RATES THE OF CHEMICAL REACTIONS
Chapter 21 Outline
HOMEWORK
Exercises (part a): 3, 5, 8, 16(1st. part only), 21, 22, 23
Problems:
None
Supplementary Questions below.
Sect.
Title and Comments
Required?
1.
Experimental Techniques
YES
2.
The Rates of Reactions
YES
3.
Integrated Rate Laws
Note: You are NOT responsible for equations that are first order in
both [A] and [B] {Eqns. 21.18 and 21.19 on pg.794)
YES
4.
Reactions Approaching Equilibrium
NO - Well just comment on Reversible First-Order Reactions.
YES - Instead, well discuss Competitive First-Order Reactions.
5.
The Temperature Dependence of Reaction Rates
In addition to giving the Arrhenius Theory, we will also present the
Transition-State Theory of rate constants, given in Chapter 22 of the
text (Sects. 22.4-22.5).
YES
6.
Elementary Reactions
YES
7.
Consecutive Elementary Reactions
You do NOT have to memorize the equations for the concentrations
of the species.
YES
8.
Unimolecular Reactions
YES
9.
Polymerization Kinetics
NO
10.
Photochemistry
NO and YES
YES
Chapter 21
Supplementary Home Work Questions
S21.1 The rate law for the reaction, A + B Products, is of the form, r = k[A]x[B]y. From the initial
rate data for this reaction given below, determine the reaction orders, x and y, and the rate
constant, k (give units).
[Ao]
0.10 M
0.30
0.30
[Bo]
2.0 M
2.0
3.0
ro
8.50 Ms-1
2.83
7.80
S21.2 Consider a second order reaction, A Products. When [A]o = 1.30 M, the half life
of the reaction is 42 seconds.
(a) What is the rate constant, k, for this reaction?
(b) When [A]o = 1.30 M, what will the concentration be 60 seconds after the start of the
the reaction.
(c) When [A]o = 1.30 M, how long will it take for the concentration of A to decrease to
0.80 M?
S21.3 Consider a reaction, A Products, which is of order 3/2; i.e. Rate
d [ A]
k[ A]3/ 2
dt
(a) Develop an equation relating [A] to [A]o, k and t
(b) Develop an expression for the half-life, t1/2, of a 3/2 order reaction, in terms of [A]o and k.
(c) If a 3/2 order reaction has a rate constant, k = 0.03 M-1/2s-1. If the initial concentration of
the reactant is 0.50 M, what is the half-life of the reaction?
(d) If a 3/2 order reaction has a rate constant, k = 0.03 M-1/2s-1. If the initial concentration of
the reactant is 0.50 M, what will the concentration of A be after 25 s?
(e) If a 3/2 order reaction has a rate constant, k = 0.03 M-1/2s-1. If the initial concentration of
the reactant is 0.50 M, how long will it take for the concentration to decrease to 0.20 M?
S21.4 The reaction, A P, is of order x; i.e. Rate = k[A]x. When [Ao]= 0.2 M, the half-life of the
reaction is 200 s. When [A]o= 0.4 M, the half-life of the reaction is 25 s. What is the order of
this reaction (i.e. what is x)?
S21.5 The reaction, A P, is of order x; i.e. Rate = k[A]x. When [Ao]= 0.1 M, the half-life of the
reaction is 200 s. When [A]o= 0.25 M, the half-life of the reaction is 126 s. What is the order of
this reaction (i.e. what is x)?
S21.6 Consider the competitive first-order reactions,
k1
A B
k2
A C
(a) If one begins an experiment with [A]o = 1.20 M, it is found that [B] = 0.90 M at the
conclusion of the experiment.
What is the ratio of the two rate constants, k1/k2?
(b) It is found that the rate constant for the first reaction is k1 = 0.60 s-1. what is the
concentration, [C], 2.0 seconds after the start of the reaction?
S21.7 The rate constant for a first order reaction is 1.5x10-3 s-1 at 40 oC and 8.6x10-2 s-1 at 80 oC.
(a) Calculate the Arrhenius parameters, A and Ea, for this reaction.
(b) Calculate the rate constant of this reaction at 130 oC.
(c) Calculate the temperature at which the half-life of this reaction is 200 s.
S21.8 The Transition State Theory Equation for the rate constant is:
S
H
RT R RT
k
ee
R, NA and h are universal constants:
N Ah
R
8.31
2.3 x107
34
23
N A h (6.02 x10 )(6.63x10 )
For a given kinetics experiment, a plot of ln(k/T) vs. 1/T was a straight line with
Slope = -5450 K and Intercept = +6.80
Calculate the reactions Activation Enthalpy, H (in kJ/mol), and the
Activation Entropy, S (in J/mol-K).
S21.9 One theory of rate constants for bimolecular gas phase reactions is Collision Theory. The
collision theory form for the rate constant is: k C T e
threshold energy for reaction.
Eo
RT
, where C is a constant, and Eo is the
Develop and expression relating the threshold energy, Eo, to the Arrhenius Activation Energy.
S21.10 The reaction, 2 A + B P (P is the product) proceeds by the following mechanism.
k1
A + B F AB
k-1
k2
AB + A P
AB is an intermediate present in steady-state concentration. Use the steady-state
approximation on [AB] to develop an expression for the rate of formation of P as a function of
[A], [B], k1, k-1 and k2.
S21.11 In a pulsed laser fluorescence experiment on Napthalene (dissolved in hexane), the fluorescence
intensity 35 ns after the experiment begins is 65% of the intensity at the beginning of the
experiment.
In a separate experiment, it was determined that the fluorescence rate constant is:
kF = 4.8x106 s-1.
Calculate (a) the singlet state lifetime, 0 (in ns), and (b) the fluorescence quantum yield of
napthalene.
Chapter 21 Supplementary Homework Answers (Solutions on Web Site)
S21.1 x = -1 , y = 5/2 , k = 0.15 M-1/2s-1
S21.2 (a) k = 0.0183 M-1s-1
(b) [A] = 0.54 M
(c) t = 26.3 s
S21.3 (a)
1
1
k
t
1/2
1/2
[ A]
[ A]o
2
(b) t1/2
2
0.828
2 1
1/2
o
k[ A]
(c) t1/2 = 39 s
(d) [A] = 0.31 M
(e) t = 54.8 s 55 s
S21.4 Fourth order; i.e. x = 4
k[ A]1/2
o
S21.5 3/2 order; i.e. x = 3/2
S21.6 (a) k1/k2 = 3.00
(b) [C] = 0.24 M
S21.7 (a) Ea = 93.0 kJ/mol , A = 5.0x1012 s-1
(b) k = 4.4 s-1 at 130 oC
(c) At T = 48 oC, t1/2 = 200 s
S21.8 H = 45.3 kJ/mol
S = -84.3 J/mol-K
S21.9 Ea
1
RT Eo
2
S21.10 Rate
d [ P] k1k2 [ A]2 [ B ]
dt
k1 k2 [ A]
S21.11 (a) 0 = 81 ns
(b) F = 0.40
Chapter 21
21
The Rates of
Chemical Reactions
1
Spontaneous Reactions Dont Always Occur
Consider:
H2(g) + O2(g) H2O(l)
At 298 K, Go = -237.1 kJ/mol
K = 4x1041
Therefore, this reaction proceeds 100% to completion.
But, how long does it take?
Forever!!!
Without a catalyst, the formation of water from
hydrogen and oxygen proceeds infinitesimally slowly.
Thus, we see that thermodynamics tells us only whether a
reaction can occur. It cannot tell us whether it will occur, or
It
it
if it will, how fast is the reaction.
That is the subject of Chemical Kinetics.
2
1
The Reaction Rate
RP
Rate
[P2 ] [P1 ] [P]
t 2 t1
t
OR
P
conc.
R
[P]
t
[R ] [ R1 ]
[R]
Rate 2
t 2 t1
t
time
Example: Rate = (0.1 M - 0.0 M) / (10 s - 0 s) = 0.01 Ms-1
or: Rate = - (0.0 M 0.1 M) / (10 s - 0 s) = 0.01 Ms-1
Note: the text uses "v" for velocity of a reaction. In accordance
with common notation, we will use "Rate" or "R" to denote the
reaction rate.
3
Rates actually change with time
RP
R
P
conc.
d[P]
Rate =
dt
The derivative, d[P]/dT, is the
tangent of the curve.
OR
Rate
time
d[R]
dt
4
2
One must consider stoichiometry when writing
rates using different species.
Consider: A B + 2C
In t = 1 sec: [A] = -0.1 M
Rate = -
[A]
= -(-0.1 M/1 s) = 0.1 Ms-1
t
[B] = +0.1 M
Rate = +
[B]
= +(+0.1 M/1 s) = 0.1 Ms-1
t
[C] = +0.2 M
Rate = +
[C]
= +(+0.2 M/1 s) = 0.2 Ms-1
t
Rate = +
1 [C]
= 0.1 Ms-1
0.1
2 t
General Rule: Divide by Stoichiometric Coefficients when comparing
rates of change of different species.
