16 chemical kinetics. 2 chapter goals 1.the rate of a reaction factors that affect reaction rates...
TRANSCRIPT
16 Chemical Kinetics
2
Chapter Goals
1. The Rate of a Reaction
Factors That Affect Reaction Rates
2. Nature of the Reactants
3. Concentrations of the Reactants: The Rate-Law Expression
4. Concentration Versus Time: The Integrated Rate Equation
5. Collision Theory of Reaction Rates
3
Chapter Goals
6. Transition State Theory
7. Reaction Mechanisms and the Rate-Law Expression
8. Temperature: The Arrhenius Equation
9. Catalysts
4
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.
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 be imperceptible.
OUSINSTANTANE
kJ -79=G OHOH+H
SLOW VERY
kJ 396G COO C
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:
7
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100
150
200
250
300
350
Time
Con
cent
rati
ons
of
Rea
ctan
ts &
Pro
duct
s
[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 reactants disappear or products appear.
• This movie is an illustration of a reaction rate.
9
The Rate of Reaction
• Mathematically, the rate of a reaction can be written as:
t d
D+
t c
C+or
t b
B-
t a
A-= Rate
10
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)
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 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.
12
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
13
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 laws
2. rate equations
3. 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.order first and
ONin order first isreaction This
ONk= Rate
O + NO4ON 2
52
52
g2g2g52
15
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
16
The Rate of Reaction
• A final example of the order of a reaction is:
ALLYEXPERIMENT DETERMINED
ARE SEXPRESSION RATE ALL REMEMBER,
overallorder third and ,Oin order first
NO,in order second isreaction This
Ok[NO]=Rate
NO 2O+NO 2
2
2
2
g2g2g
17
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
18
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.
19
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 The
reaction. rapid and violent a is This
HNaOH 2OH 2Na 2
2
g2aq2s
20
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
21
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
22
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
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 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
25
Concentrations of Reactants: The Rate-Law Expression
• Example 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
Number
Initial [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 Expression
.Bignorecan can weThus,
k[A]=Rateor BAk=Rate
012
constant. remains rate initial that theNotice
2.by increases [B] that theandconstant is [A] the
thatsee we3 and 1 sexperiment compare weIf
BAk=Rate
:form theof bemust law rate The
0 xx
y
yx
y
27
Concentrations of Reactants: The Rate-Law Expression
You do it!
xx
reaction? for this
k of units and value theisWhat
overall.order 1 andA to
respect order with 1 isreaction This
k[A]=Rateor k[A]= Rate
122
2.by increases rate theand
2by increases [A] that theNotice
2. and 1 sexperiment compare Next,
st
st
1
28
Concentrations of Reactants: The Rate-Law Expression
[A] 10 x 2.0=Rate
as written becan law rate theThus
10 x 2.0 10.0
10 x 2.0=k
1 experiment from [A] and Rate of values theUsing
A
Rate=k
law.rate thefromk of value thefindcan We
s13-
s13-s
4-
M
M
29
Concentrations of Reactants: The Rate-Law Expression• Example 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)
Experiment
Initial [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
30
Concentrations of Reactants: The Rate-Law Expression
yxzyx
z z
BAk=Rateor CBAk=Rate
013
constant. remains rate but the 3by increases C The
constant.remain B and A that Notice
3. and 1 sexperiment Compare
31
Concentrations of Reactants: The Rate-Law Expression
BAk=Rateor BAk=Rate
133
3.by increases rate theand 3by increases B The
constant. remains A The
2. and 1 sexperiment compare Next,
1 xx
y y
32
Concentrations of Reactants: The Rate-Law Expression
overall.order 2 and B, respect to order with1
A, respect to order with 1 isreaction This
BAk=Rateor BAk= Rate
133
3.by increases rate theand 3by increases A The
constant. remains B The
4. and 2 sexperiment compare Next,
ndst
st
11
xx
33
Concentrations of Reactants: The Rate-Law Expression
BA100.1 = Rate
as written becan law-rate theThus,
100.1
0.10 0.20
100.2
BA
Rate=k
4.or 3, 2, 1, experiment from data theuseCan
k. of units and value thedetermine Finally,
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 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!
