Homework #4 Chapter 5 Chemical Kinetics 8. Arrhenius Equation Therefore, k depends only on temperature. The rate of the reaction depends on all of these items (a d). 4. a) d) b) c) e) 5. Rate has units of therefore, the units of the rate constant must compensate for any missing/additional units. 8. a) General Rate Law Experiment [NO] o (M) [Cl 2 ] o (M) Initial Rate 0.0 0.0 0.8 2 0.0 0.20 0.36 3 0.20 0.20.45 From experiments and 2 ([NO] o is kept constant) as the initial concentration of Cl 2 is doubled, the initial rate of the reaction doubles..; therefore, the reaction is first. order with respect to Cl 2. From experiments 2 and 3 ([Cl 2 ] o is kept constant) as the initial concentration of NO is doubled, the initial rate of the reaction quadruples..; therefore, the reaction is. second order with respect to NO. b) 0.8 0.0 80 9. a) General Rate Law Experiment [I ] o (M) [OCl ] o (M) Initial Rate 0.2 0.8 7.9 0 2 2 0.060 0.8 3.95 0 2 3 0.030 0.090 9.88 0 2 4 0.24 0.090 7.9 0 2 From experiments and 2 ([OCl ] o is kept constant) as the initial concentration of I is doubled, the initial rate of the reaction doubles...; therefore, the reaction is first order with respect to I.
You must find the order of OCl the long way. Divide the 2 equations (in which the [OCl ] changes) by each other. 7.9 0 0.20.8 9.88 0 0.0300.090 2.0 2.0 b) 7.9 0 0.2 0.8 3.7 c) 3.7 0.5 0.5 0.082 20. General Rate Law Experiment [N 2 O 5 ] o (M) Initial Rate 0.0750 8.90 0 4 2 0.90 2.26 0 3 3 0.275 3.26 0 3 4 0.40 4.85 0 3 You must solve this problem the long way. 8.90 0 0.0750 2.26 0 0.90 Divide the 2 equations by each other. 8.90 0 0.0750 2.26 0 0.90 0.394 0.395 8.90 0 0.0750 0.09 2. a) General Rate Law Experiment [NOCl] o Initial Rate 3.0 0 6 5.98 0 4 2 2.0 0 6 2.66 0 4 3.0 0 6 6.64 0 3 4 4.0 0 6.06 0 5 From experiments 2 and 3, as the initial concentration of NOCl is doubled, the initial rate of the reaction quadruples...; therefore, the reaction is second order with respect to NOCl. b) 5.98 0 3.00 2
6.60 c) 6.60. 4.00 22. a) General Rate Law Experiment [Hb] o o Initial Rate 2.2.00 0.69 2 4.42.00.24 3 4.42 3.00 3.7 From experiments and 2 ([CO] o is kept constant) as the initial concentration of Hb is doubled, the initial rate of the reaction doubles..; therefore, the reaction is. first order with respect to Hb. You cannot use the shortcut to calculate the order with respect to [CO] Find 2 equations where the concentration of Hb is constant.24 3.7 4.42.00 4.42 3.00 Divide the 2 equations by each other. 4.42.00.24 3.7 0.33 4.42 3.00.00 0.33 3.00 The reaction is st order with respect to CO b) c) 2.2.00 0.69 0.280 d) 0.280 0.280 3.36 2.40 2.26 23. a) General Experiment [ClO 2 ] o (M) [OH ] o (M) Initial Rate 0.0500 0.00 5.75 0 2 2 0.00 0.00 2.30 0 3 0.00 0.0500.5 0 From experiments and 2 ([OH ] o is kept constant) as the initial concentration of ClO 2 is doubled, the initial rate of the reaction quadruples...; therefore, the reaction is second order with respect to ClO 2. From experiments 2 and 3 ([ClO 2 ] o is kept constant) as the initial concentration of OH is doubled, the initial rate of the reaction doubles...; therefore, the reaction is first order with respect to OH. 3
Find k 5.75 0 0.0500 0.00 230. 230. b) 230. 0.75 0.0844 0.594 30. a) Because the plot of verse time was linear the rate law is a second order rate law. 0 (integrated rate law) The rate constant is the slope of the line 3.60 0 3.60 0 (rate law) b) The half life is the amount of time it takes for ½ of the original concentration to react. 0.52.80 0 3.60 0 2.80 0 c) 9920 7.00 0 3.60 0 29,800 2.80 0 3. a) Because the plot of ln[a] verses t is linear, this is a st order rate law. 0 (integrated rate law) The rate constant is the negative slope of the line 2.97 0 2.97 0 b) (rate law) 4
