Chem 116 POGIL Worksheet - Week 6 Kinetics - Part 2

Transcription

1 Chem 116 POGIL Worksheet - Week 6 Kinetics - Part 2 Why? A different form of the rate law for a reaction allows us to calculate amounts as a function of time. One variation on this gives us the concept of half-life, which applies to chemical reactions and nuclear decay. Chemical reaction rates are affected by temperature, and the magnitude of the rate increase with rising temperature is related to the energy barriers associated with steps in the mechanism of the reaction. The mechanism of the reaction, the set of molecular-level steps by which the reaction proceeds, is related to the experimentally observed differential rate law. Knowing the rate law, we can determine whether or not a proposed mechanism is or is not reasonable for the reaction. Learning Objectives Know and understand the meaning of first- and second-order integrated rate law expressions Understand the concept of half-life Understand the Arrhenius concept of the relationship between rate and temperature Understand the connection between the observed differential rate law of a reaction and its mechanism Success Criteria Be able to distinguish between first- and second-order reactions on the basis of how concentration changes with time Be able to carry out simple half-life calculations for first-order reactions Be able to sketch an Arrhenius reaction profile and calculate activation energies and reaction enthalpies Be able to carry out calculations with the Arrhenius equation Be able to determine whether or not a proposed simple mechanism is consistent with kinetic data Prerequisite Have read all of Chapter 14

2 Key Question (Review) 1. Consider the following kinetic data for the reaction 2 NO(g) + O 2(g) 2 NO 2(g). Exp. [NO] (M) [O 2] (M) Initial Rate (M/s) a. What is the rate law for the reaction? b. What are the units of the rate constant? c. What is the value of the rate constant? Information (Integrated Rate Laws) The differential rate laws we have been looking at show the dependence of rate on concentration. By using calculus, such rate laws can be converted into equations that show the dependence of concentration on time. These expressions are called the integrated rate laws. The form of the integrated rate law equation varies with the order of the differential rate law from which it is derived. We will look at first- and second-order reactions. For a reaction, A products, which is first-order in A, we can write Suppose we consider the change in concentration of A from its initial value [A] o to its value [A] at some later time t. It can be shown by integrating the differential rate law (a process in calculus) that the dependence on time will be given by This expression, in any of the forms above, is the first-order integrated rate law equation. We can more easily see how concentration of A varies with time by rearranging to give the following form: ln[a] = kt + ln[a] o

3 This is a straight-line equation of the form y = mx + b, where ln[a] is the y variable, t is the x variable, k is the slope, and ln[a] o is the y intercept. If we plot the natural logarithm of [A] versus t we should get a straight line whose slope is -k and whose intercept is ln[a]. If a reaction is second order in one reactant, say A, then its differential rate law has the form o If we integrate this equation, this time the result is where [A] is the concentration of A at some time t, k is the rate constant, and [A] o is the initial concentration of A at t = 0. This expression is the second-order integrated rate law equation. This, too, is a straight line equation of the form y = mx + b, where 1/[A] is the y variable, k is the slope, t is the x variable, and 1/[A] is the y intercept. Key Question o 2. The reaction AB 2 AB + B might be first order or second order in AB 2. Describe how you might go about determining which order it is from a table of data for concentrations of AB 2 at various times throughout the course of the reaction. [Hint: How might you plot the data to see the difference between first- and second-order?] Depending on which order it is, how could you calculate k in either case, using the appropriate integrated rate law? Information (Half Life) For any reaction we can define the half-life (symbolized t ½) as that time required for the concentration of a reactant to diminish to half of what its value was at some initial time. Thus, at t = t ½ we have [A] = ½[A] o. If we consider only first order reactions, we can obtain an expression for the relationship between the t ½ value and the value of k for the reaction rate law by substituting into the first-order integrated rate law expression. ln 2 = kt or = kt ½ ½ This shows that the half life, t ½, is a characteristic constant of any first-order reaction, and that it is inversely proportional to the rate constant, k, of the reaction. The half-life concept means that after each t ½ period only half as much A will remain from what was present at the beginning of the period. If we let the number of half-lives that have elapsed be h = t/t ½, then the concentration of A at any time is given by

