Chemical Kinetics for iit jee

CONTENTS S.NO.

TOPIC

PAGE NO

1.

INTRODUCTION

2

2.

IMPORTANT TERMSAND DEFINITIONS

2

3.

INSTANTANEOUS RATE OFTHE REACTION

4

4.

FACTORSAFFECTING THE RATE OFREACTION

4

5.

SPECIFIC REACTION RATE

5

6.

DISTINCTION BETWEEN RATE OF REACTIONAND RATE CONSTANT 5

7.

RATE LAW

6

8.

ORDER OF REACTION

6

9.

MOLECULARITYOFAREACTION

7

10.

ELEMENTARY REACTIONS

7

11.

COMPLEX REACTIONS

7

12.

EQUILIBRIUMAPPROACH

7

13.

STEADYSTATEAPPROXIMATION

8

14.

TRUEANDAPPARENT RATE CONSTANT

9

15.

PSEUDO UNIMOLECULAR REACTION

9

16.

TYPES OF REACTIONS BASED ON ORDER OF REACTION

10

17.

MISCELLENEOUS REACTIONS

15

18.

DETERMINATION OF ORDER OF REACTION

19

19.

TEMPERATURE EFFECT

20

20.

TEMPERATURE DEPENDENCE OFRATE CONSTANTS

20

21.

MECHANISM BASED ON INTERMEDIATE FORMATION

24

22.

ACTIVATION ENERGYAND THE RATE OF REACTION

24

23.

ACTIVATION ENERGIES OFAREVERSIBLE REACTION

25

24.

ACTIVATION ENERGIESAND ENERGY CHANGE DURING REACTION

26

25.

MAXWELL'S- BOLTZMANN DISTRIBUTION CURVE

26

26.

EFFECT OFCATALYST

27

27.

COLLISION THEORYOFCHEMICALREACTIONS

27

28.

THRESHOLD ENERGYAND ORIENTATION OFCOLLIDING MOLECULES 28

1

Chemical Kinetics

CHEMICALKINETICS 1. INTRODUCTION : Chemical Kinetics means study of rate of chemical reactions. Rate of reaction depends on amount of change in concentration of reactants per unit time. Rate of reaction is influenced by various factors like nature of substance, physical state of substance, temperature, concentration, presence of catalyst, etc. Classification of reactions : [In terms of rates] : (i) Fast reactions  too fast e.g. Detonation of explosives, acid-base neutraliztion, precipitation of AgCl by NaCl andAgNO 3. (ii) Moderate reaction  Neither too fast nor too slow e.g. combination of H2 and Cl2 in presence of light, hydroysis of ethyl acetate catalyzed by acid, decomposition of azomethane. (iii) Very Slow reaction There are certain reactions which are too slow e.g. rusting ofiron, weathering of rocks.

2. IMPORTANT TERMS AND DEFINITIONS : 2.1 RATE OF REACTION It is defined as the change in concentration of reactant (or product) in a particular time interval. Unit of rate of reaction is mol L–1 s–1. Change in the concentration of reactan ts or (Products) Time 2.2 AVERAGE RATE OF REACTION Change in the concentration of reactants or products measured over bigger interval of time is called average reaction rate. If c is the change in the concentration of reactants and product in t time, then Change in the concentration of reactan ts or (Products) Average Rate = Time Rate of reaction =

C2 – C1 C =  =± t –t t 2 1

Unit of average velocity =

Note: For the reaction Rate of reaction = –

Unit of concentration gram mo le / Litre = Unit of time Second

= gram mole litre–1 second–1 n1A+ n2B  m1C + m2D 1 [A] 1 [B] 1 [C] 1 [D] =– =+ =+ n1 t n 2 t m1 t m 2 t

2

Chemical Kinetics

Note: For the reaction N2 + 3H2  2NH3 (i) Rate of formation of ammonia = +

 [NH3 ] t

(ii) Rate of disappearance of nitrogen = –

 [N2 ] t

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

Rate = +

1  [NH3 ]  [N2 ] 1  [H 2 ] =– =– 3 t t t 2

Thus, Rate = –

 [N2 ] 1  [NH3 ] = t 2 t

or rate of formation of ammonia = Twice the rate of disappearance of nitrogen i.e. Ex.1

2   [H 2 ]   [NH3 ] – = t  t 3 

For the reaction R  P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Sol.

Average rate = –

[R ] t

=–

=–

[R]2 – [R ]1 t 2 – t1

0.02M – 0.030M – 0.01M =– 25minutes 25minutes

= 4 × 10–4 mol litre–1 minutes–1 = Ex.2

– 0.01M = 6.66 × 10–6 mol litre–1 second–1. 25 60 s

In a reaction, 2A Products, the concentration ofAdecreases form 0.5 mol L–1 to 0.04 mol L–1 in 10 minutes. Calculate the rate during this interval.

Sol.

Average rate = –

1 [A] 1 [A]2 – [A]1 =– 2 t 2 – t1 2 t

× 0.4 M – 0.5 M = – 1 × – 0.1M 10 minutes 10 minutes 2 2 –3 = 5 × 10 mole minutes–1. =–

1

3

Chemical Kinetics

3. INSTANTANEOUS RATE OF THE REACTION : It is the rate of reaction when the average rate is taken over a very short interval of time or rate of reaction at a particular time is known as instantaneous rate. dc c = t 0 t dt

Instantaneous rate = ± Lim

n1A + n2B  m1C + m2D

Note: For the reaction

Instantaneous Rate of reaction = – Ex.3 Sol.

Ex.4 Sol.

1 d[A] 1 d[B] 1 d[C] 1 d[D] =– =+ =+ n1 dT n 2 dT m1 dT m 2 dT

For the rate of reaction : N2 + 3H2  2NH3 in terms of the concentrations of N2, H2 and NH3 can be expressed as 1 d[ NH3 ] d[N 2 ] 1 d[H2 ] Rate = – =+ =– 2 dt 3 dt dt What should be (a) the rate of disappearance of B and (b) the rate of formation of C, if the rate of disappearance ofAfor the reactionA+ B  2C is 10–2 mole/litre/second at a particular temperature ? (a) Rate of disappearance ofA= Rate of disappearance of B = 10–2 mole/litre/second Rate of disappearance of A=

1

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

4. FACTORS AFFECTING THE RATE OF REACTION : (i) Concentration : Law of mass action enunciates that greater is the conc. of the reactants, the more rapidly the reaction proceeds. (ii) Pressure (Gaseous reaction) : On increasing the pressure, volume decreases and conc. increases hence the rate of reaction in creases. (iii) Temperature : It is generally observed that rise in temperature increases the reaction rate. It has been found that rate is either doubled or tripled for every 10º rise in temperature. Temp. coefficient of reaction rate k T  10 kT

 2 or 3

WhereT + 10 and kT are rate constants at two temp. differing by 10º. (iv) Nature of the reactants : The rate depends upon specific bonds involved and hence on the nature of reactants. (v) Surface area of the reactants : In heterogeneous reactions, more powered is the form of reactants, more is the velocity. [as more active centres are provided] (vi) Catalyst : Affects the rate immensely. Presence of positive catalyst increases the rate of reaction by decreasing the activation energy and presence of negative catalyst decreases the rate of reaction by increasing the activation enrgy. Acatalyst mainly affects the activation energy of reaction and hence, the rate constant and rate of reaction changes. (vii) Intensity of Radiation : The rate of photochemical reactions normally increases with increase in intensity of radiation. (viii) pH of the medium

4

Chemical Kinetics

5. SPECIFIC REACTION RATE : Applying law of mass action to the reaction : n1A + n2B  m1C + m2D Rate  [A]n1 [B]n2 or r = k [A]n1 [B]n2 This equation is known as rate law. Where k is the proportionality constant and is called Rate constant or Rate coefficient (Specific reaction Rate) On putting [A] = [B] = 1, we have : r = k Hence, specific reaction rate is the rate of the reaction when the concentration of each reactant is taken as unit. Note: Unit of Specific Reaction Rate conc. = k [conc.]n1 +n 2 r = k [A]n1 [B]n 2  time –1 k = [conc.][1( n1  n 2 )] × [time] [1-(n 1+ n 2 )]

 mole  or k =   litre 

–1

.[second]

where n1 + n2 = n which is order of reaction 1–n

 mole  K=    litre  when n = 0 n=1 n=2 Ex.5

Sol.

  

–1

second

k = mol.lit–1.sec–1 k = sec–1 k = mol–1.lit1.sec–1

Identify the reaction order from each of the following rate constant. (i) k = 2.3 × 10–5 L mol–1 s–1 (ii) k = 3 × 10–4 s–1 Order for k = 2.3 × 10–5 L mol–1 s–1 is 2 and order for k = 3 × 10–4 s–1 is 1 from units.

6. DISTINCTION BETWEEN RATE OF REACTION AND RATE CONSTANT: Rate of a reaction : (i) The rate of reaction at any instant of time depends upon the molar concentrations of the reactants at that time. (ii) Its units are always mole litre–1 time–1. Rate constant : (i) The rate constant is constant for a particular reaction at a particular temperature and does not depend upon the concentration of the reactants. (ii) Its unit depends upon the order of reaction.

5

Chemical Kinetics

7. RATE LAW : (i) It may also not depend upon the concentration of each reactant or product of the reaction. Suppose,mA + nB  Product R  [A]m m[B]n (ii) Rate of a chemical reaction is directly proportional to the concentration of reactants. (iii) The rate law represents the experimentally observed rate of reaction which depends upon the slowest step of the reaction. (iv) Rate law cannot be deduce from the equation for a given reaction. It can be find by experiments only. (v) The rate law may not bear a simple relationship for the stoichiometric equation. (vi) It may not depend upon the concentration of species, which do not appear in the equation for the over all reaction.

