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Chemical Kinetics Sri Vyshnavi

Chapter5

Chemical KineticsIntroduction

Chemical Kinetics is the branch of science that deals with rate of reaction, factors affectingthe rate of reaction and reaction mechanism.

Different reactions occur at different rate. In fact a chemical reaction involves

redistribution of bonds breaking in the reactant molecule(s) and making of bonds in the

product molecule(s). he rate of a chemical reaction actuall! depends upon the strength

of the bond(s) and number of bonds to be broken during the reaction. It takes longer time

for the reactant molecules to ac"uire higher amount of energ! which the! do b! collision.

#ence reactions involving strong bond breaking occur at relativel! slower rate while

those involving weak bond breaking occur at relativel! faster rate. $n the basis of rate,

reactions are classified as.

(i) Instantaneous or e%tremel! fast reaction i.e. reactions with half&life of the order of

fraction of second.

(ii) '%tremel! slow reactions i.e. reactions with half&life of the order of !ears.

(iii) eactions of moderate or measurable rate.

Ionic reactions are instantaneous. If a drop of silver nitrate solution is added to a

solution of the chloride salt of an! metal or solution of #Cl, a white precipitate of silver

chloride appears within twinkling of e!e. his is because of the fact that in a"ueous

solution an ionic compound e%ists as its constituent ions. o bond needs to be brokenduring the reaction. #ence reaction takes no time to complete. he half life period of an

ionic reaction is of the order of *+*+s.

3 3Na Cl Ag NO AgCl Na NO+ + + + + + + +

ree radicals being ver! unstable (reactive) due to the presence of unpaired electron,

reactions involving free radicals also occur instantaneousl!. hus, the reactions, are

instantaneous.

3 2 3CH Cl CH Cl Cl + +

3 3 3 3CH CH H C CH

+

5.1 Rate of Reaction

-ith the progress of reaction the concentration of reactants decreases while that

(those) of product(s) increases. If the reaction occurs fastl!, the concentration (mole *) of

reactant will decrease rapidl! while that of product will increase rapidl!. hus, the rate

of reaction is defined as the rate at which the concentration of reactant decreases or

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Sri Vyshnavi Chemical Kinetics

alternativel!, the rate at which concentration of product increases. hat is the change in

concentration of an! of the substance (reactant/product) per unit time during the reaction

is called rate of reaction. If C be the change in concentration of an! reactant or an!

product during the time interval t then the rate ma! given as

ate 0 1C

t

( ve sign applies in the case of reactant whose concentration goes on decreasing with

time and 2 ve sign applies in the case of product whose concentration goes on increasing

with time).

#owever, the rate of reaction is not uniform. -ith the passage

of time the concentration(s) of reactant(s) goes on decreasing

and hence according to aw of 3ass 4ction, the rate of

reaction goes on decreasing. Infact, the rate of reaction

decreases moment to moment. his is shown in the graph of

rate vs time curve of a reaction.

ate varies from moment to moment so rate of reaction has to

be specified at a given instant of time called instantaneous

rate or rate at an! time t. his is as defined below.

T im e

Rate

rinst or tdc

rdt

=

-here dc is the infinitesimal change in concentration during infinitesimal time

interval dt after time t i..e. between t and t 2 dt. he time interval dt being infinitesimal

small, the rate of reaction ma! be assumed to be constant during the interval.

he rate of reaction e%pressed as 1 C/t is actuall! the average rate with time interval

considered.

or the reaction5 66$7 8$62 $6

he rate of reaction at an! time t ma! be e%pressed b! one of the following

2 5 2 2d[N O ] d[NO ] [O ]

, or ddt dt dt

+ +

-here s"uare bracket terms denote molar concentration of the species enclosed.

he above three rates are not e"ual to one another as is evident from the

stoichiometr! of the reaction. or ever! mole of 6$7decomposed, 6 moles $6and */6mole of $6will be formed. $bviousl!, the rate of formation of $6will be four times that of

$6and it is twice the rate of consumption of 6$7. hus, their three rates are interrelated

as,

2 52 2d[N O ]d[NO ] d[O ]

2 4dt dt dt

+ = =

Dividing through out b! 8,

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2 5 2 2d[N O ] d[NO ] d[O ]1 1

2 dt 4 dt dt

= + = +

hus, rate of reaction e%pressed in terms of the various species involved in a

reaction will be e"ual to one another if each of there is divided b! the stoichiometric

constituent of the species concerned. ate divided b! stoichiometric coefficient ma! be

called as rate per mole.he kinetic e%periment has shown that the rate of reaction mentioned above

increases same number of times as the number of times the concentration of 6$7 is

increased. hat is, rate is doubled b! doubling the concentration of 6$7. his ma! be

mathematicall! e%pressed as

ate 96$7:

or ate 0 k 96$7: ;(*)

-here k is the proportionalit! constant called rate constant, velocit! coefficient or

specific reaction rate of the reaction, and is a constant for a given reaction at a given

temperature. '"uation * showing the concentration dependence of rate is called rate law

of the reaction. '%pressing rate in terms of change of concentration of various species,

e"uation ma! be put as

2 5d[N O ]

dt 0 k96$7:

5d[NO ]

dt+ 0 k9$6:

2d[O ]

dt+ 0 k9$6:

when k, kand kare the rate constants of the reaction. hese three rate constants are

inter&related

k ' k ''k '''

2 4= =

hus, for a reaction represented b! the general e"uation

a4 2 b< cC 2 dD

cA B DdCdC dC dC1 1 1 1

a dt b dt c dt d dt

= = + = +

-e have,

!k k k k

a b c d= = =

where

kI, kII, kIIIand kI=

are the rate constants of the reaction when its rate is e%pressed in terms 4,


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he rate of reaction at an! time t is determined in the following wa!,

(i) Concentration of an! of the reactants or products which ever ma! be convenient is

determined at various time intervals.

(ii) hen concentration vs time curve is drawn.

(iii) 4 tangent is drawn at the point p of the curve which corresponds to the time t at

which rate is to be determined.

(iv) he slope of the tangent gives the rate of reaction at the re"uired time as shown

below.

1"# %

&

CRA

Time

R ' ta( ' % ta()1"# % * ' %t OA

OB

t

r ' t a( 't

O AO B

C&

A

tim e

B1

OB

t

(Cand C>denote concentration of reactant and product respectivel!)

?nit of rate 0+(it o co(ce(tratio(

+(it o time0 Concentration time*i.e. mole *@*

5.2 Molecularity

4 chemical reaction that takes place in one and onl! one step i.e., all that occurs in a

single step is called elementar! reaction while a chemical reaction occurring in the

se"uence of two or more steps is called complicated reaction. he se"uence of steps

through which a complicated reaction takes place is called reaction mechanism. 'ach

step in a mechanism is an elementar! step reaction.

he molecularit! of an elementar! reaction is defined as the minimum number of

molecules, atoms or ions of the reactants(s) re"uired for the reaction to occur and is e"ual

to the sum of the stoichiometric coefficient of the reactants in the chemical e"uation of

the reaction. hus, the molecularit! of some elementar! reactions are as mentioned

below.

'lementar! reaction 3olecularit!

>Cl7 A AB A A >Cl 2 Cl6 *

#62 I6 A AB A A 6#I * 2 * 0 6

eactions with molecularit! e"ual to one, two, three etc., are called unimolecular,

bimolecular, termolecular, etc., respectivel!.

4 complicated reaction has no molecularit! of its own but molecularit! of each of the

steps (elementar! reactions) involved in its mechanism.



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or e%ample, consider the reactionE 6$ 2 6#662 6#6$, which is complicated

reaction and takes place in the se"uence of following three steps.

(i) $ 2 $ A AB A A 6$6(fast and reversible)

(ii) 6$62 #6 r-d-.- 6$ 2 #6$(slow)

(iii) 6$ 2 #6 62 #6$ (fast)

he molecularit! of each step in the mechanism is two, so what we sa! is that the

reaction takes in the se"uence of three steps each of which is bimolecular. here is

another view also. 4ccording to which molecularit! of a complicated reaction is taken to

be e"ual to the molecularit! of the slowest step i.e. rate determining step (r.d.s.) in the

mechanism.

or e%ample, the reaction

2/(Cl

2ROH HCl RCl H O+ +

is said to be unimolecular nucleophilic substitution (@*) since the reaction occurs in the

se"uence of the following three steps and the slowest step i.e. r.d.s. is unimolecular.

(i)2R OH HCl R OH Cl )a.t*

+ +A AB A A

(ii)2 2R O H R H O ).lo0*

+A AB A A

(iii) R Cl RCl )a.t*

+

eactions of higher molecularit! (molecularit! F ) are rare. his is because a

reaction takes place b! collision between reactant molecules and as number of reactant

molecules i.e. molecularit! increases the chance of their coming together and collidingsimultaneousl! decreases.

