CBSE Class 12 Chemistry Chapter 4 Important Questions – Free PDF Download
Free PDF download of Important Questions for CBSE Class 12 Chemistry Chapter 4 – Chemical Kinetics 5 Marks Questions prepared by expert Chemistry teachers from latest edition of CBSE(NCERT) books, On CoolGyan.Org to score more marks in CBSE board examination.
You can also Download Chemistry Revision Notes Class 12 to help you to revise complete Syllabus and score more marks in your examinations.
Important Questions Class 12 Chemistry Chapter 4 – Chemical Kinetics
CBSE Class 12 Chemistry Important Questions Chapter 4 – Chemical Kinetics
5 Marks Questions
1. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate 
.
Ans. It is given that T1 = 298 K
Therefore, 
= 308 K
We also know that the rate of the reaction doubles when temperature is increased by 
.
Therefore, let us take the value of 
and that of 
Also, 
Now, substituting these values in the equation:
We get:
2. The activation energy for the reaction 
is 
at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Ans. In the given case:
T = 581 K
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:
In 
=18.8323
Now, x = Anti log (18.8323)
= Anti log 
3. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 
(ii) 
(iii) 
(iv) 
Ans. (i) Given rate = 
Therefore, order of the reaction = 2
Dimension of 
(ii) Given rate = 
Therefore, order of the reaction = 2
Dimension of 
(iii) Given rate = 
Therefore, order of reaction = 
Dimension of 
(iv) Given rate = 
Therefore, order of the reaction = 1
Dimension of 
4. The decomposition of 
on platinum surface is zero order reaction. What are the rates of production of 
and 
if 
?
Ans. The decomposition of on platinum surface is represented by the following equation.
Therefore, 
However, it is given that the reaction is of zero order.
Therefore, 
Therefore, the rate of production of is
And, the rate of production of is
5. The decomposition of dimethyl ether leads to the formation of 
,and CO and the reaction rate is given by Rate = 
. The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., 
If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?
Ans. If pressure is measured in bar and time in minutes, then
Unit of rate = 
Therefore, unit of rate constants 
6. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Ans. The rate constant is nearly doubled with a rise in temperature by for a chemical reaction.
The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,
where, k is the rate constant,
A is the Arrhenius factor or the frequency factor,
R is the gas constant,
T is the temperature, and
is the energy of activation for the reaction
7. In a pseudo first order hydrolysis of ester in water, the following results were obtained:
| t/s | 0 | 30 | 60 | 90 |
[Ester]  | 0.55 | 0.31 | 0.17 | 0.085 |
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Ans. (i) Average rate of reaction between the time interval, 30 to 60 seconds,
(ii) For a pseudo first order reaction,
For t= 30 s, 
For t= 60 s, 
For t= 90 s,
Then, average rate constant, 
8. In a reaction between A and B, the initial rate of reaction 
was measured for different initial concentrations of A and B as given below:
 | 0.20 | 0.20 | 0.40 |
 | 0.30 | 0.10 | 0.05 |
 |  | |  |
What is the order of the reaction with respect to A and B?
Ans. Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, 
]
……….(i)
……….(ii)
…………(iii)
Dividing equation (i) by (ii), we obtain
y = 0
Dividing equation (iii) by (ii), we obtain
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.
10. The following results have been obtained during the kinetic studies of the reaction:
| Experiment | | | Initial rate of formation of  |
| I | 0.1 | 0.1 |  |
| II | 0.3 | 0.2 |  |
| III | 0.3 | 0.4 |  |
| IV | 0.4 | 0.1 |  |
Determine the rate law and the rate constant for the reaction.
Ans. Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
According to the question,
………(i)
……….(ii)
……….(iii)
……….(iv)
Dividing equation (iv) by (i), we obtain
x = 1
Dividing equation (iii) by (ii), we obtain
y = 2
Therefore, the rate law is
From experiment I, we obtain
From experiment II, we obtain,
From experiment III, we obtain
From experiment IV, we obtain
Therefore, rate constant, k =
11. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment | | | Initial rate /  |
| I | 0.1 | 0.1 |  |
| II | – | 0.2 |  |
| III | 0.4 | 0.4 | – |
| IV | – | 0.2 | |
Ans. The given reaction is of the first order with respect to A and of zero order with respect to B.
Therefore, the rate of the reaction is given by,
Rate = k [A]
From experiment I, we obtain
From experiment II, we obtain
From experiment III, we obtain
From experiment IV, we obtain
12. The experimental data for decomposition of 
in gas phase at 318K are given below:
| t(s) | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |
| 102×[N2O5]molL−1102×[N2O5]molL−1 | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |
(i) Plot 
against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log and t.
(iv) What is the rate law?
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).
Ans.
