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Std. XII Sci.

PERFECT

CHEMISTRY (Vol. I)

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Written as per the latest textbook prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune.

Written as per the new textbook Subtopic-wise segregation for powerful concept building Complete coverage of Textual Exercise Questions, Intext Questions and Numericals Includes relevant board questions of February 2020 Extensive coverage of New Type of Questions ‘Solved Examples’ guide you through every type of problem ‘Apply Your Knowledge’ section for application of concepts ‘Quick Review’ at the end of every chapter facilitates quick revision A compilation of all ‘Important Formulae’ in relevant chapters ‘Competitive Corner’ presents questions from prominent competitive examinations Reading Between the Lines, Enrich Your Knowledge, Gyan Guru, Connections,

NCERT Corner are designed to impart holistic education Topic Test at the end of each chapter for self-assessment Video/pdf links provided via QR codes for boosting conceptual retention QR Code to access the Latest Board Question Papers

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This reference book is transformative work based on textbook Chemistry; First edition: 2020 published by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

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Perfect Chemistry Std. XII, Vol. I is intended for every Maharashtra State Board aspirant of Std. XII, Science. The scope, sequence, and level of the book are designed to match the new textbook issued by the Maharashtra State board. At this crucial juncture in their lives, when the students are grappling with the pressures of cracking a career-defining board examination, we wanted to create a book that not only develops the necessary knowledge, tools, and skills required to excel in the examination, but also enables students to appreciate the beauty of the subject and piques their curiosity. We believe that students respond favourably to meaningful content, if it is presented in a way that is easy to read and understand, rather than being mired down with facts and information. Consequently, we have always placed the highest priority on writing clear and lucid explanations of fundamental concepts. Moreover, special care has been taken to ensure that the topics are presented in a logical order. The coherent Question/Answer approach helps students expand their horizon of understanding of the concepts. The primary purpose of this book is to assist the students in preparing for the board examination. However, this is closely linked to other goals: to exemplify how important and how incredibly interesting chemistry is, and to help the student become an expert thinker and problem solver. Solving numericals is essential for success in chemistry! To help the students hone their problem-solving skills, this book amalgamates numericals that are rich in both variety and number which provides the student with ample practice, ensuring mastery of each concept. The scope of the book extends beyond the State Board examination as it also offers a plethora of Multiple Choice Questions (MCQs) in order to familiarize the students with the pattern of competitive examinations. In addition, the chapter-test have been carefully crafted to focus on concepts, thus providing the students with a quick opportunity for self-assessment and giving them an increased appreciation of chapter-preparedness. We believe that the study of chemistry helps in the understanding of many fascinating and important phenomena. In this vein, we have put an effort to relate chemistry to real-world events in order to show students that chemistry is a vibrant, constantly evolving science that has relevance in our modern world. We hope this book becomes a valuable tool for you and helps you to not only understand the concepts of chemistry but also to see the world from a molecular point of view. Our Perfect Chemistry Std. XII, Vol. I adheres to our vision and achieves several goals: building concepts, developing competence to solve numerical, recapitulation, self-study, self-assessment and student engagement—all while encouraging students toward cognitive thinking. - Publisher

Edition: Second

PREFACE

Sample Content

KEY FEATURES

Reading between the lines provideselaboration or missing fragments ofconcept which is essential for completeunderstanding of the concept.

NCERT Corner covers information fromNCERT textbook relevant to topic.

Connections enable students to interlinkconcepts covered in different chapters.

QR code provides: i. Access to a video/PDF in order to

boost understanding of a concept oractivity

ii. The Latest Board Question Papers

Connections

QR Codes

Enrich Your Knowledge presents fascinating information about the concept covered.

Enrich Your

Knowledge

Reading between the lines

NCERT Corner

Gyan Guru illustrates real lifeapplications or examples related to theconcept discussed.

GG‐Gyan Guru

Apply Your Knowledge includeschallenging questions.

Apply Your

Knowledge

Quick review includes tables/ flow chartto summarize the key points in chapter.

Quick Review

Important Formulae includes all of the key formulae in the chapter.

Important Formulae

Competitive Corner includes selectivequestions from prominent [NEET (UG),JEE (Main), NEET (ODISHA), MHTCET] competitive exams based entirely onthe syllabus covered in the chapter.

Competitive Corner

Sample Content

There will be one single paper of 70 Marks in Chemistry. Duration of the paper will be 3 hours.

Section A: (18 Marks) This section will contain Multiple Choice Questions and Very Short Answer(VSA) type of questions. There will be 10 MCQs and 8 VSA type of questions, each carrying One mark. Students will have to attempt all the questions.

Section B: (16 Marks) This section will contain 12 Short Answer (SA-I) type of questions, each carrying Two marks. Students will have to attempt any 8 questions.

Section C: (24 Marks) This section will contain 12 Short Answer (SA-II) type of questions, each carrying Three marks. Students will have to attempt any 8 questions.

Section D: (12 Marks) This section will contain 5 Long Answer (LA) type of questions, each carrying Four marks. Students will have to attempt any 3 questions.

Distribution of Marks According to the Type of Questions

Type of Questions MCQ 1 Mark each 10 Marks VSA 1 Mark each 8 Marks SA - I 2 Marks each 16 Marks SA - II 3 Marks each 24 Marks LA 4 Marks each 12 Marks

PAPER PATTERN

Sample Content

Chapter No.

Chapter Name Marks without

option Marks with

option Page No.

1 Solid State 3 5 1

2 Solutions 4 6 48

3 Ionic Equilibria 4 6 93

4 Chemical Thermodynamics 6 8 133

5 Electrochemistry 5 7 184

6 Chemical Kinetics 4 6 236

7 Elements of Groups 16, 17 and 18 6 8 276

8 Transition and Inner Transition Elements 6 8 321

9 Coordination Compounds 5 7 362

Modern Periodic Table 405

Electronic Configuration of Elements 406

Logarithms and Antilogarithms 407

Scan the given Q.R. Code to access the Latest Board Question Papers.

[Reference: Maharashtra State Board of Secondary and Higher Secondary Education, Pune ‐ 04]

Note: 1. * mark represents Textual question.2. # mark represents Intext question.3. + mark represents Textual examples.4. symbol represents textual questions that need external reference for an

answer.

CONTENTS

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236

Std. XII Sci.: Perfect Chemistry (Vol. I)

Q.1. Can you recall? (Textbook page no. 120)i. What is the influence of particle size of reacting solid on rate of a chemical reaction?ii. Why is finely divided nickel used in hydrogenation of oil?iii. What is effect of change of temperature on the rate of a chemical reaction?Ans:i. Smaller the particle size of reacting solid, greater is the rate of reaction.ii. Finely divided nickel larger surface area. Greater the surface area, greater is the adsorption and hence, finely

divided nickel used in hydrogenation of oiliii. Increase in temperature increases the rate of a chemical reaction. Q.2. State TRUE or FALSE. Correct the false statement.i. Equilibrium constants predict the feasibility of reaction.ii. Thermodynamics predicts extent of reaction.iii. Chemical kinetics predict rate of reaction.Ans:i. False

Equilibrium constants predict the extent of reaction.ii. False

Thermodynamics predicts feasibility of reaction.iii. True Q.3 Define chemical kinetics.Ans: Chemical kinetics is a branch of chemistry which deals with the rates of chemical reactions and the factors

those affect them.

Q.4. Write the applications of chemical kinetics?Ans:i. It predicts the rates of reactions for different reasons.ii. It predicts how rapidly the reaction approaches equilibrium.iii. It gives information on the mechanism of chemical reactions.

Q.5. Explain: Rate of reactionAns:i. The rate of reaction describes how rapidly the reactants are consumed or the products are formed.ii. The rate of reaction represents a decrease in concentration of the reactant per unit time or increase in

concentration of product per unit time.iii. The SI unit of rate of reaction is mol dm–3 sec–1.

6 Chemical Kinetics

6.1 Introduction 6.2 Rate of reactions 6.3 Rate of reaction and reactant concentration 6.4 Molecularity of elementary reactions

6.5 Integrated rate law 6.6 Collision theory of bimolecular reactions 6.7 Temperature dependence of reaction 6.8 Effect of catalyst on the rate of reaction

CONTENTS AND CONCEPTS

6.2 Rate of reactions

6.1 Introduction

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237

Chapter 6: Chemical Kinetics

Q.6. Define: Average rate of reactionAns: The average rate of a reaction is the change in concentration of reactant or product divided by time interval

over which the change occurs.

Average rate = changeinconcentrationof aspecieschangein time

= Ct

Q.7. How is average rate calculated? (NCERT) Ans: Consider a simple reaction A B, in which A is consumed and B is produced.

Average rate of consumption of A = [A]t

, Average rate of formation of B = [B]

t

Hence, average rate of reaction = [A]t

= [B]

t

*Q.8. How is instantaneous rate of reaction determined?Ans:i. To determine the instantaneous rate of a reaction the progress of a reaction is followed by measuring the

concentrations of a reactant or product for different time intervals.ii. The concentration of a reactant or a product is plotted against time.iii. A tangent drawn to the curve at time t1 gives the rate of the reaction. The slope thus obtained gives the

instantaneous rate of the reaction at time t1.

iv. Mathematically the instantaneous rate is expressed by replacing by derivative dc/dt in the expression ofaverage rate.

Instantaneous rate = d[C]dt

v. For a reaction A B

Instantaneous rate of consumption of A at time t = d[A]dt

Instantaneous rate of the formation of B at time t = d[B]dt

Therefore, instantaneous rate of the reaction at time t = d[A]dt

= d[B]dt

Q.9. Express the rate of reaction in terms of change in concentration of each constituent for a reaction,A + 3B 2C.

Ans: Rate of reaction = d[A] 1 d[B] 1 d[C]dt 3 dt 2 dt

time t1

(a)

reac

tant

con

cent

ratio

n

time t1

(b)

prod

uct c

once

ntra

tion

Determination of instantaneous rate

When one mole of A is consumed three moles of B are consumed and two moles of C are formed. Thestoichiometric coefficients of the three species are different. Thus the rate of consumption of B is three times therate of consumption of A. Likewise the rate of formation of C is twice the rate of consumption of A. We write,

d[B] d[A] d[C] d[A]3 and 2dt dt dt dt

READING BETWEEN THE LINES

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#Q.10. Write the expression for 2N2O(g) 4NO2(g) + O2(g).

Ans: Rate of reaction = – 2 2 21 d[N O] 1 d[NO ] d[O ]2 dt 4 dt dt

*Q.11. For the reaction, N2(g) + 3H2(g) 2NH3(g), what is the relationship among 2 2 3d[N ] d[H ] d[NH ], anddt dt dt

?

Ans: The relationship among 2 2 3d[N ] d[H ] d[NH ], anddt dt dt

is 2 2 3d[N ] 1 d[H ] 1 d[NH ]dt 3 dt 2 dt

=- =

+Q.12. For the reaction 2N2O5(g) 4NO2(g) + O2(g) in liquid bromine, N2O5 disappears at a rate of0.02 mol dm–3 sec–1. At what rate NO2 and O2 are formed? What would be the rate of reaction?

Solution:

Given: Rate of disappearance of N2O5 = 2 5d[N O ]dt

= 0.02 moles dm–3 sec–1

To find: i. Rate of formation of NO2 ii. Rate of formation of O2 iii. Rate of reaction

Calculation: Rate of reaction = 2 51 d[N O ]2 dt

= 21 d[NO ]4 dt

= 2d[O ]dt

Rate of formation of O2 = 2d[O ]dt

= 2 51 d[N O ] 1 0.022 dt 2

mol dm–3 sec–1

= 0.01 mol dm–3 sec–1

Rate of formation of NO2 = 2d[NO ]dt

= 2 54 d[N O ]2 dt

= 2 × 0.02 mol dm3 sec–1

= 0.04 mol dm3 sec–1

Rate of reaction = 2 51 d[N O ]2 dt

= 2 51 d[N O ] 1 0.022 dt 2

mol dm–3 sec–1

= 0.01 mol dm–3 sec–1

Ans: i. Rate of formation of NO2 = 0.04 mol dm3 sec–1

ii. Rate of formation of O2 = 0.01 mol dm–3 sec–1

iii. Rate of reaction = 0.01 mol dm–3 sec–1

Q.13. Try this (Textbook page no. 121)For the reaction,

2 2(aq) 2 8(aq) 3(aq) 4(aq)3 I + S O I + 2 SO

Calculate the rate of formation of 3I , the rates of consumption of I– and 2

2 8S O and the overall rate of reaction if the rate of formation of 2

4SO is 0.022 mol dm–3 sec–1. Solution: Given: Rate of formation of 2

4SO = 0.022 mol dm–3 sec–1.

To find: i. Rate of formation of 3I

ii. Rate of consumption of I– and 22 8S O

iii. Overall rate of reactionCalculation: For the reaction:

2 2(aq) 2 8(aq) 3(aq) 4(aq)3 I S O I 2 SO

Overall rate of reaction 2 2

2 8 3 41 d[I ] d[S O ] d[I ] 1 d[SO ]3 dt dt dt 2 dt

SOLVED EXAMPLES

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To see complete chapter buy Target Notes or Target E‐Notes

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258


*Q.79. Write Arrhenius equation and explain the terms involved in it.Ans: Arrhenius equation: k = ( )E /RTaAe-

where, k is the rate constant, Ea is the activation energy, R molar gas constant, T temperature in kelvin and A is the pre-exponential factor.

Note: i. The pre exponential factor A and the rate constant have same unit in case of the first order reactions. Besides A is found to be related to frequency of collisions.

ii. The fraction of successful collisions is given by f = e-Ea/RT.*Q.80. How will you determine activation energy:i. graphically using Arrhenius equationii. from rate constants at two different temperatures?Ans:i. Graphical representation of activation energy

Arrhenius equation isE /RTak Ae-=

Taking logarithm of both sides of eqn. we obtain

ln k = aERT

- + ln A

Converting natural base to base 10 we write

log10k = aE 12.303 R T

- + log10A

This equation is of the form of straight line y = mx + c.

The Arrhenius plot of log10k versus 1T

gives a straight line as shown in the diagram. A slope of the line is

aE2.303 R

- with its intercept being log10A.

From a slope of the line, the activation energy can be determined.

y m x c

Variation of log10k with 1T

1T

log 1

0k

NCERT Corner

The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).

Effective collisions are those collisions in which molecules collide with sufficient kinetic energy (i.e. threshold energy) and proper orientation to facilitate breaking of bonds between reacting species and formation of new bonds to form products. Where, Threshold energy = Activation energy + energy possessed by reacting species

Collision frequency (Z):

Effective collision:

6.7 Temperature dependence of reaction rates

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259


ii. Determination of activation energy from rate constants at two different temperatures.Arrhenius equation isk = E / RTaAe-

From two different temperatures T1 and T2

log10k1 = log10A a

1

E2.303RT

- …(1)

log10k2 = log10A a

2

E2.303RT

…(2)

where k1 and k2 are the rate constants at temperatures T1 and T2 respectively. Subtracting Eq. (1) from Eq. (2),

log10k2 log10k1 = a a

2 1

E 1 E 12.303R T 2.303R T

- +

Hence, log102 a

1 1 2

k E 1 1k 2.303 R T T

æ ö÷ç ÷= -ç ÷ç ÷çè ø = a 2 1

1 2

E T T2.303 R T T

æ ö- ÷ç ÷ç ÷ç ÷çè ø

*Q.81. Explain with the help of Arrhenius equation, how does the rate of reaction changes withi. temperature and ii. activation energy.

Ans:

i. Arrhenius equation E /RTak e-= shows that as the temperature rises, aERT

decreases, this causes an increase in

aERT

. This increases k and the rate of reaction.

ii. The decrease in energy of activation (Ea) decreases aERT

, hence, increases aERT

and the rate of reaction.

*Q.82. Explain graphically the effect of temperature on the rate of reaction.Ans:i. The average kinetic energy of molecules is proportional to temperature.ii. At a given temperature, the fraction of molecules with their kinetic energy equal to or greater than Ea may

lead to the product.iii. With an increase of temperature, the fraction of molecules having their energies (Ea) would increases. The

rate of the reaction thus would increase.iv. This is depicted by plotting a fraction of molecules with given kinetic energy versus kinetic energy for two

different temperatures T1 and T2 (T2 > T1). The area between the curve and the horizontal axis isproportional to the total number of molecules. The total area is the same at T1 and T2.

v. The entire shaded area which represents a fraction of molecules with kinetic energy exceeding Ea is larger atT2 than at T1 (since T2 > T1).

vi. This indicates that a fraction of molecules possessing energies larger than Ea increase with temperature. Therate of reaction increases accordingly.

Kinetic Energy

Frac

tion

of m

olec

ules

w

ith g

iven

kin

etic

ene

rgy

T1

T2

Comparison of fraction of molecules activated at T1 and T2

It has been observed that the rates of most of the chemical reactions usually increase with temperature. Ineveryday life we see that the fuels such as oil, coal are inert at room temperature but burn rapidly at highertemperatures. Many foods spoil rapidly at room temperature and lasts longer in freezer.

ENRICH YOUR KNOWLEDGE

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264


a10 1 1

0.2851 E 308 K 298 Klog0.1043 2.303 8.314 J K mol 308 K 298 K

0.4366 = a1

E 308 2982.303 8.314 J mol 308 298

Ea = 76728 kJ mol–1= 76.728 kJ mol–1

Now,

log10k = log10(4 × 1010 ) 1

1 176728 J mol

2.303 8.314 J K mol 318 K

log10 k =10.6020 – 12.6052

k = antilog (-2.0032) k = 9.92× 10–3 s-1

Ans: The activation energy is 76.728 kJ mol–1 and the rate constant at 318 K is 9.92× 10–3 s-1.

Q.93. What is a catalyst? Give an example.Ans: A catalyst is a substance added to the reactants that increases the rate of the reaction without itself being

consumed in the reaction. e.g. MnO2 is the catalyst in the reaction; 2KClO3(s) + 3O2(g) MnO2 2KCl(s)

*Q.94. How catalyst increases the rate of reaction? Explain with the help of potential energy diagram forcatalyzed and uncatalyzed reactions.

Ans:i. A catalyst provides alternative pathway associated with lower activation energy.

E 0

E 0

E0

E0

Energy

reaction coordinatePotential energy barriers for

catalyzed and uncatalyzed reactions

6.8 Effect of catalyst on the rate of reaction

Connections

In std. XI , chapter 11, you studied in details on catalyst.

Catalytic converters!

Emissions from internal combustion engines of automobiles are hazardous due to production of toxic byproducts like nitrogen oxide, carbon monoxides and hydrocarbons. To control these emissions, catalytic converters are used which convert these byproducts in less hazardous carbon dioxide, water vapour and nitrogen gas.

GG - Gyan Guru

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APPLY YOUR KNOWLEDGE

QUICK REVIEW

Q.97. One of the nuclear waste products is 90Sr with a half life of 6.93 years. If 1 μg of 90Sr was absorbed inbones of newly born baby in place of calcium, how much amount of time, in years, is required to reduce it by 90 %, if it is not lost metabolically.

Solution: Given: t1/2 = 6.93 years

[A]0= 1.0 g[A]t = 1.0 – 0.9 = 0.01 g

To find: Time for 90Sr to reduce by 90 %.

