Chemical Kinetics

Reconstruct Your Chemistry With Prince Sir

By Princ

e Sir

Chemical Kinetics is the branch of chemistry which deals with the rate of change of conecentration fromintial state to final state under non-equilibrium conditions and the factors affecting the rate of reaction.

RATE OF REACTIONThe rate of reaction is defined as the change in conecentration of reactants or products in unit time.Rate of reaction =

takenTimeProductsreactants/ofionconcentratinChange

Rate can be defined in two ways :(i) Average Rate:- The rate of reaction measured over a time interval isdefined as the average rate of reaction.

Rav. = tC

ttCC

12

12

(ii) Instanteneous Rate :- The rate of reaction at any given instant is defined asthe instanteneous rate .

“It is simply the slope of concentration time graph at that particular instant”

r =av r =inst– [R] d[P]–[C – C ]2 1

[t – t ]2 1t

[R]0[P]

C1

C2

t2 time time

conc

entr

atio

n

conc

entr

atio

n

t1

slope =dt

=

(a) (b)

Dependence of rate on StoichiometryFor any Reaction n1A +n2 B n3C + n4D

dtDd

ndtCd

ndtBd

ndtAd

n][1][1][1][1ReactionofRate

4321

For reactants negative sign is used because the concentration of reactantsdecreases with time.

For products positive sign is used because the concentration of productsincreases with time.

Specific care should be taken while dealing with this expressionhere ,


By Princ

e Sir

dtAd

n][1

1 = rate of reaction where as dt

Ad ][ = rate of disappearence of A

dtCd

n][1

1 = rate of reaction where as dt

Cd ][ = rate of appearence of C

Example 1. For the reaction:4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

The concentration of NO changes from 1.8M to 5.4 M in 100 secondsthan calculate -(i) Rate of disappearence of Ammonia (ii) Rate of reaction

Solution - Rate of appearence of NO = 11-2 secLmol106.3100)8.14.5(][

dt

NOd

Now from the rate expression

dtOHd

dtNOd

dtOd

dtNHd ][

61][

41][

51][

41ReactionofRate 223

Rate of disappearence of NH3 =11-23 secLmol106.3][][

44][ dt

NOddtNOd

dtNHd

(ii) Rate of reaction = 11-32 secLmol109106.341][

41 dt

NOd

LAW OF MASS ACTIONAccording to this law the rate of a chemical reaction is directlyproportional to the product of active masses of reactants raised to thepower their stochiometric coefficients.For any reaction n1A + n2B n3C + n4D

21

21

][][][][

nn

nn

BAKRateBARate

This expression is called rate expression or rate law.Active Mass - For aqeous phase active mass can be replaced byconcentration (Molarity).For gaseous reactants active mass can be replaced by either concentrationor partial pressure.

Here ‘K’ is called the rate constant or specific rate of the reaction.Greateris the value of rate constant more will be the rate of reaction.

Note :-1. The reagent which is present in excess is not included in therate law, since the change in its concentration is insignificant.2. Intermediates are not involved in the rate law they has to beeleminated if are present in the rate law.


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e Sir

ORDER AND MOLECULARITYMOLECULARITY :- The number of reactant molecules which collidewith each other in a chemical reaction to form products is called themolecularity of the reaction.On the basis of Molecularity reactions can beclassified as

Br2 2Br - UnimolecularH2 + I2 2HI - Bimolecular2NO + O2 2NO2- trimolecular

Note :- 1. Molecularity can niether be zero , negative , fractional andinfinite.2. Moleuclarity is a theoretical quantity.3. Molecularity of a reaction cannot be greater than 3 becausemore than 3 molecules can’t collide effectively to form products.4. Molecularity is independent of pressure and temperature.5. Molecularity of a single step reaction i.e. simple reaction is thenumber of molecules participating in the reaction.

6.But for a complex multistep reaction the molecularity of each stepis different and there is no meaning of overall molecularity of thereaction

ORDER It is defined as -

“The sum of the powers of all the concentration terms included in the rate law”n1A + n2 B products

21 ][][ nn BAkrate Order of reaction = (n1+n2)

Note:- 1. Order of a reaction is an experimentally determined quantity.2. Order can be zero, negative and fractional but cannot be infinity.3. Order depends on pressure and temperature.4. For a multistep complex reaction the overall order of thereaction is determined by RDS(Rate Determinind Step).

Example 2 : For the reaction 2H2 + 2NO N2 + 2H2O, the followingmechanism has been suggested:(i) 2NO N2O2 equilibrium constant K1 (fast)(ii) N2O2 + H2 2k N2O+ H2O (slow)(iii) N2O + H2 3k N2 + H2O (fast)Determine the order of the reaction

Sol:- The slowest step is the rate RDS(Rate Determining Step)N2O2 + H2 2k N2O+ H2O (slow)

the rate law for RDS rate = k[N2O2] [H2] ............(1)


By Princ

e Sir

But since N2O2 is an inetermediate it cannot be included in rate law so ithas to be elemintaedSo for the equilibrium reaction 2NO N2O2 equilibrium constant K1

21222

221 ][][][

][ NOKONNOONK

replacing [N2O2] in equation ..(1)Rate = KK1[NO]2[H2] = K’ [NO]2[H2]So the order of the reaction = 2+1 = 3

The molecularity of each step is as follows:(i) 2 (ii) 2 (iii) 2

Significance of the Order Of ReactionThe order of a reaction is different with repect to each reactant. The orderwith respect to a particular reactant tells that how the rate of reaction willchange on changing the concentration of that reactant. For example considerthe rate law

rate = k [A]2[B]3So if concentration of A is doubled than rate of reaction will become fourtimes as the order with respect to A is 2.if concentration of B is doubled than rate of reaction will become 8 times asthe order with respect to B is 3.

Types of reactions on the basis of orderZERO ORDER REACTIONS

Consider a general reactionA Products

at t=0 [A0]at t=t [At]Since the order iszero so the rate of reaction can be written as

Rate = dt]d[At =k[At]0

]d[At =kdtOn integrating [At] = -kt + c

At t=0, [At] = [A0] , So c = [A0]

So for zero order reaction [A]t = [A]0 – kt

Half Life“the time taken for the concentration of reactants to decrease to half ofits initial value”

t1/2 = kA2

][ 0


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e Sir

[A]

conc.

time

Rate

time

Examples :(i) Photochemical Reactions

H2(g) + Cl2 (g) h 2HCl (g)(ii) Decomposition of substances over metal surface

N2O(g) SurfacePthot . N2 (g) + 2

1 O2 (g)

2NH3 (g) surfaceWorMo N2 + 3H2

(iii) Enzyme catalysed reactionSubstrate(S) )(EEnzyme product(P).

FIRST ORDER REACTIONConsider a general reaction

A Productsat t=0 [A0] = aat t=t [At]= (a-x)

Since the reaction is of first order the rate of reaction can bewritten as

Rate = dt]d[At =k[At]1

On integrating we get , ][][ln1 0

tAA

tk

][][log303.2 0

10tA

Atk or )(log303.2

10 xaa

tk

The Exponential form of 1st order equation is [A]t = kteA 0][

Half -life : t1/2 = k693.0

For a first order reaction half life is independent of initial concentrationof reactant.


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e Sir

t1/2

a

The equation ][][ln1 0

tAA

tk , can be written as

ln[Ao] - ln[At] = kt or ln[At] = -kt + ln[Ao]The graph between ln[At] and t will be a straight line with negative

slope, andslope = rate constant (k)

In [A]t

t

log [A]10 t

t

If the graph is plotted between ln[At] and t the slope will be equal to k andif the graph is plotted between log10[At] and ‘t’ the slope will be 303.2

k .

Examples:(i) Radioactive disintegration is a first order reaction.(ii) C12H22O11 + H2O Inversion

hydrolysiscatalysedH

C6H12O6 + C6H12O6. (glucose) (fructose)

(iii) Mineral acid catalyzed hydrolysis of esters.(iv) Decomposition of H2O2 in aqueous solution.

Methods To Calculate the rate constant of First Order Reaction1. By Titration:- In this method an unknown solution (say ‘A’) is titratedagainst a standard solution of known concentration(Say ‘B’). The volume ofstandard solution required at different time intervals is measured fromwhich the concentration of unknown solution can be calculate .

