NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

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# NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

Edited By Vishal kumar | Updated on Mar 06, 2023 12:05 PM 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.

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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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 -

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

Answer :

According to the formula of an average rate

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

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

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

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

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

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

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

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)

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

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

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)

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

Answer :

Given pieces of information

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

so the order of the reaction is 2

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

Answer :

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

therefore the order of the reaction is 2

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

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

Answer :

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

therefore the order of the reaction is 3/2

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

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

Answer :

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

so the order of the reaction is 1

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

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

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) = $k[A^{'}][B^{'}]$

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

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

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

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 $/(bar)^{3/2}$

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

Answer :

The following factors that affect the rate of reaction-

• the concentration of reactants
• temperature, and
• presence of catalyst

Answer :

Let assume the concentration of reactant be x

So, rate of reaction, $\dpi{100} R =k[x]^{2}$

Now, if the concentration of reactant is doubled then $x\rightarrow 2x$ . So the rate of reaction would be $\dpi{100} R = k[2x]^{2} = 4kx^{2}=4R$

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

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.

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

(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.

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.

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.

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.

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)

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 $=6\ L^{2}\ mol^{-2}\ min^{-1}$

 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

Answer :

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

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)

Answer :

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

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

Plot $[N_2O_5]$ against t.

Answer :

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

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.

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

Answer :

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

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,

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

Answer :

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

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 $=4.6 \times 10^{-2} s$

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,

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

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.

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)

Calculate the rate constant.

Answer :

Decomposition is represented by equation-

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

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;

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 $\times 10^{-4}\ atm\ s^{-1}$

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 2.83 $10^{5}*K/S^{-}$ 0.0787 1.7 25.7 178 2140 $ln\ K$ -7.147 -4.075 -1.359 -0.577 3.063

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,

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

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)

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

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,
............................(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

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)

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

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

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

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

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

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

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

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

$T_2=6.1K$

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,

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

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,

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

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
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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.

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### 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.

Also check NCERT Exemplar Class 12 Solutions

### Frequently Asked Question (FAQs)

1. What are the important topics of this chapter?

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.

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

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.

3. What is the weightage of NCERT class 12 Chemistry chapter 4 in NEET?

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

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

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

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

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Hope this information helps you.

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

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.

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.

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.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

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 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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##### Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
##### AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available
##### Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.

4 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available
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