6.3 Order of a Reaction

he speed of a chemical reaction, in general depends on the concentration of reacting

species of the reaction. 4n earl! generaliGation in this regard is due to Hulberg and

-aage. his generaliGation is known as law of mass action and is stated as follows.

he rate of a chemical reaction is proportional to the product of effective

concentration (acting masses) of the reacting species, each raised to a power that is e"ual

to the corresponding stoichiometric number of the substance appearing in the chemical

reaction.

hus, for a general reaction a4 2 b< cC 2 dD ;(6a)

-e have

94:a9


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If the rate of a reaction is determined e%perimentall!, it is found that e"uation (6b) is

not alwa!s applicable. #owever, the e%perimental results can be fitted to satisf! a relation

of the t!pe of e"uation (6b) where the e%ponents ma! or ma! not be e"ual to the

respective stoichiometric coefficients. In general, we ma! write the rate as

0 k 94:a9


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hese facts clearl! indicate that the rate e"uation cannot be predicted from the form of

the stoichiometric e"uation for the overall reaction. #ence, the rate e"uation must be

determined e%perimentall!. In some cases, the order of the reaction is meaningless. or

e%ample, for the h!drogen&bromine reaction, the reaction is of first&order with respect to

h!drogen gas but it would be impossible to assign the order with respect to bromine and

to h!drogen bromide. hus, the concept of order of reaction has no meaning if a rate law

does not have the form as given in e"uation (6c).

In the reaction between h!pochlorite and iodide ions, though $#does not appear in the

overall reaction, !et it appears in the denominator of the rate law. his indicates that $#

acts as an inhibitor. @imilarl!, the reaction between acetone and iodine does not involve

#2in the overall reaction, but it appears in the numerator of the rate law. his shows that

#2acts as an accelerator or a catal!st.

5.4 Difference Between Order and Molecularity

(i

) $rder is an e%perimentall! propert! while molecularit! is the theoretical propert!.(ii) $rder concerns with kinetics (rate&law) while molecularit! concerns with mechanism.(iii) $rder ma! be an! number, fractional, integral or even Gero whereas molecularit! is

alwa!s an integer e%cepting Gero.Example

eactions ate aw $rder

3 4CH CHO CH CO + ate 9C#C#$:/6 *.7

3 2 2

1 3NH N H

2 2 +

ate 9#:+ +

2 22H H + ate 9#I:+

i.e. ate 0 k

+

ote that a Gero order reaction is one in which concentration of reactant remainsthroughout constant and as such rate of reaction remains throughout constant e"ualto the rate constant.

(iv) $rder ma! change with change in e%perimental condition while molecularit! cant.

Example

i.omeriatio(H

2C CH3

he reaction follows *st

order kinetics at high gas pressure and 6nd

order kinetics atlow gas pressure of c!clopropane.

5.5 Kinetics of First Order Reaction

4 first order reaction is one whose rate varies as *stpower of the concentration of the

reactant i.e. the rate increases as number of times as the number of times the

concentration of reactant is increased.

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et us consider a unimolecular first order reaction represented b! the general e"uation.

4 LLLLLLLLLLL >roduct

4 % 0 + ; t 0 +

4 % % 0 % ; t 0 +

he initial concentration of 4 is a mole *and its concentration after an! time t is (a

%) mole *. his means during the time interval t, % mole *of 4 has reacted.

he rate of reaction at an! time t is given b! the following first order kinetics.

d)a *)a *

dt

or

d)*

dt(a %)

or d

dt0 k (a %)

(da

#dt

= M a has a given value for a given e%pt.)

where k is the rate constant of the reaction.

d kdta

=

his is differential rate e"uation and can be solved b! integration.

dk dt

a =

or ln (a %) 0 k.t 2 C ;(*)

he constant C can be evaluated b! appl!ing the initial condition of the reaction i.e. when

t 0 +, % 0 +. >utting these in e"uation *, we get

C 0 lna

>utting the value of C in e"uation *, we get

ln (a %) 0 k.t. lna or1 a 2-3#3 a

k ( logt a t a

= =

l ;(6)

If 94+: and 94: be the concentrations of reactant at Gero time and time t, respectivel! then

'". 6 ma! be put as

#[A ]1k (t [A]

= l

his is the integrated rate e%pression for first order reaction. ?nit of the rate constant

1 ak (t a = l

kt 0 lna ln(a %)

a % 0 aekt

% 0 a(* ekt)

he differential rate e%pression for nth order reaction is as follows.

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(d k)a *dt

= or ( (d co(ce(tratio(

k)a * dt )co(ce(tratio(* time

= =

0 conc. *n. time*

If concentration be e%pressed in mole *and time in minute, then

k 0 (mole *)*nmin*

or Gero order reactionE n 0 + and hence k 0 mole *min*

or *

st

order reactionE n 0 * and hencek 0 (mole *)+min*0 min*

or 6ndorder reactionE n 0 6 and hence

K 0 (mole *)*min*0 mole** min*

he rate constant of a first reaction has onl! time unit. It has no concentration unit. his

means the numerical value of k for a first order reaction is independent of the unit in

which concentration is e%pressed. If concentration unit is changed the numerical value of

k for a first order reaction will not change. #owever, it would change with change in time

unit. @a!, k is N.+ *+ min then it ma! also be written as * *+8 s* i.e. numerical

decrease N+ times if time unit is changed from however to minute or from minute to

second.

#alf&time or half&life period of a first order reaction

he half&time of a reaction is defined as the time re"uired to reduce the concentration of

the reactant to half of its initial value. It is denoted b! the s!mbol t */6. hus,

-hena

,2

= t 0 t*/6

>utting these in e"uation 6 mentioned above, we get

1 2 1 2 1 2

2-3#3 a 2-3#3 2-3#3k log log 2 #-3#1#3at t t

a2

= = =

( log 6 0 +.+*+)

t*/60#-3

k;()

@ince k is a constant for a given reaction at a given temperature and the e%pression

lacks an! concentration term so from e"uation it is evident that half&time of a * storder

reaction is a constant independent of initial concentration of reactant. his means if we

start with 8 mole

*

of a reactant reacting b! first&order kinetics and after 6+ minute it isreduced to 6 mole *will also be 6+ minute. hat is, after 6+ minutes from the start of

reaction the concentration of the reactant will be 6 mole *, after 8+ minutes from the

start of reaction of concentration is * mole *. 4fter N+ minutes from the start of reaction

the concentration of the reactant will be reduced to +.7 mole *. In order words, if during

6+ minutes 7+O of the reaction completes, then in 8+ minute P7O, in N+ minute Q7.7O of

the reaction and on will complete as shown with following plot.

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raction left after n half&lives 0(

1

2

Concentration left after n lives(

( #

1a a

2

=

It is also to be noted that e"uation helps to calculatet*/6 or k with the knowledge of k or t*/6. 4 general

e%pression for t*/6is as follows 1 2 ( 11ta

1##

65

5#

25

# 2# 4# # "#

C)7*

t )mi(*

-here n 0 order of reaction.

5.6 ra!"ical Re!resentation@ince for nth order reaction

(d k)a * ,dt

= or (d

k-cdt

=

@o, from this it is evident that a plot ofd

dtvs(a %)nwill be straight line passing

through the origin and will have its slope e"ual to k, the rate constant of reaction.hus, for a first order reaction one will get

straight line passing through the origin ofd

dti.e.

rate of reaction be plotted against a % as showbelow.

aking logarithm of the above e"uation

8 ta(

ddt

)a % *

dlog

dt0 n log (a %) 2 log k

his e"uation demands that a plot ofd

logdt

vs log (a

%) will be straight line of the slope e"ual to n, order ofreaction and intercept e"ual to log k. or first order reactionthis slope will be * as shown.

B1

d

d tlo g

B O lo g )a % *(

# 45 9 i ( 1 ,

i -e - OA O B

'"uation 6 ma! be rearranged as

log (a %) 0k

t loga2-3#3

+

hus, a plot of (a %) vs. t will be straight line with slope e"ual tok

2-3#3 and

intercept e"ual to log a, if the reaction is of first order.

IllustrationsIllustration 1

In a reaction with initiall! +.*6 3, the concentration of reactant is reduced to +.+N 3 in *+

hour and to +.+ 3 in 6+ hour.

(a) -hat is the order of reactionR

(b) -hat is rate constant R

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Solution

4ssuming *storder,2-3#3 a

8 log)a *

=

or case *2-3#3 #-12

8 log1# #-#

=

0 +.+NS hr*.

eaction is of *storder and rate constant 0 +.+NS hr*.

