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Chemical KineticsChemical Kinetics

Chapter 12Chapter 12

Chapter 12 Slide 2

Reaction Rates 01Reaction Rates 01

• Reaction Rate: The change in the concentration of a

reactant or a product with time (M/s).

• change in concentration divided by the change in time

Reactant → ProductsA → B

Rate = −∆[A]

∆t Rate =

∆[B]

∆t2 HI(g) H2(g) + I2(g)

Chapter 12 Slide 3

Reaction Rates and StoichiometryReaction Rates and Stoichiometry

• What is the general rate of the following reaction ?

2 HI(g) → H2(g) + I2(g)

Rate = − 12

∆[HI]∆t

= ∆[I2]∆t

Chapter 12 Slide 4

Reaction Rates and StoichiometryReaction Rates and Stoichiometry

• To generalize, for the reaction

aA + bB cC + dD

Rate* = −1a

∆[A]∆t

= −1b

∆[B]∆t

=1c

∆[C]∆t

1d

∆[D]∆t

=

*: General rate of reaction

+

2 2 2HCO H(aq) + Br (aq) 2H (aq) + 2Br (aq) + CO (g)

red colorless

−→

Which of the expressions below,corresponding to the reaction ofbromine with formic acid, isincorrect?

2 2d[CO ] d[Br ]1. =

dt dt−

+d[Br ] d[H ]2. =

dt dt

−

+

2d[HCO H] 1 d[H ]4. =

dt 2 dt

2d[Br ] d[Br ]5. = 2

dt dt

−

− ×2 2d[HCO H] d[Br ]

3. = dt dt

Chapter 12 Slide 7

How Do we study Rate of a reaction?How Do we study Rate of a reaction?

• Consider the decomposition of N2O5 to give NO2

and O2: 2 N2O5(g) 4 NO2(g) + O2(g)

Brown Colorless

Chapter 12 Slide 8

Reaction Rates: concentration versus time curve 03Reaction Rates: concentration versus time curve 03

Average Rate = Rate between two points in time

•2 N2O5(g) 4 NO2(g) + O2(g)

The slope of each triangle Between two points

Reaction RatesReaction Rates

2N2O5(g) 4NO2(g) + O2(g)

Chapter 12 Slide 10

Instantaneous rate:Instantaneous rate:Rate for specific instance in time

Slope of the tangent to a concentration versus time curve

Initial Rate

Chapter 12 Slide 11

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

time

393 nmlight

Detector

∆[Br2] α ∆Absorption3

93

nm

Br2 (aq)

Chapter 12 Slide 12

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

average rate = -∆[Br2]

∆t= -

[Br2]final – [Br2]initial

tfinal - tinitial

slope oftangent

slope oftangent

slope oftangent

instantaneous rate = rate for specific instance in timeThe slope of a line tangent to the curve at any point is the instantaneous rate at that time

Chapter 12 Slide 13

←

←

←

slope oftangent

slope oftangent

Chapter 12 Slide 14

rate α [Br2]

rate = k [Br2]

k = rate

[Br2]= rate constant

The Rate Law; rate = 3.50 x 10-3 s-1 [Br2]

Chapter 12 Slide 13

k = 3.50 x 10-3 s-1

Chapter 12 Slide 16

The Rate Law and Reaction Order

The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.

aA + bB cC + dD

Rate = k [A]x[B]y

reaction is xth order in A

reaction is yth order in B

reaction is (x +y)th order overall

Chapter 12 Slide 17


are Experimentally Determined


are Experimentally Determined

Chapter 12 Slide 18

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2]x[ClO2]y

Double [F2] with [ClO2] constant

Rate doubles x = 1

Quadruple [ClO2] with [F2] constant

Rate quadruples y = 1

rate = k [F2][ClO2]

The instantaneous rate at the beginning of a reaction is called initial rate

----

Determine the reaction order for:

1 vs 3

1 vs 2

Chapter 12 Slide 19

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2][ClO2]

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.

1

Chapter 12 Slide 20

Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)

Experiment [S2O82-] [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 [S2O82-]x[I-]y

Double [I-], rate doubles (experiment 1 & 2)

y = 1

Double [S2O82-], rate doubles (experiment 2 & 3)

x = 1

k = rate

[S2O82-][I-]

=2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

rate = k [S2O82-][I-]

Chapter 12 Slide 21

Determine the Rate Law and Reaction OrderDetermine the Rate Law and Reaction Order

Comparing Experiments 1 and 2, when [NH4+] doubles, the initial

rate doubles.

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

�̳

Chapter 12 Slide 22

Likewise, comparing Experiments 5 and 6, when [NO2−]

doubles, the initial rate doubles.