(ii) Time corresponding to the concentration, 
is the half life. From the graph, the half life is obtained as 1450 s.
(iii)
| t(s) |  |  |
| 0 | 1.63 | –1.79 |
| 400 | 1.36 | –1.87 |
| 800 | 1.14 | –1.94 |
| 1200 | 0.93 | –2.03 |
| 1600 | 0.78 | –2.11 |
| 2000 | 0.64 | –2.19 |
| 2400 | 0.53 | –2.28 |
| 2800 | 0.43 | –2.37 |
| 3200 | 0.35 | –2.46 |
(iv) The given reaction is of the first order as the plot, v/s t, is a straight line. Therefore, the rate law of the reaction is 
(v) From the plot, 
v/s t, we obtain
Again, slope of the line of the plot v/s t is given by
Therefore, we obtain,
13. During nuclear explosion, one of the products is 
with half-life of 28.1 years. If 
of was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Ans. Here, 
It is known that, 
Therefore, of will remain after 10 years.
Again, 
Therefore, 
of will remain after 60 years.
14. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Ans. For a first order reaction, the time required for 99% completion is
For a first order reaction, the time required for 90% completion is
Therefore, 
Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.
15. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
| t(sec) | P(mm of Hg) |
| 0 | 35.0 |
| 360 | 54.0 |
| 720 | 63.0 |
Calculate the rate constant.
Ans. The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.
After time, t, total pressure, 
Therefore, 
For a first order reaction,
When t = 360 s, 
When t = 720 s, 
Hence, the average value of rate constant is
16. The following data were obtained during the first order thermal decomposition of 
at a constant volume. 
| Experiment |  | Total pressure/atm |
| 1 | 0 | 0.5 |
| 2 | 100 | 0.6 |
Calculate the rate of the reaction when total pressure is 0.65 atm.
Ans. The thermal decomposition of at a constant volume is represented by the following equation.
After time, t, total pressure,
Therefore,
For a first order reaction,
When t= 100 s, 
When 
,
= 0.65 – 0.5
= 0.15 atm
Therefore, when the total pressure is 0.65 atm, pressure of 
is
= 0.5 – 0.15
= 0.35 atm
Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,
17. The rate constant for the decomposition of 
at various temperatures is given below:
 | 0 | 20 | 40 | 60 | 80 |
 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph between ln k and 1/T and calculate the values of A and . Predict the rate constant at 
and 
.
Ans. From the given data, we obtain
 | 0 | 20 | 40 | 60 | 80 |
| T/K | 273 | 293 | 313 | 333 | 353 |
 |  |  |  |  |  |
 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
| In K | – 7.147 | – 4.075 | – 1.359 | – 0.577 | 3.063 |
Slope of the line,
According to Arrhenius equation,
Slope = 
= 
Again,
When T = 273 K
In k = –7.147
Then, In A = 
=37.911
Therefore, A = 
When, T = 30 + 273K = 303 K
Then, at 
,
In k = –2.8
Therefore, k = 
Again, when T = 50 + 273K = 323K,
18. The rate constant for the decomposition of hydrocarbons is 
at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Ans. 
T= 546 K
According to the Arrhenius equation,
= (0.3835 – 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 
(approximately)
19. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with
. What fraction of sample of sucrose remains after 8 hours?
Ans. For a first order reaction,
It is given that,
Therefore, 
= 
Then,
(approx)
= 0.158
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
20. The decomposition of hydrocarbon follows the equation 
. Calculate 
.
Ans. The given equation is……………..(i)
Arrhenius equation is given by,
……………….(ii)
From equation (i) and (ii), we obtain
21. The rate constant for the first order decomposition of 
is given by the following equation: 
Calculate for this reaction and at what temperature will its half-period be 256 minutes?
Ans. Arrhenius equation is given by,
The given equation is
……………………..(ii)
From equation (i) and (ii), we obtain
(approximately)
Also, when
It is also given that, 
= 668.95 K
= 669 K (approximately)
22. The decomposition of A into product has value of k as 
at 
and energy of activation 
. At what temperature would k be 
?
Ans. From Arrhenius equation, we obtain 
Also,
=273 + 10 = 283 K
Then, 
(approximately)
= 297 K
= 
Hence, k would be at .
23. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 
. Calculate k at 318 K and .
Ans. For a first order reaction, 
At 298 K, 
At 308 K,
According to the question, t = t’
From Arrhenius equation, we obtain
To calculate k at 318 K,
It is given that, 
Again, from Arrhenius equation, we obtain
=(0.6021+10) – 12.5876
= –1.9855
Therefore, k = Antilog (–1.9855)
=
24. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Ans. From Arrhenius equation, we obtain 
It is given that, 
Therefore, 
Hence, the required energy of activation is 
.