Illustration 2

or a reaction 4 >roducts, it is found that the rate of reaction doubles ifconcentration of 4 is increased four times, calculate order of reaction.

Solution

ate 0 K 9eactant:n if 9eactant: 0 "

rate 0 r*

r*0 K9a:n ;(*) if 9eactant: 0 8 "

6r*0 K98a:n;(6) rate 0 6r*

Dividing

(

1 12 4

=

, 1(2

=

Practice Exercise*. he rate constant for the reaction

2 3CO OH HCO +

In the a"ueous solution is 8 *+litre mol*sec*. Calculate the n umber of mole ofC$6and $#used up per second when 9C$6: and 9$#: are *+Nand *+* mol litre*.

4lso predict the mole of #C$formed per second.

6. or a reaction 4 >roducts, it is found that the rate of reaction doubles ifconcentration of 4 is increased four times, calculate order of reaction.

Answers

1. 8 *+*+mol *sec*. 2.1

(2

=

5.# $alf%&ife of a nt"order Reaction

et us venture on to find out the t*/6for a nthorder reaction where ( 1 .

#1 2

#

[A ]t2

(

( (

[A ] #

d[A] d[A] d[A]k[A] kdt k dt

dt [A] [A]

= = =

##

# # 2

A[A ] 1 (( (

1 2 1 2

[A ] A2

[A][A] d[A] kt kt

1 (

= =



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1 ( 1 (1 (

1 ( # ## 1 2 1 2

[A] [A]1 1[A] kt 1 kt

1 ( 2 1 ( 2

= =

( 1

1 2( 1#

1[1 2 ] kt

)1 (*[A ]

=

( 1

1 2( 1#

)2 1*kt

)( 1* [A ]

=

(order n *)

herefore for a nth order reaction, the half life is inversel! related to the initial

concentration raised to the power of (n *).

oteE It can be noted that for a Gero order reaction #1 2[A]

t2k

.

5.' ()a*!les Of First+Order Reaction4 large number of reactions e%hibiting first&order kinetics are known. 4 few e%amples are

listed below

(i) 3 3 3 3 3 3 2 )CH * COOC)CH * 2CH COCH C H +

(ii) 2 5 2 21

N O )g* 2NO )g* O )g*2

+

(iii) 3 2 3 2 2 14)CH * CHN NCH)CH * N C H= +

(iv) 2 2 2 2:O Cl )g* :O )g* Cl )g* +

(v)

NO2NO2

NO2

NO2NO2

NO2

; CO2)g*

COOH

(vi) 2 2 2 21

H O ).o l( -* H O) * O )g*2

+l

(vii) 4ll radioactive deca!s.

here are reactions in which more than one species is involved in the rate

determining step, but the order of the reaction is one. @uch reactions are known as

pseudo&unimolecular reactions and the! involve solvent molecules or a catal!st as

one of the reacting species.

Examples of this type of reactions are:

(viii) 4cid h!drol!sis of an esterE C#C$$C6#72 #6$ H+

C#C$$# 2 C6#7$#

(ix) Inversion of cane sugarE C*6#66$**2 #6$ H+

CN#*6$N(glucose) 2 CN#*6$N(fructose)

(x) Decomposition of benGene diaGoniumE 5 2 5 2C H N NCl H O C H OH N HCl

+ + + +

5., -oncentrations Re!laced By Ot"er uantities In First%

Order Inte/rated Rate &aw



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he rate constant for a first&order reaction, it is necessar! to determine the ratio of the

concentration at two different times. Muantities proportional to the concentration terms

ma! be substituted in these e"uations, since the proportionalit! constants cancel. -e

describe below a few t!pical cases where concentration terms are replaced b! other easil!

measurable "uantities.

Case I:

et there be a firstorder reaction of the t!pe, 4 < 2 C. et us assume that all the

three species are gases. -e are re"uired to calculate the value of rate constant based on

the following data

ime + t >artial pressure of 4 >+ >t

Strategy:

-e know that for a first&order reaction,

#

t

A1k (

t A=

4t t 0 +, the partial pressure of 4 is > +and at time t, it is > t. 4ccording to the ideal gas law

(& RT

!= or

( &

! RT=

# t

t # t t

& &( (+, it must also be the partial pressure of 4 at t 0 +,

since in the beginning onl! the reactant will be present. et the decrease in pressure of 4

due to reaction till time t be %. hus, at time t, the total pressure would be

(>+ %) 2 % 2 % 0 >+2 %

>+2 % 0 >t

% 0 >t >+

(>+ %) 0 >+ (>t >+) 0 6>+ >t

#

# t

&1k (

t 2& &=

Case III:

ime + t >artial pressure of 4 2 < 2

C

>t >

Calculate the e%pression of rate constant.

Strategy:

et the initial pressure of 4 be >+and the pressure of 4 reacted till time t be %.

( ) ( ) ( )

#

#

# #

A g B g C g

At time # & # #

At time t &

At time # & &

+==

= is the time when the reaction is complete.

4t time 0 t, the total pressure will be

(>+ %) 2 % 2 % 0 >t

>+2 % 0 >t

% 0 >t >+

4t time 0 , the total pressure will be

>0 + 2 >+2 >+0 6>+ 5 6>+0 >5 #&

&

2

=

% 0 >t >+0 >t&

2

# t t

& &)& * & & &

2 2

= =

t

&1k (

t 2)& & *

=



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Case IV:

ime + t >artial pressure of < 2 C >t >

Calculate the e%pression of rate constant.

Strategy:

et the initial pressure of 4 be p+and the amount of pressure reacted till time t be %.

4t time infinit!, the total pressure of < and C will be

>+2 >+0 6>+0 >

#

&&

2

=

4t time t, the total pressure of < and C will be

% 2 % 0 6% 0 >t

% 0 t&

2

t

& 21k (t & 2 & 2

=

t

&1k (

t & &

=

.

Case V:

A B C +

ime >artial pressure of

6 >

ind k.

ow let us assume that 4, < and C are substances present in a solution. rom a solution of

4, a certain amount of the solution (small amount) is taken and titrated with a suitable

reagent that reacts with 4. he volume of the reagent used is = *at t 0 + and =6at t 0 t.

ime + =olume of reagent =* =6

he reagent reacts onl! with 4. ind k.

It can be understood that the volume of the reagent consumed is directl! proportional to

the concentration of 4. herefore the ratio of volume of the reagent consumed against 4

at t 0 + and t 0 t is e"ual to 94: +/94:t

# 1t 2

[A] !1 1k ( (

t [A] t != = .


4 < 2 C



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ime + =olume of reagent =* =6

he reagent reacts with 4, < and C. ind k.

Soltion-e can use a similar logic for solving this one. hat is

1

1

1 1

A B C

At t # ! # #

At t t !

At t # ! !

+== =

where % is the volume of the reagent for those moles of a which have been converted into< and C =*2 % 0 =6 % 0 =6 =* =* % 0 6=* =6

#t 1 2

[A]1 1 !k ( (

t [A] t )2! ! *= =

his is true onl! if we make an assumption in the beginning of this problem. he

assumption is that the Tn factor of 4, < and C with the reagent is same.his can be made clear as follows. et ! moles of 4 be converted to < and C. If !moles of 4 re"uire % ml of reagent and if the Tn factor of 4 is n *with the reagent then thee"uivalents of 4 converted are n*! and in % ml of reagent n*! e"uivalents of the reagentare present. ow since the same ! moles of < and C are formed and we find the samevolume of reagent used for < and C, it can be calculated that ! moles of < and C alsocontain n*! e"uivalent each, which implies that Tn factor of < and C are also Tn *.

5.10 euential Or -onsecutie Reaction

eaction in which a reactant gives product, which further goes to give another product,

are called se"uential or consecutive reaction.

1 2k kA B C

he differential rate e"uation are

ate of disappearance of 4, 1d[A]

k [A]dt

= ;(*)

ate of change of


17/57


rom e"uation (Qe), we get

1k t2 1 #

d[B]k [B] k [ A] e

dt

+ =

Integration of this e"uation is not possible as we are unable to separate the two

variables, 9


18/57


1 2)k k *t1

2

k1 1 e

k

=

tma%01

1 2 2

k1(

)k k * k ;(S)

3a%imum concentration of assive molecules A AB A A 4ctive molecules, # 0 2ve

(iv) Concept of energ! of activation ('a)

he e%tra amount of energ! which the reactant molecules (having energ! less than

the threshold) must ac"uire so that their mutual collision ma! lead to the breaking of

bond(s) and hence the reaction, is known as energ! of activation of the reaction. It is

denoted b! the s!mbol 'a. hus,

'a0 hreshold energ! 4ctual average energ!

