NCERT solutions for class 12 chemistry chapter 3 Chemical Kinetics

NCERT Solutions for class 12 Chemistry Chapter 3 (Exercise Questions)

Question 3.1(i) From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

$3NO(g)\to N_{2}ORate=k{\left[NO\right]}^2$

Answer :

Given pieces of information

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

so the order of the reaction is 2

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

$\begin{array}{rl}& ={\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}/{\mathrm{mol}}^2{L}^{-2}\\ & =L{\mathrm{mol}}^{-1}{\mathrm{s}}^{-1}\end{array}$

Question 3.1(ii) From the rate expression for the following reactions determine their order of reaction and the dimensions of the rate constants.

(ii) $H_2{O}_2(aq)+3I^{-}(aq)+2H^{+}\to 2H_{2}O(I)+I_3^{-}$ $Rate=k\left[H_2{O}_2\right]\left[I^{-}\right]$

Answer :

Given rate = $k[H_2{O}_2][I^{-}]$

therefore the order of the reaction is 2

Dimension of k = $rate/[H_2{O}_2][I^{-}]$

$$\begin{gathered} =mol{L}^{-1}{s}^{-1}/(mol{L}^{-1})(mol{L}^{-1}) \\ =Lmo{l}^{-1}{s}^{-1} \end{gathered}$$

Question 3.1(iii) From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

$C{H}_{3}CHO(g)\to C{H}_4(g)+CO(g)Rate=k{\left[C{H}_{3}CHO\right]}^{3/2}$

Answer :

Given $Rate=k[C{H}_{3}CHO{]}^{3/2}$

therefore the order of the reaction is 3/2

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

$$\begin{gathered} =mol{L}^{-1}{s}^{-1}/(mol{L}^{-1}{)}^{3/2} \\ =mol{L}^{-1}{s}^{-1}/mo{l}^{3/2}{L}^{-3/2} \\ =L^{1/2}mo{l}^{-1/2}{s}^{-1} \end{gathered}$$

Question 3.1(iv) From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

$C_2{H}_{5}Cl(g\to )C_2{H}_4(g)+HCL(g)Rate=k\left[C_2{H}_{5}Cl\right]$

Answer :

$rate=k[C_2{H}_{5}Cl]$

so the order of the reaction is 1

and the dimension of k = $rate/[C_2{H}_{5}Cl]$

$$\begin{gathered} =mol{L}^{-1}{s}^{-1}/mol{L}^{-1} \\ =s^{-1} \end{gathered}$$

Question 3.2 For the reaction:
$2A+B\to A_{2}B$
the rate = $k\left[A\right]{\left[B\right]}^2$ with $k=2.0\times {10}^{-6}mo{l}^{-2}{L}^2{s}^{-1}$ . Calculate the initial rate of the reaction when $\left[A\right]=0.1mol{L}^{-1},\left[B\right]=0.2mol{L}^{-1}$ . Calculate the rate of reaction after $\left[A\right]$ is reduced to $0.06mol{L}^{-1}$ .

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 $\times {10}^{-9}mo{l}^{-2}{L}^2{s}^{-1}$

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

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 $[B^{{}^{\prime }}=0.18]$

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

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

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

Question 3.3 The decomposition of $N{H}_3$ on platinum surface is zero order reaction. What are the rates of production of $N_2$ and $H_2$ if $k=2.5\times {10}^{-4}mo{l}^{-1}L{s}^{-1}$ ?

Answer :

The decomposition of $N{H}_3$ on the platinum surface reaction

$2N{H}_3(s)\stackrel{Pt}{\to }{N}_2(g)+3H_2(g)$

therefore,

Rate = $-\frac{1}{2}\frac{d[N{H}_3]}{dt}=\frac{d[N_2]}{dt}=\frac{1}{3}\frac{d[H_2]}{dt}$

For zero order reaction rate = k

therefore, $-\frac{1}{2}\frac{d[N{H}_3]}{dt}=\frac{d[N_2]}{dt}=\frac{1}{3}\frac{d[H_2]}{dt}=k$

