Chapter 5
Chemical Kinetics
Introduction
Chemical reactions require varying lengths of time for
completion, depending on the characteristics of the
reactants and products and the conditions under which the
reaction is run. May reactions are over in a fraction of a
second, whereas others can take much longer. If we add
barium ion to an aqueous solution of sulphate ion, a
precipitate of BaS04 forms almost immediately. On the
other hand, the reactions that occur in a cement mixture as
it hardens to concrete require years for completion.
The study of the rate, or speed, of a reaction has
important applications. In the manufacture of ammonia
from N2 and H2 we may wish to know what conditions will
help the reaction to precede faster.
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Or, we may wish to know if the nitric oxide, No, in the
exhaust gases of supersonic transports will destroy ozone
in the stratosphere faster than the ozone is produced.
A further reason for studying reaction rates is to understand
how chemical reactions occur (mechanism of reaction). By
nothing how the rate of a reaction is affected by changing
conditions, we can sometimes learn the details of what is
happening at the molecular level.
Two factors control the outcome of chemical reactions:
1. Chemical Thermodynamics
2. Chemical Kinetics
•
Chemical Kinetics: study of rates of chemical
reactions and mechanisms by which they occur
• Rate of reaction describes how fast
reactants are
used up and products are formed
• There are 4 basic factors that affect reaction rates
(i) Concentration
(ii)
Physical state
(iii)
Temperature
(iv)
Catalysts
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• The rate of chemical reaction increases with:
1. increases in concentration
2. increase in surface area of particles (physical
nature)
3. increase in temperature
• Reaction Rates and their Measurement:
• Rate of reaction is typically measured as the change in
concentration with time
•
This change may be a decrease or an increase
in [products]
in [reactants]
Rate = ______________
= ______________
change in time
change in time
• Rate has units of moles per liter per unit time
M s-1,
M h-1
•
Consider the hypothetical reaction
aA + bB  cC + dD
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•
We can write
• Note the use of the negative sign
- rate is defined as a positive quantity
- rate of disappearance of a reactant is negative
2N2O5(g)  4NO2(g) + O2(g)
Rate of = - 1 [N2O5] = 1 [NO2]
reaction
2
4
t
t
=
[O2]
t
The rate (or speed) of any
chemical reaction can be
expressed as the ratio of the
change in the concentration of
a reactant (or product) to a
change of time. The rate is
usually expressed in moles /
dm3.s (moles/Ls).
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Example: 5.1
4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
Suppose
reaction
that at a
particular
the ammonia
is
moment during
reacting
at the
the
rate of
0.24moi/L.s.
a) What is the rate at which O2 is reacting?
b) What is the rate at which H2O is being formed?
Solution: The coefficient in the equation:
6 mole H2O
5 mole O2
and
4 mole NH 3
4moleNH 3
a)
For the rate at which O2 is consumed
Rate (for O2) =
b)
0.24 mol NH 3
5 mol O2
0.3 mol O2
x

L.S
4 mole NH 3
L.S
For the rate at which H2O is formed
Rate (for H2O) =
0.24 mol NH3
6 mol H2O 0.36 mol H2O
x