8. ORDER OF REACTION : The order of a reaction may be defined as the sum of the powers to which conc. terms must be raised in an experimentally determined differential rate equation : For the reaction : mA + nB  product Experimental rate equation : r = k [A]p [B]q order with respect to A= p order withe respect to B = q Total order = p + q Note: (i) Order may be zero, fractional, integer or negative. (ii) p and q may be equal to mAand mB. Ex.6

Ex.7

Ex.8 Sol. Ex.9 Sol.

Reaction Exp. rate equ. order 0 0 r = k [H ] [Cl ] 0 H2 + Cl2  2HCl 2 2 1/2 3/2 r = k [H2] [Br2] H2 + Br2  2HBr 2 r = k [H2] [I2] H2 + I2  2HI Reaction : CO(g) + Cl2(g)  COCl2(g) r = k [CO]2 [Cl2]1/2 order = 2.5 Reaction : COCl2(g)  CO(g) + Cl2(g) 3/2 r = k [COCl2] order = 1.5 For a reaction, A+ B  Product; the rate law is given by, r = k[A]1/2 [B]2. What is the order of the reaction ? 1 1 Order of the reaction = + 2 = 2 or 2.5. 2 2 The conversion of the molecules X toYfollows second order kinetics. If the concentration X is increased to three times, how will its affect the rate of formation ofY ? For the reaction, X Y, as it follows second order kinetics, the rate law equation will be Rate = k [X]2 If concentration of X is increased to three times, then Rate = k [3X]2 or Rate = 9k [X]2 Thus, the rate of reaction will become 9 times. Hence, the rate of formation ofYwill increase 9 times.

6

Chemical Kinetics

9. MOLECULARITY OF A REACTION : ‘‘Molecularity is defined as the number of molecules, atoms, or radicals that must collide simultaneously in order for the reaction to take place.’’It is always a whole number and cannot be negative. In the elementary processes : Participating species Molecularity One species participates..... unimolecular, 1 Two species participates..... bimolecular, 2 Three species participates....trimolecular, 3 Note: Molecularity cannot be zero, fractional or more than three. Ex.10 N2O4  2NO2 ...... unimolecular (molecularity = 1) H2 + I2  2HI ...... bimolecular (molecularity = 2) 2FeCl3 + SnCl2  2FeCl2 + SnCl4 ...... trimolecular (molecularity = 3) Note : If the reaction takes place in two or more steps then the overall molecularity of the reaction is monitored by the slow or rate determining step. TYPE OFREACTIONS On the basis of number of steps involved in the reaction, they are classified as elementary or complex reactions.

10. ELEMENTARY REACTIONS : These are single step reactions. For such reactions, order and molecularity are always same. In another words, rate law and the law of mass action will have same expression.

11. COMPLEX REACTIONS : These are multi-step reactions. For such reactions, order and molecularity may or may not be same.  In such reactions, some intermediates are formed. Intermediates are the speciesdifferent than reactants as well as products.  Each step of such reaction is elementary reaction.  The overall rate of reaction will be equal to the rate of slowest step. This is why, the slowest step is called the rate determining step (RDS) of reaction. Areaction can have more than one RDS,  The overall molecularity of reaction is the molecularity of the RDS. However, it has no significance.  The order ofreaction may or maynot be the overall order of reaction. If depends on the concentration terms involved in the rate law expression of RDS. If it is not overall order, it may be determined by equilibrium approach or be steady state approximation.

12. EQUILIBRIUM APPROACH : The concentration ofintermediate can be determined from equilibrium constant of the reaction involved. For example, let the mechanism of reactionA+ 2 B  C + D is Step I:

A+ B

k1 k–1

I

k2 I + B  C+D  r = rate of step II = k2[I][B] Now, the overall rate of reaction,

Step II:

...(i)

7

Chemical Kinetics

It cannot be the correct rate law for the overall reaction because the overall rate of reaction should be in terms ofconcentrations ofAand B(reactants). The concentration ofI can be related with the concentrations ofAand B with the help of first equilibrium. For step I, equilibrium constant, Keq =

k1 [I ] = k –1 [A][B]

k1 [I] = k [A][B] –1



k1 2 Putting this value in equation (i), r = k2 [A][B][B] = k[A][B] k–1 , k1 where, k = k –1 Hence, the overall rate of reaction is 1 + 2 = 3

13. STEADY STATE APPROXIMATION : In this method, we assume that the intermediates formed are so reactive that after some time from initiation of reaction (called induction period), the net rate of their formation becomes zero. They react with the same rate of their formation. For example, let the mechanism of reaction A + 2 B  C + D is k1

I

Step I:

A+ B

Step II:

k I + B 2  C + D

k–1

The rate of reaction may be given as r = +

d[C] = k2[I][B] dt

...(ii)

d[I ] =0 dt k1[A][B] – k–1[I] – k2[I][B] = 0

Now, from steady state approximation on the intermediate, I + or, or,

[I] =

k1[A][B] k –1  k 2 [B]

k1[A][B] k k [A][B]2 [B] = 1 2 k–1  k 2 [B] k –1  k 2 [B]  Steady state approximation gives better result when the intermediate is less stable while equilibrium approach gives better result when the intermediate is more stable. More stable intermediate reacts very less and hence, the concentration of intermediate at any time remains nearly equal to its equilibrium concentration.  When conditions of equilibrium approach are applied on the result obtained from steady state approximation, the same rate law expression will come. For the above reaction, step I is faster than step II.As step I is at equilibrium, r1 = r–1 >> r2 or, k–1[I] >> k2[I][B] Putting this value in equation (ii), r = k2

or, 

k–1 >> k2 [B] k–1 + k2 [B]  k–1

k 2k 1[A][B]2  [A][B]2, k –1  k 2 [B] which is exactly same expression obtained from equilibrium approach. Now, the rate expression obtained from steady state is r =

8

Chemical Kinetics

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

H RCCOR' + H2O  RCOOH + R'OH the rate of reaction, r = k[RCOOR'][H2O][H+ ], where k is the true rate constant of the reaction, Now, H+ is the catalyst of reaction and hence, throughout its concentration will remain constant. H2O is solvent for the reaction and taken in large excess and hence, its concentration will also remain constant. Hence, k[H2O][H+] = constant, called apparent rate constant of the reaction.

15. PSEUDO UNIMOLECULAR REACTION : Consider the reaction : CH3COOC2H5 + H2O H CH3COOH + C2H5OH Since water is present in large excess, its concentration hardly changes during the course of the reaction. And as such rate depends only on the concentration of ester. The order is one but the molecular is two. Such reactions are called pseudo unimolecular reaction. Ex.11 The reaction 2NO + Br2  2NOBr, obeys the following mechanism : Fast Step I: NO + Br2 NOBr2 Slow Step II: NOBr2 + NO 2NOBr Suggest the rate expression. Sol. Step II is the rate determining step of the reaction and hence, r = k[NOBr2] [NO] However, NOBr2 is an intermediate and thus its concentration should be determined from the equilibrium of step I. [NOBr2 ] For step I, equilibrium constant, keq = [NO][Br2 ] [NOBr2] = keq [NO][Br2] ... (ii)  2 Thus, by equations (i) and (ii), r = k keq[NO] [Br2] r = k' [NO]2[Br2]. where k' = kk eq or, Ex.12 The following mechanisms are proposed for the reaction CO + NO2  CO2 + NO at low temperature: (a) 2NO2 N2O4 (fast) (slow) N2O4 + 2CO  2CO2 + 3NO (slow) (b) 2NO2  NO3 + NO (fast) NO3 + CO  NO2 + CO2 d[CO 2 ] 2 = k[NO2] ? which of the above mechanism are consistent with the observed rate law : + dt Sol. For mechanism (a), r = rate of step II 2 ...(i) = k[N2O4][CO] [ N 2O 4 ] Now, from step I : keq = [NO ]2 22 or, [N2O4] = keq[NO2] 2 2 Hence from (i), r = k.keq [NO2] [CO] , but it is not the given rate law Hence, the mechanism is not consistent with the rate law. r = rate of step, I = k[NO2]2 , which is the given rate law. For mechanism (b), Ex.13 For a hypothetical reaction A+ B  products, the rate law is, r = k [B] [A]º, the order and molecularity of reaction is : Sol. Order = 1 + 0 = 1, molecularity = 1 + 1 = 2

9

Chemical Kinetics

16. TYPES OF REACTIONS BASED ON ORDER OF REACTION :

Concentration of R

(I) Zero Order Reactions : Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants. Consider the reaction, RP d[R] Rate = – = k [R]0 dt As any quantity raised to power zero is unity d[R] Rate = – =k×1 dt d[R] = – k dt Integrating both sides [R] = – k t + I ...(1) where, I is the constant of integration. At t = 0, the concentration of the reactant R = [R]0 , where [R]0 is initial concentration of the reactant. Substituting in equation (1) [R]0 = –k × 0 + I [R]0 = I Substituting the value of I in the equation (1) [R] = –kt + [R]0 ...(2)

O

K = - slope

Time (t)

Comparing with equation of a straight line, y = mx + c, if we plot [R] against t, we get a straight line with slope = –k and intercept equal to [R]0. Further simplifying equation (2), we get the rate constant, k as [R ]0 – [R ] k= ...(3) t Zero order reactions are relatively uncommon but they occur under special conditions. Some enzyme catalysed reactions and reactions with occure on metal surfaces are a few examples of zero order reactions. The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure. 2NH3 (g)

1130 K Pt catalyst

N2 (g) + 3H2 (g)

Rate = k [NH3]0 = k In this reaction, platinum metal acts as a catalyst. At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration. The thermal decomposition of HI on gold surface is another example of zero order reaction.