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

�̳

Chapter 12 Slide 23

• This meansRate ∝ [NH4

+]Rate ∝ [NO2

−]Rate ∝ [NH+] [NO2

−]or

Rate = k [NH4+] [NO2

−]• This equation is called the rate law, and k is the rate constant.

Chapter 12 Slide 24

Rate LawsRate Laws

• The exponents tell the order of the reaction with respect to each reactant.

• This reaction isFirst-order in [NH4

+]First-order in [NO2

−]

ㄠҊ

Chapter 12 Slide 25

Rate Law & Reaction OrderRate Law & Reaction Order

• The reaction of nitric oxide with hydrogen at

1280°C is: 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g)

• From the following data determine the rate law and

rate constant.

Experiment [NO] [H2] Initial Rate (M/s)

1 5.0 x 10–3 2.0 x10–3 1.3 x 10–5

2 10.0 x 10–3 2.0 x 10–3 5.0 x 10–5

3 10.0 x 10–3 4.0 x 10–3 10.0 x 10–5

k = 1/3(250+250+260) = 250 M-2.s-1Second order in NO, First order in H2

ᦀ

Chapter 12 Slide 26

First-Order ReactionsConcentration and Time Equation

01

First-Order ReactionsConcentration and Time Equation

01

• First Order: Reaction rate depends on the reactant concentration raised to first power.

Rate = k[A]

Rate = -∆ A[ ]

∆t∆[A]

∆t= k [A]-

A product

Chapter 12 Slide 27

Concentration and Time Equation For A First-Order Reactions

k = rate[A]

= 1/s or s-1M/sM

=

∆[A]= K ∆t-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t = 0

[A] = [A]0exp(-kt)

[A]-(ln[A] -ln[A]0) = ktln[A] = ln[A]0 - kt

[A] = [A]0exp(-kt)

la

w

de

ca

y

lex

po

ne

nt

ia

e[A

]

[A

]

kt0t−=

ln[A]0

[A]= k t

See next slide for proof of the formula

匠Ҋ

Chapter 12 Slide 28

Integration:

�Ҕ

Chapter 12 Slide 29

First-Order Reactions


[A]0 is the concentration of A at time t = 0ln[A] = ln[A]0 - kt

ln[A] = ln[A]0 - kt

ln[A]0

[A]= k t

Chapter 12 Slide 30

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 ?

t = = 66 s

[A]0 = 0.88 M

[A] = 0.14 M

ln[A]0

[A]

k

ln0.88 M

0.14 M

2.8 x 10-2 s-1=

ln[A]0

[A]= k t

t = ?

厐Ҋ

Chapter 12 Slide 31

What is Half- Life ?

The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t½ = t when [A] = [A]0/2

ln[A]0

[A]0/2

k=t½

ln2

k=

0.693

k=ln

[A]0

[A]= k t

Half Life For the First Order Reaction

�Ҕ

Chapter 12 Slide 32

Reaction OrdersReaction Orders

Units of Rate Constants vs Reaction Orders

Zeroth Order Reaction:

Rate = K [A]0 = K

叠Ҋ

Chapter 12 Slide 33

What is the order of decomposition of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?What is the half life of decomposition of N2O5 ?

t½ln2

k=

0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

Therefore, decomposition is first order?

units of k (s-1)

2N2O5(g) →→→→ 4NO2(g) + O2(g)

䖀Ҙ

Chapter 12 Slide 34

Half life of a First Order ReactionHalf life of a First Order Reaction

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Chapter 12 Slide 35

A product

First-order reaction

# of half-lives [A]

1

2

3

4

½ [A]0

1/4 [A]0

1/8 [A]0

1/16 [A]0

[A] = [A]0 x (1/2)n

�Ҕ

Chapter 12 Slide 36

First-Order ReactionFirst-Order Reaction

• Show that the decomposition of N2O5 is first order and calculate the rate constant and Half life.

k = 1.7 x 10-3 s-1

2N2O5(g) →→→→ 4NO2(g) + O2(g)

t1/2 = 408 S

�Ҕ

Chapter 12 Slide 37

Second-Order Reactions

A product rate = -∆[A]

∆trate = k [A]2

k = rate[A]2

= 1/M•s or M-1 s-1M/sM2=

∆[A]

∆t= k [A]2-


[A]0 is the concentration of A at time t = 0

1[A]

=1

[A]0+ kt

What is Unit of K ?

What is Conc. Vs time equation?

Chapter 12 Slide 38

Second-Order Reactions

So if a process is second-order in A,

a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k.

