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CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 15 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Chemical Kinetics
Quiz 15
For the following first-order competing reaction:
A
+
R
e
a
g
e
n
t
→
p
r
o
d
u
c
t
B
+
R
e
a
g
e
n
t
→
P
r
o
d
u
c
t
the ratio of
K
1
/
K
2
, if only
50
% of
B
will have been reacted when
94
% of
A
has been reacted is
(
log
2
=
0.3
,
log
3
0.48
)
Report Question
0%
4.06
0%
0.246
0%
8.33
0%
0.12
In the given first-order sequential reactions:
A
K
1
→
B
K
2
→
C
K
3
→
D
, what is the ratio of number of atoms of
A
to the number of atoms of
B
after long time interval starting with pure
A
?
(
K
1
ln
2
1200
;
K
2
=
ln
2
30
)
Report Question
0%
0.67
0%
10
0%
20
0%
40
Two reactions: (I)
A
→
p
r
o
d
u
c
t
s
and (II)
B
→
p
r
o
d
u
c
t
s
, follow first-order kinetics. The rate of reacion-I is doubled when temperature is raised from
300
to
310
K
. The half-life for this reaction at
310
K
is
30
m
i
n
. At the same temperature,
B
decomposes twice as fast as
A
. If the energy of activation for the reaction II is half that of reaction I, the rate constant of reaction II at
300
K
is
Report Question
0%
0.0233
m
i
n
−
1
0%
0.0327
m
i
n
−
1
0%
0.0164
m
i
n
−
1
0%
0.0654
m
i
n
−
1
Explanation
For reaction (1)
A
→
P
r
o
d
u
c
t
s
T
1
= 300K ,
T
2
= 310K
Given :
(
k
2
(
310
)
k
1
(
300
)
)
A
=
2
Using Arrhenius equation for reaction (1):
l
o
g
(
k
2
(
310
)
k
1
(
300
)
)
A
=
E
a
(
A
)
2.303
R
(
T
2
−
T
1
T
2
T
1
)
l
o
g
2
=
f
r
a
c
E
a
(
A
)
2.303
R
(
T
2
−
T
1
T
2
T
1
)
(For reaction 2)
B
→
Products
Since at 310 K, B decomposes twice as fast as A.
\because [K_{2}(310K)]_{A} = 0.0231 min^{-1}
\because [K_{2}(310K)]_{B} = 2 \times 0.0231 min^{-1}
Using Arrhenius equation for reaction (2)
log (\frac{k_{2}(310)}{k_{1}(300)})_{B} = \frac{E_{a}(B)}{2.303R} (\frac{T_{2}-T_{1}}{T_{2}T_{1}})
Given :
(E_{a})_{B} = \frac{1}{2}(E_{a})_{A}
Substittute the value of above equation and Comparing equations , we get
(\frac{k_{2}(310)}{k_{1}(300)})_{B} = \frac{1}{2} \times log 2
(\frac{0.0462}{k_{1}(300)})_{B} = 1.414
\therefore {k_{1}(300)} = 0.0327 min^{-1}
Surface-catalysed reactions that are incorporated by the product, obey the differential rate expression,
\cfrac { dy }{ dt } =\cfrac { k\left[ { C }_{ 0 }-y \right] }{ 1+by }
, where
{ C }_{ 0 }=
initial concentration and
k
and
b
are constants. The half-life of reaction is
Report Question
0%
\left( 1+{ C }_{ 0 }b \right) \ln { 2 } -\cfrac { { C }_{ 0 }b }{ 2 }
0%
\left( 1-{ C }_{ 0 }b \right) \ln { 2 } +\cfrac { { C }_{ 0 }b }{ 2 }
0%
\left( 1-{ C }_{ 0 }b \right) \ln { 2 } -\cfrac { { C }_{ 0 }b }{ 2 }
0%
\left( 1+{ C }_{ 0 }b \right) \ln { 2 } +\cfrac { { C }_{ 0 }b }{ 2 } \quad
A substance undergoes first-order decomposition. The decomposition follows two parallel first-order reaction with
{ K }_{ 1 }=1.26\times { 10 }^{ -4 }{ s }^{ -1 }
for the formation of
B
and
{ K }_{ 2 }=3.15\times { 10 }^{ -4 }{ s }^{ -1 }
for the formation of
C
. The percentage distribution of
B
and
C
are
Report Question
0%
29
%
B
,
71
%
C
0%
75
%
B
,
25
%
C
0%
90
%
B
,
10
%
C
0%
60
%
B
,
40
%
C
Rate of an uncatalyzed first-order reaction at
T
K
is half of the rate of a catalyzed reaction at
0.5T
T
. If the catalyst lowers the threshold energy by
20\ kcal
, what is the activation energy of the uncatalyzed reaction?
