NARAYANA Class Xii Chem Chemical Kinetics

IIT-JEE

CHEMICAL KINETICS &NUCLEAR CHEMISTRY

C H EM I S T R YS T U D Y M A T E R I A L

PHASE - I

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CONTENTSCONTENTSCONTENTSCONTENTSCONTENTS

CHEMICAL KINETICS& NUCLEAR CHEMISTRY

1. Theory

2. Solved Problems

(i) Subjective Type Problems

(ii) Single Choice Problems

(iii) Multiple Choice Problems

(iv) Miscellaneous Problems

Comprehension Type Problems

Matching Type Problems

Assertion-Reason Type Problems

3. Assignments

(i) Subjective Questions

(ii) Single Choice Questions

(iii) Multiple Choice Questions

(iv) Miscellaneous Questions

Comprehension Type Questions

Matching Type Questions

Assertion-Reason Type Questions

(v) Problems Asked in IIT-JEE

4. Answers

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CHEMICAL KINETICS & NUCLEAR CHEMISTRY

IIT-JEE Syllabus Rates of chemical reactions; Order of reactions; Rate constant; First order reactions; Temperature dependence of rate constant (Arrhenius equation).

CONTENTS

Rate of reaction

Elementary reaction

Integrated rate laws

Activation energy

Kinetics of some complex first order reactions

Methods of determining order

Nuclear chemistry

Kinetics of radioactive disintegration

INTRODUCTION Some chemical reaction occurs within few micro seconds or milli seconds due to formation of reaction intermediate (carbocation, carboanion or free radicals). But some reaction occur in few year or months because breaking of strong ionic or metallic bond is a slow process. Under chemical kinetics we would study that reaction which occurs at measurable rate.

In Chemical Kinetics we would be able to understand the velocity as well as different factors which would effect the chemical reaction. Under this we will be studying the mechanism of the reaction. Here mechanism of reaction means how reactants are converted into product or in how many intermediate steps reactant is converted into product. And which intermediate step is rate determining step.

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NARAYANA INSTITUTE OF CORRESPONDENCE COURSES

Chemistry : Chemical Kinetics & Nuclear Chemistry

1. RATE OF REACTION

Rate of a reaction is defined as the increase in molar concentration of a product per unit time or decrease in molar concentration of a reactant per unit time. Molar concentration is normally measured in moles per litre, the rate of a reaction is specified in mole per litre per time.

1.1 AVERAGE RATE

The quantity xt

is known as average rate, where is the change in concentration in

time. As we have discussed in chemical equilibrium rate of change of concentration of reactant decreases as the reaction proceeds. This means that rate of change of concentration is not constant. If the time interval is quite large then average rate will show large deviations from the actual rate.

x

t

1.2 INSTANTANEOUS RATE In order to precisely define rate time interval is made smaller ti.e. instantaneous rate = average rate as approaches zero. t

= t 0

[A]t

= d[A]

dt

Consider the hypothetical reaction

A 2B 3C D+ + Rate of disappearance of A = d[A]

dt

Rate of disappearance of B = d[B]dt

Rate of appearance of C = d[C]dt

+

Rate of appearance of D = d[D]dt

+ But from the stoichiometry it is apparent that when one mole of A is reacted, two moles of B are also consumed.

i.e. rate of disappearance of B = 2 rate of disappearance of A d[B] d[A]2

dt dt =

or d[A] 1 d[B]dt 2 dt

=


2



Similarly we can prove that

d[A] 1 d[C]dt 3 dt

+ =

So, rate of reaction = d[A] 1 d[B] 1 d[C] d[D]dt 2 dt 3 dt dt

= = = We can also generalize our statement i.e. for a general reaction

m A 1 2 1 2m B n C n D+ +

Rate of reaction = 1 2 1 2

1 d[A] 1 d[B] 1 d[C] 1 d[D]m dt m dt n dt n dt

= = + = +

Illustration 1 : Dinitrogen pentaoxide decomposes as follows:

2 5 2 2122

N O NO O + . If, 2 5 2 5[ ] [ ]d N O K N Odt =

2 2 5[ ] [ ]O K N O

dt=d N

2 2 5[ ] [ ]K N Odt

=d O Derive a relation in, , '' '''K K and K

Solution : For the given change 2 5 2 2d[N O ] 1 d[NO ] d[O ]2dt 2 dt dt

= + = +

On substituting values, 2 5 2 5 2 51[N O ] K [N O ] 2K [N O ]2

= =K or 2K K K = =

EXERCISE 1

1. For the reaction; 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g), the rate of reaction in terms of disappearance of NH3 is 3

d[NH ] ,dt

then write rate expression in terms of O2, NO and H2O.

