Class: 12
Subject: Chemistry
Topic: Chemical kinetics
No. of Questions: 27
1.
Why we prefer instantaneous rate of reaction over average rate of reaction?
The rate of reaction decreases continuously with time except for a zero order reaction.
Therefore, average rate of reaction has no significance for the reaction. But
instantaneous rate of reaction for a given instant of time does not change with time.
2.
Define rate of a reaction.
Solution:
Rate of a reaction may be defined as the change in the concentration of any one of the
reactants or products per unit time.
3.
Define specific rate constant.
Solution:
It is defined as the rate of a chemical reaction when the concentration of each reactant
appearing in the rate equation is taken as unity.
4.
 2Cl  I2 .
The following reaction was carried out in water. Cl2  2I 
IT
ia
ns
Solution:
1
and concentration after 10 minutes was
kI
The initial concentration of I was 0.50 mol
1
0.46 mole lit . Calculate the rate of disappearance of I and rate of appearance of iodine.
 I   0.46  0.50  0.04 mol
1
t  10.0  0.0  10 min
as
Solution:

Rate of disappearance of I = 
= 0.004 mol
1
t

  0.04 
10
min1
Rate of appearance of iodine =
=
 I 
1
 0.004  0.002 mol
2
1
1
(Rate of disappearance of I)
2
min1
 AB2 , the rate constant is 1.26  103 L mol1s1. What is
For the reaction A  2B 
5.
the order of the reaction?
Solution:
The units of rate constant is L mol1s1 or (mol L1)1s1. Equate this with general
1n
expression of (mol L1) s1.
mol L 
1
1

s1  mol L1

1n
s1
 1 = 1  n or n = 2
The order of reaction = 2
6
For a reaction,
2NO2 + F2 
 2NO2 F
The experimental rate law is r  K NO2 F2  . Propose the mechanism of the reaction.
The rate law is r  K NO2 F2  since rate law is proportional to single power of NO2 and
ns
Solution:
ia
single power of F2, it implies that only one molecule of NO2 and one molecule of F2 are
involved in the slow step. Thus the various steps are:
Step 1
NO2  F2 
NO2F  F..............(slow)
NO2  F 
NO2F..............(fast)
Step 2
The experimental data for the reaction
2A  B2 
 2AB is
kI
7.
IT
2NO2  F2 
 2NO2F
Solution:
as
Experiment
[A] mole/lit
1.
0.50
2.
0.50
3.
1.00
Write the rate law equation.
[B2] mol/lit
0.50
1.00
1.00
Let the rate law equation be
r  K  A  B2 
x
y
Now,
r1  1.6  104  K  0.50  0.50 
x
y
r2  3.2  104  K  0.50  1.00 
x
r3  3.2  10
4
 K 1.00  1.00 
x
…(ii)
y
….(iii)
Divide (ii) by (i)
r2 K  0.50  1.00 
3.2  104


x
y
r1 K  0.50   0.50 
1.6  104
x
y
y
 1.00 
y
 0.50   2  2  2, y  1


…(i)
y
Initial rate
1.6  104
3.2  104
3.2  104
Now divide (iii) by (ii)
r3 K 1.00  1.00 
3.2  104


r2 K  0.50 x 1.00 y 3.2  104
x
y
x
 1.00 
 0.50   1


or 2x  1
x
2 =1
x=0
 Rate law = r  K  A  B2 
0
Solution:
(i)
Pt
2NH3  g 
N2  g  3H2  g
r = k[NH3]
N2  g  2H2O 
(ii) NH4NO2  s 
9

IT
r = k[NH4NO2]
ns
Give any one example of
(i) zero order reaction
(ii) first order reaction
ia
8.
1
For decomposition of N2O5 in CCl4 solution at 320 K.
2N2O5 
 4NO2  O2
Solution:
as
kI
Show that the reaction is of first order and also calculate the rate constant:
Time in min
10
15
20
25

