Chapter 3 Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions for Class 12 Chemistry Chapter 3 (Chemical Kinetics) play a crucial role in helping students build a strong conceptual foundation for both CBSE board exams and competitive exams like JEE Main, JEE Advanced, and NEET. This chapter gives students the mathematical foundations for finding the Rate of Reaction, understanding the Order and Molecularity, and calculating the time required for a reaction to complete using Integrated Rate Equations. Understanding these concepts is essential for students as they lead to many of the systems that utilize this technology, including industrial catalyst design and pharmaceutical shelf-life determination.

The NCERT Solutions for Class 12 Chemistry Chapter 3 (Chemical Kinetics) will help students learn how to use both concentration-time graphs and the Arrhenius Equation to calculate the temperature dependence of reaction rates. Well-explained NCERT solutions enable students to strengthen fundamentals, practice important numerical questions, and perform better in term exams, board exams, and national-level competitive tests.

1.0Class 12 Chemistry Chapter 3 Chemical Kinetics: Key Concepts

Chemical Kinetics deals with the rate of chemical reactions and the factors that affect how quickly or slowly a reaction takes place. The main topics covered in this chapter are:

Rate of Reaction: Learn how to calculate the average and instantaneous rate of a chemical reaction and understand how the balanced chemical equation is related to the rate.
Factors Affecting Reaction Rates: Understand how concentration, temperature, and catalysts affect the rate of a chemical reaction.
Order and Molecularity: Learn the difference between the order of a reaction, which is found from experimental data, and molecularity, which refers to the number of reacting species involved in an elementary reaction.
Integrated Rate Equations: Study the integrated rate equations for zero-order and first-order reactions and learn how to use them to solve numerical problems.
Half-Life (t₁/₂): Understand half-life as the time required for the concentration of a reactant to become half of its initial value.
Pseudo First Order Reactions: Learn how some reactions involving more than one reactant can show first-order behaviour when one reactant is present in a large amount. The inversion of cane sugar is a common example.
Temperature Dependence and Arrhenius Equation: Learn how temperature affects the rate constant of a reaction and use the Arrhenius equation to calculate activation energy (Eₐ):
ln(k₂/k₁) = (Eₐ/R) [1/T₁ − 1/T₂]
Collision Theory: Understand that a chemical reaction occurs when particles collide with enough energy and the correct orientation. The minimum energy needed for an effective reaction is called activation energy.

2.0NCERT Solutions Class 12 Chemistry Chapter 3 Chemical Kinetics: Detailed Solutions

INTEXT QUESTIONS

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. Sol. Average rate of reaction

$= - \frac{Δ [ R ]}{Δ t} = - \frac{{ [ R _{2} ] - [ R _{1} ] }}{t _{2} - t _{1}} = - \frac{{ 0.02 - 0.03 }}{25} M min^{- 1} = 4 \times 1 0^{- 4} M Min^{- 1} = \frac{4 \times 1 0 ^{- 4}}{60} Ms^{- 1} = 6.67 \times 1 0^{- 6} M s^{- 1}$

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? Sol. Average rate $= - \frac{1}{2} \frac{Δ [ A ]}{Δ t} = - \frac{1}{2} \frac{{ [ A ] _{2} - [ A ] _{1} }}{t _{2} - t _{1}} =$

$- \frac{1}{2} \frac{{[ 0.4 - 0.5 ]}}{10} = - \frac{1}{2} \frac{{ - 0.1 }}{10} = 0.005 mol L^{- 1} min^{- 1} = 5 \times 1 0^{- 3} M min^{- 1}$

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? Sol. The order of the reaction, $\frac{1}{2} + 2 = 2 \frac{1}{2} = 2.5$ .
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? Sol. 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 $= k [a]^{2} = ka^{2}$ If the concentration of X is increased to three times, then $[X] = 3 amol L^{- 1}$ Now, the rate equation will be : Rate $= k [3 a]^{2} = 9 (ka^{2})$ Hence, the rate of formation will increase by 9 times.

A first order reaction has a rate constant $1.15 \times 1 0^{- 3} s^{- 1}$ . How long will 5g of this reactant take to reduce to 3g? Sol. We know that for a $1^{st}$ order reaction,

$t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ]} = \frac{2.303}{1.15 \times 1 0 ^{- 3}} lo g \frac{5}{3} = \frac{2.303}{1.15 \times 1 0 ^{- 3}} \times 0.2219 = 444.38 s$

or 444 s (approx)

6. Time required to decompose $SO_{2} Cl_{2}$ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. Sol. We know that for a $1^{st}$ order reaction,

$t_{1/2} = \frac{0.693}{k} .$

It is given that $t_{1/2} = 60 min$

$∴ k = \frac{0.693}{t _{1/2}} = \frac{0.693}{60} = 0.01155 min^{- 1} = 1.155 \times 1 0^{- 2} min^{- 1} = \frac{1.155 \times 1 0 ^{- 2}}{60} sec^{- 1} = 0.01925 \times 1 0^{- 2} s^{- 1} = 1.925 \times 1 0^{- 4} s^{- 1}$

What will be the effect of temperature on rate constant? Sol. The rate constant of a reaction is nearly doubled with a $1 0^{\circ} C$ rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation, $k = Ae^{- Ea / RT}$

Where, A is the Arrhenius factor or the frequency factor T is the temperature R is the gas constant $E_{a}$ is the activation energy

