Reaction order
• The rate law can be written in a generalized form:
v = k [A]a[B]b….
where a is the order of the reaction with respect to the
species A, and b is the order of the reaction with respect to the
reagent B.
• The reaction order is (a + b +… . ).
• The order of a chemical reaction needs not to be an integral. Certain
reactions do not have an overall order !!!
Example 1: v = k [A]1/2[B]1
Example 2: v = k
(zero order reaction, such as ……)
• How to determine the unit of k?
Determination of the rate law
• Isolation method:
v = k [A]a[B]b
----->
v = k’[B]b
• Method of initial rates (often used in conjunction with the isolation
method):
v = k [A]a
at the beginning of the reaction
v0 = k [A0]a
taking logarithms gives:
logv0 = log k + a log[A0]
therefore the plot of the logarithms of the initial rates against the
logarithms of the initial concentrations of A should be a straight line
with the slope a (the order of the reaction).
Self-test 22.3: The initial rate of a reaction depended on the
concentration of a substance B as follows:
[B]0/(mmol L-1)
5.0
8.2
17
30
v0/(10-7 mol L-1s-1)
3.6
9.6
41
130
Determine the order of the reaction with respect to B and
calculate the rate constant.
Solution:
Log([B]0)
-2.30
-2.086
-1.770
-1.523
Log(v0)
-6.444
-6.018
-5.387
-4.886
0
-2.5
-2
-1.5
-1
-0.5
0
-1
-2
-3
Series1
-4
-5
-6
-7
22.3 Integrated rate law
• First order reaction:
A  Product
d [ A]
  k[ A]
dt
The solution of the above differential equation is:
 [ A] 
   kt
ln 
[
A
]
0 

or:
[A] = [A]0e-kt
• In a first order reaction, the concentration of reactants decreases
exponentially in time.
Self-test 22.4: In a particular experiment, it was found that the concentration
of N2O5 in liquid bromine varied with time as follows:
t/s
0
200
400
600
1000
[N2O5]/(mol L-1)
0.110
0.073
0.048
0.032
0.014
confirm that the reaction is first-order in N2O5 and determine the rate constant.
Solution: To confirm that a reaction is first order, plot ln([A]/[A]0) against
time and expect a straight line:
t/s
0
200
400
600
1000
ln([A]/[A]0)
0
-0.410 -0.829 -1.23
-2.06
0
0
200
400
600
800
1000
1200
-0.5
-1
Series1
-1.5
-2
-2.5
Half-lives and time constant
• For the first order reaction, the half-live equals:
t1 / 2

ln 2
k
therefore, is independent of the initial concentration.

• Time constant, , the time required for the
concentration of a reactant to fall to 1/e of its initial
value.
1
 
k
for the first order reaction.
Second order reactions
• Case 1: second-order rate law: d [ A]   k[ A] 2
(e.g. A → P)
dt
• Can one use A + A → P to represent the above process?
• The integrated solution for the above function is:
1
1

 kt
[ A] [ A]0
or
[ A] 
[ A]0
1  kt[ A]0
• The plot of 1/[A] against t is a straight line with the slope k.
t1 / 2 
1
k[ A]0
• Case 2: The rate law
d [ A]
  k[ A][ B ]
dt
(e.g. A + B → Product)
• The integrated solution (to be derived on chalk board) is :
 [ B] /[ B]0 
  ([ B]0  [ A]0 )kt
ln 
[
A
]
/[
A
]
0 

22.4 Reactions approaching equilibrium
Case 1: First order reactions:
A
→
B
B
→
A
the net rate change for A is therefore
v = k [A]
v = k’ [B]
d [ A]
  k[ A]  k '[ B]
dt
if [B]0 = 0, one has [A] + [B] = [A]0 at all time.
d [ A]
  k[ A]  k '([ A]0  [ A])   (k
dt
 k ' )[ A]  k '[ A]0
the integrated solution for the above equation is
[A] =  k '  ke ( k  k ') t 

[ A]0
 k  k' 
As t → ∞, the concentrations reach their equilibrium values:
[A]eq = k '[ A]0
[B]eq = [A]0 – [A]eq = k[ A]
k  k'
0
k  k'
• The equilibrium constant can be calculated as K =
thus:
K
[ B ]eq
[ A]eq
k
k'
• In a simple way, at the equilibrium point there will be no net change
and thus the forward reaction will be equal to the reverse reaction:
k[A]eq = k’ [B]eq
thus
[ B]eq k
[ A]eq

k'
the above equation bridges the thermodynamic quantities and
reaction rates through equilibrium constant.
• For a general reaction scheme with multiple reversible steps:
K
k1 k2
 '  ...
'
k1 k2
Determining rate constants with
relaxation method
• After applying a perturbation, the system (A ↔ B) may have a new
equilibrium state. Assuming the distance between the current state and the
new equilibrium state is x, one gets
[A] = [A]eq + x; [B] = [B]eq - x;
d [ A]
  k a ( x  [ A]eq )  kb ([ B]eq  x)  ( k a  kb ) x
dt
Because
d [ A] dx one gets dx/dt = - (ka + kb)x

dt
dt
therefore
x  x0 e t /
 is called the relaxation time
1

 ka  kb
Example 22.4: The H2O(l) ↔ H+(aq) + OH-(aq) equilibrium relaxes in 37 μs at
298 K and pKw = 14.0. Calculate the rate constants for the forward and
backward reactions.
Solution: the net rate of ionization of H2O is
d [ H 2 O]
 k 1 [ H 2 O]  k 2 [ H  ][OH  ]
dt
we write
[H2O] = [H2O]eq + x;
and obtain: dx

