552 CHAPTER 14 Chemical Kinetics Recall t ha t when an exponent is 1, it need not be shown [ ~ Section 3.3]. we mu st compare two experiments in which one reactant concentration changes, and the other remains constant. For example, we first compare the data from experiments 1 and 3: Experiment [F 2 ] (M) [Cl0 2 ] (M) Initial Rate (Mis) 1 0.10 0.010 1.2 X 10- 3 2 3 [F, I doubles 0.10 0.20 ICIO,] unchanged 0.040 0.010 Rate doubles 4.8 X 10- 3 2.4 X 10- 3 When the concentration of fluorine doubles, with the chlorine dioxide concentration held constant, the rate doubles. [F2h = 0.20 M = 2 [F 2 ] 1 O.lOM rate3 2.4 X 10- 3 Mis = 2 rate I 1.2 X 10- 3 Mis This indicates that the rate is directly proportional to the concentration of fluorine and the value of x is 1. rate = k [F 2 ] [Cl0 2 Y Similarl y, we can compare experiments 1 and 2: Experiment 1 2 3 [F 2 ] (M) [F, I unchanged { 0.10 0.10 0.20 [CI0 2 ] (M) [CIO,] quadruples { 0.01O 0.040 0.010 Initial Rate (Mis) { 1.2 X 10- 3 Rate quadruples 3 4.8 X 10- 2.4 X 10- 3 We find that the rate quadruples when the concentration of chlorine dioxide is quadrupled, but the fluorine concentration is held con stant. [Cl0 2 h = 0.040 M = 4 [Cl0 2 ]l 0.010M 4.8 X 10- 3 Mis = 4 1.2 X 10- 3 Mis This indicates that the rate is also directly proportional to the concentration of chlorine dioxide, so the value of y is also 1. Thus, we can write the rate law as follows: rate = k[F 2 ][Cl0 2 ] Because the concentrations of F2 and Cl0 2 are each raised to the first power, we say that the reac- tion is first order in F2 and first order in Cl0 2 . The reaction is second order overall. Knowing the rate law, we can then use the data from anyone of the experiments to calculate the rate constant. Using the data for the first experiment in Table 14.2, we can write k = rate 1.2 X 10- 3 Mis = 12M-I. - I [F 2 ] [Cl0 2 ] (0.10 M)(0.01O M)' s Table 14.3 contains initial rate data for the hypothetical reaction aA + bB +. cC + dD which has the general rate law rate = k[A Y [BY Comparing experiments 1 and 2, we see that when [A] doubles, with [B] unchanged, the rate also doubles. Experiment 1 2 3 Thus, x = 1. Experiment 1 2 3 [A] (M) [A] doubles { OJ 0 0.20 0.10 [A] (M) 0.10 0.20 0.10 [B] (M) [B] unchanged {0.o15 0.015 0.010 [B] (M) 0.015 0.015 0.030 Initial Rate (Mis) { 1.2 X 10- 3 Rate doubles 3 4.8 X 10- 2.4 X 10- 3 Initial Rate (Mis) 2.1 X 10- 4 4.2 X 10- 4 8.4 X 10- 4 SECTION 14.2 Dependence of Reaction Rate on Reactant Concentration 553 Overall Reaction Order Sample Rate Law Units of k o rate = k 1 rate = k[A] or rate = k[B] 2 rate = k[A]2, rate = k[Bf, or rate = k[A][B] 3* rate = k[A]2[B] orrate = k[A][Bf M - 1 . S S-1 M -1 -I . S ~2 . S- 1 * Another possibility for a third- or der reaction is rate = k [Aj(Bj(C], although such reactions are very rare. Comparing experiments 1 and 3, when [B] doubles with [A] unchanged, the rate quadru- pie s. Experiment [A] (M) [B] (M) Initial Rate (Mis) 1 0.10 2 [A I unchanged 0.20 3 0.10 0.015 I BI double s 0.015 0.030 2.1 X 10- 4 Rate quadrupl es 4.2 X 10- 4 8.4 X 10- 4 Thus, the rate is n ot directly proportional to [B] to the first power, but rather it is directly propor- tional to [B] to the second power (i.e., y = 2): 7 rate ex [B]- The overall rate law is rate = k[A][Bf This reaction is therefore first order in A, second order in B, and third order overall. Once again, knowing the rate law, we can use data from any of the experiments in the table to calculate the rate constant. Using the data from experiment 1, we get rate k= - [A][Bf 2.1 X 10- 4 Mis = 9.3 M - 2 . S- 1 (0.10 M)(0.015 M)2 Note that the units of this rate constant are different from those for the rate constant we calculated for the F T CI0 2 reaction and the bromine reaction. In fact, the units of a rate constant depend on the overall order of the reaction. Table 14.4 compares the units of the rate constant for reactions that are zeroth-, first-, secondo, and third-order overall. The following are three important things to remember about the rate law: 1. The exponents in a rate law must be determined from a table of experimental data-in general, they are not related to the stoichiometric coefficients in the balanced chemical equation. 