Formulae: i. t1/2 = 0.693

kii. 0

10t

2.303 [A]k = logt [A]

Calculation: Radioactive disintegration follows first order reaction. The half-life of first order reaction is,

t1/2 = 0.693

k

k = 0.6936.93 year

= 0.1 year–1

Time required for 90Sr to reduce by 90 % 0

10t

2.303 [A]k = logt [A]

0.1 = 102.303 1.0log

t 0.01

t = 102.303 1.0log0.1 0.01

t = 23.03 × (log10 [1.0] – log10 [0.01]) t = 46 years.

Ans: The time require to reduce 90Sr by 90 % is 46 years. Q.98. With reference to the following plot for a reaction, calculate the value of activation energy (Ea).

Ans: From the slope of the graph of ln k versus 1/T, activation energy can be calculated.

Slope = aER

Ea = (Slope) (R) = ( 5.5 103 K) (8.314 J K1 mol1) Ea = 45727 4.6 104 J mol1

ln k

Slope = 5.5 103 K

1T

Average Rate of reaction

r = ct

Instantaneous rate of reaction

r = dcdt

Rate of reaction

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1. Average rate of reaction:

(rav) = Totalchange inconcentrationTimetakenfor thechange

= Ct

where, [C] = Concentration of reactants or products

2. For the reaction, aA + bB cC + dD:i. Instantaneous rate of reaction

= 1a

d Adt

= 1b

d Bdt

= + 1c

d Cdt

= + 1d

d Ddt

ii. Rate of disappearance of A = d Adt

iii. Rate of disappearance of B = d Bdt

iv. Rate of formation of C = + d Cdt

v. Rate of formation of D = + d Ddt

3. Differential rate lawRate of the reaction = k[A]x[B]y

where, x and y are experimentally determinedexponents of concentrations on which the rate ofreaction depends.Order of reaction = x + y

4. Integrated rate Law:i. For zero order reaction:

k = 0 tA A

t

ii. For first order reaction:

k = 2.303

tlog10

0

t

AA

= 2.303

tlog10

aax

where [A]0 = a = Initial concentration of the reactants, [A]t = (a x) = Concentration of reactants attime t, k = Rate constant, t = time

5. Half life period:

i. For zero order reaction t1/2 = 0[A]2k

ii. For first order reaction t1/2 =0.693

k

Types of reaction

Elementary reaction

Occurs in single step.e.g. O3 O2 + O

Complex reaction

Occurs in multiple step. e.g. NO2Cl NO2 + Cl NO2Cl + Cl NO2 + Cl

2NO2Cl 2NO2 + Cl2

Unimolecular reaction

One reactant involved. e.g. C2H5I C2H4 + HI

Biomolecular reaction

Two reactants involved. e.g. O3 O2 + O

Rate Law

Differential rate law

Rate = k[A]x[B]y

Integrated rate law

First order

k =

0

t

A2.303 logt A

Zero order kt = [A]0 [A]t

IMPORTANT FORMULAE

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6. Arrhenius equation:

k = A E /RTae or log10 k = log10 A aE2.303RT

where k = Rate constant, A = Frequency factor or pre-exponential factor,

Ea = Activation energy, R = Gas constant = 8.314 J K1 mol1

7. The fraction of successful collisions:f = e-Ea/RT

where, f = fraction of successful collisions8. Temperature dependence of reaction rates:

For two different temperatures T1 and T2.

log10 2

1

kk

= aE2.303R 1 2

1 1T T

where k1 and k2 are rate constants at

temperatures T1 and T2 respectively.

6.1 Introduction 1. What does thermodynamics tell us about the

reaction?Ans: Thermodynamics predicts feasibility of reaction. 2. What is chemical kinetic?Ans: Refer Q.3. 6.2 Rate of reactions 3. Explain: Instantaneous rate of reaction.Ans: Refer Q.8. 4. Express the rate of reaction in terms of change

in concentration of each constituent for areaction: 2A + 3B C

Ans: Refer Q.9. 5. For the reaction,

N2(g) + 3H2(g) 2 NH3(g), what is the

relationship among 2 3d[N ] d[NH ] anddt dt

?

Ans: Refer Q.11. 6. Ammonia and oxygen react at high temperature

as:4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) In an experiment, rate of formation of NO(g) is3.6 103 mol L1 s1. Calculate:

i. Rate of disappearance of ammonia.ii. Rate of formation of water.Ans:i. 3.6 10–3 mol L–1 s –1

ii. 5.4 10–3 mol L–1 s–1 6.3 Rate of reaction and reactant concentration 7. Explain the term ‘rate law’.Ans: Refer Q.15.

8. i. How does change in concentration affect the rate of reaction?

ii. What is overall order of the reaction?Ans: i. Refer Q.16. ii. Refer Q. 20. (i) 9. For hypothetical reaction reaction,

A + B CA rate law is rate = k[A][B]

i. How does reaction rate changes if [B] isdecreased by a factor of 5?

ii. What is change in rate if concentrations of bothreactants are doubled?

