lecture13 chemical kinetics
TRANSCRIPT
1
Chemical Kinetics
2
Chapter Goals
1. The Rate of a ReactionFactors That Affect Reaction Rates
2. Nature of the Reactants3. Concentrations of the Reactants: The
Rate-Law Expression4. Concentration Versus Time: The
Integrated Rate Equation5. Collision Theory of Reaction Rates
2
3
Chapter Goals
6. Transition State Theory7. Reaction Mechanisms and the Rate-Law
Expression8. Temperature: The Arrhenius Equation9. Catalysts
4
The Rate of a Reaction
• Kinetics is the study of rates of chemicalreactions and the mechanisms by whichthey occur.
• The reaction rate is the increase inconcentration of a product per unit time ordecrease in concentration of a reactantper unit time.
• A reaction mechanism is the series ofmolecular steps by which a reactionoccurs.
3
5
The Rate of a Reaction• Thermodynamics (Chapter 15) determines if a reaction
can occur.• Kinetics (Chapter 16) determines how quickly a reaction
occurs.– Some reactions that are thermodynamically
feasible are kinetically so slow as to beimperceptible.
OUSINSTANTANE
kJ-79=GOHOH+H
SLOWVERY
kJ396GCOOC
o2982
-aq
+aq
o298g2g2diamond
l
6
The Rate of Reaction
• Consider the hypothetical reaction,aA(g) + bB(g) cC(g) + dD(g)
• equimolar amounts of reactants, A and B,will be consumed while products, C and D,will be formed as indicated in this graph:
4
7
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100
150
200
250
300
350
Time
Con
cent
ratio
ns o
fR
eact
ants
& P
rodu
cts
[A] & [B][C] & [D]
• [A] is the symbol for the concentration of A in M ( mol/L).• Note that the reaction does not go entirely to completion.
– The [A] and [B] > 0 plus the [C] and [D] < 1.
8
The Rate of Reaction
• Reaction rates are the rates at which reactantsdisappear or products appear.
• This movie is an illustration of a reaction rate.
5
9
The Rate of Reaction
• Mathematically, the rate of a reaction canbe written as:
tdD+
tcC+or
tbB-
taA-=Rate
10
The Rate of Reaction• The rate of a simple one-step reaction is directly
proportional to the concentration of the reactingsubstance.
• [A] is the concentration of A in molarity ormoles/L.
• k is the specific rate constant.– k is an important quantity in this chapter.
Ak=RateorARate
C+BA (g)(g)(g)
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11
The Rate of Reaction
• For a simple expression like Rate = k[A]– If the initial concentration of A is doubled, the initial
rate of reaction is doubled.• If the reaction is proceeding twice as fast, the amount of
time required for the reaction to reach equilibrium wouldbe:
A. The same as the initial time.B. Twice the initial time.C. Half the initial time.
• If the initial concentration of A is halved theinitial rate of reaction is cut in half.
12
The Rate of Reaction
• If more than one reactant molecule appearsin the equation for a one-step reaction, wecan experimentally determine that thereaction rate is proportional to the molarconcentration of the reactant raised to thepower of the number of molecules involved inthe reaction.
22
ggg
Xk=RateorXRate
Z+YX2
7
13
The Rate of Reaction
• Rate Law Expressions must be determinedexperimentally.
– The rate law cannot be determined from thebalanced chemical equation.
– This is a trap for new students of kinetics.• The balanced reactions will not work because most
chemical reactions are not one-step reactions.
• Other names for rate law expressions are:1. rate laws2. rate equations3. rate expressions
14
The Rate of Reaction
• Important terminology for kinetics.• The order of a reaction can be expressed in
terms of either:1 each reactant in the reaction or2 the overall reaction. Order for the overall reaction is the sum of the orders
for each reactant in the reaction.
• For example:
overall.orderfirstandONinorderfirstisreactionThis
ONk=Rate
O+NO4ON2
52
52
g2g2g52
8
15
The Rate of Reaction
• A second example is:
overall.orderfirstand,OHinorderzero
CBr,CHinorderfirstisreactionThis]CBrCHk[=Rate
BrCOHCHOHCBrCH
-33
33
-aqaq33
-aqaq33
16
The Rate of Reaction
• A final example of the order of a reaction is:
ALLYEXPERIMENTDETERMINEDARESEXPRESSIONRATEALLREMEMBER,
overallorderthirdand,OinorderfirstNO,inordersecondisreactionThis
Ok[NO]=RateNO2O+NO2
2
2
2
g2g2g
9
17
The Rate of Reaction
• Given the following one step reaction and itsrate-law expression.– Remember, the rate expression would have to be
experimentally determined.