2. Comparing changes in individual reactant concentrations with changes in rate shows how the rate depends on each reactant concentration. 3. Reaction order is always defined in terms of reactant concentration s, never product concentrations. Sample Problem 14.3 shows how to use initial rate data to determine a rate law. The gas-phase reaction of nitric oxide with hydrogen at 1280°C is 2NO(g) + 2H ig ) N 2 (g) + 2H 2 0(g) From the following data collected at 1280°C, determine (a) the rate law, (b) the rate constant, including unit s, and (c) the rate of the reaction when [NOl = 4.8 X 10- 3 M and [H2l = 6.2 X 10- 3 M. Experiment 1 2 3 [NOl (M) 5.0 X 10 - 3 1.0 X 10 - 2 1.0 X 10- 2 [H2l (M) 2.0 X 10- 3 2.0 X 10- 3 4.0 X 10- 3 Initial Rate (Mis) 1.3 X 10- 5 5.0 X 10- 5 1.0 X 10- 4 (Continued) 554 CHAPTER 14 Chemical Kinetics A quotient of numbers, each raised to the same power, is equal to the quotient raised to that power: xn /y n = {xly) n Think About It The exponent for the concentration of Hz in the rate law is 1, whereas the coefficient for Hz in the balanced equation is 2. It is a common error to try to write a rate law using the stoichiometric coefficients as the exponents. Remember that, in general, the exponents in the rate law are not related to the coefficients in the balanced equation. Rate laws must be determined by examining a table of experimental data. Strategy Compare two experiments at a time to determine how the rate depends on the concentration of each reactant. Setup The rate law is rate = k[NOn H zY Comparing experiments 1 and 2, we see that the rate increases by approximately a factor of 4 when [NO] is doubled but [Hz] is held constant. Comparing experiments 2 and 3 shows that the rate doubles when [H z] doubles but [NO] is held constant. Solution (a) Dividing the rate from experiment 2 by the rate from experiment 1, we get rat~ = 5.0 X 10 - 5 ~ = 4 = k(1.0 X 10 - 2 M)"(2.0 X 10- 3 M)Y ratel 1.3 X 10 - 5 ~ k(5 .0 X 10 - 3 M)X(2 .0 X 10- 3 M)Y . . . . . . . . . Canceling identical terms in the numerator and denominator gives (1.0 X 10- 2 M)x . -' = 2" = 4 (5.0 X 10- 3 M)x Therefore, x = 2. The reaction is second order in NO. Dividing the rate from experiment 3 by the rate from experiment 2, we get rate3 = 1.0 X 10- 4 ~ = 2 = k( 1.0 X 1O - 2 .M/( 4.0 X 10 - 3 M)Y ratez 5.0 X 10 - 5 ~ k(1.0 X lO - z M)X(2 .0 X 1O - 3 .M)Y Canceling identical terms in the numerator and denominator gives (4.0 X 10- 3 M)Y , :: = 2) = 2 (2.0 X 10 - 3 M)Y Therefore, y = 1. The reaction is first order in Hz. The overall rate law is (b) We can use data from any of the experiments to calculate the va lue and units of k. Using the data from experiment 1 gives k = rate [NO f[ H 2 J 1.3 X 10 - 5 Mis = 2.6 X 10 2 M - 2 • S-I (5.0 X 10- 3 M) 2(2 .0 X 1O - 3 .M) (c) Using the rate constant determined in part (b) and the concentrations of NO and Hz given in the problem statement, we can determine the reaction rate as follows: rate = (2.6 X 10 2 M - 2 • s-I)(4.8 X 10 - 3 M)\6.2 X 10 - 3 M) = 3.7 X 10 - 5 Mis Practice Problem A The reaction of peroxydisulfate ion (SzO t) with iodide ion (1-) is From the following data collected at a certain temperature, determine the rate law and calculate the rate constant, including its units. Experiment [ S20~ - J (M) WJ (M) Initial Rate (Mis) 1 0.080 0.034 2.2 X 10 - 4 2 0.080 0.017 1.1 X 10 - 4 3 0.16 0.017 2.2 X 10 - 4 Practice Problem B For the following general reaction, rate = k[Af and k = 1.3 X lO- z M - 1 • S -I : A + B ·2C Use this information to fill in the mi ss ing table entries. Experiment 1 2 3 [AJ (M) 0.013 0.026 [BJ (M) 0.250 0.250 0.500 Initial Rate (Mis) 2.20 X 10 - 6 2.20 X 10- 6 SECTION 14 .3 Dependence of Reactant Concentrat ion on Ti me 555 Dependence of Reaction Rate on Reactant Concentration Answer questions 14.2.1 through 14.2.4 using the table of initial rate data for the reaction Ex~eriment [A](M) [B](M) A + 2B • 2C + D 1 0.12 0.010 2 0.36 0.010 3 0.12 0.020 Initial Rate (Mis ) 2.2 X 10- 3 6.6 X 10- 3 2.2 X 10- 3 14.2.1 What is the rate law for the reaction? 