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CHEMISTRY 12 Chapter 6 Rates of Chemical Reactions Solutions for Practice Problems Student Textbook page 272 1. Problem Cyclopropane, C3H6 , is used in the synthesis of organic compounds and as a fast- acting anesthetic. It undergoes rearrangement to form propene. If cyclopropane disappears at a rate of 0.25 mol/s, at what rate is propene being produced? What Is Required? Calculate the rate of propene production. What Is Given? The rate of rearrangement of cyclopropane is 0.25 mol/s. Plan Your Strategy Cyclopropane and propene both have the formula C3H6 . The balanced chemical equation is C3H6 (cyclopropane) → C3H6 (propene) Determine the mole to mole ratio between cyclopropane and propene. Use this to determine the rate of formation of propene. Act on Your Strategy 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 0.25 mol/s. Check Your Solution The 1:1 stoichiometry of the reaction indicates that the rate of reaction of cyclopropane is equivalent to the rate of formation of propene. 2. Problem Ammonia, NH3, reacts with oxygen to produce nitric oxide, NO, and water vapour. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) At a speciﬁc time in the reaction, ammonia is disappearing at a rate of 0.068 mol/(L·s). What is the corresponding rate of production of water? What Is Required? Calculate the rate of production of water. What Is Given? You have the balanced chemical equation and the rate of disappearance of ammonia. Plan Your Strategy Determine the mole to mole ratio between NH3 and H2O. Since 6 mol of H2O are produced for every 4 mol of NH3 that react, the rate of appearance of H2O is 6 or 3 4 2 the rate of reaction of NH3. Chapter 6 Rates of Chemical Reactions • MHR 84 CHEMISTRY 12 Act on Your Strategy Rate of appearance of H2O = 3 × (rate of disappearance of NH3) 2 = 3 × 0.068 mol/(L·s) 2 = 0.10 mol/(L·s) Check Your Solution 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) + O2(g) → 2Br2(g) + 2H2O(g) How does the rate of decomposition of HBr (in mol/(L·s)) compare with the rate of formation of Br2 (also in mol/(L·s))? Express your answer as an equation. What Is Required? Determine the rate of decomposition of HBr relative to the rate of formation of Br2 , and write your answer as an equation. What Is Given? You are given the balanced chemical equation. Plan Your Strategy 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 Br2 . Act on Your Strategy Since 4 mol HBr are consumed for every 2 mol of Br2 produced, the rate of decomposition of HBr is equal to 2 times the rate of formation of Br2 . Rate of decomposition of HBr = 2 × (rate of formation of Br2 ) or, ∆[Br2] = −0.5∆[HBr] ∆t ∆t Check Your Solution 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 Br2 . 4. Problem Magnesium metal reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Over an interval of 1.00 s, the mass of Mg(s) changes by −0.011 g. (a) What is the corresponding rate of consumption of HCl(aq) (in mol/s)? (b) Calculate the corresponding rate of production of H2(g) (in L/s) at 20˚C and 101 kPa. What Is Required? First, calculate the rate of consumption of HCl(aq) , in mol/s. Then calculate the rate of production of H2 , in L/s at 20˚C and 101 kPa. What Is Given? You have the balanced chemical equation, and you know the rate of consumption of Mg, in g/s. Chapter 6 Rates of Chemical Reactions • MHR 85 CHEMISTRY 12 Plan Your Strategy From the balanced chemical equation, you can obtain the relative rates of consump- tion 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 H2 are both 1. This means that the rate of production of H2 , in mol/s, is equal to the rate of disappearance of Mg, in mol/s. For H2 , you then need to convert mol/s to L/s at 20˚C, 101 kPa. Recall that 1.00 mol of any gas occupies a volume of 24.0 L at 20˚C, 101 kPa. Act on Your Strategy 0.011 g/s (a) Rate of disappearance of Mg = 24.3 g/mol = 4.5 × 10−4 mol/s Rate of consumption of HCl = 2 × (rate of disappearance of Mg) = 2 × 