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20

Chemical Kinetics

C H A P T E R
C O N T E N T S
CHEMICAL KINETICS
REACTION RATE
UNITS OF RATE
RATE LAWS
ORDER OF A REACTION
ZERO ORDER REACTION
MOLECULARITY OF A REACTION
MOLECULARITY VERSUS ORDER
OF REACTION
PSEUDO-ORDER REACTIONS
INTEGRATED RATE EQUATIONS
ZERO ORDER REACTIONS
FIRST ORDER REACTIONS
SECOND ORDER REACTIONS
THIRD ORDER REACTIONS
UNITS OF RATE CONSTANT
HALF-LIFE OF A REACTION
HOW TO DETERMINE THE ORDER
OF A REACTION
COLLISION THEORY OF
REACTION RATES
EFFECT OF INCREASE OF
TEMPERATURE ON REACTION
RATE
LIMITATIONS OF THE COLLISION
THEORY
TRANSITION STATE THEORY
ACTIVATION ENERGY AND
CATALYSIS

S

o far we have studied equilibrium reactions
...
But most chemical reactions are
spontaneous reactions
...
A spontaneous
reaction may be slow or it may be fast
...
The precipitate of AgCl is formed as fast as AgNO3
solution is added to NaCl solution
...

The branch of Physical chemistry which deals with the rate
of reactions is called Chemical Kinetics
...

(2) The factors as temperature, pressure, concentration and
catalyst, that influence the rate of a reaction
...

The knowledge of the rate of reactions is very valuable to
731

732

20

PHYSICAL CHEMISTRY

understand the chemical of reactions
...


REACTION RATE
The rate of a reaction tells as to what speed the reaction occurs
...
The rate of
reactions is defined as the change in concentration of any of reactant or products per unit time
...

Thus
rate of reaction = rate of disappearance of A
= rate of appearance of B
or

rate = –

d [ A]
dt

d [ B]
dt
where [ ] represents the concentration in moles per litre whereas ‘d’ represents infinitesimally small
change in concentration
...

=+

UNITS OF RATE
Reactions rate has the units of concentration divided by time
...
Therefore, the units of reaction rates may be
mole/litre sec
or
mol 1–1 s
mole/litre min
or
mol 1–1 min–1
mole/litre hour
or
mol 1–1 h–1 and, so on
Average Rate of Reaction is a Function of Time
Let us consider the reaction between carbon monoxide (CO) and nitrogen dioxide
...
The results of such an
experiment are listed below
...
of CO
0
...
067
0
...
040
0
...
20
...

As the reaction proceeds the concentration of CO decreases rapidly in the initial stages of
the reaction
...
Obviously the rate of
reaction is a function of time
...
067 – 0
...
033
=
=
= 0
...
100

[CO] = 0
...
080

0
...
040

[CO] = 0
...
020

0

10

20

30

40

Time/s
Figure 20
...
The average rate
is equal to the slope of the curve
...


In the time interval between 30 and 40 seconds, the average rate is much smaller
...
033 – 0
...
007
=
=
= 0
...

We shall see that average rates are not always useful
...
So, a
better way to estimate the rate of reaction, we need to
0
...


d [ ]t
dt
where [ ]t is the concentration at time t
...
080

[CO] Mol/Litre

Instantaneous Rate of Reaction
The average rates obtained by finding the slope
of the curve are not always useful
...
So, a better way to estimate the rate of a
reaction is to make the time interval as small as
possible
...
060

0
...
020

0

10

20

Time/s

30

Figure 20
...


40

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20

PHYSICAL CHEMISTRY

to the slope of a straight line drawn tangent to the curve at that time
...
20
...
0022 mol l–1 s–1
...

The exact relation between concentration and rate is determined by measuring the reaction
rate with different initial reactant concentrations
...

Thus for a substance A undergoing reaction,
rate
rate ∝ [A]n
or
rate = k [A]n

...
Thus the rate of reaction may be expressed as
rate = k [A]m [B]n

...

An expression which shows how the reaction rate is related to concentrations is called the rate
law or rate equation
...
The proportionality constant k is called the rate constant for the reaction
...
That
is [H2]1 = [H2]
...
It cannot be
written by merely looking at the equation with a background of our knowledge of Law of Mass Action
...
But usually the powers of concentration in the rate law are different from
coefficients
...
For NO the rate is proportional to [NO]2 and power ‘2’ corresponds
to the coefficient
...

Let us consider the example of a reaction which has the rate law
rate = k [A]m [B]n

...


CHEMICAL KINETICS

735

The order of a reaction can also be defined with respect to a single reactant
...
The overall order of reaction (m + n) may range
from 1 to 3 and can be fractional
...
If in the rate law (1) above
m + n = 1, it is first order reaction
m + n = 2, it is second order reaction
m + n = 3, it is third order reaction

ZERO ORDER REACTION
A reactant whose concentration does not affect the reaction rate is not included in the rate
law
...
Thus [A]0 = 1
...
For example, the
rate law for the reaction
NO2 + CO ⎯⎯ NO + CO2
→
at 200° C is
rate = k [NO2]2
Here the rate does not depend on [CO], so this is not included in the rate law and the power of
[CO] is understood to be zero
...
The reaction is second
order with respect to [NO2]
...


MOLECULARITY OF A REACTION
Chemical reactions may be classed into two types :
(a) Elementary reactions
(b) Complex reactions
An elementary reaction is a simple reaction which occurs in a single step
...

Molecularity of an Elementary Reaction
The molecularity of an elementary reaction is defined as : the number of reactant molecules
involved in a reaction
...
, according as one, two or three
reactant molecules are participating in the reaction
...
Thus we have :
(a) Unimolecular reactions : (molecularity = 1)
A ⎯⎯ product
→
→
Examples are : (i)
Br2 ⎯⎯ 2Br
H

(ii)

C

COOH

H

C

COOH

maleic acid

⎯⎯
→

H

C

COOH

COOH

C

H

fumaric acid

(b) Bimolecular reactions : (molecularity = 2)
→ products
A + B ⎯⎯
A + A ⎯⎯
→ products
Examples are :
→ CH3COOH + C2H5OH
(i) CH3 COOC2H5 + H2O ⎯⎯
Ethyl acetate

acetic acid

ethyl alcohol

→ H2 + I2
(ii)
2HI ⎯⎯
(c) Termolecular reactions : (molecularity = 3)
→ products
A + B + C ⎯⎯
Examples are :
→ 2NO2
2NO + O2 ⎯⎯
→ 2NOCl
2NO + Cl2 ⎯⎯

Why High Molecularity Reactions are Rare ?
Most of the reactions involve one, two or at the most three molecules
...
The rarity of reactions with high molecularity can be explained on
the basis of the kinetic molecular theory
...
The chances of
simultaneous collision of reacting molecules will go on decreasing with increase in number of molecules
...
For a reaction of molecularity 4, the four molecules must come closed and collide with one
another at the same time
...
Hence the reactions involving many molecules proceed through a series of
steps, each involving two or three or less number of molecules
...


