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9 Chapter 6 Rates of Chemical Reactions Solutions for Practice Problems Student Textbook page 7 1. Problem Cyclopropane, C 3 H 6, is used in the synthesis of organic compounds and as a fastacting anesthetic. It undergoes rearrangement to form propene. If cyclopropane disappears at a rate of 0.5 mol/s, at what rate is propene being produced? Calculate the rate of propene production. The rate of rearrangement of cyclopropane is 0.5 mol/s. Cyclopropane and propene both have the formula C 3 H 6. The balanced chemical equation is C 3 H 6 (cyclopropane) C 3 H 6 (propene) Determine the mole to mole ratio between cyclopropane and propene. Use this to determine the rate of formation of propene. The mole to mole ratio between cyclopropane and propene is 1:1. The rate of formation of propene is the same as the rate of reaction of cyclopropane, or 0.5 mol/s. The 1:1 stoichiometry of the reaction indicates that the rate of reaction of cyclopropane is equivalent to the rate of formation of propene.. Problem Ammonia, NH 3, reacts with oxygen to produce nitric oxide, NO, and water vapour. 4NH 3(g) + 5O (g) 4NO (g) + 6H O (g) At a specific time in the reaction, ammonia is disappearing at a rate of mol/(l s). What is the corresponding rate of production of water? Calculate the rate of production of water. You have the balanced chemical equation and the rate of disappearance of ammonia. Determine the mole to mole ratio between NH 3 and H O. Since 6 mol of H O are produced for every 4 mol of NH 3 that react, the rate of appearance of H O is 6 4 or 3 the rate of reaction of NH 3. 84

10 Rate of appearance of H O = 3 (rate of disappearance of NH 3) = mol/(l s) = 0.10 mol/(l s) The rate of production of water is greater than the rate of disappearance of ammonia, which makes sense, given the mole ratio. 3. Problem Hydrogen bromide reacts with oxygen to produce bromine and water vapour. 4HBr (g) + O (g) Br (g) + H O (g) How does the rate of decomposition of HBr (in mol/(l s)) compare with the rate of formation of Br (also in mol/(l s))? Express your answer as an equation. Determine the rate of decomposition of HBr relative to the rate of formation of Br, and write your answer as an equation. You are given the balanced chemical equation. First check that the equation is balanced. Then use the molar coefficients in the balanced equation to determine the relative rates of decomposition of HBr and formation of Br. Since 4 mol HBr are consumed for every mol of Br produced, the rate of decomposition of HBr is equal to times the rate of formation of Br. Rate of decomposition of HBr = (rate of formation of Br ) or, [Br ] t = 0.5 [HBr] t From the coefficients in the balanced equation, you can see that the rate of decomposition of HBr is twice that of the rate of formation of Br. 4. Problem Magnesium metal reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas. Mg (s) + HCl (aq) MgCl (aq) + H (g) Over an interval of 1.00 s, the mass of Mg (s) changes by g. (a) What is the corresponding rate of consumption of HCl (aq) (in mol/s)? (b) Calculate the corresponding rate of production of H (g) (in L/s) at 0 C and 101 kpa. First, calculate the rate of consumption of HCl (aq), in mol/s. Then calculate the rate of production of H, in L/s at 0 C and 101 kpa. You have the balanced chemical equation, and you know the rate of consumption of Mg, in g/s. 85

11 From the balanced chemical equation, you can obtain the relative rates of consumption or production of reactants and products, in mol/s. (a) You must first convert the rate of disappearance of Mg from g/s to mol/s. From the balanced chemical equation you can see that the rate of reaction of HCl (aq) is twice the rate of reaction of Mg. (b) From the balanced chemical equation you can see that the coefficients of Mg and of H are both 1. This means that the rate of production of H, in mol/s, is equal to the rate of disappearance of Mg, in mol/s. For H, you then need to convert mol/s to L/s at 0 C, 101 kpa. Recall that 1.00 mol of any gas occupies a volume of 4.0 L at 0 C, 101 kpa g/s (a) Rate of disappearance of Mg = 4.3 g/mol = mol/s Rate of consumption of HCl = (rate of disappearance of Mg) = mol/s = mol/s (b) Rate of appearance of H = rate of disappearance of Mg = ( mol/s) (4.0 L/mol) = L/s at 0 C, 101 kpa You have correctly used the coefficients in the balanced equation. The units are correct. Solutions for Practice Problems Student Textbook page Problem When heated, ethylene oxide decomposes to produce methane and carbon monoxide. C H 4 O (g) CH 4(g) + CO (g) At 415 C, the following initial rate data were recorded. Experiment 1 3 [C H 4 O] 0 (mol/l) Initial rate (mol/(l s)) Determine the rate law equation and the rate constant at 415. You need to find the value of m and k in the general rate law equation for the reaction. Rate = k[c H 4 O] m You know the initial concentration of the reactant and the initial rates for three experiments. Since there is only one reactant, simply compare the initial rates of two experiments with different initial [C H 4 O]. Compare the change in initial rate with the change in initial concentration. 86

12 Begin with experiments 1 and. Set up a ratio of the two rates as shown. Rate = k [ ] m mol/l Rate 1 k [ mol/l ] m = mol/(l s) mol/(l s) Because k is constant, it can be cancelled. Rate = k [ ] m mol/l Rate 1 k [ mol/l ] m = mol/(l s) mol/(l s) m = m = 1 (by inspection) Similarly, you can set up a ratio involving experiments 1 and 3 or experiments and 3 to obtain m = 1. Therefore, the reaction is first order with respect to ethylene oxide. To determine the value of k, substitute m = 1 into the rate law equation. You can use data from any of experiments 1, or 3 and solve for k. Substituting data from experiment 1 into the rate law equation gives: mol/(l s) = k [ mol/l ] 1 k = mol/(l s) mol/l = s 1 (at 415 C) The rate law equation for the reaction is: Rate = ( s 1 )[C H 4 O] 1 (at 415 C) The value of m = 1 can be checked by inspection. When the concentration of ethylene oxide doubles, so will the initial rate. The value of k can be checked by using data from one of the other experiments. The units of k are consistent with a first order reaction. The number of significant digits for k are appropriate for the given data. 6. Problem Iodine chloride reacts with hydrogen to produce iodine and hydrogen chloride. ICl + H I + HCl At temperature T, the following initial rate data were recorded. Experiment [ICI] 0 (mol/l) [H ] 0 (mol/l) Initial rate (mol/(l s)) Determine the rate law equation and the rate constant at temperature T. You need to find the values of m, n, and k in the general rate law equation for the reaction. Rate = k[icl] m [H ] n You know the initial concentrations of the reactants and the initial rates for three experiments. 87

13 Step 1 Find two experiments in which [ICl] remains constant and [H ] changes. Compare the rates and concentrations to solve for n. Then find two experiments in which [H ] remains constant and [ICl] changes. Compare the rates and concentrations to solve for m. Step Use the data and your calculated values for m and n to solve for k, using the following equation. Rate = k[icl] m [H ] n Step 1 Begin with experiments 1 and 3, where [ICl] remains constant, to determine the order with respect to H. Set up a ratio for the two rates as shown. Notice that you can cancel [ICl] because you chose experiments where [ICl] did not change. Rate 3 = k [ ] m[0.00] 0.00 mol/l n Rate 1 k [ 0.00 mol/l ] m [0.050] n = mol/(l s) mol/(l s) Because k is a constant, you can cancel it out. Rate 3 = k [ ] m[0.00] 0.00 mol/l n Rate 1 k [ 0.00 mol/l ] m [0.050] n = mol/(l s) mol/(l s) (4) n = 4 n = 1 (by inspection) Therefore, the reaction is first order with respect to H. Set up a ratio for experiments 1 and, where [H ] is constant. You can cancel [H ] because it is the same in both experiments. Rate = k [ ] m[0.050] mol/l 1 Rate 1 k [ 0.00 mol/l ] m [0.050] 1 = mol/(l s) mol/(l s) Because k is a constant, you can cancel it out. Step Rate = k [ ] m[0.050] mol/l 1 Rate 1 k [ 0.00 mol/l ] m [0.050] 1 () m = m = 1 (by inspection) The reaction is first order with respect to ICl. = mol/(l s) mol/(l s) Overall, the rate law equation is: Rate = k[icl] 1 [H ] 1 To find the value of the rate constant, substitute data from any of the three experiments into the rate law equation. Using the data from experiment 1 gives mol/(l s) = k [ 0.0 mol/l ] 1[ ] mol/l k = mol/(l s) [ 0.0 mol/l ] 1 [ mol/l ] 1 = 0.15 L/(mol s) (at temperature T ) The overall rate law equation for the reaction is: Rate = ( 0.15 L/(mol s) ) [ICl][H ] Check the values of m and n by inspection. When [ICl] doubles (when [H ] is constant), the rate also doubles. When [H ] quadruples (when [ICl] is constant), the rate also quadruples. To check the value for k, substitute data from experiments or 3 into the equation and solve for k. Check that the units for k are consistent with an overall second order reaction. Also check the number of significant digits for k. 88

