Loading...

PPT – Chemical Kinetics Chapter 13 PowerPoint presentation | free to download - id: 77bddc-YjQ3M

The Adobe Flash plugin is needed to view this content

Chemical Kinetics Chapter 13

Topics 6 and 16

1.

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.

2.

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.

Mechanisms How the reaction rate depends on the sequence of molecular scale processes.

3.

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.

4.

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. - ?A vs ?t

5.

Example 1 Reaction Rates

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

C4H9Cl M

- In this reaction, the concentration of butyl

chloride, C4H9Cl, was measured at various times,

t.

?C4H9Cl ?t

Rate

6.

Reaction Rates Calculation

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl (aq)

Average Rate, M/s

- The average rate of the reaction over each

interval is the change in concentration divided

by the change in time

7.

Reaction Rate Determination

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.

8.

Reaction Rates

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

- 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.

9.

Reaction Rates

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

- The reaction slows down with time because the

concentration of the reactants decreases.

10.

Reaction Rates and Stoichiometry

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

- In this reaction, the ratio of C4H9Cl to C4H9OH

is 11. - Thus, the rate of disappearance of C4H9Cl is the

same as the rate of appearance of C4H9OH.

11.

Reaction Rates Stoichiometry

- Suppose that the mole ratio is not 11?

Example H2(g) I2(g) ??? 2 HI(g)

2 moles of HI are produced for each mole of H2

used.

The rate at which H2 disappears is only half of

the rate at which HI is generated

12.

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 kAxBy
- 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 - To determine the rate law we measure the rate at

different starting concentrations.

13.

Concentration and Rate

NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)

- Compare Experiments 1 and 2when NH4 doubles,

the initial rate doubles.

14.

Concentration and Rate

NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)

- Likewise, compare Experiments 5 and 6 when

NO2- doubles, the initial rate doubles.

15.

Concentration and Rate

NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)

This equation is called the rate law, and k is

the rate constant.

16.

The Rate Law

- A rate law shows the relationship between the

reaction rate and the concentrations of

reactants. - For gas-phase reactants use PA instead of A.
- k is a constant that has a specific value for

each reaction. - The value of k is determined experimentally.
- Rate K NH4 NO2-
- Constant is relative here-
- the rate constant k is unique for each reaction
- and the value of k changes with temperature

17.

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 NH4 1NO2- 1

18.

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.

19.

The First Order Rate Equation

Consider a simple 1st order reaction A ? B Rate

kA

How much reactant A is left after time t? The

rate equation as a function of time can be

written as

Where At the concentration of

reactant A at time t Ao the concentration of

reactant A at time t 0

K the rate constant

At Ao e-kt Ln At - LnAo - kt

20.

First-Order Processes

- Consider the process in which methyl isonitrile

is converted to acetonitrile.

How do we know this is a first order reaction?

21.

First-Order Processes

- This data was collected for this reaction at

198.9C.

Does ratekCH3NC for all time intervals?

22.

First-Order Processes

- When ln P is plotted as a function of time, a

straight line results. - The process is first-order.
- k is the negative slope 5.1 ? 10-5 s-1.

23.

Half-Life of a Reaction

- Half-life is defined as the time required for

one-half of a reactant to react. - Because A at t1/2 is one-half of the original

A, - At 0.5 A0.

24.

Half-Life of a First Order Reaction

- For a first-order process, set At0.5 A0 in

integrated rate equation

NOTE For a first-order process, the half-life

does not depend on the initial concentration,

A0.

25.

First Order Rate Calculation

Example 1 The decomposition of compound A is

first order. If the initial A0 0.80 mol

dm-3. and the rate constant is 0.010 s-1, what

is the concentration of A after 90 seconds?

26.

First Order Rate Calculation

Example 1 The decomposition of compound A is

first order. If the initial A0 0.80 mol

dm-3. and the rate constant is 0.010 s-1, what

is the concentration of A after 90 seconds?

LnAt lnAo -kt

LnAt Ln0.80 - (0.010 s-1 )(90 s) LnAt

- (0.010 s-1 )(90 s) ln0.80 LnAt

-0.90 - 0.2231 lnAt -1.1231 At

0.325 mol dm-3

27.

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.

28.

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/t1/2 0.693/120s 0.005775

s-1 LnA Ln(2.00) -0.005775 s-1 (80 s)

-0.462 ln A - 0.462 0.693 0.231 A

1.26 mol dm-3

29.

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?

30.

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/t1/2 0.693/28.8 yr 0.02406

yr-1 Ln1 ln(100) - (0.02406 yr-1)t -

4.60 t - 4.60 . - 0.0241

yr-1 t 191 years

31.

Second-Order Processes

- Similarly, integrating the rate law for a process

that is second-order in reactant A - Rate kA2

1 Ao

1 At

kt

Where At the concentration of reactant A at

time t Ao the concentration of reactant A at

time t 0 K the rate constant

32.

Second-Order Rate Equation

- So if a process is second-order in A, a graph of

1/A vs. t will yield a straight line with a

slope of k.

33.

