Kinetics ppt (hl)

Chemical Kinetics

Chapter 14

Chemistry, The Central Science, 7th& 8theditions

What do we mean by

kinetics?

Kinetics refers to



the

rate

at which chemical

reactions occur.



The

reaction mechanism

or

pathway through which a reaction

proceeds.

Topics for study in kinetics

Reaction Rates How we measure rates.

Rate Laws How the reaction rate depends on amounts of reactants. Molecularity and order.

Integrated Rate Laws How to calculate amount left or time to reach a given amount of reactant or product.

Half-life How long it takes for 50% of _{reactants to react} Arrhenius Equation How rate constant changes with _temperature.

Factors That Affect Reaction Rates

 The Nature of the Reactants

 Chemical compounds vary considerably in their

chemical reactivities.

 Concentration of Reactants

 As the concentration of reactants increases, so does

the likelihood that reactant molecules will collide.  Temperature

 At higher temperatures, reactant molecules have

more kinetic energy, move faster, and collide more often and with greater energy.

 Catalysts

 Change the rate of a reaction

by changing the mechanism.

Reaction Rates

The rate of a chemical reaction can be determined by monitoring the change in

concentration of either reactants or the products as a function of time.

Example 1: Reaction Rates

In this reaction, the concentration of

butyl chloride,

C₄H₉Cl, was

measured at various times, t.

C₄H₉Cl(aq) + H₂O(l)  C₄H₉OH(aq) + HCl(aq)

[C₄H₉Cl] M

Rate =



[C

4

H

9

Cl]



t

Reaction Rates Calculation

The average rate of the reaction over each

interval is the change in concentration divided by the change in time:

C₄H₉Cl(aq) + H₂O(l)  C₄H₉OH(aq) + HCl (aq)

 Note that the average

rate decreases as the reaction proceeds.

 This is because as the

reaction goes forward, there are fewer

collisions between the reacting molecules.

C

₄

H

₉

Cl(

aq

) + H

₂

O(

l

)



C

₄

H

₉

OH(

aq

) + HCl(

aq

)

Reaction Rate

Determination

Reaction Rates

 A plot of concentration vs. time for this reaction yields a curve like this.  The slope of a line

tangent to the curve at any point is the

instantaneous rate at that time.

Reaction Rates



The reaction slows

down with time

because the

concentration of the

reactants decreases.

C

₄

H

₉

Cl(

aq

) + H

₂

O(

l

)



C

₄

H

₉

OH(

aq

) + HCl(

aq

)

Reaction Rates and

Stoichiometry



In this reaction, the ratio

of C

₄

H

₉

Cl to C

₄

H

₉

OH is

1:1.



Thus, the rate of

disappearance

of C

₄

H

₉

Cl

is the same as the rate of

appearance

of C

₄

H

₉

OH.

C

₄

H

₉

Cl(

aq

)

+ H

₂

O(

l

)



C

₄

H

₉

OH

(

aq

) + HCl(

aq

)

Rate = -[C4H9Cl]

t =

[C₄H₉OH]

Reaction Rates &

Stoichiometry

Suppose that the mole ratio is not 1:1?

Example

H

₂(g)

+ I

₂(g)



2 HI

(g)

2 moles of HI are produced for each mole of H

₂

used.

Concentration and Rate

Each reaction has its own equation that gives its rate as a function of reactant concentrations.

This is called its Rate Law

The general form of the rate law is

Rate = k[A]x[B]y

Where k is the rate constant, [A] and [B] are the

concentrations of the reactants. X and y are exponents known as rate orders that must be determined

experimentally

Concentration and Rate

Compare Experiments 1 and 2:

when [NH

₄+

]

doubles

, the initial rate

doubles.

NH

₄+

(aq)

+ NO

2-

(aq)



N

2

(g) + 2H

2

O (l)

Concentration and Rate

Likewise, compare Experiments 5 and 6:

when [NO

₂-

]

doubles

, the initial rate

doubles.

NH₄+ (aq) + NO

Concentration and Rate

This equation is called the

rate law

, and

k

is the

rate constant.

NH

₄+

(aq)

+ NO

2-

(aq)



N

2

(g) + 2H

2

O (l)

The Rate

Law

 A rate law shows the relationship between the

reaction rate and the concentrations of reactants.  For gas-phase reactants use P

A instead of [A].

 k is a constant that has a specific value for each

reaction.

 The value of k is determined experimentally.

Rate = K [

NH

₄+

][NO

2-

]

“Constant” is relative here-

The Rate

Law

 Exponents tell the order of the reaction

with respect to each reactant.

