NCERT Solutions Class 12

NCERT SolutionsMathematics
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 1)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 2)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 3)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 4)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 5)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 1)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 2)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 3)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 1)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 2)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 3)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 4)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 5)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 1)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 2)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 3)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 4)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 5)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 6)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 7)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 1)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 2)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 3)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 4)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 5)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 6)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 7)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 8)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 9)

NCERT SolutionsChemistry
 Aldehydes, Ketones and Carboxylic Acids : NCERT Solutions – Class 12 Chemistry
 Alcohols, Phenols and Ethers : NCERT Solutions – Class 12 Chemistry
 Amines : NCERT Solutions – Class 12 Chemistry
 Biomolecules : NCERT Solutions – Class 12 Chemistry
 Chemical Kinetics : NCERT Solutions – Class 12 Chemistry
 Chemistry in Everyday Life : NCERT Solutions – Class 12 Chemistry
 Coordination Compounds : NCERT Solutions – Class 12 Chemistry
 Electrochemistry : NCERT Solutions – Class 12 Chemistry
 General Principles and Processes of Isolation of Elements : NCERT Solutions – Class 12 Chemistry
 Haloalkanes and Haloarenes : NCERT Solutions – Class 12 Chemistry
 Polymers : NCERT Solutions – Class 12 Chemistry
 Surface Chemistry : NCERT Solutions – Class 12 Chemistry
 The dand fBlock Elements : NCERT Solutions – Class 12 Chemistry
 The pBlock Elements : NCERT Solutions – Class 12 Chemistry
 The Solid State : NCERT Solutions – Class 12 Chemistry
 Solutions : NCERT Solutions – Class 12 Chemistry