Vo = Volume of standard solution used at time t = 0Vt = Volume of standard solution used at time tV = Volume of standard solution used at completion of titration i.e.

at t =


By Princ

e Sir

So, V - Vo Initial amount of A present

V - Vt Amount of A present at any time t

][][ln1 0

tAA

tk

k =

t

oVVVV

t log303.2

2. By Measuring angle of rotation:-In this method the total angle of rotation of the sample is measured atdifferent time intervals from which we can calculate the amount present atany given time. Consider a reaction:

A BIfr0 = initial angle of rotation of the sample (when only A is present)rt = angle of rotation after time t (when both A and B are present)r = angle of rotation after completion of reaction (when only B is present)

r0- r Initial amount of A presentr - rt Amount of A present at any time t

k =

rrrr

t t

0log303.2

3. By Measuring Pressure

A(g) B (g) + C(g)at t=0 Poat t=t Po- x x xTotal pressure at time t Pt = Po- x + x + x = Po+ x ot PPx

k = xPP

t 0

0log303.2

k =TPP

Pt 0

02log303.2

whereP0 = Initial partial pressure of APT = Total pressure of gaseous system at time 't'


By Princ

e Sir

Kinetics Of Some Miscelleneous Reactions(i) Parallel Reactions :- here both the side reactions are of first order.

at any time t , AB

C(A - x )X

XLet us assume the initial concentration of A is ‘a’ and after time t ‘x’ of itdecomposes to form B and C.So we can write

Rate of diappearence of A = Rate of formation of B + rate of formation of C

)()( 21 xakxakdtdx

=(k1+k2) (a-x)

)()( 21 kkxadx

t

)(ln1)()()(ln 2121 xaa

tkktkkxaa

Here , (K1 + K2) = Kav. and2

1][][

kk

CB

So, % Yield of B = 10021

1 KKK

% Yield of C = 10021

2 KKK

[A] = tkKo eA )+( 21][ ; [B] = )1(+

][ )+

21

01 21 tkkekkAk

-(- ; [C] = )(+][ )+

21

02 21 tkkekkAk

(--1

(ii) Reversible reactionConsider the reaction where both the forward and backward reac-

tions are of first order.A

kf

kb

BAt time t = 0 a bAt time t = t a – x xAt time t = teq a – xeq xeqRate of the forward reaction = Kf(a-x)Rate of the backward reaction = Kbxat equilibrium ,Rate of forward reaction = Rate of backward reaction

Kf(a-x) = Kbx eq

eq

xax =

b

fkk


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e Sir

Now net rate of formation of B dtdx = Kf(a-x)- Kbx

Substituting the value of Kb and on integrating we get

(kf + kb) =

xxx

t eq

eqln1

(iii) Consecutive Reaction

A 1k B 2k C

Here we can write that , Rate of formation of B = Rate of disappearence of A - rate of formation of C

[A]t = tkeA 10][ ; [B]t = )(

][12

01kk

Ak tktk ee 21

[C]t = [A0] – ( [A] + [B] )

tmax =

2

1

21l)(

1kknkk ; [B]max = [A0].

122

2

1kk

k

kk

[C]

[B][A]

•conc.

time

In this reaction the concentration of B will first reach to a mximum andthan decreases as initially the rate of formation of B is greater than rate ifits decomposition.

(iv) Pseudo First Order Reactions: These are the reactions which doesn’tappear to be first order but are actually of first order. in these types ofreaction one of the reactants is present in excess, so it is not included in ratelaw and the rate of reaction does not depend on the concentration of thatreactant.Consider the hydrolysis of ester

CH3COOC2H5 + H2O H CH3COOH + C2H5OH

Here water is taken in excess so its concentration doesn’t changesmuch and therefore it is not included in the rate law .

Rate = k [CH3COOC2H5]Intially the reaction appears to be of the second order but the reaction

actually is of first order and therefore is called a pseudo first order reaction.For a pseudo first order reaction we can apply the kinetics of the first order.


By Princ

e Sir

METHODS OF DETERMINATION OF ORDER OF REACTIONS :

A few methods commonly used are given below :1. Hit & Trial Method : It is method of using integrated rate equations,where the experimentalvalues of a, x & t are put into these equations. Onewhich gives a constant value of k for different sets of a, x & t correspond tothe order of the reaction.2. Graphical Method : In this method graphs are plotted between theconcentration of reactants and time(i) A plot of [At] Vs ‘t’ gives a straight line for a zero order reaction.

[A]

conc.

time

(i) A plot of log [At] versus 't' gives a straight line for the First orderreaction.

In [A]t

t

(ii) A plot of (a – x)– (n–1) versus 't' gives a straight line any reactionof the order n (except n = 1).3 . Half Life Method : The half life of nth order reaction is

nat 1

21 )( , Where n is the order of the reaction

So,n

aa

tt

1

2

1

22/1

12/1)()(

By experimental observation of the dependence of half life on initialconcentration we can determine n, the order of reaction.

n = 1 +2010

12/122/1)log()log(

)log()log(aatt

.


By Princ

e Sir

4. By the unit of rate constant :

The unit of rate constant for an ‘nth’ order reaction is 1)1(

sec

n

Litremol

For zero order reaction the unit will be mol L-1sec-1.For first order reaction the unit of rate constant is sec-1.For second order reaction the unit of rate constant is Lmol-1sec-1.

DEPENDENCE OF RATE CONSTANT ON TEMPERATURE(Arrehnius Equation)According to Arrhenius hypothesis :(i) Among all the reactant molecules only a certain number of molevulesparticipate in a chemical reaction which are known as active molecules andthe rest are known as passive molecules.(ii) there exist an equillibrium between active and passive molecules andwhen temperature is increased the active molecules are converted into passivemolecules by absorbing energy.

Active molecules Passive molecules

The temperature dependence of rate constant can be explained on the basisof Arrhenius equation:

k = A. RTEae

Where A is Arrhenius constant or frequency factor or pre-exponential factorand is characteristicof a reaction.R Universal gas constantEa Activation energy (in J/Mol-1)

Straight from NCERT

* In the Arrhenius equation RTEae is the fraction of molecules that have

kinetic energy greater than activation energy .

Using k = A. RTEae

ARTEk a lnln

Thus the plot of lnk Vs 1/T will be a straight line


By Princ

e Sir

let at temperature T1 rate constant be K1 and temperature T2 rate constantbe K2, than

ARTEkAek aRT

Ea

lnln1

111

& ARTEkAek aRT

E a

lnln2

22 1

on solving we get

log

211

2 11303.2 TTRE

kk a

Temperature coefficient()The temperature coefficient of a chemical reaction is defined as the ratio ofthe reat of reaction two temperatures differing by 10°C.

Temperature coefficient() =t

tk

k 10 .Generally for every 100C rise in temperature the rate of reaction becomesby twice or thrice , therefore the value of temperature coefficient lies between2 & 3.

This change in reaction rate can be explained on the basis ofdistribution curve proposed by Boltzman and Maxwell. They plotted thekinetic energy with fraction of molecules

T

ENN

( Where NE number of molecules with energy E and

NT total number of molecules).


By Princ

e Sir

Let the temperature coefficient of a reaction be ‘’ when temperature isincreased from T1 to T2 than the ratio of rtae constants can be calculated as

10)( 12

2

2 )(TT

T

T

Kk

COLLISION THEORYAccording to this theory , the reactant molecules are assumed to be hardspheres and a reaction will occurr only if the molecules collide with eachother. Not all collisions result in product formation, those collisions which resultin product formation are known as Effective Collisions.For a collision to beeffective following two barrier has to be crossed

(i) Energy Barrier (ii) Orientation Barrier(i) Energy Barrier :- For an effective collision to occurr the colliding reactantmolecules must have kinetic energy greater than the activation energy.

Frac

tion

ofm

olec

ules

Energy

Fraction of molecules havingenergy greater than activation energy and can prdouceeffective collisions

A collision between high energy molecules overcomes the force of repulsionand forms an unstable activated complex which has maximum energy andis highly unstable.This activated than breaks into products or reactants.

Thershold Energy(ET) :- The minimum energy which the reactant moleculesmust have in order to form products in a chemical reaction is called ThresholdEnergy .Activation Energy (Ea) :- The extra energy which the reactant moleculesrequire to collide effectively and form products in a chemical reaction iscalled activation energy . Here,Eaf Activation energy of forward reactionEab Activation energy of backward reaction


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e Sir

E a Eb

E T

E R E p

ReactantsProduct

Reaction Co-ordinates

Ener

gy

Overall activation energy of reaction = Eaf-Eab = H

(ii) Orientation Barrier :- For a collision to be effective the reactant moleculesmust collide with proper orientation, such that they cover the maximumsurface area of each other.ADD DIAGRAM FROM NCERTThus in collision theory Activation energy and proper orientation togetherdetermine the criteria for an effective collision and the rate of chemicalreaction. There are two factors on which the rate of reaction depends:(i) Collision Frequency (ii) Activation energyCollision Frequency(Z) :- The number of collisions per second per unit volumeof the reaction mixture is known as collision frequency.

For a reaction A + B Products

The rate of reaction can be written as rate = ZAB RT

Ea

e

Where, ZAB is the collision frequency and RT

Ea

e

is the fraction of moleculeshaving energy equal to or greater than Ea.

REACTION MECHANISMThe mechanism of a reaction is the path through which the

conversion of reactants take place into products.The reaction mechanism could consist of several steps . these steps can befast and slow reactions. We are concerned with the slowest reaction whichwe call RDS (Rate Determining Step) , because we can observe the changein concentration of reactants and products only for slow reactions.

To determine the order of the reaction we will first write the rate lawfor the RDS and if the RDS consist of any intermediate specie we willeleminate it .


By Princ

e Sir

following two methods can be used to determine the rate law for amultistep reaction(i) By equilibrium Consideration

consider the following mechanism for the reaction2O3(g) 3O2(g)Mechanism :

Step-1 O3(g)kf

kb

O2(g) + O(g) fast

step-2 O(g) + O3(g) 2k 2O2(g) slow

Since step-2 is the RDS so the rate law will be :Rate = k [O] [O3]but here [O] is the intermediate and it can not be included in the

rate law so we have to eleminate it therefore for the step -1

equilibrium constant (Kc) =b

fkk

= ][][

][][]][[

2

3

3

2OOK

OOOO c

So the rate law will be rate = 12

23

'

2

23 ][][][][. OOkO

Okk c

Hence the order of the reaction is = 2+(-1) = 1

(ii) By Steady State Approximation Here we assume that the concentration of the intermediate is constantunder steady state condition so its rate of formation/ disappearence iszero.Consider the reaction NO2(g) + CO (g) CO2(g) + NO(g)

The reaction mechanism can be given asstep 1 : NO2 + NO2 1k NO + NO3 (fast )step 2 : NO3 + CO 2k CO2 + NO2 ( slow)

Since , NO3 is the intermediate , from steady state approximation

we can say that dtNOd ][ 3 = o

But dtNOd ][ 3 = k1[NO2]2 -k2[NO3][CO] = 0


By Princ

e Sir

So, [NO3] = ][][ 2

2

2

1CO

NOkk

rate of reaction = rate of disappearence of NO2

rate = dtNOd ][ 2 = k1[NO2]2 + k2[NO3][CO]

= k1[NO2]2 + k2 ][][ 2

2

2

1CO

NOkk [CO]

rate = 2k1[NO2]2So, the order of the reaction is 2.

