Ch part 2.notebook. November 30, Ch 12 Kinetics Notes part 2

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1 Ch 12 Kinetics Notes part 2 IV. The Effect of Temperature on Reaction Rate Revisited A. According to the kinetic molecular theory of gases, the average kinetic energy of a collection of gas molecules is proportional to the absolute temperature. B. At a given temperature, T 1, a definite fraction of the collisions have sufficient energy, E a, to react to form product molecules upon collision. C. At a higher temperature, T 2, a greater percentage of the collisions possess the necessary activation energy, and the reaction A. proceeds at a higher rate. D. What two factors work to increase the rate of reaction when the temperature is increased? At a hig necessar V. Rate Laws: Dependence of the Rate of Reaction on Concentrations of Reactants A. We are now aware that the higher the reactants concentrations, the greater the number of collisions per second between reactant molecules. 1. Consider the hypothetical one step gaseous reaction A + B AB. In the reaction vessel pictured below (I) there is one atom of A and one atom of B. 4. This same reaction vessel shown in (II) has one atom of A and two atoms of B in it. 3. Since these are gas molecules, they move at high velocities, and collide with each other and the walls of the container. 5. How does the number of collisions per second of the A atom with a B atom in vessel I compare with the number of collisions per second of A with in vessel II? 6. Based on the collisions, the rate of the reaction is directly proportional to the concentration of B. 1

2 7. This reaction vessel shown in (III) contains two atoms of A and one atom of B. B. The relationship between the rate of reaction and the concentrations of A and B can be expressed as Rate = k[a][b], where k is a proportionality constant called a rate constant. C. Rate = k[a][b] is called a rate law, which describes how reaction rate depends on reactant concentrations. The rate law for any step of a reaction mechanism is equal to the step s rate constant times the product of the reactant concentrations in that step. 8. How does the number of collisions per second of the B atom with an A atom in vessel I compare with the number of collisions per second of B with in vessel III? 9. The rate of the reaction is directly proportional to the concentration of A If the hypothetical chemical reaction AB + AB AB 2 + A goes by the mechanism AB + AB 2A + B 2 (step 1) and A + B 2 AB 2 (step 2), the rate law for the mechanism s first step is rate = k 1 [AB][AB] or rate = k 1 [AB] 2. Problem: What is the rate law for the second step of the mechanism? 1. The rate law for an entire chemical reaction that has reactants A and B can be expressed as rate = k[a] x [B] y. 2. The value of x is said to be the order of the reaction with respect to A. 3. The value of y is the order of the reaction with respect to B. 4. The overall order of the reaction is x + y. E. Let s explore how the rate law for a reaction is obtained from experimental data: 1. The data in the table below represents experimentally measured data for the chemical reaction A 2 + 2B 2AB at a specific temperature. Trial Run Initial [A2] Initial [B] Initial Rate of Formation of AB (Ms 1) M M 4.0 x M M 4.0 x M M 8.0 x 10 5 (a) We begin with the general form, rate = k[a 2 ] x [B] y. (b) Here s one way to find the values of x and y: 1. Examine the data in trials 1 and 2. Which concentration changes, a by how much does it change? 2. What does the exponent (y) on [B] have to be in order for the rate formation of AB to remain unchanged between trials 1 and 2? Examine the data in trials 2 and 3. Which concentration changes and by how much? 2

