Chemical Kinetics

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INTRODUCTION

IMPORTANT TERMSAND DEFINITIONS

INSTANTANEOUS RATE OFTHE REACTION

FACTORSAFFECTINGTHE RATE OFREACTION

SPECIFIC REACTION RATE

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DISTINCTION BETWEENRATE OFREACTIONAND RATE CONSTANT

RATE LAW

ORDER OF REACTION

MOLECULARITYOFAREACTION

ELEMENTARYREACTIONS

COMPLEX REACTIONS

EQUILIBRIUMAPPROACH

STEADYSTATEAPPROXIMATION

TRUEANDAPPARENT RATE CONSTANT

PSEUDO UNIMOLECULAR REACTION

TYPES OFREACTIONS BASED ON ORDER OF REACTION

MISCELLENEOUSREACTIONS

DETERMINATION OFORDER OFREACTION

TEMPERATURE EFFECT

TEMPERATURE DEPENDENCE OFRATE CONSTANTS

MECHANISM BASED ON INTERMEDIATEFORMATION

ACTIVATION ENERGYANDTHE RATE OFREACTION

ACTIVATION ENERGIES OFAREVERSIBLE REACTION

ACTIVATION ENERGIESAND ENERGYCHANGE DURINGREACTION

MAXWELL'S- BOLTZMANN DISTRIBUTION CURVE

EFFECT OFCATALYST

COLLISIONTHEORYOFCHEMICALREACTIONS

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THRESHOLD ENERGYAND ORIENTATION OFCOLLIDING MOLECULES

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CONTENTS

S.NO. TOPIC PAGE NO

Chemical Kinetics

1. INTRODUCTION :Chemical Kinetics means study of rate of chemical reactions. Rate of reaction depends on amount of

change in concentration of reactants per unit time. Rate of reaction is influenced by various factors like

nature ofsubstance, physical state ofsubstance, temperature, concentration, presence of catalyst, etc.

Classification of reactions : [In terms of rates] :

(i) Fast reactions too fast e.g. Detonation of explosives, acid-base neutraliztion, precipitation ofAgCl by NaCl andAgNO .3

(ii) Moderate reaction Neither too fast nor too slow e.g. combination of H and Cl in presence of2

light, hydroysis ofethyl acetate catalyzed by acid, decomposition of azomethane.2

(iii) VerySlow reaction There are certain reactions which are too slow e.g. rusting ofiron, weatheringof rocks.

IMPORTANT TERMS AND DEFINITIONS :2.1 RATE OFREACTION

It is defined as the change in concentration of reactant (or product) in a particular time interval. Unit

of rate of reaction is mol L–1 s–1.

Change in the concentration of reactants or (Products)

2.

Rate of reaction =Time

2.2AVERAGE RATE OFREACTIONChange in the concentration of reactants or products measured over bigger interval of time is called average reaction rate. If c is the change in the concentration of reactants and product int time, then

Change in the concentration of reactan ts or (Products)Average Rate =

Time

C2 – C1

C= = ± t – tt 2 1

Unit of concentration gram mole / LitreUnit of average velocity = =Unit of time Second

= gram mole litre–1 second–1

n1A+ n2B m1C + m2DNote: For the reaction

1 [A] [B] [C] [D]1 1 1Rate of reaction = – = – = + = +

t t t tn1 n 2 m1 m2

2

CHEMICALKINETICS

Chemical Kinetics

Note: For the reaction N2 + 3H2 2NH3

[NH3 ](i) Rate of formation of ammonia = +

t

[N2 ](ii) Rate of disappearance of nitrogen = –

t

[H 2 ](iii) Rate of disappearance of hydrogen = –

t

1 [NH3 ] [N2 ] [H 2 ]1Rate = + = – = –

2 t t 3 t

[N2 ]=

1 [NH3 ]Thus, Rate = –

t 2 t

or rate of formation of ammonia =Twice the rate ofdisappearance of nitrogen

– [H2 ] [NH3 ] 2i.e. = tt 3

Ex.1 For the reaction R P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes.

Calculate the average rate of reaction using units of time both in minutes and seconds.

[R ]Average rate = –

[R] – [R]= – 2 1Sol.

t t 2 – t1

= –0.02M – 0.030M – 0.01M

= –25minutes 25minutes

= 4 × 10–4 mol litre–1 minutes–1

– 0.01M= = 6.66 × 10–6 mol litre–1 second–1.

2560s

Ex.2 In a reaction, 2A Products, the concentration ofAdecreases form0.5 mol L–1 to 0.04 mol L–1 in 10

minutes. Calculate the rate during this interval.

Average rate = –1 [A] = –

1 [A]2 – [A]1Sol.2 t2 – t12 t

0.4 M – 0.5M= –1

× = –1 – 0.1M

×10minutes2 2 10 minutes

= 5 × 10–3 mole minutes–1.

3

Chemical Kinetics

3. INSTANTANEOUS RATE OF THE REACTIONIt is the rate of reaction when the average rate is taken over avery short interval of time or rate of reaction at a particular timeis knownas instantaneous rate.

:

Limc dc

Instantaneous rate = ± =dtt0 t

Note: For the reaction n1A + n2B m1C + m2D

1 d[A] 1 d[B] 1 d[C] 1 d[D]Instantaneous Rate of reaction = – = – = + = +

n1 dT n 2 dT m1 dT m 2 dT

Ex.3Sol.

For the rate of reaction : N2 + 3H2 2NH3

in terms of the concentrations of N2, H2 and NH3 can be expressedas d[N 2 ]

= –1

= –d[H2 ] 1 d[NH3 ]

Rate = +2dt 3 dt dt

Ex.4 What should be (a) the rate of disappearance of B and (b) the rate of formation of C, if the rate of

disappearance ofAfor the reactionA+ B 2C is 10–2 mole/litre/second at a particular temperature ?

(a) Rate of disappearance ofA= Rate of disappearance of B = 10–2 mole/litre/second

1Rate of disappearance ofA= × Rate of formation of C

Sol.

2(b) Rate of formation of C = 2 × Rate of disappearance ofA= 2 × 10–2 mole/litre/second

4. FACTORS AFFECTING THE RATE OF REACTION :(i) Concentration : Law of mass action enunciates that greater is the conc. of the reactants, the more

rapidly the reaction proceeds.

(ii) Pressure (Gaseous reaction) : On increasing the pressure, volume decreases and conc. increaseshence the rate of reaction in creases.

(iii) Temperature : It is generally observed that rise in temperature increases the reaction rate. It hasbeen found that rate is either doubled or tripled for every 10º rise in temperature. Temp. coefficientof reaction rate

kT 10 2 or 3

k T

WhereT + 10 and kT are rate constants at two temp. differing by 10º.

(iv) Nature of the reactants : The rate depends upon specific bonds involved and hence on the natureof reactants.

(v) Surface area ofthe reactants : In heterogeneous reactions, more powered is the formof reactants,more is the velocity. [as more active centres are provided]

(vi) Catalyst : Affects the rate immensely. Presence of positive catalyst increases the rate of reaction bydecreasing the activation energy and presence ofnegative catalyst decreases the rate of reaction byincreasing the activation enrgy.Acatalyst mainlyaffects the activation energy of reaction and hence,the rate constant and rate of reaction changes.

(vii) Intensity of Radiation : The rate of photochemical reactions normally increases with increase inintensity of radiation.

(viii) pH of the medium

4

Chemical Kinetics

5. SPECIFIC REACTION RATE :Applying law of mass action to the reaction :

n1A+ n2B m1C + m2D

Rate [A]n1 [B]n2 or r = k [A]n1 [B]n2

This equation is known as rate law. Where k is the proportionality constant and is called Rate constant

or Rate coefficient (Specific reaction Rate)

On putting [A] = [B] = 1, we have : r = k

Hence, specific reaction rate is the rate of the reaction when the concentration of each reactant is taken

as unit.

Unit of Specific Reaction RateNote:

conc.r = k [A]n1 [B]n 2 = k [conc.]n1 +n 2

time

× [time]–1k = [conc.][1( n1n 2 )]

[1-(n + n )]mole 1 2

litre –1

or k = .[second]

where n1 + n2 = n which is order of reaction

mole 1–n

–1 K = second litre

k = mol.lit–1.sec–1

k = sec–1

k = mol–1.lit1.sec–1

when n = 0

n = 1

n = 2

Ex.5 Identify the reaction order from each of the following rate constant.

(i) k = 2.3 × 10–5 L mol–1 s–1

(ii) k = 3 × 10–4 s–1

Order for k = 2.3 × 10–5 L mol–1 s–1 is 2 and order for k = 3 × 10–4 s–1 is 1 from units.Sol.

6. DISTINCTION BETWEEN RATE OF REACTION AND RATE CONSTANT:Rate of a reaction :

(i) The rate of reaction at any instant of time depends upon the molar concentrations of the reactants at

that time.

(ii) Its units are always mole litre–1 time–1.

Rate constant :

(i) The rate constant is constant for a particular reaction at a particular temperature and does not

depend upon the concentration of the reactants.

(ii) Its unit depends upon the order of reaction.

5

Chemical Kinetics

7. RATE LAW :(i) It may also not depend upon the concentration of each reactant or product of the reaction.

Suppose,mA+ nB Product

(ii) Rate ofa chemical reaction is directly proportional to the concentration of reactants.(iii) The rate lawrepresents the experimentallyobserved rate of reaction which dependsupon the slowest

step of the reaction.

(iv) Rate law cannot be deduce from the equation for a given reaction. It can be find by experiments

only.

(v) The rate law may not bear a simple relationship for the stoichiometric equation.

(vi) It may not depend upon the concentration of species, which do not appear in the equation for the

over all reaction.

8. ORDER OF REACTION :The order ofa reaction may be defined as the sumof the powers to which conc. terms must be raised in

an experimentally determined differential rate equation :

For the reaction : mA + nB product Experimental rate equation : r = k [A]p [B]q

order with respect toA= p

order withe respect to B = q

Total order = p + q

Note: (i) Order maybe zero, fractional, integer or negative.(ii) p and q may be equal to mAand mB.

Ex.6 Reaction

H2 + Cl2 2HCl

H2 + Br2 2HBr

H2 + I2 2HI

Exp. rate equ.r = k [H2] [Cl2]

order0

3/2

2

0 0

1/2r = k [H2] [Br2]

r = k [H2] [I2]

Ex.7 Reaction : CO(g) + Cl2(g) COCl2(g)

r = k [CO]2 [Cl ]1/2 order = 2.52

COCl2(g) CO(g) + Cl2(g)Reaction :3/2

r = k [COCl2] order = 1.5

For a reaction,A+ B Product; the rate law is given by, r = k[A]1/2 [B]2. What is the order of thereaction ?

Ex.8

1 1Sol.

Ex.9

Order of the reaction = + 2 = 22

or 2.5.2

The conversion of the molecules X toYfollows second order kinetics. If the concentration X is increasedto three times, how will its affect the rate of formation ofY?For the reaction, X Y, as it follows second order kinetics, the rate law equation will be

Rate = k [X]2

If concentration ofX is increased to three times, then

Rate = k [3X]2 or Rate = 9k [X]2

Thus, the rate of reaction will become 9 times. Hence, the rate of formation ofYwill increase 9 times.

Sol.

6

R [A]m m[B]n

Chemical Kinetics

9. MOLECULARITY OFA REACTION :‘‘Molecularity is defined as the number ofmolecules, atoms, or radicals that must collide simultaneously

in order for the reaction to take place.’’It is always a whole number and cannot be negative.

In the elementary processes :

Participating species Molecularity

One species participates..... unimolecular, 1

Two species participates..... bimolecular, 2

Three species participates....trimolecular, 3

Molecularity cannot be zero, fractional or more than three.Note:

Ex.10 N2O4 2NO2 ......

H2 + I2 2HI ......

unimolecular

bimolecular

(molecularity = 1)

(molecularity = 2)

(molecularity = 3)2FeCl3 + SnCl2 2FeCl2 + SnCl4 ...... trimolecular

Note : If the reaction takes place in two or more steps then the overall molecularity of the reaction is

monitored by the slow or rate determining step.

TYPE OFREACTIONS

On the basis of number of steps involved in the reaction, they are classified as elementary or complex

reactions.

10. ELEMENTARY REACTIONS :

These are single step reactions. For such reactions, order and molecularity are always same. In another

words, rate law and the law of mass action will have same expression.

11. COMPLEX REACTIONS :

These are multi-step reactions. For such reactions, order and molecularity may or may not be same.

In suchreactions, some intermediates are formed. Intermediates are the speciesdifferent than reactants

as well as products.

Each step of such reaction is elementary reaction.

The overall rate of reaction will be equal to the rate of slowest step. This is why, the slowest step is

called the rate determining step (RDS) of reaction.Areaction can have more than one RDS,

The overallmolecularity of reaction is the molecularity of the RDS. However, it has no significance.

The order ofreaction may or maynot be the overallorder ofreaction. If depends on the concentration

terms involved in the rate law expression of RDS. If it is not overall order, it may be determined by

equilibrium approach or be steady state approximation.

12. EQUILIBRIUM APPROACH :The concentration ofintermediate can be determined from equilibrium constant of the reaction involved.

For example, let the mechanism of reactionA+ 2 B C + D is

k1

Step I: A + B Ik–1

k2

Step II: I + B C + D

Now, the overall rate of reaction, r = rate of step II = k2[I][B] ...(i)

7

Chemical Kinetics

It cannot be the correct rate law for the overall reaction because the overall rate of reaction should be interms ofconcentrationsofAand B(reactants). The concentration ofI canberelatedwith theconcentrations ofAand B with the help of first equilibrium.

k1 [I ]For step I, equilibrium constant, Keq = =

k [A][B]–1

k1 [I] = [A][B]k–1

k1 2Putting this value in equation (i), r = k2 k

[A][B][B] = k[A][B]

,–1

k1where, k =

k –1

Hence, the overall rate of reaction is 1 + 2 = 3

13. STEADY STATE APPROXIMATION :In this method, we assume that the intermediates formed are so reactive that after some time frominitiation of reaction (called induction period), the net rate of their formation becomes zero. Theyreactwith the same rate of their formation. For example, let the mechanism of reaction A+ 2 B C + Dis

k1

Step I: A + B Ik–1

k2 Step II: I + B C + D

d[C]The rate of reaction may be given as r = + = k2[I][B] ...(ii)

dt

d[I ]Now, from steady state approximation on the intermediate, I + = 0

dtor, k1[A][B] – k–1[I] – k2[I][B] = 0

k [A][B]1or, [I] =k –1 k 2[B]

k k [A][B]2k1[A][B]1 2Putting this value in equation (ii), r = k2 k

[B] =[B] k k k [B]–1 2 –1 2

Steady state approximation gives better result when the intermediate is less stable while equilibriumapproach gives better result when the intermediate is more stable. More stable intermediate reactsvery less and hence, the concentration of intermediate at any time remains nearly equal to its equilibriumconcentration.

