Chemical Kinetics

Transcript Chemical Kinetics

Chemical Kinetics

http://www.chem1.com/acad/webtext/dynamics/ dynamics-3.html

http://ibchem.com/IB/ibnotes/brief/kin-sl.htm

http://www.docbrown.info/page03/3_31rates.ht

m#TheeffectofaCatalyst http://www.practicalchemistry.org/experiments/ the-rate-of-reaction-of-magnesium-with hydrochloric-acid,100,EX.html

What do we mean by kinetics?

Kinetics is the study of

• the

rate

occur.

at which chemical reactions • The

reaction mechanism

or pathway through which a reaction proceeds.

How an Airbag Works: http://www.youtube.com/watch?v=dZfLOnXoVOQ • When sodium azide, NaN 3, is ignited by a spark, it releases nitrogen gas which can instantly inflate an airbag.(one 25 th of a second) 10NaN 3 (s) + 2KNO 3 SiO 2 (s)=> K 2 O (s) + 5 Na 2 O(s) + 16N 2 (g) Chemical Kinetics

•

Reaction Rate

The change in the concentration of a reactant or a

product with time (M/s).

Reactant → Products A → B • Since reactants go away with time: Rate = - Δ[A]/Δt = Δ[B]/Δt mol dm -3 s -1 Simulation: http://www.chm.davidson.edu/vce/kinetics/ReactionRates.html

Chemical Kinetics

A => B Rate = - Δ[A]/Δt = Δ[B]/Δt until completion Chemical Kinetics

http://www.avogadro.co.uk/kinetics/rate_equation.htm

When the rate does not depend on the concentration: Chemical Kinetics

Reaction Mechanism

• Most chemical reactions consist of a sequence of two or more simpler reactions.

For example, recent evidence indicates that the reaction 2O 3 ---> 3O 2 occurs in three steps after intense ultraviolet radiation from the sun liberates chlorine atoms from certain compounds in Earth’s stratosphere. Chemical Kinetics

For example:

1. Chlorine atoms decompose ozone according to the equation Cl + O 3 --- > O 2 + ClO 2. Ultraviolet radiation causes the decomposition reaction O 3 --- > O 2 + O 3. ClO produced in the reaction in step 1 reacts with O produced in step 2 according to the equation ClO + O --- > Cl + O 2 Chemical Kinetics

• A complex reaction is one that consists of two or more elementary steps. The complete sequence of elementary steps that make up a complex reaction is called a reaction mechanism.

Chemical Kinetics

Rate Determining Step • Chemical reactions, too, have a “weakest link” in that a complex reaction can proceed no faster than the slowest of its elementary steps. In other words, the slowest elementary step in a reaction mechanism limits the instantaneous rate of the overall reaction.

Chemical Kinetics

Reaction Rates

C 4 H 9 Cl(aq) + H 2 O(l)



C 4 H 9 OH(aq) + HCl(aq)

•

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

Chemical Kinetics

11.

•

Collision Theory

http://www.saskschools.ca/curr_content/chem30_05/2_kinetics/kinetics2_1.htm

• For a reaction to occur: Particles must collide with each other and these collisions must be effective: • Sufficient energy(Activation energy) and proper orientation.

Chemical Kinetics

Factors That Affect Reaction Rates http://www.docbrown.info/page03/3_31rates.htm

http://www.blackgold.ab.ca/ict/Divison3/reactionrates/reactionrates.htm

• • • •

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.

– –

Surface Area http://www.physchem.co.za/OB12-che/rates.htm

Chemical Kinetics

• Maxwell Boltzman As you increase the temperature the rate of reaction increases. As a rough approximation, for many reactions happening at around room temperature, the rate of reaction doubles for every 10°C rise in temperature.

Chemical Kinetics

Catalyst and Ea • A catalyst provides an alternative route for the reaction. That alternative route has a lower activation energy.

