Kinetics - Concluded - Solutions

Chem Worksheet - Week 6 Kinetics - Concluded - Solutions Key Questions 1. Consider the first-order decomposition reaction AB(s) ÷ A(s) + B(g), for which t½ = 10 s at 1000 K. If a 16.0-mol sample of AB(s) is allowed to decompose at 1000 K, how much material will remain after (a) 10 s, (b) 20 s, (c) 30 s, (d) 40 s? (a) A = (½)hAo = (½)1(16.0 mol) = 8.0 mol (b) A = (½)hAo = (½)2(16.0 mol) = 4.0 mol (c) A = (½)hAo = (½)3(16.0 mol) = 2.0 mol (d) A = (½)hAo = (½)4(16.0 mol) = 1.0 mol 2. A first-order reaction X2 6 2X initially has a reactant concentration of 4.80 mol/L. After running for 96 s, X2 concentration is 0.600 mol/L. What is the half-life period of the reaction? What is the value of the rate constant, k? First determine what fraction of the initial concentration of X2 remains after 96 s. Then, by solving [A] = (½)h[A]o, determine how many half-lives, h, have elapsed. Knowing that h = t/t½, calculate t½. The value of k is obtained from rearranging and solving t½ = 0.693/k. [X2]/[X2]o = 0.600/4.80 = 0.125 = c t½ = t/h = (96 s)/3 = 32 s 3. Y (½)h = c Y h = 3 Y k = 0.693/t½ = 0.693/(32 s) = 0.02166 s–1 = 0.022 s–1 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? h = t/t½ = 54.0 s/ 36.0 s = 1.50 [A] = (½)1.50 (12.0 mol/L) = (0.35355)(12.0 mol/L) = 4.24 mol/L 4. The reaction, 2 NOCl(g) ÷ 2 NO(g) + Cl2(g), is believed to proceed by a one-step mechanism. Its standard enthalpy is ÄHo = +75.5 kJ/mol. The activation energy of the forward reaction is 98.2 kJ/mol. Sketch the Arrhenius plot for the reaction. What is the value of Ear, the activation energy of the reverse reaction? The Arrhenius plot would look something like the following: The activation energy of the reverse reaction would be Ear = Eaf – ÄH = 98.2 kJ – 75.5 kJ = 22.7 kJ 5. 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? A plot of ln k vs. 1/T should give a downward sloping straight line, whose slope is -Ea/R and whose intercept is ln A. 6. For the reaction 2 NOCl(g) ÷ 2 NO(g) + Cl2(g) the experimentally determined rate law is Rate = d[Cl2]/dt = k[NOCl]2 The rate constant, k, is 2.6 x 10–8 L/molAs at 300 K and 4.9 x 10-4 L/molAs at 400 K. What is the activation energy, Ea, of the reaction? Plug into the two-point Arrhenius equation and solve for activation energy: 7. Answer the following questions about elementary step molecularity and the rate expression for the step: a. Consider the elementary reaction 2 NO2 ÷ N2O4. What is the molecularity of this step? What is the rate law expression for this step? The reaction is bimolecular. Therefore, rate = k[NO2]2. b. Consider the elementary reaction Cl• + H2 ÷ HCl + H•. What is the molecularity of this step? What is the rate law expression for this step? The reaction is bimolecular. The rate law will be first order in each of the reactants and second order overall; i.e., rate = k[Cl•][H2]. c. Consider the elementary reaction H+ + Br– + H2O2 ÷ HOBr + H2O. What is the molecularity of this step? What is the rate law expression for this step? The reaction is termolecular. The rate law will be first order in each of the reactants and third order overall; i.e., rate = k[H+][Br–][H2O2]. d. The highest conceivable molecularity is termolecular. Why? Termolecular implies that three species meet simultaneously to react. This is a low probability event. Expecting even more species to meet simultaneously to react is even less plausible. 8. 