When conditions of equilibrium approach are applied on the result obtained from steady stateapproximation, the same rate law expression will come. For the above reaction, step I is faster thanstep II.As step I is at equilibrium,

r1 = r–1 >> r2

or, k–1[I] >> k2[I][B]

k–1 >> k2 [B]k–1 + k2 [B] k–1

or,

k k [A][B]2

[A][B]2,2 1Now, the rate expression obtained from steady state is r =k –1 k 2[B]

which is exactly same expression obtained from equilibrium approach.

8

Chemical Kinetics

14. TRUE AND APPARENT RATE CONSTANT :For acidic hydrolysis of ester:

H

RCCOR' + H2O RCOOH + R'OH+the rate of reaction, r = k[RCOOR'][H2O][H ], where k is the true rate constant of the reaction,

Now, H+ is the catalyst of reaction and hence, throughout its concentration will remain constant. H2O is

solvent for the reaction and taken in large excess and hence, its concentration will also remain constant.Hence, k[H2O][H ] = constant, called apparent rate constant of the reaction.

15. PSEUDO UNIMOLECULAR REACTION :

+

HConsider the reaction : CH3COOC2H5 + H2O CH3COOH + C2H5OH

Since water is present in large excess, its concentration hardlychanges during the course of the reaction.And as such rate depends only on the concentration of ester. The order is one but the molecular is two. Such reactions are called pseudo unimolecular reaction.

The reaction 2NO + Br2 2NOBr, obeys the following mechanism :Ex.11Fast

Step I: NO + Br2 NOBr2SlowStep II: NOBr2 + NO 2NOBr

Suggest the rate expression.Step II is the rate determining step of the reaction and hence, r =

k[NOBr2] [NO]

However, NOBr2 is an intermediate and thus its concentration should be determined from the equilibriumof step I.

[NOBr2 ]

Sol.

For step I, equilibrium constant, keq =[NO][Br ]2

[NOBr2] = keq [NO][Br2] ... (ii)2

Thus, by equations (i) and (ii), r = k keq[NO] [Br2]r = k' [NO]2[Br ]. where k' = kkor,

2 eq

Ex.12 The following mechanisms are proposed for the reactionCO + NO2 CO2 + NO at low temperature:

(a) 2NO2 N2O4 (fast)(slow)(slow)(fast)

N2O4 + 2CO 2CO2 + 3NO(b) 2NO2 NO3 + NO

NO3 + CO NO2 + CO2

d[CO 2 ] 2which of the above mechanism are consistent with the observed rate law : +

For mechanism (a), r = rate of step II

= k[NO2] ?dt

Sol.2

= k[N2O4][CO] ...(i)[ N2O4 ]

Now, from step I : keq = ]2[NO22

or, [N2O4] = keq[NO2]2 2

Hence from (i), r = k.keq [NO2] [CO] , but it is not the given rate lawHence, the mechanism is not consistent with the rate law.

2For mechanism (b), r = rate of step, I = k[NO2] , which is the given rate law.

Ex.13 For a hypothetical reactionA+ B products, the rate law is, r = k [B] [A]º, the order and molecularity of reaction is :Order = 1 + 0 = 1, molecularity = 1 + 1 = 2Sol.

9

Chemical Kinetics

16. TYPES OF REACTIONS BASED ON ORDER OF REACTION :(I) Zero Order Reactions :

Zero order reaction means that the rate of the reaction isproportional to zero power of theconcentration of reactants. Consider the reaction,

R Pd[R]

Rate = – = k [R]0

dtAs anyquantity raised to power zero is unity

d[R]Rate = – = k × 1

dtd[R] = – k dt

Integrating both sides[R] = – k t + I

where, I is the constant of integration....(1)

At t = 0, the concentration of the reactant R = [R] , where [R] is initial concentration of the reactant.0 0

Substituting inequation (1)[R] = –k × 0 + I

0

[R] = I0

Substituting the value of I in the equation (1)[R] = –kt + [R] ...(2)

0

O Time (t)

Comparing withequation of a straight line, y= mx + c, if we plot [R] against t, we get a straight line withslope = –k and intercept equal to [R] .

0

Further simplifying equation (2), we get the rate constant, k as[R ]0 – [R]

k = ...(3)t

Zero order reactions are relatively uncommon but they occur under special conditions. Some enzymecatalysed reactions and reactions with occure on metal surfaces are a few examples of zero orderreactions. The decomposition of gaseous ammonia on a hot platinumsurface is a zero order reaction athigh pressure.

1130K2NH (g) N (g) + 3H (g)

3 Pt catalyst 2 2

Rate = k [NH ]0 = k3

In this reaction, platinum metal acts as a catalyst.At high pressure, the metal surface gets saturated withgas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia onthe surface of the catalyst making rate of the reaction independent of its concentration. The thermaldecomposition of HI on gold surface is another example of zero order reaction.

10

Con

cent

rati

ono

fR

K = - slope

Chemical Kinetics

Note: (i) Reaction betweenAcetone and Bromine(ii)Dissociation of HI on gold surface

(A) Unit of Rate Constant –[R]0 – [R]

k =t

Unit of rate of reaction = Unit of rate constant.

(B) Half-life period (t1/2) - The time in which half reaction is

completed R 0At t = t1/2 ; R =

2

t1 / 2 R 0or1/ 2

The half life period is directly proportional to the initial concentration of the reactants.

Ex.14 The reaction 2A + B + C D + 2E; is found to be first order inA; second order in B and zero order in C.(i)(ii)

(i)

Give the rate law for the above reaction in the form of a differential equation.What is the effect on the rate of increasing the concentration of A, B and C two times?

The rate law according to given information maybe given as,Sol.

dx K[A]1[B]2[C]0

dt

(ii) When concentration ofA, B and C are doubled then rate will be

dx K[2A][2B]2[C]0

dt

8K[A][B]2[C]0

i.e., rate becomes 8 fold, the original rate.

(II) First Order Reactions :

In this class of reactions, the rate of the reaction is proportional to the first power of the concentration of

the reactant R. For example,

R P

d[R] d[R]Rate = – = k [R] or – = – kdt

dt [R]Integrating thisequation, we get ln [R] = – kt + I ...(4)

Again, I is the constant of integration and its value can be determined easily.When t = 0, R = [R] , where [R] is the initial concentration of the reactant.

0 0

Therefore, equation (4) can be written asln [R] = –k × 0 + I

0

ln [R] = I0

Substituting the value of I in equation (4)ln [R] = – kt + ln[R] ...(5)

0

Rearranging this equation

[R]ln = – kt

[R]0

[R]01or k = ln ...(6)

t [R]

11

t R0

2K

K mol lit-1 sec-1

Chemical Kinetics

At time t fromequation (4)1

ln[R] = –kt + ln[R] ...(7)1 1 0

At time t2

ln[R] = –kt + ln[R] ...(8)2 2 0

where, [R] and [R] are the concentrations of the reactants at time t and t respectively.1 2 1 2

Subtracting (8) from (7)ln[R] – ln[R] = – kt – (–kt )

1

[R]1ln

2 1 2

= k (t – t )[R]2

2 1

[R ]11k = ln ...(9)

(t 2 – t1 ) [R ]2

Equation (5) can also be written as

[R ]1ln = –kt

[R ]2

Taking antilog of both sides[R] = [R] e–kt

0

Comparing equation (5) with y= mx + c, if we plot ln [R] against t (Fig. 4.4) we get a straight line withslope = –k and intercept equal to ln [R]

0

The first order rate equation (6) can also be written in the form[R]02.303

k = log ...(10)[R]t

[R]0 ktlog =

[R] 2.303If we plot a graph between log [R] / [R] vs t,

0

the slope = k/2.303Hydrogenation of ethene is an example of first order reaction.

C H (g) + H (g) C H (g)2 4

Rate = k [C H ]2 2 6

2 4

All natural and artificial radioactive decay of unstable nuclei take place byfirst order kinetics.

O t

Fig. :Aplot between ln[R] and tfor a first order reaction

Fig. : Plot of log[R] /[R] vs time for a first0

order reaction

226Ra 4He 222RnNote: (i) 88 2 86

Rate = k [Ra](ii)(A)

Decomposition of N O and N O are some more examples of first order reactions.2 5

Unit of Rate Constant :2

2.303 [R 0 ]logK =

t [R]K = sec–1

12

ln[R

]

ln[R0]

k = – slope

Chemical Kinetics

R 0(B) Half-life period (t1/2) : At t = t1/2 , R = ,

2

=2.303 R 0 2.303 0.693

log 2K = log =t1 / 2 R 0 / 2 t1 / 2 t1 / 2

0.693t =

K1/2

Some Examples of First Order reaction andTheir Rate constant1. For Gas Phase ReactionLet us consider a typical first order gas phase reaction

A(g) B(g) + C(g)Let p be the initial pressure ofAand p the total pressure at time 't'. Integrated rate equation for such ai

reaction can be derived ast

Total pressure p = p + p + p (pressure units)t A B C

p , p and p are the partial pressures ofA, B and C, respectively.A B C

If x atm be the decrease in pressure ofAat time t and one mole each of B and C is being formed, theincrease in pressure of B and C will also be x atm each.

A(g)p atm

B(g)0 atmxatm

+ C(g)0 atmxatm

At t = 0At time t

i

(p –x) atmi

where, p is the initial pressure at time t = 0.i

p = (p – x) + x + x = p + xt i

x = (p – p )i

t i

where, p = p – x = p – (p – p ) = 2p – pA i

2.303

i t i i t

p log i k =

...(11)

t pA pi2.303

= log (2p – p )i tt

2. Pseudo Unimolecular Reaction : Inversion of cane sugar

H C H O + H O C H O + C H O12 22 11 2

Sucrose6 12 6 6 12 6

Glucose FructoseThe progressof the reaction canbe studied using apolarimeter. Cane sugar and glucose are dextrorotatorywhile fructose is laevorotataryIf is polarimetric reading at zero time.0

is polarimetric reading after time t.t

and is polarimetric reading after infinite time

2.303log

0 – k =t t –

3. (a) Decomposition of H O2 2

1H O Pt H O + O2 2 2 2 2

The progress of the reaction can be studied either bymeasuring the volume of oxygen gas after differentintervals of times or by titrating a definite amount of reaction mixture with standard KMnO at different

4

intervals of time.If V and V represent the volumes of KMnO used at the start of the reactions and at any time t,

0

respectivelyt 4

R V , R V0 0 t

2.303 V0Therefore, k = log

Vt t

13

Chemical Kinetics

(b) Decomposition of ammonium nitrite : NH NO N + 2H O4 2 2 2

IfV is the volume ofN evolved at anytime t and V is the volume ofN evolved when the decompositiont

is complete, then,2 2

V2.303logk =

V – Vtt4. Hydrolysis of ethyl acetate (ester)

H CH COOC H + H O CH COOH + C H OH3 2 5 2 3 2 5

Kinetics of this reaction is studied by titrating a definite volume of the reaction mixture with standardalkali solution.If V , V and V are volumes of standard alkali needed to neutralise a definite amount of x will be0 t

proportional to V – V and a will be proportional to V – V .t 0 0

2.303log

V – V0Hence, k = .t V – V t

5. Oxide layer formation

1 maxk = lnt max –

where

= Thickness of oxide layer after timesmax

= Thickness of oxide layer at time 't'.

6. Bacterial Growth :

1ln

a xk =

t aSecond order reactions: When the rate of a reaction is determined by variation of two concentration terms, thereaction is said to be of second order. For a general reaction,

A + B products, the rate law may be,

dx = k [A]2 [B]0, or ... (x)dt

dx = k [A]0 [B]2, or ... (xi)dt

dx= k [A] [B] ... (xii)

dt

Thus, the rate of a second order reaction varies directly as the square of the concentration of reactant.

Unit of the second order rate constant :From equation (x) we get

dx Mol / litre –1 –1k = = = lit. mol time .dt[A]2Time interval Mol.2 /lit.2

The unit of second order rate constant involves concentration as well as time.

Examples of Second Order Reactions :373 K

1.

2.

3.

4.

5.

Conversion of ozone into oxygen at 373 K : 2O3 3O2

Thermal decomposition of nitrous oxide : 2N2O — 2N2 + O2

Thermal decomposition of chlorine monoxide at 200º C (473 K) : 2Cl2O 2Cl2 + O2200ºC

Alkaline hydrolysis of an ester (saponification) : CH3 COOC2H5 + NaOH — CH3COONa + C2H5OHKCN alc.

Benzoin condensation, C6H5CHO + OHC.C6H5 C6H5CH(OH)CO.C6H5reflux

14

Chemical Kinetics

6. Isomerisation of ammonium cyanate into urea.

O

heat NH2 – C – NH2NH4CNO

Second order rate equation: All the second order reactions obey the following kinetic equation,

1 x

t a(a x)k = .... (xiii)

where k = second order rate constanta = initial concentration of reactant (s)x = concentration of reactant converted into products after time ‘t’t = time elapsed

Characteristics of a Second Order Reaction :1

(i) When a graph is plotted between ‘t’ and a straight line is obtained. On rearranging, equation(a x)1 1 1(xiii).t = The slope of the line . From this ‘k’ an be evaluated.k(a x) ka k

l/(a-x)

(ii) The value of second order rate constant depends upon the unit in which concentration of the reactant(s)x

is expressed, because the value of will be changed when the unit of concentration is changed.a(a x)

(iii) The half-life period (t1/2) of a second order reaction is inversely proportional to the initial concentration

of reactant.

From expression (xiii)

1 x

t a(a x)k = ... (xiii)

when, t = t1/2, x = a/2

1 a / 2so, k = t1/ 2

1

a(a a / 2)

or, t1/2 = ... (xv)ka

Thus, it is evident from equation (xv) that half-change time for a second order reaction is inversely proportional tothe initial concentration of reactant.

17. MISCELLENEOUS REACTIONS :(a) ParallelReactions

B

ln[A]0 [B]

k1A = (k + k )t[A]t [C] k 2

k2[A]0

1 2C

k1[A]0[A] = [A ] e–kt [B] = (1 – e–kt) [C] = (1 – e–kt)

k1 k 2 k1 k 20

15

Tim

e

Chemical Kinetics

(b) Consequtive Reaction

A k1 k2

B C

e – e k1[A]0 – k1t – k 2t[A] = [A ] e–k1t ; [B] = (k – k )t 0 t

2 1

k 2

k k 2 – k1

k

1[C] = [A ] – ([A] + [B] ) ; [B] = [A ] .t 0 t t max 0

2

conc.[C]

1 Kn 1[B]

[A]t =

max K1 – K 2 K 2

time(c) Reversible reaction

Consider the reaction

kr BA

kb

At time t = 0At time t = t

aa – x

a – x

bx

xAt time t = taq eq eq

x x keq eq fln x = (k + k )t ;– x a – x kf b eq eq b

Table of Formulae

16

[

Chemical Kinetics

Important Graphical Representation

Zero Order First Order Second Order Third Order

Ex.15 Time required to decompose SO Cl to half of its initial amount is 60 minutes. If the decomposition is a2 2

first order reaction, calculate the rate constant of the reaction.

For a first order reaction,Sol.

0.693 0.693k = = = 1.155 × 10–2 s–1.

t1 / 2 60 minutes

Ex.16 Afirst order reaction has a rate constant 1.15 × 10–3 s–1. How long will 5 g of this reactant take to reduce

to 3 g ?

