Chem Chapter 13 LEC
Transcript of Chem Chapter 13 LEC
-
8/13/2019 Chem Chapter 13 LEC
1/103
Chapter 13
Chemical
Kinetics
2008, Prentice Hall
Chemistry: A Molecular Approach, 1stEd.
Nivaldo Tro
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
-
8/13/2019 Chem Chapter 13 LEC
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
-
8/13/2019 Chem Chapter 13 LEC
3/103
Tro, Chemistry: A Molecular Approach 3
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
-
8/13/2019 Chem Chapter 13 LEC
4/103
Tro, Chemistry: A Molecular Approach 4
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
-
8/13/2019 Chem Chapter 13 LEC
5/103
Tro, Chemistry: A Molecular Approach 5
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.
-
8/13/2019 Chem Chapter 13 LEC
6/103
Tro, Chemistry: A Molecular Approach 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
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
-
8/13/2019 Chem Chapter 13 LEC
7/103
Tro, Chemistry: A Molecular Approach 7
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
-
8/13/2019 Chem Chapter 13 LEC
8/103
-
8/13/2019 Chem Chapter 13 LEC
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
-
8/13/2019 Chem Chapter 13 LEC
10/103
Tro, Chemistry: A Molecular Approach 10
Hypothetical Reaction
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
-
8/13/2019 Chem Chapter 13 LEC
11/103
Tro, Chemistry: A Molecular Approach 11
Hypothetical Reaction
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
-
8/13/2019 Chem Chapter 13 LEC
12/103
Tro, Chemistry: A Molecular Approach 12
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
-
8/13/2019 Chem Chapter 13 LEC
13/103
Tro, Chemistry: A Molecular Approach 13
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
-
8/13/2019 Chem Chapter 13 LEC
14/103
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
-
8/13/2019 Chem Chapter 13 LEC
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
Avg. Rate, M/sAvg. Rate, M/sTime (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t
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
Avg. Rate, M/sAvg. Rate, M/sTime (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t
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
-
8/13/2019 Chem Chapter 13 LEC
16/103
Tro, Chemistry: A Molecular Approach 16
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
-
8/13/2019 Chem Chapter 13 LEC
17/103
Tro, Chemistry: A Molecular Approach 17
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
-
8/13/2019 Chem Chapter 13 LEC
18/103
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
-
8/13/2019 Chem Chapter 13 LEC
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
-
8/13/2019 Chem Chapter 13 LEC
20/103
Tro, Chemistry: A Molecular Approach 20
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
-
8/13/2019 Chem Chapter 13 LEC
21/103
Tro, Chemistry: A Molecular Approach 21
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
-
8/13/2019 Chem Chapter 13 LEC
22/103
Tro, Chemistry: A Molecular Approach 22
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
-
8/13/2019 Chem Chapter 13 LEC
23/103
Tro, Chemistry: A Molecular Approach 23
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
-
8/13/2019 Chem Chapter 13 LEC
24/103
Tro, Chemistry: A Molecular Approach 24
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
Factors Affecting Reaction Rate
Temperature
-
8/13/2019 Chem Chapter 13 LEC
25/103
Tro, Chemistry: A Molecular Approach 25
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
Factors Affecting Reaction Rate
Catalysts
-
8/13/2019 Chem Chapter 13 LEC
26/103
Tro, Chemistry: A Molecular Approach 26
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)
Factors Affecting Reaction Rate
Reactant Concentration
-
8/13/2019 Chem Chapter 13 LEC
27/103
Tro, Chemistry: A Molecular Approach 27
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
-
8/13/2019 Chem Chapter 13 LEC
28/103
Tro, Chemistry: A Molecular Approach 28
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
-
8/13/2019 Chem Chapter 13 LEC
29/103
Tro, Chemistry: A Molecular Approach 29
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.
