# Chapter 13Rates of Reaction - HCC Learning Web

Chapter 13 Rates of Reaction 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. Contents and Concepts Reaction Rates We will investigate how to determine the rate of chemical reactions and identify the factors that influence this rate. Definition of Reaction Rate Experimental Determination of Rate Dependence of Rate on Concentration Change of Concentration with Time Temperature and Rate; Collision and Transition-State Theories Arrhenius Equation

Reaction Mechanisms We continue to look at reaction rate; however, the focus has shifted to an atomic and molecular perspective. 7. Elementary Reactions 8. The Rate Law and the Mechanism 9. Catalysis Learning Objectives Reaction Rates Definition of a Reaction Rate a. Define reaction rate. b. Explain instantaneous rate and average rate of a reaction. c. Explain how the different ways of expressing reaction rates are related. d. Calculate average reaction rate.

2. Experimental Determination of Rate Describe how reaction rates may be experimentally determined. 3. Dependence of Rate on Concentration Define and provide examples of a rate law, rate constant, and reaction order. Determine the order of reaction from the rate law. Determine the rate law from initial rates. 4. Change of Concentration with Time Learn the integrated rate laws for first-order, second-order, and zero-order reactions. Use an integrated rate law. Define half-life of a reaction. Learn the half-life equations for first-order, second-order, and zero-order reactions.

Relate the half-life of a reaction to the rate constant. Plot kinetic data to determine the order of a reaction. 5. Temperature and Rate; Collision and TransitionState Theories State the postulates of collision theory. Explain activation energy (Ea). Describe how temperature, activation energy, and molecular orientation influence reaction rates. State the transition-state theory. Define activated complex. Describe and interpret potential-energy curves for endothermic and exothermic reactions. 6. Arrhenius Equation

Use the Arrhenius equation. Reaction Mechanisms 7. Elementary Reactions Define elementary reaction, reaction mechanism, and reaction intermediate. Determine the rate law from initial rates. Write the overall chemical equation from a mechanism. Define molecularity. Give examples of unimolecular, bimolecular, and termolecular reactions. 7. Elementary Reactions (cont.) e. Determine the molecularity of an elementary reaction. f. Write the rate equation for an elementary reaction. 8. The Rate Law and the Mechanism

Explain the rate-determining step of a mechanism. Determine the rate law from a mechanism with an initial slow step. Determine the rate law from a mechanism with an initial fast, equilibrium step. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 9 9. Catalysis Describe how a catalyst influences the rate of a reaction. Indicate how a catalyst changes the potential-energy curve of a reaction. Define homogeneous catalysis and heterogeneous catalysis.

Explain enzyme catalysis. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 10 Chemical kinetics is the study of reaction rates, including how reaction rates change with varying conditions and which molecular events occur during the overall reaction. 13 | 11 The following questions are explored in this chapter: How is the rate of a reaction measured? What conditions will affect the rate of a

reaction? How do you express the relationship of rate to the variables affecting the rate? What happens on a molecular level during a chemical reaction? What conditions will affect the rate of a reaction? Four variables affect the rate of reaction. 1. Concentrations of the reactant 2. Concentration of the catalyst 3. Temperature at which the reaction occurs 4. Surface area of the solid reactant or catalyst 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 13

Concentrations of Reactants In certain reactions 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. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 14 Concentration of Catalyst Hydrogen peroxide, H2O2, decomposes rapidly in the presence of HBr, giving oxygen and water. Some of the HBr is oxidized to Br2, as can be seen above. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website

for classroom use. 13 | 15 Temperature at Which Reaction Occurs In most cases reactions speed up when the temperature increases. It takes less time to boil an egg at sea level than on a mountaintop, where water boils at a lower temperature. Reactions during cooking go faster at higher temperature. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 16 Surface Area of a 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. For example, a wood fire burns faster if the logs are chopped into smaller pieces. Similarly, the surface area of a solid catalyst is important to the rate of reaction. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 17 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/(Ls) or M/s. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 18 The rate of this reaction can be found by measuring the concentration of O2 at various times. Alternatively, the concentration of NO2 could be measured. Both of these concentrations increase with time. The rate could also be determined by measuring the

concentration of N2O5, which would decrease over time. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 19 2N2O5 ( g ) 4NO2 ( g ) + O2 ( g ) Molar concentrations are denoted by enclosing the substance in square brackets. [NOO 2 ] Rate of formation of O2 = t [NONO 2 ] Rate of formation of NO2 = t [NON 2 O 5 ]