Types of reactions on the basis of influence of temperature

Effect of presence of a catalyst

Rate constant in presence of catalyst RTE

pp

Aek

Rate constant in absence of catalyst RTE

aaAek

So, RTEE

a

p pa

ekk )(

Ea Activation energy in absence of catalystEp Activation energy in presence of catalyst


By Princ

e Sir

BOOST YOUR BASICSQ.1. For the equation 2NO + 2H

2N

2+ 2H

2O, write the most probable rate law from the follow

ing data. What is the order of reaction? Also calculate rate constant.Experiment [NO] [H

2] Rate (mol L-1s-1)

I 0.2 0.2 3 × 10-3

II 0.4 0.2 1.2 × 10-2

III 0.2 0.4 6 × 10-3

Q.2. For the reaction OCl- + I- + OH- OI- + Cl- + H2O kinetic data are given below. Calculate rate law

and value of rate constant.

Exp. No. [OCl–] [I–] [OH–] d[IO ]

dt

(mol dm-3) (mol dm-3) (mol dm-3) (mol dm-3) (s-1)1. 0.0017 0.0017 1.0 1.75×10-4

2. 0.0034 0.0017 1.0 3.5×10-4

3. 0.0017 0.0034 1.0 3.5×10-4

4. 0.0017 0.0017 0.5 3.5×10-4

Q.3. According to the reaction Cr2O

72- + 5H+ + 3HNO

2 2Cr3+ + 3NO3

- + 4H2O The rate of

disappearance of Cr2O

72- is found to be 2.4 × 10-4 mole L-1s-1. Find the rate of appearance of

Cr3+ during given time interval.

Q.4. Fill in the blanks in the following table which treats reaction of a compound A with acompound B, that is of the first order with respect to A and zero order with respect to B.

Expt.No.

[A](mol L-1)

[B](mol L-1)

Initial rate(mol L-1 s-1)

1. 0.1 0.1 20 × 10-2

2. ---- 0.2 4.0 × 10-2

3. 0. 4 0. 4 -----4. ---- 0.2 2.0 × 10-2

Q.5. The reaction 2NO + O2 2NO

2, follows the rate law = k[NO]2[O

2]. What is the order of the

reaction?If k = 2.0×10–6 mol–2 L2 s–1, what is the rate of the reaction when [NO] = 0.04 mol L–1 and[O

2] = 0.2 mol L–1 ?

Q.6. From the following data of initial concentrations and rates, calculate the order of reactionaA Products; and its rate constant.[A] mol L–1 0.1 0.2 0.4Rate mol L–1 s–1 9 × 10–5 36×10–5 144 × 10–5

Q.7. For a reaction aA + bB mM + nN, the rate of reaction is given as k[A]x[B]y

Calculate the order of the reaction and the rate constant from the given initial concentrations andthe corresponding rates.[A] mol L–1 0.1 0.1 0.2[B] mol L–1 0.2 0.4 0.2Rate mol L–1 s–1 4 × 104 16 × 104 8 × 104


By Princ

e Sir

Q.8. The data given in the following table pertain to the reaction,2A + B CDetermine the form of the rate equation and the value of the rate constant

Initial concentration(mol L-1)

Expt.No.

[A] [B]

Initial rate(mol L-1 s-1)

1. 1 × 10-3 1 × 10-3 7 × 10-6

2. 1 × 10-3 2 × 10-3 14 × 10-6

3. 1 × 10-3 3 × 10-3 21 × 10-6

4. 2 × 10-3 3 × 10-3 84 × 10-6

Q.9. The following initial rate data were obtained for the reaction :2NO(g) + Br

2(g) 2NOBr(g)

Run [NO]/M [Br2]/M Rate/M min-1

1. 1.0 1.0 1.30 × 10-3

2. 2.0 1.0 5.20 × 10-3

3. 4.0 2.0 4.16 × 10-2

Determine the reaction rate law and the value of the rate constant.

Q.10. For the reaction AB, it was found that the concentration of B increased by 0.3 mol L–1 in2 hours. What is the average rate of reaction ?

Q.11. In the reaction XY, the initial concentration of X is 2.5 mol L–1 and its concentration after3 hours is 0.7 mol L–1. What is the average rate of the reaction ?

Q.12. When ammonia is treated with O2 at elevated temperaturs, the rate of disappearance of

ammonia is found to be 3.5×10–2 mol dm–3s–1 during a measured time interval. Calculate therate of appearance of nitric oxide and water.

Q.13. In a reaction N2O

5 2NO

2 +12

O2, the rate of disappearance of

N2O

2 is 6.5 × 10–3 mol L–1 s–1. Compute the rates of formation of NO

2 and O

2.

Q.14. The following reaction was carried out at 44°C :N

2O

2 2NO

2 + 1/2O

2

The concentration of NO2 is 6.0 × 10–3 M after 10 minutes of the start of the reaction.

Calculate the rate of production of NO2 over the first ten minutes of the reaction.

Q.15. The rate of a particular reaction doubles when temperature changes from 27° C to 37°C.Calculate the energy of activation for such reaction. (R = 8.314 J K–1 mol–1)

Q.16. The rate constant of a reaction is 2 × 10–2 s–1 at 300 K and 8 × 102s–1 at 340 K. Calculate theenergy of activation of the reaction.

Q.17. For a chemical reaction the energy of activation is 85 kJ mol–1. If the frequency factor is4.0 × 109 L mol–1 s–1, what is the rate constant at 400 K ?

Q.18. The energy of activation of a reaction is 140 kJ mol–1. If its rate constant at 400 K is2.0 × 10–6 s–1, what is the value at 500 K ?


By Princ

e Sir

Q.19. Calculate the ratio of the catalysed and uncatalysed rate constant at 20° C if the energy ofactivation of a catalysed reaction is 20 kJ mol–1 and for the uncatalysed reaction is 75 kJmol–1.

Q.20. The decomposition of methyl iodide,2CH

3I(g) C

2H

6(g) + I

2(g)

at 273° C has a rate constant of 2.418 × 10–5 s–1. If activation energy for the reaction is+179.9 mol–1, what is the value of collision factor 'A' at 273°C ?

Q.21. The reaction,2N

2O

5 2N

2O

4 + O

2

occurs in carbon tetrachloride. The rate constant is 2.35×10–4 sec–1 at 30°C. Calculate theactivation energy of the reaction.

Q.22. For the inversion of cane sugar,C

12H

22O

11 + H

2O C

5H

12O

6 + C

6H

12O

6

the rate constant is 2.12 × 10–4 L mol–1sec–1 at 27°C. The activation energy of the reaction is1.07 × 105 J mol–1. What is the rate constant of the reaction at 37° C ?

Q.23. For a first order reaction, the rate constant is 0.1 s–1. How much time will it take to reducethe concentration from initial value of 0.6 mol L–1 to 0.06 mol L–1 ?

Q.24. A substance decompose following first order reaction. If the half life period of the reactionis 35 minutes, what is the rate constant of this reaction ?

Q25. For a certain first order reaction, it take 5 minutes for the initial concentration of 0.6 molL–1 to become 0.2 mol L–1. What is the rate constant for this reaction ? [log 3 = 0.4771]

Q.26. Find the two -thirds life (t2/3

) of a first order reaction in which k = 5.48 × 10–1 sec–1 .

Q.27. A first order reaction has a specific rate of 10–3sec–1. How much time will it take from 10 g of the reactant to reduce to 7.5 g (log 2 = 0.3010; log 4 = 0.6020 and log 6 = 0.7782) ?

Q.28. In a reaction 5 g ethyl acetate is hydrolysed per litre in presence of dilute HCl in 300 min.If the reaction is of first order and initial concentration is 22 g per litre, calculate the rateconstant for the reaction.

Q.29. Calculate the half life of the reaction A B, when the initial concentration of A is 0.01 molL–1 and initial rate is 0.00352 mol L–1 min–1. The reaction is of the first order.

Q.30. In reaction A B + C, the following data were obtained:t in seconds 0 900 1800Concentration of A 50.8 19.7 7.62Prove that it is a first order reaction.

Q.31. The first order reaction has k = 1.5 × 10–6 per second at 200°. If the reaction is allowed torun for 10 hours at the same temperature, what percentage of the initial concentration wouldhave changed into the product ? What is the half life period of this reaction ?

Q.32. A first order reaction is 20% completed in 10 minutes. Calculate (i) the specific rate constant of the reaction and (ii) the time taken for the reaction to go to 75% completion.

Q.33. The rate constant of a reaction with respect to reactant A is min–1. If we start with [A] = 0.8mol L–1, when would [A] reach the value of 0.08 mol L–1 ?


By Princ

e Sir

Q.34. What will be the initial rate of a reaction if the constant is 10–3 min–1 and the concentration is0.2 mol dm–3?How much of the reactant will be converted into the product in 200 minutes ?

