SGCh13 - Chapter Thirteen CHEMICAL KINETICS • • • •...

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Unformatted text preview: Chapter Thirteen CHEMICAL KINETICS • • • • • • The Rate of Reaction Rate Laws The Relation Between Reactant Concentration and Time Activation Energy and Temperature Dependence of Reaction Rates Reaction Mechanisms Catalysis THE RATE OF REACTION STUDY OBJECTIVES 1. 2. 3. Define the rate of reaction. Express the rate of a given reaction in terms of the change in concentration with time of a reactant or a product. Calculate the average rate of reaction from given concentration vs. time data. Expressing the Rate of Reaction. C hemical kinetics i s the area of chemistry concerned with the study of the rates of chemical reactions. The purposes of kinetic studies are to find the factors that affect reaction rates and determine the reaction mechanism. Knowledge of the factors that affect reaction rates enables chemists to control rates. Finding the reaction mechanism means we can identify the intermediate steps by which reactants are converted into products. A rate is a change in some quantity with time. The rate of population growth is the change in population per change in time. The change in a quantity such as population is always equal to the the difference, population (after) minus population (before). The symbol for "the change in" is ∆ . ∆ (population) = populationfinal – populationinitial The change in time is some appropriate time interval, ∆ t, where ∆ t = tfinal – t initial The rate of population change is rate = ∆ (population) ∆t As a chemical reaction proceeds the concentrations of reactants and products change with time. For instance, as the reaction A + B → C progresses, the concentration of C increases. The rate is expressed as the change in the concentration of C during the time interval ∆ t. rate = ∆ [C] ∆t 2 48 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 49 For a specific reaction we need to take into account the stoichiometry; that is, we need the balanced equation. For example, let's express the rate of the following reaction in terms of the concentrations of the individual reactants and products. 2NO(g) + O2 (g) → 2NO2 (g) These concentrations can be analyzed or monitored as a function of time. Notice from the balanced equation, that 2 mol NO reacts with 1 mol O2 ; therefore the concentration of NO will decrease twice as fast as that of O2 . The convention used for writing the rate of reaction is ∆ [O2 ] ∆ [NO2 ] ∆ [NO] rate = – =– = 2∆ t ∆t 2∆ t Division of each concentration change by the coefficient from the balanced equation makes all of these rates equal. Notice also a negative sign is inserted before terms involving reactants. The change in NO concentration, ∆ [NO], is negative because the concentration of NO decreases with time. Since a negative times a negative makes a positive, inserting a negative sign in the expression makes the rate of reaction a positive quantity. For a general equation aA + bB → cC the rate can be expressed in terms of any reactant or product. rate = – ∆ [A] ∆ [B] ∆ [C] =– = a ∆t b ∆t c ∆t Each rate calculation is the same and all have the same sign. Calculating an Average Rate. The average rate of reaction over any time interval is equal to the decrease in the concentration of a reactant ∆ [A] (or the increase in the concentration of a product) divided by the time interval, ∆ t, over which that change occurred. decrease in the concentration of A ∆ [A] = length of time interval ∆t [A]2 – [ A] 1 ∆ [A] average rate = – =– ∆t t2 - t 1 average rate = – The concentration term [A]2 is the concentration of A at time t2 , and [A]1 is the concentration of A at time t1 . _______________________________________________________________________________ EXAMPLE 13.1 Expressing the Rate of Reaction Write expressions for the rate of the following reaction in terms of each of the reactants and products. 2N2 O5 (g) → 4NO2 (g) + O 2 (g) •Method of Solution Recall that the rate is defined as the change in concentration of a reactant or product with time. Each "change-inconcentration" term is divided by the corresponding stoichiometric coefficient. Terms involving reactants are preceded by a minus sign. Therefore, the rate is expressed as ∆ [N2 O5 ] ∆ [NO2 ] ∆ [O2 ] rate = – = = 2∆ t 4∆ t ∆t _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 50 / Chemical Kinetics _______________________________________________________________________________ EXAMPLE 13.2 Rate of Reaction Oxygen gas is formed by the decomposition of nitric oxide: 2NO(g) → O2 (g) + N 2 (g) If the rate of formation of O2 is 0.054 M/s, what is the rate of change of NO concentration? •Method of Solution From the stoichiometry of the reaction, 2 mol NO react for each mole of O2 that forms. Therefore: ∆ [O2 ] ∆ [NO] rate = – = 2∆ t ∆t • Calculation –2∆ [O2 ] ∆ [NO] = = – 2(0.054 M/s) ∆t ∆t = – 0.108 M/s _______________________________________________________________________________ EXAMPLE 13.3 Calculation of the Average Rate Experimental data for the hypothetical reaction: A → 2B are listed in the following table. Time (s) [A] (mol/L) 0.00 1.000 10.0 0.891 20.0 0.794 30.0 0.707 40.0 0.630 a. b. Calculate the average rate of change of [A] for each 10.0 s interval from 0 to 30 s. Why does the rate decrease from one time interval to the next? •Method of Solution a. The average rate of reaction is given by: [A]2 – [ A] 1 ∆ [A] average rate = – =– ∆t t2 – t 1 For the time interval 0 – 10 s the reaction rate is average rate = – =– Back Forward 0.891 mol/L – 1.000 mol/L 10.0 s – 0 s –0.109 mol/L = – 0.0109 mol/L·s 10.0 s Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 51 For the time interval 10 – 20 s the reaction rate is 0.794 mol/L – 0.891 mol/L 20.0 s – 10.0 s average rate = – –0.097 mol/L = – 0.0097 mol/L·s 10.0 s =– For the time interval 20 – 30 s the reaction rate is 0.707 mol/L – 0.794 mol/L 30.0 s – 20.0 s –0.087 mol/L =– = – 0.0087 mol/L·s 10.0 s average rate = – b. The rate decreases as the time of reaction increases because the rate rate is proportional to the concentration of the reactant A, and the concentration of A is decreasing as the time of reaction increases. _______________________________________________________________________________ EXERCISES 1. Write expressions for the rate of reaction in terms of each of the reactants and products. N2 (g) + 3H 2 (g) → 2NH3 (g) 2. Thiosulfate ion is oxidized by iodine in aqueous solution according to the equation 2– 2– 2S 2 O3 (aq) + I2 (aq) → 2S 4 O6 (aq) + 2I– (aq) 2– If 0.025 mol of S2 O3 is consumed in 0.50 L solution per minute: 2– a. Calculate the rate of removal of S 2 O3 b. What is the rate of removal of I2 ? 