10

Chemical Kinetics

Note: (i) Reaction betweenAcetone and Bromine (ii)Dissociation of HI on gold surface (A) Unit of Rate Constant – [R]0 – [R] k= t K

mol lit -1 sec -1

Unit of rate of reaction = Unit of rate constant.

Half-life period (t1/2) - The time in which half reaction is completed R0 At t = t1/2 ; R= 2 (B)

t1/ 2 

R0 2K

or

t1 / 2  R 0

The half life period is directly proportional to the initial concentration of the reactants. Ex.14 The reaction 2A + B + C  D + 2E; is found to be first order inA; second order in B and zero order in C. (i) Give the rate law for the above reaction in the form of a differential equation. (ii) What is the effect on the rate of increasing the concentration of A, B and C two times?

Sol.

(i)

The rate law according to given information may be given as, dx  K[A]1[B]2 [C]0 dt

(ii)

When concentration ofA, B and C are doubled then rate will be dx

 K[2A][2B]2 [C]0

dt  8K[A][B] 2 [C] 0

i.e., rate becomes 8 fold, the original rate. (II) First Order Reactions : In this class of reactions, the rate of the reaction is proportional to the first power of the concentration of the reactant R. For example, RP d[R] d[R] Rate = – = k [R] or – = – kdt dt [R] Integrating this equation, we get ln [R] = – kt + I ...(4) Again, I is the constant of integration and its value can be determined easily. When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant. Therefore, equation (4) can be written as ln [R]0 = –k × 0 + I ln [R]0 = I Substituting the value of I in equation (4) ...(5) ln [R] = – kt + ln[R]0 Rearranging this equation [R] ln = – kt [R ]0 or k =

[R]0 1 ln t [R]

...(6)

11

Chemical Kinetics

At time t1 from equation (4) ...(7) ln[R]1 = –kt1 + ln[R]0 At time t2 ln[R]2 = –kt2 + ln[R]0 ...(8) where, [R]1 and [R]2 are the concentrations of the reactants at time t1 and t2 respectively. Subtracting (8) from (7) ln[R]1 – ln[R]2 = – kt 1 – (–kt 2) ln

[R ]1 = k (t2 – t 1) [R ]2

k=

[R ]1 1 ln (t 2 – t1 ) [R ]2

...(9)

Equation (5) can also be written as [R ]1 ln = –kt [R ]2 Taking antilog of both sides [R] = [R]0 e–kt Comparing equation (5) with y = mx + c, if we plot ln [R] against t (Fig. 4.4) we get a straight line with slope = –k and intercept equal to ln [R]0 The first order rate equation (6) can also be written in the form [R]0 2.303 k= log ...(10) [R] t [R]0 kt log = 2.303 [R] If we plot a graph between log [R]0 / [R] vs t, the slope = k/2.303 Hydrogenation of ethene is an example of first order reaction. C2H4 (g) + H 2 (g)  C 2H 6(g) Rate = k [C2 H4 ] All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics. ln[R0]

ln[R]

k = – slope

O

t

Fig. :Aplot between ln[R] and t for a first order reaction Note: (i) (ii) (A)

226 88 Ra

 24 He 

Fig. : Plot of log[R]0/[R] vs time for a first order reaction

222 Rn 86

Rate = k [Ra] Decomposition of N2O5 and N2O are some more examples of first order reactions. Unit of Rate Constant : [R ] 2.303 log 0 K= t [R] K = sec–1

12

Chemical Kinetics

R0 , 2 R0 2.303 2.303 0.693  log 2 = K= log = R0 / 2 t1 / 2 t1 / 2 t1 / 2 0.693 t1/2 = K Some Examples of First Order reaction and Their Rate constant 1. For Gas Phase Reaction Let us consider a typical first order gas phase reaction A(g)  B(g) + C(g) Let pi be the initial pressure ofAand pt the total pressure at time 't'. Integrated rate equation for such a reaction can be derived as Total pressure p t = p A + pB + pC (pressure units) pA, p B and p C are the partial pressures ofA, B and C, respectively. If x atm be the decrease in pressure ofAat time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each. A(g) B(g) + C(g)  At t = 0 pi atm 0 atm 0 atm At time t (pi–x) atm x atm x atm where, pi is the initial pressure at time t = 0. pt = (pi – x) + x + x = p i + x x = (pt – p i) where, pA = p i – x = p i – (p t– p )i = 2p i– p t (B)

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

p   2.303    log i  k=  pA   t  pi 2.303 = log (2p – p ) t i t 2. Pseudo Unimolecular Reaction : Inversion of cane sugar

...(11)

H C12H22O11 + H2O  C6H12O6 + C6H12O6 Glucose Fructose Sucrose The progress of the reaction can be studied using a polarimeter. Cane sugar and glucose are dextrorotatory while fructose is laevorotatary If 0 is polarimetric reading at zero time. t is polarimetric reading after time t. and  is polarimetric reading after infinite time

2.303  – log 0  t t –  3. (a) Decomposition of H 2O 2 1 Pt H2O2  H2O + O2 2 The progress of the reaction can be studied either by measuring the volume of oxygen gas after different intervals of times or by titrating a definite amount of reaction mixture with standard KMnO4 at different intervals of time. If V0 and V t represent the volumes of KMnO 4 used at the start of the reactions and at any time t, respectively R0  V0, R  V t 2.303 V log 0 Therefore, k= t Vt k=

13

Chemical Kinetics

(b) Decomposition of ammonium nitrite : NH4NO2  N2 + 2H 2O If Vt is the volume of N2 evolved at any time t and V is the volume of N2 evolved when the decomposition is complete, then, V 2.303 k= log V – Vt t 4. Hydrolysis of ethyl acetate (ester) H CH3COOC2H5 + H2O  CH3COOH + C2H5OH Kinetics of this reaction is studied by titrating a definite volume of the reaction mixture with standard alkali solution. If V0, V t and V  are volumes of standard alkali needed to neutralise a definite amount of x will be proportional to Vt – V 0 and a will be proportional to V – V 0.

Hence,

k=

2.303 V – V0 log  t V – Vt .

5. Oxide layer formation 1  k = ln max t max –  where max = Thickness of oxide layer after  times  = Thickness of oxide layer at time 't'. 6. Bacterial Growth : 1 ax k = ln t a Second order reactions: When the rate of a reaction is determined by variation of two concentration terms, the reaction is said to be of second order. For a general reaction, A + B  products, the rate law may be, dx = k [A]2 [B]0, or dt dx = k [A]0 [B]2, or dt dx = k [A] [B] dt

... (x) ... (xi) ... (xii)

Thus, the rate of a second order reaction varies directly as the square of the concentration of reactant. Unit of the second order rate constant : From equation (x) we get dx

Mol / litre

k = dt [A]2 = = lit. mol–1 time–1. Time interval Mol. 2 /lit. 2 The unit of second order rate constant involves concentration as well as time. Examples of Second Order Reactions : 373 K

1.

Conversion of ozone into oxygen at 373 K : 2O3

2.

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

3.

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

4.

Alkaline hydrolysis of an ester (saponification) : CH3 COOC2H5 + NaOH — CH3COONa + C2H5OH

5.

Benzoin condensation, C6H5CHO + OHC.C6H5

KCN alc. reflux

3O2

200ºC

2Cl2 + O2

C6H5CH(OH)CO.C6H5

14

Chemical Kinetics

6.

Isomerisation of ammonium cyanate into urea. O NH4CNO

heat

NH2 – C – NH2

Second order rate equation: All the second order reactions obey the following kinetic equation, 1

x

k = t  a(a  x) where k = second order rate constant a = initial concentration of reactant (s) x = concentration of reactant converted into products after time ‘t’ t = time elapsed Characteristics of a Second Order Reaction : (i)

.... (xiii)

1

When a graph is plotted between ‘t’ and (a  x) a straight line is obtained. On rearranging, equation 1

1

1

Time

(xiii).t = k(a  x)  ka The slope of the line . From this ‘k’ an be evaluated. k

l/(a-x)

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

is expressed, because the value of a(a  x) will be changed when the unit of concentration is changed. (iii) The half-life period (t1/2) of a second order reaction is inversely proportional to the initial concentration of reactant. From expression (xiii) x

1

k = t  a(a  x) when, t = t1/2, x = a/2 1

a/2

k = t  a(a  a / 2) 1/ 2

so, or,

... (xiii)

t1/2 =

1 ka

... (xv)

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

17. MISCELLENEOUS REACTIONS : (a)

Parallel Reactions B [A]0 A ln [A]t = (k1 + k2)t C [A] = [A0] e–kt

[B] =

k1[A]0 (1 – e–kt) k1  k 2

[B] k1  [C] k 2 k 2 [A]0 [C] =

k1  k 2

(1 – e–kt)

15

Chemical Kinetics

(b) Consequtive Reaction k2 C A k1  B   [A]t = [A0] e– k1t

;

k1[A]0 – k1t – k 2 t –e [B]t = (k – k ) e 2 1

;

 k  k 2 – k1 [B]max = [A0] .  1   k2 





k2

[C]t = [A0] – ([A] t + [B] t) conc.