NCERT SolutionsBiology

NCERT SolutionsPhysics
 Electrostatic Potential And Capacitance : NCERT Solutions – Class 12 Physics
 Electric Charges And Fields : NCERT Solutions – Class 12 Physics
 Semiconductor Electronics: Materials, Devices And Simple Circuits : NCERT Solutions – Class 12 Physics
 Ray Optics And Optical Instruments : NCERT Solutions – Class 12 Physics
 Nuclei : NCERT Solutions – Class 12 Physics
 Moving Charges And Magnetism : NCERT Solutions – Class 12 Physics
 Magnetism And Matter : NCERT Solutions – Class 12 Physics
 Electromagnetic Induction : NCERT Solutions – Class 12 Physics
 Dual Nature Of Radiation And Matter : NCERT Solutions – Class 12 Physics
 Current Electricity : NCERT Solutions – Class 12 Physics
 Communication Systems : NCERT Solutions – Class 12 Physics
 Atoms : NCERT Solutions – Class 12 Physics
 Alternating Current : NCERT Solutions – Class 12 Physics
Chemical Kinetics : NCERT Solutions – Class 12 Chemistry
4.1. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3 NO(g) → N_{2}O_{ }(g) Rate = k[NO]^{2}
(ii) H_{2}O_{2 }(aq) + 3 I^{− }(aq) + 2 H^{+} → 2 H_{2}O (l) + Rate = k[H_{2}O_{2}][I^{−}]
(iii) CH_{3}CHO(g) → CH_{4}(g) + CO(g) Rate = k [CH_{3}CHO]^{3/2}
(iv) C_{2}H_{5}Cl(g) → C_{2}H_{4}(g) + HCl(g) Rate = k [C_{2}H_{5}Cl]
Answer
(i) Given rate = k [NO]^{2}
Therefore, order of the reaction = 2
Dimension of
(ii) Given rate = k [H_{2}O_{2}] [I^{−}]
Therefore, order of the reaction = 2
Dimension of
(iii) Given rate = k [CH_{3}CHO]^{3/2}
Therefore, order of reaction =
Dimension of
(iv) Given rate = k [C_{2}H_{5}Cl]
Therefore, order of the reaction = 1
Dimension of
4.2. For the reaction:
2A + B → A_{2}B
the rate = k[A][B]^{2} with k = 2.0 × 10^{−6} mol^{−2} L^{2} 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}.
Answer
The initial rate of the reaction is
Rate = k [A][B]^{2}
= (2.0 × 10^{−6} mol^{−2} L^{2} s^{−1}) (0.1 mol L^{−1}) (0.2 mol L^{−1})^{2}
= 8.0 × 10^{−9} mol^{−2} L^{2} s^{−1}
When [A] is reduced from 0.1 mol L^{−1} to 0.06 mol^{−1}, the concentration of A reacted = (0.1 − 0.06) mol L^{−1 }= 0.04 mol L^{−1}
Therefore, concentration of B reacted = 0.02 mol L^{−1}
Then, concentration of B available, [B] = (0.2 − 0.02) mol L^{−1}
= 0.18 mol L^{−1}
After [A] is reduced to 0.06 mol L^{−1}, the rate of the reaction is given by,
Rate = k [A][B]^{2}
= (2.0 × 10^{−6} mol^{−2} L^{2} s^{−1}) (0.06 mol L^{−1}) (0.18 mol L^{−1})^{2}
= 3.89 mol L^{−1} s^{−1}
4.3. The decomposition of NH_{3} on platinum surface is zero order reaction. What are the rates of production of N_{2} and H_{2} if k = 2.5 × 10^{−4} mol^{−1} L s^{−1}?
Answer
The decomposition of NH_{3} 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 N_{2} is
And, the rate of production of H_{2} is
= 7.5 × 10^{−4} mol L^{−1} s^{−1}
4.4. The decomposition of dimethyl ether leads to the formation of CH_{4}, H_{2} and CO and the reaction rate is given by
Rate = k [CH_{3}OCH_{3}]^{3/2}
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?
Answer
If pressure is measured in bar and time in minutes, then
Unit of rate = bar min^{−1}
Therefore, unit of rate constants
4.5. Mention the factors that affect the rate of a chemical reaction.
Answer
The factors that affect the rate of a reaction are as follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii) Presence of a catalyst
4.6. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled (ii) reduced to half?
Answer
Letthe concentration of the reactant be [A] = a
Rate of reaction, R = k [A]^{2}
= ka^{2}
(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
= 4ka^{2}
= 4 R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e. , then the rate of the reaction would be
Therefore, the rate of the reaction would be reduced to
4.7. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Answer
The rate constant is nearly doubled with a rise in temperature by 10° 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
E_{a} is the energy of activation for the reaction
4.8. In a pseudo first order hydrolysis of ester in water, the following results were obtained:
t/s 
0 
30 
60 
90 
[Ester]mol L^{−1} 
0.55 
0.31 
0.17 
0.085 
Answer
(i) Average rate of reaction between the time interval, 30 to 60 seconds,
= 4.67 × 10^{−3} mol L^{−1} s^{−1}
(ii) For a pseudo first order reaction,
For t = 30 s,
= 1.911 × 10^{−2} s^{−1}
For t = 60 s,
= 1.957 × 10^{−2} s^{−1}
For t = 90 s,
= 2.075 × 10^{−2} s^{−1}
Then, average rate constant,
4.9. A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
Answer
(i) The differential rate equation will be
(ii) If the concentration of B is increased three times, then
Therefore, the rate of reaction will increase 9 times.
(iii) When the concentrations of both A and B are doubled,
Therefore, the rate of reaction will increase 8 times.
4.10. In a reaction between A and B, the initial rate of reaction (r_{0}) was measured for different initial concentrations of A and B as given below:

A/ mol L^{−1}
0.20
0.20
0.40
B/ mol L^{−1}
0.30
0.10
0.05
r_{0}/ mol L^{−1} s^{−1}
5.07 × 10^{−5}
5.07 × 10^{−5}
1.43 × 10^{−4}
What is the order of the reaction with respect to A and B?
Answer
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
Dividing equation (i) by (ii), we obtain
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.
4.11. The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D

Experiment
A/ mol L^{−1}
B/ mol L^{−1}
Initial rate of formation of D/mol L^{−1} min^{−1}
I
0.1
0.1
6.0 × 10^{−3}
II
0.3
0.2
7.2 × 10^{−2}
III
0.3
0.4
2.88 × 10^{−1}
IV
0.4
0.1
2.40 × 10^{−2}
Determine the rate law and the rate constant for the reaction.
Answer
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,
Dividing equation (iv) by (i), we obtain
Dividing equation (iii) by (ii), we obtain
Therefore, the rate law is
Rate = k [A] [B]^{2}
From experiment I, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
From experiment II, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
From experiment III, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
From experiment IV, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
Therefore, rate constant, k = 6.0 L^{2} mol^{−2} min^{−1}
4.12. 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:

Experiment
A/ mol L^{−1}
B/ mol L^{−1}
Initial rate/mol L^{−1} min^{−1}
I
0.1
0.1
2.0 × 10^{−2}
II
—
0.2
4.0 × 10^{−2}
III
0.4
0.4
—
IV
—
0.2
2.0 × 10^{−2}
Answer
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]^{1} [B]^{0}
⇒ Rate = k [A]
From experiment I, we obtain
2.0 × 10^{−2} mol L^{−1} min^{−1} = k (0.1 mol L^{−1})
⇒ k = 0.2 min^{−1}
From experiment II, we obtain
4.0 × 10^{−2} mol L^{−1} min^{−1} = 0.2 min^{−1} [A]
⇒ [A] = 0.2 mol L^{−1}
From experiment III, we obtain
Rate = 0.2 min^{−1} × 0.4 mol L^{−1}
= 0.08 mol L^{−1} min^{−1}
From experiment IV, we obtain
2.0 × 10^{−2} mol L^{−1} min^{−1} = 0.2 min^{−1} [A]
⇒ [A] = 0.1 mol L^{−1}
4.13. Calculate the halflife of a first order reaction from their rate constants given below:
(i) 200 s^{−1} (ii) 2 min^{−1} (iii) 4 years^{−1}
Answer
(i) Half life,
= 3.47 ××10 ^{3} s (approximately)
(ii) Half life,
= 0.35 min (approximately)
(iii) Half life,
= 0.173 years (approximately)
4.14. The halflife for radioactive decay of ^{14}C is 5730 years. An archaeological artifact containing wood had only 80% of the ^{14}C found in a living tree. Estimate the age of the sample.
Answer
Here,
It is known that,
= 1845 years (approximately)
Hence, the age of the sample is 1845 years.
4.15. The experimental data for decomposition of N_{2}O_{5}
in gas phase at 318K are given below:

t(s)
0 400 800 1200 1600 2000 2400 2800 3200 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35
(i) Plot [N_{2}O_{5}] against t.
(ii) Find the halflife period for the reaction.
(iii) Draw a graph between log [N_{2}O_{5}] and t.
(iv) What is the rate law?
(v) Calculate the rate constant.
(vi) Calculate the halflife period from k and compare it with (ii).
Answer
(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,
(vi) Halflife is given by,
This value, 1438 s, is very close to the value that was obtained from the graph.
4.16. 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?
Answer
It is known that,
Hence, the required time is 4.6 × 10^{−2} s.
4.17. During nuclear explosion, one of the products is ^{90}Sr with halflife of 28.1 years. If 1μg of ^{90}Sr 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.
Answer
Here,
It is known that,
Therefore, 0.7814 μg of ^{90}Sr will remain after 10 years.
Again,
Therefore, 0.2278 μg of ^{90}Sr will remain after 60 years.
4.18. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer
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, t_{1} = 2t_{2}
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.
4.19. A first order reaction takes 40 min for 30% decomposition. Calculate t_{1/2}.
Answer
For a first order reaction,
Therefore, t_{1/2} of the decomposition reaction is
= 77.7 min (approximately)
4.20. 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.
Answer
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.
After time, t, total pressure,
= 2P_{0} − P_{t}
For a first order reaction,
When t = 360 s,
= 2.175 × 10^{−3} s^{−1}
When t = 720 s,
= 2.235 × 10^{−3} s^{−1}
Hence, the average value of rate constant is
= 2.21 × 10^{−3} s^{−1}
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
4.21. The following data were obtained during the first order thermal decomposition of SO_{2}Cl_{2} at a constant volume.

Experiment
Time/s^{−1}
Total pressure/atm
1
0
0.5
2
100
0.6
Calculate the rate of the reaction when total pressure is 0.65 atm.
Answer
The thermal decomposition of SO_{2}Cl_{2} at a constant volume is represented by the following equation.
After time, t, total pressure,
Therefore,
= 2 P_{0} − P_{t}
For a first order reaction,
When t = 100 s,
= 2.231 × 10^{−3} s^{−1}
When P_{t} = 0.65 atm,
P_{0} + p = 0.65
⇒ p = 0.65 − P_{0}
= 0.65 − 0.5
= 0.15 atm
Therefore, when the total pressure is 0.65 atm, pressure of SOCl_{2} is
= P_{0} − p
= 0.5 − 0.15
= 0.35 atm
Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,
Rate = k()
= (2.23 × 10^{−3} s^{−1}) (0.35 atm)
= 7.8 × 10^{−4} atm s^{−1}
4.22. The rate constant for the decomposition of N_{2}O_{5} at various temperatures is given below:

T/°C 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 E_{a}.
Predict the rate constant at 30º and 50ºC.
Answer
From the given data, we obtain

T/°C
0
20
40
60
80
T/K
273
293
313
333
353
3.66×10^{−3}
3.41×10^{−3}
3.19×10^{−3}
3.0×10^{−3}
2.83 ×10^{−3}
0.0787
1.70
25.7
178
2140
ln k
−7.147
− 4.075
−1.359
−0.577
3.063
Slope of the line,
According to Arrhenius equation,
Again,
When ,
Then,
Again, when ,
Then, at ,
4.23. The rate constant for the decomposition of hydrocarbons 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 preexponential factor.
Answer
k = 2.418 × 10^{−5} s^{−1}
T = 546 K
E_{a} = 179.9 kJ mol^{−1} = 179.9 × 10^{3} J mol^{−1}
According to the Arrhenius equation,
= (0.3835 − 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 10^{12} s^{−1} (approximately)
4.24. 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}.
Answer
k = 2.0 × 10^{−2} s^{−1}
T = 100 s
[A]_{o} = 1.0 moL^{−1}
Since the unit of k is s^{−1}, the given reaction is a first order reaction.
Therefore,
= 0.135 mol L^{−1} (approximately)
Hence, the remaining concentration of A is 0.135 mol L^{−1}.
4.25. 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?
Answer
For a first order reaction,
It is given that, t_{1/2} = 3.00 hours
Therefore,
= 0.231 h^{−1}
Then, 0.231 h^{−1}
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
4.26. The decomposition of hydrocarbon follows the equation
k = (4.5 × 10^{11 }s^{−1}) e^{−28000 }^{K}^{/}^{T}
Calculate E_{a}.
Answer
The given equation is
k = (4.5 × 10^{11 }s^{−1}) e^{−28000 }^{K}^{/}^{T} (i)
Arrhenius equation is given by,
(ii)
From equation (i) and (ii), we obtain
= 8.314 J K^{−1} mol^{−1} × 28000 K
= 232792 J mol^{−1}
= 232.792 kJ mol^{−1}
4.27. The rate constant for the first order decomposition of H_{2}O_{2} is given by the following equation:
log k = 14.34 − 1.25 × 10^{4 }K/T
Calculate E_{a} for this reaction and at what temperature will its halfperiod be 256 minutes?
Answer
Arrhenius equation is given by,
The given equation is
From equation (i) and (ii), we obtain
= 1.25 × 10^{4} K × 2.303 × 8.314 J K^{−1} mol^{−1}
= 239339.3 J mol^{−}1 (approximately)
= 239.34 kJ mol^{−1}
Also, when t_{1/2} = 256 minutes,
= 2.707 × 10^{−3} min^{−1}
= 4.51 × 10^{−5} s^{−1}
It is also given that, log k = 14.34 − 1.25 × 10^{4} K/T
= 668.95 K
= 669 K (approximately)
4.28. The decomposition of A into product has value of k as 4.5 × 10^{3} s^{−1} at 10°C and energy of activation 60 kJ mol^{−1}. At what temperature would k be 1.5 × 10^{4} s^{−1}?
Answer
From Arrhenius equation, we obtain
Also, k_{1} = 4.5 × 10^{3} s^{−1}
T_{1} = 273 + 10 = 283 K
k_{2} = 1.5 × 10^{4} s^{−1}
E_{a} = 60 kJ mol^{−1} = 6.0 × 10^{4} J mol^{−1}
Then,
= 297 K
= 24°C
Hence, k would be 1.5 × 10^{4} s^{−1} at 24°C.
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
4.29. 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
4 × 10^{10 }s^{−1}. Calculate k at 318 K and E_{a}.
Answer
For a first order reaction,
At 298 K,
At 308 K,
According to the question,
From Arrhenius equation, we obtain
To calculate k at 318 K,
It is given that,
Again, from Arrhenius equation, we obtain
4.30. 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.
Answer
From Arrhenius equation, we obtain
Hence, the required energy of activation is 52.86 kJmol^{−1}.