1. Chapter Twelve:
CHEMICAL KINETICS

2. Chemical Kinetics
Once thermodynamics determines that a reaction will occur…
Chemical kinetics is the study of the rates of spontaneous reactions.
12.1
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3. Reaction Rate
Change in concentration (conc.) of a given reactant or product per unit time.
Or… if no reactant or product is specified, it is the rate of the species with a coefficient of “1”.
AP almost ALWAYS specifies which reactant a given rate refers to.
12.1
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4. Reaction Rate For 2A + B  C + 3D the rate with respect to A is:
Δt (reaction rates are shown as positive values)
Rate =
conc of A
at time 2 1 t -
12.1
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5. [ ] [ ] Reaction Rate For 2A + B  C + 3D
But if = 0.24 mol L-1 min-1, then -
= 0.12 mol L-1 min-1 [ ] D A t [ ] D B t
12.1
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6. Reaction Rate
Reaction Rates can also be defined as the rate per reaction as written.
In this case, there is only one value for the reaction rate…
Which is equal to each of the individual reaction rates divided by their coefficients
12.1
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7. Reaction Rate Per Rxn. As Written
For 2A + B  C + 3 D
Rxn rate = - 1 Δ[A] = - Δ[B] = Δ[C] = 1 Δ[D]
2 Δt Δt Δt Δt
12.1
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8. Reaction Rate Rxn rates decrease over time
(So you can’t use Δ[ ]/Δtime to predict concentrations over a given time period  )
12.1
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9. Instantaneous Reaction Rate
Note: by striking a tangent at any point along the curve, one can determine the rate of the reaction at that point.
12.1
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10. Instantaneous Reaction Rate
This is accomplished by finding the slope of that tangent line, since slope is change in y divided by change in x… in other words:
Rate= slope = D[N2O5]
Dtime
12.1
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11. Instantaneous Reaction Rate
This must be done “old school”… meaning you must choose two “convenient” points anywhere on the tangent line and find the slope by Dy Dx
12.1
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12. Reaction Rate A graph representing the following reaction:
2NO2  2NO + O2
Notice how the stoichiometry is evident
12.1
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13. Homework
A “reactant concentration vs. time” graph was plotted for a certain reaction. Determine the “instantaneous” rate of the reaction at 10 seconds elapsed time (do your best to estimate values on the axes)
12.2
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14. “Differential” Rate Laws
For the decomposition of nitrogen dioxide:
2NO2(g) → 2NO(g) + O2(g)
Rate = k[NO2]n
k = rate constant
Depends on temperature and “collision factors”
n = order of the reactant
Shows how “sensitive” the reaction’s rate is to changes in concentration
12.2
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15. “Differential” Rate Laws
Rate = k[NO2]n
The concentrations of the products do not appear in the rate law
12.2
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16. “Differential” Rate Laws
Rate = k[NO2]n
The value of the exponent n must be determined by experiment; it cannot be determined from the balanced equation (unless it’s a single-step reaction)
12.2
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17. Determining Differential Rate Law Exponents - Method of Initial Rates
Rate Law exponents must be determined from actual experimental data.
12.2
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18. Determining Differential Rate Law Exponents - Method of Initial Rates
Rate Law exponents must be determined from actual experimental data.
12.2
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19. Method of Initial Rates
The basic idea is to look at how the initial rate of a reaction changes due to a change in a reactant concentration.
12.2
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20. Method of Initial Rates
The basic idea is to look at how the initial rate of a reaction changes due to a change in a reactant concentration.
If the concentration is tripled(3x), and the rate increases by 9x, then the exponent for that reactant is 2 (since 32=9)
12.2
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21. Method of Initial Rates
Example: Using the following data table, determine the exponents for the reactants NH4+ and NO2-
NO2-
12.2
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22. Method of Initial Rates
Note that between trials 1 and 2, NH4+ remains constant and NO2- doubles and the rate doubles. The exponent for NO2- must therefore be 1.
NO2-
12.2
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23. Method of Initial Rates
Note that between trials 2 and 3, NO2- remains constant and NH4+ doubles and the rate doubles. The exponent for NH4+ also must be 1.