-
8/13/2019 Chem Chapter 13 LEC
30/103
Tro, Chemistry: A Molecular Approach 30
Reactant Concentration vs. Time
A Products
-
8/13/2019 Chem Chapter 13 LEC
31/103
Tro, Chemistry: A Molecular Approach 31
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
-
8/13/2019 Chem Chapter 13 LEC
32/103
Tro, Chemistry: A Molecular Approach 32
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
-
8/13/2019 Chem Chapter 13 LEC
33/103
Tro, Chemistry: A Molecular Approach 33
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
-
8/13/2019 Chem Chapter 13 LEC
34/103
Tro, Chemistry: A Molecular Approach 34
ln[A]0
ln[A]
time
H lf Lif f Fi t O d R ti
-
8/13/2019 Chem Chapter 13 LEC
35/103
Tro, Chemistry: A Molecular Approach 35
Half-Life of a First-Order Reaction
Is Constant
R D f
-
8/13/2019 Chem Chapter 13 LEC
36/103
Tro, Chemistry: A Molecular Approach 36
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
-
8/13/2019 Chem Chapter 13 LEC
37/103
Tro, Chemistry: A Molecular Approach 37
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
)
-
8/13/2019 Chem Chapter 13 LEC
38/103
Tro, Chemistry: A Molecular Approach 38
C4H9Cl + H2O C4H9OH + 2 HCl
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)
-
8/13/2019 Chem Chapter 13 LEC
39/103
Tro, Chemistry: A Molecular Approach 39
C4H9Cl + H2O C4H9OH + 2 HCl
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
-
8/13/2019 Chem Chapter 13 LEC
40/103
Tro, Chemistry: A Molecular Approach 40
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
-
8/13/2019 Chem Chapter 13 LEC
41/103
Tro, Chemistry: A Molecular Approach 41
l/[A]0
1/[A]
time
-
8/13/2019 Chem Chapter 13 LEC
42/103
R D G h F
-
8/13/2019 Chem Chapter 13 LEC
43/103
Tro, Chemistry: A Molecular Approach 43
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
-
8/13/2019 Chem Chapter 13 LEC
44/103
R D G h F
-
8/13/2019 Chem Chapter 13 LEC
45/103
Tro, Chemistry: A Molecular Approach 45
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)
-
8/13/2019 Chem Chapter 13 LEC
46/103
Tro, Chemistry: A Molecular Approach 46
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
-
8/13/2019 Chem Chapter 13 LEC
47/103
Tro, Chemistry: A Molecular Approach 47
Practice - Complete the Table and
-
8/13/2019 Chem Chapter 13 LEC
48/103
Tro, Chemistry: A Molecular Approach 48
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
-
8/13/2019 Chem Chapter 13 LEC
49/103
Tro, Chemistry: A Molecular Approach 49
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
-
8/13/2019 Chem Chapter 13 LEC
50/103
Tro, Chemistry: A Molecular Approach 50
[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
-
8/13/2019 Chem Chapter 13 LEC
51/103
Tro, Chemistry: A Molecular Approach 51
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)
-
8/13/2019 Chem Chapter 13 LEC
52/103
Tro, Chemistry: A Molecular Approach 52
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
-
8/13/2019 Chem Chapter 13 LEC
53/103
Tro, Chemistry: A Molecular Approach 53
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
-
8/13/2019 Chem Chapter 13 LEC
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
-
8/13/2019 Chem Chapter 13 LEC
55/103
Tro, Chemistry: A Molecular Approach 55
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
-
8/13/2019 Chem Chapter 13 LEC
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
Ex 13.2Determine the rate law and rate constant for the
-
8/13/2019 Chem Chapter 13 LEC
57/103
Tro, Chemistry: A Molecular Approach 57
reaction NO2(g)+ CO(g)NO(g)+ CO2(g)
given the data below.
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
Ex 13.2Determine the rate law and rate constant for the
-
8/13/2019 Chem Chapter 13 LEC
58/103
Tro, Chemistry: A Molecular Approach 58
reaction NO2(g)+ CO(g)NO(g)+ CO2(g)
given the data below.
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
Ex 13.2Determine the rate law and rate constant for the
-
8/13/2019 Chem Chapter 13 LEC
59/103
Tro, Chemistry: A Molecular Approach 59
reaction NO2(g)+ CO(g)NO(g)+ CO2(g)
given the data below.
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
Ex 13.2Determine the rate law and rate constant for the
-
8/13/2019 Chem Chapter 13 LEC
60/103
Tro, Chemistry: A Molecular Approach 60
reaction NO2(g)+ CO(g)NO(g)+ CO2(g)
given the data below.
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
Ex 13.2Determine the rate law and rate constant for the
-
8/13/2019 Chem Chapter 13 LEC
61/103
Tro, Chemistry: A Molecular Approach 61
reaction NO2(g)+ CO(g)NO(g)+ CO2(g)
given the data below.