[C] [[B] [A]

tmax =

K 1 n 1 K1 – K 2 K 2

time (c) Reversible reaction Consider the reaction A At time t = 0 At time t = t At time t = taq

kr kb

a a–x a – xeq

 x eq    ln  x – x  = (kf + kb)t eq  

B b x xeq x eq ;

a – x eq



kf kb

Table of Formulae

16

Chemical Kinetics

Important Graphical Representation Zero Order First Order

Second Order

Third Order

Ex.15 Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. Sol. For a first order reaction, k=

0.693 0.693 = = 1.155 × 10–2 s–1. t1 / 2 60 minutes

Ex.16 Afirst order reaction has a rate constant 1.15 × 10–3 s–1. How long will 5 g of this reactant take to reduce to 3 g ? Sol. Given : [A]0 = 5 g, [A] = 3 g, k = 1.15 × 10–3 s–1. Applying first order kinetics equation and substituting the values, we get t=

=

[A]0 2.303 log [A] t

2.303 5g log = 2.00 × 103 (log 1.667) s –3 3g 1.15 10

= 2.0 × 103 × 0.2219 s = 443.8 s

17

Chemical Kinetics

Ex.17 A first order reaction has a specific reaction rate of 10–2 s–1. How much time will it take 10 g of its reactant to be reduced to 2.5 g ? Sol. Rate constant, k = 10–2 s–1 Initial reactant conc., [A]0 = 10 g Final reactant conc., [A]t = 2.5 g Time required, t = ? For a first order reaction, t=

2.303 [A]0 2.303 10 g log = – 2 –1 log = 2.303 × 102 log 4 s = 230.3 × 0.6020 s = 138.6 s. [A]t k 2.5 g 10 s

Ex.18 Areaction that is of first order with respect to reactant Ahas rate constant 6 min–1. If we start with [A] = 5.0 mol L–1, when would [A] reach the value of 0.5 mol L–1 ? Sol. Rate constant = 6 min–1 [A]0 = 5.0 mol L–1 [A]t = 0.5 mol L–1 t=? For a first order reaction, t=

=

5 mol L–1 2.303 [A]0 2.303 log log = [A]t 6 min –1 0.5 mol L–1 k

2.303 log 10 min = 0.39 min. 6

Ex.19 For a first order reactions, it takes 5 minutes for the initial concentration of 0.6 mol L–1 to become 0.4 mol L–1. How long in all will it take for the inital concentration to become 0.3 mol L–1 ? Sol. For a first order reaction, k=

2.303 k

log

[A]0 [A]t [A]t = 0.4 mol L–1

We have, [A] 0 = 0.6 mol L t = 5 min So, 2.303  0.1761 2.303 0.6 2.303 k= × log = log 1.5 min–1 = min–1 = 8.1 × 10–2 min–1 0.4 5 5 5 min For, [A]t = 0.3 mol L–1 –1

t=

2.303 2.303 2.303 [A]0 0.6 log = = log 2 min = 8.5 min. – 2 log [A]t 8.110 0.3 8.110– 2 k

Ex.20 The following data were obtained during the first order thermal decomposition of N2 O5 (g) at constant volume : 2N2O5 (g)  2N2O4 (g) + O2 (g) S.No. Total Pressure/(atm) Time /s 1. 0.5 0 2. 0.512 100 Calculate the rate constant.

18

Chemical Kinetics

Sol.

Let the pressure of N2O5 (g) decrease by 2x atm.As two moles of N2O 5 decompose to give two moles of N2O4 (g) and one mole of O 2 (g) the pressure of N 2O 4 (g) increases by 2x atm and that of O 2 (g) increases by x atm. 2N2O4 (g) + O2 (g) 2N2O5 (g)  Start t = 0 0.5 atm 0 atm 0 atm At time t (0.5 – 2x) atm 2x atm x atm pt = p N 2O 5 + p N 2 O 4 + pO 2 = (0.5 – 2x) + 2x + x = 0.5 + x x = pt – 0.5 = 0.5 – 2 (p t – 0.5) = 1.5 – 2p t At t = 100 s; pt = 0.512 atm p N 2O 5 = 1.5 – 2 × 0.512 = 0.476 atm Using equation (11) k=

=

pi 2.303 2.303 0.5atm log = log t pA 100s 0.476atm

2.303 × 0.0216 = 4.98 × 10 –4s –1. 100s

Ex.21 For a reaction 2A  Product, it is found that the rate of reaction becomes 2.25 when the concentration ofAis increased 1.5 times, calculate the order of reaction. r = k [A]n Sol. Rate, .... (1) When concentration is increased 1.5 times, the rate of reaction become 2.25. 2.25 r = k [1.5A]n .... (2)  Divide (2) by (1)

2.25r r

k[1.5A]n

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

18. DETERMINATION OF ORDER OF REACTION : (I)

Integration Method In this method, value of K is determined by putting values of initial concentration ofreactants andchange in concentration with time in kinetic equation of first, second and third order reactions. The equation by which constant value of K is obtained is called order of that reaction. R0

K1 =

2.303 t

K2 =

x 1 a(a - x) t 1

log

R

(For first order reaction)

(For second order reaction)

x(2a - x)

K3 = 2t a2 (a - x)2 (For third order reaction)

19

Chemical Kinetics

(II)

van’t Hoff Differential Method van’t Hoff gave the following relationship between velocity V of nth order reaction and concentration of reactants, C. n=

log(V1 / V2 ) log(C1 / C2 )

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

Graphical Method If a straight line is obtained on drawing a graph between log (a – x) and time or dx , then it is first order dt reaction. If a straight line is obtained on drawing a graph between (a – x)2 and

dx dt

, then it is second order

reaction. (IV)

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

t1/2 

1 an 1

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

Ostwald Isolation Method This method is used to find out the order of complex reactions. If nA, nB and nC molecules of substance A, B and C, respectively, are present in a reaction, then nA+ nB + nC will be the order of reaction. When B and C are in excess, the order of reaction will be nA. WhenAand B are in excess, the order of reaction will be nC. WhenAand C are in excess, the order of reaction will be nB.

19. TEMPERATURE EFFECT : The rate of reaction is dependent on temperature. This is expressed in terms of temperature coefficient which is a ratio of two rate constants differing by a temperature of 100. Generally the temperature selected are 298K and 308K. It is mathematically expressed as, Temperatur e coefficien t 

rate constant at 308K rate constant at 298K

According to collision theory the reaction rate depends on collision frequency and effective collisions. For a molecule to have effective collision it should fulfill two conditions : (i) Proper orientation (ii) Sufficient energy

20. TEMPERATURE DEPENDENCE OF RATE CONSTANTS : Arrhenius Equation the temperature dependence of the rate of a chemical reaction can be accuratly explain byArrhenius eq. k = AeE a / RT ...(i) where k =Arrhenius factor, E a =Activation energy, R = Gas Constant, T = Temperature

20

Chemical Kinetics

For most reactions, the rate constant increases as the temperature increases. These rate constants in most of the cases vary with temperature according to theArrehenius equation : The parametersAand E a for a given reaction are collectively calledArrhenius parameters. Ea is called the 1 activation energy. Value of E a is determined from the graph for and ln k determined experimentally. T Value ofAin turn is calculated once E ais known. In theArrhenius equation (i) the factor e Ea / RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. Taking natural logarithm of both sides of equation (i) we get Ea ...(ii) RT The plot of ln k vs 1/T gives a straight line according to the equation (ii) as shown in (Figure below). ln k = lnA–

Intercept = ln A slope = –Ea/R ln k 0 1/T A plot between ln k vs 1/T

Thus, it has been found fromArrhenius equation (i) that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant. Ea In the plot, slope = – and intercept = lnA. So we can calculate E and Ausing these R values. a At temperature T1, equation (ii) is

Ea ln k1 = – RT + lnA 1

...(iii)

At temperature T2, equation (ii) is ln k2 = –

Ea + lnA RT2

...(iv)

(sinceAis constant for a given reaction) k1 and k2 are the values of rate constants at temperatures T1 and T2 respectively. Subtracting equation (iii) from (iv), we obtain ln k2 – ln k1 =

Ea E – a RT1 RT2

1 Ea  1 k2   =  R  T1 T2  k1 Ea  1 1 k2   log =  2.303R  T1 T2  k1

ln

21

Chemical Kinetics

Ex.22 What will be the effect of temperature on rate constant ? Sol. Rate constant of a reaction is nearly doubled with rise in temperature by 10°. The dependece of therate constant on temperature is given byArrhenius equation, k =Ae – Ea / RT , whereAis called frequency factor and Ea is the activation energy of the reaction. Ex.23 The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea. Sol. Here we are given that When T1 = 298 K, k 1 = k (say) When T2 = 308 K, k 2 = 2k k2 E a  T2 – T1    = k1 2.303 R  T1T2 

log

Substituting these values in the equation, we get log

Ea 308 K – 298K 2k = 2.303  8.314 J K –1mol – × 298 K  308K k 1

Ea 10 × 2.3038.314 298 308

or

log 2 =

or

0.3010 =

or

Ea =

or

Ea = 52897.8 J mol–1 = 53.6 kJ mol–1.