Rate of decomposition of N2O5 = t 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 20 Relative rates We can relate these expressions by taking into account the reaction stoichiometry. Rate of decomposition of N2O5 = N2O5 t 1 Rate of formationof O 2 = rate of decomposition of N2O5

2 O2 t 1 N2O5 = 2 t 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 21 These equations give the average rate over the time interval Dt. As Dt decreases and approaches zero, the equations give the instantaneous rate.

The next slides illustrate this relationship graphically for the increase in concentration of O 2. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 22 The average rate is the slope of the hypotenuse of the triangle formed. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 23 As Dt 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. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 24 Consider the reaction of nitrogen dioxide with fluorine to give nitryl fluoride, NO2F. 2NO2 (g ) +F2 ( g ) 2NO2F( g ) How is the rate of formation of NO2F related to the rate of reaction of fluorine?

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 25 Rate of formation of NO 2F = NO2F t F2 Rate of reaction of F2 = t Divide each rate by the corresponding coefficient and then equate them: F2

1 NO2F = 2 t t 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 26 Calculate the average rate of decomposition of N2O5, [NON2O5]/t, by the reaction 2N2O5 (g ) 4NO2 (g ) + O2 (g ) During the time interval from t = 600s to t = 1200s (regard all time figures as significant). Use the following data:

Time 600s 1200s [N2O5] 1.2410 2 M 0.9310 2 M 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 27 Solution Average rate of decomosition of N2O5 =

2 = 0.93 1.24 10 1200 600 s M N2O5 t 0.31 10 2 M = 600 s = 5.210 6 M / s Note that this rate is twice the rate of formation of

O2 in the same time interval. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 28 Shown here is a plot of the concentration of a reactant D versus time. a. How do the instantaneous rates at points A and B compare? b. Is the rate for this reaction constant at all points in time?

a. The slope at point A is greater than the slope at point B, so the instantaneous rate at point A is greater than the instantaneous rate at point B. b. No. If it were, the graph would be linear. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 29 Rates are determined experimentally in a variety of ways. For slow reactions, samples can be taken and analyzed from the reaction at several different time intervals. Continuously following the reaction is more convenient. This can be done by measuring pressure change, as shown on the next slide, or by measuring light absorbance and color change.

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 30 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 31 Experimentally, it has been found that a reaction rate depends on the concentration of one or more reactants as well as the concentration of catalyst (if any). This information is captured in the rate law, an equation that relates the rate of a reaction to the concentration of a reactant (and catalyst) raised to

various powers. The proportionality constant, k, is the rate constant. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 32 For the generic reaction aA + bB C dD + eE C = catalyst the rate law can be written in the following manner: Rate = k A m

n B C p The exponents m, n, and p are frequently integers and must be experimentally determined. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 33 The reaction order with respect to a specific reactant is the exponent of that species in the experimentally determined rate law.

The overall order of a reaction is the sum of the orders of the reactant species from the experimentally determined rate law. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 34 Nitrogen monoxide, NO, reacts with hydrogen according to the equation The experimentally determined rate law is 2 Rate = k [NONO] [NOH2 ] Thus, the reaction is second order in NO, first order in H2, and third order overall.