Common

Less common

Rare

Figure 20
...


CHEMICAL KINETICS

737

Molecularity of a Complex Reaction
Most chemical reactions are complex reactions
...
Each step is an
elementary reaction
...
In any mechanism, some of the steps will be fast, others will
be slow
...
Thus the slowest step is the ratedetermining step of the reaction
...
It occurs by the following steps :
→ 2NO2 + 2NO3
(slow)
Step 1
2N2O5 ⎯⎯
→ NO + NO2 + O2
Step 2
NO2 + NO3 ⎯⎯
(slow)
→ 2NO2
Step 3
NO + NO3 ⎯⎯
(fast)
Overall reaction

2N2O5

⎯⎯
→

4NO2 + O2

Each elementary reaction has its own molecularity equal to the number of molecules or atoms
participating in it
...
At best could be thought
of as : the number of molecules or atoms taking part in the rate-determining step
...


MOLECULARITY VERSUS ORDER OF REACTION
The term molecularity is often confused with order of a reaction
...

The sum of the powers to which the concentrations are raised in the rate law is known as the
order of reaction
...
The number of collisions in turn is proportional to the concentration of each
reactant molecule (or atom)
...

→
2A + B ⎯⎯ products
rate ∝ [A] [A] [B]
or
rate = k [A]2 [B]
(rate law)
Two molecules of A and one molecule of B are participating in the reaction and, therefore,
molecularity of the reaction is 2 + 1 = 3
...
Thus the molecularity and order for an elementary reaction are equal
...
1
...

Reactions

Molecularity

A

⎯⎯ products
→

1

Rate law
rate = k [A]

Order
1

A+A

⎯⎯ products
→

2

rate = k

A+B

⎯⎯ products
→

2

rate = k [A] [B]

2

A + 2B

⎯⎯ products
→

3

rate = k[A] [B]2

3

A+B+C

⎯⎯ products
→

3

rate = k [A] [B] [C]

3

[A]2

2

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20

PHYSICAL CHEMISTRY

Differences Between Order and Molecularity
Order of a Reaction

Molecularity of a Reaction

1
...

2
...

3
...

4
...

5
...


1
...

2
...

3
...

4
...

5
...


PSEUDO–ORDER REACTIONS
A reaction in which one of the reactants is present in a large excess shows an order different from
the actual order
...
Since for elementary reactions molecularity and order are identical, pseudo-order reactions may
also be called pseudo molecular reactions
...
Since it is an elementary reaction, its rate law can
be written as
rate = k [A] [B]
As B is present in large excess, its concentration remains practically constant in the course of
reaction
...
Thus the actual order of the reaction is second-order but in
practice it will be first-order
...

Examples of Pseudo-order Reactions
(1) Hydrolysis of an ester
...

⎯⎯
→
CH3COOH + C2H5OH
CH3COOC2H5 + H2O
ethyl acetate
(excess)
acetic acid
ethyl alcohol
Here a large excess of water is used and the rate law can be written as
rate = k [CH3COOH] [H2O]
= k′ [CH3COOH]
The reaction is actually second-order but in practice it is found to be first-order
...

(2) Hydrolysis of sucrose
...

⎯⎯
→
C12H22O11 + H2O
C6H12O6 + C6H12O6
sucrose
(excess)
glucose
fructose
If a large excess of water is present, [H2O] is practically constant and the rate law may be written
as

CHEMICAL KINETICS

739

rate = k [C12H22O11] [H2O]
= k [C12H22O11]
The reaction though of second-order is experimentally found to be first-order
...


ZERO ORDER REACTIONS
In a zero order reaction, rate is independent of the concentration of the reactions
...

a
0
Final conc
...

In zero order reaction, the rate constant is equal to the rate of reaction at all concentrations
...
If
after time t, x moles of A have changed, the concentration of A is a – x
...
Thus,
dx
= k (a – x)
dt
dx
= k dt
a – x

or


...
(2)
where I is the constant of integration
...

Thus,
I = – 1n a
Substituting for I in equation (2)
1n

or

a
= kt
a–x
1
a
k = 1n
t a– x


...
303
a
log
t
a – x


...

Sometimes the integrated rate law in the following form is also used :
k =

(a – x1 )
2
...

Examples of First order Reactions
Some common reactions which follow first order kinetics are listed below :
(1) Decomposition of N2O5 in CCl4 solution
...
20
...
The progress of the reaction is monitored
by measuring the volume of oxygen evolved from time to time
...
4
An apparatus for monitoring the volume of O2 evolved in the
decomposition of N2O5 dissolved in carbon tetrachloride
...
Thus,
k =

V∞
2
...
Thus it is a reaction of the first order
...
From the following data for the decomposition of N2O5 in CCl4 solution at
48°C, show that the reaction is of the first order
t (mts)
10
15
20
∞
6
...
95
11
...
75
Vol of O2 evolved
SOLUTION
For a first order reaction the integrated rate equation is
V∞
1
log
=k
t
V∞ – Vt

In this example, V∞ = 34
...
75
log
= 0
...
45
1
34
...
80
= 0
...
80
1
34
...
35
= 0
...
35
Since the value of k is fairly constant, it is a first order reaction
...
45

(2) Decomposition of H2O2 in aqueous solution
...

Pt

H 2 O2 ⎯⎯ H 2 O + O
→
The progress of the reaction is followed by titrating equal volumes of the reaction mixture against
standard KMnO4 solution at different time intervals
...
A solution of H2O2 when titrated against KMnO4 solution at different time
intervals gave the following results :
t (minutes)
0
10
20
Vol KMnO4 used
for 10 ml H2SO4
23
...
7 ml
9
...

SOLUTION
The integrated rate equation for first order reaction is
2
...
8 ml
(a – x) = 14
...
1
when t = 20 mts
Substituting these values in the rate equation above, we have
2
...
8
log
10
14
...
2303 (log 23
...
7)

k =

742

20

PHYSICAL CHEMISTRY

= 0
...
3766 – 1
...
04820
2
...
8
and
k =
log
20
9
...
10165 (log 23
...
1)
= 0
...
3766 – 0
...
04810
Since the value of k is almost constant, the decomposition of H2O2 is a first order reaction
...
The hydrolysis of ethyl acetate or methyl acetate in the presence of
a mineral acid as catalyst, is a first order reaction
...
At various intervals of time, a known volume of the
reaction mixture is titrated against a standard alkali solution
...
Therefore as the reaction proceeds, the volume of alkali required for titration goes on increasing
...
The following data was obtained on hydrolysis of methyl acetate at 25°C in
0
...
Establish that it is a first order reaction
...
36
29
...
72
47
...
303
a
k =
log
t
a– x
At any time, the volume of alkali used is needed for the acid present as catalyst and the acid
produced by hydrolysis
...