2 2.00 0 2.97 0 2.00 0 23.3 c) 2.50 0 2.97 0 2.00 0 70.0 32. a) Because the plot of [C 2 H 5 OH] verses t is linear, this is a 0 th order rate law 0 (integrated rate law) The rate constant equals the negative slope of the line; 4.000 4.00 0 (rate law) b).25 0 4.000.25 0 56 c) Since this is a 0 th order rate law (which means that the rate is independent of concentration), the amount of time that it takes to decompose the total amount of C 2 H 5 OH is just double the half life. 256 32 Or 04.000.25 0 32 33. Plot [H2O2] [H 2 O 2 ] vs. t, ln[h 2 O 2 ] vs. t, and [H 2 O 2 ] vs. t.5 0.5 0 0 000 2000 3000 4000 time (s) ln[h2o2] 0.5 25 0.5 20.5 5 2.5 0 5 3.5 0 0 2000 4000 0 2000 4000 time (s) time (s) Since the plot of ln[h 2 O 2 ] vs. t results in a linear relationship, the reaction is st order. [H2O2]^ 5
If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order. Time (s) [H 2 O 2 ] ln[h 2 O 2 ] [H 2 O 2 ] 0.00 0.59.00= 0.4 0.00 0.53 0.00= 0.53.00.7.00=0.7 600 0.59 0.22 0.53 0.46.7.0 200 0.37 0.5 0.99 0.5 2.7.8 800 0.22 0.09.5 0.5 4.5 3.3 2400 0.3 0.05 2.0 0.5 7.7 4.3 3000 0.082 2.5 2 Since the all the differences between ln[h 2 O 2 ] are all approximately 0.5 this will be a straight line and the reaction is st order. Differential Rate Law Integrated Rate Law The slope of the line will equal k... 8.3 0, 8.30 Calculate the concentration of H 2O 2 after 4000. s. 8.3 0 4,000.00 0.036 35. Plot [NO2] 0.6 0.5 0.4 0.3 0.2 0. 0 [A] vs. t, ln[a] vs. t, and [A] vs. t 0 5000 0000 5000 time (s) ln[no2] 0.5.5 2 0 5000 0000 5000 time (s) [NO2]^ 8 6 4 2 0 0 5000 0000 5000 time (s) Since the plot of [A] vs. t is linear, this is a 2 nd order rate law If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order. You must have at least 3 points. Time (s) [NO 2 ] ln[no 2 ] [NO 2 ] 0 0.500 0.25 0.50= 0.25 0.693.39 0.693= 0.69 2.00 4.00 2.00=2.00 9.00 0 3 0.250 0.076.39 0.36 4.00.75.80 0 4 0.74.75 5.75 This problem it is a little hard to determine. (I will not give you something this hard on an exam.) The difference between the difference between the [NO 2 ] points is 0.7 over a concentration range of 0.326 making the difference between differences 52%. 00% 52% of the. 6
range. The difference between the difference between the ln[no 2 ] is 0.33 over a ln[no 2 ] range of.06 making the difference between differences 3%. 00% 33% of the range. The. difference between the difference between [NO 2 ] is 0.25 over a range of 3.75 making the difference between differences 6%. 00% 6%of the range. The graph [NO 2] has the. smallest difference between differences, when you take the size of the range into account, the order is 2 nd order. Differential Rate Law Integrated Rate Law The slope of the line is the rate constant... 2.080 2.08 0 2.08 0 2.70 0 0.500 7.62 0.3 36. a) Plot [O] vs. t, ln[o] vs. t, and [O] vs. t [O] 6.E+09 4.E+09 2.E+09 0.E+00 0 0.0 0.02 0.03 0.04 time ln[o] 23 22 2 20 9 0 0.02 0.04 time [O]^ 5.E 09 4.E 09 3.E 09 2.E 09.E 09 0.E+00 0 0.02 0.04 time Since ln[o] vs. t is linear, the rate expression is st order with respect to O. If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order. Time (s) [O] ln[o] [O] 0 5.0 0 9.9 0 9 5.0 0 9 = 3. 0 9 22 2 22=.0 2.0 0 0 5.2 0 0 2.0 0 0 = 3.2 0 0 0.00.9 0 9.2 0 9 2.0 5.2 0 0 9.8 0 0 0.020 6.8 0 8 4.3 0 8 20..0.5 0 9 2.5 0 9 0.030 2.5 0 8 9 4.0 0 9 Since the all the differences between ln[h 2 O 2 ] are all approximately.0 this will be a straight line and the reaction is st order. Differential Rate Law b) Overall Rate Law Because the concentration of [NO 2 ] >>[O] this can be turned into a pseudo rate law 7
Determine k' using the st order integrated rate law Therefore, the slope of the line is k'....0 0 Find k.00.0 0.00 38. a) The order of the reaction with respect to A is 2 nd order because the plot of [A] vs. t is linear. The integrated rate law is. Therefore, the [A] 0 is the y intercept. [A] 0 = 0 b) 0 0 9 0 00 0.0 c) Calculate the time for the st half life ([A] 0. 0.05) 39. a) Plot [NO] 8.E+08 6.E+08 4.E+08 2.E+08 0.E+00 0.05 0 0. Calculate the time for the 2 nd half life 0.05 2 Calculate the time for the 3 rd half life 0.025 0 0.025 4 0 500 000 500 time 0.025 0 [NO] vs. t, ln[no] vs. t, and, [NO] vs. t ln[no] 20.5 20 9.5 9 8.5 8 0 500 000 500 and [A] 0 = 0. M Since the plot of ln[no] vs. t is linear, the rate law is st order with respect to NO. time [NO]^ 2.E 08.E 08 5.E 09 0.E+00 0 500 000 500 time 8