Q.35.. A substance A decomposes by the first order reaction. Starting initially with [A] = 2.00 M,after 200 minutes [A] = 0.25M. For this reaction what is t

1/2 and k ?

Q.36. In this case we have2A3B + CTime t Total pressure of A + B+C P2 P3Find k.

Q.37. A2B + 3CTime t

Total pressure of ( B+C) P2 P3Find k.

Q.38. A2B + 3CTime t

Volume of reagent V2 V3Reagent reacts with all A, B and C. Find k. [Assuming n-factor of A, B & C are same]

Q.39. k1 = x hr–1; k1 : k2 = 1 : 10. Calculate ]A[

]C[ after one hour from the start of

the reaction. Assuming only A was present in the beginning.

Q.40. How much time would be required for the B to reach maximum concentration for thereaction .

A 1k B 2k C. Given k1 = 4

2nl, k2 = 2

2nl.


By Princ

e Sir

Insight to the concepts

Q.1. The reaction

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 k1is 3.3 × 10 4s 1. In an experiment starting with the pure cis form, how long would it take for halfthe equilibrium amount of the trans isomer to be formed ?

Q.2. Two reations (i) A products (ii) B products, follow first order kinetics. The rate of thereaction (i) is doubled when the temperature is raised from 300 K to 310K. The half life for thisreaction at 310K is 30 minutes. At the same temperature B decomposes twice as fast as A. If theenergy of activation for the reaction (ii) is half that of reaction (i), calculate the rate constant ofthe reaction (ii) at 300K.

Q.3. The half life for a reaction between fixed concentration of reactants veries with temperature asfollows :

t°C 520 533 555 574t1/2

sec 1288 813 562 477Calculate the activation energy of this reaction.

Q.4. What percentage of reactant molecules will crossover the energy barrier at 325 K ? Heat ofreaction is 0.12 kcal and activation energy of backward reaction is 0.02 kcal.

Q.5. A decomposition reaction has following mechaism2N

2O

5 4NO

2+ O

2(overall)

(i) N2O

5 NO

2 + NO

3(Fast decomposition)

(ii) NO2 + NO

3 NO + NO

2 + O

2(slow)

(iii) NO + NO32NO

2(fast)

Determine rate law and the order of reaction.

Q.6. For the system A(g) B(g), H for the forward reaction is –33 kJ/mol (Note : H = E in this

case). Show that equilibrium constant K =[B][A] = 5.572 × 105 at 300 K. If the activation energies

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

Q.7. The half life period for the reaction N2O

5 2NO2+

1

2O

2is 2.4 hrs. a 30oC. What time would be

required to reduce 5 × 1010 molecules of N2O

5to 108 molecules ?

.

Q.8. 10 gram atoms of an -active radio isotope are disintegating in a sealed container. In one hourthe helium gas collected at STP is 11.2 cm3. Calculate the half-life of the radio-isotope.

Q.9. In a reaction, aA products, the rate is doubled when the concentration of A is increased 4times. If 50 % of the reaction occurs in 1414s, how many seconds would it take for the completionof 75% reaction?


By Princ

e Sir

Q.10.Mechanism for the reaction 2A C is given as follows.

(i) A + A 1k A* + AA (Activation)

(ii) A* + A 2k A + AA (Deactivation)

(iii) A* 3k C (Reaction)Deduce general rate law for the reaction, using steady state approximation.

Q.11 (a) The decomposition of HI to I2at 508oC has a half-life of 135 minutes when P

HI= 0.1 atm

(initial) which comes down to one-tenth of that value when the initial pressure is 1 atm. Calculatethe rate constant.

Q.12 Above 500oC, the reaction NO2+ CO CO

2+ NO obeys the rate law rate = k[NO

3][CO]. Below

500oC the rate law for this reaction is rate = k[NO2]2. Suggest mechanism for each of the cases

Q.13. Decomposition of non-volatile solute 'A' into another non-volatile solute B and C, when dis-solved in water follows first order kinetics as:

A 2B + C

When one mole of A is dissolved in 180 gm of water and left for decomposition, the vapour

pressure of solution was found to be 20 mm Hg after 12 hrs. Determine the vapour pressure of

the solution after 24 hrs. Assume constant temperature of 25°C, throughout. The vapour pressure

of pure water at 25°C is 24 mm Hg. [Give your answer by adding all the digits ]

Q.14. Consider a reversible reaction :

Ak1

k2

B

Which is a first order in both the directions (k1 = 3

38.1 × 10–2 min–1). The variation in concentration

is plotted with time as shown below.

0.1

0

0.2

0.3

[A]

[B]

time

conc.(M)

Calculate the time (in minute) at which 25 % of A would be exhausted. [n 2 = 0.69 ]


By Princ

e Sir

OBJECTIVESQ.1 A zero order reaction is one-

(A) In which reactants do not react (B) In which one of the reactants is in large excess(C) Whose rate does not change with time (D) Whose rate increase with time

Q.2 If concentration are measured in mole/li9tre and time in minutes, the unit for the rate constant ofa 3rd order reaction are(A) mol lit-1min-1 (B) lit2mol-2min-1 (C) lit mol-1min-1 (D) min-1

Q.3 Following reaction was carried out at 300 K. 2SO2(g) + O

2(g)2SO

3(g) How is the rate of

formation of SO3related to the rate of disappearance of O

2?

(A) – 2[O ]

t

= +1

23[SO ]

t

(B) – 2[O ]

t

= 3[SO ]

t

(C) – 2[O ]

t

= –1

23[SO ]

t

(D) None of these

Q.4 According collision theory of reaction-(A) Every collision between reactiont molecules leads to a chemical reaction.(B) Rate of reaction is proportional to the velocity of the molecules(C) Rate of reaction is proportional to the average energy of the molecules(D) Rate of reaction is proportional to the number of collision per second.

Q.5 Which of the following rate laws has an overall order of 0.5 for the reaction, A + B + Cproduct-(A) R = k[A].[B].[C] (B) R = k[A]0.5[B]0.5[C]0.5

(C) R = k[A]1.5[B]-1[C]0 (D) R = k[A] [B]0[C]0.5

Q.6 For the system A2(g) + B

2(g) 2AB(g), H =–160 kJ If the activation energy for the forword

step is 100 kcal/mol. What is the ratio of temperature at which the forword and backward reactionshows the same % change of rate constant per degree rise of temperature ? (1 cal = 4.2 J)

(A) 0.72 (B) 0.84 (C) 0.42 (D) 1

Q.7 The t1/2

of first order reaction is 10 minutes. Starting with 100 grams/lit of the reaction, the amountremaining after one hour is(A) 25.0 g (B) 3.130 g (C) 12.50 g (D) 1.563 g

Q.8 For the reaction, 2NON2+ O

2, the experssion -

1

2

d[NO]

dt represents

(A) The rate of formation of NO (B) The average rate of the reaction(C) The instantaneous rate reaction (D) All the above

Q.9 For a reaction of the type 2A + B 2C, the rate of the reaction is given by k[A]2[B]. When thevolume of the reaction vessel is reduced to 1/4 th of the original volume, the ratev of reaction bya factor of(A) 0.25 (B) 16 (C) 64 (D) 4


By Princ

e Sir

Q.10 What is order withrespect to A, B, C respectively[A] [B] [C] rate(M/sec.)0.2 0.1 0.02 8.08×10-3

0.1 0.2 0.02 2.01×10-3

0.1 1.8 0.18 6.03×10-3

0.2 0.1 0.08 6.464×10-2

(A) –1,1, 3/2 (B) –1,1, 1/2 (C) 1, 3/2, –1 (D) 1, –1, 3/2

Q.11 For reaction, rate = k[A] [B]2/3 the order of reaction is-(A) 1 (B) 2 (C) 5/3 (D) Zero

Q.12 The elementary reaction A + B products, has k = 2 × 10-5 M-1 s-1 at a temperature of 27oC. Severlexperimentary runs carried out using stoichiometric proportion. The reaction has a temperaturecorfficient value of 2.0. At what temperature should the reaction be carried out if inspite of halvingthe concentrations, the rate of reaction is desired to be 50% higher than a previous run. (Given

n6

n2

= 2.585)

(A) 47oC (B) 53oC (C) 57oC (D) 37oC

Q.13 According to collision theory-(A) All collisions are sufficiently violent(B) All collisions are responsible formation.(C) All colisions are effective.(D) Only a fraction of collisions are effective which have enough energy to form products.

Q.14 When the temperature of a reaction increases from 270C to 370C, the rate increases by 2.5 times,the activation energy in the temperature range is(A) 53.6 kJ (B) 12.61 kJ (C) 7.08 kJ (D) 70.8 kJ

Q.15 For the reaction R – X + OH– ROH + X–The Rate is given as Rate = 5.0 × 10-5 [R – X][OH–] + 0.20 × 10-5[R–X] what percentage of R – X reacted by SN2 mechanism when [OH-] = 1.0 × 10-

2 M(A) 96.1% (B) 3.9% (C) 80% (D) 20%

Q.16 Which of the following curves represents a Ist order reaction-

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

Q.17 The rate of a reaction increases 4-fold when conecentration of reaction is increased 16 times. Ifthe rate of reaction is 4 × 10-6 mole L-1 when concentration of the reaction is 4 × 10-4 mole L-1, therate constant of the reaction will be(A) 2 × 10-4mole1/2L-1/2sec-1 (B) 1 × 10-2sec-1

(C) 2 × 10-4mole-1/2,L1/2sec-1 (D) 25 mole-1Lmin-1

Q.18 If 'a' is the intial concentration of a substance which reacts according to zero order kinetic and krate conatant, the time for the reaction to go to completion is-(A) a/k (B) 2/ka (C) k/a (D) Infinite


By Princ

e Sir

Q.19 The following data are obtained from the decomposition of a gaseous compound

Initial pressure, (atm) 1.6 0.8 0.4Time for 50% reaction,(min) 80 113 160

The order of the reaction is(A) 1.0 (B) 1.5 (C) 2.0 (D) 0.5

Q.20 For a given reaction the concentration of the reactant plotted against time gave a straight linenegative slope. The order of the reaction is-(A) 3 (B) 2 (C) 1 (D) 0

Q.21 The rate constant of a first order reaction is generally determined from a plto of(A) Concentration of reactant vs time t (B) log (concentration of reaction) vs time t

(C) 1

concentration of reaction vs time t (D) Concentration of reactant vs log time t

Q.22 Which of the following correctly represents the variation of the rate of the reaction with tempera-ture ?

