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NCERT Solutions for Class 12 Chemistry Chapter 3 – Chemical Kinetics

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Intext Questions with Solutions
Exercise Questions with Solutions

Intext Questions with Solutions of Class 12 Chemistry Chapter 3 – Chemical Kinetics

3.1

📌 Key Points to Note from Question

Given Data:

Initial concentration of A: 0.03 M
Final concentration of A: 0.02 M
Time interval (Δt): 25 minutes = 1500 seconds

Key Concept:

Average Rate of Reaction, R = – Δ[R]/Δt

What to Calculate:

Change in concentration Δ[R]
Average rate in M/min (minutes) and M/s (seconds)

Ans – The average reaction rate can be obtained by relating the change in the concentration of the reactant with time taken. The values given are:

R₁ = 0.03 M
R₂ = 0.02 M
t₂ – t₁ = 25 min
Substituting the values, we obtain:

Thus, the average rate will be 4 × 10⁻⁴ M min⁻¹
To determine the average rate in seconds, we must divide the above result by 60. Therefore, it will be:

∴ Average rate in seconds = 6.66 × 10⁻⁶ M s⁻¹

3.2

📌 Key Points to Note from Question

Given Data:

Initial concentration of A: 0.5 mol L⁻¹
Final concentration of A: 0.4 mol L⁻¹
Time interval (t): 10 minutes = 600 seconds

Key Concept:

Rate of Reaction, R = – Δ[A]/Δt

What to Calculate:

Change in concentration Δ[A]
Rate of reaction (R) during the interval.

Ans –

3.3

📌 Key Points to Note from Question

Given Data:

Rate law: r = k[A]^1/2 [B]²

Key Concept:

Order of Reaction: The sum of the powers of the concentrations of reactants in the rate law.

What to Calculate:

Order of reaction.

Ans – The reaction order can be determined by summing the stoichiometric coefficients of the reactants in the specified reaction rate.
The rate can be expressed as r = k [A]^1/2 [B]²
⇒ reaction order = 1/2 + 2 = 2.5

3.4

📌 Key Points to Note from Question

Given Data:

Reaction: X → Y
Rate law: r = k[X]²

Key Concept:

Effect of Concentration Change on Rate

Ans –

3.5

📌 Key Points to Note from Question

Given Data:

Rate constant, k = 1.15 × 10^-3 s^-1
Initial amount ([R]₀): 5 g
Final amount ([R]): 3 g

Key Concept:

First-Order Reaction Formula

What to Calculate:

Time (t) required for the reactant to reduce from 5 g to 3 g.

Ans –

3.6

📌 Key Points to Note from Question

Given Data:

Half-life t_1/2 = 60 minutes.
Reaction: First-order decomposition.

What to Calculate:

Rate constant (k) of the reaction.

Ans – Given t_1/2 = 60 minutes and the breakdown of SO₂Cl₂ is classified as a first-order reaction. Thus, we can write:

3.7

Ans – A reaction’s rate constant nearly doubles in magnitude as the temperature rises by 10°. In any event, the precise temperature dependence of a chemical reaction rate is provided by the formula of Arrhenius.
The following is the Arrhenius calculation:
k = Ae^−E_a/RT
Wherein T is the temperature, R represents the gas constant, Ea denotes the activation energy, and A indicates the Arrhenius aspect, also known as the frequency factor.

3.8

📌 Key Points to Note from Question

Given Data:

Temperature increase: T₁ = 298 K, T₂ = 308 K.
Rate doubles: k₂/k₁ = 2.

Key Concept:

Arrhenius Equation (Ratio Form)

What to Calculate:

Activation energy (E_a).

Ans – Given T₁ = 298 K and T₂ = 308 K
Also k₁ = k and k₂ = 2k

3.9

📌 Key Points to Note from Question

Given Data:

Activation energy (E_a): 209.5 kJ/mol = 209500 J/mol
Temperature (T): 581 K

What to Calculate:

Fraction of molecules (f) with energy equal to or greater than E_a.