Thus,
a = 47
...
36 = 22
...
15 – 29
...
83 ml
(a – x) after 7140 sec = 47
...
72 = 15
...
303
22
...
00005455
4500
17
...
303
22
...
0000546
7140
15
...

k=

(4) Inversion of Cane sugar (sucrose)
...
The progress of the reaction is followed by noting the optical rotation
of the reaction mixture with the help of a polarimeter at different time intervals
...
The change in rotation is proportional to the amount of sugar decomposed
...

The concentration at time t, (a – x) is ∝ (rt – r∝)
Substituting in the first order rate equation,
k =

2
...
303
log10 0 ∝
t
(rt – r∝ )
If the experimental values of t (r0 – r∝) and (rt – r∝) are substituted in the above equation, a
constant value of k is obtained
...
The optical rotation of sucrose in 0
...

time (min)
0
7
...
1
∞
rotation (degree)
+24
...
4
+17
...
74
Show that inversion of sucrose is a first order reaction
...

r –r
2
...
09 – (– 10
...
83 for all time intervals
...

k =

time (t)

rt = r∝

k =

(r – r )
1
log 0 ∞
t
(rt – r∞ )

7
...
14

k =

1
34
...
0047
7
...
14

18

28
...
83
log
= 0
...
44

1
34
...
0048
27
...
74
Since the value of k comes out to be constant, the inversion of sucrose is a first order reaction
...
1

25
...
If after time t, x moles of A have reacted,
the concentration of A is (a – x)
...
Thus,
dx
= k (a – x) 2
dt


...
I can be evaluated by putting x = 0 and t = 0
...
(2)


...
(4)

Substituting for I in equation (3)
1
1
= kt +
a– x
a
kt =

1
1
–
a–x a

1
x
k =
...


Thus

Examples of Second order Reaction
Hydrolysis of an Ester by NaOH
...

→
CH3COOC2H5 + NaOH ⎯⎯ CH3COONa + C2H5OH
ethyl acetate

ethyl alcohol

The reaction is carried in a vessel at a constant temperature by taking
...
Measured volumes of the reaction mixture (say, 25 ml) are withdrawn at various
times and titrated against a standard acid
...
Thus the volume of the acid used when t = 0, gives the initial concentration (a) of the
reactants
...
The value of x can be
calculated
...

SOLVED PROBLEM
...
From the data given below, establish that this is a second order reaction
...
00
10
...
13
4
...


...

Therefore,
a, initial concentration = 16
...
24
and
x = 5
...
13
and
x = 9
...
32
and
x = 11
...
76
k =

...
0070
16 × 5 10
...
85

...
0067
16 × 15 6
...
68

...
00675
16 × 25 4
...

k =

THIRD ORDER REACTIONS
Let us consider a simple third order reaction of the type
→
3A ⎯⎯ products
Let the initial concentration of A be a moles litre–1 and after time t, x, moles have reacted
...
The rate law may be written as :
dx
= k ( a – x )3
dt
Rearranging equation (1), we have

dx
( a – x)3
On integration, it gives


...
(2)

1

= kt + I
2 (a – x) 2
where I is the integration constant
...
Thus,
I =


...
2
t 2a ( a – x ) 2

This is the integrated rate equation for a third order reaction
...
A few of the known examples are :
→
(i)
2FeCl3(aq) + SnCl2(aq) ⎯⎯ 2FeCl2 + SnCl4
(ii)

→
2NO(g) + O2(g) ⎯⎯ 2NO2(g)

(iii)

→
2NO(g) + Cl2(g) ⎯⎯ 2NOCl(g)

UNITS OF RATE CONSTANT
The units of rate constant for different orders of reactions are different
...

Units of First order Rate constant
The rate constant of a first order reaction is given by
[A]0
2
...
It has
the unit
time–1
k =

Units of Second order Rate constant
The rate constant for a second order reaction is expressed as
1
x
k= ×
t [A]0 ([A]0 – x)
concentration
1
=
×
or
concentration × concentration time
1
1
=
×
concentration time
1
1
=
×
mole/litre time
= mol–1 l time–1
Thus the units for k for a second order reactions are
mol–1 1 time–1
Units of Third order Rate constant
The rate constant for a third order reaction is
1 x (2a – x)
k =
...
It is defined as : the
time required for the concentration of a reactant to decrease to half its initial value
...
It is
represented by the symbol t1/2 or t0
...


Concentration of A

[A]0

[A]0 /2

[A]0 /4
[A]0 /8
0
0

1

2

3

4

Time

Figure 20
...
Concentration of a reactant A as a function of time for a first-order
reaction
...
For a
first-order reaction, each half-life represents an equal amount of time
...
303
log
t
[A]
where [A]0 is initial concentration and [A] is concentration at any time t
...
e
...
303
2
...
303
2
...
3010
k
k
0
...

(2) it is inversely proportional ot k, the rate-constant
...
For illustration, let us calculate the time in which two-third of the reaction is completed
...
303
log
k
[A]
2
1
Here, the initial concentration has reacted reducing it to
...
303
2
...
303
× 0
...
Compound A decomposes to form B and C the reaction is first order
...
450 s–1
...
693
k
k = rate constant

t1/ 2 =

where

Substituting the value of k = 0
...
693
0
...
54 s

Thus half-life of the reaction A → B + C is 1
...

SOLVED PROBLEM 2
...
Calculate
the rate constant
...
693
k
Putting t1/2 = 15 min in the expression and solving for k, we have
t1/ 2 =

k=

0
...
693
=
= 4
...
For the reaction

→
2N2O5 ⎯⎯ 4NO2 + O2
the rate is directly proportional to [N2O5]
...
Find the
value of the rate constant k
...
The integrated rate equation is
k =

[N 2 O 5 ]0
2
...
303
log
1
3600
[N 2 O5 ] 0
10

2
...
303
log 10 =
×1
3600
3600
2
...
40 × 10 –4 s –1
3600
=

Thus

SOLVED PROBLEM 4
...
22 × 10–4 sec–1
...
100 M to drop to 0
...

