Chapter 12 Chemical Kinetics

2ClO2 (aq) + 2OH– (aq)  ClO3– (aq) + ClO2– (aq) + H2O (l) QUESTION Chlorine dioxide (ClO2) is a disinfectant used in municipal water-treatment plants. It dissolves in basic solution producing ClO3– and ClO2–: 2ClO2 (aq) + 2OH– (aq)  ClO3– (aq) + ClO2– (aq) + H2O (l) Of the following, which would not be a proper expression to relate information about the rate of the reaction? 1. DClO2/Dt = 2DClO3–/Dt 2. DClO2/Dt = DOH–/Dt 3. DClO2/Dt = DClO2–/Dt 4. DOH–/Dt = 2DClO2–/Dt Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 3 is not a proper expression for the relative rates of this reaction. The stoichiometry of the reaction indicates that ClO2 disappears twice as fast as ClO2– forms; so to determine the rate for ClO2 , the rate of ClO2– would have to be doubled. Section 12.1: Reaction Rates Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION The average rate for the disappearance of NO2 between the time intervals given in the table shows a gradually slowing rate. Which statement correlates with the data? HMClassPrep Figure 12.2 Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION (continued) 1. The rates represent the average within 50 second intervals. These are not the same even though the times are the same because instantaneous rates were not used. 2. If the rate were taken from 0  250 seconds, the rate obtained would be more meaningful. 3. As NO2 decomposes the rate at which it does so over a 50 second interval is constant. 4. The instantaneous rates would decrease with time, but could be taken within a 50 second interval. Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 supplies correct information. The instantaneous rates would decrease over time (as average rates do), would have different values than those listed, and could be taken at any time. Section 12.1: Reaction Rates Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION A reaction is shown to be first order with respect to one reactant, with a second order dependency on another reactant. Which statement makes an accurate comment about these two observations? Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION (continued) 1. Assuming no temperature change, doubling the concentration of both reactants would cause the reaction rate to double. 2. Assuming no temperature change, doubling the concentration of both reactants would cause the reaction rate to increase by a factor of four. 3. Assuming no temperature change, doubling the concentration of both reactants would cause the reaction rate to increase by a factor of five. 4. Assuming no temperature change, doubling the concentration of both reactants would cause the reaction rate to increase by a factor of eight. Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 provides the correct response for how doubling the concentration in this case would affect the reaction rate. Doubling the concentration for a first order dependency would double the rate. However, doubling the concentration when there is a second order dependency would quadruple the rate. The combined effect would then be an eight-fold increase. Section 12.3: Determining the Form of the Rate Law Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION While studying the shelf life of a particular antibiotic a chemist found that when the initial concentration was 0.0036 M the rate of decay was 1.5  10–4 mol/L s. In another experiment under the same conditions a concentration of 0.0013 M decayed at a rate of 1.9  10–5 mol/L s. What is the order of this decay reaction? 1. More information is needed. This cannot be determined from only two experiments. 2. Zero order 3. First order 4. Second order Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 shows the correct order. (rate 2/rate 1) = (conc2/conc1)m. To solve for m, take the ln of both sides of the equation. This yields ln (rate 2/rate 1) = m ln (conc2/conc1); solving for m = 2 Section 12.3: Determining the Form of the Rate Law Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION A specific ingredient in a nutritional supplement can slowly decompose while still in the package. The specific rate constant for the first order kinetics of this decay, at 25.0°C, is 4.2  10–5 hr–1. If the initial concentration were considered to be 100.0% how long would it take to be reduced to 75.0% of its potency? 1. 6,800 hours 2. 32,000 hours 3. –6,800 hours 4. 2,900 hours HMClassPrep Table 12.6 Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 1 provides the correct answer. As the reaction is known to be first order the integrated rate law would be ln(Ao/A) = kt; Ao = 100.0%; A = 75.0%; k = 4.2  10–5 hr–1. In this case t = 6,800 hours. Note two significant digits. Section 12.4: The Integrated Rate Law Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION Two reactions have the same rate. One reaction is first order whereas the other is second order. Which comment about the half lives of both reactions is accurate? 1. As their rates are equal, their half lives will be equal. 2. The half life of the second order reaction will be twice that of the other. 3. The half life of the second order reaction will be 1/2 that of the other. 4. No direct comparison can be made with only that information. HMClassPrep Table 12.6 Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 correctly points out that the comparison cannot be made. The half life equation for a first order equation is t1/2 = 0.693/k. The half life for a second order reaction has an initial concentration dependency; t1/2 = 1/(k [Ao]) Section 12.4: The Integrated Rate Law Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION In a pseudo-first order reaction experiment involving two reactants where component A is used in small concentrations compared to the other reactant, the ln [A] versus time plot will yield a straight line. The value for the specific rate constant can be obtained by… Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION (continued) 1. determining the slope of the resulting straight line that is the pseudo-rate constant, and then multiplying that by the nearly constant concentration of the other component. 