Chemical Kinetics Chapter 14

Chemical Kinetics Chapter 14

Chemistry, The Central Science, 7th& 8theditionsTheodore L. Brown; H. Eugene LeMay, Jr.; and

Bruce E. Bursten1.

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 kineticsReaction Rates How we measure rates.

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

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

Half-lifeHow long it takes for 50% of reactants to react

Arrhenius EquationHow rate constant changes with temperature.

MechanismsHow 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

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 =[C4H9Cl]

t

6.

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)

Average Rate, M/s

7.

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.


Reaction Rate Determination

8.

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.


9.

Reaction Rates

The reaction slows down with time because the concentration of the reactants decreases.


10.

Reaction Rates and Stoichiometry

In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1.

Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.


Rate =-[C4H9Cl]

t=

[C4H9OH]t

11.

Reaction Rates & Stoichiometry

Suppose that the mole ratio is not 1:1?

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 = 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

To determine the rate law we measure the rate at different starting concentrations.

13.


Compare Experiments 1 and 2:when [NH4

+] doubles, the initial rate doubles.

NH4+ (aq) + NO2

- (aq) N2 (g) + 2H2O (l)

14.


Likewise, compare Experiments 5 and 6: when [NO2

-] doubles, the initial rate doubles.

NH4+ (aq) + NO2

- (aq) N2 (g) + 2H2O (l)

15.


This equation is called the rate law, and k is the rate constant.

NH4+ (aq) + NO2

- (aq) N2 (g) + 2H2O (l)

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 reactionand 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+ ]1[NO2

- ]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.

[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

0.20 0.25 2.0 x 10-4

19.

The First Order Rate EquationConsider 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 constant 20.

First-Order ProcessesConsider the process in which methyl isonitrile is converted to acetonitrile.

CH3NC CH3CN

How do we know this is a first order reaction?

21.

First-Order Processes

This data was collected for this reaction at 198.9°C.

CH3NC CH3CN

Does rate=k[CH3NC] 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], [A]t = 0.5 [A]0.

24.

Half-Life of a First Order Reaction

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

NOTE: For a first-order process, the half-life does not depend on the initial concentration, [A]0.

25.

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] after 90 seconds?

First Order Rate Calculation

27.

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] after 90 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.

28.


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

Ln[A] – 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.


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.


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

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

t = - 4.062 . - 0.02406 yr-1

t = 168.8 years

31.

Second-Order ProcessesSimilarly, 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 = 0K = 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 300°C is described by the equation:

NO2 (g) NO (g) + 1/2 O2 (g)

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 0.0038034.

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 OrderDistinguishing Between 1st and 2nd Order

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 [A]t=0.5 [A]0 in 2nd order equation.

In this case the half-life depends on the initial concentration of the reactant A.

37.

Sample Problem 1: Second OrderAcetaldehyde, 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 + 133.330.334 t = 533.4 t = 1600 seconds

38.

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

1 ..00150 = 0.334 mol-1dm3s-1 t + 1 .

.00750

Sample Problem 2: Second OrderAcetaldehyde, 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

39.

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-1dm3

= 0.00111 mol dm-3

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.

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 distribution of kinetic energies.

47.


As the temperature increases, the curve flattens and broadens.

Thus at higher temperatures, a larger population of molecules has higher energy.

48.


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.

Maxwell–Boltzmann DistributionsThis fraction of molecules can be found through the expression:

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

50.

Arrhenius EquationSvante 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 J mol-1 K-1

51.

Arrhenius Equation

Taking the natural logarithm of both sides, the equation becomes

1RT

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:

53.

Write the above equation twice, once for each of the twoTemperatures and then subtract the lower temperature conditions from the higher temperature. The equation then becomes:

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 EaLn ---------------- = - ------------------------ 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[ ]

Overview of Kinetics EquationsFirst 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

The rate law for this reaction is found experimentally to be

Rate = k [NO2]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.

NO2 (g) + CO (g) NO (g) + CO2 (g)

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 NOBrBy 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

k1

k−1

[NO] [Br2] = [NOBr2]

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

69.

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

CatalystsOne 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.

CatalystsSome catalysts help to lower the energy for formation for the activated complex or provide a new activated complex with a lower activation energy

71.

AlCl3 + Cl2 Cl+ + AlCl4-

Cl+ + C6H6 C6H5Cl + H+

H+ + AlCl4- HCl + AlCl3

Overall reaction

C6H6 + Cl2 C6H5Cl + HCl

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.

Ozone can also be split up again into ordinary oxygen and an oxygen radical by absorbing ultraviolet light.

Catalysts & Stratospheric Ozone

73.

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 itBack to oxygen. It is then regenerated to repeat theProcess. The result is that each chlorine free radical can Repeat this process many many times. The result is thatOzone is destroyed faster than it is formed, causing its level to drop

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.

Chemical Kinetics Chapter 14

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