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Chapter 16. Chemical Kinetics 化學 動力學. Outline. The Rate of a Reaction Nature of the Reactants Concentrations of the Reactants: The Rate-Law Expression Concentration Versus Time: The Integrated Rate Equation Collision Theory of Reaction Rates Transition State Theory - PowerPoint PPT Presentation
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1
Chapter 16Chemical Kinetics 化學動力學
2
Outline1.The Rate of a Reaction2.Nature of the Reactants3.Concentrations of the Reactants: The
Rate-Law Expression4.Concentration Versus Time: The
Integrated Rate Equation5.Collision Theory of Reaction Rates6.Transition State Theory7.Reaction Mechanisms and the Rate-
Law Expression8.Temperature: The Arrhenius Equation9.Catalysts
3
The Rate of a Reaction• Kinetics is the study of rates of chemical
reactions and the mechanisms by which they occur.
• The reaction rate is the increase in concentration of a product per unit time or decrease in concentration of a reactant per unit time.
• A reaction mechanism is the series of molecular steps by which a reaction occurs.
4
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 be imperceptible.
OUSINSTANTANE
kJ -79=G OHOH+H
SLOW VERY
kJ 396G COO C
o2982
-aq
+aq
o298g2g2diamond
l
5
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:
6
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.
The Rate of Reaction
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The Rate of Reaction• Reaction rates are the rates at which
reactants disappear or products appear.
• This movie is an illustration of a reaction rate.
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The Rate of Reaction• Mathematically, the rate of a reaction can
be written as:
- A - B + C + DRate =
a t b t c t d t
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The Rate of Reaction• The rate of a simple one-step reaction is
directly proportional to the concentration of the reacting substance.
• [A] is the concentration of A in molarity or moles/L.
• k is the specific rate constant.– k is an important quantity in this
chapter.
Ak = Rateor ARate
C + BA (g)(g)(g)
10
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 would be:
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 the initial rate of reaction is cut in half.
11
The Rate of Reaction• If more than one reactant molecule
appears in the equation for a one-step reaction, we can experimentally determine that the reaction rate is proportional to the molar concentration of the reactant raised to the power of the number of molecules involved in the reaction.
22
ggg
Xk = Rateor XRate
Z+ YX 2
12
The Rate of Reaction• Rate Law Expressions must be determined
experimentally.– The rate law cannot be determined from
the balanced 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 equations or rate expressions
13
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.order first and ONin order first isreaction This
ONk= Rate
O + NO4ON 2
52
52
g2g2g52
14
The Rate of Reaction• A second example is:
overall.order first and ,OHin order zero
CBr,CHin order first isreaction This ]CBrCHk[= Rate
BrCOHCHOHCBrCH
-
33
33
-aqaq33
-aqaq33
15
The Rate of Reaction• A final example of the order of a
reaction is:
ALLYEXPERIMENT DETERMINEDARE SEXPRESSION RATE ALL REMEMBER,
overallorder third and ,Oin order firstNO,in order second isreaction This
Ok[NO]=Rate NO 2O+NO 2
2
2
2
g2g2g
16
The Rate of Reaction• Given the following one step reaction and
its rate-law expression.– Remember, the rate expression would
have to be experimentally determined.
• Because it is a second order rate-law expression:– 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
CBA 2
17
Factors That Affect Reaction Rates
• 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.
18
Nature of Reactants• This is a very broad category that
encompasses the different reacting properties of substances.
• For example sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide.
burns. and ignites H Thereaction. rapid and violent a is This
HNaOH 2OH 2Na 2
2
g2aq2s
19
Nature of Reactants• Calcium reacts with water only slowly at
room temperature to liberate hydrogen and form calcium hydroxide.
reaction. slowrather a is This
HOHCaOH 2Ca g2aq22s
20
Nature of Reactants• The reaction of magnesium with water at
room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.
reaction No OH Mg 2s
21
Nature of Reactants• However, Mg reacts with steam rapidly
to liberate H2 and form magnesium oxide.
• The differences in the rate of these three reactions can be attributed to the changing “nature of the reactants”.
g2sC100
)g(2s HMgOOHMg o
22
Concentrations of Reactants: The Rate-Law
Expression• This movie illustrates how changing
the concentration of reactants affects the rate.