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 3.4 The decomposition of dimethyl ether leads to the formation of CH₄, H_2, and CO and the reaction rate is given by Rate $=k{\left[{\mathrm{CH}}_3{\mathrm{OCH}}_3\right]}^{3/2}$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 $=k{\left[P_{{\mathrm{CH}}_3{\mathrm{OCH}}_3}\right]}^{3/2}$ If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Answer :

Given that

$rate=k(P_{C{H}_{3}OC{H}_3}{)}^{3/2}$

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

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

$$\begin{gathered} =barmi{n}^{-1}/(bar{)}^{3/2} \\ =ba{r}^{-1/2}mi{n}^{-1} \end{gathered}$$

Question 3.5 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 3.6(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 $x\to 2x$ . So the rate of reaction would be $R=k[2x{]}^2=4k{x}^2=4R$

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

Question 3.6(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 $x\to \frac{x}{2}$ . So the rate of reaction would be $R=k[\frac{x}{2}{]}^2=\frac{k{x}^2}{4}=\frac{R}{4}$

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

Question 3.7 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,

$K=A{e}^{-E_a/RT}$ where k is rate constant

A is Arrhenius factor

R is gas constant

T is temperature and

$E_a$ is the activation energy

Question 3.8 In pseudo first order hydrolysis of ester in water, the following results were obtained:

(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-

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

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

Question 3.9(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 3.9(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 3.9(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 3.10 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:

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

$$\begin{gathered} 5.07\times {10}^{-5}=k[0.2{]}^x[0.3{]}^y \\ 5.07\times {10}^{-5}=k[0.2{]}^x[0.1{]}^y \\ 1.43\times {10}^{-4}=k[0.4{]}^x[0.05{]}^y \end{gathered}$$ 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,

$x=\frac{\log 2.821}{\log 2}$
= 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 3.11 The following results have been obtained during the kinetic studies of the reaction:
2A + B $\to$ 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,
$$\begin{gathered} 6\times {10}^{-3}=k[0.1{]}^x[0.1{]}^y \\ 7.2\times {10}^{-2}=k[0.3{]}^x[0.2{]}^y \\ 2.88\times {10}^{-2}=k[0.3{]}^x[0.4{]}^y \\ 2.4\times {10}^{-2}=k[0.4{]}^x[0.1{]}^y \end{gathered}$$ 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, $k=rate/[A][B{]}^2$
$$\begin{gathered} =6\times {10}^{-3}/(0.1)\times (0.1{)}^2 \\ =6L^{2}mo{l}^{-2}mi{n}^{-1} \end{gathered}$$

Hence the rate constant of the reaction is $=6L^{2}mo{l}^{-2}mi{n}^{-1}$

Question 3.12 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:

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] = $0.2mol/L$

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

from the experiment 4th,

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

Question 3.13 (1) Calculate the half-life of a first order reaction from their rate constants given below:
$200s^{-1}$

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 3.13 (2) Calculate the half-life of a first order reaction from their rate constants given below:
$2mi{n}^{-1}$

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

Answer :

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

Question 3.14 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 3.15 (1) The experimental data for decomposition of

$2N_2{O}_5\to 4N{O}_2+O_2$
in gas phase at 318K are given below:

Plot $[N_2{O}_5]$ against t.

Answer :

On increasing time, the concentration of $N_2{O}_5$ gradually decreasing exponentially.

Question 3.15 (2) The experimental data for decomposition of $2N_2{O}_5\to 4N{O}_2+O_2$ 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 3.15 (3) The experimental data for decomposition of $N_2{O}_5[2N_2{O}_5\to 4N{O}_2+O_2]$ in gas phase at 318K are given below:

Draw a graph between $\log [N_2{O}_5]$ and t.

