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Chapter 11

Chemical Kinetics

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Chemical Kinetics is the study of the rates and mechanisms of reactions.

Reaction Rates are determined through the changes in concentration of the reactants and the products over time

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Factors that Influence the Reaction Rate:

1. Concentration: The higher the

concentration of the reactants, the more

collisions that occur.

2. Physical State: The reactants have to be

able to mix in order for them to collide.

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Consider the following phases and discuss how

each may influence the collisions of reactants:

Aqueous Gaseous Liquid Solid

Consider the various physical states of sugar

Granular Powdered Rock candy

Which of these states has the greatest surface area

and how might this affect the collisions of

reactants?

Example: Top photo is a hot nail in oxygen gas.

Bottom photo is hot steel wool in oxygen gas.

Why does the steel wool burn so much more

rapidly than the nail?

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3. Temperature: Reactant molecules must

collide AND collide with enough energy to

react.

The higher the temperature, the higher the kinetic

energy.

The higher the kinetic energy, the more collisions.

The higher the kinetic energy, the more collisions

with sufficient energy to react.

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Mathematical Expressions of Reaction Rates

Consider the reaction: 2 N2O5(g) � 4 NO2(g) + O2(g)

Average Rate of Decomposition: Rate = - ∆ [N2O5]

∆ t

Where ∆t = tf – ti and

∆[N2O5] = [N2O5] f - [N2O5] i

Average Rate of O2 Formation: Rate = ∆ [O2]

∆ t

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Note: - ∆ [N2O5] does not equal ∆ [O2]

∆ t ∆ t

WHY?

So…. - ∆ [N2O5] equals ∆ [O2]

2 ∆ t ∆ t

and

- ∆ [N2O5] equals ∆ [NO2]

2 ∆ t 4 ∆ t

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General Equation:

a A + b B � c C + d D

R = -∆[A] = -∆[B] = ∆[C] = ∆[D]

a ∆t b ∆t c ∆t d ∆t

Rates of reactions are expressed in the following units: M/s or mol/L-s

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Example: Express the rate of the reaction below in terms of the changes in concentrations with respect to (wrt) time.

4NH3(g) + 7O2(g) � 4NO2(g) + 6H2O(g)

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Reaction Rates Do Not Remain

Constant Over Time

Why?

[ ]’s change, thus the number of

collisions change

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Rates can be determined in 3 ways:

1. Average Rate

2. Instantaneous Rate

3. Initial Rate

C2H4 + O3 � C2H4O + O2

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Note the slope of the previous graph is the rate of the reaction wrt the disappearance of O3: rise/run or -∆[O3]/∆t

Example Problem: Calculate the average rate for the reaction below wrt [N2O5]

2 N2O5 � 4 NO2 + O2

Time [N2O5]

600 s 1.24 X 10-2 M

1200 s 0.93 X 10-2 M

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Methods for Determining Rates of Chemical Reactions

Spectrometric

Gas Chromatography

pH

Titration

Pressure Changes

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Rate Law Expressions

General Equation: a A + b B � c C + d D

Rate = k [A]m [B]n

Where k is the rate constant and is temp dependent

m and n are the reaction orders wrt each

reactant.

Note: Rate constants and reaction orders are

experimentally determined

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Example: Consider the reaction below and determine the orders of the reaction from the data.

A + B + C � D

The rate doubles when [A] is doubled while holding [B] and [C] constant.

The rate quadruples when [B] doubles while holding [A] and [C] constant.

The rate does not change when [C] triples while holding [A] and [B] constant.

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Important: The reaction coefficients do not always agree with the order of the reaction wrt the particular reactant.

Here’s another one: Write the rate law expression for the reaction below. The rate doubles when [NO2] doubles while [F2] remains constant. The rate also doubles when [F2] is doubled when [NO2] is held constant.

2 NO2 + F2 � 2 NO2F

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Consider the data above for the reaction below:

O2 + 2 NO � 2 NO2

Write the rate law expression and determine the value of k for run # 1.

Another Example: Determine the Rate Law Expression.

H+

2 Fe2+ + Cl2 � 2 Fe3+ + 2 Cl-

Run [Fe2+] [Cl2] [H+] Rate (M/s)

1 0.0020M 0.0020M 1.0 M 1.0 X 10-5

2 0.0040M 0.0020M 1.0 M 2.0 X 10-5

3 0.0020M 0.0040M 1.0 M 2.0 X 10-5

4 0.0020M 0.0020M 0.5 M 2.0 X 10-5

Additional Problems in Lab Manual: Part A of Problem Solving with Rates of

Reactions

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Integrated Rate Laws

Changes in concentration of reactant(s) over time.