L.S
4 mol NH3
L.S
Measurement of reaction rates
An accurate, quantitative estimate of the rate of reaction
at any given moment during the reaction can be obtained
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from the slope of the tangent to concentration -time curve at
that particular instant. This is shown in figure 2. from the
tangent to the curve for the reaction
A → B we can write
rate =
 [B]
t
Figure 2: Estimation of the rate of reaction based on the
change in concentration of B with time
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• Rate Laws:
For any given reaction, one of the most important
controlling
influences
is
the
concentration
of the
reactants. Generally, if we follow a chemical reaction over
a period of time, we find that its rate gradually decreases as
the reactants are consumed. In fact, the rate is nearly
always directly proportional to the concentration of the
reactants raised to some power.
A→B
Rate ∝ [A]x
Where, x is called the order of the reaction. When:
• x = 1 we have first order reaction,
When the reaction rate is doubled by doubling the
concentration of a reactant the order with respect to that
reactant is 1.
• x = 2 we have second order reaction,
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If the rate is increased by a factor of four when the
concentration of a reactant is doubled, the reaction is
second order with respect to that compound.
• x = 3 we have third order reaction,
If the rate undergoes an eightfold increases, when the
concentration is doubled (23 = 8), the reaction is third
order with respect to that component.
• There are also example of zero - order reactions,
where x = 0. For a zero - order reaction the rate is
constant and does not depend on the concentration of
reactant.
A very important fact is that there is not necessarily any
direct relationship between the coefficient in the chemical
equation for a reaction and the order of the reaction. The
value of x can only be determined from experiment.
For example, A + B → products
The rate is usually depends on the concentrations of
both A and B.
Normally, the rate is proportional to the product of the
concentrations of A and B, each raised to same power.
Rate ∝ [A]x [B]y
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rate = K [A]x [B]y
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We say that the order of the reaction with respect to A
is x, that the order with respect to B is y and that the overall
order (i.e., the sum of the individual orders) is x + y and
where K is the proportionality constant, which we call the
rate constant which varies with temperature. The resulting
equation, termed the rate law for the reaction is, rate = K
[A]x [B]y
The following examples illustrate how can use these
ideas to obtain the rate law for a reaction by varying the
concentrations of reactants.
Example 5.2. Below are some data collected in a series
of experiments on the reaction of nitric oxide
with bromine
2 NO(g) + Br2(g) → 2 NOBr(g)
Experiment
1
2
3
4
5
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at 273°C
Initial concentration
Initial Rate
mol / L
(mol.l-1s-1)
NO
0.1
0.1
0.1
0.2
0.3
Br2
0.1
0.2
0.3
0.1
0.1
12
24
36
48
108
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Determine the rate law for the reaction and compute the
value of the rate constant.
Solution: The rate law for the reaction will have the form
r = rate = K [NO]x [Br2]y
r2  [A 2 ] 


r1  [A 1 ] 
X
we will study how the rate changes when the
concentration of one reactant varies while that of the other
reactant stay the same.
y
r  [0.2] 
24
 2 y  21
For Br2 2  
 
r1  [0.1]  12
 y for Br2 = 1
first order respect to Br2
X
r
48  [0.2] 

For NO 4 
 ,4  2 2  2 X
r1 12  [0.1] 
 X for NO = 2
second order respect to NO
There fore Rate = k [NO]2 [Br2]
12 mol/L.s = K (0.1 mol/l)2 (0.1 mol/l)
12 mol/L.s = K (0.001 mol3/l3)
K=
12 mol /l.s
 1.2x10 4 l2 /mol 2 .s
3
3 3
1.0x10 mol /l
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Example 5.3: The following data were collected for the
reaction of t- butyl bromide, (CH3)3 CBr, with
hydroxide ion at 55°C
(CH3)3 CBr + OH- (CH3)3 COH + BrExperiment Initial concentration (M) Initial rate formation
1
2
3
4
5
(CH3)3 CBr
OH-
of (CH3)3COH mol/L.s
0.1
0.2
0.3
0.1
0.1
0.1
0.1
0.1
0.2
0.3
0.001
0.002
0.003
0.001
0.001
What
is the rate law and rate constant for this reaction?
ii
Solution: we expect a rate law of the form
rate = K [(CH3)3 CBr]x [OH]y
For (CH3)3 CBr
X
r2  [0.2] 
0.002

