#### Transcript Chemical Kinetics - Dr. Parvin Carter

**Chemical Kinetics**

# Chapter 12

### Reaction 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 A B Products Rate [

*t A*

] 2 HI (

*g*

) Rate H 2 (

*g*

) + I Chapter 12 2 (

*g*

) [

*B*

*t*

] Slide 2

### Reaction Rates and Stoichiometry

• What is the general rate of the following reaction ?

2 HI (

*g*

) H 2 (

*g*

) + I 2 (

*g*

) Rate = − 1 2 [HI]

*t*

= [I 2 ]

*t*

Chapter 12 Slide 3

### Reaction Rates and Stoichiometry

• To generalize, for the reaction

*a*

A +

*b*

B

*c*

C +

*d*

D Rate* = − 1

*a*

[A]

*t*

= − 1

*b*

[B]

*t*

= 1

*c*

[C]

*t*

= 1

*d*

[D]

*t*

***: General rate of reaction**

Chapter 12 Slide 4

HCO H(aq) + Br (aq) 2 2 + 2H (aq) + 2Br (aq) + CO (g) 2 red colorless Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?

1. d[CO ] 2 = dt d[Br ] 2 dt 2. dt = + d[H ] dt 3. d[HCO H] 2 = dt d[Br ] 2 dt 4. dt 5. d[Br ] = dt + 1 d[H ] = 2 dt d[Br ] dt 2

HCO H(aq) + Br (aq) 2 2 + 2H (aq) + 2Br (aq) + CO (g) 2 red colorless Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?

1. d[CO ] 2 = dt d[Br ] 2 dt 2. dt = + d[H ] dt 3. d[HCO H] 2 = dt d[Br ] 2 dt 4 . d [ H C O H ] 2 d t 5. d[Br ] = dt = 2 + 1 d [ H ] d t d[Br ] 2 dt

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

• Consider the decomposition of N 2 O 5 and O 2 : 2 N 2 O 5

*(g)*

4 NO 2

*(g)*

Brown to give NO 2 + O 2

*(g)*

Colorless Chapter 12 Slide 7

• 2 N 2 O 5

*(g)*

4 NO 2

*(g)*

+ O 2

*(g)*

Reaction Rates:

**concentration versus time curve**

### 03

Average Rate = Rate between two points in time

**The slope of each triangle Between two points**

Chapter 12 Slide 8

### Reaction Rates

2N 2 O 5 (

*g*

) 4NO 2 (

*g*

) + O 2 (

*g*

)

### Instantaneous rate:

**Rate for specific instance in time **

Slope of the tangent to a concentration versus time curve Initial Rate Chapter 12 Slide 10

Br 2 (

*aq*

) + HCOOH (

*aq*

) 2Br (

*aq*

) + 2H + (

*aq*

) + CO 2 (

*g*

)

*time*

393 nm light Detector [Br 2 ] a Absorption Chapter 12 Br 2 (

*aq*

) Slide 11

Br 2 (

*aq*

) + HCOOH (

*aq*

) 2Br (

*aq*

) + 2H + (

*aq*

) + CO 2 (

*g*

) slope of tangent slope of tangent slope of tangent

*average rate*

= [Br 2 ]

*t*

= [Br 2 ] final

*t*

final – [Br -

*t*

2 initial ] initial

*instantaneous rate*

= rate for specific instance in time

**The slope of a line tangent to the curve at any point is the instantaneous rate at that time**

Chapter 12 Slide 12

slope of tangent slope of tangent Chapter 12 Slide 13

rate a [Br 2 ] rate =

*k*

[Br 2 ]

*k*

rate = [Br 2 ] =

*rate constant*

The Rate Law; rate = 3.50 x 10 -3 s -1 [Br 2 ] Chapter 12 Slide 14

k = 3.50 x 10 -3 s -1 Chapter 12 Slide 13

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.