14.2.3 What is the overall order of the reac ti on? a) rate = k[A][B] 2 a) a b) rate = k[A] 2[ B] b) 1 c) rate = k[A ]3 c) 2 d) rate = k[A] 2 d) 3 e) rate = k[A] e) 4 14.2.2 Calculate the rate constant. 14.2.4 Determine the rate when [A] = 0.50 M a) 0.15 M- 1 . S- l and [B] = 0.25 M. b) 0.15M·s- 1 a) 9.2 X 10- 3 Mis c) 0.15 S- I b) 2.3 X 10- 3 M is d) 0.018 S- I c) 4.5 X 10- 3 M is e) 0.018 M- 1 . S- l d) 5.0 X 10- 3 Mis e) 1.3 X 10- 2 Mis Dependence of Reactant Concentration on Time We can use the rate law to determine the rate of a reaction using the rate constant and the reactant concentrations: Rate law 1 I J rate = k[Ay[BF ! \ Rate Rate constant A rate law can also be used to determine the remaining concentration of a reactant at a specific time during a reaction. We will illustrate this use of rate laws using reactions that are first order overall and reactions that are second order overall. First-Order Reactions Afirst-order reaction is a reaction whose rate depends on the concentration of one of the reac- tants raised to the first power. Two examples are the decomposition of ethane (C 2 H 6 ) into highly reactive fragments ca lled methyl radicals ('CH 3 ), and the decomposition of dinitrogen pentoxide (N 2 0 S ) into nitrogen dioxide (N0 2 ) and molecular oxygen (0 2 ): C 2 H 6 - _. 2 ·CH 3 2N 2 0 S (g) - _. 4N0 2 (g) + 0 2(g) In a first-order reaction of the type A • product rate = k[C?H 6 J rate = k[N 2 0 SJ the rate can be expressed as the rate of change in reactant concentration, as well as in the form of the rate law: Ll[A J rate = - - - I1t rate = k[AJ 556 CHAPTER 14 Chemical Kinetics It is not necessary for you to be ab le to do the calculus required to arrive at Equation 14.3, but it is very important that you know how to use Equation 14.3. The inverse of In x is f! [ ~ Appendix 1) . Think About It Don't forget the minus sign in Equation 14.3. If you calculate a concentration at time t that is greater than the concentration at time 0 (or if you get a negative time required for the concentration to drop to a specified level), check your solution for this common error. Becau se p res su re is proportional to concentration, for gaseous reactions [ H4 Section 11.4) Equat i ons 14.3 and 14.4 can be w ri tten as and Pi In- = - kt Po In P, = - kt + In Po re spectively, where Po and P, are the pressures of reactant A at times 0 and 1. respectively. Setting these two expressions of the rate equal to each other we get _ ,1[AJ = k[A] ,1t Applying calculus to the preceding equation, we can show that Equation 14.3 1 [A]t = -kt n [AJo where In is the natural logarithm, and [A]o and [A]t are the concentrations of A at times 0 and t, respectively. In general, time 0 refers to any specified time during a reaction-not necessarily the beginning of the reaction. Time t refers to any specified time after time O. Equation 14.3 is some- times called the integrated rate law. In Sample Problem 14.4 we apply Equation 14.3 to a specific reaction. Sample The decomposition of hydrogen peroxide is first order in HzOz. The rate constant for this reaction at 20 0 e is 1.8 X 10- 5 s - I. If the starting concentration of H 2 0 2 is 0.75 M, determine (a) the concentration of HzOz remaining after 3 h and (b) how long it will take for the H 2 0 2 concentration to drop to 0.10 M. - Strategy Use Equation 14.3 to find [HzOzJ " where t = 3 h, and then solve Equation 14.3 for t to determine how much time must pass for [HzOzJ, to equal 0.10 M. Setup [H 2 0 2 JO = 0.75 M; time t for part (a) is (3 h)(60 minJh)(60 s/min) = 10,800 s. Solution (a) In [HzOzJ, = -k t [HzO zJo In [HzOzJ, = -(1.8 X 10- 5 s-I)(10,800 s) = -0.1944 0.75 M Take the inverse natural logarithm of both sides of the equation to get [H 2 0 2 J, = e- O . 