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 35 Bromide ion is oxidized by bromate ion in acidic solution. 5Br (aq ) +BrO3 (aq ) + 6H+ (aq ) 3Br2 (aq ) + 3H2O( l ) The experimentally determined rate law is + Rate = k Br BrO3 H

2 What is the order of reaction with respect to each reactant species? What is the overall order of the reaction? 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 36 Solution 5Br (aq )+BrO3 (aq ) + 6H+ (aq ) 3Br2 (aq ) + 3H2O( l ) + Rate = k Br BrO3 H

2 a. The reaction is first order with respect to Br . The reaction is first order with respect to BrO3 . The reaction is second order with respect to H +. b. The reaction is fourth order overall (1+1+2). 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 37 Consider the reaction Q +R S + T and the rate law for the reaction:

0 Rate = k Q R a. 2 You run the reaction three times, each time starting with [NOR] = 2.0 M. For each run you change the starting concentration of [NOQ]: run 1, [NOQ] = 0.0 M; run 2, [NOQ] = 1.0 M; run 3, [NOQ] = 2.0 M. Rank the rate of the three reactions using each of these concentrations. b. The way the rate law is written in this problem is not typical for expressions containing reactants that are zero order in the rate law. Write the rate law in the more typical fashion. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website

for classroom use. 13 | 38 a. b. [NOQ] = 0.0 M is the slowest (no reaction). The other two have equal rates because they are zero order with respect to [NOQ]. Rate = k R 2 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website

for classroom use. 13 | 39 The rate law for a reaction must be determined experimentally. We will study the initial rates method of determining the rate law. This method measures the initial rate of reaction using various starting concentrations, all measured at the same temperature. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 40 Returning to the decomposition of N2O5, we have

the following data: 2N2O5 ( g ) 4NO2 ( g ) + O2 ( g ) Initial N2O5 Concentration Initial Rate of Disappearance of N2O5 Experiment 1 1.0 102 mol/L 4.8 106 mol/(Ls) Experiment 2 2.0 102 mol/L

9.6 106 mol/(Ls) Note that when the concentration doubles from experiment 1 to 2, the rate doubles. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 41 Consider a rate law in terms of reactant A Rate1 = k[NOA]1 Rate2 = k[NOA]2 x x

Therefore.. Rate2/Rate1 = ([A]2/[A]1) x Where x is the order of reaction with respect to A Examining the effect of doubling the initial concentration gives us the order in that reactant. In this case, when the concentration doubles, the rate doubles, so m = 1. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 43 Iodide ion is oxidized in acidic solution to triiodide ion, I3, by hydrogen peroxide. H2O2 (aq ) + 3 I (aq ) + 2H+ (aq ) I3 (aq ) + 2H2O( l ) A series of four experiments was run at different concentrations, and the initial rates of I3 formation were determined. (see table in the next slide) a. + From these data, obtain the reaction orders with respect to H2O2, I , and H . b. Find the rate constant.

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 44 Initial Concentrations (mol/L) H2O2 I H+ Initial Rate [NOmol/(Ls)] Exp.1 0.010 0.010 0.00050

1.1510 6 Exp.2 0.020 0.010 0.00050 2.3010 6 Exp.3 0.010 0.020 0.00050 2.3010 6 Exp.4 0.010 0.010 0.00100 1.1510 6 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 45 To find the reaction orders of the reactants, know the rate law. Using the reactants in the chemical equation, assume that the rate law has the following form: m n + Rate = k H2O2 I H p Each of the reaction orders can be determined by

choosing experiments in which all concentrations of reactants, except one, are held constant. To determine the rate constant, use data from one of the experiments and substitute them into the rate law. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 46 Solution Write the rate law for two experiments (the subscript denotes the experiments). n + p

n + p m I H 1 1 m I H 2 2

Rate 1 = k H2O2 1 Rate 2 = k H2O2 2 Divide the second equation by the first. m n + p

Rate 2 = k H2O2 2 I 2 H 2 m n + p Rate 1 = k H2O2 1 I 1 H 1 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 47 Group the terms. Rate 2 Rate 1 H2O2 2

H2O2 1 m I 2 I 1 n + H

+ 2 H 1 p Substitute the values from Experiments 1 and 2. 6 m n 2.3010 0.020 0.010 0.00050 = 6

1.1510 0.010 0.010 0.00050 2=2 p m

m =1 The reaction is first order in H2O 2 . 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 48 Comparing Experiments 1 and 3, it is clear that with the doubling of the I concentration, the rate doubles. n=1 Comparing Experiments 1 and 4, it is clear that with the doubling of the H+ concentration, the rate is not affected. p=0 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website

for classroom use. 13 | 49 Since [NOH+]0 = 1, the rate law is: Rate =k H2O2 I The reaction orders with respect to H2O2, I, and H+ are 1, 1, and 0, respectively. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 50 The rate constant can be calculated by substituting values from any of the experiments into the rate law. By using Experiment 1:

1.1510 6 mol mol mol = k 0.010 0.010 L s L L 1.1510 6 / s k= 0.0100.010mol / L k = 1.210 2 L / mol s 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 51 Note that the units on the rate constant are specific to the overall order of the reaction. For a zero-order reaction, the unit is M/s or mol/ (L s). For a first-order reaction, the unit is 1/s or s 1. For a second-order reaction, the unit is 1/(M s) or L/(mol s). If you know the rate constant, you can deduce the overall rate of reaction. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 52 Rate laws are not restricted to chemical systems; they are used to help describe many everyday events. For example, a rate law

for tree growth might look something like this: Rate of growth = soil type w x y temperature light fertilizer z In this equation, like chemical rate equations, the exponents need to be determined by experiment. (Can you think of some other factors?) a.

Say you are a famous physician trying to determine the factors that influence the rate of aging in humans. Develop a rate law, like the one above, that would take into account at least four factors that affect the rate of aging. b. c. Explain what you would need to do in order to determine the exponents in your rate law. Consider smoking to be one of the factors in your rate law. You conduct an experiment and find that a person smoking two packs of cigarettes a day quadruples (4) the rate of aging over that of a one-pack-a-day smoker. Assuming that you could hold all other factors in your rate law constant, what would be the exponent of the smoking term in your rate law? 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 53 a. Rate of aging = diet

w x y exercise sex occupation z b. You would need to run controlled studies quantifying each of the factors and measures of aging. c. When smoking doubles, aging increases fourfold. That is a second-order reaction. Therefore, m = 2. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website

for classroom use. 13 | 54 The rate law tells us the relationship between the rate and the concentrations of reactants and catalysts. To find concentrations at various times, we need to use the integrated rate law, the mathematical relationship between concentration and time. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 55 First-Order Rate Law Consider the equation where A is the substance

that reacts to give products. aA products If this equation has a first-order rate law, Rate = A t = k A Using calculus, you derive the following integrated rate law equation: A t In = A 0

kt first - order integrated rate law 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 56 Second-Order Rate Law Consider the following reaction: aA products Suppose it has the second-order rate law. A 2 Rate = = k A t

Using calculus, you can obtain the following relationship between the concentrations of A and the time. 1 1 second - order integrated rate law = kt + A t A 0 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 57 Zero-Order Reactions Consider the following reaction: A B+C

Suppose it has a zero-order rate law of Rate = k Relationship between concentration and time would be: A t = kt + A 0 zero - order integrated rate law 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 58 The decomposition of N2O5 to NO2 and O2 is first order, with a rate constant of 4.8010 4/s at 45C.

a. If the initial concentration is 1.6510 2 mol/L, what is the concentration after 825 s? b. How long would it take for a concentration of N2O5 to decrease to is 1.0010 2 mol/L from its initial value, given in a?

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 59 Solution a. The equation for first order reaction is A t In = A 0 kt

Substituting the appropriate values: In N2O5 t 1.6510 2 mol / L = - 4.8010 4 / s825 s = 0.396 To solve for [NON2O5]t, take the antilongarithm (antiln) of both sides. This removes the 0.396 In from the left and gives the antiln(0.396), or e , on the right which equals 0.673. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 60 N2O5 t 2 1.6510 mol / L = 0.673 Hence, N2O5 t = 1.6510 2 mol / L 0.673 N2O5 t = 0.0111mol / L 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 61

b. Substitute into the same first-order equation relating concentration to time. 1.0010 2 mol / L 4 In = 4.8010 / s t 2 1.6510 mol / L 0.501= 4.8010 4 / s t t = 1.04103 s 17.4 min

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 62 The half-life of a reaction, t, is the time it takes for the reactant concentration to decrease to onehalf of its initial value. By substituting [NOA]0 for [NOA]t, we solve the integrated rate law for the special case of t = t. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 63 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 64 Sulfuryl chloride, SO2Cl2, is a colorless, corrosive liquid whose vapor decomposes in a first-order reaction to sulfur dioxide and chlorine. SO2Cl2 ( g ) SO2 ( g ) + Cl2 ( g ) 5 At 320C, the rate constant is 2.2010 /s. a. What is the half-life of SO2Cl2 vapor at this temperature? b. How long (in hours) would it take for 50.0% of the SO2Cl2 to decompose?

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 65 Solution a. 0.693 0.693 t1/2 = = k 2.2010 5 / s t1/2 = 3.1510 4 s b. The half-life time required to decompose one

half of SO2Cl2 is 3.15104 s or 8.75 hours. The time required for two half-lives is 28.75 hours = 17.5 hours 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 66 Lets look again at the integrated rate laws. y = mx + b y = mx + b y = mx + b In each case, the rate law is in the form of y = mx + b, allowing us to use the slope and

intercept to obtain the rate constant for a reaction. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 67 Zero Order : [A]t kt [A]0 For a zero-order reaction, a plot of [NOA]t versus t is linear. The y-intercept is [NOA]0. First Order : ln [A]t kt ln [A]0 For a first-order reaction, a plot of ln[NOA]t versus t is linear. The graph crosses the origin (b = 0). 1 1 Second Order : kt

[A]t [A]0 For a second-order reaction, a plot of 1/[NOA]t versus t is linear. The y-intercept is 1/[NOA]0. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 68 Graphs of concentration and time can also be used to determine the order of reaction in that reactant. For zero-order reactions, [NOA] versus t is linear. For first-order reactions, ln[NOA] versus t is linear. For second order reactions, 1/[NOA] versus t is linear. This is illustrated on the next slide for the

decomposition of NO2. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 69 Left: The plot of ln[NONO2] versus t is not linear, so the reaction is not first order. Right: The plot of 1/[NONO2] versus t is linear, so the reaction is second order in NO2. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 70 A reaction believed to be either first or second order has a half-life of 20 s at the beginning of the reaction but a half-life of 40 s sometime later. What is the order of the reaction?

[NOA]0 Zero Order : t1/2 2k 0.693 First Order : t1/2 k 1 Second Order : t1/2 k [NOA]t The initial concentration decreases in each time interval. The only equation that results in a larger value for t is the second-order equation. The reaction is second order.

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 71 Each test tube contains potassium permanganate, KMnO4, and oxalic acid, H2C2O4, at the same concentrations. Permanganate ion oxidizes oxalic acid to CO2 and H2O. Top: One test tube was placed in a beaker of warm water (40C); the other was kept at room temperature (20C). Bottom: After 10 minutes, the test tube at 40C showed a noticeable reaction, whereas the other did not. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 72 Rate Constant and Temperature The rate constant depends strongly on temperature. How can we explain this relationship? Collision theory assumes that reactant molecules must collide with an energy greater than some minimum value and with the proper orientation. The minimum energy is called the activation energy, Ea. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 73 We will now explore the effect of a temperature

increase on each of the three requirements for a reaction to occur. The rate constant can be given by the equation k = Zfp Where Z = collision frequency f = fraction of collisions with the minimum energy p = orientation factor 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 74 Certainly, Z will increase with temperature, as the average velocity of the molecules increases with temperature.

However, this factor alone cannot explain the dramatic effect of temperature changes on reaction rates. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 75 The fraction of molecular collisions having the minimum energy required is given by f. f = e Ea / RT Now we can explain the dramatic impact of temperature. f decreases with increasing values of Ea. Since k depends on f, reactions with larger activation energies have small rate constants and vice versa.

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 76 The reaction rate also depends on p, the proper orientation for the collision. This factor is independent of temperature changes. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 77 Importance of Molecular Orientation in the Reaction of NO and Cl2

a. NO approaches with its N atom toward Cl2, and an NCl bond forms. Also, the angle of approach is close to that in the product NOCl. b. NO approaches with its O atom toward Cl2. No NCl bond can form, so NO and Cl2 collide and then fly apart. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 78

Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex (transition state), an unstable grouping of atoms that can break up to form products. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 79 The potential energy diagram for a reaction visually illustrates the changes in energy that occur. The next slide shows the diagram for the reaction NO + Cl2 NOCl2 NOCl + Cl where NOCl2 indicates the activated complex.

