1 Chapter 16 Chemical Kinetics 化學動力學. 2 Outline 1.The Rate of a Reaction 2.Nature of the Reactants 3.Concentrations of the Reactants: The Rate- Law Expression

1

Chapter 16

Chemical Kinetics 化學動力學

2

Outline1.The Rate of a Reaction2.Nature of the Reactants3.Concentrations of the Reactants: The

Rate-Law Expression4.Concentration Versus Time: The

Integrated Rate Equation5.Collision Theory of Reaction Rates6.Transition State Theory7.Reaction Mechanisms and the Rate-

Law Expression8.Temperature: The Arrhenius Equation9.Catalysts

3

The Rate of a Reaction• Kinetics is the study of rates of chemical

reactions and the mechanisms by which they occur.

• The reaction rate is the increase in concentration of a product per unit time or decrease in concentration of a reactant per unit time.

• A reaction mechanism is the series of molecular steps by which a reaction occurs.

4

The Rate of a Reaction• Thermodynamics (Chapter 15) determines if

a reaction can occur.• Kinetics (Chapter 16) determines how

quickly a reaction occurs.– Some reactions that are

thermodynamically feasible are kinetically so slow as to be imperceptible.

OUSINSTANTANE

kJ -79=G OHOH+H

SLOW VERY

kJ 396G COO C

o2982

-aq

+aq

o298g2g2diamond

l

5

The Rate of Reaction• Consider the hypothetical reaction,

aA(g) + bB(g) cC(g) + dD(g)

• equimolar amounts of reactants, A and B, will be consumed while products, C and D, will be formed as indicated in this graph:

6

0

0.2

0.4

0.6

0.8

1

1.2

0 50 100

150

200

250

300

350

Time

Con

cent

rati

ons

ofR

eact

ants

& P

rodu

cts

[A] & [B]

[C] & [D]

• [A] is the symbol for the concentration of A in M ( mol/L).

• Note that the reaction does not go entirely to completion.– The [A] and [B] > 0 plus the [C] and [D] < 1.

The Rate of Reaction

7

The Rate of Reaction• Reaction rates are the rates at which

reactants disappear or products appear.

• This movie is an illustration of a reaction rate.

8

The Rate of Reaction• Mathematically, the rate of a reaction can

be written as:

- A - B + C + DRate =

a t b t c t d t

9

The Rate of Reaction• The rate of a simple one-step reaction is

directly proportional to the concentration of the reacting substance.

• [A] is the concentration of A in molarity or moles/L.

• k is the specific rate constant.– k is an important quantity in this

chapter.

Ak = Rateor ARate

C + BA (g)(g)(g)

10

The Rate of Reaction• For a simple expression like Rate = k[A]

– If the initial concentration of A is doubled, the initial rate of reaction is doubled.• If the reaction is proceeding twice as fast,

the amount of time required for the reaction to reach equilibrium would be:

A.The same as the initial time.B.Twice the initial time.C.Half the initial time.

• If the initial concentration of A is halved the initial rate of reaction is cut in half.

11

The Rate of Reaction• If more than one reactant molecule

appears in the equation for a one-step reaction, we can experimentally determine that the reaction rate is proportional to the molar concentration of the reactant raised to the power of the number of molecules involved in the reaction.

22

ggg

Xk = Rateor XRate

Z+ YX 2

12

The Rate of Reaction• Rate Law Expressions must be determined

experimentally.– The rate law cannot be determined from

the balanced chemical equation.– This is a trap for new students of

kinetics.• The balanced reactions will not work

because most chemical reactions are not one-step reactions.

• Other names for rate law expressions are:1. rate laws2. rate equations or rate expressions

13

The Rate of Reaction• Important terminology for kinetics.• The order of a reaction can be

expressed in terms of either:1 each reactant in the reaction or2 the overall reaction.

Order for the overall reaction is the sum of the orders for each reactant in the reaction.

