1. Chapter 14 Chemical Kinetics

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14.1 Reaction Rates
Kinetics: the study of how fast reactions take place
Some reactions are fast (photosynthesis)
Some reactions are slow (conversion of diamond to graphite)
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3. Importance of Studying Reaction Rates
Speed up desirable reactions
Minimize damage by undesirable reactions
Useful in drug design, pollution control, and food processing
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Rate of Reaction
Expressed as either:
Rate of disappearance of reactants (decrease or negative)‏ OR
Rate of appearance of products (increase or positive)
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Average Reaction Rate
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Average Reaction Rate
Equation A  B
rate =
Why the negative on [A]?
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Average Reaction Rate
Br2(aq) + HCOOH(aq)2Br(aq) + 2H+(aq) + CO2(g)‏
Note: Br2 disappears over time
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Average Reaction Rate
Br2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)‏
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9. Calculate Average Rate
Avg. rate =
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Average Rate
Average rate depends on time interval
Plot of [Br2] vs time = curve
Plot of Rate vs [Br2] = straight line
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Instantaneous Rate
Instantaneous: rate at a specific instance in time (slope of a tangent to the curve)
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Rate Constant
Using data from Table what can you conclude?
250 50
Time (s)
1.75 x 105
3.52 x 105
0.0101
Rate (M/s)
[Br2]
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Rate Constant
Answer:
When the [Br2] is halved; the rate is halved
Rate is directly proportional to [Br2]
rate = k [Br2]
k = proportionality constant and is constant as long as temp remains constant
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Rate Constant
Calculate the value of the rate constant for any set of data and get basically the same answer!
k = rate / [Br2]
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15. Stoichiometry and Reaction Rate
When stoichiometric ratios are not 1:1 rate of reaction is expressed as follows
General equation:
aA + bB  cC + dD
rate =
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16. Stoichiometry and Reaction Rate
Write the rate expression for the following reaction
2NO(g) + O2(g)  2NO2(g)
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17. Stoichiometry and Reaction Rate
4PH3(g)  P4(g) H2(g)
If molecular hydrogen is formed at a rate of M/s, at what rate is P4 being produced?
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18. Stoichiometry and Reaction Rate
4PH3(g)  P4(g) H2(g)
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19. 14.2 Dependence of Reaction Rate on Reactant Concentration
Rate law expression
For the general equation:
aA + bB  cC + dD
rate law = k[A]x[B]y
k = proportionality constant
x and y = the order of the reaction with respect to each reactant
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Order
Exponents represent order
Only determined via experimental data
1st order - rate directly proportional to concentration
2nd order - exponential relationship
0 order - no relationship
Sum of exponents (order) indicates overall reaction order
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21. Experimental Determination of Rate Law
Method of initial rates - examine instantaneous rate data at beginning of reaction
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rate = k[F2]x [ClO2]y
Find order (exponents) by comparing data
Exp. 1 and 3: [ClO2] is held constant
1st order with respect to [F2] (rate and M directly related)
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rate = k[F2]1[ClO2]y
Find order (exponents) by comparing data
Exp. 1 and 2: [F2] is held constant
1st order with respect to [ClO2] (rate and M directly related)
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rate = k[F2]1[ClO2]1 overall order = 2
Find k (use any set of data)
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Determining Rate Law
8.4 x 102
0.030
0.10 3
4.2 x 104
0.015
0.20 2
2.1 x 104 1
Initial Rate (M/s)
[B] (M)‏
[A] (M)‏
Exp.
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What is Different?
In experiment 1 and 2; [B] is constant; [A] doubles and rate doubles - the reaction is 1st order with respect to [A]
In experiment 1 and 3; [A] is constant; [B] doubles but the rate quadruples! This means that the reaction is 2nd order with respect to [B]
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27. Calculate the Rate Constant
Rate = k[A] [B]2
The rxn is 1st order w/ respect to [A]
The rxn is 2nd order w/ respect to [B]
The rxn is 3rd order overall (1 + 2)
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28. 14.3 Dependence of Reactant Concentration on Time
First-Order reactions may be expressed in several ways
Example: A  products
rate = k[A]
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29. Integrated Rate Law (First order)‏
When the two expressions are set equal to each other,we get an expression that can be rearranged in the form of a straight line.
y = mx b
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Graphical Methods
Given concentration and time data, graphing can determine order
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31. Integrated Rate Law (First order)‏
For a 1st order reaction, a plot of ln [A] vs time yields a straight line
The slope = k (the rate constant)
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Try Graphing
132
300
150
250
170
200
193
220
100
284
P (mmHg)‏
Time (s)‏
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Graphing
Plot ln [Pressure] on y-axis and time on x-axis
If the plot is a straight line, then the integrated rate law equation can be used to find the rate constant, k, or the slope of the line can be calculated for the rate constant.
