Chapter 13 Outline

Chapter 13 Chemical Kine1cs

Kine%cs •  kine%cs is the study of the factors that affect the _____ of a reac%on and the __________ by which a reac%on proceeds.

•  experimentally it is shown that there are 4 factors that influence the speed of a reac%on: – nature of the reactants, –  temperature, – catalysts, – concentra1on

2

Defining Rate

•  _____ is how much a quan%ty changes in a given period of %me

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

3

Defining Reac%on Rate

•  the rate of a chemical reac%on is generally measured in terms of how much the concentra%on of a reactant decreases in a given period of %me –  or product concentra%on increases

•  for reactants, a nega%ve sign is placed in front of the defini%on

4

Reac%on Rate Changes Over Time

•  as %me goes on, the rate of a reac%on generally __________ – because the concentra%on of the reactants decreases.

•  at some %me the reac%on stops, either because the reactants run out or because the system has reached equilibrium.

5

6

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

Rate = !" A[ ]"t

= !A[ ] 2

! A[ ]1( )t2 ! t1( )

Rate =Rate = !

" X[ ]"t

= !X[ ] 2

! X[ ]1( )t2 ! t1( )

Rate =

7

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

Rate =! C[ ]!t

=C[ ] 2

" C[ ]1( )t2 " t1( )

Rate =

Rate =! Z[ ]!t

=Z[ ] 2

" Z[ ]1( )t2 " t1( )

Rate =

8

Rate = !" X[ ]"t

= !X[ ] 2

! X[ ]1( )t2 ! t1( )

Rate =

Rate = !" A[ ]"t

= !A[ ] 2

! A[ ]1( )t2 ! t1( )

Rate =

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

9

Rate =! C[ ]!t

=C[ ] 2

" C[ ]1( )t2 " t1( )

Rate =

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

Rate =! Z[ ]!t

=Z[ ] 2

" Z[ ]1( )t2 " t1( )

Rate =

10

Rate = !" X[ ]"t

= !X[ ] 2

! X[ ]1( )t2 ! t1( )

Rate =

Rate = !" A[ ]"t

= !A[ ] 2

! A[ ]1( )t2 ! t1( )

Rate =

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

11

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

Rate =! C[ ]!t

=C[ ] 2

" C[ ]1( )t2 " t1( )

Rate =

Rate =! Z[ ]!t

=Z[ ] 2

" Z[ ]1( )t2 " t1( )

Rate =

Hypothe%cal Reac%on Red → Blue

12

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 turns into one molecule of Blue

the number of molecules will always total 100

the rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time


13

100

84

71

59

50

42

35 30

25 21

18

0

16

29

41

50

58

65 70

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

Nu

mb

er o

f M

olec

ule

s

Time (sec)

Concentration vs Time for Red -> Blue

Number Red

Number Blue


14

Rate of Reaction Red -> Blue

5, 3.2

10, 2.615, 2.4

20, 1.825, 1.6

30, 1.4

35, 1 40, 145, 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, Δ

[Blu

e]/Δ

t

Reac%on Rate and Stoichiometry •  in most reac%ons, the coefficients of the balanced equa%on are not all the same

•  for these reac%ons, the change in the number of molecules of one substance is a mul%ple of the change in the number of molecules of another –  for the above reac%on, 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 concentra%on of each substance is mul%plied by 1/coefficient

15

a) 0.0014 Ms–1

b) 0.00070 Ms–1

c) 0.0028 Ms–1

d) None of the above are correct

H2O2 can be used as a disinfectant; it decomposes as:

2 H2O2 → 2 H2O + O2

If the rate of appearance of O2 is 0.0014 Ms–1, what is the rate of disappearance of H2O2?

Copyright © 2011 Pearson Education, Inc.

a) 3.4 x 10–4 M/s

b) 5.9 x 10–4 M/s

c) 4.1 x 10–4 M/s

d) 1.6 x 10–4 M/s

e) 2.1 x 10–4 M/s

Determine the rate of disappearance of NO in the first 100 seconds.

[NO] time (s)

0.100 0 0.078 50 0.059 100 0.043 150 0.031 200


Average Rate

•  the average rate is the change in measured __________ in any par%cular %me period –  linear approxima%on of a curve

•  the larger the %me interval, the more the average rate deviates from the instantaneous rate

18


19

Avg. 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

20

H2 I2 HI

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

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


Time (s) [H2], M [HI], M -Δ[H2]/Δt 1/2 Δ[HI]/Δt 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 moles At 60 s, we used 0.699 moles of H2. Therefore the amount of HI made is 2(0.699 moles) = 1.398 moles

21

H2 I2 HI


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



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

Avg. Rate, M/s

Time (s) [H2], M [HI], M -Δ[H2]/Δt 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.

