1 CHAPTER 11 Rates of Reactions. 2 Chemical Kinetics Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur. Reaction

1

CHAPTER 11

Rates of Reactions

2

Chemical Kinetics

• Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur.

• Reaction rate– increase in concentration of a product per unit time or– decrease in concentration of a reactant per unit time

• Reaction mechanism– series of steps by which a reaction occurs

3

Chemical Kinetics

• Thermodynamics tells us if a reaction can occur• Kinetics tells us how quickly the reaction occurs

– some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible

OUSINSTANTANE

kJ -79=G OHOH+H

SLOW VERY

kJ396G COO C

o2982

-aq

+aq

o298g2g2diamond

l

4

The Rate of Reaction

• Consider the hypothetical reaction,

A(g) + B(g) C(g) + D(g)

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

5

0

0.2

0.4

0.6

0.8

1

1.2

Time

Con

cen

trat

ion

s of

R

eact

ants

& P

rodu

cts

[A] & [B][C] & [D]

• [A] = concentration of A in M ( mol/L)• Note that rxn does not go to completion

6

R =

- A

t

- B

t or

+ C

t

+ D

t

• Reaction rates– rates at which reactants disappear or products appear


7


• Rate of a simple one-step reaction is directly proportional to the concentration of the reacting substance[X] is concentration of X in molarity or moles/L

k = specific rate constant

A(g) B(g) + C(g)

R A or R = k A

8


• For a simple expression like R = k[A]

• doubling the initial concentration of A doubles the initial rate of reaction

• halving the initial concentration of A halves the initial rate of reaction

9


• Rate Law Expressions must be determined experimentally– cannot be determined from balanced equations– trap for new students of kinetics– most chemical reactions are not one-step reactions

• Rate law expressions are also called:– rate laws– rate equations– rate expressions

10


• Order of a reaction– expressed in terms of either:1 each reactant2 overall reaction

• For example:

2 4 N O NO + O R = k N O

first order in N O and first order overall

2 5 g 2 g 2 g 2 5

2 5

11


CH CBr OH CH COH Br R = k[ CH CBr

first order in CH CBr, zero order in OH , and first order overall

3 aq aq-

3 aq aq-

3

3-

3 3 3

3

]

12


CH CBr OH CH COH Br R = k[ CH CBr

first order in CH CBr, zero order in OH , and first order overall

2 NO + O 2 NO R = k[NO] O

second order in NO, first order in O and third order overall

ALL RATE EXPRESSIONS ARE

DETERMINED EXPERIMENTALLY

3 aq aq-

3 aq aq-

3

3-

g 2 g 2 g2

2

2

3 3 3

3

]

,

13


• Look at the following one step reaction and its experimentally determined rate-law expression

2 A B C Rate = k Ag g g 2

14


• Look at the following one step reaction and its experimentally determined rate-law expression

• because it is a second order rate-law expression– doubling the [A] increases the rate of reaction by a factor of 4

22 = 4– halving the [A] decreases the rate of reaction by a factor of 4

(1/2)2 = 1/4

2 A B C Rate = k Ag g g 2

15

Factors That Affect Reaction Rates

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

nature of the reactants concentrations of reactants temperature presence of a catalyst we will look at each factor individually

16

Nature of Reactants

• Broad category that includes the different reacting properties of substances. For example:

1 Sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide.

2 Na H O NaOH H

violent reaction

H ignites and burns

s 2 aq 2 g

2

2 2l

17

Nature of Reactants

2 Calcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.

Ca H O Ca OH H

slow reaction

s 2 aq 2 g 22l

18

Nature of Reactants

3 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.

Mg H O No reactions 2 l

19

Nature of Reactants

• However, Mg reacts with steam rapidly to liberate H2 and form magnesium oxide.

• Differences due to “nature of the reactants”

Mg H O MgO Hs 2C

s 2 g l

o100

20

Concentrations of Reactants

• Simplified representation of effect of different numbers of molecules in the same volume.– 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



21


• Example: 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)

Experiment Initial [A] Initial [B] Initial Rate ofFormation of C

1 0.10 M 0.10 M 2.0 x 10-4 M s-1

2 0.20 M 0.30 M 4.0 x 10-4 M s-1

3 0.10 M 0.20 M 2.0 x 10-4 M s-1

22


Bignorecan k[A]=Ror BAk=R

012constant remains rate

2by increases [B] constant is [A]

3 & 1 sexperiment compare

BAk=R

0

xx

y

yx

y

23


k[A]=Ror k[A]=R

122 2by increases rate

2by increases [A]

2 & 1 sexperiment compare

1

xx

24


k =

R

A

From experiment 1

k =2.0 x 10

2.0 x 10

R = 2.0 x 10 [A]

-4s -3

s

-3s

M

M0101

1

.