\left( T=300K,\ln { 2 } =0.7 \right)
Report Question
0%
39.58\ kcal/mol
0%
19.58\ kcal/mol
0%
40.42\ kcal/mol
0%
20.42\ kcal/mol
Explanation
Given ,
\frac{k_{2}}{k_{1}} = 2
T_1 = 300\ K
T_{2} = \frac{T_{1}}{2}
As we know,
log(\frac{k_{2}}{k_{1}}) = \frac{E_{2}-E_{1}}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})
Putting values in the above equation, we get;
log(2) = \frac{20-E_{1}}{1.987}(\frac{2}{T_1}-\frac{1}{T_1})
0.7 = \frac{20-E_{1}}{1.987}(\frac{2}{T_1}-\frac{1}{T_1})
After solving, we get ;
Activation energy of the uncatalyzed reaction,
E_{1} = 39.58\ kcal/mol
Hence , option A is correct .
A first-order reaction:
A\rightarrow B
, activation energy is
4.8kcal/mol
. When a
20
% solution of
A
was kept at
{ 27 }^{ o }C
for
21.6min
,
75
% decomposition took place. What will be the percent decomposition in
8.0min
in a
30
% solution maintained at
{ 47 }^{ o }C
? Assume that activation energy remains constant in this range of temperature.
(e=2.7)
Report Question
0%
25
%
0%
50
%
0%
75
%
0%
87.5
%
Decomposition of a non-volatile solute
A
into another non-volatile solute
B
and
C
, in aqueous solution follows first-order kinetics as:
A\rightarrow 2B+C
When one mole of
A
is dissolved in
180g
water and left for decomposition, the vapour pressure of solution was found to be
20mm
Hg
after
12h
. What is the vapour pressure of solution after
24h
?
Assume constant temperature of
{ 25 }^{ o }C
, throughout. The vapour pressure of water at
{ 25 }^{ o }C
is
24mm
Hg
Report Question
0%
18mm
Hg
0%
19.2mm
Hg
0%
10mm
Hg
0%
16mm
Hg
What is overall activation energy of reaction, if steps (1) to (4) are much faster than (5)?
Report Question
0%
\cfrac { { E }_{ { a }_{ 5 } }.{ E }_{ { a }_{ 3 } }.{ E }_{ { a }_{ 1 } }^{ 1/2 } }{ { E }_{ { a }_{ 4 } }.{ E }_{ { a }_{ 2 } }^{ 1/2 } }
0%
{ E }_{ { a }_{ 5 } }+{ E }_{ { a }_{ 3 } }+\cfrac { 1 }{ 2 } { E }_{ { a }_{ 1 } }-{ E }_{ { a }_{ 2 } }- { E }_{ { a }_{ 4 } }
0%
{ E }_{ { a }_{ 1 } }+{ E }_{ { a }_{ 3 } }+{ E }_{ { a }_{ 5 } }-{ E }_{ { a }_{ 2 } }-{ E }_{ { a }_{ 4 } }\quad
0%
-{ E }_{ { a }_{ 5 } }-{ E }_{ { a }_{ 3 } }-\cfrac { 1 }{ 2 } { E }_{ { a }_{ 1 } }+{ E }_{ { a }_{ 2 } }+\cfrac { 1 }{ 2 } { E }_{ { a }_{ 4 } }
Arrhenius equation gives the change in rate constant (and hence rate of reaction) with temperature. If the activation energy of the reaction is found to be equal to
RT
, then
Report Question
0%
The rate of reaction does not depend upon initial concentration
0%
The rate constant becomes about
35
% of the Arrhenius constant
A
0%
The rate constant becomes equal to
73
% of the Arrhenius constant
A
0%
The rate of the reaction becomes infinite or zero
Explanation
k = Ae^{\dfrac{-E_{a}}{RT}}
when
E_{a} = RT
k = \dfrac{A}{e}
\Rightarrow k=0.37 \times A
Hence the rate constant becomes approx. 35% of A.