2. For the decomposition reaction: N2O4(g) 2NO2(g); the initial pressure of N2O4 falls from 0.46 atm to 0.28 atm in 30 minute. What is the rate of appearance of NO2?

3. A chemical reaction 2A 4B + C; in gaseous phase shows an increase in concentration of B by 5 103 M in 10 second. Calculate:

a) rate of appearance of B, b) rate of the reaction,

c) rate of disappearance of A


3



2. ELEMENTARY REACTION

An elementary reaction is a single molecular event, such as collision of molecules resulting in a reaction. The set of elementary reactions whose overall effect is given by the net chemical equation is called reaction mechanism. Let us consider an example for the reaction of nitrogen dioxide with carbon monoxide.

(i) 2(g) (g) (g) 2(g)NO CO NO CO [net chemical equation]+ +Suppose this chemical reaction takes place in two steps

(ii) 2(g) 2(g) 3(g) (g)NO NO NO NO [Elementary reaction]+ + (iii) 3(g) (g) 2(g) 2(g)NO CO NO CO [Elementary reaction]+ +Thus net reaction (i) is obtained by the combination of two elementary reaction (ii) and (iii) and is a reaction intermediate as it is produced and consumed during the course of reaction.

3NO

2.1 MOLECULARITY It is defined as the number of molecules taking part on the reactant side of an elementary reaction. A unimolecular reaction is an elementary reaction involves one reactant molecule, a bimolecular reaction is an elementary reaction that involves two reactant molecules.

2.2 DIFFERENTIAL RATE LAW It is a relationship that relates the variation of rate of reaction with the concentration of reactants.

Consider the reaction

2A 3B C+ +Das we have seen

m n1 d[A] d[C] [A] [B]2 dt dt

= [law of mass action states that rate of reaction is directly proportional to the product of the active masses of reactants raised to some powers]

So, m nd[C] k[A] [B]dt

= k is known as velocity constant or rate constant or specific reaction rate. m is known as order with respect to A, n is known as order with respect to B. Sum m + n is known as overall order of the reaction.

2.3 ORDER OF REACTION It is defined as the sum of exponents or powers which are raised to concentration terms in the rate law expression of a reaction.


4



Difference between order and Molecularity : Molecularity of a reaction can be used to describe only an elementary process and molecularity can be predicted just be viewing the elementary reaction. Whereas order of a reaction refers to the overall reaction and can be determined experimentally, however for elementary process the order and molecularity are same. Molecularity has got no meaning for an overall reaction and order cannot be predicted from a balanced chemical equation.

Note : If a reaction can be written as a combination of several elementary reaction then its the slowest step that governs the rate of reaction i.e. the slowest step is the rate determining step.

Differential Rate Law : Consider reaction P 2Q+ RThe differential rate law is written as

m nd[P] 1 d[Q]Rate k[P] [Q]dt 2 dt

= = = Value of m and n can be determined by performing the reaction in laboratory i.e. order w.r.t P is m and order w.r.t. Q is n and overall order will be m + n.

4.4 UNITS OF k In general rate law for a nth order reaction can be written as

ndC kCdt

= Where k is rate constant, k is characteristic of a reaction at a given temperature. It changes only when temperature changes and n is the order of reaction.

ndC / dtk

C=

Units of k: (concentration)1ntime1

For a zero order reaction (n = 0) Units of k = mol/L/s

For a first order reaction (n = 1) Units of k = time1

For a second order reaction (n = 2) Units of k = (mol/l)1time1 = l/mol/s


5



Illustration 2 :The data given below are for the reaction of NO and Cl2 to form NOCl at 25K

[Cl2] [NO] Initial rate 103 (mol litre1sec1)

0.05 0.05 1

0.15 0.05 3

0.05 0.15 9

a) What is the order with respect to NO and Cl2 in the reaction?

b) Write the rate expression.

c) Calculate the rate constant

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

Solution : For the reaction; 2NO + Cl2 2NOCl Rate = k[Cl2]m[NO]n (1)

Where, m and n are order of reaction w.r.t Cl2 and NO, respectively.