SOLUTION
Calculation of half-life
t1/ 2 =

0
...
693
=
= 1
...
22 × 10 –4 sec –1

Calculation of time in seconds for drop of [N2O5] from 0
...
0100 M
From first order integrated rate equation,
t=

or

[N O ]
2
...
303
log 2 5 0
[N 2 O5 ] t
k

Substituting values
t=

=

2
...
22 × 10

–4

2
...
22 × 10 –4

log

×1

= 3
...
100
0
...
For a certain first order reaction t0
...
How long will it take for the
reaction to be completed 75% ?
SOLUTION
Calculation of k
For a first order reaction
0
...
693
or
100 =
k
0
...
00693 sec –1
100
Calculation of time for 75% completion of reaction
The integrated rate equation for a first order reaction is
t1/ 2 =

[A] 0
2
...
303
t=
log
k
[A]

k=
or

3
1
initial concentration has reacted, it is reduced to
4
4
Substituting values in the rate equation

When

[A] 0
2
...
00693
[A] 0
4
2
...
00693

t3/ 4 =

SOLVED PROBLEM 6
...
Calculate the
time required for its 100% completion
...
303
log
t
[A]

After 40 mts, the initial concentration is reduced to

4
That is,
5

4
[A] 0
5
Substituting values in the equation above
[A] =

[A] 0
2
...
303
k =
log 5 – log 4 = 0
...
303
log
t
[A]

[A] 0
2
...
Thus,
t=

t1 =

[A] 0
2
...
00558
0

SOLVED PROBLEM 7
...
Calculate the time
required to complete 90% of the reaction
...
693
k
0
...
693
k =
=
= 0
...
5
23

t0
...
303
log
t
[A]

[A] 0
2
...
That is
...
303
2
...
0301304
0
...
303
=
= 76
...
0301304

t=

Half-life for a Second order Reaction
For the simple second order reaction 2A → Products, the integrated rate equation is
1
1
–
[A] [A] 0
where [A]0 is the initial concentration and [A] is the concentration when time t has elapsed
...

kt=

[A] =

1
[A] 0
2


...
(2)

752

20

PHYSICAL CHEMISTRY

and we have
k t1/ 2 =

k t1/ 2

or

1

–

1
[A] 0

1
[A] 0
2
2
1
=
–
[A] 0 [A] 0

Solving for t1/2 we find that
1
k [A] 0
As in case of a first order reaction, half-life for a second order reaction is inversely proportional
to rate constant k
...
This fact can be used to distinguish
between a first order and a second order reaction
...
6
Second-order half-life
...
Note that each half-life is twice as long as the preceding
one because t ½ = 1/k[A]0 and the concentration of A at the beginning of each
successive half-life is smaller by a factor of 2
...

(1) Using integrated rate equations
The reaction under study is performed by taking different initial concentrations of the reactant (a)
and noting the concentration (a – x) after regular time intervals (t)
...
The rate equation which yields a constant value of k corresponds to the correct order of the
reaction
...
It is still used extensively to find the order of simple reactions
...

In case of First order
We have already derived the integrated rate equation for first order as
a
1n
= kt
a–x
Simplifying, it becomes
1n (a – x) = – kt + 1n a
↑
↑
↑
y
= mx +
b
Thus the two variables in the first order rate equation are :
a
1n
and t
a–x
a
is plotted against t and straight line results (Fig
...
7), the corresponding
Hence, if 1n
a–x
reaction is of the first order
...


t

t

Figure 20
...


Figure 20
...


In case of Second order
We have already shown that second order rate equation can be written as
1
1
= kt +
a−x
a
↑
↑
↑
y
= mx +
b
This is the equation of a straight line, y = mx + b
...
20
...
In case a curve is obtained, the reaction is not second order
...

The progress of the reaction in each case is recorded by analysis
...
Let the initial concentrations in the two experiments be [A1] and
[A2], while times for completion of half change are t1 and t2 respectively
...
We know that half-life period for a first order reaction is
independent of the initial concentration, [A]
...
In the reduction of nitric oxide, 50% of reaction was completed in 108 seconds
when initial pressure was 336 mm Hg and in 147 seconds initial pressure was 288 mm Hg
...

SOLUTION
We know that
n –1

t2 ⎡ A1 ⎤
=

...

Taking logs of the expression (1), we have

n =1+

log [t2 / t1 ]
log [A1 / A 2 ]

Substituting values in expression (2),
n =1+

log 108 /147
log 288 / 336


...
1339
=1+ 2 = 3
0
...

=1+

(4) The Differential method
This method was suggested by van’t Hoff and, therefore, it is also called van’t Hoff’s differential
method
...

dC
–
= k Cn
dt
where C = concentration at any instant
...
(1)
–
= k C1n
dt
dC2
n

...
(3)

...
(5)

⎛ dC ⎞
To find n,⎜− ⎟ in the two experiments is determined by plotting concentrations against time (t)
...
Using the values of
⎝ dt ⎠
⎛ dC2 ⎞
⎛ dC1 ⎞
slopes ⎜ –
⎟ in the equation (5), n can be calculated
...
Suppose the reaction under
consideration is :
→
A + B + C ⎯⎯ products
The order of the reaction with respect to A, B and C is determined
...
The order of the reaction is then determined by using any of the
methods described earlier
...
If
nA, nB and nC are the orders of the reaction with respect to A, B and C respectively, the order of the
reaction n is given by the expression
...
But not all collisions are effective
...

The two main conditions for a collision between the reacting molecules to be productive are :
(1) The colliding molecules must posses sufficient kinetic energy to cause a reaction
...

Now let us have a closer look at these two postulates of the collision theory
...
The energy for the breaking of bonds comes from the
kinetic energy possessed by the reacting molecules before the collision
...
20
...

Energy barrier

Energy

Ea

A–A
B–B
Reactants
2A – B
Product

Reaction progress
Figure 20
...
The activation energy Ea provides the energy barrier
...
20
...
Only the molecules that collide with a kinetic energy
greater than Ea, are able to get over the barrier and react
...
The collisions between them are unproductive and the
molecules simply bounce off one another
...
The correct
orientation is that which ensure direct contact between the atoms involved in the breaking and
forming of bonds
...
20
...


CHEMICAL KINETICS

Effective collision

757

Ineffective collision

Figure 20
...


Collision Theory and Reaction Rate Expression
Taking into account the two postulates of the collision theory, the reaction rate for the elementary
process
...


EFFECT OF INCREASE OF TEMPERATURE ON REACTION RATE
It has been found that generally an increase of temperature increases the rate of reaction
...
Thus the ratio of rate constants of
a reaction at two different temperatures differing by 10 degree is known as Temperature Coefficient
...
e
...
As
a rule, an increase of temperature by 10°C doubles the reaction rate
...
Thus as the
temperature of a system is increased, more and more molecules will acquire necessary energy greater
that Ea to cause productive collisions
...

In 1889, Arrhenius suggested as simple relationship between the rate constant, k, for a reaction
and the temperature of the system
...
(1)
k = A e– Ea / RT
This is called the Arrhenius equation in which A is an experimentally determined quantity, Ea is

758

20

PHYSICAL CHEMISTRY

the activation energy, R is the gas constant, and T is Kelvin temperature
...
303 RT
1n k = –


...
(3)

If k1 and k2 are the values of rate constants at temperatures T1 and T2 respectively, we can derive
Ea ⎡ T2 – T1 ⎤
k2
=

...
303 R ⎢ T1 T2 ⎥
⎣
⎦
Arrhenius equation is valuable because it can be used to calculate the activation energy, Ea if the
experimental value of the rate constant, k, is known
...
(3)

↑
+

b

You can see that the equation (3) is that of a straight line, y = mx + b
...

Thus if we plot the natural logarithm of k against 1/T, we get a straight line (Fig
...
11)
...

