Chemical Kinetics
Unit 11
Overview
 Factors affecting reactions
 Collision Model
 Activation energy, activated complex
 Exothermic/endothermic reactions
 Energy of reactions ∆E
 Reaction Rate
 Average reaction rate
 Instantaneous reaction rate
 Kinetics Stoichiometry
 Reaction Rate and Concentration
 (Differential) Rate laws
 Specific rate constant (k)
 Writing rate laws
 Reaction Order
 Rate laws
 Integrated Rate Laws
 Graphing
 Solving (0, 1st, 2nd order)
 Half life equations
 Rate Constant and Temperature
 Arrhenius equation
 Catalysis
 Homogeneous, heterogeneous,
enzymes
 Reaction Mechanisms
 Identifying intermediates
 Rate determining step
Chemical Kinetics
 Thermochemistry = does a reaction take place?
 Kinetics = how fast does the reaction happen?
 Kinetics: The area of chemistry concerned with
the speeds (rates) of reactions
Factors Affecting Reaction Rates
 Physical state of reactants (surface area)
 Concentration of reactants
 Temperature at which reaction occurs
 Presence of a catalyst
 Pressure of gaseous reactants or products
Collision Model
Molecules must collide to react
Only a small fraction of collisions produces
a reaction. Why?
Collision Model
 Collisions must have enough energy to produce the
reaction (must equal or exceed the activation
energy).
 Reactants must have proper orientation to allow the
formation of new bonds.
Collision Model
Activation Energy
 Activation Energy (Ea)– minimum energy required to transform
reactants into the activated complex
 (The minimum energy required to produce an effective collision)
 Example: flame, spark, high temperature, radiation
 The lower the Ea, the faster the reaction
Activation Energy
 Activated Complex – transitional structure
 All bonds have been broken but no new bonds have formed
 Point in reaction of highest energy but lowest stability
Exothermic Processes
Processes in which energy is released as it proceeds, and
surroundings become warmer
Reactants  Products + energy
Endothermic Processes
Processes in which energy is absorbed as it proceeds, and
surroundings become colder
Reactants + energy  Products
Energy of Reactions (∆E)
 Same as ∆H
 “Heat” and “energy” can be used interchangeably
 ∆E = ∆Eproducts- ∆Ereactants
 - ∆E = exothermic (energy lost)
 +∆E = endothermic (energy gained)
Reaction Rate
Reaction Rate: the change in concentration of reactants per
unit time as a reaction proceeds (M/s)
A
B
D[A]
rate = Dt
D[A] = change in concentration of A over
time period Dt
D[B]
rate =
Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
A
B
time
D[A]
rate = Dt
rate =
D[B]
Dt
13.1
Reaction Rate
 Average reaction rate: average rate throughout the course of
the reaction
 Instantaneous reaction rate: rate of reaction at a specific
point in time during the reaction
 Think of a car trip where you drove 100 miles in 2 hours
 Average speed = 50 mi/hr
 Instantaneous speed is speed at specific
point – you don’t go exactly 50 mi/hr
the entire trip
Br2 (aq) + HCOOH (aq)
2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
D[Br2]
average rate = Dt
=-