3. in M/s. N2 O5 is an unstable compound that decomposes according to the following equation. 2N2 O5 → 4NO2 + O2 The following data was obtained at 50°C. [N2 O5 ] (M) Time (s) 1.00 0.88 0.78 0.69 0.61 0.54 0.48 0.43 0 200 400 600 800 1000 1200 1400 a. What is the average rate of N2 O5 disappearance in the time interval 200-400 s? b. What is the average rate of N2 O5 disappearance in the time interval 800-1000 s? c. What is the rate of O 2 production in the time interval 800-1000 s? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 52 / Chemical Kinetics RATE LAWS STUDY OBJECTIVES 1. 2. Describe all the terms in "the rate equation," and define the order of reaction. Apply the isolation method to derive the rate law for a reaction. Effect of Concentration. The rate of a reaction is proportional to the reactant concentrations. For the reaction 1 NO + 2 O 2 → NO2 the rate is proportional to the concentrations of NO and O2 : rate α [NO]x [O2 ]y where x and y are exponents. The rate law or rate equation for the reaction is: rate = k[NO]x [O2 ]y The proportionality constant k is called the r ate constant . The value of k depends on the reaction and the temperature. The exponents x and y determine how strongly the concentration affects the rate. The exponent x is called the o rder with respect to NO , and y is the o rder with respect to O 2 . The sum x + y is the overall order. The values of x and y must be determined from experiment, and cannot be derived by any other means. We will discuss how to determine the order of reaction in the next section. For now we will just use the results. Experiments show that x = 2 and y = 1 for the NO reaction with O2 . Therefore, the rate law for this reaction is: rate = k[NO]2 [O2 ] This reaction is second order in nitric oxide, and first order in oxygen. It is third order overall. The fact that the reaction is first order in O2 means that the rate is directly proportional to the O2 concentration. If [O2 ] doubles or triples, the rate will double or triple also. We can show this mathematically. Consider two experiments. In expt 1 the concentration of O2 is c. In expt 2 the concentration of O2 is doubled from c to 2c. If the concentration of NO is the same in both experiments, it will have no effect on the rate. Use of the rate law allows us to write the ratio of the two rates: rate (expt 2) k[NO]2 (2c) = =2 rate (expt 1) k[NO]2 (c) As discussed, we see that doubling the concentration of a reactant that is first order should cause the rate to double. If the concentration of O2 is held constant in two experiments and the concentration of NO doubles (from c to 2c), the rate law predicts that the rate will quadruple. k[O2 ] (2c)2 rate (expt 2) = = 22 = 4 rate (expt 1) k[O2 ] (c)2 The fact that the reaction is second order in NO means that the rate is proportional to the square of the concentration of NO. Doubling or tripling of [NO] causes the rate to increase four- or nine-fold, respectively. In general, if the concentration of one reactant doubles while the other reactant concentration is unchanged, and the rate is: 1. 2. 3. Back unchanged, the order of the reaction is zero order with respect to the changing reactant. doubled, the order of the reaction is first order with respect to the changing reactant. quadrupled, the order of the reaction is second order with respect to the changing reactant. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 53 The Isolation Method. One procedure used to determine the rate law for a reaction involves the isolation method. In this method the concentration of all but one reactant is fixed, and the rate of reaction is measured as a function of the concentration of the one reactant whose concentration is varied. Any variation in the rate is due to the variation of this reactant's concentration. In practice the experimenter observes the dependence of the initial rate on the concentration of the reactant. To determine the order with respect to A in the following chemical reaction 2A + B → C the initial rate would be measured in several experiments in which the concentration of A is varied and the concentration of B is held constant. To determine the order with respect to B, the concentration of A must be held constant and the concentration of B is varied in several experiments. The application of this method is illustrated in Example 13.5. _______________________________________________________________________________ EXAMPLE 13.4 Concentration Effect on the Rate The reaction A + 2B → products was found to have the rate law rate = k[A][B]3 . By what factor will the rate of reaction increase if the concentration of B is increased from c to 3c, while the concentration of A is held constant? •Method of Solution Write a ratio of the rate law expressions for the two different concentrations of B. rate (expt 2) k[A][3c]3 33 c3 = =3 3 rate (expt 1) k[A][c] c = 27 The rate will increase 27-fold when [B] is increased three-fold. _______________________________________________________________________________ EXAMPLE 13.5 Finding the Rate Law The following rate data were collected for the reaction: 2NO + 2H 2 → N2 + 2H2 O __________________________________________________ Experiment [NO]0 [H2 ]0 ∆ [N2 ]/∆ t (M) (M) (M/h) __________________________________________________ 1 0.60 0.15 0.076 2 0.60 0.30 0.15 3 0.60 0.60 0.30 4 1.20 0.60 1.21 __________________________________________________ a. b. Determine the rate law. Calculate the rate constant. •Method of Solution a. We want to determine the exponents in the equation rate = k[NO]x [H2 ]y Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 54 / Chemical Kinetics To determine the order with respect to H2 , first find two experiments in which [NO] is held constant. This can be done by comparing the data of experiments 1 and 2. When the concentration of H2 is doubled, the reaction rate doubles. Thus, the reaction is first order in H2 . When the NO concentration is doubled (experiments 3 and 4) the reaction rate quadruples. Thus the reaction is second order in NO. The rate law is rate = k[NO]2 [H2 ] b. Rearrange the rate law from part (a). k= rate [NO]2 [H2 ] Then substitute data from any one of the experiments. Using experiment 1: k= 0.076 M/h = 1.4/M2 h [0.60 M] 2 [0.15 M] •Comment You should get the same value of k from all four experiments. Note that the units of k are those of a third-order rate constant. Taking a closer look at proving that x = 2. Write the ratio of rate laws for experiments 4 and 3. rate4 k [NO]x [ H 2 ] = rate3 k [NO]x [ H 2 ] 121 k (1.20)x ( 0.60) = = 0.30 k (0.60)x ( 0.60) 1.20 x 0.60 4.0 = 2.0x Therefore x = 2 _______________________________________________________________________________ EXERCISES 4. The rate law for the reaction 2A + B → C was found to be rate = k[A][B]2 . If the concentration of B is tripled and the concentration of A is unchanged, by how many times will the reaction rate increase? 