Ans – The fraction of molecules with energy either equal or surpassing the activation energy is stated as:

Exercise Questions with Solutions of Class 12 Chemistry Chapter 3 – Chemical Kinetics

3.1

Ans – (i) Order = 2

(ii) Order = 2

(iii) Order = 3/2

(iv) Order = 1

3.2

📌 Key Points to Note from Question

Given Data:

Reaction: 2A + B → A₂B
Rate law: Rate = k[A][B]²
Rate constant (k) = 2.0 x 10^-6 mol^-2 L² s^-1
Initial concentrations of [A] = 0.1 mol L^-1 and [B] = 0.2 mol L^-1
Reduced concentration of [A] = 0.06 mol L^-1

What to Calculate:

Initial rate of reaction (Rate_initial)
New rate of reaction (Rate_new) when [A] = 0.06 mol L⁻¹

Ans – The reaction rate is given as follows:
Rate = k[A][B]²
Substituting the values gives the rate as follows:
Rate = 2.0 × 10^-6 × 0.1 × (0.2)² = 8.0 × 10^-9 mol L⁻¹ s⁻¹
When [A] decreases from 0.10 mol L⁻¹ to 0.06 mol L⁻¹. The quantity of [A] that reacted will be:
= 0.10 − 0.06 = 0.04 mol L⁻¹
Thus, the concentration of reacted B will be:
= 1/2 × 0.04 = 0.02 mol L⁻¹
∴ new [B] = 0.2 – 0.02 = 0.18 mol L⁻¹
The new rate of reaction will be:
Rate = 2.0 × 10^-6 × 0.06 × (0.18)² = 3.89 × 10^-9 mol L⁻¹ s⁻¹
⇒ Rate of reaction = 3.89 × 10^-9 mol L⁻¹ s⁻¹.

3.3

📌 Key Points to Note from Question

Given Data:

Rate law (zero-order): Rate = k
Rate constant k = 2.5 × 10^-4 mol^-1 L s^-1

What to Calculate:

Rate of N₂ production.
Rate of H₂ production.

Ans – Reaction will be
NH₃ → 1/2 N₂ + 3/2 H₂

3.4

Ans – If pressure is measured in bar and time in minutes, the unit of rate will be: bar min^-1
The reaction rate is given as: Rate = k[CH₃OCH₃]^3/2

Thus, we can define the units of the rate constant as:

3.5

Ans – The following variables affect how quickly a response occurs.

Reactant type: The type of reactant has an impact on the velocity of the reaction. Ionic substance reactions, for instance, happen more quickly than covalent compound events.
The reactants’ states are as follows: gas reactions happen extremely quickly, liquid reactions happen swiftly, and solid reactions happen slowly.
Temperature: The reaction rate is also significantly impacted by temperature. The rate of response doubles for each 10°C as the temperature rises.
Catalyst availability: The involvement of a catalyst in an action also influences its pace. Expanding the reaction region, creating fragile intermediaries with the base, and providing an alternate route with less activation energy, speed up the process.

3.6

📌 Key Points to Note from Question

Given Data:

Reaction order: Second order with respect to the reactant.
Rate law: Rate = k[Reactant]².

Key Concept:

Effect on Rate: If the concentration of the reactant is changed, the rate of the reaction is proportional to the square of the concentration: Rate ∝ [Reactant]²

Ans –

3.7

📌 Key Points to Note from Question

Key Concept:

Effect of Temperature on Rate Constant (k):
- As temperature increases, the rate constant (k) increases exponentially.
- This is because higher temperatures provide more molecules with sufficient energy to overcome the activation energy barrier.
Arrhenius Equation: k=Ae^−E_a/RT

What to Highlight:

Rate constant (k) increases exponentially with temperature.
Represented by the Arrhenius equation.

Ans – A reaction’s rate constant (k) rises with temperature and about doubles with roughly every 10° increase in temperature. Arrhenius equation helps one to express the effect.

3.8

Ans – (i) During the interval 30 – 60 sec,

(ii) Initial concentration of A, [A₀] = 0.55 M

When t = 30 sec,

When t = 60 sec,

When t = 90 sec,

∴ Average

3.9

📌 Key Points to Note from Question

Given Data:

Reaction is first order in A and second order in B.