Drive the formula for half life of a second order reaction

t = t1/2

=t1/2

[A]2

[A]0

Chapter 12 Slide 39

= kt +[A]0

1[A]t

1t = t1/2

=t1/2

[A]2

[A]0

Half-life for a second-order reaction

[A]0

1= kt1/2 +

[A]0

2=t1/2

k[A]0

1

�Ҕ

Chapter 12 Slide 40

Second-Order ReactionsSecond-Order Reactions

=t1/2k[A]0

1

For a second-order reaction, the half-life is dependent on the initial concentration.

Each successive half-life is twice as long as the preceding one.

Chapter 12 Slide 42

2 NO2(g) →→→→ 2NO(g) + O2(g)

a) Is the following reaction first or second order ?

b) What is the value of k?

Example:

議҈

Chapter 12 Slide 43

議҈

Chapter 12 Slide 44

k = 0.54 M-1 . S-1 Second-Order Reactions

Chapter 12 Slide 45

Zero Order Reaction:

Rate = k

議҈

Chapter 12 Slide 46

Reaction Mechanisms 01Reaction Mechanisms 01

• A reaction mechanism

is a sequence of

molecular events, or

reaction steps, that

defines the pathway

from reactants to

products.

議҈

Chapter 12 Slide 47


• Single steps in a mechanism are called elementary steps (reactions).

• An elementary step describes the behavior of individual molecules.

• An overall reaction describes the reaction stoichiometry.

Chapter 12 Slide 48

Reaction MechanismsNO2(g) + CO(g) → NO(g) + CO2(g)

Reaction MechanismsNO2(g) + CO(g) → NO(g) + CO2(g)

• NO2(g) + NO2(g) → NO(g) + NO3(g) Elementary

• NO3(g) + CO(g) → NO2(g) + CO2(g) Elementary

• NO2(g) + CO(g) → NO(g) + CO2(g) Overall

• The chemical equation for an elementary reaction is a description of an individual molecular event that involves the breaking and/or making of chemical bonds.

NO3(g) is called reaction intermediate.

議҈

Chapter 12 Slide 49


• Molecularity: is the number of molecules (or atoms) on the reactant side of the chemical equation.

• Unimolecular: Single reactant molecule.

議҈

Chapter 12 Slide 50


• Bimolecular: Two reactant molecules.

• Termolecular: Three reactant molecules.

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Chapter 12 Slide 51


• Determine individual steps , the reaction intermediates, and the molecularity of each individual step.

2N2O2N2 + O2

Chapter 12 Slide 52

Rate Laws and Reaction Mechanisms 01Rate Laws and Reaction Mechanisms 01

• Rate law for an overall reaction must be determined experimentally.

• Rate law for elementary step follows from its molecularity.

텐ҍ

Chapter 12 Slide 53

Rate Laws and Reaction Mechanisms 02Rate Laws and Reaction Mechanisms 02

• The rate law of each elementary step follows its molecularity.

• The overall reaction is a sequence of elementary steps called the reaction mechanism.

텐ҍ

Chapter 12 Slide 54

Rate-Determining StepRate-Determining Step

• The slowest elementary step in a multistep reaction is called the rate-determining step.

• The overall reaction cannot occur faster than the speed of the rate-determining step.

• The rate of the overall reaction is therefore determined by the rate of the rate-determining step.

텐ҍ

Chapter 12 Slide 55

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

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

텐ҍ

Determining Reaction Mechanism From The Rate LawDetermining Reaction Mechanism From The Rate Law

텐ҍ

Chapter 12 Slide 57

Activation Energy (Ea):

텐ҍ

Chapter 12 Slide 58

The Arrhenius EquationThe Arrhenius Equation

Is it hidden in k?

2N2O5(g) 4NO2(g) + O2(g)

rate = k[N2O5]Where is temperature dependence?

Typically, as the temperature increases, the rate of reaction increases.

텐ҍ

Chapter 12 Slide 59

The Arrhenius Equation 01The Arrhenius Equation 01

• Collision Theory: A bimolecular reaction occurs when two correctly oriented molecules collide with sufficient energy.

• Activation Energy (Ea): The potential energy barrier that must be surmounted before reactants can be converted to products.

텐ҍ

Chapter 12 Slide 60

Temperature dependence of Rate

Constat

Temperature dependence of Rate

Constat

• Adding more of the reactants speeds up a reaction by increasing the number of collisions that occur.

Collision rate = Z [A][B]

• The fraction of collisions with an energy equal or more than activation energy( Ea) :

f = e-Ea/RT

• Raising the temperature speeds up a reaction by providing the energy of activation to more colliding molecules.

Z is a constant related to collision frequency

텐ҍ

Chapter 12 Slide 61

• Only the fraction of collisions having proper orientation can result to products.

