Chapter 14: Chemical Kinetics: Reactions in the Air We Breathe

Transcription

1 Chapter 4: Chemical Kinetics: Reactions in the Air We Breathe Problems: , 4.-4., 4.4, , 4.37, 4.4 Consider the formation of rust (or oxidation of iron). a. It can occur over a period of several years (e.g., the rusty winch below). b. It can occur quite rapidly (e.g., the combustion of iron in pure oxygen). 4. CARS, TRUCKS, AND AIR QUALITY The increased use of automobiles in the last 50 years has created air quality problems, often in the form of photochemical smog (smoke fog). The brownish gas we often equate with smog (e.g. see layer over Los Angeles in the photo at the right) is actually NO gas, which has a reddish-brown color at high concentrations: NO forms from the following series of reactions: step : N (g) + O (g) NO(g) ΔH =80.6 kj step : NO(g) + O (g) NO (g) ΔH = -4. kj Although N and O are abundant in air, the formation of NO requires very high temperatures, so it only forms inside internal combustion engines and fossil fuel power plants or when lightning strikes. When NO escapes into the atmosphere, it reacts with more O to form NO, usually with the help of energy provided by sunlight hence, the name photochemical smog. page

2 CHEMICAL KINETICS AND COLLISION THEORY chemical kinetics: study of the factors that influence reaction rates i.e., how quickly reactions occur collision theory (or collision model): molecules must collide to react Now, consider the reverse reaction or decomposition of NO : NO (g) NO(g) + O (g). System before any reactions occur System after a few NO molecules react Ex. : a. As the reaction proceeds, the [NO ]. stays the same b. As the reaction proceeds, the [NO]. stays the same c. As the reaction proceeds, the [O ]. stays the same d. As the [NO ] decreases, is the reaction more likely or less likely to occur? Explain. e. As the [NO] and [O ] increase, is the reaction more likely or less likely to occur? Explain. page

3 4. REACTION RATES We can define the rate of a reaction quantitatively as the change in concentration of a reactant or product per unit time: change inreactant (or product) concentration rate = change intime Δ [reactant] = or Δt Δ [product] Note: A reaction rate must be positive since it expresses how quickly a reaction occurs. For example, consider the decomposition of NO, a brown gas responsible for smog: NO (g) NO(g) + O (g) The reaction rate can be expressed in terms of the formation of a product: rate Δ [NO] = or rate Δt = Δ The reaction rate can also be expressed in terms of the consumption of a reactant: [O Δt ] However, because the concentration of NO decreases, Δ[NO ] is negative, so a negative sign is used since reaction rates are positive. rate = Δ [NO Δt ] Consider the following data on reactant and product concentrations for the reaction at 300 C: NO (g) NO(g) + O (g) Ex. : Consider the data at the right to calculate the rate of the reaction with respect to NO, NO, and O in parts a and b below: a. Calculate the reaction rate with respect to NO i.e., the rate of production of NO between t=0 s and t=50 s. Show the final answer with the correct units and sig figs. Δ [NO] [NO] f - [NO] i rate = = = Δt t - t f i page 3

4 b. Calculate the reaction rate with respect to O (i.e., the rate of production of O ) and with respect to NO (i.e., rate of consumption of NO ) between t=0 s and t=50 s. c. Fill in the boxes below: rate of consumption of NO rate of production of NO rate of production of O d. Write the chemical equation for the reaction then explain any differences in the rates above. e. Use the data on the previous page to calculate the reaction rate with respect to NO, NO, and O between t=50 s and t=50 s. rate of consumption of NO rate of production of NO rate of production of O f. How do these reaction rates compare to those calculated in part a? Thus, the reaction rate must be defined with respect to a specific reactant or product or a general reaction rate can be written accounting for the stoichiometric coefficients, as follows: reaction rate = Δ [NO] = Δt Δ [O Δt ] = Δ [NO Δt ] page 4

5 Ex. Consider the following reaction: A + 3 B C Write the reaction rate in terms of all the reactants and products in the reaction. Average Reaction Rates versus Instantaneous Rates Thus, as the reaction proceeds, the concentration of any reactants decreases while the concentration of any products increases. Consider the following plot of the concentrations of reactants and products over time for the reaction, NO (g) NO(g) + O (g), and the reaction rate data with respect to the [O ]: Time period (in s) 0 s 50 s 50 s 00 s 00 s 50 s 50 s 00 s 00 s 50 s Δ [O ] Δt. 0-5 M/s M/s M/s M/s M/s As we calculated previously, the reaction rate decreases over time. Note that the reaction rates in the table above are average reaction rates i.e., they are the average rates over 50- second intervals. By calculating the average reaction rate over shorter and shorter intervals of time, we can determine the rate for a specific instant in time. an instantaneous rate page 5