1944 = 0.823 0.75 M [HzOzJ, = (0.823)(0.75 M) = 0.62 M The concentration of HzOz after 3 his 0.62 M. (b) In ( ~.1O M) = -2.015 = -(1.8 X 10- 5 S-I)t .75M 2.015. =t=1.12 X 10 5 s 1.8 X 10 -) S-1 The time required for the peroxide concentration to drop to 0.10 M is 1.1 X 105 s or about 31 h. Practice Problem A The rate constant for the reaction 2A • B is 7.5 X 10- 3 S-1 at 110°C. The reaction is first order in A. How long (in seconds) will it take for [AJ to decrease from 1.25 M to 0.71 M? Practice Problem B Refer again to the reaction 2A • B, for which k = 7.5 X 10- 3 S- I at llO o e. With a starting concentration of [AJ = 2.25 M, what will [AJ be after 2.0 min? Equation 14.3 can be rearranged as follows: Equation 14.4 In [AJt = -kt + In [AJ o Equation 14.4 has the form of the linear equation y = mx + b: In [AJt = (-k)(t) + In [AJo y m x + b SECTION 14.3 Dependence of Reactant Concentration on Time 557 k ln [Ala '-< slope = -k t (a) (b) Figure 14.8(a) shows the decrease in concentration of reactant A during the course of the reaction. As we saw in Section 14.1, the plot of reactant concentration as a function of time is not a straight line. For a first-order reaction, however, we do get a straight line if we plot the natural log of reac- tant concentration (In [A]t) versus time (y versus x). The slope of the line is equal to -k [Figure 14.8(b)], so we can determine the rate constant from the slope of this plot. Sample Problem 14.5 shows how a rate constant can be determined from experimental data. I ! The rate of decomposition of azomethane is studied by monitoring the partial pressure of the reactant as a function of time: The data obtained at 300°C are listed in the following table: Time (s) Pazomethane (mmHg) 0 284 100 220 150 193 200 170 250 150 300 132 Determine the rate constant of the reaction at this temperature. Strategy We can use Equation 14.3 only for first-order reactions, so we must first determine if the decomposition of azomethane is first order. We do this by plotting In P against time. If the reaction is first order, we can use Equation 14.3 and the data at any two of the times in the table to determine the rate constant. Setup The table expressed as In P is Time (s) o 100 150 200 250 300 InP 5.649 5.394 5.263 5.136 5.011 4.883 Plotting these data gives a straight line, indicating that the reaction is indeed first order. Thus, we can use Equation 14.3 expressed in terms of pressure. PI . In- = -kt Po P t and Po can be pressures at any two times during the experiment. Po need not be the pressure at Os-it need only be at the earlier of the two times. (Continued) Figure 14.8 First-order reaction characteristics: (a) Decrease of reactant concentration with time. (b) A plot of In [All versus t. The slope of the line is equal to -k. Th is grap h ica l determin at i on is an alternative to using the method of initial rates to determi ne the value of k. 558 CHAPTER 14 Chemical Kinetics Think About It We could equally well have determined the rate constant by calculating the slope of the plot of In P versus t. Using the two points labeled on the plot, we get I 5.011 - 5.394 s ope = 250 - 100 = -2.55 X 10- 3 S-I Remember that slope = -k, so k = 2.55 X 10- 3 S-I . Solution Using data from times 100 sand 250 s of the original table (Pazomethane versus t), we get l50mmHg In 220 mmHg = -k(150 s) In 0.682 = -k(1 50 s) k = 2.55 X 10- 3 S- I Practice Problem A Ethyl iodide (C2HsI) decomposes at a certain temperature in the gas phase as follows: From the following data, determine the rate constant of this reaction. Time (min) [C~HsI] (M) 0 0.36 15 0.30 30 0.25 48 0.19 75 0. 