slope of
tangent
slope of
tangent
[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time (slope of tangent)
Kinetics Stoichiometry
 When mole ratios are not 1:1, stoichiometry is used to
compare rates of reactions
aA + bB  cC + dD
1 ∆[A]
1 ∆[B]
Rate = - a ∆t = - b ∆t =
1 ∆[C]
1 ∆[D]
c ∆t = d ∆t
Kinetics Stoichiometry
2HI  H2 + I2
If the rate at which H2 appears, ∆[H2]/∆t, is 6.0×10-5 M/s at a
particular instant, at what rate is HI disappearing at this same
time, - ∆[HI]/∆t?
1 ∆[HI]
Rate = =
2 ∆t
Therefore, -
1 ∆[H2]
1 ∆t
1
1 ∆[HI]
= 1 (6.0×10-5 M/s )
2 ∆t
- ∆[HI] = 1.2×10-4 M/s
∆t
Reaction Rate and Concentration
 The initial rate of a reaction depends on the concentration of
the reactants
 Rate changes based on the concentration of reactants
 (Differential) Rate Law: an equation that relates reaction
rate and concentrations of reactants
Rate a [reactants]
a means “proportional to”
Rate Law
R = k[A]n[B]m
R = reaction rate
k = specific rate constant
[A] and [B] = concentrations of reactants
n and m = the order of the reactant (usually 0, 1, 2)
(not related to moles of reactants from the equation)
Overall order of reaction = n + m
Rate Laws
 Rate laws are always determined experimentally.
 Reaction order is always defined in terms of reactant (not
product) concentrations.
 The order of a reactant is not related to the stoichiometric
coefficient of the reactant in the balanced chemical equation.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2]1
13.2
Specific Rate Constant (k)
1.
2.
3.
4.
5.
6.
Once the reaction orders are known, the value of k must be
determined from experimental data
Value of k is for a specific reaction; k has a different value
for other reactions, even at the same conditions
Units of k depend on the overall order of the reaction
Value of k does not change for different concentrations of
reactants or products
Value of k is for reaction at a specific temperature; if
temperature increases, k increases
Value of k changes if a catalyst is present
Writing a (differential) Rate Law
Problem -Write the rate law, determine the value of the rate
constant, k, and the overall order for the following reaction:
2 NO(g) + Cl2(g)  2 NOCl(g)
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
11.4 x 10-6
Writing a Rate Law
Step 1 – Determine the values for the exponents in the rate law:
R = k[NO]n[Cl2]m
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
1.14 x 10-5
In experiment 1 and 2, [Cl2] is constant while [NO] doubles.
The rate quadruples, so the reaction is second order with
respect to [NO]
 R = k[NO]2[Cl2]m
Writing a Rate Law
R = k[NO]2[Cl2]m
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
1.14 x 10-5
In experiment 2 and 4, [NO] is constant while [Cl2] doubles.
The rate doubles, so the reaction is first order with
respect to [Cl2]
 R = k[NO]2[Cl2]
Writing a Rate Law
Part 2 – Determine the value for k, the rate constant, by using any
set of experimental data:
R = k[NO]2[Cl2]
Experiment
1
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
0.250
0.250
1.43 x 10-6
2
mol
mol  
mol 