4.5 × 10−4 mol/s = 9.0 × 10−4 mol/s (b) Rate of appearance of H2 = rate of disappearance of Mg = (4.5 × 10−4 mol/s) × (24.0 L/mol) = 0.011 L/s at 20˚C, 101 kPa Check Your Solution You have correctly used the coefﬁcients in the balanced equation. The units are correct. Solutions for Practice Problems Student Textbook page 284 5. Problem When heated, ethylene oxide decomposes to produce methane and carbon monoxide. C2H4O(g) → CH4(g) + CO(g) At 415˚C, the following initial rate data were recorded. Experiment [C 2H 4O] 0 (mol/L) Initial rate (mol/(L • s)) 1 0.002 85 5.84 × 10−7 2 0.004 28 8.76 × 10−7 3 0.005 70 1.17 × 10−6 Determine the rate law equation and the rate constant at 415˚. What Is Required? You need to ﬁnd the value of m and k in the general rate law equation for the reaction. Rate = k[C2H4O]m What Is Given? You know the initial concentration of the reactant and the initial rates for three experiments. Plan Your Strategy Since there is only one reactant, simply compare the initial rates of two experiments with different initial [C2H4O]. Compare the change in initial rate with the change in initial concentration. Chapter 6 Rates of Chemical Reactions • MHR 86 CHEMISTRY 12 Act on Your Strategy Begin with experiments 1 and 2. Set up a ratio of the two rates as shown. m Rate2 = k 0.002 85 mol/L = 5.84 × 10−7 mol/(L·s) Rate1 k 0.004 28 mol/L m 8.76 × 10−7 mol/(L·s) Because k is constant, it can be cancelled. m Rate2 = k 0.002 85 mol/L = 5.84 × 10−7 mol/(L·s) Rate1 k 0.004 28 mol/L m 8.76 × 10−7 mol/(L·s) 0.666m = 0.666 m = 1 (by inspection) Similarly, you can set up a ratio involving experiments 1 and 3 or experiments 2 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, 2 or 3 and solve for k. Substituting data from experiment 1 into the rate law equation gives: 1 5.84 × 10−7 mol/(L·s) = k 0.002 85 mol/L −7 k = 5.84 × 10 mol/(L·s) 0.002 85 mol/L = 2.05 × 10−4 s−1 (at 415˚C) The rate law equation for the reaction is: Rate = (2.05 × 10−4 s−1)[C2H4O]1 (at 415˚C) Check Your Solution 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. 2ICl + H2 → I2 + 2HCl At temperature T, the following initial rate data were recorded. Experiment [ICI] 0 (mol/L) [H 2 ] 0 (mol/L) Initial rate (mol/(L • s)) 1 0.20 0.050 0.0015 2 0.40 0.050 0.0030 3 0.20 0.200 0.0060 Determine the rate law equation and the rate constant at temperature T. What Is Required? You need to ﬁnd the values of m, n, and k in the general rate law equation for the reaction. Rate = k[ICl]m[H2]n What Is Given? You know the initial concentrations of the reactants and the initial rates for three experiments. Chapter 6 Rates of Chemical Reactions • MHR 87 CHEMISTRY 12 Plan Your Strategy Step 1 Find two experiments in which [ICl] remains constant and [H2 ] changes. Compare the rates and concentrations to solve for n. Then ﬁnd two experiments in which [H2 ] remains constant and [ICl] changes. Compare the rates and concentrations to solve for m. Step 2 Use the data and your calculated values for m and n to solve for k, using the following equation. Rate = k[ICl]m[H2]n Act on Your Strategy Step 1 Begin with experiments 1 and 3, where [ICl] remains constant, to determine the order with respect to H2 . 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. m Rate3 k 0.020 mol/L [0.200]n = m = 0.0060 mol/(L·s) Rate1 k 0.020 mol/L [0.050]n 0.0015 mol/(L·s) Because k is a constant, you can cancel it out. m Rate3 k 0.020 mol/L [0.200]n = m = 0.0060 mol/(L·s) Rate1 k 0.020 mol/L [0.050]n 0.0015 mol/(L·s) (4)n = 4 n = 1 (by inspection) Therefore, the reaction is ﬁrst order with respect to H2 . Set up a ratio for experiments 1 and 2, where [H2 ] is constant. You can cancel [H2 ] because it is the same in both experiments. m 1 Rate2 = k 0.040 mol/L [0.050] = 0.0030 mol/(L·s) m Rate1 k 0.020 mol/L [0.050]1 0.0015 mol/(L·s) Because k is a constant, you can cancel it out. m 1 Rate2 = k 0.040 mol/L [0.050] = 0.0030 mol/(L·s) m Rate1 k 0.020 mol/L [0.050]1 0.0015 mol/(L·s) (2) = 2 m m = 1 (by inspection) The