22

H2 I2 HI



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

23

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)

conc

entr

atio

n, (M

)

[H2], M

[HI], M

average rate in a given time period = - slope of the line connecting the _____ points; and ½ +slope of the line for _____

the average rate for the first 10 s is 0.0181 M/s

the average rate for the first 40 s is 0.0150 M/s

Instantaneous Rate

•  the instantaneous rate is the change in concentra%on at any one par%cular %me – slope at __________ of a curve

•  determined by taking the slope of a line ______ to the curve at that par%cular point – _______________ of the func%on

24

H2 (g) + I2 (g) → 2 HI (g)

25

Using [H2], the instantaneous rate at 50 s is:

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

Example -‐ For the reac%on given, the [I-‐] changes from 2.000 M to 0.735 M in the first 25 s. Calculate the average rate in the

first 25 s and the Δ[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

Rate = ! 13"

#$%

&'([I! ](t

=

Rate =

Rate = ![H+ ]!t

=

![H+ ]!t

=

Measuring Reac%on Rate •  in order to measure the reac%on rate you need to be able to measure the concentra%on of at least one component in the mixture at many points in %me

•  there are two ways of approaching this problem •  (1) for reac%ons that are complete in less than 1 hour, it is best to use __________ monitoring of the concentra%on,

•  (2) for reac%ons that happen over a very long %me, _________ of the mixture at various %mes can be used –  when sampling is used, o`en the reac%on in the sample is stopped by a quenching technique

27

Factors Affec%ng Reac%on Rate Nature of the Reactants

•  nature of the reactants means what kind of reactant molecules and what physical condi%on they are in. –  small molecules tend to react _____ than large molecules; –  gases tend to react _____ than liquids which react _____ than solids;

–  powdered solids are _____ reac%ve than “blocks” •  more surface area for contact with other reactants

–  certain types of chemicals are more reac%ve than others •  e.g., the ac%vity series of metals

–  ions react faster than molecules •  no bonds need to be broken

28

Factors Affec%ng Reac%on Rate Temperature

•  increasing temperature _________ reac%on rate – chemist’s rule of thumb -‐ for each 10°C rise in temperature, the speed of the reac%on doubles

•  for many reac%ons

•  there is a mathema%cal rela%onship between the absolute temperature and the speed of a reac%on discovered by Svante Arrhenius which will be examined later

29

Factors Affec%ng Reac%on Rate Catalysts

•  catalysts are substances which affect the speed of a reac%on _______________consumed. – homogeneous = present in _____ phase – heterogeneous = present in ______ phase

•  how catalysts work will be examined later

30

Factors Affec%ng Reac%on Rate Reactant Concentra%on

•  generally, the larger the concentra%on of reactant molecules, the _____ the reac%on –  increases the __________ of reactant molecule contact

– concentra%on of gases depends on the par%al pressure of the gas • higher pressure = higher concentra%on

•  concentra%on of solu%ons depends on the solute to solu%on ra%o (__________)

31

The Rate Law •  the Rate Law of a reac%on is the mathema%cal rela%onship

between the rate of the reac%on and the concentra%ons of the reactants –  and homogeneous catalysts as well

•  the rate of a reac%on is _______________ to the concentra%on of each reactant raised to a power

•  for the reac%on nA + mB → products the rate law would have the form given below –  n and m are called the _____ for each reactant –  k is called the __________

32

The Integrated Rate Law

•  Rela%onship between the concentra%on of reactants and %me

•  Obtained from integra%on of differen%al rate laws

•  It has the form of an equa%on for a straight line y = mx + b

33

Reac%on Order •  the exponent on each reactant in the rate law is called the _____ with respect to that __________

•  the sum of the exponents on the reactants is called the _______________

•  The rate law for the reac%on:

34

2 NO(g) + O2(g) → 2 NO2(g)

is Rate = The reaction is

second order with respect to [NO], first order with respect to [O2],

and third order overall

Reactant Concentra%on vs. Time A → Products

35

Half-‐Life •  the half-‐life, ___, of a reac%on is the length of %me it takes for the concentra%on of the reactants to fall ____ its ini%al value

•  the half-‐life of the reac%on depends on the _____ of the reac%on

36

Zero Order Reac%ons •  Rate = =

– constant rate reac1ons •  Integrated Rate Law: [A] = •  graph of [A] vs. %me is straight line with slope = ___ and y-‐intercept = ____

