NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

Edited By Vishal kumar | Updated on Mar 6, 2023 - 12:05 p.m. IST | #CBSE Class 12th

NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics- Students appearing in the Class 12 board exam must check these NCERT Class 12 Chemistry Chapter 4 solutions. NCERT solutions for Class 12 Chemistry chapter 4 Chemical Kinetics covers all the questions from NCERT books for Class 12 Chemistry.

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In NCERT Class 12 Chemistry solutions chapter 4, there are questions and solutions of some important topics like average and instantaneous rate of a reaction, factors affecting the rate of reaction, integrated rate equations for zero and first-order reactions, etc. Read further to know all the Chemical Kinetics Class 12 exercise solutions.

Also Read,

Check NCERT Class 12 solutions for other subjects.

Rate of reaction- It is defined as the rate of change in concentration of reactant or product. Unit of rate is $mole\:L^{-1}S^{-1}$ .

Reactants, R $\rightarrow$ Products, P

nA+mB $\rightarrow$ pC+qD

$Rate=-\frac{1}{n}\frac{\Delta \left [ A \right ]}{\Delta t}=-\frac{1}{m}\frac{\Delta \left [ B \right ]}{\Delta t}=+\frac{1}{p}\frac{\Delta \left [ C \right ]}{\Delta t}=+\frac{1}{n}\frac{\Delta \left [ D \right ]}{\Delta t}$

NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics -

Question 4.1 For the reaction $R\rightarrow P$ , the concentration of a reactant change from $0.03M$ to $0.02M$ in $25$ minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Answer :

We know that,

The average rate of reaction = $\frac{-\Delta [R]}{\Delta t}$

= $-\frac{ [R]_{2}-[R]_{1}}{t_{2}-t_{1}}$

= $\frac{0.03-0.02}{25}Mmin^{-1}$

= $4\times 10^{-4}Mmin^{-1}$

In seconds we need to divide it by 60. So,

= $\frac{4\times 10^{-4}}{60}Msec^{-1}$

= 6.67 $\times 10^{-6}Msec^{-1}$

Question 4.2 In a reaction, $2A\rightarrow Products$ P, 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?

Answer :

According to the formula of an average rate

= $-\frac{1}{2}\frac{\Delta[ A]}{\Delta t}$ (final concentration - initial conc.)/time interval

= $-\frac{1}{2}\frac{[ A]_{2}-[A]_{1}}{t_{2}-t_{1}}$

= $-\frac{1}{2}\frac{[ 0.4]-[0.5]}{10}$

= $\frac{0.1}{20}Mmin^{-1}$

= $5\times 10^{-3}Mmin^{-1}$

Question 4.3 For a reaction, $A+B\rightarrow product$ ; the rate law is given by, $r=k\left [ A \right ]^{1/2}\left [ B \right ]^{2}$ . What is the order of the reaction?

Answer :

Order of reaction = Sum of power of concentration of the reactant in the rate law expressions

So, here the power of A = 0.5

and power of B = 2

order of reaction = 2+0.5 =2.5

Question 4.4 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 ?

Answer :

The order of a reaction means the sum of the power of concentration of the reactant in rate law expression.

So rate law expression for the second-order reaction is $R=k[x]^{2}$ here R = rate

if the concentration is increased to 3 times means $x^{'}_{new}=3x$

new rate law expression = $R^{'}=k[3x]^{2}$ = $9k[x]^{2}$ = 9R

the rate of formation of Y becomes 9 times faster than before

Question 4.5 A first order reaction has a rate constant $1.15\times 10^{-3}s^{-1}$ . How long will $5g$ of his reactant take to reduce to $3g$ ?

Answer :

Given data,

initial conc. = 5g

final conc. = 3g

rate const. for first-order = $1.15\times 10^{-3}s^{-1}$

We know that for the first-order reaction,

$t=\frac{2.308}{k}\log\frac{[R]_{0}}{[R]}$

$=\frac{2.308}{1.15\times 10^{-3}}\log\frac{5}{3}$ [log(5/3)= 0.2219]

= 444.38 sec (approx)

Question 4.6 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.

Answer :

We know that t(half ) for the first-order reaction is $\frac{0.693}{k} = t_{1/2}$

and we have given the value of half time $t_{1/2} = 60 min$

thus, $k = \frac{0.693}{60}min^{-1}$

= 0.01155 /min

OR = 1.1925 $\times 10^{-4}sec$

Alternative method

we can also solve this problem by using the first-order reaction equation.

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

put $[R]=[R_{0}]/2$

Question 4.7 What will be the effect of temperature on rate constant ?

Answer :

The rate constant of the reaction is nearly doubled on rising in 10-degree temperature.

Arrhenius equation depicts the relation between temperature and rates constant.