5
General Rule: Divide by Stoichiometric Coefficients when comparing
rates of change of different species.
Thus, for the reaction: aA + bB cC + dD
The rate is:
Rate = -
1 d[A]
1 d[B]
1 d[C]
1 d[D]
==+
=+
a dt
b dt
c dt
d dt
6
3
Monitoring the Concentration
In order to study the rate of a reaction, one must be able to measure
the concentration of one of the reactants or products as a function of time.
There are a number of ways to accomplish this depending upon the
nature of the reaction. These methods include:
Chemical titration:
e.g. if Cl- is formed, one may determine the
concentration by titration with AgNO3(aq).
pH measurement:
Good for reactions in which H+ or OH- is
produced or consumed.
Pressure:
Good if number of gas phase moles changes;
e.g. 2N2O5(g) 4NO2(g) + O2(g)
7
Optical Rotation:
Good if reactant or product is optically active.
Spectrophotometry:
Useful if reactant or product has characteristic
absorption band;
e.g. IR C=O absorption in ketone decomposition.
IR C=O absorption in ketone decomposition
Vis. absorption of Br2 in: H2 + Br2 2HBr.
Gas Chromatography
NMR Spectra
Mass Spectrometry
Spectrometry
8
4
Experimental Methods
The experimental procedures used in the laboratory are
dependent upon the time scale of the reaction.
Hours, days, weeks, etc.
Extract aliquots (~10) and analyze.
aliquots
and analyze
1/2 Hour
Reaction will continue during analysis of aliquot.
Extract aliquots, quench (cool or dilute) and analyze.
1 - 5 minutes
Insufficient time to extract and quench a suitable number (10)
of samples to analyze.
In situ concentration measurements; e.g. spectroscopy, pH, etc.
9
1 millisecond - 1 minute
If reactants mixed in normal fashion by pouring together, the
reaction would be over by the time you finished mixing.
Use rapid mixing, flow methods.
Standard Flow
Stopped Flow
Requires less reactant.
10
5
1 microsecond - 1 millisecond
No time to mix reactants.
Relaxation Methods: Use a sudden shock to perturb a system
from equilibrium and measure the relaxation
to the new equilibrium concentrations
Temperature Jump (T-Jump)
-
+
Let reactants and products reach
reach equilibrium in aqueous solution.
Add electrolyte [e.g. NaCl(aq)] and
capacitor.
Discharge capacitor quickly (1 s).
Temperature jumps.
Measure relaxation to new equilibrium.
Analogous Methods: P-Jump, E-Jump
+
A+ B
+-
AB
H 0
11
1 picosecond - 1 microsecond
Flash Photolysis
Create excited state reactant photochemically
with ultrashort laser pulse (< 1 ns).
Measure decay of reactant or production of product.
12
6
Rate Laws: Order of a Reaction
aA + bB Products
Rate = k[A]x[B]y[C]z
[A], [B], [C] can be reactants, products or catalysts
x = order of reaction w.r.t. [A]
y = order of reaction w.r.t. [B]
z = order of reaction w.r.t. [C]
n = x + y + z = overall order of reaction
x, y, z are not necessarily equal to the stoichiometric coefficients
x, y, z can be:
positive integers, negative integers, non-integers
Not all rate laws are of the form above
e.g. Rate = k1[A][B] / (k2 + [B]) (rate law for Enzyme Cat. Rxns.)
The rate changes with time because the concentrations
change with time
13
Units of the Rate Constant
Rate = k[A]x[B]y[C]z
or Rate = k[conc]n
k=
n=x+y+z
1
1 [conc]
Rate =
n
[conc]
[conc]n (time)
k
11
1
1
n-1
n-1
M
s
[conc]
time
(Unit Analysis Only)
If
[conc] = M
and time = s
n=1
k = 1 / (M1-1s) = s-1
(M
n=2
k = 1 / (M2-1s) = M-1s-1
n = 5/2
k = 1 / (M5/2-1s) = M-3/2s-1
14
7
Determination of Reaction Order
The Initial Rate Method
A Products
R=-
d[A]
= k[A]n
dt
t
The rate changes with time because
[A] changes with time.
Measure for about
5% of reaction
conc.
It is often convenient to measure the
initial rate [Ro] for only a small amount
of time after the reaction begins.
time after the reaction begins
[A]
Ro = k[A]on
time
15
Principle of the Method
Experiment 1: (Ro)1 = k([A]on )1
Experiment 2: (Ro)2 = k([A]on )2
n
(Ro )2 k([A]o )n ([A]o )2
2
=
=
n
(Ro )1 k([A]o )1 ([A]o )1
When [A]o is doubled, Ro is doubled.
n
2(Ro )1 2([A]o )1
=
(Ro )1 ([A]o )1
2 = 2n
n=1
16
8
When [A]o is doubled, Ro is quadrupled.
n
4(Ro )1 2([A]o )1
=
(Ro )1 ([A]o )1
4 = 2n
n=2
17
When ([A]o)1 = 0.5 M ,
(Ro)1 = 0.10 Ms-1
When ([A]o)2 = 1.25 M , (Ro)2 = 1.56 Ms-1
n
(Ro )2 ([A]o )2
=
(Ro )1 ([A]o )1
n
1.56 Ms-1 1.25 M
=
0.10 Ms-1 0.5 M
15.6 = (2.5)n
Method of Logarithms
ln(15.6) = ln(2.5)n =nln(2.5)
[ln(a)b = bln(a)]
n = ln(15.6) / ln(2.5) = 2.75 / 0.92 = 2.99
n=3
Note: Can use either natural or common logs.
18
9
Generalization to Multiple Concentrations
A+B C
Ro = k [A]ox [B]oy [C]oz
x
y
z
(Ro )2 k([A]o )2 ([B]o )2 ([C]o )2
=
x
y
z
(Ro )1 k([A]o )1([B]o )1([C]o )1
x
y
([A]o )2 ([B]o )2 ([C]o )2
=
([A]o )1 ([B]o )1 ([C]o )1
z
Can isolate a given species by holding other concentrations
constant.
e.g. Double [A]o while holding [B]o and [C]o constant.
19
Example: Consider the reaction, 2A + B 2C
Ro = k [A]ox [B]oy [C]oz
Use the experimental data below to determine
x, y, z and the rate constant, k.
Expt. [A]o
#1 0.50 M
[B]o
0.10 M
[C]o
0.80 M
Ro
78 Ms-1
#2
0.75
0.05
0.20
176
#3
0.75
0.10
0.80
176
#4
0.75
0.05
0.80
352
x=2
y = -1
z = 1/2
Ro = k
2
[A]o [C]1/2
o
[B]o
k = 35 M-1/2s-1
20
10
Determination of Reaction Order
Use of the Integrated Rate Equation
A Products
and
R=-
d[A]
= k[A]n
dt
One may integrate the rate equation to obtain
th
[A] as a function of k, [A]o and t.
A = A([A]o, k, t)
The form of the function depends upon the order
of the reaction, n.
The order, n, and the rate constant, k, can be
th
calculated by determining which order equation
fits the experimental data.
21
First Order Reactions
A Products
R=-
d[A]
= k[A]1
dt
1
d[A] = -k dt
[A]
[A]
[A]o
t
1
d[A] = -k dt
0
[A]
[A]
ln
= -kt
[A]o
e
[A]
ln
[A]o
[A]
= e-kt
[A]o
[A] = [A]o e-kt
x2
x1
x2
x1
1
dx = ln(x 2 / x1 )
x
Cdx = C(x 2 - x1 )
= e-kt
22
11
[A] = [A]o e-kt
At t = 0
, [A] = [A]o
As t , [A] 0
[A]
[A]o
0
time
23
A Linear Equation
It is not straightforward to use plots of [A] vs. t to determine k.
It would be better to have a linear relation.
[A]
ln
= -kt
[A]o
Lets start again at this point.
ln([A]) = ln([A o ]) - k t
ln([A])o
Slope = -k
ln([A])
ln([A]) - ln([A o ]) = -kt
ln([A]) = ln([A o ]) - k t
y
x
t
24
12
A Products
0.0
The following data were obtained:
-0.2
ln([A])
-0.33
15
0.63
-0.47
25
0.55
-0.60
35
0.48
-0.73
45
0.42
ln([A])
t
[A]
5 min 0.72 M
-0.87
-0.4
-0.6
-0.8
Is this reaction first order?
0
Yes! Because ln([A]) vs. t is a straight line.
20
40
t (min)
What is k?
Slope =
-0.80 - (-0.40)
= -0.0133 min-1 = -k
(40 -10) min
k = 0.0133 min-1
25
Half-Life of a First Order Reaction
The Half-Life (t1/2) of a reaction (any order) is defined by:
t = t1/2 when [A] = [A]o
[A]
= -kt
[A]o
For a first order reaction: ln
1/ 2[A]o
ln
= -kt1/2
[A]o
1
ln = -kt1/2
2
-0.693 = -kt1/2
t1/2 =
0.693
k
Notes: (a) t1/2 1/k
(b) t1/2 is independent
of [A]o
26
13
1.0
[A]o
t1/2
10 - 0 = 10 s
0.5
20 - 10 = 10 s
0.25
30 - 20 = 10 s
10
Note that t1/2 is
independent of [A]o
[A] (M)
1.0 M
0.50
0.25
0.125
0
k = 0.693 / t1/2
10
20
30
time (sec)
= 0.693 / 10 s
= 0.0693 s-1
27
Example: For a first order reaction, A Products, the half-life
is 150 s.