25

Concentration vs. Time: The Integrated Rate Equation

• Integrated rate equation relates time and concentration for a chemical or nuclear reaction

• First Order Reactions1st order in reactant A & 1st order overall

• For example:

a A products

common for many chemical reactions and all simple radioactive decays

26


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

Aln 0

27


• Solve the first order integrated rate equation for t.

A

Aln

k a

1t 0

28


• 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

29


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

k a

693.0t

2lnk a

1

A1/2

Aln

k a

1t

1/2

0

01/2

30


• Example: 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

CH2

CH2CH3

CH

( g ) ( g )

31


• Example: 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

CH2

CH2CH3

CH

( g ) ( g )

s 075.0s 2.9

693.0

k

693.0t

1-1/2

32


• Example: 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)

33


• Example: 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)

1-

1/21/2

hr 00101.0hr 688

0.693k

t

0.693k

k

693.0t

34


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

k tAlnAlnk tA

Aln

1-

00

35


unreacted 97%or g 9.2g86.2eA

1.0521.1)--(0.048ln(A)

0.048ln(A)-1.1

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

k tAlnAlnk tA

Aln

1.052

1-

00

36


• For reactions that are second order with respect to a particular reactant and second order overall, the rate equation is

1 1

0A Ak t

37


• At t1/2 [A] = 1/2[A]0, so

11 2

1

0 0

0

/ A Ak t

which has a common denominator of A

1/2

38


• Use the common denominator to derive:

2 1

10 0

0

A Ak t or

Ak t

1/2

1/2

39


• Solve for t1/2:

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

t

k A1/2 1

0

40


• Example: 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 5270C.

(a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 5270C?

CH CHO CH + CO3 g 4 g g

41


tk A

hr

hr

1/2

-1

1

1

2 0 10 010

5 0 10

0

2 1

2

. .

.

M M

42


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

43



1 1

1 1010

2 0 10 200

110 4 0

0

2 1

1 1

A Ak t

A hr hr

A

-1

..

.

MM

M M

44



114

1

14

0 071

0 071 mol

11A

A

A

mol = 1.0 L x 0.071 molL

M

M

M.

? .

45


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

46


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

% unreacted =0.071 mol0.10 mol

unreacted & 29% reacted

100%

71%

47

Rate equations

• Table 11.2 is a review of the Rate Equations!

48

Enrichment - Derivation of Integrated Rate Equations

• For the first order reaction

a A products

the rate can be written as:

rate =1a

At

49


• We can use calculus to rearrange the rate equation and get the integrated rate equation

-lnA

Aa k t (first order)

t

0

50


• 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

51


the second order integrated rate equation

k t aA

1

A

1

0

52


• For a zero order reaction the rate expression is:

k

t a

A

d

d

53


• Which gives this relationship:

the zeroeth order integrated rate equation

k t a-AA

or

k t -aAA

0

0

54

Enrichment -Rate Equations to Determine Reaction Order

• Plots of the integrated rate equations can help us determine the order of a reaction.

• Rearrange the first-order integrated rate equation– laws of logarithms give ln (x/y) = ln x - ln y

0

0

Alnk t aAln

or

k t aAlnAln

55


• The equation for a straight line is:

• Compare this equation to the rearranged rate-lawbmy x

56


• Now we can interpret the parts of the equation as follows:– y = ln[A] plot on y-axis

– m = -ak slope of line

– x = t plot on x-axis

– b = ln[A]0 y-intercept

0Alnk t aAln

b m y

x

57


• Example: Concentration-versus-time data for the thermal decomposition of ethyl bromide are given in the table below. Use the graphs to determine the rate of the reaction and the value of the rate constant

700Kat HBrHCBrHC gg42g52

58


t(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 1.00 1.2 1.5 1.8 2.2 2.7

59


• Make three different graphs

1 [C2H5Br] vs. t

– if linear then reaction is zero order

2 ln [C2H5Br] vs time

– if linear then reaction is first order

3 1/ [C2H5Br] vs. time

– if linear then reaction is second order

60


• Plot of [C2H5Br] versus time

0

1

0 1 2 3 4 5

Time, minutes

61


• Plot of ln [C2H5Br] versus time

-1-0.5

0

0 1 2 3 4 5

Time, minutes

62


• Plot of 1/[C2H5Br] versus time

123

0 1 2 3 4 5

Time, minutes

63


• Note that the only graph which is linear is the plot of ln[C2H5Br] vs. time.

• Thus this is a First Order Reaction.• Determine the value of the rate constant from the slope of the line

on the graph of ln[C2H5Br] vs. time

64


• Note that the only graph which is linear is the plot of ln[C2H5Br] vs. time.