$k = Ae^{-\frac{E_{a}}{RT}}$

A= Arrhenius factor

Ea = Activation energy

R = gas constant

T = temperature

Question 4.8 The rate of the chemical reaction doubles for an increase of $10K$ in absolute temperature from $298K$ Calculate $E_{a}$ .

Answer :

Given data

$T_{1}$ (initial temperature) = 298K and $T_{2}$ (final temperature)= 308K

And we know that rate of reaction is nearly doubled when temperature rise 10-degree

So, $\frac{k_{2}}{k_{1}}=2$ and R = 8.314 J/mol/K

now, $\log\frac{k_{2}}{k_{1}}=\frac{ E_{a}}{2.303}[\frac{T_{2}-T_{1}}{T_{1}T_{2}}]$

On putting the value of given data we get,

$\log2=\frac{ E_{a}}{2.303}[\frac{10}{298\times 308}]$

Activation energy ( $E_{a}$ ) = $\frac{2.303\times 8.314\times 298\times 308\times \log2}{10}Jmol^{-1}$

=52.9 KJ/mol(approx)

Question 4.9 The activation energy for the reaction $2HI(g)\rightarrow H_{2}+I_{2}(g)$ is $209.5 KJ mol^{-1}$ at $518 K$ . Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Answer :

We have

Activation energy = 209.5KJ/mol

temperature= 581K

R = 8.314J/mol/K

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

$x= e^{-E_{a}/RT}$

taking log both sides we get

$\log x = -\frac{E_{a}}{RT}$

$=\frac{209500Jmol^{-1}}{2.303\times 8.314Jmol^{-1}K^{-1}\times 581}$

= 18.832

x = antilog(18.832)

= 1.471 $\times 10^{-19}$

NCERT Solutions for class 12 chemistry chapter 4

Question 4.1(i) From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

$3NO(g)\rightarrow N_{2}O\: \: Rate= k\left [ NO \right ]^{2}$

Answer :

Given pieces of information

Rate = $k[NO]^{2}$

so the order of the reaction is 2

The dimension of k = $Rate/[NO]^{2}$

$\\=molL^{-1}s^{-1}/mol^{2}L^{-2}\\ =L\ mol^{-1}\ s ^{-1}$

Question 4.1(ii) From the rate expression for the following reactions determine their order of reaction and the dimensions of the rate constants.

(ii) $H_{2}O_{2}(aq)+3I^{-}(aq)+2H^{+}\rightarrow 2H_{2}O(I)+I_{3}^{-}$ $Rate=k\left [ H_{2}O_{2}\right ]\left [ I^{-} \right ]$

Answer :

Given rate = $k[H_{2}O_{2}][I^{-}]$

therefore the order of the reaction is 2

Dimension of k = $rate/[H_{2}O_{2}][I^{-}]$

$\\=mol\ L^{-1}\ s^{-1}/(mol\ L^{-1})(mol\ L^{-1})\\ =L\ mol^{-1}\ s^{-1}$

Question 4.1(iii) From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

$CH_{3}CHO(g)\rightarrow CH_{4}(g)+CO(g)\: \: Rate=k\left [ CH_{3} CHO\right ]^{3/2}$

Answer :

Given $Rate = k[CH_{3}CHO]^{3/2}$

therefore the order of the reaction is 3/2

and the dimension of k $=Rate/[CH_{3}CHO]^{3/2}$

$\\=molL^{-1}s^{-1}/(molL^{-1})^{3/2}\\ =molL^{-1}s^{-1}/mol^{3/2}L^{-3/2}\\ =L^{1/2}\ mol^{-1/2}\ s^{-1}$

Question 4.1(iv) From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

$C_{2}H_{5}Cl(g\rightarrow )C_{2}H_{4}(g)+HCL(g) \: Rate=k\left [ C_{2}H_{5}Cl \right ]$

Answer :

$rate = k[C_{2}H_{5}Cl]$

so the order of the reaction is 1

and the dimension of k = $rate/[C_{2}H_{5}Cl]$

$\\=molL^{-1}s^{-1}/mol L^{-1}\\ =s^{-1}$

Question 4.2 For the reaction:
$2A+B\rightarrow A_{2}B$
the rate = $k\left [ A \right ]\left [ B \right ]^{2}$ with $k=2.0\times 10^{-6}mol^{-2}L^{2}s^{-1}$ . Calculate the initial rate of the reaction when $\left [ A \right ]=0.1 mol L^{-1}\: \: ,\left [ B \right ]=0.2 mol L^{-1}$ . Calculate the rate of reaction after $\left [ A \right ]$ is reduced to $0.06molL ^{-1}$ .