23
Concentrations of Reactants: The Rate-Law
Expression• This is a simplified representation of the
effect of different numbers of molecules in the 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 B BA B
A BA BA B
4 different possible A-B collisions
6 different possible A-B collisions
9 different possible A-B collisions
24
Concentrations of Reactants: The Rate-Law
ExpressionExample 16-1: The following rate data were obtained at 25oC for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction?
2 A(g) + B(g) 3 C(g)Experiment
NumberInitial [A]
(M)Initial [B]
(M)
Initial rate of formation 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
25
Concentrations of Reactants: The Rate-Law
ExpressionExperiment
NumberInitial [A]
(M)Initial [B]
(M)
Initial rate of formation 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
26
Concentrations of Reactants: The Rate-Law
ExpressionExample 16-2: The following data were obtained for the following reaction at 25oC. What are the rate-law expression and the specific rate constant for the reaction?
2 A(g) + B(g) + 2 C(g) 3 D(g) + 2 E(g)
ExperimentInitial [A]
(M)Initial [B]
(M)Initial [C]
(M)
Initial rate of formation of 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
27
Concentrations of Reactants: The Rate-Law
Expression
28
Concentrations of Reactants: The Rate-Law
ExpressionExample 16-3: consider a chemical reaction between compounds A and B that is first order with respect to A, first order with respect to B, and second order overall. From the information given below, fill in the blanks.
You do it!
29
Concentrations of Reactants: The Rate-Law
Expression
ExperimentInitial Rate
(M/s)Initial [A]
(M)Initial [B]
(M)1 4.0 x 10-3 0.20 0.0502 1.6 x 10-2 ? 0.0503 3.2 x 10-2 0.40 ?
30
Concentration vs. Time: The Integrated Rate
Equation• The integrated rate equation relates time
and concentration for chemical and nuclear reactions.– From the integrated rate equation we can
predict the amount of product that is produced in a given amount of time.
• Initially we will look at the integrated rate equation for first order reactions.These reactions are 1st order in the
reactant and 1st order overall.
31
Concentration vs. Time: The Integrated 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 many chemical reactions and all simple radioactive decays.
• Two examples of this type are:2 N2O5(g) 2 N2O4(g) + O2(g)
238U 234Th + 4He
32
Concentration vs. Time: The Integrated 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 reactions is:
k t aAAln 0
33
Concentration vs. Time: The Integrated Rate
Equation
• Solve the first order integrated rate equation for t.
• Define the half-life, t1/2, of a reactant as the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A]0.
AAln
k a1t 0
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Concentration vs. Time: The Integrated Rate
Equation• At time t = t1/2, the expression becomes:
k a693.0t
2lnk a
1t
A1/2Aln
k a1t
1/2
1/2
0
01/2
35
Concentration vs. Time: The Integrated Rate
EquationExample 16-4: Cyclopropane, an anesthetic, decomposes to propene according to the following equation.
The reaction is first order in cyclopropane with k = 9.2 s-1 at 10000C. Calculate the half life of cyclopropane at 10000C.
CH2 CH2
CH2CH2CH3
CH
(g) (g)
36
Concentration vs. Time: The Integrated Rate
EquationExample 16-5: Refer to Example 16-4. How much of a 3.0 g sample of cyclopropane 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.
37
Concentration vs. Time: The Integrated Rate
EquationExample 16-6: The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours.
CS2(g) CS(g) + S(g)
You do it!
38
Concentration vs. Time: The Integrated Rate
Equation
hr) 48)(hr 00101.0(Aln-ln(3.0)
k tAlnAlnk tAAln
1-
00
unreacted 97%or g 9.2g86.2eA
1.0521.10)--(0.048Aln0.048Aln-1.10
hr) 48)(hr 00101.0(Aln-ln(3.0)
k tAlnAlnk tAAln
1.052
1-
00
39
Concentration vs. Time: The Integrated Rate
Equation• For reactions that are second order with
respect to a particular reactant and second order overall, 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 since beginning of reaction.
a = stoichiometric coefficient of A in balanced overall equation.
k t aA1
A1
0
40
Concentration vs. Time: The Integrated Rate
Equation• Second order reactions also have a
half-life.– Using the second order integrated rate-
law as a starting point.• At the half-life, t1/2 [A] = 1/2[A]0.
