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. 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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3. 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.
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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. Suppose that at one point in the reaction, the concentration of A is 0
Suppose that at one point in the reaction, the concentration of A is M and 125s later the concentration has fallen to M. During this time, what is the average rate of reaction?
Kinetics classwork problems 1
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12. 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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13. 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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14. 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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15. 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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16. 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]
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17. 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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18. Given the reaction, 2A + B  2C + D At some point in the reaction, [C] = M and 153s later, the concentration of C is M. What is the average rate of reaction during this time period? What is the average rate of formation of D during this time?
Kinetics classwork problems 2
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19. Given the reaction, 2A + B  3C + D If –Δ[A]/Δt is 2
Given the reaction, 2A + B  3C + D If –Δ[A]/Δt is 2.10 E-5 M/s, What is the rate of formation of C?  
Kinetics classwork problems 3
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20. Given A + 2B  C + 3D If the rate of disappearance of B is -6
Given A + 2B  C + 3D If the rate of disappearance of B is -6.2E-4 M/s, what is the rate of reaction, the rate of disappearance of A and the rate of formation of D?
Kinetics classwork problems 4
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21. If the initial concentration of hydrogen peroxide is 0
If the initial concentration of hydrogen peroxide is M and the initial rate of reaction is 9.32 E-4M/s. What will be the concentration of hydrogen peroxide remaining from its decomposition at t=35s. (reaction given for 1 mole of hydrogen peroxide decomposition)  
Kinetics classwork problems 5
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22. Given the same reaction, the initial concentration of hydrogen peroxide is M and 12s later the concentration is M. What is the initial rate of this reaction in both M/s and M/min.  
Kinetics classwork problems 6
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23. 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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24. 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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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  [NH+] [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. aA + bB  cC + dD
The rate law: Rate = k [A]m[B]n k = rate constant m = experimentally determined #, the value shows the order of A n = experimentally determined #, the value shows the order of B m + n = overall order of the reaction
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30. Rate = k[A]o[B]1 m = 0; rxn is 0th order in A
n = 1; rxn is 1st order in B
m + n = 0 + 1; overall rxn is 1st order
Since anything [ ]0 = 1  simplify
Rate = k[B]1
= k[B]
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31. H2SeO3 + 6I- + 4H+  Se + 2I3- + 3H2O Rate = k[H2SeO3] [I3-]3 [H+]2
Does the coefficient always correspond to the order of the reaction?
Kinetics classwork problems 7
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32. Units of k Rxn order Units of k 0th order 1st order 2nd order
3rd order
nth order
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33. A + B  C Given rate = k[A]0[B] What does this mean?
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34. A + B  C Given rate = k[A]0[B]2 What does this mean? Double A
Double B
Half B
Triple B
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35. Kinetics classwork problems 8
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36. A  B
We find that when [A] has fallen to half its initial value, the reaction proceeds at the same rate. What order is the reaction?
Kinetics classwork problems 9
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37. H2O2  H2O + ½ O2
For the decomposition of 1 mole of hydrogen peroxide, use k = 3.66 E-3 1/s to determine the instantaneous rate of the 1st order decomposition of 2.05M H2O2?
Kinetics classwork problems 10
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38. Determine the reaction order through experimental data
2NO + Cl2  2NOCl
By comparison write the rate laws (differential rate law)
Experiment
[NO]
[Cl2]
Initial Rate (M/s) 1
0.0125
0.0255
2.27E-5 2
0.0510
4.55E-5 3
0.0250
9.08E-5
Kinetics classwork problems 11

39. Determine the reaction order through experimental data
2NO + Cl2  2NOCl
By comparison write the rate laws (differential rate law)
Experiment
[NO]
[Cl2]
Initial Rate (M/s) 1
0.0125
0.0255
2.27E-5 2
0.0510
4.55E-5 3
0.0250
9.08E-5
Suppose there is a hypothetical experiment 4, which has [NO]=0.0500M and [Cl2]=0.0255M what is the rate in this experiment?
Kinetics classwork problems 12

40. Experiment [S2O8-2] [l-] Initial Rate (M/s) 1 2 3
Determine the reaction order through experimental data
S2O I-  2SO I3-
By comparison write the rate laws.
What is the order for each reactant and overall?
Solve for the rate constant.
What would be the initial rate of reaction if [S2O8-2] = 0.083M and [I-] =0.115M
Experiment
[S2O8-2]
[l-]
Initial Rate (M/s) 1
0.038
0.060
1.4E-5 2
0.076
2.8E-5 3
0.120
5.6E-5
Kinetics classwork problems 13
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41. Concentration and Reaction Rate
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42. The rate expression will always have the form Initial rate = k [A]m[B]n[C]p k = rate constant [A] = conc. of reactant A [B] = conc. of reactant B [C] = conc. of catalyst (rare on AP/IB) m = order of rxn for reactant A n = order of rxn for reactant B p = order of rxn for catalyst
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43. Initial rate = k [A]m[B]n[C]p Must be determined by experimentation
Initial rate = k [A]m[B]n[C]p
Exponents can be zero, whole numbers, or fractions (rare)
Must be determined by experimentation
The rate constant, k
Temperature dependent
Determined experimentally
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44. 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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45. Zeroth order half-life
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46. Zeroth order half-life
[A]t = -kt +[A]o (not in calculus – trust me)
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47. Zeroth order half-life
[A]t = -kt +[A]o
This looks like the equation for what???