Answer :

The initial rate of reaction =

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

substitute the given values of [A], [B] and k,

rate = $2\times 10^{-6}\times 0.1\times (0.2)^{2}$

=8 $\times 10^{-9}mol^{-2}\ L^{2}\ s^{-1}$

When [A] is reduced from 0.1 mol/L to 0.06 mol/L $[A^{'}=0.06]$

So, conc. of A reacted = 0.1-0.06 = 0.04 mol/L

and conc. of B reacted = 1/2(0.04) = 0.02mol/L

conc. of B left = (0.2-0.02) = 0.18 mol/L $[B^{'}=0.18]$

Now, the rate of the reaction is (R) = $k[A^{'}][B^{'}]$

$=2\times 10^{-6}\times 0.06\times (0.18)^{2}$

= $3.89\times 10^{-6}$ $mol L^{-1}s^{-1}$

Question 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 $k=2.5\times 10^{-4}mol^{-1}L s^{-1}$ ?

Answer :

The decomposition of $NH_{3}$ on the platinum surface reaction

$2NH_{3}(s)\overset{Pt}{\rightarrow}N_{2}(g)+3H_{2}(g)$

therefore,

Rate = $-\frac{1}{2}\frac{d[NH_{3}]}{dt}=\frac{d[N_{2}]}{dt}=\frac{1}{3}\frac{d[H_{2}]}{dt}$

For zero order reaction rate = k

therefore, $-\frac{1}{2}\frac{d[NH_{3}]}{dt}=\frac{d[N_{2}]}{dt}=\frac{1}{3}\frac{d[H_{2}]}{dt}=k$

So $\frac{d[N_{2}]}{dt}= 2.5\times 10^{-4}mol\ L^{-1}\ s^{-1}$

and the rate of production of dihydrogen $(H_{2})$ = 3 $\times$ (2.5 $\times 10^{-4}$ ) $mol\ L^{-1}\ s^{-1}$

= 7.5 $\times 10^{-4}$ $mol\ L^{-1}\ s^{-1}$

Question 4.4 The decomposition of dimethyl ether leads to the formation of $CH_{4}$ , $H_{2}$ and and the reaction rate is given by $Rate=k\left [ CH_{3}OCH_{3} \right ]^{3/2}$ 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., $Rate=k\left [ P_{CH_{3}OCH_{3} \right ]^{3/2}$ If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Answer :

Given that

$rate = k(P_{CH_{3}OCH_{3}})^{3/2}$

So, the unit of rate is bar/min .( $bar\ min^{-1}$ )

And thus the unit of k = unit of rate $/(bar)^{3/2}$

$\\=bar\ min^{-1}/(bar)^{3/2}\\ =bar^{-1/2}min^{-1}$

Question 4.5 Mention the factors that affect the rate of a chemical reaction.

Answer :

The following factors that affect the rate of reaction-

the concentration of reactants
temperature, and
presence of catalyst

Question 4.6(i) A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is doubled

Answer :

Let assume the concentration of reactant be x

So, rate of reaction, $R =k[x]^{2}$

Now, if the concentration of reactant is doubled then $x\rightarrow 2x$ . So the rate of reaction would be $R = k[2x]^{2} = 4kx^{2}=4R$

Hence we can say that the rate of reaction increased by 4 times.

Question 4.6(ii) A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half ?

Answer :

Let assume the concentration of reactant be x

So, rate of reaction, R = $k[x]^{2}$

Now, if the concentration of reactant is doubled then $x\rightarrow \frac{x}{2}$ . So the rate of reaction would be $R = k[\frac{x}{2}]^{2} = \frac{kx^{2}}{4}=\frac{R}{4}$

Hence we can say that the rate of reaction reduced to 1/4 times.

Question 4.7 What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

Answer :

The rate constant is nearly double when there is a 10-degree rise in temperature in a chemical reaction.

effect of temperature on rate constant be represented quantitatively by Arrhenius equation,

$K =Ae^{-E_{a}/RT}$ where k is rate constant

A is Arrhenius factor

R is gas constant

T is temperature and

$E_{a}$ is the activation energy

Question 4.8 In pseudo first order hydrolysis of ester in water, the following results were obtained:

1594744535986

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

Answer :

The average rate of reaction between the time 30 s to 60 s is expressed as-

$R = \frac{d[ester]}{dt}$

$\\=(0.31-0.17)/60-30\\ =0.14/30=4.67\times 10^{-3} mol\ L^{-}\ s^{-}$

Question 4.9(i) A reaction is first order in A and second order in B.

(i)Write the differential rate equation.

Answer :

the reaction is first order in A and second order in B. it means the power of A is one and power of B is 2

The differential rate equation will be-

$-\frac{d[R]}{dt}=k[A][B]^{2}$

Question 4.9(ii) A reaction is first order in A and second order in B.

(ii) How is the rate affected on increasing the concentration of B three times?

Answer :

If the concentration of [B] is increased by 3 times, then

$-\frac{d[R]}{dt}=k[A][3B]^{2}$

$=9k[A][B]^{2}$

Therefore, the rate of reaction will increase 9 times.

Question 4.9(iii) A reaction is first order in A and second order in B.