Given : [A] = 5 g, [A] = 3 g, k = 1.15 × 10–3 s–1.Sol.0

Applying first order kinetics equation and substituting the values, we get

[A]02.303t = log

[A]t

2.303log

5g= = 2.00 × 10 (log 1.667) s3

1.1510 –3 3g

= 2.0 × 103 × 0.2219 s = 443.8 s

17

Chemical Kinetics

Ex.17 A first order reaction has a specific reaction rate of 10–2 s–1. How much time will it take 10 g of its

reactant to be reduced to 2.5 g ?

Rate constant, k = 10–2 s–1

Initial reactant conc., [A] = 10 g

Sol.

0

Final reactant conc., [A] = 2.5 gt

Time required, t = ?

For a first order reaction,

2.303log

[A]0 2.303 log

10 gt = = = 2.303 × 102 log 4 s = 230.3 × 0.6020 s = 138.6 s.

10 – 2 s –1k [A]t 2.5g

Ex.18 Areaction that is of first order with respect to reactant Ahas rate constant 6 min–1. If we start with

[A] = 5.0 mol L–1, when would [A] reach the value of 0.5 mol L–1 ?

Rate constant = 6 min–1

[A] = 5.0 mol L–1

Sol.

0

[A] = 0.5 mol L–1t

t = ?

For a first order reaction,

–12.303

log[A]0 2.303 5mol L

logt = =[A]t 6 min –1 0.5 mol L–1k

2.303= log 10 min = 0.39 min.

6

Ex.19 For a first order reactions, it takes 5 minutes for the initial concentration of 0.6 mol L–1 to become 0.4

mol L–1. How long in all will it take for the inital concentration to become 0.3 mol L–1 ?

For a first order reaction,

2.303log

[A]0

Sol.

k =k [A]t

We have, [A] = 0.6 mol L–1 [A] = 0.4 mol L–1 t = 5 min0 t

So,

2.303 0.6 2.303 2.303 0.1761k =

For,

× log = log 1.5 min =–1 min = 8.1 × 10– min–1 2 –1

5min 0.4 5 5[A] = 0.3 mol L–1

t

2.303 2.3032.303log

[A]0 0.6logt = = = log 2 min = 8.5 min.

[A]t 8.110–2 0.3 8.110–2k

Ex.20 The following data were obtained during the first order thermal decomposition of N O (g) at constant2 5

volume:

2N O (g) 2N O (g) + O (g)2 5 2 4

Time /s

0

100

2

S.No.

1.

2.

Total Pressure/(atm)

0.5

0.512

Calculate the rate constant.

18

Chemical Kinetics

Sol. Let the pressure of N O (g) decrease by 2x atm.As two moles of N O decompose to give two moles2 5 2 5

of N O (g) and one mole of O (g) the pressure of N O (g) increases by 2x atm and that of O (g)2 4

increases byx atm.2 2 4 2

2N O (g) 2N O (g) + O (g)2 5

0.5 atm

(0.5 – 2x) atm

2 4

0 atm

2x atm

2

0 atm

xatm

Start t = 0

At time t

pN O2 5p = + p + pN2O4 O2t

= (0.5 – 2x) + 2x + x = 0.5 + x

x = p – 0.5t

= 0.5 – 2 (p – 0.5) = 1.5 – 2pt

At t = 100 s; p = 0.512 atmt

t

pN O2 5= 1.5 – 2 × 0.512 = 0.476 atm

Using equation (11)

2.303 pi 2.303 0.5atmk = log

p =

100slog

0.476atmt A

2.303= × 0.0216 = 4.98 × 10 s .–4 –1

100s

Ex.21 For a reaction 2A Product, it is found that the rate of reaction becomes 2.25 when the

concentration ofAis increased 1.5 times, calculate the order of reaction.

r = k [A]nSol. Rate, .... (1)

When concentration is increased 1.5 times, the rate of reaction become 2.25.

2.25 r = k [1.5A]n .... (2)

k[1.5A]n2.25rDivide (2) by (1) =

k[A]nr2.25 = (1.5)n

(1.5)2 = (1.5)n

n = 2

18. DETERMINATION OF ORDER OF REACTION :(I) Integration Method

In this method, value of Kis determined by putting values of initial concentration ofreactants andchange

in concentration with time in kinetic equation of first, second and third order reactions. The equation by

which constant value of K is obtained is called order of that reaction.

R02.303K1 = log (For first order reaction)

t R

x1K2 = (For second order reaction)

a(a - x)t

x(2a - x)1K3 = (For third order reaction)

a2 (a - x)22t

19

Chemical Kinetics

(II) van’t Hoff Differential Method

van’t Hoff gave the following relationship between velocityV of nth order reaction and concentration of

reactants, C.

log(V1 / V2 )n =

log(C1 / C2 )

C1 and C2 are two different concentrations, while V1 and V2 are their velocities.

Graphical Method

If a straight line is obtained on drawing a graph between log (a – x) and time ordt

, then it is first order

(III)

dx

reaction.

If a straight line is obtained on drawing a graph between (a – x)2 anddx

, then it is second orderdt

reaction.

Half-life Method(IV)

1t1/2

Relation between half-life period of a reaction and initial concentration is as follows :an 1

For first order reaction (Half life a)

For second order reaction (Half life 1/a) For third order reaction (Half life 1/a2)

Ostwald Isolation Method

This method is used to find out the order ofcomplex reactions. If nA, nB and nC molecules of substance

A, B and C, respectively, are present in a reaction, then nA+ nB + nC will be the order of reaction.

When B and C are in excess, the order of reaction will be nA.

WhenAand B are in excess, the order of reaction will be nC.

WhenAand C are in excess, the order of reaction will be nB.

TEMPERATURE EFFECT :The rate of reaction is dependent on temperature. This is expressed in terms of temperature coefficient

which is a ratio of two rate constants differing by a temperature of 100. Generally the temperature

selected are 298K and 308K. It is mathematically expressed as,

Temperatur e coefficien t rate constant at 308K

(V)

19.

rate constant at 298K

According to collision theory the reaction rate depends on collision frequencyand effective collisions.

For a molecule to have effective collision it should fulfill two conditions :

(i) Proper orientation (ii) Sufficient energy

20. TEMPERATURE DEPENDENCE OF RATE CONSTANTS :Arrhenius Equation the temperature dependence of the rate of a chemical reaction can be

accuratly explain byArrhenius

AeE a / RTeq. k = ...(i)

where k =Arrhenius factor, E =Activation energy, R = Gas Constant, T = Temperaturea

20

Chemical Kinetics

For most reactions, the rate constant increases as the temperature increases. These rate constants inmost of the cases vary with temperature according to theArrehenius equation :

The parametersAand E for a given reaction are collectively calledArrhenius parameters. E iscalled thea a

1activation energy. Value of E is determined from the graph for and ln k determined experimentally.

Ta

Value ofAin turn is calculated once E is known.a

In theArrhenius equation (i) the factor eEa / RT corresponds to the fraction of molecules that have

kinetic energygreater than E .a

Taking natural logarithm of both sides of equation (i) we get

Ealn k = lnA– ...(ii)RT

The plot of ln k vs 1/T gives a straight line according to the equation (ii) as shownin (Figure below).

slope = –Ea/R

ln k

0 1/T

A plot between ln k vs 1/T

Thus, it has been found fromArrhenius equation (i) that increasing the temperature or decreasing theactivation energy will result in an increase in the rate of the reaction and an exponential increase in the

rate constant.

EaIn the plot, slope = – and intercept = lnA. So we can calculate E andAusing thesevalues.

aR

At temperature T , equation (ii) is1

Ealn k = – + lnA ...(iii)RT11

At temperature T , equation (ii) is2

Ealn k = – + lnA ...(iv)2 RT2

(sinceAis constant for a given reaction)

k and k are the values of rate constants at temperatures T and T respectively.1 2 1 2

Subtracting equation (iii) from (iv), we obtain

Ea Ealn k – ln k = –2 1 RT RT1 2

Ea 1 1k 2ln = R T1 T2 k1

Ea 1 k 2log =

1 k1 2.303R T1 T2

21

Intercept = ln A

Chemical Kinetics

Ex.22

Sol.

What will be the effect of temperature on rate constant ?

Rate constant ofa reaction is nearlydoubled with rise in temperature by10°. The dependece of therate

constant on temperature is given byArrhenius equation, k =Ae– Ea / RT , whereAis called frequency

factor and E is the activation energy of the reaction.a

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K.

Calculate E .

Ex.23

a

Here we are given that

When T = 298 K, k = k (say)

Sol.

1 1

When T = 308 K, k = 2k2 2

E T – T log k 2 a 2 1 =

2.303 R T T k1 1 2

Substituting these values in the equation, we get

Ea 308K – 298Klog

2k= ×2.303 8.314 J K –1mol –

1

298K308Kk

E a 10or log 2 = ×

2.3038.314 298308

Ea 10or 0.3010 =

2.303 8.314 298308

0.3010 2.3038.314 298308or E =

a 10

or E = 52897.8 J mol–1 = 53.6 kJ mol–1.a

Ex.24 How the value of activation energy is calculated fromthe rate constants at two different temperatures. Ifthe value of activation energy is 50 kJ/mol show that by increasing the temperature from 300 K to 310

K rate constant becomes nearly double.

We know thatArrhenius equation can be written as

Ea 1

Sol.

log k = logA–2.303 R T

If k and k are the rate constants at two different temperature T and T thenArrhenius equation at both1 2 1 2

the temperatures can be written as

Ea 1log k = logA–

1 2.303 REa

T1

1log k = logA–

2 2.303 R T2

Substracting eq. (i) from (ii), we get

1 Ea 1–log k – log k = 2 1 2.303 R T1 T2

k 2 E a 1 1–or log = k1 2.303 R T1 T2

22

Chemical Kinetics

By knowing the values of k and k at temperature T and T value of E can be calculated.1 2 1 2 a

Given T = 300 K, T = 310 K, E = 50 kJ or 50,000 J1 2 a

Put these values in eq. (iii)

k 2 50000 1 1 log = –

k1 2.303 8.314 300 310

50000 310 – 300 = 310 300 2.3038.314

logk 2 = 0.2808ork1

=Antilog 0.2808 = 1.91 2k 2

k1

It is clear that by increasing the temperature from 300 K to 310 K rate constant becomes double.

Afirst order reaction is 50% complete in 30 minutes at 27°C and in 10 minutes at 47°C. Calculate the

rate constants at 27°C and 47°C, and the energy of activation of the reaction in kJ/mol.

Time for the completion of 50% reaction means t .

Ex.25

Sol.1/2

It means t of the reaction at 27°C is 30 minutes and at 47°C is 10 min.1/2

0.693We know that k =

t1 / 2

0.693or k = = 0.0231 min–1 (at 27°C)

300.693

and

We know that

klog 2

k = = 0.0693 min–1 (at 47°C)10

1 Ea 1–= 2.303 R T1 T2 k1

k = 0.0693 min–1, k = 0.0231 min–1,2 1

T = 27 + 273 = 30 K, T = 47 + 273 = 300 K,1 2

R = 8.314 JK–1 mol–1, E = ?a

log0.0693 E a 1 1

–= 0.0231 2.303 8.314 300 320

E a 320 – 300 or log 3 =

2.303 8.314 300 320

E a 20 0.4771 =

2.303 8.314 300 320

0.4771 2.303 8.314 300 320 or E = a 20

= 43848.49 J mol–1

E = 43.85 kJ mol–1a

23

Chemical Kinetics

21. MECHANISM BASED ON INTERMEDIATE FORMATION :For understanding it let us use the following simple reaction

H (g) + I (g) 2HI (g)2 2

H I H I H I++

H I H I H I

Formation of HI through the intermediate

For understanding the product formation in the above reaction we highlight the following points:

According toArrhenius, this reaction can take place only when a molecule of hydrogen and a molecule

of iodine collide to form an unstable intermediate (Figure). It exists for a very short time and then breaks

up to form two molecules of hydrogen iodide.

The energy required to form this intermediate, called activated complex (C), is known as activationenergy (E ).

a

Figure is obtained by plotting potential energy v/s reaction coordinate. The reaction coordinaterepresents the profile of energy change when reactants change into products.

Activated

2HI

Reaction coordinate

Diagram showing plot of potential energy vs reaction coordinate.

Some energy is released when the complex decomposes to form products. So, the final heat of thereaction depends upon the nature of reactants and products.All the molecules is the reacting species do not have the same kinetic energy.

Kinetic energy

Distribution curve showing energies among gaseous molecules

22. ACTIVATION ENERGYAND THE RATE OF REACTION :The molecules of the reacting substances must be promoted to the top of the energy barrier before theseare able to react. For this to happen, the molecules must absorb energy equal to the activation energy.Depending upon the magnitude of the activation energy, the following three cases are possible:(i) When the activation energy is small. If the activation energy is low, then larger number of the reactant

molecules will be able to cross over the top of the energy-barrier. As a result, the reaction will befaster. Thus, if the activation energy is low, then the reaction is fast.

(ii) When the activation energy is high, then only a few molecules would be able to cross the top of theenergy barrier. As a result, the reaction will be slow. Thus, if the activation energy of a reaction ishigh, then the reaction is slow.

24

Pot

enti

alen

ergy

Most probablekinetic energy

complex

C

Activation energy

A

H2 + I2 B

Chemical Kinetics

(iii) When the activation energy is zero, then each molecule will be able to cross the top of the energybarrier.As a result, the reaction will be instantaneous, and almost explosive. Thus, if the activationenergy of a reaction is zero, then the reaction is instantaneous, (very very fast).

23. ACTIVATION ENERGIES OF A REVERSIBLE REACTION :In a reversible reaction, the reactants react to form products, and the products react back to give thereactant molecules. Thus, in a reversible reaction, there are two reactions proceeding in the oppositedirections. Each reaction has its own characteristic activation energy. So, in a reversible reaction there aretwo activation energies: one for the forward reaction, and the other for the backward reaction. These arecommonly called as the energyof activation for the forward reaction (E ) and the energyofactivation

af

for the backward reaction (E ). Depending upon the relative energies of the reactants and products, thea,b

following two cases are possible.(i) When the energy of the products is lower than that of reactantWhen the energy of products is lower than that of the reactant, [Fig. (a)], thenActivation energy for the forward reaction < Energy of activation for the backward reactionor, E < E

af a,b

Therefore, the reaction in the forward direction is faster than that in the backward direction.

Er

ER

Ep

Reaction coordinate

(Exothermic reaction)

(a)

(ii) When the energy of products is higher than that of the reactants

When the energyof products is higher than that of the reactants [Fig (b)], then

Activation energy for the forward reaction >Activation energy for the backward reaction

or, E > Eaf a,b

Er

Ep

ER

Reaction coordinate

(Endothermic reaction)

(b)

Thus, the reaction in the forward direction is slower than the reaction in the backwarddirection.

It must be remarked here that the activation energy of a reaction in any direction determines the speed

(rate) of reaction in that direction. It does not say anything about the extent of reaction in that direction.