(A) (B)

(C) (D)

Q.23 The plot of log K versus1

Tis linear with a slope of

(A) aE

R(B) aE

R

(C) aE

2.303R(D) aE

2.303R

Q.24 According to the collision theory, the rate of reaction increases with temperature due to-(A) Greater number of collision(B) Higher velocity of reacting molecules(C) Greater number of molecules having the activation energy(D) Decrease in the activation energy

Q.25 If a reaction A + B C is exothermic to the extent of 30 kJ/mol and the forward reaction has anactivation energy 70 kJ/mol, the activation energy for the reverse reaction is (A) 30 kJ/mol (B) 40 kJ/mol (C) 70 kJ/mol (D) 100 kJ/mol

Q.26 A first order reaction takes 40 min for 30% decomposition. Calculate t1/2

.(A) 77.7 min. (B) 27.2 min. (C) 55.3 min. (D) 67.3 min.


By Princ

e Sir

Q.27 For a certain reaction involving a single reaction, it is found that C0 T is constant where C

0is the

initial concentration of the reaction and T is the half-life. What is the order the reaction ?(A) 1 (B) Zero (C) 2 (D) 3

Q.28 The high temperature ( 1200K) decomposition of CH3COOH(g) occurs as follows as per simul-taneous 1st order reactions.

CH3COOH k1 CH

4+ CO

2

CH3COOH k 2 CH

2CO + H

2O

What would be the % of CH4 by mole in the product mixture (excluding CH3COOH) ?

(A) 1

1 2

50k

k k (B) 1

1 2

100k

k k (C) 1

1 2

200k

k k (D) it depends on time

Q.29 The number of half-lives (t1/2

) required to bring the ratio to0

N

N = 0.125 from

0

N

N = 1 of a

radioactive element is(A) 2 (B) 4 (C) 3 (D) 1

Q.30 The reaction, A(g) + 2B(g)C(g) + D(g) is an elementary process. In an experiment, the initialpartial pressure of A and B are P

A= 0.60 and P

B= 0.80 atm. When P

C= 0.2 atm the rate of reaction

relative to the initial rate is(A) 1/48 (B) 1/24 (C) 9/16 (D) 1/6

Q.31 Half life period for a first order reaction is 20 minutes. How much time is required to change theconcentration of the reactants from 0.08M to 0.01 M(A) 20 minutes (B) 60 minutes (C) 40 minutes (D) 50 minutes

Q.32 The kinetic datas for the reaction: 2A + B22AB are as given below:

[A] [B] Ratemol L-1 mol L-1 mol L-1min-1

0.5 1.0 2.5 × 10-3

1.0 1.0 5.0 × 10-3

0.5 2.0 1 × 10-2

The order of reaction w.r.t. A and B2are, respectively,

(A) 1 and 2 (B) 2 and 1 (C) 1 and 1 (D) 2 and 2

Q.33 If for any reaction rate constant is equal to the rate of the reaction at all conaentrations, the orderis-(A) 0 (B) 2 (C) 1 (D) 3

Q.34 For the irreversible unimolecular type reaction A k products, in a batch reactor, 80% reactantA(C

A0= 1 mole/lit.) is converted in a 480 second run and conversion is 90% after 18 mintue. The

order of this reaction is-(A) 1 (B) 2 (C) 1/2 (D) 3/2

Q.35 The rate constant, the activation energy and the Arrhenius parmeter of a chemical reaction at 25oCa r e3.0 × 10-4s-1, 104.4 kJ mol-1 and 6.0 × 1014s-1 respectively.The value of the rate constant as T

is:


By Princ

e Sir

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

Q.36 For the first order reaction t99%

= x × t90%

the value of 'x' will be-(A) 10 (B) 6 (C) 3 (D) 2

Q.37 The gas phase decomposition 2N2O

54NO2+ O

2follows the first order rate law with rate con-

s t a n tK = 7.5 × 10-3 sec-1. The initial pressure of N

2O

5is 0.1atm. The time of decomposition of N

2O

5so

that total pressure becomes 0.15 atm will be-(A) 54 sec (B) 5.4 sec (C) 3.45 sec (D) 34.55 sec

Q.38 If in the fermentation of sugar in an enzymatic solution that is 0.12 M, the concentration of thesugar is reduced to 0.06 M in 10h and to 0.03 M in 20h, what is the order of the reaction-(A) 1 (B) 2 (C) 3 (D) 0

Q.39 In a Ist order reaction, A products, the concentration of the reactant decreases to 6.25% of itsinitial value in 80 minutes. What is the rate constant reaction 100 minutes after the starts, if theinitial concentration is 0.2 mole/litre ?(A) 2.17 × 10-2min-1 (B) 3.465 × 10-2min-1

(C) 3.465 × 10-3min-1 (D) 2.166 × 10-3min-1

Q.40 For the first order reaction A(g) 2B(g)

+ C(g)

, the initial pressure is PA = 90 mm Hg, the pressureafter 10 minutes is found to be 180 mm Hg. The rate constant of the reaction is(A) 1.15 × 10-3sec-1 (B) 2.3 × 10-3sec-1

(C) 3.45 × 10-3sec-1 (D) 6 × 10-3sec-1

Q.41 For a certain reaction the variation of the rate constant with temperature is given by the equation

In kt= 1n k

0+

1n3

10 t (t0oC) The value of the temperature coefficient of the reaction rate is

therefore-(A) 4 (B) 3 (C) 2 (D) 10

Q.42 A catalyst lowers the activation energy of a reaction from 20 kJ mole-1 to 10 kJ mole-1. The tem-perature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27oCis(A) –123oC (B) 327oC (C) 1200oC (D) +23oC

Q.43 The rate of reaction triples when temperature changes from 20oC to 50oC. Calculate energy ofactivation for the reaction.(A) 28.81 kJ mol-1 (B) 38.51 kJ mol-1

(C) 18.81 kJ mol-1 (D) 8.31 kJ mol-1

Q.44 The half life time for the decomposition of a substance dissolved in CC14is 2.5 hour at 30oC. How

much of the substance will be left after 10 hours, if the initial weight of the substance is 160 gm ?(A) 20 gm (B) 30 gm (C) 40 gm (D) 10 gm

Q.45 For the reaction [Cr(H2O)

4C1

2]+

(aq.)1k [Cr(H

2O)

5C1]2+

(aq.)2k [Cr(H

2O)

6C1]3+

(aq.)

K1= 1.78 × 10-3s-1 and k

2= 5.8 × 10-5s-1

If initial concentration of [Cr(H2O)

4C1

2]+ is 0.0174 M at 273 K. Calculate time at which concen-

tration of [Cr(H2O)

5C1]2+ is maximum.

(A) 1990 seconds (B) 1090 seconds(C) 2100 seconds (D) 2110 seconds.


By Princ

e Sir

Q.46 What is the activation energy for the decomposition of N2O

5 as N

2O

5 2NO

2+ (1/2) O

2If the

value of the rate constant are 3.45 × 10-5 and 6.9 × 10-3 at 27oC and 67oC respectively.(A) 102 × 102 kJ (B) 488.5 kJ (C) 92 kJ (D) 112.3 kJ

Q.47 If for a first order reaction, rate constant varies with temperature according to the

graph given below. At 27oC, 1.5 × 10-4 percent of the reaction molecules areable to cross-over the potential barrier. At 52oC, the slope of this graph isequal to 0.2 K-1 sec-1, calculate the value of rate constant at 52oC, assumingthat activation energy does not change in this temperature range.(A) 3.14×10-2min-1 (B) 1.35 × 10-2min-1

(C) 0.75 × 10-2min-1 (D) 8.75 × 10-2min-1

Q.48 the rate constant for a first order reaction is 60s-1. How much time will it take to reduce the initialconcentration of the reactant to its 1/16th value-(A) 0.046 s (B) 0.46 s (C) 0.124 s (D) 2.123 s

Q.49 1 mole of gas changes linearly from initial state (2 atm, 10 lt) to final state (8 atm, 4 lt). Find thevalue of rate constant, at the maximum temperature, that the gas can attain. Maximum rate con-stant is equal to 20 sec-1 and value of activation energy is 40 kJ mole-1, assuming that activationenergy does not change in this temperature range.(A) 0.56 × 10-3 sec-1 (B) 3.16 × 10-3 sec-1

(C) 1.56 × 10-3 sec-1 (D) 5.12 × 10-3 sec-1

Q.50 A first order reaction was started with a decimolar solution of the reactant. After 8 minutes and 20seconds, its concentration was found to be M/100. Determine the rate constant of the reaction.(A) 4.6 × 10-3 sec-1 (B) 16.6 × 10-3 sec-1

