Chemical Kinetics Question Answers

Chemical Kinetics Question Answers: NCERT Class 12 Chemistry

Current NCERT chapter

Academic session 2026-27

2 Solutions · 0 Topics · 0 Notes resources

This chapter is part of the verified current curriculum mapping for the academic session shown below.

Current NCERT PDF
Read textbook resource Current Exemplar
Practice Exemplar problems

Chapter Details

Chapter
3 — Chemical Kinetics

NCERT Book
Chemistry Parts I and II

Language
English

Edition
Academic session 2026-27

Use this Class 12 Chemistry chapter hub to study Chemical Kinetics from one place. SaralStudy currently provides 2 visible NCERT solution entries on this page.

NCERT Solutions

Review the SaralStudy solution entries currently available for this chapter.

Exercise 1

Q 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.

Q 2:

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

Q 3:

For a reaction, A + B → Product; the rate law is given by, r = k [A]^½ [B]². What is the order of the reaction?

The order of the reaction = 1/2 + 2

= 2.5

Q 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?

The order of reaction is defined as the sum of the powers of concentrations In the rate law.

The rate of second order reaction can be expressed as rate = k [A]²

The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k [X]^{2 ________________________}(1)

Let [X] = a mol L^-1, then equation (1) can be written as:

Rate = k [a]²

= ka²

If the concentration of X is increased to three times, then [X] = 3a mol L^-1

Now, the rate equation will be:

Rate = k [3a]²

= 9 (ka²)

Hence, the rate of formation will increase by 9 times.

Q 5:

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

From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10 ^{- 3}s^{- 1}

We know that for a 1^storder reaction,

= 444.38 s

= 444 s (approx)

Q 6:

Time required to decompose SO₂Cl₂to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

We know that for a 1^storder reaction,

t½ = 0.693 / k

It is given that t_1/2= 60 min

k = 0.693 / t½

= 0.693 / 60

= 0.01155 min^-1

= 1.155 min^-1

k = 1.925 x 10^-2 s^-1

Q 7:

What will be the effect of temperature on rate constant?

The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

k = Ae^{- E_a / RT}

Where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

E_ais the activation energy

Q 8:

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

It is given that T₁ = 298 K

∴T2 = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k₁ = k and that of k₂ = 2k

Also, R = 8.314 J K ^{- 1} mol ^{- 1}

Now, substituting these values in the equation:

= 52897.78 J mol^{- 1}

= 52.9 kJ mol^{- 1}

Q 9:

The activation energy for the reaction 2HI_(g) → H₂ + I_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?

In the given case:

E_a = 209.5 kJ mol^{- 1} = 209500 J mol^{- 1}

T = 581 K

R = 8.314 JK^{- 1} mol^{- 1}

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as

Exercise 2

Q 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) → N₂O(g) Rate = k[NO]²

(ii) H₂O₂ (aq) + 3 I^- (aq) + 2 H⁺→ 2 H₂O (l) + I₃^- Rate = k[H₂O₂][I^-]

(iii) CH₃CHO(g) → CH₄(g) + CO(g) Rate = k [CH₃CHO]^3/2

(iv) C₂H₅Cl(g) → C₂H₄(g) + HCl(g) Rate = k [C₂H₅Cl]

(i) Given rate = k [NO]²

Therefore, order of the reaction = 2

Dimension of k = Rate / [NO]²

= mol L^-1 s^-1 / (mol L^-1)²

= mol L^-1 s^-1 / mol² L^-2

= L mol^-1 s^-1

(ii) Given rate = k [H₂O₂][I^-]

Therefore, order of the reaction = 2

Dimension of k = Rate / [H₂O₂][I^-]

= mol L^-1 s^-1/ (mol L^-1) (mol L^-1)

= L mol^-1 s^-1

(iii) Given rate =k [CH₃CHO]^3/2

Therefore, order of reaction = 3/2

Dimension of k = Rate / [CH₃CHO]^3/2

= mol L^-1 s^-1/ (mol L^-1)^3/2

= mol L^-1 s^-1/ mol^3/2 L^-3/2

= L^½ mol^-½s^-1

(iv) Given rate = k [C₂H₅Cl]

Therefore,order of the reaction = 1

Dimension of k = Rate / [C₂H₅Cl]

= mol L^-1 s^-1/ mol L^-1

= s^-1

Q 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:

A/ mol L^{- 1}	0.20	0.20	0.40
B/ mol L^{- 1}	0.30	0.10	0.05
r₀/ 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?

