Chemical Kinetics
Tro, Chemistry: A Molecular Approach 1
Tro, Chemistry: A Molecular Approach 2
• 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
Tro, Chemistry: A Molecular Approach 3
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
Tro, Chemistry: A Molecular Approach 4
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
Tro, Chemistry: A Molecular Approach 5
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.
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
25.0
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
Tro, Chemistry: A Molecular Approach 7
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
Tro, Chemistry: A Molecular Approach 8
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
Tro, Chemistry: A Molecular Approach 10
Hypothetical ReactionRed Blue
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
Tro, Chemistry: A Molecular Approach 11
Hypothetical ReactionRed 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
Tro, Chemistry: A Molecular Approach 12
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
Tro, Chemistry: A Molecular Approach 13
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
Time (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t
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
Time (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t
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
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)
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
Tro, Chemistry: A Molecular Approach 17
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
Tro, Chemistry: A Molecular Approach 20
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
Tro, Chemistry: A Molecular Approach 21
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
Tro, Chemistry: A Molecular Approach 22
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
Tro, Chemistry: A Molecular Approach 23
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
Tro, Chemistry: A Molecular Approach 24
• 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
Tro, Chemistry: A Molecular Approach 25
• 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
Tro, Chemistry: A Molecular Approach 26
• 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
Tro, Chemistry: A Molecular Approach 27
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
Tro, Chemistry: A Molecular Approach 28
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
Tro, Chemistry: A Molecular Approach 29
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.
Tro, Chemistry: A Molecular Approach 30
Reactant Concentration vs. TimeA Products
Tro, Chemistry: A Molecular Approach 31
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
Tro, Chemistry: A Molecular Approach 32
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
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 = -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
Tro, Chemistry: A Molecular Approach 34
ln[A]0
ln[A]
time
slope = −k
Tro, Chemistry: A Molecular Approach 35
Half-Life of a First-Order ReactionIs Constant
Tro, Chemistry: A Molecular Approach 36
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
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)
conc
entr
atio
n, (
M)
Tro, Chemistry: A Molecular Approach 38
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
)
Tro, Chemistry: A Molecular Approach 39
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
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 = 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
Tro, Chemistry: A Molecular Approach 41
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
Tro, Chemistry: A Molecular Approach 43
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)
Tro, Chemistry: A Molecular Approach 44
Rate Data Graphs For2 NO2 2 NO + O2
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)
Tro, Chemistry: A Molecular Approach 45
Rate Data Graphs For2 NO2 2 NO + O2
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
)
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 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
Tro, Chemistry: A Molecular Approach 47
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
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
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)
con
cen
trat
ion
, M
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
(co
nce
ntr
atio
n)
Tro, Chemistry: A Molecular Approach 52
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
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 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
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 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
Tro, Chemistry: A Molecular Approach 57
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.
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
Tro, Chemistry: A Molecular Approach 58
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.
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
Tro, Chemistry: A Molecular Approach 59
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.
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
Tro, Chemistry: A Molecular Approach 60
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.
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
Tro, Chemistry: A Molecular Approach 61
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.
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
Tro, Chemistry: A Molecular Approach 62
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
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 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
Tro, Chemistry: A Molecular Approach 65
Tro, Chemistry: A Molecular Approach 66
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
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
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
Tro, Chemistry: A Molecular Approach 69
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
Tro, Chemistry: A Molecular Approach 70
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 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)
Tro, Chemistry: A Molecular Approach 72
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
Tro, Chemistry: A Molecular Approach 73
Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the
following data:
use a spreadsheet to graph ln(k) vs. (1/T)
Tro, Chemistry: A Molecular Approach 74
Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the
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
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 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
Tro, Chemistry: A Molecular Approach 77
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.
Tro, Chemistry: A Molecular Approach 78
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
Tro, Chemistry: A Molecular Approach 79
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
Tro, Chemistry: A Molecular Approach 80
Effective CollisionsOrientation Effect
Tro, Chemistry: A Molecular Approach 81
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
Tro, Chemistry: A Molecular Approach 82
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
Tro, Chemistry: A Molecular Approach 83
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
Tro, Chemistry: A Molecular Approach 84
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
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)
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
Tro, Chemistry: A Molecular Approach 86
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
Tro, Chemistry: A Molecular Approach 87
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]
Tro, Chemistry: A Molecular Approach 88
Rate Laws of Elementary Steps
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 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
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 its activation energy is larger.
The rate law of the first step is the same as the rate law of the overall reaction.
Tro, Chemistry: A Molecular Approach 91
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
Tro, Chemistry: A Molecular Approach 92
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
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
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
Tro, Chemistry: A Molecular Approach 94
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
Tro, Chemistry: A Molecular Approach 95
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
Tro, Chemistry: A Molecular Approach 96
Ozone Depletion over the Antarctic
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
Tro, Chemistry: A Molecular Approach 98
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
Tro, Chemistry: A Molecular Approach 99
Types of Catalysts
Tro, Chemistry: A Molecular Approach 100
Catalytic HydrogenationH2C=CH2 + H2 → CH3CH3
Tro, Chemistry: A Molecular Approach 101
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
Tro, Chemistry: A Molecular Approach 102
Enzyme-Substrate BindingLock and Key Mechanism
Tro, Chemistry: A Molecular Approach 103
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
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