E a 10 2.303 8.314  298 308

0.3010  2.303 8.314  298 308 10

Ex.24 How the value of activation energy is calculated from the rate constants at two different temperatures. If the value of activation energy is 50 kJ/mol show that by increasing the temperature from 300 K to 310 K rate constant becomes nearly double. Sol. We know that Arrhenius equation can be written as Ea 1 log k = logA– 2.303 R T If k1 and k2 are the rate constants at two different temperature T1 and T2 thenArrhenius equation at both the temperatures can be written as Ea 1 2.303 R T1 1 Ea log k2 = log A– 2.303 R T2 Substracting eq. (i) from (ii), we get log k1 = log A–

log k2 – log k1 = or

log

Ea 1 1  – 2.303 R  T1 T2 

k2 Ea 1 1  – = k1 2.303 R  T1 T2 

22

Chemical Kinetics

By knowing the values of k1 and k2 at temperature T1 and T 2 value of Ea can be calculated. Given T 1 = 300 K, T 2 = 310 K, E a = 50 kJ or 50,000 J Put these values in eq. (iii) log

or

log

k2 k1

=

1   1 – 2.303  8.314  300 310 

=

50000  310 – 300  2.3038.314  310  300

50000

k2 = 0.2808 k1

k2 = Antilog 0.2808 = 1.91  2 k1 It is clear that by increasing the temperature from 300 K to 310 K rate constant becomes double. Ex.25 Afirst order reaction is 50% complete in 30 minutes at 27°C and in 10 minutes at 47°C. Calculate the rate constants at 27°C and 47°C, and the energy of activation of the reaction in kJ/mol. Sol. Time for the completion of 50% reaction means t1/2 . It means t1/2 of the reaction at 27°C is 30 minutes and at 47°C is 10 min. We know that k = or

k=

0.693 t1 / 2 0.693

= 0.0231 min–1 (at 27°C) 30 0.693 k= = 0.0693 min–1 (at 47°C) 10

and We know that

Ea k2 1 1  – = 2.303 R  T1 T2  k1 k2 = 0.0693 min–1, k 1 = 0.0231 min–1, T1 = 27 + 273 = 30 K, T 2 = 47 + 273 = 300 K, R = 8.314 JK–1 mol–1, E a = ? log

log

or

1  Ea 0.0693  1 – =  0.0231 2.303  8.314  300 320 

log 3 =

Ea 2.303  8.314

0.4771 =

or

320 – 300   300  320 

Ea  20   2.303  8.314  300  320

2.303  8.314  300  320  Ea = 0.4771   20 = 43848.49 J mol–1 Ea = 43.85 kJ mol–1

23

Chemical Kinetics

21. MECHANISM BASED ON INTERMEDIATE FORMATION : For understanding it let us use the following simple reaction H (g) + I (g)  2HI (g) 2

2

I

H

H

I

H

I +

+

I I I H H H Formation of HI through the intermediate

For understanding the product formation in the above reaction we highlight the following points: According toArrhenius, this reaction can take place only when a molecule of hydrogen and a molecule of iodine collide to form an unstable intermediate (Figure). It exists for a very short time and then breaks up to form two molecules of hydrogen iodide.  The energy required to form this intermediate, called activated complex (C), is known as activation energy (Ea).  Figure is obtained by plotting potential energy v/s reaction coordinate. The reaction coordinate represents the profile of energy change when reactants change into products.

Potential energy

Activated complex C Activation energy A H2 + I2

B 2HI

Reaction coordinate

Diagram showing plot of potential energy vs reaction coordinate.

 

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

Most probable kinetic energy

Kinetic energy

Distribution curve showing energies among gaseous molecules

22. ACTIVATION ENERGY AND THE RATE OF REACTION : The molecules of the reacting substances must be promoted to the top of the energy barrier before these are able to react. For this to happen, the molecules must absorb energy equal to the activation energy. Depending upon the magnitude of the activation energy, the following three cases are possible: (i) When the activation energy is small. If the activation energy is low, then larger number of the reactant molecules will be able to cross over the top of the energy-barrier. As a result, the reaction will be faster. Thus, if the activation energy is low, then the reaction is fast. (ii) When the activation energy is high, then only a few molecules would be able to cross the top of the energy barrier. As a result, the reaction will be slow. Thus, if the activation energy of a reaction is high, then the reaction is slow.

24

Chemical Kinetics

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

23. ACTIVATION ENERGIES OF A REVERSIBLE REACTION : In a reversible reaction, the reactants react to form products, and the products react back to give the reactant molecules. Thus, in a reversible reaction, there are two reactions proceeding in the opposite directions. Each reaction has its own characteristic activation energy. So, in a reversible reaction there are two activation energies: one for the forward reaction, and the other for the backward reaction. These are commonly called as the energy of activation for the forward reaction (E ) and the energy of activation af for the backward reaction (Ea,b ). Depending upon the relative energies of the reactants and products, the following two cases are possible. (i) When the energy of the products is lower than that of reactant When the energy of products is lower than that of the reactant, [Fig. (a)], then Activation energy for the forward reaction < Energy of activation for the backward reaction or, Eaf < Ea,b Therefore, the reaction in the forward direction is faster than that in the backward direction. Er

Ea,f

Energy

Ea,b

ER

Reactant

Ep

Product

Reaction coordinate (Exothermic reaction) (a)

(ii) When the energy of products is higher than that of the reactants When the energy of products is higher than that of the reactants [Fig (b)], then Activation energy for the forward reaction >Activation energy for the backward reaction or,

Eaf > Ea,b Er Ea,b

Ea,f

Energy

Ep Product

ER

Reactant

Reaction coordinate (Endothermic reaction) (b)

Thus, the reaction in the forward direction is slower than the reaction in the backward direction. It must be remarked here that the activation energy of a reaction in any direction determines the speed (rate) of reaction in that direction. It does not say anything about the extent of reaction in that direction. Extent of reaction in any direction is governed by the equilibrium constant of the reaction.

25

Chemical Kinetics

24. ACTIVATION ENERGIES AND ENERGY CHANGE DURING REACTION: In a reaction, the energy change is given by, Energy change in a reaction = Energy of products – Energy of reactants or, (i) E = EP – ER According to the concept of activation energy, the reactant molecules in the forward reaction, and the products molecules in the backward reaction must pass over the top of the energy-barrier (ET). Then, one can write Eq.(i), in the following form E = EP – ER + ET – ET = (ET – ER) – (ET – EP) or,

E = Ea,f – Ea,b

(ii) (iii)

Under constant pressure conditions, E =H, So, H = Ea,f – Ea,b (a) When Ea,f < Ea,b E = – ve H = – ve

and, Thus, when the activation energy for the forward reaction is less than that for the backward reaction, energy is released during the course of reaction. (b) When

Ea,f > Ea,b E = + ve H = + ve

and, Thus, when the activation energy for the forward reaction is more than that for the backward reaction, energy is absorbed during the course of reaction.

25. MAXWELL'S- BOLTZMANN DISTRIBUTION CURVE : Since it is difficult to predict the behavior of any one molecule with precision, Ludwig Boltzmann and James Clark Maxwell used statistics to predict the behaviour of large number of molecules.According to them, the distribution of kinetic energy may be described by plotting the fraction of molecules (NE /NT ) with a given kinetic energy (E) vs kinetic energy (Figure). Here, NE is the number of molecules with energy E and NT is total number of molecules. t (t+10)

Energy of activation

This area shows fraction of molecules reacting at t

This area show s fraction of additional mole cules which react at (t + 10

Kinetic energy

Distribution curve showing temperature dependence of rate of a reaction

26

Chemical Kinetics

The peak of the curve corresponds to the most probable kinetic energy, i.e., kinetic energy of maximum fraction of molecules. There are decreasing number of molecules with energies higher or lower than this value. When the temperature is raised, the maximum of the curve moves to the higher energy value (Figure) and the curve broadens out, i.e., spreads to the right such that there is a greater proportion of molecules with much higher energies. The area under the curve must be constant since total probability must be one at all times. We can mark the position of E on Maxwell Boltzmann distribution curve. a Increasing the temperature of the substance increases the fraction of molecules, whichcollide withenergies greater than Ea. It is clear from the diagram that in the curve at (T + 10), the area showing the fraction of molecules having energy equal to or greater than activation energy gets doubled leading to doublingthe rate of a reaction.

26. EFFECT OF CATALYST : A catalyst is a substance which alters the rate of a reaction without itself undergoing any permanent chemical change. For example, MnO2 catalyses the following reaction so as to increase itsrate considerably. MnO2 2KClO3   2KCl + 3O2 Catalytic Mechanism : The action of the catalyst can be explained by intermediate complex theory. According to this theory, a catalyst participates in a chemical reaction by forming temporary bonds with the reactants resultingin an intermediate complex. This has a transitory existence and decomposes to yield products and the catalyst. It is believed that the catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy between reactants and products and hence lowering the potential energy barrier as shown in the figure.

Potential energy

Reaction path without catalyst

Reactants

Reaction path with catalyst

Energy of activation with catalyst

Energy of activation without catalyst

Products

Reaction coordinate Effect of catalyst on activation energy

It is clear fromArrhenius equation, k =Ae Ea / RT , that the lower the value of activation energy faster will be the rate of a reaction. Note: for catalysts  Asmall amount of the catalyst can catalyse a large amount of reactants.  Acatalyst does not alter Gibbs energy, G of a reaction.  It catalyses the spontaneous reactions but does not catalyse non-spontaneous reactions.  It is also found that a catalyst does not change the equilibrium constant of a reaction rather, it helps in attaining the equilibrium faster, that is, it catalyses the forward as well as the backward reaction to the same extent so that the equilibrium state remains same but is reached earlier.

27. COLLISION THEORY OF CHEMICAL REACTIONS : Though Arrhenius equation is applicable under a wide range of circumstances, collision theory, which was developed by Max Trautz and William Lewis in 1916-18, provides a greater insight into the energetic and mechanistic aspects of reactions. It is based on kinetic theory of gases.

27

Chemical Kinetics

Collision Theory : According to this theory, the reactant molecules are assumed to be hard spheres and reaction is postulated to occur when molecules collide with each other. Various noteworthy points of collision theory are:  Collision Frequency - The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).  Activation Energy - Another factor which affects the rate of chemical reactions is activation energy (as we have already studied).  Bimolecular Elementary Reaction For a bimolecular elementary reaction A + B  Products rate of reaction can be expressed as Rate = ZAB e– Ea / RT ...(vi) where ZAB represent the collision frequency of reactants, A and B and e– Ea / RT represents the fraction of molecules with energies equal to or greater than Ea .