(C) 24.6 × 10-3 sec-1 (D) 40.6 × 10-3 sec-1

Q.51 87.5% of a radioactive substance disintegrates in 40 minutes. What is the half life of the substance?(A) 13.58 min (B) 135.8 min (C) 1358 min (D) None of these

Q.52 A gaseous substance AB3decomposes irreversibly according to the overall equation AB

31

2A

2

+3

2B

2. Starting with pure AB

3, the partial pressure of the reaction varies time for which the data

are given below.Time in hours 0 5.0 15.0 35.0

3ABP mm Hg 660 330 165 82.5

What is the order of the reaction ?(A) 2 (B) 0.5 (C) 1 (D) 1.5

Q.53 The inversion of cane sugar proceeds with half-life of 500 minute at pH 5 for any concentration ofsugar. However if pH = 6, the half-life changes to 50 minute. The rate law ecpression for the sugarinversion can be written as(A) r = K[sugar]2[H]6 (B) r = K[sugar]1[H]0

(C) r = K[sugar]0[H+]6 (D) r = K[sugar]0ss[H+]1

Q.54 The decomposition of A into product has value of k as 4.5 × 103s-1 at 10oC and energy of activation


By Princ

e Sir

60 kJ mol-1. At what temperature would k be 1.5 × 104s-1 ?(A) 12oC (B) 24oC (C) 48oC (D) 36oC

Q.55 Decomposition on NH3 on heated tungsten yields the following data :Initial pressure (mm) 65 105 y 185Half-life 290 x 670 820What are the values of x and y in that order ?(A) 420 s, 110 mm (B) 500 s, 160 mm(C) 520 s, 170 mm (D) 460 s, 150 mm

Q.56 The half life period of gaseous substance undergoing thermal decomposition was measured forvarious initial pressure 'P' with the following result.P(mm) 250 300 400 450t1/2

(min) 136 112.5 85 75.5Calculate the order of reaction.(A) 2 (B) 4 (C) 6 (D) 10

Q.57 For the reaction 2N2O

5(g) 4NO

2(g) + O

2(g), the concentration of NO

2increases by 2.4 × 10-2

Mollit-1 in 6 second. What will be the rate of appearance of NO

2and the rate of disappearance of N

2O

5.

(A) 2 × 10-3 mol L-1sec-1, 4 × 10-3 mol L-1 sec-1 (B) 2 × 10-3 mol L-1sec-1, 1 × 10-3 mol L-1 sec-1

(C) 2 × 10-3 mol lit-1sec-1, 2 × 10-3 mol lit.-1 sec-1 (D) 4 × 10-3 mol lit.-1sec-1, 2 × 10-3 mol lit.-1 sec-1

Q.58 The rate constant of a particular reaction has the dimensions of a frequency. What is the order ofthe reaction ?(A) Zero (B) First (C) Second (D) Fractional

Q.59 The reaction of iodomethane with sodium ethoxide proceeds as : EtO + MeI EtOMe + 1 .

A plot of log MeI

EtO

on the Y-axis against 's' on the X-axis gives a straight line with a positive

slope. What is the order of the reaction ?(A) Second (B) First (C) Third (D) Fractional

Q.60 For the reaction,C2H

5I + OH-C

2H

5OH + I- the rate constant was found to have a value of

5.03×10-2 mol-1dm3s-1 at 289 K and 6.71 mol-1dm3s-1 at 333 K. What is the rate constant at 305 K ?(A) 1.35 mol-1dm3s-1 (B) 0.35 mol-1dm3s-1

(C) 3.15 mol-1dm3s-1 (D) 7.14 mol-1dm3s-1

Q.61 A plot of ln rate Vs ln C for the nth order reaction gives-(A) a straight line with slope n and intercept ln k

n.

(B) a straight line with slope (n - 1)(C) a straight line with slope ln k

nand intercept 'n'

(D) a straight line with slope -n and intercept kn.

Q.62 The first order rate constant for a certain reaction increases from 1.667×10-6 s-1 at 727oC to 1.667×10-

4 s-1 at 1571oC. The rate constant at 1150oC, assuming constancy of activation energy over to giventemperature range is(A) 3.911 × 10-5sec-1 (B) 1.139 × 10-5sec-1

(C) 3.318 × 10-5sec-1 (D) 1.193 × 10-5sec-1


By Princ

e Sir

Q.63 The solvolysis of 2-chloro-2-methyl propane is aqueous acetone : H2O + (CH

3)

3C - Cl HO -

C(CH3)

3+ H+Cl- has a rate equation. Rate = K[(CH

3)

3C-Cl]. From this it may be inferred that the

energy profile of the reaction leading from reactants to products is

(A) (B)

(C) (D)

Q.64 Thermal decomposition of a compound is of first order. If 50 % of a sample of the compound isdecompound in 120 minutes, show how long will it take for 90 % of the compound to decompose?(A) 399 min (B) 410 min (C) 250 min (D) 120 min

Q.65 The rate constant of a certain first order reaction increases by 11.11% per degree rise of tempera-ture at 27oC. By what % will it increases at 127oC, assuming constancy of activation energy overthe given temperature range ?(A) 5.26% (B) 5.62% (C) 6.25% (D) 7.33%

Q.66 The half-life for radioactive decay of 14C is 5730 y. An archaeogical artefact contained wood hadonly 80% of the 14C found in a living tree. Estimate the age of the sample.[Radioactive decays follow the first order kinetic](A) 1657.3 y (B) 1845.4 y (C) 1512.4 y (D) 1413.1 y

Q.67 Two substance A and B are present such that [A0] = 4[B

0] and half-life of A.is 5 minute and that of

B is 15 minute. If they start decaying at the same time following first order kinetics how much timelater will the concentration of both of them would be same.(A) 15 minute (B) 10 minute (C) 5 minute (D) 12 minute

Q.68 The following data were obtained during the first order thermal decomposition of SO2Cl

2at a

constant volume. SO2Cl

2(g) SO

2(g) + Cl

2(g)

Experiment Time/s Total pressure/atm1 0 0.52 100 0.6

Calculate the rate of the reaction when total pressure is 0.65 atm.(A) 7.8 × 10-4s-1 atm. (B) 0.8 × 10-4s-1 atm.(C) 2.4 × 10-2s-1 atm. (D) 6.1 × 10-8s-1 atm.


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SUBJECT APTITUDEOne or more than one choice may be correct :Q.1 Bicyclo hexane was found to undergo two paeallel first order rearrangements, as given below-

Bicyclo hexane

Choose the correct options-(A) % of cyclohexane = 77 (B) % of methylcyclopentance = 23(C) % of methylcyclopentance = 77 (D) % of cyclohexane = 23

Q.2 Consider the following case of competing 1st order reaction.

After the start of the reaction at t = 0 with only A, the [C] is equal to the [D] at all times. The timein which all three concentrations will be equal is given by

(A) t =1

1n3

2k (B) t =

2

1n3

2k (C) t =

1

1n3

3k (D) t =

2

1n3

3k

Q.3 At 25oC, the second order rate constant for the reaction I- + ClO- IO3-+Cl- is 0.0606 M-1sec-1. Ifa solution is initially 3.5 × 10-3 M with respect to each reactants. Choose the correct options –(A) concentration of A = 3.29 × 10-3 M after 300 sec.(B) concentration of B = 3.29 × 10-3 M after 300 sec.(C) concentration of A = 0.19 × 10-3 M after 300 sec.(D) None of these

Q.4 Which is correct graph :

(A) (B)dc

ndt

(C)dc

ndt

(D) 0.75

0.5

t

t

Q.5 In a pseudo first order hydrolysis of ester in water, the following results were obtained :t/s 0 30 60 90[Ester]/M 0.55 0.31 0.17 0.085


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(A) Average rate of reaction between the time interval 30 to 60 seconds is 4.7 × 10-3 mol L-1s-1

(B) The pseudo first order rate constant for the hydrolysis of ester is 1.98 × 10-2s-1

(C) The pseudo first order rate constant for the hydrolysis of ester is 4.7 × 10-3s-1

(D) Average rate of reaction between the time interval 30 to 60 seconds is 1.98 × 10-2 mol L-1s-1

Q.6 The zone layer in the earth's upper atmosphere is important in shielding the earth from very harm-ful ultravilet radiation. The ozone, O

3, decomposes according to the equation 2O

3(g) 3O

2(g).

The mechanism of the reaction is through to proceed through an initial fast, reversible step.

Step 1 : Fast, reversible O3(g) O2

(g) + O(g)

Step 2 : Slow O3(g) + O(g) 2O

2(g)

The which of the following is correct ?

(A) Step 2 is rate determining step (B) The rate expression for step 2 is-d

dt [O

3] = k[O

3][O]

(C) For Step 1, molecularity is 1 (D) For Step 2, molecularity is 2

Q.7 The substance undergoes first order decomposition. The decomposition follows two parallel firstorder reactions as:

K1= 1.26 × 10-4sec-1 and K

2= 3.8 × 10-5sec-1

The percentage distribution of B and C(A) 76.83% B (B) 23.17% C (C) 90% B (D) 40% C

Q.8 For a certain reaction A products, the t1/2

as a function of [A]0is given as below :

[A]0(M) : 0.1 0.025

t1/2

(min) : 100 50Which of the following is true :

(A) The order is 1/2 (B) t1/2

would be 100 10 min for [A]0= 1 M

(C) The order is 1 (D) t1/2

would be 100 min for [A]0= 1 M

Q.9 For the reaction 2A + BC with the rate lawd[C]

dt = k[A]1[B]-1 and started with A and B in

stoichimetric proportion which is/are teur ?(A) unit of k is Mole litre-1s-1 (B) [A], [B] and [C] all will be linear functions of time(C) [C] = 2kt (D) [C] = kt

Q.10 The half-period T for the decomposition of ammonia on tungsten wire, was measured for differentinitial pressure P of ammonia at 25oC. ThenP (mm Hg) 11 21 48 73 120T (sec) 48 92 210 320 525(A) zero order reaction(B) First order reaction(C) rate constant for reaction is 0.114 seconds.(D) rate constant for reaction is 1.14 seconds.