6 The instantaneous rate can be determined by calculating the slope of a line tangent to the concentration curve at that point in time. Ex. : a. The plot above shows the tangent drawn at t=00 s for the NO curve and at t=50 s for both NO and O. Calculate the instantaneous rate at each of these points. b. How do the rates compare? Is this expected? Ex. : Consider the following reaction, N O 5 (g) 4 NO (g) + O (g). a. Write the rate expression in terms of the consumption of N O 5 and the formation of NO and O. b. If oxygen is forming at a rate of 0.05M/s at a given instant, calculate the rate of consumption of N O 5 and the rate of formation of NO at that same instant. c. How do the rates compare? Is this expected? page 6

7 4.3 EFFECT OF CONCENTRATION ON REACTION RATE Consider the following plot of reactant concentration over time. Note the tangents drawn for three different times: (a), (b), and (c). Point (a) defines the initial rate i.e, the rate as soon as the reactants are mixed at t=0. Point (b) is the rate about half-way to equilibrium, and point (c) is the rate when the reaction almost reaches equilibrium. Ex. a. Use the plot to explain if the reaction rate increases, decreases, or stays the same as the reaction progresses. b. Use collision theory to explain the trend described in part a. Thus, experimental observations and theory suggest that reaction rates depend on reactant concentrations but do not indicate the extent of that dependence. This information is given by the concept of reaction order. REACTION ORDER AND RATE CONSTANTS We can express the rate of a general reaction, a A + b B c C + d D, in terms of the initial reactant concentrations as a rate law: rate = k [A] x [B] y where k=rate constant and x is the reaction order with respect to [A], y is the reaction order with respect to [B], and the sum of x and y is the overall reaction order. Consider the following: The rate law is written in terms of the reactant concentration because the rate is based only on the initial concentration of the reactants, before appreciable products are made. The values of x and y must be determined experimentally. They are NOT simply the coefficients in the balanced chemical equation. page 7

8 Example: Consider the following reactions and their corresponding rate laws: a. N O 5 (g) 4 NO (g) + O (g) rate = k [N O 5 ] b. CH 3 Cl(g) + Cl (g) CCl 4 (g) + HCl(g) rate = k [CH 3 Cl] [Cl ] / c. HCO H(aq) + Br (aq) H + (aq) + Br - (aq) + CO (g) rate = k [Br ] d. C H 4 Br (aq) + 3 KI(aq) C H 4 (g) + KBr(aq) + KI 3 (aq) rate = k [C H 4 Br ] [KI] Indicate the overall reaction order for examples above. DETERMINING THE FORM OF THE RATE LAW Because reactions are reversible, appreciable accumulation of the products will result in the reverse reaction occurring, which will change the concentration of reactants. To avoid this complication, we focus on the moment when the reactants are first combined and the contributions to the reactant concentration by the reverse reaction are negligible. Thus, the rate depends only on the initial concentration of reactants = initial rate. Consider the following examples that show how increasing concentrations of NO and O increases the number of possible collisions (shown as double arrows) and the number of potential reaction events, given by the experimentally determined: Rate [NO] [O ] Ex. Calculate the relative rate for each example above. page 8

9 Note again that the rate law is based on experimental data it CANNOT be determined by simply using at the coefficients in the chemical equation! Initial Rates Method for Determining a Rate Law Several experiments are carried out where the initial concentrations are changed and the reaction rates are measured. We can determine the effects of initial concentrations on the initial rates. Ex. : Consider the following reaction, a. Determine the rate law for the reaction. F (g) + ClO (g) FClO (g), and data: [F ] (in M) [ClO ] (in M) Initial Rate (in M/s) rate law: b. What is the overall reaction order? c. Solve for the rate constant, k. page 9

10 Ex. : Consider the reaction of nitric oxide with H, NO(g) + H (g) N (g) + H O(g) and the following experimental data: a. Determine the rate law for the reaction. [NO] (in M) [H ] (in M) Initial Rate (in M/s) rate law: b. What is the overall reaction order? c. Solve for the rate constant, k. Differential Rate Law versus Integrated Rate Law These last two examples have shown how experimental data on initial reactant concentrations can be used to determine the rate law for a reaction. These are considered differential rate laws (often simply called rate laws) since they express how the rate depends on the concentrations. A second type of rate law, called the integrated rate law, expresses how the concentration depends on time. page 0