If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order. Time (s) [NO] ln[no] [NO] 0 6.0 0 8 2.4 0 8 6.0 0 8 = 3.6 0 8 20 9 20=.0.7 0 9 4.2 0 9.7 0 9 = 2.5 0 9 500 2.4 0 8.4 0 8 9.0 4.2 0 9 5.8 0 9 000 9.9 0 7 8.0 0 8 Since the all the differences between ln[no] are all approximately.0 this will be a straight line and the reaction is st order. Plot [O 3 ] vs. t, ln[o 3 ] vs. t, and [O 3 ] vs. t 2.E+0 23.5 4.E 0 [O3].E+0 5.E+09 ln[o3] 23 22.5 22 [O3]^ 3.E 0 2.E 0.E 0 0.E+00 0 200 400 Time 2.5 0 200 400 Time 0.E+00 0 00 200 300 400 Time Since plot of ln[o 3 ] vs. t is linear, the rate law is st order with respect to O 3 If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order. Time (s) [O 3 ] ln[o 3 ] [O 3 ] 0.0 0 0 7.0 0 9.0 0 0 = 3.0 0 9 23.0 22.7 23.0= 0.3.0 0 0.4 0 0.0 0 0 = 4.0 0 00 7.0 0 9 2. 0 9 22.7 0.4.4 0 0 6.0 0 200 4.9 0 9.5 0 9 22.3 0.4 2.0 0 0 9.0 0 300 3.4 0 9 2.9 2.9 0 0 Since the all the differences between ln[o 3 ] are all approximately 0.4 this will be a straight line and the reaction is st order. b) Rate Law c) When the [NO] data was taken the concentration of [O 3 ]>>[NO]. Therefore, a pseudo rate law can be written. Since the reaction was first order with respect to [NO] the slope of the line equals k'... 0.008.8 When the [O 3 ] data was taken the concentration of [NO]>>[O 3 ]. Therefore, a pseudo rate law can be written. Since the reaction was first order with respect to [O 3 ] the slope of the line equals k'' 9
3.4 0.00 300. 0 0.0036 3.6 d) You can solve this using either of the two data sets. In both data sets the reactions are pseudo st order system. Overall rate = pseudo st order rate Using data set [O 3 ] held constant.8.80.0 0.80 Using data set 2 [NO] held constant 3.6 2.0 0.80 43. 0.250 20..00 0.06 0.06 0.350 0.06 2.00 5.80 47. a) Calculate k 0.250 320. 0.250 320. 0.00433 Calculate time for st half life 0.00433 0.00433 59 Calculate time for the 2 nd half life 0.00433 0.00433 59 b) 0.00 0.00433 0.00 0.00433 0
532 48. Since the time of the half life doubles every successive half life, the reaction is a 2 nd order reaction a) 0.0 0.05 0.0.0.0 80.0 0.0 90. 0.0 b).0 30.0 40.. 0.025 52. The problem wants you to find the time when 4.00. Known 4.500 3.700 Both st order reactions Initially Plug in for and 4.00 forand solve for 4.00 4.00 4.00 Plug into equation for and solve for t 4.00 4.00 4.00 4.00 4.00 4.00 4.00 3.70 0 4.500 427 54. The concentrations of B and C are so much larger than A, therefore, during the course of the reaction they do not effectively change making this a pseudo st order reaction. The rate of the pseudo st order reaction is equal to the rate of the overall reaction times. a) 3.8 0 8.0.0 0 0.2
0.2.0 0 0.002 0.002.0 0 3.0 0.03 b) 0.2 0.0050 0.2.0 0 5.8 c) 0.2 0.2 3.0.0 0 0.002 d) The concentration of C is the 2.00 0.002 0.008 2 2.0 20.008 2.0 This should not be a surprising answer because in order to have a pseudo st order reaction with respect to A, the concentration of C must be essentially constant. 56. a) Elementary Step: An individual reaction in a proposed reaction mechanism. The rate law can be written from the coefficients in the balanced equation b) Molecularity: The number of reactant molecules (or free atoms) taking part in an elementary reaction. This is also the number of species that must collide in order to produce the reaction represented by an elementary reaction. c) Reaction Mechanism: The pathway that is proposed for an overall reaction and accounts for the experimental rate law. d) Intermediate: A species that is produced and consumed during a reaction but does not appear in the overall chemical equation. e) Rate Determining Step: The elementary reaction that governs the rate of the overall reaction. This is the slowest elementary reaction that occurs in a mechanism. 59. For elementary reactions the rate is just equal to the rate constant times the concentration of the reactants. a) b) c) d) e) 60. The rate law found in problem 33 was In order for this to be the rate, the first step would have to be the rate determining step. Overall Reaction 2H 2 O 2 2H 2 O + O 2 6. Overall Reaction C 4 H 9 Br + 2H 2 O Br + C 4 H 9 OH + H 3 O + 2