Experiment
Initial 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 ?
35
Concentrations of Reactants: The Rate-Law Expression
BA40.0 Rate theThus
40.0
0.050 0.20
100.4
BA
Rate=k
k. of value thedeterminecan we1 experiment From
BAk=Rate
s 1
s 1
s3
M
M
M
MM
36
Concentrations of Reactants: The Rate-Law Expression
M
MM
M
80.0]A[
050.0s0.40
s106.1]A[
k[B]
Rate[A]
2 experiment from data andk of value theUse
1-1
1-2
37
Concentrations of Reactants: The Rate-Law Expression
M
MM
M
20.0[B]
40.0s0.40
s102.3[B]
k[A]
R[B]
determinecan we3 experiment from Similarly,
1-1
1-2
38
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.
39
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
40
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 aA
Aln 0
41
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.
A
Aln
k a
1t 0
42
Concentration vs. Time: The Integrated Rate Equation• At time t = t1/2, the expression becomes:
k a
693.0t
2lnk a
1t
A1/2
Aln
k a
1t
1/2
1/2
0
01/2
43
Concentration vs. Time: The Integrated Rate Equation
• Example 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)
s 075.0s 2.9
693.0
k
693.0t
1-1/2
44
Concentration vs. Time: The Integrated Rate Equation• Example 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.
45
Concentration vs. Time: The Integrated Rate Equation
remains 1%or g 03.0eA
5.310.16.4Aln
6.4Aln 10.1
s 50.0s 2.9Aln -3.0ln
k t AlnAln
.logarithms of laws theusemust We
reaction for thisk t k t aA
Aln
3.5-
1-
0
0
46
Concentration vs. Time: The Integrated Rate Equation• Example 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!
47
Concentration vs. Time: The Integrated Rate Equation
1-
1/2
1/2
hr 00101.0hr 688
0.693k
t
0.693k
k
693.0t
1. areaction For this
48
Concentration vs. Time: The Integrated Rate Equation
unreacted 97%or g 9.2g86.2eA
1.0521.10)--(0.048Aln
0.048Aln-1.10
hr) 48)(hr 00101.0(Aln-ln(3.0)
k tAlnAlnk tA
Aln
1.052
1-
0
0
49
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 aA
1
A
1
0
50
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 aA
1
A2/1
1
1/20
1/200
k t aA
1
or k t aA
1
A
2
51
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 a
1t
52
Concentration vs. Time: The Integrated Rate Equation• Example 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
53
Concentration vs. Time: The Integrated Rate Equation
tk A
hr
hr
1/2
-1
1
1
2 0 10 010
5 0 10
0
2 1
2
. .
.
M M
54
Concentration vs. Time: The Integrated Rate Equation• (b) How many moles of CH3CHO remain after
200 hours?
1 1
1 1010
2 0 10 200
110 4 0
0
2 1
1 1
A Ak t
A hr hr
A
-1
..
.
MM
M M
Concentration vs. Time: The Integrated Rate Equation
55
114
1
14
0 071
0 071 mol
11A
A
A
mol = 1.0 L x 0.071 molL
M
M
M.
? .
56
Concentration vs. Time: The Integrated Rate Equation• (c) What percent of the CH3CHO remains
after 200 hours?
reacted 29% and unreacted %71
%100mol 0.10
mol 0.071= unreacted %
57
Concentration vs. Time: The Integrated Rate Equation• Example 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!
58
Concentration vs. Time: The Integrated Rate Equation
tk A
hr
hr
note the time has increased by 10
over Example 16 - 7:
1/2
-1
1
1
2 0 10 0 010
5 0 10
0
2 1
3
. .
.
M M
59
Concentration vs. Time: The Integrated Rate Equation• (b) How many moles of CH3CHO remain
after 200 hours?
You do it!
