1. Chapter 14 Chemical Kinetics
Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 14 Chemical Kinetics
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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2. Aim: What goes into the kinetics of a chemical reaction?
Do Now: How do reactions take place on the molecular level?
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3. Kinetics
In kinetics we study the rate at which a chemical process occurs.
Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).
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4. Turn and Talk
Turn to your neighbor, and see if you can come up with a complete list of factors that affect reaction rates in chemical reactions. Write down your list in your notes. I will stop you in 2 minutes.
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5. Factors That Affect Reaction Rates
Physical State of the Reactants
In order to react, molecules must come in contact with each other.
The more homogeneous the mixture of reactants, the faster the molecules can react.
For heterogeneous mixtures, surface area plays a role. Larger the surface area, the quicker the rate
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6. Think about it…
Why is medicine found in form of a fine powder in caplets?
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7. Factors That Affect Reaction Rates
Concentration of Reactants
As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.
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8. Factors That Affect Reaction Rates
Temperature
At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.
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9. Factors That Affect Reaction Rates
Presence of a Catalyst
Catalysts speed up reactions by changing the mechanism of the reaction.
Catalysts are not consumed during the course of the reaction.
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10. Reaction Rates
Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.
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11. Stop and Think
Can anyone think of a rate that is used in most daily life?
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12. Sample Exercise 14.1 (Pg. 587)
AB is a hypothetical reaction If we know that at 20s the concentration of A is .54 M, and at 40s the concentration is .30M, can we calculate average rate of reaction over this time interval?
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13. Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times.
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14. Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
The average rate of the reaction over each interval is the change in concentration divided by the change in time:
Average rate =
[C4H9Cl] t
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15. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
Note that the average rate decreases as the reaction proceeds.
This is because as the reaction goes forward, there are fewer collisions between reactant molecules.
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16. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
A plot of [C4H9Cl] vs. time for this reaction yields a curve like this.
The slope of a line tangent to the curve at any point is the instantaneous rate at that time.
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17. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
All reactions slow down over time.
Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction.
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18. Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1.
Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.
Rate =
-[C4H9Cl] t =
[C4H9OH]
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19. Reaction Rates and Stoichiometry
What if the ratio is not 1:1?
2 HI(g)  H2(g) + I2(g)
In such a case,
Rate = − 1 2
[HI] t =
[I2]
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20. Reaction Rates and Stoichiometry
To generalize, then, for the reaction
aA + bB
cC + dD
Rate = − 1 a
[A] t
= − b
[B] = c
[C] d
[D]
* Use this for the rate of reaction when you aren’t specifying a particular reactant or product
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21. Aim: How does concentration affect rates, and what are rate laws?
Do Now: At a certain time in a reaction, substance A is disappearing at a rate of 4.0 x 10-2 M/s, substance B is appearing at a rate of 2.0 x 10-2 M/s, and substance C is appearing at a rate of 6.0 x 10-2 M/s. Which of the following could be the stoichiometry for the reaction being studied?
2A + B  3C
A 2B + 3C
2A  B + 3C
4A 2B + 3C
A + 2B  3C
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22. Sample Exercise 14.3 (Pg.581)
a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2O3(g)3O2(g)? b) If the rate at which O2 appears, ∆[O2]/∆t is 6.0 x 10-5 M/s at a particular instant, at what rate is O3 disappearing at this same time, -∆[O3]/∆t.
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23. Concentration and Rate
One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.
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24. Concentration and Rate
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
If we compare Experiments 1 and 2, we see that when [NH4+] doubles, the initial rate doubles.
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25. Concentration and Rate
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Likewise, when we compare Experiments 5 and 6, we see that when [NO2−] doubles, the initial rate doubles.
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26. Concentration and Rate
This means
Rate  [NH4+]
Rate  [NO2−]
Rate  [NH4+] [NO2−]
which, when written as an equation, becomes
Rate = k [NH4+] [NO2−]
This equation is called the rate law, and k is the rate constant.
Therefore,
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27. Rate Laws
A rate law shows the relationship between the reaction rate and the concentrations of reactants.
The exponents tell the order of the reaction with respect to each reactant.
Since the rate law is
Rate = k [NH4+] [NO2−]
the reaction is
First-order in [NH4+] and
First-order in [NO2−].
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28. Rate Laws Rate = k [NH4+] [NO2−]
The overall reaction order can be found by adding the exponents on the reactants in the rate law.
This reaction is second-order overall.
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29. Practice with Rate Laws Sample Exercise 14.4 (Pg. 584)
Consider a reaction A + B C for which the rate=k[A][B]2. Each of the following boxes represents a reaction mixture in which A is shown in red spheres and B as purple ones. Rank the mixtures in order of increasing rate of reaction.