(iii) How is the rate affected when the concentrations of both A and B are doubled?

Answer :

If the concentration of [A] and[B] is increased by 2 times, then

$-\frac{d[R]}{dt}=k[2A][2B]^{2}$

$=8k[A][B]^{2}$

Therefore, the rate of reaction will increase 8 times.

Question 4.10 In a reaction between A and B, the initial rate of reaction (r0) was measure for different initial concentrations of A and B as given below:

1649395389830

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

Answer :

we know that
rate law ( $r_{0}$ ) = $k[A]^{x}[B]^{y}$
As per data

$\\5.07\times 10^{-5} =k[0.2]^{x}[0.3]^{y}\\ 5.07\times 10^{-5}=k[0.2]^{x}[0.1]^{y}\\ 1.43\times 10^{-4}=k[0.4]^{x}[0.05]^{y}$ these are the equation 1, 2 and 3 respectively

Now, divide eq.1 by equation2, we get
$1= (0.3/0.1)^{y}$
from here we calculate that y = 0

Again, divide eq. 2 by Eq. 3, we get
Since y =0 also substitute the value of y
So,
= $(\frac{1.43}{0.507})= (\frac{0.4}{0.2})^{x}$
= $2.821=2^{x}$

taking log both side we get,

$x = \frac{\log2.821}{\log2}$
= 1.496
= approx 1.5
Hence the order of reaction w.r.t A is 1.5 and w.r.t B is 0(zero)

Question 4.11 The following results have been obtained during the kinetic studies of the reaction:
2A + B $\rightarrow$ C + D

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

Answer :

Let assume the rate of reaction wrt A is $x$ and wrt B is $y$ . So, the rate of reaction is expressed as-
Rate = $k[A]^{x}[B]^{y}$

According to given data,
$\\6\times 10^{-3}=k[0.1]^{x}[0.1]^{y}\\ 7.2\times 10^{-2}=k[0.3]^{x}[0.2]^{y}\\ 2.88\times 10^{-2} =k[0.3]^{x}[0.4]^{y}\\ 2.4\times 10^{-2} =k[0.4]^{x}[0.1]^{y}$ these are the equation 1, 2, 3 and 4 respectively

Now, divide the equation(iv) by (i) we get,
$4 = (0.4/0.1)^{x}$
from here we calculate that $x =1$

Again, divide equation (iii) by (ii)
$4 =(0.4/0.2)^{y}$
from here we can calculate the value of y is 2

Thus, the rate law is now, $Rate = k[A][B]^{2}$
So, $k = rate/[A][B]^{2}$
$\\= 6\times 10^{-3}/(0.1)\times (0.1)^{2}\\ =6\ L^{2}\ mol^{-2}\ min^{-1}$

Hence the rate constant of the reaction is $=6\ L^{2}\ mol^{-2}\ min^{-1}$

Question 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]/molL^-1	[B]/molL^-1	Initial rate/ mol L^-1 min^-1
I	0.1	0.1	2*10^-2
II	-	0.2	4*10^-2
III	0.4	0.4	-
IV		0.2	2*10^-2

Answer :

The given reaction is first order wrt A and zero order in wrt B. So, the rate of reaction can be expressed as;
$Rate = k[A][B]^{0}$
Rate = k[A]

from exp 1,
$2\times 10^{-2}=k(0.1)$
k = 0.2 per min.

from experiment 2nd,
$4\times 10^{-2}=0.2[A] \\$
[A] = $0.2\ mol/L$

from experiment 3rd,
$rate =(0.2\ min^{-1})\times (0.4\ mol/L)$
$= 0.08\ mol\ L^{-1}\ min^{-1}$

from the experiment 4th,

$2\times 10^{-2}=0.2*[A]$
from here [A] = 0.1 mol/L

Question 4.13 (1) Calculate the half-life of a first order reaction from their rate constants given below:
$200 s^{-1}$

Answer :

We know that,
half-life ( $t_{1/2}$ ) for first-order reaction = $0.693/k$
= $0.693/200$
$\approx 3.4\times 10^{-3}s$

Question 4.13 (2) Calculate the half-life of a first order reaction from their rate constants given below:
$2\: \: min ^{-1}$

Answer :

the half-life for the first-order reaction is expressed as ;

$t_{1/2}=0.693/k$
= 0.693/2
= 0.35 min (approx)

Question 4.13 (3) Calculate the half-life of a first order reaction from their rate constants given below:
$4 years ^{-1}$

Answer :

The half-life for the first-order reaction is $t_{1/2}= 0.693/k$
= 0.693/4
= 0.173 year (approximately)

Question 4.14 The half-life for radioactive decay of $^{14} C$ 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.