NO2-
12.2
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24. Method of Initial Rates
So far, we have the differential rate law as:
rate = k[NH4+][NO2-]
All that’s left is to determine the rate constant, “k”
NO2-
12.2
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25. Method of Initial Rates
This is done by picking a trial (several, if this is actual lab data and not “book” data) and plugging in the concentration and rate values, along with the determined exponents.
NO2-
12.2
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26. Method of Initial Rates
Using trial 1:
1.35x10-7mol/L∙s = k(0.100mol/L)1(0.0050mol/L)1
2.70x10-4 L/mol∙s = k (Take note of the units!)
NO2-
12.2
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27. The Problem with Differential Rate Laws
“differential” rate laws like
rate = k[A]2
…cannot relate time elapsed to concentration remaining, since the rate constantly changes during the reaction.
12.4
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28. Integrated Rate Laws
The only way to predict how much of a reactant will remain over a given time period is to do hundreds (or thousands) of individual calculations, each of which takes into account the changed rate… the shorter the time period for each calculation, the more precise the result will be.
NOT FUN…
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29. Integrated Rate Laws
The only way to predict how much of a reactant will remain over a given time period is to do hundreds (or thousands) of individual calculations, each of which takes into account the changed rate… the shorter the time period for each calculation, the more precise the result will be.
NOT FUN… or easy!!!!
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30. Integrated Rate Laws
But that is EXACTLY what the calculus process of “integration” does… it takes into account the function of the curve… which takes into account the constantly changing reaction rate!
12.4
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31. First-Order Reactant Integrated Rate Law
Differential rate law:
Rate = k[A]
12.4
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32. First-Order Reactant Integrated Rate Law
Differential rate law:
Rate = k[A]
Integrated rate law:
ln[A]remaining - ln[A]initial = -k ∙telapsed
12.4
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33. First-Order Reactant Integrated Rate Law
Differential rate law:
Rate = k[A]
Integrated rate law:
ln[A]remaining - ln[A]initial = -k ∙telapsed or
ln [A]rem = -k ∙telapsed
[A]i
12.4
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34. First-Order Reactant Integrated Rate Law
Differential rate law:
Rate = k[A]
Integrated rate law:
ln[A]remaining - ln[A]initial = -k ∙telapsed or
ln [A]rem = -k ∙telapsed
[A]I
1st order is MOST common reactant order!!
12.4
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35. Exercise
A first order reaction is 35% complete at the end of 55 minutes. What is the value of k?
k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1.
12.4
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36. Exercise ln [A]rem = - k ∙ t elapsed [A]initial
A first order reaction is 35% complete at the end of 55 minutes. What is the value of k?
ln [A]rem = - k ∙ t elapsed
[A]initial
k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1.
12.4
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37. Second-Order Integrated Rate Law
Differential Rate Law:
rate = k[A]2
12.4
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38. Second-Order Integrated Rate Law
Differential Rate Law:
rate = k[A]2
Integrated Rate Law:
= kt
[A]rem [A]i
12.4
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39. Zero-Order Rate = k[A]0 = k Integrated: [A]rem - [A]i = -ktelapsed
12.4
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40. Another Method (Besides “Initial Rates”) to Determine Reactant Order: Graphical Technique
Given experimental concentrations at given times, we can use graphing as a diagnostic tool to determine reactant order.
12.4
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41. Graphing Technique to Determine IF a reactant is 1st order
1st order Integrated rate law:
ln[A]remaining - ln[A]initial = -k ∙telapsed
12.4
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42. Graphing Technique to Determine IF a reactant is 1st order
1st order Integrated rate law:
ln[A]remaining - ln[A]initial = -k ∙telapsed
Rearranged:
ln[A]remaining = -k ∙telapsed + ln[A]initial
y = mx b
12.4
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43. Graphing Technique to Determine IF a reactant is 1st order
ln[A]remaining = -k ∙telapsed + ln[A]initial
y = mx b
So IF a graph of the ln[A] vs. time is linear, THEN the reactant is first order.