• For example:

overall.order first and

ONin order first isreaction This

ONk= Rate

O + NO4ON 2

52

52

g2g2g52

14

The Rate of Reaction• A second example is:

overall.order first and ,OHin order zero

CBr,CHin order first isreaction This

]CBrCHk[= Rate

BrCOHCHOHCBrCH

-

33

33

-aqaq33

-aqaq33

15

The Rate of Reaction• A final example of the order of a

reaction is:

ALLYEXPERIMENT DETERMINED

ARE SEXPRESSION RATE ALL REMEMBER,

overallorder third and ,Oin order first

NO,in order second isreaction This

Ok[NO]=Rate

NO 2O+NO 2

2

2

2

g2g2g

16

The Rate of Reaction• Given the following one step reaction and

its rate-law expression.– Remember, the rate expression would

have to be experimentally determined.

• Because it is a second order rate-law expression:– If the [A] is doubled the rate of the

reaction will increase by a factor of 4. 22 = 4

– If the [A] is halved the rate of the reaction will decrease by a factor of 4. (1/2)2 = 1/4

2ggg

Ak = Rate

CBA 2

17

Factors That Affect Reaction Rates

• There are several factors that can influence the rate of a reaction:

1. The nature of the reactants.2. The concentration of the reactants.3. The temperature of the reaction.4. The presence of a catalyst.• We will look at each factor individually.

18

Nature of Reactants• This is a very broad category that

encompasses the different reacting properties of substances.

• For example sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide.

burns. and ignites H The

reaction. rapid and violent a is This

HNaOH 2OH 2Na 2

2

g2aq2s

19

Nature of Reactants• Calcium reacts with water only slowly at

room temperature to liberate hydrogen and form calcium hydroxide.

reaction. slowrather a is This

HOHCaOH 2Ca g2aq22s

20

Nature of Reactants• The reaction of magnesium with water at

room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.

reaction No OH Mg 2s

21

Nature of Reactants• However, Mg reacts with steam rapidly

to liberate H2 and form magnesium oxide.

• The differences in the rate of these three reactions can be attributed to the changing “nature of the reactants”.

g2sC100

)g(2s HMgOOHMg o

22

Concentrations of Reactants: The Rate-Law

Expression• This movie illustrates how changing

the concentration of reactants affects the rate.

23


Expression• This is a simplified representation of the

effect of different numbers of molecules in the same volume.– The increase in the molecule numbers is

indicative of an increase in concentration.A(g) + B (g) Products

A B

A B

A B BA B

A BA BA B

4 different possible A-B collisions



24


ExpressionExample 16-1: The following rate data were obtained at 25oC for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction?

2 A(g) + B(g) ® 3 C(g)

ExperimentNumber

Initial [A](M)

Initial [B](M)

Initial rate of formation of C

(M/s)

1 0.10 0.10 2.0 x 10-4

2 0.20 0.30 4.0 x 10-4

3 0.10 0.20 2.0 x 10-4

25


Expression

ExperimentNumber

Initial [A](M)

Initial [B](M)

Initial rate of formation of C

(M/s)

1 0.10 0.10 2.0 x 10-4

2 0.20 0.30 4.0 x 10-4

3 0.10 0.20 2.0 x 10-4

26


Expression

Example 16-2: The following data were obtained for the following reaction at 25oC. What are the rate-law expression and the specific rate constant for the reaction?

2 A(g) + B(g) + 2 C(g) ® 3 D(g) + 2 E(g)

ExperimentInitial [A]

(M)Initial [B]

(M)Initial [C]

(M)

Initial rate of formation of D (M/s)

1 0.20 0.10 0.10 2.0 x 10-4

2 0.20 0.30 0.20 6.0 x 10-4

3 0.20 0.10 0.30 2.0 x 10-4

4 0.60 0.30 0.40 1.8 x 10-3

27


Expression

28


ExpressionExample 16-3: consider a chemical reaction between compounds A and B that is first order with respect to A, first order with respect to B, and second order overall. From the information given below, fill in the blanks.

You do it!