11 The Integrated Rate Law or Deriving a Rate Law from a Single Experiment Consider the following general decomposition reaction: a A products which would have the following rate expression and rate law: rate Δ [A] = rate = k [A] x Δt Setting these two equal to each other yields rate Δ [A] = = k [A] x Δt We will consider the integrated rate laws for the reaction orders x=0,, and. The integrated rate laws will allow us to determine the concentration of reactants or products as a given time or the time required for the concentrations of reactants or products to reach a specified target. Zero-Order Rate Laws The decomposition of nitrous oxide, N O, N O (g) N (g) + O (g) occurs on a platinum surface. Because the platinum surface can only accommodate a limited number of N O molecules, increasing the concentration of N O molecules has no effect on the reaction rate. Pt Pt Thus, this is an example of a zero-order reaction. For a zero-order reaction, x=0, so the rate law is rate = k [A] 0 = k so rate Δ [A] = = k Δt Δ [A] = Δt [A] t [A] 0 t 0 = [A] [A] 0 t t = k where [A] t = the concentration of A at time t and [A] 0 = the initial concentration of A Thus, the integrated rate law for a zero-order reaction is [A] t [A] 0 = kt page

12 First-Order Rate Laws For a first-order reaction, x=, so the rate law is rate = k [A] so rate Δ [A] = = k [A] and using calculus and integrating yields the following. Δt [A] The integrated rate law for a first-order reaction is ln t = [A] and using algebra, this can be written as 0 kt ln [A] t ln [A] 0 = kt Example: The decomposition of hydrogen peroxide in dilute sodium hydroxide is as follows, H O (aq) H O(l) + O (g) The reaction is first-order in H O. The rate constant for the decomposition of H O at 0 C is s -. a. If the initial concentration of H O is 0.50M, what is the concentration of H O after 5.0 hours? b. How many hours will it take for the concentration of H O to drop from 0.40M to 0.0M? c. How many hours will it take for 5% of the H O to decompose? page

13 Half-Life of a First-Order Reaction half-life (t / ): the time required for the concentration of a reactant to decrease by half Consider the following first-order reaction: N O(g) N (g) + O (g). Note that after each half life, only half of the original amount is present. Next, consider the st - order decomposition of N O 5, N O 5 (g) 4 NO (g) + O (g). The following plot of [N O 5 ] versus time shows how the concentration drops from its initial concentration, [N O 5 ] 0 =0.000M. Example: What is the half life for the decomposition of N O 5? page 3

14 At the half life, t = t / and [A] t = [A] 0. For a first-order reaction, we can substitute into [A] ln [A] 0 0 = ln = = k t / Thus, for a first-order reaction, t / = k Ex. : The decomposition of hydrogen peroxide is H O (aq) H O(l) + O (g). The rate constant for this first-order reaction at 0 C is s -. Calculate the half-life in hours, then calculate the time required for a sample of hydrogen peroxide to decompose to.5% (or one-eighth) of its original concentration. Ex. : The thermal decomposition of phosphine, 4 PH 3 (g) P 4 (g) + 6 H (g), is a first-order reaction. The half-life for the reaction is 35.0 s at 680 C. a. Calculate the rate constant for the reaction. b. Calculate the time in minutes required for 95.0% of the original sample to decompose. page 4

15 Second-Order Rate Laws For a second-order reaction of the form, a A products so x=, so the rate law is rate = k [A] so rate Δ [A] = = k [A] and using calculus and integrating yields the following. Δt The integrated rate law for a second-order reaction is = kt + [A] [A] t 0 Ex.: The decomposition of hydrogen iodide is as follows, HI(aq) H (g) + I (g). The reaction is second-order in HI. The rate constant for the decomposition of HI at 40 C is 0.03 M - min -, and the initial concentration of HI is 0.500M. a. What is the concentration of HI after.5 hours? b. How many hours does it take for the HI concentration to drop from 0.50M to 0.00M? Half-Life of a Second-Order Reaction For example, for a general second-order reaction, A products, [A] At the half life, t = t / and [A] t = 0. For a second-order reaction, we can substitute into the integrated rate law: = k t / + [A] 0 [A] 0 = k t/ + [A] 0 [A] 0 = = k t/ [A] 0 [A] 0 [A] 0 page 5