• Because it is a second order rate-lawexpression:– If the [A] is doubled the rate of the reaction will
increase by a factor of 4. 22 = 4– If the [A] is halved the rate of the reaction will
decrease by a factor of 4. (1/2)2 = 1/4
2ggg
Ak=Rate
CBA2
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Factors That Affect ReactionRates• There are several factors that can
influence the rate of a reaction:1. The nature of the reactants.2. The concentration of the reactants.3. The temperature of the reaction.4. The presence of a catalyst.• We will look at each factor individually.
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Nature of Reactants
• This is a very broad category that encompassesthe different reacting properties of substances.
• For example sodium reacts with waterexplosively at room temperature to liberatehydrogen and form sodium hydroxide.
burns.andignitesHThereaction.rapidand violentaisThis
HNaOH2OH2Na2
2
g2aq2s
20
Nature of Reactants
• Calcium reacts with water only slowly atroom temperature to liberate hydrogenand form calcium hydroxide.
reaction.slowratheraisThis
HOHCaOH2Ca g2aq22s
11
21
Nature of Reactants
• The reaction of magnesium with water atroom temperature is so slow that that theevolution of hydrogen is not perceptible tothe human eye.
reactionNoOHMg 2s
22
Nature of Reactants
• However, Mg reacts with steam rapidly toliberate H2 and form magnesium oxide.
• The differences in the rate of these threereactions can be attributed to the changing“nature of the reactants”.
g2sC100
)g(2s HMgOOHMg o
12
23
Concentrations of Reactants:The Rate-Law Expression• This movie illustrates how changing the
concentration of reactants affects the rate.
24
Concentrations of Reactants:The Rate-Law Expression• This is a simplified representation of the
effect of different numbers of molecules inthe same volume.– The increase in the molecule numbers is
indicative of an increase in concentration.A(g) + B (g) Products
A B
A B
A BB
A B
A BA BA B
4 different possibleA-B collisions
6 different possibleA-B collisions
9 different possibleA-B collisions
13
25
Concentrations of Reactants:The Rate-Law Expression
• Example 16-1: The following rate data were obtained at25oC for the following reaction. What are the rate-lawexpression and the specific rate-constant for thisreaction?
2 A(g) + B(g) 3 C(g)
ExperimentNumber
Initial [A](M)
Initial [B](M)
Initial rate offormation of
C (M/s)1 0.10 0.10 2.0 x 10-4
2 0.20 0.30 4.0 x 10-4
3 0.10 0.20 2.0 x 10-4
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Concentrations of Reactants:The Rate-Law Expression
.Bignorecancan weThus,k[A]=RateorBAk=Rate
012
constant.remainsrateinitial that theNotice2.byincreases[B] that theandconstantis[A]the
thatsee we3and1sexperimentcompare weIfBAk=Rate
:form theofbemustlawrateThe
0 xx
y
yx
y
14
27
Concentrations of Reactants:The Rate-Law Expression
You do it!
xx
reaction?for thiskofunitsand value theisWhat
overall.order1andAtorespectorder with1isreactionThis
k[A]=Rateork[A]=Rate122
2.byincreasesrate theand2byincreases[A] that theNotice2.and1sexperimentcompareNext,
st
st
1
28
Concentrations of Reactants:The Rate-Law Expression
[A]10 x2.0=Rateas writtenbecanlawrate theThus
10 x2.010.010 x2.0=k
1experimentfrom[A]andRateof values theUsingA
Rate=k
law.rate thefromkof value thefindcanWe
s13-
s13-s
4-
M
M
15
29
Concentrations of Reactants:The Rate-Law Expression• Example 16-2: The following data were obtained
for the following reaction at 25oC. What are therate-law expression and the specific rate constantfor the reaction?