Answer :

Given ,
half-life of radioactive decay = 5730 years
So, $t_{1/2}= 0.693/k$
$k = 0.693/5730$ per year

we know that, for first-order reaction,
$t = \frac{2.303}{k} \log\frac{[R_{0}]}{[R]}$
$t = \frac{2.303}{.693/5730} \log\frac{100}{80}$
= 1845 years (approximately)

Thus, the age of the sample is 1845 years

Question 4.15 (1) The experimental data for decomposition of

$2 N_2 O_5 \rightarrow 4 NO_2 + O_2$
in gas phase at 318K are given below:

1594744756099

Plot $[N_2O_5]$ against t.

Answer :

On increasing time, the concentration of $N_{2}O_{5}$ gradually decreasing exponentially.

1594744892475

Question 4.15 (2) The experimental data for decomposition of $2 N_2O_5 \rightarrow 4 NO_2 + O_2$ in gas phase at 318K are given below:

Find the half-life period for the reaction.

Answer :

The half-life of the reaction is-
The time corresponding to the $1.63 \times 10^{2}/2$ mol/ L = 81.5 mol /L is the half-life of the reaction. From the graph, the answer should be in the range of 1400 s to 1500 s.

Question 4.15 (3) The experimental data for decomposition of $N_2O_5 [2N_2O_5 \rightarrow 4NO_2 + O_2]$ in gas phase at 318K are given below:

Draw a graph between $\log [ N_2 O_5 ]$ and t.

Answer :

$t/S$	$10^{2}\times [N_{2}O_{5}]$	$\log[N_{2}O_{5}]$
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

1594745470627

Question 4.15 (4) The experimental data for decomposition of $N_2O_5 [2N_2O_5 \rightarrow 4NO_2 + O_2]$ in gas phase at 318K are given below:

What is the rate law ?

Answer :

Here, the reaction is in first order reaction because its log graph is linear.
Thus rate law can be expessed as $Rate = k[N_{2}O_{5}]$

Question 4.15 (5) The experimental data for decomposition of $N_2O_5 [2N_2O_5 \rightarrow 4NO_2 + O_2]$ in gas phase at 318K are given below:

Calculate the rate constant.

Answer :

From the log graph,

the slope of the graph is = $\frac{-2.46-1.79}{3200}$
= -k/2.303 ..(from log equation)

On comparing both the equation we get,

$-k/2.303 = -0.67/3200$
$k= 3200\times (0.67/3200)$
$k= 4.82 \times 10^{-4}\ s^{-1}$

Question 4.15 (6) The experimental data for decomposition of $N_2O_5 [2N_2O_5 \rightarrow 4NO_2 + O_2]$ in gas phase at 318K are given below:

Calculate the half-life period from k and compare it with (ii).

Answer :

The half life produce = $0.693/k$
= $=0.693/4.82 \times 10^{-4}$
$1.438\times 10^{3} s$

Question 4.15(7) 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 1/16th value?

Answer :

We know that,
for first order reaction,
$t=\frac{2.303}{k}\log\frac{[R]_{0}}{[R]}$
$\\=\frac{2.303}{60}\log\frac{1}{1/16}\\ =\frac{2.303}{60} \log2^{4}$
$=4.6 \times 10^{-2} s$ (nearly)

Hence the time required is $=4.6 \times 10^{-2} s$

Question 4.17 During nuclear explosion, one of the products is $^{90} Sr$ with half-life of 28.1 years. If $1 \mu g$ 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.

Answer :

Given,
half life = 21.8 years
$\therefore\ k=0.693/t_{1/2}$
= 0.693/21.8

and, $t = \frac{2.303}{k}\log\frac{[R]_{0}}{[R]}$

by putting the value we get,

$10= \frac{2.303}{0.693/21.8}\log\frac{1}{[R]}$
$\log[R] = -\frac{10\times 0.693}{2.303\times 21.8}$
taking antilog on both sides,
[R] = antilog(-0.1071)
= 0.781 $\mu g$

Thus 0.781 $\mu g$ of ${Sr}^{90}$ will remain after given 10 years of time.

Again,

1594745092603
Thus 0.2278 $\mu g$ of ${Sr}^{90}$ will remain after 60 years.

Question 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 :

case 1-

for 99% complition,
$t^{1} = \frac{2.303}{k}\log\frac{100}{100-99}$
$= \frac{2.303}{k}\log100$
$=2\times (\frac{2.303}{k})$

CASE- II
for 90% complition,

$t^{2}=\frac{2.303}{k}\log\frac{100}{100-90}$
$=\frac{2.303}{k}\log10$
$=(\frac{t^{1}}{2})$

$t^{1}=2t^{2}$
Hence proved.

Question 4.19 A first order reaction takes 40 min for 30% decomposition. Calculate $t_{1/2}$

Answer :

For the first-order reaction,

$t =\frac{2.303}{k}\log \frac{[R]_{0}}{[R]}$
$k =\frac{2.303}{40}\log \frac{100}{100-30}$ (30% already decomposed and remaining is 70%)

$=8.918\times 10^{-3} min^{-1}$

therefore half life = 0.693/k
= $0.693/8.918\times 10^{-3}$
= 77.7 (approx)

Question 4.20 For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.