 This reaction is

First-order in [NH₄+]

First-order in [NO₂−]

 The overall reaction order can be found by

adding the exponents on the reactants in the rate law.

 This reaction is second-order overall.

Rate = K [NH

₄+

]

1

[NO

2-

]

1

Determining the Rate constant

and Order

The following data was collected for the reaction of substances A and B to produce products C and D.

Deduce the order of this reaction with respect to A and to B. Write an expression for the rate law in this reaction and

calculate the value of the rate constant.

[NO] mol dm-3 [O2] mol dm-3 Rate mol dm-3 s-1

0.40 0.50 1.6 x 10-3

0.40 0.25 8.0 x 10-4

The First Order Rate

Equation

Consider a simple 1st order reaction: A



B

Rate = k[A]

How much reactant A is left after time t? The rate equation as a function of time can be written as

[A]

_t

=[A]

_o

e

-kt

Ln [A]

_t

- Ln[A]

_o

= - kt

Where

[A]_t = the concentration of reactant A at time t [A]_o = the concentration of reactant A at time t = 0

K = the rate

First-Order Processes

Consider the process in which methyl isonitrile is converted to acetonitrile.

CH

₃

NC

CH

₃

CN

First-Order Processes

This data was

collected for this reaction at

198.9°C.

CH

₃

NC

CH

₃

CN

Does

rate=k[CH

₃

NC]

for all time intervals?

First-Order Processes

 When Ln P is plotted as a function of time, a straight line results.

 The process is first-order.

Half-Life of a Reaction

 Half-life is defined

as the time

required for one-half of a reactant to react.

 Because [A] at t_1/2

is one-half of the original [A],

[A]_t = 0.5 [A]₀.

Half-Life of a First Order

Reaction

For a first-order process, set [A]_t=0.5 [A]₀ in integrated rate equation:

First Order Rate

Calculation

26.

Example 1: The decomposition of compound A is first order. If the initial [A]₀ = 0.80 mol dm-3. and the rate

constant is 0.010 s-1, what is the concentration of [A]

First Order Rate

Calculation

Ln[A]_t – Ln[A]_o = -kt

Ln[A]_t – Ln[0.80] = - (0.010 s-1 _{)(90 s)}

Ln[A]_t = - (0.010 s-1 _{)(90 s) + Ln[0.80]}

Ln[A]_t = -0.90 - 0.2231 Ln[A]_t = -1.1231

[A]_t = 0.325 mol dm-3

Example 1: The decomposition of compound A is first order. If the initial [A]₀ = 0.80 mol dm-3. and the rate

constant is 0.010 s-1, what is the concentration of [A]

First Order Rate

Calculations

Example 2: A certain first order chemical reaction required 120 seconds for the concentration of the reactant to drop from 2.00 M to 1.00 M. Find the rate constant and the concentration of reactant [A] after 80 seconds.

First Order Rate

Calculations

Example 2: A certain first order chemical reaction required 120 seconds for the concentration of the reactant to drop from 2.00 M to 1.00 M. Find the rate constant and the concentration of reactant [A] after 80 seconds.

Solution

k =0.693/t_1/2 =0.693/120s =0.005775 s-1

Ln[A] – Ln(2.00) = -0.005775 s-1 (80 s)= -0.462

Ln A = - 0.462 + 0.693 = 0.231

First Order Rate

Calculations

Example 3: Radioactive decay is also a first order process. Strontium 90 is a radioactive isotope with a half-life of 28.8 years. If some strontium 90 were accidentally released, how

long would it take for its concentration to fall to 1% of its original concentration?

First Order Rate

Calculations

Example 3: Radioactive decay is also a first order process. Strontium 90 is a radioactive isotope with a half-life of 28.8 years. If some strontium 90 were accidentally released, how

long would it take for its concentration to fall to 1% of its original concentration?

Solution

k =0.693/t_1/2 =0.693/28.8 yr =0.02406 yr-1

Ln[1] – Ln(100) = - (0.02406 yr-1)t = - 4.065

t = - 4.062 .

- 0.02406 yr-1

Second-Order Processes

Similarly, integrating the rate law for a

process that is second-order in reactant A:

Rate = k[A]

2

1

[A]

_t

= kt +

1

[A]

_o

Where

[A]_t = the concentration of reactant A at time t [A]_o = the concentration of reactant A at time t = 0

K = the rate constant

Second-Order Rate

Equation

So if a process is second-order in A,

a graph of

1/[A] vs. t

will yield a

Determining Reaction

Order

Distinguishing Between 1

st

and 2

nd

Order

The decomposition of NO2 at 300°C is described by the equation:

NO

₂ (g)

NO

(g)

+ 1/2 O

₂ (g)

A experiment with this reaction yields this data:

Time (s) [NO₂], M

Graphing ln [NO₂] vs. t yields:

Time (s) [NO₂], M ln [NO₂]

0.0 0.01000 -4.610 50.0 0.00787 -4.845 100.0 0.00649 -5.038 200.0 0.00481 -5.337 300.0 0.00380 -5.573

• The graph is not a straight

line, so this process cannot be first-order in [A].