16 Thus, for a second-order reaction, t / = = k [A] 0 Example: The decomposition of hydrogen iodide is as follows, HI(aq) H (g) + I (g). The reaction is second-order in HI. The rate constant for the decomposition of HI at 40 C is 0.03 M - min -, and the initial concentration of HI is 0.500M. a. Calculate the half-life (in hours) for the sample of HI described above. b. Calculate the half-life (in hours) for the remaining sample of HI after the first half-life described in part a. c. Calculate the half-life (in hours) for the remaining sample of HI after the two half-lives described in part b. d. Are the half-lives for a second order reaction constant like the half-lives for a first-order reaction? If not, explain if each successive half-life is longer or shorter than the previous half-life. page 6

17 Thus, for a second-order reaction, t / =, so the half-life is not constant, and it is k [A] 0 dependent on the initial reactant concentration. After each half-life, the original concentration is cut in half [A] i.e., for each subsequent half life, [A] 0 becomes 0 t/ = = k [A] 0 k [A] 0 Thus, for second-order reactions each subsequent half-life is always twice as long as the previous half-life, as shown on the plot of concentration versus time for a second-order reaction below.. Half-Life of a Zero-Order Reaction For example, for a general zero-order reaction, A products, Example: Starting with the integrated rate law for a zero-order reaction, [A] t [A] 0 = kt, calculate the general formula for the half-life for a zero-order reaction. page 7

18 But what if you don t know the reaction order for a reaction? Using experimental data of concentration versus time, you can plot the data and determine which plot gives you a straight line. Plots of Zero-Order Reactions Rearranging the integrated rate law for a zero-order reaction yields [A] t = kt + [A] 0 which has the form of the linear equation, y = mx + b. y = m x + b Plotting concentration versus time for a zero-order reaction will yield a straight line with a slope=-k. Thus, if one plots the experimental data for reactant concentration versus time for a reaction of unknown order, a straight line with a negative slope indicates the reaction is zero order. [A] slope = -k Plots of First-Order Reactions Rearranging the integrated rate law for a first-order reaction yields ln[a] t = k t + ln[a] 0 which has the form of the linear equation, y = mx + b. y = m x + b Plotting experimental data for ln [A] versus time will yield a straight line with a negative slope for all first-order reactions. Consider experimental data at 45 C for the following reaction, N O 5 (g) 4 NO (g) + O (g). time (in min) [N O 5 ] (in M) ln [N O 5 ] Consider the two plots below, and note that only a plot of ln [N O 5 ] versus time is linear. time [NO5] (in M) ln [NO5] - slope = -k Time (in min.) - Time (in min.) page 8

19 Plots of Second-Order Reactions Rearranging the integrated rate law for a nd -order reaction yields = kt + [A] [A] t 0 which has the form of the linear equation, y = mx + b. y = m x + b Plotting experimental data of versus time will yield a straight line for a nd -order reaction. [A] Consider experimental data at 40 C for the following reaction, HI(g) H (g) + I (g). time (in min) [HI] (in M) ln [HI] /[HI] Consider the three plots below, [HI] 0.3 ln [HI] Time (in min.) -.7 Time (in min.) slope = k /[HI] Time (in min.) Thus, for second-order reactions, only plots of versus time are linear, where slope=k. [A] page 9

20 Note that plotting the data can be time-consuming, and a much easier way to determine reaction order is by comparing slopes. Any linear plot will result in the same slope (or similar slopes) for any two sets of points. If the set of points is not linear, then the slope for one pair of points will not be the same (or similar) to the slope for a second pair of point. This difference in the slopes is often greater for exponential decay, especially if the pairs of points compared are near the start and near the end of the curve. Ex. Consider experimental data at 0 C for the reaction, NOBr(g) NO(g) + Br (g). a. Compare the slopes using the first two and the last two sets of points for a plot of [NOBr] versus time. If the slopes are similar, the reaction is what order? time (in min) [NOBr] (in M) b. Compare the slopes using the first two and the last two sets of points for a plot of ln [NOBr] versus time. If the slopes are similar, the reaction must be what order? c. Finally, compare the slopes using the first two and the last two sets of points for a plot of versus time. If the slopes are similar, the reaction must be what order? [NOBr] d. Thus, the following, NOBr(g) NO(g) + Br (g), is a -order reaction. e. Calculate the rate constant, k, for the reaction. page 0