reaction is ﬁrst order with respect to ICl. Step 2 Overall, the rate law equation is: Rate = k[ICl]1[H2]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 1 1 0.0015 mol/(L·s) = k 0.20 mol/L 0.050 mol/L k= 0.0015 mol/(L·s) 1 1 0.20 mol/L 0.050 mol/L = 0.15 L/(mol·s) (at temperature T ) The overall rate law equation for the reaction is: Rate = 0.15 L/(mol·s) [ICl][H2] Check Your Solution Check the values of m and n by inspection. When [ICl] doubles (when [H2 ] is constant), the rate also doubles. When [H2 ] quadruples (when [ICl] is constant), the rate also quadruples. To check the value for k, substitute data from experiments 2 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. Chapter 6 Rates of Chemical Reactions • MHR 88 CHEMISTRY 12 7. Problem Sulfuryl chloride (also known as chlorosulfuric acid and thionyl chloride), SO2Cl2 , 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. SO2Cl2(g) → SO2(g) + Cl2(g) [SO2Cl 2 ] (mol/L) Initial rate (mol/(L • s)) 0.150 3.3 × 10−6 0.300 6.6 × 10−6 0.450 9.9 × 10−6 (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. What Is Required? You need to ﬁnd the values of m and k in the general rate law equation for the reaction. Rate = k[SO2Cl2]m What Is Given? You know the initial concentration of the reactant and the initial rates for three experiments. Plan Your Strategy Since there is only one reactant, simply compare the initial rates of two experiments with different initial [SO2Cl2]. Compare the change in initial rate with the change in initial concentration. Act on Your Strategy (a) Begin with the ﬁrst and second experiments. Set up a ratio of the two rates as shown. m −6 Rate2 = k 0.150 mol/L m = 3.3 × 10−6 mol/(L·s) Rate1 k 0.300 mol/L 6.6 × 10 mol/(L·s) Because k is constant, it can be cancelled. m Rate2 = k 0.150 mol/L = 3.3 × 10−6 mol/(L·s) m Rate1 k 0.300 mol/L 6.6 × 10−6 mol/(L·s) 0.5m = 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[SO2Cl2] (b) To determine the value of k, substitute m = 1 into the rate law equation. You can use data from any of experiments 1, 2 or 3 and solve for k. Substituting data from the ﬁrst experiment into the rate law equation gives: 3.3 × 10−6 = k[0.15]1 −6 k = 3.3 × 10 mol/(L·s) 0.15 mol/L = 2.2 × 10−5 s−1 The rate law equation for the reaction is: Rate = (2.2 × 10−5 s−1)[SO2Cl2]1 Chapter 6 Rates of Chemical Reactions • MHR 89 CHEMISTRY 12 Check Your Solution 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. 2A + 3B + C → products This reaction was found to obey the following rate law equation. Rate = k[A]2[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. Initial [A] Initial [B] Initial [C] Initial rate Experiment (mol/L) (mol/L) (mol/L) (mol/(L • s)) 1 0.10 0.20 0.050 0.40 2 0.10 (a) 0.10 0.40 3 0.20 0.050 (b) 0.20 4 (c) 0.025 0.040 0.45 5 0.10 0.010 0.15 (d) What Is Required? You need to calculate the missing values in the table: (a) = Initial [B] for experiment 2 (b) = Initial [C] for experiment 3 (c) = Initial [A] for experiment 4 (d) = Initial rate for experiment 5 What Is Given? You know the rate law equation, and you are given some incomplete rate data. Plan Your Strategy You know that the reaction is ﬁrst order in B and C. This means that when the con- centration 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 22 , or 4 times. If [A] is tripled, the rate will increase by a factor of 32 , 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 2 to 5. Act on Your Strategy Substituting data from experiment 1 into the rate law equation gives: 0.40 mol/(L·s) = k[0.10 mol/L]2[0.20 mol/L]1[0.050 mol/L]1 k= 0.40 mol/(L·s) [0.10 mol/L]2[0.20 mol/(L·s)]1[0.050 mol/L]1 = 4.0 × 103 L3/(s·mol3) Chapter 6 Rates of Chemical Reactions • MHR 90 CHEMISTRY 12 (a) To solve for (a) in experiment 2, use the value of k and the given data. Substitute into the rate equation. 0.40 mol/(L·s) = 4.0 × 103 L3/(s·mol3) × [0.10 mol/L]2 × (a) × [0.10 mol/L]1 (a) = 0.40 mol/(L·s) 4.0 × 103 L3/(s·mol3)[0.10 mol/L]2 × [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. 