Calculate the rate constant.

Answer :

Decomposition is represented by equation-

1594745741332
After t time, the total pressure $p_{T}$ = $p_{0}-p+(p+p) = p_{0}+p$

So, $p = p_{t}-p_{0}$

thus, $p_{0}-p = 2p_{0}-p_{t}$

for first order reaction,

$k= \frac{2.303}{t}\log\frac{p_{0}}{p_{0}-p}$
$= \frac{2.303}{360}\log\frac{p_{0}}{2p_{0}-p_{t}}$
now putting the values of pressures,

when t =360sec

$= \frac{2.303}{360}\log\frac{35}{2*35-54}$
$= 2.175\times 10^{-3}s^{-1}$

when t = 270sec

$= \frac{2.303}{270}\log\frac{35}{2*35-54}$
$= 2.235\times 10^{-3}s^{-1}$

So, $k_{avg}=k_{1}+k_{2}/2$
$=2.21\times 10^{-3}\ s^{-1}$

Question 4.21 The following data were obtained during the first order thermal decomposition of $SO_2 Cl_2$ at a constant volume.
$( SO_2Cl_2 g) \rightarrow SO_2 (g) + Cl_2 (g)$

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

Answer :

The thermal decomposition of $SO_{2}Cl_{2}$ is shown here;
1594745223164

After t time, the total pressure $p_{t}$ = $p_{0}-p+(p+p) = p_{0}+p$

So, $p = p_{t}-p_{0}$

thus, $p_{0}-p = 2p_{0}-p_{t}$

for first order reaction,

$k= \frac{2.303}{t}\log\frac{p_{0}}{p_{0}-p}$
$= \frac{2.303}{360}\log\frac{p_{0}}{2p_{0}-p_{t}}$
now putting the values of pressures, when t = 100s
$k = \frac{2.303}{100}\log\frac{0.5}{2*0.5-0.6}$
$= 2.231 \times 10^{-3}\ s^{-1}$

when $p_{t} = 0.65\ atm$

$p = p_{t}-p_{0}$
= 0.65 - 0.5
= 0.15 atm

So, $p(_{SO_{2}Cl_{2}}) = p_{0}-p$
= 0.5 - 0.15
= 0.35 atm

Thus, rate of reaction, when the total pressure is 0.65 atm
rate = k( $p(_{SO_{2}Cl_{2}})$ )
= $2.31\times 10^{-3}\times 0.35$
= 7.8 $\times 10^{-4}\ atm\ s^{-1}$

Question 4.22 The rate constant for the decomposition of N2O5 at various temperatures
is given below:

Draw a graph between ln k and 1/T and calculate the values of A and
$E_a$ . Predict the rate constant at 30° and 50°C.

Answer :

From the above data,

T/ $C^{0}$	0	20	40	60	80
T/K	273	293	313	333	353
$1/T$ $/$ $K^{-1}$ ( $\times 10^{-3}$ )	3.66	3.41	3.19	3.0	2.83
$10^{5}*K/S^{-}$	0.0787	1.70	25.7	178	2140
$ln\ K$	-7.147	-4.075	-1.359	-0.577	3.063

1594745258069
Slope of line = $\frac{y2-y2}{x2-x1} = -12.30\ K$

According to Arrhenius equations,

Slope = $-E_{a}/R$
$E_{a}=$ 12.30 $\times$ 8.314
= 102.27 $KJ mol^{-1}$

Again,
1594745277635
When T = 30 +273 = 303 K and 1/T =0.0033K
$\ln k= -2.8$

$\therefore$ k = $6.08\times 10^{-2}\ s^{-1}$

When T = 50 + 273 = 323 K and 1/T = 3.1 $\times 10^{-3}$ K
$\ln k = -0.5$
$\therefore$ k = 0.607 per sec

Question 4.23 The rate constant for the decomposition of hydrocarbons is $2.418 \times 10 ^{-5} s ^{-1}$ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Answer :

Given that,
k = $2.418 \times 10 ^{-5} s ^{-1}$
$E_{a}$ = 179.9 KJ/mol
T(temp) = 546K

According to Arrhenius equation,

$k=Ae^{-E_{a}/RT}$
taking log on both sides,
$\log k = \log A - \frac{E_{a}}{2.303 RT}$

$\log A =\log k + \frac{E_{a}}{2.303 RT}$

$=\log (2.418\times 10^{-5}) + \frac{179.9\times 10^{3}}{2.303 \times 8.314 \times 546}$

= (0.3835 - 5) + 17.2082
= 12.5917
Thus A = antilog (12.5917)
A = 3.9 $\times 10^{12}$ per sec (approx)