(a) What is the rate constant, k?
k = 4.62x10-3 s-1
(b) If [A]o = 0.40 M, what is [A] after 240 s?
[A] = 0.13 M
(c) If [A]o = 0.40 M, how long does it take for
[A] to decrease to 0.08 M?
t = 350 s
28
14
Second Order Reactions
A Products
R=-
-
1
1
=
+ kt
[A] [A]o
d[A]
= k[A]2
dt
1
d[A] = k dt
[A]2
1
1
+ kt
[A]o
[A] =
t
1
d[A] = k dt
2
[A]o [A]
0
-
[A]
1
1
-
= kt
[A]o [A]
11
= kt
[A] [A]o
x2
x1
x2
x1
1
11
dx = 2
x
x1 x 2
Cdx = C(x 2 - x1 )
29
At t = 0
1
1
+ kt
[A]o
[A]o
, [A] = [A]o
As t , [A] 0
[A]
[A] =
time
It is not straightforward to use the above curve to determine
the rate constant, k.
It would be better to have a linear relation.
30
15
A Linear Equation
We ALREADY have a linear relation!!!
1
1
=
+k t
[A] [A]o
x
1/[A]
y
This is the step just before
solving for [A].
Slope = +k
1/[A]o
If you believe a reaction may be
second order, plot 1/[A] vs. t.
order plot 1/[A] vs
t
If the plot is a straight line, you have
verified the order. The rate constant can
be obtained from the slope.
31
Half-Life of a Second Order Reaction
As before, the Half-Life (t1/2) of a reaction (any order) is defined by:
t = t1/2 when [A] = [A]o
For a second order reaction:
11
= kt
[A] [A]o
1
1
= kt1/2
1/ 2 [A]o [A]o
2
1
= kt1/2
[A]o [A]o
Notes: (a) t1/2 1/k
(b) t1/2 1/[A]o
Not Independent of [A]o
1
= kt1/2
[A]o
t1/2 =
1
k[A]o
32
16
1.0
[A]o
t1/2
10 - 0 = 10 s
0.5
30 - 10 = 20 s
0.25
70 - 30 = 40 s
40
Note that t1/2 1/[A]o
[A] (M)
1.0 M
0.50
0.25
0.125
0
10
30
70
time (sec)
k = 1/t1/2[A]o = 1/(10 s1 M) = 0.1 M-1s-1
or k = 1/t1/2[A]o = 1/(20 s0.5 M) = 0.1 M-1s-1
or k = 1/t1/2[A]o = 1/(40 s0.25 M) = 0.1 M-1s-1
33
Third Order Reactions
A Products
R=-
d[A]
= k[A]3
dt
What should we plot to get a straight line?
1
1
= 2 + 2kt
2
[A]
[A]o
1/[A]2
In Class
Slope = +2k
1/[A]o2
In Class
t
t1/2 =
3
2 Notes: (a) t1/2 1/k
2k[A]o
(b) t1/2 1/[A]o2
Not Independent of [A]o
34
17
Example: The reaction, A Products, is third order;
i.e. -d[A]/dt = k[A]3
(a) When [A]o = 0.40 M, it takes 75 s for the
concentration to decrease to 0.10 M.
What is the rate constant, k?
k = 0.625 M-2s-1
(b) When [A]o = 0.40 M, what is the concentration
of A after 315 s?
[A] = 0.05 M
35
Zeroth Order Reactions
A Products
R=-
d[A]
= k[A]0
dt
R=-
d[A]
=k
dt
What???
How can the rate of a reaction be independent
of the concentration of reactant??
the concentration of reactant??
Photochemical Reactions
Rate limited by photon flux
Surface Catalyzed Reactions
Rate limited by number of active sites on surface
Enzyme Catalyzed Reactions
At high substrate concentration, rate limited
by enzyme concentration
36
18
A Products
R=-
d[A]
= k[A]0 = k
dt
What should we plot to get a straight line?
[A]o
[A]
Integrate
[A] = [A]o - kt
Slope = -k
t
t1/2 =
[A]o
2k
Notes: (a) t1/2 1/k
(b) t1/2 [A]o
Not Independent of [A]o
37
Generalization: Linear Plots
n=1
[A]
ln([A])
n=0
Ignore
n=1
t
t
vs. t
n=2
ln([A]) vs. t
1/[A]2
1/[A]
[A]
(=1/[A]0-1)
n=3
t
1 / [A] (=1 / [A]2-1) vs. t
t
1/
[A]2
(=1 / [A]3-1) vs. t
38
19
The Trend in Plots
For any order (n) except n = 1, a plot of
yields a straight line.
1
vs. t
[A]n-1
For example, if you believe that the reaction order might
be n = 3/2, then plot:
1
1
= 1/2 vs. t
3/2-1
[A]
[A]
A straight line would verify the assumed reaction order.
39
Generalization: Half-Lives
n=0
=
n=2
[A]o
2k
t1/2 =
n=1
1
1
2k [A]0-1
o
t1/2 =
=
1
k[A]o
t1/2 =
=
n=3
1
1
2-1
k [A]o
Trend: For all orders (including n = 1),
0.693
k
0.693 1
1-1
k
[A]o
t1/2 =
3
2
2k[A]o
t1/2 =
3
1
3-1
2k [A]o
t1/2
1
[A]n-1
o
40
20
Determining Reaction Order: Trial and Error
A Products
-d[A]/dt = k[A]n
-2.4
ln([A])
t
[A]
10 s 0.239 M -1.43
0.153
50
0.122
-2.10
70
0.104
-2.26
90
0.092
-2.39
-2.2
-1.88
Is this reaction first order?
No Way!!!
ln([A])
30
-2.0
-1.8
-1.6
-1.4
0
40
80
t (s)
41
A Products
-d[A]/dt = k[A]n
14
1/[A]
t
[A]
10 s 0.239 M 4.18 M-1
0.153
6.54
50
0.122
8.20
70
0.104
9.62
90
0.092
10.87
Is this reaction second order?
Close, but no cigar!!
12
1 / [A] (M-1)
30
10
8
6
4
0
40
80
t (s)
42
21
A Products
-d[A]/dt = k[A]n
1/[A]2
t
[A]
10 s 0.239 M 17.5 M-2
0.153
50
0.122
67.5
70
0.104
92.5
90
0.092
105
42.5
118
1 / [A]2 (M-2)
30
130
80
55
30
Is this reaction third order?
Yes!!!
5
What is k?
0
40
80
t (s)
1
1
= 2 + 2kt
2
[A]
[A]o
Slope =
(130 - 30) M-2
= 1.25 M-2 s-1 = 2 k
(100 - 20) s
k = 0.625 M-2s-1
43
Direct Determination of Reaction Order: Half-Life Method
1.0
1.0
n=1
t1/2
[A] (M)
[A] (M)
t1/2
n=2
1
constant
[A]11
o
0.50
1
1
2
[A] o1 [A] o
0.50
0.25
0.25
0.125
0.125
0
10
20
0
30
time (sec)
10
30
70
time (sec)
Chapter 1: Slide 3
Chapter 1: Slide 4
In contrast to trial and error, there are a number of direct methods
to determine the order of a reaction.
determine the order of reaction
One of these is the Half-Life Method.
One can determine the reaction order by learning how the
half-life depends upon the initial concentration, [A]o
44
22
t1/2
1
[A]n-1
o
When [A]o = 0.2 M, t1/2 = 60 s.
When [A]o = 0.4 M, t1/2 = 15 s.
What is the order, n?
Note that when [A]o is doubled, t1/2 is reduced by a factor of four.
Therefore,
Hence, 2 = n-1
t1/2
1
1
2
[A]o [A]n-1
o
n=3
The proportionality between [A]o and t1/2 is not always obvious
from inspection.
One may use a mathematical method to determine n from the data.
45
Mathematical Procedure to Determine n
(t1/2 )1
1
([A]n-1 )1
0
and
(t1/2 )2
1
([A]n-1 )2
0
(t1/2 )2 1/ ([A]n-1 )2
0
=
(t1/2 )1 1/ ([A]n-1 )1
0
(t1/2 )2 ([A]o )1
=
(t1/2 )1 ([A]o )2
When [A]o = 0.2 M, t1/2 = 60 s.
[A]
60
When [A]o = 0.4 M, t1/2 = 15 s.
n-1
What is the order, n?
15 s 0.2 M
=
60 s 0.4 M
n-1
46
23
0.25 = (0.50)n-1
ln(0.25) = ln(0.50)n-1 = (n-1)ln(0.50)
n-1 = ln(0.25) / ln(0.50) = (-1.39) / (-0.69) = 2.0
ln(0.25) ln(0.50)
2.0
n=3
47
Reactions Approaching Equilibrium
(Reversible First-Order Reactions
We'll just discuss this material briefly. You are not responsible for it.
Reversible reactions (first order and more complex reactions) are very
important, and can be well studied by relaxation methods (introduced
earlier).
Consider the reversible reaction, A F B, in the case that the rates
of the forward and reverse reactions are both first order:
AB
RF = -d[A]/dt = kf[A]
BA
RR = -d[B]/dt = +d[A]/dt= kR[B]
With a bit of algebra, it can be shown that:
[ A] [ A]o e ( kF kB ) t where
[ A] [ A] [ A]eq
[ A]o [ A]o [ A]eq
Thus, the deviation of [A] from its equilibrium value decreases
exponentially with a rate constant equal to the sum of the forward
and reverse first-order rate constants.
48
24
[ A] [ A]o e ( kF kB ) t where
[ A] [ A] [ A]eq
[ A]o [ A]o [ A]eq
Thus, the deviation of [A] from its equilibrium value decreases
exponentially with a rate constant equal to the sum of the forward
and reverse first-order rate constants.
Therefore, measurement of [A] as a function of time (from a relaxation
experiment) allows one to determine the sum of the two rate
constants, kF + kB.
The ratio of the two constants, kF/kB, can be determined from
the equilibrium concentrations:
K
[ B ]eq
[ A]eq
kF
kB
Thus, measurement of the decay kinetics, and the equilibrium
concentrations at infinite time permit determine of both
the forward and reverse rate constants for the reversible reaction.
49
Competitive First Order Reactions
When a synthetic chemist performs a reaction, (s)he will often
obtain more than one product (e.g. lovely white crystals + ugly
black gunk).
This is an example of multiple reaction pathways for the given
reactant. This can be studied using a "Competitive" first order
reaction mechanism.
Consider the two first order reactions:
k1
A B
(Product #1)
2
A k C
(Product #2)
Below, we will develop expressions for [A], [B] and [C]
as a function of time.
50
25
k1
A B
k2
A C
[A] vs. time
Both reactions cause [A] to decrease with time.
d[A]
= -k1[A] - k 2 [A] = - k1 + k 2 [A] = -k'[A]
dt
This is a simple first-order rate law, with the effective rate constant,
k' = k1 + k2
It may be integrated directly to obtain the following equation
for [A].
[A]
[ A] [ A]o e ( k1 k2 )t [ A]o e k ' t
Notice that the rate constant for the disappearance of [A] is the
sum of the rate constants for the two competing reactions.
51
k1
A B
[ A] [ A]o e ( k1 k2 )t [ A]o e k ' t
k2
A C
[B] vs. time
[B] is formed only from the first of the two rate equations.
d [ B]
k1[ A]
dt
To integrate this equation, we insert the above expression
for [A].
d [ B]
k1[ A]o e ( k1 k2 )t k1[ A]o e k 't
dt
52
26
d [ B]
k1[ A]o e ( k1 k2 )t k1[ A]o e k 't
dt
With the initial condition, [B]o = 0, we can integrate (in class) to get
the following equation for [B] vs. time.
[ B]
k1
k
[ A]o 1 e ( k1 k2 )t 1 [ A]o 1 e k 't
k1 k2
k'
[C] vs. time
Using the identical procedure for [C], we have:
d [C ]
k2 [ A]
dt
This yields: [C ]
k2
k
[ A]o 1 e ( k1 k2 )t 2 [ A]o 1 e k 't
k1 k2
k'
53
[ A] [ A]o e ( k1 k2 ) t
[ B]
k1
[ A]o 1 e ( k1 k2 )t
k1 k2
[C ]
k2
[ A]o 1 e ( k1 k2 ) t
k1 k2
As one would expect, [A] decreases exponentially with a rate
constant equal to the sum, k1 + k2
Perhaps surprisingly, [B] and [C] both increase exponentially,
with rate constants equal to the sum, k1 + k2
However, the relative amounts of the two products depend upon
their respective rate constants.
54
27
[ A] [ A]o e ( k1 k2 ) t
[ B]
k1
[ A]o 1 e ( k1 k2 )t
k1 k2
[C ]
k2
[ A]o 1 e ( k1 k2 ) t
k1 k2
Let's consider the relative concentrations of the two products,
[B]/[C].
k1
[ A]o 1 e ( k1 k2 ) t
k
[ B ] k1 k2
1
k2
[C ]
[ A]o 1 e ( k1 k2 ) t k2
k1 k2
55
k1
[ A]o 1 e ( k1 k2 ) t
k
[ B ] k1 k2
1
k2
[C ]
[ A]o 1 e ( k1 k2 ) t k2
k1 k2
Thus, we see that the relative yields of two different products
in a reaction is a measure of their relative rate constants.
Consider a reactant, [A] which undergoes two first-order reactions to
form the products, [B] and [C]
(A) If one begins with an initial concentration of the reactant, 0.90 M.
At the conclusion of the experiment, the concentration of C was
0.55 M. What is the value of the ratio, k1/k2 ?
k1/k2 = 0.64
56
28
k1
[ A]o 1 e ( k1 k2 ) t
k
[ B ] k1 k2
1
k2
[C ]
k2
( k1 k2 ) t
[ A]o 1 e
k1 k2
Thus, we see that the relative yields of two different products
in a reaction is a measure of their relative rate constants.
Consider a reactant, [A] which undergoes two first-order reactions to
form the products, [B] and [C]
(B) The rate constant, k1, for the first reaction was found to be
k1 = 0.050 s-1. If one begins with an initial concentration of the
reactant of 0.90 M, what will be the concentration of [C]
10 s after the start of the reaction?
[C] = 0.40 M
57
Temperature Dependence of the Rate Constant
k
It is observed for most reactions that the rate constant, k, increases
exponentially with rising temperature.
T
58
29
Fraction with Energy (E)
Energy
Ea
Rcts
Low T
High T
Prods
Reaction Coordinate
Energy (E)
Ea
Fraction: E Ea
In order for molecules to react, they
must overcome an energy barrier,
called the Activation Energy (Ea).
At low temperature, only a small
fraction of collisions have E Ea
At high temperature, a larger
fraction of collisions have E Ea
59
The Arrhenius Equation
k Ae
Ea
RT
k
Svante Arrhenius (1889)
Matches observed k vs. T
A = Pre-Exponential Factor
T
Ea = Activation Energy
Units: kJ/mol
R = 8.31 J/mol-K
Energy
Units: Same as k
Ea
Rcts
Prods
T = Temperature (K)
Reaction Coordinate
60
30
Relation Between Ea and Temperature
Dependence of k
Ea
RT
ln(k ) ln( A)
Ea
RT
ln(k)
k Ae
E
d ln( k )
a2
dT
RT
T (K)
This equation predicts that a plot of ln(k) vs. T will NOT be
a straight line. Rather the slope will become smaller at
straight line Rather the slope will become smaller at
higher temperatures.
We will use the above expression for dln(k)/dT in a later section.
However, for now let's determine how to obtain a straight
line plot
61
E
d ln(k )
d ln(k ) d (1/ T )
1 d ln(k )
a2
2
dT
RT
d (1/ T
dT
T d (1/ T
Therefore:
E
d ln(k )
a
d (1/ T
R
ln(k)
Thus, we expect that if ln(k) is plotted vs. 1/T, we should get a straight
we expect that if ln(k) is plotted vs 1/T we should get straight
line with Slope = -Ea/R
-1
1/T (K )
62
31
Determination of the Arrhenius Parameters
Ea
RT
Ea
ln(k ) ln Ae RT
ln(A)
Ea
ln(k ) ln( A) ln e RT
ln(k) ln(A)
Slope = -Ea/R
ln(k)
k Ae
Ea 1
RT
1/T
ln(k) ln(A)
y
Ea 1
RT
x
63
A rate constant was measured as a function of temperature, and the
following Arrhenius plot [ln(k) vs. 1000/T] was obtained.
Calculate A and Ea for this reaction.
ln(k) = ln(A) 8
ln(k)
NOT Int.
Slope
6
ln(k)
0.0 4.0
(1/T) (4.2 3.0)x103 K 1
Note
E
Slope 3330 K a
R
4
2
Ea = -R(-3330 K)
K)
= -8.31 J/mol-K(-3330 K)
0
-2
2.6
Ea 1
RT
3.4
1000/T (K-1)
4.2
= +27690 J/mol
Ea = 27.7 kJ/mol
64
32
ln(k) = ln(A) 8
Int ln(A) ln(k1 )
6
ln(k)
NOT Int.
Ea 1
RT
4.0
4
2
Ea 1
R T1
27,690 J/mol
3.0x103 K 1
8.314 J/mol-K
ln(A) = 14.0
0
A = 1.2x106 s-1
-2
2.6
3.4
4.2
1000/T (K-1)
65
Two Point Analysis
ln(k1 ) ln(A)
ln(k 2 ) ln(A)
Ea 1
R T1
Ea 1
R T2
ln(k 2 )-ln(k1 )
Ea 1 Ea 1
R T2 R T1
ln(k 2 /k1 )
Ea
R
1 1
T2 T1
For a first order reaction, the measured rate constant was
5. s-1 at 25 oC and 15. s-1 at 35 oC.