X 1
 
r1  [0.1] 
0.001
so the reaction is first order with respect to (CH3)3 CBr
In experiment 1,4,5, the [(CH3)3 CBr] concentration is the
same.
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Changing the [OH-] has no effect on the rate. This means
that the reaction is zero - order with respect to OH-.
Therefore
Rate = k [(CH3)3 CBr]1 [OH]°
:. rate = k [(CH3)3 CBr]
0.01 mol/L.s = K = (0.1mol/L)
K=
0.001 mol /l.s
 0.01 S -1
0.1 mol /l
Example 5.4: For the reaction
OH-
I- + ClO- →
IO- + Cl-
In order to use the initial - rate method to obtain the rate law
for this reaction, the following experiments were run:
Experiment Initial concentration (M)
I-
CIO-
OH-
Initial rate
(mol/L.s)
1
2
0.01
0.02
0.01
0.01
0.01
0.01
6.1 x 10-4
12.2 x 10-4
3
0.01
0.02
0.01
12.2 x I0-4
4
0.01
0.01
0.02
3.0 x 10-4
What is the rate and the rate constant for this reaction?
Solution: 1- By comparing Exp. 1 and 2
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X
r2  [0.2] 
12.2 x 10 - 4
1


2
 
r1  [0.1] 
6.1 x 10 - 4
X 1
The reaction is first order in I2- by comparing Exp 1 and 3
y
r3  [0.2]  12.2 x 10 - 4

 21
 
-4
r1  [0.1] 
6.1 x 10
y 1
The reaction is first order in CIO3- by comparing Exp 1 and 4
Z
r4  [0.2] 
3x 10 - 4

 1  2Z
 
-4
2
r1  [0.1] 
6.1 x 10
Z  1
 The rate with respect to [OH-] is inversely proportional.
The rate law is
[I- ][ClO  ]
Rate = K
[OH - ]
6.1 x 10-4
mol
0.01 mol/L x 0.01 mol/L
K
L.s
0.01 mol/L
Therefore K = 6.1 x 10-2S-1
The overall order of the reaction = 1 + 1 – 1 = 1
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• Concentration and Time: Half - Lives:
The rate law for a reaction tells us how the rate of a
reaction is related to the concentration of the reactants. By
applying calculus to the rate law which we won't attempt to
go through here - an expression relating the concentration
to time can be derived table 1.
Half - lives: t½
The length of time required for the concentration of the
reactant to be decreased to half of its initial value. At this
point
t½, and [A]t = ½ [A]0.
Table 1: Relation of concentration and time for various
orders of reactions, their half lives and k units.
Order of the Rate law
Relation of
reaction
concentration and
time
Zero-order
r=k
First-order
r = k [B]
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CBo - CB = kt
log
CBo
Kt

CB 2.303
t½
K units
mol/L.s
C Bo
2K
0.693
-1
K S
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Second-order r = k [B]2
Third – order r = k [B]3
1
1

 Kt
CB CBo
1
1

 - 2Kt
CBo  CB 
L/mol.s
1
KCBo
3 L/mol.s
2KCBo2
Bo initial concentration
B concentration at time t.
Example 5.5: In an investigation of the decomposition of
N2O5(g) at 35°C,
2 N2O5(g)  4 NO2(g) + O2(g).
an initial concentration of N2O5(g)/ [N2O5]0, of 0.03 mol/L
was used. The rate constant was found to be 1.35 x 10-4S-1
a) What will the concentration of N2O5(g) be after 30.0
minutes?
b) How many minutes will it take for the concentration of
N2O5(g) to drop to 0.02 mol/L.
c) How many minutes will it take for 90% of N2O5(g) to
decompose?
d) What is the half - life for the reaction?
Solution: Since units of K in S-1
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 The reaction is first order reaction.
Since the questions are asked in terms of minutes, it is
convenient to convert the value of K into units of min-1 from
S-1.
1.35 x 10 -4 60 s
x
 8.10 x 10 - 3 min -1
K=
1s
1 min
a) Log
[N2O5 ]O
kt

[N2O5 ] 2.303
0.03 mol/L 8.10 x 10-3 min 1 30.0 min
log

 0.1055
[N2O5 ]
2.303
0.03 mol/L
 antilog 0.1055  1.275
[N2O5 ]
[N2O5 ] 
0.03 mol/L
 0.0253 mol/L
1.275
0.03 mol/L 8.10 x 10 -3 min 1x t

b) Log
0.02 mol/L
2.303
2.303 log 1.5 = (8.10 x 10-3 min-1) t
t
2.303log1. 5
2.303x0.17 6

