# Class 12 NCERT Solutions Chemistry Chapter 4 – Chemical Kinetics

## Chemical Kinetics Class 12 Chemistry NCERT Solutions Chapter 4

Chemical Kinetics Class 12 Chemistry NCERT Solutions Chapter 4 is a topic that defines chemical reactions, variables, and mechanisms. It includes both the chemical reaction and the physical process. This chapter describes in detail all the aspects that students need to understand the concept and fair well in their examination. The Chemical Kinetics Class 12 Chemistry NCERT Solutions Chapter 4 provided by Extramarks will help students with that.

Chemical Kinetics deals with chemical reactions, factors, and mechanisms. It is closely related to the chemical reaction and physical process. Chemical kinetics Class 12 is categorized into swift, prolonged, and moderate reactions based on their varying rate.

Students learn various subtopics related to chemical kinetics like dependence on the rate of concentration, integrated equations, collision theory, catalyst and Arrhenius equation under class 12 Chemistry chapter 4 solutions.

### The key topics covered in NCERT solutions class 12 Chemistry chapter 4 include

 Exercise Topic 4.1 Introduction 4.2 Rate of Chemical reaction 4.3 Factor influencing rate of Reaction 4.4 Integrated Rate equation 4.5 Pseudo-first-order Reaction 4.6 Arrhenius Equation 4.7 CollisionTheory of chemical Reaction 4.8 FAQ

Students can refer to these exercises by clicking on the respective topic. Given below is a brief of all the exercises under NCERT solutions class 12 Chemistry chapter 4.

4.1 Introduction

Chemical kinetics, also known as reaction kinetics, helps us understand the rates of reactions and how certain conditions influence them. It further helps to define the characteristics of a chemical reaction by gathering and analyzing information about the mechanism of the reaction.

4.2Rate of Chemical Reaction

Under this exercise of NCERT solutions class 12 Chemistry chapter 4, students will learn about

The Rate Law: The rate law (also called the rate equation) for a chemical reaction is an expression that shows a relationship between the rate of the reaction and the concentrations of the reactants participating in it.

Expression: For a  chemical reaction given by aA + bB → cC + dD

a, b, c, and d are the stoichiometric coefficients of the reactants or products.

The rate equation for the reaction is obtained as below.

Rate: ∝ [A][B]y ⇒ Rate = k[A][B]y

Where

• [A] & [B] represents the concentrations of the reactants A and B.
• x & y represent the partial reaction orders for reactants A & B (that may or may not be equal to their stoichiometric coefficients a & b).
•  ‘proportionality constant is the rate constant of the reaction.

It’s key to understanding that the expression of the rate law for a specific reaction can only be determined experimentally. The rate law expression cannot be achieved from the balanced chemical equation (since the partial orders of the reactants are not necessarily equal to the stoichiometric coefficients). Students can refer to Chemical Kinetics Class 12 Chemistry NCERT Solutions Chapter 4 by Extramarks for more information on the ‘rate of chemical reaction’.

### Reaction Orders: The combination of the partial orders of the reactants in the rate law expression equals the total order of the reaction.

If Rate = k[A]x[B]y, the total order of the reaction (n) = x+y

The order of a reaction gives insight into the change in the rate of the reaction that can be anticipated by increasing the concentration of the reactants.

For example,

• If this reaction is a zero-order reaction, doubling the reactant concentration will not affect the rate of reaction.
• If it is a first-order reaction, doubling the reactant concentration will double the rate of reaction.
• In the case of second-order reactions, doubling the concentration of the reactants will quadruple the overall rate of reaction.
• For third-order reactions, the tot rate increases by eight times when the reactant concentration is doubled.

Rate Constants: Rearranging the rate of the equation, the value of the rate constant ‘k’ is

shown as below k = Rate/[A]x[B]y

Hence, the units of k (assuming that concentration is represented in mol.L-1 or M and time are represented in seconds) can be calculated via the following equation.

k = (M.s-1)*(M-n) = M(1-n).s-1

#### 4.3 Factors affecting the rate of reaction

Under exercise 4.3 of NCERT solutions class 12 Chemistry chapter 4, students will learn about

The Nature of reactant: The nature of bonding in the reactants determines the rate of a reaction. The ionic compounds react faster than covalent compounds due to the energy requirement in covalent compounds to cleave the existing binds.

The reaction between ionic compounds:

AgNO3 + NaCl –> AgCl + NaNO3

Precipitation of AgCl

Temperature: The rate of reaction increases with the increase in temperature due to an increase in average kinetic energy, increasing the number of molecules having more incredible energy than the threshold energy and consequently increasing the number of effective collisions. The rate of a reaction is increased by 100% or doubled with a 10oC rise in temperature.

Pressure: An increase in partial pressure increases the number of collisions. Therefore, the rate of reactions involving gaseous reactants increases with the increase in partial pressures.

Catalyst: As per Chemical Kinetics Class 12 Chemistry NCERT Solutions Chapter 4, a catalyst increases the rate of reaction by giving an alternative path with lower activation energy (Ea’) for the reaction to proceed.

The concentration of reactants: Increasing concentration increases the number of collisions and the activated collisions between the reactant molecules. According to the collision theory, the rate is directly proportional to the collision frequency. Consequently, the rate of a reaction increases with the rise in the concentration of reactant.

Surface area: According to Chemical Kinetics Class 12 Chemistry NCERT Solutions Chapter 4, the rate of reaction increases with an increase in the surface area of a solid reactant.

4.4 Integrated Rate of Equations

The integrated rate of equations expresses the concentration of the reactants in a chemical reaction as a function of time. Thus, such rate of equations can be employed to check how long it would take for a given percentage of the reactants to be consumed in a chemical reaction. It is essential to observe that different order reactions have different integrated rates of equations.

Integrated Rate of Equation for Zero-Order Reactions: With reference to Chemical Kinetics Class 12 Chemistry NCERT Solutions Chapter 4, Integrated Rate of Equation for Zero-Order Reactions is given by:

kt = [R]0 – [R] (or) k = ([R]0 – [R])/t

[R]0 is the initial reactant concentration ( t = 0)

[R] is the reactant concentration at a time ‘t.’

and the Rate constant is k

The Integrated Rate Equation for First-Order Reactions: The integrated rate law for first-order reactions is:

kt = 2.303log([R]0/[R]) (or) k = (2.303/t)log([R]0/[R])

Integrated Rate Equation for Second-Order Reactions: The integrated rate of the equation is:

kt = (1/[R]) – (1/[R]0)

More details of the Integrated Rate of reaction are described in NCERT Solutions Class 12 Chemistry Chapter 4.

What is Half-Life Reaction?As per Chemical Kinetics Class 12 Chemistry NCERT Solutions Chapter 4, the half-life of a chemical reaction can give as the time taken for the concentration of a given reactant to reach 50% of its initial concentration (i.e. the time taken for the reactant concentration to get half of its initial value). It is denoted by the symbol ‘t1/2‘ and is usually represented in seconds.

Half-Life Formula: It is important to note that the formula for the half-life of a reaction varies with the order of the reaction.

• For a zero-order reaction, the mathematical expression that can be employed to determine the half-life is: t1/2 = [R]/2k
• For a first-order reaction, the half-life is given by: t1/2 = 0.693/k
• For a second-order reaction, the formula for the half-life of the reaction is 1/k[R]0

Where

• t1/2 is the half-life of the reaction (unit: seconds)
• [R]0 is the initial reactant concentration (unit: mol.L-1 or M)
• k is the rate constant of the reaction (unit: M(1-n)s-1 where ‘n’ is the reaction order)

Students may refer to Chemical Kinetics Class 12 NCERT Solutions for more information on this exercise.

4.5 Pseudo order reaction

The reaction that appears to be an nth order reaction but belongs to some different order is called Pseudo order reaction.

Referring to Chemical Kinetics Class 12 NCERT Solutions, a pseudo-first-order reaction is a chemical reaction between two reactants participating in a chemical reaction and, therefore, should be a second-order reaction. But it resembles a first-order reaction due to the presence of reactants in negligible quantities.

Let R` + R“ –> P

Rate = k[A]1[B]1

Order of reaction = 2.

Let us consider another reaction,

CH3Br + OH−→ CH3OH+Br−

The rate law for this reaction is

Rate = k [OH−][CH3Br]

Rate = k [OH−][CH3Br] = k(constant)[CH3Br]=k′[CH3Br]

As only the concentration of CH3Br would change during the reaction, the rate would solely depend upon the changes in the CH3Br reaction.

4.6 Arrhenius equation

Under this exercise of Chemical Kinetics NCERT solutions class 12 Chemistry chapter 4, students will learn about the Arrhenius equation. The formula used to calculate the energy of activation and justify the effect of temperature on the rate of reaction is called Arrhenius Equation.

The formula is,

K = A e-Ea/RT

Where

k = Rate constant

A= Frequency factor

e = mathematical quantity

Ea= activation energy

R = gas constant

T = kelvin temperature

The above relation was created by Swedish chemist Svante Arrhenius and hence named after him.

ln K = ln A – Ea/(2.303RT)

Equation of a straight line with slope = –Ea /R.

When Ea = 0 , Temperature = Infinity

e-Ea/RT =Boltzmann factor.

For a chemical reaction, the rate constant gets doubled for a rise of 10° temperature due to the Arrhenius Equation.

The NCERT solutions class 12 Chemistry chapter 4 exercise 4.6 offers more examples.

K = Ae-Ea/RT

Taking log on both sides of the equation

Ln k = ln A – Ea/RT

Compared with the equation of a straight line

y= mx+c,

[m= slope of the line

c= y-intercept]

So we have

y = ln k

x = 1/T

m = -Ea / R

c = ln A

Image:   Plotting k Vs (1/T)

4.7 Collision theory

As per NCERT solutions class 12 Chemistry chapter 4 Exercise 4.7 collision theory, the molecules collide with significant kinetic energy to create a chemical reaction. The molecules of the reacting species collide through space in a rectilinear motion. The rate of a chemical reaction is proportional to the number of collisions between the molecules of the reacting species. Chemical Kinetics Solutions provided by Extramarks is very easy to understand.

The molecules must be correctly oriented. Rate of successful collisions ∝ Fraction of successful collisions X Overall collision frequency. The number of collisions per second per unit volume of the molecules in a chemical reaction is called collision frequency (Z).

Let A+B –> C + D

Rate = ZABe-Ea/RT

Here ZAB = collision frequency of A and B.

In many reactions Rate = P ZABe-Ea/RT

Where p= steric factor, which considers the proper orientation of the molecules participating in a chemical reaction. More information is provided in Chemical Kinetics NCERT Solutions class 12 Chemistry chapter 4, provided by Extramarks.

### Chemical Kinetics Class 12 NCERT Solutions Article Link

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Q.1 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3 NO(g) → N2O (g); Rate = k[NO]2
(ii) H2O2 (aq) + 3 I (aq) + 2 H+ → 2 H2O (l) + I3 ; Rate = k[H2O2][I]
(iii) CH3CHO(g) → CH4 (g) + CO(g); Rate = k [CH3CHO]3/2
(iv) C2H5Cl(g) → C2H4(g) + HCl(g); Rate = k [C2H5Cl]

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{Given Rate=k}{\left[\text{NO}\right]}^{\text{2}}\\ \text{Therefore, order of the reaction=2}\\ \text{k=}\frac{\text{Rate}}{{\text{[NO]}}^{\text{2}}}\\ \text{Dimension\hspace{0.17em}of k}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{{\text{(mol L}}^{\text{-1}}\text{)}}^{\text{2}}}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{\text{mol}}^{\text{2}}{\text{L}}^{\text{-2}}}\\ {\text{=L mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ \\ \left(\text{ii}\right){\text{Given rate=k[H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{][I}}^{\text{–}}\text{]}\\ \text{Therefore, order of the reaction=2}\\ \text{k=}\frac{\text{Rate}}{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]{\text{[I}}^{\text{–}}\text{]}}\\ \text{Dimension of k}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{\left({\text{mol L}}^{\text{-1}}\right){\text{(mol L}}^{\text{-1}}\text{)}}\\ {\text{=L mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ \\ \left(\text{iii}\right)\text{Given rate = k}{\left[{\text{CH}}_{\text{3}}\text{CHO}\right]}^{\frac{\text{3}}{\text{2}}}\text{=}\frac{\text{3}}{\text{2}}\\ \text{Therefore,order of reaction=\hspace{0.17em}k=}\frac{\text{Rate}}{{{\text{[CH}}_{\text{3}}\text{CHO]}}^{\frac{\text{3}}{\text{2}}}}\\ \text{Dimension of k}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{{\text{(mol L}}^{\text{-1}}\text{)}}^{\frac{\text{3}}{\text{2}}}}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{\text{mol}}^{\frac{\text{3}}{\text{2}}}{\text{L}}^{\frac{\text{3}}{\text{2}}}}\\ {\text{=L}}^{\frac{\text{1}}{\text{2}}}{\text{mol}}^{\frac{\text{-1}}{\text{2}}}{\text{s}}^{\text{-1}}\\ \\ \left(\text{iv}\right){\text{Given rate=k[C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl]}\\ \text{Therefore, order of the reaction=1}\\ \text{k=}\frac{\text{Rate}}{{\text{[C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl]}}\\ \text{Dimension of k}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{\text{mol L}}^{\text{-1}}}\\ {\text{=s}}^{\text{-1}}\end{array}$

Q.2 For the reaction:
2A + B → A2B
The rate = k[A][B]2 with k = 2.0 × 10-6 mol-2 L2 s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 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 the reaction is

Rate = k [A][B]2

= (2.0 × 10-6 mol-2 L2 s-1) (0.1 mol L-1) (0.2 mol L-1) 2

= 8.0 × 10-9 mol L-1 s-1

When [A] is reduced from 0.1 mol L-1 to 0.06 mol-1, the concentration of A reacted = (0.1 − 0.06) mol L-1 = 0.04 mol L-1

Therefore, concentration of B reacted

$=\frac{1}{2}×0.04mo{l}^{-1}$

= 0.02 mol L-1

Then, concentration of B available, [B] = (0.2 − 0.02) 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 [A][B]2

= (2.0 × 10-6 mol-2 L2 s-1) (0.06 mol L-1) (0.18 mol L-1) 2

= 3.89×10-9mol L-1 s-1

Q.3 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10-4 mol-1 L s-1?

Ans. Q.4 The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]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.,

$\text{Rate}=k\text{}{\left({\text{P}}_{{\text{CH}}_{3}{\text{OCH}}_{3}}\right)}^{3/2}$

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

Ans.

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

Unit of rate = bar min-1 Q.5 Mention the factors that affect the rate of a chemical reaction.

Ans.

The factors that affect the rate of a reaction are as follows.

(i) Concentration of reactants (pressure in case of gases)

(ii) Temperature

(iii) Nature of reactants

(iv) Presence of a catalyst

(v) Surface area

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
(i) doubled
(ii) reduced to half?

Ans.

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

Rate of reaction, R = k [A]2

= ka2

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

R’= k(2a)2

= 4ka2

=4R

Therefore, the rate of the reaction would increase by 4 times.

$\begin{array}{l}\begin{array}{l}\left(\text{ii}\right)\text{If the concentration of the reactant is reduced to half, i.e.}\left[\text{A}\right]\text{=}\frac{\text{1}}{\text{2}}\text{a then the rate}\\ \text{of the reaction would be}\end{array}\\ {\text{R}}^{\text{*}}\text{=k}{\left(\frac{\text{1}}{\text{2}}\text{a}\right)}^{\text{2}}\\ \text{=}\frac{\text{1}}{\text{4}}\\ \text{=}\frac{\text{1}}{\text{4}}\text{R}\\ \text{Therefore, the rate of the reaction would be reduced to =}\frac{{\text{1}}^{\text{th}}}{\text{4}}\end{array}$

Q.7 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

${\mathrm{SO}}_{2}{\mathrm{Cl}}_{2}\left(\mathrm{g}\right)\to {\mathrm{SO}}_{2}\left(\mathrm{g}\right)+\mathrm{}{\mathrm{Cl}}_{2}\left(\mathrm{g}\right)$

 Experiment Time/s-1 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. Q.8 The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.34 − 1.25 × 104 K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

Ans. Q.9 What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?

Ans.

The rate constant is nearly doubled with a rise in temperature by 10°C for a chemical reaction.

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

k = Ae-Ea/RT

where, k is the rate constant,

A is the Arrhenius factor or the frequency factor,

R is the gas constant,

T is the temperature, and

Ea is the energy of activation for the reaction

Q.10 In a pseudo first order hydrolysis of ester in water, the following results were obtained:

 t/s 0 30 60 90 [A]/ mol L-1 0.55 0.31 0.17 0.085

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

Ans.

Average rate of reaction between the time interval, 30 to 60 seconds,

$\begin{array}{l}=\frac{\text{d}\left[\text{Ester}\right]}{\text{dt}}\\ =\frac{0.31-0.17}{60-30}\\ =\frac{0.14}{30}\\ =\mathrm{}\text{4}.\text{67}×\text{1}{0}^{-\text{3}}{\text{molL}}^{-\text{1}}{\text{s}}^{-\text{1}}\\ \end{array}$

Q.11 A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{The differential rate equation will be}\\ \text{–}\frac{\text{d[R]}}{\text{dt}}\text{=k}\left[\text{A}\right]{\text{[ B]}}^{\text{2}}\\ \\ \left(\text{ii}\right)\text{If the concentration of B is increased three times then}\\ \text{–}\frac{\text{d[R]}}{\text{dt}}\text{=k}\left[\text{A}\right]{\text{[3B]}}^{\text{2}}\\ \text{=9.k}\left[\text{A}\right]{\text{[B]}}^{\text{2}}\\ \text{Therefore, the rate of reaction will increase 9 times.}\\ \\ \left(\text{iii}\right)\text{When the concentrations of both A and B are doubled,}\\ \text{–}\frac{\text{d[R]}}{\text{dt}}\text{=k}\left[\text{A}\right]{\text{[B]}}^{\text{2}}\\ \text{=k}\left[2\text{A}\right]{\text{[2B]}}^{\text{2}}\\ \text{=8.k}\left[\text{A}\right]{\text{[B]}}^{\text{2}}\\ \text{Therefore, the rate of reaction will increase 8 times.}\end{array}$

Q.12 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 r0/ mol L-1 s-1 5.07 × 10-5 5.07 × 10-5 1.43 × 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.

Therefore,

$\begin{array}{l}{\mathrm{r}}_{0}=\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{x}}{\left[\mathrm{B}\right]}^{\mathrm{y}}\\ 5.07×{10}^{-5}=\mathrm{k}{\left[0.20\right]}^{\mathrm{x}}{\left[0.30\right]}^{\mathrm{y}}----\left(\mathrm{i}\right)\\ 5.07×{10}^{-5}=\mathrm{k}{\left[0.20\right]}^{\mathrm{x}}{\left[0.10\right]}^{\mathrm{y}}----\left(\mathrm{ii}\right)\\ 1.43×{10}^{-4}=\mathrm{k}{\left[0.40\right]}^{\mathrm{x}}{\left[0.05\right]}^{\mathrm{y}}----\left(\mathrm{iii}\right)\\ \text{Dividing equation}\left(\mathrm{i}\right)\text{by}\left(\mathrm{ii}\right)\text{, we obtain}\\ \frac{5.07\mathrm{}×\mathrm{}{10}^{-5}}{5.07\mathrm{}×\mathrm{}{10}^{-5}\text{}}=\frac{\mathrm{k}{\left[0.20\right]}^{\mathrm{x}}{\left[0.30\right]}^{\mathrm{y}}}{\mathrm{k}{\left[0.20\right]}^{\mathrm{x}}{\left[0.10\right]}^{\mathrm{y}}}\\ ⇒1=\frac{{\left[0.30\right]}^{\mathrm{y}}}{{\left[0.10\right]}^{\mathrm{y}}}\\ ⇒{\left(\frac{0.30}{0.10}\right)}^{0}={\left(\frac{0.30}{0.10}\right)}^{\mathrm{y}}\\ ⇒\mathrm{y}=0\\ \text{Dividing equation}\left(\mathrm{iii}\right)\text{by}\left(\mathrm{ii}\right)\text{, we obtain}\\ \frac{1.43\mathrm{}×\mathrm{}{10}^{-4}}{5.07\mathrm{}×\mathrm{}{10}^{-5}}=\frac{\mathrm{k}{\left[0.40\right]}^{\mathrm{x}}{\left[0.05\right]}^{\mathrm{y}}}{\mathrm{k}{\left[0.20\right]}^{\mathrm{x}}{\left[0.30\right]}^{\mathrm{y}}}\\ ⇒\frac{1.43\mathrm{}×\mathrm{}{10}^{-4}}{5.07\mathrm{}×\mathrm{}{10}^{-5}}=\frac{{\left[0.40\right]}^{\mathrm{x}}}{{\left[0.20\right]}^{\mathrm{x}}}\mathrm{}\left[\begin{array}{c}\mathrm{Since}\mathrm{}\mathrm{y}=0,\\ {\left[0.05\right]}^{\mathrm{y}}={\left[0.30\right]}^{\mathrm{y}}=1\end{array}\right]\\ ⇒2.821=\mathrm{}{2}^{\mathrm{x}}\\ ⇒\mathrm{log}2.821=\mathrm{xlog}\mathrm{}2\left(\mathrm{Takinglogon}\mathrm{}\mathrm{both}\mathrm{}\mathrm{sides}\right)\\ ⇒\mathrm{x}=\frac{\mathrm{log}2.821}{\mathrm{log}2}\\ =1.496\\ =1.5\left(\text{approximately}\right)\\ \text{Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.}\end{array}$

Q.13 The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

 Experiment [A]/ mol L-1 [B]/ mol L-1 Initial rate of formation of D/mol L-1 min-1 I 0.1 0.1 6.0 × 10-3 II 0.3 0.2 7.2 × 10-2 III 0.3 0.4 2.88 × 10-1 IV 0.4 0.1 2.40 × 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, the rate of the reaction is given by,

$\begin{array}{l}\mathrm{Rate}=\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{x}}{\left[\mathrm{B}\right]}^{\mathrm{y}}\\ \text{According to the question,}\\ {\text{6.0×10}}^{-3}=\mathrm{k}{\left[0.1\right]}^{\mathrm{x}}{\left[0.1\right]}^{\mathrm{y}}----\left(\mathrm{i}\right)\\ 7.2×{10}^{-2}=\mathrm{k}{\left[0.3\right]}^{\mathrm{x}}{\left[0.2\right]}^{\mathrm{y}}----\left(\mathrm{ii}\right)\\ 2.88×{10}^{-1}=\mathrm{k}{\left[0.3\right]}^{\mathrm{x}}{\left[0.4\right]}^{\mathrm{y}}----\left(\mathrm{iii}\right)\\ \text{Dividing equation}\left(\mathrm{iv}\right)\text{by}\left(\mathrm{i}\right)\text{, we obtain}\\ \frac{2.40\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}}{6.0\mathrm{}\mathrm{x}\mathrm{}{10}^{-3}}=\frac{\mathrm{k}{\left[0.4\right]}^{\mathrm{x}}{\left[0.1\right]}^{\mathrm{y}}}{\mathrm{k}{\left[0.1\right]}^{\mathrm{x}}{\left[0.1\right]}^{\mathrm{y}}}\\ ⇒4=\frac{{\left[0.4\right]}^{\mathrm{x}}}{{\left[0.1\right]}^{\mathrm{x}}}\\ ⇒4={\left(\frac{0.4}{0.1}\right)}^{\mathrm{x}}\\ ⇒{\left(4\right)}^{1}={\left(4\right)}^{\mathrm{x}}\\ ⇒\mathrm{x}=1\\ \text{Dividing equation}\left(\mathrm{iii}\right)\mathrm{by}\left(\mathrm{ii}\right),\text{we obtain}\\ \frac{2.88\mathrm{}\mathrm{x}\mathrm{}{10}^{-1}}{7.2\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}}=\frac{\mathrm{k}{\left[0.3\right]}^{\mathrm{x}}{\left[0.4\right]}^{\mathrm{y}}}{\mathrm{k}{\left[0.3\right]}^{\mathrm{x}}{\left[0.2\right]}^{\mathrm{y}}}\\ ⇒4={\left(\frac{0.4}{0.2}\right)}^{\mathrm{y}}\\ ⇒4={2}^{\mathrm{y}}\\ ⇒{2}^{2}={2}^{\mathrm{y}}\\ ⇒\mathrm{y}=2\\ \text{Therefore, the rate law is rate = k}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^{2}\\ ⇒\mathrm{k}=\frac{\mathrm{Rate}}{\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^{2}}\\ \text{From experiment I, we obtain}\\ \text{k}=\frac{6.0\text{x}{10}^{-3}{\text{mol L}}^{-1}{\text{min}}^{-1}}{\left(0.1{\text{mol L}}^{-1}\right){\left(0.1{\text{mol L}}^{-1}\right)}^{2}}\\ =6.0\mathrm{}{\mathrm{L}}^{2}\mathrm{}{\mathrm{mol}}^{-2}\mathrm{}{\mathrm{min}}^{-1}\\ \text{From experiment II, we obtain}\\ \mathrm{k}=\frac{7.2\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\mathrm{}{\mathrm{min}}^{-1}}{\left(0.3\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\right){\left(0.2\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\right)}^{2}}\\ =6.0\mathrm{}{\mathrm{L}}^{2}\mathrm{}{\mathrm{mol}}^{-2}\mathrm{}{\mathrm{min}}^{-1}\\ \text{From experiment III, we obtain}\\ \mathrm{k}=\frac{2.88\mathrm{}\mathrm{x}\mathrm{}{10}^{-1}\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\mathrm{}{\mathrm{min}}^{-1}}{\left(0.3\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\right){\left(0.4\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\right)}^{2}}\\ =6.0\mathrm{}{\mathrm{L}}^{2}\mathrm{}{\mathrm{mol}}^{-2}\mathrm{}{\mathrm{min}}^{-1}\\ \text{From experiment IV, we obtain}\\ \mathrm{k}=\frac{2.40\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\mathrm{}{\mathrm{min}}^{-1}}{\left(0.4\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\right){\left(0.1\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\right)}^{2}}\\ =6.0\mathrm{}{\mathrm{L}}^{2}\mathrm{}{\mathrm{mol}}^{-2}\mathrm{}{\mathrm{min}}^{-1}\\ \mathrm{Therefore},\mathrm{rate}{\text{constant, k=6.0 L}}^{2}{\mathrm{mol}}^{-2}{\mathrm{min}}^{-1}\end{array}$

Q.14 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/ mol L-1 B/ mol L-1 Initial rate/mol L-1 min-1 I 0.1 0.1 2.0 × 10-2 II — 0.2 4.0 × 10-2 III 0.4 0.4 — IV — 0.2 2.0 × 10-2

Ans.

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

Therefore, the rate of the reaction is given by,

Rate = k [A]1 [B]0

⇒ Rate = k [A]

From experiment I, we obtain

2.0 × 10-2 mol L-1 min-1 = k (0.1 mol L-1)

k = 0.2 min-1

From experiment II, we obtain

4.0 × 10-2 mol L-1 min-1 = 0.2 min-1 [A]

⇒ [A] = 0.2 mol L-1

From experiment III, we obtain

Rate = 0.2 min-1 × 0.4 mol L-1

= 0.08 mol L-1 min-1

From experiment IV, we obtain

2.0 × 10-2 mol L-1 min-1 = 0.2 min-1 [A]

⇒ [A] = 0.1 mol L−1

Q.15 Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s-1 (ii) 2 min-1 (iii) 4 years-1

Ans.

$\begin{array}{l}\left(\mathrm{i}\right){\text{Half life, t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ =\frac{0.693}{200{\text{s}}^{-1}}\\ =3.47×{10}^{–5}\mathrm{s}\left(\text{approximately}\right)\\ \\ \left(\mathrm{ii}\right){\text{Half life, t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ =\frac{0.693}{2{\text{min}}^{-1}}\\ =0.35\text{min}\left(\text{approximately}\right)\\ \\ \left(\mathrm{iii}\right){\text{Half life, t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ =\frac{0.693}{4\text{\hspace{0.17em}}{\mathrm{years}}^{-1}}\\ =0.173\text{\hspace{0.17em}}\mathrm{years}\text{\hspace{0.17em}}\left(\mathrm{approximately}\right)\end{array}$

Q.16 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.

Ans.

$\begin{array}{l}\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}\\ \text{Here,}\\ \text{=}\frac{0.693}{5730}{\mathrm{years}}^{-1}\\ \mathrm{It}\text{is known that,}\\ \text{t=}\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{{\left[\mathrm{R}\right]}_{0}}{\left[\mathrm{R}\right]}\\ =\frac{2.303}{\frac{0.693}{5730}}\mathrm{log}\frac{100}{80}\\ =1845\text{years}\left(\mathrm{approximatelly}\right)\\ \mathrm{Hence},\text{\hspace{0.17em}}\mathrm{the}\text{age of the sample is 1845 years.}\end{array}$

Q.17 The experimental data for decomposition of N2O5

[2N2O5 → 4NO2 + O2]

In gas phase at 318K are given below:

 t(s) 0 400 800 1200 1600 2000 2400 2800 3200 102×[N2O5]mol L-1 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35

(i) Plot [N2O5] against t.

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

(iii) Draw a graph between log [N2O5] and t.

(iv) What is the rate law?

(v) Calculate the rate constant.

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

Ans.

(i) (ii) Time corresponding to the concentration, 1.630×102/2=mol L-1 =81.5 mol L-1 is the half life. From the graph, the half life is obtained as 1450 s.

(iii) t(s) 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 Q.18 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.

$\begin{array}{l}\text{It is known that,}\\ \text{t=}\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{{\left[\mathrm{R}\right]}_{0}}{\left[\mathrm{R}\right]}\\ =\frac{2.303}{60{\mathrm{s}}^{-1}}\mathrm{log}\frac{1}{1/16}\\ =\frac{2.303}{60{\mathrm{s}}^{-1}}\mathrm{log}16\\ =4.6×{10}^{-2}\mathrm{s}\left(\text{approximately}\right)\\ {\text{Hence, the required time is 4.6×10}}^{-2}\mathrm{s}\end{array}$

Q.19 During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1µg of 90Sr 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.

$\begin{array}{l}\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1}{2}}}=\frac{0.693}{28.1}{\mathrm{y}}^{-1}\\ \text{Here, It is known that,}\\ \mathrm{t}=\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{{\left[\mathrm{R}\right]}_{0}}{\left[\mathrm{R}\right]}\\ ⇒10=\frac{2.303}{\frac{0.693}{28.1}}\mathrm{log}\frac{1}{\left[\mathrm{R}\right]}\\ ⇒10=\frac{2.303}{\frac{0.693}{28.1}}\left(-\mathrm{log}\left[\mathrm{R}\right]\right)\\ ⇒\mathrm{log}\left[\mathrm{R}\right]=-\frac{10\mathrm{}\mathrm{x}\mathrm{}0.693}{2.303\mathrm{}\mathrm{x}\mathrm{}28.1}\\ ⇒\left[\mathrm{R}\right]=\mathrm{antilog}\left(-0.1071\right)\\ =\mathrm{antilog}\left(\stackrel{-}{1}.8929\right)\\ =0.7814\mathrm{\mu g}\\ \text{Therefore, 0.7814}\mathrm{\mu }{\text{g of}}^{90}\mathrm{Sr}\text{will remain after 10 years.}\\ \text{Again,}\\ \text{t=}\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{{\left[\mathrm{R}\right]}_{0}}{\left[\mathrm{R}\right]}\\ \mathrm{}⇒60=\frac{2.303}{\frac{0.693}{28.1}}\mathrm{log}\frac{1}{\left[\mathrm{R}\right]}\\ ⇒\mathrm{log}\left[\mathrm{R}\right]=-\frac{60\mathrm{}\mathrm{x}\mathrm{}0.693}{2.303\mathrm{}\mathrm{x}\mathrm{}28.1}\\ ⇒\left[\mathrm{R}\right]=\mathrm{antilog}\left(-0.6425\right)\\ =\mathrm{antilog}\left(\stackrel{-}{1}.3575\right)\\ =0.2278\mathrm{\mu g}\\ \text{Therefore, 0.2278}\mathrm{\mu }{\text{g of}}^{90}\mathrm{Sr}\text{will remain after 60 yeras.}\end{array}$

Q.20 For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

Ans. Q.21 A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.

Ans.

$\begin{array}{l}\mathrm{For}\text{a first order reaction,}\\ \text{t=}\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{{\left[\mathrm{R}\right]}_{0}}{\left[\mathrm{R}\right]}\\ \mathrm{k}=\frac{2.303}{40\mathrm{min}}\mathrm{log}\frac{100}{100-30}\\ =\frac{2.303}{40\mathrm{min}}\mathrm{log}\frac{10}{7}\\ =8.918×{10}^{-3}{\mathrm{min}}^{-1}\\ {\text{Therefore, t}}_{1/2}\text{of the decomposition reaction is}\\ {\text{t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ =\frac{0.693}{8.918×{10}^{-3}}\mathrm{min}\\ =77.7\mathrm{min}\left(\text{approximately}\right)\\ \end{array}$

Q.22 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:

$\begin{array}{l}\underset{\begin{array}{l}\mathrm{At}\text{t=0}\\ \text{At t=t}\end{array}}{{\left({\mathrm{CH}}_{3}\right)}_{2}\mathrm{CHN}}=\underset{\begin{array}{l}{\mathrm{P}}_{0}\\ {\mathrm{P}}_{0-\mathrm{p}}\end{array}}{\mathrm{NCH}{\left({\mathrm{CH}}_{3}\right)}_{2\left(\mathrm{g}\right)}}\to \underset{\begin{array}{l}0\\ \mathrm{p}\end{array}}{{\mathrm{N}}_{2\left(\mathrm{g}\right)}}+\underset{\begin{array}{l}0\\ \mathrm{p}\end{array}}{{\mathrm{C}}_{6}{\mathrm{H}}_{14\left(\mathrm{g}\right)}}\\ \mathrm{After}\mathrm{time},\mathrm{t},\mathrm{total}\text{}\mathrm{pressure},{\text{P}}_{t}=\left({\mathrm{P}}_{0}-\mathrm{p}\right)+\mathrm{p}+\mathrm{p}\\ ⇒{\mathrm{P}}_{t}={\mathrm{P}}_{0}+\mathrm{p}\\ ⇒{\mathrm{P}}_{}={\mathrm{P}}_{t}-{\mathrm{P}}_{0}\\ {\text{Therefore, P}}_{0}-\mathrm{p}={\mathrm{P}}_{0}-\left({\mathrm{P}}_{\mathrm{t}}-{\mathrm{P}}_{0}\right)\\ =2{\mathrm{P}}_{0}-{\mathrm{P}}_{\mathrm{t}}\\ \text{for a first order reaction,}\\ \text{k=}\frac{2.303}{\mathrm{t}}\mathrm{log}\frac{{\mathrm{P}}_{0}}{{\mathrm{P}}_{0}-\mathrm{p}}\\ =\frac{2.303}{\mathrm{t}}\mathrm{log}\frac{{\mathrm{P}}_{0}}{2{\mathrm{P}}_{0}-{\mathrm{p}}_{t}}\\ \text{When t= 360 s, k=}\frac{2.303}{360\mathrm{s}}\mathrm{log}\frac{35.0}{2×35.0-54.0}\\ =2.175×{10}^{-3}{\mathrm{s}}^{-1}\\ \mathrm{k}=\frac{2.303}{720\mathrm{s}}\mathrm{log}\frac{35.0}{2×35.0-63.0}\\ \text{When t = 720s,}\\ \text{=2}..{\text{235×10}}^{-3}{\mathrm{s}}^{-1}\\ \text{Hence, the average value of rate constant is}\\ \text{k=}\frac{\left(2.175×{10}^{-3}\right)+\left(2.235×{10}^{-3}\right)}{2}{\mathrm{s}}^{-1}\\ =2.21×{10}^{-3}{\mathrm{s}}^{-1}\end{array}$

Note: There is a slight variation in this Solution and the one given in the NCERT textbook.

Q.23 The rate constant for the decomposition of N2O5 at various temperatures is given below:

 T/°C 0 20 40 60 80 105×k/s-1 0.0787 1.7 25.7 178 2140

Draw a graph between ln k and 1/T and calculate the values of A and Ea.
Predict the rate constant at 300 and 500C.

Ans.

From the given data, we obtain

 T/°C 0 20 40 60 80 T/K 273 293 313 333 353 1/T 3.66×10−3 3.41×10−3 3.19×10−3 3.0×10−3 2.83 ×10−3 k (s-1) 0.0787 1.70 25.7 178 2140 ln k −7.147 − 4.075 −1.359 −0.577 3.063 $\begin{array}{l}\text{Slope of the line,}\\ \frac{{\mathrm{y}}_{2}-{\mathrm{y}}_{1}}{{\mathrm{x}}_{2}-{\mathrm{x}}_{1}}=–12.301\mathrm{}\mathrm{K}\\ \text{According to Arrhenius equation,}\\ \mathrm{Slope}=-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\\ ⇒{\mathrm{E}}_{\mathrm{a}}=-\mathrm{Slope}\mathrm{}×\mathrm{}\mathrm{R}\\ =-\left(-12.301\mathrm{}\mathrm{K}\right)×\mathrm{}\left(8.314\mathrm{}{\mathrm{JK}}^{-1}\mathrm{}{\mathrm{mol}}^{-1}\right)\\ \text{Again,}\\ \mathrm{ln}\mathrm{}\mathrm{k}=\mathrm{ln}\mathrm{}\mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\\ \mathrm{ln}\mathrm{}\mathrm{A}=\mathrm{ln}\mathrm{}\mathrm{K}+\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\\ \text{When T=273k,}\\ \text{ln k =-7.147}\\ \text{Then, In A=-7.147+}\frac{102.27×{10}^{3}}{8.314×273}\\ =37.911\\ {\text{Therefore, A=2.91×10}}^{6}\\ \text{When T=30+273k=303k,}\\ \frac{1}{\mathrm{T}}=0.00333\mathrm{k}=3.3×{10}^{-3}\mathrm{K}\\ \text{Then, at}\frac{1}{\mathrm{T}}=3.3×{10}^{-3}\mathrm{k}\\ \mathrm{lnk}=\mathrm{}-2.8\\ {\text{Therefore\hspace{0.17em}k=6.08×10}}^{-2}{\mathrm{s}}^{-1}\\ \text{Again when t=50+273k=323k,}\\ \frac{1}{\mathrm{T}}=0.0031\mathrm{}\mathrm{K}=3.1\mathrm{}×\mathrm{}{10}^{-3}\mathrm{}\mathrm{K}\\ \mathrm{Then},\mathrm{at}\frac{1}{\mathrm{T}}=3.1\mathrm{}×\mathrm{}{10}^{-3}\mathrm{}\mathrm{K}\\ \mathrm{lnk}=\mathrm{}-0.5\\ \mathrm{Therefore},\mathrm{K}=0.607{\mathrm{s}}^{-1}\end{array}$

Q.24 The rate constant for the decomposition of hydrocarbons is 2.418 × 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.

$\begin{array}{l}\mathrm{K}=\text{2}.\text{418}×\text{1}{0}^{-\text{5}}{\text{s}}^{-\text{1}}\\ \mathrm{T}\mathrm{}=\mathrm{}\text{546K}\\ {\mathrm{E}}_{\text{a}}=\text{179}.{\text{9 kJmol}}^{-\text{1}}\mathrm{}=\text{179}.\text{9}×\mathrm{}\text{1}{0}^{\text{3}}{\text{Jmol}}^{-\text{1}}\\ \mathrm{According}\text{to the Arrhenius equation,}\\ \mathrm{k}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}\\ ⇒\mathrm{lnk}=\mathrm{lnA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\\ ⇒\mathrm{logk}=\mathrm{logA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{RT}}\\ ⇒\mathrm{logA}=\mathrm{logK}+\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{RT}}\\ =\mathrm{log}\left(2.418\mathrm{}×\mathrm{}{10}^{-5}{\mathrm{s}}^{-1}\right)+\frac{179.9\mathrm{}×\mathrm{}{10}^{3}\mathrm{}\mathrm{J}\mathrm{}{\mathrm{mol}}^{-1}}{2.303\mathrm{}×\mathrm{}8.314\mathrm{}{\mathrm{JK}}^{-1}\mathrm{}{\mathrm{mol}}^{-1}\mathrm{}×\mathrm{}546\mathrm{}\mathrm{K}}\\ =\mathrm{}\left(0.\text{3835}-\mathrm{}\text{5}\right)+\mathrm{}\text{17}.\text{2}0\text{82}\\ =\mathrm{}\text{12}.\text{5917}\\ \text{Therefore},\text{A}=\text{antilog}\left(\text{12}.\text{5917}\right)\\ =\text{3}.\text{9}×\text{1}{0}^{\text{12}}{\text{s}}^{-\text{1}}\mathrm{}\left(\text{approximately}\right)\end{array}$

Q.25 Consider a certain reaction A → Products with k = 2.0 × 10-2 s-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L-1.

Ans.

k = 2.0 × 10-2 s-1

T = 100 s

[A]o = 1.0 moL-1

Since the unit of k is s-1, the given reaction is a first order reaction.

$\begin{array}{l}\mathrm{K}=\frac{2.303}{\mathrm{t}}\mathrm{log}\frac{{\left[\mathrm{A}\right]}_{0}}{\left[\mathrm{A}\right]}\\ \text{Therefore,}\\ ⇒2.0\mathrm{}×\mathrm{}{10}^{-2}{\mathrm{s}}^{-1}=\frac{2.303}{100\mathrm{}\mathrm{s}}\mathrm{log}\frac{1.0}{\left[\mathrm{A}\right]}\\ ⇒2.0\mathrm{}×\mathrm{}{10}^{-2}{\mathrm{s}}^{-1}=\frac{2.303}{100\mathrm{}\mathrm{s}}\left(-\mathrm{log}\left[\mathrm{A}\right]\right)\\ ⇒-\mathrm{log}\left[\mathrm{A}\right]=\frac{2.0\mathrm{}×\mathrm{}{10}^{-2}\mathrm{}×\mathrm{}100}{2.303}\\ ⇒\left[\mathrm{A}\right]=\mathrm{antilog}\left(-\frac{2.0\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}\mathrm{}×\mathrm{}100}{2.303}\right)\end{array}$

= 0.135 mol L-1 (approximately)

Hence, the remaining concentration of A is 0.135 mol L-1.

Q.26 Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Ans. Q.27 The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s-1) e-28000 K/T Calculate Ea.

Ans.

The given equation is

k = (4.5 × 1011 s-1) e-28000 K/T (i)

Arrhenius equation is given by,

$\begin{array}{l}\mathrm{k}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}----\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right),\text{we obtain}\\ \frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}=\frac{28000\mathrm{}\mathrm{K}}{\mathrm{T}}\\ ⇒{\mathrm{E}}_{\mathrm{a}}=\mathrm{R}\mathrm{}×\mathrm{}28000\mathrm{}\mathrm{K}\\ =\mathrm{}\text{8}.{\text{314JK}}^{-\text{1}}{\text{mol}}^{-\text{1}}×\text{28}000\text{K}\\ =\mathrm{}{\text{232792Jmol}}^{-\text{1}}\\ =\mathrm{}\text{232}.{\text{792kJmol}}^{-\text{1}}\end{array}$

Q.28 The decomposition of A into product has value of k as 4.5 × 103 s-1 at 100C and energy of activation 60 kJ mol-1. At what temperature would k be 1.5 × 104 s-1?

Ans. 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 × 1010 s-1. Calculate k at 318 K and Ea.

Ans.

$\begin{array}{l}\mathrm{For}\text{a first order reaction,}\\ \mathrm{t}=\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\\ \text{At 298k,}\mathrm{t}=\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{100}{90}\\ =\frac{0.1054}{\mathrm{k}}\\ \text{At 308k,}{\mathrm{t}}^{1}=\frac{2.303}{{\mathrm{k}}^{1}}\mathrm{log}\frac{100}{75}\\ =\frac{2.2877}{\mathrm{k}}\\ \text{According to the question,}\\ \mathrm{t}=\mathrm{}{\mathrm{t}}^{1}\\ ⇒\frac{0.1054}{\mathrm{k}}=\frac{0.2877}{{\mathrm{k}}^{1}}\\ ⇒\frac{{\mathrm{k}}^{1}}{\mathrm{k}}=2.7296\\ \text{From Arrhenius equation, we obtain}\\ \mathrm{log}\frac{{\mathrm{k}}^{1}}{\mathrm{k}}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{R}}\left(\frac{{\mathrm{T}}^{1}-\mathrm{T}}{{\mathrm{TT}}^{1}}\right)\\ \text{log}\left(2.7296\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}×\mathrm{}8.314}\left(\frac{308-298}{298\mathrm{}×\mathrm{}308}\right)\\ {\mathrm{E}}_{\mathrm{a}}=\frac{2.303\mathrm{}×\mathrm{}8.314\mathrm{}×\mathrm{}298\mathrm{}×\mathrm{}308\mathrm{}×\mathrm{log}\left(2.7296\right)}{308-298}\\ =76640.096\mathrm{}\mathrm{J}\mathrm{}{\mathrm{mol}}^{-1}\\ =76.64\mathrm{}\mathrm{kJ}\mathrm{}{\mathrm{mol}}^{-1}\\ \mathrm{To}\text{calculate k at 318 k,}\\ {\text{It is given that, A=4×10}}^{10}{\mathrm{s}}^{-1},\mathrm{T}=318\mathrm{K}\\ \text{Again, from Arrhenius equation, we obtain}\\ \mathrm{logk}=\mathrm{logA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{RT}}\\ =\mathrm{log}\left(4\mathrm{}×\mathrm{}{10}^{10}\right)-\frac{76.64\mathrm{}×\mathrm{}{10}^{3}}{2.303\mathrm{}×\mathrm{}8.314\mathrm{}×\mathrm{}318}\\ =\left(0.6021+10\right)-12.5876\\ = –1.9855\\ \mathrm{Therefore}\text{, k = Antilog}\left(-1.9855\right)\\ =1.034×{10}^{-2}{\mathrm{s}}^{-1}\end{array}$

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.

$\begin{array}{l}\text{From Arrhenius equation, we obtain}\\ \mathrm{log}\frac{{\mathrm{k}}_{2}}{{\mathrm{k}}_{1}}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{R}}\left(\frac{{\mathrm{T}}_{2}-{\mathrm{T}}_{1}}{{\mathrm{T}}_{1}{\mathrm{T}}_{2}}\right)\\ {\text{It is given that, k}}_{2}=4{\mathrm{k}}_{1}\\ {\text{T}}_{\text{1}}\mathrm{}=\text{293K}\\ {\text{T}}_{\text{2}}\mathrm{}=\mathrm{}\text{313 K}\\ \text{Therefore,}\mathrm{log}\frac{{\mathrm{k}}_{1}}{{\mathrm{k}}_{2}}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}×\mathrm{}8.314}\left(\frac{313-293}{293\mathrm{}×\mathrm{}313}\right)\\ ⇒0.6021=\frac{20\mathrm{}×\mathrm{}{\mathrm{E}}_{\mathrm{a}}\mathrm{}}{2.303×\mathrm{}8.314\mathrm{}×\mathrm{}293\mathrm{}×313\mathrm{}}\\ ⇒{\mathrm{E}}_{\mathrm{a}}=\frac{0.6021\mathrm{}×\mathrm{}2.303\mathrm{}×\mathrm{}8.314\mathrm{}×\mathrm{}293\mathrm{}×\mathrm{}313\mathrm{}}{20}\\ =52863.33\mathrm{}\mathrm{J}\mathrm{}{\mathrm{mol}}^{-1}\\ =52.86\mathrm{}{\mathrm{kJmol}}^{-1}\\ {\text{Hence, the required energy of activation is 52.86 kJmol}}^{-1}.\end{array}$