The plot of concentration of a reactant vs time for a chemical reaction is shown below :
The order of this reaction with respect to the reactant is:
Report Question
0%
0
0%
1
0%
2
0%
not possible to determine from this plot
Explanation
The order of this reaction with respect to the reactant is zero-order reaction is reaction in which the rate does not vary with the increase or decrease in the concentration of the reactants.
Therefore the rate of these reactions is always equal to the rate constant (
K
) of the specific reactions (since the rate of these reactions is proportional to the 0th power of reactants concentration).
The Differential form of a zero-order reaction can be written as
Rate
=\dfrac{dA}{dt}=k[A]^{0}=k
Hence option A is correct.
Under what condition the order of the reaction,
2HI(g)\rightarrow H_2(g)+I_2(g)
, is zero.
Report Question
0%
At high temperature
0%
At high partial pressure of HI
0%
At low partial pressure of HI
0%
At high partial pressure of
H_2
0%
At high partial pressure of
I_2
Explanation
The reaction,
2HI(g)\overset{\Delta}{\longrightarrow}H_2(g)+I_2(g)
is of zero-order in which HI is present at high partial pressure.
Desorption of a gas from metal surface follows first-order kinetics. The rate constant of desorption can be given by Arrhenius equation. If the desorption of hydrogen on manganese is found to increase
10
times on increasing the temperature from
600
to
1000 K
, the activation energy of desorption is :
Report Question
0%
6.0\ kcal/mol
0%
6.9\ kcal/mol
0%
3.0\ kcal/mol
0%
57.4\ kcal/mol
For a first-order reaction,
Report Question
0%
The degree of dissociation is equal to
\left(1-{e}^{-kt}\right)
0%
A plot of reciprocal concentration of the reactant vs time gives a straight line
0%
The time taken for completion of
75\%
reaction is thrice the
{t}_{1/2}
of the reaction
0%
The pre-exponential factor in the Arrhenius equation has the dimension of time,
{T}^{-1}
Explanation
(a) is correct because degree of dissociation =
1-e^{-kt}
at any time t.
(b) is wrong because plot of
\log{\left[A\right]}
vs
t
is a straight line
(c) is wrong because time taken for
75\%
reaction is two half life.
(d) is correct because in
k = Ae^{{-E_{a}}/{RT}}
,
E_{a/RT}
is dimensionless hence
A
has the unit of
K
.
Option A and D are correct.
For the combustion of carbon,
\quad \Delta H=-ve
and
\Delta S=+ve
and hence, thermodynamically the process is spontaneous at all temperatures. But coal stored in coal depots does not burn automatically because of
Report Question
0%
very high threshold energy barrier
0%
thermodynamically stability of coal
0%
lower energy of activation needed for burning
0%
low temperature in coal depots
Explanation
\Delta G = \Delta H - T \Delta S
As
\Delta H
= -ve
As
\Delta S
= +ve
since it is given the process is spontaneous
\Delta G
will be more -ve.
\therefore \Delta G
must be -ve.
\Delta G = -nFE
Hence it requires very high threshold energy barrier.
Which of the following curves represents a zero order reaction?
Report Question
0%
0%
0%
0%
Which of the following curves represents a first order reaction?
Report Question
0%
0%
0%
0%
Explanation
t_{1/2}
is independent of
a
for the first order reaction. So, it will be a straight line parallel to concentration axis.
The reaction
A+B\to C+D;\ \Delta H=25\ kJ/ mole
should have an activation energy :
Report Question
0%
-25\ kJ/ mole
0%
< 25\ kJ/ mole
0%
> 25\ kJ/ mole
0%
either answer
(B)
or
(C)
depending upon experiment
Explanation
Option C is correct
Which of the following statements is wrong about reactions?
