Chapter 12 chemical kinetics2

General Chemistry
Presented by: Dr Chehardoli
Department of Medicinal Chemistry,
School of Pharmacy,
Hamadan University of Medical Sciences

References:
Main Ref: Chemistry, Ninth Edition Steven
Zumdahl and Susan Zumdahl
Other Refs: 1- Chang General Chemistry
2- Silberberg Chemistry
3- Mortimer Chemistry

Section 12.1
Reaction Rates
5
Reaction Rate is:
- Change in concentration of a reactant or product per
unit time. Reaction: A B
Rate =
D[B]
Dt
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Rate = -
D[A]
Dt
D[A] = change in concentration of A over
time period Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
[A] and [B] means concentration of A and B in mol/L; A is
the reactant and B is the product.

Section 12.1
Reaction Rates
A B
13.1
rate = -
D[A]
Dt
time
rate = -
D[A]
Dt
rate =
D[B]
Dt
time

Section 12.1
Reaction Rates
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
time
13.1
time
Reaction between bromine and formic acid: An example

Br2 (aq) + HCOOH (aq) 2Br– (aq) + 2H+ (aq) + CO2 (g)
= –
[Br2]final – [Br2]initial
tfinal - tinitial
13.1
slope of
tangent
slope of
tangent slope of
tangent
average rate = –
D[Br2]
Dt
= –
[Br2]final – [Br2]initial
tfinal - tinitial
Instantaneous rate

Section 12.1
Reaction Rates
9
rate a [Br2]
rate = k [Br2]
k =
rate
[Br2]
13.1
= rate constant
= 3.50 x 10–3 s–1

Section 12.1
Reaction Rates
10
Thus:
Rate = k [Reactant]

Section 12.2
Rate Laws: An Introduction
Rate Law
Decomposition of nitrogen dioxide: An example
For the decomposition of nitrogen dioxide:
2NO2(g) → 2NO(g) + O2(g)
Rate = k[NO2]n:
 k = rate constant
 n = order of the reactant
11

Section 12.2
Types of Rate Laws
 Differential Rate Law– shows how the rate of a reaction
depends on concentrations.
 Example:
 Integrated Rate Law – shows how the concentrations of
species in the reaction depend on time.
 Example:
Copyright © Cengage Learning. All rights reserved 12
D[A]
Dt
= k [A]2–
   0
1 1
= +
A A
kt

Section 12.2
Rate Law
2NO2(g) → 2NO(g) + O2(g)
Rate = k[NO2]n
 The concentrations of the products do not appear in the
rate law. [NO] [O2]
13

Section 12.2
Rate Law
Rate = k[NO2]n
 The value of the exponent n must be determined by
experiment; it cannot be written from the balanced
equation.
14

Section 12.2
×
 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

Section 12.3
Determining the Form of the Rate Law
Overall Reaction Order is:
The sum of the exponents in the reaction rate equation.
Rate = k[A]n[B]m
reaction is nth order in A
reaction is mth order in B
Overall reaction order = n + m
k = rate constant
[A] = concentration of reactant A
[B] = concentration of reactant B
16
aA + bB cC + dD

17
Determine the rate law and calculate the rate constant for
the following reaction from the following data:
S2O8
2– (aq) + 3I– (aq) 2SO4
2– (aq) + I3
– (aq)
Experiment [S2O8
2 – ] [I – ]
Initial Rate
(M/s)
1 0.08 0.034 2.2 x 10–4
2 0.08 0.017 1.1 x 10–4
3 0.16 0.017 2.2 x 10–4
rate = k [S2O8
2–]x[I–]y
Double [I–], rate doubles (experiment 1 & 2) 2y=2 y = 1
y = 1
Double [S2O8
2–], rate doubles (experiment 2 & 3) 2x=2 x = 1
x = 1
k =
rate
[S2O8
2–][I–]
=
2.2 x 10–4 M/s
(0.08 M)(0.034 M)
= 0.08/M•s
rate = k [S2O8
2–][I–]
×2 ×2
×2 ×2

Section 12.3
Determining the Form of the Rate Law
c. What is the overall reaction order? = 3
×2 ×4
×2 ×2
Rate= k[NO]2[Cl2]
0.18= k[0.1]2[0.1]= 180

1) First-Order Reaction:
 Rate = k[A]1
 Integrated
 ln[A] = –kt + ln[A]o
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
19
Relation Between Reactant Concentrations and Time
D[A]
Dt
= k [A]Differential: –
A product
[A] = [A]0exp(–kt)