0
1/200
A ofr denominatocommon a haswhich
k t aA1
A2/11
1/20
1/200
k t aA1
or k t aA1
A2
41
Concentration vs. Time: The Integrated Rate
Equation• If we solve for
t1/2:
• Note that the half-life of a second order reaction depends on the initial concentration of A.
01/2 Ak a1t
42
Concentration vs. Time: The Integrated Rate
EquationExample 16-7: Acetaldehyde, CH3CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide.
The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC?
CH CHO CH + CO3 g 4 g g
43
Concentration vs. Time: The Integrated Rate
EquationExample 16-7:
The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC?
CH CHO CH + CO3 g 4 g g
44
Concentration vs. Time: The Integrated Rate
Equation(b) How many moles of CH3CHO remain after 200 hours?
45
Concentration vs. Time: The Integrated Rate
Equation(c) What percent of the CH3CHO remains after 200 hours?
46
Concentration vs. Time: The Integrated Rate
EquationExample 16-8: Refer to Example 16-7. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 10.0 L vessel at 527oC?– Note that the vessel size is increased by a
factor of 10 which decreases the concentration by a factor of 10!
You do it!
47
Concentration vs. Time: The Integrated Rate
Equation• (b) How many moles of CH3CHO remain after
200 hours? You do it!
48
Concentration vs. Time: The Integrated Rate
Equation• (c) What percent of the CH3CHO remains
after 200 hours?You do it!
49
Concentration vs. Time: The Integrated Rate
Equation• Let us now summarize the results
from the last two examples.
Initial Moles
CH3CHO
[CH3CHO]0
(M)
[CH3CHO]
(M)
Moles of CH3CH
O at 200 hr.
% CH3CHO remainin
gEx. 16-
70.10 0.10 0.071 0.071 71%
Ex. 16-8
0.010 0.010 0.0096 0.096 96%
50
Enrichment - Derivation of Integrated Rate
Equations• For the first order reaction
a A productsthe rate can be written as:
tA
a1-=Rate
51
Enrichment - Derivation of Integrated Rate
Equations• For a first-order reaction, the rate is
proportional to the first power of [A].
-1a
At
k A
52
Enrichment - Derivation of Integrated Rate
Equations• In calculus, the rate is defined as the
infinitesimal change of concentration d[A] in an infinitesimally short time dt as the derivative of [A] with respect to time. -1
aAt
k Add
53
Enrichment - Derivation of Integrated Rate
Equations• Rearrange the equation so that all of
the [A] terms are on the left and all of the t terms are on the right.
- AA
a k td d
54
Enrichment - Derivation of Integrated Rate
Equations• Express the equation in integral form.
- A
Aa k t
A
A td d0 0
55
Enrichment - Derivation of Integrated 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
56
Enrichment - Derivation of Integrated Rate
Equations• Which rearranges to the integrated
first order rate equation.
k t aAAln
t
0
57
Enrichment - Derivation of Integrated Rate
Equations• Derive the rate equation for a
reaction that is second order in reactant A and second order overall.
• The rate equation is:
2Ak t a
A
dd
58
Enrichment - Derivation of Integrated Rate
Equations• Separate the variables so that the A
terms are on the left and the t terms on the right.
tk A aA
2 dd
59
Enrichment - Derivation of Integrated Rate
Equations• Then integrate the equation over the
limits as for the first order reaction.
t
0
A
A2 tk a
A A
0
dd
60
Enrichment - Derivation of Integrated Rate
Equations• Which integrates the second order
integrated rate equation.
k t aA1
A1
0
61
Enrichment - Derivation of Integrated Rate
Equations• For a zero order reaction the rate
expression is:
k t a
A
dd
62
Enrichment - Derivation of Integrated Rate
Equations• Which rearranges to:
tk aA dd
63
Enrichment - Derivation of Integrated Rate
Equations• Then we integrate as for the other
two cases:
t
0
A
A
tk aA0
dd
64
Enrichment - Derivation of Integrated Rate
Equations• Which gives the zeroeth order
integrated rate equation.