Report Question
0%
There can be only three values of molecularity, that is
1, 2,
and
3
.
0%
There can be only four values of order, that is ,
0, 1, 2,
and
3
.
0%
There can be infinite number of values for order.
0%
The order involves rate while molecularity does not
Explanation
The order can be anything
-ve
fractional,
O
or
+ve
. Its maximum value is
3
But as we know between any two real no. infinite no are possible & since order can be fractional / any real no. so it has infinite values.
50\%
of a zero order reaction completes in
10
minutes.
100\%
of the same reaction shall complete in:
Report Question
0%
5\ min
0%
10\ min
0%
20\ min
0%
\infty
time
Explanation
For zeroth order reaction:
At=Ao-Kt
;
After
50\%
completion, At
=\dfrac{Ao}{2}
A/q, \dfrac{Ao}{2}=Ao-K\times 10\Rightarrow K=\dfrac{Ao}{20}
Now for
100\%
completion
At=0
0=Ao-\dfrac{Ao}{20}\times t\Rightarrow t=20\min
The rate for a first order reaction is
0.6932\times 10^{-2}\ mol\ L^{-1}\ min^{-1}
and the initial concentration of the reactant is
0.1\ M. t_{\tfrac{1}{2}}
is equal to:
Report Question
0%
0.6932\times 10^{-2}\ min
0%
0.6932\times 10^{-3}\ min
0%
10\ min
0%
6.932\ min
Explanation
Rate
=0.6932\times 10^{-2}mole/lit-\min =KA_o
A_{0}=0.1\ M, t/1/2=\dfrac{0.693}{K}=\dfrac{0.693}{\dfrac{0.693\times 10^{2}}{10^{-1}}}=10\min
Hence, option C is correct.
For a zero order reaction with the initial reactant concentration a, the time for completion of the reaction is
Report Question
0%
k/a
0%
a/k
0%
2k/a
0%
a/2k
Explanation
For zero order reaction .
dQ/dt=k
Q= concentration of reactant at time t .
And k = rate. Constant.
On integrating above equation
Q - Qo = k ( t-to)
Qo = initial concentration and to = start time
So at t= 0; Q= a
At end of the reaction, reactant Concentration=0
a = kt
So, t = a/ k
Which of the following curves represent(s) a zero-order reaction ?
Report Question
0%
0%
0%
0%
In which of the following reactions of the following orders the molecularity and order can never be same?
Report Question
0%
Zero order
0%
First order
0%
Second order
0%
Third order
Explanation
Option A is correct because molecularity can never be zero.
So, for zero-order reaction, molecularity and order can never be same.
A plot of reactant concentration versus time for a reaction is a straight line with a negative slope giving the rate constant, and the intercept, giving the initial concentration of the reactant. The order of the reactant is:
Report Question
0%
0
0%
1
0%
2
0%
None of these
For a reaction of the order of
0.5
, when the concentration of the reactant is doubled the rate
Report Question
0%
doubles
0%
increases four times
0%
decrease four times
0%
increases
\sqrt{2}
times
Explanation
The expression for the rate of the reaction is
r=k[A]^n
now,
r=k[A]^{1/2}......(1)
When the concentration of A is doubled, then, rate:
r′=k[2A]^{1/2}......(2)
Dividing (2) by (1), we get
\dfrac{k[2A]^{1/2}}{k[A]^{1/2}}=\sqrt2
The rate constant of a first-order reaction is
6.93 \times 10^{-2} min^-
. The half-life of the reaction is
Report Question
0%
10 \min
0%
100 \min
0%
1 \min
0%
1000 \min
Explanation
As we know that Half-life of first order reaction is given by:
t_{1/2} = \cfrac{0.693}{k}
\implies t_{1/2} = \cfrac{0.693}{6.93 \times 10^{-2}}
\implies t_{1/2} = 10 \ min
Hence, Option "A" is the correct answer.
