# 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

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

xth order

in A reaction is

yth 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

   

## 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

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

2

g

g

g

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

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

-Ea/RT

### ( p.Z )= A

Chapter 12 Slide 62

### The Arrhenius Equation

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

k = A .e

-Ea/RT k = A .e

(-Ea/R)(1/T)

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

## 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