Determining Reaction

Order

Distinguishing Between 1

st

and 2

nd

Order

Second-Order Reaction Kinetics

A graph of 1/[NO2] vs. t gives this plot.

Time (s) [NO₂], M 1/[NO₂]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• This is a straight line.

Therefore, the process is second-order in [NO₂].

• The slope of the line is

the rate constant, k.

Half-Life for 2nd Order

Reactions

For a second-order process, set

[A]_t=0.5 [A]₀ in 2nd order equation.

Sample Problem 1: Second

Order

Acetaldehyde, CH₃CHO, decomposes by second-order

kinetics with a rate constant of 0.334 mol-1dm3s-1 at 500oC.

Calculate the amount of time it would take for 80% of the acetaldehyde to decompose in a sample that has an initial concentration of 0.00750 M.

666.7 = 0.334 t + 133.33 0.334 t = 533.4

t = 1600 seconds

38.

The final concentration will be 20% of the original 0.00750 M or = 0.00150

1 .

Sample Problem 2: Second

Order

Acetaldehyde, CH₃CHO, decomposes by second-order

kinetics with a rate constant of 0.334 mol-1dm3s-1 at 500oC. If

the initial concentration of acetaldehyde is 0.00200 M. Find the concentration after 20 minutes (1200 seconds)

= 0.334 mol-1dm3 s-1 (1200s) + 500 mol-1dm3

= 900.8 mol-1dm3 Solution

1 .

[A]_t = 0.334 mol-1dm3s-1 (1200s) + 1 .

0.00200 mol dm-3

1 . [A]_t

[A]_t = 1 _____.

900.8 mol-1dm = 0.00111 mol3 dm

Summary of Kinetics

Equations

First order Second order Second order

Rate Laws

Integrat ed Rate

Laws complicated

Half-life complicated

Temperature and Rate



Generally speaking, the

reaction rate increases

as the temperature

increases.



This is because

k

is

temperature dependent.



As a rule of thumb a

The Collision Model



In a chemical reaction, bonds are

broken and new bonds are formed.



Molecules can only react if they

collide with each other.



These collisions must occur with

sufficient energy and at the

appropriate orientation.

The Collision Model

Furthermore, molecules must collide with

the correct

orientation

and with enough

Activation Energy

 In other words, there is a minimum amount of energy

required for reaction: the activation energy, Ea.

 Just as a ball cannot get over a hill if it does not roll

up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

Reaction Coordinate

Diagrams

It is helpful to visualize

energy changes throughout a

process on a reaction

coordinate diagram like this one for the rearrangement of methyl

Reaction Coordinate Diagrams

 It shows the energy of

the reactants and products (and,

therefore, E).

 The high point on the

diagram is the

transition state.

•

The species present at the transition state is

called the

activated complex

.

•

The energy gap between the reactants and

the activated complex is the activation

energy barrier.

Maxwell–Boltzmann

Distributions

 Temperature is defined as a

measure of the average kinetic energy of the molecules in a sample.

•

At any temperature there is a wide

Maxwell–Boltzmann Distributions

 As the temperature increases, the

curve flattens and broadens.

 Thus at higher temperatures, a

larger population of molecules has

higher energy.

Maxwell–Boltzmann

Distributions

 If the dotted line represents the activation energy,

as the temperature increases, so does the fraction of molecules that can overcome the activation

energy barrier.

Maxwell–Boltzmann

Distributions

This fraction of molecules can be found through the expression:

where R is the gas constant and T is the temperature in Kelvin .

Arrhenius Equation

Svante Arrhenius developed a mathematical relationship between k and Ea:

where A is the frequency factor, a number that

represents the likelihood that collisions would occur with the proper orientation for reaction. E_a is the

activation energy. T is the Kelvin temperature and R is the universal thermodynamics (gas) constant.

Arrhenius Equation

Taking the natural logarithm of both sides, the equation becomes

1

RT

y

=

mx

+

b

When

k

is determined experimentally at

several temperatures,

E

_a

can be calculated

from the slope of a plot of

ln

k

vs. 1/

T

.