14 7. Problem Sulfuryl chloride (also known as chlorosulfuric acid and thionyl chloride), SO Cl, is used in a variety of applications, including the synthesis of pharmaceuticals, rubber-based plastics, dyestuff, and rayon. At a certain temperature, the rate of decomposition of sulfuryl chloride was studied. SO Cl (g) SO (g) + Cl (g) [SO Cl ] (mol/l) Initial rate (mol/(l s)) (a) Write the rate law equation for the decomposition of sulfuryl chloride. (b) Determine the rate constant, k, for the reaction, with the appropriate units. You need to find the values of m and k in the general rate law equation for the reaction. Rate = k[so Cl ] m You know the initial concentration of the reactant and the initial rates for three experiments. Since there is only one reactant, simply compare the initial rates of two experiments with different initial [SO Cl ]. Compare the change in initial rate with the change in initial concentration. (a) Begin with the first and second experiments. Set up a ratio of the two rates as shown. Rate = k [ ] m mol/l Rate 1 k [ mol/l ] m Because k is constant, it can be cancelled. Rate = k [ ] m mol/l Rate 1 k [ mol/l ] m = mol/(l s) mol/(l s) = mol/(l s) mol/(l s) 0.5 m = 0.5 m = 1(by inspection) Similarly, you can set up a ratio involving other experiments to obtain m = 1. Therefore, the reaction is first order with respect to sulfuryl chloride. The rate law equation is: Rate = k[so Cl ] (b) To determine the value of k, substitute m = 1 into the rate law equation. You can use data from any of experiments 1, or 3 and solve for k. Substituting data from the first experiment into the rate law equation gives: = k[0.15] 1 k = mol/(l s) 0.15 mol/l = s 1 The rate law equation for the reaction is: Rate = ( s 1 )[SO Cl ] 1 89

15 The value m = 1 can be checked by inspection. When the concentration of sulfuryl chloride doubles, so will the initial rate. The value of k can be checked by using data from one of the other experiments. The units of k are consistent with a first order reaction. The number of significant digits for k are appropriate for the given data. 8. Problem Consider the following reaction. A + 3B + C products This reaction was found to obey the following rate law equation. Rate = k[a] [B][C] Copy the following table into your notebook. Then use the given information to predict the blank values. Do not write in this textbook. Experiment You need to calculate the missing values in the table: (a) = Initial [B] for experiment (b) = Initial [C] for experiment 3 (c) = Initial [A] for experiment 4 (d) = Initial rate for experiment 5 You know the rate law equation, and you are given some incomplete rate data. You know that the reaction is first order in B and C. This means that when the concentration of B is doubled, if the concentration of other reactants are kept constant, the rate will also double. The rate also doubles when C is doubled, if all else is kept constant. The reaction is second order in A. When the concentration of A is doubled, if [B] and [C] are kept constant, the rate will increase by a factor of, or 4 times. If [A] is tripled, the rate will increase by a factor of 3, or 9 times. You can use the complete set of data, given in experiment 1, along with the rate law equation, to solve for k, the rate constant. This will allow you to determine any unknown concentrations in experiments to 5. Substituting data from experiment 1 into the rate law equation gives: 0.40 mol/(l s) = k[0.10 mol/l] [0.0 mol/l] 1 [0.050 mol/l] 1 k = Initial [A] (mol/l) (c) 0.10 Initial [B] (mol/l) 0.0 (a) mol/(l s) [0.10 mol/l] [0.0 mol/(l s)] 1 [0.050 mol/l] 1 = L 3 /(s mol 3 ) Initial [C] (mol/l) (b) Initial rate (mol/(l s)) (d) 90