13 Practice Problem B Use the calculated k from Practice Problem A to fill in the missing values in the following table: Time (min) o 10 20 30 40 0.45 We often describe the rate of a reaction using the half-life. The half-life (t1/2) is the time required for the reactant concentration to drop to half its original value. We obtain an expression for tll2 for a first-order reaction as follows: According to the definition of half-life, t = t1/2 when [A], = i [A]o, so 1 [A]o t 1/2 = - In ., ~ k i[A]o Because [A]of i [A]o = 2, and In 2 = 0.693, the expression for t ll2 simplifies to Equation 14.5 t - 0.693 1I2 - k According to Equation 14.5, the half-life of a first-order reaction is independent of the initial concentration of the reactant. Thus, it takes the same time for the concentration of the reactant to decrease from 1.0 M to 0.50 M as it does for the concentration to decrease from 0.10 M to 0.050 M (Figure 14.9). Measuring the half-life of a reaction is one way to determine the rate constant of a first-order reaction. The half-life of a first-order reaction is inversely proportional to its rate constant, so a sh ort half-life corresponds to a large rate constant. Consider, for example, two radioactive isotopes used in nuclear medicine: 24Na (t1/2 = 14.7 h) and 60 CO (t1/2 = 5.3 yr). Sodium-24, with the shorter half-life, decays faster. If we started with an equal number of moles of each isotope, most of the sodium-24 would be gone in a week whereas most of the cobalt-60 would remain unchanged. SECTION 14.3 Dependence of Reactant Concentration on Time 559 [Ala [Al a 12 -t+- ' [Ala/ 4 I· [Ala/ 8 O ~ ~~~~~ ~~~~ ' r r ~ o 1 2 3 4 Time (min) Sample Problem 14.6 shows how to calculate the half-life of a first-order reaction, given the rate constant. The decomposition of ethane (C 2 H 6 ) to meth yl radicals ( CH 3 ) is a first-order reaction with a rate constant of 5.36 X 10- 4 S- 1 at 7 00 °C: Calculate the half-life of the reaction in minutes. Strategy Use Equation 14.5 to calculate t 1/2 in seconds, and then conv elt to minutes. Setup Solution seconds X (1 minute/60 seconds) = minut es 0.693 tl/2 = k 1293 s X 1 min = 21.5 min 60 s The half-life of ethane decomposition at 700°C is 21.5 min. Practice Problem A Calculate the half-life of the decomposition of azomethane, discuss ed in Sample Problem 14.5. Practice Problem B Calculate the rate constant for the first-order deca y of 24Na (t 1/2 = 14.7 h). Figure 14.9 A plot of [AJ versus time for the first -order reaction A • products. The half-life ofthe reaction is 1 min. The concentration of A is halv ed ev er y half-life. Think About It Half-lives and rate constants can be express ed using any units of time and reciprocal time, respectively. Track units carefully w hen you convert from one unit of time to another. , 560 CHAPTER 14 Chemical Kinetics • Th is equa tion can also be wri tt en as 2 1( g) • I,(g). Second-Order Reactions A second-order reaction is a reaction whose rate depends on the concentration of one reactant raised to the second power or on the product of the concentrations of two different reactants (each raised to the first power). For simplicity, we will consider only the first type of reaction: where the rate can be expressed as or as A • product MA] rate = - I::: t rate = k[A] 2 As before, we can combine the two expre ss ions of the rate: _ I::: [A] = k[Af I::: t Again, using calculus, we obtain the following integrated rate law: Equation 14.6 1 = kt + 1 [A], [A] o As before, we can obtain the expression for the half-life by setting [A] , = i [A]o in Equation 14.6. Solving for tl/2, we obtain Equation 14,7 1 1 -, ' ' = kt 