Chapter13 chemical kinetics

Chemical Kinetics

Tro, Chemistry: A Molecular Approach 1


• kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds.

• experimentally it is shown that there are 4 factors that influence the speed of a reaction: nature of the reactants, temperature, catalysts,concentration

Kinetics


Defining Rate

• rate is how much a quantity changes in a given period of time

• the speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour)so the rate of your car has units of mi/hr

time

distance Speed


Defining Reaction Rate• the rate of a chemical reaction is generally measured in

terms of how much the concentration of a reactant decreases in a given period of timeor product concentration increases

• for reactants, a negative sign is placed in front of the definition

time

[reactant]

time

[product] Rate

time

ionconcentrat Rate


Reaction Rate Changes Over Time

• as time goes on, the rate of a reaction generally slows down because the concentration of the reactants decreases.

• at some time the reaction stops, either because the reactants run out or because the system has reached equilibrium.


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

25.0

061

04Rate

tt

CC

t

CRate

12

12

25.0

061

84Rate

tt

AA

t

ARate

12

12

0625.0

061

01Rate

tt

ZZ

t

ZRate

12

12

0625.0

061

87Rate

tt

XX

t

XRate

12

12


125.0

061

46Rate

tt

CC

t

CRate

12

12

0625.0

061

76Rate

tt

XX

t

XRate

12

12

125.0

061

42Rate

tt

AA

t

ARate

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


0625.0

061

65Rate

tt

XX

t

XRate

12

12

125.0

061

20Rate

tt

AA

t

ARate

12

12

at t = 32[A] = 2[B] = 2[C] = 6

at t = 32[X] = 6[Y] = 6[Z] = 2

at t = 48[A] = 0[B] = 0[C] = 8

at t = 48[X] = 5[Y] = 5[Z] = 3

125.0

061

68Rate

tt

CC

t

CRate

12

12

0625.0

061

23Rate

tt

ZZ

t

ZRate

12

12

9

Hypothetical ReactionRed Blue

Time (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 65

35 30 70

40 25 75

45 21 79

50 18 82

in this reaction, one molecule of Red turnsinto one molecule of Blue

the number of moleculeswill always total 100

the rate of the reaction canbe measured as the speed ofloss of Red molecules over time, or the speed of gain of Blue molecules over time



Concentration vs Time for Red -> Blue100

84

71

59

50

4235

3025

2118

0

16

29

41

50

5865

7075

7982

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20 25 30 35 40 45 50

Time (sec)

Num

ber

of M

olec

ules

Number Red

Number Blue



Rate 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)

Rat

e,

[Blu

e]/

t


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

equation are not all the same

H2 (g) + I2 (g) 2 HI(g)

• for these reactions, the change in the number of molecules 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 H2 used, 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 concentration of each substance is multiplied by 1/coefficient

t

HI][

2

1

t

]I[

t

]H[ Rate 22


Average Rate

• the average rate is the change in measured concentrations in any particular time periodlinear approximation of a curve

• the larger the time interval, the more the average rate deviates from the instantaneous rate

14

Hypothetical Reaction Red BlueAvg. Rate Avg. Rate Avg. Rate

Time (sec)

Number Red

Number Blue

(5 sec intervals)

(10 sec intervals)

(25 sec intervals)

0 100 0

5 84 16 3.2

10 71 29 2.6 2.9

15 59 41 2.4

20 50 50 1.8 2.1

25 42 58 1.6 2.3

30 35 65 1.4 1.5

35 30 70 1

40 25 75 1 1

45 21 79 0.8

50 18 82 0.6 0.7 1

15

H2 I2

HIAvg. Rate, M/s Avg. Rate, M/s

Time (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t

0.000 1.000

10.000 0.819

20.000 0.670

30.000 0.549

40.000 0.449

50.000 0.368

60.000 0.301

70.000 0.247

80.000 0.202

90.000 0.165

100.000 0.135

Avg. Rate, M/s Avg. Rate, M/s


0.000 1.000 0.000

10.000 0.819 0.362

20.000 0.670 0.660

30.000 0.549 0.902

40.000 0.449 1.102

50.000 0.368 1.264

60.000 0.301 1.398

70.000 0.247 1.506

80.000 0.202 1.596

90.000 0.165 1.670

100.000 0.135 1.730

Stoichiometry tells us that for every 1 mole/L of H2 used, 2 moles/L of HI are made.