•  t ½ = •  when Rate = M/sec, k = M/sec

37

[A]0

[A]

time

Half Life from Rate Law

38

First Order Reac%ons •  Rate = •  Integrated Rate Law: ln[A] = •  graph ln[A] vs. %me gives straight line with slope = ___ and y-‐

intercept = ______ –  used to determine the rate constant

•  t½ = •  the half-‐life of a first order reac1on is constant •  the when Rate = M/sec, k = sec-‐1

39

ln[A]0

ln[A]

time

Second Order Reac%ons •  Rate = •  Integrated Rate Law: 1/[A] = •  graph 1/[A] vs. %me gives straight line with slope = __ and

y-‐intercept = ____ –  used to determine the rate constant

•  t½ = •  when Rate = M/sec, k = M-‐1·sec-‐1

40

l/[A]0

1/[A]

time

Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions

Order Rate Law Concentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

ln[A] = ln[A]0 - kt

1 [A]

= 1

[A]0 + kt

[A] = [A]0 - kt

t½ ln2 k

=

t½ = [A]0 2k

t½ = 1 k[A]0

13.3

Ex. 13.4 – The reac%on 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 condi%ons. 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

for a 1st order process:

ln[SO2Cl2 ]=ln[SO2Cl2 ]=ln[SO2Cl2 ]=[SO2Cl2 ]=

What is the order of a reaction that has the following time and concentration data?

Time (s) [Concentration] 0 0.01000 50 0.00887 100 0.00797 150 0.00723 200 0.00662 250 0.00611

a)  Zero order b)  First order c)  Second order d)  None of the above


a) Second order

b) Third order

c) Fourth order

d) Fifth order e) Sixth order

Determine the overall order for a reaction with the following rate law.

Rate = k[A]2[B][C]3


Prac%ce -‐ Determine the Rate Equa%on for the Reac%on A → 2 Prod

45

[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

46

[A] vs. Time

0

0.02

0.04

0.06

0.08

0.1

0.12

0 50 100 150 200 250

time, (s)

conc

entr

atio

n, M

47

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(c

once

ntra

tion)

48

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

conc

entr

atio

n, M

-1

Ini%al Rate Method •  another method for determining the order of a reactant is to see the effect on the ini%al rate of the reac%on when the ini%al concentra%on of that reactant is changed –  for mul1ple reactants, keep ini1al concentra1on of all reactants constant except one

–  zero order = changing the concentra%on has _____ on the rate

–  first order = the rate changes by the _____ _____as the concentra%on

•  doubling the ini%al concentra%on will double the rate –  second order = the rate changes by the _____ of the factor the concentra%on changes

•  doubling the ini%al concentra%on will quadruple the rate

49

Ex 13.2 – Determine the rate law and rate constant for the reac%on NO2(g) + CO(g) → NO(g) + CO2(g)

given the data below.

50

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

Rate =



51

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

[NO2 ]expt 2

[NO2 ]expt 1

=Rateexpt 2

Rateexpt 1

=



52

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

[NO2 ]expt 2

[NO2 ]expt 1

!

"##

$

%&&

n

=Rateexpt 2

Rateexpt 1



53

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

[CO]expt 3

[CO]expt 2

=Rateexpt 3

Rateexpt 2

=

[CO]expt 3

[CO]expt 2

!

"##

$

%&&

m

=Rateexpt 3

Rateexpt 2

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



54

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 2=n = 2, m = 0

Rate = Rate =



55

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

Rate = k[NO2 ]2

for expt 1

a) 1 b) 0 c) 2 d) 3 e) 4

What is the order of the reaction with respect to OH–? 2 ClO2 (aq) + 2 OH– (aq) → ClO3

– (aq) + ClO2– (aq) + H2O (l)

[ClO2] [OH–] Initial Rate (M/s)

0.060 0.030 0.0248 0.020 0.030 0.00276 0.020 0.090 0.00828


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

•  Svante Arrhenius inves%gated this rela%onship and showed that:

57

R is the gas constant in energy units, _______________ where T is the temperature in _____

A is a factor called the frequency factor Ea is the _______________ , the extra energy needed to start the molecules reacting

58

The Arrhenius Equa%on: The Exponen%al Factor

•  the exponen%al factor in the Arrhenius equa%on is a number between 0 and 1

•  it represents the frac%on of reactant molecules with sufficient energy so they can make it over the energy barrier –  the higher the energy barrier (larger ac%va%on energy), the fewer

molecules that have sufficient energy to overcome it •  that extra energy comes from conver%ng the kine%c energy of

mo%on to poten%al energy in the molecule when the molecules collide –  increasing the temperature increases the average kine%c 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 reac%on rate