We will consider three types of reaction rates:

First, Second, and Zero Order wrt ONE reactant.

First, Second, and Zero Orders of Reaction

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Can you write equations for each of the examples in the previous slide?

These equations simplified, I’ll let you do the math:

First Order: ln [A]o/[A]t = kt

Second Order: 1/[A]t – 1/[A]0 = kt

Zero Order: [A]t – [A]0 = -kt

The decomposition of N2O5. What order of reaction is

this decomposition?

See next slide for assistance.

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Lab Manual: Problem Solving with Rates of Reactions Part B

First Order Reaction. Cyclopropane is used as an anesthetic. Its isomerization to propene is a first order reaction. It has a rate constant of 9.2 /s at 1,000oC. If 6.00 M cyclopropane is used initially, what would be its concentration after just 1.0 second?

First Order: ln ([6.00 M]0/[A]t) = (9.2/s)(1.0s)

[A] = 6.1 X 10-4 M

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Half-Life Reactions

Half-Life: The amount of time it takes for the reactant concentration to reach one half of its initial value.

0.0000

0.0100

0.0200

0.0300

0.0400

0.0500

0.0600

0 24 48 72Time (min)

[N2O

5]

0.0000

0.0100

0.0200

0.0300

0.0400

0.0500

0.0600

0 24 48 72Time (min)

[N2O

5]

t0

0.0000

0.0100

0.0200

0.0300

0.0400

0.0500

0.0600

0 24 48 72Time (min)

[N2O

5]

After t1/2

t1/2

t0

0.0000

0.0100

0.0200

0.0300

0.0400

0.0500

0.0600

0 24 48 72Time (min)

[N2O

5]

After 2t1/2

After t1/2

t1/2 t1/2

t0

0.0000

0.0100

0.0200

0.0300

0.0400

0.0500

0.0600

0 24 48 72Time (min)

[N2O

5]

After 2t1/2

Time (min)

After t1/2

After 3t1/2

t1/2 t1/2 t1/2

t0

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First Order Half Life

Recall: ln ([A]0/[A]t) = kt

ln ([A]0/1/2[A]0) = kt1/2

ln 2 = kt1/2

Lab Manual Problem II. Half life using First Order Rxn.

The following reaction is a first order reaction wrt

nitrite ion and its rate constant is 0.0030 /s. In a

solution that is 10.0 M NO2- , what is the half-life of the

reaction?

NH4+ + NO2

- � N2 + 2 H2O

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Oftentimes we need to know how much of a substance is remaining after a certain period of time.

Given the half-life we can calculate the rate constant for the reaction. Then, using the rate constant in the original equation, we can solve for the amount of substance remaining after specific period of time.

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Problem

A first order reaction has a half-life of 35.0 seconds. What is the concentration of a reactant remaining after 2.0 minutes if the original concentration is 6.5 M?

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Second Order Half-Life Reactions

Recall: 1/[A]t – 1/[A]0 = kt

1/(1/2[A]0) – 1/[A]0 = kt1/2

2/[A]0 - 1/[A]0 = kt1/2

1/[A]0 = kt1/2

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Zero Order Half Life Reactions

Recall: [A]t – [A]0 = -kt

1/2[A]0 – [A]0 = - kt1/2

½[A]o = kt1/2

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What is this “Rate Constant?”

The rate constant (k) is the product of…

collision frequency (Z),

the fraction of collisions with the minimum

amount of energy (f), and

the fraction of collisions with the proper

orientation (p).

k = Zfp or

since A = Zp, then k = Af

where both Z and f are temperature dependent.

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The Arrhenius Equation: Relationship between the reaction rate constant, temperature, and activation energy.

k = Ae-Ea/RT

where R = 8.314 J/mol-K

Let’s look at a reaction at two different temperatures.

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Effect of Temperature on Reaction

Rates

General Rule of Thumb: For every 10oC increase in temperature the reaction rate will double.

Two different temperatures for comparison of

reaction rates

Activation Energy (Ea) is the minimum amount of energy the reactants must have to form the products.

The Arrhenius Equation can be mathematically manipulated in the following format:

ln k = -Ea/R(1/T) + ln A

y = m (x) + b

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Using mathematical manipulation, the equation

can be used in the following format:

ln (k2/k1) = -(Ea/R)(1/T2 – 1/T1)

Problem:

The rate constant for the rxn of hydrogen

reacting with iodine to form hydrogen iodide is

2.7 X 10-4 L/mol-s at 600 K and 3.5 X 10-3

L/mol-s at 650 K. (a) Determine the activation

energy for this reaction. (b) Determine the

rate constant at 700. K.

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Two Theories

I. Collision Theory – Assumes in order for a reaction to occur:

(a) molecules must collide with a

minimum amount of energy (Ea).