Chemical Kinetics

http://www.gcsescience.com/rc4-sodium-thiosulfate-hydrochloric.htm

• investigate the effect that the temperature rate of antacid tablets.

of water has on the reaction • investigate the change in reaction rate when varying sizes of zinc particles are reacted with sulfuric acid. • explore the relationship between concentration and reaction rate.

• observing how quickly reactants are used up or how quickly products are forming. This can be calculated in three ways: of precipitation - When the products of the reaction is a precipitate, which will cloud the solution. Observing a marker through the solution and timing how long it takes for this marker to disappear will determine the rate of reaction.

Hydrochloric Acid and Sodium Thiosulphate HCl (aq) + Na 2 S 2 O 3(aq) NaCl (aq) + SO 2(g) + S (s) + H 2 O (l) Chemical Kinetics

• The graph lines W, X, original, Y and Z on the left diagram are

typical of when a gaseous product is being collected

X , a catalyst was added, forming the same amount of product, but faster.

• Z could represent taking half the

amount of reactants or half a

concentration. The reaction is slower and only half as much gas is formed • W might represent taking double the quantity of reactants, forming twice as much gas e.g. same volume of reactant solution but doubling the concentration, so producing twice as much gas, initially at double the speed (gradient twice as steep).

•

W > X > original > Y > Z See SG graphs

Chemical Kinetics

Measuring the rate of a reaction The change in concentration can be measured by using any property that changes during conversion of reactants to products: CaCO 3 (s) + HCl(aq) => CaCl 2 (aq) + H 2 O(l) + CO 2 (g) 1. mass and volume changes for gaseous reactions 2. change in pH for acids and bases 3. change in conductivity for metals 4. colorimetry for color changes Chemical Kinetics

Measuring the volume of gas produced with time Chemical Kinetics

Mass Loss Method

Chemical Kinetics

• Graphs The graph below shows the volume of carbon dioxide gas produced against time when excess calcium carbonate is added to x cm 3 of 2.0 mol dm −3 hydrochloric acid.

(i) (ii) Write a balanced equation for the reaction.

State and explain the change in the rate of reaction with time. Outline how you would determine the rate of the reaction at a particular time.

(iii) Sketch the above graph on an answer sheet. On the same graph, draw the curves you would expect if: I.

II.

the same volume (x cm 3 ) of 1.0 mol dm −3 HCl is used.

double the volume (2x cm 3 ) of 1.0 mol dm −3 HCl is used.

Label the curves and explain your answer in each case.

Chemical Kinetics

Graph mass x time

Chemical Kinetics

SG page 36

Chemical Kinetics

Example 1: Reaction Rates

C 4 H 9 Cl

(

) + H 2 O (

)  C 4 H 9 OH (

) + HCl (

)

[C 4 H 9 Cl] M In this reaction, the concentration of butyl chloride, C 4 H 9 Cl , was measured at various times, t .

Rate = 

[C 4 H 9 Cl]



t 25.

Reaction Rates Calculation

C 4 H 9 Cl(aq) + H 2 O(l)



C 4 H 9 OH(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:

26.

Reaction Rate Determination

C 4 H 9 Cl(aq) + H 2 O(l)



C 4 H 9 OH(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.

27.

Reaction Rates

• •

C 4 H 9 Cl(aq) + H 2 O(l)



C 4 H 9 OH(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.

28.

Reaction Rates

C 4 H 9 Cl(aq) + H 2 O(l)



C 4 H 9 OH(aq) + HCl(aq)

•

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

29.

Reaction Rates and Stoichiometry

C 4 H 9 Cl(aq) + H 2 O(l)



C 4 H 9 OH (aq) + HCl(aq)

• •

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.

Rate =  [C 4 H 9 Cl] 

= 

[C 4 H 9 OH]



30.

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.

The rate at which H 2 disappears is only half of the rate at which HI is generated

31.

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.

32.

Concentration and Rate

NH 4 + (aq) + NO 2 (aq)



N 2 (g) + 2H 2 O (l) Compare Experiments 1 and 2: when [NH 4 + ] doubles , the initial rate doubles.

33.