Assuming a 1 L container, at 10 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/s

Time (s) [H2], M [HI], M -[H2]/t

0.000 1.000 0.000

10.000 0.819 0.362 0.0181

20.000 0.670 0.660 0.0149

30.000 0.549 0.902 0.0121

40.000 0.449 1.102 0.0100

50.000 0.368 1.264 0.0081

60.000 0.301 1.398 0.0067

70.000 0.247 1.506 0.0054

80.000 0.202 1.596 0.0045

90.000 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

s 10.000

M 181.0

Avg. Rate, M/s Avg. Rate, M/s


0.000 1.000 0.000

10.000 0.819 0.362 0.0181 0.0181

20.000 0.670 0.660 0.0149 0.0149

30.000 0.549 0.902 0.0121 0.0121

40.000 0.449 1.102 0.0100 0.0100

50.000 0.368 1.264 0.0081 0.0081

60.000 0.301 1.398 0.0067 0.0067

70.000 0.247 1.506 0.0054 0.0054

80.000 0.202 1.596 0.0045 0.0045

90.000 0.165 1.670 0.0037 0.0037

100.000 0.135 1.730 0.0030 0.0030


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)

con

cen

trat

ion

, (M

)

[H2], M

[HI], M

average rate in a given time period = slope of the line connecting the [H2] points; and ½ +slope of the line for [HI]

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


Instantaneous Rate

• the instantaneous rate is the change in concentration at any one particular timeslope at one point of a curve

• determined by taking the slope of a line tangent to the curve at that particular pointfirst derivative of the function

for you calculus fans

18

H2 (g) + I2 (g) 2 HI (g) Using [H2], the instantaneous rate at 50 s is:

s

M 0.0070 Rate

s 40

M 28.0 Rate

Using [HI], the instantaneous rate at 50 s is:

s

M 0.0070 Rate

s 40

M 56.0

2

1 Rate

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 of the 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

s

M3-104.40 Rate

s 10

M 000.1M 868.0

3

1

t

]I[

3

1 Rate

s

M3-s

M3- 108.80 104.402t

]H[

Rate2t

]H[

t

]H[

2

1 Rate


Measuring Reaction Rate• in order to measure the reaction rate you need to be

able 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) for reactions 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 usedwhen sampling is used, often the reaction in the sample is

stopped by a quenching technique


Continuous Monitoring• polarimetry – measuring the change in the degree of

rotation of plane-polarized light caused by one of the components over time

• spectrophotometry – measuring the amount of light of a particular wavelength absorbed by one component over time the component absorbs its complimentary color

• total pressure – the total pressure of a gas mixture is stoichiometrically related to partial pressures of the gases in the reaction


Sampling• gas chromatography can 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 aliquots from the mixture and doing quantitative analysis titration for one of the componentsgravimetric analysis


Factors Affecting Reaction RateNature of the Reactants

• nature of the reactants means what kind of reactant molecules 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 reactantscertain 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


• increasing temperature increases reaction ratechemist’s rule of thumb - for each 10°C rise in

temperature, the speed of the reaction doublesfor many reactions

• there is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius which will be examined later

Factors Affecting Reaction RateTemperature


• catalysts are substances which affect the speed of a reaction without being consumed.

• most catalysts are used to speed up a reaction, these are called positive catalysts catalysts 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

Factors Affecting Reaction RateCatalysts


• generally, the larger the concentration of reactant 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)

Factors Affecting Reaction RateReactant Concentration


The Rate Law• the Rate Law of a reaction is the mathematical relationship

between the rate of the reaction and the concentrations of the reactantsand homogeneous catalysts as well

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

• for the reaction aA + bB products the rate law would have the form given belown and m are called the orders for each reactantk is called the rate constant

mnk [B][A] Rate


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

called the order with 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


Reaction Rate Law CH3CN CH3NC Rate = k[CH3CN]

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

2 N2O5 4 NO2 + O2 Rate = k[N2O5]

H2 + I2 2 HI Rate = k[H2][I2]

Tl+3 + Hg2+2 Tl+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.