Extent of reaction in any direction is governed by the equilibrium constant of the reaction.

25

Ene

rgy

Ene

rgy

Ea,fEa,b

Product

Reactant

Ea,f

Ea,b

Reactant

Product

Chemical Kinetics

24. ACTIVATION ENERGIES AND ENERGY CHANGE DURING REACTION:

In a reaction, the energy change is given by,

Energy change in a reaction = Energy of products – Energy of reactants

or, E = E – E (i)P R

According to the concept of activation energy, the reactant molecules in the forward reaction, and the

products molecules in the backward reaction must pass over the top of the energy-barrier (E ). Then,T

one can write Eq.(i), in the following form

E = E – E + E – E = (E – E ) – (E – E ) (ii)

(iii)P R T T T R

E = E – ET P

or,a,f a,b

Under constant pressure conditions, E =H,

So,

(a) When

H = E – Ea,f a,b

E < Ea,f a,b

E = – ve

H = – veand,

Thus, when the activation energy for the forward reaction is less than that for the backward reaction,

energy is released during the course of reaction.

(b) When E > Ea,f a,b

E = + ve

H = + veand,

Thus, when the activation energy for the forward reaction is more than that for the backward reaction,

energy is absorbed during the course of reaction.

25. MAXWELL'S- BOLTZMANN DISTRIBUTION CURVE :Since it is difficult to predict the behavior of any one molecule with precision, Ludwig Boltzmann andJames Clark Maxwellused statistics to predict the behaviour of large number of molecules.Accordingto them, the distribution of kinetic energy maybe described by plotting the fraction ofmolecules (N /N )

E T

with a given kinetic energy (E) vs kinetic energy (Figure). Here, N is the number of molecules withE

energy E and N is total number of molecules.T

s nal h

Kinetic energy

Distribution curve showing temperature dependence of rate of a reaction

26

t

(t+10) Energy ofactivation This area show

fraction of additio

This area cules whic

shows fraction t at (t + 10

of moleculesreacting at t

mole reac

Chemical Kinetics

The peak of the curve corresponds to the most probable kinetic energy, i.e., kinetic energy of maximumfraction ofmolecules. There are decreasing number of molecules withenergies higher or lower than thisvalue. When the temperature is raised, the maximum of the curve moves to the higher energy value(Figure) and the curve broadens out, i.e., spreads to the right such that there is a greater proportion ofmolecules with much higher energies. The area under the curve must be constant since total probabilitymust be one at all times. We can mark the position of E on Maxwell Boltzmann distribution curve.

a

Increasing the temperature of the substance increases the fraction ofmolecules, whichcollide withenergiesgreater than E . It is clear from the diagram that in the curve at (T + 10), the area showing the fraction of

a

molecules having energy equal to or greater than activation energygets doubled leading to doublingtherate of a reaction.

26. EFFECT OF CATALYST :Acatalyst is a substance which alters the rate of a reaction without itself undergoing any permanent chemical change. For example, MnO catalyses the followingreactionsoas to increase itsrate considerably.

2

2KCl + 3O2KClO MnO2 3 2

Catalytic Mechanism :The action of the catalyst can be explained by intermediate complex theory.According to this theory, acatalyst participates in a chemical reaction by forming temporarybonds with the reactants resultingin anintermediate complex. This has a transitoryexistence and decomposes to yield products and the catalyst.It is believed that the catalyst provides an alternate pathway or reaction mechanism by reducing theactivation energy between reactants and products and hence lowering the potential energy barrier asshownin the figure.

Energy of activation without catalyst

catalystpath with

Products

Reaction coordinate

Effect of catalyst on activation energy

It is clear fromArrhenius equation, k =AeEa / RT ,

that the lower the value of activation energy faster will be the rate of a reaction.Note: for catalysts

Asmallamount of the catalyst can catalyse a large amount of reactants.Acatalyst does not alter Gibbs energy, G of a reaction.

It catalyses the spontaneous reactions but does not catalyse non-spontaneous reactions.

It is also found that a catalyst does not change the equilibrium constant of a reaction rather, it helps

in attaining the equilibrium faster, that is, it catalyses the forward as well as the backward reaction to

the same extent so that the equilibrium state remains same but is reached earlier.

27. COLLISION THEORY OF CHEMICAL REACTIONS :

Though Arrhenius equation is applicable under a wide range of circumstances, collision theory, which

was developed by Max Trautz and William Lewis in 1916-18, provides a greater insight into the energetic

and mechanistic aspects of reactions. It is based on kinetic theory of gases.

27

Po

ten

tial

ener

gy

Reaction pathwithout catalyst

Energy of activation

Reactants Reaction with

catalyst

Chemical Kinetics

Collision Theory :

According to this theory, the reactant molecules are assumed to be hard spheres and reaction is postulated

to occur whenmolecules collide with each other. Various noteworthy points ofcollision theory are: Collision Frequency - The number of collisions per second per unit volume of the reaction mixture

is known as collision frequency (Z).Activation Energy - Another factor which affects the rate ofchemical reactions is activation energy(as we have already studied). Bimolecular Elementary Reaction -For a bimolecular elementary reaction A+ B Productsrate of reaction can be expressed as

e– Ea / RTRate = Z ...(vi)AB

e– Ea / RTwhere Z represent the collision frequency of reactants, Aand B and represents theAB

fraction of molecules with energies equal to or greater than E .a

28. THRESHOLD ENERGY AND ORIENTATION OF COLLIDING MOLECULES :The equation (vi) predicts the value of rate constants fairly accurately for the reactions that involveatomic species or simple molecules but for complex molecules significant deviations are observed.The reason could be that all collisions do not lead to the formation of products. The collisions in whichmolecules collide with sufficient kinetic energy (called threshold energy*) and proper orientation, so asto facilitate breaking of bonds between reacting species and formation of new bonds to form productsare called as effective collisions.

Note: The formation of methanol from bromoethane depends upon the orientation of reactant molecules as

showninthe figure. The proper orientation ofreactant molecules lead to bond formationwhereas

improper orientation makes them simply bounce back and no products are formed.CH3Br + OH CH3OH + Br

HImproper

Br + OH No productsH CH H

H C Br + OHH

C

HH

HO C

H

Br

H

OH H + BrOrientation

HIntermediate

Steric Factor

To account for effective collisions, another factor P, called the probability or steric factor is introduced.It takes into account the fact that in a collision molecules must be properly oriented i.e.,

e–Ea / RTRate = PZAB

Note: for collision theory

In collision theoryactivation energy and proper orientation of the molecules together determine thecriteria for an effective collision and hence the rate of a chemical reaction.

Collision theory also has certain drawbacks as it considers atoms/ molecules to be hard spheres and

ignores their structural aspect.

Comparing the equation (vi) withArrhenius equation, we can saythatA(ofArrhenius Equation) is

related to collision frequency.

28

Orientation

proper

Chemical Kinetics

SOLVED EXAMPLES

Ex.1 The half-life period of a first order reaction is 30 minutes. Calculate the specific reaction rate of thereaction. What fraction of the reactant remains after 70 minutes ?

0.6932 0.6932Sol. k = = = 0.231 min–1.

1 t 1 302

Let the reaction be

a

A

0

Product

x

Initial concentration

(a – x) Concentration after 70 minutes

(a – x) fraction of the reactant remained unreacted = .

a

Now,

2.303log

ak =

1 t a – x

2.303log

aor 0.0231 =

70 a – x

a 0.0231 70log = = 0.7021.

a – x 2.303

Taking antilog, we get

a= 5.036.

a – x

a – x 1 = 0.2.

a 5.036

Ex.2 The specific reaction rate of a first-order reaction is 0.02 s–1.The intial concentration of the reactant is 2moles/litre. Calculate (a) initial rate, and (b) rate after 60 seconds.

We know that for a first-order reaction :

Rate of a reaction = k × molar concentration of the reactant.

(a) inital rate = k × initial concentration

= 0.02 × 2 = 0.4 mole/litre/second.

(b) Now, to calculate rate after 60 seconds, let us first calculate concentration of the reactant after 60 seconds.

2

Sol.

2.303 logk = = 0.02.concentration after 60seconds60

concentration of the reactant after 60 seconds = 0.60 M.

rate after 60 seconds = k × concentration of the reactant after 60 seconds.

= 0.02 × 0.60

= 0.012 mole /litre/second.

29

Chemical Kinetics

Ex.3 The rate constant is numerically the same for three reactions of first, second and third order respectively,the unit of concentration being in moles per litre. Which reaction should be the fastest and is thistrue for all ranges of concentrations ?

Suppose R , R and R are the rates of three reactions of first, second and third order respectively andSol.1 2 3

k is the rate constant, which is the same for the three reactions,

R = k[A]11

R = k[A]22

R = k[A]33

[A] being the concentration of the reactantAin moles per litre.

Now if, [A] = 1,

[A] < 1,

[A] > 1,

R = R = R ;1 2 3

R > R > R ;1 2 3

R < R < R .and1 2 3

Ex.4 For the reaction 2NO + Cl 2NOCl, it is found that doubling the concentration of both reactants2

increases the rate by a factor of 8, but doubling the Cl concentration alone, only doubles the rate. What2

is the order of the reaction with respect to NO and Cl ?2

Sol. Rate = k [NO]m [Cl ]n2

Let the concentrations of NO and Cl be x and y respectively.2

According to the question,

R = k xm yn1

R = k (2x)m (2y)nand,2

= k . xm yn . 2m + n

R 2

= 2m + n = 8 = 23. (given)R1

Again,

m + n = 3

R = k (x)m (2y)n3

= k xm yn . 2n.

R3

= 2n = 2. (given)R1

n = 1

m = 3 – 1 = 2.

Ex.5 In a reaction 2N O 4NO + O , the rate can be expressed as2 5 2 2

–d[ N 2O5 ]

(i) = k [N O ]1 2 5dt

d[NO2 ](ii) = k [N O ]

2 2 5dt

d[O2 ](iii) = k [N O ]

3 2 5dt

How are k , k and k related ?1 2 3

30

Chemical Kinetics

Sol. The rate law of the given reaction is

1

d[N2O5 ] 1

d[NO2 ] d[O2 ]rate = –

2= = = k[N O ]

2 5dt 4 dt dt

d[N 2O5 ] – = 2k [N O ] = k [N O ]

2 5 1 2 5dt

d[ NO2 ]= 4k [N O ] = k [N O ]

dtd[O2 ]

2 5 2 2 5

= k [N O ] = k [N O ]dt 2 5 3 2 5

k1 2k

k2 4k

k3 k

k1 k 2or k = = = k

32 4

or 2k = k = 4k .1 2 3

Ex.6 The reaction, N O 4NO + O , is forming NO at the rate of 0.0072 mol/L/s at some time.2 5 2 2 2

(a) What is the rate of change of [O ] at this time ?2

(b) What is the rate of change of [N O ] at this time ?2 5

(c) What is the rate of reaction at this time ?

The rate of the reaction is expressed asSol.

1 d[N 2O5 ] 1 d[NO2 ] d[O2 ]rate = –

2= +

4= +

dt dt dt

d[ NO2 ]and given that = 0.0072 mole/L/s.

dt

1(a) Rate of appearance of O = × rate of appearance of NO

42 2

d[O2 ] 1 d[ NO2 ]= ×

dt 4 dt

1= × 0.0072 = 0.0018 mole/L/s.

4

1(b) Rate of disappearnace of N O = × rate of appearance of NO

2

1 d[NO2 ]

2 5 2

d[N 2O5 ]– = ×

dt 2 dt

d[N 2O5 ] 1= – × 0.0072 = –0.0036 mole/L/s.

dt 2

1 d[ NO2 ] 1= × 0.0072 = 0.0018 mole / L/s.(c) Rate of reaction = ×

4 dt 4

31

Chemical Kinetics

Ex.7 The rate of a first order reaction is 0.04 mole/L/s at 10 minutes and 0.03 mole/L/s at 20 minutes afterinitiation. Find the half-life of the reaction.

Let the concentrations of the reactant after 10 min and 20 min be C and C respectively.Sol.1 2

rate after 10 min = k.C = .04 × 601

and rate after 20 min = k.C = .03 × 60.2

c1 4 =

c2 3

Suppose the reaction starting after 10 minutes

2.303 2.303c1 4k = log =

clog = 0.02878

310 102

0.6932 0.6932 t = = = 24.086 min.

k 0.028781/2

Ex.8 For a reaction at 800° C, 2NO + 2H N + 2H O, the following data were obtained :2 2 2

d[NO]/dt × 10–4

mole litre–1 min–1

4.4

2.2

0.24

[NO] × 10–4

mole / litre

(i) 1.5

(ii) 1.5

(iii) 1.5

[H ] × 10–32

mole / litre

7.0

3.5

2.0

What is the order of this reaction with respect to NO and H ?2

From the data (i) and (ii), we see that when the concentration of H is halved, the rate is also halved atSol.2

constant concentration of NO. Hence the reaction is of first order with respect to H . Let us now2

consider the data (ii) and (iii) to determine the order with respect to NO as [H ] is constant.2

The rate law of the above reaction is

–1

d[NO]

rate = = k[NO] [H ]2

m 1

2 dt

where m is the order with respect to NO

–d[NO]

or = 2k [NO] [H ]2

m

dt

Substituting data (ii) and (iii), we get

2.2 × 10–4 = 2k (1.5 × 10–4)m . (2 × 10–3)

0.24 × 10–4 = 2k (0.5 × 10–4)m . (2 × 10–3)

Dividing (1) by(2),

...(1)

...(2)

(1.5 10 –4 )m2.2 220

= = 3m or = 3m

(0.510 – 4 )m0.24 24

Taking log, log 220 – log 24 = m log 3

2.4324 – 1.3802 = m × 0.4771

0.9622 = 0.4771 m

0.9622

or

or m = = 2.0.4771

Hence the reaction is of second and first order with respect to NO and H respectively.

32

Chemical Kinetics

For a chemical reactionA+ B Product, the order is 1 with respect to each ofAand B. Find x and yfrom the given data.

Ex.8

Rate (moles/ L/s)

0.10

0.40

0.80

[A]

0.20 M

x

0.40 M

[B]

0.05 M

0.05 M

y

Sol. The rate law may be written as

rate = k [A] [B]

Substituting the first set of data in the rate law, we get,

0.10 = k × 0.20 × 0.05

k = 10.

Now substituting the second and third sets of data, we get,

0.40 = 10 × x × 0.05

x = 0.80 M.

And, 0.80 = 10 × 0.40 × y

y = 0.20 M.

Ex.9 The energy of activation and specific rate constant for a first-order reaction at 50° C

2N O 2N O + O2 5

(in CCl )2 4 2

(in CCl )4 4

are 100 kJ/mole and 3.46 × 10–5 s–1 respectively. Determine the temperature at which the half-life of thereaction is 2 hours.

Sol. Let us calculate the rate constant (say k ) at a temperature (say T ) at which t is given to be2 2 1/2

2 × 60 × 60 seconds.

The rate constant at a temperature 298 K (say T ) is given as 3.46 × 10–5 s–1 (say k )1 1

0.6932k = = 9.62 × 10–5 s–1.