'ais e%pressed in kcals mole*or kV mole*

he essence of 4rrhenius heor! of reaction

rate is that e%ists an energ! barrier in the

reaction path between reactant(s) and

product(s) and for reaction to occur the reactant

molecules must climb over the top of the barrier

which the! do b! collision. he e%istence of

=a1=a2

Reacta(t H ' H % H ' =a ' =a& R 1 2

HR

H&

=(t>al?@ )H*

&rodct.

T>re.>old e(t>al?@ or e(erg@

&rogre.. o, Reactio()or reactio( coordi(ate*

Sri Vyshnavi SIIT CHEMISTRY 1$


20/57


energ! barrier and concept of 'a can be

understood from the following diagram.

RH 0 @ummation of enthalpies of reactants

&H 0 @ummation of enthalpies of products# 0 'nthalp! change during the reaction

'a*0 'nerg! of activation of the forward reaction ()

'a60 'nerg! of activation of the backward reaction (

#off e"uation of thermod!namics which is as follows

?

2

d(8 H

dT RT

=

If k*and k6be the rate constant of and utting these in the above e"uation we get,

1 2a a1 2

2 2

= =d(k d(k

dT dT RT RT =

@plitting into two parts



21/57


1a1

2

=d(k/

dT RT= + (or )

2a2

2

=d(k/

dT RT= + (or


22/57


@ince a=

2-3#3Rand log4 both are constants for a given reaction. @o from e"uation P it is

evident that a plot of log k vs.1

Twill be a straight line of the slope e"ual to

a=

2-3#3R and

intercept e"ual to log4 as shown below.

a=

2-3#3R 0 tan0 tan (*Q+ ) 0 OA

OB

aOA

= 2-3#3R OB

=

log4 0 $4

1 " # %

A

log 8

O

1 T

hus, from this plot 'aand 4 both can be determined accuratel!.

If k* be the rate constant of a reaction at two different temperature * and 6

respectivel! then from e"uation (8), we ma! write

log k*0a

2

= 1- log A

2-3#3R T+

log k60a

2

= 1- log A

2-3#3R T+

@ubtracting

a2

1 1 2

=k 1 1log

k 2-3#3R T T

=

;(7)

-ith the help of this e"uation it is possible to calculate ' aof a reaction provided, rate

constants of reaction at two different temperatures are known. 4lternativel! one can

calculate rate constant of a reaction at a given temperature provided that rate constant ofthe reaction at some other temperature and also 'aof the reaction is known.


he half time of first of decomposition of nitramide is 6.* hour at *7UC.

#6$6(a".) 6$(g) 2 #6$(l)

If N.6 g of #6$6is allowed to decompose, calculate.

(i) time taken for #6$6is decompose SSO

(ii) volume of dr! 6$ produced at this point measured at @>.

Solution

M 1#2-3#3 a

t log8 )a *

=



23/57


If1 a

t , resence of a catal!st lowers the

energ! of activation b! P7O. -hat will be effect on rate of reaction at 6+UC5 other

things being e"ualR

Answers

3. 1.308 102minute1 4. 5.206 kt 103minute1

5. 43.85 kJ mol1, 93.85 kJ mol1 6. 2.35 1013



26/57


5.12 -atalyst4 catal!st is a substance, which increases the rate of a reaction without itself being

consumed at the end of the reaction, and the phenomenon is called catal!sis. here are

some catal!sts which decrease the rate of reaction and such catal!sts are called negative

catal!st. $bviousl!, the catal!st accelerating the rate will be positive catal!st. #owever,

the term positive is seldom used and catal!st itself implies positive catal!st.

Catal!st are generall! foreign substances but sometimes one of the product formedma! act as a catal!st and such catal!st is called Xauto catal!stY and the phenomenon is

called auto catal!sis. hermal decomposition of KCl$ is found to be accelerated b! the

presence of 3n$6. #ere 3n$6(foreign substance) acts as a catal!st.

6KCl$ 2 93n$6: 6KCl 2 $6 2 93n$6:3n$6can be received in the same composition and mass at the end of the reaction.

In the permanganate titration of o%alic acid in the presence of bench #6@$8(acid medium),

it is found that the titration in the beginning there is slow discharge of the colour of

permanganate solution but after sometime the discharge of the colour become faster. his

is due to the formation of 3n@$8during the reaction which acts as a catal!st for the same

reaction. hus, 3n@$8is an Xauto catal!stY for this reaction. his is an e%ample of auto

catal!st.

6K3n$82 #6@$82 7#6C6$6 K6@$82 Q#6$ 2 *+C$6

,eneral characteristics of catalyst(i) 4 catal!st does not initiate the reaction. It simpl! fastens it.

(ii) $nl! a small amount of catal!st can catal!se the reaction.

(iii) 4 catal!st does not alter the position of e"uilibrium i.e. magnitude of e"uilibrium

constant HU. It simpl! lowers the time needed to attain e"uilibrium. his means if areversible reaction in absence of catal!st completes to go to the e%tent of P7O till

attainment of e"uilibrium, and this state of e"uilibrium is attained in 6+ minutes thenin presence of a catal!st also the reaction will go to P7O of completion before the

attainment of e"uilibrium but the time needed for this will be less than 6+ minutes.

(iv) 4 catal!st drives the reaction through a different route

for which energ! barrier is of shortest height and

hence 'ais of lower magnitude. hat is, the function of

the catal!st is to lower down the activation.

'a0 'nerg! of activation in absence of catal!st.

'a 0 'nerg! of activation in presence of catal!st.

'a 'a 0 lowering of activation energ! b! catal!st.

&rodct.

=a

a&-=-

HR

H&

Reactio( Coordi(ate

Reacta(t.

If k and kcatbe the rate constant of a reaction at a given temperature , and 'aand 'aare the activation energies of the reaction in absence and presence of catal!st,

respectivel!, the

a

='aRTcat

= RT

k Ae

k Ae

=

a a) = = ' * RTcatk

Aek

=

2! SIIT CHEMISTRY SriVyshnavi


27/57


@ince 'a, 'a is 2 so kcatF k. the ratio catk

kgives the number of times the rate of

reaction will increase b! the use of catal!st at a given temperature and this depends upon

'a 'a*. Hreater the value of 'a 'a*, more number of times kcatis greater than k.

he rate of reaction in the presence of catal!st at an! temperature *ma! be made

e"ual to the rate of reaction in absence of catal!st but for this sake we will have to raise

the temperature. et this temperature be 6this= RTa 2

a 1= ' RT ee =

or a a

1 2

= ' =

T T= ;(N)

5.13 Radioactiity

4ll radioactive deca! follow *storder kinetics and this is where the similarit! ends. his will

be e%plained in a short while.

-e measured the rate of reaction in chemical kinetics based on the rate of change of

concentration of reactant or products.


28/57


1 2

#-3t =

Note:et us start with *+ nuclei. If the half life is 7 minutes, then at the end of first 7

minutes, number of nuclei would be 7. ow what would be the number of nuclei after ne%t

minutesR -ill it be 6.7 or 6 or R -e can clearl! see that it cannot be called as half life.

his dilemma can be overcome b! understanding that all formula relating to kinetics are

onl! valid when the sample siGe is ver! large and in such a large sample siGe, a small

difference of +.7 will be insignificant.

he fact that radioactive deca! follows the e%ponential law implies that this

phenomenon is statistical in nature. 'ver! nucleus in a sample of a radionuclide has a

certain probabilit! of deca!ing, but there is no wa! to know in advance which nuclei will

actuall! deca! in a particular time span. If the sample is large enough that is, if man!

nuclei are present the actual fraction of it that deca!s in a certain time span will be ver!

close to the probabilit! for an! individual nucleus to deca!. o sa! that, a certain

radioisotope has a halflife of 7 hr., then, signifies that ever! nucleus of this isotope has a

7+ percent chance of deca!ing in ever! 7 hr. period. his does not mean a *++ percent

probabilit! of deca!ing in *+ hr. 4 nucleus does not have a memor!, and its deca!

probabilit! per unit time is constant unit it actuall! does deca!. 4 half life of 7 hr. implies a

P7O probabilit! of deca! in *+ hr., which increases to QP.7O in *7 hr, to S.P7O in 6+ hr,

and so on, because in ever! 7 hr. interval the probabilit! of deca! is 7+ percent.