(b) molecules must have the proper

orientation.

This figure illustrates the importance of

concentration.

Illustration of the importance of the

appropriate orientation.

Second Theory

II. Transition State Theory: A high energy

species forms when an effective collision

takes place.

A – B + C – D �A-----B-----C-----D

bonds beginto break

Bond beginsto form

This is an example of a high energy species

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Let’s see how this theory can be explained through graphs and diagrams.

BrCH3 + OH–

Reaction progress

Po

ten

tial en

erg

y

BrCH3 + OH–

Br– + CH3OH

∆∆∆∆Hrxn

Reaction progress

Po

ten

tial en

erg

y

BrCH3 + OH–

Br– + CH3OH

∆∆∆∆Hrxn

Reaction progress

Br C O

Overlap changes

as reaction proceeds

Po

ten

tial en

erg

y

Reactants

C

H

HH

OH–+Br

BrCH3 + OH–

Br– + CH3OH

∆∆∆∆Hrxn

Reaction progress

Br C O Br C O

Before transition state

H

H H

C OHBr

Overlap changes

as reaction proceeds

Po

ten

tial en

erg

y

Reactants

C

H

HH

OH–+Br

BrCH3 + OH–

Br– + CH3OH

∆∆∆∆Hrxn

Reaction progress

Br C O Br C O Br C O

Before transition state

H

H H

C OHBr

Transition state

H

HH

C OHBr

Overlap changes

as reaction proceeds

Po

ten

tial en

erg

y

Reactants

C

H

HH

OH–+Br

BrCH3 + OH–

Br– + CH3OH

Ea(fwd)

Ea(rev)

∆∆∆∆Hrxn

Reaction progress

Br C O Br C O Br C O

Before transition state

H

H H

C OHBr

Transition state

H

HH

C OHBr

Overlap changes

as reaction proceeds

Po

ten

tial en

erg

y

Reactants

C

H

HH

OH–+Br

BrCH3 + OH–

Br– + CH3OH

Ea(fwd)

Ea(rev)

∆∆∆∆Hrxn

Reaction progress

Br C O Br C O Br C O Br C O

Before transition state

H

H H

C OHBr

Transition state

H

HH

C OHBr

After transition state

H

H

C OHBr

H

Overlap changes

as reaction proceeds

Po

ten

tial en

erg

y

Reactants

C

H

HH

OH–+Br

BrCH3 + OH–

Br– + CH3OH

Ea(fwd)

Ea(rev)

∆∆∆∆Hrxn

Reaction progress

Br C O Br C O Br C O Br C O Br C O

Po

ten

tial en

erg

y

Reactants

C

H

HH

OH–+Br

Before transition state

H

H H

C OHBr

Transition state

H

HH

C OHBr

After transition state

H

H

C OHBr

H

Products

C OHBr–

H

HH

+

Overlap changes

as reaction proceeds

Exothermic and Endothermic Reaction Energy Diagrams with Transition States

Reaction Mechanisms

A reaction mechanism is a sequence of single

step reactions that when added together give

the overall or net reaction.

Example:

A + Bk1

k-1C

C + B D + Ek2

Ek3 F + G

A + 2B D + F + G

bimolecular

bimolecular

unimolecular

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Write rate law expressions for each of the reaction steps of the previous reaction.

R1 = k1[A][B] also, R1 = k-1 [C]

R2 = k2 [C][B]

R3 = k3 [E]

Note: One of these steps is slower than the others. This is called the rate determining step (rds) and represents the Rate Law expression.

In addition, species C & E are reaction intermediates and not actual substances. How do you know this?

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If the rds for the reaction above is step 2, write a rate law expression for this reaction.

R2 = k2[C][B]

However, C is not an actual substance, its an intermediate.

We need a rate law expression in terms of actual substances.

So… through mathematical substitutions we can derive one.

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k1[A][B] = k-1[C]

Therefore, k1[A][B] = [C]

k-1

Substitute

R2 = k2k1 [A][B]2

k-1

R2 = k[A][B]2 where k = k2k1/k-1

Another example: What is the overall rxn and rate law expression if the second step is the rds?

NO + O2 NO3

NO3 + NO 2 NO2

Another Example: What is the overall rxn

and rate law expression if the first step is the rds.

2 NO N2O2 fast

N2O2 + H2 N2O + H2O slow

N2O + H2 N2 + H2O fast

Energy diagram illustrating two transition states

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Catalysts

Catalysts are used to speed up a reaction by lowering the activation energy.

Note: The catalyst is not used up in a reaction.

Demos: (1) Hydrogen Peroxide with KI

(2) Cobalt (II) chloride with sodium potassium tartrate.