12.4
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44. Graphing Technique to Determine IF a reactant is 1st order
ln[A]remaining = -k ∙telapsed + ln[A]initial
y = mx b
And if it is linear… what does the slope give us??
12.4
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45. Graphing Technique to Determine IF a reactant is 1st order
ln[A]remaining = -k ∙telapsed + ln[A]initial
y = mx b
And if it is linear… what does the slope give us??
The rate constant, k k = |m|
12.4
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46. Graphing Technique to Determine IF a reactant is 2nd order
2nd order Integrated rate law:
= kt
[A]rem [A]i
12.4
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47. Graphing Technique to Determine IF a reactant is 2nd order
2nd order Integrated rate law:
= kt
[A]rem [A]i
Rearranged:
= kt
[A]rem [A]i
y = mx b
12.4
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48. Graphing Technique to Determine IF a reactant is 2nd order
= kt
[A]rem [A]i
y = mx b
So IF a graph of 1/[A] vs. time is linear, THEN the reactant is 2nd order,
with k = |m|
If it’s not linear, then it will be zero or 1st order
12.4
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49. Zero-Order Rate = k[A]0 = k 12.4
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50. Zero-Order Rate = k[A]0 = k Integrated: [A]rem - [A]i = -ktelapsed
Graphical “diagnostic”:
If zero order, a graph of [A] vs. time will be linear, and k = |m|
12.4
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51. Half Life Amt. of time for ½ of the reactants to react away.
So [A]remaining = ½ [A]initial
Most important one is 1st order
All radioactive decay is 1st order!!
12.6
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52. 1st Order Half Life Equation
ln ½ [A]initial = -k∙t½
[A]initial
12.6
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53. 1st Order Half Life Equation
ln ½ [A]initial = -k∙t½
[A]initial
So ln 0.5 = -k∙t½
= -k∙t½
0.693 = k∙t½
12.6
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54. 1st Order Half Life Equation
ln ½ [A]initial = -k∙t½
[A]initial
So ln 0.5 = -k∙t½
= -k∙t½
0.693 = k∙t½ it’s worth your while
to memorize this eqn.
12.6
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55. 1st Order Half Life Equation
ln ½ [A]initial = -k∙t½
[A]initial
So ln 0.5 = -k∙t½
= -k∙t½
0.693 = k∙t½
NOTE that for 1st order, the half-life is independent of concentration!
12.6
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56. Half Life
The Half-life equations of the other orders can be derived in the same fashion, but they are rarely used… their half-lives are concentration dependent.
12.6
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57. Half Life
You will often be presented with the half-life of a 1st order reactant as a means to determine the rate constant, k, to use in a related problem.
12.6
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58. The “Problem” with Integrated Rate Laws
They only work for “one-reactant” systems.
For two-reactant systems, either one reactant must already be zero order…
…Or we must “trick” the reaction into behaving like one of the reactants is zero order.
12.6
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59. “Flooding”
This is done by “flooding” the reaction with such a high concentration of that reactant that its concentration changes negligibly during the course of the reaction.
This will result in a negligible change to the rate due to the flooded reactant… it will appear zero order!
12.6
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60. “Flooding”
Only one reactant affects the rate, now, so the integrated rate laws will work for that reactant.
The actual order of the “flooded” reactant will be found afterward by method similar to method of initial rates once the rate constant is known.
12.6
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61. Flooding Mathematics For A + F  Products
Where “F” will be the flooded reactant and A is the reactant affecting the rate
12.6
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62. Flooding Mathematics For A + F  Products Rate = kreal [A]x [F]y
Where “F” will be the flooded reactant and A is the reactant affecting the rate
Rate = kreal [A]x [F]y
12.6
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63. Flooding Mathematics For A + F  Products Rate = kreal [A]x [F]y
Where “F” will be the flooded reactant and A is the reactant affecting the rate
Rate = kreal [A]x [F]y
= k′[A]x
Where k′ = kreal[F]y
12.6
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64. Flooding Mathematics For A + F  Products
Where “F” will be the flooded reactant and A is the reactant affecting the rate
Rate = kreal [A]x [F]y = k′[A]x
Where k′ = kreal[F]y
“x” is determined by graphical “diagnostic”, and k′ = |m| linear graph
12.6
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65. Flooding Mathematics
To determine the order of reactant “F”, do the experiment twice, each with different initial concentrations of “F”.