1) First-Order Reaction: Plots
20
ln[A] = –kt + ln[A]o
[A] = [A]0exp(–kt)

Compare: ln[A] = –kt + ln[A]o With Y=mX + b
1) First-Order Reactions:
 Straight line
 Negative slope
Tangent = -k =
Δy
Δx

13.3
2N2O5 4NO2 (g) + O2 (g)
k = 5.714 X 10–4 s–1
Example:
ln[N2O5] = –kt + ln[N2O5]o
Tangent = -k =
Δy
Δx

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]t = ln[A]0 – kt
kt = ln[A]0 – ln[A]
t =
ln[A]0 – ln[A]
k
= 66 s
[A]0 = 0.88 M
[A]t = 0.14 M
ln
[A]0
[A]
k
=
ln
0.88 M
0.14 M
2.8 x 10–2 s–1
=

1) First-Order Reaction:
 Time required for a reactant to reach half its original concentration. Its
application is used in medicine to predict the concentration of a substance
over time. It means that even though we can calculate and predict the
probability that any drug may decrease in the target during a certain interval of
time (sec, minute, day,etc.).
 Half–Life for a first-order reaction:
k = rate constant
 For a first-order reaction: Half–life does not depend on the
concentration of reactants.
24
1
2
0.693
=t
k
Half-Life (t½ ) of First-Order Reactions
t½ = t when [A] = [A]0/2
ln
[A]0
[A]0/2
k
=t½
ln2
k
=
0.693
k
=

Half-Life of a First-Order Reaction
The half-life of a first-order reaction stays the same.
As you can see, the time to reach one-half the starting
concentration in a first-order reaction does not depend
on what that starting concentration is.
1
2
0.693
=t
k

What is the half-life of N2O5 if it decomposes with a rate
constant of 5.7 x 10–4 s–1?
t½
ln2
k
=
0.693
5.7 x 10–4 s–1
=
How do you know decomposition is first order?
= 1200 s = 20 minutes
units of k = s-1

2) Second-Order Reaction:
 Rate = k[A]2
k = rate constant
t = time
27
   0
1 1
= +
A A
kt
A product
D[A]
Dt
= k [A]2Differential: – Integrated:

Compare: With Y=mX + b
2) Second-Order Reactions:
 Straight line
 Positive slope
Tangent = k =
Δy
Δx
   0
1 1
= +
A A
kt

2) Second-Order reaction:
 Half–Life:
k = rate constant
 Half–life gets longer as the reaction progresses and the
concentration of reactants decrease.
 Each successive half–life is double the preceding one.
29
 1
2
0
1
=
A
t
k
Half-Life (t½ ) of Second-Order Reactions

Half-Lives of Second-Order Reactions
Each half-life is double the time of the previous half-life.

The combination of iodine atoms to form molecular iodine is a
second-order kinetics and has the high rate constant
7.03×109/M.s at room temperature : I(g) + I(g) I2(g)
a) If the initial concentration of I was 0.068 M, calculate the
concentration after 3.5 min.
b) Calculate the half-life of the reaction if the initial
concentration of I is 0.53 M. c) and if it is 0.39 M.
1
[A]
=
1
0.068 M
+ 7.03×109/M.s (3.5 min×60 s/1 min)
1
[A]
=
1
[A]0
+ kt
[A]= 6.8×10-13 M
a)
b) For [I]0= 0.53 M
 1
2
0
1
=
A
t
k
=
1
7.03×109/M.s (0.53 M)
= 2.7×10-10 s
c) For [I]0= 0.39 M = 3.7×10-10 s

3) Zero-Order Reaction:
 Rate = k[A]0 = k
Integrated: [A] = –kt + [A]o
k = rate constant
t = time
32
A product
D[A]
Dt
= kDifferential: –

Compare: [A] = –kt + [A]o With Y=mX + b
3) Zero-Order Reactions:
 Straight line
 Negative slope
Tangent = -k =
Δy
Δx

3) Zero-Order Reaction:
34
Half-Life (t½ ) of Second-Order Reactions
 Half–Life:
k = rate constant
 Half–life gets shorter as the reaction progresses and the
concentration of reactants decrease.
 0
1
2
A
=
2
t
k

Half-Lives of a Zero-Order Reaction
Each half-life is ½ the time of the previous half-life.