Vol. of O2
6.30
8.95
11.40
13.50
34.75
Evolved (in ml)
If the reaction is of first order, it must obey the equation
2.303
a
K
log
t
ax
In the above reaction, NO2 remains in solution and oxygen is liberated and collected at
different intervals of time.
Therefore
Vt  x , V  a
Substituting these values in the first order equation.
V
2.303
K
log
t
V  Vt
Time t = 10
Time t = 15
2.303
34.75
log
 0.0198
10
34.75  6.30
2.303
34.75
K
log
 0.0198
15
34.75  8.95
K
Time t = 20
Time t = 25
2.303
34.75
log
 0.0198
20
34.75  11.40
2.303
34.75
K
log
 0.0198
25
34.75  13.50
K
Since the value of K comes out to be constant, the reaction is therefore is of first order.
Differentiate between the rate of reaction and the rate constant.
Sol.
(i)
Rate of reaction
It is defined as the change in
concentration of the reactant or
product with time, each divided by its
stoichimetric coefficient.
Rate constant
It is defined as the rate of
chemical reaction when the
concentration of each reactant
appearing in the rate equation is
taken as unity.
Its units depends on the order of
th
a reaction. For n order reaction
the units of rate constant
1n
= (conc) time1
It is independent of the initial
concentration of the reactants.
(i)
It always has a unit of (conc)/time.
(ii)
(iii)
It
depends
on
the
initial
concentration of the reactants.
(iii)
ia
(ii)
ns
10
Define order of a reaction. Can it be a fractional value? If yes then give an example of a
fractional order reaction.
Sol.
The order of a reaction is defined as the sum of the exponents to which each concentration
terms is raised in the experimently derived rate equation. For example in the reaction
A + B + C  Products
and the rate law expression is
r  k  A  B C
x
y
z
kI
IT
11
as
Then, overall order of reaction = x + y + z.
Order of a reaction can be fractional value. An example of a fractional order reaction is gasphase decomposition of CH3CHO.
CH3CHO g 
 CO g  CH4 g and r  k CH3CHO
3/2
Differentiate between order and molecularity of a reaction.
(i)
(ii)
(iii)
(iv)
It does not give an ideal of the
mechanism of the reaction.
(v)
(iii)
(iv)
(ii)
(iii)
(iv)
Chemical Reactions
This involves making and breaking of
bonds in order to rearrange the
atoms.
Only electrons in the atomic orbitals
are involved in the making and
breaking of bonds.
It involves the absorption or release
of relatively small amount of energy.
IT
(i)
Sol.
(ii)
It can be obtained by adding the
molecules of the slowest step.
Overall
molecularity
of
a
complex
reaction
has
no
significance; only slowest step is
significant.
It explains the mechanism of the
reaction.
How would you compare chemical reactions and the nuclear reactions?
Sol.
14
It need not be whole number it can
be factional or zero also.
It can only be experimentally
determined.
It is for the overall reaction and no
separate steps are written to obtain it.
kI
13
Molecularity of a reaction
In simple reactions, it is equal to
the number of molecules of the
reactants while in complex
reactions, it is number of
molecules involved in rate
determining step.
Always a whole number.
(i)
ia
(v)
Order of a reaction
It is sum of the exponents to which
the concentration terms in the rate
law expression are raised to express
observed rate of reaction.
ns
Sol.
as
12
Rates of reaction are influenced by
temperature,
pressure,
concentrations and catalyst.
(i)
(ii)
(iii)
(iv)
Nuclear Reactions
This involves the conversion of
one element into another.
It involves protons, neutrons,
electrons, and other elementary
particles.
These
reactions
are
accompanied by absorption or
release of tremendous amounts
of energy.
Rates of reaction normally are
not affected by temperature,
pressure, and catalyst.
Calculate the rate constant of a reaction at 293 K when the energy of activation is
103 KJ mol1 and the rate constant at 273 K is 7.87 ×10 -7 sec -1 .
Ea  T2  T1 
K2



K1 2.303R  T1T2 
K1  7.87  107 S1; Ea  103KJ mol1
log10
T1 = 273K, T2 = 293K
K2
103  20
log10

7
7.87  10
2.303  8.314  293  273  103
K2
log10
 1.345
7.87  107
K 2  1.74  105 sec 1
15
The rate law for the reaction,
2Cl 2 O 
 2Cl 2 + O2 at 200C is found to be rate = K Cl 2 O  ,
2
(a)How would the rate change if [Cl2O] is reduced to one third of its original value?
(b)How should the [Cl2O] be charged in order to double the rate?
(c)How would the rate change if [Cl2O] is raised to three fold of its original value?
Sol.
(a) Rate equation for the reaction
r  K Cl2O
2
Let new rate be r; so
2
ns
1
 Cl O 
r'  K 2   r
9
 3 
(b) In order to have the rate = 2r, let the concentration of Cl2O be x.
2
so 2r = Kx
…(i)
2
we know that r = K Cl2O
…(ii)
2
2
2 Cl2O
or x =
IT
x = 2 Cl2O
ia
Dividing equation (i) and (ii)
2r
kx 2
x2