- line
Graph [A] vs. t will produce a straight line, with slope = -k and y-intercept of [A]0
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48. First-Order Processes
So how do we straighten the line???
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49. First order half-life
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50. 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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51. First-Order Processes
ln [A]t = -kt + ln [A]0
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52. In the first order decomposition of hydrogen peroxide, k = 3
In the first order decomposition of hydrogen peroxide, k = 3.66E-3 1/s and [H2O2]0 = 0.882M
Determine the time at which [H2O2] = 0.600M
Determine the [H2O2] after 225s
Kinetics classwork problems 14
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53. The decomposition of NH2NO2 has a rate law of rate = k[NH2NO2]. k = 5
The decomposition of NH2NO2 has a rate law of rate = k[NH2NO2]. k = 5.62E-3 1/min at 15oC and [NH2NO2]o = 0.105
At what time is [NH2NO2] = M
What is [NH2NO2] after 6.00hrs
What is the rate of reaction after 35 mins, if the [NH2NO2]0=0.0750M
Kinetics classwork problems 15
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54. 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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55. 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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56. What is the rate constant of the first order decomposition of N2O5 if the t1/2 = 120s
Kinetics classwork problems 16
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57. Determine the time for 90% of a N2O5 sample to undergo decomposition.
How long will it take for an initial [N2O5] to reach 1/16 of its value?
Find the mass of dinitrogen pentoxide remaining after a 4.80g sample has decomposed for 10mins.
Determine the time for 90% of a N2O5 sample to undergo decomposition.
Kinetics classwork problems 17-19
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58. The half-life of the first order decomposition of SO2Cl2 at 320oC is 8
The half-life of the first order decomposition of SO2Cl2 at 320oC is 8.75 hr. What is the value of the rate constant?
What is the pressure of sulfuryl chloride 3.00 hr after the start of the reaction, if the initial pressure of sulfuryl chloride is 722mmHg?
Kinetics classwork problems 20, 21
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59. Second-Order Processes
Similarly, integrating the rate law for a process that is second-order in reactant A, we get (the math is a little more complicated) 1
[A]t
= kt +
[A]0
also in the form
y = mx + b
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60. 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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61. 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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62. 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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63. Second-Order Processes1
[NO2]
Graphing ln 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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64. 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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65. Find the rate constant, and the half-life of the decomposition of 1
Find the rate constant, and the half-life of the decomposition of 1.00M NO2
Kinetics classwork problems 22
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66. (2nd order) A  B, takes 55s for the [A] to fall to 0.40 from [A]o = 0.80M What is the k for this reaction? How long would it take for [A] to fall to 0.20M? to 0.10M?
Kinetics classwork problems 23, 24
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67. In a process where Cl atoms combine to form Cl2­, the reaction follows a second order rate law. The first half-life is 12ms (initial concentration is .500 M). How long will it take for the [Cl] to drop to 1/8 of its initial value?
Kinetics classwork problems 25
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68. Integrated Rate Laws
Use graphical methods to determine the order of a given reactant. The value of the rate constant k is equal to the absolute value of the slope of the best fit line which was decided by performing 3 linear regressions and analyzing the regression correlation coefficient r.
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69. Integrated Rate Law: [ ] / time
Create 3 graphs
Time always goes on the x-axis
Graph 1: [A] on y-axis
Graph 2: ln[A] on y-axis
Graph 3: 1/[A] on y-axis [A]-1
Alphabetic order:
Concentration = 0 order
Natural log = 1st order
Reciprocal = 2nd order
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70. Integrated Rate Law: [ ] / time
Alphabetic order:
Concentration = 0 order
Natural log = 1st order
Reciprocal = 2nd order
Choose the graph that forms a straight line.
(r value closest to ±1)
The equation formed by line allows you to predict either [ ] at a time or a time for a specific [ ].
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71. Integrated Rate Law: [ ] / time
y = mx + b Zero order: [A] = -kt + [A]o First order: ln[A] = -kt + ln[Ao] Second order 1/[A] = kt + 1/[Ao] *** |slope| = k *** Rate expression =Rate law Rate k[A]order determined from graph
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72. Kinetics classwork problems 26
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73. Determine the order of this reaction using the given experimental data.
Determine k.
[A]
Time(min)
0.80
0.60 8
0.35 24
0.20 40
1st order k=0.0345
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74. Given the following information, determine the order of the reaction and the value of k, the reaction constant.
Concentration (M)
Time (s)
1.0  10
0.50 20
0.33 30
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75. Temperature and Rate
Generally, as temperature increases, so does the reaction rate.
This is because k is temperature dependent.
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76. 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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77. 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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78. 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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79.
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80. 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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81. 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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82. 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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83. 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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84. 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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85. 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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86. 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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87. 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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88. 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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89. 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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90. Reaction Mechanisms
The molecularity of a process tells how many molecules are involved in the process.
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91. 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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92. 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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93. 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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94. 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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95. 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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96. 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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97. 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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98. 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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99. 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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100. 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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101. 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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102. 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.
 2009, Prentice-Hall, Inc.