2 A(g) + B(g) + 2 C(g) 3 D(g) + 2 E(g)
Experiment
Initial [A](M)
Initial [B](M)
Initial[C](M)
Initial rateof
formationof D (M/s)
1 0.20 0.10 0.10 2.0 x 10-4
2 0.20 0.30 0.20 6.0 x 10-4
3 0.20 0.10 0.30 2.0 x 10-4
4 0.60 0.30 0.40 1.8 x 10-3
30
Concentrations of Reactants:The Rate-Law Expression
yxzyx
z z
BAk=RateorCBAk=Rate
013
constant.remainsratebut the3byincreasesCTheconstant.remainBandA thatNotice
3.and1sexperimentCompare
16
31
Concentrations of Reactants:The Rate-Law Expression
BAk=RateorBAk=Rate
133
3.byincreasesrate theand3byincreasesBTheconstant.remainsAThe
2.and1sexperimentcompareNext,
1 xx
y y
32
Concentrations of Reactants:The Rate-Law Expression
overall.order2andB,respect toorder with1A,respect toorder with1isreactionThis
BAk=RateorBAk=Rate
133
3.byincreasesrate theand3byincreasesATheconstant.remainsBThe
4.and2sexperimentcompareNext,
ndst
st
11
xx
17
33
Concentrations of Reactants:The Rate-Law Expression
BA100.1=Rateas writtenbecanlaw-rate theThus,
100.10.100.20
100.2BA
Rate=k
4.or3,2,1,experimentfromdata theuseCank.ofunitsand value thedetermineFinally,
s12
s12
s4
M
M
M
MM
34
Concentrations of Reactants:The Rate-Law Expression• Example 16-3: consider a chemical reaction
between compounds A and B that is first orderwith respect to A, first order with respect to B,and second order overall. From the informationgiven below, fill in the blanks.
You do it!ExperimentInitial Rate
(M/s)Initial [A]
(M)Initial [B]
(M)1 4.0 x 10-3 0.20 0.050
2 1.6 x 10-2 ? 0.050
3 3.2 x 10-2 0.40 ?
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35
Concentrations of Reactants:The Rate-Law Expression
BA40.0Rate theThus40.0
0.0500.20100.4
BARate=k
k.of value thedeterminecan we1experimentFromBAk=Rate
s1
s1
s3
M
M
M
MM
36
Concentrations of Reactants:The Rate-Law Expression
M
MMM
80.0]A[050.0s0.40s106.1]A[
k[B]Rate[A]
2experimentfromdataandkof value theUse
1-1
1-2
19
37
Concentrations of Reactants:The Rate-Law Expression
M
MMM
20.0[B]40.0s0.40s102.3[B]
k[A]R[B]
determinecan we3experimentfromSimilarly,
1-1
1-2
38
Concentration vs. Time: TheIntegrated Rate Equation• The integrated rate equation relates time
and concentration for chemical andnuclear reactions.– From the integrated rate equation we can
predict the amount of product that is producedin a given amount of time.
• Initially we will look at the integrated rateequation for first order reactions.These reactions are 1st order in the reactant
and 1st order overall.
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39
Concentration vs. Time: TheIntegrated Rate Equation• An example of a reaction that is 1st order
in the reactant and 1st order overall is:a A products
This is a common reaction type for manychemical reactions and all simpleradioactive decays.
• Two examples of this type are:2 N2O5(g) 2 N2O4(g) + O2(g)
238U 234Th + 4He
40
Concentration vs. Time: TheIntegrated Rate Equation
where:[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.k = specific rate constant. t = time elapsed since
beginning of reaction.a = stoichiometric coefficient of A in balanced overall
equation.
• The integrated rate equation for first order reactionsis:
k taAAln 0
21
41
Concentration vs. Time: TheIntegrated Rate Equation
• Solve the first order integrated rate equation for t.
• Define the half-life, t1/2, of a reactant as the timerequired for half of the reactant to be consumed,or the time at which [A]=1/2[A]0.
AAln
ka1t 0
42
Concentration vs. Time: TheIntegrated Rate Equation• At time t = t1/2, the expression becomes:
ka693.0t
2lnka
1t
A1/2Aln
ka1t
1/2
1/2
0
01/2
22
43
Concentration vs. Time: TheIntegrated Rate Equation
• Example 16-4: Cyclopropane, ananesthetic, decomposes to propeneaccording to the following equation.
The reaction is first order in cyclopropanewith k = 9.2 s-1 at 10000C. Calculate the halflife of cyclopropane at 10000C.
CH2 CH2CH2
CH2CH3CH
(g) (g)
s075.0s2.9
693.0k693.0t 1-1/2
44
Concentration vs. Time: TheIntegrated Rate Equation• Example 16-5: Refer to Example 16-4.
How much of a 3.0 g sample ofcyclopropane remains after 0.50 seconds?– The integrated rate laws can be used for any
unit that represents moles or concentration.– In this example we will use grams rather than
mol/L.
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45
Concentration vs. Time: TheIntegrated Rate Equation
remains1%org03.0eA
5.310.16.4Aln6.4Aln10.1
s50.0s2.9Aln-3.0lnk tAlnAln
.logarithmsoflaws theusemustWe
reactionfor thisk tk taAAln
3.5-
1-
0
0
46
Concentration vs. Time: TheIntegrated Rate Equation• Example 16-6: The half-life for the
following first order reaction is 688 hoursat 10000C. Calculate the specific rateconstant, k, at 10000C and the amount ofa 3.0 g sample of CS2 that remains after48 hours.