29


Expression

ExperimentInitial Rate

(M/s)Initial [A]

(M)Initial [B]

(M)1 4.0 x 10-3 0.20 0.050

2 1.6 x 10-2 ? 0.050

3 3.2 x 10-2 0.40 ?

30

Concentration vs. Time: The Integrated Rate

Equation• The integrated rate equation relates time

and concentration for chemical and nuclear reactions.– From the integrated rate equation we can

predict the amount of product that is produced in a given amount of time.

• Initially we will look at the integrated rate equation for first order reactions.These reactions are 1st order in the

reactant and 1st order overall.

31


Equation• An example of a reaction that is 1st order in

the reactant and 1st order overall is:a A ® products

This is a common reaction type for many chemical reactions and all simple radioactive decays.

• Two examples of this type are:2 N2O5(g) 2 N2O4(g) + O2(g)

238U 234Th + 4He

32


Equation

where:[A]0= mol/L of A at time t=0. [A] = mol/L of

A at time t.k = specific rate constant. t = time elapsed

since beginning of

reaction.a = stoichiometric coefficient of A in balanced

overall equation.

• The integrated rate equation for first order reactions is:

k t aA

Aln 0

33


Equation

• Solve the first order integrated rate equation for t.

• Define the half-life, t1/2, of a reactant as the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A]0.

A

Aln

k a

1t 0

34


Equation• At time t = t1/2, the expression becomes:

k a

693.0t

2lnk a

1t

A1/2

Aln

k a

1t

1/2

1/2

0

01/2

35


EquationExample 16-4: Cyclopropane, an anesthetic, decomposes to propene according to the following equation.

The reaction is first order in cyclopropane with k = 9.2 s-1 at 10000C. Calculate the half life of cyclopropane at 10000C.

CH2 CH2

CH2CH2CH3

CH

(g) (g)

36


EquationExample 16-5: Refer to Example 16-4. How much of a 3.0 g sample of cyclopropane remains after 0.50 seconds?– The integrated rate laws can be used for any

unit that represents moles or concentration.– In this example we will use grams rather

than mol/L.

37


EquationExample 16-6: The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours.

CS2(g) ® CS(g) + S(g)

You do it!

38


Equation

hr) 48)(hr 00101.0(Aln-ln(3.0)

k tAlnAlnk tA

Aln

1-

00

unreacted 97%or g 9.2g86.2eA

1.0521.10)--(0.048Aln

0.048Aln-1.10

hr) 48)(hr 00101.0(Aln-ln(3.0)

k tAlnAlnk tA

Aln

1.052

1-

0

0

39


Equation• For reactions that are second order with

respect to a particular reactant and second order overall, the rate equation is:

• Where:[A]0= mol/L of A at time t=0. [A] = mol/L of

A at time t.k = specific rate constant. t = time

elapsed since beginning of reaction.

a = stoichiometric coefficient of A in balanced overall equation.

k t aA

1

A

1

0

40


Equation• Second order reactions also have a

half-life.– Using the second order integrated rate-

law as a starting point.

• At the half-life, t1/2 [A] = 1/2[A]0.

0

1/200

A ofr denominatocommon a haswhich

k t aA

1

A2/1

1

1/20

1/200

k t aA

1

or k t aA

1

A

2

41


Equation• If we solve for

t1/2:

• Note that the half-life of a second order reaction depends on the initial concentration of A.

01/2 Ak a

1t

42


Equation

Example 16-7: Acetaldehyde, CH3CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide.

The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC?

CH CHO CH + CO3 g 4 g g

43


Equation

Example 16-7:

The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC?

CH CHO CH + CO3 g 4 g g

44


Equation

(b) How many moles of CH3CHO remain after 200 hours?

45


Equation(c) What percent of the CH3CHO remains after 200 hours?

46


EquationExample 16-8: Refer to Example 16-7. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 10.0 L vessel at 527oC?– Note that the vessel size is increased by a

factor of 10 which decreases the concentration by a factor of 10!