Key Concept:

Differential Rate Equation, Rate = k[A]¹[B]²

Ans – (i) Differential rate equation will be:

(ii) When B = 3B, then the rate equation will be:

∴ rate will be increased by 9 times

(iii) When A = 2A and B = 2B, then the rate equation will be:

∴ rate will be increased by 8 times

3.10

Ans – Assume x for the reaction order for A and y for the reaction order concerning B.
Therefore, we can say:

⇒ x = 1.496 = 1.5

3.11

Ans – Assume the reaction order with respect to A is x and with respect to B is y.
Consequently, we can express:

3.12

Ans –

3.13

📌 Key Points to Note from Question

Given Data:

Reaction is a first-order reaction and their different rate constants.

What to Calculate:

Half life (t_1/2)

Ans – The half-life of a first-order reaction is,

3.14

📌 Key Points to Note from Question

Given Data:

Half-life of ¹⁴C: t_1/2 = 5730 years
Initial ¹⁴C content (N₀): 100%
Remaining ¹⁴C content (N): 80%

Ans – Decay in radioactive form follows first-order kinetics.

3.15

Ans –

3.16

📌 Key Points to Note from Question

Given Data:

Rate constant (k): 60 s⁻¹
Reduction in concentration: [A]/[A]₀ = 1/16

Key Concept:

For a first-order reaction: t = [ln([A]₀/[A])]/k

Ans – We know,

The initial concentration of the reactant has reduced to to 1/16^th. Substituting the values, we get:

3.17

📌 Key Points to Note from Question

Given Data:

Initial mass of ⁹⁰Sr: 1 μg
Half-life (t_1/2): 28.1 years
Time (t): 10 years, 60 years

What to Calculate:

Mass remaining after 10 and 60 years.

Ans – Radioactive decay follows to first-order kinetics. So we can calculate decay constant as,

To find the amount that will remain after 10 years,

To find the amount that will remain after 60 years,

3.18

📌 Key Points to Note from Question

Given Data:

Reaction is first order.

What to Calculate:

Time Required for 99% vs 90% Completion

Ans – For a first order reaction, we can write,

99% completion indicates that x = 99% of a, or x = 0.99 a.

90% completion indicates that x = 90% of a, or x = 0.90 a.

We can now find the ratio as,

Therefore, the time needed for 99% completion of a first-order reaction is two times that needed for 90% completion of the reaction.

3.19

📌 Key Points to Note from Question

Given Data:

Time for 30% decomposition = 40 min

Key Concept:

Formula for first-order reactions:

What to Calculate:

t_1/2

Ans – 30% decomposition indicates that x = 30% of a, or x = 0.30 a
Given that the reaction is of first order, we can write:

Given t = 40 min

Given the rate constant value we can now calculate the half-life period

3.20

Ans – The decomposition of azoisopropane into hexane and nitrogen at 543 K can be written as

The total pressure after time t will be:

Applying the rate constant formula of first order reaction

At t = 360 s

At t = 720 s

∴ Average value will be

3.21

Ans – Thermal decomposition of SO₂Cl₂at constant volume will be,

After time t, total pressure we will be,

Substitute the variable p with the pressure of the reactant at time t.

Applying the rate constant formula of first order reaction

At t = 100 s

When P_t = 0.65 atm

When total pressure is 0.65 atm rate of equation is given by

3.22

Ans – From the given data we can write:

T/^oc	0	20	40	60	80
T/K	273	293	313	333	353
(1/T)/k-1	3.66 × 10^-3	3.41 × 10^-3	3.19 × 10^-3	3.0 × 10^-3	2.83 × 10^-3
10⁵ × k/s	0.0787	4.075	25.7	178	2140
ln K	-7.147	-4.075	-1.359	-0.577	3.063

3.23

📌 Key Points to Note from Question

Given Data:

k = 2.418 × 10⁻⁵ s⁻¹
T = 546 K
E_a = 179.9 kJ/mol

Key Concept:

Arrhenius equation: k=Ae^−E_a/RT

What to Find/Calculate:

Pre-exponential factor A

Ans – Given K = 2.418 × 10^-5 s^-1
T = 546K
E_a = 179.9 kJ mol^-1 = 179.9 × 10³ J mol⁻¹

∴ A = antilog (12.5917)

A = 3.912 × 10¹² s^-1

3.24

📌 Key Points to Note from Question

Given Data:

k = 2.0 × 10⁻² s⁻¹
Initial concentration, [A]₀ = 1.0 mol L⁻¹
Time, t = 100 s

Key Concept:

First-order reaction formula: [A] = [A]₀⋅e^−kt

What to Calculate:

[A], the concentration of A remaining after t = 100 s.