Intermediates C 4 H 9 + and C 4 H 9 OH 2 + 63. The rate of a reaction is dependent on the slowest step. a) The mechanism is not consistent with rate law. b) NO 3 is an intermediate, therefore, it needs to be eliminated from the rate equation. Use equilibrium to eliminate The mechanism is consistent with rate law. c) The mechanism is not consistent with rate law. d) In order for a mechanism to be plausible the elementary reactions must add up to the overall reaction. These elementary reactions do not add up to the overall reaction. In fact the middle reaction is not even balanced. 65. H 2 BrO 3 + is an intermediate therefore, needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate. HBrO 3 is an intermediate therefore, needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate. 68. a) COCl is an intermediate, therefore, it needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate. Cl is an intermediate, therefore, it needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate. 3
b) COCl and Cl are intermediates 69. a) MoCl 5 is an intermediate b) Use the steady state approximation to eliminate the intermediate, rate of formation = rate of consumption 70. Need to calculate the rate of decomposition of O 3 Note: Rate of decomposition=ways O 3 used up ways O 3 forms O is an intermediate therefore need to remove it from the rate law. Use the same steady state approximation to solve for [O] 2 72. a) As the activation energy decreases the rate of the reaction goes up. b) As the temperature increases the molecules/atoms move faster and more collisions occur, therefore, the faster the rate of the reaction. c) As the frequency of collisions increases the rate of the reaction goes up. d) As the orientation of collisions is more specific the rate of the reaction goes down. 4
79. 3.52 0 9.750 80. 7.2 0.7 0 8.345 20,000.7 0 86 0.008345 20,000 8.345 555 645 720. 660. 720. 598.30 After each half life the number of moles of iodoethane is halved. Therefore after 3 half lives there is 8 the number of moles of iodoethane that there was initially. For a gas, PV=nRT, therefore, since the volume, gas constant, and temperature are constant 894 2 8. 7 5.40 0 8.345 324 5 295 83. a) Therefore, when the ln(k) is plotted vs. the slope of the line equals 8.345 9.50.0 0 9.5 b) When the ln(k) is plotted vs. T the y intercept of the line equals ln A 33.5 3.54 0 c) 9,50 8.345 3.54 0 3.43 0.0324 5
84. Therefore if you plot the ln(k) vs. the slope of the line will be 0 2 4 6 8 0 2 0.0028 0.003 0.0032 0.0034 T^ 3.5 0 4.90 298 2,400 338 2,400 0.008345 03 ln(k) 86. a) b) c) 6
87. The second graph is a two step problem. st : reactants intermediates 2 nd : intermediates products. Since the second reaction has the greater activation energy it will be the slower reaction and will determine the rate of the reaction. Therefore, we label the activation energy of the 2 nd step. 88. The activation energy for the reverse reaction is the ΔE + E a. 26 25 34 90. A catalyst increases the reaction rate because it provides another pathway with a lower activation energy for the reaction to proceed by. Homogeneous catalysts are present in the same phase as the reactants and heterogeneous catalyst are present in a different phase from the reactants. The uncatalyzed and catalyzed reactions most likely will have different rate laws because they have different pathways that they occur by. 94. a) The catalyst is the species that is initially added to the reaction but not used in the overall reaction. Therefore NO is the catalyst. b) An intermediate is a species that is formed in one of the steps of the reaction but not one of the reactants or products. Therefore NO 2 is an intermediate. c) 0.848 8.345 298,900 4,000 0.848 7
2.3 8