16 (a) To solve for (a) in experiment, use the value of k and the given data. Substitute into the rate equation mol/(l s) = L 3 /(s mol 3 ) [0.10 mol/l] (a) [0.10 mol/l] 1 (a) = 0.40 mol/(l s) L 3 /(s mol 3 )[0.10 mol/l] [0.10 mol/l] 1 = 0.10 mol/l (b) To solve for (b) in experiment 3, use the value of k and the given data. Substitute into the rate equation. 0.0 mol/(l s) = L 3 /(s mol 3 ) [0.0 mol/l] [0.050] 1 (b) (b) = 0.0 mol/(l s) L 3 /(s mol 3 ) [0.0 mol/l] [0.050 mol/l] 1 = 0.05 mol/l (c) To solve for (c) in experiment 4, use the value of k and the given data. Substitute into the rate equation mol/(l s) = L 3 /(s mol 3 ) (c) [0.05] 1 [0.040] 1 (c) = 0.45 mol/(l s) L 3 /(s mol 3 )[0.05 mol/l] 1 [0.040 mol/l] 1 = 0.11 mol /L (c) = 0.34 mol/l (d) To solve for (d) in experiment 5, use the value of k and the given data. Substitute into the rate equation. (d) = L 3 /(s mol 3 ) [0.10 mol/l] [0.010] 1 [0.15] 1 (d) = mol/(l s) Check your calculations. The units and significant digits are correct. You can also check by inspection, using the fact that the reaction is second order in A and first order in B and C. Solutions for Practice Problems Student Textbook page Problem Cyclopropane, C 3 H 6, has a three-membered hydrocarbon ring structure. It undergoes rearrangement to propene. At 1000 C, the first-order rate constant for the decomposition of cyclopropane is 9. s 1. (a) Determine the half-life of the reaction. (b) What percent of the original concentration of cyclopropane will remain after 4 half-lives? First, you need to find the half-life of the reaction. Next, you need to find the percentage of the original concentration of cyclopropane remaining after 4 half-lives. You know that the reaction is first order in cyclopropane, with k = 9. s 1. (a) To find t 1/, use the following equation. t 1/ = k 91

17 (b) Each half-life reduces the initial amount of cyclopropane by 1, or 50%. Let the original concentration of cyclopropane be A. After 1 t 1/, [cyclopropane] = 1 A After t 1/, [cyclopropane] = 1 1 A = 1 4 A After 3 t 1/, [cyclopropane] = A = 1 8 A After 4 t 1/, [cyclopropane] = A = 1 16 A After 4 half-lives, there is 1 16, or ( 1 ) 4 of the original amount of cyclopropane remaining. (a) Solve for t 1/. t 1/ = s 1 = s The half-life of the reaction is s. ) 4 = 1 16 (b) After 4 half-lives, there will be ( 1 of the original amount present. If we take the original concentration of cyclopropane to be 100, this corresponds to = 6. Therefore, 6.% of the original amount of cyclopropane will remain after 4 half-lives. The units and number of significant digits for the half-life are correct. The relatively large rate constant corresponds to a short half-life. 4 half-lives corresponds to = 1 of the original amount, or about 6.% remaining Problem Peroxyacetyl nitrate (PAN), H 3 CCO ONO, is a constituent of photochemical smog. It undergoes a first-order decomposition reaction with t 1/ = 3 min. (a) Calculate the rate constant in s 1 for the first-order decomposition of PAN. (b) 18 min after a sample of PAN began to decompose, the concentration of PAN in the air is molecules/l. What was the concentration of PAN when the decomposition began? First, you must find the first order rate constant, in s 1, for the decomposition of PAN. Next, find the concentration of PAN originally present if [PAN] is molecules/l after decomposing for 18 minutes. (a) The half-life of the reaction is 3 minutes. (b) [PAN] = molecules/l at t = 18 minutes. (a) Rearrange the following equation and solve for k. t 1/ = k 9

