chem chapter 13 lec

8/13/2019 Chem Chapter 13 LEC

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Chapter 13

Chemical

Kinetics

2008, Prentice Hall

Chemistry: A Molecular Approach, 1stEd.

Nivaldo Tro

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA


2/103

Tro, Chemistry: A Molecular Approach 2

kinetics is the study of the factors that affectthe speed of a reaction and the mechanismby which a reaction proceeds.

experimentally it is shown that there are 4factors that influence the speed of areaction:

nature of the reactants,

temperature,

catalysts,

concentration

Kinetics


3/103


Defining Rate

rateis how much a quantity changes in a givenperiod of time

the speed you drive your car is a ratethe distance

your car travels (miles) in a given period of time (1hour)

so the rate of your car has units of mi/hr

time

distanceSpeed


4/103


Defining Reaction Rate

the rate of a chemical reaction is generally measured interms of how much the concentration of a reactant

decreases in a given period of time

or product concentration increases

for reactants, a negative sign is placed in front of thedefinition

time

[reactant]

time

[product]Rate

time

ionconcentrat

Rate


5/103


Reaction Rate Changes Over Time

as time goes on, the rate of a reaction generally

slows downbecause the concentration of the reactants decreases.

at some time the reaction stops, either because thereactants run out or because the system has

reached equilibrium.


6/103


at t = 0

[A] = 8[B] = 8

[C] = 0

at t = 0

[X] = 8[Y] = 8

[Z] = 0

at t = 16

[A] = 4

[B] = 4

[C] = 4

at t = 16

[X] = 7

[Y] = 7

[Z] = 1

25.0061

04Rate

tt

CC

tCRate

12

12

25.0061

84Rate

tt

AA

tARate

12

12

0625.0061

01Rate

tt

ZZ

tZRate

12

12

0625.0061

87Rate

tt

XX

tXRate

12

12


7/103


125.0061

46Rate

tt

CC

t

CRate

12

12

0625.0061

76Rate

tt

XX

tXRate

12

12

125.0061

42Rate

tt

AA

tARate

12

12

at t = 16

[A] = 4

[B] = 4[C] = 4

at t = 16

[X] = 7

[Y] = 7[Z] = 1

at t = 32

[A] = 2

[B] = 2

[C] = 6

at t = 32

[X] = 6

[Y] = 6

[Z] = 2

0625.0

061

12Rate

tt

ZZ

t

ZRate

12

12


8/103


9/103

9

Hypothetical Reaction

RedBlueTime

(sec)

Number

Red

Number

Blue

0 100 0

5 84 16

10 71 29

15 59 41

20 50 50

25 42 58

30 35 6535 30 70

40 25 75

45 21 79

50 18 82

in this reaction,one molecule of Redturns

into one molecule of Blue

the number of moleculeswill always total 100

the rate of the reaction can

be measured as the speed ofloss of Red molecules

over time, or the speed of

gain of Blue molecules

over time


10/103



RedBlueConcentration vs Time for Red -> Blue

100

84

71

59

50

42

3530

2521

18

0

16

29

41

50

58

6570

75 79

82

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20 25 30 35 40 45 50

Time (sec)

NumberofMolecules

Number Red

Number Blue


11/103



RedBlueRate of Reaction Red -> Blue

5, 3.2

10, 2.6

15, 2.4

20, 1.8

25, 1.6

30, 1.4

35, 1 40, 1

45, 0.8

50, 0.6

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 10 20 30 40 50

time, (sec)

Rate,[

Blue]/t


12/103


Reaction Rate and Stoichiometry in most reactions, the coefficients of the balanced

equation are not all the sameH2 (g)+ I2 (g)2 HI(g)

for these reactions, the change in the number ofmolecules of one substance is a multiple of the change in

the number of molecules of another for the above reaction, for every 1 mole of H2used, 1 mole of I2

will also be used and 2 moles of HI made

therefore the rate of change will be different

in order to be consistent, the change in the concentrationof each substance is multiplied by 1/coefficient

t

HI][

2

1

t

]I[

t

]H[Rate 22


13/103


Average Rate

the average rate is the change in measuredconcentrations in any particular time period

linear approximation of a curve

the larger the time interval, the more the averagerate deviates from the instantaneous rate