$/erage(ife 0ime4verage life time is defined as the life time of a single isolated nucleus. et us imagine a

single nucleus which deca!s in * second. 4ssuming * second time interval to be ver! small

the rate of change of nuclei would be */* (because d 0 * and dt 0 *). -e can also see

that sincedN

Ndt

= , for a single isolated nucleus 0 *,

dN

dt

= . herefore in this present

case 0 *.

ow let us assume the same nucleus deca!s in 6 seconds, we can see thatdN

dt

i.e.,

is e"ual to */6. Zou will also notice that in the * stcase the nucleus survived for * second

and in the second case it survived for 6 second. herefore the life time of a single isolated

nucleus is1

.

a1

t =

.

2# SIIT CHEMISTRY SriVyshnavi


29/57


5.14 ctiity

4ctivit! b! definition is the rate of deca! of a radioactive element. It is represented as T4

and is e"ual to .

nuclei represented b!dN

dt

. his is because

dN

dt

talks about the overall change in the

number of nuclei in a given instant of time while activit! onl! talks about that change

which is deca!. or e%ample if !ou go out to a market with s. 7+ in !our pocket and !ou

spend s. 6+ in 7 minutes then !our rate of change of mone! in the wallet is s. 8/ min

and in fact the rate of spending the mone! is also s. 8/min. #ere !ou can see both are

same.


30/57


0 * curie (M * Ci 0 .P *+*+dps)

0 .P *+8utherford ( * rd 0 *+Ndps)

he @i unit of activit! is dps of

4a

Example 2

he rate of a reaction increases 8&fold when concentration of reactant is increased *N times. If therate of reaction is 8 *+Nmole *@*mole *when concentration of the reactant is 8 *+8, therate constant of the reaction will be(a) 6 *+8mole*/6*/6@* (b) * *+6@*

(c) 6 *+8mole*/6, */6@* (d) 67 mole* min*

Solution

ate co(ce( , ate 0 k co(ce(

1 2 4 1 2 2

Rate 4 1# 4 1#k

)co(ce(* )4 1# * 2 1#

= = =

0 6 *+8mole*/6*/6@*

4a

Example 34 catal!st lowers the activation energ! of a reaction from 6+ kV mole* to *+ kV mole*. hetemperature at which the uncatal!sed reaction will have the same rate as that of the catal!sed at6PUC is(a) *6 UC (b) 6P UC (c) 66P UC (d) 2 6 UC

Solution

a

1 2 2

== 'a 1# 2#

T T 3## T= = =

60 N++ K 0 6PUC 4"

Example 4

4 first order reaction is QP.7O complete in an hour. he rate constant of the reaction is(a) +.+8N min* (b) +.+NS h* (c) +.+NS min* (d) +.+8N h*

Solution

or QP.7O change there will be three half&changes. #ence half&time of the reaction will be1

3h 0

#2#

3= min, since half&time of a first order reaction is a constant independent of initial

concentration.



31/57


*++O 2# mi( 2# mi( 2# mi(

)( reacted*5#7 257 12-57

1 2

#-3 #-3k

t 2#= = 0 +.+8N min*

4a

Example 5he half&life a first order reaction is 68 hours. If we start with *+ 3 initial concentration of the

reactant then conc. 4fter SN hours will be(a) N.67 3 (b) *.67 3 (c) +.*67 3 (d) +.N67 3

Solution

umber of half&lives 0

424

=

Concentration remaining after SN hrs 0 41#

)2*0 +.N67 3

46

Example 6During the particular reaction *+O of the reactant decompose in one hour 6+O in two hours +O in

three hours and so on. he unit of the rate constant is(a) #our* (b) mol*hour* (c) mol *hour* (d) mol hour*

Solutionhis reaction is a Gero order reaction. 4c

Example 7he temperature coefficient of a reaction is 6, b! what factor the rate of reaction increases whentemperature is increased from +UC to Q+UC.(a) *N (b) 6 (c) N8 (d) *6Q

Solution

"#

3#

k

k

0 670 6

4"

Example 8

he rate constant, the activation and 4rrhenius parameter of a chemical reaction at 67UC are *+8s*, *+8.8 kV mol*and N *+*8s*respectivel!. he value of the rate constant at is(a) 6 *+*Qs* (b) N *+*8 (c) (d) .N *++s*

Solution

K 0 4e'a/when value is N *+*8s*

4"

Example 9If t*/6does not change with initial concentration, then order of the reaction is(a) Gero (b) second (c) first (d) third

Solution

1 2 ( 1

1t

a K 0 4e'a/when

4c



32/57


Example 104 substance undergoes a first order decomposition. he decomposition follows two parallel first

order reaction as

A

81

82

B

Che percentage distribution of < and C are(a) Q+O < and 6+O C (b) P7O < and 67O C (c) S+O < and *+O C (d) N+O < and 8+O C

Solution

or parallel reaction1

2

8 [B]

8 [C]=

4

5

[B] 1-14 1# 3

[C] 13-" 1#

= =

O of < 0 P7 and percentage of C 0 67 4"

Example 11

4 tangent drawn on the curve obtained b! plotting concentration of product (mole *) of a firstorder reaction vs. time (min) at the point corresponding to time 6+ minute makes an angle to +Uwith concentration a%is. #ence the rate of formations of product after 6+ minutes will be(a) +.7Q+ mole *min* (b) *.P6 mole *min* (c) +.6S+ mole *min* (d) +.QNN mole *

min*

Solutionangent makes an angle of +U with concentration a%is so it must make an angle of N+U with thetime a%is(ve direction). he slope of the tangent will be tan N+U i.e., *.P6. 4"Example 12

or reaction 4 products, if is found that the rate of reaction increases 8fold when

concentration of 4 is increased *N times keeping the temperature constant. he order of reactionisR(a) 6 (b) * (c) * (d) +.7Solution

he rate data suggests the rate law as follows

ate A i.e., r 4*/6

order 0 +.7 46

Example 13

he thermal decomposition of acetaldeh!deE C#C#$ C#82 C$, has rate constant of *.Q

*+mole*/6*/6min*at a given temperature. #ow would 3d[CH CHO]

dt

will change if concentration

of acetaldeh!de is doubled keeping the temperature constantR(a) will increase b! 6Q6Q times (b) will increase b! **.*times(c) will not change (d) will increase b! 8 times

Solution?nit of the rate constant mole */6*/6. min*/6 suggests that the reaction obe!s kinetics of *.7

order.ate 0 k 9C#C#$:/6



33/57


$r1 1

3 2 1 3 2

3

Rate mole mi(k

[CH CHO] )mole *

= = 0 3ole*/6, */6. min*

@o, b! doubling the concentration of acetaldeh!de the rate will increase 6*/7i.e., 6.Q6Q times. 4a

Example 14

he reaction 5 6$ $6, is assigned the following mechanism.

(I) $ A AB A A $62 $(II) $2 $ .lo0 6$6

he rate law of if the reaction will, therefore be(a) r 9$:69$6: (b) r 9$:69$6:* (c) r 9$: (d) r 9$: 9$6:6

Solution@tep II, being r.d.s.ate of overall reaction 0 rate of @tep II 0 KII9$: 9$:>utting the value of 9$: from the e"uilibrium of @tep I,

ate 02

C 3

2

8 8 [O ]

[O ]

4"

Example 15

or an endothermic reaction where # represents the enthalp! of the reaction, the minimum valuefor the energ! of activation will be(a) ess than # (b) Gero (c) more than # (d) e"ual to #

Solution

=

=R

H=

&

Reactio( coordi(ate

4c

Example 16he rate law for a reaction between the substances 4 and < is given b!ate k 94:n9


34/57


he half&life of a reaction is halved as the initial concentration of the reactant is doubled. he orderof reaction is

(a) +.7 (b) * (c) 6 (d) +

Solution

*/6( 1

( 1

1 2 1 # 2 # 2

( 1 ( 1

# 1 2 2 # 1 # 1

)t * [A ] [A ]1,

[A ] )t * [A ] [A ]

= =

( 1

t 2a

t 2 a

=

or 6 0 (6)n*

or n * 0 * or n 0 6. 4c

Example 18he rate constant of the reaction at temperature 6++ K is *+ times less than the rate constant at8++ K. -hat is the activation energ! of the reactionR(a) *Q86.8 (b) S6*.6 (c) 8N+.N (d) 6+.

Solution

4t *0 6++ K, if k*0 k, thenat 60 8++ K, k60 *+ k

a=1# k 4## 2##

logk 2-3#3 R 4## 2##

= or 'a0 S6*.6 4"

Example 19

In the reaction 64 2 < 46


35/57


46



36/57


S"3ecti/e 0ype

Example 1

4 h!drogenation reaction is carried out at 7++ K. If the same reaction is carried out in the presenceof a catal!st at the rate, the temperature re"uired is 8++ K. Calculate the activation energ! of thereaction if the catal!st lowers the activation energ! b! 6+ kV mol*.