1.43 x 10
 k  0.250
  0.250

Ls
L  
L 

6
2
 1.43 x106   mol   L3 
L
5
k 

9.15
x
10


3
3 
2
0.250
L

s
mol
mol
s




Reaction Order
Given the rate law
R = k[NO]2[Cl2]
We say that the reaction is
 2nd order with respect to NO
 1st order with respect to Cl2
Overall the order of the reaction is a 3rd order reaction
2+1=3
Example 2
Run # Initial [A]
([A]0)
Initial [B]
([B]0)
Initial Rate (v0)
1
1.00 M
1.00 M
1.25 x 10-2 M/s
2
1.00 M
2.00 M
2.5 x 10-2 M/s
3
2.00 M
2.00 M
2.5 x 10-2 M/s
What is the order with respect to A?
0
What is the order with respect to B?
1
What is the overall order of the reaction?
1
Example 3
[NO(g)] (mol
dm-3)
[Cl2(g)] (mol
dm-3)
Initial Rate
(mol dm-3 s-1)
0.250
0.250
1.43 x 10-6
0.250
0.500
2.86 x 10-6
0.500
0.500
1.14 x 10-5
What is the order with respect to Cl2?
1
What is the order with respect to NO?
2
What is the overall order of the reaction?
3
Reaction Order
Zero
Order
Rate Law R = k
1st
Order
2nd
Order
3rd Order
R = k[A]3
R = k[A][B]
R = k[A]
R = k[A][B][C]
2
R = k[A]
R = k[A] 2[B]
Concentration VS Time
 Integrated Rate Law
 Graph data to solve for order of reaction
Zero Order = time vs concentration is linear
1st Order = time vs ln[concentration] is linear
2nd
1
Order = time vs concentration is linear
Zero Order Reactions
13.3
First Order Reactions
13.3
Second Order Reactions
Solving an Integrated Rate Law
Time (s)
[H2O2] (mol/L)
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Problem: Find the integrated
rate law and the value for the
rate constant, k
Time vs. [H2O2]
Time
[H2O2]
Regression results:
y = ax + b
a = -2.64 x 10-4
b = 0.841
r2 = 0.8891
r = -0.9429
Time (s)
[H2O2]
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Time vs. ln[H2O2]
ln[H2O2]
Time (s)
ln[H2O2]
0
0
120
-0.0943
300
-0.2485
600
-0.5276
1200
-0.9943
1800
-1.514
Regression results:
2400
-2.04
y = ax + b
a = -8.35 x 10-4
b = -.005
r2 = 0.99978
r = -0.9999
3000
-2.501
3600
-2.996
Time
Time vs. 1/[H2O2]
Time
1/[H2O2]
Regression results:
y = ax + b
a = 0.00460
b = -0.847
r2 = 0.8723
r = 0.9340
Time (s)
1/[H2O2]
0
1.00
120
1.0989
300
1.2821
600
1.6949
1200
2.7027
1800
4.5455
2400
7.6923
3000
12.195
3600
20.000
Integrated Rate Law
 The graph is linear for “time vs ln[H2O2]”
 The reaction is 1st order
VS
VS
Concentration VS Time
Rate laws for concentration of reactants versus time are based on
the linear equation y = ax + b
 Zero Order Reactions (time vs concentration)
[A] = -kt + [A]0
 1st Order (time vs ln[concentration])
ln[A] = -kt + ln[A]0
 2nd Order (time vs 1/concentration)
1
1
 kt 
[ A]
[ A]0
The reaction 2A
B is first order in A with a rate constant
of 2.8 x 10-2 s-1 at 800C. How long will it take for A to
decrease from 0.88 M to 0.14 M ?
ln[A] = -kt + ln[A]0
ln[A] - ln[A]0 = - kt
[A]0 = 0.88 M
[A] = 0.14 M
ln[A]0 - ln[A] = kt
ln[A]0 – ln[A]
t=
k
ln
=
[A]0
[A]
k
ln
=
0.88 M
0.14 M
2.8 x
10-2 s-1
= 66 s
Half-life t½
 Time it takes for concentration of reactant to equal ½ of its
initial value
 A fast reaction will have a short half-life
 In a first order reaction, concentration of reactant decreases by
½ each time interval of t½
[A] = ½ [A]0
Half-life t½
 For zero and second order reactions, half-life changes based on
concentration of the reactant
 To find equation for half-life, place ½[A] 0 in for [A] in each
integrated rate law and put t½ in for t
When [A] = ½[A]0 and t = t½ …
Zero Order:
[A] = -kt + [A]0
[ A]0
t1/ 2 
2k
1st Order
ln[A] = -kt + ln[A]0
0.693
t1/ 2 
k
2nd
1
1
 kt 
[ A]
[ A]0
1
t1/ 2 
k [ A]0
Order
Half-life t½
What is the half-life of N2O5 if it decomposes with
a rate constant of 5.7 x 10-4 s-1?
t½ =
0.693
= 1200 s = 20 minutes
-4
-1
5.7 x 10 s
How do you know decomposition is first order?
units of k are (s-1)
SUMMARY
Order
Rate Law
Concentration-Time
Equation
0
rate = k
[A] - [A]0 = - kt
1
rate = k [A]
ln[A] - ln[A]0 = - kt
2
[A]2
rate = k
1
1
= kt
[A] [A]0
Half-Life
[A]0
2k
Ln 2
t½ =
k
1
t½ =
k[A]0
t½ =
Rate Constant and Temperature
 Reaction rate increases with temperature
 Molecules move faster
 More molecules collide to cause a reaction in a time period
 Molecules have greater kinetic energy at higher temperatures
 More molecules will have enough energy to reach the activation
energy of the reaction at higher temperatures
Boltzmann Distribution Curve