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14

Hypothetical Reaction RedBlueAvg. Rate Avg. Rate Avg. Rate

Time(sec) NumberRed NumberBlue (5 secintervals) (10 secintervals) (25 secintervals)

0 100 05 84 16 3.2

10

71

29

2.6

2.9

15 59 41 2.420 50 50 1.8 2.125 42 58 1.6 2.330

35

65

1.4

1.5

35 30 70 140 25 75 1 145 21 79 0.850 18 82 0.6 0.7 1


15/103

15

H2 I2HI

Avg. Rate, M/sAvg. Rate, M/sTime (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t

0.000 1.00010.000 0.81920.000 0.67030.000 0.54940.000 0.44950.000 0.36860.000 0.30170.000 0.24780.000 0.20290.000 0.165

100.000 0.135


0.000 1.000 0.00010.000 0.819 0.36220.000 0.670 0.66030.000 0.549 0.90240.000 0.449 1.10250.000 0.368 1.26460.000 0.301 1.39870.000 0.247 1.50680.000 0.202 1.59690.000 0.165 1.670

100.000 0.135 1.730

Stoichiometry tells us that for

every 1 mole/L of H2used,

2 moles/L of HI are made.

Assuming a 1 L container, at10 s, we used 0.181 moles of

H2. Therefore the amount of

HI made is 2(0.181 moles) =

0.362 molesAt 60 s, we used 0.699 moles

of H2. Therefore the amount

of HI made is 2(0.699 moles)

= 1.398 moles

Avg. Rate, M/sTime (s) [H2], M [HI], M -[H2]/t

0.000 1.000 0.00010.000 0.819 0.362 0.018120.000 0.670 0.660 0.014930.000 0.549 0.902 0.012140.000 0.449 1.102 0.010050.000 0.368 1.264 0.008160.000 0.301 1.398 0.006770.000 0.247 1.506 0.005480.000 0.202 1.596 0.004590.000 0.165 1.670 0.0037

100.000 0.135 1.730 0.0030

The average rate is

the change in the

concentration in a

given time period.

In the first 10 s, the

[H2] is -0.181 M,

so the rate is

s

M0181.0

s10.000

M181.0


0.000 1.000 0.00010.000 0.819 0.362 0.0181 0.018120.000 0.670 0.660 0.0149 0.014930.000 0.549 0.902 0.0121 0.012140.000 0.449 1.102 0.0100 0.010050.000 0.368 1.264 0.0081 0.008160.000 0.301 1.398 0.0067 0.006770.000 0.247 1.506 0.0054 0.005480.000 0.202 1.596 0.0045 0.004590.000 0.165 1.670 0.0037 0.0037

100.000 0.135 1.730 0.0030 0.0030


16/103


Concentration vs. Time for H2+ I2--> 2HI

0.000

0.200

0.400

0.600

0.800

1.000

1.200

1.400

1.600

1.800

2.000

0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000

time, (s)

concentration,

(M

[H2], M

[HI], M

average rate in a given

time period =

slope ofthe line connecting the

[H2] points; and +slope

of the line for [HI]

the average rate for thefirst 10 s is 0.0181 M/sthe average rate for thefirst 40 s is 0.0150 M/sthe average rate for thefirst 80 s is 0.0108 M/s


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Instantaneous Rate

the instantaneous rate is the change inconcentration at any one particular time

slope at one point of a curve

determined by taking the slope of a line tangentto the curve at that particular point

first derivative of the functionfor you calculus fans


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18

H2 (g)+ I2 (g)2 HI(g) Using [H2], theinstantaneous rate at50 s is:

s

M0.0070Rate

s40

M28.0Rate

Using [HI], the

instantaneous rate at

50 s is:

s

M0.0070Rate

s40

M56.0

2

1Rate

E 13 1 F h i i h [I ] h f


19/103

Ex 13.1 - For the reaction given, the [I] changes from

1.000 M to 0.868 M in the first 10 s. Calculate the

average rate in the first 10 s and the [H+].