3 4. What does the exponent (x) on [A 2 ] have to be in order for the rate of formation of AB to double between trials 2 and 3? 5. What is the rate law for this reaction? (c) The rate law can also be determined algebraically, by using the general rate equation, rate = k[a 2 ] x [B] y, and solving for x and y. (e) The rate constant (k) is easily determined by inserting the data for any trial into the rate law. For trial 1, k = 4.0 x 10 5 Ms 1 / ( M) = s 1. (f)what rate constant would be obtained using data from trial 2? 1. Putting the data into this equation and dividing trial one s data by trial two s data, we get 4.0 x 10 5 = k(0.0010) x (0.0010) y or 1 = (1/2) y. y = x 10 5 k(0.0010) x (0.0020) y 2. Dividing the data from trial 2 by the data from trial 3, we get 4.0 x 10 5 = k(0.0010) x (0.0020) y or ½ = (1/2) x. x = x 10 5 k(0.0020) x (0.0020) y (b) So, as before the rate law is rate = k[a 2 ]. F. As another example, kinetic data on the reaction H 2 (g) + Cl 2 (g) 2HCl(g) are given in the table below. Trial Initial [H2] (mole/l) Initial [Cl2] (mole/l) Initial Rate of Formation of HCl(mole/Ls) x x x Problem: What is the rate law for this reaction and what is the value of the rate constant? Specify units in your answers VI. Reaction Mechanisms and the Rate Law A. Most reaction mechanisms involve more than one step. 1. The slow step of a multistep mechanism is called the rate determining step. 2. The speed with which the slow step occurs determines the rate at which the overall reaction occurs. 3. Using a combination of experimental data and our prior experience with reactions, it is possible to postulate a mechanism by which a reaction occurs. B. A chemical reaction goes by the multistep mechanism, AC + AC A 2 C 2 (slow) A 2 C 2 + B 2 AB 2 C + AC (fast) 3

4 1. What is the chemical equation for this reaction? 2. What is the rate law for this reaction? 3. Generalizing from the example shown above, what two requirements must be met by any reaction mechanism? C. Any mechanism that meets these two criteria is valid, and additional experimental evidence is needed to determine which mechanism describes the actual reaction process. D. Consider the reaction NO 2 (g) + CO(g) NO(g) + CO 2 (g) below 225 o C. Rate = k[no 2 ] 2 1. The following two step mechanism has been proposed: NO 2 + NO 2 N 2 O 4 (slow) N 2 O 4 + CO NO + CO 2 + NO 2 (fast) 2. Does this mechanism meet the two criteria for an acceptable mechanism? Explain. 3. There is another mechanism that has been proposed for this reaction: NO 2 + NO 2 NO 3 + NO (slow) NO 3 + CO NO + CO 2 (fast) E. Dinitrogen oxide, N 2 O, decomposes by the chemical reaction 2N 2 O 2N 2 + O 2 with the rate law, rate = k[n 2 O]. How would we write a mechanism for this reaction that meets the two criteria for all acceptable mechanisms? 4. Does this mechanism also meet the two criteria for an acceptable mechanism? Explain. 5. This latter mechanism is thought to be the correct one because nitrogen trioxide, NO 3, has been identified from experimental data as a short lived reaction intermediate. 4

5 F. We have seen that the rate determining step of a multistep mechanism determines the experimentally observed rate of reaction. 1. When the rate determining step is the first step of the mechanism, the experimentally determined rate law is just the rate law for the first step. 2. But, let s look at a mechanism in which the first step is not the rate determining step. 3. Consider the chemical reaction 2ClO 2 (g) + F 2 (g) 2ClO 2 F (g). 4. Suppose this reaction goes by the following mechanism: 1. ClO 2 + F 2 ClO 2 F 2 (fast) k 2 k 3 ClO 2 F 2 ClO 2 F + F (slow) ClO 2 + F ClO 2 F (fast) 5. What is the rate law for the slow step of this reaction? (a) Does ClO 2 F 2 appear in the overall reaction equation? (b) Would you characterize ClO 2 F 2 as a catalyst or a reaction intermediate? (c) Do you think [ClO 2 F 2 ] would be easy to measure experimentally? Justify your answer. (d) Because ClO 2 F 2 is difficult to measure, we need to express the rate law in terms of the overall reaction s reactants, whose initial concentrations are known. 6. Let s now eliminate ClO 2 F 2 from the rate law. 7. Since the first step is faster than the second step, a point is reached where the rate of the forward reaction in the first step is equal to the rate of the reverse reaction in the first step. k 1 [ClO 2 ][F 2 ] = k 2 [ClO 2 F 2 ] 8. Solving for [ClO 2 F 2 ], we get [ClO 2 F 2 ] = (k l /k 2 )[ClO 2 ][F 2 ] 9. Substituting this expression into the rate law gives rate = (k 3 k 1 /k 2 ) [ClO 2 ][F 2 ], or rate = k [ClO 2 ][F 2 ] Problem: The chemical reaction 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g) is thought to go by the following mechanism NO + NO k 2 N 2 O 2 (fast) k 3 N 2 O 2 + H 2 N 2 O + H 2 O (slow) N 2 O + H 2 N 2 + H 2 O (fast) What is the rate law for this reaction? 10. The experimentally determined rate law shows that it is first order in ClO 2 and first order in F 2. 5