112 + :: ":::- i [A] o [A]o t - 1 112 - k[A] o Note that unlike the half-life of a first-order reaction, which is independent of the starting con- centration, the halfclife of a second-order reaction is inversely proportional to the initial reactant concentration. Determining the half-life at several different initial concentrations is one way to distinguish between first-order and second-order reactions, Sample Problem 14.7 shows how to use Equations 14,6 and 14,7 to calculate reactant con- centrations and the half-life of a second-order reaction, Iodine atoms combine to form molecular iodine in the gas phase: • I(g) + I(g) +. 1 2 (g) This reaction is second order and has a rate constant of7.0 X 10 9 M- 1 , S-1 at 23 ° C. (a) If the initial concentration of I is 0.086 M, calculate the concentration after 2.0 min. (b) Calculate the half-life of the reaction when the initial concentration of I is 0.60 M and when the initial concentration of I is 0.42 M. Strategy Use Equation 14.6 to detenrune [I] , at t = 2,0 min; use Equation 14.7 to determine t1/2 when [1]0 = 0.60 M and when [1] 0 = 0.42 M. Setup t = (2.0 min)(60 s/ min) = 120 s. I I SECTION 14.3 Dependence of Reactant Concentration on Time 561 Solution () I_k+l a [All - t [Alo = (7.0 X 10 9 M - I • s-I)(120 s) + 0.0~6 M = 8.4 X 1011 M- I [All = 1 = 1.2 X 10 - 12 M 8.4 X 1011 M- I The concentration of atomic iodine after 2 min is 1.2 X 10- 12 M. (b) When [110 = 0.60 M, t - 1 l iZ - k[Alo When [110 = 0.42 M, t - 1 1/2 - k[Alo 1 =2.4 X lO - lO s (7.0 X 10 9 M- I • sl)(0.60 M) • ___ -:- _ :.1_-:- ___ = 3.4 X 10 - 10 s (7.0 X 10 9 M - I • s- I)(0.42 M) Practice Problem The reaction 2A • B is second order in A with a rate constant of 32 M- I • S- I at 25°C. (a) Starting with [Alo = 0.0075 M, how long will it take for the concentration of A to drop to 0.0018 M? (b) Calculate the half-life of the reaction for [Alo = 0.0075 M and for [Alo = 0.0025 M. First- and second-order reactions are the most common reaction types. Reactions of overall order zero exist but are relatively rare. For a zeroth-order reaction A • product the rate law is given by rate = k[AJ o = k Thus, the rate of a zeroth-order reaction is a constant, independent of reactant concentration. Third- order and higher-order reactions are quite rare and too complex to be covered in this book. Table 14.5 summarizes the kinetics for first-order and second-order reactions of the type A • product. Order Rate Law o rate = k 1 rate = k[AJ 2 rate = k[AJ2 Kinetics ReaCtions Integrated Rate Law [AJI = -kt + [AJo [AJt In [AJo = -kt 1 =kt+ 1 [AJI [AJo Half-Life [AJo 2k 0.693 k 1 k[AJo Think About It (a) Iodine, like the other halogens, exists as diatomic molecules at room temperature. It makes sense, therefore, that atomic iodine would react quickly, and essentially completely, to form I z at room temperature. The very low remaining concentration of I after 2 min makes sense. (b) As expected, the half-life of this second-order reaction is not constant. (A constant half-life is a characteristic of first-order reactions .) [...]... collision that does result in a reaction is called an effective collision A molecule in motion possesses kinetic energy; the faster it is moving, the greater its kinetic energy, When molecules collide, part of their kinetic energy is converted to vibrational energy If the initial kinetic energies are large, then the colliding molecules will vibrate Is the Shroud of Turin Really the Burial Cloth of Christ?... carbon-14 isotope and hydrogen: 4COz, which The carbon-14 atoms eventually find their way into 1 mixes with the ordinary carbon dioxide ( 12C0 2) in the atmosphere As the carbon-14 isotope decays, it emits {3 particles (electrons) The rate of decay (as measured by the number of electrons emitted per second) obeys first-order kinetics It is customary in the study of radioactive