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30. Using the data, determine The rate law for the reaction
The rate constant
The rate of the reaction when [A] = .050M and [B] =.100M
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31. Aim: How can we understand rate laws using concentration changes with time?
Do Now: Take out your homework from before the break.
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32. Integrated Rate Laws
Using calculus to integrate the rate law for a first-order process gives us ln
[A]t
[A]0
= −kt
Where
[A]0 is the initial concentration of A, and
[A]t is the concentration of A at some time, t, during the course of the reaction.
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33. Integrated Rate Laws Manipulating this equation produces… ln [A]t [A]0
= −kt
ln [A]t − ln [A]0 = − kt
ln [A]t = − kt + ln [A]0
…which is in the form
y = mx + b
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34. First-Order Processes
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k.
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35. First-Order Processes
Consider the process in which methyl isonitrile is converted to acetonitrile.
CH3NC
CH3CN
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36. First-Order Processes
CH3NC
CH3CN
This data was collected for this reaction at °C.
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37. First-Order Processes
When ln P is plotted as a function of time, a straight line results.
Therefore,
The process is first-order.
k is the negative of the slope: 5.1  10-5 s−1.
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38. Sample Exercise 14.7(Pg. 588)
The decomposition of a certain insecticide in water at 12∘C follows first-order kinetics with a rate constant of 1.45 yr A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10-7 g/cm3. Assume that the temperature of the lake is constant.
What is the concentration of the insecticide on June 1 of the following year?
How long will it take for the insecticide concentration to decrease to 3.0 x 10-7 g/cm3 ?
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39. Second-Order Processes
Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1
[A]t
= kt +
[A]0
also in the form
y = mx + b
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40. Second-Order Processes1
[A]t
= kt +
[A]0
So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k. 1
[A]
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41. Second-Order Processes
The decomposition of NO2 at 300°C is described by the equation
NO2 (g)
NO (g) O2 (g) 1 2
and yields data comparable to this:
Time (s)
[NO2], M
0.0
50.0
100.0
200.0
300.0
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42. Practice (Sample Exercise 14.8 Pg. 590)
Using the data table for the reaction:
Is the reaction first or second order in NO2?
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43. Aim: How can we calculate Half-Life for first order and second order reactions?
Do Now: Take out your homework graphs.
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44. Second-Order Processes
Plotting ln [NO2] vs. t yields the graph at the right.
The plot is not a straight line, so the process is not first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
−4.610
50.0
−4.845
100.0
−5.038
200.0
−5.337
300.0
−5.573
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45. Second-Order Processes1
[NO2]
Graphing vs. t, however, gives this plot.
Because this is a straight line, the process is second-order in [A].
Time (s)
[NO2], M
1/[NO2]
0.0
100
50.0
127
100.0
154
200.0
208
300.0
263
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46. Zero Order Process
In some reactions, the rate is apparently independent of the reactant concentration. The rates of these zero-order reactions do not vary with increasing nor decreasing reactants concentrations. This means that the rate of the reaction is equal to the rate constant, k , of that reaction.
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47. Half-Life
Half-life is defined as the time required for one-half of a reactant to react.
Because [A] at t1/2 is one-half of the original [A],
[A]t = 0.5 [A]0.
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48. Half-Life For a first-order process, this becomes 0.5 [A]0 [A]0 ln
= −kt1/2
ln 0.5 = −kt1/2
−0.693 = −kt1/2
= t1/2
0.693 k
NOTE: For a first-order process, then, the half-life does not depend on [A]0.
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49. Half-Life For a second-order process, 1 0.5 [A]0 = kt1/2 + [A]0 2 [A]0
2 − 1
[A]0
= kt1/2 1 =
= t1/2 1
k[A]0
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50. Sample Exercise 14.9 (Pg. 592)
The reaction of C4H9Cl with water is a first-order reaction. a) Use the graph to estimate the half life of the reaction. b) Use the half life to calculate the rate constant.
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51. Summary
Half-Life for a first order reaction will decrease the concentration of the reactant by one half in each of a regularly spaced time interval, each interval equal to t1/2
Half life of a second order reaction depends on initial concentration of reactant. The lower the initial concentration, the longer the half life
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52. Practice Problems
14.39:
Define the following symbols that are encountered in rate equations for the generic reaction AB. : [A]0, t1/2[A]t, k
What quantity, when graphed versus time, will yield a straight line for a first order reaction?
How can you calculate the rate constant for a first order reaction from the graph in part b?