Let the order of the reaction with respect to A be xand with respect to B be y.

Therefore,

= 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.

Q 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.

Q 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}

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 x 10^-2mol L^-1min^-1= k (0.1 mol L^-1)

⇒ k = 0.2 min^-1

From experiment II, we obtain

4.0 x 10^-2mol L^-1min^-1= 0.2 min^-1[A]

⇒ [A] = 0.2 mol L^-1

From experiment III, we obtain

Rate = 0.2 min^-1 x 0.4 mol L^-1

= 0.08 mol L^-1min^-1

From experiment IV, we obtain

2.0 x 10^-2mol L^-1min^-1= 0.2 min^-1[A]

⇒ [A] = 0.1 mol L^-1

Q 13:

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

(i) 200 s^-1

(ii) 2 min^-1

(iii) 4 years^-1

(i) Half life, t_½ = 0.693 / k

= 0.693 / 200 s^-1

= 3.47 ×10 ^-3 s (approximately)

(ii) Half life, t_½ = 0.693 / k

= 0.693 / 2 min^-1

= 0.35 min (approximately)

(iii) Half life,t_½ = 0.693 / k

= 0.693 / 4 years^-1

= 0.173 years (approximately)

Q 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.

Here,

K = 0.693 / t_½

= 0.693 / 5730 years^-1

It is known that,

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Q 15:

The experimental data for decomposition of N₂O₅

[2N₂O₅ → 4NO₂ + O₂]

in gas phase at 318K are given below:

t/s	0	400	800	1200	1600	2000	2400	2800	3200
10² × [N₂O₅] mol L^-1	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

(i) Plot [N₂O₅] against t.

(ii) Find the half-life period for the reaction.

(iii) Draw a graph between log[N₂O₅] 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).

(i)

(ii) Time corresponding to the concentration, 1630x10² / 2 mol L^-1 = 81.5 mol L^-1 is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

t/s	10² × [N₂O₅] mol L^-1	Log [N₂O₅]
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, log[N₂O₅] v/s t, is a straight line. Therefore, the rate law of the reaction is

Rate = k [N₂O₅]

(v) From the plot, log[N₂O₅]

v/s t, we obtain

Slope = -2.46 -(1.79) / 3200-0

= -0.67 / 3200

Again, slope of the line of the plot log[N₂O₅] v/s t is given by

- k / 2.303

.Therefore, we obtain,

- k / 2.303 = - 0.67 / 3200

⇒ k = 4.82 x 10^-4 s^-1

(vi) half life is given by,

t_½ = 0.693 / k

= 0.639 / 4.82x10^-4s

=1.438 x 10³s

Or we can say

1438 S

Which is very near to what we obtain from graph.

Q 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?

Q 17:

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

Therefore, 0.7814 μg of ⁹⁰Sr will remain after 10 years.

Again,

Therefore, 0.2278 μg of ⁹⁰Sr will remain after 60 Years

Q 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.

For a first order reaction, the time required for 99% completionis

t₁= 2.303/k Log 100/100-99

= 2.303/k Log 100

= 2x 2.303/k

For a first order reaction, the time required for 90% completion is

t₂= 2.303/k Log 100/100-90

= 2.303/k Log 10

= 2.303/k

Therefore, t₁= 2t₂

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.

Q 19:

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

For a first order reaction,

t= 2.303/k Log [R] _º/ [R]

k = 2.303/40min Log 100/ 100-30

= 2.303/40min Log 10/ 7

= 8.918 x 10^-3 min^-1

Therefore, t_1/2 of the decomposition reaction is

t_1/2= 0.693/k

= 0.693 / 8.918 x 10^-3 min

= 77.7 min (approximately)

Q 2:

For the reaction:

2A + B → A₂B

the rate = k[A][B]²with k= 2.0 x 10^-6mol^-2L²s^-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.