decay to write the rate law as... , which is linen, a cloth made from the flax plant In 1955, American chemi st Willard F Libby suggested 4C: that the 1 12C ratio could be used to estimate the length of time the carbon-14 isotope in a particular specimen has been decaying without replenishment Rearranging Equation 14.3 , we can write No InN=kt I where No and Nt are the number of 1 nuclei present at t = 0 and 4C t = t, respectively Because... shroud, independently showed by carbon-14 dating that the shroud dates from between A.D 1260 and A.D 1390 Thus , the shroud could not have been the burial cloth of Christ Libby received the Nobel Prize in Chemistry in 1960 for his work on radiocarbon dating so strongly as to break some of the chemical bonds This bond breaking is the first step toward product formation If the initial kinetic energies are... molecules, vary greatly Normally, only a small fraction of the colliding molecules-the fastest-moving ones have sufficient kinetic energy to exceed the activation energy These molecules can therefore take part in the reaction The relationship between rate and temperature should now make sense According to kinetic molecular theory, the average kinetic energy of a sample of molecules increases as the temperature... constant at one temperature, we can determine the value of the rate constant at any other temperature Sample Problems 14.9 and 14.10 show how to use Equation 14.11 Sample Problem 14.9 The rate constant for a particular first-order reaction is given for three different temperatures: T(K) 400 450 500 2.9 X 10- 3 6.1 X 10- 2 7.0 X 10- 1 Using these data, calculate the activation energy of the reaction Strategy... reaction has an activation energy of 83 kJ/mol If the rate constant for this reaction is 2.1 X 10- 2 s -I at 150°C, what is the rate constant at 300°C? Strategy Use Equation 14.11 to solve for k 2• Pay particular attention to units in this type of problem · Setup Solving Equation 14.11 for k2 gives · · • · • • k2 = • Note that E, is converted to joules so that the units will cancel properly Alternatively,... reaction, because it does not occur in a single step If it did, the reaction would be second order in H20 2 (as a result of the collision of two H2 0 2 molecules) What's more, the r- ion, which is not even part of the overall equation, would not appear in the rate law expression How can we reconcile these facts? First, we can account for the observed rate law by assuming that the reaction takes place in... the catalyst are in different phases The catalyst is usually a solid, and the reactants are either gases or liquids Heterogeneous catalysis is by far the most important type of catalysis in industrial chemistry, especially in the synthesis of many important chemicals Heterogeneous catalysis is also used in the catalytic converters in automobiles At high temperatures inside a car's engine, nitrogen and... atmospheric conditions, thus reducing production costs and minimizing the decomposition of products at high temperatures In addition, homogeneous catalysts can be designed to function selectively for particular types of reactions, and homogeneous catalysts cost less than the precious metals (e.g., platinum and gold) used in heterogeneous catalysis Enzymes: Biological Catalysts Of all the intricate . M - 2 • S-I (5.0 X 10- 3 M) 2(2 .0 X 1O - 3 .M) (c) Using the rate constant determined in part (b) and the concentrations of NO and Hz given in the problem statement, we can determine. how much time must pass for [HzOzJ, to equal 0.10 M. Setup [H 2 0 2 JO = 0.75 M; time t for part (a) is (3 h)(60 minJh)(60 s/min) = 10,800 s. Solution (a) In [HzOzJ, = -k t [HzO zJo. experimental data. I ! The rate of decomposition of azomethane is studied by monitoring the partial pressure of the reactant as a function of time: The data obtained at 300°C are listed