chapter 14 chemical kinetics. copyright mcgraw-hill 2009 14.1 reaction rates kinetics: the study of...
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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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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]?
[A] [B]ort t
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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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Calculate Average Rate
• Avg. rate =
• Avg. rate =
2 2[Br ] [Br ]
final initial
t tfinal initial
= 3.80 0.0101 0.0120
50 0 M M
s sM / s
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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 14.1 - what can you conclude?
250
50
Time (s)
1.75 x 1050.00596
3.52 x 1050.0101
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]
53.52 x 10 /s 33.5x 10 /s0.0101
MkM
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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 = 1 [A] 1 [B] 1 [C] 1 [D]a t b t c t d t
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Stoichiometry and Reaction Rate
• Write the rate expression for the following reaction
2NO(g) + O2(g) 2NO2(g)
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Stoichiometry and Reaction Rate
4PH3(g) P4(g) + 6H2(g)
If molecular hydrogen is formed at a rate of 0.168 M/s, at what rate is P4 being produced?
3 4 21 [PH ] 1 [P ] 1 [H ]4 1 6t t t
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Stoichiometry and Reaction Rate
4PH3(g) P4(g) + 6H2(g)
21 [H ] 1 (0.168 /s) 0.028 /s6 6
M Mt
41 [P ] 0.028 /s1
Mt
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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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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)
2
2
[F ]0.203 2
[F ] 0.101
MM
[rate] 32.4 10 /s3 2
3[rate] 1.2 10 /s1
M
M
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rate = k[F2]1[ClO2]y
Find order (exponents) by comparing data Exp. 1 and 2: [F2] is held constant
2
2
[ClO ]0.0402 4
[ClO ] 0.0101
MM
1st order with respect to [ClO2] (rate and M directly related)
[rate] 34.8 10 /s2 43[rate] 1.2 10 /s
1
M
M
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rate = k[F2]1[ClO2]1 overall order = 2
Find k (use any set of data)
4[rate] 2.1 10 /s 2 19.3 s2 2[A] [B] (0.10 ) (0.015 )
Mk MM M
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Determining Rate Law
8.4 x 1020.0300.103
4.2 x 1040.0150.202
2.1 x 1040.0150.101
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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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)
4[rate] 2.1 10 /s 2 19.3 s2 2[A] [B] (0.10 M) (0.015 )
Mk MM
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14.3 Dependence of Reactant Concentration on Time
• First-Order reactions may be expressed in several ways
• Example: A products
rate = k[A]
[ ]ratet
A
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Integrated Rate Law (First order)
y = mx + b
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.
0ln lnA kt A
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Graphical Methods
• Given concentration and time data, graphing can determine order
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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
132300
150250
170200
193150
220100
2840
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 103 s1 at 110C. 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
[A]ln
[A]0
t kt
(0.71 M) 3 1ln 7.5 10 s ( )(1.25 M)
75 s
t
t
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Your Turn!
• Consider the same first order reaction
2A B, for which k = 7.5 x 103 s1 at 110C. 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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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
t1/ 2 .693
k
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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 104 s1 at 700 C.
C2H6 2CH3
Calculate the half-life in minutes.
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Half-Life
1/2 4 1
0.693t 1293 s
5.36 10 s
1 min1293 s 21.5 min
60 s
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Dependence of Reactant Concentration on Time
Second-order reactions may be expressed in several ways
• Example: A product
rate = k[A]2
[A]ratet
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Integrated Rate Law for Second Order Reactions
Again, the relationships can be combined to yield the following relationship in the form of astraight line
0
1 1kt
A A
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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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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
1/20
1t
[A]k
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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 M1 s1 at 23C. a) If the initial [I] is 0.086 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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a) Use integrated rate equation for 2nd order
9 1 1
11 1
1211 1
1 1(7.0 10 s ) 120s
A 0.086
8.4 10
1[A] 1.2 10
8.4 10
MM
M
MM
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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!)
101/2 9 1 1
101/2 9 1 1
12.4 10 s
(7.0 10 s ) (0.60 )
13.4 10 s
(7.0 10 s ) (0.42 )
tM M
tM M
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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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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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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
a /E RTk Ae
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Arrhenius Equation
In the form of a straight line…what plot will give a straight line?
a 1ln ln
Ek A
R t
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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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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
3180.332
3080.184
2980.101
2880.0521
T (Kelvin) k (M1 s1)
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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)
1 a
2 12
1 1ln k E
T Tk R
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Arrhenius Another Way!
• Use the data to calculate activation energy of the reaction mathematically
7.0 x 101500
6.1 x 102450
2.9 x 103400
k (s1)T (Kelvin)
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Arrhenius
1
2a
2 1
a
ln
1 1
32.9 10ln
26.1 108.3145J/Kmol
1 1450 400
91 kJ/mol
kkE R
T T
E
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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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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]
k1¾ ¾k2¾ ¾
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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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Potential Energy Diagram
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Reaction Mechanism • Given overall equation:
H2(g) + I2(g) 2HI(g)
Step 1: I2 2I (fast)
Step 2: H2 + 2I 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
k2¾ ¾
k1
k
¾ ¾¬ ¾¾
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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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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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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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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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Reaction Pathway without and with Enzyme-Substrate
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Enzyme-Substrate Complex
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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