You do it!

47


Equation• (b) How many moles of CH3CHO remain after

200 hours? You do it!

48


Equation• (c) What percent of the CH3CHO remains

after 200 hours?You do it!

49


Equation• Let us now summarize the results

from the last two examples.

Initial Moles

CH3CHO

[CH3CHO]

0

(M)

[CH3CHO]

(M)

Moles of CH3CH

O at 200 hr.

% CH3CHO remainin

g

Ex. 16-7

0.10 0.10 0.071 0.071 71%

Ex. 16-8

0.010 0.010 0.0096 0.096 96%

50

Enrichment - Derivation of Integrated Rate

Equations• For the first order reaction

a A productsthe rate can be written as:

t

A

a

1-=Rate

51


Equations• For a first-order reaction, the rate is

proportional to the first power of [A].

-1a

At

k A

52


Equations• In calculus, the rate is defined as the

infinitesimal change of concentration d[A] in an infinitesimally short time dt as the derivative of [A] with respect to time. -

1a

At

k Add

53


Equations• Rearrange the equation so that all of

the [A] terms are on the left and all of the t terms are on the right.

-A

Aa k t

dd

54


Equations• Express the equation in integral form.

-

AA

a k tA

A tdd

0 0

55


Equations• This equation can be evaluated as:

-ln A a k t or

-ln A A a k t - a k 0

which becomes

-ln A A a k t

t0t

t

t

0

0

0

ln

ln

56


Equations• Which rearranges to the integrated

first order rate equation.

k t aA

Aln

t

0

57


Equations• Derive the rate equation for a

reaction that is second order in reactant A and second order overall.

• The rate equation is:

2Ak t a

A

d

d

58


Equations• Separate the variables so that the A

terms are on the left and the t terms on the right.

tk A a

A2 d

d

59


Equations• Then integrate the equation over the

limits as for the first order reaction.

t

0

A

A2 tk a

A

A

0

dd

60


Equations• Which integrates the second order

integrated rate equation.

k t aA

1

A

1

0

61


Equations• For a zero order reaction the rate

expression is:

k

t a

A

d

d

62


Equations• Which rearranges to:

tk aA dd

63


Equations• Then we integrate as for the other

two cases:

t

0

A

A

tk aA0

dd

64


Equations• Which gives the zeroeth order

integrated rate equation.

k t a-AA

or

k t -aAA

0

0

65

Enrichment - Rate Equations to Determine

Reaction Order • Plots of the integrated rate equations

can help us determine the order of a reaction.

• If the first-order integrated rate equation is rearranged. – This law of logarithms, ln (x/y) = ln x - ln y,

was applied to the first-order integrated rate-equation.

0

0

Alnk t aAln

or

k t aAlnAln

66


Reaction Order • The equation for a straight line is:

• Compare this equation to the rearranged first order rate-law.

bmy x

67


Reaction Order

b m y x

• Now we can interpret the parts of the equation as follows: y can be identified with ln[A] and plotted

on the y-axis. m can be identified with –ak and is the

slope of the line. x can be identified with t and plotted on

the x-axis. b can be identified with ln[A]0 and is the

y-intercept.

0Alnk t aAln

68


Reaction Order• Example 16-9: Concentration-versus-

time data for the thermal decomposition of ethyl bromide are given in the table below. Use the following graphs of the data to determine the rate of the reaction and the value of the rate constant. 700Kat HBrHCBrHC gg42g52

69


Reaction Order

Time(min) 0 1 2 3 4 5

[C2H5Br] 1.00 0.82 0.67 0.55 0.45 0.37

ln [C2H5Br] 0.00 -0.20 -0.40 -0.60 -0.80 -0.99

1/[C2H5Br] 1.0 1.2 1.5 1.8 2.2 2.7

70


Reaction Order• We will make three different graphs of the

data.1 Plot the [C2H5Br] (y-axis) vs. time (x-axis)

– If the plot is linear then the reaction is zero order with respect to [C2H5Br].