Ans: Refer Q.18. 10. Give example of a reaction in which order of the

reaction is fractional. Ans: Refer Q.23. 11. For hypothetical reaction A + B products,

find the rate law from the following data. [A]/M [B]/M Rate/M s–1

0.3 0.01 0.10 0.6 0.01 0.20 0.6 0.04 0.80

Ans: Rate = k[A][B] 12. The rate of the reaction for reaction

A + B product is 0.460 mol dm–3 s–1 when[A]= [B] = 0.2 mol dm–3. If the reaction is offirst order in A and zeroth order in B, what isthe rate constant?

Ans: 2.3 s–1

13. Consider the reaction A + B products. If

the concentration of A and B are halved, therate of the reaction decreases by a factor of 8.If the concentration of A is increased by afactor of 2.5, the rate increases by the factor of2.5. What is the order of the reaction? Writethe rate law.

Ans: rate = k[A][B]2 6.4 Molecularity of elementary reactions

14. What is elementary reaction?Ans: Refer Q.32. (i) 15. Give an example of complex reaction.Ans: Refer Q. 32. (ii-e.g.) 16. Identify the molecularity and write the rate law

for each of the following elementary reactions.i. 2NO2(g) 2NO2(g) + O2(g)ii. C2H5I(g) C2H4(g) + HI(g)Ans: Refer Q.34. 17. What is the relationship between coefficients of

reactants in a balanced equation for an overallreaction and exponents in rate law.

Ans: Refer Q.35.

EXERCISE

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271


1. For the reaction, the rate of reaction at a giveninstant is represented as _______.+ d A d B d C1

A 2 dt dt

Identify the chemical reaction. (A) A + B C (B) C A + 2B

(C) C A +12

B (D) A + 2B C

2. Which of the following expressions is correctfor the rate of reaction given below?

(aq)aq aq5Br BrO 6H 2 ( )2 aqBr 3H O l

(A) Br H

5t t

(B) Br H6

t 5 t

(C) Br H5

t 6 t

(D) Br H

6t t

3. In the reaction A + 3B 2C, the rate offormation of C is _______.(A) the same as rate of consumption of A(B) the same as the rate of consumption of B(C) twice the rate of consumption of A(D) 3/2 times the rate of consumption of B

*4. The rate law for the reaction aA + bB P israte = k[A] [B]. The rate of reaction doubles if _______. (A) concentrations of A and B are both

doubled.(B) [A] is doubled and [B] is kept constant(C) [B] is doubled and [A] is halved(D) [A] is kept constant and [B] is halved.

5. Rate law for the reaction A + 2B C isfound to beRate = k [A]2 [B]

Concentration of reactant A is doubled, keepingconcentration of ‘B’ constant, the value of rateconstant will be _______.(A) the same (B) doubled(C) quadrupled (D) halved

6. Rate law for a reaction is found to beRate = [A]–x [B]y. Which of the followingstatement is correct?(A) Rate increases with increase in

concentration of A.(B) Rate increases with increase in

concentration of B.

(C) Rate decreases with increase intemperature.

(D) Rate is independent of concentration of Aor B and temperature.

*7. Rate law for the reaction,2NO + Cl2 2 NOCl is rate = k[NO2]2[Cl2]. Thus of k would increase with _______. (A) increase of temperature(B) increase of concentration of NO(C) increase of concentration of Cl2 (D) increase of concentrations of both Cl2 and

NO

8. In any unimolecular reaction _______.(A) only one reacting species is involved in

rate determining step(B) The order and molecularity of slowest

step is equal to one(C) The molecularity of reaction is one and

border is zero(D) both molecularity and order are one

9. In a reaction: A + B C + D, themolecularity of the reaction is _______.(A) 0 (B) 1(C) 2 (D) 3

*10. The elementary reactionO3(g) + O(g) 2 O2(g) is _______. (A) unimolecular and second order(B) bimolecular and first order(C) bimolecular and second order(D) unimolecular and first order

11. For a complex reaction ______________.(A) order of overall reaction is same as

molecularity of the slowest step.(B) order of overall reaction is less than the

molecularity of the slowest step.(C) order of overall reaction is greater than

molecularity of the slowest step.(D) molecularity of the slowest step is never

zero or non integer.

*12. The reaction, 3ClO– ClO3– +2Cl– occurs in

two steps, (i) 2ClO– ClO2

–

(ii) ClO2– + ClO– ClO3

– + Cl–

The reaction intermediate is _______.(A) Cl– (B) ClO2

–

(C) ClO3– (D) ClO–

13. Rate law can be determined from balancedchemical equation if _______.(A) reverse reaction is involved(B) it is an elementary reaction(C) it is a sequence of elementary reactions(D) any of the reactants is in excess

MULTIPLE CHOICE QUESTIONS

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272


*14. Slope of the graph ln[A]t versus t for first orderreaction is _______. (A) –k (B) k(C) k/2.303 (D) –k/2.303