6 VII. Integrated Rate Equations A. Up to this point, we have studied kinetics in terms of initial reactant concentrations. However, concentrations change as the reaction proceeds over time. Thus, the rate also changes over time. Using calculus to integrate the rate law expressions, we obtain new expressions that relate concentration as a function of time. B. Zero order reactions: [A] = kt + [A] 0 C. First order in a particular reactant and first order overall: ln [A] = kt + ln [A] 0 as natural logarithms, or as log [A] = kt log [A] 0 or log [A] 0 /[A] = kt/2.303 for base 10 logarithms. Problem: The reaction N 2 O 5 (g) N 2 O 4 (g) + ½O 2 (g) obeys the rate law Rate = k[n 2 O 5 ] in which the specific rate constant is 1.68 x 10 2 s 1 at a certain temperature. (a) What is the half life of N 2 O 5 at this temperature? (b) If 2.50 moles of N 2 O 5 (g) are placed in a 5.00 liter container at that temperature, how many moles of N 2 O 5 would be left after 1.00 minute? 1. [A] o represents the initial concentration of A and [A] represents the concentration at time, t. The half life (t 1/2 ) of a first order reaction is the time it takes for one half of a reactant to be used up. At t = t 1/2, [A] = ½[A] o. kt 1/2 /2.303 = log [A] 0 /(1/2[A] 0 t 1/2 = (2.303 log 2)/k t 1/2 = 0.693/k D. Second order in a particular reactant and second order overall: 1/[A] = kt + 1/[A] The relationship between the half life and the rate constant can be obtained by substituting t 1/2 for t and ½[A] 0 for [A] in this equation. 1/1/2[A] 0 1/[A] 0 = kt 1/2 Problem: The recombination of iodine atoms to form molecular iodine in the gas phase I (g) + I (g) I 2 (g) is second order in I (g) and second order overall. The rate constant (k) for this reaction is 7.0 x 10 9 M 1 s 1 at 23 o C. (a) If the initial concentration of I was M, calculate the concentration after 2.0 min. 1/1/2[A] 0 (1/2)/(1/2[A] 0 ) = kt 1/2 (1 ½)/(1/2[A 0 ] = kt 1/2 1/[A] 0 =kt 1/2 t 1/2 = 1/k[A 0 ] (b) Calculate the half life of the reaction if the initial concentration of I is 0.60 M. 6

7 3. The table below shows integrated rate equations for zero order, firstorder and second order, expressed in y = mx + b format, to facilitate graphing as straight lines. Problem: The hypothetical molecule A 2 decomposes by the reaction A 2 2A. Kinetic data were taken on this reaction by measuring the concentration versus time. Three graphs on the data were made and are shown below. What is the order of this reaction? Order Integrated Rate Equation y x Graph Slope (m) SlopeUnits t1/2 Zero [A]= kt+[a] 0 [A] t [A] vs t k Ms 1 [A] 0 /2k First ln[a]= kt+ln[a] 0 ln [A] t Second 1/[A]=kt+1/[A] 0 1[A] t ln [A] vs k s t /k 1/[A] vs t k M 1 s 1 1/k[A] 0 7