H2O2(aq)+ 3 I

(aq)+ 2 H+

(aq)I3

(aq)+ 2 H2O(l)

Solve the equation

for the Rate (in

terms of the change

in concentration ofthe Given quantity)

Solve the equation

of the Rate (in terms

of the change in the

concentration for the

quantity to Find) for

the unknown value

sM3-104.40Rate

s10

M000.1M868.0

3

1

t

]I[

3

1Rate

s

M3-

s

M3- 108.80104.402

t

]H[

Rate2t

]H[

t

]H[

2

1Rate


20/103


Measuring Reaction Rate

in order to measure the reaction rate you need to beable to measure the concentration of at least one

component in the mixture at many points in time

there are two ways of approaching this problem (1) forreactions that are complete in less than 1 hour, it is best

to use continuous monitoring of the concentration, or

(2) for reactions that happen over a very long time,

sampling of the mixture at various times can be used

when sampling is used, often the reaction in the sample is

stopped by a quenching technique


21/103


Continuous Monitoring polarimetrymeasuring the change in the degree of

rotation of plane-polarized light caused by one of thecomponents over time

spectrophotometrymeasuring the amount of light ofa particular wavelength absorbed by one componentover time the component absorbs its complimentary color

total pressurethe total pressure of a gas mixture isstoichiometrically related to partial pressures of thegases in the reaction


22/103


Sampling gas chromatographycan measure the concentrations

of various components in a mixture for samples that have volatile components

separates mixture by adherence to a surface

drawing off periodic aliquotsfrom the mixture anddoing quantitative analysis

titration for one of the components

gravimetric analysis


23/103


Factors Affecting Reaction Rate

Nature of the Reactants

nature of the reactants means what kind of reactantmolecules and what physical condition they are in.small molecules tend to react faster than large molecules;

gases tend to react faster than liquids which react faster

than solids;

powdered solids are more reactive than blocks

more surface area for contact with other reactants

certain types of chemicals are more reactive than others

e.g., the activity series of metals

ions react faster than molecules

no bonds need to be broken


24/103


increasing temperature increases reaction ratechemists rule of thumb - for each 10C rise in

temperature, the speed of the reaction doublesfor many reactions

there is a mathematical relationship between

the absolute temperature and the speed of areaction discovered by Svante Arrhenius

which will be examined later


Temperature


25/103


catalysts are substances which affect the speed ofa reaction without being consumed.

most catalysts are used to speed up a reaction,

these are called positive catalystscatalysts used to slow a reaction are called negative

catalysts.

homogeneous = present in same phase heterogeneous = present in different phase

how catalysts work will be examined later


Catalysts


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generally, the larger the concentration ofreactant molecules, the faster the reaction

increases the frequency of reactant molecule

contactconcentration of gases depends on the partial

pressure of the gas

higher pressure = higher concentration concentration of solutions depends on the

solute to solution ratio (molarity)


Reactant Concentration


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The Rate Law

the Rate Law of a reaction is the mathematical relationshipbetween the rate of the reaction and the concentrations ofthe reactants

and homogeneous catalysts as well

the rate of a reaction is directly proportional to theconcentration of each reactant raised to a power

for the reactionaA + bB products the rate lawwould have the form given below

nand mare called the orders for each reactant

kis called the rate constant

mnk [B][A]Rate


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Reaction Order the exponent on each reactant in the rate law is

called the orderwith respect to that reactant the sum of the exponents on the reactants is

called the order of the reaction

The rate law for the reaction:2 NO(g) + O2(g) 2 NO2(g)

is Rate = k[NO]2[O2]

The reaction is

second order with respect to [NO],

first order with respect to [O2],

and third order overall


29/103


Reaction Rate Law

CH3CN CH3NC Rate = k[CH3CN]

CH3CHO CH4+ CO Rate = k[CH3CHO]3/2

2 N2O54 NO2+ O2 Rate = k[N2O5]

H2+ I22 HI Rate = k[H2][I2]

Tl+3+ Hg2+2Tl+1+ 2 Hg+2 Rate = k[Tl

+3][Hg2

+2][Hg

+2]-1

Sample Rate Laws

The reaction is autocatalytic, because a product affects the rate.

Hg2+is a negative catalyst, increasing its concentration slows the reaction.