The initial rate of the reaction is

Rate = k [A][B]²

= (2.0 × 10^{- 6}mol^{- 2}L²s^{- 1}) (0.1 mol L^{- 1}) (0.2 mol L^{- 1})²

= 8.0 × 10^{- 9}mol^{- 2}L²s^{- 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= 1/2 x 0.04 mol L^-1 = 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}L²s^{- 1}) (0.06 mol L^{- 1}) (0.18 mol L^{- 1})²

= 3.89 mol L^{- 1}s^{- 1}

Q 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

The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

After time, t, total pressure, P_t = (P_º - p) + p + p

⇒ P_t = (P_º + p)

⇒ p = P_t - P_º

therefore, P_º - p = P_º- P_t - P_º

= 2 P_º- P_t

For a first order reaction,

k = 2.303/t Log P_º/ P_º- p

= 2.303/t Log P_º/ 2 P_º- P_t

When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0 - 54.0

= 2.175 × 10 ^{- 3} s ^{- 1}

When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0 - 63.0

= 2.235 × 10 ^{- 3} s ^{- 1}

Hence, the average value of rate constant is

k = (2.175 × 10 ^{- 3}+ 2.235 × 10 ^{- 3} ) / 2 s ^{- 1}

= 2.21 × 10 ^{- 3}s^{- 1}

Q 21:

The following data were obtained during the first order thermal decomposition of SO₂Cl₂at a constant volume.

SO₂Cl₂₍g) → SO₂₍g) + Cl₂(g)

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.

The thermal decomposition of SO₂Cl₂at a constant volume is represented by the following equation.

After time, t, total pressure,P_t = (P_º - p) + p + p

⇒ P_t = (P_º + p)

⇒ p = P_t - P_º

therefore, P_º - p = P_º- P_t - P_º

= 2 P_º- P_t

For a first order reaction,

k = 2.303/t Log P_º/ P_º- p

= 2.303/t Log P_º/ 2 P_º- P_t

When t= 100 s,

k = 2.303 / 100s log 0.5 / 2x0.5 - 0.6

= 2.231 × 10^{- 3}s^{- 1}

When Pt= 0.65 atm,

P₀+ p= 0.65

⇒ p= 0.65 - P₀

= 0.65 - 0.5

= 0.15 atm

Therefore, when the total pressure is 0.65 atm, pressure of SOCl₂ is

p_SOCL2 = P₀ - 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(p_SOCL2)

= (2.23 × 10 ^{- 3}s ^{- 1}) (0.35 atm)

= 7.8 × 10 ^{- 4}atm s^{- 1}

Q 22:

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

T/°C	0	20	40	60	80
10⁵ X K /S^-1	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 30 º and 50 ºC.

From the given data, we obtain

T/°C	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 X K /S^-1	0.0787	1.70	25.7	178	2140
In k	-7.147	-4.075	-1.359	-0.577	3.063

Slope of the line,

In k= - 2.8

Therefore, k = 6.08x10^-2s^-1

Again when T = 50 + 273K = 323K,

1/T = 3.1 x 10^-3 K

In k = - 0.5

Therefore, k = 0.607 s^-1

Q 23:

The rate constant for the decomposition of hydrocarbons is 2.418 x 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.

k= 2.418 × 10^-5 s^-1

T= 546 K

E_a= 179.9 kJ mol^{- 1} = 179.9 × 10³J mol^{- 1}

According to the Arrhenius equation,

= (0.3835 - 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10¹² s^{- 1}(approximately)

Q 24:

Consider a certain reaction A → Products with k = 2.0 x 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.

k= 2.0 × 110^-2 s^-1

T= 100 s

[A]_o= 1.0 moL ^{- 1}

Since the unit of kis s^{- 1}, the given reaction is a first order reaction.

Therefore, k = 2.303/t Log [A]_º/ [A]

⇒2.0 × 110^-2 s^-1= 2.303/100s Log 1.0/ [A]

⇒2.0 × 110^-2 s^-1= 2.303/100s ( - Log [A] )

⇒ - Log [A] = - (2.0 x 10^-2 x 100) / 2.303

⇒ [A] = antilog [- (2.0 x 10^-2 x 100) / 2.303]

= 0.135 mol L^{- 1}(approximately)

Hence, the remaining concentration of A is 0.135 mol L ^{- 1}.

Q 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?

For a first order reaction,

k = 2.303/t Log [R]_º/ [R]

It is given that, t_1/2= 3.00 hours

Therefore, k = 0.693 / t_1/2

= 0.693 / 3 h^-1

= 0.231 h^{- 1}

Then, 0.231 h^{- 1}= 2.303 / 8h Log [R]_º/ [R]

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

Q 26:

The decomposition of hydrocarbon follows the equation

k = (4.5 x 10¹¹ s^-1) e^{-28000 K/T}

Calculate E_a.