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53. Practice Problems
14.41
The gas-phase decomposition of SO2Cl2, SO2Cl2 SO2 + Cl2, is first order for SO2Cl2. At 600K the half life for this process is 2.3 x 105s. What is the rate constant at this temperature?
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54. Aim: How does the rate of a reaction change with temperature change?
Do Now: How does temperature increase the rate of a chemical reaction?
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55. Temperature and Rate
Generally, as temperature increases, so does the reaction rate.
This is because k is temperature dependent.
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56. The Collision Model
In a chemical reaction, bonds are broken and new bonds are formed.
Molecules can only react if they collide with each other.
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57. Turn and Talk
When H2 and I2 are at ordinary temperatures and pressures, each molecule undergoes about 1010 collisions per second. Why then does the reaction proceed very slowly at room temperature?
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58. The Collision Model
Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.
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59. Activation Energy
In other words, there is a minimum amount of energy required for reaction: the activation energy, Ea.
Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.
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60. Reaction Coordinate Diagrams
It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.
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61. Reaction Coordinate Diagrams
The diagram shows the energy of the reactants and products (and, therefore, E).
The high point on the diagram is the transition state.
The species present at the transition state is called the activated complex.
The energy gap between the reactants and the activated complex is the activation energy barrier.
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62. Give It Some Thought
Suppose you could measure the rates for both the forward and reverse reactions of this process. In which direction would the rate be larger?
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63. Maxwell–Boltzmann Distributions
Temperature is defined as a measure of the average kinetic energy of the molecules in a sample.
At any temperature there is a wide distribution of kinetic energies.
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64. Maxwell–Boltzmann Distributions
As the temperature increases, the curve flattens and broadens.
Thus at higher temperatures, a larger population of molecules has higher energy.
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65. Maxwell–Boltzmann Distributions
If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier.
As a result, the reaction rate increases.
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66. Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the expression
where R is the gas constant and T is the Kelvin temperature.
-Ea RT
f = e
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67. Arrhenius Equation
Svante Arrhenius developed a mathematical relationship between k and Ea:
k = A e
where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.
-Ea RT
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68. Arrhenius Equation
Taking the natural logarithm of both sides, the equation becomes
ln k = ( ) + ln A Ea R 1 T
y = m x b
Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. . 1 T
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70. Aim: What are reaction mechanisms?
Do Now: Turn and Talk Do most reactions happen in one step or multiple steps? If it’s a multi-step reaction, how is rate determined?
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71. Reaction Mechanisms
The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.
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72. Reaction Mechanisms
Reactions may occur all at once or through several discrete steps.
Each of these processes is known as an elementary reaction or elementary process.
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73. Reaction Mechanisms
The molecularity of a process tells how many molecules are involved in the process.
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75. Multistep Mechanisms
In a multistep process, one of the steps will be slower than all others.
The overall reaction cannot occur faster than this slowest, rate-determining step.
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76. Slow Initial Step NO2 (g) + CO (g)  NO (g) + CO2 (g)
The rate law for this reaction is found experimentally to be
Rate = k [NO2]2
CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.
This suggests the reaction occurs in two steps.
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77. Slow Initial Step A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
The NO3 intermediate is consumed in the second step.
As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.
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79. Fast Initial Step The rate law for this reaction is found to be
2 NO (g) + Br2 (g)  2 NOBr (g)
The rate law for this reaction is found to be
Rate = k [NO]2 [Br2]
Because termolecular processes are rare, this rate law suggests a two-step mechanism.
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80. Fast Initial Step A proposed mechanism is Step 1: NO + Br2
NOBr2 (fast)
Step 2: NOBr2 + NO  2 NOBr (slow)
Step 1 includes the forward and reverse reactions.
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81. Fast Initial Step
The rate of the overall reaction depends upon the rate of the slow step.
The rate law for that step would be
Rate = k2 [NOBr2] [NO]
But how can we find [NOBr2]?
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82. Fast Initial Step NOBr2 can react two ways:
With NO to form NOBr
By decomposition to reform NO and Br2
The reactants and products of the first step are in equilibrium with each other.
Therefore,
Ratef = Rater
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83. Fast Initial Step Because Ratef = Rater , k1 [NO] [Br2] = k−1 [NOBr2]
Solving for [NOBr2] gives us k1
k−1
[NO] [Br2] = [NOBr2]
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84. Fast Initial Step
Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives
k2k1
k−1
Rate =
[NO] [Br2] [NO]
= k [NO]2 [Br2]
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86. Catalysts
Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction.
Catalysts change the mechanism by which the process occurs.
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87. Catalysts
One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.
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88. Enzymes Enzymes are catalysts in biological systems.
The substrate fits into the active site of the enzyme much like a key fits into a lock.
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