NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

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.

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.

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Chemical Kinetics Class 12 NCERT Notes
Chemical Kinetics Class 12 NCERT Exemplar Solutions

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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 1594744535593

(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:

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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

1594744682490 1594744682077

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 1594744755760

Plot $[N_2O_5]$ against t.

Answer :

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

1594744892475 1594744892159

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:

1594744778959 1594744778214

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:

1594744983681 1594744983358

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 1594745470023

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:

1594744804544 1594744803921

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:

1594744815280 1594744814878

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:

1594744831806 1594744831445

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 1594745092246
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.

1594745551536 1594745548735

Calculate the rate constant.

Answer :

Decomposition is represented by equation-

1594745741332 1594745740908
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.
$( SO_2Cl_2 g) \rightarrow SO_2 (g) + Cl_2 (g)$

1594745212870 1594745210169

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 1594745222570

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:

1594745242910 1594745242519

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 1594745257653
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 1594745277231
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 1594745326105
= 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 1594745378630

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

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