 50.0 min
-3
1
-3
1
8.10 x 10 min
8.10 x 10 min
c) Since 90% of N2O5 has decomposed, [N2O5] is equal
to 10.0% of the original concentration [N2O5]0
[N2O5] = 0.100 [N2O5]°
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a) = 0.1x0.03mol/L
= 0.003 mol/L
Log
[N2O5 ]O
kt

[N2O5 ] 2.303
8.10 x 10 -3 min 1x t
Log 10 =
2.303
t
2.303 log 10
 284 min
8.10 x 10 -3 min 1
d) t½ =
0.693
0.693

 85.6 min
K
8.10 x 10 -3 min 1
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• THEORIES OF REACTION RATE
1. Collision Theory
According to this theory, a chemical reaction takes place
only by collisions between the reacting molecules. But not all
collisions are effective. Only a small fraction of the collisions
produce a reaction. The two main conditions for a collision
between the reacting molecules to be productive are:
1. The colliding molecules must possess sufficient kinetic
energy to cause a reaction.
2. The
reacting molecules must come
with
proper
orientation
Now let us have a closer look at these two postulates of
the collision theory.
1. The molecule must collide with sufficient kinetic
energy
Let us consider a reaction
A-A + B-B → 2 A-B-B → 2A-B
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A chemical reaction occurs by breaking bonds between
the atoms of the reacting molecules and forming new bonds
in the product molecules. They energy for the breaking of
bonds comes from the kinetic energy possessed by the
reacting molecules before the collision. Next figure shows
the energy of molecules A2 and B2. as the reaction A2 + B2
→2AB progress.
The energy of the colliding molecules as the reaction A2
+ B2 →2AB proceeds. The activation energy Ea provides the
energy barrier.
The Fig. also shows the activation energy Ea. That is the
minimum energy necessary to cause a reaction between the
colliding molecules. Only the molecules that collide with a
kinetic energy greater than Ea, are able to get over the barrier
and react. The molecules colliding with kinetic energies less than
Ea fail to surmount the barrier. The collisions
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between them
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are
unproductive
and the molecules simply bounce off one
another.
2. The molecules must collide with correct orientation
The reactant molecules must collide with favorable orientation
(relative position). The correct orientation is that which ensures
direct contact between the atoms involved in the breaking and
forming of bonds. From the above discussion it is clear that:
only the molecules colliding with kinetic energy greater than
Ea and with correct orientation can cause reaction.
• Not all collisions leads to a reaction
• For effective collisions proper orientation of
the molecules must be possible
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Effect of Temperature on Reaction Rate;
In the last section we interpreted the activation energy
Ea to be the minimum kinetic required for a collision to be
effective. All molecules having kinetic energies equal or
higher than Ea are, therefore, capable of reacting. An
increase in temperature increases the fraction of molecules
with energies higher than the energy of activation. As a
result, the rate of the reaction increases.
In 1889, Arrhenius suggested a simple
relationship between the rate constant, K,
for a reaction and the temperature of the
system.
The rate constant, k, varies with in a
manner described by the following equation (Arrhenius
equation).
Where A (frequency
factor) is a constant
that is characteristic
of the reaction being
studied, e is the base of natural logarithms, Ea is the
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energy of activation for the reaction (in J / mol), R is the
molar gas constant [8.314 J/K. mol], and T is the absolute
temperature. For a single - step reaction, the factor e-Ea/RT
represents the fraction of molecules that has the energy of
activation needed for a successful reaction.
If we take the logarithms of the Arrhenius equation we get
Log K = log A -
Ea
2.303 RT
For a given reaction, there are two variables in this
equation, k and T. If we rearrange it into
Log K = -
Ea
 1
    log A
2.303 R  T 
Y
m
X
b
The values of Ea and A for a reaction can be found from the
rate constants for two different temperatures. If the rate
constant at T1 is k1 and at T2 is K2, then
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K 
Ea  1 1 
Log  2  
T  T 
K
2.303
R
 1
 1
2 
K 
Ea  T2  T1 
Log  2  