Section 12.4
The Integrated Rate Law
Summary of the Rate Laws
36

Section 12.5
Reaction Mechanisms
Reaction Mechanism
 Most chemical reactions occur by a series of elementary
steps. The sequence of elementary steps that leads to
product formation is the reaction mechanism.
A D
Elementary steps: A B C D
 An intermediate is formed in one step and used up in a
subsequent step and thus is never seen as a product in
the overall balanced reaction.
Intermediate: B and C
37

Section 12.5
Reaction Mechanisms
A Molecular Representation of the Elementary Steps in the Reaction
of NO2 and CO
NO2(g) + CO(g) → NO(g) + CO2(g)
38
NO2 NO2 NO3 NO
NO3
NO2
CO2CO
Intermediate

Section 12.5
Reaction Mechanisms
Elementary Steps (Molecularity)
 Unimolecular – reaction involving one molecule; first
order.
 Bimolecular – reaction involving the collision of two
species; second order.
 Termolecular – reaction involving the collision of three
species; third order. This type is very rare.
39
A products rate = k [A]
A + A products
A + B products
rate = k [A]2
rate = k [A][B]

Section 12.5
Reaction Mechanisms
Rate-Determining Step
 A reaction is only as fast as its slowest step.
 The rate-determining step (slowest step) determines the
rate law and the molecularity of the overall reaction.
A C
1) A B Rate-Determining Step: Rate=k[A]
2) B C
40
Fast
Slow

The experimental rate law for the reaction between NO2
and CO to produce NO and CO2 is rate = k[NO2]2. The
reaction is believed to occur via two steps:
Step 1: NO2 + NO2 NO + NO3
Step 2: NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO NO + CO2
What is the intermediate?
NO3 (unstable)
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
13.5
Step 1: Stable molecules
converted to unstable particle
Step 2: Unstable particle
converted to stable products

Section 12.6
A Model for Chemical Kinetics
Collision Model
 Molecules must collide to react.
 Main Factors:
 Activation energy, Ea
 Temperature
 Molecular orientations
42

Section 12.6
Activation Energy, Ea
 Energy that must be overcome to produce a chemical
reaction.
43

Section 12.6
Transition States and Activation Energy
44

Section 12.6
Change in Potential Energy
45
Example: 2BrNO 2NO + Br2

Section 12.6
For Reactants to Form Products
 Collision must involve enough energy to produce the
reaction (must equal or exceed the activation energy).
 Relative orientation of the reactants must allow
formation of any new bonds necessary to produce
products.
46

Section 12.6
13.4
Importance of Orientation

Arrhenius Equation
k = rate constant
A = frequency factor (represents the frequency of
collisions between reactant molecules)
Ea = activation energy (J/mol)
R = gas constant (8.3145 J/K·mol)
T = temperature (in K)
48
/
=
 a
E RT
k Ae
Temperature Dependence of the Rate Constant

Section 12.6
Linear Form of Arrhenius Equation
49
 a
E 1
ln( ) = ln
R T
 
  
 
k + A
For Two Temperatures:
= ( - )
Ea
R
1
T2
1
T1
Ln( )
k2
k1Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T1 is the initial temperature
T2 is the final temperature
k1 is the rate constant in T1
k2 is the rate constant in T2

At 550 0C the rate constant for the below reaction is 1.1 L/mol.s, and at
625 0C the rate constant is 6.4 L/mol.s. Using these values, calculate
Ea for this reaction.
CH4(g) + 2S2(g) CS2(g) + 2H2S(g)
= ( - )
Ea
R
1
T2
1
T1
Ln( )
k2
k1
( - ) = 1.4 × 105 J/molLn( ) =
6.4
1.1
Ea
8.314 J/K.mol
1
898 k
1
823 K

Section 12.7
Catalysis
Catalyst
 A substance that speeds up a reaction without being
consumed itself.
 Catalyst provides a new pathway for the reaction with a
lower activation energy.
51

Section 12.7
Catalysis
Energy Plots for a Catalyzed and an Uncatalyzed Pathway
for a Given Reaction
52
ratecatalyzed > rateuncatalyzed
Ea < Ea

Section 12.7
Catalysis
Heterogeneous Catalyst
Example: Haber synthesis of ammonia
53
Catalyst Types:
Homogeneous Catalyst
Example: Acid catalyzed reaction

Section 12.7
Catalysis
Heterogeneous Catalyst
 Most often involves gaseous reactants being adsorbed
on the surface of a solid catalyst.
54

Section 12.7
Catalysis
Homogeneous Catalyst
 Exists in the same phase as the reacting molecules.
 Enzymes are nature’s homogeneous catalysts.
55

Chapter 12 chemical kinetics2

Chapter 12 chemical kinetics2

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Chapter 12 chemical kinetics2