Chapter 12 Chemical Kinetics - An-Najah National University · Section 12.7 Catalysis Copyright ©2017 Cengage Learning. All Rights Reserved. Title: PowerPoint Presentation Author:

Chapter 12

Chemical Kinetics

Copyright ©2017 Cengage Learning. All Rights Reserved.

Chapter 12Table of Contents


(12.1) Reaction rates

(12.2) Rate laws: An introduction

(12.3) Determining the form of the rate law

(12.4) The integrated rate law

(12.5) Reaction mechanisms

(12.6) A model for chemical kinetics

(12.7) Catalysis

Section 12.1Reaction Rates

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2 22NO ( ) 2NO( ) + O ( )g g g

Reaction Rate Change in concentration of a

reactant or product per unit time

[A] = Concentration of the reactant or product in mol/L

Δ = Change in a given quantity

2 1

2 1

[A] at time – [A] at time Rate =

–

[A] =

t t

t t

t



Since the concentration of NO2 decreases with time, rate is a negative quantity

2[NO ]Rate = –

t

Since the concentration of NO increases with time, rate is a positive quantity

[NO]Rate =

t

Instantaneous Rate and Initial rate



Determining the Rate of Reaction in Terms of Products

Rate of consumption of NO2 = Rate of production of NO = 2(rate of production of O2)

2 2[NO ] [O ][NO] = = 2

t t t

[A] Rate of rxn =

a t

1



Section 12.2Rate Laws: An Introduction


Rate Law

k = rate constant

x and y are the orders of the reaction

The concentrations of the products do not appear in the rate law

The value of the exponent x, and y must be determined by experiment

aA bB cC dD

x yRate=k[A] [B]

Section 12.2Rate Laws: An Introduction


Types of Rate Laws

Differential rate law (rate law)

Integrated rate law

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Section 12.3Determining the Form of the Rate Law


Method of Initial Rates


Section 12.3Determining the Form of the Rate Law


Example 12.1– – +

3 2 2BrO ( ) + 5Br ( ) + 6H ( ) 3Br ( ) + 3H O( )aq aq aq l l

Section 12.4The Integrated Rate Law


Integrated Rate Law

11

Zero Order Reactions

𝐴 − 𝐴0 = −𝑘𝑟𝑡



Integrated Rate Law

12

1st Order Reactions

𝑙𝑛𝐴 − 𝑙𝑛𝐴0 = −𝑘𝑟𝑡



Integrated Rate Law

13

2nd Order Reactions

1/𝐴 − 1/𝐴0 = 𝑘𝑟𝑡



Summary of the Kinetics for Reactions of Type aA → Products



Calculate [N2O5] at 150 s after the start of the following reaction:

Use the following information, verify that the rate law is first order in [N2O5], and calculate the value of the rate constant

Example 12.3

2 5 2 22N O ( ) 4NO ( ) + O ( ) g g g



Using the data given in Example 12.2, calculate [N2O5] at

150 s after the start of the reaction.



Example 12.4

A certain first-order reaction has a half-life of 20.0 minutes

a. Calculate the rate constant for this reaction

b. How much time is required for this reaction to be 75% complete?



Section 12.5Reaction Mechanisms


Reaction Mechanism Most chemical reactions occur by a series of steps

Example - The reaction between nitrogen dioxide and carbon monoxide involves the following steps:

Where k1 and k2 are the rate constants of the individual reactions

OBSERVED RATE LAW


2 2 3

3 2 2

2

NO (g) + NO ( ) NO ( ) + NO( )

NO ( ) + CO( ) NO ( ) + CO ( )

--------------------------------------------------------

NO ( ) + CO( ) NO( ) + CO ( )

k

k

g g g

g g g g

g g g g

1

2

2

2

2Rate=k[NO ]



Reaction Mechanism

In the reaction, NO3 is an intermediate

Each of the reactions is called an elementary step

Elementary step: A reaction whose rate law can be written from its molecularity


2 2 3

3 2 2

NO (g) + NO ( ) NO ( ) + NO( )

NO ( ) + CO( ) NO ( ) + CO ( )

k

k

g g g

g g g g

1

2



Molecularity Number of species that must collide to produce the reaction

represented by an elementary step

Unimolecular: Reaction that involves one molecule

Bimolecular: Reaction that involves the collision of two species

Termolecular: Reaction that involves the collision of three species

Requirements of a Reaction Mechanism

The sum of the elementary steps must give the overall balanced equation for the reaction

The mechanism must agree with the experimentally determined rate law



Table 12.7 - Examples of Elementary Steps



Example 12.6

The balanced equation for the reaction of the gases nitrogen dioxide and fluorine is

Experimentally determined rate law is

Suggested mechanism for this reaction is

Is this an acceptable mechanism?

2 2 22NO ( ) + F (g) 2NO F( )g g

2 2Rate = [NO ] [F ]k

1

2

2 2 2

2 2

NO + F NO F + F (slow)

F + NO NO F (fast)

k

k

Section 12.6A Model for Chemical Kinetics


Collision Model (Transition State Theory(

States that a reaction requires the collision of molecules

Activation energy: Threshold energy that must be overcome to produce a chemical reaction

24

22BrNO( ) 2NO( ) + Br ( )g g g



Relation between Activation Energy and Number of Collisions

Ea - Activation energy R - Universal gas constant

T - Temperature in Kelvin

e–Ea/RT - Fraction of collisions with energy Ea or greater at T


a– /

Number of collisions with the activation energy

= (total number of collisions) E RT

e

Requirements to be Satisfied for Reactants to Collide Successfully

The collision energy must equal or exceed the activation energy

The relative orientation of the reactants must allow the formation of any new bonds necessary to produce products



Rate Constant in Molecular Collision

Represented as k = zpe–Ea/RT

z - Collision frequency

p - Steric factor

Always lesser than 1

Reflects the fraction of collisions with effective orientations

e–Ea/RT - Fraction of collisions with sufficient energy to produce a reaction



Arrhenius Equation

A modification of the rate constant equation for collision of molecules

A - Frequency factor for the reaction

Taking the natural logarithm of each side gives

a– / =

E RTk Ae

a 1ln( ) = – + ln( )

Ek A

R T



Example 12.7 The following reaction was studied at several temperatures

Values of k were obtained

Calculate the value of Ea for this reaction

2 5 2 22N O ( ) 4NO ( ) + O ( )g g g





Rate constants at different temperatures

a1

1

ln ( ) = – + ln( )E

k ART

a2

2

ln ( ) = – + ln( )E

k ART

a2

1 1 2

1 1 ln =

Ek–

k R T T



Example 12. The gas-phase reaction between methane and diatomic sulfur is

given by the equation

At 550°C the rate constant for this reaction is 1.1 L/mol • s

At 625°C the rate constant is 6.4 L/mol • s

Using these values, calculate Ea for this reaction

4 2 2 2CH ( ) + 2S ( ) CS ( ) + 2H S ( ) g g g g

Section 12.7Catalysis


Catalyst

A substance that speeds up a reaction without being consumed itself

Provides a new pathway with a lower activation energy for the reaction


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Chapter 12 Chemical Kinetics - An-Najah National University · Section 12.7 Catalysis Copyright ©2017 Cengage Learning. All Rights Reserved. Title: PowerPoint Presentation Author: