# NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics

## NCERT Solutions For Class 12 Chemistry Chapter 4 PDF Free Download

NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics is one study material which will help students in getting tuned in with the concepts involved in chemical kinetics. Chemical kinetics class,12 NCERT solutions pdf is crucial for the students of CBSE  class 12th. Students aspiring to make a career in the medical or engineering field must practice the class 12 chemistry chapter 4 questions to score well in board exam as well as various competitive entrance exams.

This Chapter presented here is to assist the students to understand the lesson in an easy and interesting way. The class 12 NCERT Chemistry Chapter 4 solutions are created by subject experts as per the latest CBSE syllabus (2018-19). Students must practice the solutions regularly to prepare effectively for their examinations.

## Class 12 NCERT Solutions for Chemical Kinetics

Chemical Kinetics is a branch of chemistry. It deals with the rate of chemical reaction, the factors affecting it, the mechanism of the reaction. Based on the rate of reaction we have 3 types: Instantaneous reactions, Slow reactions and moderately slow reactions. Any chemical reaction that completes in less than 1ps time is called fast reaction. Any chemical reaction happening for some minutes to some years is called a slow reaction. Intermediate chemical reactions which occur between fast and slow chemical reactions are called moderately slow reactions. This was brief on Chemical Kinetics.

### Topics involved in class 12 chemistry chapter 4 Chemical Kinetics

1. The rate of a Chemical Reaction
2. Factors Influencing the Rate of a Reaction
1. Dependence of Rate on Concentration
2. Rate Expression and Rate Constant
3. Order of a Reaction
4. Molecularity of a Reaction
3. Integrated Rate Equations
1. Zero Order Reactions
2. First Order Reactions
3. Half-Life of a Reaction
4. Pseudo First Order Reaction
5. Temperature Dependence of the Rate of a Reaction
1. Effect of Catalyst Ex 4.6 – Collision Theory of Chemical Reactions.

### NCERT Class 12 Chemical Kinetics Important Questions

Q 1. Find out the rate of expression for the given reaction, also find the dimensions of the rate constant and their order of reaction.

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

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

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

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

Ans:

(a) Given rate = $k\left [ NO \right ]^{2}$

Therefore, order of the reaction = 2

Dimensions of $k = \frac{Rate}{\left [ NO \right ]^{2}}$ $\\ = \frac{mol \; L^{-1} s^{-1}}{\left ( mol \; L^{-1} \right )^{2}} \\ \\ = \frac{mol \; L^{-1} s^{-1}}{mol^{2}\; L^{-2}} \\ \\ = L \; mol^{-1} s^{-1}$

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

Therefore, order of the reaction = 2

Dimensions of  $k = \frac{Rate}{\left [ H_{2}O_{2} \right ]\left [ I^{-} \right ]}$ $\\ = \frac{mol \; L^{-1} S^{-1}}{\left ( mol \; L^{-1} \right ) \left ( mol \; L^{-1} \right )} \\ \\ = L \; mol^{-1} s^{-1}$

(c) Given rate = $= k \left [ CH_{3} CHO \right ]^{\frac{3}{2}}$

Therefore, the order of reaction = $\frac{3}{2}$

Dimensions of $k = \frac{Rate}{\left [ CH_{3} CHO \right ]^{\frac{3}{2}}} \\ \\ = \frac{mol \; L^{-1}s^{-1}}{\left (mol \; L^{-1} \right )^{\frac{3}{2}}} \\ \\ = \frac {mol\; L^{-1} s^{-1}}{mol^{\frac{3}{2}}\; L^{\frac{3}{2}}} \\ \\ L^{\frac{1}{2}}\; mol^{ -\frac{1}{2}} \; s^{-1}$

(d) Given rate = $k = \left [ C_{2}H_{5}Cl \right ]$

Therefore, order of the reaction = 1

Dimension of $k = \frac{Rate}{\left [ C_{2}H_{5}Cl \right ]} \\ \\ = \frac{mol\; L^{-1} s^{-1}}{mol \; L^{-1}} \\ \\ = s^{-1}$

Q 2.  The rate is given for a reaction $2A + B \rightarrow A_{2}B$ is $k\left [ A \right ]\left [ B \right ]^{2}$ 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 [A] is reduced to $0.06 \; mol \; L^{-1}$.

Ans:

The initial rate of reaction is

Rate = $k\left [ A \right ]\left [ B \right ]^{2} \\ \\ = \left ( 2.0 \times 10^{-6} mol^{-2} L^{2} s^{-1} \right )\left ( 0.1 \; mol \; L^{-1} \right )\left ( 0.2 \; mol \; L^{-1} \right )^{2} \\ \\ = 8.0 \times 10^{-9} mol^{-2}L^{2} s^{-1}$

When [A] is reduced from $0.1 \; mol\; L^{-1} \; to \; 0.06 \; mol\; L^{-1}$, the concentration of A reacted = $\left (0.1 – 0.06 \right ) \; mol\; L^{-1} = 0.04 \; mol\; L^{-1}$

Therefore, concentration of B reacted $= \frac {1}{2} \times 0.04 \; mol \; L^{-1} = 0.02 \; mol \; L^{-1}$

Then, concentration of B available, $\left [ B \right ] = \left ( 0.2 – 0.02 \right ) mol \; L^{-1} = 0.18\; mol \; L^{-1}$

After [A] is reduced to $0.06 \; mol \; L^{-1}$, the rate of the reaction is given by,

Rate = $k \left [ A \right ]\left [ B \right ]^{2} \\ \\ = \left ( 2.0 \times 10^{6} mol^{-2}L^{2}s^{-1} \right )\left ( 0.06\; mol L^{-1} \right )\left ( 0.18 \; mol \; L^{-1} \right )^{2} \\ \\ = 3.89\; mol \; L^{-1} s^{-1}$

Q 3. The decomposition of $NH_{3}$ on platinum surface is zero order reaction. What are the rates of production of $N_{2} \; and\; H_{2} \; if \; k = 2.5 \times 10^{-4} mol^{-1} L\, s^{-1}$?