4 [A] = [A] o(½) h Key Questions 3. A first-order reaction X 2 2X initially has a reactant concentration of 4.80 mol/l. After running for 96 s, X 2 concentration is mol/l. What is the half-life period of the reaction? What is the value of the rate constant, k? 4. A first-order reaction A B initially has a reactant concentration of 12.0 mol/l. If the reaction has a half life of 36.0 seconds, how much reactant will remain after 54.0 seconds? Information (Rate and Temperature) All chemical reactions speed up as temperature increases. For example, in a typical chemical o reaction it is not unusual for the rate to double for every 10 C increase in temperature. In terms of the differential rate law, this means that the rate constant, k, increases with temperature. The first successful interpretation of the relationship between k and T was made by Arrhenius in Arrhenius reasoned that any reaction process must proceed through a transition state, involving formation of a high-energy species, called the activated complex, which then breaks apart to form products. Suppose we have the hypothetical reaction A + B C + D which proceeds by a single-step mechanism in which A and B molecules come together to form products. We might imagine the reaction proceeding through the following process A + B [AB ] C + D in which [AB ] is the transition state species, the activated complex, which decomposes to give the products C and D. This can be illustrated with an Arrhenius plot, which is a semiquantitative representation of the energy of the system in the process of converting from reactants to products. For this exothermic, one-step reaction, the Arrhenius plot might look like the following.

5 For an endothermic one-step reaction, the Arrhenius plot might look like the following: f r In both cases, E a is the activation energy of the forward reaction, and E a is the activation energy for the reverse reaction. In either direction, the activation energy represents an energy barrier that must be overcome for reaction to occur. Note: If E < E, then ÄH < 0 (exothermic) f r a a rxn If E > E, then ÄH > 0 (endothermic) f r a a rxn

6 Key Question 5. The reaction, 2 NOCl(g) 2 NO(g) + Cl 2(g), is believed to proceed by a one-step o mechanism. Its standard enthalpy is ÄH = kj/mol. The activation energy of the forward reaction is 98.2 kj/mol. Sketch the Arrhenius plot for the reaction. What is the r value of E, the activation energy of the reverse reaction? a Information (Arrhenius Equation) Arrhenius was able to take this model and develop an equation for the relationship between a reaction s rate constant, k, and the absolute temperature, T. The Arrhenius equation is where A is a constant (called the Arrhenius constant) and E a is the activation energy. The value of A for a reaction is determined by the characteristics of the reactants, including possible orientation requirements for effective reaction. If we take logarithms of both sides of the Arrhenius equation we obtain ln k = ln A - E a/rt Rearranging, This is a straight line equation, in which ln k is the y variable, 1/T is the x variable, and ln k. Because only two points are needed to define a straight line, only two pairs of ln k and 1/T data are needed to define the slope. This being the case, we can obtain a single equation that relates the two pairs of ln k and 1/T data to the value of E a. At any two temperatures, T 1 and T 2, we can write

7 Subtracting the second equation from the first, we obtain This allows us to calculate E a from kinetic data that give the rate constant for the reaction at two temperatures. Moreover, proceeding in this manner we do not need to know A, the Arrhenius constant. Key Questions 6. Sketch a plot of ln k vs. 1/T. What is the value of the slope? What information can be obtained from the y intercept? 7. For the reaction 2 NOCl(g) 2 NO(g) + Cl 2(g) the experimentally determined rate law is Rate = d[cl 2]/dt = k[nocl] The rate constant, k, is 2.6 x 10 L/mol s at 300 K and 4.9 x 10 L/mol s at 400 K. What is the activation energy, E, of the reaction? a

8 Information (Mechanisms, Elementary Reactions, and Rate Law) The mechanism of a reaction is composed of a series of molecular-level steps, called elementary reactions or elementary steps. In representing the mechanism, these elementary reactions are written in the same way that we write overall reaction equations. However, in this context, they are given a special meaning: they are representations of what is presumed to occur on the molecular level. Unlike reaction equations in general, we can make a direct connection between the stoichiometry of the elementary reaction and the order with respect to each species in differential rate law expression for that step. On the molecular level, we can conceive of three plausible kinds of elementary reactions, depending on whether one, two, or three molecules are involved. Hence, every elementary reaction in a mechanism can be categorized as either unimolecular, bimolecular, or termolecular. These terms refer to the molecularity of the elementary reaction. We can relate the molecularity of an elementary reaction to the rate law expression for that step by applying the Law of Mass Action: For any elementary reaction, the rate law expression for that step is proportional to the concentration of each reactant such that the molecularity determines the overall order. Suppose we consider the elementary process of chlorine molecules, Cl 2, dissociating to form chlorine atoms (free radicals), Cl : Cl 2 2Cl The more Cl 2 we have, the more Cl will form, or in other words the rate is directly proportional to the concentration of Cl 2 and nothing else. Thus, for this elementary reaction, occurring at the molecular level, we can write its rate as rate = k[cl 2] Note that this unimolecular process has a first-order rate law. [We will use rate (not capitalized) to refer to the rate of an elementary reaction, and Rate (capitalized) to refer to the rate of an overall reaction; the two are not generally the same.] 8. Answer the following questions about elementary step molecularity and the rate expression for the step: a. Consider the elementary reaction 2 NO 2 N2O 4. What is the molecularity of this step? What is the rate law expression for this step? b. Consider the elementary reaction Cl + H 2 HCl + H. What is the molecularity of this step? What is the rate law expression for this step? + c. Consider the elementary reaction H + Br + H2O 2 HOBr + H2O. What is the molecularity of this step? What is the rate law expression for this step?