*a*

A +

*b*

B

*c*

C +

*d*

D Rate =

*k*

[A]

*x*

[B]

*y*

reaction is

*x*th order

in A reaction is

*y*th order

in B reaction is

**(x +y)th order overall**

Chapter 12 Slide 16

**The Rate Law and Reaction Order are Experimentally Determined**

Chapter 12 Slide 17

**Determine the reaction order for:**

F 2 (

*g*

) + 2ClO 2 (

*g*

) 2FClO 2 (

*g*

)

** ---**

rate =

*k*

[F 2 ]

*x*

[ClO 2 ]

*y*

**1 vs 3**

Double [F 2 ] with [ClO 2 ] constant Rate doubles

*x*

= 1

**1 vs 2**

Quadruple [ClO 2 ] with [F 2 ] constant rate =

*k*

[F 2 ][ClO 2 ] Rate quadruples

*y*

= 1

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

Chapter 12 Slide 18

**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.

F 2 (

*g*

) + 2ClO 2 (

*g*

) 2FClO 2 (

*g*

) rate =

*k*

[F 2 ][ClO 2 ] 1 Chapter 12 Slide 19

Experiment 1 2 3 Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2 (

*aq*

) + 3I (

*aq*

) 2SO 4 2 (

*aq*

) + I 3 (

*aq*

) [S 2 O 8 2 ] 0.08

0.08

0.16

[I ] 0.034

0.017

0.017

Initial Rate (

*M*

/s) 2.2 x 10 -4 1.1 x 10 -4 2.2 x 10 -4 rate =

*k y = 1 x = 1*

rate =

*k*

[S 2 O 8 2 ]

*x*

[I ]

*y*

[S 2 O 8 2 ][I ] Double [I ], rate doubles (experiment 1 & 2) Double [S 2 O 8 2 ], rate doubles (experiment 2 & 3)

*k*

rate = [S 2 O 8 2 ][I ] 2.2 x 10 -4

*M*

/s = (0.08

*M*

)(0.034

*M*

) = 0.08/

*M*

• s Chapter 12 Slide 20

### Determine the Rate Law and Reaction Order

NH 4 + (

*aq*

) + NO 2 − (

*aq*

) N 2 (

*g*

) + 2 H 2 O (

*l*

) Comparing Experiments 1 and 2, when [NH 4 + ] doubles, the initial rate doubles.

Chapter 12 Slide 21

NH 4 + (

*aq*

) + NO 2 − (

*aq*

) N 2 (

*g*

) + 2 H 2 O (

*l*

) Likewise, comparing Experiments 5 and 6, when [NO 2 doubles, the initial rate doubles.

− ] Chapter 12 Slide 22

• • This means Rate Rate Rate [NH 4 + ] [NO 2 − ] [NH + ] [NO 2 − ] Rate =

*k*

or [NH 4 + ] [NO 2 − ] This equation is called the rate law , and

*k*

the rate constant .

is Chapter 12 Slide 23

### Rate Laws

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

of the reaction • This reaction is First-order in [NH 4 + ] First-order in [NO 2 − ] Chapter 12 Slide 24

### Rate Law & Reaction Order

• The reaction of nitric oxide with hydrogen at 1280 °C is: 2 NO (g) + 2 H 2

*(g)*

N 2

*(g)*

+ 2 H 2 O

*(g)*

• From the following data determine the rate law and rate constant.

Experiment 1 2 3 [NO] 5.0 x 10 –3 10.0 x 10 –3 10.0 x 10 –3 [H 2 ] 2.0 x10 –3 2.0 x 10 –3 4.0 x 10 –3 Initi al Rate (M/s) 1.3 x 10 –5 5.0 x 10 –5 10.0 x 10 –5 Second order in NO, First order in H 2 k = 1/3(250+250+260) = 250 M Chapter 12 -2 .s

-1 Slide 25

### First-Order Reactions

Concentration and Time Equation

### 01

•

**First Order: **

Reaction rate depends on the reactant concentration raised to first power.