0.20 mol/(L·s) = 4.0 × 103 L3/(s·mol3) × [0.20 mol/L]2[0.050]1 × (b) (b) = 0.20 mol/(L·s) 4.0 × 103 L3/(s·mol3) × [0.20 mol/L]2[0.050 mol/L]1 = 0.025 mol/L (c) To solve for (c) in experiment 4, use the value of k and the given data. Substitute into the rate equation. 0.45 mol/(L·s) = 4.0 × 103 L3/(s·mol3) × (c)2 × [0.025]1[0.040]1 (c)2 = 0.45 mol/(L·s) 4.0 × 103 L3/(s·mol3)[0.025 mol/L]1[0.040 mol/L]1 = 0.11 mol2/L2 (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) = 4.0 × 103 L3/(s·mol3) × [0.10 mol/L]2[0.010]1[0.15]1 (d) = 0.060 mol/(L·s) Check Your Solution 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 287 9. Problem Cyclopropane, C3H6 , 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 9.2 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? What Is Required? 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. What Is Given? You know that the reaction is ﬁrst order in cyclopropane, with k = 9.2 s−1 . Plan Your Strategy (a) To ﬁnd t1/2 , use the following equation. t1/2 = 0.693 k Chapter 6 Rates of Chemical Reactions • MHR 91 CHEMISTRY 12 1 (b) Each half-life reduces the initial amount of cyclopropane by , or 50%. 2 Let the original concentration of cyclopropane be A. 1 After 1 t1/2 , [cyclopropane] = 2 A 1 1 1 After 2 t1/2 , [cyclopropane] = 2 × 2 A = 4 A After 3 t1/2 , [cyclopropane] = 1 ×2 1 4 A = 1 8 A After 4 t1/2 , [cyclopropane] = 1 ×2 1 8 A = 1 16 A 4 After 4 half-lives, there is 16 , or 1 1 2 of the original amount of cyclopropane remaining. Act on Your Strategy (a) Solve for t1/2 . t1/2 = 0.693 9.2 s−1 = 7.5 × 10−2 s The half-life of the reaction is 7.5 × 10−2 s. 1 4 1 (b) After 4 half-lives, there will be = 16 of the original amount present. If we 2 take the original concentration of cyclopropane to be 100, this corresponds to 1 16 × 100 = 6.2 Therefore, 6.2% of the original amount of cyclopropane will remain after 4 half-lives. Check Your Solution 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 2 1 × 1 × 1 × 1 = 16 of the original amount, or about 6.2% remaining. 2 2 2 10. Problem Peroxyacetyl nitrate (PAN), H3CCO2ONO2 , is a constituent of photochemical smog. It undergoes a ﬁrst-order decomposition reaction with t1/2 = 32 min. (a) Calculate the rate constant in s−1 for the ﬁrst-order decomposition of PAN. (b) 128 min after a sample of PAN began to decompose, the concentration of PAN in the air is 3.1 × 1013 molecules/L . What was the concentration of PAN when the decomposition began? What Is Required? 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 3.1 × 1013 molecules/L after decomposing for 128 minutes. What Is Given? (a) The half-life of the reaction is 32 minutes. (b) [PAN] = 3.1 × 1013 molecules/L at t = 128 minutes. Plan Your Strategy (a) Rearrange the following equation and solve for k. t1/2 = 0.693 k Chapter 6 Rates of Chemical Reactions • MHR 92 CHEMISTRY 12 (b) Begin by determining the number of half-lives in 128 minutes. Use this to determine the original [PAN]. We know that 1 n 2 × (original concentration) = (concentration after n half-lives) or, (original concentration) = (concentration after n half-lives) 1 n 2 Act on Your Strategy (a) 32 min = 32 min × 60 s/min = 1.9 × 103 s k = 0.693 t1/2 = 0.693 1.9 × 103 s = 3.6 × 10−4 s−1 (b) Number of half-lives = 128 min 32 min/half-life = 4 half-lives After 4 half-lives, [PAN] = 3.1 × 1013 molecules/L. Substituting into the equation gives (original concentration) = concentration after n half-lives 1 n 2 = 3.1 × 10 1molecules/L 13 4 2 = 5.0 × 10 molecules/L 14 The original [PAN] is 5.0 × 1014 molecules/L . Check Your Solution 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 3.1 × 1013 molecules/L . 