Reactant Concentration vs. TimeA Products


Half-Life• the half-life, t1/2, of a

reaction is the length of time it takes for the concentration of the reactants to fall to ½ its initial value

• the half-life of the reaction depends on the order of the reaction


Zero Order Reactions• Rate = k[A]0 = k

constant rate reactions

• [A] = -kt + [A]0

• graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0

• t ½ = [A0]/2k• when Rate = M/sec, k = M/sec

[A]0

[A]

time

slope = - k


First Order Reactions• Rate = k[A]

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

slope = -k and y-intercept = ln[A]0

used to determine the rate constant

• t½ = 0.693/k• the half-life of a first order reaction is

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


ln[A]0

ln[A]

time

slope = −k


Half-Life of a First-Order ReactionIs Constant


Rate Data for C4H9Cl + H2O C4H9OH + HCl

Time (sec) [C4H9Cl], M

0.0 0.1000

50.0 0.0905

100.0 0.0820

150.0 0.0741

200.0 0.0671

300.0 0.0549

400.0 0.0448

500.0 0.0368

800.0 0.0200

10000.0 0.0000


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)

conc

entr

atio

n, (

M)


C4H9Cl + H2O C4H9OH + 2 HClRate 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)

Rat

e, (

M/s

)


C4H9Cl + H2O C4H9OH + 2 HClLN([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(c

once

ntra

tion)

slope = -2.01 x 10-3

k =2.01 x 10-3 s-1

s 345s 1001.2

693.0

693.0t

1-3

21

k


Second Order Reactions

• Rate = k[A]2

• 1/[A] = kt + 1/[A]0

• graph 1/[A] vs. time gives straight line with slope = k and y-intercept = 1/[A]0

used to determine the rate constant

• t½ = 1/(k[A0])

• when Rate = M/sec, k = M-1∙sec-1


l/[A]0

1/[A]

time

slope = k

42

Rate Data For2 NO2 2 NO + O2

Time (hrs.)

Partial Pressure

NO2, mmHg ln(PNO2) 1/(PNO2)

0 100.0 4.605 0.01000

30 62.5 4.135 0.01600

60 45.5 3.817 0.02200

90 35.7 3.576 0.02800

120 29.4 3.381 0.03400

150 25.0 3.219 0.04000

180 21.7 3.079 0.04600

210 19.2 2.957 0.05200

240 17.2 2.847 0.05800


Rate Data Graphs For2 NO2 2 NO + O2

Partial 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)

Pres

sure

, (m

mH

g)



ln(PNO2) vs. Time

2.4

2.6

2.8

3

3.2

3.4

3.6

3.8

4

4.2

4.4

4.6

4.8

0 50 100 150 200 250

Time (hr)

ln(p

ress

ure)



1/(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)

Inve

rse

Pres

sure

, (m

mH

g-1

)


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 rate

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

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

is 2, rate constant = slope

• initial ratesby comparing effect on the rate of changing the initial

concentration of reactants one at a time



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


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 -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


[A] vs. Time

0

0.02

0.04

0.06

0.08

0.1

0.12

0 50 100 150 200 250

time, (s)

con

cen

trat

ion

, M


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

(co

nce

ntr

atio

n)


1/([A]) vs. Timey = 0.1x + 10

0

5

10

15

20

25

30

35

40

0 50 100 150 200 250

time, (s)

inve

rse

con

cen

trat

ion

, M

-1


Practice - Complete the Table and Determine the Rate Equation for the 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 with a rate constant of 2.90 x 10-4 s-1 at 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 10-4 s-1

[SO2Cl2]

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

[SO2Cl2][SO2Cl2]0, t, k

0ln[A]tln[A] :processorder 1st afor k

M 0.0175 ]Cl[SO

4.043.790.251]Clln[SO0.0225lns 865s 102.90]Clln[SO

]Clln[SOt]Clln[SO

(-4.04)22

22

1-4-22

02222

e

k


Initial Rate Method• another method for determining the order of a

reactant is to see the effect on the initial rate of the reaction when the initial concentration of that reactant is changed for multiple reactants, keep initial concentration of all

reactants constant except onezero order = changing the concentration has no effect on

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

concentrationdoubling the initial concentration will double the rate

second order = the rate changes by the square of the factor the concentration changesdoubling the initial concentration will quadruple the rate

56

Ex 13.2 – Determine the rate law and rate constant for the 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 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

Write a general rate law including all reactants

Examine the data and find two 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 0.0082

3. 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

mnk [CO]][NO Rate 2




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 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