Question 4.24 Consider a certain reaction A $\rightarrow$ Products with $k = 2.0 \times 10 ^{-2 } s ^{-1}$ . Calculatethe concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol $L^{-1}$

Answer :

Given that,
k = $=2\times 10^{-2}$
t = 100 s
$[A]_0= 1\ mol\ L^{-1}$
Here the unit of k is in per sec, it means it is a first-order reaction.
therefore,
$k = \frac{2.303}{t}\log\frac{[A]_o}{[A]}$
$\\2\times 10^{-2} = \frac{2.303}{100}\log\frac{1}{[A]}\\ 2\times 10^{-2} = -\frac{2.303}{100}\log[A]\\ \log[A]=-\frac{2}{2.303}\\$
$\\A = anti\log\frac{-2}{2.303}\\ A= 0.135 mol\ L^{-1}$

Hence the concentration of rest test sample is 0.135 mol/L

Question 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 first order reaction,

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

given that half life = 3 hrs ( $t_{1/2}$ )

Therefore k = 0.693/half-life
= 0.231 per hour

Now,

$\\0.231 = \frac{2.303}{8}\log\frac{[R]_{0}}{[R]}\\ \log\frac{[R]_{0}}{[R]} = 0.231\times \frac{8}{2.203}$
= antilog (0.8024)
= 6.3445

$[R]_{0}/[R] = 6.3445$

$[R]/[R]_{0} = 0.157$ (approx)

Therefore fraction of sample of sucrose remains after 8 hrs is 0.157

Question 4.26 The decomposition of hydrocarbon follows the equation k=(4.51011s^-1)e^-28000K/T . Calculate $E_{a}$

Answer :

The Arrhenius equation is given by
$k=Ae^{-E_{a}/RT}$ .................................(i)
given equation,
$k = (4.5 × 1011s^{-1}) e^{-28000K/T}$ ............................(ii)

by comparing equation (i) & (ii) we get,

A= 4.51011 per sec
$E_{a}/RT =28000/T$
Activation energy = 28000 $\times$ (R = 8.314)
= 232.792 KJ/mol

Question 4.27 The rate constant for the first order decomposition of $H_2O_2$ is given by the following equation:
$\log k = 14.34 - 1.25 \times 10 ^ 4K/T$ .

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

Answer :

The Arrhenius equation is given by
$k=Ae^{-E_{a}/RT}$

taking log on both sides,

$\log k = \log A -\frac{E_{a}}{2.303RT}$ ....................(i)
given equation,

$\log k = 14.34 - 1.25 \times 10 ^ 4K/T$ .....................(ii)

On comparing both equation we get,

$E_{a}/2.303R=1.25 \times 10^{4}$

activation energy
$\\=1.25 \times 10^{4} \times 2.303 \times 8.314\\ =239.34\ KJ/mol$

half life ( $t_{1/2}$ ) = 256 min

k = 0.693/256
$k = 4.51\times 10^{-5} s^{-1}$

With the help of equation (ii),

$\log4.51\times 10^{-5} s^{-1} = 14.34-\frac{1.25\times 10^{4}}{T}$

$\frac{1.25\times 10^{4}}{T} = 18.686$
T = $\frac{1.25\times 10^{4}}{18.686}$

= 669 (approx)

Question 4.28 The decomposition of A into product has value of k as $4.5 \times 10 ^3 s ^{-1}$ at 10°Cand energy of activation 60 kJ mol–1. At what temperature would k be $1.5 \times 10 ^4 s ^{-1}$ ?

Answer :

The decomposition of A into a product has a value of k as $4.5 \times 10 ^3 s ^{-1}$ at 10°C and energy of activation 60 kJ mol–1.

K1 = $4.5 \times 10 ^3 s ^{-1}$
K2 = $1.5 \times 10 ^4 s ^{-1}$

$E_a$ = 60 kJ mol–1

K2 = $1.5 \times 10 ^4 s ^{-1}$

$log\frac{K_2}{K_1}=\frac{E_a(T_2-T_1)}{2.303RT_1T_2}$

$log(\frac{1.5\times 10^{4}}{4.5\times 10^{3}})=\frac{60(T_2-283)}{2.303\times R\times 283\times T_2}$

$log(\frac{150}{45})=\frac{60(T_2-283)}{5418.61\times T_2}$

$log150-log45=\frac{60T_2-16980}{5418.61\times T_2}$

$2.176-1.653=\frac{60T_2-16980}{5418.61\times T_2}$

$0.5229=\frac{60T_2-16980}{5418.61\times T_2}$

$T_2=\frac{16980}{2733.4}$

$T_2=6.1K$

Question 4.29 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is $4 \times 10 ^{10} s ^{-1}$ . Calculate k at 318K and Ea.