30/103


Reactant Concentration vs. Time

A Products


31/103


Half-Life the half-life, t1/2, of a

reaction is the lengthof time it takes for theconcentration of the

reactants to fall to itsinitial value

the half-life of thereaction depends on

the order of thereaction

Z O d R i


32/103


Zero Order Reactions Rate = k[A]0= kconstant rate reactions

[A] = -kt + [A]0 graph of [A] vs. time is straight line with

slope = -kand y-intercept = [A]0

t= [A0]/2k when Rate = M/sec, k= M/sec

[A]0

[A]

time


33/103


First Order Reactions

Rate = k[A] ln[A] = -kt + ln[A]0 graph ln[A] vs. time gives straight line with

slope = -kand y-intercept = ln[A]0

used to determine the rate constant

t= 0.693/k the half-life of a first order reaction is

constant

the when Rate = M/sec, k= sec-1


34/103


ln[A]0

ln[A]

time

H lf Lif f Fi t O d R ti


35/103


Half-Life of a First-Order Reaction

Is Constant

R D f


36/103


Rate Data for

C4H9Cl + H2O C4H9OH + HCl

Time (sec) [C4H9Cl], M0.0 0.1000

50.0 0.0905

100.0 0.0820

150.0 0.0741

200.0 0.0671

300.0 0.0549

400.0 0.0448500.0 0.0368

800.0 0.0200

10000.0 0.0000


37/103


C4H9Cl + H2O C4H9OH + 2 HCl

Concentration vs. Time for the Hydrolysis of C4H9Cl

0

0.02

0.04

0.06

0.08

0.1

0.12

0 200 400 600 800 1000

time, (s)

concentration,(M

)


38/103



Rate vs. Time for Hydrolysis of C4H9Cl

0.0E+00

5.0E-05

1.0E-04

1.5E-04

2.0E-04

2.5E-04

0 100 200 300 400 500 600 700 800

time, (s)

Rate,

(M/s)


39/103



LN([C4H9Cl]) vs. Time for Hydrolysis of C4H9Cl

y = -2.01E-03x - 2.30E+00

-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0 100 200 300 400 500 600 700 800

time, (s)

LN(concentration)

slope =

-2.01 x 10-3

k =

2.01 x 10-3s-1

s345

s1001.2

693.0

693.0t

1-3

21

k


40/103


Second Order Reactions

Rate = k[A]2

1/[A] = kt + 1/[A]0

graph 1/[A] vs. time gives straight line with

slope = kand y-intercept = 1/[A]0used to determine the rate constant

t= 1/(k[A0])

when Rate = M/sec, k= M-1sec-1


41/103


l/[A]0

1/[A]

time


42/103

R D G h F


43/103


Rate Data Graphs For

2 NO22 NO + O2Partial Pressure NO2, mmHg vs. Time

0.0

10.0

20.0

30.0

40.0

50.0

60.0

70.0

80.0

90.0

100.0

0 50 100 150 200 250

Time, (hr)

Pressure,

(mmHg


44/103

R D G h F


45/103


Rate Data Graphs For

2 NO22 NO + O21/(PNO2) vs Time

1/PNO2= 0.0002(time) + 0.01

0.00000

0.01000

0.02000

0.03000

0.04000

0.05000

0.06000

0.07000

0 50 100 150 200 250

Time, (hr)

In

versePressure,

(mm

Hg-1)


46/103


Determining the Rate Law can only be determined experimentally

graphically rate = slope of curve [A] vs. time

if graph [A] vs time is straight line, then exponent on A in ratelaw is 0, rate constant = -slope

if graph ln[A] vs time is straight line, then exponent on A in ratelaw is 1, rate constant = -slope

if graph 1/[A] vs time is straight line, exponent on A in rate lawis 2, rate constant = slope

initial ratesby comparing effect on the rate of changing the initial

concentration of reactants one at a time


47/103


Practice - Complete the Table and


48/103


Practice Complete the Table and

Determine the Rate Equation for the

Reaction A 2 Prod[A], (M) Prod M Time sec ln([A]) 1/[A]

0.100 0 0

0.067 50

0.050 100

0.040 150

0.033 200

0.029 250



49/103


Practice Complete the Table and


Reaction A 2 Prod[A], (M) Prod M Time sec ln([A]) 1/[A]

0.100 0 0 -2.3 10

0.067 0.066 50 -2.7 15

0.050 0.100 100 -3.0 20

0.040 0.120 150 -3.2 25

0.033 0.134 200 -3.4 30

0.029 0.142 250 -3.5 35


50/103


[A] vs. Time

0

0.02

0.04

0.06

0.08

0.1

0.12

0 50 100 150 200 250

time, (s)

concent

ration,

M


51/103


LN([A]) vs. Time

-3.8

-3.6

-3.4

-3.2

-3

-2.8

-2.6

-2.4

-2.2

-2

0 50 100 150 200 250

time, (s)