The rate of the chemical reaction doubles for an increase of 10 K in an absolute temperature from 298 K. Calculate $E_{a}$ . Sol. It is given that $T_{1} = 298 K$ $T_{2} = (298 + 10) K = 308 K$ We also know that the rate of the reaction doubles when temperature is increased by 10°. Therefore, let us take the value of $k_{1} = k$ and that $k_{2} = 2 k$ Now, substituting these values in the equation : $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} [\frac{T _{2} - T _{1}}{T _{1} T _{2}}]$ We get : $lo g \frac{2 k}{k} = \frac{E _{a}}{2.303 \times 8.314} [\frac{10}{298 \times 308}]$

$\Rightarrow \Rightarrow lo g 2 = \frac{E _{a}}{2.303 \times 8.314} [\frac{10}{298 \times 308}] E_{a} = \frac{2.303 \times 8.314 \times 298 \times 308 \times l o g 2}{10} = 52897.78 J mol^{- 1} = 52.9 kJ mol^{- 1}$

The activation energy for the reaction $2 HI (g) \to H_{2} (g) + I_{2} (g)$ is $209.5 kJ mol^{- 1}$ at 581 K. Calculate the fraction of molecules of reactant having energy equal to or greater than activation energy? Sol. The fraction of molecules of reactants having energy equal to or greater than activation energy is given as : $x = e^{- Ea / RT}$

$\Rightarrow \Rightarrow \Rightarrow ln x = - Ea / RT lo g x = \frac{- E _{a}}{2.303 RT} lo g x = \frac{- 209500 J mol ^{- 1}}{2.303 \times 8.314 JK ^{- 1} mol ^{- 1} \times 581} = - 18.8323 Now, x = Anti lo g (- 18.8323) = 1.471 \times 1 0^{- 19}$

NCERT EXERCISE

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants : (i) $3 NO (g) ⟶ N_{2} O (g) + NO_{2}$ ; Rate $= k [NO]^{2}$ (ii) $H_{2} O_{2} (aq) + 3 I^{-} (aq) + 2 H^{+} ⟶ 2 H_{2} O (l) + I_{3}^{-}$ ; Rate $= k [H_{2} O_{2}] [I^{-}]$ (iii) $CH_{3} CHO (g) \to CH_{4} (g) + CO (g)$ ; Rate $= k [CH_{3} CHO]^{3/2}$ (iv) $C_{2} H_{5} Cl (g) ⟶ C_{2} H_{4} (g) + HCl (g)$ ; Rate $= k [C_{2} H_{5} Cl]$

Sol.

(i) Given rate $= k [NO]^{2}$ Therefore, order of the reaction $= 2$ Dimension of $k = \frac{Rate}{[ NO ] ^{2}}$ $= \frac{molL ^{- 1} s ^{- 1}}{( molL ^{- 1} ) ^{2}} = mol^{- 1} L s^{- 1}$ (ii) Give rate $= k [H_{2} O] [I]$ Therefore, order of the reaction $= 2$ . Dimension of $k = \frac{Rate}{[ H _{2} O _{2} ] [ I ^{-} ]}$ $= \frac{molL ^{- 1} s ^{- 1}}{( molL ^{- 1} ) ( molL ^{- 1} )} = L mol^{- 1} s^{- 1}$ (iii) Given rate $= k [CH_{3} CHO]^{3/2}$ Therefore, order of reaction $= \frac{3}{2}$ Dimension of $k = \frac{Rate}{[ CH _{3} CHO ] ^{\frac{3}{2}}}$ $= \frac{molL ^{- 1} s ^{- 1}}{( molL ^{- 1} ) ^{\frac{3}{2}}} = \frac{molL ^{- 1} s ^{- 1}}{mol ^{\frac{3}{2}} L ^{- \frac{3}{2}}} = L^{\frac{1}{2}} mol^{- \frac{1}{2}} s^{- 1}$ (iv) Given rate = $k [C_{2} H_{5} Cl]$ Therefore, order of the reaction $= 1$ . Dimension of $k = \frac{Rate}{[ C _{2} H _{5} Cl ]} = \frac{molL ^{- 1} s ^{- 1}}{molL ^{- 1}} = s^{- 1}$

For the reaction : $2 A + B ⟶ A_{2} B$ the rate $= k [A] [B]^{2}$ with $k = 2.0 \times 1 0^{- 6} mol^{- 2} L^{2} s^{- 1}$ . 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 $0.06 mol L^{- 1}$ . Sol. The initial rate of the reaction is

$Rate = k [A] [B]^{2} = (2.0 \times 1 0^{- 6} mol^{- 2} L^{2} s^{- 1})$

$(0.1 mol L^{- 1}) (0.2 mol L^{- 1})^{2} = 8.0 \times 1 0^{- 9} mol$ $L^{- 1} s^{- 1}$ 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

$= \frac{1}{2} \times 0.04 mol L^{- 1} = 0.02 mol L^{- 1}$

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.

$Rate = k [A] [B]^{2} = (2.0 \times 1 0^{- 6} mol^{- 2} L^{2} s^{- 1})$

$(0.06 mol L^{- 1}) (0.18 mol L^{- 1})^{2}$

$= 3.88 \times 1 0^{- 9} mol L^{- 1} s^{- 1} .$

The decomposition of $NH_{3}$ on platinum surface is zero order reaction. What are the rate of production of $N_{2}$ and $H_{2}$

$if k = 2.5 \times 1 0^{- 4} mol litre^{- 1} s^{- 1} .$

Sol. $2 NH_{3} \to N_{2} + 3 H_{2}$ . Rate of reaction (r)

$= - \frac{1}{2} \frac{d [ NH _{3} ]}{dt} = \frac{d [ N _{2} ]}{dt} = \frac{1}{3} \frac{d [ H _{2} ]}{dt}$

