Kinetics
Chemical Chemical
Chemical Kinetics: Ch. 14
Bradley J. Siwick
Siwick
Departments of Physics and Chemistry
50
T = -1 ps
T=-1ps
T = +6 ps (x4)
T = +50 ps (x4)
B
40
4.05
30
H(r)
20
Al-O
10
14-1 The Rate of a Chemical Reaction
14-2 Measuring Reaction Rates
14-3 Effect of Concentration on
Reaction Rates: The Rate Law
14-4 Zero-Order Reactions
Reactions
14-5 First-Order Reactions
14-6 Second-Order Reactions
14-7 Reaction Kinetics: A Summary
0
-10
-20
-30
0
2
4
6
8
10
Chemical Kinetics
12
What is Chemical Kinetics?
14-8 Theoretical Models for Chemical Kinetics
14-9 The Effect of Temperature on Rates
14-10 Reaction Mechanisms
Mechanisms
14-11 Catalysis
Concerned with the speed (or rate) at
which chemical reactions take place and
hi
the mechanisms by which they occur.
A reaction mechanism is the step by step
th
sequence of chemical events by which
reactant molecules are turned into product
reactant molecules are turned into product
molecules.
The rate at which a reaction proceeds
rate at which reaction proceeds
depends on the mechanism, so by
measuring rates we can hope to better
measuring rates we can hope to better
understand mechanisms!
14-1 The Rate of a Chemical Reaction
Rates of Chemical Reactions
Rate of change of concentration with
respect to time.
2 Fe3+(aq) + Sn2+(aq) 2 Fe2+(aq) + Sn4+(aq)
Fe
Sn
Fe
Sn
t = 38.5 s
t = 38.5 s
[Fe2+] = 0.0010 M
[Fe2+] = (0.0010 0) M
Average Rate of Fe2+ [Fe2+]
0.0010 M
=
= 2.610-5 M s-1
=
Formation
t
38.5 s
2 Fe3+(aq) + Sn2+(aq) 2 Fe2+(aq) + Sn4+(aq)
1 [Fe3+]
[Sn4+]
1 [Fe2+]
==
t
t
2 t
2
Rate of disappearance is a negative quantity
di
Rate of formation is a positive quantity
What number should we take for the rate??
number should we take for the rate??
General Rate of Reaction
14-2 Measuring Reaction Rates
aA+bBcC+dD
Rate of reaction = -rate of disappearance/consumption of reactants
=-
1 [B]
1 [A]
=b t
a t
H2O2(aq) H2O(l) + O2(g)
Or, by titration:
= rate of appearance/production of products
2 MnO4-(aq) + 5 H2O2(aq) + 6 H+
2 Mn2+ + 8 H2O(l) + 5 O2(g)
Mn
1 [D]
1 [C]
=
=
d t
c t
A very simple device for measuring the
very simple device for measuring the
rate of a reaction that produces gas
14-2 Measuring Reaction Rates
Determining and Using an Initial Rate of Reaction.
H2O2(aq) H2O(l) + O2(g)
H2O2(aq) H2O(l) + O2(g)
-(-2.32 M / 1360 s) = 1.7 10-3 M s-1
Rate =
-[H2O2]
t
-(-1.7 M / 2800 s) = 6 10-4 M s-1
2800 s)
2 MnO4-(aq) + 5 H2O2(aq) + 6 H+
2 Mn2+ + 8 H2O(l) + 5 O2(g)
Mn
A very simple device for measuring the
very simple device for measuring the
rate of a reaction that produces gas
Determining and Using an Initial Rate of Reaction.
H2O2(aq) H2O(l) + O2(g)
Rate =
-(-1.7 M / 2800 s) = 6 10-4 M s-1
2800 s)
-[H2O2]
t
Reaction progress
-(-2.32 M / 1360 s) = 1.7 10-3 M s-1
Ultrafast
Spectroscopy:
Spectroscopy:
Watching
Chemistry as it
Happens
Happens
Transient states
14-3 Effect of Concentration on
Reaction Rates: The Rate Law
a A + b B . g G + h H .
Rate of reaction = k[A]m[B]n .
of reaction
Method of Initial Rates
A method that can be used to determine the rate
law for reaction
law for a reaction
Requires that the initial rate be determined: At
this stage the concentration of the reactants is
known (i.e. under our control)
Rate constant = k
Overall order of reaction = m + n + .
In general, m a and n b!
general
and
b!
With the rate law determined, we can calculate the
rate for known concentrations and derive an
equation for concentration as a function of time!