Arrhenius Equation for 2

Temperatures

When measurements are taken for two different temperatures the Arrhenius equation can be

symplified as follows:

Write the above equation twice, once for each of the two Temperatures and then subtract the lower temperature

Arrhenius Equation Sample

Problem 1

The rate constant for the decomposition of

hydrogen iodide was determined at two different temperatures

2HI  H₂ + I₂.

At 650 K, k₁ = 2.15 x 10-8 dm3 mol-1s-1

At 700 K, k₂ = 2.39 x 10-7 dm3 mol-1s-1

Find the activation energy for this reaction.

54.

2.39 x 10-7 Ea

Ln --- = - --- x 2.15 x 10-8 (8.314 J mol-1 K-1)

E_a = 180,000 J mol-1 = 180 kJ mol-1

1 1 -- ---700K 650K

Overview of Kinetics Equations

First order Second order Second order

Rate Laws

Integrated

Rate Laws complicated

Half-life complicated

Reaction Mechanisms

The sequence of events that describes

the actual process by which reactants

become products is called the

reaction mechanism

.

Reaction Mechanisms



Reactions may occur all at once or

through several discrete steps.



Each of these processes is known as

Reaction Mechanisms

• _{The molecularity of a process tells how many}

molecules are involved in the process.

• _{The rate law for an elementary step is written}

directly from that step.

Multistep Mechanisms



In a multistep process, one of the steps will

be slower than all others.



The overall reaction cannot occur faster

Slow Initial Step

 The rate law for this reaction is found experimentally

to be

Rate =

k

[NO

₂

]

2

 CO is necessary for this reaction to occur, but the

rate of the reaction does not depend on its concentration.

 This suggests the reaction occurs in two steps.

NO

₂ (g)

+ CO

(g)



NO

(g)

+ CO

₂ (g)

Slow Initial Step

 A proposed mechanism for this reaction is

Step 1: NO2 + NO2  NO3 + NO (slow) Step 2: NO3 + CO  NO2 + CO2 (fast)

 The NO₃ intermediate is consumed in the second step.

 As CO is not involved in the slow, rate-determining step, it

Fast Initial Step



The rate law for this reaction is

found (experimentally) to be



Because termolecular (= trimolecular)

processes are rare, this rate law

suggests a two-step mechanism.

Fast Initial Step



A proposed mechanism is

Step 1 is an

equilibrium

-

Fast Initial Step



The rate of the overall reaction depends

upon the rate of the slow step.



The rate law for that step would be



But how can we find [NOBr

₂

]?

Fast Initial Step



NOBr

2

can react two ways:



With NO to form NOBr



By decomposition to reform NO and Br

2



The reactants and products of the first step

are in equilibrium with each other.



Therefore,

Fast Initial Step



Because Rate

_f

= Rate

_r

,

k

₁

[NO] [Br

₂

] =

k

₋₁

[NOBr

₂

]

Solving for [NOBr

₂

] gives us

k

₁

k

₋₁

[NO] [Br

2

] = [NOBr

2

]

Fast Initial Step

Catalysts

 Catalysts increase the rate of a reaction by

decreasing the activation energy of the reaction.

 Catalysts change the mechanism by which the

process occurs.

 Some catalysts also make atoms line up in the

correct orientation so as to enhance the reaction rate

Catalysts

Catalysts may be

either homogeneous or heterogeneous

A homogeneous

catalyst is in the same phase as the

substances reacting. A heterogeneous

Catalysts

One way a catalyst can speed up a

reaction is by holding the reactants

together and helping bonds to break.

Heterogeneous catalysts often act in this way

Catalysts

Some catalysts help to lower the energy for formation for the activated complex or

provide a new activated

complex with a lower activation energy

AlCl₃ + Cl₂  Cl+ + AlCl 4

-Cl+ + C

6H6  C6H5Cl + H+

H+ + AlCl

4-  HCl + AlCl3

Overall reaction

Catalysts & Stratospheric Ozone

72.

In the stratosphere, oxygen molecules absorb ultraviolet light and break into individual oxygen atoms known as free radicals

The oxygen radicals can then combine with ordinary oxygen molecules to make ozone.

Catalysts & Stratospheric Ozone

The presence of chlorofluorcarbons in the

stratosphere can catalyze the destruction of ozone. UV light causes a Chlorine free radical to be released

The chlorine free radical attacks ozone and converts it Back to oxygen. It is then regenerated to repeat the

Enzymes

 Enzymes are

catalysts in biological systems.

 The substrate fits

into the active

site of the enzyme much like a key

fits into a lock.