Concentration and Rate

NH 4 + (aq) + NO 2 (aq)



N 2 (g) + 2H 2 O (l) Likewise, compare Experiments 5 and 6: when [NO 2 ] doubles , the initial rate doubles.

34.

Concentration and Rate

NH 4 + (aq) + NO 2 (aq)



N 2 (g) + 2H 2 O (l) This equation is called the rate law , and rate constant.

is the 35.

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 constant k is unique for each reaction and the value of k changes with temperature 36.

The Rate

Law

• • • •

Exponents tell the order each reactant.

of the reaction with respect to This reaction is

First-order First-order

in [NH 4 + ] in [NO 2 − ] 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 +

]

[NO

2 -

]

1 37.

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

0.40

0.20

[O2] mol dm -3

0.50

0.25

Rate mol dm -3 s -1

1.6 x 10 -3 8.0 x 10 -4 2.0 x 10 -4

38.

Graphical Representation

• http://www.chem.purdue.edu/gchelp/howtos olveit/Kinetics/IntegratedRateLaws.html



Chemical Kinetics

Zero Order

• For a zero order reaction: [A] versus t (linear for a zero order reaction)

rate = k

(k = - slope of line) Chemical Kinetics

First Order

For a 1 st order reaction:

rate = k[A]

(k = - slope of line) Chemical Kinetics

First-Order Processes

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

CH 3 NC CH 3 CN

How do we know this is a first order reaction?

42.

First-Order Processes

CH 3 NC

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

Does rate=k[CH 3 NC] for all time intervals?

CH 3 CN

43.

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 .

44.

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]

= 0.5 [A] 0 .

45.

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 .

46.

First Order Rate Calculation

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?

47.

First Order Rate Calculation

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?

Ln[A] t – Ln[A] o = -kt Ln[A] t – Ln[0.80] = - ( 0.010 s -1 Ln[A] t = - ( 0.010 s -1 )(90 s) )(90 s) + Ln[0.80] Ln[A] t = -0.90 - 0.2231

Ln[A] t = -1.1231

[A] t = 0.325 mol dm -3

48.

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.

49.

First Order Rate Calculations

Example 2:

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

A = 1.26 mol dm -3 50.

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?

51.

First Order Rate Calculations

Example 3:

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 t = 168.8 years 52.

• [A] versus t (linear for a zero order reaction) • ln [A] versus t (linear for a 1 st order reaction) • 1 / [A] versus t (linear for a 2 nd order reaction) Chemical Kinetics

Second Order

For a 2 nd order reaction:

rate = k[A]

2 (k = slope of line) Chemical Kinetics

Determining Reaction Order

Distinguishing Between 1 st and 2 nd Order The decomposition of NO 2 equation: at 300

C is described by the

NO 2 (

) NO (

) + 1/2 O 2 (

)

A experiment with this reaction yields this data:

Time (s) 0.0

50.0

100.0

200.0

300.0

[NO 2 ], M 0.01000

0.00787

0.00649

0.00481

0.00380

55.

Determining Reaction Order

Distinguishing Between 1 st and 2 nd Order

•

Graphing ln [NO 2 ] vs. t yields: The graph is not a straight line, so this process cannot be first-order in [A].

Time (s)

0.0

50.0

100.0

200.0

300.0

[NO 2 ], M

0.01000

0.00787

0.00649

0.00481

0.00380

ln [NO 2 ]

-4.610

-4.845

-5.038

-5.337

-5.573

56.

Second-Order Reaction Kinetics

A graph of 1/[NO 2 ] vs. t gives this plot.

Time (

) [NO 2 ],

0.0

0.01000

50.0

0.00787

100.0

200.0

300.0

0.00649

0.00481

0.00380

1/[NO 2 ]

100 127 154 208 263 • •

This is a straight line. Therefore, the process is second-order in [NO 2 ].

The slope of the line is the rate constant, k.

57.

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.

58.

Sample Problem 1: Second Order

Acetaldehyde, CH 3 CHO, decomposes by second-order kinetics with a rate constant of 0.334

mol

-1 dm 3 s -1 at 500 o C. 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

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

1 .