k t a-AAor
k t -aAA
0
0
65
Enrichment - Rate Equations to Determine
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 t aAlnor
k t aAlnAln
66
Enrichment - Rate Equations to Determine
Reaction Order • The equation for a straight line is:
• Compare this equation to the rearranged first order rate-law.
bmy x
67
Enrichment - Rate Equations to Determine
Reaction Order b m y x
• Now we can interpret the parts of the equation as 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 t aAln
68
Enrichment - Rate Equations to Determine
Reaction Order• Example 16-9: Concentration-versus-
time data for the thermal decomposition of ethyl bromide are given in the table below. Use the following graphs of the data to determine the rate of the reaction and the value of the rate constant. 700Kat HBrHCBrHC gg42g52
69
Enrichment - Rate Equations to Determine
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
70
Enrichment - Rate Equations to 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 zero order 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 second order with respect to [C2H5Br].
71
Enrichment - Rate Equations to 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
5Br
]
72
Enrichment - Rate Equations to 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]
73
Enrichment - Rate Equations to 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]
74
Enrichment - Rate Equations to 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
min 20.0min 3
60.0slope
min 14)20.0(80.0
x-xy-y slope
75
Enrichment - Rate Equations to Determine
Reaction Order• From the equation for a first order
reaction we know that the slope = -a k.– In this reaction a = 1.
.min 0.20kconstant rate theThus
-k-0.20slope1-
76
Enrichment - Rate Equations to Determine
Reaction Order• The integrated rate equation for a reaction
that is second order in reactant A and second order overall.
• This equation can be rearranged to: k t a
A1
A1
0
0A1k t a
A1
77
Enrichment - Rate Equations to 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 equation as 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.
b m y x
0A1k t a
A1
78
Enrichment - Rate Equations to Determine
Reaction OrderExample 16-10: Concentration-versus-time data for the decomposition of nitrogen dioxide are given in the table below. Use the graphs to determine the rate of the reaction and the value of the rate constant
500Kat ONO 2NO 2 g2gg2
79
Enrichment - Rate Equations to 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
80
Enrichment - Rate Equations to 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.
81
Enrichment - Rate Equations to 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]
82
Enrichment -Rate Equations to 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]
83
Enrichment - Rate Equations to 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
O 2]
84
Enrichment - Rate Equations to 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 on the graph of 1/[NO2] vs. time.
85
Enrichment - Rate Equations to Determine
Reaction Order
• From the equation for a second order reaction we know that the slope = a k– In this reaction a = 2.
1-1 min 0.45kconstant rate theThus
k 20.90slope
M
2 1
2 1
11
min
1( )y -y (5.50 1.90)slope x -x (5 1) min
3.60 slope 0.904 min
MM
M
86
Collision Theory of Reaction Rates
• Three basic events must occur for a reaction to occur the atoms, molecules or ions must:
1. Collide.2. Collide with enough energy to break and
form bonds.3. Collide with the proper orientation for a
reaction to occur.
87
Collision Theory of Reaction Rates
• One method to increase the number of collisions and the energy necessary to break and reform bonds is to heat the molecules.
• As an example, look at the reaction of methane and 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.
kJ 891 OH CO O CH (g)22(g)2(g)4(g)
88
Collision Theory of Reaction Rates
• Illustrate the proper orientation of molecules that is necessary for this reaction.
X2(g) + Y2(g) 2 XY(g)• Some possible ineffective collisions are :
X
XY Y Y
YX X X X Y Y
89
Collision Theory of Reaction Rates
• An example of an effective collision is:
X Y
X Y
X Y
X Y
X Y +
X Y
90
Collision Theory of Reaction Rates
• This picture illustrates effective and ineffective molecular collisions.
91
Transition State Theory• Transition state theory postulates that
reactants form a high energy intermediate, the transition state, which then falls apart into the products.
• For a reaction to occur, the reactants must acquire sufficient energy to form the transition state.– This energy is called the activation energy
or Ea.• Look at a mechanical analog for activation
energy
92
Transition State Theory
Epot = mgh
Cross section of mountain
Boulder
Eactivation
h
h2
h1
Epot=mgh2
Epot=mgh1
Height
93
Transition State Theory
PotentialEnergy
Reaction Coordinate
X2 + Y2
2 XY
Eactivation - a kinetic quantity
E Ha thermodynamic quantity
Representation of a chemical reaction.