Rate $(r) = k [NH_{3}]^{\circ} = k$

$(∵ zero order reaction) = 2.5 \times 1 0^{- 4}$

$∴ \frac{d [ N _{2} ]}{dt} = r = 2.5 \times 1 0^{- 4} mol lit^{- 1} s^{- 1}$

$\frac{d [ H _{2} ]}{dt} = 3 r$

$= 3 \times 2.5 \times 1 0^{- 4}$

$= 7.5 \times 1 0^{- 4} mol lit^{- 1} sec^{- 1}$

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 presure of dimethyl ether i.e.,

$Rate = k [P_{CH_{3} OCH_{3}}]^{3/2}$

If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants? Sol. $CH_{3} OCH_{3} ⟶ CH_{4} + CO + H_{2}$ Rate $= k [CH_{3} OCH_{3}]^{3/2} = k [P_{CH_{3} OCH_{3}}]^{3/2} &$ unit of rate = bar min $^{- 1}$ Unit of $K = \frac{Rate}{[ P _{CH_{3} OCH_{3}} ] ^{3/2}} = \frac{bar min ^{- 1}}{bar ^{3/2}}$ $= bar^{- 1/2} min^{- 1}$

5. Mention the factors that affect the rate of a chemical reaction. Sol. The important factors on which the rate of a chemical reaction depends are (i) Nature of the reacting species. (ii) Concentration of the reacting species. (iii) Temperature at which a reaction proceeds. (iv) Surface area of the reactants. (v) Presence of a catalyst.

6. A reaction is of 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 ? Sol. Given rate $(r_{0}) = K [A]^{2}$ Therefore, the rate of reaction would increase by 4 times. (i) If [A] is doubled : $r_{1} = k [2 A]^{2} ∴ r_{1} = 4 r_{0}$ Therefore, the rate of reaction would increase by 4 times. (ii) If [A] is reduced to half: $r_{2} = k [\frac{A}{2}]^{2} ∴ r_{2} = \frac{1}{4} r_{0}$

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature effect on rate constant be represented quantitatively? Sol. The rate constant is nearly doubled with a rise in temperature by 100°C for a chemical reaction. The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation, $k = Ae^{- E_{a} / RT}$

$or 2.303 lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} [\frac{T _{2} - T _{1}}{T _{1} T _{2}}]$

where, k is the rate constant at temperature A is the Arrhenius parameter, R is the gas constant, T is the temperature, And $E_{a}$ is the energy of activation which is always positive.

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

t/s	[Ester] / mol $L^{- 1}$
0	0.55
30	0.31
60	0.17
90	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.

Sol.

(i) Average rate of reaction between the time interval

$30 to 60 seconds = \frac{d [ Ester ]}{dt} = - \frac{( 0.17 - 0.31 )}{60 - 30} = \frac{- ( - 0.14 )}{30} = 4.67 \times 1 0^{- 3} mol L^{- 1} s^{- 1}$

(ii) For a pseudo first order reaction,

$k = \frac{2.303}{t} lo g \frac{[ R _{0} ]}{[ R ]}$

$For t = 30 s, k_{1} = \frac{2.303}{30} lo g \frac{0.55}{0.31} = 1.911 \times 1 0^{- 2} s^{- 1} For t = 60 s, k_{2} = \frac{2.303}{60} lo g \frac{0.55}{0.17} = 1.957 \times 1 0^{- 2} s^{- 1} For t = 90 s, k_{3} = \frac{2.303}{90} lo g \frac{0.55}{0.085} = 2.075 \times 1 0^{- 2} s^{- 1}$

Then, average rate constant, $k = \frac{k _{1} + k _{2} + k _{3}}{3}$

$= \frac{( 1.911 \times 1 0 ^{- 2} ) + ( 1.957 \times 1 0 ^{- 2} ) + ( 2.075 \times 1 0 ^{- 2} )}{3} = 1.98 \times 1 0^{- 2} s^{- 1}$

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 concentration of both A and B are doubled ?

Sol.

(i) Rate $= k [A]^{1} [B]^{2}$ (ii) $r_{0} = k [A]^{1} [B]^{2}$

$r_{1} = k [A]^{1} [3 B]^{2} r_{1} = 9 \times r_{0}$

(iii) $r_{0} = k [A^{1}] [B]^{2}$

$r_{2} = k [2 A]^{1} [2 B]^{2} r_{2} = 8 \times r_{0}$

In a reaction between A and B, the initial rate of reaction $(r_{0})$ was measured for different initial concentrations A and B as given below :

A/mol $L^{- 1}$	$B / mol L^{- 1}$	$r_{0} / mol L^{- 1} s^{- 1}$
0.20	0.30	$5.07 \times 1 0^{- 5}$
0.20	0.10	$5.07 \times 1 0^{- 5}$
0.40	0.05	$1.43 \times 1 0^{- 4}$

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

Sol. Let the order of the reaction with respect to A be x and with respect to B be y . Therefore, $r_{0} = k [A]^{x} [B]^{y}$

$5.07 \times 1 0^{- 5} = k [0.20]^{x} [0.30]^{y}$

$5.07 \times 1 0^{- 5} = k [0.20]^{x} [0.10]^{y}$

$1.43 \times 1 0^{- 4} = k [0.40]^{x} [0.05]^{y}$

Dividing equation (i) by (ii), we obtain

$\frac{5.07 \times 1 0 ^{- 5}}{5.07 \times 1 0 ^{- 5}} = \frac{k [ 0.20 ] ^{x} [ 0.30 ] ^{y}}{k [ 0.20 ] ^{x} [ 0.10 ] ^{y}}$