5. Use the following data to determine (a) the rate law and (b) the rate constant for the reaction 2A + B → C Experiment [A]0 [B]0 Rate (M/s) __________________________________ 1 0.25 0.10 0.012 2 0.25 0.20 0.048 3 0.50 0.10 0.024 __________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 55 THE RELATION BETWEEN REACTANT CONCENTRATION AND TIME STUDY OBJECTIVES 1. 2. 3. Determine for first- and second-order reactions the concentration of a reactant at any time after the reaction has started, or the time required for a given fraction of the sample to react. Determine the half-life of a first-order reaction. Apply rate law equations and graphs to determine whether a reaction is first order or second order. First-Order Reactions. One of the most widely encountered kinetic forms is the first-order rate equation. In this case the exponent of [A] in the rate law is 1. A → products rate = – ∆ [A] = k[A] ∆t For a first-order reaction the unit of the rate constant is reciprocal time, 1/t. Convenient units are s–1 , h –1 , etc. The equation that relates the concentration of A remaining to the time since the reaction started is [A]0 ln = kt [A] This is a very useful equation called the i ntegrated first-order equation . Here [A]0 is the concentration of A at time = 0, and [A] is the concentration of A at time = t. The rate constant k is the first-order rate constant. The concentration [A] decreases as the time increases. This equation allows the calculation of the rate constant k when [A]0 is known, and [A] is measured at time t. Also, once k is known, [A] can be calculated for any future time. To determine whether a reaction is first order, we rearranged the first-order equation into the form: ln [A] = – kt + ln [A]0 corresponding to the linear equation y = mx + b Here m is the slope of the line and b is the intercept on the y axis. Comparing the last two equations, we can equate y and x to experimental quantities. y = ln [A] and x = t Therefore, the intercept b = ln [A]0 , and the slope of the line m = –k. Thus a plot of ln [A] versus t for a firstorder reaction gives a straight line with a slope of –k as shown in Figure 13.1 below. If a plot of ln [A] versus t yields a curved line rather than a straight line, the reaction is not a first-order reaction. This graphical procedure is the method used by most chemists to determine whether or not a given reaction is first order. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 56 / Chemical Kinetics [ A ]0 ln [ A ] Concentration of A ln [ A ] 0 0 0 time time Figure 13.1 A plot of [A] versus time for a first-order reaction gives a curved line. A plot of ln [A] versus time gives a straight line for a first-order reaction. Half life. The half-life of a reaction, t 1/2 , is a useful concept. For a first-order reaction, the half-life is given by: t1/2 = 0.693 k where 0.693 is a constant and k is the rate constant. Knowledge of the half-life allows the calculation of the rate constant k. The h alf-life of a reaction is the time required for the concentration of a reactant to decrease to half of the initial value. After one half-life, the ratio [A]/[A]0 is equal to 0.5. If the reaction continues, then [A] will drop by 1/2 again during the second half-life period as shown in Figure 13.2. After two half-life periods the fraction of the original concentration of A remaining, [A]/[A]0 , will be 1/2 of the concentration remaining after the first half-life, so [A]/[A]0 = 0.5 × 1/2 = 0.25. Concentration of A [ A ]0 1 [A ] 0 2 1 [ A ]0 4 1[ ] A0 8 0 1 2 3 Number of half-lives elapsed Figure 13.2 A plot of [A] versus time for a first-order reaction gives a curved (exponential) line. Over each half-life period, [A] drops in half. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 57 Second-Order Reactions. In a second-order reaction, the rate is proportional either (1) to the square of the concentration of one reactant, A → product rate = – ∆ [A] = k[A] 2 ∆t or (2) to the product of the concentrations of two reactants, each raised to the first power: A + B → product rate = – ∆ [A] = k[A][B] ∆t This reaction is first order in A and first order in B, and so it is second order overall. For a second-order reaction, the rate constant has units 1/(molarity × time) which is 1/M·s or M–1 s –1 . An important equation called t he integrated second-order equation is 1 1 = + kt [A] [A]0 where [A]0 , [A], and t have their usual meaning, and k is the second-order rate constant. This equation allows the calculation of the concentration of A at any time t after the reaction has begun. Alternatively, if [A] 0 , [A], and t are known, the rate constant can be calculated. Example 13.6 in the text shows an example calculation using this equation. The half-life for a second-order reaction is given by t1/2 = 1 k[A]0 Here we see that the half-life is inversely proportional to the initial concentration, [A]0 . This situation is different from that for a first-order reaction, where t1/2 is independent of [A]0 . Zero Order Reactions. A zero-order reaction is one in which the rate does not depend on the concentrations of reactants. A → products rate = –[A] = k[A]0 = k ∆t A plot of [A] versus time is a straight line for a zero order reaction. The rate does not slow down as the reactant is used up.. Summary. The integrated rate equations and graphical methods allow us to distinguish between the various overall orders of reaction. 1. 2. 3. Back The reaction is first order when a plot of ln [A] versus t is a straight line. A plot of [A] versus time will be curved. The reaction is zero order when a plot of [A] versus t is a straight line and plots of [A] versus t, and ln[A] versus t are curved. The half-life equations also provide a way to distinguish between first- and second-order reactions. The halflife of a first-order reaction is independent of starting concentration, whereas the half-life of a second-order reaction is inversely proportional to the initial concentration. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 58 / Chemical Kinetics _______________________________________________________________________________ EXAMPLE 13.6 First-Order Reaction Methyl isocyanide undergoes a first-order isomerization reaction to form methyl cyanide. CH3 NC(g) → CH 3 CN(g) The reaction was studied at 199°C. The initial concentration of CH3 NC was 0.0258 mol/L, and after 11.4 min analysis showed the concentration of product was 1.30 × 10 –3 mol/L. a. What is the first-order rate constant? b. What is the half-life of methyl isocyanide? c. How long will it take for 90 percent of the CH3 NC to react? •Method of Solution for (a) This problem illustrates the use of the integrated first-order rate equation [CH3 NC]0 ln = kt [CH3 NC] Where k is the rate constant and [CH3 NC] is the reactant concentration at time t. The initial concentration is [CH3 NC]0 = 0.0258 M. After 11.4 min the product concentration is 1.30 × 10–3 M. This implies that the concentration of CH3 NC remaining must be 0.0258 – 0.0013 = 0.0245 M. •Calculation for (a) Substitution gives 0.0258 M = k (11.4 min) 0.0245 M ln 1.05 k= 11.4 min ln k = 4.54 × 10–3 min –1 •Method of Solution for (b) For a first-order reaction the half-life equation is t1/2 = 0.693 k • Calculation for (b) t1/2 = 0.693 = 153 min 4.54 × 1 0 –3 m in –1 •Method of Solution for (c) If 90 percent of the initial CH3 NC is consumed, then 10 percent remain. Therefore: [CH3 NC] = 0.10 [CH 3 NC]0 •Calculation for (c) Substitution into the rate equation gives [CH3 NC]0 ln = (4.54 × 10–3 min –1 )t 0 .10[CH 3 NC]0 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 59 Solving for t, we get: ln 10 4.54 × 1 0 –3 m in –1 t= = 2.303 4.54 × 1 0 –3 m in –1 t = 507 min _______________________________________________________________________________ EXAMPLE 13.7 First-Order Reaction At 230°C the rate constant for methyl isocyanide isomerization is 9.25 × 10–4 s –1 . CH3 NC → CH 3 CN a. b. What fraction of the original isocyanide will remain after 60.0 min? What is the half-life of methyl isocyanide at this temperature? •Method of Solution for (a) a. Again applying the first-order equation: [CH3 NC]0 ln = kt [CH3 NC] The fraction of methyl isocyanide remaining is given by [CH 3 NC]/[CH3 NC]0 •Calculation for (a) [CH3 NC]0 [CH3 NC] ln = – ln = kt [CH3 NC] [CH3 NC]0 – ln [CH3 NC] 60 s = 9.25 × 10–4 s –1 (60 min) × [CH3 NC]0 1 min = –1.45 [CH3 NC] = 10 –1.45 [CH3 NC]0 = 0.035 •Calculation for (b) The half-life is t1/2 = 0.693 9.25 × 1 0 –4 s –1 = 749 s or 12.4 min •Comment Note the much shorter half-life at 230°C than at 199°C (Example 13.6). _______________________________________________________________________________ EXERCISES 6. Back A certain first-order reaction A → B is 40% complete (40% of the reactant is used up) in 75 s. What is (a) the rate constant and (b) the half-life of this reaction? Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 60 / Chemical Kinetics 7. a. Write the integrated rate equation for a first-order reaction? Define the variables. b. How can the equation be plotted to give a straight line? Define x, y, and the slope. c. What information do you need in order to calculate the first-order rate constant? 8. The half-life of a certain first order reaction is 112 min. What percent of the initial concentration of reactant will remain after 89 min? 9. Consider the reaction in exercise 3 above for the reaction: 2N2 O5 → 4NO2 + O2 a. Use the data to calculate the first-order rate constant. b. How long will it take for the concentration of N2 O5 to fall to 0.25 M? c. What is the half-life? 10. What are the units of a first-order and a second-order rate constant? 11. A decompositon reaction has a rate constant of 0.12 y–1 at a certaian temperature. a. What is the half-life of the reaction at the same temperature? b. How long will it take for the concentration of the reactant to reach 20% of its initial value? ACTIVATION ENERGY AND TEMPERATURE DEPENDENCE OF REACTION RATES STUDY OBJECTIVES 1. 2. 3. 4. Describe three fundamental tenents to the collision theory. Calculate the activation energy for a reaction when given rate constants at several different temperatures. Describe the concept of activation energy and show how it explains the variation of reaction rate with temperature. Describe a reaction energy profile, including the activation energy and energy change of reaction. Collision Theory. The collision theory of chemical reactions provides a general explanation of how reaction rates are affected by reactant concentrations, and temperature. The basic ideas of the theory are: 1. 2. 3. Back In order for atoms, molecules, or ions to react they must first collide with each other. The rate of reaction is proportional to the rate of collisions, called the c ollision frequency . The more concentrated the reactants, the greater the collision frequency, and the reaction rate. For molecules to react they must come together in the proper orientation. Molecules can be complex and often it is just one atom in a molecule that reacts upon collision with another molecule. When reactant molecules collide, they must possess a minimum amount of kinetic energy in order for an effective collision —a reaction— to occur. Without this necessary energy two molecules will just bump each other and bounce back without reacting. The minimum amount of energy required to initiate a chemical reaction is called the a ctivation energy . Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 61 Effect of Temperature. The temperature of a reaction system is an important variable because of its strong effect on reaction rates. As a rough rule, reaction rates approximately double with a 10°C rise in temperature. In general, the rate equation is rate = k[A]x [B]y . Since the concentrations [A] and [B] are unaffected by temperature, it is the rate constant that changes with temperature. In 1889, S. Arrhenius found that a plot of the natural logarithm of the rate constant (ln k) versus the reciprocal of the absolute temperature 1/T gave a straight line. See Figure 13.16 (text). Arrhenius identified the slope of the line as being related to an energy term Ea m=– R where R is the ideal gas constant (in units of joules) and Ea is the activation energy. The logarithmic form of the Arrhenius equation is ln k = – Ea + ln A RT where ln A is the intercept. Both the Arrhenius A and Ea are constants for a particular reaction. Taking the antilog of both sides gives the Arrhenius equation. k = A e–Ea/RT The Meaning of A and Ea . The parameter A is called the f requency factor . It is related to the frequency of molecular collisions, and the fraction of collisions that have the correct orientation. As discussed above, the collision frequency is important because molecules must collide in order to react. The activation energy Ea is related to the formation of the activated complex. The a ctivated