From the given data:

1 103 = k[0.05]m [0.05]n (2)

3 103 = k [0.15]m [0.05]n (3)

9 103 = k[0.05]m [0.15]n (4)

By equations (2) and (3),

m = 1 By equations (2) and (4)

n = 1

a) order with respect to NO is 2 and w.r.t. to Cl2 is 1. b) Also, rate expression r = k[Cl2]1 [NO2]2

c) And rate constant.,3

2 12

r 1 10k[Cl ][NO] [0.05] [0.05]

= = 2 = 8 litre2 mol2 sec1

d) Further, r = k[Cl2]1 [NO]2 = 8 [0.2]1 [0.4]2

= 0.256 mol litre1 sec1


6



EXERCISE 2 :

1. The reaction; 2A + B + C D + 2E; is found to be I order in A II order in B and zero order in C.

a) Write the rate expression b) What is the effect on rate on increasing the conc. of A, B and C two times.

2. The reaction; 2NO + Br2 2NOBr, is supposed to follow the following mechanism, i) NO + Br2 NOBr

fastZZZXYZZZ 2 ii) NOBr2 + NO 2NOBr slow Suggest the rate law expression 3. The thermal decomposition of N2O5 occurs in the following steps: Step I: N2O5 NOslow 2 + NO3 Step II N2O5 + NO3 3NOfast 2 + O2 Overall reaction 2N2O5 4NO2 + O2- Suggest the rate expression

3. INTEGRATED RATE LAWS The differential rate laws show how the rates of reaction depend on the concentrations of reactants. It is also useful to known how the concentration is depend on time, this information can be obtained from the differential rate law by integration.

3.1 FIRST ORDER REACTION A reaction is said to be first order if its rate is determined by the change of one concentration term only. Consider a first order reaction

A Products The differential rate law equation will be

d[A] k[A]dt

=

Integrating both sides taking limits [ , which is the concentration at t = 0 and [A , the

concentration at time t

0A] t]t

0

[A]

[A]

d[A]dt

=t

0

k dt


7



0t

[A]ln kt[A]

= or 0t

[A]2.303k logt [A

=]

A Products t = 0 a

t = t a x

2.303 ak logt a

= x

3.2 CHARACTERISTICS OF A FIRST ORDER REACTION A first order reaction must follow above form of rate law for all time instants. This means if we are given value of a i.e. initial concentration and values of x at different time instants i.e. a x is known to us. If values of K are calculated for different time instants by using above expression, if the reaction is following a first order kinetics then all values of K will approximately be equal to each other.

3.3 HALF LIFE PERIOD It is defined as the time in which half of the reactants are reacted for a first order reaction.

2.303 ak logt a

= x

at 1/ 2at t ; x =2

=

1/ 2

2.303 ak logt a

=/ 2

1/ 20.6932t

k=

In fact we can define any life suppose we want to define 7/8 life.

2.303 ak logt a

= x

at 7187at t ; x=8

=

at 718

2.303 ak log 7at a8

=

7182.303 2.303t log8 3

k k= = log 2

7 /8 1/ 2t 3 t=


8



Important thing is that in a first order reaction time required for the completion of a definite fraction is independent of initial concentration of reactants.

3.4 HALF LIFE FOR NTH ORDER REACTION Let A products is following nth order kinetics. i.e. nA] k[A] , it can be shown that

dt=d[

n 1

1/ 2 n 10

2 1tk(n 1)a

=

1/ 2 n 10

1i.e.t (n 2)a

Expression for first order reaction is

2.303 ak logt a

= x

or 2.303 at logK a

= x = 2.303[log a log(a x)]

k

2.303 2.303t log a log(a x)k k

= Comparing this equation with Y = mx + C

So a plot of t vs log (a x) will be straight line with a intercept of 2.303 log ak

and a slope

of 2.303k

2.303 logaK

2.303slopeK

=

log(a x)

time

3.5 EXAMPLES OF FIRST ORDER REACTION

3.5.1 Decomposition of H2O2 2 2 2 22H O 2H O O + 2 2 2H O H O O [slow i.e. rate determining step] + [fast] 2O O O+


9



So 12 2Rate [H O ]or 2 2R k[H O= ]

2

x

Order = 1

Kinetics of this reaction can be studied by withdrawing a definite volume of reaction mixture after regular interval of time and titrating it against KM solution in presence of dil.