Answer :

Question 3.15 (4) The experimental data for decomposition of $N_2{O}_5[2N_2{O}_5\to 4N{O}_2+O_2]$ 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 3.15 (5) The experimental data for decomposition of $N_2{O}_5[2N_2{O}_5\to 4N{O}_2+O_2]$ 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 3.15 (6) The experimental data for decomposition of $N_2{O}_5[2N_2{O}_5\to 4N{O}_2+O_2]$ 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 3.15(7) The rate constant for a first order reaction is $60s^{-1}$ . 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,
$t=\frac{2.303}{k}\log \frac{[R{]}_0}{[R]}$
$$\begin{gathered} =\frac{2.303}{60}\log \frac{1}{1/16} \\ =\frac{2.303}{60}\log 2^4 \end{gathered}$$
$=4.6\times {10}^{-2}s$ (nearly)

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

Question 3.17 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,

$\begin{array}{rl}& t=\frac{2.303}{k}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}\\ & \Rightarrow 60=\frac{2.303}{\frac{0.693}{28.1}}\log \frac{1}{[\mathrm{R}]}\\ & \Rightarrow \log [\mathrm{R}]=-\frac{60\times 0.693}{2.303\times 28.1}\\ & \begin{array}{rl}\Rightarrow [\mathrm{R}]& =\mathrm{antilog}(-0.6425)\\ & =\mathrm{antilog}(\bar{1}.3575)\\ & =0.2278\mu \mathrm{g}\end{array}\end{array}$
Thus 0.2278 $\mu g$ of ${Sr}^{90}$ will remain after 60 years.

Question 3.18 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,
$t^1=\frac{2.303}{k}\log \frac{100}{100-99}$
$=\frac{2.303}{k}\log 100$
$=2\times (\frac{2.303}{k})$

CASE- II
for 90% complition,

$t^2=\frac{2.303}{k}\log \frac{100}{100-90}$
$=\frac{2.303}{k}\log 10$
$=(\frac{t^1}{2})$

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

Question 3.19 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%)

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

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

Question 3.20 For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.

Calculate the rate constant.

Answer :

$$\begin{array}{lcc}{\left({\mathrm{CH}}_3\right)}_2\mathrm{CHN}=\mathrm{NCH}{\left({\mathrm{CH}}_3\right)}_{2g}\to {\mathrm{N}}_{2g}+{\mathrm{C}}_6{\mathrm{H}}_{14g}& & \\ \text{ Anitialpressure }& P_0& 0& 0\\ \text{ Pressureaftertimet }& P_0-P& P& P\\ \text{ Totalpressureaftertimet }\left(P_t\right)=(P_0-P)+P+P=P_0+P& & & \\ \text{ or }P=P_t-P_0& & & \\ R_0\propto P_0\text{ and }R\propto P_0-P& & \end{array}$$

On substituting the value of $p$,

$$R\propto P_0(P_t-P_0)i.eR\propto 2P_0-P_t$$

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

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

when t = 270sec

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

So, $k_{avg}=k_1+k_2/2$
$=2.21\times {10}^{-3}{s}^{-1}$

Question 3.21 The following data were obtained during the first order thermal decomposition of $S{O}_{2}C{l}_2$ at a constant volume.
$(S{O}_{2}C{l}_{2}g)\to S{O}_2(g)+C{l}_2(g)$

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

Answer :

The thermal decomposition of $S{O}_{2}C{l}_2$ is shown here;

${\mathrm{SO}}_2{\mathrm{Cl}}_2\to {\mathrm{SO}}_2+{\mathrm{Cl}}_2$

Initial pressure $p_{0}00$
Pressure at time $t{p}_0-ppp$
Let initial pressure, $P_0\propto R_0$
Pressure at time $t,p_t=p_0-p+p+p=p_0+p$
Pressure of reactants at time $t,p_0-p=2p_0-p_t\propto R$

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
$k=\frac{2.303}{100}\log \frac{0.5}{2\ast 0.5-0.6}$
$=2.231\times {10}^{-3}{s}^{-1}$

when $p_t=0.65atm$

$p=p_t-p_0$
= 0.65 - 0.5
= 0.15 atm

So, $p{(}_{S{O}_{2}C{l}_2})=p_0-p$
= 0.5 - 0.15
= 0.35 atm

Thus, rate of reaction, when the total pressure is 0.65 atm
rate = k( $p{(}_{S{O}_{2}C{l}_2})$ )
= $2.31\times {10}^{-3}\times 0.35$
= 7.8 $\times {10}^{-4}atm{s}^{-1}$