Ln(concentration)


52/103


1/([A]) vs. Time

y = 0.1x + 10

0

5

10

15

20

25

30

35

40

0 50 100 150 200 250

time, (s)

inverseconcentration,

M-1


53/103




Reaction A 2 Prod

the reaction is second order, Rate =-[A]

t

= 0.1 [A]2

Ex. 13.4 The reaction SO2Cl2(g) SO2(g) + Cl2(g) is first order


54/103

Ex. 13.4 The reaction SO2Cl2(g)SO2(g) Cl2(g)is first order

with a rate constant of 2.90 x 10-4s-1at a given set of conditions.

Find the [SO2Cl2] at 865 s when [SO2Cl2]0= 0.0225 M

the new concentration is less than the original, as

expected

[SO2Cl2]0= 0.0225 M, t = 865, k= 2.90 x 10-4s-1

[SO2Cl2]

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

[SO2Cl2][SO2Cl2]0, t, k

0ln[A]tln[A]:processorder1stafor k

M0.0175]Cl[SO

4.043.790.251]Clln[SO

0.0225lns865s102.90]Clln[SO

]Clln[SOt]Clln[SO

(-4.04)22

22

1-4-

22

02222

e

k

I iti l R t M th d


55/103


Initial Rate Method another method for determining the order of a

reactant is to see the effect on the initial rate of thereaction when the initial concentration of thatreactant is changed

for multiple reactants, keep initial concentration of allreactants constant except one

zero order = changing the concentration has no effect onthe rate

first order = the rate changes by the same factor as theconcentration

doubling the initial concentration will double the rate second order = the rate changes by the square of the factor

the concentration changes

doubling the initial concentration will quadruple the rate

Ex 13.2Determine the rate law and rate constant for the


56/103

56

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

given the data below.

Expt.

Number

Initial

[NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.00823. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

Write a general

rate lawincluding all

reactants

Examine the

data and findtwo experiments

in which the

concentration of

one reactant

changes, but the

other

concentrations

are the same

Expt.

Number

Initial

[NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.00823. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

Comparing Expt #1 and Expt #2, the

[NO2] changes but the [CO] does not

mn

k [CO]][NORate 2



57/103




Determine by

what factor the

concentrations

and rates change

in these two

experiments.

Expt.

Number

Initial

[NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.00823. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

2M10.0

M20.0

][NO

][NO

1expt2

2expt2 4

0021.0

0082.0

Rate

Rate

sM

sM

1expt

2expt



58/103




Determine to

what power the

concentration

factor must be

raised to equal

the rate factor.

Expt.

Number

Initial

[NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.00823. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

2M10.0

M20.0

][NO

][NO

1expt2

2expt2

40021.0

0082.0

Rate

Rate

sM

sM

1expt

2expt

2

42

RateRate

][NO][NO

1expt

2expt

1expt2

2expt2

n

n

n



59/103




Repeat for the

other reactants

Expt.

Number

Initial

[NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.00823. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

2M10.0M20.0

[CO][CO]

2expt

3expt 10082.00083.0

RateRate

sM

sM

2expt

3expt

0

12

RateRate

[CO][CO]

2expt

3expt

2expt

3expt

m

m

m

Expt.

Number

Initial

[NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.00823. 0.20 0.20 0.0083

4. 0.40 0.10 0.033



60/103




Substitute the

exponents into

the general rate

law to get the

rate law for the

reaction

Expt.

Number

Initial

[NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.00823. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

mn

k [CO]][NORate 2n= 2, m= 02

2

022

][NORate[CO]][NORate

kk



61/103




Substitute the

concentrations

and rate for any

experiment into

the rate law and

solve for k

Expt.

Number

Initial

[NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.00823. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

1-1-

2

sM

2s

M

22

sM21.0M01.0

0.0021

M10.00.0021

1exptfor][NORate

k

k

k

Practice - Determine the rate law and rate constant for the


62/103


reaction NH4+1+ NO2

-1 2 2 H

2O


Expt.No.