Rate

*= k*

[A] A product

## Rate = -

## A t

[A]

*t*

=

*k*

[A] Chapter 12 Slide 26

[A] [A] ln [A] 0 [A] Concentration and Time Equation For A First-Order Reactions = K

**Δt**

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

*kt*

ln[A] = ln[A] 0 -

*kt*

See next slide for proof of the formula

**= k t**

[A] = [A] 0 exp(

*-kt*

)

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

*k*

= rate [A] =

*M*

/

*s M*

= 1/s or s -1

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

[A] = [A] 0 exp(

*-kt*

) [A] t [A] 0 e kt exponentia l decay law Chapter 12 Slide 27

Integration: Chapter 12 Slide 28

ln [A] 0

**= k t**

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

*kt*

First-Order Reactions ln[A] = ln[A] 0 -

*kt*

[A] is the concentration of A at any time

*t*

[A] 0 is the concentration of A at time

*t *

= 0 Chapter 12 Slide 29

The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88

*M*

to 0.14

*M *

?

*t*

= ln [A] 0 [A] ln [A] 0 [A]

*k*

**= k t**

[A] 0 = 0.88

*M*

[A] = 0.14

*M*

**t = ?**

= 0.88

*M*

ln 0.14

*M 2.8 x 10 -2 s -1*

= 66 s Chapter 12 Slide 30

What is Half- Life ?

The

*half-life*, *t* ½ ,

is the time required for the concentration of a reactant to decrease to half of its initial concentration.

*t ½*

=

*t*

when [A] = [A] 0 /2 ln [A] 0 [A]

**= k t**

*t*

½ Half Life For the First Order Reaction [A] 0 ln = [A] 0 /2

*k*

= ln2

*k*

= 0.693

*k*

Chapter 12 Slide 31

### Reaction Orders

Zeroth Order Reaction: Rate = K [A] 0 = K

### Units of Rate Constants vs Reaction Orders

Chapter 12 Slide 32

What is the order of decomposition of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ?

What is the half life of decomposition of N 2 O 5 ?

**2N 2 O 5 (**

*g*

**) **

**4NO 2 (**

*g*

**) + O 2 (**

*g*

**)**

units of

*k*

(s -1 ) Therefore, decomposition is first order?

*t*

½ = ln2

*k*

= 0.693

5.7 x 10 -4 s -1 = 1200 s = 20 minutes Chapter 12 Slide 33

### Half life of a First Order Reaction

Chapter 12 Slide 34

Chapter 12 First-order reaction A product # of half-lives 1 2 3 4 [A] ½ [A] 0 1/4 [A] 0 1/8 [A] 0 1/16 [A] 0 [A] = [A] 0 x (1/2)

**n**

Slide 35

### First-Order Reaction

**2N 2 O 5 (**

*g*

**) **

**4NO 2 (**

*g*

**) + O 2 (**

*g*

**)**

• Show that the decomposition of N 2 O 5 is first order and calculate the rate constant and Half life.

Chapter 12

**k = 1.7 x 10 -3 s -1**

t 1/2 = 408 S Slide 36

Second-Order Reactions A product rate = [A]

*t*

rate =

*k*

[A] 2 What is Unit of k ?

*k*

= rate [A] 2 =

*M*

/

*s M 2*

= 1/

*M*

•

*s or M -1 s -1*

What is Conc. Vs time equation?

[A]

*t*

=

*k*

[A] 2 1 [A] = 1 [A] 0 +

*kt*

[A] is the concentration of A at any time

*t*

[A] 0 is the concentration of A at time

*t *

= 0 Chapter 12 Slide 37

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*

=

*t*

**1/2**

[A]

*t*

**1/2**

= [A]

**0**

2 Chapter 12 Slide 38

Half-life for a second-order reaction 1 [A]

*t*

=

*kt*

+ 1 [A]

**0**

2 [A]

**0**

=

*kt*

**1/2**

+ 1 [A]

**0**

Chapter 12

*t*

=

*t*

**1/2**

[A]

*t*

**1/2**

= [A]

**0**

2 1

*t*

**1/2**

=

*k*

[A]