complex is the high-energy intermediate species that dissociates into the products. Figure 13.3 shows that the activation energy is the difference in energy between the reactants and the activated complex. The activation energy is provided by the kinetic energy of rapidly moving molecules during collisions. Reactants must "get over the barrier" before they become products. The factor e–Ea/RT that appears in the Arrhenius equation is the fraction of molecules with energies equal to or greater than the activation energy. This factor changes significantly with temperature. As temperature increases a greater fraction of molecules has an energy equal to the activation energy. Energy Activated complex Ea A+B C+D Reaction Progress Figure 13.3 Molecules of A and B with average energies must first acquire an energy Ea before they can react. From the Arrhenius equation we can point out that reactions for which E a is large will be much slower than those for which Ea is small. As Ea increases, the negative exponent increases, and so k decreases. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 62 / Chemical Kinetics A convenient equation that can be used to calculate the activation energy is ln k1 = Ea T1 – T 2 R T1 T2 k2 where k1 is the rate constant at temperature T1 , and k2 is the rate constant at temperature T2 . Use of this equation requires that rate constants k1 and k2 be measured at two temperatures T1 and T2 , respectively. _______________________________________________________________________________ EXAMPLE 13.8 Calculating Activation Energy For the reaction NO + O 3 → NO2 + O2 the following rate constants have been obtained: ____________________________ Temperature, °C k(M–1 s –1 ) ____________________________ 10.0° 9.30 × 106 30.0° 1.25 × 107 ____________________________ Calculate the activation energy for this reaction. •Method of Solution Here we are given two rate constants corresponding to two temperatures. The activation energy (Ea) can be calculated by substituting into the equation given earlier that shows the temperature dependence of the rate constant. ln Let k1 = Ea T1 – T 2 k2 R T1 T2 k1 = 9.3 × 106 M –1 s–1 at T1 = 273 + 10.0 = 283 K k2 = 1.25 × 107 M –1 s–1 at T2 = 273 + 30.0 = 303 K Recall R = 8.31 J/mol·K. Substitute into the preceding equation to solve for Ea: ln Ea 283 K – 303 K 9.3 × 1 0 6 M – 1 s–1 = 7 M – 1 s–1 8.31 J/mol K (283 K) (303 K) 1.25 × 1 0 Solving for Ea yields: (ln 0.744)(8.31J/mol K) = Ea –20 K 85749 K 2 (–0.296)(8.31 J/mol K) = Ea(–2.33 × 10–4 K–1 ) Ea = 1.06 × 104 J/mol (or 10.6 kJ/mol) _______________________________________________________________________________ EXAMPLE 13.9 Reaction Energy Profile Draw a reaction energy profile for the following endothermic reaction: 2HI(g) → H2 (g) + I 2 (g) o ∆ Hrxn = 12.5 kJ Given the activation energy Ea = 185 kJ/mol. Label the activation energy and the activated complex. What is the activation energy for the reverse reaction? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 63 •Method of Solution The reaction is endothermic so the products are at a higher level than the reactants. Energy Activated complex E a = 185 kJ H2 + I 2 ∆ H = 12.5 kJ 2HI Reaction Progress Using the diagram we can see that the activation energy for the reverse reaction is 185 kJ – 12.5 kJ = 173 kJ. _______________________________________________________________________________ EXERCISES 12. Calculate the activation energy for a reaction given that the rate constant is 4.60 × 10–4 s –1 at 350°C and 1.87 × 10–4 s –1 at 320°C. 13. The rate constant for a first-order reaction is 4.60 × 10–4 s –1 at 250°C. If the activation energy is 100.0 kJ/mol, calculate the rate constant at 300°C. 14. A substance decomposes according to first-order kinetics, the rate constants at various temperatures being as follows: Temp (°C) 15.0 25.0 30.0 k (s–1 ) 4.41 × 10 –6 1.80 × 10 –5 2.44 × 10 –5 a. If one were to make an Arrhenius plot what would its slope be equal to according to the Arrhenius theory? b. Estimate the numerical value of the slope by using the first and third points. c. Calculate the activation energy from your value for the slope. 15. The reaction 2NOCl → 2NO + Cl2 has an E a of 102 kJ/mol and a ∆ Hrxn of 75.5 kJ. Sketch the reaction energy diagram and determine the activation energy for the reverse reaction. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 64 / Chemical Kinetics REACTION MECHANISMS STUDY OBJECTIVES 1. 2. Describe what is meant by a reaction mechanism and distinguish between elementary steps and balanced chemical reactions. Derive the rate law for a reaction from its proposed mechanism. Elementary Steps. The purpose of a reaction mechanism is to indicate how the reactants are converted into products. A reaction mechanism consists of a set of so-called elementary steps. A mechanism indicates just what molecules must collide with each other and in what sequence. Each elementary step marks the progress of the conversion of molecules of reactants into products. A mechanism points out intermediates that are formed and consumed along the way, as well as, which steps are fast and which are slow. This information is not supplied by the balanced chemical equation, because it is intended only to indicate the number of moles of one reactant that are consumed per mole of the other reactant. Elementary steps have a property called molecularity, which pertains to the number of molecules that must collide in a single step. An elementary step that involves three molecules is called termolecular. A step that involves two molecules is bimolecular, and a step in which one molecule decomposes or rearranges is called unimolecular. Thus the equation A→B represents a unimolecular reaction in which a single molecule of A reacts to form a single molecule of B. The following equation represents a bimolecular reaction. A+B →C That is one in which a molecule of A collides with a molecule of B, and as a result a molecule of C is formed. Reaction Order for Elementary Steps. The rate of a unimolecular step A→B will depend on the number of molecules of A per unit volume of the container. Thus unimolecular reactions follow first-order rate laws. rate = k[A] The rate of a bimolecular step will depend on the number of molecules of A per unit volume times the number of molecules of B per unit volume. rate = k[A][B] Therefore, bimolecular steps follow second-order rate laws. This can be understood also by realizing that A and B must collide with each other if they are to react, and so the rate of bimolecular process depends on the rate of collisions of A and B. Doubling the concentration of A will double the probability of collision between A and B. Similarly, doubling the concentration of B will also double the rate of collisions of A and B. See Figure 13.14 in the textbook. Reasoning further along these lines, we conclude that termolecular reactions are third order. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 65 Rate-Determining Step. It often turns out that one of the elementary steps in a mechanism is much slower than all the rest. This step determines the overall rate of reaction much as the slowest person ahead of you a cafeteria line determines how fast you and others move through the line. The slowest step in a sequence of elementary steps is called the r ate-determining step . This allows us to say that the overall law predicted by a mechanism will be the one corresponding to the rate-determining step. Testing a Mechanism. The sequence of events in a kinetic study of a reaction that leads to a proposed mechanism is as follows: 1. 2. 3. Determine the rate law experimentally. Propose a mechanism for the reaction. Test the mechanism. To test a proposed mechanism, assume one step to be the rate-determining step. Then establish the rate law for that step. This gives the rate law predicted by the mechanism. If the mechanism is adequate, then when the predicted rate law is compared with the experimental rate law, the two will match. On the other hand, if the predicted and experimental rate laws do not match, the mechanism has been proved inadequate. Only those mechanisms consistent with all the data can be considered adequate. _______________________________________________________________________________ EXAMPLE 13.10 Reaction Mechanism The rate law for the following reaction has been experimentally determined to be third order: 2NO(g) + O2 (g) → 2NO2 (g) rate = k[NO]2 [O2 ] Which of the two proposed mechanisms that follow is more satisfactory? k1 NO + NO → N2 O2 slow k2 N2 O2 + O2 → 2NO2 a. k1 fast b. NO + NO N 2 O2 fast k –1 k2 N 2 O2 + O2 → 2NO2 slow •Method of Solution Testing mechanism (a) first, we note that the first step is the rate-determining step. The rate law for this bimolecular step is rate = k1 [NO]2 I This mechanism predicts a rate law that is second order in NO concentration and zero order in O2 . By comparison, the experimental rate law is second order in NO and first order in O2 . These do not match, and so mechanism (a) is not satisfactory. Testing mechanism (b), we find that the rate-determining step is the second step. Since it is bimolecular, its rate law should be: rate = k2 [N2 O2 ][O2 ] II It is not possible to compare this predicted rate law directly with the experimental rate law because of the unique term, which is the concentration of N 2 O2 . In this mechanism, N2 O2 is an intermediate. Being formed in step 1 and consumed in step 2, its concentration is always small and usually not measurable. The way around this situation is to find a mathematical substitution for [N2 O2 ]. Note that step 1 is reversible and equilibrium can be established. This means that the rates of the forward and reverse reactions are equal: k1 [NO]2 = k –1 [N2 O2 ] Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 66 / Chemical Kinetics where k1 is the rate constant for the forward reaction, and k–1 for the reverse. Rearranging terms gives us: [N2 O2 ] = k1 [NO]2 k–1 We now have an expression for [N2 O2 ], which we can substitute into equation II. This yields: rate = k2 k1 [NO]2 [O2 ] k–1 Whenever a reaction step is reversible, we can use the equality of the forward and reverse rates to form an expression to substitute for the concentration of an intermediate. Now we compare this rate law with the experimental rate law, which is rate = k[NO]2 [O2 ] We see that this mechanism predicts that the reaction will be second order in NO and first order in O2 , just as observed. Also, the collection of constants k 2 k1 /k–1 will equal the rate constant. k= k2 k1 k–1 Thus, the second mechanism predicts a rate law that matches the experimental order of reaction. _______________________________________________________________________________ EXERCISES 16. Which of the following species can be isolated from a chemical reaction? product, activated complex, intermediate 17. Carbon monoxide can be converted to carbon dioxide by the following overall reaction. NO2 (g) + CO(g) → NO(g) + CO 2 (g) The experimentally determined rate law is rate = k [NO2 ]2 . The suggested mechanism involves two bimolecular elementary steps. NO2 (g) + NO2 (g) NO3 (g) + NO(g) NO3 (g) + CO(g) → NO2 (g) + CO 2 (g) step 1 step 2 a. What is the rate law for each step? b. Derive the predicted rate laws when step 1 is rate determining, and when step 2 is rate determining. 18. The reaction of nitric oxide and chlorine has the following rate law as determined by experiment: 2NO(g) + Cl 2 (g) → 2NOCl(g) rate = k [NO] [Cl2 ] Is the following proposed mechanism consistent with the experimental rate law? NO(g) + Cl 2 (g) → NOCl 2 (g) slow NO(g) + NOCl 2 (g) → 2NOCl(g) fast Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 67 CATALYSIS STUDY OBJECTIVES 1. 2. Describe in general how a catalyst increases the rate of a reaction without being consumed. Describe the essentials of heterogeneous, homogeneous, and enzyme catalysis. Mechanism of a Catalyzed Reaction. A c atalyst is a substance that increases the rate of a chemical reaction without being consumed in the reaction. For this reason a catalyst does not appear in the stoichiometric equation. Chromium(III) oxide, Cr2 O3 , is the catalyst present in catalytic converters in American automobiles. This compound accelerates the reaction of carbon monoxide with oxygen. By converting CO to CO2 in the exhaust stream, CO emissions are reduced. Cr 2 O3 1 CO(g) + 2 O2 (g) → CO2 (g) The lifetime of catalytic converters is quite long; the Cr 2 O3 and other catalysts do not need to be replaced. All catalysts accelerate reaction rates by providing a new reaction pathway (mechanism) which has a lower activation energy. See Figure 13.17 (textbook). In the catalytic converter both CO and O 2 are chemically adsorbed on the catalyst's surface. Adsorption of O2 by Cr 2 O3 weakens the O—O bond enough so that oxygen atoms can react with adsorbed CO. The reaction path involving a weakened O—O bond has a significantly lower activation energy than that of the reaction occurring purely in the gas phase. Recall that as the activation energy is lowered the rate constant for a reaction increases. After CO2 is formed, it desorbs from the surface, leaving the Cr2 O3 catalyst chemically unaltered. The catalyst is not