CS2(g) CS(g) + S(g)
You do it!
24
47
Concentration vs. Time: TheIntegrated Rate Equation
1-
1/2
1/2
hr00101.0hr688
0.693k
t0.693k
k693.0t
1.areactionFor this
48
Concentration vs. Time: TheIntegrated Rate Equation
hr)48)(hr00101.0(Aln-ln(3.0)
k tAlnAlnk tAAln
1-
00
unreacted97%org9.2g86.2eA
1.0521.10)--(0.048Aln0.048Aln-1.10
hr)48)(hr00101.0(Aln-ln(3.0)
k tAlnAlnk tAAln
1.052
1-
00
25
49
Concentration vs. Time: TheIntegrated Rate Equation• For reactions that are second order with respect
to a particular reactant and second orderoverall, the rate equation is:
• Where:[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.
k = specific rate constant. t = time elapsed sincebeginning of reaction.
a = stoichiometric coefficient of A in balanced overall equation.
k taA1
A1
0
50
Concentration vs. Time: TheIntegrated Rate Equation• Second order reactions also have a half-life.
– Using the second order integrated rate-law as astarting point.
• At the half-life, t1/2 [A] = 1/2[A]0.
0
1/200
Aofrdenominatocommonahaswhich
k taA1
A2/11
1/20
1/200
k taA1
ork taA1
A2
26
51
Concentration vs. Time: TheIntegrated Rate Equation
• If we solve for t1/2:
• Note that the half-life of a second orderreaction depends on the initialconcentration of A.
01/2 Aka1t
52
Concentration vs. Time: TheIntegrated Rate Equation• Example 16-7: Acetaldehyde, CH3CHO, undergoes
gas phase thermal decomposition to methane andcarbon monoxide.
The rate-law expression is Rate = k[CH3CHO]2,and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) Whatis the half-life of CH3CHO if 0.10 mole isinjected into a 1.0 L vessel at 527oC?
CH CHO CH + CO3 g 4 g g
27
53
Concentration vs. Time: TheIntegrated Rate Equation
tk A
hr
hr
1/2
-1
1
12 0 10 010
5 0 10
0
2 1
2
. .
.
M M
54
Concentration vs. Time: TheIntegrated Rate Equation• (b) How many moles of CH3CHO remain after
200 hours?
1 1
1 1010
2 0 10 200
1 10 4 0
0
2 1
1 1
A Ak t
A hr hr
A
-1
..
.
MM
M M
1 14 114
0 071
0 071 mol
11A
A
A
mol = 1.0 L x 0.071 molL
M
MM.
? .
28
55
Concentration vs. Time: TheIntegrated Rate Equation• (c) What percent of the CH3CHO remains
after 200 hours?
reacted29%andunreacted%71
%100mol0.10mol0.071=unreacted%
56
Concentration vs. Time: TheIntegrated Rate Equation• Example 16-8: Refer to Example 16-7. (a)
What is the half-life of CH3CHO if 0.10mole is injected into a 10.0 L vessel at527oC?– Note that the vessel size is increased by a
factor of 10 which decreases theconcentration by a factor of 10!
You do it!
29
57
Concentration vs. Time: TheIntegrated Rate Equation
tk A
hr
hrnote the time has increased by 10over Example 16 - 7:
1/2
-1
1
12 0 10 0 010
5 0 10
0
2 1
3
. .
.
M M
58
Concentration vs. Time: TheIntegrated Rate Equation• (b) How many moles of CH3CHO remain
after 200 hours?You do it!
30
59
Concentration vs. Time: TheIntegrated Rate Equation
1 1
1 10 010
2 0 10 200
1 100 4 0
0
2 1
1 1
A Ak t
A hr hr
A
-1
..
.
MM
M M
1 104 1104
0 0096
0 096 mol
11A
A
A
mol = 10.0 L x 0.0096 molL
M
MM.
? .
60
Concentration vs. Time: TheIntegrated Rate Equation• (c) What percent of the CH3CHO remains
after 200 hours?You do it!
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61
Concentration vs. Time: TheIntegrated Rate Equation
% unreacted = 0.096 mol0.100 mol
unreacted & 4% reacted
100%
96%
62
Concentration vs. Time: TheIntegrated Rate Equation• Let us now summarize the results from the last
two examples.
InitialMoles
CH3CHO
[CH3CHO]0
(M)
[CH3CHO]
(M)
Moles ofCH3CH
O at 200hr.