21 Ex. Consider data on the decomposition of ethyl iodide, C H 5 I(g) C H 4 (g) + HI(g). a. Determine the reaction order by comparing the slopes using the first two and the last two sets of points for a plot of [C H 5 I] versus time, ln [C H 5 I] versus time, and versus time. [C H I] 5 time (in s) [C H 5 I] (in M) Thus, this is a -order reaction. b. Write the rate law for the reaction: c. Using the data given, can you determine an approximate half-life for the reaction? Is the halflife constant or close to constant? (Consider what reaction order has a constant half-life.) d. Determine the rate constant and the half-life for the reaction. page

22 Ex. 3 Consider data on the decomposition of NO, NO (g) NO(g) + O (g). a. Determine the reaction order by comparing the slopes using two sets of points for a plot of [NO ] versus time, ln [NO ] versus time, and versus time. [NO ] time (in s) [NO ] (in M) Thus, this is a -order reaction. b. Write the rate law for the reaction: c. Using the data given, can you determine an approximate half-life for the reaction? Is the halflife constant or close to constant? Consider what reaction has a constant half-life. d. Determine the rate constant and the half-life for the reaction. page

23 4.4 REACTION RATES, TEMPERATURE, AND THE ARRHENIUS EQUATION collision theory (or collision model): molecules must collide to react Consider the following molecular-level images for the reaction: NO (g) NO(g) + O (g). System before any reactions occur System after a few NO molecules react Experiments show that virtually all rate constants increase exponentially with an increase in absolute (Kelvin) temperature. Ex. : Using kinetic molecular theory, explain why this observation is expected. However, the reaction rate is generally much smaller than the calculated frequency in a collection of gas particles. Increasing the number of collisions is not sufficient to account for the increase in the rate constant and the reaction rates. Every collision does not necessarily result in a reaction. An additional requirement must be met for a reaction to occur. In 880s, Svante Arrhenius proposed the existence of a threshold energy, called the activation energy, that reactants must have for the reaction to occur. The reactants kinetic energy is converted to vibrational energy that causes bonds to break. Not only must the molecules collide, but they must have sufficient energy to collide and vibrate strongly enough to break and make chemical bonds. page 3

24 Consider the following plot showing how the number of particles with higher energy increase with temperature: Thus, as temperature increases, the number of reactant molecules with the required activation energy, E a, increases. More collisions and more reactants with the required E a have the vibrational energy resulting in bonds breaking. More reactions The reaction rate increases. Potential Energy Diagrams and Covalent Bond Formation When two H atoms come near one another, the nucleus of one atom is attracted to the electrons of the other atom. Together each H atom has electrons like a He atom has electrons. They become more stable like He (indicated by the drop in energy in the diagram below). Note: In potential energy diagrams like the one below, the more negative the energy, the stronger the bond(s) and the more stable the compound. However, the atoms cannot come too close together, or their nuclei and electrons get too close, resulting in nucleus-nucleus and electron-electron repulsions The optimal distance between nuclei = bond length. Thus, the energy associated with the formation of any molecule can be shown in an energy diagram like the one above. The energy associated with molecules reacting involving bonds breaking and bonds forming can be shown with a reaction energy profile or reaction energy diagram. page 4

25 Consider the following reaction energy profile for the reaction: O 3 + NO NO + O In this reaction energy profile the energy of the reactants and products is plotted on the y-axis and the progress of the reaction is shown on the x-axis. In order for the reactants to be converted to products, they must collide with sufficient energy to form the activated complex (the arrangement of atoms where the bonds between the reactants are half broken and the bonds between products are half formed). The energy state when the activated complex exists is called the transition state and corresponds to the highest energy state on the reaction energy diagram. Reactants must collide with sufficient energy to break and form bonds to make the activated complex, and this energy requirement is called the activation energy (E a ) for the reaction. Note that the difference in energy between the reactants and products is equal to the change in energy, which is often equal to the change in enthalpy, ΔH. The reaction energy profile shows if a reaction is exothermic or endothermic. Ex. : Consider the reaction energy profile above to answer the following questions: a. What is the activation energy (E a ) for the forward reaction? b. What is the activation energy (E a ) for the reverse reaction? c. Given their relative energies, the bonds are stronger in the. (Circle one) reactants products d. So when the reaction occurs, it heat. absorbs releases e. Thus, the reaction is. endothermic exothermic page 5