Method of Initial Rates: Example
Reactant Concentration vs.
Time: Three examples
Zero order reactions
First order reactions
order reactions
Second order reactions
The equation that describes the time-dependence of reactant
concentration is different in each case. However, we will see
that the rate constant k for the process can be always be
that the rate constant k for the process can be always be
obtained from the slope of a particular straight-line plot.
14-4 Zero-Order Reactions
A products
Rrxn = k [A]0
Rrxn = k
[k] = mol L-1 s-1= M s-1
k
[ A]0
tf
Integrated Rate Law
The rate law can be used with a little
littl
calculus to determine the concentration of a
reactant as a function of time
as
-[A]
t
=k
Move to the
-d[A]
infinitesimal
infinitesimal
dt
=k
And integrate from 0 to time t
[A]t
-
[A]0
t
d[A]
= k dt
0
-[A]t + [A]0 = kt
[A]t = [A]0 - kt
14-5 First-Order Reactions
First-Order Reactions
H2O2(aq) H2O(l) + O2(g)
Rate of reaction = k[H2O2]
d[H2O2 ]
= -k[H2O2]
dt
[A]t
[A]0
ln
[k] = s-1
t
d[H2O2 ]
= - k dt
[H2O2]
0
[A]t
= -kt
[A]0
ln[A]t = -kt + ln[A]0
ln[A]
Half-Life
Half-Life
t is the time taken for one-half of a
reactant to be consumed
reactant to be consumed.
ln
ln
[A]t
[A]0
= -kt
[A]0
[A]0
ButOOBut(g) 2 CH3CO(g) + C2H4(g)
= -kt
- ln 2 = -kt
t =
ln 2 0.693
=
k
k
Half-Life
ButOOBut(g)
2 CH3CO(g) + C2H4(g)
14-6 Second-Order Reactions
Rate law where sum of exponents m +
n + = 2.
A products
Rate of Reaction = k[A]2
of Reaction k[A]
d[A]
dt
= -k[A]2
[A]t
[A]0
d[A]
[A]2
[k] = M-1 s-1 = L mol-1 s-1
mol
t
= - k dt
0
1
1
= kt +
[A]t
[A]0
Second-Order Reaction
Consider kinetic data from 3 reactions
(blue, green red). What order is each
(blue, green red). What order is each
reaction?
91%
1. second, first, zero
first zero
2. first, second, zero
3. zero, first, second
first second
3%
1
6%
2
3
Pseudo First-Order Reactions
Pseudo First-Order Reactions
A second order reaction can sometimes be
made to behave like first order reaction by
made to behave like a first order reaction by
holding one reactant concentration constant:
pseudo first order reaction.
A second order reaction can sometimes be
made to behave like first order reaction by
made to behave like a first order reaction by
holding one reactant concentration constant:
pseudo first order reaction.
Example
CH3CO2C2H5 + H2O CH3CO2H + C2H5OH
CH
Example
CH3CO2C2H5 + H2O CH3CO2H + C2H5OH
CH
If the concentration of water does not
change appreciably during the reaction
change appreciably during the reaction.
Rate law appears to be first order. The rxn.
Behaves as if it were zero order in [H2O]
Simplify the kinetics of complex reactions
Rate laws become easier to work with.
14-7 Reaction Kinetics: A Summary
Calculate the rate of a reaction from a known
rate law using:
rate law using:
Rate of reaction = k [A]m[B]n .
Determine the instantaneous rate of the
the instantaneous rate of the
reaction by:
Finding the slope of the tangent line of [A] vs t or,
the slope of the tangent line of [A] vs
Evaluate [A]/t, with a short t interval.
If the concentration of water does not
change appreciably during the reaction.
Rate law appears to be first order.
Simplify the kinetics of complex reactions
Rate laws become easier to work with.
Summary of Kinetics
Determine the order of reaction by:
Using the method of initial rates.
Find the graph that yields a straight line.
Test for the half-life to find first order
reactions.
Substitute data into integrated rate laws to
find the rate law that gives a consistent
value of
value of k.
14-8 Theoretical Models for
Chemical Kinetics
Summary of Kinetics
Find the rate constant k by:
Determining the slope of a straight line graph.
Evaluating k the with integrated rate law.
with the integrated rate law.
Measuring the half life of first-order reactions.
Find reactant concentrations or times for
certain conditions using the integrated rate
law after determining k.