Ans:

The decomposition of NH3 on platinum surface is represented by the following equation.

$2NH^{3(g)} \overset{Pt}{\rightarrow} N_{2(g)} + 3H_{2(g)}$

Therefore,

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

However, it is given that the reaction is of zero order.

Therefore,

$-\frac{1}{2}\frac{d\left [ NH_{3} \right ]}{dt} = \frac{d\left [ N_{2} \right ]}{dt} = \frac{1}{3}\frac{d\left [ H_{2} \right ]}{dt} = k \\ \\ = 2.5 \times 10^{-4}\; mol L^{-1} s^{-1}$

Therefore, the rate of production of $N_{2}$ is

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

And, the rate of production of $H_{2}$  is

$\frac{d\left [H_{2} \right ]}{dt} = 3 \times 2.5 \times 10^{-4} mol\;L^{-1}s^{-1} \\ \\ = 7.5 \times 10^{-4} \; mol \;L^{-1}s^{-1}$

Q 4.  The decomposition of dimethyl ether leads to the formation of $CH_{4}, H_{2}, \; and \; CO$ and the reaction rate is given by $Rate = k\left [ CH_{3} O\, C\! H_{3} \right ]^{\frac{3}{2}}$

The rate of reaction is followed by an 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.,

$Rate = k\left ( P_{CH_{3} O\, C\! H_{3}} \right )^{\frac{3}{2}}$

If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Ans:

If pressure is measured in bar and time in minutes, then

Unit of rate = $bas \; min^{-1}$ $Rate = k\left ( P_{CH_{3} O\, C\! H_{3}} \right )^{\frac{3}{2}} \\ \\ \Rightarrow k = \frac{Rate}{k\left ( P_{CH_{3} O\, C\! H_{3}} \right )^{\frac{3}{2}}}$

Therefore, unit of rate constants  $(k) = \frac{bar\; min^{-1}}{bar^{\frac{3}{2}}} \\ \\ = bar^{\frac{-1}{2}}min^{-1}$

Q 5. List the reason which affects the chemical reaction’s rate.

Ans:

The factors which are responsible for the effect in chemical reaction’s rate are:

(a) Temperature

(b) Presence of a catalyst

(c) The concentration of reactants (pressure in case of gases)

Q 6.  A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (a) doubled (b) reduced to half?

Ans:

Let the concentration of the reactant be [A] = a

Rate of reaction, $R = k [A]^{2}\\ \\ = ka^{2}$

(a) If the concentration of the reactant is doubled, i.e [A] = 2a, then the rate if the reaction would be

$R’ = k\left ( A \right )^{2} \\ \\ = 4 ka^{2} \\ \\ = 4 \; R$

Therefore, the rate of the reaction now will be 4 times the original rate.

(b) If the concentration of the reactant is reduced to half, i.e $\left [ A \right ] = \frac{1}{2}a$, then the rate of the reaction would be

$R” = k\left ( \frac{1}{2} a \right )^{2} \\ \\ = \frac{1}{4} k a \\ \\ = \frac{1}{4} R$

Therefore, the rate of the reaction will be reduced to $\frac{1}{4} ^{th}$

Q 7. What change would happen in the rate constant of a reaction when there is a change in its temperature? How can this temperature effect on rate constant be represented quantitatively?

Ans:

When a temperature of $10^{ \circ }$ rises for a chemical reaction then the rate constant increases and becomes near to double of its original value.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,

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

Where,

k = rate constant,

A = Frequency factor / Arrhenius factor,

R = gas constant

T = temperature

$E_{a}$ = activation energy for the reaction.

Q 8. The given is the result obtained from pseudo first order hydrolysis of ester in water.:

 t/s 0 30 60 90 [Ester]mol / L 0.55 0.31 0.17 0.085

(a) Find the avg rate of the reaction between the time intervals of 30 to 60 seconds.

(b) Calculate the pseudo-first-order rate constant for the hydrolysis of the ester.

Ans:

(a) Avg rate of reaction between the time intervals, 30 to 60 seconds,

$= \frac{d\left [ Ester \right ]}{dt} \\ \\ = \frac{0.31 – 0.17}{60 – 30} \\ \\ = \frac{0.14}{30} \\ \\ = 4.67 \times 10^{-3}\; mol \; l^{-1}\; s^{-1}$

(b) For a pseudo first order reaction,

$\\k = \frac{2.303}{t} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]}\\ \\ For \; t = 30\; s \\ \\ k_{1} = \frac{2.303}{30} \log \frac{ 0.55}{ 0.31} \\ \\ = 1.911 \times 10^{-2} s^{-1} \\ \\ For \; t = 60\; s \\ \\ k_{2} = \frac{2.303}{60} \log \frac{ 0.55}{ 0.17} \\ \\ = 1.957 \times 10^{-2} s^{-1} \\ \\ For \; t = 90\; s \\ \\ k_{3} = \frac{2.303}{90} \log \frac{ 0.55}{ 0.085} \\ \\ = 2.075 \times 10^{-2} s^{-1}$

Then, avg rate constant, $k = \frac{k_{1} + k_{2} + k_{3}}{3} \\ \\ = \frac{\left (1.911 \times 10^{-2} \right ) + \left (1.957 \times 10^{-2} \right ) + \left (2.075 \times 10^{-2} \right )}{3} \\ \\ = 1.98 \times 10^{-2}\;s^{-1}$

Q 9. A reaction is first order in A and second order in B.