18 (b) Begin by determining the number of half-lives in 18 minutes. Use this to determine the original [PAN]. We know that ( 1 ) n (original concentration) = (concentration after n half-lives) or, (original concentration) = (concentration ( after n half-lives) 1 ) n (a) 3 min = 3 min 60 s/min = s k = t 1/ = s = s 1 (b) Number of half-lives = 18 min 3 min/half-life = 4 half-lives After 4 half-lives, [PAN] = molecules/l. Substituting into the equation gives (original concentration) = concentration ( after n half-lives 1 ) n = molecules/l ( 1 ) 4 = molecules/l The original [PAN] is molecules/l. The units and number of significant digits for the half-life and original concentration of PAN are correct. Dividing the original [PAN] in half four times gives molecules/l. 11. Problem In general, a reaction is essentially over after 10 half-lives. Prove that this generalization is reasonable. Show that only a negligible percentage of the original amount of a substance remains after 10 half-lives. You are given the time of 10 half-lives. Use the following equation: Amount remaining after n half-lives = (original amount) ( 1 ) n For simplicity, take the original mass of the substance to be 100 g. Amount remaining after 10 half-lives = (100 g) ( 1 ) 10 Amount remaining after 10 half-lives = g After 10 half-lives, a 100 g sample would be reduced to g, or 0.098% of the original amount. This is a negligible amount. 93

19 Take 100 and divide it by ten times to get the same answer. Less than one-tenth of one percent of a substance remains after 10 half-lives. 1. Problem The half-life of a certain first-order reaction is 10 s. How long do you estimate that it will take for 90% of the original sample to react? Estimate how long it will take for 90% of the sample to react. You know that the half-life of the reaction is 10 s. Assume a 100 g sample size. When 90% of the sample has reacted, 10 g will remain. To estimate how long this will take, divide 100 g by several times, until you reach approximately 10 g. Alternatively, use the equation below and solve for n, the number of half-lives. This will give a more accurate solution, but will require the use of logarithms. Amount remaining after n half-lives = (original amount) ( 1 ) n Dividing 100 g by several times gives 100 = 50 (1 t 1/ ) 50 = 5 ( t 1/) 5 = 1.5 (3 t 1/) 1.5 = 6.5 (4 t 1/ ) It takes between 3 and 4 half-lives for 90% of the sample to react. If the half-life is 10 s, this will take about 400 s. Alternate Solution (using logarithms) Amount remaining after n half-lives = Original amount ( 1 ) n or, ( 1 ) n Amount remaining after n half-lives = Original amount = = 0.10 Taking the logarithm of both sides gives: log ( 1 ) n = log(0.10) n log ( 1 ) = log(0.10) n = log(0.10) log ( 1 ) = 3.3 It will take 3.3 half-lives, or s, for 90% of the sample to react. The estimated answer agrees with the calculated value. 94

20 Solutions for Practice Problems Student Textbook page Problem The following reaction is exothermic. ClO (g) Cl (g) + O (g) Draw and label a potential energy diagram for the reaction. Propose a reasonable activated complex. Solution Since the reaction is exothermic, the products should be lower than the reactants on the potential energy diagram. The activated complex exists between the reactants and products, and is of higher energy than the reactants. possible activated complex Cl... Cl... O O Potential Energy (kj) ClO (g) H Cl (g) + O (g) Reaction Progress A reasonable activated complex might be a species where the bonds between Cl and O in the reactants are breaking, while new bonds between two Cl atoms and between two O atoms are forming. 14. Problem Consider the following reaction. AB + C AC + B H =+65 kj, E a(rev) = 34 kj Draw and label a potential energy diagram for this reaction. Calculate and label E a(fwd). Include a possible structure for the activated complex. Solution The reaction is endothermic, since H > 0. The products will be of higher energy than the reactants. 95