2 60602

E T2 – T1 k 2 2.303 R T T

Thus, log k = 1 2 1

T – 298 9.62 10–5

log3.4610–5

100000 2 = 29T2.303 8.314 2

or T = 310 K.2

Ex.10 InArrhenius's equation for a certain reaction, the value ofAand E (activation energy) are 4 × 1013 s–1

and 98.6 kJ mol–1 respectively. If the reaction is of first order, at what temperature will its half-life period be ten minutes ?

We have,

k = Ae–E/RT

E

Sol.

ln k = lnA–RT

E2.303 log k = 2.303 logA–

RT

33

Chemical Kinetics

Eor log k = logA– ...(i)2.303RT

Given that A = 4 × 1013 s–1, E = 98.6 kJ mol–1

t = 10 × 60 s.1/2

0.6932 0.6932For first-order reaction k = = = s–1

t1/ 2 600

Thus (i) becomes,

98.60.6932log = log (4 × 10 ) –13 [R = 8.314× 10 kJ/K/mol]–3

2.3038.31410–3 T600

T = 311.2 K.

Ex.11 The activation energyfor the reaction,

O (g) + NO (g) NO (g) + O (g)3 2 2

is 9.6 kJ/ mole. Prepare an activation energy plot it H° for this reaction is –200 kJ/mole. What is the

energy of activation for the reverse reaction ?

Energy of activation for reverse reaction

= 9.6 + 200 kJ

= 209.6 kJ.

Sol.

Reaction coordinate

Ex.12 Fromthe followingdata for thedecompositionofdiazobenzene chloride, showthat the reactionisoffirst order

Time:

Vol. of N :

20

10

50

25

70

33

162

(min)

(mL)2

Sol. C H N Cl C H Cl + N6 5 2 6 5 2

Initial concentration

Concentration after time t

a

(a – x) x

At time, i.e, when the reaction is complete, the whole of C H N Cl converts into N . Hence volume6 5 2 2

of N at time corresponds to the inital concentration 'a' while volumes of N at different time intervals2

correspond to x as shown above.2

Inserting the given data in Equation of first order reaction, we get the following results.

2.303 162For t = 20 min, k = log = 0.0032 min–1

20 162 –101

34

En

ergy

9.6 kJ

Ereverse

200 kJ

Chemical Kinetics

2.303 162For t = 50 min, k = log = 0.0033 min–1

1 50

2.303

162 – 25

162For t = 70 min, k = log = 0.0032 min–1

1 70 162 – 33

The constancy of k shows tha the decomposition of C H N Cl is a first order reaction.1 6 5 2

Ex.13 1 mL ofmethyl acetate was added to 20 mLof 0.5 N HCland 2 mL of the mixture was withdrawn fromtime t time during the progress of hydrolysis of the ester and titrated with a solution ofalkali. The amount

c

of alkali needed for titration at various intervals is given below :

42.03

Time:

Alkali used :

0

19.24

20

20.73

119

26.6

(min)

(mL)

Establish that the reaction is of first order.

CH COOCH + H O HCl[H+]Sol. CH COOH + CH OH

3 3 2 3 3

Initial concentration

Concentration after time t

a

(a – x) x

HClacts as a catalyst. Since in every titration the amount of HCl is the same, the alkaliused against HClis subtracted from the total alkali used (given in the data) to get the volume of alkali used only againstCH COOH.At zero time no CH COOH is formed, so alkali used at zero time is only for HCl.

3

we thus have,

Time (min) :

Vol. of alkali

(mL) used

against CH COOH

3

0 20 119

: 19.24 20.73 26.6 42.03

3

–19.24 = 0 –19.24 =

1.49

(x)

–19.24 =

7.36

(x)

–19.24 =

22.79

(a)

Now following the same way as inExample 40, we get the following results :

2.303 22.79For t =20 ; k = log = 0.0033 min–1

20

2.303

22.79 –1.49

22.79

1

= 0.0032 min–1For t =119 ; k = log119 22.79 – 7.361

The constancy of k shows that the reaction is of first order.1

Ex.14 The optical rotation of can sugar in 0.5 N lactic acid at 25° C at various time intervals are givenbelow :

Time (min) :

Rotation(°) :

0

34.50°

1435

31.10°

11360

13.98°

–10.77°

Show that the reaction is offirst order.

Lactic acidSol. C12H22O11 + H2O C6H12O6 + C6H12O6

Sucrose (excess) GlucoseDextro rotatory

FructoseLaevo rotatory

Dextrorotatory

35

Chemical Kinetics

Since in this reaction dextro formchanges to laevo form, the optical rotation decreases with the progressof the reaction. Thus change in rotation is proportional to the amount of sugar remaind after differenttime intervals. We now have,

Substituting the data inEquation,

for t = 1435 min

2.303 45.27k = log = 5.442 × 10–5

1 1435 41.87

and for t = 11360 min

2.303 45.27k = log = 5.311 × 10–5

1 11360 24.75

The values of k are fairly constant and so the reaction is of first order.1

Ex.15 Bicyclohexane was found to undergo two parallel first order rearrangements.At 730 K, the first orderrate constant for the formation of cyclohexene was measurred as 1.26 × 10–4 s–1, and for the formationof methyl cyclopenten the rate constant was 3.8 × 10–5 s–1. What is the percentage distribution of therearrangement products?

Sol.

k1Percentage of cyclohexene = × 100k k1 2

1.2610–4

= × 1001.2610– 4 3.810–5

= 77%.

percentage ofmethylcyclopenten = 23%.

Ex.16 For the displacement reaction

[Co(NH ) Cl]2+ + H O [Co(NH ) (H O)]3+ + Cl–3 5 2 3 5 2

the rate constant is given by

–11067 K

ln [k / (min–1)] = + 31.33.T

evaluate k, E andAfor the chemical reaction at 25° C.

Substituting T = 298 K in the given equation, we get,

k = 3.06 × 10–3 min–1 = 5.10 × 10–5 s–1.

Sol.

36

Time (min) 0 1435 11360

Change in rotation (°)

34.50 –(–10.77)= 45.27

(a)

31.10 –(–10.77)= 41.87 (a – x)

13.98 –(–10.77)= 24.75 (a – x)

–10.77 –(–10.77)

= 0

Chemical Kinetics

Further, we have,

k = Ae–E/RT

Eor ln k = lnA–

RT

Compairing this equation with the given equation, we get,

lnA= 31.33

or 2.303 logA= 31.33 ; A = 4.04 × 1013 min–1

= 6.73 × 1011 s–1

Eand, –

RT = –

11067 K

T

E = (11067 K) (8.314 J K–1 mol–1)

= 92011 J/mole = 92.011 kJ/ mole.

The complexation of Fe2+ with the chelating agent dipyridyl has been studied kinetically in both forward and reverse directions.

Fe2+ + 3 dipy Fe(dipy) 2+

Ex.17

3

Rate (forward) = (1.45 × 1013) [Fe2+][dipy]3

and rate (reverse) = (1.22 × 10–4) [Fe (dipy) 2+].3

Find the stability constant for the complex.

At dynamic equilibrium,

rate of formation of complex = rate of its decomposition

(1.45 × 1013) [Fe2+][dipy]3 = (1.22 × 10–4) [Fe(dipy) 2+]

Sol.

3

[Fe(dipy )2 ] 1.4510133

K = = = 1.19 × 1017.2 3 1.2210–4s [Fe ][dipy]

Ex.18 The approach to the following equilibrium was observed kinetically fromboth direction.

PtCl 2– + H O Pt (H O)Cl – + Cl– at 25° C.4 2 2 3

d[PtCl2– ]4–It was found that = (3.9 × 10–5) [PtCl 2–)] – (2.1 × 10–3) [Pt(H O)Cl –] [Cl–]dt 4 2 3

Calculate the equilibrium constant for the complexation of the fourth Cl– byPt (II).

d[PtCl2– ]4Sol. At equilibrium, = 0dt

Hence, 3.9 × 10–5 [PtCl2– ] = 2.1 × 10–3 [Pt(H O)Cl2– ][Cl– ]2 34

[PtCl2– ] –32.1104

or K = = = 53.85.[Pt(H O)Cl– ][Cl– ] –53.9102 3

37

Chemical Kinetics

Ex.19 For the reaction

[Cr(H O) Cl ]+ (aq) k1

[Cr(H O) Cl ]2+ (aq) k2

[Cr(H O) ]3+ (aq)2 4 2 2 5 2 2 6

k = 1.78 × 10–3 s–1 and k = 5.8 × 10–5 s–1 for the initial concentration of [Cr(H O) Cl ]+ is 0.01741 2 2 4 2

mole/litre at 0°C. Calculate the value of t at which the concentration of [Cr(H O) Cl ]2+ is maximum.2 5 2

Sol. We have,

2.303(log k1 – log k 2 ) 2.303(log1.7610–3 – log 5.810–5 )t = = = 1990 seconds.k – k –3 –51.7610 – 5.8101 2

Ex.20 From the following reaction scheme, write the rate law for the disappearance ofA, B and C.

(1) A+ B k1C + D (2) C + D k A+ B (3) B + C k3 E + D

Sol. The reactant Ais removed in Step 1 and produced in Step 2

d[A] – = k [A] [B] – k [C] [D]

1 2dt

Similarly,

d[B]– = k [A] [B] + k [B] [C] – k [C] [D]

1 3 2dt

d[C]and – = k [C] [D] + k [B] [C] – k [A] [B].

2 3 1dt

38

Chemical Kinetics

EXERCISE - I

Q.1 Asolution ofAis mixed with an equal volume of a solution of B containing the same number of moles,and the reaction A+B=C occurs. At the end of 1h, A is 75 % reacted. How much of A will be leftunreacted at the end of 2 h if the reaction is (a) first order inAand zero order in B; (b) first order in both AandB ; and (c) zero order in both Aand B ?

The reaction CH3CH2NO2 + OH– CH3CHNO2 + H2O obeys the rate law for pseudo first

order kinetics in the presence of a large excess of hydroxide ion. If 1% ofnitro ethane undergoes reactionin halfa minute when the reactant concentration is0.002 M, What is the pseudo first order rate constant?

The decomposition of a compound P, at temperature T according to the equation2P(g) 4Q(g) + R(g) + S(l)

is the first order reaction. After 30 minutres from the start of decomposition in a closed vessel, the totalpressure developed is found to be 317 mm Hg and after a long period of time the total pressure observed to be617 mm Hg. Calculate the total pressure of the vessel after 75 mintute, if volume of liquid S is supposed tobe negligible.Also calculate the time fraction t7/8.Given : Vapour pressure of S (l) at temperature T = 32.5 mm Hg.

A certain reactant Bn+ is getting converted to B(n+4)+ in solution. The rate constant of this reaction ismeasured by titrating a volume of the solution with a reducing reagent which only reacts with Bn+ andB(n+4)+. In this process, it converts Bn+ toB(n2)+ and B(n+4)+ toB(n1)+. At t=0, the volume of the reagentconsumed is 25 ml and at t = 10 min, the volume used up is 32 ml. Calculate the rate constant of theconversion of Bn+ to B(n+4)+ assuming it to be a first order reaction.

Decomposition of H2O2 is a first order reaction. A solution of H2O2 labelled as 20 volumes was left open.Due to this, some H2O2 decomposed. To determine the new volume strength after 6 hours, 10 mL of thissolutionwas diluted to 100mL. 10mL of this diluted solution was titrated against 25mLof0.025MKMnO4 solution under acidic conditions. Calculate the rate constant for decomposition of H2O2.

A metal slowly forms an oxide film which completely protects the metal when the film thickness is3.956 thousand ths of an inch. If the film thickness is 1.281 thou. in 6 weeks, how much longer willit bebefore it is 2.481 thou.? The rate of film formation follows first order kinetics.

An optically active compound Aupon acid catalysed hydrolysis yield two optically active compound B and Cby pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while aftercompletion of the reaction it was – 20°. If optical rotation per mole of A, B & C are 60°,40° & – 80°. Calculate half life of the reaction.

A vessel contains dimethyl ether at a pressure of 0.4 atm. Dimethyl ether decomposes asCH3OCH3(g) CH4(g) + CO(g) + H2(g). The rate constant of decomposition is 4.78×103 min1.

Calculate the ratio of initial rate ofdiffusionto rateofdiffusionafter 4.5hoursof initiationofdecomposition.

At room temperature (20°C) orange juice gets spoilt in about 64 hours. In a referigerator at 3°C juice can bestored three times as long before it gets spoilt. Estimate (a) the activation energy of the reaction that causesthe spoiling of juice. (b) How long should it take for juice to get spoilt at 40°C?

Afirst order reaction, A B, requires activation energy of 70 kJ mol1. When a 20% solution ofAwas kept at25°C for 20 minutes, 25% decomposition took place. What will be the percent decomposition in the sametime in a 30% solution maintianed at 40°C? Assume that activation energy remains constant in this range oftemperature.

Two reations (i) A products (ii) B products, follow first order kinetics. The rate of the reaction (i) is

doubled when the temperature is raised from 300 K to 310K. The half life for this reaction at 310K is 30

minutes. At the same temperature B decomposes twice as fast as A. If the energy of activation for the reaction

(ii) is half that of reaction (i), calculate the rate constant of the reaction (ii) at 300K.

Q.2

Q.3

Q.4

Q.5

Q.6

Q.7

Q.8

Q.9

Q.10

Q.11

39

Chemical Kinetics

Q.12 Acertain organic compoundAdecomposes by two parallel first order mechanism

If k1 : k2 = 1 : 9 and k1 = 1.3 × 10–5 s–1.

Calculate the concentration ratio of C toA, if an experiment is started withonlyAand allowed torun for one hour.

The reactionQ.13

cisCr(en)2(OH)+2 transCr(en)2(OH)+

2

is first order in both directions. At 25°C the equilibrium constant is 0.16 and the rate constant k1 is3.3 × 10 4s 1. In an experiment starting with the pure cis form, how long would it take for half theequilibrium amount of the trans isomer to be formed ?

Q.14 For a reversible firstorder reaction A B

k1 = 10 2 s 1 and [B]eq /[A] eq = 4. If [A]0 = 0.01 mole L 1 and [B]0 = 0, what will be the concentration

of B after 30 s ?

For the systemA(g)lB(g), H for the forward reaction is –33 kJ/mol (Note : H = E in this case).

[B]Show that equilibrium constant K = [A] = 5.572 × 105 at 300 K. If the activation energies Ef & Eb are

Q.15

in the ratio 20 : 31, calculate Ef and Eb at this temperature.Assume that the pre-exponential factor is thesame for the forward and backward reactions.

The complex [Co(NH3)5F]2+ reacts with water according to the equation. [Co(NH3)5F]2+ + H2O [Co(NH3)5(H2O)]3+ + F

The rate of the reaction = rate const. x[complex]ax[H+]b. The reaction is acid catalysed i.e. [H+] doesnot change during the reaction. Thus rate = k[Complex]a where k’ = k[H+]b, calculate ‘a’ and ‘b’ giventhe following data at 250C.