28. THRESHOLD ENERGY AND ORIENTATION OF COLLIDING MOLECULES : The equation (vi) predicts the value of rate constants fairly accurately for the reactions that involve atomic species or simple molecules but for complex molecules significant deviations are observed. The reason could be that all collisions do not lead to the formation of products. The collisions in which molecules collide with sufficient kinetic energy (called threshold energy*) and proper orientation, so as to facilitate breaking of bonds between reacting species and formation of new bonds to form products are called as effective collisions. Note: The formation of methanol from bromoethane depends upon the orientation of reactant molecules as shown inthe figure. The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no products are formed. CH3Br + OH

H H H

CH3OH + Br Improper Orientation

C

Br + OH

H H H

C

Br + OH

No products H

H proper Orientation

HO H

C

Br H

Intermediate

OH

C

H + Br

H

Steric Factor To account for effective collisions, another factor P, called the probability or steric factor is introduced. It takes into account the fact that in a collision molecules must be properly oriented i.e., Rate = PZAB e– Ea / RT Note: for collision theory  In collision theory activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of a chemical reaction.  Collision theory also has certain drawbacks as it considers atoms/ molecules to be hard spheres and ignores their structural aspect.  Comparing the equation (vi) withArrhenius equation, we can say that A(ofArrhenius Equation) is related to collision frequency.

28

Chemical Kinetics

SOLVED EXAMPLES Ex.1

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

Sol.

k1 =

0.6932 0.6932 = = 0.231 min–1. t1 30 2

Let the reaction be a 0 Initial concentration A  Product (a – x) Concentration after 70 minutes x  fraction of the reactant remained unreacted =

(a – x) . a

Now, k1 = or

2.303 a log t a–x

0.0231 = log

2.303 a log 70 a–x

0.0231 70 a = = 0.7021. a–x 2.303

Taking antilog, we get a = 5.036. a–x  Ex.2 Sol.

a–x 1 =  0.2. a 5.036

The specific reaction rate of a first-order reaction is 0.02 s–1. The intial concentration of the reactant is 2 moles/litre. Calculate (a) initial rate, and (b) rate after 60 seconds. We know that for a first-order reaction : Rate of a reaction = k × molar concentration of the reactant. (a)  inital rate = k × initial concentration = 0.02 × 2 = 0.4 mole/litre/second. (b) Now, to calculate rate after 60 seconds, let us first calculate concentration of the reactant after 60 seconds. k=

2 2.303 log concentration after 60 seconds = 0.02. 60

 concentration of the reactant after 60 seconds = 0.60 M.  rate after 60 seconds = k × concentration of the reactant after 60 seconds. = 0.02 × 0.60 = 0.012 mole /litre/second.

29

Chemical Kinetics

Ex.3

Sol.

Ex.4

Sol.

The rate constant is numerically the same for three reactions of first, second and third order respectively, the unit of concentration being in moles per litre. Which reaction should be the fastest and is thistrue for all ranges of concentrations ? Suppose R1, R2 and R3 are the rates of three reactions of first, second and third order respectively and k is the rate constant, which is the same for the three reactions, R1 = k[A]1  R2 = k[A]2 R3 = k[A]3 [A] being the concentration of the reactantAin moles per litre. Now if, [A] = 1, R1 = R2 = R3 ; [A] < 1, R1 > R2 > R3 ; and [A] > 1, R1 < R2 < R3 . For the reaction 2NO + Cl2  2NOCl, it is found that doubling the concentration of both reactants increases the rate by a factor of 8, but doubling the Cl2 concentration alone, only doubles the rate. What is the order of the reaction with respect to NO and Cl2 ? Rate = k [NO]m [Cl2]n Let the concentrations of NO and Cl2 be x and y respectively. According to the question, R1 = k xm yn R2 = k (2x)m (2y)n and, = k . xm yn . 2m + n R2 m+n = 8 = 23. (given) R1 = 2



m+n=3

 Again,

R3 = k (x)m (2y)n = k xm yn . 2n. R3

Ex.5



n R 1 = 2 = 2.



n=1 m = 3 – 1 = 2.

(given)

In a reaction 2N2O5  4NO2 + O2, the rate can be expressed as d[ N 2 O5 ] = k1 [N2O5] dt d[NO 2 ] = k2 [N 2O5] (ii) dt

(i) –

(iii)

d[O 2 ] dt

= k3 [N 2O5]

How are k1, k2 and k 3 related ?

30

Chemical Kinetics

Sol.

The rate law of the given reaction is 1 d[N 2 O5 ] 1 d[NO 2 ] d[O 2 ] = k[N2O5] =  = rate = –  2 dt 4 dt dt

 –

d[N 2O5 ] = 2k [N2O5] = k1 [N2O5] dt

d[ NO2 ] = 4k [N2O5] = k 2 [N2O 5] dt d[O 2 ] = k [N2O5] = k 3 [N2O 5] dt



k1  2k  k 2  4k k  k  3

or k =

k1 k2 = = k3 2 4

or 2k1 = k2 = 4k3. Ex.6

Sol.

The reaction, N2O5  4NO2 + O2, is forming NO 2 at the rate of 0.0072 mol/L/s at some time. (a) What is the rate of change of [O2] at this time ? (b) What is the rate of change of [N2O5] at this time ? (c) What is the rate of reaction at this time ? The rate of the reaction is expressed as 1 d[N 2O5 ] 1 d[NO2 ] d[O2 ] =+ =+ 2 4 dt dt dt d[ NO2 ] and given that = 0.0072 mole/L/s. dt 1 (a) Rate of appearance of O2 = × rate of appearance of NO2 4 d[O2 ] 1 d[ NO2 ] = × dt 4 dt 1 = × 0.0072 = 0.0018 mole/L/s. 4

rate = –

(b) Rate of disappearnace of N2 O5 = –

1 × rate of appearance of NO2 2

1 d[NO 2 ] d[N 2O5 ] = × dt 2 dt

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

(c) Rate of reaction =

d[ NO2 ] 1 1 × = × 0.0072 = 0.0018 mole / L/s. 4 dt 4

31

Chemical Kinetics

Ex.7 Sol.

The rate of a first order reaction is 0.04 mole/L/s at 10 minutes and 0.03 mole/L/s at 20 minutes after initiation. Find the half-life of the reaction. Let the concentrations of the reactant after 10 min and 20 min be C1 and C2 respectively.  rate after 10 min = k.C1 = .04 × 60 and rate after 20 min = k.C2 = .03 × 60. 

4 c1 = 3 c2

Suppose the reaction starting after 10 minutes k=

2.303 2.303 c 4 log 1 = log = 0.02878 10 10 c2 3

 t1/2 = Ex.8

Sol.

0.6932 0.6932 = = 24.086 min. k 0.02878

For a reaction at 800° C, 2NO + 2H2  N2 + 2H2O, the following data were obtained : d[NO]/dt × 10–4 [NO] × 10–4 [H2] × 10–3 mole / litre mole litre–1 min–1 mole / litre 4.4 7.0 (i) 1.5 3.5 2.2 (ii) 1.5 0.24 2.0 (iii) 1.5 What is the order of this reaction with respect to NO and H2 ? From the data (i) and (ii), we see that when the concentration of H2 is halved, the rate is also halved at constant concentration of NO. Hence the reaction is of first order with respect to H2. Let us now consider the data (ii) and (iii) to determine the order with respect to NO as [H2] is constant. The rate law of the above reaction is 1 d[NO] = k[NO]m[H2]1 rate = –  2 dt where m is the order with respect to NO –

or

d[NO] dt

= 2k [NO]m [H2]

Substituting data (ii) and (iii), we get 2.2 × 10–4 = 2k (1.5 × 10–4)m . (2 × 10–3)

...(1) ...(2)

0.24 × 10 = 2k (0.5 × 10 )m . (2 × 10 ) Dividing (1) by (2), –4

–4

–3

(1.5 10 –4 )m 2.2 = = 3m (0.5 10 – 4 )m 0.24 Taking log,

or

or

log 220 – log 24 = m log 3 2.4324 – 1.3802 = m × 0.4771 0.9622 = 0.4771 m

or

m=

220 = 3m 24

0.9622

= 2. 0.4771 Hence the reaction is of second and first order with respect to NO and H respectively.

32

Chemical Kinetics

Ex.8

For a chemical reactionA+ B  Product, the order is 1 with respect to each ofAand B. Find x and y from the given data. Rate (moles/ L/s) [A] [B] 0.10 0.20 M 0.05 M 0.40 x 0.05 M 0.80 0.40 M y

Sol.

The rate law may be written as rate = k [A] [B] Substituting the first set of data in the rate law, we get, 0.10 = k × 0.20 × 0.05 k = 10. Now substituting the second and third sets of data, we get,

And,

Ex.9

Sol.

0.40 = 10 × x × 0.05 x = 0.80 M. 0.80 = 10 × 0.40 × y y = 0.20 M.

The energy of activation and specific rate constant for a first-order reaction at 50° C 2N2O5  2N2O 4 + O 2 (in CCl4) (in CCl4) are 100 kJ/mole and 3.46 × 10–5 s–1 respectively. Determine the temperature at which the half-life of the reaction is 2 hours. Let us calculate the rate constant (say k2) at a temperature (say T2) at which t 1/2 is given to be 2 × 60 × 60 seconds. The rate constant at a temperature 298 K (say T1) is given as 3.46 × 10–5 s–1 (say k 1) k2 = Thus,

0.6932 = 9.62 × 10–5 s–1. 2 60  60

k2 E  T2 – T1  log k = 2.303 R  T T  1  1 2  100000 9.62 10 –5 = log –5 2.303 8.314 3.46 10

or

 T2 – 298     29T2 

T2 = 310 K.