%CH3CHOremainin
gEx. 16-
70.10 0.10 0.071 0.071 71%
Ex. 16-8
0.010 0.010 0.0096 0.096 96%
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63
Enrichment - Derivation ofIntegrated Rate Equations• For the first order reaction
a A productsthe rate can be written as:
tA
a1-=Rate
64
Enrichment - Derivation ofIntegrated Rate Equations• For a first-order reaction, the rate is proportional
to the first power of [A].
-1a
At
k A
33
65
Enrichment - Derivation ofIntegrated Rate Equations• In calculus, the rate is defined as the
infinitesimal change of concentration d[A] in aninfinitesimally short time dt as the derivative of[A] with respect to time.
-1a
At
k Add
66
Enrichment - Derivation ofIntegrated Rate Equations• Rearrange the equation so that all of the [A]
terms are on the left and all of the t terms are onthe right.
- AA
a k td d
34
67
Enrichment - Derivation ofIntegrated Rate Equations• Express the equation in integral form.
- A
Aa k t
A
A td d0 0
68
Enrichment - Derivation ofIntegrated Rate Equations• This equation can be evaluated as:
-ln A a k t or
-ln A A a k t - a k 0which becomes-ln A A a k t
t0t
t
t
0
0
0
ln
ln
35
69
Enrichment - Derivation ofIntegrated Rate Equations• Which rearranges to the integrated first order
rate equation.
k taAAln
t
0
70
Enrichment - Derivation ofIntegrated Rate Equations• Derive the rate equation for a reaction that is
second order in reactant A and second orderoverall.
• The rate equation is:
2Ak ta
A
dd
36
71
Enrichment - Derivation ofIntegrated Rate Equations• Separate the variables so that the A terms are
on the left and the t terms on the right.
tkAaA
2 dd
72
Enrichment - Derivation ofIntegrated Rate Equations• Then integrate the equation over the limits as for
the first order reaction.
t
0
A
A2 tka
AA
0
dd
37
73
Enrichment - Derivation ofIntegrated Rate Equations• Which integrates the second order integrated
rate equation.
k taA1
A1
0
74
Enrichment - Derivation ofIntegrated Rate Equations• For a zero order reaction the rate expression is:
k ta
A
dd
38
75
Enrichment - Derivation ofIntegrated Rate Equations• Which rearranges to:
tkaA dd
76
Enrichment - Derivation ofIntegrated Rate Equations• Then we integrate as for the other two cases:
t
0
A
A
tkaA0
dd
39
77
Enrichment - Derivation ofIntegrated Rate Equations• Which gives the zeroeth order integrated rate
equation.
k ta-AAor
k t-aAA
0
0
78
Enrichment - Rate Equations toDetermine Reaction Order• Plots of the integrated rate equations can help us
determine the order of a reaction.• If the first-order integrated rate equation is
rearranged.– This law of logarithms, ln (x/y) = ln x - ln y, was applied
to the first-order integrated rate-equation.
0
0
Alnk taAlnor
k taAlnAln
40
79
Enrichment - Rate Equations toDetermine Reaction Order• The equation for a straight line is:
• Compare this equation to the rearranged firstorder rate-law.
bmy x
80
Enrichment - Rate Equationsto Determine Reaction Order
bmy x
• Now we can interpret the parts of the equationas follows:– y can be identified with ln[A] and plotted on the y-axis.– m can be identified with –ak and is the slope of the
line.– x can be identified with t and plotted on the x-axis.– b can be identified with ln[A]0 and is the y-intercept.
0Alnk taAln
41
81
Enrichment - Rate Equations toDetermine Reaction Order• Example 16-9: Concentration-versus-time data
for the thermal decomposition of ethyl bromideare given in the table below. Use the followinggraphs of the data to determine the rate of thereaction and the value of the rate constant.
700KatHBrHCBrHC gg42g52
82
Enrichment - Rate Equations toDetermine Reaction Order
Time(min) 0 1 2 3 4 5
[C2H5Br] 1.00 0.82 0.67 0.55 0.45 0.37
ln [C2H5Br] 0.00 -0.20 -0.40 -0.60 -0.80 -0.99
1/[C2H5Br] 1.0 1.2 1.5 1.8 2.2 2.7
42
83
Enrichment - Rate Equationsto Determine Reaction Order• We will make three different graphs of the
data.1 Plot the [C2H5Br] (y-axis) vs. time (x-axis)
– If the plot is linear then the reaction is zeroorder with respect to [C2H5Br].