21 possible activated complex A... C... B Potential Energy (kj) AB + C E a(fwd) = 99 kj AC + B H =+65 kj Reaction Progress For an endothermic reaction, E a(fwd) = H + E a(rev) = 65 kj + 34 kj = 99 kj The activation energy for the forward reaction is 99 kj. A possible structure for the activated complex could involve simultaneous bond breaking in the reactants and the formation of new bonds in the products. 15. Problem Consider the reaction below. C + D CD H = 13 kj, E a(fwd) = 61 kj Draw and label a potential energy diagram for this reaction. Calculate and label E a(rev). Include a possible structure for the activated complex. Solution Since this reaction is exothermic, the potential energy diagram will be similar to Figure 6.1. possible activated complex C... D Potential Energy (kj) E a(rev) = 193 kj C + D H = 13 kj CD Reaction Progress 96

22 For an exothermic reaction, E a(rev) = H + E a(fwd) = 13 kj + 61 kj = 193 kj (Note: Use the positive value of H.) A possible structure of the activated complex might involve partial bond formation between C and D. 16. Problem In the upper atmosphere, oxygen exists in other forms other than O (g). For example, it exists as ozone, O 3(g), and as single oxygen atoms, O (g). Ozone and atomic oxygen react to form two molecules of oxygen. For this reaction, the enthalpy change is 39 kj and the activation energy is 19 kj. Draw and label a potential energy diagram. Include a value for E a(rev). Propose a structure for the activated complex. Solution The reaction is O 3(g) + O (g) O (g) H = 39 kj; E a(fwd) = 19 kj Since this reaction is exothermic, the potential energy diagram will be similar to Figure 6.1. Potential Energy (kj) O 3 + O H = 39 kj E a(rev) = 411 kj O Reaction Progress For an exothermic reaction, E a(rev) = H + E a(fwd) = 39 kj + 19 kj = 411 kj A possible structure of the activated complex might involve partial bond breakage in the O 3 molecule coupled with partial bond formation to produce two O molecules. Solutions for Practice Problems Student Textbook page Problem NO (g) and F (g) react to form NO F (g). The experimentally determined rate law for the reaction is written as follows: Rate = k[no ][F ] A chemist proposes the following mechanism. Determine whether the mechanism is reasonable. Step 1 NO (g) + F (g) NO F (g) + F (g) (slow) Step NO (g) + F (g) NO F (g) (fast) 97

23 You need to determine whether the proposed mechanism is reasonable. To do this, you need to answer the following questions: - Do the steps add up to give the overall reaction? You need to write the balanced chemical equation since it was not given. - Are the steps reasonable in terms of their molecularity? - Is the mechanism consistent with the experimentally determined rate law? You know the proposed mechanism and the rate law for the overall reaction. First, write the balanced chemical equation. Next, add the two reactions and cancel out reaction intermediates. Check the molecularity of the steps. Determine the rate law equation for the rate determining step, and compare it to the overall rate law equation. The balanced chemical equation is NO (g) + F (g) NO F (g) Add the two steps. Step 1 NO (g) + F (g) NO F (g) + F (g) Step NO (g) + F (g) NO F (g) NO (g) + F (g) + F (g) NO F (g) + F (g) or, NO (g) + F (g) NO F (g) The two steps add up to give the overall reaction. Both steps are bimolecular, which is chemically reasonable. The first step of the mechanism is rate-determining, and its rate law equation is Rate 1 = k 1 [NO ][F ]. This rate law equation matches the overall reaction. Based on the steps of the proposed mechanism, the molecularity of these steps, and the rate law equation of the rate-determining step, the proposed mechanism seems reasonable. The reaction intermediate in the proposed mechanism is atomic fluorine, F. When the steps were added, you were able to cancel F. 18. Problem A researcher is investigating the following overall reaction. C + D E The researcher claims that the rate of law equation for the reaction is written as follows: Rate = k[c][d] (a) Is the rate law equation possible for the given reaction? (b) If so, suggest a mechanism that would match the rate of law. If not, explain why not. Parts (a) and (b) can be answered together. You need to determine if the given rate law equation is possible, and provide a possible mechanism. You know the chemical equation and the rate law equation. 98