Calculate A and Ea for this reaction.
A = 2.5x1015 s-1
Ea = 83.8 kJ/mol
66
33
A second order reaction has an activation energy of 60 kJ/mol.
The rate constant is 3.0 M-1s-1 at 25 oC.
What is the value of k at 50 oC?
k = 19.5 M-1s-1
19
A first order reaction has an activation energy of 45 kJ/mol.
The half-life is 50 s at 25 oC.
At what temperature (in oC) is the half-life equal to 10 s?
T = 54 oC
54
67
Transition State Theory
The material on Transition State Theory can be found in Chapter 22
of the text (Sects. 22.4 and 22.5)
Deficiencies of the Arrhenius Theory
The Arrhenius Equation is basically empirical. Whereas the activation
energy, Ea, can be interpreted as the energy barrier to reaction, there
can be interpreted as the energy barrier to reaction there
is no interpretation of the pre-exponential factor, A.
Furthermore, it is not possible to predict the parameters theoretically.
Transition State Theory (aka Activated Complex Theory)
In 1935, Henry Eyring applied the theoretical methods of equilibrium
statistical mechanics to determine the rate constants for elementary
reactions.
He assumed that the reactants are in a quasi-equilibrium with a
transition state (or activated complex)
K
A B AB
68
34
AB
K
A B AB
[AB] is related to the reaction concentrations
by the constant, equilibrium K:
K
[ AB ]
[ A][ B]
or
[ AB ] K [ A][ B]
A+ B
One particular vibration of the activated
complex, , leads to conversion of AB to
products.
The rate of the reaction is then proportional
to the frequency of the vibration, , and
the concentration of activated complexes, [AB].
concentration of activated complexes [AB
Rate [ AB ] K [ A][ B]
Note: Many treatments of TST include a transmission coefficient, ,
representing the fraction of complexes that proceed to products.
It is often assumed that 1, as we have done here.
69
Rate [ AB ] K [ A][ B] kr [ A][ B]
kr is the reaction rate constant, given by: kr K
is the frequency of the vibration of the activated complex which leads
to dissociation into products
K is the equilibrium constant between reactants and activated complex.
Using statistical mechanics formulae for vibrational frequencies, it can
be shown that:
k BT
RT
h
N Ah
kB is Boltzmann's constant, and is related to the gas constant, R,
by kB = R/NA, and h is Planck's Constant (6.63x10-34 J-s)
R/N and is Planck Constant (6
Thus, we have the TST expression for the rate constant:
k
k BT RT
K
K
h
N Ah
70
35
k
k BT RT
K
K
h
N Ah
One advantage of Transition State Theory over the Arrhenius Theory is
that Statistical Mechanical methods have been well studied to predict
equilibrium constants.
Thus, one can use the equation above to predict values for the rate
constants of elementary reactions.
However, a big advantage of TST for experimental kineticists is that,
as we shall see the. TST expression for the rate constant, like the
Arrhenius Equation, has two parameters and both are interpretable.
71
Thermodynamic Formulation of TST
k
RT
K
N Ah
The equilibrium constant, K, may be related to the Gibbs Activation
Energy, G, and to the Activation Enthalpy, H and
Activation Entropy, S, by the standard relations:
H S
ln( K )
G RT ln( K ) H T S
RT
R
Therefore: K e
H S
RT
R
e
S
R
e
H
RT
Thus, the TST equation for the rate constant is:
S
H
RT R RT
k
ee
N Ah
72
36
S
H
RT R RT
k
ee
N Ah
Note: The equation in the text (Eqn. 22.43) differs from the one above:
S
H
RT RT R RT
k
o e e
N Ah p
Text Eqn. 22.43 after minor manipulation
A comparison shows that the text equation has the additional
factor, RT/po.
That term arises from the conversion from Kp to Kc. These are
different by that factor for bimolecular gas phase reactions.
How for reactions in solution and for unimolecular gas phase reactions,
Kp and Kc are the same. The form of the equation that we present is
the correct one for these cases.
73
It is useful to compare the TST and Arrhenius equations for the
rate constant:
k
S
H
RT R RT
ee
N Ah
Transition State
Theory
k Ae
Ea
RT
Arrhenius
Theory
Note that TST has two parameters (H and S) just like the
Arrhenius Theory (Ea and A). However, both TST parameters
have a mechanistic interpretation.
H, the Activation Enthalpy, has a meaning qualitatively similar
to Ea. It represents the barrier which the colliding molecules must
overcome in order to react to form products.
S, the Activation Entropy, represents the relative amount of disorder
of the activated complex compared to reactants.
This parameter is often very useful in determining the mechanism of
the reaction.
74
37
S, the Activation Entropy, represents the relative amount of disorder
of the activated complex compared to reactants.
This parameter is often very useful in determining the mechanism of
the reaction.
Metal Carbonyl Substitution
L + M(CO)6 ML(CO)5 + CO
Associative: L + M(CO)6 ML(CO)6 ML(CO)5 + CO S < 0
Dissociative: L + M(CO)6 L + M(CO)5 + CO
ML(CO)5 + CO
S > 0
Ring Opening Reaction
If S 0, the ring structure is preserved in the transition state.
the ring structure is preserved in the transition state
If S > 0, the ring has opened in the transition state
75
Determination of the TST Parameters
A linearized form of the TST equation can be developed in the
following manner.
S
H
RT R RT
k
ee
N Ah
T ln NRh e
ln k
A
k
S
R
e
H
RT
T
S
H
R R RT
ee
N Ah
S H C H
ln R N h
A
R
RT
RT
S
where C ln R N h
A
R
One expects a plot of ln(k/T) vs. 1/T will be a straight line
76
38
S
T ln R N h R
ln k
A
S
C ln R
N h
A
R
H
H
C
RT
RT
Slope
T
ln k
C
y
H 1
RT
x
Int
ln(k/T)
S
Int = C = ln R
N h +
A
R
Slope = -H/R
1/T
77
Relation Between TST and Arrhenius Parameters
Relation between Ea and H
Recall that we showed that the Arrhenius Equation leads to an
expression relating Ea to dln(k)/dT.
k Ae
Ea
RT
E
d ln(k )
a2
dT
RT
Let's use this to relate Ea to H.
k
S
H
RT R RT
ee
N Ah
Therefore:
Then:
S H
ln(k ) ln R
N h ln T
A
R
RT
d ln(k )
1
H H 1
0 0
dT
T
RT 2 RT 2 T
Ea
H 1
RT 2 RT 2 T
Ea H RT
or H Ea RT
78
39
Ea H RT
or H Ea RT
The difference between the TST Activation Enthalpy
and the Arrhenius Activation Energy is not especially
large for reactions around room temperature.
For example, if H = 50.0 kJ/mol, then:
3
At 300 K: Ea 50.0 kJ (8.31x10 kJ / mol K )(300 K ) 52.5 kJ / mol
i.e. approximately 5% higher
However, there is a major interest in high temperature kinetics
(e.g. in combustion chemistry), in which reactions occur at
1500 K to 2000 K or higher.
to 2000 or higher
3
At 1500 K: Ea 50.0 kJ (8.31x10 kJ / mol K )(1500 K ) 62.5 kJ / mol
i.e. approximately 25% higher.
Thus, one observes a very significant deviation between the
two parameters at elevated temperatures.
79
Relation between A and S
H Ea RT
Let's substitute this relation into the TST Equation:
k
Ea
Ea
S
H
S
S
RT
S
RT R RT
RT R EaRTRT
RT R RT RT
RTe R RT
ee
ee
eee
ee
N Ah
N Ah
N Ah
N Ah
Thus: k
Ea
E
S
a
RTe R RT
Ae RT
ee
N Ah
S
This gives: A RTe e R
N Ah
Therefore, we see that A
S
RTe R
e
showing that the Arrhenius
N Ah
pre-exponential factor is an indirect measure of the Activation Enthalpy.
80
40
It is instructive to evaluate A (at room temperature) for S = 0:
A
S
RTe R
(8.31 J / mol K )(298 K )(2.72) 0
e
e 1.7 x1013 s 1
N Ah
(6.02 x1023 mol 1 )(6.63x1034 J s )
Therefore, as a rule of thumb, if an experimental activation energy is:
A < 1.7x1013 s-1 S < 0, and
A > 1.7x1013 s-1 S > 0
S
A
1x1016
1x1015
1x1014
1x1013
1x1012
+52 J/mol-K
+34
+15
-4
-24
81
Catalysis
Energy
k Ae
Ea
Ea
Reaction Coordinate
Ea
RT
We learned that k, and hence the
reaction rate, can be increased by
raising the temperature.
th
At higher temperatures, a greater
fraction of collisions have an
energy greater than the
activation energy, Ea.
A second way to increase the rate of a reaction is to add a catalyst.