2

2
2
r
K Cl2O
Cl2O
(c) New rate = K 3Cl2O  9K Cl2O  9r nine times of original rate.
2
2
Explain in brief the collision theory of reaction rates.
Sol.
According to this theory, the reaction takes places as a result of the collision between the
reactant molecules. The number of collisions per unit volume per unit time is known as collision
frequency ZAB. But all collisions are not effective. The collision which actually converts the
reactants into products are called effective collisions. These collisions takes the molecules to
the top of the energy barrier and finally results in the formation of products.
There are two conditions for the effective collision:
(i) Energy barrier: The reactant molecules must posses the minimum energy known as
threshold energy before they could react and form products.
(ii) Orientation barrier: The reactant molecules must be properly oriented in order to have an
effective collision.
The rate constant is given by
k  PZAB eEa / RT
Where ZAB = collision frequency
P = orientation factor and eEa / RT = fraction of total collision which are effective.
as
kI
16.
0
1
e and 01 ?
What is the difference between
Sol.
The symbol 01 e represents the electron in or from an atomic orbital. The symbol 01
represents an electron, that although physically identical to any other electron, comes from a
nucleus and not form an atomic orbital.
The emission of 01 involves the conversion of a neutron into a proton.
18
How are the radioactive decay series distinguished? Which one of the decay series is not
natural but artificial?
Sol.
A radioactive decay series is a sequence of nuclear reactions that ultimately result in the
formation of stable isotope.
There are four decay series distinguished by whether the mass number are
(i) divided by four (4n series) or
(ii) divided by four with remainder equals to one (4n + 1)
(iii) divided by four with remainder equals to two (4n + 2)
(iv) divided by four with remainder equals to three (4n + 3)
237
The (4n + 1) series in not natural but artificial series and it starts with 93
Np and ends with
.
19.
Explain the following:
(i) Mass defect (ii) Binding energy
209
83
Bi
ia
ns
17
(i) Mass defect is the difference between the actual mass of an isotope of an element and the
sum of the masses of protons, neutrons and electrons present in it.
(ii) Binding energy is the energy required to hold the nucleus together. Binding energy of a
nucleus is generally quoted as energy in million volts (MeV) per nucleons.
The binding energy per nucleon is a measure of the stability of the nucleus. The greater the
binding energy per nucleon, the more stable is the nucleon.
20.
Decomposition of N2O5 is expressed by the equation, N 2 O5 
 2NO 2 +
kI
1
O
2 2
If during a certain time interval, the rate of decomposition of N 2O5 is 1.8  103 mol lit 1 min 1 ,
as
Sol.
IT
Sol.
what will be the rates of formation of NO2 and O2 during the same interval?
The rate expression for the decomposition of N2O5 is
 N2O5  1  NO2  2. O2 


t
2 t
t
 NO2  2. N2O5 

 2  1.8  103
So
t
t
 3.6  103 mol lit 1 min1
1  N2O5  1

  1.8  103
t
2
t
2
3
1
1
 0.9  10 mol lit min
and
 O2 
21.
The conversion of molecules X to Y follows the second order of kinetics. If concentration
of X is increased 3 times, how will it affect the rate of formation of Y.
Sol:
Rate
22.
The rate law for a reaction is
[ ][ ]
Rate
Can the reaction be an elementary process? Explain.
No, an elementary process would have a rate law with orders equal its molecularities and
therefore must be integral form.
ns
If the decomposition of nitrogen oxide is represented as
2 N2O5  4 NO2  O2
follows a first order kinetics.
ia
23.
(i)
Calculate the rate constant for a 0.05 M solution if the instantaneous rate is 1.5 ×
10-6 mol/l/s.
Sol:
Rate
[
]
What concentration of
as
(ii)
]
kI
[
IT
Sol:
[ ]
[ ]
[ ]
The rate of formation will become nine times.
would give a rate of 2.45 ×10-5 mol L-1 S-1
Rate
24.
Write the difference between order and molecularity of reaction.
ORDER
It is the sum of the powers of
concentration terms in the rate law
expression.
It is determined experimentally
Order of reaction need not be a whole
number
Order of reaction can be zero.
MOLECULARITY
It is the number of reacting species
undergoing simultaneous collision in a
reaction.
It is a theoretical concept
It is whole no. only
It can’t be zero or fractional
25.
A first order reaction takes 69.3 min for 50% completion. Set up an equation for
determining the time needed for 80% completion.
Sol:
min-1
[
]
[ ]
in
The decomposition of NH3 on platinum surface is zero order reaction.
What are the rate of production of N2 and H2? The rate constant is
2.5 104 mol1 sec1
Sol:
2NH3  N2 +3H2
d  NH3 
 Rate  k x  NH3 
0
1 d[NH3 ]
2
dt
as

kI
dt
2.5 104 mol1 sec1
d  N2 
ia
1 d[NH3 ] d  N 2  1 d  H 2 

=
2
dt
dt
3 dt
IT

ns
26.
dt
1
  2.5 104 molL1 sec1
2
d  H2 
3 d  NH3
0
 k x  NH3
dt
2
dt
4
 3.75 10 molL1 sec1
d  NH3
0
 k X  NH3
Rate 
dt
4
2.5 10 molL1 sec1
Rate of production of N2  2.5 104 molL1 sec1

27.
Time required to decompose
to half of its initial amount is 60 minutes. If the
decomposition is a first order reaction, calculate the rate constant of the reaction.
Sol:
We know that for a 1st order reaction,
0.693
t1/2 
k
It is given that t1/2  60 min
0.693
∴k 
t1/2
0.693

60
ns
Min-1
Min-1
as
kI
IT
ia
Or