*15. The rate constant for the reaction2 N2O5(g) 2 N2O4(g) + O2(g) is 4.98 × 10–4 s–1. The order of reaction is _______. (A) 2 (B) 1 (C) 0 (D) 3

*16. Time required for 90 % completion of a certainfirst order reaction is t. The time required for 99.9 % completion will be _______. (A) t (B) 2t (C) t/2 (D) 3t

*17. What is the half life of a first order reaction iftime required to decrease concentration of reactant from 0.8 M to 0.2 M is 12 h? (A) 12 h (B) 3 h(C) 1.5 h (D) 6 h

18. Which of the following process follows firstorder kinetics?(A) Decomposition of nitrous oxide(B) Hydrolysis of cane sugar(C) Radioactive decay of 90Sr(D) Decomposition of HI

19. Which among the following reactions is anexample of a zero order reaction?(A) H2(g) + I2(g) 2HI(g)(B) 2H2O2(l) 2H2O(l) + O2(g) (C) C12H22O11(aq) + H2O(l) C6H12O6(aq)

+ C6H12O6(aq)

(D) 2NH3(g) Pt N2(g) + 3H2(g)

*20. The order of the reaction for which the units ofrate constant are mol dm–3 s–1 is _______. (A) 1 (B) 3 (C) 0 (D) 2

21. The half-life period of zero order reactionA product is given by _______.

(A) 0Ak

(B) 0.693

k

(C) 0A2k

(D) 02 Ak

22. The kinetic order for the following reaction is_______.2N2O(g) Pt 2N2(g) + O2(g)(A) zero (B) first(C) second (D) third

23. Which of the following statement is INCORRECTabout the collision theory of chemical reaction?(A) Number of collisions determines the rate

of reaction.(B) Collision of atoms or molecules with

energy equal to greater than activationenergy leads to product formation.

(C) Molecules should collide with properorientation for collision to be effective.

(D) Increase in temperature increases thecollision frequency, which increases therate of product formation.

24. Which of the following is TRUE?(A) The potential energy of the transition state

is greater than the potential energy ofeither reactants or products.

(B) The potential energy of the transition stateis less than the potential energy of eitherreactants or products.

(C) The potential energy of the transition stateis greater than the potential energy ofreactants but less than the potentialenergy of products.

(D) The potential energy of the transition stateis less than the potential energy ofreactants but more than the potentialenergy of products.

*25. For an endothermic reaction, X Y . If Ef isactivation energy of the forward reaction and Er that for reverse reaction, which of the following is correct? (A) Ef = Er(B) Ef < Er(C) Ef > Er(D) ΔH = Ef – Er is negative

26. What happens when the temperature of a reaction

mixture is increased from 25 C to 52 C?(A) The rate of reaction remains unchanged

and rate constant increases.(B) The rate of reaction increases and rate

constant decreases.(C) The rate of reaction decreases and so the

rate constant decreases.(D) Both rate and rate constant of the reaction

increase.27. On which factor does the rate constant of the

reaction depends upon?(A) Concentration (B) Catalyst(C) Temperature (D) Time

28. In the presence of a catalyst, the heat evolved orabsorbed during the reaction ___________.(A) increases.(B) decreases.(C) remains unchanged.(D) may increase or decrease.

29. Which of the following statement is TRUEabout a catalyst?(A) It alters the enthalpy change of the reaction.(B) It is used for altering the velocity of the

reaction.(C) It initiates a reaction.(D) It is formed during the reaction mechanism.

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273


COMPETITIVE CORNER

1. (B) 2. (C) 3. (C) 4. (B)5. (C) 6. (B) 7. (A) 8. (D)9. (C) 10. (C) 11. (A) 12. (B)13. (B) 14. (A) 15. (B) 16. (D)17. (D) 18. (C) 19. (D) 20. (C)21. (C) 22. (A) 23. (A) 24. (A)25. (C) 26. (D) 27. (C) 28. (C)29. (A)

16. Refer Q.74. 17. concentration is reduced to 25%. It means it

takes two half lives to decrease concentration ofreactant from 0.8 M to 0.2 M. Hence, half life is12/2 = 6 hours.

1. If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of thereaction. is given by _______. [NEET (UG) 2019](A) t = 6.909/k (B) t = 4.606/k (C) t = 2.303/k (D) t = 0.693/k

Hint: For 1st order reaction, ln

0

t

AA

= kt

99% of the reaction is complete, so if [A]0 = 100, then [A]t = 100 – 99 = 1 ln100 = kt

t = 1k 2.303 log102

t = 1k 2.303 2 log10

t =4.606

k2. For the chemical reaction, N2(g) + 3H2(g) 2NH3(g), the CORRECT option is _______.