(a) What is the equation of differential rate?

(b) What would be the change in rate if the concentration of B is increased 3 times?

(c) When the concentration for both A and B is doubled then what would be the change in the rate?

Ans:

(a) The differential rate equation will be

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

(b) If the concentration of B is increased three times, then

$-\frac{d\left [ R \right ]}{dt} = k\left [ A \right ]\left [ 3B \right ]^{2} \\ \\ = 9. k\left [ A \right ]\left [ B \right ]^{2}$

Therefore, the reaction rat will be increased by 9 times.

(c) When the concentrations of both A and B are doubled,

$-\frac{d\left [ R \right ]}{dt} = k\left [ 2 \right ]\left [ 2B \right ]^{2} \\ \\ = 8. k\left [ A \right ]\left [ B \right ]^{2}$

Therefore, the rate of reaction will increase 8 times.

Q10.  In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:

 $A/mol\; L^{-1}$ 0.20 0.20 0.40 $B/mol\; L^{-1}$ 0.30 0.10 0.05 $r_{0}/mol\; L^{-1}\; s^{-1}$ $5.07 \times 10^{-5}$ $5.07 \times 10^{-5}$ $1.43 \times 10^{-4}$

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

Ans:

Let the order of the reaction with respect to A be x and with respect to B be y.

Then,

$r_{0} = k\left [ A \right ]^{x} \left [ B \right ]^{y} \\ \\ 5.07 \times 10^{-5} = k\left [ 0.20 \right ]^{x}\left [ 0.30 \right ]^{y} \;\;\;\; (i) \\ \\ 5.07 \times 10^{-5} = k\left [ 0.20 \right ]^{x}\left [ 0.10 \right ]^{y} \;\;\;\; (ii) \\ \\ 1.43 \times 10^{-4} = k\left [ 0.40 \right ]^{x}\left [ 0.05 \right ]^{y} \;\;\;\; (iii)$

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

$\frac{5.07 \times 10^{-5} }{5.07 \times 10^{-5} } = \frac{k\left [ 0.20 \right ]^{x}\left [ 0.30 \right ]^{y} }{k\left [ 0.20 \right ]^{x}\left [ 0.10 \right ]^{y} } \\ \\ \Rightarrow 1 = \frac{\left [ 0.30 \right ]^{y}}{\left [ 0.10 \right ]^{y}} \\ \\ \Rightarrow \left ( \frac{0.30}{0.10} \right )^{0} = \left ( \frac{0.30}{0.10} \right )^{y} \\ \\ \Rightarrow y = 0$

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

$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} = \frac{k\left [ 0.40 \right ]^{x}\left [ 0.05 \right ]^{y}}{k\left [ 0.20 \right ]^{x}\left [ 0.30 \right ]^{y}} \\ \\ \Rightarrow \frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} = \frac{\left [ 0.40 \right ]^{x}}{\left [ 0.20 \right ]^{x}} \;\;\;\;\; \begin{bmatrix} Since\; y = 0,\\ \left [ 0.05 \right ]^{y} = \left [ 0.30 \right ]^{y} = 1 \end{bmatrix} \\ \\ \Rightarrow 2.821 = 2^{x} \\ \\ \Rightarrow \log 2.821 = x \log 2 \;\;\;\;\; (taking\; log \; on \; both\; sides) \\ \\ \Rightarrow x = \frac{\log 2.821}{\log 2} \\ \\ = 1.496 \\ \\ = 1.5 \; (Approximately)$

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

Q 11. The following results have been obtained during the kinetic studies of the reaction:

2A + B $\rightarrow$ C + D

 Exp. $\frac{A}{mol L^{-1}}$ $\frac{B}{mol L^{-1}}$ Initial rate of formation of $\frac{D}{mol\; L^{-1} \;min^{-1}}$ 1 0.1 0.1 $6.0 \times 10^{-3}$ 2 0.3 0.2 $7.2 \times 10^{-2}$ 3 0.3 0.4 $2.88 \times 10^{-1}$ 4 0.4 0.1 $2.4 \times 10^{-2}$

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

Ans:

Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, rate of the reaction is given by,

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

According to the question,

$6.0 \times 10^{-3} = k\left [ 0.1 \right ]^{x} \left [ 0.1 \right ]^{y}$ ——- (1)

$7.2 \times 10^{-2} = k\left [ 0.3 \right ]^{x} \left [ 0.2 \right ]^{y}$ ——–(2)

$2.88 \times 10^{-1} = k\left [ 0.3 \right ]^{x} \left [ 0.4 \right ]^{y}$ ——–(3)

$2.4 \times 10^{-2} = k\left [ 0.4 \right ]^{x} \left [ 0.1 \right ]^{y}$ ——–(4)

