Chemical Kinetics
CK-1
Chemical kinetics is the study of time dependence
of the change in the concentration of reactants and
products.
“The field of chemical kinetics has not yet matured
to a point where a set of unifying principles has
been identified…There are many different
theoretical models for describing how chemical
reactions occur.” (M&S, 1137)
Generally, we want to understand the rate of reaction:
d [reactant ]
Rate  
Reactant  Product
dt
Recall from Slide CEq-2…
CK-2
vA A( g )  vB B( g )  vY Y ( g )  vZ Z ( g )
Reactants
Products
nA (t )  nA (0)  vA (t )
nY (t )  nY (0)  vY  (t )
nB (t )  nB (0)  vB (t )
nZ (t )  nZ (0)  vZ  (t )
For each constituent…
1 dn A d [ A]
v A d (t )


V dt
dt
V dt
Rate of reaction, v(t)
CK-3
v(t), the rate of reaction, is defined as the rate of change in ξ(t) with
time per unit volume
1 d (t )
1 d [ A]
1 d [ B] 1 d [Y ] 1 d [ Z ]
v(t ) 




V dt
v A dt
vB dt
vY dt
vZ dt
Note all quantities are positive.
Examples:
What are units of v(t)?
2NO(g) + O2 (g)  2NO2 (g)
1 d (t )
v(t ) 

V dt
H2(g) + I2 (g)  2HI (g)
1 d (t )
v(t ) 

V dt
The integrated rate law
CK-4
The rate can be expressed as a function of the reactant concentrations.
Most common function is of form:
1 d
v(t ) 
 k[ A]m [ B]n
V dt
Rate laws must be determined experimentally and, generally,
cannot be deduced from the balanced reaction!!
General
vA A( g )  vB B( g )  vY Y ( g )  vZ Z ( g )
v(t )  k[ A] [ B]
m
n
Example
2NO(g) + O2 (g)  2NO2 (g)
v(t )  k[ NO]2 [O2 ]
Determined experimentally!
The
Order:
mth order in A
nth order in B
(m+n)th order overall
Finding rate laws experimentally
CK-5
There are two common methods for determining rate laws:
Method of isolation
Set up reaction so one reactant is in excess. Any change in rate
will be due to changes in other reactant. Repeat for other reactant.
v  k [ B]
n
where
k   k[ A]
m
Method of initial rates
Measure concentration change as a function of time, ~v(t), for a
series of experimental conditions. (Conditions must include sets
where the reactant A has the same initial concentration but B
changes and vice versa).
Units of k, rate constant
CK-6
1 d
m
n
v(t ) 
 k[ A] [ B]
V dt
(concentration)
n
concentration
time
Rate Law
(concentration)m
Order
0
Units of k
v=k
1
v =k[A]
2
v = k[A]2
v = [A][B] 1 in [A], [B]
2 overall
v = k[A]1/2
1/2
v(t )
(concentrat ion )1mn
k

m
n
[ A] [ B]
time
Rate laws can be complicated
H2(g) + I2 (g)  2HI (g)
CK-7
v(t )  k[H2 ][I2 ]
2k [H 2 ][ Br2 ]1/ 2
H2(g) + Br2 (g)  2HBr (g) v(t ) 
1  k [HBr ][ Br2 ]1
These rate laws suggest that these two reactions occur via
different mechanisms (sets of individual steps).
The first may be a elementary reaction (one step) whereas the
latter is certainly a multistep process.
We will soon explore how to obtain complicated rate laws
from suggested mechanisms.
Elementary Rxns vs. Complex Reactions
CK-8
Complex Reactions
Elementary Reactions
Reactants  Intermediates  Products
Reactants  Products
Mechanism of a complex
reaction is a sequence of
elementary reactions.
Molecularity of Elementary Reactions:
Unimolecular
A  products
Bimolecular
A  B  products
Termolecular
A  B  C  products
First order reactions
The reaction: A  B  products
CK-9
d [ A]
has rate law: v(t )  
 k[ A]
dt
Let’s integrate…
t
1
[ A]0 [ A] d[ A]  0 kdt
d [ A]
 kdt
[ A]
[ A ]t
Solution:
[ A]t
ln
  kt
[ A]0
ln[ A]t  ln[ A]0  kt
[ A]t  [ A]0 e
 kt
First order reactions decay exponentially.
Ozone decays via first order kinetics
CK-10
O3 ( g )  O2 ( g )  O( g )
k = 1.078 × 10-5 s-1 at 300 K
[O3 ]t  [O3 ]0 e  kt
[O 3 ]t
ln
  kt
[O 3 ]0
What is slope?
What happens as k increases?
[ A]t  [ A]0 e  kt
k = 0.0125 s-1
k = 0.0250 s-1
k = 0.0500 s-1
k = 0.1000 s-1
[ A]t
ln
  kt
[ A]0
CK-11
Half-life of a first order reaction
CK-12
The half-life, t1/2, is the time it
takes to fall to ½ of the starting
concentration:
At
t  t1/ 2
, [ A]t1 / 2
1
 [ A]0
2
[ A]1/ 2 1
  e  kt1 / 2
[ A]0
2
Figure 28.3
Other order reactions…
Second order reaction:
Second order rate:
2 A  products
1 d [ A]
v(t )  
 k[ A]2
2 dt
Integrated rate law:
1
1