26 Ex. : Consider the following reaction energy profile: a. Give the number corresponding to each of the following: i. E a for the forward reaction ii. E a for the reverse reaction i. ΔH for the forward reaction c. This reaction is. endothermic exothermic Arrhenius postulated that the number of collisions with energy greater than or equal to E a is given by the following: # of collisions with E a = (total # of collisions) e RT where E a =activation energy, R=ideal gas constant (8.345 J/mol K), and T is temperature (in K). - E a The factor, e RT, represents the fraction of collisions with energy E a at temperature T. However, experimental results showed that the observed reaction rate was lower than the rate of collisions with the required activation energy. Many collisions with the required activation energy do not result in a reaction. Why not?!!! - E a Consider the following reaction: NO(g) + O 3 (g) NO (g) + O (g) correct collision geometry reaction occurs incorrect collision geometry no reaction page 6

27 Thus, the reaction only occurs when the molecular orientations can lead to effective bond formation for the products. To summarize, the two requirements for a reaction to occur are:. The collision energy must equal or exceed the activation energy, E a.. Only the correct orientation of the reactant molecules results in the formation of new bonds to make the products and the breaking of old bonds in the reactants. Taking these two factors into account, the rate constant - E a can be represented as follows in the Arrhenius equation: k = A e RT - E a RT where A=frequency factor and e = the fraction of collisions with the energy E a and A = z p where z=collision frequency and p=steric factor (<) that reflects the fraction of collisions with the correct orientation/geometry Determining the Activation Energy (E a ) from Experimental Data Taking the natural log of the Arrhenius equation and rearranging so it s in the form of a linear equation yields the following: E ln k = ( ) a + ln A R T A plot of ln k versus E for a reaction at several temperatures gives a slope equal to a T R and a y-intercept equal to ln A. For example, consider the following plot for the reaction: N O 5 N O 4 + O. Ex. : Use the plot to calculate E a (in kj/mol) for the reaction, using temperature (T) is in Kelvins and R=8.345 J. mol K page 7

28 page 8 Note at temperature T where the rate constant is k, ln k = T R E a + ln A And at temperature T where the rate constant is k, ln k = T R E a + ln A Subtracting the first from the second equation yields ln k ln k = + A ln T R a E + A ln T R a E = + T R a E T R a E This is the two-point form of the Arrhenius equation: = T T R a E k k ln Ex. : Consider the following reaction: N O 5 4 NO + O. Calculate the activation energy (in kj/mol) if the rate constants for the reaction are s - at 5 C and s - at 55 C.

29 Ex. : The first-order rate constant for the reaction of methyl chloride (CH 3 Cl) with water to produce methanol (CH 3 OH) and hydrochloric acid is s - at 5 C. Calculate the rate constant at 45 C if the activation energy for the reaction is 6 kj/mol. Ex. 3: The rate constant for a first-order reaction is s - at 350 C. At what temperature is the rate constant s - if the activation energy for the reaction is 04 kj/mol. page 9

30 4.5 REACTION MECHANISMS Most chemical reactions occur by a series of steps called the reaction mechanism. A single step in a reaction mechanism is called an elementary step (or elementary reaction). For example, consider the following reaction, which occurs by a two-step mechanism: NO (g) + CO(g) NO(g) + CO (g), Step : NO (g) + NO (g) NO(g) + NO 3 (g) first elementary step Step : NO 3 (g) + CO(g) NO (g) + CO (g) second elementary step NO (g) + CO(g) NO(g) + CO (g) overall reaction Note that the elementary steps describe the individual molecular events that involve bonds breaking and bonds being made, as shown below. In contrast, the overall reaction describes the stoichiometry of the overall process. An intermediate is a substance that is formed and consumed during the reaction sequence. Example: Identify the intermediate in the reaction shown above. page 30

31 Each elementary step is classified on the basis of its molecularity. The molecularity is defined as the number of species/molecules that collide/react in the elementary step. I. A unimolecular step involves only a single reactant molecule breaking up. e.g., the decomposition of O 3 is a unimolecular step. (Note the asterisk in O 3 indicates an excited form of ozone that is unstable, causing one of the O-O bonds to break.) II. A bimolecular step involves two reactant molecules/atoms colliding in order to react. e.g., the reaction between O 3 and O atom in the upper atmosphere is an example of a bimolecular step/reaction. III. A termolecular step involves three reactant molecules/atoms colliding in order to react. e.g., the reaction between two O atoms in the upper atmosphere often involving a third molecule, M, is an example of termolecular reaction. Note: Termolecular reactions are extremely rare! What is the probability of three molecules colliding simultaneously with the correct orientation? This is highly improbable! Most reactions occur as a series of unimolecular and bimolecular steps. page 3

