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Published on: 13/10/2020
12th Standard Chemistry Chemical Kinetics English Medium Free Online Test One Mark Questions with Answer Key 2020 - 2021
Download Tamil Nadu 12th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
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1.
The excess energy which a molecule must possess to become active is known as _____.
kinetic energy
threshold energy
potential energy
activation energy
2.
For the second order reaction \({ t }_{ \frac { 1 }{ 2 } }\alpha \) _____.
\(\frac { 1 }{ { a }^{ 2 } } \)
\(\frac { 1 }{ { a } } \)
Constant
a
3.
Concentration is expressed in?
\(\frac { number\ of\ moles/litre }{ time\ in\ sec } \)
\(\frac { time\ in\ sec }{ number\ of\ moles/litre } \)
\(\frac { number\ of\ moles/litre }{ volume } \)
\(\frac { volume }{ number\ of\ moles/litre } \)
4.
During decomposition of an activated complex.
energy is always released
energy is always absorbed
energy does not change
products may be formed.
5.
Which of the above graphs is correct for first order reactions?
I, II
II, III
I, IV
I, III
6.
How much time will be taken for 20 gm to reduce 5 g? [R = 2 x 10-3s-1 (I order reaction)]
693.1 s
693.1 s-1
6.931 s
6.931 s-1
7.
A reaction having equal activation energies for forward and reverse reactions has ______.
ΔG = 0
ΔH = 0
ΔH = ΔG = ΔS = 0
ΔS = 0
8.
Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide
2NO(g) + O2(g) ⟶ 2NO2(g)
If the mechanism of reaction is ______.
NO + O2 \(\overset { K }{ \rightleftharpoons } \) NO3 (fast)
NO3 + NO \(\overset { { K }_{ 1 } }{ \rightleftharpoons } \) NO2 + NO2 (slow)
Rate = K' [NO] [O2]
Rate = K' [NO] [O2]2
Rate = K' [NO]2 [O2]
Rate = K' [NO]3 [O2]
9.
For an exothermic chemical process occurring in 2 steps as
(i) A +B ⟶ X (slow) ;
(ii) X ⟶ AB (fast)
The progress of the reaction can be best described by (x- intermediate).
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None of these
10.
For the reaction, 2N2O5 ⟶4 NO2+O2, select the correct statement.
Rate of formation of O2 is same as rate of formation of NO2
Rate of disappearance of N2O5 is two times the rate of formation of NO2.
Rate of formation of O2 is 0.5 times rate of disappearance of N2O5
Rate of formation of NO2 is equal to rate of disappearance of N2O5
11.
What would be the rate of disappearance of oxygen, if the rate of formation of nitric oxide (NO) is 3.6 x 10-3mol L-1 s-1?
4 x 10-3mol L-1s-1
4 x 10-3mol-1 L-1s-1
4.5 x 10-3mol L-1s-1
4.5 x 10-3mol-1 L-1s-1
12.
After 2 hours, a radioactive substance becomes \(\left( \frac { 1 }{ 16 } \right) ^{ th }\) of original amount Then the half life (in min) is _______.
60 minutes
120 minutes
30 minutes
15 minutes
13.
The half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to
\(\left( \frac { 1 }{ 2 } \right) g\)
\(\left( \frac { 1 }{ 4 } \right) g\)
\(\left( \frac { 1 }{ 8 } \right) g\)
\(\left( \frac { 1 }{ 16 } \right) g\)
14.
15.
For a first order reaction, the rate constant is 6.909 min-1 the time taken for 75% conversion in minutes is _______.
\(\left( \frac { 3 }{ 2 } \right) { \log 2 }\)
\(\left( \frac { 2 }{ 3 } \right) \log2\)
\(\left( \frac { 3 }{ 2 } \right) \log\left( \frac { 3 }{ 4 } \right) \)
\(\left( \frac { 2 }{ 3 } \right) \log\left( \frac { 4 }{ 3 } \right) \)
16.
In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively −x kJ mol-1 and y kJ mol-1. Therefore, the energy of activation in the backward direction is _______.
(y-x) kJ mol-1
(x+y) J mol-1
(x-y) KJ mol-1
(x+y) x 103J mol-1
17.
The addition of a catalyst during a chemical reaction alters which of the following quantities?
Enthalpy
Activation energy
Entropy
Internal energy
18.
The decomposition of phosphine (PH3) on tungsten at low pressure is a first order reaction. It is because the _____.
rate is proportional to the surface coverage
rate is inversely proportional to the surface coverage
rate is independent of the surface coverage
rate of decomposition is slow
19.
For a first order reaction A ⟶ product with initial concentration x mol L-1, has a half life period of 2.5 hours. For the same reaction with initial concentration \(\left( \frac { x }{ 2 } \right) \) mol L-1 the half life is
(2.5 x 2) hours
\(\left( \frac { 2.5 }{ 2 } \right) \) hours
2.5 hours
Without knowing the rate constant, t1/2 cannot be determined from the given data
20.
A zero order reaction X ⟶ Product, with an initial concentration 0.02M has a half life of 10 min. if one starts with concentration 0.04M, then the half life is
10 s
5 min
20 min
cannot be predicted using the given information
1.
(d)
activation energy
2.
(b)
\(\frac { 1 }{ { a } } \)
3.
(a)
\(\frac { number\ of\ moles/litre }{ time\ in\ sec } \)
4.
(a)
energy is always released
5.
(c)
I, IV
6.
(a)
693.1 s
7.
(b)
ΔH = 0
8.
(c)
Rate = K' [NO]2 [O2]
9.
(c)
10.
(c)
Rate of formation of O2 is 0.5 times rate of disappearance of N2O5
11.
(c)
4.5 x 10-3mol L-1s-1
12.
13.
In 140days ⇒ initial concentration reduced to (1/2) g
In 280 days ⇒ initial concentration reduced to (1/4) g
In 420 days ⇒ initial concentration reduced to (1/8) g
In 560 days ⇒ initial concentration reduced to (1/16) g
14.
(a)
15.
\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A0] = 100: [A] = 25
\(6.909=\frac { 2.303 }{ t } \log\frac { \left[ {100 } \right] }{ \left[ 25\right] } \)
\(t =\frac { 2.303 }{ 6.909 } \log(4)\)
\(t =\frac { 1 }{ 3 } \log(2^2)\)
\(= \left( \frac { 2 }{ 3 } \right) \log2\)
16.
17.
A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.
18.
Given:
At low pressure the reaction follows first order therefore,
Rate α [reactant]1
Rate α (surface area)
At high pressure due to the complete coverage of surface area, the reaction follows zero order.
Rate α [reactant]0
Therefore the rate is independent of surface area.
19.
For a first order reaction
t1/2 = \(\frac { 0.693 }{ { k }}\)
t1/2 does not depend on the initial concentration and it remains constant (whatever may be the initial concentration)
t1/2 = 2.5 hrs
20.
for n ≠ 1 t1/2 = \(\frac{2^{n-1} -1}{(n- 1) k[A_{0}]^{-1}}\)
for n = 0; t1/2 = \(\frac{1}{2 k[A_{0}]^{-1}}\)
t1/2 = \(\frac{[A_{0}]}{2 k}\)
t1/2 α [A0] ...(1)
Given [A0] = 0.002 M; t1/2 = 10 min
[A0] = 0.04M; t1/2 = ?
Substitute in (1)
10 min α 0.02 M...(2)
t1/2 α 0.04M ....(3)
(3)(2)
⇒ t1/2 / 10 min
= 0.04 M/0.02 M
t1/2 = 2 x 10 min = 20 min
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