Slope = –

Ea
R

ln k
1/T

ln k

ln k

Slope =

1/T

1/T
Figure 20
...
The slope of
line Δ In k /Δ I/T gives Ea using the expression given above
...
The values of the rate constant (k) for the reaction 2N2O5(g) ⎯ 4NO2(g) +
O2(g) were determined at several temperatures
...
2 × 104 K
...
314 JK–1 mol–1) (– 1
...
0 × 105 J mol–1
Thus the activation energy for the reaction is 1
...
Ea, is then calculated using the formula
that can be derived as follows from equation (3) above
...
(1)
RT1
At temperature T2, where the rate constant is k2,
E
1n k2 = – a + 1n A

...
303 R ⎢ T1 T2 ⎥
⎣
⎦
⎝
Thus the values of k1 and k2 measured at T1 and T2 can be used to find Ea
...
The gas-phase reaction between methane (CH4) and diatomic sulphur (S2) is
given by the equation
→
CH4 (g) + 2S2 (g) ⎯⎯ CS2 (g) + 2H2S (g)
At 550°C the rate constant for this reaction is 1
...
4 1 mol–1 sec
...

SOLUTION
Here

k 1 = 1
...
T1 = 550 + 273 = 823 K
k 2 = 6
...
T2 = 625 + 273 = 898 K
Substituting the values in the equation
1 ⎞
⎛k ⎞ E ⎛ 1
1n ⎜ 2 ⎟ = a ⎜ –
⎟
R ⎝ T1 T2 ⎠
⎝ k1 ⎠

760

20

PHYSICAL CHEMISTRY

Ea
⎛ 6
...
1 ⎠ 8
...
4 ⎞
(8
...
1 ⎠
Ea =
1 ⎞
⎛ 1
⎜ 823K – 898 K ⎟
⎝
⎠
5 J/mol
= 1
...
Such reactions
are known as complex reactions as these do not take place in a single step
...
Generally, following types of
complications occur
...
These are also known as sequential reactions
...
Various step reactions can be written for the overall reaction as shown below :

Initial conc
...
after time t

A
[A]o
[A]

k1

B
0
[B]

k2

C
0
[C]

In the above reaction the product C is formed from the reactant A through intermediate B
...

The net or overall rate of reaction depends upon the magnitude of these two rate constants
...

It is clear that
[A]0 = [A] + [B] + [C]
The differential rate expressions are
– d [A]
= k1 [A]
dT

d [B]
= k1 [A] – k2 [B]
dT

and

d [C]
= k2 [B]
dT

During the course of the reaction the concentration of A, B and C vary as shown in the Fig
...
12

CHEMICAL KINETICS

761

[A]

Concentration

[C]

[B]

Time
Figure 20
...


From the Fig 20
...

Examples of First Order Consecutive Reactions
(a) Decomposition of dimethyl ether is gaseous phase
k1

CH3 COCH3

CH4 + HCHO

k2

H2 + CO

(b) Decomposition of Ethylene oxide
CH2

CH2

k1

(CH3CHO)*

k2

CH4 + CO

O

(c) Any radioactive decay of the type
218
Po
84

214
Pb
82

214
Bi
83

214
Po
84

(2) Parallel or Side Reactions
In these reactions the reacting substance follows two or more paths to give two or more products
...
The reaction in which the maximum yield of the products is obtained is called the main or
major reaction while the other reaction (or reactions) are called side or parallel reactions
...
If k1 > k2 the reaction A ⎯ B will be the major
→
reaction and A ⎯ C will be the side or parallel reaction
...
The differential rate expressions are

762

20

PHYSICAL CHEMISTRY

r1 =

– d [A]
= k1 [A]
dT


...
(ii)

– d [A]
= r1 + r2 = k1 [A] + k2 [A]
dT
= (k1 + k2) [A]
= k ′ [A]

...
It is equal to the sum of the two constants k1 and k2 of two
side reactions
...
(iv)

where [A]0 is the initial concentration of the reactant A and [A]t is concentration of A at time t
...


Examples of Parallel or Side Reactions
(a) Reaction of ethyl bromide with potassium hydroxide
CH2
CH3CH 2Br

+

KOH

CH 2

KBr

+

ethylene

ethyl bromide

CH 3CH 2OH

+

KBr

ethyl alcohol

(b) Dehydration of 2-methyl–2–butanol
CH 3
CH3

CH 3
CH3

C

CH 2

OH
2-methyl-2-butanol

CH 3

C

CH

2-methyl-2-butene

CH3

CH3
CH 2

C

CH2

2-methyl-1-butene

CH3

+

H 2O


...

Initially, the rate of forward reaction is very large which decreases with passage of time and the rate of
backward or reverse reaction is zero which increases with passage of item
...
This situation is called the chemical equilibrium
...
e
...
A reaction of this type may be
represented as
kf
A
B
kb
Initial concentration
[A] 0
0
Conc
...

The overall rate of reaction is given by
Rate of Reaction = Rate of forward reaction – Rate of backward reaction
– d [A] d [B]
=
= k f [A] – kb [B]
dt
dt
If [A]0 is the initial concentration of A and x moles of it have reacted in time t
then
[A]t = [A]0 – x
and
[B] = x
Substituting these in equation (i), we get

i
...
,

dx
= k f ([A]0 – x) – kb x
dt

At equilibrium


...
(ii)

dx
=0
dt

Hence
kf ([A]0 – xeq) = kb xeq
where xeq
...
From equation
(iii) we have
⎛ [ A]0 – xeq ⎞
kb = k f ⎜
⎟
⎜
⎟
xeq
⎝
⎠
Substituting the value of kb in equation (ii), we get


...
All these
quantities can be measured easily
...

⎛ [A 0 ] – xeq ⎞
kb = k f ⎜
⎟
⎜
⎟
xeq
⎝
⎠
Examples of Opposing Reactions
(a) Dissociation of hydrogen iodides
kf
2HI
H 2 + I2
kb
(b) Isomerisation of cyclopropane into propene

kf
cyclopropane

kb

CH 3

CH

CH 2

propene

(c) Isomerisation of ammonium cyanate into urea in aqueous solution
...
cyanate

O

kf
kb

NH 2

C

NHR

alkylurea

(e) Reaction between gaseous CO and NO2
kf
CO(g) + NO 2 (g)

kb

CO 2 (g) + NO(g)

Limitations of the Collision Theory
The collision theory of reaction rates is logical and correct
...

(1) The theory applies to simple gaseous reactions only
...


CHEMICAL KINETICS

765

(2) The values of rate constant calculated from the collision theory expression (Arrhenius
equation) are in agreement with the experimental values only for simple bimolecular
reactions
...

(3) There is no method for determining the steric effect (p) for a reaction whose rate constant has
not been determined experimentally
...
There is no reason
why the rotational and vibrational energies of molecules should be ignored
...

The various drawbacks in the simple collision theory do not appear in the modern transition-state
theory
...
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
...
During the collision, the reactant molecules
form a transition state or activated complex which decomposes to give the products
...