60
Concentration vs. Time: The Integrated Rate Equation
1 1
1 10 010
2 0 10 200
1100 4 0
0
2 1
1 1
A Ak t
A hr hr
A
-1
..
.
MM
M M
Concentration vs. Time: The Integrated Rate Equation
61
1104
1
104
0 0096
0 096 mol
11A
A
A
mol = 10.0 L x 0.0096 molL
M
M
M.
? .
62
Concentration vs. Time: The Integrated Rate Equation• (c) What percent of the CH3CHO remains
after 200 hours?
You do it!
63
Concentration vs. Time: The Integrated Rate Equation
% unreacted =0.096 mol0.100 mol
unreacted & 4% reacted
100%
96%
64
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
g
Ex. 16-7
0.10 0.10 0.071 0.071 71%
Ex. 16-8
0.010 0.010 0.0096 0.096 96%
65
Enrichment - Derivation of Integrated Rate Equations• For the first order reaction
a A products
the rate can be written as:
t
A
a
1-=Rate
66
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
67
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.
-1a
At
k Ad
d
68
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.
-A
Aa k t
dd
69
Enrichment - Derivation of Integrated Rate Equations• Express the equation in integral form.
-
AA
a k tA
A tdd
0 0
70
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 0
which becomes
-ln A A a k t
t0t
t
t
0
0
0
ln
ln
71
Enrichment - Derivation of Integrated Rate Equations• Which rearranges to the integrated first order
rate equation.
k t aA
Aln
t
0
72
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
d
d
73
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 a
A2 d
d
74
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
75
Enrichment - Derivation of Integrated Rate Equations• Which integrates the second order integrated
rate equation.
k t aA
1
A
1
0
76
Enrichment - Derivation of Integrated Rate Equations• For a zero order reaction the rate expression is:
k
t a
A
d
d
77
Enrichment - Derivation of Integrated Rate Equations• Which rearranges to:
tk aA dd
78
Enrichment - Derivation of Integrated Rate Equations• Then we integrate as for the other two cases:
t
0
A
A
tk aA0
dd
79
Enrichment - Derivation of Integrated Rate Equations• Which gives the zeroeth order integrated rate
equation.
k t a-AA
or
k t -aAA
0
0
80
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 aAln
or
k t aAlnAln
81
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
82
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
83
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
84
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
85
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].
86
Enrichment - Rate Equations to Determine Reaction Order• Plot of [C2H5Br] versus time.
– Is it linear or not?
[C2H5Br] vs. time
0
0.20.4
0.60.8
11.2
0 1 2 3 4 5
Time (min)
[C2
H5
Br]
87
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
0
0 1 2 3 4 5
Time (min)
ln [
C2H
5B
r]
88
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 5
Time (min)
1/[C
2H5B
r]
89
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-x
y-y slope
90
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-
91
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 aA
1
A
1
0
0A
1k t a
A
1
92
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
0A
1k t a
A
1
93
Enrichment - Rate Equations to Determine Reaction Order• Example 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
94
Enrichment - Rate Equations to Determine Reaction Order
Time
(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
95
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.
96
Enrichment - Rate Equations to Determine Reaction Order• Plot of [NO2] versus time.
– Is it linear or not?
[NO2] vs. time
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5
Time (min)
[NO
2]
97
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
0
0 1 2 3 4 5
Time (min)
ln [
NO
2]
98
Enrichment - Rate Equations to Determine Reaction Order• Plot of 1/[NO2] versus time.
– Is it linear or not?1/[NO2] vs.time
01
23
45
6
0 1 2 3 4 5
Time (min)
1/[
NO
2]
99
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.
100
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
min 1
1
1
12
12
90.0min 4
60.3slope
min 15
)90.1(50.5
x-x
y-y slope
MM
M
101
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.