12.6
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66. Flooding Mathematics
To determine the order of reactant “F”, do the experiment twice, each with different initial concentrations of “F”.
You will get different k′ values corresponding to the different [F]initial values
12.6
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67. Flooding Mathematics
To determine the order of reactant “F”, do the experiment twice, each with different initial concentrations of “F”.
You will get different k′ values corresponding to the different [F]initial values
k′rxn 1 = kreal [F]yrxn 1 and
k′rxn 2 = kreal [F]yrxn 2
Where the [F] values are known
12.6
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68. Flooding Mathematics Substituting, rearranging k′rxn 1 = [F]yrxn 1
12.6
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69. Flooding Mathematics Substituting, rearranging k′rxn 1 = [F]yrxn 1
Taking the log10 of both sides:
log k′rxn 1 = y∙ log [F]rxn 1
k′rxn [F]rxn 2
12.6
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70. Flooding Mathematics Solving for “y” gives the order of reactant “F”
(we usually round to the nearest order… sometimes nearest ½ order)
12.6
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71. Flooding Mathematics Solving for “y” gives the order of reactant “F”
Substitute back into
k′rxn 1 = kreal [F]yrxn 1 and
k′rxn 2 = kreal [F]yrxn 2
to determine kreal for each trial, and average them.
12.6
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72. Reaction Mechanism Most chemical reactions occur by a series of steps.
12.6
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73. Reaction Mechanism Most chemical reactions occur by a series of steps.
A “Reaction Mechanism” is a listing of these steps.
12.6
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74. Reaction Mechanism Most chemical reactions occur by a series of steps.
A Reaction Mechanism is a listing of these steps.
Each step is called an “elementary reaction” and occurs literally as the reaction equation states.
12.6
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75. Reaction Mechanism Each step is a literal reaction:
the reactants of a given step collide
12.6
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76. Reaction Mechanism Each step is a literal reaction:
the reactants of a given step collide
an old bond breaks while a new bond simultaneously is formed.
12.6
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77. Reaction Mechanism Each step is a literal reaction:
the reactants of a given step collide
an old bond breaks while a new bond simultaneously is formed.
A-B  C  A--B--C  A + B-C
12.6
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78. Reaction Mechanism Each step is a literal reaction:
the reactants of a given step collide
an old bond breaks while a new bond simultaneously is formed.
A-B  C  A--B--C  A + B-C
This “in-between” substance is called the activated complex
It is also known as the “transition state” of the reaction
12.6
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79. Reaction Mechanism
An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product.
12.6
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80. Reaction Mechanism
An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product.
Catalysts are substances that are reacted away, but subsequently are reformed.
12.6
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81. A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO NO2 + CO  NO +CO2
12.6
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82. A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO NO2 + CO  NO +CO2
12.6
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83. A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO NO2 + CO  NO +CO2
12.6
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84. Elementary Steps (Molecularity)
Unimolecular – reaction involving one molecule ; first order
12.6
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85. Elementary Steps (Molecularity)
Unimolecular – reaction involving one molecule ; first order (no collision… just a decomposition… often initiated by some form of energy (hn))
12.6
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86. Elementary Steps (Molecularity)
Unimolecular – reaction involving one molecule ; first order (no collision… just a decomposition… often initiated by some form of energy (hn))
Bimolecular – reaction involving the collision of two molecules… could be the same type of molecule; second order
12.6
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87. Elementary Steps (Molecularity)
Unimolecular – reaction involving one molecule ; first order (no collision… just a decomposition… often initiated by some form of energy (hn))
Bimolecular – reaction involving the collision of two molecules… could be the same type of molecule; second order
Termolecular – reaction involving the simultaneous collision of three species; third order (rare)
12.6
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88. Rate-Determining Step
A reaction is only as fast as its slowest step.