dt

[H+] = [H+]eq – x; [OH-] = [OH-]eq – x

  k1  k 2 ([ H ]eq  [OH  ]eq ) x  k1 [ H 2 O]eq  k 2 [ H  ]eq [OH  ]eq  k 2 x 2
Because x is small, k2x2 can be ignored,
so 1





 k1  k 2 ([ H ]eq  [OH ]eq )
Because k1[H2O]eq = k2[H+]eq[OH-]eq at equilibrium condition


k1 [ H ]eq [OH ]eq

k2
[ H 2 O ]eq
1

hence
=
K w (mol L1 ) 2
[ H 2 O ]eq
=
Kw
mol L1
55.6
 k 2 ( K  [ H  ]eq  [OH  ]eq )  k 2 ( K  K w1 / 2  K w1 / 2 )
k2 = 1.4 x 1011 L mol-1 s-1
k1 = 2.4 x 10-5 s-1
• Self-test 22.5: Derive an expression for the relaxation time of a
concentration when the reaction A + B ↔ C + D is second-order in
both directions.
To be demonstrated on in class
22.5 The temperature dependence
of reaction rates
• Arrhenius equation:
k  A e  Ea / RT
A is the pre-exponential factor; Ea is the activation energy. The
two quantities, A and Ea, are called Arrhenius parameters.
• In an alternative expression
E
lnk = lnA - RT
one can see that the plot of lnk against 1/T gives a
straight line.
a
Example: Determining the Arrhenius parameters from the following data:
T/K
300
350
400
450
500
k(L mol-1s-1)
7.9x106
3.0x107
7.9x107
1.7x108
3.2x108
Solution:
1/T (K-1)
lnk (L mol-1s-1)
0.00333
15.88
0.00286
17.22
0.0025
18.19
0.00222
18.95
0.002
19.58
25
20
15
Series1
10
5
0
0
0.0005
0.001
0.0015
0.002
0.0025
0.003
0.0035
The slope of the above plotted straight line is –Ea/R, so Ea = 23 kJ mol-1.
The intersection of the straight line with y-axis is lnA, so A = 8x1010 L mol-1s-1
The interpretation of the Arrhenius
parameters
• Reaction coordinate: the collection of
motions such as changes in
interatomic distance, bond angles, etc.
• Activated complex
• Transition state
• For bimolecular reactions, the
activation energy is the minimum
kinetic energy that reactants must
have in order to form products.
Applications of the Arrhenius principle
Temperature jump-relaxation method:
consider a simple first order reaction:
A↔B
at equilibrium: ddt[ A]  0
d [ A]
  k a '[ A]  k b' [ B ]
dt
ka '[ A]  k [ B]
'
eq
'
b
'
eq
After the temperature jump the system has a new
equilibrium state. Assuming the distance between the
current state and the new equilibrium state is x, one gets
[A] = [A]eq + x; [B] = [B]eq - x;
22.6 Elementary reactions
• Elementary reactions: reactions which involves only a small
number of molecules or ions.
A typical example:
H + Br2 → HBr
+ Br
• Molecularity: the number of molecules coming together to react in
an elementary reaction.
• Molecularity and the reaction order are different !!! Reaction
order is an empirical quantity, and obtained from the experimental
rate law; molecularity refers to an elementary reaction proposed as
an individual step in a mechanism. It must be an integral.
• An elementary bimolecular reaction has a second-order rate law:
A + B → P
d [ A]
  k[ A][ B ]
dt
• If the reaction is an elementary bimolecular process, then it has
second-order kinetics; However, if the kinetics are second-order,
then the reaction might be complex.
22.7 Consecutive elementary
reactions
•
An example:
239U
→ 239Np →239Pu
• Consecutive unimolecular reaction
A
→ B → C
The rate of decomposition of A is:
d [ A]
  k 1 [ A]
dt
• The intermediate B is formed from A, but also decays to C. The net
rate of formation of B is therefore:
d [ B]
 k1 [ A]  k 2 [ B]
dt
• The reagent C is produced from the unimolecular decay of B:
d [C ]
 k 2 [ B]
dt
• Integrated solution for the first order reaction (A) is:
[ A]  [ A]0 e k1t
• Then one gets a new expression for the reactant B:
d [ B]
 k1 [A]0 e -k1t  k 2 [ B]
dt
the integrated solution for the above equation is:
[ B] 
k1
(e k1t  e k2t )[ A]0
k 2  k1
when assuming [B]0 = 0.
• Based on the conservation law [A] + [B] + [C] = [A]0

k1e  k2t  k 2e  k1t 
[C ]  1 
[ A]0
k 2  k1


Example. In an industrial batch process a substance A produces the desired
compound B that goes on to decay to a worthless product C, each step of
the reaction being first-order. At what time will B be present in the greatest
concentration?
Solution: At the maximum value of B
d[ B]
0
dt
Using the equation 25.7.6 and taking derivatives with respect to t:
k1 [ A]0 (k1e  k1t  k 2 e  k2t )
d [ B]

dt
k 2  k1
In order to satisfy
d[ B]
0
dt
k1e  k1t  k2e  k2t
tmax =
=0
k
1
ln 1
k1  k 2 k 2
The maximum concentration of B can be calculated by plugging the tmax into
the equation.