• Thus this is a First Order Reaction.• Determine the value of the rate constant from the slope of the line

on the graph of ln[C2H5Br] vs. time

20.03

60.0slope

14

)20.0(80.0

x-x

y-y slope

12

12

65


• From the equation for a first order reaction we know that:

slope = -a k• In this reaction a = 1

1-min 0.20k

-k-0.20slope

66


• The integrated rate equation for a reaction that is second order in reactant A and second order overall.

• This equation rearranges to: k t aA

1

A

1

0

67


• Compare this equation with an equation for a straight line

0A

1k t a

A

1

bmy x

68


• Now we can interpret the parts of the equation as follows:– y = 1/[A] plot on y-axis

– m = a k slope of line

– x = t plot on x-axis

– b = 1/[A]0 y-intercept

0A

1k t a

A

1

b m y

x

69


70


• Note that the only graph which is linear is the plot of 1/[NO2] vs. time.

• Thus this is a Second Order Reaction.• Determine the value of the rate constant from the slope of the line

on the graph of 1/[NO2] vs. time

71


• Note that the only graph which is linear is the plot of 1/[NO2] vs. time.

• Thus this is a Second Order Reaction.• You can now determine the value of the rate constant from the slope of

the line on the graphSolve in the same way as the previous example

72

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

73


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

• Look at the reaction of methane and oxygen:

• Must start reaction with a match.– Provides the initial energy necessary to break the first few bonds.– Afterwards reaction is self-sustaining.

kJ891OH 2 COO 2 CH g2g2g2g4

74


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

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

• Some ineffective possible collisions are :

X

X

Y YY

Y

X X X X Y Y

75


• An example of an effective collision is:

X Y

X Y

X Y

X Y

X Y

+X Y

76

Transition State Theory

• 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.– Activation energy or Ea

• Mechanical analogy for activation energy

77


Cross section of mountain

Boulder

Eactivation

h

h2

h1

Epot=mgh2

Epot=mgh1

Epot = mgh

Height

78


PotentialEnergy

Reaction Coordinate

X2 + Y2

2 XY

Eactivation - a kinetic quantity

E Ha thermodynamic quantity

79


• The relationship between the activation energy for forward and reverse reactions is– Forward reaction = Ea

– Reverse reaction = Ea + E

– difference = E

80


• The distribution of molecules possessing different energies at a given temperature may be represented as

81

Reaction Mechanisms & the Rate-Law Expression

• Use experimental rate-law to postulate a mechanism.• The slowest step in a reaction mechanism is the rate determining step.• Elementary step: any process that occurs in a single step.• Elementary steps must add to give the balanced chemical equation.• Intermediate: a species which appears in an elementary step which is not a reactant or

product.

82


• Molecularity: the number of molecules present in an elementary step.– Unimolecular: one molecule in the elementary step,

– Bimolecular: two molecules in the elementary step, and

– Termolecular: three molecules in the elementary step.

• It is not common to see termolecular processes (statistically improbable).

83


• Consider the iodide ion catalyzed decomposition of hydrogen peroxide to water and oxygen.

2 H O 2 H O + O2 2I

2 2 g

-

l l

84


• Reaction is known to be first order in H2O2 , first order in I- , and second order overall.

• Mechanism is thought to be:

-22

2222

-2222

-

2--

22

IOHk=R law rate alExperiment

O+OH 2OH 2 rxn. Overall

I+O+OHOH+ IO stepFast

OH+IOI+OH step Slow

85


• Important notes:one hydrogen peroxide molecule and one iodide ion

are involved in the rate determining stepthe iodide ion catalyst is consumed in step 1 and

produced in step 2 in equal amountshypoiodite ion has been detected in reaction mixture

as a short-lived reaction intermediate

86


• Ozone, O3, reacts very rapidly with nitrogen oxide, NO, in a reaction that is first order in each reactant and second order overall.

O + NO NO + O

Experimental rate - law R = k O NO

3 g g 2 g 2 g

3

87


• A possible mechanism is:

Slow step O + NO NO + O

Fast step O + NO NO + O

Overall rxn. O + NO NO + O

3 3

3 2 2

3 2 2

88


• A mechanism that is inconsistent with the rate-law expression is:

Slow step O O + O

Fast step O + NO NO

Overall rxn. O + NO NO O

Rate - law from this mechanism

R = k O cannot be correct

3 2

2

3 2 2

3

89


• Experimentally determined reaction orders indicate the number of molecules involved in:

the slow step only

or

the slow step and the equilibrium steps preceding the slow step.

90

Temperature: The Arrhenius Equation

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

k = Ae

or

ln k = ln A -ERT

-E RT

a

a

91


• Illustrate the effect of temperature on a reaction, with all other variables in Arrhenius equation remaining constant.

Write the Arrhenius equations for two temperatures (T2 >T1 )

92


• Illustrate the effect of temperature on a reaction, with all other variables in Arrhenius equation remaining constant.