2 Plot the ln [C2H5Br] (y-axis) vs. time (x-axis)– If the plot is linear then the reaction is first

order with respect to [C2H5Br].3 Plot 1/ [C2H5Br] (y-axis) vs. time (x-axis)

– If the plot is linear then the reaction is second order with respect to [C2H5Br].

71


Reaction Order• Plot of [C2H5Br] versus time.

– Is it linear or not?

[C2H5Br] vs. time

0

0.20.4

0.60.8

11.2

0 1 2 3 4 5

Time (min)

[C2

H5

Br]

72


Reaction Order• Plot of ln [C2H5Br] versus time.


ln [C2H5Br] vs. time

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0 1 2 3 4 5

Time (min)

ln [

C2H

5B

r]

73


Reaction Order• Plot of 1/[C2H5Br] versus time.

– Is it linear or not?1/[C2H5Br] vs. time

0

1

2

3

0 1 2 3 4 5

Time (min)

1/[C

2H5B

r]

74


Reaction Order• Note that the only graph which is linear is

the plot of ln[C2H5Br] vs. time.– Thus this is a first order reaction with

respect to [C2H5Br].• Next, we will determine the value of the

rate constant from the slope of the line on the graph of ln[C2H5Br] vs. time.– Remember slope = (y2-y1)/(x2-x1).

1-

12

12

min 20.0min 3

60.0slope

min 14

)20.0(80.0

x-x

y-y slope

75


Reaction Order• From the equation for a first order

reaction we know that the slope = -a k.– In this reaction a = 1.

.min 0.20kconstant rate theThus

-k-0.20slope1-

76


Reaction Order• The integrated rate equation for a reaction

that is second order in reactant A and second order overall.

• This equation can be rearranged to:

k t aA

1

A

1

0

0A

1k t a

A

1

77


Reaction Order • Compare the equation for a straight line and

the second order rate-law expression.

• Now we can interpret the parts of the equation as follows: y can be identified with 1/[A] and plotted

on the y-axis. m can be identified with a k and is the

slope of the line. x can be identified with t and plotted on

the x-axis b can be identified with 1/[A]0 and is the

y-intercept.

b m y x

0A

1k t a

A

1

78


Reaction OrderExample 16-10: Concentration-versus-time data for the decomposition of nitrogen dioxide are given in the table below. Use the graphs to determine the rate of the reaction and the value of the rate constant

500Kat ONO 2NO 2 g2gg2

79


Reaction Order

Time(min) 0 1 2 3 4 5[NO2] 1.0 0.53 0.36 0.27 0.22 0.18

ln [NO2] 0.0 -0.63 -1.0 -1.3 -1.5 -1.7

1/[NO2] 1.0 1.9 2.8 3.7 4.6 5.5

80


Reaction Order• Once again, we will make three

different graphs of the data.1. Plot [NO2] (y-axis) vs. time (x-axis).

– If the plot is linear then the reaction is zero order with respect to NO2.

2. Plot ln [NO2] (y-axis) vs. time (x-axis).• If the plot is linear then the reaction is

first order with respect to NO2.

3. Plot 1/ [NO2] (y-axis) vs. time (x-axis).– If the plot is linear then the reaction is

second order with respect to NO2.

81


Reaction Order• Plot of [NO2] versus time.


[NO2] vs. time

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5

Time (min)

[NO

2]

82

Enrichment -Rate Equations to Determine

Reaction Order

• Plot of ln [NO2] versus time.– Is it linear or not?

ln [NO2] vs. time

-2

-1.5

-1

-0.5

0

0 1 2 3 4 5

Time (min)

ln [

NO

2]

83


Reaction Order

• Plot of 1/[NO2] versus time.– Is it linear or not?

1/[NO2] vs.time

01

23

45

6

0 1 2 3 4 5

Time (min)

1/[

NO

2]

84


Reaction Order• Note that the only graph which is linear is

the plot of 1/[NO2] vs. time.• Thus this is a second order reaction with

respect to [NO2].• Next, we will determine the value of the

rate constant from the slope of the line on the graph of 1/[NO2] vs. time.