**0**

Slide 39

### Second-Order Reactions

1

*t*

**1/2**

=

*k*

[A]

**0**

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 40

**Example: a) Is the following reaction first or second order ?**

**b) What is the value of k?**

**2 NO 2 (g) **

**2NO**

*(g)*

**+ O 2**

*(g)*

Chapter 12 Slide 42

Chapter 12 Slide 43

**k = 0.54 M -1 . S -1**

Second-Order Reactions Chapter 12 Slide 44

**Zero Order Reaction: Rate = k**

Example of Zeroth Order Reaction: Decomposition of N 2 O on hot platinum surface: N 2 O → N 2 Rate [N d[N 2 2 O] 0 + 1/2 O = k[N O]/dt = k 2 2 O] 0 = k Chapter 12 Slide 45

### Reaction Mechanisms

• A

**reaction mechanism**

is a sequence of molecular events, or reaction steps, that defines the pathway from reactants to products.

Chapter 12

### 01

Slide 46

### Reaction Mechanisms 02

• 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 47

### Reaction Mechanisms NO

2

### (

*g*

### ) + CO(

*g*

### )

### NO(

*g*

### ) + CO

2

### (

*g*

### )

• NO 2 (

*g*

) + NO 2 (

*g*

) NO(

*g*

) + NO 3 (

*g*

) • NO 3 (

*g*

) + CO(

*g*

) NO 2 (

*g*

) + CO 2 (

*g*

) • NO 2 (

*g*

) + CO(

*g*

) NO(

*g*

) + CO 2 (

*g*

) Elementary Elementary 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.

NO 3 (

*g*

) is called reaction intermediate.

Chapter 12 Slide 48

### Reaction Mechanisms

•

**Molecularity: **

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

•

**Unimolecular: **

Single reactant molecule.

### 04

Chapter 12 Slide 49

### Reaction Mechanisms

•

**Bimolecular: **

Two reactant molecules.

### 05

•

**Termolecular: **

Three reactant molecules.

Chapter 12 Slide 50

### Reaction Mechanisms

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

**2N 2 O 2N 2 + O 2**

### 06

Chapter 12 Slide 51

### Rate Laws and Reaction Mechanisms

• Rate law for an

*overall reaction*

must be determined experimentally.

• Rate law for

*elementary step*

follows from its molecularity.

### 01

Chapter 12 Slide 52

### Rate Laws and Reaction Mechanisms

• The

*rate law*

of each

*elementary step*

follows its molecularity.

• The

*overall reaction*

is a sequence of elementary steps called the

*reaction mechanism*

.

### 02

Chapter 12 Slide 53

*Rate-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*

.

Chapter 12 Slide 54

The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate =

*k*

[NO 2 ] 2 . The reaction is believed to occur via two steps: Step 1: Step 2: NO 2 + NO 2 NO + NO 3 NO 3 + CO NO 2 + CO 2 What is the equation for the overall reaction?

NO 2 + CO NO + CO 2 What is the intermediate?

NO 3 What can you say about the relative rates of steps 1 and 2?

rate =

*k*

[NO 2 ] 2 is the rate law for step 1 so step 1 must be slower than step 2 Chapter 12 Slide 55

Determining Reaction Mechanism From The Rate Law

**Activation Energy (**

*E*

**a):**

Chapter 12 Slide 57

### The Arrhenius Equation

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

2N 2 O 5 (

*g*

) 4NO 2 (

*g*

) + O 2 (

*g*

) rate =

*k*

[N 2 O 5 ] Where is temperature dependence?

Is it hidden in k?

Chapter 12 Slide 58

### The Arrhenius Equation 01

•

**Collision Theory:**

A bimolecular reaction occurs when two correctly oriented molecules collide with sufficient energy.

•

**Activation Energy ( E a ):**

The potential energy barrier that must be surmounted before reactants can be converted to products.