Initial[NH4

+], MInitial[NO2

-], MInitial Rate,(x 10-7), M/s

1 0.0200 0.200 10.8

2 0.0600 0.200 32.3

3 0.200 0.0202 10.8

4 0.200 0.0404 21.6

Practice - Determine the rate law and rate constant for the


63/103

63

reaction NH4+1+ NO2

-1 2 2 H

2O


Expt.No.

Initial[NH4

+], MInitial[NO2

-], MInitial Rate,(x 10-7), M/s

1 0.0200 0.200 10.8

2 0.0600 0.200 32.3

3 0.200 0.0202 10.8

4 0.200 0.0404 21.6

orderfirst1,

33][NHFactorRate

3108.10

103.32

1Expt

2ExptRate,

30200.0

0600.0

1Expt

2Expt],[NHFor

4

4

7

7

n

nn

Rate = k[NH4+]n[NO2

]m

orderfirst1,

22][NOFactorRate

2108.10

106.21

3Expt

4ExptRate,

20202.0

0404.0

3Expt

4Expt],[NOFor

2

2

7

7

m

mm

1-1-423-s

M7-s

M7-

24

sM1070.2M1000.4

1010.8

M.2000M.020001010.8

1exptfor

][NO][NHRate

k

k

k


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The Effect of Temperature on Rate changing the temperature changes the rate

constant of the rate law

Svante Arrhenius investigated this relationshipand showed that:

RT

Ea

eAk

Ris the gas constant in energy units, 8.314 J/(molK)

where Tis the temperature in kelvins

Ais a factor called the frequency factor

Eais the activation energy, the extra energy needed to start

the molecules reacting


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A ti ti E d th


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Activation Energy and the

Activated Complex

energy barrier to the reaction

amount of energy needed to convert reactants

into the activated complexaka transition state

the activated complex is a chemical species with

partially broken and partially formed bondsalways very high in energy because partial bonds


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Isomerization of Methyl Isonitrile

methyl isonitrile rearranges to acetonitrile

in order for the reaction to occur,

the H3C-N bond must break; and

a new H3C-C bond form

Energy Profile for the


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68

Energy Profile for the

Isomerization of Methyl Isonitrile

As the reaction

begins, the C-N

bond weakens

enough for the

CN group to

start to rotate

the collision frequency

is the number of

molecules thatapproach the peak in a

given period of time

the activation energy

is the difference in

energy between thereactants and the

activated complex

the activated complex

is a chemical species

with partial bonds

The Arrhenius Equation:


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The Arrhenius Equation:

The Exponential Factor

the exponential factor in the Arrhenius equation is a numberbetween 0 and 1 it represents the fraction of reactant molecules with sufficient

energy so they can make it over the energy barrier

the higher the energy barrier (larger activation energy), the fewer

molecules that have sufficient energy to overcome it

that extra energy comes from converting the kinetic energy ofmotion to potential energy in the molecule when the moleculescollide

increasing the temperature increases the average kinetic energy of the

molecules

therefore, increasing the temperature will increase the number ofmolecules with sufficient energy to overcome the energy barrier

therefore increasing the temperature will increase the reaction rate


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Arrhenius Plots


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Arrhenius Plots the Arrhenius Equation can be algebraically

solved to give the following form:

AR

Ek a ln

T

1)ln(

this equation is in the formy= mx+ bwhere y= ln(k) andx= (1/T)

a graph of ln(k) vs. (1/T) is a straight line

(-8.314 J/molK)(slope of the line) =Ea,(in Joules)

ey-intercept=A, (unit is the same as k)

Ex. 13.7 Determine the activation energy and frequency


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factor for the reaction O3(g)O2(g)+ O(g)given the

following data:

Temp, K k,M-1s-1 Temp, K k,M-1s-1

600 3.37 x 103 1300 7.83 x 107

700 4.83 x 104 1400 1.45 x 108

800 3.58 x 105

1500 2.46 x 108

900 1.70 x 106 1600 3.93 x 108

1000 5.90 x 106 1700 5.93 x 108

1100 1.63 x 107 1800 8.55 x 108

1200 3.81 x 107 1900 1.19 x 109



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following data:

use a

spreadsheet

to graph

ln(k) vs.(1/T)



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following data:

Ea= m(-R)

solve forEa

mol

kJ

mol

J4

Kmol

J4

1.93

1031.9314.8K1012.1

a

a

E

E

A = ey-intercept

solve forA 11-11

118.26

sM1036.4

1036.4

A

eA


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Arrhenius Equation:

Two-Point Form if you only have two (T,k) data points, the

following forms of the Arrhenius Equation canbe used:

TT

k

k

xTTR

TT

TT

k

k

R x

TT

k

k

R xEa

21

2

1

21

21

21

2

1

12

2

1

lnln

11

ln

211

2

T1

T1ln

RE

kk a

Ex. 13.8The reaction NO2(g) + CO(g)CO2(g)+ NO(g)has a rate

f 2 57 M 1 1 701 K d 567 M 1 1 895 K Fi d


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a

a

a

E

E

E

mol

kJ

mol

J5

1-4

Kmol

J

Kmol

JsM

sM

1451045.1

K10290.3314.8

5639.5

K895

1

K701

1

314.857.2

567ln

1

1-1-

-1-1

constant of 2.57 M-1s-1at 701 K and 567 M-1s-1at 895 K. Find

the activation energy in kJ/mol

most activation energies are tens to hundreds of

kJ/molso the answer is reasonable

T1= 701 K, k1= 2.57 M-1

s-1

, T2= 895 K, k2= 567 M-1

s-1

Ea, kJ/mol

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

EaT1, k1, T2, k2

211

2

T1

T1ln

RE

kk a

C lli i Th f Ki ti


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Collision Theory of Kinetics

for most reactions, in order for a reaction to takeplace, the reacting molecules must collide intoeach other.

once molecules collide they may react togetheror they may not, depending on two factors -1. whether the collision has enough energy to

"break the bonds holding reactant molecules

together";

2. whether the reacting molecules collide in theproper orientation for new bonds to form.

Eff ti C lli i


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Effective Collisions collisions in which these two conditions are

met (and therefore lead to reaction) are calledeffective collisions

the higher the frequency of effective

collisions, the faster the reaction rate

when two molecules have an effectivecollision, a temporary, high energy (unstable)

chemical species is formed - called anactivated complex ortransition state

Effective Collisions


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Kinetic Energy Factor

for a collision to lead

to overcoming the

energy barrier, thereacting molecules

must have sufficient

kinetic energy so that

when they collide it

can form the

activated complex

Eff ti C lli i


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Orientation Effect

Collision Theory and


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Collision Theory and

the Arrhenius Equation

Ais the factor called the frequency factorandis the number of molecules that can approach

overcoming the energy barrier

there are two factors that make up the frequencyfactorthe orientation factor (p) and the

collision frequency factor (z)

RTE

RTE aa

pzeeAk


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Orientation Factor

the proper orientation results when the atoms arealigned in such a way that the old bonds can break andthe new bonds can form

the more complex the reactant molecules, the lessfrequently they will collide with the proper orientation

reactions between atoms generally havep= 1 reactions where symmetry results in multiple orientations

leading to reaction havepslightly less than 1

for most reactions, the orientation factor is less than 1

for many,p 1 in which an electron

is transferred without direct collision

Reaction Mechanisms


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Reaction Mechanisms we generally describe chemical reactions with an

equation listing all the reactant molecules and productmolecules

but the probability of more than 3 molecules collidingat the same instant with the proper orientation andsufficient energy to overcome the energy barrier is

negligible most reactions occur in a series of small reactions

involving 1, 2, or at most 3 molecules

describing the series of steps that occur to produce the

overall observed reaction is called a reactionmechanism

knowing the rate law of the reaction helps usunderstand the sequence of steps in the mechanism


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Elements of a Mechanism


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H2(g) + 2 ICl(g)

2 HCl(g) + I2(g)1) H2(g) + ICl(g) HCl(g) + HI(g)

2) HI(g) + ICl(g) HCl(g) + I2(g)

Intermediates

notice that the HI is a product in Step 1, but then areactant in Step 2

since HI is made but then consumed, HI does notshow up in the overall reaction

materials that are products in an early step, but then areactant in a later step are called intermediates


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Molecularity

the number of reactant particles in an elementarystep is called its molecularity

a unimolecular step involves 1 reactant particle

a bimolecular step involves 2 reactant particlesthough they may be the same kind of particle

a termolecular step involves 3 reactant particlesthough these are exceedingly rare in elementary steps