Dividing equation (4) by (1), we get

$\frac{2.4 \times 10^{-2}}{6.0 \times 10^{-3}} = \frac{k\left [ 0.4 \right ]^{x} \left [ 0.1 \right ]^{y}}{k\left [ 0.1 \right ]^{x} \left [ 0.1 \right ]^{y}}$

$4 = \frac{\left [ 0.4 \right ]^{x}}{\left [ 0.1 \right ]^{x}}$

$4 = \left (\frac{0.4}{0.1 } \right )^{x}$

$\left (4 \right )^{1} = \left (4 \right )^{x}$

x = 1

Dividing equation (3) by (2), we get

$\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \frac{k\left [ 0.3 \right ]^{x} \left [ 0.4 \right ]^{y}}{k\left [ 0.3 \right ]^{x} \left [ 0.2 \right ]^{y}}$

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

$4 = 2^{y}$

$2^{2} = 2^{y}$

y = 2

Hence, the rate law is

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

From experiment 1, we get

$k = \frac{6.0 \times 10^{-3} mol\; L^{-1} \;min^{-1}}{\left (0.1 \;mol \;L^{-1} \right ) \left (0.1 \;mol \;L^{-1} \right )^{2}}$

= 6.0 $L^{2} \;mol^{-2} \;min^{-1}$

From experiment 2, we get

$k = \frac{7.2 \times 10^{-2} mol\; L^{-1} \;min^{-1}}{\left (0.3 \;mol \;L^{-1} \right ) \left (0.2 \;mol \;L^{-1} \right )^{2}}$

= 6.0 $L^{2} \;mol^{-2} \;min^{-1}$

From experiment 1, we get

$k = \frac{2.88 \times 10^{-1} mol\; L^{-1} \;min^{-1}}{\left (0.3 \;mol \;L^{-1} \right ) \left (0.4 \;mol \;L^{-1} \right )^{2}}$

= 6.0 $L^{2} \;mol^{-2} \;min^{-1}$

From experiment 1, we get

$k = \frac{2.4 \times 10^{-2} mol\; L^{-1} \;min^{-1}}{\left (0.4 \;mol \;L^{-1} \right ) \left (0.1 \;mol \;L^{-1} \right )^{2}}$

= 6.0 $L^{2} \;mol^{-2} \;min^{-1}$

Thus, rate constant, k = 6.0 $L^{2} \;mol^{-2} \;min^{-1}$

Q 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:

 Exp. $\frac{A}{mol L^{-1}}$ $\frac{B}{mol L^{-1}}$ Initial rate $mol\; L^{-1} \;min^{-1}$ 1 0.1 0.1 $2.0 \times 10^{-2}$ 2 — 0.2 $4.0 \times 10^{-2}$ 3 0.4 0.4 — 4 — 0.2 $2.0 \times 10^{-2}$

Ans:

The given reaction is of the first order with respect to A and of zero order with respect to B.

Thus, the rate of the reaction is given by,

Rate = $k \left [ A \right ]^{1} \left [ B \right ]^{0}$

Rate = $k \left [ A \right ]$

From experiment 1, we get

$2.0 \times 10^{-2}\, mol\; L^{-1} min^{-1} = k\left ( 0.1 \; mol \; L^{-1} \right ) \\ \\ \Rightarrow k = 0.2\; min^{-1}$

From experiment 2, we get

$4.0 \times 10^{-2}\, mol\; L^{-1} min^{-1} = 0.2 min^{-1} \left [ A \right ] \\ \\ \Rightarrow \left [ A \right ] = 0.2\; mol\; L^{-1}$

From experiment 3, we get

Rate = $0.2 \; min^{-1} \times 0.4 \; mol \; L^{-1}\\ \\ = 0.08\; mol\; L^{-1}min^{-1}$

From experiment 4, we get

$2.0 \times 10^{-2}\; mol \; L^{-1} min^{-1} = 0.2\; min^{-1} \left [ A \right ] \\ \\ \Rightarrow \left [ A \right ] = 0.1\; mol \; L^{-1}$

Q 13. Calculate the half-life of a first order reaction from their rate constants given below:

(a) $200 \; s^{-1}$

(b) $2 \; min^{-1}$

(c) $4 \; years^{-1}$

Ans:

(a) Half life, $t_{ \frac{1}{2}} = \frac{0.693}{k} \\ \\ = \frac {0.693}{200\; s^{-1}} \\ \\ = 3.47\; s$  (Approximately)

(b) $t_{ \frac{1}{2}} = \frac{0.693}{k} \\ \\ = \frac {0.693}{2\; min^{-1}} \\ \\ = 0.35\; min$  (Approximately)

(c) $t_{ \frac{1}{2}} = \frac{0.693}{k} \\ \\ = \frac {0.693}{4\; years^{-1}} \\ \\ = 0.173\; years$   (Approximately)

Q 14. The half-life for radioactive decay of $\,^{14}C$ is 5730 years. An archaeological artefact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.

Ans:

Here, $k = \frac{0.693}{t_{\frac{1}{2}}} \\ \\ = \frac{0.693}{5730}years^{-1}$

It is known that,

$t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ = \frac{2.303}{0.693} \log \frac{100}{80} \\ \\ = 1845\; years$      (approximately)

Hence, the age of the sample is 1845 years.