consumed in the reaction. Catalysts are regenerated in one of the last steps of the mechanism. The textbook describes three types of catalysts, which we will review here. Heterogeneous Catalysis. In heterogeneous catalysis the reactants are in one phase and the catalyst is in another phase. The catalyst is usually a solid, and the reactants are gases or liquids. Ordinarily the site of the reaction is the surface of the solid catalyst. Many industrially important reactions involve gases and are catalyzed by solid surfaces. For example, the Haber process for the synthesis of ammonia is catalyzed by iron plus a few percent of the oxides of potassium and aluminum. The famous Ziegler-Natta catalyst for polymerizing ethylene gas (C2 H4 ) into polyethylene polymer contains triethylaluminum and titanium or vanadium salts. Homogeneous Catalysis. In homogeneous catalysis the reactants, products, and catalysts are all in the same phase, which is usually the gas or the liquid phase. Many reactions are catalyzed by acids. The decomposition of formic acid (HCO2 H) is an example. HCO2 H → H2 O + CO Formic acid is normally stable and lasts a long time on the shelf. However, when sulfuric acid is added, bubbles of carbon monoxide gas can be observed immediately. The hydrogen ions from the sulfuric acid initiate a new reaction path. The mechanism is + HCO2 H + H + →HCO2 H2 + HCO2 H2 → H2 O + HCO+ HCO+ → CO + H+ _____________________________ HCO2 H → CO + H2 O Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 68 / Chemical Kinetics Note that the H + ion is consumed in the first step, but, as for all catalysts, H+ is regenerated. The net reaction is + the sum of the three steps. The intermediate HCO 2 H2 cancels out as does the catalyst. Enzyme Catalysis. Nearly all chemical reactions that occur in living organisms require catalysts. Enzymes are biological catalysts. All enzymes are protein molecules with molecular masses well over 10,000 amu. All enzyme molecules are highly specific. That is, they can only affect the reaction rates of a few specific reactant molecules. Typically, an enzymes catalyzes a single reaction, or a group of closely related reactions. The molecule on which an enzyme acts is called a substrate. The simplest mechanism which explains enzyme activity, and is consistent with these trends is one in which the substrate S and the enzyme E form an enzyme-substrate complex, ES. This complex has a lower energy than the activated complex without the enzyme (Figure 13.26 of the text). The enzyme-substrate complex can either dissociate back into E and S, or break apart into the products P and the regenerated enzyme E. The simplest mechanism has the two steps shown below. E+S ES ES → E + P _______________________________________________________________________________ EXAMPLE 13.11 Intermediates and Catalysts Given the following mechanism for the decomposition of ozone in the stratosphere, identify the intermediate and the catalyst: O3 + C l → O2 + ClO ClO + O → C l + O 2 •Method of Solution Adding the two steps gives the overall reaction O3 + O → 2O2 An intermediate is formed in an early step and removed in a later step and does not appear in the overall equation; therefore ClO is the intermediate. A catalyst is consumed in an early step and is later regenerated. Atomic chlorine, Cl, is the catalyst. _______________________________________________________________________________ EXERCISES 19. Define a catalyst. Why is a catalyst not comsumed in the reaction? 20. Identify the catalyst, any intermediates, and the overall equation for the following mechanism. O3 + NO → O2 + NO 2 NO2 + O → NO + O2 ______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 69 CONCEPTUAL QUESTIONS 1. Explain why the rate of reaction increases rapidly with increasing temperature. 2. Cooking an egg involves the denaturation of a protein called albumin from a soluble globular protein to an insoluble fibrous protein. This reaction has an activation energy of roughly 75 kJ/mol. At sea level it takes three minutes to prepare a "soft-boiled egg." However on top of Pikes Peak at 445 m elevation it takes 7.5 minutes to cook a "soft-boiled egg." Explain why its takes longer to cook an egg on the tops of mountains. 3. Describe the difference between an overall reaction (balanced chemical equation) and an elementary reaction. PRACTICE TEST 1. The following experimental data were obtained for the reaction 2A + B → products. What is the rate law for this reaction? Expt. Initial Concentration (M) Rate [A]0 [B]0 M/s ____________________________________________ 1 0.40 0.20 5.6 × 10–3 2 0.80 0.20 5.5 × 10–3 3 0.40 0.40 22.3 × 10–3 2. For the reaction 30CH3 OH + B10 H14 → 10B(OCH3 )3 + 22H2 express the rate in terms of the change in concentration with time for each of the reactants and each of the products. 3. The hydrolysis of sucrose (C12 H22 O11 ) yields the simple sugars, glucose (C6 H12 O6 ) and fructose (C6 H12 O6 ), which just happen to be isomers. C 12 H22 O11 + H2 O → C 6 H12 O6 + C 6 H12 O6 The rate follows the rate equation, rate = k[C12 H22 O11 ]. At 27°C the rate constant is 2.1 × 10–6 /s. a. Starting with the sucrose solution with a concentration of 0.10 M at 27°C, what would the concentration of sucrose be 24 hours later? (The solution is kept at 27°C). b. What is the half-life of sucrose at 27°C? 4. N2 O5 decomposes according to first-order kinetics: 2N2 O5 (g) → 4NO2 (g) + O 2 (g) a. At a certain temperature 20.0% of the initial N2 O5 decomposes in 2.10 h. Determine the rate constant. b. What fraction of the initial N2 O5 will remain after 13.0 h? c. What is the half-life? d. If the initial conc of N 2 O5 was 0.222 M, then what conc remains after 24 h? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 70 / Chemical Kinetics 5. The following data were obtained on the rate of disintegration of a pesticide in soil at 30°C. Time (days) % Pesticide remaining96 11 80 60 71 93 a. What is the order of reaction with respect to pesticide concentration? b. What is the value of the rate constant? 6. a. It takes 30.0 min for the concentration of a reactant in a second-order reaction to drop from 0.40 M to 0.30 M. What is the value of the rate constant for this reaction? b. How long will it take for the concentration to drop from 0.40 M to 0.20 M? 7. The reaction A + 2B → C + D has for its experimental rate law r = k[A]2 [B]. By what factor will the rate increase if the concentration of A is doubled and the concentration of B is tripled? 8. The rate of the reaction of hydrogen with iodine H2 + I2 → 2HI has rate constants of 1.41 × 10–5 /M·s at 393°C and 1.40 × 10–4 /M·s at 443°C. Calculate the activation energy for this reaction. 