2M 10.0

M 20.0

][NO

][NO

1expt 2

2expt 2 4 0021.0

0082.0

Rate

Rate

sM

sM

1expt

2expt




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 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

2M 10.0

M 20.0

][NO

][NO

1expt 2

2expt 2 4 0021.0

0082.0

Rate

Rate

sM

sM

1expt

2expt

242

Rate

Rate

][NO

][NO

1expt

2expt

1expt 2

2expt 2

n

n

n




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 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

2M 10.0

M 20.0

[CO]

[CO]

2expt

3expt 1 0082.0

0083.0

Rate

Rate

sM

sM

2expt

3expt

012

Rate

Rate

[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 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033




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 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

mnk [CO]][NO Rate 2n = 2, m = 0

22

022

][NO Rate

[CO]][NO Rate

k

k




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 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

1-1-

2s

M

2s

M

22

sM 21.0M 01.0

0.0021M 10.0 0.0021

1expt for ][NO Rate

k

k

k


Practice - Determine the rate law and rate constant for the reaction NH4

+1 + NO2-1

given the data below. 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

63

Practice - Determine the rate law and rate constant for the reaction NH4

+1 + NO2-1

given the data below. 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

orderfirst 1,

3 3 ][NH Factor Rate

3108.10

103.32

1Expt

2Expt Rate,

30200.0

0600.0

1Expt

2Expt ],[NHFor

4

4

7

7

n

nn

Rate = k[NH4+]n[NO2

]m

orderfirst 1,

2 2 ][NO Factor Rate

2108.10

106.21

3Expt

4Expt Rate,

20202.0

0404.0

3Expt

4Expt ],[NOFor

2

2

7

7

m

mm

1-1-4

23-s

M7-s

M7-

24

sM 1070.2M 1000.4

1010.8

M .2000M .02000 1010.8

1expt for ][NO][NH Rate

k

k

k


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

constant of the rate law

• Svante Arrhenius investigated this relationship and showed that:

RT

Ea

eAk

R is the gas constant in energy units, 8.314 J/(mol∙K)where T is the temperature in kelvins

A is a factor called the frequency factorEa is the activation energy, the extra energy needed to start the molecules reacting



Activation Energy and theActivated 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


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

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 that approach the peak in a given period of time

the activation energy is the difference in energy between the reactants and the activated complex

the activated complex is a chemical species with partial bonds


The Arrhenius Equation:The Exponential Factor

• the exponential factor in the Arrhenius equation is a number between 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 of motion to potential energy in the molecule when the molecules collide increasing the temperature increases the average kinetic energy of the

molecules therefore, increasing the temperature will increase the number of

molecules with sufficient energy to overcome the energy barrier therefore increasing the temperature will increase the reaction rate



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 form y = mx + bwhere y = ln(k) and x = (1/T)

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

(-8.314 J/mol∙K)(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 factor for the reaction O3(g) O2(g) + O(g) given the

following data:

Temp, K k, M-1∙s-1 Temp, K k, M-1∙s-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



following data:

use a spreadsheet to graph ln(k) vs. (1/T)



following data:

Ea = m∙(-R)

solve for Ea

molkJ

molJ4

KmolJ4

1.93

1031.9314.8K 1012.1

a

a

E

E

A = ey-intercept

solve for A 11-11

118.26

sM 1036.4

1036.4

A

eA


Arrhenius Equation:Two-Point Form

• if you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used:

T T

kk

x T TR

TT TT

kk

R x

TT

kk

R x

Ea21

2

121

21

21

2

1

12

2

1 lnln

11

ln

211

2

T

1

T

1ln

R

E

k

k a

a

a

a

E

E

E

mol

kJ

mol

J5

1-4

Kmol

J

Kmol

JsM

sM

1451045.1

K 10290.3 314.8

5639.5

K 895

1

K 701

1

314.857.2

567ln

1

1-1-

-1-1

Ex. 13.8 – The reaction NO2(g) + CO(g) CO2(g) + NO(g) has a rate constant of 2.57 M-1∙s-1 at 701 K and 567 M-1∙s-1 at 895 K. Find the

activation energy in kJ/mol

most activation energies are tens to hundreds of kJ/mol – so 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

T

1

T

1ln

R

E

k

k a


Collision Theory of Kinetics

• for most reactions, in order for a reaction to take place, the reacting molecules must collide into each other.

• once molecules collide they may react together or 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 the proper orientation for new bonds to form.