Q 15. The experimental data for decomposition of $N_{2} O_{5}$

$\left [2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2} \right ]$

In gas phase at 318K are given below:

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

(a) Plot $\left [N_{2} O_{5} \right ]$ against t.

(b) Find the half-life period for the reaction.

(c) Draw a graph between log $\left [N_{2} O_{5} \right ]$ and t.

(d) What is the rate law?

(e) Calculate the rate constant.

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

Ans:

(a)

(b) Time corresponding to the concentration, $\frac{1.630 \times 10^{2}}{2}\; mol\; L^{-1} = 81.5 mol\; L^{-1}$ is the half life. From the graph , the half life obtained as 1450 s.

(c)

 t(s) $10^{2} \times \left [N_{2} O_{5} \right ] mol \; L^{-1}$ $\log \left [ N_2 O_{5} \right ]$ 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

(d) The given reaction is of the first order as the plot, $\log \left [ N_2 O_{5} \right ]$ v/s t, is a straight line.

Therefore, the rate law of the reaction is

Rate = $k \left [ N_2 O_{5} \right ]$

(e) From the plot, $\log \left [ N_2 O_{5} \right ]$ v/s t, we obtain

$Slope = \frac{-2.46 – \left ( -1.79 \right )}{3200 – 0} \\ \\ = \frac{-0.67}{3200}$

Again, slope of the line of the plot $\log \left [ N_2 O_{5} \right ]$ v/s t is given by

$- \frac{k}{2.303}$.

Therefore, we obtain,

$- \frac{k}{2.303} = – \frac{0.67}{3200} \\ \\ \Rightarrow k = 4.82 \times 10^{-4}s^{-1}$

(f) Half – life is given by,

$t_{\frac{1}{2}} = \frac{0.639}{k} \\ \\ = \frac{0.693}{4.82 \times 10^{-4}}S \\ \\ = \frac{1.483}{10^{3}}s \\ \\ = 1438 s$

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

Q 16. The rate constant for a first order reaction is $60\; s ^{-1}$. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Ans:

It is known that,

$t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ = \frac{2.303}{60\; s^{-1}} \log \frac{1}{1/6} \\ \\ = \frac{2.303}{60\; s^{-1}} \log 16 \\ \\ = 4.6 \times 10^{-2} \left ( approximately \right )$

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

Q 17. During the 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.

Ans:

$k = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{28.1}\; y^{-1}$

Here,

It is known that,

$t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ \Rightarrow 10 = \frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{\left [ R \right ]} \\ \\ \Rightarrow 10 = \frac{2.303}{\frac{0.693}{28.1}}\left ( – \log \left [ R \right ] \right ) \\ \\ \Rightarrow \log \left [ R \right ] = – \frac{10 \times 0.693}{2.303 \times 28.1}\\ \\ \Rightarrow \left [ R \right ] = antilog \left ( – 0.1071 \right )\\ \\ = antiog \left ( 1.8929 \right ) \\ \\ = 0.7814 \mu g$

Therefore, $0.7814 \; \mu g$ of $^{90}Sr$ will remain after 10 years.

Again,

$t = \frac{2.303}{k} \;log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ \Rightarrow 60 = \frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{\left [ R \right ]}\\ \\ \Rightarrow log \left [ R \right ] = – \frac{60 \times 0.693}{2.303 \times 28.1}\\ \\ \left [ R \right ] = antilog \; \left ( – 0.6425 \right ) \\ \\ = antilog \left ( 1.3575 \right )\\ \\ = 0.2278 \mu g$

Therefore, $0.2278 \mu g \; o\!f \; ^{90}Sr$ will remain after 60 years.

Q 18. For a first-order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of the reaction.

Ans:

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

$t_{1} = \frac{2.303}{k} \log \frac{100}{100 – 99} \\ \\ = \frac{2.303}{k} \log 100 \\ \\ = 2 \times \frac{2.303}{k}$

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

$t_{1} = \frac{2.303}{k} \log \frac{100}{100 – 90} \\ \\ = \frac{2.303}{k} \log 10 \\ \\ = \frac{2.303}{k}$

Therefore, $t_{1} = 2 \; t_{2}$

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

Q 19. A first order reaction takes 40 min for 30% decomposition. Calculate $t_{ \frac{1}{2} }$

Ans:

For a first order reaction,

$t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ k = \frac{2.303}{40\; min} \log \frac{100}{100 – 30} \\ \\ = \frac{2.303}{40 \; min} \log \frac{10}{7} \\ \\ = 8.918 \times 10^{-3} \; min^{-1}$

Therefore, $t _{\frac{1}{2}}$ of the decomposition reaction is

$t _{\frac{1}{2}} = \frac{0.693}{k} \\ \\ = \frac{0.693}{8.918 \times 10^{-3}} min \\ \\ = 77.7 \; min \; \left ( approximately \right )$

Q 20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

 t(sec) P(mm of Hg) 0 35.0 360 54.0 720 63.0

Calculate the rate constant.