12.6
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89. Rate-Determining Step
A reaction is only as fast as its slowest step.
So the “slow” step in a reaction mechanism is called the “rate-determining” step
12.6
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90. Rate-Determining Step
A reaction is only as fast as its slowest step.
So the “slow” step in a reaction mechanism is called the “rate-determining” step
The rate-determining step, therefore, is the step that determines the differential rate law!
12.6
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91. Rate-Determining Step
A reaction is only as fast as its slowest step.
So the “slow” step in a reaction mechanism is called the “rate-determining” step
The rate-determining step, therefore, is the step that determines the differential rate law!
This is yet ANOTHER way to determine a rate law.
12.6
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92. Using the Slow Step to Determine the Differential Rate Law
For a given OVERALL reaction
mA + nB  Products
The general form of the Differential Rate law would be
Rate = k[A]x[B]y
12.6
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93. Using the Slow Step to Determine the Differential Rate Law
For a given OVERALL reaction
mA + nB  Products
The general form of the Differential Rate law would be
Rate = k[A]x[B]y
Recall that the differential rate law exponents (x and y) are NOT determined from the balanced equation coefficients (m and n)
12.6
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94. Using the Slow Step to Determine the Differential Rate Law
That is because the rate relationship between reactants and rate is not determined by the overall equation, but by the slow step of the reaction.
12.6
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95. Using the Slow Step to Determine the Differential Rate Law
That is because the rate relationship between reactants and rate is not determined by the overall equation, but by the slow step of the reaction.
So… the coefficients of the slow step reactants ARE the exponents in the rate law!
12.6
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96. Using the Slow Step to Determine the Differential Rate Law
If you know the mechanism… you know the rate law!
12.6
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97. Using the Slow Step to Determine the Differential Rate Law
Example:
for the Reaction 2A + B  C + D
Proposed Mechanism:
12.6
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98. Using the Slow Step to Determine the Differential Rate Law
Example:
for the Reaction 2A + B  C + D
Proposed Mechanism
A + B  C + E slow
A + E  D fast
12.6
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99. Using the Slow Step to Determine the Differential Rate Law
Example:
for the Reaction 2A + B  C + D
Proposed Mechanism
A + B  C + E slow
A + E  D fast
The rate law would be rate = k[A][B]
Since 1 A and 1 B appear as reactants in the slow step.
12.6
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100. Using the Slow Step to Determine the Differential Rate Law
Different Example:
for the Reaction 2A + B  C + D
Proposed Mechanism
A + A  C + E slow
B + E  D fast
12.6
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101. Using the Slow Step to Determine the Differential Rate Law
Different Example:
for the Reaction 2A + B  C + D
Proposed Mechanism
A + A  C + E slow
B + E  D fast
The rate law would be rate = k[A]2 since 2 “A”s appear as reactants in the slow step.
12.6
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102. Using the Slow Step to Determine the Differential Rate Law
Different Example:
for the Reaction 2A + B  C + D
Proposed Mechanism
A + A  C + E slow
B + E  D fast
The rate law would be rate = k[A]2 since 2 “A”s appear as reactants in the slow step. Note that “B” is a zero order reactant because it doesn’t “appear” in the slow step!
12.6
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103. Using the Slow Step to Determine the Differential Rate Law
So…. If you were asked to predict which of the previous 2 proposed mechanisms was most likely, you would have to compare the PREDICTED rate laws (from the mechanisms) to the REAL rate law (determined by experiment)
12.6
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104. Reaction Mechanism Requirements
The sum of the elementary steps must give the overall balanced equation for the reaction.
12.6
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105. Reaction Mechanism Requirements
The sum of the elementary steps must give the overall balanced equation for the reaction.
The mechanism must agree with the experimentally determined rate law.
12.6
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106. Reaction Mechanism Requirements
The sum of the elementary steps must give the overall balanced equation for the reaction.
The mechanism must agree with the experimentally determined rate law.
It must not include more than 3 species colliding… and it’s even “suspect” if 3 species collide.