The given equation is

k = (4.5 x 10¹¹ s^-1) e^{-28000 K/T} (i)

Arrhenius equation is given by,

k= Ae^-E_a/RT(ii)

From equation (i) and (ii), we obtain

E_a/ RT = 28000K / T

⇒ E_a= R x 28000K

= 8.314 J K^{- 1}mol^{- 1}× 28000 K

= 232792 J mol ^{- 1}

= 232.792 kJ mol^{- 1}

Q 27:

The rate constant for the first order decomposition of H₂O₂is given by the following equation:

log k = 14.34 - 1.25 x 10⁴K/T

Calculate E_afor this reaction and at what temperature will its half-period be 256 minutes?

Arrhenius equation is given by,

k= Ae ^-E_a/RT

⇒In k = In A - E_a/RT

⇒In k = Log A - E_a/RT

⇒ Log k = Log A - E_a/2.303RT (i)

The given equation is

Log k = 14.34 - 1.25 10⁴ K/T (ii)

From equation (i) and (ii), we obtain

E_a/2.303RT = 1.25 10⁴ K/T

⇒ E_a=1.25 × 10⁴K × 2.303 × R

= 1.25 × 10⁴K × 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,

k = 0.693 / t_1/2

= 0.693 / 256

= 2.707 × 10^{- 3}min ^{- 1}

= 4.51 × 10^{- 5}s^{- 1}

It is also given that, log k= 14.34 - 1.25 × 10⁴K/T

= 668.95 K

= 669 K (approximately)

Q 28:

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

From Arrhenius equation, we obtain

log k₂/k₁= E_a / 2.303 R (T₂ - T₁) / T₁T₂

Also, k₁ = 4.5 × 10³ s^{- 1}

T₁ = 273 + 10 = 283 K

k₂ = 1.5 × 10⁴ s^{- 1}

E_a = 60 kJ mol ^{- 1} = 6.0 × 10⁴ J mol ^{- 1}

Then,

= 297 K = 24°C

Hence, k would be 1.5 × 10⁴ s^{- 1} at 24°C.

Q 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 x 10¹⁰ s^-1. Calculate k at 318 K and E_a.

For a first order reaction,

t = 2.303 / k log a / a - x

At 298 K,

t = 2.303 / k log 100 / 90

= 0.1054 / k

At 308 K,

t' = 2.303 / k' log 100 / 75

= 2.2877 / k'

According to the question,

t = t'

⇒ 0.1054 / k = 2.2877 / k'

⇒ k' / k = 2.7296

From Arrhenius equation,we obtain

To calculate k at 318 K,

It is given that, A = 4 x 10¹⁰ s^-1, T = 318K

Again, from Arrhenius equation, we obtain

Therefore, k = Antilog (-1.9855)

= 1.034 x 10^-2 s ^-1

Q 3:

The decomposition of NH₃on platinum surface is zero order reaction. What are the rates of production of N₂and H₂if k = 2.5 x 10^-4mol^-1L s^-1?

The decomposition of NH₃on platinum surface is represented by the following equation.

= 7.5 × 10^{- 4}mol L^{- 1}s^{- 1}

Q 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.

From Arrhenius equation, we obtain

Hence, the required energy of activation is 52.86 kJ mol^{- 1}.

Q 4:

The decomposition of dimethyl ether leads to the formation of CH₄, H₂and CO and the reaction rate is given by

Rate = k [CH₃OCH₃]^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.,

Rate = k (P_CH3OCH3)^3/2

If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?

If pressure is measured in bar and time in minutes, then

Unit of rate = bar min^{- 1}

Rate = k (P_CH3OCH3)^3/2

⇒ k = Rate / (P_CH3OCH3)^3/2

Therefore, unit of rate constants (k) = bar min^-1/ bar^3/2

= bar^-½ min^{- 1}

Q 5:

Mention the factors that affect the rate of a chemical reaction.

The factors that affect the rate of a reaction areas follows.

(i) Concentration of reactants (pressure in case of gases)

(ii) Temperature

(iii) Presence of a catalyst

Q 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?

Let the concentration of the reactant be [A] = a

Rate of reaction, R = k [A]² = ka²

(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

R = k(2a)²

= 4ka²

= 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. [A] = 1/2 a, then the rate of the reaction would be

R = k(1/2a)²

= 1/4 Ka²

= 1/4 R

Q 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?

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,

K = Ae^{-E_a/ RT}

where, kis 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

Q 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

(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.