Ans – Given,
k = 2.0 × 10^-2 s^-1
t = 100 s
[A]_o=1.0 mol L^-1
Since the units of k are expressed in s^-1, the reaction is first order.

3.25

📌 Key Points to Note from Question

Given Data:

Half-life t_1/2 = 3.00 hours
Time t = 8 hours

What to Calculate:

Fraction of sucrose remaining after 8 hours.

Ans –

Sucrose undergoes decomposition in accordance with first-order rate law,

3.26

📌 Key Points to Note from Question

Given Data:

Rate constant equation: k = (4.5 × 10¹¹ s^-1)e^-280000K/T

Key Concept:

Arrhenius equation: k = Ae^−E_a/RT

What to Calculate:

Activation energy E_a

Ans –

⇒ E_a = 28000 × R
E_a = 28000 × 8.314 = 232.79 kJ mol^-1

3.27

📌 Key Points to Note from Question

Given Data:

Rate constant equation, log k = 14.34 – 1.25 × 10⁴ K/T
t_1/2 = 256minutes

What to Calculate:

Activation energy E_a.
Temperature at which t_1/2 = 256 minutes.

Ans –

3.28

📌 Key Points to Note from Question

Given Data:

k₁ = 4.5 × 10³ s⁻¹ at T₁ = 10^∘C = 283 K
k₂ = 1.5 × 10⁴ s⁻¹
Activation energy E_a = 60 kJ/mol = 60000 J/mol

Key Concept:

Arrhenius equation (for two temperatures):

What to Calculate:

Temperature T₂

Ans – Given,
k₁ = 4.5 × 10³
T₁ = 10 + 273 = 283 K
k₂ = 1.5 × 10⁴
E_a = 60 kJ mol^-1
Applying Arrhenius equation,

3.29

📌 Key Points to Note from Question

Given Data:

A = 4 × 10¹⁰ s⁻¹
Time for 10% completion at 298 K = Time for 25% completion at 308 K

Key Concept:

Arrhenius equation (for two temperatures):

What to Calculate:

Rate constants k at 318 K and activation energy E_a.

Ans –

3.30

📌 Key Points to Note from Question

Given Data:

Temperature change: 293 K to 313 K
Rate quadruples (rate increases by a factor of 4)

Key Concept:

Arrhenius equation (for two temperatures):

What to Calculate:

Activation energy E_a using the given temperature change and rate change.

Ans – Given that,
k₂ = 4 k₁
T₁ = 293 K
T₂ = 313 K
According to Arrhenius equation,

Related Study Resources of Chapter 3 – Chemical Kinetics

Students can use the links below to get extra study materials for Class 12 Chemistry Chapter 3: Chemical Kinetics

Sl No.	Related Links
1	Class 12 Chemistry Chapter 2 Chemical Kinetics – Important Questions

NCERT Solutions for Class 12 Chemical Kinetics

In Class 12 Chemistry, Chemical Kinetics is a quite crucial chapter covering factors influencing chemical reaction rates. While the board and competitive tests should address important ideas such rate laws, order of the reaction, and activation energy, Our NCERT Solutions for Class 12 Chemical Kinetics offer textbook question step-by-step responses. This improves problem-solving abilities and strengthens the basis on reaction kinetics, thereby helping the pupils to clearly understand fundamental ideas.
One must practise if one truly wants to grasp the ideas of chemical kinetics. Work through Class 12 Chemical Kinetics NCERT answers and consult these for direction. Knowing the fundamental ideas and using them in several situations will confirm your knowledge and increase your ability to solve problems.
All things considered, chemical kinetics is a discipline that truly accomplishes amazing things to teach about the how and what occurs during a chemical reaction. If you review the Class 12 NCERT solutions through Cogniks to improve your learning, they will truly add taste to your studies.