2 Plot the ln [C2H5Br] (y-axis) vs. time (x-axis)– If the plot is linear then the reaction is first order
with respect to [C2H5Br].3 Plot 1/ [C2H5Br] (y-axis) vs. time (x-axis)
– If the plot is linear then the reaction is secondorder with respect to [C2H5Br].
84
Enrichment - Rate Equationsto Determine Reaction Order• Plot of [C2H5Br] versus time.
– Is it linear or not?[C2H5Br] vs. time
00.20.40.60.8
11.2
0 1 2 3 4 5
Time (min)
[C2H
5B
r]
43
85
Enrichment - Rate Equationsto Determine Reaction Order• Plot of ln [C2H5Br] versus time.
– Is it linear or not?
ln [C2H5Br] vs. time
-1.2-1
-0.8-0.6-0.4-0.2
00 1 2 3 4 5
Time (min)
ln [C
2H5B
r]
86
Enrichment - Rate Equationsto Determine Reaction Order
• Plot of 1/[C2H5Br] versus time.– Is it linear or not?
1/[C2H5Br] vs. time
0
1
2
3
0 1 2 3 4 5Time (min)
1/[C
2H5B
r]
44
87
Enrichment - Rate Equationsto Determine Reaction Order• Note that the only graph which is linear is the plot of
ln[C2H5Br] vs. time.– Thus this is a first order reaction with respect
to [C2H5Br].• Next, we will determine the value of the rate constant
from the slope of the line on the graph of ln[C2H5Br] vs.time.– Remember slope = y2-y1/x2-x1.
1-
12
12
min20.0min3
60.0slope
min14)20.0(80.0
x-xy-yslope
88
Enrichment - Rate Equations toDetermine Reaction Order• From the equation for a first order reaction we
know that the slope = -a k.– In this reaction a = 1.
.min0.20kconstantrate theThus-k-0.20slope
1-
45
89
Enrichment - Rate Equationsto Determine Reaction Order• The integrated rate equation for a reaction that is
second order in reactant A and second orderoverall.
• This equation can be rearranged to: k ta
A1
A1
0
0A1k ta
A1
90
Enrichment - Rate Equationsto Determine Reaction Order• Compare the equation for a straight line and the
second order rate-law expression.
• Now we can interpret the parts of the equationas follows:– y can be identified with 1/[A] and plotted on the y-axis.– m can be identified with a k and is the slope of the
line.– x can be identified with t and plotted on the x-axis– b can be identified with 1/[A]0 and is the y-intercept.
bmy x
0A1k ta
A1
46
91
Enrichment - Rate Equationsto Determine Reaction Order• Example 16-10: Concentration-versus-
time data for the decomposition of nitrogendioxide are given in the table below. Usethe graphs to determine the rate of thereaction and the value of the rate constant
500KatONO2NO2 g2gg2
92
Enrichment - Rate Equationsto Determine Reaction OrderTime(min) 0 1 2 3 4 5[NO2] 1.0 0.53 0.36 0.27 0.22 0.18
ln [NO2] 0.0 -0.63 -1.0 -1.3 -1.5 -1.7
1/[NO2] 1.0 1.9 2.8 3.7 4.6 5.5
47
93
Enrichment - Rate Equationsto Determine Reaction Order• Once again, we will make three different graphs
of the data.1. Plot [NO2] (y-axis) vs. time (x-axis).• If the plot is linear then the reaction is zero order with
respect to NO2.2. Plot ln [NO2] (y-axis) vs. time (x-axis).• If the plot is linear then the reaction is first order with
respect to NO2.3. Plot 1/ [NO2] (y-axis) vs. time (x-axis).• If the plot is linear then the reaction is second order
with respect to NO2.
94
Enrichment - Rate Equationsto Determine Reaction Order• Plot of [NO2] versus time.
– Is it linear or not?
[NO2] vs. time
00.20.40.60.8
11.2
0 1 2 3 4 5
Time (min)
[NO
2]
48
95
Enrichment -Rate Equationsto Determine Reaction Order• Plot of ln [NO2] versus time.
– Is it linear or not?ln [NO2] vs. time
-2
-1.5
-1
-0.5
00 1 2 3 4 5
Time (min)
ln [N
O2]
96
Enrichment - Rate Equationsto Determine Reaction Order• Plot of 1/[NO2] versus time.
– Is it linear or not?
1/[NO2] vs.time
0123456
0 1 2 3 4 5
Time (min)
1/[N
O2]
49
97
Enrichment - Rate Equationsto Determine Reaction Order• Note that the only graph which is linear is
the plot of 1/[NO2] vs. time.• Thus this is a second order reaction with
respect to [NO2].• Next, we will determine the value of the
rate constant from the slope of the line onthe graph of 1/[NO2] vs. time.