$\Rightarrow 1 = \frac{[ 0.30 ] ^{y}}{[ 0.10 ] ^{y}}$

$\Rightarrow (\frac{0.30}{0.10})^{0} = (\frac{0.30}{0.10})^{y}$

$\Rightarrow y = 0$

Dividing equation (iii) by (i), we obtain

$\frac{1.43 \times 1 0 ^{- 4}}{5.07 \times 1 0 ^{- 5}} = \frac{k [ 0.40 ] ^{x} [ 0.05 ] ^{y}}{k [ 0.20 ] ^{x} [ 0.30 ] ^{y}}$

$\Rightarrow \frac{1.43 \times 1 0 ^{- 4}}{5.07 \times 1 0 ^{- 5}} = \frac{[ 0.40 ] ^{x}}{[ 0.20 ] ^{x}} [Since y = 0 [0.05]^{y} = [0.30]^{y} = 1]$

$\Rightarrow 2.821 = 2^{x}$

$\Rightarrow lo g 2.821 = x lo g 2$

(Taking $lo g$ on both sides)

$\Rightarrow x = \frac{l o g 2.821}{l o g 2} = 1.496 = 1.5$

Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.

The following results have been obtained during the kinetic studies of the reaction :

$2 A + B \to C + D$

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	$6.0 \times 1 0^{- 3}$
II	0.3	0.2	$7.2 \times 1 0^{- 2}$
III	0.3	0.4	$2.88 \times 1 0^{- 1}$
IV	0.4	0.1	$2.40 \times 1 0^{- 2}$

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

Sol. 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,

$Rate = k [A]^{x} [B]^{y}$

According to the question,

$6.0 \times 1 0^{- 3} = k [0.1]^{x} [0.1]^{y}$

$7.2 \times 1 0^{- 2} = k [0.3]^{x} [0.2]^{y}$

$2.88 \times 1 0^{- 1} = k [0.3]^{x} [0.4]^{y}$

$2.40 \times 1 0^{- 2} = k [0.4]^{x} [0.1]^{y}$

Dividing equation (iv) by (i), we obtain

$\frac{2.4 \times 1 0 ^{- 2}}{6.0 \times 1 0 ^{- 3}} = \frac{k [ 0.4 ] ^{x} [ 0.1 ] ^{y}}{k [ 0.1 ] ^{x} [ 0.1 ] ^{y}}$

$\Rightarrow 4 = \frac{[ 0.4 ] ^{x}}{[ 0.1 ] ^{x}} \Rightarrow 4 = (\frac{0.4}{0.1})^{x}$

$\Rightarrow \Rightarrow (4)^{1} = 4^{x} x = 1$

Dividing equation (iii) by (ii), we obtain

$\frac{2.88 \times 1 0 ^{- 1}}{7.2 \times 1 0 ^{- 2}} = \frac{k [ 0.3 ] ^{x} [ 0.4 ] ^{y}}{k [ 0.3 ] ^{x} [ 0.2 ] ^{y}}$

$\Rightarrow 4 = (\frac{0.4}{0.2})^{y}$

$\Rightarrow \Rightarrow \Rightarrow 4 = 2^{y} 2^{2} = 2^{y} y = 2$

Therefore, the rate law is

$Rate = k [A] [B]^{2}$

$\Rightarrow k = \frac{Rate}{[ A ] [ B ] ^{2}}$

From experiment I, we obtain

$k = \frac{6.0 \times 1 0 ^{- 3} mol L ^{- 1} min ^{- 1}}{( 0.1 mol L ^{- 1} ) ( 0.1 mol L ^{- 1} ) ^{2}} = 6.0 L^{2} mol^{- 2} min^{- 1}$

From experiment II, we obtain

$k = \frac{7.2 \times 1 0 ^{- 2} molL ^{- 1} min ^{- 1}}{( 0.3 molL ^{- 1} ) ( 0.2 molL ^{- 1} ) ^{2}} = 6.0 L^{2} mol^{- 2} min^{- 1}$

From exp. III, we obtain,

$k = \frac{2.88 \times 1 0 ^{- 1} molL ^{- 1} min ^{- 1}}{( 0.3 molL ^{- 1} ) ( 0.4 molL ^{- 1} ) ^{2}} = 6.0 L^{2} mol^{- 2} min^{- 1}$

From exp. IV, we obtain,

$k = \frac{2.40 \times 1 0 ^{- 2} molL ^{- 1} min ^{- 1}}{( 0.4 molL ^{- 1} ) ( 0.1 molL ^{- 1} ) ^{2}} = 6.0 L^{2} mol^{- 2} min^{- 1}$

Therefore, rate constant,

$k = 6.0 L^{2} mol^{- 2} min^{- 1} .$

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	$2.0 \times 1 0^{- 2}$
II	......	0.2	$4.0 \times 1 0^{- 2}$
III	0.4	0.4	......
IV	......	0.2	$2.0 \times 1 0^{- 2}$

Sol. The given reaction is of the first order with respect to A and 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
2.0 × 10^{–2} mol L^{–1} min^{–1}

= k (0.1 mol L^{–1}

k = 0.2 min^{–1}

From experiment II

$= [A] = 4.0 \times 1 0^{- 2} mol L^{- 1} min^{- 1} 0.2 min^{- 1} [A] [0.2] mol L^{- 1}$

From experiment III

$2.0 \times 1 0^{- 2} mol L^{- 1} min^{- 1} = 0.2 min^{- 1} [A] [A] = 0.1 mol L^{- 1}$

From experiment IV

2.0 × 10^{–2} mol L^{–1} min^{–1}

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 year $^{- 1}$

Sol.

