NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics

Chemistry NCERT Solutions for Class 12 Chapter 4 Chemical Kinetics includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 12 Chemistry Chapter 4 Chemical Kinetics NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time. Students can also download Chemistry NCERT Solutions for Class 12 Chapter 4 Chemical Kinetics PDF to access them even in offline mode.

Chemistry NCERT Solutions for Class 12 Chapter 4 Chemical Kinetics

NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

Chemistry NCERT Solutions for Class 12 Chapter 4 INTEXT Questions

Question 1.
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Solution:

Question 2.
In a reaction, 2A → products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this interval.
Solution:

Question 3.
For a reaction, A + B → product; the rate law is given by, r = k[A]^1/2 [B]2. What is the order of the reaction?
Solution:

Question 4.
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Solution:
The reaction, X → Y, follows second order kinetics hence the rate law equation will be
Rate = kC², where C = [X]
If concentration of X increases three times, now, [X] = 3C mol L^-1
∴ Rate = k(3C)² = 9kC²
Thus the rate of reaction will become 9 times. Hence, the rate of formation of Y will increase 9 times.

Question 5.
A first order reaction has a rate constant 1.15 × 10^-3 s^-1. How long will 5 g of this reactant take to reduce to 3 g?
Solution:

Question 6.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Solution:

Question 7.
What will be the effect of temperature on rate constant?
Solution:
It has been found that for a chemical reaction, with rise in temperature by 10°, the rate constant is nearly doubled.

The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation.
K = Ae^-Ea/RT
Where A is the Arrhenius factor or the frequency factor. It is also called pre exponential factor. It is a constant specific to a particular reaction. R is gas constant and E_a is activation energy measured in joules/mole (J mol^-1).

Question 8.
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_a.
Solution:

Question 9.
The activation energy for the reaction 2Hl_(g) → H_2(g) + l_2(g) is 209.5 kJ mol^-1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Solution:
Fraction of molecules having energy equal to or greater than activation energy is given as

Chemistry NCERT Solutions for Class 12 Chapter 4 NCERT Exercises

Question 1.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

Solution:

Question 2.
For the reaction : 2A + B → A2B the rate = k[A][B]2 with k = 2.0 × 10^-6 mol^-2 L² s^-1 Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1 [B] = 0.2 mol L^-1 Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.
Solution:

Question 3.
The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol L^-1 s^-1 ?
Solution:

Question 4.
The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by
Rate = k[CH₃OCH₃]^3/2
The rate of reaction is followed by increase in pressure in closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
Rate = K(pCH₃OCH₃)^3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants ?
Solution:
In terms of pressure,
Units of rate = bar min^-1

Question 5.
Mention the factors that affect the rate of a chemical reaction.
Solution:
Following are the factors on which rate of reaction depends.
(i) Nature of the reactant : Rate of reaction depends on nature of reactant.
Example : Reactions of ionic compounds are faster than that of covalent compounds.

(ii) State of reactants : Solid reactions are slow, reactions of liquids are fast whereas that of gases are very fast.

(iii) Temperature : Rate of reaction largely depends on temperature. It has been observed that every 10°C rise in temperature increases rate of reaction by 2-3 times.
\(\frac { { r }_{ t }+10 }{ { r }_{ t } } \) = 2 – 3 . This ratio is called temperature coefficient.
There are two reasons for increasing rate of reaction with increasing temperature.

(a) Increase in temperature increases average kinetic energy of reactant molecules. Hence, rate of collision increases.
(b) With increase in temperature number of molecules having threshold energy also increases i.e. number of active molecules increases. As a result, number of effective collisions increases. Hence, rate of reaction increases.

(iv) Concentration : Rate of reaction also depends on concentration of reactants.
Rate = k × C^”, where n = order of reaction, C = concentration of reactant.

(v) Presence of catalyst : Rate of reaction also depends on presence of catalyst. Catalyst increases rate of reaction by any of the following ways:
(a) Increasing surface area of reaction.
(b) Adsorbing the reactants on its surface and thus increasing chance of collision.
(c) By forming unstable intermediate with the substrate.
(d) By providing alternate path of lower activation energy.

Question 6.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

1. doubled
2. reduced to half ?

Solution:

Question 7.
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Solution:
The rate constant increases with increase in temperature and becomes almost double for every 10° increase in temperature. Swedish chemist, Arrhenius derived a quantitative relation between rate of reaction and temperature. According to Arrhenius,

Question 8.
In a pseudo first order hydrolysis of ester in water, the following results were obtained :

1. Calculate the average rate of reaction between the time interval 30 to 60 seconds.
2. Calculate the pseudo first order rate constant for the hydrolysis of ester.

Solution:

Question 9.
A reaction is first order in A and second order in B.

1. Write the differential rate equation.
2. How is the rate affected on increasing the concentration of 6 three times?
3. How is the rate affected when the concentrations of both A and B are doubled?

Solution:
(i) Reaction is first order in A and second order in B, hence differential rate equation is

Question 10.
In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below :

What is the order of the reaction with respect to A and B ?
Solution:

Question 11.
The following results have been obtained during the kinetic studies of the reaction :
2 A + B → C + D

Determine the rate law and the rate constant for the reaction.
Solution:

Question 12.
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 :

Solution:

Question 13.
Calculate the half-life of the first order reaction from their rate constants given below:

1. 200 s^-1
2. 2 min^-1
3. 4 year^-1

Solution:

Question 14.
The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
Solution:
Radioactive decay follows first order kinetics. Therefore,

Question 15.
The experimental data for the decomposition of N₂O₅
[2N₂O₅ → 4NO₂ + O₂]
in gas phase at 318K are given below :

1. Plot [N₂O₅] against t.
2. Find the half-life period for the reaction.
3. Draw a graph between log [N₂O₅] and t.
4. What is the rate law?
5. Calculate the rate constant.
6. Calculate the half-life period from k and compare it with (ii).

Solution:

Question 16.
The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16 th value ?
Solution:

Question 17.
During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1 µg of ⁹⁰Sr 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 ?
Solution:

Question 18.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution:

Question 19.
A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Solution:

Question 20.
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant
Solution:

Chemistry NCERT Solutions for Class 12 Chapter 4 Question 21.
The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.

Calculate the rate of the reaction when total pressure is 0.65 atm.
Solution:

Question 22.
The rate constant for the decomposition of N₂O₅ at various temperatures is given below :

Draw a graph between In k and 1/7 and calculate the value of A and Ea. Predict the rate constant at 30°C and 50°C.
Solution:
The values of rate constants for the decomposition of N₂O₅ at various temperatures are given below :

Question 23.
The rate constant for the decomposition of a hydrocarbon is 2.418 × 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Solution:

Question 24.
Consider a certain reaction A → Products with k = 2.0 × 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.
Solution:

Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?
Solution:
Sucrose decomposes according to first order rate law, hence

Question 26.
The decomposition of a hydrocarbon follows the equation

Solution:

Question 27.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.34 – 1.25 × 104 K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Solution:

Question 28.
The decomposition of A into product has value of k as 4.5 × 10³ s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 × 10⁴ s^-1 ?
Solution:

Question 29.
The time required for 10% completion of the first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s^-1, calculated at 318 K and E_a.
Solution:

Question 30.
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.
Solution:


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NCERT Solutions for Class 12 Chemistry