4nO

2 4H SO

4 2 4 2 4 4 22KMnO 3H SO K SO 2MnSO 3H O 5O+ + + + 5 2 2 2 2[H O O H O O ]+ + 4 2 4 2 2 2 4 4 22KMnO 3H SO 5H O K SO 2MnSO 8H O 5O+ + + + +Since H2O2 is undergoing self decomposition so quantity of H2O2 present in a definite volume will also go on decreasing, KMnO4 is reacting with H2O2 so titre value i.e. volume of KMnO4 used will also go on decreasing.

Let and be the volumes of KM used at zero time and after t time respectively. 0V tV 4nO

then [a = initial conc. of ] 0V a 2 2H O [a x = conc. of after t time] tV a 2 2H O 0

t

V aV a

= x

0t

V2.303 ogt V

=k l

Illustration 3 :From the following data show that the decomposition of hydrogen peroxide in aqueous solution is a first - order reaction. What is the value of the rate constant ?

Time in minutes 0 10 20 30 40

VolumeV/(ml) 25.0 20.0 15.7 12.5 9.6

where V is the number of ml of potassium permanganate required to decompose a definite volume of hydrogen peroxide solution.

Solution : The equation for a first order reaction is

12.303 ak log

t a=

-x

The volume of KMnO4 used, evidently corresponds to the undecomposed hydrogen peroxide.


10



Hence the volume of KMnO4 used, at zero time corresponds to the initial concentration a and the volume used after time t, corresponds to (a - x) at that time. Inserting these values in the above equation, we get

when t = 10 min. k1 = 2.303

10 log 25

20.0 = 0.022287 min-1 = 0.000314 s-1

when t = 20 min. k1 = -1 -12 .303 25 log 0.023230 min 0.0003871 s

10 15.7= =

when t = 30 min. k1 = -1 -12.303 25log 0.02369 min 0.0003948 s

30 12.5= =

when t = 40 min. k1 = -1 -12.303 25log 0.023897 min 0.0003983 s

40 9.6= =

The constancy of k, shows that the decomposition of H2O2 in aqueous solution is a first order reaction.

The average value of the rate constant is 0.0003879 s-1.

3.5.2 Decomposition of NH4NO2 in aqueous solution 4 2 2 2NH NO N 2H O + 14 2Rate [NH NO ] 4 2Rate K[NH NO ]= order = 1 Kinetics of this reaction may be studied by collecting the nitrogen evolved and measuring its volume after regular intervals of time Let and be the volumes of liberated after t time and at the end of reaction. tV V 2N

4 2 2 2NH NO N 2H O +Initial conc. a 0 Conc. after t time a x x Conc. at t = 0 a tV x V a

xor tV V a

t

V aV V a x

=

t

V2.303k logt V

= V


11



Note: t = indicates that sufficient time is provided to the reaction i.e., the reaction has practically gone to completion.

Illustration 4 : From the following data for the decomposition of diazobenzene chloride, show that the reaction is of first order:

Time (min) 20 50 70 Volume of N2 (mL) 10 25 33 162

Solution : C6H5N2Cl C6H5Cl + N2 initial concentration a Concentration of after time t (a x) x

At time, i.e., when the reaction is complete, the whole of C6H5N2Cl converts into N2. Hence volume of N2 at time corresponds to the initial concentration a while volumes of N2 at different time intervals correspond to x as shown above.

For t = 20 min, 112.303 162k log 0.0032min

20 162 10= =

For t = 50 min, 112.303 162k log 0.0033min

50 162 25= =

For t = 70 min, 112.303 162k log 0.0032min

70 162 33= =

The constant of k1 shows that the decomposition of C6H5N2Cl; is a first order reaction.

3.5.3 Conversion of N-Chloroacetinalide into p-chloro acetanialide

NCl

CH3

O

rearrangementisomerism

NH

CH3

O

ClN-chloro-N-phenylacetamide

N-(4-chlorophenyl)acetamide 1Rate [N-chloroacetinalide]

order = 1 Kinetics of this reaction may be studied by withdrawing a definite volume of the reaction mixture after regular intervals of time, adding excess of KI and titrating the liberated

against standard sodium thiosulphate solution. 2I

N-chloroacetinalide reacts with KI and liberates but p-chloroacetinalide does not. 2I

I 2 2 2 3 2 4 62Na S O Na S O 2NaI+ +


12



N-chloroacetinalide P chloroacetinalide t = 0 a 0 t = t a x x Let and be the volumes of sodium thiosulphate used at t = 0 and t = t respectively, then

0V tV

0V a tV a x 0

t

V aV a x

=

0t

V2.303k logt V

=

Illustration 5 : Acetochloracetanilide (N-chloroacetanilide) is converted into p-chloroacetanilide, it is followed by the addition of KI which acts only on the former compound. The progress of reactionis studied by titrating the iodine liberated with standard hypo solution and the following results are obtained.