32 While the rate law for a reaction cannot be determined simply knowing the overall chemical reaction based on the stoicihiometry, the rate law for each elementary step follows the molecularity of the step. Elementary Step Molecularity Rate Law A products A products A + B products A + B products A + B + C products unimolecular bimolecular bimolecular termolecular termolecular rate = k [A] rate = k [A] rate = k [A] [B] rate = k [A] [B] rate = k [A] [B] [C] Thus, if a reaction mechanism is valid if meets the following two requirements:. The sum of the elementary steps must give the overall balanced equation for the reaction.. The mechanism must agree with the experimentally determined rate law. Consider again the following reaction: NO (g) + CO(g) NO(g) + CO (g) Step : NO (g) + NO (g) NO(g) + NO 3 (g) first elementary step Step : NO 3 (g) + CO(g) NO (g) + CO (g) second elementary step NO (g) + CO(g) NO(g) + CO (g) overall reaction Thus, the first requirement is met since canceling the intermediate (NO 3 ) and the extra NO molecule results in the overall reaction. Now, consider that experimental results indicate that the rate law is rate = k [NO ]. This matches the molecularity of the first elementary step: Step : NO (g) + NO (g) NO(g) + NO 3 (g) first elementary step Thus, if only the first elementary step in the mechanism determines the reaction rate, then this mechanism is valid. When a reaction occurs in two or more elementary steps, the slowest elementary step in the reaction mechanism is the rate-determining step. Similar to traffic, the bottleneck that results from the slowest moving cars will regulate the overall rate for all of the cars. Thus, the following mechanism would match the experimental rate law, rate = k [NO ]. Step : NO (g) + NO (g) NO(g) + NO 3 (g) slow (rate-determining step) Step : NO 3 (g) + CO(g) NO (g) + CO (g) fast page 3

33 Ex. : Consider the following reaction, NO(g) + Cl (g) NOCl(g), which has the following mechanism: k Step : NO(g) + Cl (g) k Step : NOCl (g) + NO(g) NOCl (g) slow NOCl(g) fast a. Determine the rate law for the reaction: b. What is NOCl? Explain. c. If ΔH=-77. kj for the overall reaction, sketch the reaction energy diagram for the reaction accounting for the number of steps, the relative heights of the activation energies for each step, and the relative energies of the reactants and products. Ex. : The gas-phase decomposition of nitrous oxide, N O(g) N (g) + O (g), is believed to occur via two elementary steps: k Step : N O(g) N (g) + O(g) k Step : N O(g) + O(g) N (g) + O (g) a. If the experimentally determined rate law is rate = k [N O], indicate the ratedetermining step, and explain your choice. b. If ΔH=-64. kj for the overall reaction, sketch the reaction energy diagram for the reaction accounting for the number of steps, the relative heights of the activation energies for each step, and the relative energies of the reactants and products. page 33

34 Ex. 3: Consider the following reaction, NO(g) + O (g) NO (g), which has the following mechanism: k Step : NO(g) N O (g) k Step : N O (g) + O (g) fast NO (g) slow Determine the rate law for the reaction: The experimentally determined rate law for the reaction, NO(g) + O (g) NO (g), is actually rate = k [NO] [O ]. Does this match the rate law you determined above? Yes No The difference is because your rate law includes the concentration of the intermediate, [N O ]. Because intermediates are produced and consumed in the overall reaction, the concentration of an intermediate should not appear in the rate law. The rate law can only include the concentrations of reactants in the overall reaction. The concentration of an intermediate usually appears in a proposed rate law when the slow, rate-determining elementary step is not the first elementary step in the proposed mechanism. To eliminate the intermediate, we make the assumption that the fast elementary step reaches equilibrium (where the forward and reverse rates are equal). Thus, we can rewrite the mechanism as follows: k- Step : NO(g) k N O (g) Step : N O (g) + O (g) k fast NO (g) slow Ex. a. Write the rate law determined in Ex. 3 above? b. Write the rate law for the forward reaction in the first elementary step, then write the rate law for the reverse reaction in the first elementary step. c. Because the first elementary step is fast enough to achieve equilibrium (where the forward and reverse rates are equal), set the two rate laws written in part b equal to one another, then solve for [N O ]. page 34

35 d. Substitute the expression you determined in part c for [N O ] in part a. e. Does this new rate law match the experimentally determined rate law? Yes No Ex. : Consider the following reaction, CHCl 3 (g) + Cl (g) HCl(g) + CCl 4 (g), which has the following mechanism: k- Step : Cl (g) Step : Cl(g) + CHCl 3 (g) k Step 3: Cl(g) + CCl 3 (g) 3 k Cl(g) fast k HCl(g) + CCl 3 (g) CCl 4 (g) slow Consider that the first step occurs so quickly that the system reaches equilibrium. Determine the rate law for the reaction that contains only concentrations of reactants in the overall reaction. fast page 35