Ans:

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

$\left ( CH_{3} \right )_{2} CHN \! = \! NCH \left ( CH_{3} \right )_{2 \left ( g \right )} \rightarrow N_{2\left ( g \right )} + C_{6}H_{14\left ( g \right )}$

At t = 0                                                $P_{0}$                                         0          0

At t = t                                     $P_{0} – P$                                                p          p

After time, t, total pressure, $P_{1} = \left ( P_{0} – p \right ) + p + p \\ \\ \Rightarrow P_{1} = P_{0} + p \\ \\ \Rightarrow p = P_{1} – P_{0} \\ \\ Therefore, P_{0} – p = P_{0} – \left ( P_{t} – P_{0} \right ) \\ \\ = 2P_{0} – P_{t}$

For the first order reaction,

$k = \frac{2.303}{t} \log \frac{P_{0}}{P_{0} – p} \\ \\ = \frac{2.303}{t} \log \frac{P_{0}}{2P_{0} – P_{t}}$

When t = 360 s, $k = \frac{2.303}{360 \; s} \log \frac{35.0}{2 \times 35.0 – 54.0} \\ \\ = 2.175 \times 10^{-3} s^{-1}$

When t = 720 s,

$k = \frac{2.303}{720 \; s} \log \frac{35.0}{2 \times 35.0 – 63.0} \\ \\ = 2.235 \times 10^{-3} s^{-1}$

Hence the average value of rate constant is.

$k = \frac{2.21 \times 10^{-3} + 2.235 \times 10^{-3} }{2} \; s^{-1} \\ \\ = 2.21 \times 10^{-3} s^{-1}$

Q 21. The following data were obtained during the first order thermal decomposition of $SO_{2} CL_{2}$  at a constant volume.

$SO_{2} CL_{2 \left ( g \right )} \rightarrow SO_{2 \left ( g \right )} CL_{2}\left ( g \right )$

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

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

Ans:

The thermal decomposition of $SO_{2} CL_{2}$  at a constant volume is represented by the following equation.

$SO_{2} CL_{2 \left ( g \right )} \rightarrow SO_{2 \left ( g \right )} CL_{2}\left ( g \right )$

At t = 0                        $P_{0}$                                         0                      0

At t = t             $P_{0} – p$                                                0                      0

After time t, total pressure, $P_{t} = \left ( P_0 – p \right ) + p + p \\ \\ \Rightarrow P_t = P_0 + p \\ \\ \Rightarrow p = P_t – P_0 \\ \\∴ P_0 – p = P_o – \left ( P_t – P_0 \right ) \\ \\ = 2 P_0 – P_t$

For a first order reaction,

$k = \frac{2.303}{t} \log \frac{P_0}{P_0 – p} \\ \\ = \frac{2.303}{t} \log \frac{P_0}{2P_0 – P_t}$

When t = 100s,

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

When $P_t = 0.65$ atm,

$P_0 + p = 0.65 \\ \\ \Rightarrow p = 0.65 – P_0 \\ \\ = 0.65 – 0.5 \\ \\ = 0.15 \; atm$

Therefore, when the total pressure is 0.65 atm, pressure of $SO_{2} CL_{2}$ is

$P_{SOCL_{2} } = P_0 – p \\ \\ = 0.5 – 0.15 \\ \\ = 0.35 \; atm$

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

$Rate = k \left ( P_{SOCL_{2}} \right ) \\ \\ = \left ( 2.23 \times 10^{-3} s^{-1} \right )\left ( 0.35 \right ) \; atm \\ \\ = 7.8 \times 10^{-4} \; atm \; s^{-1}$

Q 22. The rate constant for the decomposition of $N_2 O_5$ at various temperatures is given below:

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

Draw a graph between In k and 1/T and calculate the values of A and $E_a$. Predict the rate constant at $30^{\circ} \; and \; 50^{\circ}\; C$.

Ans:

From the given data, we obtain

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

Slope of the line,

$\frac{y_2 – y_1}{x_2 – x_1} = – 12.301\, K$

According to Arrhenius equation,

$Slope = – \frac{E_a}{R} \\ \\ \Rightarrow E_a = – Slope \times R \\ \\ = – \left ( -12.301 \, K \right ) \times \left ( 8.314 \, JK^{-1}\, mol^{-1} \right ) \\ \\ = 102.27 \; kJ \; mol^{-1}$

Again,

In k = In $A – \frac{E_a}{RT}$

In A = In $k + \frac{E_a}{RT}$

When T = 273 k,

In k = – 7.147

Then, In A = $-7.147 + \frac{102.27 \times 10^3}{8.314 \times 273} \\ \\ = 37.911 \\ \\∴ A = 2.91 \times 10^6$

When T = 30 + 273 K = 303 K

$\frac{1}{T} = 0.0033 K = 3.3 \times 10^{-3} K$

Then, at $\frac{1}{T} = 3.3 \times 10^{-3} K \\ In\; k = -2.8$

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

Again, when T = 50 + 273 K = 323 K

$\frac{1}{T} = 0.0031 K = 3.1 \times 10^{-3} \; K \\ \\ Then, \;at \frac{1}{T} = 3.1 \times 10^{-3} \; K \\ \\ In\; k = -0.5$

Therefore, k = 0.607 /s

Q 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?