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Q.11 The reaction given below, involving the gases is boserved to be first order with rate constant 7.48× 10-3 sec-1. The time required for the total pressure in a system cosntaining A at an initial pressureof 0.1 atm to rise to 0.145 atm is t

0and the total pressure after 100 sec is x atm. 2A(g) 2B(g)

+ C(g)(A) t

0= 47.7 sec. (B) x = 0.180 atm (C) x = 0.03 atm (D) t

0= 22.7 sec.

Q.12 In a reaction bteween A and B, the initial rate of reaction was measured for different initial concen-trations of A and B as given below :A/M 0.20 0.20 0.40B/M 0.30 0.10 0.05R

0/M s-1 5.07 × 10-5 5.07 × 10-5 7.6 × 10-5

(A) The order of reaction with respect to A is 0.5(B) The order of reaction with respect to B is 0.5(C) The order of reaction with respect to B is 0(D) The order of reaction with respect to A is 1.5

Q.13 The polarimeter readings in an experiment to measure the rate of inversion of cane suger (1storder reaction) were as follows

Time (min) : 0 30 Angle (degree) : 30 20 -15

Identify the true statement (s) log 2 = 0.3, log3 = 0.48, log7 = 0.84, loge10 = 2.3

(A) The half life of the reaction is 82.7 min(B) The solution is optically inactive at 131.13 min.(C) The equimolar mixture of the products is dextrorotatory(D) The angle would be 7.5o at half time

Q.14 A certain reaction obeys the rate equation (in integeated form) [C(1 - n) - C0

(1 - n)] = (n - 1) kt whereC

0is the initial concentration and C is the concentration aftert time t, then

(A) the unti of k for n = 1 is sec-1 (B) the unti of k for n = 2 is litre mol-1sec-1

(C) the unti of k for n = 3 is mol litre-1 sec-1 (D) the unti of k for n = 3 is litre mol-2 sec-1

ASSERTION AND REASON QUESTIONS :Note : Each question contains STATMENT-1 (Assertion) and STATMENT-2 (Reason). Eachquestion has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.(A) Statment-1 is Ture, Statment-2 is Ture; Statment-2 is a correct explanation for Statment-1.

(B) Statment-1 is Ture, Statment-2 is Ture; Statment-2 is NOT a correct explanation forStatment-1.(C) Statment-1 is Ture, Statment-2 is False.(D) Statment-1 is Ture, Statment-2 is True.

Q.15 Statment-1 : A lump of coal burns at a moderate rate in air while coal dust burns explosively.Statment-2 : Coal dust contains very fine particles of carbon.

Q.16 Statment-1 : Liquid bromine reacts slowly as compared to bromine vapour.Statment-2 : In liquid bromine, the bromine molecules are held together by a force which is muchweaker than the force existing between the two molecules of bromine in the vapour phase.


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Q.17 Statment-1 : The reaction, N2(10atm) + 3H

2(10atm) 2NH

3(g) is faster..

Statment-2 : Catalyst lowers the activation energy of the reaction.

Q.18 Statment-1 : Molecularity of a reaction cannot be determined experimentally.Statment-2 : Molecularity is assigned to the reactions on the basis of mechanism.

Q.19 Statment-1 : For a first order reaction, the degree of dissocation is equal to (1 - e-kt)Statment-2 : For a first order reaction, the pre-exponential factor in the Arrhenius equation hasthe dimension of time-1.

Q.20 Statment-1 : Molecularity does not influence the rate of reaction.Statment-2 : The overal kinetics of a reaction is governed by the slowest step in the reactionscheme.

MATCH THE COLUMN TYPE QUESTIONS :Each question constant statments given in two columns which have to be matched. Statements(A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II.

Q.21 Match the following :Coiumn-I Coiumn-II(A) Slope of graph c vs t (abscissa) for zero order (P) unity(B) Slope of graph log c vs (abscissa) for first order (Q) zero

(C) Slope of dc

dt

vs c for zero order (R) -k

(D) Slope ofdc

dt

vs In c first order (S) -k

2.303

Q.22 Match the following :Coiumn-I Coiumn-II(A) Half life of first order reaction (P) Active mass

(B) Arrhenius equation (Q) k = aE / RTAe

(C) Molar concentration (R) t1/2

=0.693

k(D) half life period of zero order reaction (S) a/2k

Q.23 Match the following :Coiumn-I Coiumn-II

(A) Zero (P) k = x

at a x

(B) One (Q) k =x

t

(C) Two (R) k = 22

x 2a x

t.2a a x


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(D) Three (S) k =2.303

t log

a

a x

Q.24 Match the following :Coiumn-I Coiumn-II(A) If the activation energy is 65 kJ then how much time (P) 5 faster a reaction proceed at 25oC than 0oC(B) Rate constant of a first-order reaction is 0.0693 min-1. (Q) 11 If we start with 20 mol L-1, it is reduced to 2.5 mol L-1 in how many minutes(C) Half-lives of first - order and zeroth order reaction are (R) 30 same. Ratio of rates at the start of reaction is how many times of 0.693(D) the half-life periods are given, (S) 1/4

[A]0(M) 0.0677 0.136 0.272 (T) 2

t1/2

(sec) 240 480 960 (U) 0order of the reaction is

PASSAGE BASED QUESTIONS :.

PASSAGE -1Consider the inter conversion of nitrosotriacetoamine into nitrogen phorone and water.

N

CCH (aq)3 N (g) + H O( )2 2

CH3H C3

H C3

NO

O C

O

H C–C3 C–CH3

CH3 CH3

+(aq)-20kJ/mol

The reaction is 1st order in each dieection, with an equilibrium constant of 104, the activationenergy for the forward reaction is 57.45 kJ/ mol. Assuming Arrhenius pre-exponential factor of

1012s-1..

Q.25 What is the expected forward constant at 300K, if we initiate this reaction starting with onlyreactant(A) 102 (B) 106 (C) 108 (D) 104

.

Q.26 If the change in entropy of the reaction is 0.07 kJ K-1mol-1 at 1 atm pressure. Calculate up to whichtemperature the reaction would not be spontaneous. (For forward reaction)

(A) T < 285.7 K (B) T > 250 K (C) T < 340.2 K (D) T > 200 K

.

Q.27 Calculate Kpof the reaction at 300 K

(A) 2.4 × 104 atm-1 (B) 104 atm (C) 24.6 × 104 atm (D) 2.82 × 102 atm-1

.

.


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PASSAGE-2The reaction between A and B is of first order with respect toA and of zero order with respect toB. Fill in the blanks in the following table :Experiment [A]/M [B]/M Initial Rate/M min-1

I 0.1 0.1 2.0 × 10-2

II × 0.2 4.0 × 10-2

III 0.4 0.4 yIV z 0.2 2.0 × 10-2

.

Q.28 The value of x is-(A) 0.2 (B) 0.8 (C) 0.1 (D) 0.5

.

Q.29 The value of y is-(A) 0.2 (B) 0.08 (C) 0.1 (D) 0.5

.

Q.30 The value of z is-(A) 0..2 (B) 0.8 (C) 0.1 (D) 0.5

..

PASSAGE-3For the reaction, A + B productFollowing initial rate data is given :Exp. No. [A]

0(m/1) [B]

0(m/1) Initial rate (m/1/s)

1. 1.5 4.5 3 × 10-3

2. 3 4.5 6 × 10-3

3. 1.5 9 3 × 10-3

.

Q.31 Order of reaction with respect to A and B is:(A) 1, 1 (B) 1, 2 (C) 2, 1 (D) 1, 0

.

Q.32 Rate constant of reaction is:(A) 2 × 10-3sec-1 (B) 10-3sec-1 (C) (1/2) × 10-3sec-1 (D) 2.25 × 10-3sec-1

.

Q.33 Time required to decrease concentration of A form 16 M to 1M is:

(A) 346.5 sec (B) 639 sec (C) 1386 sec (D) 6930 sec


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FEEL THE HEAT(Previous Year’s JEE Questions)

Q.1 The degree of dissociation is 0.4 at 400K & 1.0 atm for the gasoeus reactionPCl5 PCl3 + Cl2(g). Assuming ideal behaviour of all gases. Calculate the density of equilibriummixture at 400K & 1.0 atm pressure. [JEE 1999]

Q.2 When 3.06g of solid NH4HS is introduced into a two litre evacuated flask at 27°C, 30% of thesolid decomposes into gaseous ammonia and hydrogen sulphide.

(i) Calculate KC & KP for the reaction at 27°C.(ii) What would happen to the equilibrium when more solid NH4HS is introduced into the flask?