94
Transition State Theory
95
Transition State Theory• The relationship between the activation
energy for forward and reverse reactions is– Forward reaction = Ea
– Reverse reaction = Ea + E– difference = E
96
Transition State Theory• The distribution of molecules
possessing different energies at a given temperature is represented in this figure.
97
Reaction Mechanisms and the Rate-Law
Expression• Use the experimental rate-law to
postulate a molecular mechanism.• The slowest step in a reaction
mechanism is the rate determining step.
98
Reaction Mechanisms and the Rate-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 to water and oxygen.
g22I
22 O + OH 2 OH 2-
99
Reaction Mechanisms and the Rate-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=R law rate alExperiment
O+OH 2OH 2reaction Overall
I+O+OHOH+ IO stepFast
OH+IOI+OH step Slow
100
Reaction Mechanisms and the Rate-Law
Expression• Important notes about this reaction:1. One hydrogen peroxide molecule and one
iodide ion are involved in the rate determining step.
2. The iodide ion catalyst is consumed in step 1 and produced in step 2 in equal amounts.
3. Hypoiodite ion has been detected in the reaction mixture as a short-lived reaction intermediate.
101
Reaction Mechanisms and the Rate-Law
Expression• Ozone, O3, reacts very rapidly with
nitrogen oxide, NO, in a reaction that is first order in each reactant and second order overall.
NOOk=Rate is law-rate alExperiment
O+NONO+O
3
g2g2gg3
102
Reaction Mechanisms and the Rate-Law
Expression• One possible mechanism is:
223
223
33
O+NONO+Oreaction Overall
O+NONO+O stepFast O+NONO+O step Slow
103
Reaction Mechanisms and the Rate-Law
Expression• A mechanism that is inconsistent
with the rate-law expression is:
correct. becannot mechanism thisproveswhich
Ok=Rate is mechanism thisfrom law-rate TheONONO+Oreaction Overall
NONO+O stepFast O+OO step Slow
3
223
2
23
104
Reaction Mechanisms and the Rate-Law
Expression• Experimentally determined reaction orders
indicate the number of molecules involved in:
1. the slow step only, or2. the slow step and the equilibrium steps
preceding the slow step.
105
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
106
Temperature: The Arrhenius Equation
• This movie illustrates the effect of temperature on a reaction.
107
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
108
Temperature: The Arrhenius Equation
1. Subtract one equation from the other.ln k k A - ln A - E
RTERT
ln k k ERT
- ERT
2 1a
2
a
1
2 1a
1
a
2
ln ln
ln
109
Temperature: The Arrhenius Equation
2. 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
110
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
,
.
.
111
Temperature: The Arrhenius Equation
• For reactions that have an Ea50 kJ/mol, the rate approximately doubles for a 100C rise in temperature, near room temperature.
• Consider:2 ICl(g) + H2(g) I2(g) + 2 HCl(g)
• The rate-law expression is known to be R=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
112
Catalysts• Catalysts change reaction rates by
providing an alternative reaction pathway with a different activation energy.
113
Catalysts• Homogeneous catalysts exist in same
phase as the reactants.• Heterogeneous catalysts exist in
different phases than the reactants.– Catalysts are often solids.
114
Catalysts• Examples of commercial catalyst
systems include:
NiO and Pt8 218 g 2 g 2 g g
NiO and Ptg 2 g 2 g
NiO and Ptg 2 g 2 g
2 C H +25 O 16 CO 18 H O
2 CO +O 2 CO
2 NO N O
Automobile catalytic converter system
115
Catalysts• This movie shows catalytic
converter chemistry on the Molecular Scale
116
Catalysts• A second example of a catalytic
system is:
npreparatio acid Sulfuric
SO 2OSO 2 g3NiO/Ptor OV
g2g252
117
Catalysts• A third examples of a catalytic system
is:
ProcessHaber
NH 2H 3 N g3OFeor Fe
g2g232
118
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) 2 O(surface)
• ReactionCO(surface) +O(surface) CO2(surface)
• DesorptionCO2(surface) CO2(g)
119
Catalysts