Ans:

$k = 2.418 \times 10^{-5} s^{-1}$

T = 546 K

$E_a = 179.9 kJ\; mol^{-1} = 179.9 \times 10^3\; J\; mol^{-1}$

According to the Arrhenius equation,

$k =Ae^{-E_a /RT}\\ \\ \Rightarrow I\!n \;k = I\!n A – \frac{E_a}{RT} \\ \\ \Rightarrow \log k = \log A – \frac{E_a}{2.303 \; RT} \\ \\ \Rightarrow \log A = \log k + \frac{E_a}{2.303 \; RT} \\ \\ = \log \left ( 2.418 \times 10^{-5} \; s^{-1} \right ) + \frac{179.9 \times 10^3 J mol^{-1}}{2.303 \times 8.314 \; Jk^{-1} \; mol^{-1} \times 546\; K} \\ \\ = \left ( 0.3835 – 5 \right ) + 17.2082 \\ \\ = 12.5917$

Therefore, A = antilog (12.5917)

$= 3.9 \times 10^{12 } \; s^{-1}$  (approximately)

Q 24. Consider a certain reaction $A \rightarrow$ products with $k = 2.0 \times 10^{-2} \; s^{-1}$. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.o mol /L .

Ans:

$k = 2.0 \times 10^{-2} \; s^{-1} \; T = 100 \; s$ $\left [ A \right ]_0 = 1.0 \; mol^{-1}$

Since the unit k is $s^{-1}$, the given reaction is a first order reaction.

Therefore, $k = \frac{2.303}{t} \log \frac{\left [ A \right ]_0}{\left [ A \right ]} \\ \\ \Rightarrow 2.0 \times 10^{-2}\; s^{-1} = \frac{2.303}{100 \; s} \log \frac{1.0}{\left [ A \right ]} \\ \\ \Rightarrow 2.0 \times 10^{-2} \; s^{-1} = \frac{2.303}{100 \; s}\left ( – \log \left [ A \right ] \right ) \\ \\ \Rightarrow – \log \left [ A \right ] = \frac{2.0 \times 10 ^{-2} \times 100}{2.303} \\ \\ \Rightarrow \left [ A \right ] = antilog \left ( – \frac{2.0 \times 10 ^{-2} \times 100}{2.303} \right ) \\ \\ = 0.135 \; mol \; L^{-1}$                    (approximately)

Hence, the remaining concentration of A is $0.135 \; mol \; L^{-1}$

Q 25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with $t_{\frac{1}{2}} = 3.00 \; hours$. What fraction of sample of sucrose remains after 8 hours?

For the first order reaction,

$k = \frac{2.303}{t} \log \frac{\left [ R \right ]_0}{\left [ R \right ]}$

It is given that , $t_{\frac{1}{2}} = 3.00 \; hours$.

Therefore, $k = \frac{0.693}{t_{\frac{1}{2}}}$ $\frac{0.693}{3} h^{-1} \\ \\ = 0.231 \; h^{-1}$

Then, $0.231 \; h^{-1} = \frac{2.303}{8h} \log \frac{\left [ R \right ]_0}{\left [ R \right ]} \\ \\ \Rightarrow \log \frac{\left [ R \right ]_0}{\left [ R \right ]} = \frac{0.231\; h^{-1} \times 8\; h}{2.303} \\ \\ \Rightarrow \frac{\left [ R \right ]_0}{\left [ R \right ]} = antilog \left ( 0.8024 \right ) \\ \\ \Rightarrow \frac{\left [ R \right ]_0}{\left [ R \right ]} = 6.3445 \\ \\ \Rightarrow \frac{\left [ R \right ]}{\left [ R \right ]_0} = 0.1576$     (approx)

= 0.158

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

Q 26.  The decomposition of hydrocarbon follows the equation

$k = \left ( 4.5 \times 10_{11} S – 1 \right ) e_{-28000} K/T$

Calculate $E_{a}$.

Ans:

The given equation is $k = \left ( 4.5 \times 10_{11} S – 1 \right ) e_{-28000} K/T$      ….(i)

The Arrhenius equation is given by,

$k = Ae^{-E_a /RT}$          ….(ii)

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

$\frac{E_a }{RT} = \frac{28000\; K}{T} \\ \\ \Rightarrow E_a = R \times 28000 \; K \\ \\ = 8.314\; J \; K^{-1} mol^{-1} \times 28000 \; K \\ \\ 232791\; J \; mol^{-1}\\ \\ = 232.791\; kJ\; mol^{-1}$

Q 27.  The rate constant for the first order decomposition of $H_2 O_2$ is given by the following equation:

$log \; k = 14.34 – 1.25 \times 10^4 \; K/T$

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

Ans:

Arrhenius equation is given by,

$k = Ae^{-E_a/RT} \\ \\ \Rightarrow I\!n\; k = I\!n\; A – \frac{E_a}{RT} \\ \\ \Rightarrow I\!n\; k = \log A – \frac{E_a}{RT} \\ \\ \Rightarrow \log k = \log A – \frac{E_a}{2.303\; RT} \;\;\;\;\;\;\;\; . . . (i)$

The given equation is

$\log k = 14.34 – 1.25 \times 10^{4}\; K/T \;\;\;\;\;\;\;\; . . . (ii)$

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

$\frac{E_a}{2.303\; RT} = \frac{1.25 \times 10^4\; K}{T}\\ \\ \Rightarrow E_a = 1.25 \times 10^4\; K \times 2.303 \times R \\ \\ = 1.25 \times 10^4\; K \times 2.303 \times 8.314 \; J \; K^{-1} mol^{-1} \\ \\ = 239339.3\; J\; mol^{-1}\;\;\;\; (approximately) \\ \\ = 239.34\; kJ\; mol^{-1}$

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

$k = \frac{0.693}{t_{\frac{1}{2}}} \\ \\ = \frac{0.693}{256} \\ \\ = 2.707 \times 10^{-3} \; min^{-1}\\ \\ = 4.51 \times 10^{-5}\; s^{-1}$