[JEE 2000]

Q.3 When two reactants A and B are mixed to give products C and D, the reaction quotient Q, at theinitial stages of the reaction :[JEE 2000](A) is zero (B) decrease with time(C) independent of time (D) increases with time

Q.4 For the reversible reaction : [JEE 2000]N2(g) + 3H2(g) 2NH3(g) at 500°C. The value of Kp is 1.44 × 10–5, when partial pressure ismeasured in atmospheres. The corresponding value of Kc with concentration in mol L–1 is :(A) 1.44 × 10–5 /(0.082 × 500)2 (B) 1.44 × 10–5 /(8.314 × 773)2

(C) 1.44 × 10–5 /(0.082 × 500)2 (D) 1.44 × 10–5 /(0.082 × 773)2

Q.5 At constant temperature, the equilibrium constant (KP) for the decomposition reaction.N2O4 2NO2 is expressed by KP = 4x2P/(1 – x2) where P is pressure, x is extent of decomposition.Which of the following statement is true ? [JEE 2001](A) KP increases with increase of P(B) KP increases with increase of x(C) KP increases with decrease of x(D) KP remains constant with change in P or x

Q.6 When 1-pentyne (A) is treated with 4N alcoholic KOH at 175°C, it is converted slowly into anequilibrium mixture of 1.3% 1-pentyne (A), 95.2% 2–pentyne (B) & 3.5% of 1, 2,–pentadiene(C). The equilibrium was maintained at 1750C. Calculate G° for the following equilibria.

B = A G10 = ?

B = C G20 = ?

From the calculated value of G10 & G2

0 indicate the order of stability of A, B & C. Write areasonable reaction mechanism sharing all intermediate leading to A, B & C.[JEE 2001]

Q.7 Consider the following equilibrium in a closed container : N2O4(g) 2NO2(g).At a fixed temperature, the volume of the reaction container is halved. For this change, which ofthe following statements holds true regarding the equilibrium constant (KP) and degree ofdissociation () : [JEE 2002](A) Neither KP nor changes


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(B) Both KP and change(C) KP changes, but does not change(D) KP does not change, but changes

Q.8 N2O4(g) 2NO2(g)This reaction is carried out at 298 K and 20 bar. 5 mol each of N2O4 and NO2 are taken initially.

Given: oON 42

G = 100 kJ mol–1; oNO2

G = 50 kJ mol–1

(i) Find G for reaction at 298 K under given condition.(ii) Find the direction in which the reaction proceeds to achieve equilibrium. [JEE 2004]

Q.9 If Ag+ + NH3 [Ag(NH3)]+ ; K

1 = 1.6 × 103 and [JEE 2006]

[Ag(NH3)]+ + NH

3 [Ag(NH

3)

2]+ ; K

2 = 6.8 × 103 .

The formation constant of [Ag(NH3)

2]+ is :

(A) 6.08 × 10–6 (B) 6.8 × 10–6 (C) 1.6 × 103 (D) 1.088 × 107

Q.10 N2 + 3H

2l 2NH

3

Which is correct statement if N2 is added at equilibrium condition?(A) The equilibrium will shift to forward direction because according to II law of thermodynamics

the entropy must increases in the direction of spontaneous reaction.(B) The condition for equilibrium is

322 NHHN G2G3G where G is Gibbs free energy per mole

of the gaseous species measured at that partial pressure. The condition of equilibrium isunaffected by the use of catalyst, which increases the rate of both the forward and backwardreactions to the same extent.

(C) The catalyst will increase the rate of forward reaction by and that of backward reaction by .(D) Catalyst will not alter the rate of either of the reaction. [JEE 2006]

Q.11 STATEMENT-1 : For every chemical reaction at equilibrium, standard Gibbs energy ofreaction is zero

andSTATEMENT-2 : At constant temperature and pressure, chemical reactions are spontaneousin the direction of decreasing Gibbs energy.(A) STATEMENT-1 is True, STATEMENT-2 is True ; STATEMENT-2 is a correct explanationfor STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True ; STATEMENT-2 is NOT a correctexplanation for STATEMENT-1(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is True [JEE 2008]

Q.12. For a first order reaction A P, the temperature (T) dependent rate constant (k) was found tofollow the equation

log k = -(2000)T1

+ 6.0

The pre-exponential factor A and the activation energy Ea, respectively,are(A) 1.0 × 106 s-1 and 9.2 kJmol-1 (B) 6.0 s-1 and 16.6 kJmol-1


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(C) 1.0 × 106 s-1 and 16.6 kJmol-1 (D) 1.0 × 106 s-1 and 38.3 kJmol-1 [JEE 2009]Q.13. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The

plot that follows Arrhenius equation is [JEE 2010]

(A*)

K

T

(B)

K

T

(C)

K

T

(D)

K

T

Q.14. The concentration of R in the reaction R P was measured as a function of time and thefollowing data is obtained:[R] (molar) 1.0 0.75 0.40 0.10t (min.) 0.0 0.05 0.12 0.18The order of the reaction is? Sol. 0

Q.15. For the first order reaction2N

2O

5(g) 4NO

2 g 2O

2(g)

(A) the concentration of the reactant decreases exponentially with time(B) the half-life of the reaction decreases with increasing temperature(C) the half-life of the reaction depends on the initial concentration of the reactant(D) the reaction proceeds to 99.6% completion in eight half-life duration

Q.16. An organic compound undrgoes first order decomposition. The time taken for itsdecomposition to 1/8 and 1/10 of its initial concentration are t

1/8 and t

1/10 respectively. What is

the value of 100×][

][

10/1

8/1

t

t? [JEE-2012]


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BOOST YOUR BASICSAnswer Key1. Order =3, k = 0.375 mol-2L2sec-1 2. Rate = [OCl-]1[I-]1[OH-]-1 3. rate = 4.8×10-4

4. 0.2, 8×10-2, 0.1 5. rate = 6.4×10-10 molL-1 sec-1 6. k = 9×10-3mol-1Lsec-1

7. order =3 8. Rate = k[A]2[B] 9. Rate = k[NO]2[Br2] 10. 0.15 mol L–1 hr–1

11. 6.0 ×10–1 mol L–1 hr–1 12. 2[NO] 3.5 10dt

mol dm–3 s–1, 22[H O] 5.3 10

dt

mol dm–3 s–1

13. 22d[NO ] 1.25 10dt

mol L–1 s–1, 32d[O ] 3.125 10dt

mol L–1 s–1 14. 6.0 × 10–4 molL–1 min–1

15. 53.59 kJ mol–1 16.29.39 kJ mol–1 17. k = 3.19 × 10–2 s–1 18. 9.06 × 10–3 s–1

19. 6.4 × 109 20. 3.9 × 1012s–1 21. 100.343 kJ mol–1 22. 8.46 Msec–1

23. 23.0 second 24.1.98 × 10–2 min–1 25. 0.2197 min-1 26.2.005 s

27. 287.7 s 28. k =8.59 × 10–4 min–1 29. 1.969 min 31.t1/2

=128.3 hours;5.27%

32. k = 2.232 × 10–2 min–1; t = 62.12 min 33. 0.3838 min 34. Initial rate = 2 ×10–4

mol L–1 s–1; 0.036 mol 35. t1/2

= 66.66 min, k = 0.01039 min–1 36. )(2ln

1

23

3

PP

P

tk

37. )(5

4ln

1

23

3

PP

P

tk

38. k = )VV(5

V4n

t

l

23

3

l 39. ]A[

]C[ =

11

10(e11x1x – 1)

40. t = 4 min

Insight To The concepts1. 4.83 min 2. k = 0.0327 min-1 3. 86.47 kJmol-1 4. 4.810-3

5. k = 1.910-2 min-1 6. Ef = 6104,E

b = 9.3104 J 7. 21.5 hrs 8. 1.58 yrs

9. 2412 seconds 10.33

231

][

][

kAk

Akk

11. k2=710-3

12. (i) 2NO2 NO + O

3 (fast) (ii) NO

3 + CO CO

2 +NO

2

13. t1/2

= 1.386 time 14. 3.4 min

Objective exercise1. C 2.B 3.A 4.D 5.C 6.B 7.D 8.C 9.C 10.D 11.C 12.B13.D 14.D 15.D 16.B 17.A 18.A 19.B 20.D 21.B 22.B 23.D 24.C25.D 26.A 27.D 28.A 29.C 30.D 31.B 32.A 33.A 34.B 35.B 36.D37.A 38.A 39.B 40.A 41.B 42.B 43.A 44.D 45.A 46.D 47.D 48.A49.C 50.A 51.A 52.A 53.B 54.B 55.D 56.A 57.D 58.B 59.A 60.B61.A 62.C 63.B 64.A 65.C 66.B 67.A 68.A


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Subject Aptitude1. a,b 2. a,b 3. a,b 4.a,b,c 5. a,b 6. a,b,c,d 7.a,b 8. a,b 9. a,b,c 10. a,c 11.a,b 12. a,c

13. a,b,d 14. a,b,d 15. a 16. c 17. b 18. a 19. b 20. b 21. A r, B s,

Cq,D p 22. Ar, Bq Cp Ds 23.Aq, Bs Cp Dr 24. Aq, Br Ct

Du 25. b 26. a 27. c 28.a 29. b 30. c 31. d 32. a 33. c

Feel The HeatQ.1 4.54 g dm–3

Q.2 (i) Kc= 8.1 × 10–5 mol2 L2 ; Kp = 4.91 × 10–2 atm2 (ii) Noeffect;

Q.3 D Q.4 D Q.5 D

Q.6 15991 J mol–1 , 12304 J mol–1 ; B > C > A Q.7 D

Q.8 (i) 5.705 × 103 J mol–1

(ii) Since initial Gibbs free energy change of the reaction is positive, so the reverse reaction willtake place

Q.9 D Q.10 B Q.11 D Q.12. D Q.13. AQ.14. 0 Q.15. (A, B, D) Q.16. 9

Chemical Kinetics

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