The transition state theory may be summarised as follows :
(1) In a collision, the fast approaching reactant molecules (A and BC) slow down due to gradual
repulsion between their electron clouds
...

(2) As the molecules come close, the interpenetration of their electron clouds occurs which
allows the rearrangement of valence electrons
...
This leads to formation of an activated complex or transition state
...
20
...


766

20

PHYSICAL CHEMISTRY

Activated
complex
A

B

C
A

B

C

Potential energy

Ea

A + B—C
Reactants
A

B

ΔE
A—B + C
Products A

C

B

C

Reaction coordinate
Figure 20
...


Here the potential energy of the system undergoing reaction is plotted against the reaction
coordinate (the progress of the reaction)
...
The reactants must have this minimum energy
to undergo the reaction through the transition state
...
Also, the energy is obtained in passing from the transition state to the products
...
20
...
Thus such a reaction
will be exothermic
...
14
A potential energy diagram for an endothermic reaction
...


ACTIVATION ENERGY AND CATALYSIS
We know that for each reaction a certain energy barrier must be surmounted
...
20
...

Activated
complex
Activated
complex involving
catalyst

Potential energy

Ea
Ecat

Reactants
E
Products

Reaction coordinate
Figure 20
...


The catalyst functions by providing another pathway with lower activation energy, Ecat
...
Since the rate of reaction is
proportional to effective collisions, the presence of a catalyst makes the reaction go faster, other
conditions remaining the same
...


LINDEMAN’S THEORY OF UNIMOLECULAR REACTIONS
A number of unimolecular reactions, for example,
→
2N2O5(g) ⎯⎯ 4NO2(g) + O2(g)
are found to be of the first order
...
If two molecules must collide in order to provide necessary
activation energy, a second order rate law should result
...
During this time
lag, the activated molecules could either react or be deactivated
...
(1)
* ⎯⎯ A + A
→
A+A
deactivation
...
(3)
If the time lag is long, step (3) is slow, the reaction should follow first order kinetics
...


768

20

PHYSICAL CHEMISTRY

The proof of Lindeman’s theory is provided by studying the effect of change of pressure on the
reaction
...
On the other hand, at sufficiently
low pressure all the activated molecules will react before they can be deactivated
...
Several
gases are known to exhibit this behaviour
...


2
...


4
...


6
...

8
...

The first order rate constant for the decomposition of N2O2 of 0°C is 5
...
If the energy of
activation is 6200 joules per mole, calculate the rate constant at 25°C
...
7
...
Show that for first order
reactions the half life period is independent of the initial concentration
...
2 × 10–3 sec–1 at 303 K
...
12 kJ mol–1
...
1
...

(b) What is activation energy? How is it determined?
(a) For the kinetics of bimolecular reactions briefly discuss the “Collision Theory”
...
What is the order of such reactions?
(d) If the rate constant at one temperature alongwith the activation energy is given, how can the rate
constant at any other temperature be determined?
Explain Arrhenius equation
...
Give graphical representation
(Jiwaji BSc, 2000)
of activation energy diagram
...
Write rate law expression for it
...

(Allahabad BSc, 2001)
Distinguish between reaction rate and rate constant of a reaction
...

(kathmandu BSc, 2001)
(a) Explain briefly the collision theory of reaction rates
...

10
...


CHEMICAL KINETICS

769

(b) Write unit of rate constant for zero order reaction
...
(a) A reactant R is converted into product by the following mechanism :
R →x→P
where both the steps are of first order
...

(b) A solution containing equal concentrations of ethyl acetate and NaOH is 25% saponified in
5 minutes
...

(Vidyasagar BSc (H), 2002)
Answer
...
11 × 10–5
13
...
What
(Arunachal BSc (H), 2002)
is half-life period of the second order reaction?
14
...

(b) Derive an expression for the dependence of rate constant of a reaction with temperature
...

(d) What are zero order reactions? Give one example
...
If 15% of a substance decomposes in first ten minutes in a first order reaction, calculate how much of it
would remain undecomposed after one hour?
Answer
...
75%
(Vidyasagar BSc, 2002)
16
...

(Jamia Millia BSc, 2002)
17
...
Derive the rate equation for the first order reaction and show that :
(i) Half-life is independent of initial concentration
...

(MD Rohtak BSc, 2002)
19
...
25 × 10–3 at 303 K and 11
...
Calculate the
energy of activation of the reaction
...
53
...
(a) Describe the expression for the rate constant of the reaction A + B → Products
...

(b) Describe the graphical method for the determination of order of reaction
...
(HS Gaur BSc, 2002)
21
...
Calculate the specific reaction rate for the reaction
at 300 K, if the frequency factor is 2
...

Answer
...
3 × 10–3 sec–1
(Mumbai BSc, 2002)
22
...
How is the energy of activation
determined from the plot?
(b) The value of rate constant for the decomposition of nitrogen pentoxide (N2O5 → N2O4 + ½O2) is
4
...
87 × 10–3 at 65°C
...

(R = 8
...
103
...
(a) Discuss the collision theory of Bimolecular reactions
...

(Nagpur BSc, 2002)
24
...

(b) Give theory of absolute reaction rates
...
If the half-life of a first order reaction in A is 15 min
...
2990 sec
(Arunachal BSc, 2002)

770

20

PHYSICAL CHEMISTRY

26
...
70 × 10 –5 dm 3 mol –1 sec –1 at 25°C and
1
...
Calculate the activation energy of the reaction
...
314 JK–1 mol–1)
Answer
...
645 kJ
(Mizoram BSc (H), 2002)
27
...
On what factors does it depend? What is meant by collision
diameter?
(Guru Nanak Dev BSc, 2002)
28
...
54 × 10–3 sec–1
...

Answer
...
How is the order of the reaction determined by Differential method and Half-life method?
(Arunachal BSc, 2002)
30
...
How long will it take for the reaction to go to 75% completion?
Answer
...
Explain the following :
(i) If a reaction is pseudo order, the half life is always defined with respect to the species present in the
smallest amount
...

(Delhi BSc, 2002)
32
...

(b) Give one example each for pseudo unimolecular, second order, third order and zero order reactions
...
Explain, with examples, the functions of catalytic promoters and poisons in chemical reactions
...
The activation energy of a non-catalysed reaction at 37°C is 83
...
10 kJ mol–1
...

Answer
...
9975
(Kalyani BSc, 2003)
35
...
Express the value of k in cm3 mol–1 s–1
...
5 × 10–5 sec–2 and 9
...
Evaluate
the activation energy of the reaction
...
53
...
Write short notes on:
(i) Transition state theory of reaction rates
(ii) Parallel reactions
(iii) Opposing reactions
(Allahabad BSc, 2003)
37
...
The slope of the straight line was found to be –2
...

Calculate the activation energy of the reaction
...
21
...
Calculate the activation energy of a reaction whose reaction rate at 27°C gets doubled for 10° rise in
temperature (R = 8
...
53
...
What do you understand by the term: ‘rate determining step’ of a complex reaction? What is steady–state
hypothesis?
(Arunachal BSc, 2003)
40
...