Determining Reaction OrderDistinguishing Between

1st and 2nd Order

The decomposition of NO2 at 300C is described by

the equation

A experiment with this reaction yields this data

Time (s) NO2, M

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

34.

Determining Reaction OrderDistinguishing Between

1st and 2nd Order

Graphing ln NO2 vs. t yields

- The graph is not a straight line, so this process

cannot be first-order in A.

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

Does not fit the first order equation

35.

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.

36.

Half-Life for 2nd Order Reactions

- For a second-order process, set
- At0.5 A0 in 2nd order equation.

In this case the half-life depends on the initial

concentration of the reactant A.

37.

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.

The final concentration will be 20 of the

original 0.00750 M or 0.00150

1 . .00150

1 . .00750

0.334 mol-1dm3s-1 t

666.7 0.334 t 133.33 0.334 t 533.4

t 1600 seconds

38.

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)

Solution

1 . At

1 . 0.00200 mol dm-3

0.334 mol-1dm3s-1 (1200s)

1 . At

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

900.8 mol-1dm3

1 _____. 900.8 mol-1dm3

At

0.00111 mol dm-3

39.

Summary of Kinetics Equations

First order Second order Second order

Rate Laws

Integrated Rate Laws complicated

Half-life complicated

40.

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 reaction rate increases

about 10 fold for each 10oC rise in temperature

41.

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.

42.

The Collision Model

- Furthermore, molecules must collide with the

correct orientation and with enough energy to

cause bonds to break and new bonds to form

43.

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.

44.

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 isonitrile.

45.

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.

46.

MaxwellBoltzmann 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 distribution

of kinetic energies.

47.

MaxwellBoltzmann Distributions

- As the temperature increases, the curve flattens

and broadens. - Thus at higher temperatures, a larger population

of molecules has higher energy.

48.

MaxwellBoltzmann 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.

- As a result, the reaction rate increases.

49.

MaxwellBoltzmann Distributions

- This fraction of molecules can be found through

the expression - where R is the gas constant and T is the

temperature in Kelvin .

50.

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.

Ea is the activation energy. T is the Kelvin

temperature and R is the universal thermodynamics

(gas) constant. - R 8.314 J mol-1 K-1 or 8.314 x 10-3 kJ

mol-1 K-1

51.

Arrhenius Equation

- Taking the natural logarithm of both sides, the

equation becomes

y mx b

When k is determined experimentally at several

temperatures, Ea can be calculated from the slope

of a plot of ln k vs. 1/T.

52.

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 conditions from the higher

temperature. The equation then becomes

53.

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.

2.39 x 10-7 Ea Ln ---------------- -

------------------------ x 2.15 x 10-8

(8.314 J mol-1 K-1) Ea 180,000

J mol-1 180 kJ mol-1

1 1 ------ -- ------ 700K 650K

54.

Overview of Kinetics Equations

First order Second order Second order

Rate Laws

Integrated Rate Laws complicated

Half-life complicated

Rate and Temp (T)

55.

Reaction Mechanisms

- The sequence of events that describes the actual

process by which reactants become products is

called the reaction mechanism.

56.

Reaction Mechanisms

- Reactions may occur all at once or through

several discrete steps. - Each of these processes is known as an elementary

reaction or elementary process.

57.

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.

58.

Multistep Mechanisms

- In a multistep process, one of the steps will be

slower than all others. - The overall reaction cannot occur faster than

this slowest, rate-determining step.

59.

Slow Initial Step

NO2 (g) CO (g) ??? NO (g) CO2 (g)

- The rate law for this reaction is found

experimentally to be - Rate k NO22
- 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.

60.

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 does not appear in the

rate law.

61.

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.

62.

Fast Initial Step

- A proposed mechanism is

Step 1 is an equilibrium- it includes the

forward and reverse reactions.

63.

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 NOBr2?

64.

Fast Initial Step

- NOBr2 can react two ways
- With NO to form NOBr
- By decomposition to reform NO and Br2
- The reactants and products of the first step are

in equilibrium with each other. - Therefore,
- Ratef Rater

65.

Fast Initial Step

- Because Ratef Rater ,
- k1 NO Br2 k-1 NOBr2
- Solving for NOBr2 gives us

66.

Fast Initial Step

- Substituting this expression for NOBr2 in the

rate law for the rate-determining step gives

67.

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

68.

Catalysts

- Catalysts may be either homogeneous or

heterogeneous - A homogeneous catalyst is in the same phase as

the substances reacting. - A heterogeneous catalyst is in a different phase

69.

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

70.

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 AlCl4- Cl C6H6 ? C6H5Cl

H H AlCl4- ? HCl AlCl3 Overall

reaction C6H6 Cl2 ? C6H5Cl HCl

71.

Catalysts Stratospheric Ozone

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. Ozone can also

be split up again into ordinary oxygen and an

oxygen radical by absorbing ultraviolet light.

72.

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 Process. The result is

that each chlorine free radical can Repeat this

process many many times. The result is

that Ozone is destroyed faster than it is formed,

causing its level to drop

73.

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.

74.

75.