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 80 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 81 The reaction of NO with Cl2 is an endothermic reaction. The next slide shows the curve for a generic exothermic reaction. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 82 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 83 Rate constants for most chemical reactions closely follow an equation in the following form: k = Ae Ea / RT This is called the Arrhenius equation, where e is the base of natural logarithms. Ea is the activation energy. R is the gas constant, 8.31 J/(Kmol).

T is the absolute temperature. A is the frequency factor, a constant. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 84 A more useful form of the Arrhenius equation is its two-point form. k 2 Ea ln k1 R 1 1

T1 T2 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 85 The rate constant for the formation of hydrogen iodide from the elements H2 (g ) +I2 (g ) 2HI( g ) 4 3 is 2.2710 L/(mols) at 600 K and 3.510 L/(mols) at 650 K. a. Find the activation energy Ea.

b. Calculate the rate constant at 700 K. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 86 Solution a. Ea 3.510 3 1 1 In

= 4 2.710 8.31J / mol K 600 K 650 K Ea In 1.3010 = 2.56 = 1.2810 4 8.31J / mol 1 Ea = 2.568.31 J / mol 1.2810 4

Ea = 1.66105 J / mol 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 87 b. Substitute Ea = 1.66105 J/mol. k1 = 2.7104 L/(mols) (T1 = 600 K) k2 = unknown as yet (T2 = 700 K) k2 1.66105 J / mol 1 1 In =

= 4.77 4 2.710 L / mol s 8.31J / mol K 600 K 700 K k2 4.77 2 = e = 1.210 2.710 4 L / mol s Taking antilogarithms: k2 = 1.2102 2.710 4 L / mol s

k2 = 3.210 2 L / mol s 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 88 Reaction Mechanism A balanced chemical equation is a description of the overall result of a chemical reaction. However, what actually happens on a molecular level may be more involved than what is represented by this single equation. For example, the reaction may take place in several steps. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 89 Each step in the reaction mechanism is called an elementary reaction. It is a single molecular event. The set of elementary reactions whose overall effect is given by the net chemical equation is called the reaction mechanism. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 90 A reaction intermediate is a species produced during a reaction that does not appear in the net equation because it reacts in a subsequent step in the mechanism.

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 91 Elementary reactions are classified according to their molecularity, the number of molecules on the reactant side of an elementary reaction. Unimolecular reactions involve one reactant molecule. Bimolecular reactions involve two reactant

molecules. Termolecular reactions involve three reactant molecules. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 92 Carbon tetrachloride, CCl4, is obtained by chlorinating methane or an incompletely chlorinated methane such as chloroform, CHCl 3. The mechanism for the gas-phase chlorination of CHCl3 is Cl2 2Cl Cl + CHCl3 HCl + CCl3 Cl + CCl3 CCl4

(elementary reaction) (elementary reaction) (elementary reaction ) Obtain the net, or overall, chemical equation from this mechanism. Also determine the molecularity of each step. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 93 Solution Cl2 2Cl Cl + CHCl3 HCl + CCl3 Cl + CCl3 CCl4

Cl2 + CHCl3 HCl + CCl4 (overall equation) The forward part of the first step is unimolecular. The reverse part of the first step, the second step, and the third step are each bimolecular. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 94 Because an elementary reaction is an actual molecular event, the rate of an elementary reaction is proportional to the concentration of each reactant molecule. This means we can write the rate law directly from an elementary reaction.