(i)

$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{200 s ^{- 1}} = 3.4 \times 10^{- 3} sec (approx.)$

(ii)

$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{2 min ^{- 1}} = 0.35 min (approx.)$

(iii)

$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4 year ^{- 1}} = 1.73 \times 10^{- 1} year (approx.)$

The half-life 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. Sol. Here,

$k = \frac{0.693}{t _{2}} = \frac{0.693}{5730} years^{- 1}$

⇒

$t = \frac{2.303}{k} lo g \frac{a}{( a - x )} t_{1/2} of^{14} C = 5730 yr;$

⇒ Also,

$a = 100, (a - x) = 80 t = \frac{2.303 \times 5730}{0.693} lo g \frac{100}{80} = 19042.12 \times 0.097 = 1847 years$

The experimental data for decomposition of $N_{2} O_{5} [2 N_{2} O_{5} ⟶ 4 NO_{2} + O_{2}]$ in gas phase at 318 K are given below :

t/(s)	$10^{2} \times [N_{2} O_{5}] / mol L^{- 1}$
0	1.63
400	1.36
800	1.14
1200	0.93
1600	0.78
2000	0.64
2400	0.53
2800	0.43
3200	0.35

(a) Plot $[N_{2} O_{5}]$ against t. (b) Find the half - life period for the reaction. (c) Draw a graph between $lo g [N_{2} O_{5}]$ and t . (d) What is the rate law? (e) Calculate the rate constant. (f) Calculate the half - life period from k and compare it with (ii). Sol. (a)

(b) Time corresponding to the concentration, $\frac{1.630 \times 1 0 ^{2}}{2} molL^{- 1} = 81.5 mol L^{- 1}$ is the half life. From the graph the half life is obtained as 1450 s.

(c)

$t (s) 0400800120016002000240028003200 10^{2} \times [N_{2} O_{5}] / mol^{- 1} 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 log [N_{2} O_{5}] - 1.79 - 1.87 - 1.94 - 2.03 - 2.11 - 2.19 - 2.28 - 2.37 - 2.46$

Class 12 Chapter 3 Chemical Kinetics Ques 15

(d) The given reaction of the first order as the plot, $lo g [N_{2} O_{5}] v / st$ , is a straight line. Therefore, the rate law of the reaction is Rate $= k [N_{2} O_{5}]$ (e) From the plot, $lo g [N_{2} O_{5}] v / st$ , we obtain Slope $= \frac{- 2.46 - ( - 1.79 )}{3200 - 0} = \frac{- 0.67}{3200}$ Again, slope of the line of the plot log $[N_{2} O_{5}] v / s t$ is given by $- \frac{k}{2.303}$ Therefore, we obtain, $- \frac{k}{2.303} = - \frac{0.67}{3200} \Rightarrow k = 4.82 \times 1 0^{- 4} s^{- 1}$ (f) Half-life is given by,

$t_{1/2} = \frac{0.639}{k} = \frac{0.693}{4.82 \times 1 0 ^{- 4}} s = 1437.75 sec \approx 1438 sec$

This value, 1438 s, is very close to the value that was obtained from the graph.

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 $1/1 6^{th}$ value? Sol. It is known that, $t = \frac{2.303}{k} lo g \frac{a}{( a - x )}$

$If a = 1 then (a - x) = \frac{1}{16} t = \frac{2.303}{60 s ^{- 1}} lo g \frac{1}{1/16} = 0.0462 s = 4.62 \times 1 0^{- 2} sec (approx)$

17. During nuclear explosion, one of the products is $^{90} Sr$ with half-life of 28.1 years. If $1 μ g$ 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. Sol. Here, $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{28.1} y^{- 1}$ It is known that, $t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ]}$ $\Rightarrow 10 = \frac{2.303}{\frac{0.693}{28.1}} lo g \frac{1}{[ R ]}$ $\Rightarrow 10 = \frac{2.303}{\frac{0.693}{28.1}} (- lo g [R])$ $\Rightarrow lo g [R] = - \frac{10 \times 0.693}{2.303 \times 28.1}$ $\Rightarrow [R] = anti lo g (- 0.107) = 0.7816 μ g$ $= anti lo g (\overline{1} .8929) = 0.7814 μ g$

Therefore, $0.7816 μ g$ of $^{90} Sr$ will remain after 10 years. Again, $t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ]}$

$\Rightarrow 60 = \frac{2.303}{\frac{0.693}{28.1}} lo g \frac{1}{[ R ]}$

$\Rightarrow lo g [R] = - \frac{60 \times 0.693}{2.303 \times 28.1}$

$\Rightarrow [R] = antilog (- 0.6425) = 0.2278 μ g$

Therefore, $0.2278 μ g$ of $^{90} Sr$ will remain after 60 years.

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

Sol. For a first order reaction, the time required for 99% completion is

$t_{1} = \frac{2.303}{k} lo g \frac{100}{100 - 99} = \frac{2.303}{k} lo g 100 = 2 \times \frac{2.303}{k}$

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

$t_{2} = \frac{2.303}{k} lo g \frac{100}{100 - 90} = \frac{2.303}{k} lo g 10 = \frac{2.303}{k}$

Therefore, $t_{1} = 2 t_{2}$ 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.