 2kt
[ A]t [ A]0
CK-13
A  B  products
v(t )  
 [ B]0 [ A]t
1
ln 
[ A]0  [ B]0   [ A]0 [ B]t
Zero order reaction:
A  products
Zero order rate:
v(t )  
Integrated rate law:
d [ A]
 k[ A][ B]
dt
d [ A]
k
dt
[ A]t  [ A]0  kt

  kt

Pseudo-first order reactions
CK-14
You can “overload” the other reactants to determine
the order with respect to one individual reactant
(method of isolation).
For A  B  products , what happens if [B] >> [A]?
d [ A]
v(t )  
 k[ B ][ A]
dt
Similar to SN2 Lab
Reversible reactions (small DrG)
A
k1
k-1
B
CK-15
Assume first order, elementary rxn in both directions
d [ A]

 k1[ A]  k 1[ B]
dt
Rate:
Conservation of Mass:
[ A]0  [ B]0  [ A]  [ B]
[ B]  [ A]0  [ B]0  [ A]
d [ A]

 k1[ A]  k 1 [ A]0  [ B]0  [ A]
dt
Integrate:
 k1[ A]t  k1 [ A]0  [ B]0  [ A]t  
  (k1  k1 )t
ln 
k1[ A]0  k1[ B]0


At equilibrium
A
k1
k-1
B
CK-16
d [ A]

 k1[ A]  k 1[ B]
dt
d
[
A
]
At equilibrium… 
0
dt
k1[ A]eq  k 1[ B]eq
The forward rate equals the
reverse at equilibrium.
What is the equilibrium constant for this reaction?
In terms of rate constants?
k1
K eq 
k 1
K eq 
[ B]eq
[ A]eq
Temperature Dependence of k
CK-17
The rate constant can vary in different ways with T.
Svante Arrhenius
Winner of the 3rd Nobel
Prize in Chemistry
Differential form of the Arrhenius Equation:
Ea
d ln k

2
dT
RT
Arrhenius Parameters
CK-18
Integrated forms of Arrhenius equation:
Ea
ln k  ln A 
RT
k  Ae  Ea / RT
Activated (or transition) state
Ea is the activation energy. This is the
energy required to get over a barrier (at
the activated or transition state) between
the reactants and products. Ea has units
of energy and is T independent.
A is the pre-exponential or Arrhenius
factor and is T dependent. A is a
measure of rate at which collisions
occur (and takes lots of things into
acct such as orientation, molecular
size, number of molecules per
volume, molecular velocity, etc).
2HI(g)→I2(g) + H2(g)
Transition-State Theory
CK-19
AB‡ is the transition state (or activated complex.)
Transition state theory assumes that
the transition state and reactants are in
equilibrium with each other, and uses
concepts from chemical equilibrium
and statistical mechanics to find kinetic
info such as rate constants!
Eyring Equation (key to transition-state theory)
From CEq:
K e
‡
 DG ‡ / RT
Change in Gibbs energy from reactants to TS
So…
k BT
k
e
h
k BT ‡
k
KC
h
‡
 DG / RT
DG  DH  TDS
k BT DS ‡/ R  DH‡ / RT
k
e
e
Enthalpy of activation
h
Entropy of activation
Relating Ea to thermodynamics!
CK-20
Necessary Pieces…
Arrhenius Equation: ln k  ln A 
Ea
d ln k