102
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)
103
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
X
Y YY
Y
X X X X Y Y
104
Collision Theory of Reaction Rates• An example of an effective collision is:
X Y
X Y
X Y
X Y
X Y +
X Y
105
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
106
Transition State Theory
Epot = mgh
Cross section of mountain
Boulder
Eactivation
h
h2
h1
Epot=mgh2
Epot=mgh1
Height
107
Transition State Theory
PotentialEnergy
Reaction Coordinate
X2 + Y2
2 XY
Eactivation - a kinetic quantity
E Ha thermodynamic quantity
Representation of a chemical reaction.
108
Transition State Theory
109
Transition State Theory
• The relationship between the activation energy for forward and reverse reactions is– Forward reaction = Ea
– Reverse reaction = Ea + E
– difference = E
110
Transition State Theory
• The distribution of molecules possessing different energies at a given temperature is represented in this figure.
111
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.
112
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-
113
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
114
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.
115
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
116
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
117
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 The
ONONO+Oreaction Overall
NONO+O stepFast
O+OO step Slow
3
223
2
23
118
Reaction Mechanisms and the Rate-Law Expression• Experimentally determined reaction
orders indicate the number of molecules involved in:
1. the slow step only or
2. the slow step and the equilibrium steps preceding the slow step.
119
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 = Ae
or
ln k = ln A -ERT
-E RT
a
a
120
Temperature: The Arrhenius Equation• This movie illustrates the effect of temperature
on a reaction.
121
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 -E
RT
1a
1
2a
2
122
Temperature: The Arrhenius Equation1. Subtract one equation from the other.
ln k k A - ln A -E
RTERT
ln k kERT
-E
RT
2 1a
2
a
1
2 1a
1
a
2
ln ln
ln
123
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
124
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?
lnkk
ER
T - TT T
lnkk
8.314
K
lnkk
kk
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
,
.
.
125
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 s
k approximately doubles
0 -1 -1
0 -1 -1
M
M
126
Catalysts
• Catalysts change reaction rates by providing an alternative reaction pathway with a different activation energy.
127
Catalysts
• Homogeneous catalysts exist in same phase as the reactants.
• Heterogeneous catalysts exist in different phases than the reactants.– Catalysts are often solids.
128
Catalysts
• Examples of commercial catalyst systems include:
systemconverter catalytic Automobile
ONNO 2
CO 2O+CO 2
OH 18CO16O 25+HC
g2g2Pt and NiO
g
g2Pt and NiO
g2g
g2g2Pt and NiO
g2g188
129
Catalysts
• This movie shows catalytic converter chemistry on the Molecular Scale
130
Catalysts
• A second example of a catalytic system is:
npreparatio acid Sulfuric
SO 2OSO 2 g3NiO/Ptor OV
g2g252
131
Catalysts
• A third examples of a catalytic system is:
ProcessHaber
NH 2H 3 N g3OFeor Fe
g2g232
132
Catalysts
• Look at the catalytic oxidation of CO to CO2
• Overall reaction
2 CO(g)+ O2(g) 2CO2(g)
• Absorption
CO(g) CO(surface) + O2(g)
O2(g) O2(surface)
• Activation
O2(surface) O(surface)
• Reaction
CO(surface) +O(surface) CO2(surface)
• Desorption
CO2(surface) CO2(g)
133
Synthesis Question
• The Chernobyl nuclear reactor accident occurred in 1986. At the time that the reactor exploded some 2.4 MCi of radioactive 137Cs was released into the atmosphere. The half-life of 137Cs is 30.1 years. In what year will the amount of 137Cs released from Chernobyl finally decrease to 100 Ci? A Ci is a unit of radioactivity called the Curie.
134
Synthesis Question
Ci 100 reaches Chernobylat emitted Csfrom
ity radioactiv the when2426 440 1986
years440 years439 y0230.0
10.1t
t y0230.010.1
t y0230.0Ci 100
Ci 102.4ln
decay eradioactiv for this 1a andk t a A
Aln
Ci 102.4 MCi2.4
y0230.0 y1.30
693.0
t
693.0k
k
693.0t
137
1-
1-
1-6
0
6
1-
21
21
16 Chemical Kinetics