11. Problem In general, a reaction is essentially over after 10 half-lives. Prove that this generaliza- tion is reasonable. What Is Required? Show that only a negligible percentage of the original amount of a substance remains after 10 half-lives. What Is Given? You are given the time of 10 half-lives. Plan Your Strategy Use the following equation: n Amount remaining after n half-lives = (original amount) × 1 2 For simplicity, take the original mass of the substance to be 100 g. Act on Your Strategy 10 Amount remaining after 10 half-lives = (100 g) × 1 2 Amount remaining after 10 half-lives = 0.098 g After 10 half-lives, a 100 g sample would be reduced to 0.098 g, or 0.098% of the original amount. This is a negligible amount. Chapter 6 Rates of Chemical Reactions • MHR 93 CHEMISTRY 12 Check Your Solution Take 100 and divide it by 2 ten times to get the same answer. Less than one-tenth of one percent of a substance remains after 10 half-lives. 12. Problem The half-life of a certain ﬁrst-order reaction is 120 s. How long do you estimate that it will take for 90% of the original sample to react? What Is Required? Estimate how long it will take for 90% of the sample to react. What Is Given? You know that the half-life of the reaction is 120 s. Plan Your Strategy 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 2 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. n Amount remaining after n half-lives = (original amount) × 1 2 Act on Your Strategy Dividing 100 g by 2 several times gives 100 = 50 (1 t ) 1/2 2 50 = 25 (2 t ) 1/2 2 25 = 12.5 (3 t ) 1/2 2 12.5 = 6.25 (4 t ) 1/2 2 It takes between 3 and 4 half-lives for 90% of the sample to react. If the half-life is 120 s, this will take about 400 s. Alternate Solution (using logarithms) n Amount remaining after n half-lives = Original amount × 1 2 or, 1 n Amount remaining after n half-lives = 2 Original amount = 10 100 = 0.10 Taking the logarithm of both sides gives: 1 n log 2 = log(0.10) 1 n × log 2 = log(0.10) log(0.10) n= log 12 = 3.3 It will take 3.3 half-lives, or 3.9 × 102 s, for 90% of the sample to react. Check Your Solution The estimated answer agrees with the calculated value. Chapter 6 Rates of Chemical Reactions • MHR 94 CHEMISTRY 12 Solutions for Practice Problems Student Textbook page 294 13. Problem The following reaction is exothermic. 2ClO(g) → Cl2(g) + O2(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) 2ClO(g) ∆H Cl2(g) + O2(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, Ea(rev) = 34 kJ Draw and label a potential energy diagram for this reaction. Calculate and label Ea(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. Chapter 6 Rates of Chemical Reactions • MHR 95 CHEMISTRY 12 possible activated complex A ... C ... B Potential Energy (kJ) AC + B Ea(fwd) = 99 kJ ∆H = +65 kJ AB + C Reaction Progress For an endothermic reaction, Ea(fwd) = ∆H + Ea(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 = −132 kJ, Ea(fwd) = 61 kJ Draw and label a potential energy diagram for this reaction. Calculate and label Ea(rev) . Include a possible structure for the activated complex. Solution Since this reaction is exothermic, the potential energy diagram will be similar to Figure 6.12. possible activated complex C ... D Potential Energy (kJ) Ea(rev) = 193 kJ C+D ∆H = −132 kJ CD Reaction Progress Chapter 6 Rates of Chemical Reactions • MHR 96 CHEMISTRY 12 For an exothermic reaction, Ea(rev) = ∆H + Ea(fwd) = 132 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 O2(g) . For example, it exists as ozone, O3(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 −392 kJ and the activation energy is 19 kJ. Draw and label a potential energy diagram. Include a value for Ea(rev) . Propose a structure for the activated complex. Solution The reaction is O3(g) + O(g) → 2O2(g) ∆H = −392 kJ; Ea(fwd) = 19 kJ Since this reaction is exothermic, the potential energy diagram will be similar to Figure 6.12. Potential Energy (kJ) Ea(rev) = 411 kJ O3 + O ∆H = −392 kJ 2O2 Reaction Progress For an exothermic reaction, Ea(rev) = ∆H + Ea(fwd) = 392 kJ + 19 kJ = 411 kJ A possible structure of the activated complex might involve partial bond breakage in the O3 molecule coupled with partial bond formation to produce two O2 molecules. Solutions for Practice Problems Student Textbook page 301 17. Problem NO2(g) and F2(g) react to form NO2F(g) . The experimentally determined rate law for the reaction is written as follows: Rate = k[NO2][F2] A chemist proposes the following mechanism. Determine whether the mechanism is reasonable. Step 1 NO2(g) + F2(g) → NO2F(g) + F(g) (slow) Step 2 NO2(g) + F(g) → NO2F(g) (fast) Chapter 6 Rates of Chemical Reactions • MHR 97 CHEMISTRY 12 What Is Required? 