Q.16

[H+]M0.010.02

[Complex]M0.10.2

T1/2hr1

0.5

T3/4hr21

For the two parallel reactionsA k1 B andA k 2 C, show that the activation energy E forthe disappearance ofAis given in terms of activation energies E1 and E2 for the two paths by

Q.17

k1E1 k2E2E = k k1 2

k1 kQ.18 For the mechanism A+ B C ; C 3 Dk2

(a) Derive the rate law using the steady-state approximation to eliminate the concentration of C.(b)Assuming that k3 << k2, express the pre-exponential factorAand Ea for the apparent second-order rate constant in terms ofA1,A2 andA3 and Ea1, Ea2 and Ea3 for the three steps.

The reaction of formation of phosgene from CO and Cl2 is CO + Cl2 COCl2The proposed mechanism is

Q.19

(i)

(iii)

Cl2 2Cl (fast equilibrium) (ii) Cl + CO COCl (fast equilibrium)

COCl + Cl2 K3

COCl2 + Cl (slow)d[COCl 2 ]

=K[CO][Cl2]3/2.Show that the above mechanism leads to the following rate lawdt1/ 2k k 2Where K = k3. 1 .

k 2 k 1

40

Chemical Kinetics

Q.20 For the following first order gaseous reaction

The initial pressure in a container of capacity V litres is 1 atm. Pressure at time t = 10 sec is 1.4 atm andafter infinite time it becomes 1.5 atmosphere. Find the rate constant k1 and k2 for the appropriate reactions.

Afirst order reaction takes 69.3 minutes for 50% completion. How much time will be needed for 80%completion.

The specific rate constant for a reaction increases by a factor 4, of the temperature is changed from27ºC to 47ºC. Find, the activation energy for the reaction

Q.21

Q.22

Q.23 The reaction 2A+ B + C D + 2 E is of first order with respect toA, of second order with respect toB and is of zero order with respect to C(i)(ii)

Write down the rate law for the reactionWhat will be the effect ofdoubling concentration ofA, B and C.

Q.24 Afirst order reaction is 50% completed in 30 min at 27ºC and in 10 minutes at 47ºC. Calculate the rateconstant at 27ºC and the energy of activation of the reaction in kJ per mole.

The optical rotation of sucrose in 0.5 M HCl at 35ºC at different time intervals are given below. Show

that the reaction follows first order kinetics.

Q.25

Time (min.) 0 10 20 30 40

+32.4 +28.8 +25.5 +22.4 +19.6 –11.1Rotation (degrees)

Q.26 T½ of a reaction is halved as the initial concentration of the reaction is doubled. find out the order of the

reaction.

The rate constant of a reaction is 1.5 × 10–7 sec–1 at 50ºC and 4.5 × 10–7 sec–1 at 100ºC. Evaluate the

Arrthenius parameters Aand Ea.

Asubstance reacts according to the law of first order reaction the velocity constant of the reaction is1.0 × 10–2 sec–1. If initial canc. of the substance is 1.0 M.

Q.27

Q.28

(a)(b)

Find out the initial rateFind out the rate after 1 min.

What will be the initial rate of a reaction if its rate constant is 10–3 min–1 and the concentration of reactant

is 0.2 mol dm–3. How much of reactant will be converted into products in 200 minutes.

Afirst order reaction is 20% complete in 10 minutes. Calculate

Q.29

Q.30

(a)

(b)

Specific rate constant of the reaction and

Time taken for the reaction to go to 75% completion

41

Chemical Kinetics

EXERCISE - II Q.1 Ammonia and oxygen reacts at higher temperatures as

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)In an experiment, the concentration of NO increases by 1.08 ×10–2 mol litre–1 in 3 seconds. Calculate.(i) rate of reaction. (ii) rate ofdisappearance of ammonia (iii) rate offormation of water

In the following reaction 2H2O2 2H2O + O2

rate of formation of O2 is 3.6 M min–1. (a) What is rate of formation of H2O?(b) What is rate of disappearance of H2O2?

In a catalytic experiment involving the Haber process, N2 + 3H2 2NH3, the rate of reaction was measured as

Q.2

Q.3

[NH3 ]= 2 × 10–4 mol L–1 s–1.Rate =

tIf there were no sides reactions, what was the rate of reaction expressed in terms of (a) N2, (b) H2?

The reaction 2A+ B + C D + E is found to be first order in A second order in B and zero order in C. (i)Give the rate law for the reaction in the form of differential equation.(ii) What is the effect in rate of increasing concentrations ofA, B, and C two times?

At 27°C it was observed during a reaction of hydrogenation that the pressure of hydrogen gas decreasesfrom 2 atmosphere to 1.1 atmosphere in 75 minutes. Calculate the rate of reaction (in M sec–1) and rate ofreaction in terms of pressure.

For the elementary reaction 2A + B2 2AB. Calculate how much the rate of reaction will change ifthe volume of the vessel is reduced to one third of its original volume?

For the reaction 3BrO— BrO3— + 2Br— in an alkaline aquesous solution, the value of the second

Q.4

Q.5

Q.6

Q.7–[BrO ]

order (in BrO—) rate constant at 80°C in the rate law for – was found to be 0.056L mol–1s–1.t – –

[BrO3 ], (b)

[Br ]?What is the rate of constant when the rate law is written for (a)

Dinitropentaoxide decomposes as follows :1

t tQ.8

N2O5(g) 2NO2(g) +2

O2(g)

Given that –d [N2O5] / dt = k1[N2O5]d [NO2] / dt = k2[N2O5]d [O2] / dt = k3[N2O5]

What is the relation between k1, k2 and k3?

Suppose that the Sun consists entirely of hydrogen atom and releases the energy bythe nuclear reaction,Q.9

4 1H 4He with 26 MeV of energy released. If the total output power of the Sun is assumed to1 2

remain constant at 3.9 × 1026 W, find the time it will take to burn all the hydrogen. Take the mass of theSun as 1.7 × 1030 kg.The reactionA(g) + 2B(g) C(g) + D(g) is an elementary process. In an experiment, the initialpartial pressure ofA& B are PA = 0.6 and PB = 0.8 atm. Calculate the ratio of rate of reaction relative

to initial rate when PC becomes 0.2 atm.

In the following reaction, rate constant is 1.2 × 10–2 M s–1 A B. What is concentration of B after

10 and 20 min., if we start with 10 M ofA.

Q.10

Q.11

42

Chemical Kinetics

Q.12 For the following data for the zero order reactionA products. Calculate the value of k.Time (min.)0.0

1.0

2.0

[A]0.10 M

0.09 M

0.08 M

The rate constant for a zero order reaction is 2 × 10–2 mol L–1sec–1, if the concentration of the reactant after 25 sec is 0.25 M, calculate the initial concentration.

Adrop of solution (volume 0.10 ml) contains 6 × 10–6 mole of H+, if the rate constant of disappearance of H+ is 1 × 107 mole litre–1 sec–1. How long would it take for H+ in drop to disappear?

Acertain substanceAis mixed withan equimolar quantity ofsubstance B.At the end of an hourAis75%reacted. Calculate the time whenAis 10% unreacted. (Given: order of reaction is zero)

Afirst order reaction is 75% completed in 72 min.. How long time will it take for

Q.13

Q.14

Q.15

Q.16(i) 50% completion (ii) 87.5% completion

Q.17 Afirst order reaction is 20% complete in 10 min. calculate (i) the specific rate constant , (ii) the timetaken for the reactions to go to 75% completion.

Show that in case of unimolecular reaction, the time required for 99.9% of the reaction to take place in tentimes that required for half of the reaction.

A first order reaction has a rate constant is 1.5 × 10–3 sec–1. How long will 5.0 g of this reactant take toreduce to 1.25 g.

A drug is known to be ineffective after it has decomposed 30%. The original concentration of asample was 500 units/ml. When analyzed 20 months later, the concentration was found to be420 units/ml. Assuming that decomposition is of I order, what will be the expiry time of the drug?

A viral preparation was inactivated in a chemical bath. The inactivation process was found to be firstorder in virus concentration. At the beginning of the experiment 2.0 % of the virus was found to beinactivated per minute . Evaluate k for inactivation process.

If a reaction AProducts, the concentrations of reactant Aare C0, aC0, a2C0, a3C0, ............. after time

interval 0, t, 2t, 3t, ............. where a is a constant. Given 0 < a < 1. Show that the reaction is of first order.

Also calculate the relation in k, a and t.

The reaction SO2Cl2(g) SO2(g) + Cl2(g) is a first order gas reaction with k =2.2 × 10–5 sec–1 at

320°C. What % of SO2Cl2 is decomposed on heating this gas for 90 min.

Two substances A(t1/2= 5 mins) and B (t1/2 = 15 mins) follow first order kinetics are taken in such a way that

initially [A]= 4[B]. Calculate the time after which the concentration of both the substance will be equal.

Q.18

Q.19

Q.20

Q.21

Q.22

Q.23

Q.24

Q.25 At 800° C the rate of reaction

2 NO + H2 N2 + H2O

Changes with the concentration of NO and H2 are

1 d[NO]

–1[NO] in M [H2] in M

4 × 10–3

2 × 10–3

2 × 10–3

in M sec2 dt

1.5 × 10–4 4.4 × 10–4

2.2 × 10–4

8.8 × 10–4

(i)

(ii) 1.5 × 10–4

(iii) 3.0 × 10–4

43

Chemical Kinetics

(a) What is the order of this reaction?

(b) What is the rate equation for the reaction?

(c) What is the rate when

[H2] = 1.5 ×10–3 M and [NO] = 1.1 × 10–3M?

The data below are for the reaction if NO and Cl2 to form NOCl at 295 KQ.26

Initial Rate (M s–1)

1 × 10–3

3 × 10–3

9 × 10–3

Concentration of Cl2 [M]

0.05

0.15

0.05

Concentration of NO

0.05

0.05

0.15

(a) What is the order w.r.t NO and Cl2 in the reaction.

(b) Write the rate expression

(c) Calculate the rate constant

(d) Determine the reaction rate when concentration of Cl2 and NO are 0.2 M & 0.4 M respectively.

The catalytic decomposition of N2O by gold at 900°C and at an initial pressure of 200mm is 50%

complete in 53 minutes and 73% complete in 100 minutes.

(i) What is the order of the reaction?

(ii) Calculate the velocity constant.

(iii) How much of N2O willdecompose in 100 min. at the same temperature but at initial pressureof 600

mm?

The pressure ofa gas decomposing at the surface of a solid catalyst has been measured at different times and

the results are given below

Q.27

Q.28

t (sec)

Pr. (Pascal)

0

4 × 103

100

3.5 × 103

200

3 × 103

300

2.5 × 103

Determine the order of reaction, its rate constant.

The half life period of decomposition of a compound is 50 minutes. If the initial concentration is halved,

the half life period is reduced to 25 minutes. What is the order of reaction?

In this case we haveA B + C

Q.29

Q.30

TimeTotal pressure of A+ B+C Find k.

A B + C

TimeTotal pressure of ( B+C) Find k.

A B + C

TimeVolume of reagent

tP2

P3

Q.31tP2

P3

Q.320 t

V2

[Assuming n-factor ofA, B & C are same]V1

The reagent reacts withA, B and C. Find k.

44

Chemical Kinetics

Q.33 A 2B + 3CTime

Volume of reagent

t V3

[Assuming n-factor ofA, B & C are same]

V2

Reagent reacts with allA, B and C. Find k.

S G + FQ.34Time

Rotation ofGlucose & FructoseFind k.

trt

r

3The reaction AsH3(g) As(s) +

2H2(g) was followed at constant volume at 310°C by measuringQ.35

the gas pressure at intervals Show fromthe following figures that reaction is offirst order.Time (inhrs)Total pressure (in mm)

0758

5827

7.5856

10882

Q.36 The thermal decomposition of dimethyl ether as measured by finding the increase in pressure of the

(CH3)2O(g) CH4(g) + H2(g) + CO(g)reactionat 500°C is as follows:Time (sec.)

61939096

1195250

3155467Pressure increase (mm Hg)

the initial pressure of ether was 312 mm Hg. Write the rate equation for this reaction and determinethe rate constant of reaction.

From the following data show that decomposition of H2O2 in aqueous solution is first order.Q.37

Time (in minutes)Volume (in c.c. of KMnO4)

022.8

1013.3

208.25

Q.38 The following data were obtained in experiment on inversion of cane sugar.

–3.8

Time (minutes)Angle of rotation (degree)

0+13.1

60+ 11.6

120+ 10.2

180+9.0

360+5.87

Show that the reaction is of first order.After what time would you expect a zero reading in polarimeter?At 100°C the gaseous reaction A 2B + C was observed to be of first order. On starting with pureAit is found that at the end of 10 minutes the totalpressure ofsystem is 176 mm. Hg and after a long time270 mm Hg. From these data find (a) initial pressure ofA(b) the pressure ofAat the end of 10 minutes(c) the specific rate of reaction and (d) the half life period of the reaction?

The decomposition of N2O5 according to the equation 2 N2O5 (g) 4 NO2(g) + O2(g) is a firstorder reaction. After 30 min. from start of decomposition in a closed vessel the total pressuredeveloped is found to be 284.5 mm Hg. On complete decomposition, the total pressure is584.5 mm Hg. Calculate the rate constant of the reaction.

A definite volume of H2O2 under going spontaneous decomposition required 22.8 c.c. of standardpermanganate solution for titration. After 10 and 20 minutes respectively the volumes ofpermanganate required were 13.8 and 8.25 c.c.(a) Find order of reaction. How may the result be explained?(b) Calculate the time required for the decomposition to be half completed. (c) Calculate the fraction of H2O2 decomposed after 25 minutes.

Hydrogen peroxide solution was stored in a mild steel vessel. It was found, however, that the hydrogenperoxide decomposed on the walls of the vessel (a first order reaction). An experiment with 100 ml of asolution gave 10.31 ml oxygen (corrected to 1 atm & 273 K) after 5.1 days. Find how long the peroxidecan be stored before the loss of 20.00 ml oxygen occurs (per 100 ml solution) under similar storage

Q.39

Q.40

Q.41

Q.42

conditions. if complete decomposition of the sample to H2O2 gave 46.34 ml oxygen.

45

Chemical Kinetics

The reaction given below, rate constant for disappearance ofAis 7.48 × 10–3 sec–1. Calculate the timeQ.43required for the totalpressure in a system containingAat an initial pressure of0.1 atm to rise to 0.145 atmand also find the total pressure after 100 sec.

2A (g) 4B(g) + C(g)

The reaction A(aq) B (aq) + C (aq) is monitered by measuring optical rotation of reaction mixture

at different time interval. The speciesA, B and C are optically active with specific rotations 20°,30° and

– 40° respectively. Starting with pure A if the value of optical rotation was found to be 2.5° after 6.93minutes and optical rotation was –5° after infinite time. Find the rate constant for first order conversionofA into B and C.

Q.44

[x]t, calculate value of ratio, [y] [z] at any given instant

t.

Q.45 For a reaction

[C]k1 = x hr–1; k1 : k2 = 1 : 10. Calculate [A] after one hour from the start of the reaction.Q.46

Assuming onlyAwas present in the beginning.