This is a species which increases the reaction rate without being
consumed in the reaction.
It accomplishes this by providing an alternative reaction pathway
with a lower activation energy, Ea.
82
41
Accounting for the Rate Laws
Most reactions require more than a single step. The reaction
mechanism is the detailed series of individual steps required for
transformation of the reactants to products.
Elementary Reactions
Sometimes, a reaction occurs in a single step. In this case,
the rate law can be written immediately by inspection of the
reaction stoichiometry.
k
H2 + I2 2HI
R = k[H2][I2]
Note that the converse is not necessarily true;
i.e. if the experimental rate law follows the stoichiometry
of the overall reaction, the mechanism may still be more
than a single step.
83
Elementary First Order Reactions
Consider the simple first order reaction: A P
If it is an elementary reaction, the rate law is: Rate = -
d[A]
= k[A]
dt
Assuming that [A](t=0) = [A]o and [P](t=0) = 0, this equation integrates
th [A](t
[A]
[P](t
thi
to:
[A] = [A]o e kt
One can also determine [B] from the relation: [B] = [A]o - [A]
[B] = [A]o [A] [A]o [A]o e kt
or
[B] = [A]o 1 e kt
Thus we see that:
(1) [A] decreases exponentially from [A]o to 0
(2) Simultaneously, [B] increases exponentially from 0 to [A]o
84
42
Consecutive First Order Reactions
k2
Consider two consecutive first order reactions: A k1 I P
The initial reactant, A, forms an intermediate, I, when then reacts to form
the product, P.
The rate equations for each species are:
d[A]
= -k1[A]
dt
d[I]
= +k1[A] - k 2 [I]
dt
d[P]
= +k 2 [I]
dt
Reactant Concentration, [A]
This integrates fairly easily
d[A]
= -k1[A]
dt
[A] = [A]o e k1t
85
k2
1
A k I P
d[A]
= -k1[A]
dt
d[I]
= +k1[A] - k 2 [ I ]
dt
d[P]
= +k 2 [I]
dt
Intermediate Concentration, [I]
Plug in concentration, [A]
d[I]
= +k1[A] - k 2 [ I ]
dt
d[I]
= +k1[A]o e k1t - k 2 [ I ]
dt
This is an inhomogeneous first order differential equation, which can be
solved using standard (but advanced) techniques to yield:
[I] =
k1
[A]o e-k1t - e-k 2t
k 2 - k1
86
43
k2
1
A k I P
d[A]
= -k1[A]
dt
d[I]
= +k1[A] - k 2 [ I ]
dt
d[P]
= +k 2 [I]
dt
Product Concentration, [P]
d[P]
= +k 2 [I]
dt
One can plug in [I] and solve this differential equation.
However, it's easier to just use: [P] = [A]o - [A] - [I]
This yields:
k e-k 2t - k 2e-k1t
[P] = [A]o 1+ 1
k 2 - k1
Yecch!!!! However, you do NOT have to memorize these results.
87
Concentrations at t = 0 and t
[A] = [A]o e k1t
[A](t=0) = [A]o
[A](t) = 0
[I] =
k1
[A]o e-k1t - e-k 2t
k 2 - k1
k e-k2t - k 2e-k1t
[P] = [A]o 1+ 1
k 2 - k1
[I](t=0) = 0
[I](t) = 0
[P](t=0) = 0
[P](t) = [A]o
Note that the concentrations of all three species at the start and end
th th
th
th
of the reaction are the values that one expects physically.
[A] decreases monotonically towards 0
[I] first increases and then decreases back towards 0
[P] increases monotonically towards [A]o
88
44
Limiting Case: k1 >> k2
k2
1
A k I P
One expects that [A] will drop very rapidly towards 0 and [I] should rise
quickly up to almost [A]o
One then has a simple first order reaction:
[I] will drop exponentially towards 0
[P] will rise exponentially towards [A]o
89
Limiting Case: k2 >> k1
k2
1
A k I P
One expects that [I] will rise only very slightly from 0 because it
will be used up almost immediately by the second, very fast reaction.
One expects that:
[A] will drop exponentially from [A]o
[I] will rise slightly, but remain constant
[P] will rise exponentially towards [A]o
We shall consider this limiting case in
more detail soon. It represents a very
good introduction to the
Steady State Approximation.
90
45
Mechanisms and Rate Laws
We have already seen that if a reaction involves only a single
elementary step, then the rate law may be written directly from the
reaction stoichiometry.
Sometimes, a reaction occurs in a single step. In this case,
reaction occurs in single step In this case
the rate law can be written immediately by inspection of the
reaction stoichiometry.
H2 + I2 2HI
R = k[H2][I2]
However, more commonly, a reaction occurs in a series of
elementary steps, in which case the rate law may differ
significantly from the reaction stoichiometry
91
Slow Rate Determining Step (RDS)
I2H2O2(aq) 2H2O(l) + O2(g)
R k[H2O2]2
k[H
Observed Rate Law: R = k[H2O2][I-]
k
(1) H2O2 + I- H2O + IO-
Slow RDS
(2) IO- + H2O2 H2O + O2 + I-
Mechanism:
Fast
R = d[O2]/dt d[IO-]/dt = k[H2O2][I-]
92
46
Pre-Equilibrium: Hydrolysis of Sucrose
[H+]
Sucrose + H2O Glucose + Fructose
Catalyzed by [H+]
Observed Rate Law: R = k[Suc][H+][H2O]
K
Mechanism: (1) Suc + H+
Fast Pre-Equilibrium
SucH+
k1
(2) SucH+ + H2O Glu + Fru + H+
Slow RDS
R = d[Glu]/dt = k1[SucH+][H2O] = k1K[Suc][H+][H2O]
K
[SucH ]
[Suc ][H ]
[SucH+] = K[Suc][H+]
93
For the reaction, Hg22+(aq) + Tl3+(aq) 2 Hg2+(aq) + Tl+(aq)
the observed rate law is: r
2
[ Hg 2 ][Tl 3 ]
d [Tl ]
k'
dt
[ Hg 2 ]
Show that the mechanism below is consistent with the observed
that the mechanism below is consistent with the observed
rate law.
K
Mechanism:
2
2
Hg Hg Hg 2
Fast Pre-Equilibrium
k2
Hg Tl 3 Hg 2 Tl
r
Slow RDS
2
2
[ Hg 2 ][Tl 3 ]
[ Hg 2 ][Tl 3 ]
d [Tl ]
k2 K
k'
dt
[ Hg 2 ]
[ Hg 2 ]
94
47
The Steady-State Approximation
In a multi-step reaction, it will often occur that the intermediate is very
unstable, and decays rapidly to product.
In these cases, it is valid to assume that, after an initial induction period,
the concentration of the intermediate will remain approximately constant
and very low.
One may then solve for the rate law by assuming that the rate of change
of the intermediate is approximately zero.
A classic case where the steady-state approximation is valid is for
consecutive first order reactions when the rate constant for the step
removing the intermediate, I, is much greater than the rate constant for
creating the intermediate:
Limiting Case: k2 >> k1
k2
1
A k I P
95
Limiting Case: k2 >> k1
k2
1
A k I P
We will use the steady-state approximation on [I] to determine the
concentrations, [I] and [P] as a function of time.
We will then compare the result with the exact solution, which was
presented earlier.
96
48
k2
1
A k I P
d[A]
= -k1[A]
dt
[A] = [A]o e k1t
Apply the Steady-State Approximation: d[I]/dt 0
pp
d[I]
= +k1[A] - k 2 [ I ] 0
dt
[I] =
k1
k
[A] = 1 [A o ]e-k1t
k2
k2
k
d[P]
= +k 2 [I] = +k 2 1 [A]o e-k1t = k1[A]o e-k1t
dt
k2
This integrates to (in class):
[P] = [A]o 1- e-k1t
97
k2
1
A k I P
Exact Solution
Approximate Solution
[A] = [A]o e k1t
[A] = [A]o e k1t
[I] =
k1
[A]o e-k1t - e-k 2t
k 2 - k1
k e-k2t - k 2e-k1t
[P] = [A]o 1+ 1
k 2 - k1
[I] =
k1
k
[A] = 1 [A o ]e-k1t
k2
k2
[P] = [A]o 1- e-k1t
As we show in class, the exact solutions for the concentrations
of [I] and [P] reduce to the approximate solutions using the
steady-state approach in the limit that k2 >> k1
98
49
The steady-state approximation is a less restrictive mechanism than
assuming a rapid pre-equilibrium. Let's apply this method to a practical
example
Example:
2NO(g) + O2(g) 2NO2(g)
R = k[NO]2[O2]
k[NO]
at low [O2]
[O
R = k[NO]2
Observed Rate Law:
at high[O2]
k1
2NO N2O2
k-1
N2O2 2NO
k2
N2O2 + O2 2NO2
Mechanism:
Create Intermediate
Reverse of first step
Slow RDS
RDS
k1
N 2 O2
k-1
k2
N2O2 + O2 2NO2
Shorthand:
2NO
2NO
99
k1
N 2 O2
k-1
k2
N2O2 + O2 2NO2
R = k2[N2O2][O2] =
k1k 2 [NO]2 [O2 ]
k -1 + k 2 [O2 ]
Steady-State Approximation on N2O2
[N2O2]/t = 0 = + k1[NO]2 - k-1[N2O2] - k2[N2O2][O2]
[N2O2]{k-1 + k2[O2]} = k1[NO]2
[N 2O 2 ]
k 1[NO] 2
k 1 k 2 [O 2 ]
100
50
R=
k1k 2 [NO]2 [O2 ]
k -1 + k 2 [O2 ]
Limiting Cases
Low [O2]
R=
k2[O2] << k-1
k1k 2 [NO]2 [O2 ]
= (k1k2/k-1)[NO]2[O2]
k -1 + k 2 [O2 ]
= k[NO]2[O2]
Note: If we had employed the approximation of a rapid-preequilibrium
to this reaction, we would have obtained the above rate equation.