2. determining the slope of the resulting straight line, which is the negative of the pseudo-rate constant, and dividing that by the nearly constant concentration of the other component. 3. determining the slope of the resulting straight line and dividing by the initial concentration of the A component. Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 2 relates the procedure for using the slope of a pseudo-first order plot to obtain the specific rate constant. The rate constant, k, incorporates the specific rate constant and the nearly constant concentrations of other reaction components: k = k/(other components from the rate law) Section 12.4: The Integrated Rate Law Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION Of the following, which is NOT true of the rate determining step in proposed mechanisms for chemical reactions? 1. It could be the only step in a proposed mechanism. 2. The step could have a molecularity of two. 3. It must be followed by a fast step to fit the stoichiometry of the overall equation. 4. The rate of the overall reaction is essentially the same as the rate of this step. Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 3 does not apply to all rate determining steps in proposed mechanisms. It is possible that the rate determining step could be followed by a fast step, but there are many cases where this does not happen. Section 12.6: Reaction Mechanisms Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION HMClassPrep Figure 12.11 a Which of the following represents a correct conclusion based on the information presented in this figure? Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION (continued) 1. The forward reaction is endothermic. 2. The activation energy for the forward reaction is less than the activation energy of the reverse reaction. 3. The transition state is at a lower energy than the products. 4. The energy of the reactants represents a lower energy level than both the transition state and the products. Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 2 properly relates the activation energy of the forward reaction to the other observations. From the reactants to the transition state the energy change is less than from the products to the transition state. Section 12.7: A Model for Chemical Kinetics Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION In the Arrhenius activation energy equation for chemical reactions, the “A” factor represents… 1. the frequency factor, incorporating the frequency and the steric factors in collision theory. 2. the frequency factor, incorporating the relative frequency of collisions in a reaction. 3. the fraction of molecules that have effective collisions (always less than 1). 4. the first letter of Arrhenius’s, the discoverer of this equation, last name. Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 1 properly defines A in the Arrhenius equation for activation energy. To have effective collisions both energy and orientation must be adequate. Section 12.7: A Model for Chemical Kinetics Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION Use the Arrhenius activation energy equation to determine the activation energy for a reaction that has, at 25.0°C, a specific rate constant of 0.010 s–1 and at 50.0°C a specific rate constant of 0.020 s–1. (R = 8.3145 J/K mol) 1. 290 J 2. 320 J 3. 22,000 J 4. –22,000 J Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 3 provides the correct mathematical solution to the Arrhenius activation energy equation. Note: Make sure that the temperature is reported in K. Section 12.7: A Model for Chemical Kinetics Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION The second diagram in this figure shows … HMClassPrep Figure 12.16 Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION (continued) 1. that a greater fraction of molecules may encounter effective collisions because the catalyst increases the energy of the collisions. 2. that a greater fraction of molecules may encounter effective collisions because the activation energy is lowered by the catalyst. 3. that a lower fraction of molecules may encounter effective collisions because adding a catalyst decreases the number of collisions thereby increasing the opportunity for a reaction from reactant to product. 4. that a greater fraction of molecules may encounter effective collisions because a catalyst provides an alternate pathway that can raise the activation energy. Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 2 indicates the proper connection between a catalyst and the increased rate of a reaction. The alternate pathway for the reaction provided by a catalyst lowers the activation energy. The graph shows that the Ea value has shifted to the left along the x axis. Section 12.8: Catalysis Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION What role, as suggested in this figure, does the metal surface play in the reaction to add H atoms to a saturated C to C double bond? 1. The metal acts to absorb H2 and change it to 2H so that it can be added to the double bond; once this is completed the metal is ready to do this again. 2. The metal acts as a homogeneous catalyst to promote the hydrogenation of the compound with the double bond. 3. The metal has a greater attraction for C2H6 than for C2H4. This assists, as a catalyst, the completion of the reaction. HMClassPrep Figure 12.17 Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 1 provides a correct interpretation of the figure. The metal surface, through a series of steps, catalyzes the reaction. First the H2 is adsorbed to the surface of the metal. Then the separated H atoms migrate closer to the C2H4 molecules where the reaction takes place. Finally, the products can escape the surface (this is called desorption). Without the available metal surface more energy would be required to separate the H2 molecules into atoms prior to its reaction with C2H4. Section 12.8: Catalysis Copyright © Houghton Mifflin Company. All rights reserved.