Differentiate wrt T:
dT
RT 2
Ea
RT
or
d ln k
Ea  RT
dT
2
‡
From Eyring Equation:
d ln k 1 d ln K C
 
dT
T
dT
van’t Hoff Equation (for Kc):
Putting it all together…
d ln K C DU

dT
RT 2
‡
 1 d ln K C 
Ea  RT  

dT 
T
2
or
Ea  RT  DU
‡
What about A, the pre-exponential?
Ea  RT  DU ‡
so
CK-21
DU ‡  DH‡  RTD‡ng
and
‡
‡
Ea  RT  DH  RT D n
Unimolecular Gas Phase Reaction
AA‡Products
so
k BT DS
k
e
h
‡
Ea  RT  DH
‡
‡
/ R  DH / RT
e
k BT DS ‡ / R  Ea / RT
k e
e
e
h
Bimolecular Gas Phase Reaction
A+BAB‡Products so
Ea  DH  2 RT
k BT DS ‡/ R  E A / RT
k e
e
e
h
2
Same for
reactions
in solution
‡
What is A?
Transition State Theory and NMR Lab
CK-22
In the NMR/N,N-DMA Paper, Gasparro et al. found an activation
energy of 70.3 kJ/mol and a pre-factor of 1.87 × 1010 s-1. Using
these values, and a temperature of 298 K, find…
DH
‡
DS
‡
DG
‡
Why is TST important?
1. Provides details of a reaction on the molecular scale.
2. Connects quantum mechanics and kinetics.
3. Currently used for many computational studies on reaction rates.
EX-CK3
CK-23
Chapter 29: Reaction Mechanisms
Always remember….
CK-24
• One can never prove a reaction
mechanism, although evidence may
disprove a mechanism.
• Verifying proposed mechanisms requires
extensive experimental verification of each
proposed step!
Let’s examine a reaction …
CK-25
A  P
k o bs
Reaction could progress in multiple ways… How can we distinguish?
Case 1: One elementary step
A  P
k1
Case 2: Two step reaction
A  I
k1
I  P
k2
EX-CK4
Now let’s focus on the intermediate…
CK-26
A  I  P
k1
k2
How do k1 and k2 relate in case a? in case b?
(a) I forms quickly but decays slowly… k1 is fast relative to k2.
(b) I builds up to a constant, nearly negligible, concentration
until near end of reaction. … k1 is slow relative to k2.
EX-CK5
d[I ]
0
dt
Steady state approximation... Valid only if k2 > k1.
Rate Laws do not yield unique mechanisms
CK-27
An empirically determined rate law does not imply a
unique reaction mechanism!
Consider reaction:
obs
2NO( g )  O2 ( g ) k
2NO2 ( g )
Experimentally, it was determined that the rate is given by:
v(t )  kobs[ NO] [O 2 ]
2
Researchers proposed two possible mechanisms. They need
to determine if one of them is correct.
So how would researchers distinguish between the mechanisms?
EX-CK6
Remember the Chain Rxn from CK-6?
CK-28
1/ 2

2
k
[
H
][
Br
]
2
2
H2(g) + Br2 (g)  2HBr (g) v(t ) 
1
1  k [HBr ][ Br2 ]
Proposed Mechanism
Initiation:
Propagation:
k1
Br2 ( g )  M ( g ) 
2Br  ( g )  M ( g )
k2
Br  ( g )  H 2 ( g ) 
HBr ( g )  H  ( g )
k3
H  ( g )  Br2 ( g ) 
HBr ( g )  Br  ( g )
Inhibition:
k 2
HBr ( g )  H  ( g ) 
Br  ( g )  H 2 ( g )
Termination:
k 1
2Br  ( g )  M ( g ) 
Br2 ( g )  M ( g )
EX-CK7
What about the solution kinetics lab?!
CK-29
• Now that we’ve explored reaction
mechanisms and rate laws, let’s try to
derive the rate laws from the solution
kinetics lab…
EX-CK8
Catalysis
CK-30
Catalyst: A substance that participates in the chemical reaction but is
not consumed. Provides a new mechanism for reaction and can cause
reaction to occur faster.
In an experiment involving a catalyst, there are two competing reactions:
A  products
EX-CK9
A  catalyst  products  catalyst
If both reactions are elementary, overall rate is given by:
d [ A]

 k[ A]  kcat [ A][catalyst ]
dt