Rate Constant and Temperature
 At higher temperatures (T2), more molecules are able to
reach the activation energy
Rate Constant and Temperature
 Arrhenius Equation: relationship between k and T
Ea is the activation energy (J/mol)
-Ea 1
Ln k = + lnA
R T
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
A is the frequency factor
Takes into account collision
frequency and orientation
Graphing Arrhenius
 A plot of lnk versus 1 / T produces a straight line with the
familiar form y = - mx + b, where
x = 1 /T
y = lnk
m=-Ea/R
b = lnA
 The activation energy E a can be determined from the slope
m of this line: E a = - m · R
Catalysis
 Catalyst: A substance that speeds up a reaction by
lowering activation energy
 Homogeneous catalyst: Present in the same phase as the
reacting molecules.
 Heterogeneous catalyst: Present in a different phase than
the reacting molecules.
 Enzyme: A large molecule (usually a protein) that catalyzes
biological reactions.
 Avoids temperature increase in reactions of living organisms
A catalyst is a substance that increases the rate of a chemical
reaction without itself being consumed.
Ea
uncatalyzed
k
catalyzed
ratecatalyzed > rateuncatalyzed
13.6
Endothermic Reaction w/Catalyst
Exothermic Reaction w/Catalyst
Factors Affecting Reaction Rates
 Physical state of reactants (surface area)
 Increased surface area means more collisions between reacting molecules
 Concentration of reactants
 Higher concentration means more molecules (solute) to react meaning more
collisions
 Temperature at which reaction occurs
 Higher kinetic energy
 Molecules moving faster so collide more frequently
 Presence of a catalyst
 Speeds up a reaction without being used up
 Lowers activation energy
 Pressure of gaseous reactants or products
 Increased number of collisions
Rate Laws (vocabulary)
 Elementary Reaction – rate law is based directly on proportionality
 Describes an individual molecular event
 Unimolecular – overall rate law is 1st order
 Rate = k[A]
 Bimolecular – overall rate law is 2nd order
 Rate = k[A]2
 Rate = k[A][B]
 Termolecular – overall rate law is 3rd order
 Rate = k[A]3
 Rate = k[A] 2[B]
 Rate = k[A][B][C]
Reaction Mechanisms
 Some reactions occur in one step, others do not
 Reaction Mechanism: step-by-step sequence of reactions by
which the overall chemical change occurs
 A chemical equation does not tell us how reactants become products;
it is a summary of the overall process.
 The sum of the elementary steps must give the overall
balanced equation for the reaction
Reactants  Products
The  sign has represents the reaction mechanism, but
gives no indication of the steps in the mechanism
Reaction Mechanisms (Example)
For the reaction:
2NO (g) + O2 (g)
2NO2 (g)
The following steps occur:
Elementary Step 1:
NO + NO
N2O2
+ Elementary Step 2:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
N2O2 is detected during the reaction!
Reaction Intermediates
Intermediates are species that appear in a reaction mechanism but not in
the overall balanced equation.
An intermediate is always formed in an early elementary step and
consumed in a later elementary step.
Elementary step:
NO + NO
N2O2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
Energy Diagrams
 For multi-step reactions, each elementary step requires
activation energy and has its own transition state
 Peaks = transition states
 Valleys = intermediates
Energy Diagrams (multi-step)
Writing Reaction Mechanisms
•
The sum of the elementary steps must give the overall balanced
equation for the reaction.
•
The rate-determining step should predict the same rate law that is
determined experimentally.
The Rate-Determining Step
In a multi-step reaction, the slowest
step is the rate-determining step.
It therefore determines the rate of
reaction.
Step 1. NO + Cl2 <----> NOCl2 (fast)
Step 2. NOCl2+ NO ----> 2NOCl (slow)
Together they give:
2NO + Cl2 ----> 2NOCl
Step 2 is the rate determining step
Identifying the Rate-Determining Step
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
The experimental rate law is:
R = k[NO]2[H2]
Which step in the reaction mechanism is the rate-determining (slowest)
step?
Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2
N2O(g) + H2(g)  N2(g) + H2O(g)
Step #1 agrees with the experimental rate law