How much time would be required for the B to reach maximum concentration for the reaction

L n2 k ln2,.

Q.47

k1 k 2A B C. Given k = =1

2

4 2

For first order parallel reaction k1 and k2 are 8 and 2 min–1 respectively at

300 K. If the activation energies for the formation of B and C are respectively 20and 28.314 kJ/mol respectively find the temperature at which B and C will be obtained in molar ratio of 2 : 1.[Given : ln 4 = 1.4 ]

In gaseous reactions important for understanding the upper atmosphere, H2O and O react bimolecularly toform two OH radicals. H for this reaction is 72 kJ at 500 K and Ea = 77 kJ mol–1, then calculate Ea forthe biolecular recombination of 2OH radicals to form H2O & O at 500 K

The energy of activation of a first order reaction is 104.5 kJ mole–1 and pre – exponential factor (A) is5 ×1013 sec–1.At what temperature, will the reaction have a half life of 1 minute?

The specific rate constant for a reaction increases by a factor of 4, if the temperature is changed from27°C to 47°C. Find the activation energy for the reaction.

The energyof activation and specific rate constant for a first order reaction at 25°C are 100 kJ/ mole and3.46 × 10–5 sec–1 respectively. Determine the temperature at which half life of the reaction is 2 hours.

Acatalyst lowers the activation energy for a certain reaction from 75 kJ to 25 kJ mol–1. What will be the effect on the rate of reaction at 25°C, after things being equal.

Given that the temperature coefficient for the saponification of ethyl acetate by NaOH is 1.75. Calculate activation energy for the saponification ofethyl acetate.

At 380°C, the halflife period for the first order decomposition of H O is 360 min.The energy of

Q.48

Q.49

Q.50

Q.51

Q.52

Q.53

Q.54

Q.552 2

activation of the reaction is 200 kJ mol1. Calculate the time required for 75% decomposition at 4500C.

46

Chemical Kinetics

Q.56 TheArrhenius equation for two first order equationA B and C D is given by

k1 = 1012

k2 = 1011

e81.28(kJ) / RT

e43.10(kJ) / RT

At what temperature k1 becomes equal to k2. The unit of activation energy is kJ/molUse: ln 10 = 2.3 and R = 8.3 J/K/mol

The reaction 2NO + Br2 2NOBr, is supposed to follow the following mechanismQ.57

(i) NO + Br2

(ii) NOBr2 + NO

NOBr2

slow 2NOBr

Suggest the rate law expression.

For the reaction 2H2 + 2NO N2 + 2H2O, the following mechanism has been suggested:2NOl N2O2 equilibrium constant K1 (fast)

Q.58

kN2O2 + H2

N2O + H2

2

k 3

N2O+ H2O (slow)

N2 + H2O (fast)

Establish the rate law for given reaction.

Reaction between NO and O2 to form NO2 is 2NO + O2 2NO2 follows the following mechanismQ.59

k1 NO + NO N2O2 ( in rapid equilibrium)k–1

N2O2 + O2 k 2

2NO2

(slow)1 d[ NO 2 ]

= K[NO]2[O ]Show that the rate of reaction is given by 2 dt 2

Q.60 Deduce rate law expressions for the conversion of H2 and I2 to HI at 400°C corresponding to each

of the following mechanisms:(a) H2 + I2 2HI (one step)

(b) I2 2I

2I + H2 2HI (slow)

(c) I2 2I

I + H2 IH2

IH2 + I 2HI (slow)

(d) Can the observed rate law expression rate = k[H2][I2] distinguish among these mechanisms?

(e) If it is known that ultraviolet light causes the reaction of H2 and I2 to proceed at 200°C with the

same rate law expression, which of these mechanisms becomes most improbable?

47

Chemical Kinetics

EXERCISE - III

Q.1 The rate of a reaction is expressed in different ways as follows :

1 d[C]

1 d[D]

1 d[A]

d[B]

2 dt 3 dt 4 dt dt

The reaction is:(A) 4A+ B 2C + 3D (C)A+ B C + D

(B) B + 3 D 4A+ 2 C(D) B + D A+ C

For the reactionA+ B C; starting with different initial concentration ofAand B, initial rate ofreaction were determined graphically in four experiments.

Q.2

rate/(M sec )

Rate law for reaction from above data is

(A) r = k[A]2 [B]2 (B) r = k[A]2 [B] (C) r = k[A] [B]2 (D) r = k[A] [B]

Q.3 The rate law for a reaction between the substancesAand B is given byrate = k [A]n [B]m

On doubling the concentration ofAand halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as(A) 2 (n–m) (B) 1 / 2 (m+n) (C) (m + n) (D) (n – m)

Q.4 For the reaction system 2NO (g) + O2 (g) 2NO2 (g) volume is suddenly reduced to half of itsvalue by increasing the pressure on it. If the reaction is first order with respect to O2 and second order

with respect to NO, the rate of reaction will(A) increase to four times of its initial value(C) diminish to one-eight of its initial value

(B) diminish to one-fourth of its initial value(D) increase to eight times of its initial value

Q.5 In a first order reaction, the concentration of the reactant, decreases from 0.8 to 0.4 M in 15 minutes.The time taken for the concentration to change from 0.1 M to 0.025 M is(A) 30 minutes (B) 15 minutes (C) 7.5 minutes (D) 60 minutes

The rate equation for the reaction 2A+ B C is found to be : rate = k [A] [B]. The correct statementin relation to this reaction is (A) unit of k must be s–1

(B) t1/2 is a constant

(C) rate of formation of C is twice the rate of disappearance ofA(D) value ofk is independent of the initial concentrations ofAand B.

t1/4 can be taken as the time taken for the concentration of a reactant to drop to 3/4 of its value. If the

rate constant for a first order reaction is k, the t1/4 can be written as [ln2 = 0.695, ln = 1.1]

Q.6

Q.7

(A) 0.69 / k (B) 0.75 / k (C) 0.10 / k (D) 0.29 / k

Q.8 Areaction was found to be second order with respect to the concentration of carbon monoxide. If theconcentration ofcarbon monoxide is doubled, with everything else kept the same, the rate ofreaction will

(A) Double (B) remain unchanged (C) Triple (D) Increase by a factor of 4

48

S.No. [A]0/M (Initial conc.) [B]0/M (Initial conc.) –1

1 1.6 × 10–3 5 × 10–2 10–3

2 3.2 × 10–3 5 × 10–2 4 × 10–3

3 1.6 × 10–3 10–1 2 × 10–3

4 3.2 × 10–3 10–1 8 × 10–3

Chemical Kinetics

2AB are 180 kJ mol–1Q.9 The energies of activation for forward and reverse reactions forA2 + B2

and 200 kJ mol–1 respectively. The presence of catalyst lowers the activation energies ofboth (forward

and reverse) reactions by 100 kJ mol–1. The magnitude of enthalpy change

(A2 + B2 2AB) in the presence of catalyst will be (in kJ mol–1).

of the reaction

(A) 300 (B) 120 (C) 20 (D) –20

Which graph represents zero order reaction [A(g) B(g)] :Q.10

d[B]

(A) [B] dt(B)

t t

t1/2 t3/4

(C) (D)

[A]0[A]0

For a hypothetical reaction, A + 3B P

Q.11

H = –2 x kJ/mole ofAH = + x kJ/mole of M& M 2Q + R

If these reactions are carried simultaneously ina reactor such that temperature is not changing. Ifrate ofdisappearance of B is y M sec–1 then rate of formation (in M sec–1) of Q is :

2(A)

3y

3(B)

2y

4(C)

3y

3(D) y

4

Q.12 Gaseous reactionA B + C follows first order kinetics. Concentration ofAchanges from 1 M to0.25 M in 138.6 minutes. Find the rate of reaction when concentration ofAis 0.1 M.(A) 2×10–3 M min–1 (B) 10–3 M min–1 (C) 10–4 M min–1 (D) 5 × 10–4 M min–1

The initial rate of zero order reaction of the gaseous reactionA(g) 2B(g) is 10–2 M min–1. If the initial concentration ofAis 0.1 M, what would be the concentration of B after 60 sec.?

Q.13

(A) 0.09 M (B) 0.01 M (C) 0.02 M (D) 0.002 MQ.14 Consider the following first order competing reactions:

Y k 2 C + DX k1 A+ B and

if 50% of the reaction of X was completed when 96% of the reaction ofYwas completed, the ratio oftheir rate constants (k2/k1) is

(A) 4.06 (B) 0.215 (C) 1.1 (D) 4.65

Q.15 Consider the reaction A B, graph between half life (t1/2) and initial concentration (a) of the reactant

is

t1/2

a

d[A]

Hence graph between and time will bedt

49

Chemical Kinetics

d[A]

d[A]

(A) (B)dt dt

t

d[A]

d[A]

(C) (D)dt dt

t t

Q.16 At certain temperature, the half life period for the thermaldecomposition of a gaseous substance dependson the initial partial pressure of the substance as follows

P(mmHg)

t1 2 (in min.)

500

235

250

950

Find the order of reaction [Given log (23.5) = 1.37 ; log (95) = 1.97; log 2 = 0.30]

(A) 1 (B) 2 (C) 2.5 (D) 3

Q.17 The reactions of higher order are rare because(A) many body collisions involve very high activation energy

(B) many bodycollisions have a very low probability

(C) manybody collisions are not energetically favoured.

(D) many body collisions can take place only in the gaseous phase.

Consider the reaction : A B + C

Initial concentration ofAis 1 M. 20 minutes time is required for completion of20 % reaction.

d[B]

Q.18

If = k[A], then half life (t1/2 ) isdt

(A) 55.44 min. (B) 50 min (C) 62.13 min (D) None of these

Q.19 If decomposition reactionA(g) B (g) follows first order kinetics then the graph ofrate offormation(R) ofB against time t will be

(A) (B) (C) (D)

Q.20 For the first order decomposition of SO2Cl2(g), SO2Cl2(g) SO2(g) + Cl2(g)

a graph of log (a0 – x) vs t is shown in figure. What is the rate constant (sec–1)?Time (min)

2 4 6 8|

10|

(0,0)| | |

-1–

-2–

-3–

(B) 4.6 × 10–1 (C) 7.7 × 10–3 (D) 1.15 × 10–2(A) 0.2

50

log

(a0

–x)

Chemical Kinetics

The rate constant for the forward reactionA(g)l 2B(g) is 1.5 × 10–3 s–1 at 100 K. If 10–5 moles ofA

and 100 moles of B are present in a 10 litre vessel at equilibrium then rate constant for the backward reaction at this temperature is

Q.21

(A) 1.50× 104 L mol–1 s–1

(C) 1.5 × 1010 L mol–1 s–1

(B) 1.5 × 1011 L mol–1 s–1

(D) 1.5 × 10–11L mol–1 s–1

1 1

ReactionA+ B C + D follows following rate law : rate = k[A] 2 [B]2 . Starting with initial conc.

of 1 M ofAand B each, what is the time taken for concentration ofAof become 0.25 M.

Given : k = 2.303 × 10–3 sec–1.

Q.22

(A) 300 sec. (B) 600 sec. (C) 900 sec. (D) 1200 sec.

The reaction A(g) B(g) + 2C (g) is a first order reaction with rate constant 3465 × 10–6 s–1.Starting with0.1 mole ofAin 2 litre vessel, find the concentration ofAafter 200 sec., when thereaction

is allowed to take place at constant pressure and temperature.

Q.23

(A) 0.05 M (B) 0.025 M (C) 0.0125 M (D) None of these

Q.24 Decomposition of H2O2 is a first order reaction.Asolution of H2O2 labelled as "16.8 V" was left open.

Due to this, some H2O2 decomposed. To determine the new volume strength after

2.303 hours, 20 mL of this solution was diluted to 100 mL. 25 mLof this diluted solution was titrated

against 37.5 mL of 0.02 M KMnO4 solution under acidic conditions [Given : STPis 1 atm and 273 K]

The rate constant (in hr–1) for decomposition of H2O2 is :(A) 0.15 (B) 0.30 (C) 0.60 (D) 1.3

The variation of concentration ofAwith time in two experiments starting with two different initialconcentration ofAis given in the following graph. The reaction is represented asA(aq) B(aq).

What is the rate of reaction (M/min) when concentration ofAin aqueous solution was 1.8 M?

1.5

1.2

1

0.8

0.6

Experiment-1Experiment-2

5 10 15 20

time(min.)

(B) 0.036 M min–1

(D) 1 M min–1

(A) 0.08 M min–1

(C) 0.13 M min–1

Q.26 In respect of the equation k =Aexp (– Ea / RT), which one of the following statements is correct?

(A) R is Rydberg's constant

(C)Ais adsorption factor

(B) k is equilibrium constant(D) Ea is the energy of activation

Q.27 Rate of a reaction can be expressed byArrhenius equation as :

k = Ae–E/RT

In this equation, E represents

(A) The fraction of molecules with energygreater than the activation energyof the reaction

(B) The energyabove which all the colliding molecules will react

(C) The energybelow which colliding molecules will not react

(D) The total energy of the reacting molecules at a temperature, T

51

Co

ncen

trat

ion(

M)

Chemical Kinetics

Q.28 The rate constant, the activation energyand theArrhenius parameter (A) ofa chemical reaction at 25°C are3.0 × 10–4 s–1, 104.4 kJ mol–1 and 6.0 × 1014s–1 respectively. The value of the rate constant at T is

(A) 2.0 × 1018 s–1 (B) 6.0 × 1014 s–1 (C) infinity (D) 3.6 × 1030 s–1

Q.29 A first order reaction is 50% completed in 20 minutes at 27°C and in 5 min at 47°C. The energy ofactivation of the reaction is(A) 43.85 kJ/mol (B) 55.14 kJ/mol (C) 11.97 kJ/mol (D) 6.65 kJ/mol

For the first order reactionA— B + C, carried out at 27 ºC if 3.8 × 10–16 % of the reactant molecules exists in the activated state, the Ea (activation energy) of the reaction is [log 3.8 = 0.58]

Q.30

(A) 12 kJ/mole (B) 831.4 kJ/mole (C) 100 kJ/mole (D) 88.57 kJ/mole

Q.31 In a reaction carried out at 400 K, 0.0001% ofthe total number of collisions are effective. The energy ofactivation of the reaction is(A) zero (B) 7.37 k cal/mol (C) 9.212 k cal/mol (D) 11.05 k cal/mol

Q.32 The following mechanism has been proposed for the exothermic catalyzed complex reaction.

k1 AB + I k 2 P +AA+ B I AB

If k1 is much smaller than k2. The most suitable qualitative plot of potential energy (P.E.) versus reaction

coordinate for the above reaction.

Q.33 The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr :

NO(g) + Br2(g) NOBr2(g)

NOBr2(g) + NO(g) 2NOBr(g)

If the second step is the rate determining step, the order of the reaction with respect to NO(g) is

(A) 2 (B) 1 (C) 0 (D) 3

Q.34 Choose the correct set of identifications.