Solution

@uppose activation energ! in the absence of catal!st is 'aand that in the presence of catal!st it is'b. hen

K 0 a b= 5##R = 4##R A e Ae =(as k is same in both the cases).

#ence,a b= =

5## R 4## R= or

b a

4= =

5=

%is the partial pressure of [."ser/ation

!o.0ime 4inmintes

#x4in mm of-g

* + Q++6 *++ 8++ 6++ 6++

(i) -hat is the order of reaction with respect to [R

(ii) ind the time for P7O completion of the reaction(iii) ind the total pressure when pressure of [ is P++ mm of #g.

Solution

(i) as pressure of [ is changing with time, it cannot be a Gero order reaction. et us now check itfor *storder.

4t t 0 *++ min,

o

t

&2-3#3 2-3#3 "##log log

1## & 1## 4##= 0 N.S6 *+min*

4t t 0 6++ min,

2-3#3 "## 2-3#3k log log 4

2## 2## "##= =

0 N.S6 *+

min*

4s k comes out to be constant, hence it is a reaction of *storder.

(ii)657

2-3#3 1##t log

k 1## 65=

0 3 12-3#3

log4-32 1# mi(

0 2++ min.

3! SIIT CHEMISTRY SriVyshnavi


37/57


(iii)

2E )g* 3F)g* 2 /)g*

(itial "## mm # #

atr time t "## 2? 3? 2?

+

-hen pressure of [ is P++ mm,Q++ 6p 0 P++ or p 0 7+ mm

otal pressure 0 (Q++ 6 p) 2 p 2 6p0 Q++ 2 p 0 Q++ 2 7+ 0 &5+ mm.

Example 3

@how that for a first order reaction time re"uired for SSO completion is twice for the time re"uiredfor the completion of S+O of the reaction.

Solution

SSO0 1#2-3#3 1##

logk 1##

;(*)

S+O0 1#2-3#3 1##

logk 1## #

;(6)


38/57


Solutionhe unit of rate constant suggests the second order for reaction. hus,

ate of reaction 0 K9$: 9$: or +.*7 0 *.7 *+P#-6

[O]#-#"21 2" 6#

3

( & #-6,or O

! RT #-#"21 2" 6#

= = 9$: 0 6.87 *+8

n/= for $ 0 6.87 *+8

&

RTfor $ 0 6.87 *+8

>+ 0 6.87 *+8+.+Q6* 6SQ0 7.SS *+atm0 7.SS *+PN+ mm 0 8.7 mm

4lso, rate of reaction 0 2d[O ]1 1

2 dt 2= rate of formation of $6

ate of formation of $60 6 +.*7 0 +.+ mol *t*.

Example 6

(a) In a reaction with initiall! +.*6 3, the concentration of reaction is reduced to +.+N 3 in *+ hour

and to

+.+ 3 in 6+ hour.

(i) -hat is order of reactionR (ii) -hat is rate constantR

(b) he rate of a first order reaction is +.+8 mol litre*s*at *+ minute and +.+ mol litre*at 6+

minute

after initiation. ind the half life of the reaction.

Solution

(a) 4ssuming Ist order, i.e., K 0 1#2-3#3 a

log

t )a *

a 0 +.*6 3

(a %) 0 +.+N 3 7or case I : 1#2-3#3 #-12

8 log1# #-#

= 0 +.+NS hr* t 0 *+ hr

a 0 +.*6 3

(a %) 0 +.+ 3 7or Case II: 11#2-3#3 #-12

8 log #-# >r 2# #-#3

= = t 0 6+ hr

eaction is of * order and rate constant k 0 +.+NS hr*

(b) ate 0 K 94:

+.+8 0 K949*+and +.+ 0 K94:6+

1#

2#

[A] #-#4 4[A] #-#3 3

= =

4lso1#

2#

[A]2-3#3t log

8 [A]= when t 0 *+ min

2-3#3 41# log

8 3=

3# SIIT CHEMISTRY SriVyshnavi


39/57


2-3#3 48 log

1# 3= 0 +.6QQ min*

1 2

2-3 #-3t

8 #-#2""= = 0 68.+N min.

Example 7

he following data represent for the decomposition of #8$6in a"ueous solution.

ime in minute *+ *7 6+ 67 =olume of 6(in

m)

N.67 S.+ **.8

+

*.N7 .+7

(a) @how that reaction is of *st order (b) Calculate velocit! constant.

Solution

4 2).* 2)g* 2 ) *NH NO N 2H O

ole at t # a # #

ole at t t )a * 2

+==

l

the volume of 6formed at an! time is proportional to the amount of #8$6decomposed in that

time. 4t t 0 2N

! 0 .+7 m a .+7

I 4t t 0 *+2N

! 0 N.67 % N.67

II 4t t 0 *72N

! 0 S.+ m % S.+

III 4t t 0 6+2N

! 0 **.8+ m % **.8+

I= 4t t 0 672N

! 0 *.N7 m % *.N7

ow use, 1#2-3#3 a

k logt a

=

Case I

1#

2-3#3 33-#5k log

1# 33-#5 -25= k 0 6.+ *+6min*

@imilarl!, calculate k for each case. he values of k come almost constant and thus showing thatreaction is * order. 4verage of all k value gives rate constant.

Exercise I

,eneral 0ype47ill in the "lan8s9 0re or 7alse9 $ssertion : ;eason5

rue!"alse*. 3olecularit! of a reaction can be +, *, 6, , etc.

6. here is no difference between rate law and law of mass action.

. 4rrhenius e"uation is ' / (ak 4 e= .

8. $rder of reaction is an e%perimental parameter.

7. or a Gero order reaction t*/6is proportional to initial concentration

"ill in t#e $lan%s

Sri Vyshnavi SIIT CHEMISTRY 3$


40/57


*. he inversion of cane sugar is a LLLLLL reaction though its molecularit! is LLLLLL

6. 4 reaction is said to be of LLLLLL if the rate is entirel!

. he h!drol!sis of eth!l acetate in LLLLLL medium is a LLLLLL order reaction.

8. -hen 6d%

k94:9


41/57


(a) )2-3#3 3log 2*

1#

(b)

)2-3#3 2log3*

1#

(c) *+ 6.+ 6 log (d) *+ 6.+

log 6

8. or the reaction 4 2 < >roducts, it is found that the order of 4 is 6 and of < is in therate e%pression. -hen concentration of both is doubled the rate will increase b!

(a) *+ (b) N (c) 6 (d) *N

7. he rate law of the reaction 4 2 6< >roduct is given b! d9>roduct:/dt 0 K94:6

. 9


42/57


*S. or reaction 462 6< 64

(a) 6 (b) 6.* (c) *.S* (d) 6.

6. wo first order reactions have half lives in the ratio of E 6. Calculate the ratio of time intervals

t* E t6, the time t* and t6are the time period for 67O and P7O completion for the first and

second reaction respectivel!.

(a) +.** E * (b) +.86+ E * (c) +.6P E * (d) +.**S E *

68. he rate constant is numericall! the same for three reactions of first, second and third order.

-hich of the following is correctR

(a) If 94: 0 *, then r*0 r60 r (b) If 94: \ *, then r*F r6F r(c) If 94: F *, then rF r6F r* (d) all of these

67. he reaction, 6$ 2


43/57


(c) Catal!st lowers the energ! of activation of the forward reaction without affecting the energ!

of activation of the backward reaction

(d) Catal!st does not affect the overall enthalp! change of the reaction.

+. 'th!lene is produced b! the reaction

4 " 2 4C@clob ta(e

C H 2C H

he rate constant of the decomposition of c!clobutane is 6.8Q *+8sec*. In what time, will themolar ratio of the eth!lene to c!clobutane in the reaction mi%ture attain value * R

(a) 6P.67 minute (b) 6Q.67 minute (c) 67 minute (d) 6+ minute

*. he inversion of cane sugar proceeds with a constant half life of 7+ minute at p# 0 7, for an!

concentration of sugar. #owever, if p# 0 8, the half life changes to 7+ minute. he rate law

e%pression for inversion of cane sugar can be written as

(a) r 0 k 9sugar:69#2:+ (b) r 0 k 9sugar:*9#2:* (c) r 0 k 9sugar:*9#2:+ (d) r 0 k 9sugar:+

9#2:*

6. he order of a reaction and rate constant for a chemical change having log t7+Ovs log 94:+curve

as would be

log t5#7

45 9

Dog [A]#

(a) +, */6 (b) *, * (c) 6, 6 (d) +, *

. #alf life (t*) of the first&order reaction and half life (t6) of the second&order reaction are e"ual.