(i) Average rate of reaction between the time interval, 30 to 60 seconds, = d[ester] / dt

= (0.31-0.17) / (60-30)

= 0.14 / 30

= 4.67 × 10 - 3mol L^{- 1}s ^{- 1}

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]_º / [R]

For t= 30 s, k₁ = 2.303/ 30 log 0.55 / 0.31

= 1.911 × 10 ^{- 2}s^{- 1}

For t = 60 s, k₂ = 2.303/ 60 log 0.55 / 0.17

= 1.957 × 10^{- 2}s ^{- 1}

For t= 90 s, k₃ = 2.303/ 90 log 0.55 / 0.085

= 2.075 × 10 ^{- 2}s ^{- 1}

Then, average rate constant, k = k₁ + k₂+ k₃/ 3

= 1.911 × 10 ^{- 2}+ 1.957 × 10^{- 2} + 2.075 × 10 ^{- 2} / 3

= 1.981 x 10^-2s ^{- 1}

Q 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?

NCERT Textbook PDF

Use the available NCERT textbook PDF alongside the chapter solutions.

Current NCERT PDF Current Hindi NCERT PDF

NCERT Exemplar

Use the mapped NCERT Exemplar resource for additional chapter practice.

Current Exemplar

Removed / Changed Chapters from the Syllabus

Current and historical curriculum records are shown separately so removed material is not mixed into current practice.

Previous / Removed Chapters

Chapter 1: The Solid State (rationalised)
Chapter 5: Surface Chemistry (rationalised)
Chapter 6: General Principles and Processes of Isolation of Elements (rationalised)
Chapter 7: The p-Block Elements (rationalised)
Chapter 15: Polymers (rationalised)
Chapter 16: Chemistry in Everyday Life (rationalised)

Frequently Asked Questions about Chemical Kinetics - Class 12 Chemistry

- 1. Is Chemical Kinetics in the current curriculum?
- This chapter is part of the verified current curriculum mapping for the academic session shown below. Verified academic session: 2026-27.
- 2. How many NCERT solution entries are currently available for Chemical Kinetics?
- SaralStudy currently shows 2 visible solution entries on this chapter page.
- 3. Are topic mappings available for Chemical Kinetics?
- No public topic mappings are currently available for this chapter.
- 4. Are revision notes available for Chemical Kinetics?
- No verified chapter notes are currently available in this hub.

Latest Blog Posts

Stay updated with our latest educational content and study tips

How to Return to Work After a Career Break: Skills, Jobs and Preparation in 2026

It’s not about starting over if you took a break to care for children, take care of yourself, be healthy or for any other reason. This is a realistic strategy for the recovery of skills, selection of the right job, and overcoming the resume problem. One year, three years or 10 years of a career … Read more

Best Career Options After Class 12 for Average Students in 2026

There’s no need to score 95% or achieve a JEE/NEET rank. You don’t need to score 95% or rank in JEE/NEET to build a solid career. A candid and pragmatic guide to the routes that actually work for students who have an average mark and what to do when faced with the options. If … Read more

Coaching vs Self-Study: Which is Best in 2026 for Competitive Exams?

AI tools, online platforms that are much cheaper than coaching centres and the changing idea of what “structure” even means have upended this argument more in the past two years than in the decade preceding it. Here’s what is really the case for 2026 – and what has actually stayed the same. If you look … Read more

Career Change After 30: How to Transition from Non-Tech to IT or Data Science

Starting to make a new career after 30 isn’t a start from scratch. Marketing, teaching, financial, retail, healthcare, sales and other professionals can pursue careers in technology by applying the existing knowledge and skills they have acquired to new technology skills. There are various ways to enter the technology industry, such as in IT support, … Read more

View All Blog Posts

Benefits of Using Our NCERT Solutions for Class 12 Chemistry

When it comes to excelling in your studies, having a well-structured study guide can make a huge difference. Our NCERT Solutions for Class 12 Chemistry provide you with a comprehensive, easy-to-understand, and exam-focused resource that is specifically tailored to help you maximize your potential. Here are some of the key benefits of using our NCERT solutions for effective learning and high scores:

NCERT Solutions for Effective Exam Preparation

Preparing for exams requires more than just reading through textbooks. It demands a structured approach to understanding concepts, solving problems, and revising thoroughly. Here’s how our NCERT solutions can enhance your exam preparation:

Clear Understanding of Concepts: Our NCERT solutions are designed to break down complex topics into simple, understandable language, making it easier for students to grasp essential concepts in Chemistry. This helps in building a solid foundation for each chapter, which is crucial for scoring high marks.
Step-by-Step Solutions: Each solution is presented in a detailed, step-by-step manner. This approach not only helps you understand how to reach the answer but also equips you with the right techniques to tackle similar questions in exams.
Access to Important Questions: We provide a curated list of important questions and commonly asked questions in exams. By practicing these questions, you can familiarize yourself with the types of problems that are likely to appear in the exams and gain confidence in answering them.
Quick Revision Tool: Our NCERT solutions serve as an excellent tool for last-minute revision. The solutions cover all key points, definitions, and explanations, ensuring that you have everything you need to quickly review before exams.

Importance of Structured Answers for Scoring Higher Marks

In exams, it's not just about getting the right answer—it's also about presenting it in a well-structured and logical way. Our NCERT solutions for Class 12 Chemistry are designed to guide you on how to write answers that are organized and effective for scoring high marks.

Precise and Concise Answers: Our solutions are crafted to provide answers that are to the point, without unnecessary elaboration. This ensures that you don't waste time during exams and focus on delivering accurate answers that examiners appreciate.
Step-Wise Marks Distribution: We understand that exams often allot marks based on specific steps or points. Our NCERT solutions break down each answer into structured steps to ensure you cover all essential points required for full marks.
Improved Presentation Skills: By following the format of our NCERT solutions, you learn how to present your answers in a systematic and logical manner. This helps in making your answers easy to read and allows the examiner to quickly identify key points, resulting in better scores.
Alignment with NCERT Guidelines: Since exams are often set in alignment with NCERT guidelines, our solutions are tailored to follow the exact format and language that is expected in exams. This can improve your chances of scoring higher by meeting the examiner's expectations.

Quick & Popular Links

Fast access to frequently used study resources.

Quick Links

NCERT Sample Papers Study Planner and Calculators NCERT Books CBSE Syllabus Previous Year Question Papers CBSE Board

Chemical Kinetics Question Answers: NCERT Class 12 Chemistry

Chapter Details

NCERT Solutions

Exercise 1

Exercise 2

NCERT Textbook PDF

NCERT Exemplar

Removed / Changed Chapters from the Syllabus

Previous / Removed Chapters

Frequently Asked Questions about Chemical Kinetics - Class 12 Chemistry

Latest Blog Posts

How to Return to Work After a Career Break: Skills, Jobs and Preparation in 2026

Best Career Options After Class 12 for Average Students in 2026

Coaching vs Self-Study: Which is Best in 2026 for Competitive Exams?

Career Change After 30: How to Transition from Non-Tech to IT or Data Science

Benefits of Using Our NCERT Solutions for Class 12 Chemistry

NCERT Solutions for Effective Exam Preparation

Importance of Structured Answers for Scoring Higher Marks

Students Also Read

Discover More on SaralStudy

Latest Blogs

Popular Blogs

Quick & Popular Links

Quick Links

Popular Study Links

What Students Are Viewing

Chemical Kinetics Question Answers: NCERT Class 12 Chemistry

Chapter Details

NCERT Solutions

Exercise 1

Exercise 2

NCERT Textbook PDF

NCERT Exemplar

Removed / Changed Chapters from the Syllabus

Previous / Removed Chapters

Frequently Asked Questions about Chemical Kinetics - Class 12 Chemistry

Latest Blog Posts

How to Return to Work After a Career Break: Skills, Jobs and Preparation in 2026

Best Career Options After Class 12 for Average Students in 2026

Coaching vs Self-Study: Which is Best in 2026 for Competitive Exams?

Career Change After 30: How to Transition from Non-Tech to IT or Data Science

Related Resources

Exam Information

Full Form

Calculator

Glossary

Related Educational Resources

NCERT Solutions

NCERT Books PDF

NCERT Exemplar

CBSE Sample Papers

Previous Year Papers

Important Formulas

Math Tables

Educational Calculators

Full Forms Dictionary

Educational Glossary

Quick Access

Benefits of Using Our NCERT Solutions for Class 12 Chemistry

NCERT Solutions for Effective Exam Preparation

Importance of Structured Answers for Scoring Higher Marks

Students Also Read

Discover More on SaralStudy

Latest Blogs

Popular Blogs

Quick & Popular Links

Quick Links

Popular Study Links

What Students Are Viewing