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Integrated Rate Law
The rate constant for the reaction
2A  B is 7.5 x 103 s1 at 110C. The reaction is 1st order in A. How long (in seconds) will it take for [A] to decrease from 1.25 M to 0.71 M?
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Another form of the integrated rate law
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Your Turn!
Consider the same first order reaction
2A  B, for which k = 7.5 x 103 s1 at 110C. With a starting concentration of [A] = 2.25 M, what will [A] be after 2.0 minutes?
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Half-Life (1st order)
Half-life: the time that it takes for the reactant concentration to drop to half of its original value.
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38. Calculating First Order Half-life
Half-life is the time that it takes for the reactant concentration to drop to half of its original value.
The expression for half-life is simplified as
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Half-Life
The decomposition of ethane (C2H6) to methyl radicals (CH3) is a first order reaction with a rate constant of
5.36 x 104 s1 at 700 C.
C2H6  2CH3
Calculate the half-life in minutes.
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Half-Life
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41. Dependence of Reactant Concentration on Time
Second-order reactions may be expressed in several ways
Example: A  product
rate = k[A]2
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42. Integrated Rate Law for Second Order Reactions
Again, the relationships can be combined to
yield the following relationship in the form of a
straight line
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43. Integrated Rate Law for Second Order Reactions
For a 2nd order reaction, a plot of 1/[A] vs time yields a straight line
The slope = k (the rate constant)
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44. Calculating Second-Order Half-life
The expression for half-life is simplified as
Note: half-life for 2nd order is inversely proportional to the initial reaction concentration
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45. Calculating Second Order Half-life
I(g) + I(g)  I2(g)
The reaction is second order and has a rate constant of 7.0 x 109 M1 s1 at 23C.
a) If the initial [I] is M, calculate the concentration after 2.0 min.
b) Calculate the half-life of the reaction when the initial [I] is 0.60 M and when the [I] is 0.42 M.
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Use integrated rate equation for 2nd order
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b) Use the half-life formula for 2nd order
(note: half-life does not remain constant for a 2nd-order reaction!)
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Zero Order
Zero Order reactions may exist but are relatively rare
Example: A  product
rate = k[A]0 = k
Thus, a plot of [A] vs time yields a straight line.
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Summary of Orders
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50. 14.4 Dependence of Reaction Rate on Temperature
Most reactions occur faster at a higher temperature.
How does temperature alter rate?
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Collision Theory
Particles must collide in order to react
The greater frequency of collisions, the higher the reaction rate
Only two particles may react at one time
Many factors must be met:
Orientation
Energy needed to break bonds (activation energy)‏
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Collision Theory
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Collision Theory
Though it seems simple, not all collisions are effective collisions
Effective collisions: a collision that does result in a reaction
An activated complex (transition state) forms in an effective collision
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Activation Energy
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55. The Arrhenius Equation
The dependence of the rate constant of a reaction on temperature can be expressed
Ea = activation energy
R = universal gas constant
A = frequency factor T = Kelvin temp
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Arrhenius Equation
In the form of a straight line…what plot will give a straight line?
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Arrhenius Equation
A plot of ln k vs 1/T will give a straight line
The slope of line will equal Ea/R
The activation energy may be found by multiplying the slope by “R”
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58. Graphing with Arrhenius
Rate constants for the reaction
CO(g) + NO2(g)  CO2(g) + NO(g)
Were measured at four different temperatures. Plot the data to determine activation energy in kJ/mol.
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Graphing
318
0.332
308
0.184
298
0.101
288
0.0521
T (Kelvin)
k (M1 s1)‏
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Graphing
Steps:
Make a column of ln k data
Make a column of inverse temp (1/T)
Plot ln k vs 1/T
Calculate the slope
Multiply slope by R
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Arrhenius Another Way!
Another useful arrangement of the Arrhenius equation enables calculation of: 1) Ea (with k at two different temps)
2) the rate constant at a different temperature (with Ea, k and temps)
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Arrhenius Another Way!
