CHEMICAL KINETICS CHAPTER 2, Slides of Chemistry

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Advanced Physical Chemistry

CP4167/ 4539

Chemical Kinetics

Chapter 13 of SmartBook

Is the study of reaction rates, the changes in concentration

of reactants (or products) as a function of time.

Sub-topics for Chemical Kinetics

1.1 Reaction Rate

1.2 Measuring Reaction Progress and Expressing Reaction Rate

Average Reaction Rate

Instantaneous Rate

Stoichiometry and Reaction Rate

1.3 Dependence of Reaction Rate on Reactant Concentration

The Rate Law

Experimental Determination of the Rate Law

1.4 Dependence of Reactant Concentration on Time (Integrated rate law)

First-Order Reactions

Second-Order Reactions

Zero-Order Reactions

1.5 Dependence of Reaction Rate on Temperature

The Arrhenius Equation

1.6 Reaction Mechanisms

Elementary Reactions

Rate-Determining Step

Experimental Support for Reaction Mechanisms

Mechanisms with a Fast First Step 1.

Catalysis

1.1 Reaction Rate

Thermodynamics – does a reaction take place?

Kinetics – how fast does a reaction proceed?

Reaction rate is the change in the concentration of a

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

Because [A] decreases with time, ∆ [A] is negative.

A  B

 

 

 

 

ΔA changeinconcentrationofAover A changeinconcentrationof A over

ΔA changeinconcentrationofAover A

rate

timeperiod ΔA changeinconcentrationofAover

ΔA changeinconcentrationofAovert

ΔA changeinconcentrationofAover B ΔA changeinconcentrationofAover B changeinconcentrationof Bover

rate

ΔA changeinconcentrationofAovert

timeperiod ΔA changeinconcentrationofAover

t

t

A → B

1.1 Reaction Rate

 

 





ΔA changeinconcentrationofAover A

rate

ΔA changeinconcentrationofAover

ΔA changeinconcentrationofAover

rate

ΔA changeinconcentrationofAover

-

t

t

B

Time (s) [Br 2

] ( M )

0.0 0.

50.0 0.

100.0 0.

150.0 0.

200.0 0.

250.0 0.

300.0 0.

350.0 0.

400.0 0.

instantaneous rate = rate for specific instance in time

The instantaneous rate of a reaction can be found at any time by finding

the slope of the tangent at a given point in time.

1.2 Expressing Reaction Rate

         



2 2

Br aq + HCOOH aq 2Br aq + 2H aq + CO g

   

2 2 2 final initial

final initial

Br – Br ΔA changeinconcentrationofAover[Br ]

ΔA changeinconcentrationofAover t t – t

average rate

1.2 Expressing Reaction Rate

Hydrogen peroxide decomposes to form

oxygen gas. A manometer measures pressure

changes, and as the reaction proceeds, oxygen

gas causes the pressure to increase.

     

 

 

 

 



 



 

2 2 2 2

2

2

2

2H O 2H O O

O RT

1

O

ΔA changeinconcentrationofAover O 1

rate

aq l g

PV nRT

n

P RT

V

P

RT

ΔPP

ΔPt RT ΔPt

2A → B

Two moles of A disappear for each mole of B that is formed.

1.2 Stoichiometry and Reaction Rate

Also known as differential rate law

   

       

 

  

   

ΔA changeinconcentrationofAover A ΔA changeinconcentrationofAover B 1

rate – rate

2 ΔA changeinconcentrationofAover ΔA changeinconcentrationofAover

A B C D

ΔA changeinconcentrationofAover A ΔA changeinconcentrationofAover B ΔA changeinconcentrationofAover C ΔA changeinconcentrationofAover D 1 1 1 1

rate – –

ΔA changeinconcentrationofAover ΔA changeinconcentrationofAover ΔA changeinconcentrationofAover ΔA changeinconcentrationofAover

t t

a b c d

a t b t c t d t

Example 1.2.

Write the rate expression for the following reaction in

terms of the disappearance of the reactants and the

appearance of the products:

4NH

3

(g) + 5O

2

(g) → 4NO (g) + 6H

2

O (g)

Strategy

To express the rate of the reaction in terms of the change in

concentration of a reactant or product with time, we need to use the

proper sign (minus or plus) and the reciprocal of the stoichiometric

coefficient.