 K 1  2.303 R  T1T2 
If this equation is solved for the activation energy Ea , the
following relationship is obtained:
 TT 
K 
Ea  2.303 R  1 2  log  2 
 T2 - T1 
 K1 
Example 5.6 For the reaction
2NOCI → 2NO(g) + CI2(g)
the rate equation is
rate of production of CI2 = K [NOCI]2
The rate constant, k, is 2.6 x 10-8 L/mol.s at 300°K and
4.9 x 10-4 L/mol.s at 400°K.
a) what is the energy of activation, Ea for the reaction?
b) find the value of K at 500°K.
Solution: Let
T1 = 300°K
K1 = 2.6 x 10-8 L/mol .s
T2= 400°K
K2 = 4.9 x 10-4 L/mol .s
R is 8.31 J / °K. mol
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 TT 
K 
a) Ea  2.303 R  1 2  log  2 
 T2 - T1 
 K1 
-4
 4.9 x 10 
 300 x 400 
= 2.303 x 8.31 
log 
-8

400
300


 2.6 x 10 
= 19.1 x 1200 log 1.88 x 104 = 22.9 x 4.28 = 98000 J/mol
= 98.0 KJ/ mol.
b) Let
T1 = 400°K
K1 = 4.9 x 10-4 L/mol .s
T2= 500°K
K2 = unknown
Ea = 9.8 x 104 J/mol
K 
Ea  T2  T1 
Log  2  


 K 1  2.303 R  T1T2 
9.8 x 10 4  500  400 
3
5

  5.13x10 x5x10 K  2.57
2.303 x 8.31  500x400 
K2
 anti log 2.57  3.7 x102
K1
 K2 = 3.7 x 102 x K1 = 3.7 x 102 x 4.9 x 10-4= 0.18 L/mol.s
We can compute A from the equation
K = Ae –Ea/RT
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Limitation of the Collision theory
The collision theory of reaction rates is logical and
correct. However, it has been oversimplified and suffers
from the following weaknesses.
1. The theory applies to simple gaseous reactions only.
2. Arrhenius
equation
is
in
agreement
with
the
experimental values only for simple bimolecular
reactions. For reactions involving complex molecules,
the experimental rate constants are quite different
from the calculated values.
3. Cannot determine the steric effect.
4. The collision theory is silent on the cleavage and
formation of bonds involved in the reaction
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The various drawbacks in the simple collision theory do
not appear in the modem transition state theory.
2. Transition Sate Theory
The transition state or activated complex theory was
developed by Henry Erying (1935). This theory is also
called the absolute rate theory because with its help it is
possible to get the absolute value of the rate constant. The
transition state theory assumes that
• During a chemical reaction, reactants do not suddenly
convert to products
•
The formation of products is a continuous process of
bonding breaking and forming
•
At some point, a transitional species is formed
containing “partial” bonds
•
This species is called the transition state or activated
complex
• The transition state is the configuration of atoms at the
maximum of the reaction energy diagram
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•
The activation energy is therefore the energy needed to
reach the transition state
•
Note also that the transition state can go on to form
products or break apart to reform the reactants
CATALYSIS
Reaction rates are also affected by catalysts
A catalyst is a substance that increases the rate of a
reaction without being consumed; after the reaction has
ceased, it can be recovered from the reaction mixture
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chemically unchanged. A catalyst works by opening a
new path by which the reaction can take place. The
catalyzed path has a lower overall energy of activation
than the uncatalyzed path does (Figure 4); this accounts
for the more rapid reaction rate. Two additional points
can be derived from Figure 4.
1) The enthalpy change, H, for the catalyzed reaction is
the same as the H for the uncatalyzed reaction.
2) For reversible reaction, the catalyst has the same effect
on the reverse reaction that it has on the forward reaction.
The energy of activation for the reverse reaction Ear is
lowered by the catalyst to the same extent that Ea for the
forward reaction Eaf lowered.
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