The Arrhenius Equation 02The Arrhenius Equation 02

This is called steric factor, p., In the above example p = 0.5

텐ҍ

Chapter 12 Slide 62

Collision rate = Z [A][B] Where Z is a constant, related to the collision frequency .

Reaction rate = p.f.Z [A][B]

Reaction rate = k [A][B] k = p.f.Z,

Assume p.Z = A, frequency factor

A = frequency factork = A. f,f = e-Ea/RT

k = Ae-Ea/RT (p.Z )= A


텐ҍ

Chapter 12 Slide 63

K = Ae-Ea/RT

A = pZ

(steric factor)


텐ҍ

Chapter 12 Slide 64

k = A .e-Ea/RT

k = A .e(-Ea/R)(1/T)

Calculating Activation Energy

Ea = -R . (slope)

Ln k

1/T

議҈

Chapter 12 Slide 65

2HI(g) + H2(g) →→→→ I2(g) + H2(g)

Find the activation energy for the following reaction

Calculating Activation EnergyCalculating Activation Energy

텐ҍ

Chapter 12 Slide 67

Ea = - (8.314 j/K.mol) (-2.24 x 10 4 K)

Ea = 190 kj/mol

Slope = -2.24 x 10 4 K

Ea = -R . (slope)

텐ҍ

Chapter 12 Slide 68

Effect of Temperature on Fraction of Collisions with Activation energyEffect of Temperature on Fraction of Collisions with Activation energy

f = e-Ea/RT

텐ҍ

Effect of TemperatureEffect of Temperature

Collision Theory: As the average kinetic energy increases, the average molecular speed increases, and thus the collision rate increases.

f = e-Ea/RT

텐ҍ

Chapter 12 Slide 70

Change of Rate Constant with temperatureChange of Rate Constant with temperature

• If the Ea is known , we can calculate the Rate

Constant when temperature is changed:

ln

k2k1

= −Ea

R

1T2

−1

T1

The above formula could be used to

determine the rate constant at a different temperature.

Chapter 12 Slide 71

Homework:Determination the Activation Energy

Homework:Determination the Activation Energy

The second-order rate constant for the decomposition of nitrous oxide (N2O) into nitrogen molecule and oxygen atom has been measured at different temperatures:

Determine (graphically)the activation energyfor the reaction.

k (M -1s-1) t (°C)

1.87x10-3 6000.0113 6500.0569 7000.244 750

蜀҆

Chapter 12 Slide 72

Catalysis 01Catalysis 01

蜀҆

Chapter 12 Slide 73

• A catalyst is a substance that increases the rate of a reaction without being consumed in the reaction.


蜀҆

CatalysisCatalysis

Note that the presence of a catalyst does not affect the energy difference between the reactants and the products

蜀҆

Chapter 12 Slide 75


• The relative rates of the reaction A + B → AB in vessels a–d are 1:2:1:2. Red = A, blue = B, green = third substance C.

(a) What is the order of reaction in A, B, and C?(b) Write the rate law.(c) Write a mechanism that agrees with the rate law.(d) Why doesn’t C appear in the overall reaction?

1 12 2

蜀҆

Chapter 12 Slide 76

CatalysisCatalysis

Catalyst: A substance that increases the rate of a reaction without itself being consumed in the reaction. A catalyst is used in one step and regenerated in a later step.

H2O2(aq) + I1-(aq) H2O(l) + IO1-(aq)

H2O2(aq) + IO1-(aq) H2O(l) + O2(g) + I1-(aq)

2H2O2(aq) 2H2O(l) + O2(g) overall reaction

rate-determiningstep

fast step

蜀҆

Chapter 12 Slide 77


• Homogeneous Catalyst: Exists in the same phase as the reactants.

• Heterogeneous Catalyst: Exists in different phase to the reactants.

蜀҆

Chapter 12 Slide 78


• Catalytic Hydrogenation:

蜀҆

Chapter 12 Slide 79

H H C CA

B

X

YH H

Mechanism of Catalytic Hydrogenation:Mechanism of Catalytic

Hydrogenation:

蜀҆

Chapter 12 Slide 80

H

C CA

B

X

Y

H H H

Mechanism of Catalytic Hydrogenation:Mechanism of Catalytic Hydrogenation:

蜀҆

Chapter 12 Slide 81

H

H H HCC

AB

XY


蜀҆

Chapter 12 Slide 82

H

H H HCC

AB

XY


蜀҆

Chapter 12 Slide 83

H H

H

CC

AB

XY

H


蜀҆

Chapter 12 Slide 84

H HH

CC

AB

XY

H


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