98
Enrichment - Rate Equationsto Determine Reaction Order
• From the equation for a second order reactionwe know that the slope = a k– In this reaction a = 2.
1-1 min0.45kconstantrate theThusk20.90slope
M
min1
1
1
12
12
90.0min4
60.3slope
min15)90.1(50.5
x-xy-yslope
MM
M
50
99
Collision Theory ofReaction Rates• Three basic events must occur for a
reaction to occur the atoms, molecules orions must:
1. Collide.2. Collide with enough energy to break and
form bonds.3. Collide with the proper orientation for a
reaction to occur.
100
Collision Theory ofReaction Rates• One method to increase the number of collisions
and the energy necessary to break and reformbonds is to heat the molecules.
• As an example, look at the reaction of methaneand oxygen:
• We must start the reaction with a match.– This provides the initial energy necessary to break
the first few bonds.– Afterwards the reaction is self-sustaining.
kJ891OHCOOCH (g)22(g)2(g)4(g)
51
101
Collision Theory ofReaction Rates• Illustrate the proper orientation of
molecules that is necessary for thisreaction.
X2(g) + Y2(g)2 XY(g)
• Some possible ineffective collisions are :
X
XY Y Y
YX X X X Y Y
102
Collision Theory ofReaction Rates• An example of an effective collision is:
X Y
X Y
X Y
X Y
X Y+
X Y
52
103
Transition State Theory• Transition state theory postulates that
reactants form a high energy intermediate,the transition state, which then falls apart intothe products.
• For a reaction to occur, the reactants mustacquire sufficient energy to form thetransition state.– This energy is called the activation energy or Ea.
• Look at a mechanical analog for activationenergy
104
Transition State Theory
Epot = mg h
Cross sectionof mountain
BoulderEactivation
h
h2
h1
Epot=mgh2
Epot=mgh1
Height
53
105
Transition State Theory
PotentialEnergy
Reaction Coordinate
X2 + Y2
2 XY
Eactivation - a kinetic quantity
E Ha thermodynamicquantity
Representation of a chemical reaction.
106
Transition State Theory
54
107
Transition State Theory
• The relationship between the activationenergy for forward and reverse reactionsis– Forward reaction = Ea
– Reverse reaction = Ea + E– difference = E
108
Transition State Theory
• The distribution of molecules possessingdifferent energies at a given temperature isrepresented in this figure.
55
109
Reaction Mechanisms and theRate-Law Expression• Use the experimental rate-law to postulate a
molecular mechanism.• The slowest step in a reaction mechanism is the
rate determining step.
110
Reaction Mechanisms and theRate-Law Expression• Use the experimental rate-law to
postulate a mechanism.• The slowest step in a reaction mechanism
is the rate determining step.• Consider the iodide ion catalyzed
decomposition of hydrogen peroxide towater and oxygen.
g22I
22 O+OH2OH2-
56
111
Reaction Mechanisms and theRate-Law Expression• This reaction is known to be first order in H2O2 ,
first order in I- , and second order overall.• The mechanism for this reaction is thought to be:
-22
2222
-2222
-2
--22
IOHk=RlawratealExperiment
O+OH2OH2reactionOverall
I+O+OHOH+IOstepFast
OH+IOI+OHstepSlow
112
Reaction Mechanisms and theRate-Law Expression• Important notes about this reaction:1. One hydrogen peroxide molecule and one
iodide ion are involved in the rate determiningstep.
2. The iodide ion catalyst is consumed in step 1and produced in step 2 in equal amounts.
3. Hypoiodite ion has been detected in thereaction mixture as a short-lived reactionintermediate.
57
113
Reaction Mechanisms and theRate-Law Expression• Ozone, O3, reacts very rapidly with nitrogen
oxide, NO, in a reaction that is first order in eachreactant and second order overall.
NOOk=Rateislaw-ratealExperiment
O+NONO+O
3
g2g2gg3
114
Reaction Mechanisms and theRate-Law Expression• One possible mechanism is:
223
223
33
O+NONO+OreactionOverall
O+NONO+OstepFastO+NONO+OstepSlow
58
115
Reaction Mechanisms and theRate-Law Expression• A mechanism that is inconsistent with the
rate-law expression is:
correct.becannotmechanism thisproveswhich
Ok=Rateismechanism thisfromlaw-rateTheONONO+OreactionOverall
NONO+OstepFastO+OOstepSlow
3
223
2
23
116
Reaction Mechanisms and theRate-Law Expression• Experimentally determined reaction
orders indicate the number of moleculesinvolved in:
1. the slow step only or2. the slow step and the equilibrium steps
preceding the slow step.