For a first order reaction
A\xrightarrow [ ]{ k } B
, the degree of dissociation is equal to
Report Question
0%
e^{-kt}
0%
1-e^{-kt}
0%
e^{kt}
0%
1+e^{-kt}
The order w. r. t A is :
Report Question
0%
1
0%
2
0%
3
0%
-1
Explanation
R = k [A]^m [B]^n
5 \times 10^{-4} = k[2.5 \times 10^{-4}]^m [3 \times 10^{-5}]^n
.....(i)
4 \times 10^{-3} = k[5 \times 10^{-4}]^m [6 \times 10^{-5}]^n
.......(ii)
1.6 \times 10^{-2} = k[1 \times 10^{-3}]^m [6 \times 10^{-5}]^n
.......(iii)
Using equation (ii) and (iii) we get,
m = 2.
Option B is correct.
The time for the half-life period of a certain reaction
A\to
products is
1
hour. When the initial concentration of the reactant,
A
is
2.0\ mol\ L^{-1}
. How much time does it take initial concentration to make from
0.50
to
0.25\ mol\ L^{-1}
if it is a zero-order reaction?
Report Question
0%
4\ h
0%
0.5\ h
0%
0.25\ h
0%
1\ h
Explanation
The initial concentration of A is
\mathrm{2 \ M}
.
If the reaction would be first order reaction, then
\mathrm{k = \cfrac{a_o - 0.5a_o}{t} = 1}
Time taken for the reactant concentration to decrease from 0.5 to 0.25;
\mathrm{t = \cfrac{0.5 - 0.25}{k} = \cfrac{0.25}{1}}
So, the time taken for reactant concentration to decrease from 0.5 M to 0.25 M is
0.25 \ h
.
Hence, Option "C" is the correct answer.
The number of collisions depend upon
Report Question
0%
Pressure
0%
Concentration
0%
Temperature
0%
All the above
Explanation
Number of collision depend upon pressure, concentration and temperature.
Option D is correct.
The rate constant is doubled when temperature increases from
27^{o}C
to
37^{o}C
. Activation energy in
kJ
is:
Report Question
0%
34
0%
53.6
0%
100
0%
50
Explanation
\log{\dfrac{K_2}{K_1}}=\dfrac{E_a}{2.303R}\left[\dfrac{1}{T_1}-\dfrac{1}{T_2}\right]
If
\dfrac{K_2}{K_1}=2
\log{2}=\dfrac{E_a}{2.303\times8.314}\left[\dfrac{1}{300}-\dfrac{1}{310}\right]
{E}_{a}
=
.3010\times2.303\times8.314\left(\dfrac{300\times310}{10}\right)
=
53598.59 J{mol}^{-1}
=
53.6kJ
The energy of activation is
Report Question
0%
The energy associated with the activated molecules
0%
Threshold energy energy of normal molecules
0%
Threshold energy + energy of normal molecules
0%
Energy of products energy of reactants
Explanation
Activation energy is the energy needed by reactant molecules to gain threshold energy level.
Activation energy is given by the formula
Report Question
0%
\log{\frac{K_2}{K_1}}=\frac{E_a}{2.303R}\left[\frac{{T_2}-{T_1}}{{T_1}{T_2}}\right]
0%
\log{\frac{K_1}{K_2}}=-\frac{E_a}{2.303R}\left[\frac{{T_2}-{T_1}}{{T_1}{T_2}}\right]
0%
\log{\frac{K_1}{K_2}}=-\frac{E_a}{2.303R}\left[\frac{{T}_{1}-{T_2}}{{T_1}{T_2}}\right]
0%
None of these
Explanation
log_eK_2 = log_eA - \cfrac{E_A}{RT_2}
...........(1)
log_eK_1 = log_eA - \cfrac{E_A}{RT_1}
..............(2)
Eqn. (1) - (2) we get,
\log{\cfrac{K_2}{K_1}}=\cfrac{E_a}{2.303R}\left[\cfrac{{T_2}-{T_1}}{{T_1}{T_2}}\right]
Option A is correct.
A graph between
t_{1/2}
and concentration for
n^{th}-order
reaction is a straight line. The reaction of this nature is completed
50\%
in
10
minutes when concentration is
2\ mol. L^{−1}.