[NEET (UG) Odisha 2019]

(A) 32 d NHd N2

dt dt (B) z 32 d NHd N 1=

dt 2 dt

(C) 32 d NHd H3 2

dt dt (D) 32 d NHd H1 1

3 dt 2 dt

Hint: N2(g) + 3H2(g) 2NH3(g)

Rate = 32 2 d NHd N d H1 1dt 3 dt 2 dt

32 d NHd N 1=dt 2 dt

3. The rate of reaction A Products, at the initial concentration of 3.24 × 10-2 M is nine times its rate atanother initial concentrations of 1.2 × 10-3 M. The order of the reaction is ______. [MHT CET 2019](A) 1

2(B) 3

4(C) 3

2(D) 2

3

Hint: 9 = 2

3

3.24 101.2 10

or 9 = (27)n

(3)2 = (3)3n

n = 23

4. For a first order reaction, when a graph is plotted between 010

t

[A]log[A]

on y-axis and time (t) on x-axis, slope

of the graph is equal to _______. [MHT CET 2019]

(A) – k (B) k (C) –k

2.303 (D) k

2.303

ANSWERS TO MULTIPLE CHOICE QUESTIONS HINTS TO MULTIPLE CHOICE QUESTIONS

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274


5. The rate constant for a first order reaction is 4.606 × 10–3 s–1. The time required to reduce 2.0 g of thereactant to 0.2 g is _______. [NEET (UG) P-I 2020](A) 200 s (B) 500 s (C) 1000 s (D) 100 s

Hint: t = 2.303k

log100

t

[A][A]

= 3

2.3034.606 10-´

log 20.2

= 500 s 6. The half-life for a zero order reaction having 0.02 M initial concentration of reactant is

100 s. The rate constant (in mol L–1 s–1) for the reaction is _______. [NEET (UG) P-II 2020](A) 1.0 × 10–2 (B) 1.0 × 10–4 (C) 2.0 × 10–4 (D) 2.0 × 10–3

Hint: For zero order reaction,

t1/2 = 0[A]2k

⸫ k = 0.022 100

= 1.0 × 10–4 mol L–1 s–1

7. For the reaction 4NH3 + 5O2 → 4NO + 6H2O if the rate of disappearance of NH3 is 3.6 × 10–3 M/s. What is therate of formation of water? [MHT CET 2020] (A) 4.0 × 104 M/s (B) 5.4 × 10–3 M/s (C) 6.0 × 10–4 M/s (D) 3.6 × 10–3 M/s

Hint: Given reaction: 4NH3 + 5O2 → 4NO + 6H2O

Rate of disappearance of NH3 = 3d NHdt

= 3.6 × 10–3 M/s.

Rate = 14

3d NHdt

= – 15

2d[O ]dt

= 14

d[NO]dt

= 16

2d H Odt

Rate of formation of H2O = 16

2d H Odt

= 14

3d NHdt

2d H Odt

= 64

× 3.6 × 10–3 M/s = 5.4 × 10–3 M/s

Time: 1 Hour 30 Min TOPIC TEST Total Marks: 25

Q.1. Select and write the correct answer: [04] i. In the reaction A + 3B 2C, the rate of formation of C is _______.

(A) the same as rate of consumption of A (B) the same as the rate of consumption of B(C) twice the rate of consumption of A (D) 3/2 times the rate of consumption of B

ii. If the rate constant of a first order reaction is 0.5 10–4 s–1, the order of the reaction is _______.(A) zero (B) first(C) second (D) Insufficient data

iii. The elementary reaction2O2(g) O3(g) + O(g) is _______.(A) unimolecular and second order (B) bimolecular and first order(C) bimolecular and second order (D) unimolecular and first order

iv. Rate law for the reaction A + 2B C is found to beRate = k [A]2 [B]Concentration of reactant B is doubled, keeping concentration of ‘A’ constant, the value of rateconstant will be _______.(A) the same (B) doubled(C) quadrupled (D) halved

SECTION A

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275


Q.2. Answer the following: [03] i. Express the rate of reaction in terms of change in concentration of each constituent for a reaction,

A + B 4C.

ii. State rate law. iii. What are the units of first order reactions?

Q.3. Differentiate between order and molecularity of a reaction. Q.4. Derive the expression for half life of first order reaction. Q.5. A first order reaction takes 40 minutes for 30% decomposition. Calculate its half life. Q.6 State the key features of the order of the reaction. Q.7. What is a catalyst? Give an example. Q.8. Write a short note on reaction intermediate.

Q.9. What is pseudo-first order reaction? Explain with the help of an example. Q.10. Explain graphically the effect of catalyst on the rate of reaction. Q.11. The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at 25 C is

3.7 × 10–5 s–1. What is the rate constant at 30C? (R = 8.314 J/K mol)

Q.12. i. The reaction, CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) is first order in CHCl3 and 1/2 order in Cl2. Write the rate law and overall order of reaction.

ii. The concentration change affects the rate of the reaction. Explain with respect to rate law.

Q.13. How will you determine activation energy:i. graphically using Arrhenius equationii. from rate constants at two different temperatures?

Download the answers of the Topic Test by scanning the given Q.R. Code.

SECTION B

Attempt any Four: [08]

SECTION C

Attempt any Two: [06]

SECTION D

Attempt any One: [04]

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