59

60

Arrhenius Plots •  the Arrhenius Equa%on can be algebraically solved to give the following form:

61

this equation is in the form y = mx + b where y = and x = 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 ac%va%on energy and frequency factor for the reac%on O3(g) → O2(g) + O(g) given the following data:

62

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


63

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


64

Ea = m·(-R) solve for Ea

Ea =

Ea =

A = ey-intercept solve for A

A =A =

Arrhenius Equa%on: Two-‐Point Form

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

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⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

211

2T1

T1ln

RE

kk a

Ex. 13.8 – The reac%on 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 ac%va%on 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:

Ea T1, k1, T2, k2

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

211

2T1

T1ln

RE

kk a

The chirping of tree crickets has sometimes been used to predict temperatures. At 25.0 oC the rate was 179 chirps/min. At 21.7 oC it was 142 chirps/min. What is the Ea of the process?

a)  316 J b)  51,300 J c)  0.749 kJ d)  6.16 kJ


Collision Theory of Kine%cs

•  for most reac%ons, in order for a reac%on to take place, the reac%ng 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 "_______________holding reactant molecules together";

2. whether the reac%ng molecules collide in the _______________for new bonds to form.

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Effec%ve Collisions •  ____________________: collisions in which these two condi%ons are met

•  the higher the frequency of effec%ve collisions, the faster the reac%on rate

•  when two molecules have an effec%ve collision, a temporary, high energy (unstable) chemical species is formed -‐ called an ac1vated complex or _______________

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Effec%ve Collisions Orienta%on Effect

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Collision Theory and the Arrhenius Equa%on

•  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 – __________ factor (p) – ____________________factor (z)

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!

k = A e"EaRT

#

$ %

&

' ( = pze

"EaRT

Orienta%on Factor •  the proper orienta%on 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 orienta%on –  reac%ons between atoms generally have p = 1

•  for most reac%ons, the orienta%on factor is less than 1 –  for many, p << 1

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Reac%on Mechanisms •  we generally describe chemical reac%ons with an equa%on lis%ng all the reactant molecules and product molecules

•  but the probability of more than 3 molecules colliding at the same instant with the proper orienta%on and sufficient energy to overcome the energy barrier is negligible

•  most reac%ons occur in a series of small reac%ons involving 1, 2, or at most 3 molecules

•  describing the series of steps that occur to produce the overall observed reac%on is called a __________ __________

•  knowing the rate law of the reac%on helps us understand the sequence of steps in the mechanism

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An Example of a Reac%on Mechanism •  Overall reac%on:

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 __________ _____, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps

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Elements of a Mechanism Intermediates

•  no%ce 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 reac%on

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

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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)

Molecularity

•  the number of reactant par%cles in an elementary step is called its _______________

•  unimolecular step: involves __ reactant par%cle •  bimolecular step: involves __ reactant par%cles

–  though they may be the same kind of par%cle

•  termolecular step: involves __ reactant par%cles –  though these are exceedingly rare in elementary steps

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Rate Laws for Elementary Steps •  each step in the mechanism is like its own lirle reac%on – with its own ac%va%on energy and own rate law

•  the rate law for an overall reac%on must be determined experimentally

•  but the rate law of an elementary step can be deduced from the equa%on of the step

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

•  the result is that product produc%on cannot occur any faster than the _______________ – the step determines the rate of the overall reac%on

•  we call the slowest step in the mechanism the _____ _______________step –  the slowest step has the largest ac%va%on energy

•  the rate law of the rate determining step determines the rate law of the overall reac%on

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Another Reac%on Mechanism

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NO2(g) + CO(g) → NO(g) + CO2(g) Rateobs = 1)  NO2(g) + NO2(g) → NO3(g) + NO(g) Rate = .

2)  2) NO3(g) + CO(g) → NO2(g) + CO2(g) Rate = .

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.

Valida%ng a Mechanism

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

1.  the elementary steps must _____ to the overall reac%on

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

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Catalysts •  catalysts are substances that affect the rate of a reac%on without being consumed

•  catalysts work by providing an _________________ for the reac%on –  with a ____________________

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

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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

Catalysts •  ____________________ catalysts are in the same phase as the reactant par%cles – Cl(g) in the destruc%on of O3(g)

•  ____________________ catalysts are in a different phase than the reactant par%cles – solid cataly%c converter in a car’s exhaust system

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Types of Catalysts

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Documents

Chapter 13 Outline