Chapter 12 Slide 59

**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] **

Z is a constant related to collision frequency

**The fraction of collisions with an energy equal or more than activation energy( E a ) : f = e -Ea/RT**

•

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

Chapter 12 Slide 60

### The Arrhenius Equation

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

### 02

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

Chapter 12 Slide 61

### The Arrhenius Equation

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 factor k = A.

**f**

,

**f = e -Ea/RT**

### k = Ae

-Ea/RT

### ( p.Z )= A

Chapter 12 Slide 62

### The Arrhenius Equation

K = Ae -Ea/RT A = pZ (steric factor) Chapter 12 Slide 63

### Calculating Activation Energy

k = A .e

-Ea/RT k = A .e

(-Ea/R)(1/T)

**E**

**a**

**= -R . (slope)**

Chapter 12 Ln k 1/T Slide 64

**Find the activation energy for the following reaction 2HI(**

*g*

**) + H 2 (**

*g*

**) **

**I 2 (**

*g*

**) + H 2 (**

*g*

**)**

Chapter 12 Slide 65

### Calculating Activation Energy

**Slope = -2.24 x 10 4 K E a = -R . (slope) E a = - (8.314 j/K.mol) (-2.24 x 10 4 K) E a = 190 kj/mol**

Chapter 12 Slide 67

Effect of Temperature on Fraction of Collisions with Activation energy

**f = e**

**-Ea/RT**

Chapter 12 Slide 68

### Effect of Temperature

**f = e**

**-Ea/RT Collision Theory**

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

### Change of Rate Constant with temperature

• If the

**E a**

is known , we can calculate the Rate Constant when temperature is changed: ln

*k*

2

*k*

1

*E a R*

1

*T*

2 1

*T*

1 The above formula could be used to determine the rate constant at a different temperature.

Chapter 12 Slide 70

### Homework: Determination the Activation Energy

The second-order rate constant for the decomposition of nitrous oxide (N 2 O) into nitrogen molecule and oxygen atom has been measured at different temperatures: k (M -1 s -1 ) t ( °C) Determine (graphically) the activation energy 1.87x10

-3 600 for the reaction.

0.0113

650 0.0569

700 Ea = 241 KJ/mole 0.244

750 Chapter 12 Slide 71

### Catalysis 01

Chapter 12 Slide 72

### Catalysis 01

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

Chapter 12 Slide 73

### Catalysis

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

### Catalysis 02

• 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 2 Chapter 12 1 2 Slide 75

### Catalysis

**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.

H 2 O 2 (

*aq*

) + I 1 (

*aq*

) H 2 O 2 (

*aq*

) + I O 1 (

*aq*

) 2H 2 O 2 (

*aq*

) H 2 O(

*l*

) + I O 1 (

*aq*

) rate-determining step H 2 O(

*l*

) + O 2 (

*g*

) + I 1 (

*aq*

) fast step 2H 2 O(

*l*

) + O 2 (

*g*

) overall reaction Chapter 12 Slide 76

### Catalysis 03

•

**Homogeneous Catalyst: **

Exists in the same phase as the reactants. •

**Heterogeneous Catalyst:**

to the reactants.

Exists in different phase Chapter 12 Slide 77

### Catalysis

•

**Catalytic Hydrogenation :**

### 04

Chapter 12 Slide 78

### Mechanism of Catalytic Hydrogenation:

B

*H H*

C A C Y

*H H*

X Chapter 12 Slide 79

### Mechanism of Catalytic Hydrogenation:

B C A

*H H*

C Y X

*H H*

Chapter 12 Slide 80

### Mechanism of Catalytic Hydrogenation:

*H H*

A B C C Y X

*H H*

Chapter 12 Slide 81

### Mechanism of Catalytic Hydrogenation:

*H H*

A B C C Y X

*H H*

Chapter 12 Slide 82

### Mechanism of Catalytic Hydrogenation:

*H*

A B C

*H*

C Y X

*H H*

Chapter 12 Slide 83

### Mechanism of Catalytic Hydrogenation:

B A C C Y

*H H H*

X

*H*

Chapter 12 Slide 84