#ence, assuming 94:+to be same, the ratio of the rate at the start of the reaction (assuming

same initial concentration of the reactant in both the reactions) would be

(a) * (b) 6 (c) +.NS (d) *.88

8. he activation energies of two reactions are1 2a a

= a(d = . If the temperature of the reacting

s!stems is increased from *to 6, predict which of the following alternatives is correctR

(a) 1 2

2

k ' k '

k k= (b) 1 2

1 2

k ' k '

k k> (c) 1 2

1 2

k ' k '

k k< (d) 1 2

1 2

k ' k '2

k k (b)1 2a a

= =< (c)1 2a a

= == (d) none of these

N. he decomposition of a substance follows first order kinetics. Its concentration is reduced to

*/Qth

of its initial value in 68 minutes. he rate constant of the decomposition process is

(*) */68 min* (b) +.NS/68 min* (c) 12-3#3 1log mi(

24 "

(d) 12-3#3 "log mi(

24 1

P. he chemical reaction, 6$ $6proceeds as follows

$ A AB A A $62 $ ;..(fast)

$ 2 $ 6$6;.(slow)he rate law e%pression should be(a) r 0 k9$:6 (b) r 0 k9$:69$6:* (c) r 0 k9$: 9$6: (d) unpredictable



44/57


Q. or a reaction of first order, it takes 6+ minutes for the concentration to drop from * 3 to +.N3. the line re"uired for the concentration to drop from +.N 3 to +. 3 will be(a) F 6+ min (b) \ 6+ min (c) e"ual to 6+ min (d) infinit!

S. or the decomposition of 6$7at a particular temperature, according to the e"uations

66$7 8$6 2 $66$7 6$62 ^ $6the activation energies are '*and '6respectivel!. hen(a) '*F '6 (b) '*\ '6 (c) '*0 6'6 (d) '*0 '6

8+. he concentration of the reactant 4 in the reaction 4 < at different times are given belowConcentration

4=0ime 4=intes

+.+NS ++.+76 *P+.+7 8+.+*Q 7*

he rate constant of the reaction according to the correct order of reaction is

(a) +.++* 3/minute (b) +.++* minute* (c) +.++* minute/3 (d) +.++* 3*minute*

Exercise III

2"3ecti/e 0ype4=ore than 2ne Choice Correct5

*. he half&period for the decomposition of ammonia on tungsten wire, was measured for

different initial pressures > of ammonia at 67UC. hen

> (mm #g) ** 6* 8Q P *6

+ (sec) 8Q S6 6*+ 6

+

76

7(a) Gero order reaction (b) irst order reaction

(c) rate constant for reaction is +.**8 seconds (d) rate constant for reaction is *.*8seconds.

6. or an isomerisation [email protected]

A B in gaseous phase, the e"uilibrium constant at N+UC

is .N+. atm, of 4, the pressure became +.67 > atm in 8+ minutes. he values of individual rate

constants for the forward (kf) and backward (kb) reactions are

(a) kf0 .8NN *+6min* (b) kf0 S.N6Q *+min*

(c) kb0 S.N6Q *+min* (d) kb0 .8NN *+6min*

. or the reaction 64 2 < C with the rate lawd[C]

dt0 k 94:*9


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> < #< K< 'ing comprehension careflly an6 ans>er the follo>ing %estions.

#SE?=(EC($; ;E$C0I!S

here are man! reactions which obe! a first order rate e"uation although in realit! the! are bi& or

ter&molecular. 4s an e%ample of these ma! be taken the decomposition of carbon!l sulfide in water,

namel!,

C$@ 2 #6$ C$62 #6@4ccording to the law of mass action this reaction should order with the rate dependent on the

concentration of both the carbon!l sulfide and the water. 4ctuall! however, the rate is found to be

first order with respect to the carbon!l sulfide and independent of the water. eactions e%hibiting

such behaviour are said to be pseudo&molecular.

he pseudo&unimolecular nature of this reaction is e%plainable b! the fact that water is present insuch e%cess that its concentration remains practicall! constant during the course of the reaction.?nder these condition b % 0 b, and the rate e"uation becomes

2

dk

dt= (a %) b

$n integration this leads to

2 1#2-3#3 ak bk log

t a = =

which is the e"uation for a first order reaction. It is evident, however, that the new constant k is notindependent of the concentration, as is the case with true first order constants, but ma! var! with bif the latter is changed appreciabl!, -hen such is the case, the true constant k6can be obtainedfrom k b! dividing the latter b! b.

>seudo&molecular reactions are encountered whenever one or more of the reactants remainconstants during the course of an e%periment. his is the case with reactions conducted in solventswhich are themselves one of the reactants, as in the decomposition of carbon!l sulfide in water, orin the esterification of acetic anh!dride in alcohol.

Sri Vyshnavi SIIT CHEMISTRY 4"


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(C#C$)62 6C6#7$# 6C#C$$C6#72 #6$4gain, this is also true of reactions sub_ect to catal!sis, in which case the concentration of the

catal!st does not change. he decomposition of diacetone alcohol to acetone in a"ueous solution iscatal!sed b! h!dro%!l ions, with the rate proportional to the concentration of the alcohol and that ofthe base. @ince the concentration of the base does not change within an! one e%periment,however, the rate e"uation reduces to one of first order with respect to the alcohol.

ot all chemical reactions proceed to a stage at which the concentrations of the reactants become

vanishingl! small. #ere we consider the kinetics of such reactions.

et a reaction be represented in general terms b! the scheme

1

1

k

kA B

A A A B A AA

where k*and k*represent the rate constants for the forwards and reverse reactions, respectivel!.

he e"uilibrium constant for this reaction ma! be written as

K 0 9


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or d94:t/dt 0 (k*2 k*)1

t #

1 1

k[A] [A]

k k

+

;(6)

ow, from (*) we have 9


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(c) d

dt 0 k 9a %: 9b %: (k F +) (d)

d

dt 0 k 9a %:

9b %: (k F +)

6. #ow could the progress of this reaction be best monitoredR

(a) b! monitoring the color of the reaction mi%ture (b) b! titration of3 with

h!po

(c) b! precipitation of I

with 4g2

(d) b! monitoring the change in pressure. -hich mechanism is consistent with the facts given about the reaction rate e"uationR

(a) mechanism (I) (b) mechanism (II) (c) both (I) and (II) (d) neither (I) nor (II)

8. or the reaction 2 22 2 3 4 2: O : O 2

+ +

(i)2

2 32 d[: O ]d[ ] 1

dt 2 dt

= (ii)2

2 32 d[: O ]d[ ] 2dt dt

=

(iii)2

2 32 2 d[: O ]d[ ] d[ ]2dt dt dt

= (iv)2

2 d[: O ] 1 d[ ]

dt 2 dt

=

Correct option is

(a) onl! (i) (b) (i) J (iv) (c) (ii) J (iv) (d) onl! (iii)

7. Statement 4$: In the reaction, 2 22 2 3 4 2: O : O 2 + + , the rate of disappearancethiosulphate ions is twice the rate of disappearance of I6.

Statement 4B:he rate of disappearance of I6is one half the rate of disappearance of @6$6

ions.

(a) @tatement (4) J @tatement (


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(a) a=

RT

= (b) a=

( RT

= (c) a= RT e= (d) 4n! of these

*atc#in( +pe ,uestions

*. 3atch the followingColmn I Colmn II

(a) Decomposition of #6$6 (p) *+t*/6

(b) D+QK

6SQ

t

k K (") *storder

(c) 4rrhenius e"uation (r) emperature coefficient

(d) tSS.SO (s) a6 6 *

* * 6

'k log

k 6.+

=

(t) 6 to

(a) (a&"), (b&r,t), (c&s), (d&p) (b) (a&r), (b&s), (c&"), (d&p)

(c) (a&s), (b&r), (c&p), (d&") (d) (a&"), (b&r), (c&p), (d&s)

6. Colmn I Colmn II

(a) f

t

K tK+ (p) >hotosensitiGer

(b) atio of number of molecules reacting (") *.+*Nin given time to number of "uanta of lightabsorbed in same time

(c) In photochemical dissociation of #6 to # (r) emperature coefficientatoms #g is used as

(d) he increase in rate of reaction on raising (s)t/*+

(@) t/*+the temperature b! *+UC due tocollision fre"uenc!

(a) (a&"), (b&r,t), (c&s), (d&p) (b) (a&r), (b&s), (c&"), (d&p)

(c) (a&s), (b&r), (c&p), (d&") (d) (a&"), (b&r), (c&p), (d&s)

. Colmn I Colmn II

(a) #62 I66#I (p) $rder of reaction 0 *.7(b) CN#76Cl 2 #6$ CN#7$# 2 62 #Cl (") $rder of reaction 0 (c) 6eCl2 @nCl6 @nCl82 6eCl6 (r) $rder of reaction 0 *(d) #62


52/57


S"3ecti/e 0ype

*. he progress of the reaction with time is presented in the figure. Determine

4 n


53/57


he reaction is carried out in a constant volume container at 7++UC and has a half life of *8.7

minutes. Initiall! onl! dimeth!l ether is present at a pressure of +.8+ atmosphere. -hat is the

total pressure of the s!stem after *6 minutesR 4ssume ideal gas behavious.