9. At 300 K the rate constant is 1.5 × 10–5 /M·s for the reaction: 2NOCl → 2NO + Cl2 The activation energy is 90.2 kJ/mol. Calculate the value of the rate constant at 310 K. By what factor did the rate constant increase? 10. The activation energy for the reaction CO + NO2 → CO 2 + NO is 116 kJ/mol. How many times greater is the rate constant for this reaction at 250°C than it is at 200°C? 11. If the reaction 2HI → H2 + I2 has an activation energy of 190 kJ/mol and a ∆ H = 10 kJ, what is the activation energy for the reverse reaction? H2 + I2 → 2HI 12. The reaction between nitrite ion and oxygen gas is: – – 2NO2 + O 2 → 2NO3 – and proceeds at a rate that is first order in [NO2 ]. A mechanism has been proposed: – – 2NO2 + O2 → 2NO4 slow – – – NO4 + NO2 → 2NO3 fast – Assume that NO 4 is a very reactive intermediate and that the first step is the rate determining step. Show that this mechanism is consistent with the experimental rate law. 13. The rate law for the substitution of NH3 for H2 O in the reaction of a complex ion : 2+ Ni(H2 O)6 (aq) + NH3 (aq) → Ni(H2 O)5 (NH3 )2+ (aq) + H2 O(l) 2+ is first order in Ni(H2 O)6 , and zero order in NH3 . 2+ rate = k[Ni(H2 O)6 ] Show that the following mechanism is consistent with the experimental rate law. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 71 2+ → Ni(H2 O)5 2+ + NH3 → Ni(H2 O)5 (NH3 )2+ Ni(H2 O)6 Ni(H2 O)5 2+ + H2 O slow fast 14. The rate law for the net reaction 2NO + O 2 → 2NO2 is rate = k[NO2 ]2 [O2 ]. Could the mechanism of this reaction be a single termolecular process? 15. The rate law for the net reaction H2 (g) + I 2 (g) → 2HI(g) is rate = k[H2 ][I2 ]. A possible mechanism involves a bimolecular elementary step: H2 + I2 → 2HI A second possibility has also been proposed: I2 → 2I fast 2I + H 2 → 2HI slow Show that both mechanisms are consistent with the experimental rate law. 16. The following statements are sometimes made with reference to catalysts. Explain the fallacy in each one of them. a. A catalyst is a substance that accelerates the rate of a chemical reaction but does not take part in the reaction. b. A catalyst may increase the rate of a reaction going in one direction without increasing the rate of the reaction going in the reverse direction. 17. a. Identify the catalyst in the following mechanism. b. What is the overall reaction? c. Is this reaction an example of homogeneous or heterogeneous catalysis? H2 O2 (aq) + Br2 (aq) → 2H+(aq) + O2 (aq) + 2Br– (aq) H2 O2 (aq) + 2H+(aq) + 2Br– (aq) → Br2 (aq) + H2 O(l) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 72 / Chemical Kinetics ANSWERS Exercises ∆ [NH3 ] –∆ [N2 ] –∆ [H2 ] = = 2∆ t ∆t 3∆ t 2– –∆ [S2 O3 ] –∆ [I2 ] a. = 8.3 × 10 –4 M /s b. = 4.2 × 10 –4 M/s ∆t ∆t a. 5.0 × 10 –4 M /s b. 3.5 × 10 –4 M /s c. 1.8 × 10 –4 M/s 9 times a. rate = k [A] [B]2 k = 4.8 /M2 s a. k = 6.8 × 10 –3 / s b. 102 s [A]0 a. ln = kt b. Plot ln[A] vs. t where x = t, y = [A], and m =-k [A] 1. rate = 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. c. The initial concentration [A]0 and [A] again at a know time t. 58% a. b. c. first-order: s–1 ; second order: L/mol·s or M–1 s–1 a. 5.8 y b. 13 y 92 kJ/mol k2 = 1.4 s–1 Ea a. According to Arrhenius's interpretation the slope is: m = – R ln k 2 – l n k 1 b. Calculate the slope from m = m = – 1 .0 × 10 4 K 1/T2 – 1 /T 1 c. 83 kJ/mol Ea(rev) = 26.5 kJ/mol products and intermediates a. rate1 = k 1 [NO2 ]2 ; rate2 = k 2 [NO3 ] [CO] b. when step 1 is rate determining: rate = k1 [NO2 ]2 k2 k1 when step 2 is rate determining: rate = [NO2 ]2 [CO] k–1 18. yes 19. NO is the catalyst. NO2 is the intermediate. The overall equation is: O3 + O → 2O2 Conceptual Questions 1. Back Normally only a small fraction of the colliding molecules have enough energy to exceed the activation energy. As temperature increases the average kinetic energy of molecules increases and so a larger fraction of the colliding molecules have energies greater than the activation energy. The rate of reaction increases because the fraction of molecules with energy greater than the activation energy increases. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Kinetics / 2 73 2. First of all cooking an egg is a chemical reaction and as such its rate will increase with temperature in accordance with the Arrhenius equation. At sea level water (used to cook the egg) boils at 100°C, but at higher elevations such as 445 m water boils at a temperature lower than 100°C (86°C on top of Pikes Peak). Therefore the rate of cooking an egg is much slower on mountain tops than at sea level. 3. An elementary reaction represents a reaction between individual molecules. The reaction occurs when the molecules collide. An overall reaction is the net change due to the sum of all the elementary reactions. It does not represent the collsions and reactions of any particular molecules. Indeed any intermediates have canceled out of the net reaction. The balanced equation represents the proper molar amounts needed for the reactants to be completely consumed. Practice Test 1. Rate = k[B]2 2. Rate = – ∆ [B10 H14 ] 1 ∆ [CH3 OH] 1 ∆ [H2 ] 1 ∆ [B(OCH3 )3 ] =– = = 30 ∆t ∆t 22 ∆ t 10 ∆t a. [sucrose] = 0.083 M; b. t1/2 = 3.3 × 105 s a. k = 0.106 h–1 b. 0.252 c. 6.54 h d. 0.0174 M a. first order; b. 3.7 × 10–3 per day a. 2.76 × 10–2 /M . min b. 90 min Twelve-fold 182 kJ/mol k = 4.8 × 10–5 /M s; k2 /k1 = 3.21 16.8 180 kJ/mol The rate law for the rate determining step is – rate = k[NO2 ][O2 ] The concentration of dissolved O2 is a constant according to Henry's law (Chapter 12). – Therefore k[O2 ] = K (a constant), and rate = K[NO2 ]. 13. The rate law for the rate determining step is 2+ rate = k[Ni(H2 O)6 ] This rate law matches the experimental rate law. 14. Yes 15. The rate law for the rate determining step is rate = k2 [I]2 [H2 ] Obtaining a mathematical substitution for [I] from k1 [I2 ] = k–1 [I]2 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. yields rate = k2 k1 [H2 ][I2 ] k–1 This predicted rate law matches the experimental rate law given in the problem. 16. a. The catalyst reacts in one of the first few steps, but then is regenerated in the last step. b. If E a is lowered for the forward reaction it must also be lowered for the reverse. 17. a. Br2 b . 2H 2 O → 2H2 O + O 2 c. homogeneous ____________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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This note was uploaded on 09/15/2009 for the course CHEM 102 taught by Professor Bastos during the Spring '08 term at Adelphi.

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