(Delhi BSc, 2003)

CHEMICAL KINETICS

771

41
...
693
k

(ii) for a second order reaction t 1 =
2

1
k2 a

(Allahabad BSc, 2003)

42
...

44
...

46
...


48
...


50
...

52
...
What is the unit of k in the equation?
(Nagpur BSc, 2003)
Write down the rate and the differential rate expression for A + 2B → Products
...

(a) What are pseudo and true unimolecular reactions? Explain with examples
...
What is the half-life period of the reaction?
Answer
...

Also write how the equation you, derive, is employed in the graphical method to confirm second
order
...
What are the advantages of this
theory over collision theory?
(Guru Nanak Dev BSc, 2004)
(a) Derive a second order rate equation for the reaction
2 A → Product
...

(b) Calculate the activation energy of a reaction whose rate constant at 27°C gets doubled for 10°C rise
in temperature
...
(b) 12804
...

(b) 50% of a first order reaction is completed in 23 minutes
...

Answer
...
43 min
(Madras BSc, 2004)
A first order reaction is 15% complete in 20 minutes
...
112
...
052 × 10–3 sec–1
...
659 sec
(Indore BSc, 2004)
An acid solution of sucrose was hydrolysed to the extent of 54% after 67 minutes
...

Answer
...
4 min
(Allahabad BSc, 2005)

772

20

PHYSICAL CHEMISTRY

53
...
How long will it take for the reaction to go to 90% completion ?
Answer
...
For a certain reaction, it takes 5 minutes for the initial concentration of 0
...
25 mol lit–1 and another 5 minutes to became 0
...
(a) What is the order of the reaction ? (b)
What is the rate contant of the reaction ?
Answer
...
136 min–1
(Mizoram BSc, 2005)
55 If the half life of a first order in A is 2 min, how long will it take A to reach 25% of its initial concentration
...
4 min
(Delhi BSc, 2006)
56
...
If we start
[ A ] = 5
...
05 mol lit–1 ?
Answer
...
7676 min
(Mysore BSc, 2006)
57
...
Calculate the energy of activation for the reaction
...
83
...
Calculate the half life of a first order reaction where the specific rate constant is (a) 200 sec–1 (b) 2 min–1
...
(a) 0
...
3465 min–1
(Panjab BSc, 2006)

MULTIPLE CHOICE QUESTIONS
1
...
(d)
2
...
(d)
3
...
(d)
4
...
If this reaction takes place in a
sealed vessel and the partial pressure of nitric oxide is doubled, what effect would this have on the rate of
reaction?
(a) the reaction rate would triple
(b) the reaction rate would double
(c) the reaction rate would quadruple
(d) there would be no effect on the reaction rate
Answer
...


Which three factors affect the rate of a chemical reaction?
(a) temperature, pressure and humidity
(b) temperature, reactant concentration and catalyst
(c) temperature, reactant concentration and pressure
(d) temperature, product concentration and container volume
Answer
...
For first-order reactions the rate constant, k, has the unit(s)
(a) l mol–1
(b) time–1
–1 time–1
(c) (mol/l)
(d) time mol l–1
Answer
...
What are the units of the rate constant for a reaction in solution that has an overall reaction order of two?
(M is molarity, s is seconds
...
(a)
8
...


10
...


12
...


The reaction A → B is a second-order process
...
50 M, the
half-life is 8
...
What is the half-life if the initial concentration of A is 0
...
6 minutes
(b) 8
...
0 minutes
(d) 16
...
(c)
...
(a)
The first-order rate constant for the decomposition of N2O5 to NO2 and O2 at 70°C is 6
...

Suppose we start with 0
...
500 L container
...
5 min?
(a) 0
...
555 mol
(c) 0
...
162 mol
Answer
...
(d)
In a series of reactions, which is the rate-determining step?
(a) the simplest reaction
(b) the main reaction involving the major reactant
(c) the slowest reaction
(d) the fastest reaction
Answer
...
If 100 milliliters
of 0
...
(d)

774

20

PHYSICAL CHEMISTRY

14
...
0120 min–1
...
Calculate the rate
constant of the reaction at 525 K
...
11 × 10–7 min–1

(b)

202 min–1

(c) 8
...
2 min–1

Answer
...
Here is a second order reaction A → P
...
0818 M goes down 30
...
15
minutes, what is the rate constant for the reaction?
(a) 0
...
7 l mol–1 s–1

(c) 9
...
(b)
16
...
If the concentration of A is 0
...
0026 M

(b)

(c) 0
...
042 M

0
...
(d)
17
...
(a)
18
...
(d)
19
...
What is the order
of this reaction?
(a) zero order
(b) first order
(c) second order
(d) third order
Answer
...
To study the rate of a reaction, it is necessary to
(a) identify the reactants
(b) know the relative amounts of reactants used
(c) know the overall chemical equation for the reaction
(d) all of the above are necessary
Answer
...
Why do most chemical reaction rates increase rapidly as the temperature rises?
(a) the fraction of the molecules with kinetic energy greater than the activation energy increases rapidly
with temperature
(b) the average kinetic energy increases as temperature rises
(c) the activation energy decreases as temperature rises
(d) more collisions take place between particles so that the reaction can occur
Answer
...
The rate constant for a reaction depends upon each of the following, EXCEPT:
(a) solvent for solutions
(b) temperature
(c) concentration of reactants
(d) nature of reactants
Answer
...
All of the following are true statements concerning reaction orders EXCEPT:
(a) the rate of a zero-order reaction is constant
(b) after three half-lives, a radioactive sample will have one-ninth of its original concentration
(c) the units for the rate constant for first order reactions are sec–1
(d) if doubling the concentration of a reactant doubles the rate of the reaction, then the reaction is first
order in that reactant
Answer
...
The powers in the rate law are determined by
(a) the principle of detailed balance
(b) the physical states of the reactants and products
(c) experiment
(d) the coefficients in the balanced chemical reaction
Answer
...
Consider the reaction 3A → 2B
...
How is the average rate of appearance of B related
to the average rate of disappearance of A?
(a) –2[A]/3t
(b) [A]/t
(c) –[A]/t
(d) –3[A]/2t
Answer
...
Which one of the following is incorrect for the reaction A → B?
(a) the half-life of a second-order reaction depends on the initial concentration
(b) the half-life is the time for one half of the reactant to be consumed
(c) the second-order rate constant can be found by plotting 1/[A]2 versus time, where [A] is the
concentration of reactant
(d) the initial rates for a second-order reaction depend on the concentration of the reactant squared
Answer
...
Consider the reaction in which ammonia is synthesized from nitrogen and hydrogen gases :
N2(g) + 3 H2(g) → 2NH3(g)
How is the rate of formation of ammonia related to the rate of consumption of hydrogen?
(a) the rate of formation of ammonia is half the rate of consumption of hydrogen
(b) the rate of formation of ammonia is twice the rate of consumption of hydrogen
(c) the rate of formation of ammonia is equal to the rate of consumption of hydrogen
(d) the rate of formation of ammonia is two-thirds the rate of consumption of hydrogen
Answer
...
Which concentration plot is linear for a first-order equation? (A is one of the reactants)
...
(c)
29
...
(c)

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PHYSICAL CHEMISTRY

30
...

(a) decrease than increase
(c) increases
Answer
...
For the reaction 2NO2 + O3 → N2O5 + O2 the following observations are made :
Doubling the concentration of [NO2] doubles the rate, and doubling the concentration of [O3] doubles the
rate
...
(d)
32
...
(d)
33
...
This reaction is thus
(a) the molecularity cannot be determined from the given information
(b) termolecular
(c) bimolecular
(d) unimolecular
Answer
...
Why is a minimum energy needed for an effective collision?
(a) energy is needed to break bonds
(b) energy is needed to orient the particles correctly
(c) a minimum energy is needed, so that the particles will collide many times per second
(d) enough energy is needed to give off heat in a reaction
Answer
...
Species that are formed in one step of reaction mechanism and used up in another step are called
(a) catalysts
(b) intermediates
(c) inhibitors
(d) activated complexes
Answer
...
For a certain reaction, the rate = k [NO]2 [O2], when the initial concentration of NO is tripled, the initial
rate
(a) decreases by a factor of nine
(b) increases by a factor of three
(c) increases by a factor of six
(d) increases by a factor of nine
Answer
...
Which of the following statements associated with mechanisms of chemical reactions is incorrect?
(a) intermediates do not appear in the net chemical equation or overall rate law
(b) the first step in a mechanism always determines the rate of the reaction
(c) in elementary reactions, coefficients give the order with respect to reactants and products
(d) a plausible mechanism must account experimentally determined rate law
Answer
...
What happens in a steady state?
(a) product is being formed faster than reactants are regenerated
(b) heat is evolved

CHEMICAL KINETICS

777

(c) the concentration of an intermediate is constant
(d) nothing is happening
Answer
...
What happens when molecules collide with less than the activation energy needed for the reaction?
(a) they stick together but do not react
(b) they react, but more slowly
(c) they react if the bonds are arranged in the correct orientation
(d) they do not react; they simply bounce off of each other
Answer
...
Why do fine iron wires burst into flame when lighted, while an iron frying pan does not?
(a) the finely divided iron has lower internal free energy than the frying pan
(b) the frying pan has much greater mass than the finely divided wires
(c) the frying pan is more dense than the wires
(d) the wires have greater surface area, enabling more iron particles to react more quickly
Answer
...
Which of the following terms describes a process in which two particles come together to form an
activated complex?
(a) reaction mechanism
(b) elementary process
(c) rate determining step
(d) unimolecular
Answer
...
A + B + C → products is :
(a) unimolecular
(b) trimolecular
(c) bimolecular
(d) tetramolecular
Answer
...
Based on the collision model, the atoms at the top of the potential energy “hill” are called:
(a) top of the hill
(b) activation energy
(c) transition state
(d) steric factor
Answer
...
According to chemical kinetic theory, a reaction can occur
(a) if the reactants collide with the proper orientation
(b) if the reactants possess sufficient energy of collision
(c) if the reactants are able to form a correct transition state
(d) all of the above
Answer
...
What does termolecular refer to?
(b) 4 molecules colliding
(a) 2 molecules colliding
(c) 3 molecules colliding
(d) 1 molecule dissociating
Answer
...
What is the order of the absorption process for alcohol through the lining of the stomach and small
intestine?
(a) zero order

(b)

first order

(c) second order

(d)

third order

Answer
...
The decomposition of nitrogen dioxide to nitrogen and oxygen is second-order with a rate constant
k = 12
...
What is the half-life for the reaction if [NO2]0 = 0
...
0554 sec

(b)

30
...
5 sec

Answer
...
A first order reaction requires 8
...
0% of its
original value
...
48 months

(b)

2
...
96 months

(d)

17
...
(a)
49
...
34 M
...
12 M of the reactant
...
What is the average rate of the reaction?
(a) 4
...
33 ×

10–3

M/sec

(b)

2
...
(c)
50
...
If the concentration after 11,072 years is 0
...
0690 M

(b) 0
...
173 M

(d) 1
...
(b)
51
...
022
...
050 mol and 0
...
022 mol
(b) 0
...
028 mol
(d) 0
...
(d)
52
...
What is the rate constant for the decay of Bromine-87?
(a) 56 sec
(b) 6
...
24 × 10
(d) 6
...
(c)
53
...
0 hours
...
(d)
54
...
If the initial
concentration of NOBr is 0
...
0 min?
(a) 9
...
9 × 10–2 M
–2 M
(c) 4
...
(b)
55
...
What is the rate law for the reaction?
(a) rate = k [A]2 [B] [C]2
(b) rate = k [A] [C]2
2 [B] [C]
(c) rate = k [A]
(d) rate = k [A] [B] [C]
Answer
...
If we have the reaction A(g) → 2B(g) and the number of moles of A is as follows,
time
0
5 min
10 min

CHEMICAL KINETICS

779

moles A
0
...
085
0
...
030 mol
(b) 0
...
060 mol
(d) 0
...
(c)
57
...
50
5
...
50
10
...
5
[A]
0
...
389
0
...
236
0
...
143
(a) – 0
...
0164
(c) – 0
...
0184
Answer
...
A reaction requires 279 minutes in order to reach equilibrium
...
13 M to 0
...
What is the average rate of the reaction?
(a) 7
...
77 × 10–3 M/min
(c) 2
...
(b)
59
...
0620 M initially and 0
...
0 min
...
50 min
(b) 46
...
6 min
(d) 59
...
(d)
60
...
620 M initially and 0
...
0 min
...
09 × 10–3 min–1
(a) 9
...
0117 min–1
(d) 0
...
(c)
61
...
(d)
62
...
0150 min–1
...
400 M initially, what will [A] be
after 2
...
388 M
(b) 1
...
487 M
(d) 0
...
(d)
63
...
0 × 10–4 M–1 s–1
...
800 M?
(a) 1
...
671 M
(c) 0
...
300 M
Answer
...
In the reaction 2A + B → 2C + D, –Δ [A]/t is found to be 5
...
What is the rate of change of B?
(a) 2
...
0 M/min
Answer
...
The reaction A → P is a second-order process with t½ = 23
...
500 M
...
00 hour elapses?
(a) 0
...
139 M

780

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PHYSICAL CHEMISTRY

(c) 0
...
175 M
Answer
...
The thermal decomposition of N2O5(g) to form NO2(g) and O2(g) is a first-order reaction
...
1 × 10–4 s–1 at 318 K
...
9 × 103 s
(b) 2
...
4 × 103 s
(c) 1
...
(d)
67
...
(a)
68
...
(c)
69
...
(a)
70
...
(d)
71
...
46 s–1
...
(d)



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