59
117
Temperature:The Arrhenius Equation• Svante Arrhenius developed this
relationship among (1) the temperature(T), (2) the activation energy (Ea), and (3)the specific rate constant (k).
k = Aeor
ln k = ln A - ERT
-E RT
a
a
118
Temperature:The Arrhenius Equation• This movie illustrates the effect of temperature
on a reaction.
60
119
Temperature:The Arrhenius Equation• If the Arrhenius equation is written for two
temperatures, T2 and T1 with T2 >T1.
ln k ln A - ERT
and
ln k ln A - ERT
1a
1
2a
2
120
Temperature:The Arrhenius Equation1. Subtract one equation from the other.
ln k k A - ln A - ERT
ERT
ln k k ERT
- ERT
2 1a
2
a
1
2 1a
1
a
2
ln ln
ln
61
121
Temperature:The Arrhenius Equation2. Rearrange and solve for ln k2/k1.
ln kk
ER T T
or
ln kk
ER
T - TT T
2
1
a
1 2
2
1
a 2 1
2 1
1 1
122
Temperature:The Arrhenius Equation• Consider the rate of a reaction for which Ea=50
kJ/mol, at 20oC (293 K) and at 30oC (303 K).– How much do the two rates differ?
ln kk
ER
T - TT T
ln kk 8.314
K
ln kkkk
e
2
1
a 2 1
2 1
2
1
Jmol
JK mol
2
1
2
1
0.677
50 000 303 293303 293
0 677
197 2
,
.
.
62
123
Temperature:The Arrhenius Equation• For reactions that have an Ea50 kJ/mol, the
rate approximately doubles for a 100C rise intemperature, near room temperature.
• Consider:2 ICl(g) + H2(g) I2(g) + 2 HCl(g)
• The rate-law expression is known to beR=k[ICl][H2].
At 230 C, k = 0.163 s
At 240 C, k = 0.348 sk approximately doubles
0 -1 -1
0 -1 -1
M
M
124
Catalysts
• Catalysts change reaction rates by providing analternative reaction pathway with a differentactivation energy.
63
125
Catalysts
• Homogeneous catalysts exist in same phase asthe reactants.
• Heterogeneous catalysts exist in differentphases than the reactants.– Catalysts are often solids.
126
Catalysts
• Examples of commercial catalyst systemsinclude:
systemconvertercatalyticAutomobile
ONNO2
CO2O+CO2
OH18CO16O25+HC
g2g2PtandNiO
g
g2PtandNiO
g2g
g2g2PtandNiO
g2g188
64
127
Catalysts
• This movie shows catalytic converterchemistry on the Molecular Scale
128
Catalysts
• A second example of a catalytic system is:
npreparatioacidSulfuric
SO2OSO2 g3NiO/PtorOV
g2g252
65
129
Catalysts
• A third examples of a catalytic system is:
ProcessHaber
NH2H3N g3OFeorFe
g2g232
130
Catalysts• Look at the catalytic oxidation of CO to CO2• Overall reaction
2 CO(g)+ O2(g)2CO2(g)
• AbsorptionCO(g)CO(surface) + O2(g)
O2(g)O2(surface)
• ActivationO2(surface) O(surface)
• ReactionCO(surface) +O(surface)CO2(surface)
• DesorptionCO2(surface)CO2(g)
66
131
Synthesis Question
• The Chernobyl nuclear reactor accidentoccurred in 1986. At the time that thereactor exploded some 2.4 MCi ofradioactive 137Cs was released into theatmosphere. The half-life of 137Cs is 30.1years. In what year will the amount of137Cs released from Chernobyl finallydecrease to 100 Ci? A Ci is a unit ofradioactivity called the Curie.
132
Synthesis Question
Ci100reachesChernobylatemittedCsfromityradioactiv the when24264401986
years440 years439 y0230.0
10.1t
t y0230.010.1
t y0230.0Ci100
Ci102.4ln
decayeradioactivfor this1aandk taAAln
Ci102.4 MCi2.4
y0230.0 y1.30
693.0t693.0k
k693.0t
137
1-
1-
1-6
0
6
1-
21
21
67
133
Group Question
• 99mTc has a half-life of 6.02 hours and isoften used in nuclear medical diagnostictests. Patients are injected withapproximately 10 mCi of 99mTc that is thendirected to specific sites in the patient’sbody to detect gallstones, brain tumorsand function, and other medicalconditions. How long will the patient havea higher than normal radioactivity levelafter they have been injected with 10 mCiof 99mTc?
16 Chemical Kinetics