85


Reaction Order

• From the equation for a second order reaction we know that the slope = a k– In this reaction a = 2.

1-1 min 0.45kconstant rate theThus

k 20.90slope

M

2 1

2 1

11

min

1( )y -y (5.50 1.90)slope

x -x (5 1) min

3.60 slope 0.90

4 minM

M

M

86

Collision Theory of Reaction Rates

• Three basic events must occur for a reaction to occur the atoms, molecules or ions must:

1. Collide.2. Collide with enough energy to break and

form bonds.3. Collide with the proper orientation for a

reaction to occur.

87


• One method to increase the number of collisions and the energy necessary to break and reform bonds is to heat the molecules.

• As an example, look at the reaction of methane and oxygen:

• We must start the reaction with a match.– This provides the initial energy

necessary to break the first few bonds.– Afterwards the reaction is self-

sustaining.

kJ 891 OH CO O CH (g)22(g)2(g)4(g)

88


• Illustrate the proper orientation of molecules that is necessary for this reaction.

X2(g) + Y2(g) 2 XY(g)

• Some possible ineffective collisions are :

X

X

Y YY

Y

X X X X Y Y

89


• An example of an effective collision is:

X Y

X Y

X Y

X Y

X Y +

X Y

90


• This picture illustrates effective and ineffective molecular collisions.

91

Transition State Theory• Transition state theory postulates that

reactants form a high energy intermediate, the transition state, which then falls apart into the products.

• For a reaction to occur, the reactants must acquire sufficient energy to form the transition state.– This energy is called the activation energy

or Ea.• Look at a mechanical analog for activation

energy

92

Transition State Theory

DEpot = mgDh

Cross section of mountain

Boulder

Eactivation

Dh

h2

h1

Epot=mgh2

Epot=mgh1

Height

93


PotentialEnergy

Reaction Coordinate

X2 + Y2

2 XY

Eactivation - a kinetic quantity

DE »DHa thermodynamic quantity

Representation of a chemical reaction.

94


95

Transition State Theory• The relationship between the activation

energy for forward and reverse reactions is– Forward reaction = Ea

– Reverse reaction = Ea + DE– difference = DE

96

Transition State Theory• The distribution of molecules

possessing different energies at a given temperature is represented in this figure.

97

Reaction Mechanisms and the Rate-Law

Expression• Use the experimental rate-law to

postulate a molecular mechanism.• The slowest step in a reaction

mechanism is the rate determining step.

98


Expression• Use the experimental rate-law to postulate

a mechanism.• The slowest step in a reaction mechanism is

the rate determining step.• Consider the iodide ion catalyzed

decomposition of hydrogen peroxide to water and oxygen.

g22I

22 O + OH 2 OH 2-

99


Expression• This reaction is known to be first

order in H2O2 , first order in I- , and second order overall.

• The mechanism for this reaction is thought to be:

-22

2222

-2222

-

2--

22

IOHk=R law rate alExperiment

O+OH 2OH 2reaction Overall

I+O+OHOH+ IO stepFast

OH+IOI+OH step Slow

100


Expression• Important notes about this reaction:1. One hydrogen peroxide molecule and one

iodide ion are involved in the rate determining step.

2. The iodide ion catalyst is consumed in step 1 and produced in step 2 in equal amounts.

3. Hypoiodite ion has been detected in the reaction mixture as a short-lived reaction intermediate.

101


Expression• Ozone, O3, reacts very rapidly with

nitrogen oxide, NO, in a reaction that is first order in each reactant and second order overall.