24 You need to write a possible mechanism that is consistent with the given information. The rate law equation is first order in both C and D. Step 1 of the mechanism can involve a rate-determining bimolecular reaction between C and D to form an intermediate, B. In Step, B can react with another molecule of C to form the product, E. (a) and (b) The following proposed mechanism satisfies the criteria. Step 1 C + D B (slow) Step B + C E (fast) C + B + D B + E or, C + D E This is a reasonable mechanism, since both steps are bimolecular. Step 1 is ratedetermining, which is consistent with the rate law equation. The proposed reaction intermediate, B, cancels when the two steps are added. 19. Problem A chemist proposes the following reaction mechanism for a certain reaction. Step 1 A + B C (slow) Step C + B E + F (fast) (a) Write the equation for the chemical reaction that is described by this mechanism. (b) Write a rate law equation that is consistent with the proposed mechanism. First, you need to write the overall chemical equation that is described by the proposed mechanism. Next, you need to write a rate law equation that is consistent with the proposed mechanism. You know the proposed mechanism, and the rate-determining step. (a) Add the two steps and cancel any reaction intermediates. (b) Since Step 1 is rate-determining, this step can be used to write the rate law equation. (a) Adding the steps of the mechanism gives Step 1 A + B C (slow) Step C + B E + F (fast) A + B + C C + E + F or, A + B E + F (b) Step 1 is rate-determining, so the rate law equation for the overall reaction is simply the rate law for Step 1 of the mechanism. Rate = Rate 1 = k[a][b] The reaction intermediate, C, was cancelled to obtain the overall equation. The bimolecular rate law equation for the reaction is consistent with having Step 1 as the rate-determining step. 99

25 0. Problem Consider the reaction between -bromo--methylpropane and water. (CH 3 ) 3 CBr (aq) + H O (l) (CH 3 ) 3 COH (aq) + H + (aq) + Br (aq) Rate experiments show that the reaction is first order in (CH 3 ) 3 CBr, but zero order in water. Demonstrate that the accepted mechanism, shown below, is reasonable. Step 1 (CH 3 ) 3 CBr (aq) (CH 3 ) 3 C + (aq) + Br (aq) (slow) Step (CH 3 ) 3 C + + (aq) + H O (l) (CH 3 ) 3 COH (aq) (fast) + Step 3 (CH 3 ) 3 COH (aq) H + (aq) + (CH 3 ) 3 COH (aq) (fast) Show that the proposed mechanism is reasonable and that the reaction is first order in (CH 3 ) 3 CBr and zero order in water. You know the proposed reaction mechanism. You know that the reaction is first order in (CH 3 ) 3 CBr and zero order in water. From the given information, write the rate law equation. Add the steps of the mechanism to determine if they combine to give the overall equation. Ascertain that the proposed mechanism supports the rate law equation. The rate law equation is Rate = k[(ch 3 ) 3 CBr] 1 [H O] 0 = k[(ch 3 ) 3 CBr] Add the steps of the proposed mechanism. Notice that (CH 3 ) 3 C + (aq) and + (CH 3 ) 3 COH (aq) are both reaction intermediates and can be cancelled. Step 1 (CH 3 ) 3 CBr (aq) (CH 3 ) 3 C + (aq) + Br (aq) (slow) Step (CH 3 ) 3 C + + (aq) + H O (l) (CH 3 ) 3 COH (aq) (fast) + Step 3 (CH 3 ) 3 COH (aq) H + (aq) + (CH 3 ) 3 COH (aq) (fast) (CH 3 ) 3 CBr (aq) + H O (l) (CH 3 ) 3 COH (aq) + H + (aq) + Br (aq) Since Step 1 is rate-determining, the rate law equation will be Rate = Rate 1 = k[(ch 3 ) 3 CBr] The steps of the proposed mechanism combine to give the overall equation, after cancelling the two reaction intermediates, (CH 3 ) 3 C + + (aq) and (CH 3 ) 3 COH (aq). Since step 1 is rate-determining, the rate law equation for step 1 is equivalent to the rate law equation for the overall reaction. 100