High [O2]
R=
k2[O2] >> k-1
k1k 2 [NO]2 [O2 ]
= k1[NO]2
k -1 + k 2 [O2 ]
= k[NO]2
101
Unimolecular Gas Phase Reactions
A(g) Products(g)
Examples:
N2O5(g) NO2(g) + NO3(g)
Cl
Cl
Cl
C
H
H
C
C
C
H
H
Decompositon
Isomerization
Cl
Mechanism:
R = k[A]2
at low [A]
R = k[A]
Observed Rate Law:
at high [A]
A+A
k1
A* + A
k-1
A* is an activated
molecule
k2
A* Products
102
51
A+A
k1
A* + A
k-1
k2
A* Products
If one applies the Steady-State approximation to the concentration, [A*],
it can be shown (in class) that:
R = [Products]/t =
k1k 2 [A]2
k 2 + k -1[A]
Limiting Cases
low [A]: k-1[A] << k2
high [A]: k-1[A] >> k2
2
R=
k1k 2 [A]
k 2 + k -1[A]
R = k1[A]2 = k[A]2
R=
k1k 2 [A]2
k 2 + k -1[A]
R = (k1k2/k-1)[A] = k[A]
103
Photochemistry
Many important reactions are initiated photochemically; i.e. via
the absorption of a photon of light.
Unimolecular Reactions: A + h A* Products
Bimolecular Reactions: A + h A* + B Products
Two advantages of photochemical reactions are that:
1. The reaction may not occur thermally
2. The photochemically induced reaction may be more selective
than the thermal reaction of the same substrate(s)
A number of primary photochemical deexcitation processes compete
with the formation of products by the excited state.
Therefore, it is important to consider the time scales of the various
excitation and decay processes of excited state molecules.
104
52
Excited Electronic States
So
S2
S1
T1
T2
Excited State
Triplet
Excited State Excited State
Singlet
Singlet
Ground State
Singlet
Excited State
Triplet
105
Decay Processes
Non-Radiative Decay
S2
ISC
IC
S1
T2
ISC
IC
Absorption
T1
F
IC
ISC
P
IC
Internal Conversion
S S or T T
ISC Inter-System Crossing
S T or T S
Radiative Decay
F
Fluorescence
S1 S0 Emission
P
Phosphorescence
T1 S0 Emission
S0
106
53
Relation Between Absorption and Fluorescence
Emin
Abs.
min
Emax
Fluor.
max
The fluorescence spectrum occurs at lower frequency than
the UV (or visible) absorption spectrum.
They are (approximately) mirror images of each other.
107
Fluorescence and Phosphorescence Lifetimes
ISC
1
Laser Pulse
T1
S0
F
IC
ISC
P
IFluor
S1
1/e = 0.37
F time
F = Fluorescence Lifetime
F = 1 - 100 nanoseconds (ns)
P = Phosporescence Lifetime
P = 1 ms - days
ms days
P >> F because Triplet-Singlet transitions are spin forbidden.
Because phosphorescence lifetimes are so extremely long, one
rarely observes phosphorescence in aqueous solutions;
the Triplet state is depleted by collisional processes.
108
54
Transient Singlet State Kinetics
S1
Once molecules have been excited from
S0 to S1 by a transient laser pulse, they
will have three modes of decays:
T1
Laser Pulse
kF
S1 S0
1. Fluorescence:
ISC
2. Intersystem Crossing: S1 T1
k
3. Internal Conversion:
ISC
k IC
S1 S0
The overall rate of change of [S1] is
given by:
F
IC
S0
d [ S1 ]
k F [ S1 ] k ISC [ S1 ] k IC [ S1 ] k0 [ S1 ]
dt
where k0 k F k ISC k IC
1
0
109
d [ S1 ]
1
k F [ S1 ] k ISC [ S1 ] k IC [ S1 ] k0 [ S1 ] where k0 k F k ISC k IC
0
dt
This is straightforward to integrate to get:
or [ S1 ] [ S1 ]0 e
k0 t
Thus [S1] decays exponentially from
its initial value with a decay time, 0.
One can monitor [S1] as a function
of time from the fluorescence
intensity, IF.
[ S1 ]0 e
k0 t
[ S1 ]0 e
t
0
[S1]0
[S1]
[S ]
ln 1 k0t
[ S1 ]0
1/e = 0.37
0.37
0 t
110
55
Singlet State Lifetimes From Pulsed Laser Fluorescence
2. Monitor the fluorescence intensity
as a function of time.
1
IF
1. Excite molecules from S0 to S1 with
a short (<1 ns) laser pulse.
1/e = 0.37
IF = IF o e
-
t
0
-
0 t
t
0
ln IF = ln IF o + ln(e )
1
t
0
Slope = -1/0
ln(IF)
ln IF = ln IF o -
x
y
m = - 1/0
t
111
The Quantum Yield ()
The quantum yield for a process is a measure of the efficiency of absorbed
photons in inducing the process to occur.
The quantum yield can be defined either in terms of
(a) the rate of the process relative to the rate of photon absorption, or
(b) the number of moles undergoing the process relative to the
number of moles of photons absorbed.
Rate:
proc
Rate of Pr ocess
Rate of Pr ocess
Rate of photon absorption
I abs
Amount:*
proc
Moles of MoleculesUndergoing Pr ocess
Moles of Photons Absorbed
*Notes: (1) One has an equivalent definition using molecules instead
of moles.
(2) 1 mole of photons is often referred to as 1 einstein.
112
56
Rate:
proc
Rate of Pr ocess
Rate of Pr ocess
Rate of photon absorption
I abs
Amount:*
proc
Moles of MoleculesUndergoing Pr ocess
Moles of Photons Absorbed
Primary Processes
These are processes (such as fluorescence, phosphorescence,
intersystem crossing, etc.) in which 1 absorbed photon can induce
only 1 molecule to undergo the process.
For primary processes, 0 process 1
Secondary Processes
Processes
These are processes in which 1 absorbed photon can indirectly induce
the process to occur multiple times. Typically, these are reactions.
For secondary processes, 0 process
Chain reactions often have R > 1
113
A Chain Reaction Quantum Yield:
Chlorination of Methane
Reaction:
Cl2 + CH4 CH3Cl + HCl
Mechanism: (1) Cl2 + h 2 Cl
(2) Cl + CH4 HCl+ CH3
Chain Initiation
Chain Propagation
(3) CH3 + Cl2 CH3Cl + Cl
Chain Propagation
(4) Cl + Cl + M Cl2 + M
Chain Termination*
*M is an inert body to absorb excess translational energy.
R
Moles of CH 3Cl Pr oduced
1, 000 10, 000
Moles of Photons Absorbed
114
57
Reaction Quantum Yield Examples
Example 1 (like Exer. 21.21a)
In a photochemical reaction, A B + 3C, the quantum yield with
500 nm radiation is 130 mol/einstein (1 einstein = 1 mol of photons).
After exposure of a sample of A to the light for a period of time,
0.36 mol of C was formed.
How many photons of light were absorbed during the time period.
NA = 6.02x1023 mol-1
Nph = 4.5x1020
115
Reaction Quantum Yield Examples
Example 2 (Exer. 21.22b)
In an experiment to measure the quantum yield of a photochemical
reaction, the absorbing substance was exposed to 320 nm radiation from a
87.5 W source for 28.0 min. The intensity of the transmitted light was
25.7% that of the incident light.
th
th
li
As a result of irrdation, 0.324 mol of the absorbing substance decomposed.
Determine the reaction quantum yield, R.
Additional Information:
E ph h
hc
(6.63x1034 J s)(3.00 x108 m / s) 1.99 x1025 J / m
( m)
( m)
( m)
Note: This relation will be furnished on a test if needed.
R = 1.11
116
58
Steady-State Singlet Kinetics:
The Fluorescence Quantum Yield
We will perform the kinetic analysis to
solve for the steady-state concentration
and use the result to determine the
fluorescence quantum yield, F
Processes
1. Photon Absorption:
2. Fluorescence:
Fluorescence:
Absorption, Iabs
In an experiment where there is continuous S1
irradiation of the sample, the excited Singlet
will reach a steady-state concentration, [S1].