Activation Energy: Ea
Collision Theory
Theory
Kinetic-Molecular theory can be used to calculate
the collision frequency
the collision frequency in a gas.
gas
In gases at STP ~1030 collisions occur per second (in 1L).
If each collision produced a reaction, the rate would be
each collision produced reaction the rate would be
approximately 106 M s-1 (i.e. ~1030/NA)!
Actual rates in gas reactions are generally much slower
a typical value being 10-4 M s-1.
Only a fraction of all
collisions yield a
reaction!
Activation Energy
Not all collisions result in a chemical
reaction
reaction
For a reaction to occur there must be a
redistribution of energy sufficient to
redistribution of energy sufficient to
break certain bonds in the reacting
molecule(s)
molecule(s).
Activation Energy:
Ea
The minimum energy above the average
kinetic energy that molecules must bring to
th
their collisions for a chemical reaction to
occur.
Kinetic Energy Distribution
Collision Theory
If activation barrier is high, only a few
molecules will have sufficient kinetic energy
gy
to react.
Orientation of molecules is also important:
Shaded area =
-
e Ea
RT
Transition State
The activated complex
is a hypothetical (or
unstable) species lying
between reactants and
products at a point on
the reaction profile
called the transition
the transition
state.
14-9 Effect of Temperature on Reaction Rates
In 1889 Svante Arrhenius
determined experimentally
that many rate constants vary
with temperature according to
with temperature according to
the equation:
Collision Theory
A Prediction of Collision Theory:
Th
The form of the rate constant
k = Ae-Ea/RT
ln k =
-Ea 1
R
T
+ lnA
ln
14-9 Effect of Temperature on Reaction Rates
In 1889 Svante Arrhenius
determined experimentally
that many rate constants vary
with temperature according to
with temperature according to
the equation:
Collision Theory
Experimenal rate constants compared to the ones predicted by
collision theory for gas phase reactions
Reaction
A (Arrhenius
frequency factor)
2ClNO 2Cl +
2NO
9.4 10
2ClO Cl2 + O2
Cl
6.3 10
10
H2 + C2H4 C2H6
1.24 10
Br2 + K KBr + Br
KBr Br
k=
ln k =
Ae-Ea/RT
-Ea 1
R
T
+ lnA
ln
Z (collision
frequency)
10
5.9 10
7
12
9
2.5 10
10
6
Steric
factor
10
0.16
10
2.3 10
10
11
1.7 10
7.3 10
11
2.1 10
10
-3
4.3
-6
How do you determine Ea?:
From the slope of an Arrhenius Plot
th
Pl
ln k =
-Ea 1
R
T
ln k =
k = Ae-Ea/RT
+ lnA
ln
N2O5(CCl4) N2O4(CCl4) + O2(g)
(CCl
(CCl
Arrhenius Equation
-Ea
= -1.2104 K
R
-Ea = 1.0102 kJ mol-1
ln k2 ln k1 =
ln
Transition State
-Ea 1
R
k2
k1
T2
=
-Ea 1
R
+ ln A -
T
-Ea 1
R
-Ea
1
R
T2
-
+ ln A
- ln A
T1
1
T1
14-10 Reaction Mechanisms
A step-by-step description of a chemical
reaction.
reaction.
Each step is called an elementary process.
Any molecular event that significantly alters a
molecular event that significantly alters
molecules energy or geometry, or produces a
new molecule
new molecule.
Reaction mechanism must be consistent
with:
with:
Stoichiometry for the overall reaction.
The experimentally determined rate law.
experimentally determined rate law
Elementary Processes
Unimolecular or bimolecular.
Exponents for concentration terms in the
rate law for an elementary process are the
same as the stoichiometric factors.
Elementary processes are reversible.
Intermediates are produced in one
produced in one
elementary process and consumed in
another
another.
One elementary step if much slower than
all the others
all the others is known as the rate
known as the rate
determining step.
Reaction Mechanisms: Examples
Mechanisms: Examples
Example 1: 2 step reaction that has a rate
determining step
Example 2: 2 step reaction that involves a fast
pre-equilibrium
Example 3: Application of the steady-state
approximation
approximation to a 2 step reaction.
step reaction
These are common rxn mechanisms that by
are common rxn mechanisms that
their very natures provide a simplification to
the mathematics.