Ex.10 InArrhenius's equation for a certain reaction, the value ofAand E (activation energy) are 4 × 1013 s–1 and 98.6 kJ mol–1 respectively. If the reaction is of first order, at what temperature will its half-life period be ten minutes ? Sol. We have, k = Ae–E/RT ln k = lnA–

E RT

2.303 log k = 2.303 log A–

E RT

33

Chemical Kinetics

or

E log k = logA– 2.303 RT

Given that

A = 4 × 1013 s–1, E = 98.6 kJ mol–1

...(i)

t1/2 = 10 × 60 s. For first-order reaction k =

0.6932 0.6932 = = s–1 t1 / 2 600

Thus (i) becomes, log

98.6 0.6932 = log (4 × 1013) – 2.303 8.31410–3  T 600

[R = 8.314× 10–3 kJ/K/mol]

T = 311.2 K.  Ex.11 The activation energy for the reaction, O3 (g) + NO (g)  NO2 (g) + O 2 (g) is 9.6 kJ/ mole. Prepare an activation energy plot it H° for this reaction is –200 kJ/mole. What is the energy of activation for the reverse reaction ? Sol.

Energy of activation for reverse reaction = 9.6 + 200 kJ = 209.6 kJ.

9.6 kJ

Energy

Ereverse

200 kJ

Reaction coordinate

Ex.12 Fromthe followingdata for the decomposition ofdiazobenzene chloride, showthat the reactionisoffirst order Time : Vol. of N 2:

20 10

50 25

70 33

 162

(min) (mL)

C6H5N2Cl  C6H5Cl + N2

Sol. Initial concentration

a

Concentration after time t

(a – x)

x

At  time, i.e, when the reaction is complete, the whole of C6 H5 N2Cl converts into N2 . Hence volume of N2 at  time corresponds to the inital concentration 'a' while volumes of N2 at different time intervals correspond to x as shown above. Inserting the given data in Equation of first order reaction, we get the following results. For t = 20 min, k1 =

2.303 162 log = 0.0032 min–1 20 162 – 10

34

Chemical Kinetics

For t = 50 min, k1 =

2.303 162 log = 0.0033 min–1 50 162 – 25

2.303 162 log = 0.0032 min–1 70 162 – 33 The constancy of k1 shows tha the decomposition of C6 H5 N2Cl is a first order reaction. For t = 70 min, k1 =

Ex.13 1 mL of methyl acetate was added to 20 mL of 0.5 N HCl and 2 mL of the mixture was withdrawn from time tc time during the progress of hydrolysis of the ester and titrated with a solution of alkali. The amount of alkali needed for titration at various intervals is given below : Time : 0 20 119 (min)  Alkali used : 19.24 20.73 26.6 (mL) 42.03 Establish that the reaction is of first order. CH3COOCH3 + H2O

Sol.

HCl [H +]

CH3COOH + CH3OH

Initial concentration a Concentration after time t (a – x) x HCl acts as a catalyst. Since in every titration the amount of HCl is the same, the alkali used against HCl is subtracted from the total alkali used (given in the data) to get the volume of alkali used only against CH3COOH.At zero time no CH 3COOH is formed, so alkali used at zero time is only for HCl.  we thus have, Time (min) : Vol. of alkali (mL) used against CH3COOH

0 : 19.24 –19.24 = 0

20

119



20.73

26.6

42.03

–19.24 =

–19.24 =

–19.24 =

1.49 7.36 22.79 (x) (x) (a) Now following the same way as in Example 40, we get the following results : For t =20 ;

k1 =

For t =119 ;

k1 =

2.303 22.79 log = 0.0033 min–1 20 22.79 – 1.49 2.303 119

log

22.79 22.79 – 7.36

= 0.0032 min–1

The constancy of k1 shows that the reaction is of first order. Ex.14 The optical rotation of can sugar in 0.5 N lactic acid at 25° C at various time intervals are givenbelow : Time (min) : 0 1435 11360  Rotation(°) : 34.50° 31.10° 13.98° –10.77° Show that the reaction is of first order. Sol.

C12H22O11 + H2O Sucrose (excess) Dextrorotatory

Lactic acid

C6H12O6 + C6H12O6 Glucose Fructose Dextro Laevo rotatory rotatory

35

Chemical Kinetics

Since in this reaction dextro form changes to laevo form, the optical rotation decreases with the progress of the reaction. Thus change in rotation is proportional to the amount of sugar remaind after different time intervals. We now have, Time (min)

0

1435

11360

Change in rotation (°)

34.50 – (–10.77) = 45.27 (a)

31.10 – (–10.77) = 41.87 (a – x)

13.98 – (–10.77) = 24.75 (a – x)

–10.77 – (–10.77) =0

Substituting the data in Equation, for t = 1435 min k1 =

2.303 45.27 log = 5.442 × 10–5 1435 41.87

and for t = 11360 min k1 =

2.303 11360

log

45.27 = 5.311 × 10–5 24.75

The values of k1 are fairly constant and so the reaction is of first order. Ex.15 Bicyclohexane was found to undergo two parallel first order rearrangements.At 730 K, the first order rate constant for the formation of cyclohexene was measurred as 1.26 × 10–4 s–1, and for the formation of methyl cyclopenten the rate constant was 3.8 × 10–5 s–1. What is the percentage distribution of the rearrangement products?

Sol. k1 Percentage of cyclohexene = k  k × 100 1 2 1.2610 –4 = × 100 1.26 10 – 4  3.8 10 –5 = 77%.  percentage of methylcyclopenten = 23%. Ex.16 For the displacement reaction [Co(NH3) 5Cl]2+ + H 2O  [Co(NH 3) 5(H 2O)]3+ + Cl–

Sol.

the rate constant is given by 11067 K ln [k / (min–1)] = – + 31.33. T evaluate k, E and Afor the chemical reaction at 25° C. Substituting T = 298 K in the given equation, we get, k = 3.06 × 10–3 min–1 = 5.10 × 10–5 s–1.

36

Chemical Kinetics

Further, we have, k = Ae–E/RT or ln k = lnA–

E

RT Compairing this equation with the given equation, we get,

lnA= 31.33 or 2.303 logA= 31.33 ;

A = 4.04 × 1013 min–1 = 6.73 × 1011 s–1

E 11067 K =– RT T E = (11067 K) (8.314 J K–1 mol–1) = 92011 J/mole = 92.011 kJ/ mole.

and, –

Ex.17 The complexation of Fe2+ with the chelating agent dipyridyl has been studied kinetically in both forward and reverse directions. Fe2+ + 3 dipy  Fe(dipy) 2+ 3 13

Sol.

Rate (forward) = (1.45 × 10 ) [Fe2+][dipy]3 and rate (reverse) = (1.22 × 10–4) [Fe (dipy) 32+]. Find the stability constant for the complex. At dynamic equilibrium, rate of formation of complex = rate of its decomposition (1.45 × 1013) [Fe2+][dipy]3 = (1.22 × 10–4) [Fe(dipy) 32+] Ks =

[Fe(dipy )32 ] [Fe2 ][dipy]3

=

1.451013 1.22 10 – 4

= 1.19 × 1017.

Ex.18 The approach to the following equilibrium was observed kinetically from both direction. PtCl42– + H2O Pt (H 2O)Cl 3– + Cl– at 25° C. It was found that –

d[PtCl42 – ] = (3.9 × 10–5) [PtCl42–)] – (2.1 × 10–3) [Pt(H 2O)Cl 3–] [Cl–] dt

Calculate the equilibrium constant for the complexation of the fourth Cl– by Pt (II). Sol.

At equilibrium,

d[PtCl42– ] =0 dt

Hence, 3.9 × 10–5 [PtCl 2– ] = 2.1 × 10–3 [Pt(H 2O)Cl 2–3 ][Cl – ] 4 or

–3 [PtCl 24 – ] 2.110 K= = = 53.85. [Pt(H 2O)Cl 3– ][Cl – ] 3.910 –5

37

Chemical Kinetics

Ex.19 For the reaction

Sol.

k2 [Cr(H2O)4Cl2]+ (aq) k1 [Cr(H2O)5Cl2]2+ (aq)  [Cr(H2O)6]3+ (aq)  k = 5.8 × 10–5 s–1 for the initial  concentration of [Cr(H O) Cl ]+ is 0.0174 k1 = 1.78 × 10–3 s–1 and 2 2 4 2 mole/litre at 0°C. Calculate the value of t at which the concentration of [Cr(H2 O)5 Cl2 ]2+ is maximum. We have,

2.303(log k1 – log k 2 ) 2.303(log1.7610 –3 – log 5.810 –5 ) t= = k1 – k 2 1.7610 –3 – 5.810 –5

= 1990 seconds.

Ex.20 From the following reaction scheme, write the rate law for the disappearance ofA, B and C. (1) A+ B k1 C + D Sol.

k (2) C + D   A+ B

k3 (3) B + C  E + D

The reactant Ais removed in Step 1 and produced in Step 2 d[A] = k1 [A] [B] – k 2 [C] [D] –  dt Similarly, d[B] = k1 [A] [B] + k 3 [B] [C] – k 2[C] [D] dt d[C] = k2 [C] [D] + k 3 [B] [C] – k 1[A] [B]. – dt

– and

38

Chemical Kinetics

EXERCISE - I Q.1

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

Q.2

The reaction CH3CH2NO2 + OH–  CH3CHNO2 + H2O obeys the rate law for pseudo first order kinetics in the presence of a large excess of hydroxide ion. If 1% of nitro ethane undergoes reaction in halfa minute when the reactant concentration is 0.002 M, What is the pseudo first order rate constant? The decomposition of a compound P, at temperature T according to the equation 2P(g)  4Q(g) + R(g) + S(l) is the first order reaction. After 30 minutres from the start of decomposition in a closed vessel, the total pressure developed is found to be 317 mm Hg and after a long period of time the total pressure observed to be 617 mm Hg. Calculate the total pressure of the vessel after 75 mintute, if volume of liquid S is supposed to be negligible. Also calculate the time fraction t7/8. Given : Vapour pressure of S (l) at temperature T = 32.5 mm Hg.