36 4.6 CATALYSIS catalyst: A substance that speeds up a reaction by providing an alternative pathway with a lower activation energy for the reaction A catalyst is not consumed in the reaction. Homogeneous Catalysis A homogeneous catalyst is in the same phase (i.e., physical state) as the reactants. For example, consider the decomposition of formic acid, HCOOH, catalyzed by H + : HCOOH(aq) H + CO(aq) + H O(l) Without the H + catalyst, the reaction proceeds slowly because of the high activation energy. However, when the H + catalyst is present, the reaction occurs much faster because the H + catalyst provides an alternative pathway with a lower activation energy. Ex. : When H+ is present to act as a catalyst in the decomposition of formic acid, to what atom in HCOOH does it bind? Ex. : In step #, the Lewis base is, and the Lewis acid is. Ex. 3: What bond breaks or forms after H + binds to the HCOOH molecule? Ex. 4: Thus, H + effectively lowers the activation energy for the reaction by easing what process? Explain. Ex. 5: Given the reaction energy diagrams above, what is the rate-determining step when H + is present as a catalyst? Explain how this was determined. page 36

37 Heterogeneous Catalysis A heterogeneous catalyst is in a different phase than the reactants. Most often, a solid catalyst is used to increase the rate of a gas-phase or liquid-phase reaction (e.g. catalytic converters in cars catalyze the oxidation of CO and unburned hydrocarbons to CO and H O). For example, the hydrogenation of ethylene, C H 4 (g) + H (g) C H 6 (g), occurs much faster on a metal catalyst. (a) C H 4 and H are adsorbed onto the metal surface. (b) The H breaks up into H atoms, and one H atom bonds to C H 4 to form a metalbonded C H 5. (c) The second H atom bonds to the C H 5. (d) The resulting C H 6 desorbs from the metal surface. A general reaction energy profile for a reaction catalyzed by a heterogeneous catalyst is shown at the right. Note again that the activation energy, E a, is dramatically decreased by the catalyzed reaction, so more reactants have the required E a, and the reaction proceeds more quickly. Example: Identify the rate-determining step in the hydrogenation of ethylene on a metal catalyst. page 37

38 Enzymes: Biological Catalysts In living organisms, most biochemical reactions occur too slowly at room temperature or even at body temperature. Living organisms use enzymes, which are catalysts that speed up biochemical reactions. Enzymes are extremely specific, and each enzyme has a unique pocket called the active site that holds a particular molecule, like a lock that corresponds only to a specific key. Each enzyme catalyzes only the reaction involving its specific molecule. Each enzyme catalyzes only one reaction. Thus, living organisms can switch a biochemical reaction on or off by producing or not producing the enzyme that catalyzes that particular reaction. For example, when people consume table sugar (sucrose=c H O ), the human body must break down the C H O into two smaller molecules, glucose and fructose (both C 6 H O 6 ), as shown in the image below. However, this reaction occurs too slowly even at body temperature. An enzyme called sucrase is required to speed up the reaction. The sucrose binds to the sucrase at the active site, and binding to the sucrase weakens the bond between the glucose and fructose molecules. This allows the reaction to proceed more quickly. page 38

39 Another example of an enzyme is catalase, a common enzyme found in nearly all living organisms exposed to oxygen that catalyzes the decomposition of hydrogen peroxide (H O ) to water and oxygen gas. One catalase enzyme can break down 40 million H O molecules per second. However, as we age, catalase production in the human body tails off, leaving nothing to convert the H O into water and O gas the body can release. As hydrogen peroxide builds up in the body, it causes the bleaching of hair from the inside out, so we go gray, according to researchers at the University of Bradford in the United Kingdom, who published the results of a study in a 009 issue of the Federation of American Societies for Experimental Biology. ( fj v) Protein Data Bank (PDB) Rendering of Catalase Example: The conversion of O to ozone, O 3, is catalyzed by nitric oxide, NO, according to the following series of reactions: O (g) + NO(g) NO (g) NO (g) sunlight NO(g) + O(g) O (g) + O(g) O 3 (g) a. Show the overall reaction above. b. In this reaction NO is the. (Circle one) intermediate catalyst c. In this reaction NO is the. (Circle one) intermediate catalyst d. Explain the difference between NO and NO in the reaction above. page 39