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? D is higher energy than A, so the reaction is endothermic. b. If a reaction is exothermic or endothermic, must all its steps be the same (exothermic or endothermic)? No. As the plot shows, steps 1 and 2 are endothermic, but step 3 is exothermic. c. Step 2 in the hypothetical mechanism is rate determining. Why must it have the highest activation energy? The rate-determining step has the slowest individual rate of all the elementary steps. From the Arrhenius equation, we know that the higher the activation energy the smaller the k value will be. If step 2 is rate-determining, it must have the smallest k value of all the steps, which also means it must have the largest 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 rate-determining step? Step 2 is the “bottle neck” in the process. Even though B is made rapidly in step 1, the rest of the reaction depends on how quickly C is made at step 2. The formation of the final product D depends on there being C present in order for step 3 to take place. No matter how fast step 3 is, the rate of formation of D can be no faster than the rate of formation of C in step 2. 9. Consider the following chemical reaction in aqueous solution, 2I– + 2H+ + H2O2 ÷ I2 + 2H2O for which the experimentally determined rate law expression is Observed Rate = k[I–][H2O2] The following mechanism has been proposed: I– + H2O2 6 HOI + OH– HOI + I– 6 OH– + I2 H+ + OH– 6 H2O a. (slow) (fast) (fast) Show that this 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.] The third step needs to be multiplied by 2: I– + H2O2 ÷ HOI + OH– HOI + I– ÷ OH– + I2 2 H+ + 2 OH– ÷ 2 H2O Overall b. (slow) (fast) (fast) 2I– + 2H+ + H2O2 ÷ I2 + 2H2O From the molecularity, write the rate law expression for each step. rate1 = k1[I–][H2O2 ] rate2 = k2[HOI][I–] rate3 = k3[H+][OH–] c. Identify the reaction intermediates in this mechanism. Both HOI and OH– are produced and consumed in the course of the mechanism; therefore, they are reaction intermediates. d. Explain how this mechanism is consistent with the experimentally observed rate law. If the first step is slow, then it is rate determining. Thus, Rate = rate1 = k1[I–][H2O2 ]. This is the same as the experimentally observed rate if k = k1. 10. Consider the gas phase reaction 2 N2O5 ÷ 4 NO2 + O2 for which the observed rate law expression is Observed Rate = k[N2O5] The following mechanism has been proposed: N2O5 º NO2 + NO3 NO2 + NO3 ÷ NO + NO2 + O2 NO3 + NO ÷ 2NO2 k1 (÷), k–1 (²) k2 k3 fast equilibrium slow fast Carry out the following steps to show that this is a plausible mechanism. i. Show that this series of steps adds to give the overall stoichiometry of the reaction. [Hint: One step needs to be multiplied by an integer factor.] 2 (N2O5 º NO2 + NO3) NO2 + NO3 ÷ NO + NO2 + O2 NO3 + NO ÷ 2 NO2 ——————————————— 2 N2O5 ÷ 4 NO2 + O2 ii. Write the rate law for each step. The first step (fast equilibrium) is reversible, so you will need to write two rate law expressions, one for each direction. Step 1, forward: Step 1, reverse: Step 2: Step 3: rate1 = k1[N2O5] rate–1 = k–1[NO2][NO3] rate2 = k2[NO2][NO3] rate3 = k3[NO3][NO] iii. Which step is rate determining? Write its rate law expression. Step 2 is rate determining, for which rate2 = k2[NO2][NO3] iv. What species are reaction intermediates? NO3 and NO v. Your rate expression, based on the rate expression for the rate-determining step, contains a concentration term for an unobservable reaction intermediate. We need to express the rate law for the rate determining step in terms of observable starting materials. Note that the first step is a rapidly established equilibrium, which means its rate in the forward direction is exactly the same as in the reverse direction. Write an equality between your two expressions for the two directions, and rearrange it to solve for [NO3]. Setting the rate law expressions for the forward and reverse directions of step 1 equal to each other, we obtain k1[N2O5] = k-1[NO2][NO3] Solving for [NO3] gives vi. Substitute your expression for [NO3] into your expression for the rate law of the rate determining step. Does this agree with the experimentally observed rate law expression? This is the same as the experimentally observed rate expression, if kexp = k1k2/k–1. On this basis, the proposed mechanism is plausible.