Q.3

Q.4

A certain reactant Bn+ is getting converted to B(n+4)+ in solution. The rate constant of this reaction is measured by titrating a volume of the solution with a reducing reagent which only reacts with Bn+ and B(n+4)+. In this process, it converts Bn+ toB(n2)+ and B(n+4)+ toB(n1)+. At t=0, the volume of the reagent consumed is 25 ml and at t = 10 min, the volume used up is 32 ml. Calculate the rate constant of the conversion of Bn+ to B(n+4)+ assuming it to be a first order reaction.

Q.5

Decomposition of H2O2 is a first order reaction. A solution of H2O2 labelled as 20 volumes was left open. Due to this, some H2O2 decomposed. To determine the new volume strength after 6 hours, 10 mL of this solution was diluted to 100mL. 10mL of this diluted solution was titrated against 25mLof 0.025M KMnO4 solution under acidic conditions. Calculate the rate constant for decomposition of H2O2.

Q.6

A metal slowly forms an oxide film which completely protects the metal when the film thickness is 3.956 thousand ths of an inch. If the film thickness is 1.281 thou. in 6 weeks, how much longer willit be before it is 2.481 thou.? The rate of film formation follows first order kinetics. An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was – 20°. If optical rotation per mole of A, B & C are 60°, 40° & – 80°. Calculate half life of the reaction.

Q.7

Q.8

Q.9

Q.10

Q.11

A vessel contains dimethyl ether at a pressure of 0.4 atm. Dimethyl ether decomposes as CH3OCH3(g)  CH4(g) + CO(g) + H2(g). The rate constant of decomposition is 4.78×103 min1. Calculate the ratio of initial rate ofdiffusionto rate of diffusion after 4.5 hours of initiationofdecomposition. At room temperature (20°C) orange juice gets spoilt in about 64 hours. In a referigerator at 3°C juice can be stored three times as long before it gets spoilt. Estimate (a) the activation energy of the reaction that causes the spoiling of juice. (b) How long should it take for juice to get spoilt at 40°C? A first order reaction, A B, requires activation energy of 70 kJ mol1. When a 20% solution ofAwas kept at 25°C for 20 minutes, 25% decomposition took place. What will be the percent decomposition in the same time in a 30% solution maintianed at 40°C? Assume that activation energy remains constant in this range of temperature. Two reations (i) A  products (ii) B  products, follow first order kinetics. The rate of the reaction (i) is doubled when the temperature is raised from 300 K to 310K. The half life for this reaction at 310K is 30 minutes. At the same temperature B decomposes twice as fast as A. If the energy of activation for the reaction (ii) is half that of reaction (i), calculate the rate constant of the reaction (ii) at 300K.

39

Chemical Kinetics

Q.12

Acertain organic compoundAdecomposes by two parallel first order mechanism If k1 : k2 = 1 : 9 and k1 = 1.3 × 10–5 s–1. Calculate the concentration ratio of C toA, if an experiment is started with onlyAand allowed to run for one hour.

Q.13

The reaction cisCr(en)2(OH)+2

transCr(en)2(OH)+2

is first order in both directions. At 25°C the equilibrium constant is 0.16 and the rate constant k1 is 3.3 × 10 4s 1. In an experiment starting with the pure cis form, how long would it take for half the equilibrium amount of the trans isomer to be formed ? Q.14

For a reversible firstorder reaction A

B

k1 = 10 2 s 1 and [B]eq /[A] eq = 4. If [A]0 = 0.01 mole L 1 and [B]0 = 0, what will be the concentration of B after 30 s ? Q.15

Q.16

Q.17

Q.18

Q.19

For the systemA(g) l B(g), H for the forward reaction is –33 kJ/mol (Note : H = E in this case). [B] Show that equilibrium constant K = [A] = 5.572 × 105 at 300 K. If the activation energies Ef & Eb are 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. The complex [Co(NH3)5F]2+ reacts with water according to the equation. [Co(NH3)5F]2+ + H2O  [Co(NH3)5(H2O)]3+ + F The rate of the reaction = rate const. x[complex]ax[H+]b. The reaction is acid catalysed i.e. [H+] does not change during the reaction. Thus rate = k[Complex]a where k’ = k[H+]b, calculate ‘a’ and ‘b’ given the following data at 250C. [Complex]M [H+]M T1/2hr T3/4hr 0.1 0.01 1 2 0.2 0.5 1 0.02 k k1 For the two parallel reactionsA   B andA 2 C, show that the activation energy E for the disappearance ofAis given in terms of activation energies E1 and E2 for the two paths by k1E1  k 2 E2 E = k  k 1 2

For the mechanism

A+ B

k1 k2

C

;

k3 C  D

(a) Derive the rate law using the steady-state approximation to eliminate the concentration of C. (b)Assuming that k3 0)

(D)

(S)

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

(T)

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

(U)

1 (y-axis), t (x-axis) (order = 2) [A]

(V)

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

58

Chemical Kinetics

ANSWER KEY EXERCISE -I 1. 3. 5. 7. 9. 11.

(a) 6.25 ; (b) 14.3 ; (c) 0% Pt = 379.55 mm Hg, t7/8 = 399.96min k = 0.022 hr–1 20 min (a) 43.46kJmol-1, (b) 20.47 hour k = 0.0327 min–1

2. 4. 6. 8. 10. 12.

2 × 10–2 min–1 0.0207 min–1 15.13 week 0.26 : 1 % decomposition = 67.21% 0.537

13.

4.83 mins

14.

0.0025 m

15.

Ef = 6 × 104 J; E b = 9.3 × 104J

16.

a=b=1

18.

(a)

20.

0.0805

A1A3 d(D) k1k 3 (A)(B)  ; (b) E = E + E – E . A = a a1 a3 a2 k2  k3 dt A2 21.

161 minutes

22.

55.33 kJ mole–1

24.

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

25.

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

27.

Ea = 2.2 × 104

23.

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

26.

2

A = 5.42 × 1010 28.

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

29.

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

30.

K = 0.02231 min–1 t = 62.07 min

EXERCISE-II 1 d[NO]

= 9 ×10–4mol litre–1 sec–1 , (ii) 36 × 10–4mol litre–1 sec–1 , (iii) 54×10–4 mol litre–1 sec–1

1.

(i) r =

2. 3.

(i) 7.2 mol litre–1min–1, (ii) 7.2 mol litre–1 min–1 (a) 1 × 10–4 mol L–1 s–1 , (b) 3 × 10–4 mol L–1 s–1

4.

(i)

5. 7.

4

dt

dx

= k[A][B]2, (ii) rate increases by 8 times dt 8.12 × 10–6 Ms–1, 0.012 atm min–1 (a) 0.019 mol L–1 s–1, (b) 0.037 mol L–1 s–1

6. 8.

rate increase by 27 times 2k1 = k2 = 4k3

9.

8 31018 sec

10.

1/6

11. 13. 15. 17. 20.

(i) 7.2 M, (ii) Think 0.75 M 1.2 hr

12. 14. 16. 19. 21.

K = 0.01 M min–1 6 × 10–9 sec (i) 36 min., (ii) 108 min. 924.362 sec 3.3 × 104s1

(i) 0.0223 min–1, (ii) 62.17 min expiry time = 41 months

59

Chemical Kinetics

22. 24. 25. 26.

2.303 1 log 23. t a 15 min (a) Third order, (b) r = k[NO]2[H2], (c) 8.85 ×10–3 M sec–1.

k=

11.2%

27.

(a) order w.r.t NO = 2 and w.r.t Cl2 = 1, (b) r = K[NO] 2[Cl2], (c) K = 8 L2mol–2 s–1 , (d) rate = 0.256 mole L–1s1 (i) first order (ii) k = 1.308 × 102 min1 (iii) 73% 28. (i) Zero order, (ii) K = 5 Pa/s

29.

Zero order

31.

P3 l k = t ln (P  P ) 3 2

33. 35. 37. 39. 40. 41.

30.

V1 l k = ln t (2V1  V2 ) l r 4V3 l k = ln 34. k = t ln 5(V  V ) t (r  rt ) 3 2 (i) r= K[(CH3)2 O], 0.000428 sec–1 36. First order 38. First order 966 min (a) 90 mm, (b) 47 mm, (c) 6.49 × 10–2 per minutes, (d) 10.677 min. k1 = 2.605 × 103 min1 (a) first order, (b) 13.75 minutes, (c) 0.716 42. 11.45 days 0.180 atm, 47.69 sec 44. 0.1 min–1 32.

1 45. 47. 49. 51. 53. 55. 57. 60.

P3 l k = t ln 2(P  P ) 3 2

e (K1 K2 ) t 1

46.

[C] 10 = (e11x – 1) [A] 11

t = 4 min 5 kJ mol–1 55.33 kJ mole–1 rate of reaction increases 5.81× 108 times t = 20.4 minutes r = K' [NO]2[Br2] (d) No, (e) mechanism (a) is incorrect

48. 50. 52. 54. 56. 58.

0379.75 K 349.1 k 306 k 10.757 k cal mol–1 2000 K r = K [NO]2 [H2], where K = k2 × K1

EXERCISE -III 1. B 8. D 15. C 22. B 29. B 36. A,B,D 43. C 50. B 57. C

2. B 9. C 16. D 23. B 30. C 37. A,D 44. D 51. C 58. C

3. A 10. D 17. B 24. C 31. D 38. A,C,D 45. A 52. C

4. D 11. C 18. C 25. A 32. A 39. A,C 46. C 53. C

5. A 12. B 19. C 26. D 33. A 40. A,B,C 47. A 54. B

6. D 13. C 20. C 27. C 34. B 41. C,D 48. B 55. D

7. D 14. D 21. D 28. B 35. A,D 42. C 49. A 56. B

59. (A) S, (B) R, (C) P, (D) Q 60. (A) P (B) Q,S (C) R,T (D) V

60