19. A first order reaction takes 40 min for 30% decomposition. Calculate $t_{1/2}$ . Sol. For a first order reaction, $t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ]}$

$k = \frac{2.303}{40 min} lo g \frac{100}{100 - 30} = \frac{2.303}{40 min} lo g \frac{10}{7} = 8.918 \times 1 0^{- 3} min^{- 1}$

Therefore, $t_{1/2}$ of the decomposition reaction is

$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.918 \times 1 0 ^{- 3}} min = 77.7 min (approx.)$

$\Rightarrow 60 = \frac{2.303}{\frac{0.693}{28.1}} lo g \frac{1}{[ R ]}$ $\Rightarrow lo g [R] = - \frac{60 \times 0.693}{2.303 \times 28.1}$ $\Rightarrow [R] = antilog (- 0.6425) = 0.2278 μ g$

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

$(CH_{3})_{2} CHN = NCH (CH_{3})_{2} (g) ⟶ N_{2} (g) + C_{6} H_{14} (g)$ At $t = 0$ $P_{i}$ 00 At $t = t$ $P_{A} = P_{i - x}$ $P_{B} = x P_{C} = x$ After time, t , total pressure, $P_{t} = (P_{i} - x) + x + x$ $\Rightarrow P_{t} = P_{i} + x \Rightarrow x = P_{t} - P_{i}$

Therefore,

$P_{A} = P_{i} - x = P_{i} - (P_{t} - P_{i}) = 2 P_{i} - P_{t}$

For a first order reaction,

$k = \frac{2.303}{t} lo g \frac{P _{i}}{P _{i} - x} = \frac{2.303}{t} lo g \frac{P _{i}}{2 P _{i} - p _{t}}$

When $t = 360 s$ ,

$k = \frac{2.303}{360 s} lo g \frac{35.0}{( 2 \times 35.0 - 54.0 )} = 2.175 \times 1 0^{- 3} s^{- 1}$

When $t = 720 s$ ,

$k = \frac{2.303}{720 s} lo g \frac{35.0}{2 \times 35.0 - 63.0} = 2.235 \times 1 0^{- 3} s^{- 1}$

Hence, the average value of rate constant is

$k = \frac{( 2.175 \times 1 0 ^{- 3} ) + ( 2.235 \times 1 0 ^{- 3} )}{2} s^{- 1} = 2.205 \times 1 0^{- 3} s^{- 1}$

The following data were obtained during the first order thermal decomposition of $SO_{2} Cl_{2}$ at a constant volume.

$SO_{2} Cl_{2} (g) ⟶ SO_{2} (g) + Cl_{2} (g)$

Experiment	Times/s	Total pressure/atm
1	0	0.5
2	100	0.6

Calculate the rate of the reaction when total pressure is 0.65 atm. Sol. The thermal decomposition of $SO_{2} Cl_{2}$ at a constant volume is represented by the following equation.

$SO_{2} Cl_{2} (g) ⟶ SO_{2} (g) + Cl_{2} (g) At t = 0 P_{i} 00 At t = t P_{A} = P_{i} - x P_{B} = xP P_{C} = x$

After time, t, total pressure,

$P_{t} = (P_{i} - x) + x + x$

Therefore,

$P_{A} = P_{i} - x = P_{i} - (P_{t} - P_{i}) = 2 P_{i} - P_{t}$

For a first order reaction,

$k = \frac{2.303}{t} lo g \frac{P _{i}}{P _{i} - p} = \frac{2.303}{t} lo g \frac{P _{i}}{2 P _{i} - p _{t}}$

When $t = 100 s$ ,

$k = \frac{2.303}{100 s} lo g \frac{0.5}{2 \times 0.5 - 0.6} = 2.231 \times 1 0^{- 3} s^{- 1}$

When

$P_{t} = 0.65 atm, P_{i} + x = 0.65 x = 0.65 - P_{i} = 0.65 - 0.5 = 0.15 atm$

Therefore, when the total pressure is 0.65 atm, Pressure of $SOCl_{2}$ is

$p_{SO_{2} Cl_{2}} = P_{i} - x = 0.5 - 0.15 = 0.35 atm$

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,

$Rate = k (p_{SOCl_{2}}) = (2.23 \times 1 0^{- 3} s^{- 1}) (0.35 atm) = 7.8 \times 1 0^{- 4} atm s^{- 1}$

The rate constant for the decomposition of $N_{2} O_{5}$ at various temperature is given below :

T/°C	$10^{5} \times k / s^{- 1}$
0	0.0787
20	1.7
40	25.7
60	178
80	2140

Draw a graph between In k and 1/T and calculate the values of A and $E_{a}$ . Predict the rate constant at 30° and 50°C. Sol. For the given data, we obtain

T/°C	T/K	$1 / T / K^{- 1}$	$10^{5} \times k / s^{- 1}$	In k
0	273	$3.66 \times 1 0^{- 3}$	0.0787	-7.147
20	293	$3.41 \times 1 0^{- 3}$	1.7	-4.075
40	313	$3.19 \times 1 0^{- 3}$	25.7	-1.359
60	333	$3.0 \times 1 0^{- 3}$	178	-0.577
80	353	$2.83 \times 1 0^{- 3}$	2140	3.063

Class 12 Chapter 3 Chemical Kinetics Ques 22

Slope of the line, $\frac{y _{2} - y _{1}}{x _{2} - x _{1}} = 12.301 K$ According to Arrhenius equations, Slope $= - \frac{E _{a}}{R}$

$\Rightarrow E_{a} = - Slope \times R = - (- 12.301 K) \times (8.314 JK^{- 1} mol^{- 1}) = 102.27 kJ mol^{- 1}$

Again, In $k = In A - \frac{E _{a}}{RT}$ In $A = In k + \frac{E _{a}}{RT}$ When $T = 273 K$ , In $k = - 7.147$ Then, $In A = - 7.147 + \frac{102.27 \times 1 0 ^{3}}{8.314 \times 273} = 37.911$ Therefore, $A = 2.91 \times 1 0^{16}$ When $T = 30 + 273 K = 303 K$ ,