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? What Is Given? You know the proposed mechanism and the rate law for the overall reaction. Plan Your Strategy 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. Act on Your Strategy The balanced chemical equation is 2NO2(g) + F2(g) → 2NO2F(g) Add the two steps. Step 1 NO2(g) + F2(g) → NO2F(g) + F(g) Step 2 NO2(g) + F(g) → NO2F(g) 2NO2(g) + F2(g) + F(g) → 2NO2F(g) + F(g) or, 2NO2(g) + F2(g) → 2NO2F(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 Rate1 = k1[NO2][F2] . 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. Check Your Solution 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. 2C + 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. What Is Required? 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. What Is Given? You know the chemical equation and the rate law equation. Chapter 6 Rates of Chemical Reactions • MHR 98 CHEMISTRY 12 Plan Your Strategy 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 2, B can react with another molecule of C to form the product, E. Act on Your Strategy (a) and (b) The following proposed mechanism satisﬁes the criteria. Step 1 C+D→B (slow) Step 2 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 rate- determining, which is consistent with the rate law equation. Check Your Solution 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 2 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. What Is Required? 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. What Is Given? You know the proposed mechanism, and the rate-determining step. Plan Your Strategy (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. Act on Your Strategy (a) Adding the steps of the mechanism gives Step 1 A+B→C (slow) Step 2 C+B→E+F (fast) A + 2B + C → C + E + F or, A + 2B → 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 = Rate1 = k[A][B] Check Your Solution 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. Chapter 6 Rates of Chemical Reactions • MHR 99 CHEMISTRY 12 20. Problem Consider the reaction between 2-bromo-2-methylpropane and water. (CH3)3CBr(aq) + H2O( ) → (CH3)3COH(aq) + H+(aq) + Br−(aq) Rate experiments show that the reaction is ﬁrst order in (CH3)3CBr, but zero order in water. Demonstrate that the accepted mechanism, shown below, is reasonable. Step 1 (CH3)3CBr(aq) → (CH3)3C+(aq) + Br−(aq) (slow) + + Step 2 (CH3)3C (aq) + H2O( ) → (CH3)3COH2 (aq) (fast) + + Step 3 (CH3)3COH2 (aq) → H (aq) + (CH3)3COH(aq) (fast) What Is Required? Show that the proposed mechanism is reasonable and that the reaction is ﬁrst order in (CH3)3CBr and zero order in water. What Is Given? You know the proposed reaction mechanism. You know that the reaction is ﬁrst order in (CH3)3CBr and zero order in water. Plan Your Strategy 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. Act on Your Strategy The rate law equation is Rate = k[(CH3)3CBr]1[H2O]0 = k[(CH3)3CBr] Add the steps of the proposed mechanism. Notice that (CH3)3C+(aq) and (CH3)3COH2+(aq) are both reaction intermediates and can be cancelled. Step 1 (CH3)3CBr(aq) → (CH3)3C+(aq) + Br−(aq) (slow) + + Step 2 (CH3)3C (aq) + H2O( ) → (CH3)3COH2 (aq) (fast) Step 3 (CH3)3COH2+(aq) → H+(aq) + (CH3)3COH(aq) (fast) (CH3)3CBr(aq) + H2O( ) → (CH3)3COH(aq) + H+(aq) + Br−(aq) Since Step 1 is rate-determining, the rate law equation will be Rate = Rate1 = k[(CH3)3CBr] Check Your Solution The steps of the proposed mechanism combine to give the overall equation, after cancelling the two reaction intermediates, (CH3)3C+(aq) and (CH3)3COH2+(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. Chapter 6 Rates of Chemical Reactions • MHR 100