12.6
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107. Back to Using Reaction Mechanisms to determine Differential Rate laws
Many mechanisms contain at least one step that is reversible.
12.6
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108. Back to Using Reaction Mechanisms to determine Differential Rate laws
Many mechanisms contain at least one step that is reversible.
This, of course, will affect the subsequent steps.
12.6
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109. Back to Using Reaction Mechanisms to determine Differential Rate laws
Many mechanisms contain at least one step that is reversible.
This, of course, will affect the subsequent steps.
Many of these mechanisms end up with “Intermediates” in the slow step of the mechanism
12.6
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110. Back to Using Reaction Mechanisms to determine Differential Rate laws
Intermediates are NOT allowed to be included in rate laws, though, and must be “substituted out” in terms of actual reactants.
12.6
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111. Back to Using Reaction Mechanisms to determine Differential Rate laws
Intermediates are NOT allowed to be included in rate laws, though, and must be “substituted out” in terms of actual reactants.
This process of substitution can get rather “messy”, but we will only work with relatively simple cases in AP Chem
12.6
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112. Back to Using Reaction Mechanisms to determine Differential Rate laws
A VAST majority of the time, when an intermediate appears in the slow step, this is the “look” that you will see in the mechanism.
12.6
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113. Back to Using Reaction Mechanisms to determine Differential Rate laws
A VAST majority of the time, when an intermediate appears in the slow step, this is the “look” that you will see in the mechanism:
A + B ↔ I fast equilibrium
I + A  C + D slow
12.6
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114. Back to Using Reaction Mechanisms to determine Differential Rate laws
A VAST majority of the time, when an intermediate appears in the slow step, this is the “look” that you will see in the mechanism:
A + B ↔ I fast equilibrium
I + A  C + D slow
NOTE: the intermediate, I (that must be substituted out), is the ONLY product of a fast equilibrium.
12.6
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115. Back to Using Reaction Mechanisms to determine Differential Rate laws
A + B ↔ I fast equilibrium
I + A  C + D slow
The “unofficial” rate law would therefore be
Rate = k[I][A]
12.6
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116. Back to Using Reaction Mechanisms to determine Differential Rate laws
A + B ↔ I fast equilibrium
I + A  C + D slow
The “unofficial” rate law would therefore be
Rate = k[I][A]
In THIS case : [I] “=“ [A][B]
12.6
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117. Back to Using Reaction Mechanisms to determine Differential Rate laws
A + B ↔ I fast equilibrium
I + A  C + D slow
The “unofficial” rate law would therefore be
Rate = k[I][A]
so the “official” rate law is
rate = k[A][B][A] = k[A]2[B]
12.6
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118. Back to Using Reaction Mechanisms to determine Differential Rate laws
A + B ↔ I fast equilibrium
I + A  C + D slow rate = k[A]2[B]
A couple of things to note:
12.6
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119. Back to Using Reaction Mechanisms to determine Differential Rate laws
A + B ↔ I fast equilibrium
I + A  C + D slow rate = k[A]2[B]
A couple of things to note:
1. “B” is NOT in the slow step, but it is not zero order because it is in an equilibrium reaction prior to the slow step.
12.6
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120. Back to Using Reaction Mechanisms to determine Differential Rate laws
A + B ↔ I fast equilibrium
I + A  C + D slow rate = k[A]2[B]
A couple of things to note:
1. “B” is NOT in the slow step, but it is not zero order because it is in an equilibrium reaction prior to the slow step.
2. Even though the rate law has 2 As and 1 B in it, it does NOT mean that 2As and 1 B are colliding simultaneously… that can be clearly seen from the mechanism. SO… you can’t always tell what is actually colliding by just looking at the rate law.
12.6
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121. Concept Check
The reaction A + 2B  C has the following proposed mechanism:
A + B D (fast equil.)
D + B  C (slow)
Write the rate law for this mechanism.
rate = k[A][B]2
12.6
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122. Collision Model Molecules must collide to react.
Rare for more than 2 molecules to collide in a given elementary step
12.7
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123. Collision Model Molecules must collide to react.