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 95 Write rate equations for each of the following elementary reactions. a. b. Ozone is converted to O2 and NO in a single step. O3 +NO O2 +NO The recombination of iodine atoms occurs as follows: Where M is some atom or molecule that * absorbs energy from the reaction.

c. I+I+M I2 +M An H2O molecule absorbs energy; some time later enough of this energy flows into one OH bond to break it. H2O H+ O H 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 96 Solution a. Rate = k O3 NO

2 b. Rate = k I M c. Rate = k H2O 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 97 Rate Law and Reaction Mechanism A reaction mechanism cannot be directly observed. We can, however, determine the rate law by experiment and decide if the reaction mechanism is consistent with that rate law. The rate of a reaction is determined completely by the rate-determining step.

2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 98 The slowest step in the reaction mechanism is called the rate-determining step (RDS). The rate law for the RDS is the rate law for the overall reaction. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 99 For example, the reaction of NO2 with F2 is believed to occur in the following elementary

steps: NO + F k NO F + F 2 2 1 2 2 F + NO2 k NO2F (slow step) (fast step) 2NO2 + F2 2NO2F

What is the rate law for this mechanism? Rate = k1[NONO2 ][NOF2 ] 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 100 Ozone reacts with nitrogen dioxide to produce oxygen and dinitrogen pentoxide. O3 (g ) + 2NO2 ( g ) O2 ( g ) +N2O5 ( g ) The proposed mechanism is O3 +NO2 NO3 + O2 NO3 +NO2 N2O5 (slow)

(fast) What is the rate law predicted by this mechanism? Solution The rate law predicted from the first step is Rate = k O3 NO2 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 101 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 102

Nitrogen monoxide can be reduced with hydrogen gas to give nitrogen and water vapor. 2NO( g ) + 2H2 ( g ) (overall N2 ( g )equation) + 2H2O( g ) A proposed mechanism is k1 2NO N2O2 k (fast, equilibrium) 1 2

N2O2 +H2 k N2O +H2O 3 N2O +H2 k N2 +H2O (slow) (fast) What rate law is predicted by this mechanism? 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 103 Solution

According to the rate-determining step, Rate = k 2 N2O2 H2 Try to eliminate N2O2 from the rate law by considering the first step, which is fast and reaches equilibrium. At equilibrium, forward and reverse rates are equal. 2 k1 NO = k 1 N2O2 N2O2 = k1 / k 1 NO 2 k 2 k1 2 Rate = NO

H2 k 1 2 Rate = k NO H2 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 104 Catalysis It is an increase in the rate of reaction that results from the addition of a catalyst. Enzymes are marvelously selective biological catalysts. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 105 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. A catalyst increases the reaction rate by either increasing the frequency factor A, or providing an alternative reaction path with a lower activation energy. When Ea is reduced, k increases exponentially. This relationship is illustrated on the potential energy diagram for the decomposition of ozone. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.

13 | 106 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 107 Homogenous Catalysis It is the process which uses the catalyst in the same phase as the reacting species. Consider the decomposition of aqueous hydrogen peroxide to form water and oxygen. 2H2O2 (aq ) O2 (g ) + 2H2O( l ) Proposed mechanism is: Step 1: H2O2 (aq ) +I (aq ) IO (aq )+H2O( l ) (slow)

Step 2 :IO (aq ) +H2O2 (aq) H2O( l ) + O 2 ( g ) +I ( aq ) (fast) Overall : 2H2O2 (aq ) O 2 ( g ) + 2H2O( l ) 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 108 Heterogeneous Catalysis Applied in most industrial reactions, it uses a catalyst that exists in a different phase from the reacting species, usually a solid catalyst in contact with a gaseous or liquid solution of reactants. This occurs via chemical adsorption of reactants onto the catalyst surface. Chemisorption

It is the binding of a species to the surface by chemical bonding forces that break the bonds in the species. The best example is catalytic hydrogenation, illustrated in the following slide. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 109 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 110 Enzyme Catalysis Enzymes have enormous catalytic activity and are highly specific. The substance whose reaction the enzyme catalyzes is called the substrate.

When a substrate fits into the active site where catalysis occurs, an enzyme-substrate (ES) complex is formed. On binding, the bonds in the substrate may weaken or new bonds may form. This helps yield a product, P. E + S ES E +P 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 111 Notice that the formation of the ES complex provides a new pathway for products with lower Ea, as mentioned below in the illustration. 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 13 | 112

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