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
-
8/13/2019 Chem Chapter 13 LEC
62/103
Tro, Chemistry: A Molecular Approach 62
reaction NH4+1+ NO2
-1 2 2 H
2O
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
Practice - Determine the rate law and rate constant for the
-
8/13/2019 Chem Chapter 13 LEC
63/103
63
reaction NH4+1+ NO2
-1 2 2 H
2O
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
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
-
8/13/2019 Chem Chapter 13 LEC
64/103
Tro, Chemistry: A Molecular Approach 64
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
-
8/13/2019 Chem Chapter 13 LEC
65/103
Tro, Chemistry: A Molecular Approach 65
A ti ti E d th
-
8/13/2019 Chem Chapter 13 LEC
66/103
Tro, Chemistry: A Molecular Approach 66
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
-
8/13/2019 Chem Chapter 13 LEC
67/103
Tro, Chemistry: A Molecular Approach 67
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
-
8/13/2019 Chem Chapter 13 LEC
68/103
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:
-
8/13/2019 Chem Chapter 13 LEC
69/103
Tro, Chemistry: A Molecular Approach 69
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
-
8/13/2019 Chem Chapter 13 LEC
70/103
Tro, Chemistry: A Molecular Approach 70
Arrhenius Plots
-
8/13/2019 Chem Chapter 13 LEC
71/103
Tro, Chemistry: A Molecular Approach 71
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
-
8/13/2019 Chem Chapter 13 LEC
72/103
Tro, Chemistry: A Molecular Approach 72
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
Ex. 13.7 Determine the activation energy and frequency
-
8/13/2019 Chem Chapter 13 LEC
73/103
Tro, Chemistry: A Molecular Approach 73
factor for the reaction O3(g)O2(g)+ O(g)given the
following data:
use a
spreadsheet
to graph
ln(k) vs.(1/T)
Ex. 13.7 Determine the activation energy and frequency
-
8/13/2019 Chem Chapter 13 LEC
74/103
Tro, Chemistry: A Molecular Approach 74
factor for the reaction O3(g)O2(g)+ O(g)given the
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
-
8/13/2019 Chem Chapter 13 LEC
75/103
Tro, Chemistry: A Molecular Approach 75
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
-
8/13/2019 Chem Chapter 13 LEC
76/103
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
-
8/13/2019 Chem Chapter 13 LEC
77/103
Tro, Chemistry: A Molecular Approach 77
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
-
8/13/2019 Chem Chapter 13 LEC
78/103
Tro, Chemistry: A Molecular Approach 78
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
-
8/13/2019 Chem Chapter 13 LEC
79/103
Tro, Chemistry: A Molecular Approach 79
Effective Collisions
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
-
8/13/2019 Chem Chapter 13 LEC
80/103
Tro, Chemistry: A Molecular Approach 80
Effective Collisions
Orientation Effect
Collision Theory and
-
8/13/2019 Chem Chapter 13 LEC
81/103
Tro, Chemistry: A Molecular Approach 81
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
-
8/13/2019 Chem Chapter 13 LEC
82/103
Tro, Chemistry: A Molecular Approach 82
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
-
8/13/2019 Chem Chapter 13 LEC
83/103
Tro, Chemistry: A Molecular Approach 83
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
-
8/13/2019 Chem Chapter 13 LEC
84/103
Elements of a Mechanism
-
8/13/2019 Chem Chapter 13 LEC
85/103
Tro, Chemistry: A Molecular Approach 85
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
-
8/13/2019 Chem Chapter 13 LEC
86/103
Tro, Chemistry: A Molecular Approach 86
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
-
8/13/2019 Chem Chapter 13 LEC
87/103
Tro, Chemistry: A Molecular Approach 87
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
-
8/13/2019 Chem Chapter 13 LEC
88/103
Tro, Chemistry: A Molecular Approach 88
Rate Laws of Elementary Steps
Rate Determining Step
-
8/13/2019 Chem Chapter 13 LEC
89/103
Tro, Chemistry: A Molecular Approach 89
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
-
8/13/2019 Chem Chapter 13 LEC
90/103
Tro, Chemistry: A Molecular Approach 90
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.
-
8/13/2019 Chem Chapter 13 LEC
91/103
-
8/13/2019 Chem Chapter 13 LEC
92/103
Tro, Chemistry: A Molecular Approach 92
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
-
8/13/2019 Chem Chapter 13 LEC
93/103
Tro, Chemistry: A Molecular Approach 93
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
-
8/13/2019 Chem Chapter 13 LEC
94/103
Tro, Chemistry: A Molecular Approach 94
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
-
8/13/2019 Chem Chapter 13 LEC
95/103
-
8/13/2019 Chem Chapter 13 LEC
96/103
Tro, Chemistry: A Molecular Approach 96
Ozone Depletion over the Antarctic
Energy Profile of Catalyzed Reaction
-
8/13/2019 Chem Chapter 13 LEC
97/103
Tro, Chemistry: A Molecular Approach 97
Energy Profile of Catalyzed Reaction
polar stratospheric clouds
contain ice crystals that
catalyze reactions that
release Cl from
atmospheric chemicals
Catalysts
-
8/13/2019 Chem Chapter 13 LEC
98/103
Tro, Chemistry: A Molecular Approach 98
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
-
8/13/2019 Chem Chapter 13 LEC
99/103
Tro, Chemistry: A Molecular Approach 99
Types of Catalysts
Catalytic Hydrogenation
-
8/13/2019 Chem Chapter 13 LEC
100/103
Tro, Chemistry: A Molecular Approach 100
Catalytic Hydrogenation
H2C=CH
2+ H
2CH
3CH
3
-
8/13/2019 Chem Chapter 13 LEC
101/103
Tro, Chemistry: A Molecular Approach 101
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
-
8/13/2019 Chem Chapter 13 LEC
102/103
Tro, Chemistry: A Molecular Approach 102
Enzyme-Substrate Binding
Lock and Key Mechanism
-
8/13/2019 Chem Chapter 13 LEC
103/103