Rare for more than 2 molecules to collide in a given elementary step
For 3 (or more???) molecules to be reactants in a given step, they have to collide simultaneously (or awfully darn close to it.)
12.7
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124. Collision Model Molecules must collide to react.
And even if the correct collision DOES occur…
12.7
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125. Not every collision results in a reaction…
Collision Model
Molecules must collide to react.
And even if the correct collision DOES occur…
Not every collision results in a reaction…
12.7
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126. Requirements for an “Effective” Collision
#1: The correct molecules must collide
12.7
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127. Requirements for an “Effective” Collision
#1: The correct molecules must collide
#2: Activation Energy Consideration
12.7
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128. Requirements for an “Effective” Collision
Molecules must collide with enough energy
12.7
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129. Requirements for an “Effective” Collision
Molecules must collide with enough energy
Sufficient collision energy is necessary to disrupt the bonds of the reactants to form the activated complex ( A---B---C)
12.7
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130. Requirements for an “Effective” Collision
Molecules must collide with enough energy
Sufficient collision energy is necessary to disrupt the bonds of the reactants to form the activated complex ( A---B---C)
There is a “threshold energy” of collision that must be met in order for that collision to have a chance at reacting… called the activation energy.
12.7
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131. Change in Potential Energy
12.7
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132. Requirements for an “Effective” Collision
Molecules must collide with enough energy
Temperature of the reaction is the most important variable regarding collision energy… more on this to come…
12.7
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133. Requirements for an “Effective” Collision
#3: Collision Orientation Considerations
12.7
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134. Requirements for an “Effective” Collision
The molecules must be oriented correctly in space
If the transition state must be A--B--C, then the reaction will not proceed if “C” collides with “A” instead of “B”
C  A-B  No Rxn.
12.7
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135. Requirements for an “Effective” Collision
12.7
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136. Collision Model - “Effective” Collisions
So…
not only do the correct molecules need to collide….
12.7
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137. Collision Model - “Effective” Collisions
So…
not only do the correct molecules need to collide….
They need to collide with sufficient energy…
12.7
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138. Collision Model - “Effective” Collisions
So…
not only do the correct molecules need to collide….
They need to collide with sufficient energy…
And they need to be oriented “just right”
12.7
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139. Collision Model - “Effective” Collisions
So…
not only do the correct molecules need to collide….
They need to collide with sufficient energy…
AND they need to be oriented “just right”…
Or there will be no reaction… the collision would not have been “effective”
12.7
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140. Back to The Effect of Temperature on Effective Collisions
12.7
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141. The Arrhenius Equation
This formula relates the effect that temperature has on rate constants…
…and therefore reaction rates.
12.7
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142. Maxwell Boltzman Distribution
12.7
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143. The Arrhenius Equation
The fraction of collisions exceeding the activation energy is given by
e-Ea/RT
12.7
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144. The Arrhenius Equation
Fraction of collisions of correct orientation is symbolized by the Greek letter rho, r… this value is inherent to a given reaction.
12.7
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145. The Arrhenius Equation
The total # of collisions is symbolized by the letter “z”.
12.7
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146. The Arrhenius Equation
The total # of collisions is symbolized by the letter “z”.
A combination of (z∙r) is known as the “frequency factor”, symbolized by “A”
12.7
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147. The Arrhenius Equation
So…
k = z∙r∙ e-EA/RT
12.7
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148. The Arrhenius Equation
So…
k = z∙r∙ e-EA/RT
k = A e-EA/RT
12.7
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149. The Arrhenius Equation
So…
k = z∙r∙ e-EA/RT
k = A e-EA/RT
ln k = -EA ∙ ln A
R T
12.7
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150. The Arrhenius Equation
ln k = -EA ∙ ln A
R T
Looks like a graphing situation to me!!!
12.7
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151. The Arrhenius Equation – Two-Point Version
ln k2 = - EA ( )
k R ( T T1)
12.7
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152. Catalyst
A substance that speeds up a reaction without being (net) consumed itself.
12.8
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153. Catalyst
A substance that speeds up a reaction without being (net) consumed itself.