Effective Collisions• collisions in which these two conditions are

met (and therefore lead to reaction) are called effective collisions

• the higher the frequency of effective collisions, the faster the reaction rate

• when two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex or transition state


Effective CollisionsKinetic Energy Factor

for a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide it can form the activated complex


Effective CollisionsOrientation Effect


Collision Theory andthe Arrhenius Equation

• A is the factor called the frequency factor and is the number of molecules that can approach overcoming the energy barrier

• there are two factors that make up the frequency factor – the orientation factor (p) and the collision frequency factor (z)

RT

E

RT

E aa

pzeeAk


Orientation Factor• the proper orientation results when the atoms are

aligned in such a way that the old bonds can break and the new bonds can form

• the more complex the reactant molecules, the less frequently they will collide with the proper orientation reactions between atoms generally have p = 1 reactions where symmetry results in multiple orientations

leading to reaction have p slightly less than 1• for most reactions, the orientation factor is less than 1

for many, p << 1 there are some reactions that have p > 1 in which an electron

is transferred without direct collision


Reaction Mechanisms• we generally describe chemical reactions with an

equation listing all the reactant molecules and product molecules

• but the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient 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 reaction mechanism

• knowing the rate law of the reaction helps us understand the sequence of steps in the mechanism


An Example of a Reaction Mechanism• Overall reaction:

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) • Mechanism:

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

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

• the steps in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps


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)

Elements of a MechanismIntermediates

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

• since HI is made but then consumed, HI does not show up in the overall reaction

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


Molecularity• the number of reactant particles in an elementary

step 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• each step in the mechanism is like its own little

reaction – with its own activation energy and own rate law

• the rate law for an overall reaction must be determined experimentally

• but the rate law of an elementary step can be deduced 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]


Rate Laws of Elementary Steps


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

other steps• the result is that product production cannot occur any

faster than the slowest step – the step determines the rate of the overall reaction

• we call the slowest step in the mechanism the rate determining step the slowest step has the largest activation energy

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


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 its activation energy is larger.

The rate law of the first step is the same as the rate law of the overall reaction.


Validating a Mechanism

• in order to validate (not prove) a mechanism, two conditions must be met:

1. the elementary steps must sum to the overall reaction

2. the rate law predicted by the mechanism must be consistent with the experimentally observed rate law


Mechanisms with a Fast Initial Step

• when a mechanism contains a fast initial step, the rate limiting step may contain intermediates

• when a previous step is rapid and reaches equilibrium, the forward and reverse reaction rates are equal – so the concentrations of reactants and products of the step are relatedand the product is an intermediate

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


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

2212

1

[NO]]O[N

]O[N [NO]

k

k

kk

222

1

12

22

1

122

2222

]][NO[HRate

][NO][HRate

]O][N[HRate

k

kk

k

kk

k


Ex 13.9 Show that the proposed mechanism for the 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

123

1

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

k

kk

k

kk

k


Catalysts• catalysts are substances that affect the rate of a reaction

without being consumed• catalysts work by providing an alternative mechanism

for the reactionwith a lower activation energy

• catalysts are consumed in an early mechanism step, then made in a later step

mechanism without catalyst

O3(g) + O(g) 2 O2(g) V. Slow

mechanism with catalyst

Cl(g) + O3(g) O2(g) + ClO(g) Fast

ClO(g) + O(g) O2(g) + Cl(g) Slow


Ozone Depletion over the Antarctic


Energy Profile of Catalyzed Reaction

polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals


Catalysts• homogeneous catalysts are in the same phase

as the reactant particlesCl(g) in the destruction of O3(g)

• heterogeneous catalysts are in a different phase than the reactant particlessolid catalytic converter in a car’s exhaust system


Types of Catalysts


Catalytic HydrogenationH2C=CH2 + H2 → CH3CH3


Enzymes

• because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate

• protein molecules that catalyze biological reactions are called enzymes

• enzymes work by adsorbing the substrate reactant onto an active site that orients it for reaction


Enzyme-Substrate BindingLock and Key Mechanism


Enzymatic Hydrolysis of Sucrose

Chapter 13Chemical Kinetics

2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Chapter13 chemical kinetics

Education

rate of change

reaction rate changes

rate rateis

rate isavg

instantaneous rate tro

molecular approach

ms time s h

average rate deviates