It is also given that, $log\; k = 14.34 – 1.25 \times 10^{4}\; K/T \\ \\ \Rightarrow log \left ( 4.51 \times 10^{-5} \right ) = 14.34 – \frac{1.25 \times 10^4 \; K}{T}\\ \\ \Rightarrow log \left ( 0.654 – 05 \right ) = 14.34 – \frac{1.25 \times 10^4 \; K}{T}\\ \\ \Rightarrow \frac{1.25 \times 10^4\; K}{T} = 18.686 \\ \\ = 668.95\; K \\ \\ = 669\; K\;\;\;\; (approximately)$

Q 28. The decomposition of A into product has value of k as $4.5 \times 10^3 \; s^{-1} \; at \; 10^{\circ} C$ and energy of activation 60 kJ / mol. At what temperature would k be $1.5 \times 10^4 \; s^{-1}$?

Ans:

From Arrhenius equation, we obtain

$\log \frac{k_2}{k_1} = \frac{E_a}{2.303\, R}\left ( \frac{T_2 – T_1}{T_1 T_2} \right )$

Also, $k_1 = 4.5 \times 10^3\; s^{-1}$ $T_1 = 273 + 10 = 283\; K \; k_2\\ \\ = 1.5 \times 10^4\; s^{-1}\\ \\ E_a = 60\; kJ\; mol^{-1} = 6.0 \times 10^{4}\; J \; mol^{-1}$

Then,

$\log \frac{1.5 \times 10^4}{4.5 \times 10^3} = \frac{6.0 \times 10^4\; J\; mol^{-1}}{2.303 \times 8.314\; J\; K^{-1}\; mol^{-1}}\left ( \frac{T_2 – 283}{283\; T_2} \right )\\ \\ \Rightarrow 0.5229 = 3133.627 \left ( \frac{T_2 – 283}{283\; T_2} \right ) \\ \\ \Rightarrow \frac{0.5229 \times 283\; T_2}{3133.627} = T_2 – 283 \\ \\ \Rightarrow 0.9528 \;T_2 = 283\\ \\ \Rightarrow T_2 = 297.019\; K \;\;\;\;\; (approximately) \\ \\ = 297 \; K \\ \\ = 24^{\circ}$

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

Ans:

For a first order reaction,

$t = \frac{2.303}{k} \log \frac{a}{a – x}\\ \\ at \; 298\; K, \; t = \frac{2.303}{k} \log \frac{100}{90}\\ \\ = \frac{0.1054}{k}\\ \\ at\; 208\; K, \; t’ = \frac{2.303}{k’} \log \frac{100}{75} \\ \\ = \frac{2.2877}{k’}$

According to the question,

$t = t’ \\ \\ \Rightarrow \frac{0.1054}{k} = \frac{0.2877}{k’} \\ \\ \Rightarrow \frac{k’}{k} = 2.7296$

From Arrhenius equation, we get

$\log \frac{k’}{k} = \frac{E_a}{2.303\, R}\left ( \frac{T’ – T}{TT’} \right ) \\ \\ \Rightarrow \log \left ( 2.7296 \right ) = \frac{E_a}{2.303 \times 8.314}\left ( \frac{308 – 298}{298 \times 308} \right ) \\ \\ \Rightarrow E_a = \frac{2.303 \times 8.314 \times 298 \times 308 \times \log \left ( 2.7296 \right )}{308 – 298} \\ \\ = 76640.096\; J \, mol^{-1} \\ \\ = 76.64\; kJ\; mol^{-1}$

To calculate k at 318 K,

It is given that, $A = 4 \times 10^{10} s^{-1}$, T = 318 K

Again, from Arrhenius equation, we get

$\log k = \log A – \frac{E_a}{2.303\; R\, T} \\ \\ = \log \left ( 4 \times 10^{10} \right ) – \frac{76.64 \times 10^{3}}{2.303 \times 8.314 \times 318} \\ \\ = \left ( 0.6021 + 10 \right ) – 12.5876 \\ \\ = -1.9855$

Therefore, k = Antilog(-1.9855)

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

Q 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.

Ans:

From Arrhenius equation, we get

$\log \frac{k_2}{k_1} = \frac{E_a}{2.303\, R}\left ( \frac{T_2 – T_1}{T_1 T_2} \right )$

From the question we have, $K_2 = 4 K_1$ $\\ T_1 = 293\; K \\ \\ T_2 = 313\; K$

Therefore,  $\log \frac{4 K_1}{K_2} = \frac{E_a}{2.303 \times 8.314}\left ( \frac{313 – 293}{293 \times 313} \right )\\ \\ \Rightarrow 0.6021 = \frac{20 \times E_a}{2.303 \times 8.314 \times 293 \times 313} \\ \\ \Rightarrow E_a = \frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20}\\ \\ = 52863.33\; J\; mol^{-1}\\ \\ = 52.86\; kJ\; mol^{-1}$

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

 Also Access NCERT Exemplar for class 12 Chemistry Chapter 4 CBSE Notes for class 12 Chemistry Chapter 4

Chemical kinetics class 12 NCERT is very important from the point of a student’s life. This topic is basic to many of the concepts that you will study in the future. To avoid difficulty in future students should get a thorough understanding of this chapter and solve as many as chemical kinetics numerical problems to get well versed with the formulas, standard values and equations.

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