate-determining] (2) NO3(g) + CO(g) → NO2(g) + CO2(g) [fast] Observed or experimental rate law: Rate = k[NO2]2 NO3 functions as reaction intermediate. Rate law for the two elementary steps are: (1) Rate1 = k1[NO2][NO2] = k1[NO2]2 (2) Rate2 = k2[NO3][CO] If k1 = k, the rate law for the rate-determining step (step 1) becomes identical to the observed rate law. 16-75 The Validity of the Proposed Mechanism A valid mechanism must meet three criteria: 1) The rate law derived from rate determining step (or slow step) must consistent or correlate with the experimental rate law or observed rate law. (not the other way around) 2) The overall equation from the mechanism is consistent with the chemical equation of reaction. 3) The molecularity of all elementary steps must be reasonable. (involved unimolecular or bimolecular) 16-76  Mechanisms with a Slow Initial Step The overall reaction 2NO2(g) + F2(g) →2NO2F(g) has an experimental rate law: Rate = k[NO2][F2]. The accepted mechanism is (1) NO2(g) + F2(g) → NO2F(g) + F(g) [slow; rate determining] (2) NO2(g) + F(g) → NO2F(g) [fast] 2NO2(g) + F2(g) →2NO2F(g) The elementary steps sum to the overall balanced equation. (consistent with the overall reaction) Both elementary steps are bimolecular and are therefore reasonable. rate1 = k1[NO2[F2] Step 1 is the slow step, and rate1 correlates rate2 = k2[NO2][F] with the experimental rate law. The mechanism is therefore reasonable. 16-77 Figure 16.22 Reaction energy diagram for the two-step reaction of (A) NO2 and F2 and of (B) NO and O2. Each step in a multi-step reaction has its own transition state, which occurs at the energy maximum for that step. 16-78 16-79 16-80  Figure 16.23 Reaction energy diagram for a catalyzed (green) and an uncatalyzed (red) process. 16-81