ISC
T1
F
IC
I abs
S0 S1
F
S1 k S0
S0
3. Intersystem Crossing: S1 kISC T1
k IC
4. Internal Conversion: S1 S0
Note: The rate of the absorption step is:
1. Photon Absorption:
117
I abs
S0 S1
2. Fluorescence:
d [ S1 ]
I abs
dt
kF
S1 S0
3. Intersystem Crossing: S1 kISC T1
k IC
4. Internal Conversion: S1 S0
Steady-State concentration, [S1]
d [ S1 ]
0 I abs k F [ S1 ] k ISC [ S1 ] k IC [ S1 ] I abs k0 [ S1 ]
dt
1
where (as before): k0 k F k ISC k IC
0
Solving for [S1] yields [ S1 ]
I abs
I abs
k0
k F k ISC k IC
118
59
The Fluorescence Quantum Yield
[ S1 ]
I abs
I
abs
k F k ISC k IC
k0
The fluorescence quantum yield is given by:
F
Fluorescence Rate
k [S ]
F1
Photon Absorption Rate
I abs
Substituting the expression for [S1] then yields:
F
I abs
k F [ S1 ] k F
I abs
I abs kF k ISC kIC
kF
k
F k F 0
k F k ISC k IC
k0
This result is intuitively reasonable, and shows that the quantum yield
is the ratio of the fluorescence rate constant divided by the sum of
the ratio of the fluorescence rate constant divided by the sum of
rate constants for all S1 depletion processes.
119
F
I abs
k F [ S1 ] k F
kF
k
F k F 0
I abs
I abs k F k ISC k IC kF k ISC k IC
k0
Experimental determination of kF
We saw recently that the singlet state lifetime, 0, can be obtained from
a pulsed laser fluorescence decay experiment.
The lifetime so determined can be combined with the experimental
fluorescence quantum yield to determine the molecule's fluorescence
rate constant, kF.
In a pulsed laser fluorescence experiment on liquid benzene the
fluorescence intensity 150 ns after the experiment begins is 25% the
intensity at the start of the experiment.
In a separate steady-state fluorescence experiment, it was determined
separate steady
fluorescence experiment it was determined
that the fluorescence quantum yield in liquid benzene is 0.18
Determine (a) the singlet state lifetime, 0 (in ns), and (b) the fluorescence
rate constant, kF (in s-1) in liquid benzene.
(a) 0 = 110 ns
(b) kF = 1.7x106 s-1
120
60
Excited State Quenching
S1
Earlier, we discussed that once a molecule
has been excited to S1, there are three
mechanisms for deexcitation of the excited state.
1. Photon Absorption:
2. Fluorescence:
ISC
T1
I abs
S0 S1
F
S1 k S0
Iabs F IC
Q
3. Intersystem Crossing: S1 T1
k ISC
4. Internal Conversion:
k IC
S1 S0
S0
An additional deexcitation mechanism is the addition of
a solute, called a quencher (Q), which can
induce depopulation of the excited state.
Q
S1 Q S0 Q
k
121
An additional mechanism is the addition of
a solute, called a quencher (Q), which can
induce depopulation of the excited state.
S1
Q
S1 Q S0 Q
ISC
T1
k
The mechanisms of excited-state quenching
include:
1. Collisional deactivation
2. Resonance Energy Transfer
S0
3. Reaction
4. Spin-orbit coupling (triplet-state quenching)
Iabs F IC
Q
The net effects of adding a quencher are:
A. The excited state lifetime is reduced (to below 0)
B. The fluorescence (or phosphorescence) quantum yield is reduced.
Below, we will develop a relation between the fluorescence yield with
no quenching, F,0 , and the quantum yield with quencher added, F
122
61
S1
ISC
S0 S1
1. Photon Absorption:
I abs
T1
S1 S0
kF
2. Fluorescence:
ISC
3. Intersystem Crossing: S1 T1
k
Q
S1 S0
4. Internal Conversion:
Internal Conversion:
5. Quenching:
Iabs F IC
k IC
Q
S1 Q S0 Q
k
S0
We will use a procedure analogous to the one used earlier to obtain
an expression for the fluorescence quantum yield in the presence of
a quencher.
Steady-State concentration, [S1]
d [ S1 ]
0 I abs k F [ S1 ] k ISC [ S1 ] k IC [ S1 ] kQ [ S1 ][Q]
dt
123
Steady-State concentration, [S1]
d [ S1 ]
0 I abs k F [ S1 ] k ISC [ S1 ] k IC [ S1 ] kQ [ S1 ][Q]
dt
0 I abs [ S1 ]k F k ISC k IC kQ [Q] I abs [ S1 ]k0 kQ [Q]
where k0 k F k ISC k IC
Therefore: [ S1 ]
1
0
I abs
k0 kQ [Q]
The fluorescence quantum yield in the presence of Q is
given by:
F
I abs
Fluorescence Rate
k [S ] k
kF
F 1 F
k k [Q ] k k [Q ]
Photon Absorption Rate
I abs
I abs 0 Q
Q
0
124
62
The fluorescence quantum yield in the presence of Q is
given by:
F
I abs
Fluorescence Rate
k [S ] k
kF
F 1 F
Photon Absorption Rate
I abs
I abs k0 kQ [Q ] k0 kQ [Q ]
With no quencher (i.e. [Q] = 0), the quantum yield is: F ,0 k F
no quencher
[Q]
the quantum yield is:
k0
Thus, we see that the presence of a molecule which can quench
S1 reduces the fluorescence quantum yield.
The Stern-Volmer Equation
Let's calculate the ratio, F,0/F:
F ,0
F
kF
k0 kQ [Q]
kQ [Q]
k0
1 0 k Q [Q ]
1
kF
k0
k0
k0 kQ [Q ]
125
F ,0
F
1 0 kQ [Q]
Stern-Volmer Equation
We see from this equation that if we measure
F as a function of [Q], one gets a straight
],
line with: Slope = 0kQ
If the singlet state lifetime, 0 , has been measured
in a pulsed laser fluorescence experiment, then
the Stern-Volmer plot may be used to determine the quenching
rate constant, kQ.
Laser fluorescence experiments require relatively expensive equipment,
and are not available to all researchers.
A common application is to estimate kQ and then use the
plot to obtain a value for the singlet state lifetime,0.
One of the methods to estimate kQ is to use the theory of
Diffusion Controlled Reactions (Sect. 22.2 of the text)
126
63
Example: Singlet State Lifetime from Fluorescence Quenching
The fluorescence quantum yield for 2-aminopurine in water is
0.32. When a quencher is added to the solution, with [Q] = 0.02 M,
the quantum yield is reduced to 0.14
The quenching rate constant is: kQ = 2.5x109 M-1s-1
quenching rate constant is:
Calculate the singlet state lifetime of 2-aminopurine (in ns).
0 = 25 ns
127
Bimolecular Reactions from S1
We commented that one mechanism by which a second molecule
can quench the fluorescence of a molecule in the excited singlet state
is to react with it:
kR
A( S1 ) B Pr od
or
kR
S1 B Pr od
The fluorescence quantum yield from S1 will be reduced by the
presence of the second reactant, B. It is straightforward to show
that the Stern-Volmer Equation becomes:
F ,0
F
1 0 kR [ B]
Thus, if one has the measured fluorescence lifetime (in the absence
if one has the measured fluorescence lifetime (in the absence
of the second reactant), 0 , and measures the fluorescence
quantum yield as a function of [B], it is straightforward to use
the above equation to determine the bimolecular rate constant, kR,for
the reaction.
128
64
The Triplet State (T1)
S1
Triplet Lifetime, T1
They can then return to S0 via two processes.
can then return to
two processes
T1 S0
kP
Absorption
T1
After molecules are excited by light from S0
to S1, some of the molecules will
intersystem cross to T1.
1. Phosphorescence:
ISC
ISC
P
k ISC
2. Intersystem Crossing: T1 S0
The kinetic rate equation is:
S0
d [T1 ]
1
k P [T1 ] k ISC [T1 ] kT1 [T1 ] [T1 ]
dt
T1
This integrates to: [T1 ] [T1 ]0 e
kT1t
[T1 ]0 e
t
T1
where T1 is the triplet state lifetime, given by: T1
1
1
kT1 k P k ISC
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where T1 is the triplet state lifetime, given by: T1
1
1
kT1 k P k ISC
Because the T1 S0 transition is spin-forbidden, triplet state lifetimes, T1,
are generally many orders of magnitude longer than singlet lifetimes, 0
Typical Singlet State Lifetimes: 0 1 - 100 ns
Singlet State Lifetimes:
100 ns
Typical Triplet State Lifetimes: T1 1 ms - days
Phosphorescence in Liquid Solution
This is a short section. Phosphorescence is virtually never observed in
liquid phase solutions.
This is because triplet state lifetimes are so long that deexcitation
by collisions with solvent molecules will depopulate T1 very efficiently,
completely quenching the phosphorescence.
However, one does observe phosphorescence from molecules in the
gas phase and in frozen glasses at 77 K.
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