(1)E for

E + S

Ea for

E + S Ea for

(2)

Ea for

ES EPE for

E + S ES

Ea for

EP E + P

Ea for

ES EPEoverall

for S P

(3)Eoverall

for S P

Ea for

ES EPEoverall

for S P

Ea for

EP E + P

E for l

EP E + P

(4)

Ea for

EP E + Poverall

for S PE for

EP E + PEoverall

for S P

Ea for

EP E + P

(A)

ES

(B)

ES

(C)

ES Ea for

EP

(D)

E + S E for

E + S

ES

(E)

ES

52

Chemical Kinetics

SO3 gas is entering the environment at a constant rate of 6.93 × 10–6 gm/L/day due to the emission ofQ.35polluting gases fromthermal power plant, but at the same time it is decomposing &following first orderkinetics with half life of 100 days.Based onabove information select the true statement(s).(A) Concentration of SO3 in Kota is 1.25 × 10–5 M (Assume SO3 present in air reaches steady state) (B) If 103 L of air is passed through 1 L pure water (assuming all SO3 to be dissolved in it) & resulting

solution is titrated against 1 N NaOH solution, 15 ml is required to reach end point.(C)Anindustry is manufacturing H2SO4 at the rate of 980 kg per daywith the useof SO3 inair it should

use 8 × 105 Litre air /day.(D) If SO3 emission is stopped then after 1000 days its concentrations will reduce to ~ 1.2 ×10–8

M.

d[A]For the reactionA B, the rate law expression is 1/2Q.36 = k [A] . If initial concentration of [A] isd t

[A]0, then

2 1/ 2 1/ 2(A0 A )(A) The integerated rate expression is k =t

A (B) The graph of Vs t will be

Kt1 / 2 =(C) The half life period 1/ 22[A]0

t 3 / 4 =(D) The time taken for 75% completion of reaction

Consider the reaction,k

Q.37B

A

C

A, B and C all are opticallyactive compound . If optical rotation per unit concentration ofA, B and C are

60°, –72°, 42° and initial concentration ofAis 2 M then select write statement(s).

(A) Solution will be optically active and dextro after verylong time

(B) Solution will be optically active and levo after very long time

(C) Half life of reaction is 15 min

(D)After 75% conversion ofAinto B and C angle of rotation of solution will be 36°.

Select incorrect statement(s):

(A) Unit of pre-exponential factor (A) for second order reaction is mol L–1 s–1.

(B)Azero order reaction must be a complex reaction.

(C) Molecularity is defined only for RDS in a complex reaction.

(D) Decay constant () of radioactive substance is affected by temperature.

Q.38

In a consecutive reaction systemA E1B E 2 C when E is much greater than E , the yield of BQ.391 2

increase with(A) increase in temperature(C) increase in initial concentration ofA

(B) decreases in temperature(D) decrease in initial concentration ofA

53

[A]0

Chemical Kinetics

Q.40 Which of the following is/are correct statement?(A) Stoichiometry of a reaction tells about the order of the elementary reactions. (B) For a zero order reaction, rate and the rate constant are identical.(C)Azero order reaction is controlled by factors other than concentration of reactants. (D)Azero order reaction is always elementary reaction.

Which of the following statement is incorrect?(A) The order of reaction is the sum of powers of all the concentration terms in the rate equation.

(B) The order of reaction with respect to one reactant is the ratio of the change of logarithm of the rateof the reaction to the change in the logarithm of the concentration of the particular reactant, keeping theconcentrations of allother reactants constant.(C) Orders of reactions can not be fractional.(D) The order of a reaction can onlybe determined from the stoichiometric equation for the reaction.

Q.41

Statement Type Question :

Q.42 Statement-1 :Afractional order reaction must be a complex reaction.

Statement-2 : Fractional order of RDS equals to overall order of a complex reaction.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanationfor statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

Statement-1 : The time of completion of reactions of typeA product (order <1) may be

determined.

Statement-2 : Reactions with order 1 are either too slow or too fast and hence the time ofcompletion can not be determined.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanationfor statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

Statement-1 : Temperature coefficient of an one step reaction may be negative.

Statement-2 : The rate of reaction having negative order with respect to a reactant decreases

with the increase in concentration of the reactant.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanationfor statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

Statement-1 : The overall rate of a reversible reaction may decrease with the increase in

temperature.

Statement-2 : When the activation energy of forward reaction is less than that of backward

reaction, then the increase in the rate of backward reaction is more than that of

forward reaction on increasing the temperature.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanationfor statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

Q.43

Q.44

Q.45

54

Chemical Kinetics

Q.46 Statement-1 : In a reversible endothermic reaction, E of forward reaction is higher than thatact

of backward reaction

Statement-2 : The threshold energy of forward reaction is more than that of backward reaction(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanationfor statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

Statement-1 : Acatalyst provides an alternative path to the reaction in which conversion of

reactants into products takes place quickly

Statement-2 : The catalyst forms an activated complex of lower potential energy, with the

reactants by which more number of molecules are able to cross the barrier per

unit of time.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanationfor statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

Q.47

Paragraph 1 :Oxidation of metals is generally a slow electrochemical reaction involving many steps. These stepsinvolve electron transfer reactions. Aparticular type of oxidation involve overall first order kinetics withrespect to fraction of unoxidised metal (1– f ) surface thickness relative to maximum thickness (T) ofoxidised surface, when metal surface is exposed to air for considerable period of time

dfRate law : = k(1 – f ), where f = x/T,

dtx = thickness of oxide film at time 't'& T = thickness of oxide film at t = Agraph of ln(1 – f ) vs t is shown in the adjacent figure.

Q.48 The time taken for thickness to grow 50% of 'T' is(A) 23.1 hrs (B) 46.2 hrs (C) 100 hrs (D) 92.4 hrs

Q.49 The exponential variation of 'f' with t(hrs) is given by

(A) [1 e3t / 200 ]

Paragraph 2 :

(B) e3t / 200 1 (C) e3t / 200 (D) e3t / 200

k1 1

For a hypothetical elementary reaction wherek 2 2

Initially only2 moles ofAare present.Q.50 The total number of moles of A, B & C at the end of 50% reaction are

(A) 2 (B) 3 (C) 4 (D) 5

Q.51 Number of moles of B are(A) 2 (B) 1 (C) 0.666 (D) 0.333

55

Chemical Kinetics

Paragraph 3 :A reaction is said to be first order if it's rate is proportional to the concentration of reactant. Let us consider a reaction

A(g) a

a – x

B(g)0 x

+ C(g)0 x

At t = 0At time t

dxThe rate of reaction is given by the expression = k(a – x) and integrated rate equation for agiven dt

a1 reaction is represented as k = ln a x

where a = initial concentration and (a – x) = concentrationt

ofAafter time t.

Thermal decomposition of compound X is a first order reaction. If 75% of X is decomposed in 100Q.52min. How long will it take for 90% of the compound to decompose? Given : log 2 = 0.30

(D) 156.66 min(A) 190 min (B) 176.66 min (C) 166.66 minConsider a reaction A(g) 3B(g) + 2C(g) with rate constant 1.386 × 10–2 min–1. Starting with2 moles ofAin 12.5 litre vessel initially, if reaction is allowed to takes place at constant pressure & at298K then find the concentration of B after 100 min.

Q.53

(A) 0.04 M

Paragraph 4 :

(B) 0.36 M (C) 0.09 M (D) None of these

The gaseous reaction : n1A(g) n2B (g) is first order with respect toA. The true rate constant of reaction is k. The reaction is studied at a constant pressure and temperature. Initially, the moles ofAwere 'a' and no B were present.How many moles ofAare present at time, t?Q.54

(A) a · e–kt (B) a · en1kt a · en 2kt (D) a (1 en1kt )(C)

Q.55 If the initial volume of system were v0, then the volume of system after time, t, will ben1v0 n 2v0

(A) (B)n n2 1

n n n n2 · e n1kt

1 · e n1kt

2 2 2 1v v (C) (D)0 0 n1 n1 n1 n1

Q.56 What will be the concentration ofAat time t, if n1 = 1 and n2 = 2?

e kt e kt ] ] [A [A(A) [A0] · e–kt

Paragraph 5 :

–kt(B) (C) (D) [A0] (1–2 · e ) 2 e kt 1 e kt 0 0

For the given sequential reaction

A k1 B k 2 C

the concentration ofA, B & C at any time 't' is given by

k1[A]0 ek1t ek 2tk t[A]t = [A]0 e ; [B]t =1(k k )2 1

[C]t = [A0] – ( [A]t + [B]t )

56

Chemical Kinetics

Q.57 The time at which concentration of B is maximum is

k1 1 k1 1 k1 k 2ln ln(A) (B) (C) (D)k k k 2 k1 k 2 k1 k 2 k 2 k k2 1 2 1

Select the correct option if k1 = 1000 s–1 and k2 = 20 s–1.Q.58

Match the Column :

Q.59 For the reaction of typeA(g) 2B(g)Column-Icontains four entries and column-II contains four entries. Entryofcolumn-I are to bematched withonly one entry ofcolumn-II

Column I Column II

d[B] d[A](A) vs for first order (P)

dt dt

(B) [A] vs t for first order (Q)

(C) [B] vs t for first order (R)

(D) [A] vs t for zero order (S)

57

Chemical Kinetics

Q.60 Column-I and column-II. Entry of column-I are to be matched withone or more than one entries ofcolumn-II and vice versa.

Column I(Graphs reaction A Products)

Column II(Co-ordinates)

(A) (P) ln [A] (y-axis), t (x-axis) (order = 1)

(B) (Q) t1/2 (y-axis), [A0] (x-axis) (order = 1)

(C) (R) r (y-axis), t (x-axis) (order > 0)

(D) (S) r (y-axis), t (x-axis) (order = 0)

(T) t1/2 (y-axis), [A0] (x-axis) (order > 1)

1(U) (y-axis), t (x-axis) (order = 2)

[A]

(V) r (y-axis), [A] (x-axis) (order = 1)

58

Chemical Kinetics

ANSWER KEY

EXERCISE -I2.

4.

6.

8.

10.

12.

14.

16.

2 × 10–2 min–1

0.0207 min–1

15.13 week

0.26 : 1

% decomposition= 67.21%

0.537

0.0025 m

a = b = 1

1.

3.

5.

7.

9.

11.

13.

15.

(a) 6.25 ; (b) 14.3 ; (c) 0%

P = 379.55 mm Hg, t7/8 = 399.96mint

k = 0.022 hr–1

20 min

(a) 43.46kJmol-1, (b) 20.47 hour

k = 0.0327 min–1

4.83 mins

Ef = 6 × 104 J; E = 9.3 × 10 J4b

d(D) k1k3 (A)(B) A1A318. (a) ; (b) Ea = Ea1 + Ea3 – Ea2. A =dt k k A2 3 2

20.

22.

24.

25.

27.

0.0805

55.33 kJ mole–1

21. 161 minutes

23. rate = K[A] [13]2 rate will became 8 times.

K27 = 3.85 × 10–4 sec–1 K47 = 11.55 × 10–4 sec–1, E = 43.78 kJ/mol

It is first order kinetics with k = 8.64 × 10–3

Ea = 2.2 × 104

A = 5.42 × 1010

(a) = 1 × 10–2 mol l–1 s–1 (b) = 5.495 × 10–3 mol–1 l–1 s–1

Rate = 2 × 10–4 mol dm–3 min–1 x = 18.12%

K = 0.02231 min–1 t = 62.07 min

EXERCISE-II

1 d[NO]

26. 2

28.

29.

30.

–4 –1 –1 –4 –1 –1 –4 –1 –11. (i) r =4

= 9 ×10 mol litre sec , (ii) 36 × 10 mol litre sec , (iii) 54×10 mol litre secdt

(i) 7.2 mol litre–1min–1, (ii) 7.2 mol litre–1 min–1

(a) 1 × 10–4 mol L–1 s–1 , (b) 3 × 10–4 mol L–1 s–1

dx

2.

3.

(i) = k[A][B]2, (ii) rate increases by 8 times4.

5.

7.

9.

11.

13.

15.

17.

20.

dt

8.12 × 10–6 Ms–1, 0.012 atm min–1

(a) 0.019 mol L–1 s–1, (b) 0.037 mol L–1 s–1

6.

8.

10.

12.

14.

16.

19.

21.

rate increase by 27 times

2k1 = k2 = 4k3

1/6

K = 0.01 M min–1

6 × 10–9 sec

(i) 36 min., (ii) 108 min.

924.362 sec

3.3 × 104s1

8 31018 sec

(i) 7.2 M, (ii) Think

0.75 M

1.2 hr

(i) 0.0223 min–1, (ii) 62.17 min

expiry time = 41 months

59

Chemical Kinetics

2.303log

122.

24.

25.

26.

k = 23. 11.2%t a

15 min

(a) Third order, (b) r = k[NO]2[H2], (c) 8.85 ×10–3 M sec–1.

r = K[NO]2[Cl2], (c) K = 8 L2mol–2 s–1,(a) order w.r.t NO = 2 and w.r.t Cl2 = 1, (b) (d) rate = 0.256 mole L–1s1

(i) first order (ii) k = 1.308 × 102 min1 (iii) 73%27. 28. (i) Zero order, (ii) K = 5 Pa/s

lln

P329. Zero order 30. k = t 2(P3 P2 )

V1lln

P3 lk = ln31. k = t 32.

(P3 P2 )

4V3

t (2V1 V2 )

lln

lk = ln

r33.

35.

37.

39.

40.

41.

k = t 34.

36.

38.

5(V3 V2 ) t (r rt )

(i) r= K[(CH3)2 O], 0.000428 sec–1

966 min

First order

First order

(a) 90 mm, (b) 47 mm, (c) 6.49 × 10–2 per minutes, (d) 10.677 min.

k1 = 2.605 × 103 min1

(a) first order, (b) 13.75 minutes, (c) 0.716

0.180 atm, 47.69 sec

1

42.

44.

11.45 days

0.1 min–1

[C] 1011x45. 46. = (e – 1)(K1K2 ) [A]t 1 11e

47.

49.

51.

53.

55.

57.

60.

t = 4 min

5 kJ mol–1

55.33 kJ mole–1

rate of reaction increases 5.81× 108 times

t = 20.4 minutes

r = K' [NO]2[Br2]

(d) No, (e) mechanism (a) is incorrect

EXERCISE

48.

50.

52.

54.

56.

58.

0379.75 K

349.1 k

306 k

10.757 k cal mol–1

2000 K

r = K [NO]2 [H2], where K = k2 × K1

-III5. A

12. B

19. C

26. D

33. A

40. A,B,C

47. A

54. B

1. B

8. D

15. C

22. B

29. B

36. A,B,D

43. C

50. B

57. C

2. B

9. C

16. D

23. B

30. C

37. A,D

44. D

51. C

58. C

3. A

10. D

17. B

24. C

31. D

38. A,C,D

45. A

52. C

4. D

11. C

18. C

25. A

32. A

39. A,C

46. C

53. C

6. D

13. C

20. C

27. C

34. B

41. C,D

48. B

55. D

7. D

14. D

21. D

28. B

35. A,D

42. C

49. A

56. B

59. (A) S, (B) R, (C) P, (D) Q 60. (A) P (B) Q,S (C) R,T (D) V

60