Slow Step Followed by a Fast Step
d[P]
H2(g) + 2 ICl(g) I2(g) + 2 HCl(g)
Slow Step Followed by a Fast Step
= k[H2][ICl]
dt
Postulate a mechanism:
H2(g) + ICl(g)
HI(g) + ICl(g)
slow
d[HI]
HI(g) + HCl(g)
fast
= k[H2][ICl]
dt
d[I2]
I2(g) + HCl(g)
= k[HI][ICl]
dt
d[P]
H2(g) + 2 ICl(g) I2(g) + 2 HCl(g)
= k[H2][ICl]
dt
Fast Reversible Step Followed by a Slow Step
d[P]
2NO(g) + O2(g) 2 NO2(g)
dt
= -kobs[NO]2[O2]
Postulate a mechanism:
fast
2NO(g)
k1
k1[NO]2 = k-1[N2O2]
N2O2(g)
k-1
[N2O2] =
slow
N2O2(g) + O2(g)
k2
2NO2(g)
2NO(g) + O2(g) 2 NO2(g)
k1
d[NO2]
dt
[NO]2 = K[NO]2
k-1
= k2[N2O2][O2]
d[NO2]
k
= k2 1 [NO]2[O2]
dt
k-1
The Steady State Approximation
2NO(g)
k1
2NO(g)
k-1
N2O2(g) + O2(g)
N2O2(g)
N2O2(g)
k3
2NO2(g)
d[NO2]
dt
d[N2O2]
dt
2NO(g)
N2O2(g)
k1
k2
N2O2(g) + O2(g)
d[NO2]
2NO(g)
k3
dt
=
k1k3[NO]2[O2]
N2O2(g)
(k2 + k3[O2])
If k2 << k3[O2])
d[NO2]
dt
k1
k2
N2O2(g) + O2(g)
2NO2(g)
= k1[NO]2 k2[N2O2] k3[N2O2][O2] = 0
k1
[N
[N2O2] =
(k2 + k3[O2])
2NO(g)
N2O2(g)
= k3[N2O2][O2]
[NO]2
Kinetic Consequences of Assumptions
=
k1k3[NO]2[O2]
( k3[O2])
N2O2(g)
2NO(g)
2NO(
k3
2NO2(g)
= k1[NO]2
Or
If k2 >> k3[O2])
d[NO2]
dt
=
k1k3[NO]2[O2]
( k2)
=
k1k3
k2
[NO]2[O2]
Evaluating a rxn mechanism:
Summary
Is the proposed mechanism consistent with the
the proposed mechanism consistent with the
overall stoichiometry of the rxn? i.e. do the
steps add up to the overall rxn?
Does the mechanism consist of only elementary
processes?
Does the proposed mechanism account for the
experimentally determined rate law?
In general one can rule out a proposed mechanisms by
appropriate experiment but it is very difficult to prove
appropriate experiment, but it is very difficult to prove
unequivocally that a rxn mechanism is correct!
Problem Set Ch.14
Set Ch
(from 10th ed)
10
10,13,19,25,27,32,45,49,53,65,69,78,87,92,
94,97
The same set of problems can be found in
th
the 9th edition with the following question
th
numbers:
(from 9th ed)
10,13,19,25,27,32,45,49,53,65,69,76,86,91
,94, 96
14-11 Catalysis
14-5 Catalysis
A catalyst provides an alternative
reaction pathway of lower
reaction pathway of lower Ea
Catalysts participate in reactions, but do
not undergo permanent change
not undergo a permanent change.
Homogeneous catalysis.
All species in the reaction are in solution.
Heterogeneous catalysis.
The catalyst is in the solid state.
Reactants from gas or solution phase are
adsorbed
adsorbed.
Active sites on the catalytic surface are important.
Catalysis on a Surface
Enzyme Catalysis
E+S
k1
k-1
ES
k2
ES E + P
Saturation Kinetics
E+S
k1
k-1
d[P]
k2
ES E + P
dt
d[P]
dt
= k2[ES]
= k1[E][S] k-1[ES] k2[ES]= 0
k1[E][S] = (k-1+ k2 )[ES]
[E] = [E]0 [ES]
[E]
k1[S]([E]0 [ES]) = (k-1+ k2 )[ES]
[ES] =
Michaelis-Menten
d[P]
d[P]
dt
= k2[E]0
k1k2[E]0 [S]
=
dt
d[P]
dt
(k-1+ k2 ) + k1[S]
k2[E]0 [S]
=
(k-1+ k2 ) + [S]
k1
d[P]
dt
=
k2
[E]0 [S]
KM
d[P]
dt
=
k2[E]0 [S]
KM + [S]
k1[E]0 [S]
(k-1+ k2 ) + k1[S]