Homogeneous catalysts do participate in an early step in the mechanism, but are completely regenerated in one of the latter steps.
12.8
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154. Catalyst
A substance that speeds up a reaction without being (net) consumed itself.
Homogeneous catalysts do participate in an early step in the mechanism, but are completely regenerated in one of the latter steps.
Heterogeneous catalysts (usually metals) do not participate in any elementary steps, but provide a “reactive surface” for the reaction to occur on.
12.8
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155. Homogeneous Catalyst Example
Atmospheric Ozone depletion:
12.8
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156. Homogeneous Catalyst Example
Atmospheric Ozone depletion: hn
O O2 + O∙
12.8
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157. Homogeneous Catalyst Example
Atmospheric Ozone depletion: hn
O O2 + O∙
O3 + Cl∙  O2 + ∙OCl
12.8
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158. Homogeneous Catalyst Example
Atmospheric Ozone depletion: hn
O O2 + O∙
O3 + Cl∙  O2 + ∙OCl
O∙ + ∙OCl  O2 + Cl∙
12.8
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159. Heterogeneous Catalyst Example
Hydrogenation of Ethene :
Ni cat
CH2=CH2 + H2  CH3-CH3
12.8
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160. Heterogeneous Catalyst Example
Hydrogenation of Ethene :
Ni cat
CH2=CH2 + H2  CH3-CH3
12.8
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161. Heterogeneous Catalyst Example
Hydrogenation of Ethene :
Ni cat
CH2=CH2 + H2  CH3-CH3
Carbon’s electrons are attracted to the surface of the metal, thus weakening the double bond… making it “vulnerable” to attack by the hydrogen atoms (which were also attracted to the metal surface)
12.8
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162. Catalyst
The main effect that a catalyst has is to weaken (or even break) a bond in one of the reactants.
12.8
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163. Catalyst
The main effect that a catalyst has is to weaken (or even break) a bond in one of the reactants.
This lowers the activation energy needed to initiate the reaction.
12.8
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164. Catalyst
The main effect that a catalyst has is to weaken (or even break) a bond in one of the reactants.
This lowers the activation energy needed to initiate the reaction.
This “lower energy pathway” leads to a faster reaction rate.
12.8
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165. Catalyst
The main effect that a catalyst has is to weaken (or even break) a bond in one of the reactants.
This lowers the activation energy needed to initiate the reaction.
This “lower energy pathway” leads to a faster reaction rate.
Catalyst addition and temperature increase are the only factors that increase the rate constant, k.
12.8
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166. Potential Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction
12.8
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167. A “Maxwell-Boltzman” Look at the Effect of Catalyst Addition on Reaction Rate
12.8
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168. A “Maxwell-Boltzman” Look at the Effect of Catalyst Addition on Reaction Rate
12.8
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169. Factors Affecting Reaction Rates
Reactant Concentration
Higher concentration, more collisions, greater statistical chance of reaction
12.8
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170. Factors Affecting Reaction Rates
Reactant Concentration
Higher concentration, more collisions, greater statistical chance of reaction
Reaction Temperature
Higher temperature, greater average collision energy, greater number of collisions with sufficient activation energy
12.8
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171. Factors Affecting Reaction Rates
Reactant Concentration
Higher concentration, more collisions, greater statistical chance of reaction
Reaction Temperature
Higher temperature, greater average collision energy, greater number of collisions with sufficient activation energy
Catalyst Addition
Provides a lower energy “pathway” for the reaction to occur, resulting in more of the collisions being “effective”
12.8
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172. Factors Affecting Reaction Rates
Reactant Surface area (for solid reactants only)
Increased surface area leads to more of that reactant being exposed to the other reactant, thus increasing the reaction rate.
12.8
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173. Factors Affecting Reaction Rates
Reactant Surface area (for solid reactants only)
Increased surface area leads to more of that reactant being exposed to the other reactant, thus increasing the reaction rate.
(Physical Mixing)
(it is assumed that all reactions are done with physical mixing, which assures that there is not a depletion of reactant in one part of the reaction vessel)
12.8
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