Answer :

We know that,

for a first order reaction-

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

Case 1
At temp. = 298 K
$t = \frac{2.303}{k}\log\frac{100}{90}$
= 0.1054/k

Case 2
At temp = 308 K

$t' = \frac{2.303}{k}\log\frac{100}{75}$
= 2.2877/k'
As per the question
$t' = t$
K'/K = 2.7296

From Arrhenius equation,

1594745326768
= 76640.096 J /mol
=76.64 KJ/mol

k at 318 K
we have , T =318K
A= $4 \times 10^{10}$

Now $\log k = \log A- \frac{E_{a}}{2.303RT}$
After putting the calue of given variable, we get

$\log k = -1.9855$
on takingantilog we get,

k = antilog(-1.9855)

= 1.034 $\times 10^{-2}\ s^{-1}$

Question 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 the Arrhenius equation,

$\log\frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303R}(\frac{T_{2}-T_{1}}{T_{1}T_{2}})$ ...................................(i)
it is given that $k_{2}=4k_{1}$
T1= 293 K

T2 = 313 K
Putting all these values in equation (i) we get,

$\log 4 =\frac{E_{a}}{2.303 \times 8.314}(\frac{313-293}{313 \times 293})$

1594745379486

Activation Energy = 52.86 KJ/mo l
This is the required activation energy

Topics and Sub-topics of Chemical Kinetics Class 12 NCERT Chapter 4

4.1 Rate of a Chemical Reaction
4.2 Factors Influencing Rate of a Reaction
4.3 Integrated Rate Equations
4.4 Temperature Dependence of the Rate of a Reaction
4.5 Collision Theory of Chemical Reactions

NCERT Chapter-wise Solutions for Class 12 Chemistry

Chapter 1	The Solid State
Chapter 2	Solutions
Chapter 3	Electrochemistry
Chapter 4	Chemical Kinetics
Chapter 5	Surface chemistry
Chapter 6	General Principles and Processes of isolation of elements
Chapter 7	The P-block elements
Chapter 8	The d and f block elements
Chapter 9	Coordination compounds
Chapter 10	Haloalkanes and Haloarenes
Chapter 11	Alcohols, Phenols, and Ethers
Chapter 12	Aldehydes, Ketones and Carboxylic Acids
Chapter 13	Amines
Chapter 14	Biomolecules
Chapter 15	Polymers
Chapter 16	Chemistry in Everyday life

More About Class 12 Chemistry Chapter 4 Chemical Kinetics NCERT solutions

A total of 5 marks of questions will be asked in the Class 12 CBSE board exam of Chemistry from this chapter.
In this chapter, there are 30 questions in the exercise and 9 questions that are related to topics studied.
To clear doubts of students, the Chemical Kinetics NCERT solutions are prepared in a comprehensive manner by subject experts.
This chapter is vital for both CBSE Board exam as well as for competitive exams like JEE Mains, VITEEE, BITSAT, etc. so, students must pay special attention to the concepts of this chapter.
The NCERT solutions provided here are completely free and you can also download them for offline use also if you want to prepare or any other subject or any other class.
By referring to the NCERT Solutions for Class 12 chemistry chapter 4 PDF download, students can understand all the important concepts and practice questions well enough before their examination.

NCERT Solutions for Class 12 Subject wise

Benefits of NCERT solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics

First, the easy steps given in the NCERT Class 12 Chemistry solutions chapter 4 will help you to understand the chapter easily.
The revision will be so easy that you always remember the concepts and get very good marks in your class.
Homework will be easy now, all you need to do is check the detailed CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics and you are good to go.
If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get the answers with solutions that will help you score well in your exams.

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Solutions

Frequently Asked Question (FAQs) - NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

Question: What is the weightage of NCERT class 12 Chemistry chapter 4 in JEE Mains?

Answer:

Weightage of NCERT class 12 Chemistry chapter 4 in JEE Mains is 4 marks.

Question: What is the weightage of NCERT class 12 Chemistry chapter 4 in NEET?

Answer:

The weightage of NCERT class 12 Chemistry chapter 4 in NEET is 3%. Solve more previous year NEET papers to get more problems

Question: What is the weightage of NCERT class 12 Chemistry chapter 4 in CBSE board exam?

Answer:

The weightage of NCERT class 12 Chemistry chapter 4 in CBSE board exam is 5 marks. To practice questions on the chapter refer to NCERT book exercise and NCERT exemplar problems.

Question: What are the important topics of this chapter?

Answer:

Important topics of this chemical kinetics are the rate of reaction, concept of collision theory, effect of temperature in activation energy, Arrhenius equation, concept of collision theory.

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i couldnt clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

Upasana Barman 24th Aug, 2022

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Mayankjha.student2007 24th Aug, 2022

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

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if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Nuthalapati Vyshnavi 21st Aug, 2022

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

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I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Muskan Dhalwal 18th Aug, 2022

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

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I will have my boards in Feb 2023 (Class 12) and I still havent completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

khushi.thakurbbk 7th Sep, 2022

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

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NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

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