Write the Arrhenius equations for two temperatures (T2 >T1 ) ln k ln A -

ERT

and

ln k ln A -E

RT

1a

1

2a

2

93


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

94


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

95


• Consider the rate of a reaction for which Ea=50 kJ/mol, at 200C (293 K) and at 300C (303 K).

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

,

.

.

96


• For reactions that have an Ea50 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]

97


• For reactions that have an Ea50 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

98

Catalysts

• Catalysts change reaction rates by providing an alternative reaction pathway with a different activation energy.

99

Catalysts

100

Catalysts

• Homogeneous catalysts exist in same phase as the reactants.

• Catalysts can operate by increasing the number of effective collisions.

• That is, from the Arrhenius equation: catalysts increase k be increasing A or decreasing Ea.

• A catalyst may add intermediates to the reaction.• When a catalyst adds an intermediate, the activation

energies for both steps must be lower than the activation energy for the uncatalyzed reaction.

101

Catalysts

• Heterogeneous catalysts exist in different phases than the reactants.

• Typical example: solid catalyst, gaseous reactants and products (catalytic converters in cars).

• Most industrial catalysts are heterogeneous.• First step is adsorption (the binding of reactant molecules

to the catalyst surface).• Adsorbed species (atoms or ions) are very reactive.• Molecules are adsorbed onto active sites on the catalyst

surface.

102

Catalysts

• Examples of catalysts include:

ProcessHaber

NH 2H 3 N

npreparatio acid Sulfuric

SO 2OSO 2

chemistryconverter Catalytic

ONNO 2

CO 2O+CO 2

OH 18CO16O 25+HC

g3OFeor Fe

g2g2

g3NiO/Ptor OV

g2g2

g2g2Pt & NiO

g

g2Pt & NiO

g2g

g2g2Pt & NiO

g2g188

32

52

103

• Catalytic oxidation CO to CO2

• Overall reaction

2 CO(g)+ O2(g) 2CO2(g)

• Absorption

CO(g) CO(surface) + O2(g)

O2(g) O2(surface)

• Activation

O2(surface) O(surface)

• Reaction

CO(surface) +O(surface) CO2(surface)

• Desorption

CO2(surface) CO2(g)

104

Catalysts

Enzymes are biological catalysts.

• Most enzymes are protein molecules with large molecular masses (10,000 to 106 amu).

• Enzymes have very specific shapes.

• Most enzymes catalyze very specific reactions.

• Substrates undergo reaction at the active site of an enzyme.

105

Catalysts

106

Catalysts

A substrate locks into an enzyme and a fast reaction occurs.

The products then move away from the enzyme.

• Only substrates that fit into the enzyme lock can be involved in the reaction.

• If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors).

107

Catalysts

• Nitrogen gas cannot be used in the soil for plants or animals.

• Nitrogen compounds, NO3, NO2-, and NO3

- are used in the soil.

• The conversion between N2 and NH3 is a process with a high activation energy (the NN triple bond needs to be broken).

108

Catalysts

109

Catalysts

• An enzyme, nitrogenase, in bacteria which live in root nodules of legumes, clover and alfalfa, catalyses the reduction of nitrogen to ammonia.

• The fixed nitrogen (NO3, NO2-, and NO3

-) is consumed by plants and then eaten by animals.

• Animal waste and dead plants are attacked by bacteria that break down the fixed nitrogen and produce N2 gas for the atmosphere.

110

Synthesis Question

• The Chernobyl nuclear reactor accident occurred in 1986. At the time that the reactor exploded some 2.4 MCi of radioactive 137Cs was released into the atmosphere. The half-life of 137Cs is 30.1 years. In what year will the amount of 137Cs released from Chernobyl finally decrease to 100 Ci? A Ci is a unit of radioactivity called the Curie.

111

Synthesis Question

Ci 100 reaches Chernobylat emitted Csfrom

ity radioactiv the when2426 440 1986

years440 years439 y0230.0

10.1t

t y0230.010.1

t y0230.0Ci 100

Ci 102.4ln

decay eradioactiv for this 1a andk t a A

Aln

Ci 102.4 MCi2.4

y0230.0 y1.30

693.0

t

693.0k

k

693.0t

137

1-

1-

1-6

0

6

1-

21

21

112

Group Question

• 99mTc has a half-life of 6.02 hours and is often used in nuclear medical diagnostic tests. Patients are injected with approximately 10 Ci of 99mTc that is then directed to specific sites in the patient’s body to detect gallstones, brain tumors and function, and other medical conditions. How long will the patient have a higher than normal radioactivity level after they have been injected with 10 Ci of 99mTc?

Documents

1 CHAPTER 11 Rates of Reactions. 2 Chemical Kinetics Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur. Reaction