NOOk=Rate is law-rate alExperiment

O+NONO+O

3

g2g2gg3

102


Expression• One possible mechanism is:

223

223

33

O+NONO+Oreaction Overall

O+NONO+O stepFast

O+NONO+O step Slow

103


Expression• A mechanism that is inconsistent

with the rate-law expression is:

correct. becannot mechanism thisproveswhich

Ok=Rate is mechanism thisfrom law-rate The

ONONO+Oreaction Overall

NONO+O stepFast

O+OO step Slow

3

223

2

23

104


Expression• Experimentally determined reaction orders

indicate the number of molecules involved in:

1. the slow step only, or2. the slow step and the equilibrium steps

preceding the slow step.

105

Temperature: The Arrhenius Equation

• Svante Arrhenius developed this relationship among (1) the temperature (T), (2) the activation energy (Ea), and (3) the specific rate constant (k).

k = Ae

or

ln k = ln A -ERT

-E RT

a

a

106


• This movie illustrates the effect of temperature on a reaction.

107


• If the Arrhenius equation is written for two temperatures, T2 and T1 with T2 >T1.

ln k ln A -ERT

and

ln k ln A -E

RT

1a

1

2a

2

108


1. Subtract one equation from the other.

ln k k A - ln A -E

RTERT

ln k kERT

-E

RT

2 1a

2

a

1

2 1a

1

a

2

ln ln

ln

109


2. Rearrange and solve for ln k2/k1.

ln kk

ER T T

or

ln kk

ER

T - TT T

2

1

a

1 2

2

1

a 2 1

2 1

1 1

110


• Consider the rate of a reaction for which Ea=50 kJ/mol, at 20oC (293 K) and at 30oC (303 K). – How much do the two rates differ?

lnkk

ER

T - TT T

lnkk

8.314

K

lnkk

kk

e

2

1

a 2 1

2 1

2

1

Jmol

JK mol

2

1

2

1

0.677

50 000 303 293303 293

0 677

197 2

,

.

.

111


• For reactions that have an Ea»50 kJ/mol, the rate approximately doubles for a 100C rise in temperature, near room temperature.

• Consider:2 ICl(g) + H2(g) ® I2(g) + 2 HCl(g)

• The rate-law expression is known to be R=k[ICl][H2].At 230 C, k = 0.163 s

At 240 C, k = 0.348 s

k approximately doubles

0 -1 -1

0 -1 -1

M

M

112

Catalysts• Catalysts change reaction rates by

providing an alternative reaction pathway with a different activation energy.

113

Catalysts• Homogeneous catalysts exist in same

phase as the reactants.• Heterogeneous catalysts exist in

different phases than the reactants.– Catalysts are often solids.

114

Catalysts• Examples of commercial catalyst

systems include:

NiO and Pt8 218 g 2 g 2 g g

NiO and Ptg 2 g 2 g

NiO and Ptg 2 g 2 g

2 C H +25 O 16 CO 18 H O

2 CO +O 2 CO

2 NO N O

Automobile catalytic converter system

115

Catalysts• This movie shows catalytic

converter chemistry on the Molecular Scale

116

Catalysts• A second example of a catalytic

system is:

npreparatio acid Sulfuric

SO 2OSO 2 g3NiO/Ptor OV

g2g252

117

Catalysts• A third examples of a catalytic system

is:

ProcessHaber

NH 2H 3 N g3OFeor Fe

g2g232

118

Catalysts• Look at the catalytic oxidation of CO

to CO2

• Overall reaction2 CO(g)+ O2(g) ® 2CO2(g)

• AbsorptionCO(g) ® CO(surface) + O2(g)

O2(g) ® O2(surface)

• ActivationO2(surface) ® 2 O(surface)

• ReactionCO(surface) +O(surface) ® CO2(surface)

• DesorptionCO2(surface) ® CO2(g)

119

Catalysts

Documents

1 Chapter 16 Chemical Kinetics 化學 動力學. 2 Outline 1.The Rate of a Reaction 2.Nature of the Reactants 3.Concentrations of the Reactants: The Rate- Law Expression

1 Chapter 16 Chemical Kinetics 化學動力學. 2 Outline 1.The Rate of a Reaction 2.Nature of the Reactants 3.Concentrations of the Reactants: The Rate- Law Expression