$\frac{1}{T} = 0.0033 K = 3.3 \times 1 0^{- 3} K$

Then, at $\frac{1}{T} = 3.3 \times 1 0^{- 3} K$ , In $k = - 2.8$ Therefore, $k = 6.08 \times 1 0^{- 2} s^{- 1}$ Again, when $T = 50 + 273 K = 323 K$ ,

$\frac{1}{T} = 0.0031 K = 3.1 \times 1 0^{- 3} K$

Then, at $\frac{1}{T} = 3.1 \times 1 0^{- 3} K$ ,

$In k = 37.911 - \frac{102.27}{8.314} \times 3.1 \times 1 0^{- 3} = 37.911 - 0.0381 = 37.8729$

Therefore, $k = 0.38 s^{- 1}$

23. The rate constant for decomposition of hydrocarbons is $2.418 \times 1 0^{- 5} s^{- 1}$ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor. Sol. According to the Arrhenius equation,

$\Rightarrow \Rightarrow k = Ae^{- E_{a} / RT} \Rightarrow In k = In A - \frac{E _{a}}{RT} lo g k = lo g A - \frac{E _{a}}{2.303 RT} lo g A = lo g k + \frac{E _{a}}{2.303 RT} = lo g (2.418 \times 1 0^{- 5} s^{- 1}) + \frac{179.9 \times 1 0 ^{3} J mol ^{- 1}}{2.303 \times 8.314 Jk ^{- 1} mol ^{- 1} \times 546 K} = (0.3855 - 5) + 17.2082 = 12.5917$

Therefore,

$A = antilog (12.5917) = 3.9 \times 1 0^{12} s^{- 1}$

Consider a certain reaction A → Products with $k = 2.0 \times 1 0^{- 2} s^{- 1}$ . Calculate the concentration of A remaining after 100s if the initial concentration of A is $1.0 mol L^{- 1}$ . Sol. Since the unit of k is $s^{- 1}$ , the given reaction is a first order reaction. Therefore, $k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}$

$\Rightarrow 2.0 \times 1 0^{- 2} s^{- 1} = \frac{2.303}{100 s} lo g \frac{1.0}{[ A ]}$

$\Rightarrow 2.0 \times 1 0^{- 2} s^{- 1} = \frac{2.303}{100 s} (- lo g [A])$ $\Rightarrow lo g [A] = \frac{- 2.0 \times 1 0 ^{- 2} \times 100}{2.303}$ $\Rightarrow [A] = anti lo g (\frac{- 2.0 \times 1 0 ^{- 2} \times 100}{2.303}) = 0.135 mol L^{- 1}$ Hence, the remaining concentration of A is $0.135 mol L^{- 1}$ .

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? Sol. For a first order reaction, $k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}$ It is given that, $t_{1/2} = 3.00$ hours Therefore, $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{3} h^{- 1} = 0.231 h^{- 1}$ Then, $0.231 h^{- 1} = \frac{2.303}{8 h} lo g \frac{[ R ] _{0}}{[ R ]}$

$\Rightarrow lo g \frac{[ R ] _{0}}{[ R ]} = \frac{0.231 h ^{- 1} \times 8 h}{2.303} \Rightarrow \frac{[ R ] _{0}}{[ R ]} = anti lo g (0.8024) \Rightarrow \frac{[ R ] _{0}}{[ R ]} = 6.3445 \Rightarrow \frac{[ R ]}{[ R ] _{0}} = 0.1576 \approx 0.158$

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

The decomposition of hydrocarbon follows the equation $k = (4.5 \times 1 0^{11} s^{- 1}) e^{- 28000 K / T}$ . Calculate $E_{a}$ . Sol. The given equation is

$k = (4.5 \times 1 0^{11} s^{- 1}) e^{- 28000 K / T}$

Arrhenius equation is given by,

$k = Ae^{- E_{a} / RT}$

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

$\frac{E _{a}}{RT} = \frac{28000 K}{T}$

⇒

$E_{a} = R \times 28000 K^{-} = 8.314 J K^{- 1} mol^{- 1} \times 28000 K = 232792 J mol^{- 1} = 232.792 kJ mol^{- 1}$

The rate constant for the first order decomposition of $H_{2} O_{2}$ is given by the following equation :

$lo g k = 14.34 - 1.25 \times 1 0^{4} K / T$

Calculate $E_{a}$ for this reaction and at what temperature will its half - period be 256 minutes ? Sol. Arrhenius equation is given by,

$k = Ae^{- E_{a} / RT}$

$\Rightarrow In k = In A - \frac{E _{a}}{RT} \Rightarrow lo g k = lo g A - \frac{E _{a}}{2.303 RT}$

The given equation is,

$lo g k = 14.34 - 1.25 \times 1 0^{4} K / T$

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

$\frac{E _{a}}{2.303 RT} = \frac{1.25 \times 1 0 ^{4} K}{T}$

$\Rightarrow E_{a} = 1.25 \times 1 0^{4} K \times 2.303 \times R = 1.25 \times 1 0^{4} K \times 2.303 \times 8.314 J K^{- 1} mol^{- 1} = 239339.3 J mol^{- 1} = 239.34 kJ mol^{- 1}$

Also, when $t_{1/2} = 256$ minutes,

$k = \frac{0.693}{t _{1/2}} = \frac{0.693}{256} = 2.707 \times 1 0^{- 3} min^{- 1} = 4.51 \times 1 0^{- 5} s^{- 1}$