Rate Laws for Elementary Steps


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Rate Laws for Elementary Steps each step in the mechanism is like its own little

reactionwith its own activation energy and ownrate law

the rate law for an overall reaction must bedetermined experimentally

but the rate law of an elementary step can bededuced from the equation of the step

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)

1) H2(g) + ICl(g) HCl(g) + HI(g) Rate = k1[H2][ICl]2) HI(g) + ICl(g) HCl(g) + I2(g) Rate = k2[HI][ICl]

R L f El S


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Rate Laws of Elementary Steps

Rate Determining Step


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Rate Determining Step in most mechanisms, one step occurs slower than the

other steps the result is that product production cannot occur anyfaster than the slowest stepthe step determines therate of the overall reaction

we call the slowest step in the mechanism the ratedetermining step the slowest step has the largest activation energy

the rate law of the rate determining step determines therate law of the overall reaction

Another Reaction Mechanism


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Another Reaction MechanismNO2(g)+ CO(g)NO(g)+ CO2(g) Rateobs= k[NO2]

2

1) NO2(g)+ NO2(g)NO3(g)+ NO(g) Rate = k1[NO2]2 slow2) NO3(g)+ CO(g)NO2(g) + CO2(g) Rate = k2[NO3][CO] fast

The first step in this mechanism

is the rate determining step.

The first step is slower than the

second step because itsactivation energy is larger.

The rate law of the first step is

the same as the rate law of the

overall reaction.


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Mechanisms with a Fast Initial Step

when a mechanism contains a fast initial step, the ratelimiting step may contain intermediates

when a previous step is rapid and reaches equilibrium,the forward and reverse reaction rates are equalso theconcentrations of reactants and products of the step arerelated

and the product is an intermediate

substituting into the rate law of the RDS will produce arate law in terms of just reactants

An Example


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An Example

2 NO(g) N2O2(g) Fast

H2(g)+ N2O2(g) H2O(g)+ N2O(g) Slow Rate = k2[H2][N2O2]

H2(g)+ N2O(g)H2O(g)+ N2(g) Fast

k1

k-1

2 H2(g)+ 2 NO(g)2 H2O(g)+ N2(g) Rateobs= k [H2][NO]2

for Step 1 Rateforward= Ratereverse

2

1

122

221

2

1

[NO]]O[N

]O[N[NO]

k

kkk

222

1

12

221

122

2222

]][NO[HRate

][NO][HRate

]O][N[HRate

k

kk

k

k

k

k

Ex 13.9 Show that the proposed mechanism for the

reaction 2 O3( ) 3 O2( ) matches the observed rate law


94/103


reaction 2 O3(g)3 O2(g)matches the observed rate law

Rate = k[O3]2[O2]

-1

O3(g) O2(g)+ O(g) Fast

O3(g)+ O(g) 2 O2(g) Slow Rate = k2[O3][O]

k1

k-1

for Step 1 Rateforward= Ratereverse

1

231

1

2131

]][O[O[O]

][O][O][O

k

k

kk

1-2

23

1

12

1-23

1

132

32

][O][ORate

]][O[O][ORate

][O][ORate

kkk

k

kk

k


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Ozone Depletion over the Antarctic

Energy Profile of Catalyzed Reaction


97/103


Energy Profile of Catalyzed Reaction

polar stratospheric clouds

contain ice crystals that

catalyze reactions that

release Cl from

atmospheric chemicals

Catalysts


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Catalysts

homogeneous catalystsare in the same phaseas the reactant particles

Cl(g)in the destruction of O3(g)

heterogeneous catalystsare in a different phase

than the reactant particlessolid catalytic converter in a cars exhaust system

Types of Catalysts


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Types of Catalysts

Catalytic Hydrogenation


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Catalytic Hydrogenation

H2C=CH

2+ H

2CH

3CH

3


101/103


Enzymes

because many of the molecules are large andcomplex, most biological reactions require acatalyst to proceed at a reasonable rate

protein molecules that catalyze biologicalreactions are called enzymes

enzymes work by adsorbing the substratereactant onto an active site that orients it forreaction

Enzyme Substrate Binding


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Enzyme-Substrate Binding

Lock and Key Mechanism


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chem chapter 13 lec

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