P. he rate constant for the forward reaction is *.6 *+min*at +UC. he enthalpies of

activation and reaction are Q7.8Q kV mole*and *+P.* kV mol*, respectivel!. he rate

constant for the reverse reaction is *.6 ` *+8min* at 7+UC. what is its value at 8+UC.

Q. he time re"uired for *+O completion of a first order reaction at 6SQ K is e"ual to that

re"uired for its 677 completion at +Q K. If the pre e%ponential factor for the reaction is .7N `

*+Ss*, calculate its rate constant at *Q K and also the energ! of activation.

S. 4 sample of uraninite, a uranium containing mineral, was found on anal!sis to contain +.6*8 g

of >b6+Nfor ever! gram of uranium. 4ssuming that the lead all resulted from the radioactive

disintegration of the uranium since the geological formation of the uraninite and that all

isotopes of uranium other than 6Q? can be neglected, estimate the date when the mineral

was formed in the earths crust. he half life of 6Q? is 8.7 ` *+S!.

*+. In nature a deca! chain series starts with S+h66

and finall! terminates at Q6>b6+Q

. 4 thoriumore sample was found to contain Q ` *+7ml of helium at @> and 7 ` *+P g of h66. ind

the age of the ore sample assuming the source of helium to be onl! due to the deca! of h66.

4lso assume complete retention of helium within the ore. (#alf life of h660 *.S ` *+*+Z).



54/57


Exercise VI

II0.EE #ro"lems

$ S"ecti/e #ro"lems

*. he gas phase decomposition of dimeth!l ether follows first order kinetics

C# $ C#(g) C#8(g) 2 #6(g) 2 C$(g)

he reaction is carried out in a constant volume container at 7++UC and has a half&life of *8.7

minutes. Initiall! onl! dimeth!l ether is present at a pressure of +.8+ atmosphere. -hat is the

total pressure of the s!stem after *6 minutes R 4ssume ideal gas behaviour.

6. 4t Q+UC, the half&life period for the first order decomposition of #6$6is N+ min. he energ! of

activation of the reaction is 6++ kV mol*. Calculate the time re"uired for P7O decomposition at

87+UC.

. he rate constant for the first order decomposition of a certain reaction is described b! the

e"uation

log k (s*) 0 *8.8 8*.67 *+ K

(i) -hat is the energ! of a activation of this reactionR

(ii) 4t what temp. will its half&life period be 67N minutes R

8. he time re"uired for *+O completion of a first order reaction alt 6SQ K is e"ual to the re"uired

for its 67O completion at +Q K. If the pre&e%ponential factor for the reaction is .7N *+Ss*,calculate its rate constant at *Q K and also the energ! of activation.

7. he rate constant of a reaction is *.7 *+Ps*at 7+UC and 8.7 *+Ps*at *++UC. 'valuate the4rrhenius parameters Aand Ea.

N. or the reaction 6$7(g) 0 6 $6 (g) 2 +.7 $6(g), calculate the mole fraction of 6$7(g)decomposed at a constant volume and temperature, if the initial pressure is N++ mm #g and

the pressure at an! time is SN+ mm #g. 4ssume ideal gas behaviour.

P. he rate constant for an isomerisation reaction. 4 < is 8.7 *+ min*. If the initialconcentration of 4 is * 3, calculate the rate of reaction after * h.

Q. 4 h!drogenation reaction is carried out at 7++ K. If the same reaction is carried out in the

presence of a catal!st at the same rate, the temperature re"uired is 8++ K. Calculate the

activation energ! of the reaction if the catal!st lowers the activation energ! b! 6+ kV mol*.

S. or the reaction, 4 2 < >roducts, the following initial rates were obtained at various giveninitial concentrations

S. !o. D$ DB ;ate 4mol (1

sec

1

*. +.* +.* +.+7

6. +.6 +.* +.*+

. +.* +.6 +.+7

-rite rate law and find the rate constant of the above reaction.

*+. 4t constant temperature and volume, [ decomposes as



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6[ (g) Z(g) 2 6 W (g)

>%is the partial pressure of [.

"ser/ati

on !o.

0ime 4in

mintes

#x4in mm of

-g

*. + Q++

6. *++ 8++. 6++ 6++

(i) -hat is the order of reaction with respect to [ R

(ii) ind the time for P7O completion of the reaction.

(iii) ind the total pressure when pressure of [ is P++ mm of #g

$ns>ers

Exercise - rue!"alse*. alse 6. alse . alse 8. rue 7. rue

"ill in t#e $lan%s*. first order, two 6. Gero order . acidic, first 8. first, second, third7. +.NS

Assertion an& 'eason*. (a) 6. (c) . (b) 8. (b) 7. (c)

Exercise - /nl+ /ne /ption is correct

(e/el I*. (a) 6. (c) . (a) 8. (c) 7. (b)N. (b) P. (a) Q. (a) S. (b) *+. (a)**. (a) *6. (a) *. (c) *8. (b) *7. (d)*N. (c) *P. (b) *Q. (c) *S. (a) 6+. (d)

(e/el II6*. (a) 66. (c) 6. (a) 68. (8) 67. (a)6N. (b) 6P. (a) 6Q. (c) 6S. (c) +. (a)*. (b) 6. (a) . (c) 8. (b) 7. (d)N. (d) P. (b) Q. (b) S. (d) 8+. (a)

Exercise - *ore #an /ne #oice orrect*. (a, c) 6. (a, c) . (a, b, c) 8. (a, b, c) 7. (a, b)N. (a, b, d) P. (a, b, c, d) Q. (a, b) S. (a, b) *+. (a,c)**. (c, d) *6. (a, b) *. (a, b, c, d) *8. (b, c) *7. (b, c)*N. (a, d) *P. (a, b, d) *Q. (a,b,c) *S. (a,b,c) 6+. (a,b)

Exercise ) Passa(e ) 1

*. (a) 6. (b) . (a)

Sri Vyshnavi SIIT CHEMISTRY


56/57


Passa(e ) 2

*. (a) 6. (f) . (f)

Passa(e ) 3

*. (a) 6. (b) . (b) 8. (b) 7. (a)

Passa(e ) 4*. (b) 6. (c) . (a) 8. (b) 7. (b)

*atc#in( +pe ,uestions

*. (a ) 6. (c ) . (d ) 8. (b )

Exercise - Suectie +pe

1. (i) 6, (ii) +.* mol *hr*

2. (a) a 0 6, b 0 +, (b) 8 `*+6 mol * sec*, (c) *.N `*+mol *sec*

3. (i) *7+ mm/#g, (ii) S mm/#g (iii) .8*7 `*+6min*, (iv) 6+. min

F. irst order5 *.+* 2*+8sec*

5. P.SN `*+

). +.P8QQ atm

'. *.6* ` *+7 min*

*. S.66 ` *+8sec*, *Q. Kcal mol*

&. *.8 ` *+S!rs.

1+. 8.QS ` *+S!rs.

Exercise - -EE eel ProlemSection - A

*. +.P8S atm 6. 6+.N min . (i) 6S.8 kV mol* (ii) NNS K8. k 0 S. *+8s* 7. (i) 66.+* kV mol* (ii) 7.8 *+*+s*

N. +.8 P. .88 *+mol *min*

Q. *++ kV mol* S. +.7 s* *+. (i) N.S6 *+min* (ii) 6++ min. (iii) S7+ mm

2.1............................................................'ate o 'eaction........................................................................................1

2.10........................Seuential /r onsecutie 'eaction......................................................................................16

2.11...........Eect o emperature on t#e 'eaction 'ateArr#enius #eor+:.....................................................19

2.12........................................................................atal+st......................................................................................26

2.13................................................................'a&ioactiit+......................................................................................27

2.14.........................................................................Actiit+......................................................................................29

2.15..........................................................Speciic Actiit+......................................................................................29

2.2...................................................................*olecularit+........................................................................................4

2.3......................................................./r&er o a 'eaction........................................................................................5

2.4...............;ierence $etween /r&er an& *olecularit+........................................................................................7

2.5..................................


57/57


2.9.....oncentrations 'eplace& $+ /t#er ,uantities n"irst-/r&er nte(rate& 'ate aw................................12

Answers..........................................................................55#emical