.00150

= 0.334 mol -1 dm 3 s -1

+ 1 .

.00750

666.7 = 0.334

+ 133.33

0.334

t t

= 533.4

= 1600 seconds

59.

Sample Problem 2: Second Order

Acetaldehyde, CH 3 CHO, decomposes by second-order kinetics with a rate constant of 0.334

mol

-1 dm 3 s -1 at 500 o C. If the initial concentration of acetaldehyde is 0.00200 M. Find the concentration after 20 minutes (1200 seconds)

Solution

1 .

[A] t 1 .

[A] t = 0.334 mol -1 dm 3 s -1

(1200s)

+ 1 0.00200 mol dm = 0.334 mol -1 dm 3 s -1

(1200s)

+ 500 mol -1 dm 3 -3 .

= 900.8 mol -1 dm 3 [A] t = 1 _____ .

900.8

mol -1 dm 3

= 0.00111 mol dm -3 60.

Summary of Kinetics Equations

Rate Laws Integrat ed Rate Laws First order Half life Second order Second order

complicated complicated

61.

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 10 o C rise in temperature 62.

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.

63.

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 64.

Activation Energy

• •

In other words, there is a minimum amount of energy required for reaction: the activation energy , E

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.

65.

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.

66.

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.

67.

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.

68.

Maxwell

–

Boltzmann Distributions

• • As the temperature increases, the curve flattens and broadens.

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

69.

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.

•

As a result, the reaction rate increases.

70.

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 .

71.

Arrhenius Equation

Svante Arrhenius developed a mathematical relationship between

and

E a

: where

is the frequency factor , a number that represents the likelihood that collisions would occur with the proper orientation for reaction. E a energy. T is the activation 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 72.

Arrhenius Equation

Taking the natural logarithm of both sides, the equation becomes

RT y

When

is determined experimentally at several temperatures,

E a

can be calculated from the slope of a plot of ln k vs. 1/T .

73.

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:

74.

Arrhenius Equation Sample Problem 1

The rate constant for the decomposition of hydrogen iodide was determined at two different temperatures 2HI



H 2 + I 2 . At 650 K, k 1 = 2.15 x 10 -8 dm 3 At 700 K, k 2 = 2.39 x 10 -7 dm 3 mol mol -1 -1 s s Find the activation energy for this reaction.

-1 -1 2.39 x 10 -7 Ea

Ln ---------------- = - ------------------------ x

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

[

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

]

E a = 180,000 J mol -1 = 180 kJ mol -1 75.

Overview of Kinetics Equations

First order Second order Second order

Rate Laws Integrated Rate Laws

complicated

Half-life

complicated

Rate and Temp (T)

76.

Reaction Mechanisms

The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism .

77.

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 .

78.

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.

79.

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 .

80.

Slow Initial Step

NO 2 (g) + CO (g)



NO (g) + CO 2 (g)

• • •

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.

81.

Slow Initial Step

•

A proposed mechanism for this reaction is Step 1: NO 2 Step 2: NO 3 + NO 2



+ CO



NO 3 NO 2 + NO (slow) + CO 2 (fast)

• •

The NO 3 step.

intermediate is consumed in the second As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

82.

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.

83.

Fast Initial Step

•

A proposed mechanism is Step 1 is an

equilibrium

it includes the forward and reverse reactions

84.

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 ]?

85.

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, Rate

= Rate

86.

Fast Initial Step

•

Because Rate

= Rate

1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us

−1 [NO] [Br 2 ] = [NOBr 2 ] 87.

Fast Initial Step

Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives

88.

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 89.

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 90.

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 91.

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 AlCl 3 + Cl 2



Cl + + AlCl 4 Cl + + C 6 H 6



C 6 H 5 Cl + H + H + + AlCl 4 Overall reaction



HCl + AlCl 3 C 6 H 6 + Cl 2



C 6 H 5 Cl + HCl 92.

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.

93.

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 94.

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.

95.

96.