This is decomposed
50\%
in t minutes at
4
mol. L^{−1}
. Then
n
and
t
are respectively:
Report Question
0%
0, 20 min
0%
1, 10 min
0%
1, 20 min
0%
0, 5 min
Explanation
It can be observed from the given graph that
t_{1/2}
is constant and is independent of the concentration. We know that,
t_{1/2}∝(1/a_o)^{n−1}
---- 1
where,
a_o
is the initial concentration of the reactant
n is the order of the reaction.
Now for first order reaction
n=1
and
t_{1/2}
will be independent of
a_o
(i.e. concentration) as depicted in the graph.
Hence, the reaction is of
\text{first order.}
As,
t_{1/2}
is independent of initial concentration, it will be the same whether the concentration is
2\ mol /L
or
4\ mol /L
So,
n=1
and
t=10
minutes (as already given in the question)
Option B is correct.
The energy of activation for reaction
(KJ/mol)
is :
Report Question
0%
20.83
0%
13.83
0%
15.23
0%
10.23
8 gm of the radioactive isotope, cesium-137 were collected on February 1 and kept in a sealed tube. On July 1, it was found that only 0.25 gm of it remained. So the half-life period of the isotope is:
Report Question
0%
37.5 days
0%
30 days
0%
25 days
0%
50 days
Explanation
t = Feb 1 to July 1 = 28 + 31 + 30 + 31 + 30 + 31 = 150 days
\lambda =\dfrac{2.303}{150} log \dfrac{8}{0.25} = \dfrac{2.303}{150}log2^5 =\dfrac{0.693}{30}day^{-1}
t_{1/2} = \dfrac{0.693}{0.693/30} = 30 days
For a zero order reaction
Report Question
0%
The concentration of the reactant does not change during the reaction
0%
The concentration change only when the temperature changes
0%
The rate remains constant throughout
0%
The rate of the reaction is proportional to the concentration
Explanation
Rate of
\textbf {zero order}
reaction is independent of the concentration of the reactant and remains constant throughout the reaction.
hence option (C) is correct.
Half-life of a radioactive substance which disintegrates by
75\%
in 60 minutes, will be
Report Question
0%
120 min
0%
30 min
0%
45 min
0%
20 min
Explanation
1st half- Life → 50% decay
2nd half- Life → 75% decay
3rd half- Life → 87.5% decay
4th half- Life → 93.75% decay
Thus, the half-life period of the radioactive element, if
75
% of it disintegrates in
60
min, is
\dfrac{60}{2}
=
30 \;Minutes.
Hence option (B) is correct.
A zero order reaction is one whose rate is independent of_________________________
Report Question
0%
Temperature of the reaction
0%
The concentrations of the reactants
0%
The concentration of the products
0%
The material of the vessel in which the reaction is carried out
Explanation
The rate of zero order reaction is not depend on the concentration of the reactants.
Rate = K[A]^0
Certain bimolecular reactions which follow the first order kinetics are called_________________________
Report Question
0%
First order reactions
0%
Unimolecular reactions
0%
Bimolecular reactions
0%
Pseudounimolecular reactions
Explanation
It is the characteristic of pseudo-unimolecular reactions.
Half-life of 10 gm of radioactive substance is 10 days. The half-life of 20 gm is:
Report Question
0%
10 days
0%
20 days
0%
25 days
0%
Infinite
Explanation
Half-life period is independent of initial amount. Hence,
t_{1/2}
of 20 gm is 10 days.
The
\Delta H
value of the reaction
H_2 + Cl_2 \rightleftharpoons 2HCl
is
-44.12 \,kcal
. If
E_1
is the activation energy of the products, then for the above reaction
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0%
E_1 > E_2
0%
E_1 < E_2
0%
E_1 = E_2
0%
\Delta H
is not related to
E_1
and
E_2
0%
None is correct
Explanation
\Delta {H} = E_2- E_1= Negative
E_1
is the activation energy of the products
E_2
is the activation energy of the reactants.
Thus ,
E_1>E_2
Option (A) is correct.
Half-life period of a zero order reaction is:
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0%
Inversely proportional to the concentration
0%
Independent of the concentration
0%
Directly proportional to the initial concentration
0%
Directly proportional to the final concentration
Explanation
t_{1/2}
of zero-order reaction is independent of the concentration.