Use the data to calculate activation energy of the reaction mathematically
7.0 x 101
500
6.1 x 102
450
2.9 x 103
400
k (s1)
T (Kelvin)
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Arrhenius
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14.5 Reaction Mechanisms
Most reactions occur in a series of steps
The balanced equation does not tell us how the reaction occurs!
There are often a series of steps which add together to give the overall reaction
The series of steps is the reaction mechanism
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Reaction Mechanisms
Most chemical reactions occur in a series of steps
Energy of activation must be overcome to form intermediates
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Reaction Mechanism
Consider:
2NO(g) + O2(g)  2NO2(g)
The reaction cannot occur in a single step.
One proposed mechanism:
Step 1: NO + NO  N2O2
Step 2: N2O2 + O2  2NO2
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Reaction Mechanism
N2O2 is an intermediate in the reaction mechanism
Intermediate: a substance that is produced in an early step and consumed in a later step
Elementary reaction: one that occurs in a single collision of the reactant molecules
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Reaction Mechanism
Molecularity: the number of reactant molecules involved in the collision
Unimolecular: one reactant molecule
Bimolecular: two reactant molecules
Termolecular: three reactant molecules (fairly rare)
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Rate Determining Step
If the elementary reactions are known, the order can be written from the stoichiometric coefficients of the rate- determining step
Rate-determining step: the slowest step in the mechanism
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70. Rate-Determining Step
Steps of a mechanism must satisfy two requirements
Sum of elementary steps must equal the overall balanced equation
The rate law must have same rate law as determined from experimental data
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The decomposition of hydrogen peroxide (2H2O2  2H2O + O2 )‏
may occur in the following two steps Step 1: H2O2 + I H2O + IO
Step 2: H2O2 + IO H2O + O2 + I
If step 1 is the rate-determining step, then the rate law is
rate = k1[H2O2] [I]
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I does not appear in the overall balanced equation
I serves as a catalyst in the reaction - it is present at the start of the reaction and is present at the end
IO is an intermediate
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73. Potential Energy Diagram
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Reaction Mechanism
Given overall equation:
H2(g) + I2(g)  2HI(g)
Step 1: I I (fast)
Step 2: H I 2HI (slow)
rate = k2[H2] [I]2
This rate expression does not meet the requirement..I is an intermediate and should not appear in the rate expression
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Consider the first equilibrium step: the forward rate is equal to the reverse rate
k1[I2] = k-1 [I]2 k1/k-1 [I2] = [I]2
If we substitute for [I]2 , the rate law becomes
rate = k[H2] [I2]
This now matches the overall balanced equation!
(When step 2 is rate-determining this substitution is always possible)‏
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14.6 Catalysis
Catalyst - a substance that increases the rate of a chemical reaction without being used up itself
Provides a set of elementary steps with more favorable kinetics than those that exist in its absence
Many times a catalyst lowers the activation energy
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77. Reaction Pathway with Catalyst
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Types of Catalysts
Heterogeneous catalysts - reactants and catalyst are in different phases
Homogeneous catalysts - reactants and catalysts are dispersed in single phase
Enzyme catalysts - biological catalysts
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79. Heterogeneous Catalysts
Most important in industrial chemistry
Used in catalytic converters in automobiles
Efficient catalytic converter serves two purposes; oxidizes CO and unburned hydrocarbons into CO2 and H2O; converts NO and NO2 into N2 and O2
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Catalytic Converter
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81. Homogeneous Catalysts
Usually dispersed in liquid phase
Acid and base catalyses are the most important types of homogeneous catalysis in liquid solution
Advantages of homogeneous catalysts
Reactions performed at room conditions
Less expensive
Can be designed to function selectively
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Biological Catalysts
Enzymes: large protein molecule that contains one or more active sites where interactions with substrates occur
Enzymes are highly specific (lock and key)
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83. Reaction Pathway without and with Enzyme-Substrate
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84. Enzyme-Substrate Complex
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85. Biological Molecules (binding of glucose to hexokinase)
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Key Points
Rate of reaction can be determined in several ways
Instantaneous rate
Average rate
Graphing using integrated rate laws
Mechanisms
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Key Points
Write rate law expressions
Calculate rate constant with proper units
Distinguish orders: 1st, 2nd, 0 order
Calculate half-life
Collision theory and relationship to Arrhenius equation
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Key Points
Calculate activation energy graphically and mathematically
Reaction mechanisms
Elementary reactions
Molecularity
Rate law from slow step
Intermediates and catalysts
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Key Points
Catalysis
Heterogeneous
Homogeneous
Enzymes
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