9 d. The highest conceivable molecularity is termolecular. Why? Information (Proposing Mechanisms and the Rate Determining Step Concept) The mechanism of a reaction, consisting of one or more elementary steps, must add up to give the overall stoichiometry of the reaction. Moreover, the sequence of steps and the individual rates of each elementary process determine the form of the overall rate law for the reaction. The initial piece of information the chemist uses is the observed rate law, with its various orders with respect to reacting species. Then, using knowledge about the reactivity of the various species, the chemist proposes a series of steps that will add to give the overall stoichiometry and that can be shown to yield a rate law with the same orders as the observed rate law. Except in very carefully chosen cases, this process is beyond reasonable expectations for a beginning chemistry student (i.e., you re not going to be asked to propose mechanisms on a test.) However, it is important to be able to understand how a proposed mechanism is consistent with the observed rate law. In many cases, one step in the proposed mechanism may be significantly slower than others. Such a step is called a rate determining step, because the overall reaction rate can be no faster than the rate at that step. If a rate determining step occurs, the overall rate expression must be equivalent to the rate expression at that step. In terms of the Arrhenius concept, the rate determining step has the highest activation energy, E a, of all the steps in the mechanism, and therefore it has the smallest rate constant, k. For example, consider the hypothetical reaction A D, proceeding by a three-step mechanism in which the second step is significantly slower than the others and therefore rate determining. A B fast rate 1 = k 1[A] B C slow rate 2 = k 2[B] rate determining step C D fast rate = k [C] 3 3 A D Rate rate 2 The Arrhenius plot for this hypothetical mechanism is shown on the following page.

10 Because the second step is the slowest, the overall rate of the reaction is equivalent to the rate at step 2. From the Law of Mass Action, we can always write down the rate expression for each mechanism step, including the rate-determining step, from its molecularity. However, unless the rate-determining step is the first in the mechanism, its rate expression may depend upon concentrations of species that are not initially present in the reaction mixture, but rather are created temporarily and then destroyed in the course of the mechanism. Such transient species are called reaction intermediates. In the example above, both B and C are reaction intermediates, because they are produced in one step and consumed in a subsequent step. The dependence of the rate-determining step on the concentration of a reaction intermediate (here, B) means that the rate expression for the rate-determining step will not have the same form as the experimentally observed rate expression, which is cast in terms of reactants present at the start of the reaction. In order to show that the proposed mechanism is consistent with the experimental data, it is often necessary to derive from the rate-determining step's rate expression an equivalent expression in terms of observable reactants (and sometimes products). The final proof of a proposed reaction mechanism is that it predicts the rate law expression observed by experiment, which is always cast in terms of observable reactants, not unobservable intermediates.

11 Key Questions 9. Refer to the plot shown above for the three-step mechanism by which the hypothetical reaction A D proceeds. a. From the plot, is the overall reaction exothermic or endothermic? b. If a reaction is exothermic or endothermic, must all its steps be the same (exothermic or endothermic)? c. Step 2 in the hypothetical mechanism is rate determining. Why must it have the highest activation energy? d. Look at the proposed mechanism. Why would a slow rate at step 2 slow down the whole reaction? In other words, what does it do that makes its rate the ratedetermining step? 10. Consider the following chemical reaction in aqueous solution, + 2I + 2H + H2O 2 I 2 + 2H2O for which the experimentally determined rate law expression is Observed Rate = k[i ][H2O 2] The following mechanism has been proposed: I + H O HOI + OH HOI + I OH + I H + OH H O (slow) (fast) (fast) a. Show that his series of steps gives the overall stoichiometry of the reaction. [Note: sometimes it is necessary to multiply one or more steps by appropriate factors to make everything add up to the overall stoichiometry. These factors are usually omitted when writing the series of mechanism steps, as this case illustrates.] b. From the molecularity, write the rate law expression for each step. c. Identify the reaction intermediates in this mechanism. d. Explain how this mechanism is consistent with the experimentally observed rate law.