It is also given that,

$lo g k = 14.34 - 1.25 \times 1 0^{4} K / T$

$\Rightarrow lo g (4.51 \times 1 0^{- 5}) = 14.34 - \frac{1.25 \times 1 0 ^{4} K}{T} \Rightarrow 0.654 - 5 = 14.34 - \frac{1.25 \times 1 0 ^{4} K}{T}$

$\Rightarrow T = \frac{1.25 \times 1 0 ^{4} K}{18.686} = 668.95 K = 669 K$

28. The decomposition of A into product has value of k as $4.5 \times 1 0^{3} s^{- 1}$ at $1 0^{\circ} C$ and energy of activation $60 kJ mol^{- 1}$ . At what temperature would k be $1.5 \times 1 0^{4} s^{- 1}$ ? Sol. From Arrhenius equation, we obtain,

$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$

$lo g \frac{1.5 \times 1 0 ^{4}}{4.5 \times 1 0 ^{3}} = \frac{6.0 \times 1 0 ^{4} J mol ^{- 1}}{2.303 \times 8.314 JK ^{- 1} mol ^{- 1}} (\frac{T _{2} - 283}{283 T _{2}})$

$\Rightarrow 0.5229 = 3133.627 (\frac{T _{2} - 283}{283 T _{2}})$ $\Rightarrow \frac{0.5229 \times 283 T _{2}}{3133.627} = T_{2} - 283$ $\Rightarrow 0.9528 T_{2} = 283$ $\Rightarrow T_{2} = 297.019$ $K = 297$ $K = 2 4^{\circ} C$ Hence, k would be $1.5 \times 1 0^{4} s^{- 1}$ at $2 4^{\circ} C$

29. The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308 K. If the value of A is $4 \times 1 0^{10} s^{- 1}$ . Calculate k at 318 K and $E_{a}$ .

Sol. For a first order reaction,

$t = \frac{2.303}{k} lo g \frac{a}{a - x}$

At $298 K, t_{1} = \frac{2.303}{k _{1}} lo g \frac{100}{90} = \frac{0.1054}{k _{1}}$ At $308 K, t_{2} = \frac{2.303}{k _{2}} lo g \frac{100}{75} = \frac{0.2877}{k _{2}}$ According to the question, $t = t^{'}$ $\Rightarrow \frac{0.1054}{k _{1}} = \frac{0.2877}{k _{2}}$ $\Rightarrow \frac{k _{2}}{k _{1}} = 2.73$ From Arrhenius equation, we obtain

$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$

$\Rightarrow lo g (2.73) = \frac{E _{a}}{2.303 \times 8.314} (\frac{308 - 298}{298 \times 308})$

$E_{a} = \frac{2.303 \times 8.314 \times 298 \times 308 \times 0.4361}{308 - 298} = 76640.26 J mol^{- 1} = 76.64 kJ mol^{- 1}$

To calculate k at 318 K, It is given that,

$A = 4 \times 1 0^{10} s^{- 1}, T = 318 K$

Again, from Arrhenius equation, we obtain

$lo g k = lo g A - \frac{E _{a}}{2.303 RT} = lo g (4 \times 1 0^{10}) - \frac{76.6 \times 1 0 ^{3}}{2.303 \times 8.314 \times 318} = (0.6021 + 10) - 12.5876 = - 1.9855$

Therefore, $k = Antilog (- 1.9855)$

$= 1.05 \times 1 0^{- 2} s^{- 1}$

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. Sol. From Arrhenius equation, we obtain,

$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$

It is given that, $k_{2} = 4 k_{1}$

$T_{1} = 293 K; T_{2} = 313 K$

Therefore,

$lo g \frac{4 k _{1}}{k _{1}} \Rightarrow 0.6021 = \frac{E _{a}}{2.303 \times 8.314} (\frac{313 - 293}{293 \times 313}) = \frac{20 \times E _{a}}{2.303 \times 8.314 \times 293 \times 313}$

$\Rightarrow E_{a} = \frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20}$

$= 52863.33 J mol^{- 1} = 52.86 kJ mol^{- 1}$

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

3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

Access chapter-wise NCERT Solutions for Class 12 Chemistry with easy-to-follow answers to textbook questions, key concepts, important formulas, and step-by-step solutions to numerical problems.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Solutions
Chapter 2	Electrochemistry
Chapter 4	d- and f-Block Elements
Chapter 5	Coordination Compounds
Chapter 6	Haloalkanes and Haloarenes
Chapter 7	Alcohols, Phenols and Ethers
Chapter 8	Aldehydes, Ketones and Carboxylic Acids
Chapter 9	Amines
Chapter 10	Biomolecules

4.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 3

Easy Graph-Based Explanations: Understand how to find the slope and intercept from graphs such as ln[R] vs. t and 1/T vs. log k with simple explanations.
Step-by-Step Numerical Solutions: Solve Chemical Kinetics numericals with clear steps, correct formulas, unit conversions, and proper use of logarithm values. Special attention is given to units such as s⁻¹ for first-order rate constants.
Simple Explanation of Half-Life: Learn why the half-life of a first-order reaction does not depend on the initial concentration of the reactant.
Clear Understanding of Catalysts: Understand how a catalyst increases the rate of a reaction by providing an alternative reaction pathway with lower activation energy.
Important Formulas and Concepts: Revise important Chemical Kinetics formulas, reaction rates, order of reaction, molecularity, activation energy, Arrhenius equation, and half-life in one place.
Exam-Focused Practice: Get clear solutions to NCERT questions and numerical problems to strengthen concepts and prepare for Class 12 Chemistry exams.

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NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

1.0Class 12 Chemistry Chapter 3 Chemical Kinetics: Key Concepts

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