Solution

       

   

3 2 2

ΔA changeinconcentrationofAover NH ΔA changeinconcentrationofAover O ΔA changeinconcentrationofAover NO ΔA changeinconcentrationofAover H O 1 1 1 1

rate – –

4 ΔA changeinconcentrationofAover t 5 ΔA changeinconcentrationofAover t 4 ΔA changeinconcentrationofAover t 6 ΔA changeinconcentrationofAover t

Strategy

To calculate the rate of formation of N

2

O

5

and disappearance of

NO

2

, we need to express the rate of the reaction in terms of the

stoichiometric coefficients as in Example 13.1:

We are given

where the minus sign shows that the concentration of O

2

is

decreasing with time.

Example 1.2.

     

2 2 2 5

ΔA changeinconcentrationofAover NO ΔA changeinconcentrationofAover O ΔA changeinconcentrationofAover N O

rate – –

4 ΔA changeinconcentrationofAover t ΔA changeinconcentrationofAover t 2 ΔA changeinconcentrationofAover t

 

2

ΔA changeinconcentrationofAover O

ΔA changeinconcentrationofAover

M s

t

     

2 2 2 5

4NO g O g 2N O g

Solution

a) From the preceding rate expression we have

Example 1.2.

b) Here we have

   

2 2 5

ΔA changeinconcentrationofAover O ΔA changeinconcentrationofAover N O

ΔA changeinconcentrationofAover t 2 ΔA changeinconcentrationofAover t

 

 (^)   

2 5

ΔA changeinconcentrationofAover N O

• 2 – 0.024 0.

ΔA changeinconcentrationofAover

M s M s

t

     

2 2 2 5

4NO g O g 2N O g

   

2 2

ΔA changeinconcentrationofAover NO ΔA changeinconcentrationofAover O

• –

4 ΔA changeinconcentrationofAover t ΔA changeinconcentrationofAover t

 

 

2

ΔA changeinconcentrationofAover NO

ΔA changeinconcentrationofAover

M s M s

t

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.

1.3 The Rate Law

Rate , reaction orders and rate

constant (k) must be found by

experiment; they cannot be deduced

from the reaction stoichiometry !!!!!

Example 1.3.

The reaction of nitric oxide with hydrogen at 1280°C is

From the following data collected at this temperature, determine

a) the rate law

b) the rate constant

c) the rate of the reaction when

2NO  g  2H  g  N  g  2H O g 

2 2 2

  

   

–3 –

NO = 12.0 × 10 M and H2 = 6.0 × 10 M

         

–3 –3 –

–3 –3 –

–3 –3 –

M M M s

2

Experiment NO H InitialRates

Solution

(a) Experiments 1 and 2 show that when we double the concentration of

NO at constant concentration of H

2

, the rate quadruples.Taking the ratio

of the rates from these two experiments

Therefore,

or x = 2, that is, the reaction is second order in NO.

Example 1.3.

 

 

 

 

  

  



 

–3 –

2

–3 –

1

10.0 10 2.0 10 rate 5.0 10

4

rate 1.3 10

5.0 10 2.0 10

x y

x y

k M M M s

M s

k M M

 

 



 



10.0 10

2 4

5.0 10

x

x

x

M

M

         

  

  

  

–3 –3 –

–3 –3 –

–3 –3 –

1 5.0 10 2.0 10 1.3 10

2 10.0 10 2.0 10 5.0 10

3 10.0 10 4.0 10 10.0 10

M M M s 2

Experiment NO H InitialRates

Experiments 2 and 3 indicate that doubling

doubles the rate. Here we write the ratio as

Therefore,

or y = 1, that is, the reaction is first order in H

2

. Hence the rate

law is given by

which shows that it is a or third-order reaction overall.

Example 1.3.

    2

H at constant NO

   

   

  

  



 

–3 –

3

–3 –

2

10.0 10 M 4.0 10 M rate 10.0 10

2

rate 5.0 10

10.0 10 M 2.0 10 M

x y

x y

k M s

M s

k

 

 



 



4.0 10

2 2

2.0 10

y

y

y

M

M

 (^)    

2

2

rate k NO H

 2 + 1

         

  

  

  

–3 –3 –

–3 –3 –

–3 –3 –

1 5.0 10 2.0 10 1.3 10

2 10.0 10 2.0 10 5.0 10

3 10.0 10 4.0 10 10.0 10

M M M s 2

Experiment NO H InitialRates