Hence, option B is correct.
If 2.0 g of a radioactive isotope has a half-life of 20 hr, the half-life of 0.5 g of the same substance is:
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0%
20 hr
0%
80 hr
0%
5 hr
0%
10 hr
Explanation
Half-life period is a characteristic of radioactive isotope which is independent of initial concentration.
A radioactive isotope decays at such a rate that after 96 minutes only
\dfrac{1}{8}th
of the original amount remains. The half-life of this nuclide in minutes is:
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0%
12
0%
24
0%
32
0%
48
Explanation
\lambda = \dfrac{2.303}{t}log\dfrac{N}{N} = \dfrac{2.303}{96} log\dfrac{1}{1/8} = 0.0216
\therefore \ t_{1/2} = \dfrac{0.693}{\lambda} = 32.0 min
Number of
\alpha
-particles emitted per second by a radioactive element falls to 1/32 of its original value in 50 days. The half-life-period of this elements is:
Report Question
0%
5 days
0%
15 days
0%
10 days
0%
20 days
Explanation
T = 50 \ days, t_{1/2} = ?, N_0 = 1, N = \dfrac{1}{32}
,
N = N_0 \times \Big(\dfrac{1}{2}\Big)^n
or
\dfrac{1}{32} = 1 \times \Big(\dfrac{1}{2}\Big)^n
,
or
\Big(\dfrac{1}{2}\Big)^5 = \Big(\dfrac{1}{2}\Big)^n
or
n = 5
T = t_{1/2} \times 2
, or
t_{1/2} = \dfrac{50}{5} = 10 \ days
.
The radium and uranium atoms in a sample of uranium mineral are in the ratio of
1:2.8 \times 10^6
. If the half-life period of radium is 1620 years, the half-life period of uranium will be:
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0%
45.3 \times 10^9
years
0%
45.3 \times 10^{10}
years
0%
4.53 \times 10^9
years
0%
4.53 \times 10^{10}
years
Explanation
According to radioactive equilibrium:
\lambda_AN_A = \lambda_BN_B
or
\dfrac{0.693 \times N_A}{t_{1/2}(A)} = \dfrac{0.693 \times N_B}{t_{1/2}(A)} \Big[\lambda = \dfrac{0.693}{t_{1/2}}\Big]
Where
t_{1/2} (A)
and
t_{1/2} (B)
are half periods of A and B respectively
\therefore \dfrac{N_A}{t_{1/2}(A)} = \dfrac{N_B}{t_{1/2}(B)}
\therefore \dfrac{N_A}{t_{1/2}(A)} = \dfrac{N_B}{t_{1/2}(B)}
or
\dfrac{N_A}{N_B} = \dfrac{t_{1/2}(A)}{t_{1/2}(B)}
\therefore
At equilibrium A and B are present in the ratio of their half lives
\dfrac{1}{2.8 \times 10^6} = \dfrac{1620}{\text{Half Life of uranium}}
\therefore
Halflife of uranium
= 2.8 \times 10^6 \times 1620 = 4.53 \times 10^9
years.
If the order of the reaction
x+y\xrightarrow[]{hv}xy
is zero, it means that the rate of ____
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0%
Reaction is independent of temperature
0%
Formation of activated complex is zero
0%
Reaction is independent of the concentration of reacting species
0%
Decomposition of activated complex is zero
Explanation
In photochemical reactions the rate of reaction is independent of the concentration of reacting species.
8 \ gms
of a radioactive substance is reduced to
0.5 g
after 1 hour. The
t_{1 / 2}
of the radioactive substance is:
Report Question
0%
15 min
0%
30 min
0%
45 min
0%
10 min
Explanation
N_0 = 8gms, N = 0.5 g
and
t = 1 hr. = 60 min.
For
t_{1/2}
,
t = \dfrac{2.303 \times t_{1/2}}{0.693} log \dfrac{N_0}{N}
.
60 = \dfrac{2.303\times t_{1/2}}{0.693} log \dfrac{8}{0.5}
t_{1/2} = 15
min
Hence, option A is correct.
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