Time (hour) 0 1 2 6 ml of hypo solution 45 32 22.5 5.7

Show that the reaction is unimolecualr. Calculate the velocity constant of the reaction and also determine the fraction of N-chloroacetanilide which has reacted in three hours.

Solution : The velocity constant of the first order reaction is given by

2.303 ak logt a

= x Here the initial concentration a corresponds to the volume of hypo used against

liberated iodine at time 0, a = 45, (a x) corresponds to the volume of hypo used for titrations at different intervals of time, i.e. when t = 1 hour, (a x) = 32, and so on.

Substituting the values in the above equation,

(i) When t = 1 hour, (a x) = 32

2.303 45log 0.34111 32

= =k (ii) When t = 2 hours, (a x) = 22.5

2.303 45log 0.3468t 22.5

= =k (iii) When t = 6 hours, (a x) = 5.7


13



2.303 45k log 0.34436 5.7

= = Since the value of K in all the three cases is nearly same, the reaction must be

unimolecular. Calculation of fraction that has reacted in 3 hours

Let the amount reacted in 3 hours = x moles Given a = 45, t = 3 hours, (a x) = 45 x

Substituting these values in the reaction

2.303 alogt a

= k x 2.303 45300 log

t (45= 0. x)

45 x = 16.03

x = 28.97 Hence fraction of N-chloroacetanilide reacted in 3 hours = 28.97 0.643

45=

3.6 PSEUDO FIRST ORDER REACTION If molecularity of a reaction is 2 but order is 1 then the reaction is known as pseudo first-order reaction or pseudo unimolecular reaction.

Hydrolysis of tertiary butyl bromide :

3 3 3 3(CH ) CBr OH (CH ) OH Br + +

This reaction takes place in two steps.

i) slowCH3

CH3

CH3

Br CH3

CH3

CH3- Br+

ii) fast CH3

CH3

CH3

OHCH3

CH3

CH3- OH+

Since slowest step is rate determining step.

1 03 3Rate [(CH ) CBr] [OH ]

= 3 3K[(CH ) CBr]


14

Molecularity = 2



Order = 1 So the reaction is pseudo unimolecular

3.6.1 Hydrolysis of Ester catalysed by an acid

H3 2 5 2 3 6 5CH COOC H H O CH COOH C H OH++ +

3 2 5 2Rate [CH COOC H ][H O]Since water is in large excess so its concentration almost remains constant

13 2Rate [CH COOC H ] 55

t

a

x

3 2Rate k[CH COOC H ]= order = 1 Molecularity = 2

So the reaction is pseudo unimolecualr reaction. Kinetics of this reaction can be studied by withdrawing a definite volume of reaction mixture after regular intervals of time and titrating against NaOH solution. This reaction mixture is chilled during titration by adding ice cold water. Chilling checks hydrolysis of ester during titration. Chilling also checks reaction of ester with NaOH (second order and biomolecular).

Let and V0V , V be the volumes of NaOH used at 0 time, t time and at the end of reaction respectively. The volume of NaOH used at 0 time ( ) corresponds to the concentration of catalyst and volume of NaOH used after t time (V corresponds to the concentration of catalyst and acetic acid formed.

0V

t )

H3 2 5 2 3 2 5CH COOC H H O CH COOH C H OH++ +

t = 0 a 0

t = t a x x

t = 0 a V t 0 3V x [x conc. of CH COOH formed after t time] = V V 0 (V 0 t 0V ) (V V ) a x tV V a

0t

V V aV V a x

=

0t

V V2.303 ogt V

= k l V


15



Illustration 6 :5 ml of ethylacetate was added to a flask containing 100 ml of 0.1 N HCl placed

in a thermostat maintained at 30C. 5 ml of the reaction mixture was withdrawn at different intervals of time and after chilling, titrated against a standard alkali. The following data were obtained :

Time (minutes) 0 75 119 183 ml of alkali used 9.62 12.10 13.10 14.75 21.05

Show that hydrolysis of ethyl acetate is a first order reaction. Solution : The hydrolysis of ethy acetate will be a first order reaction if the above data

confirm to the equation.

01t

2.303 V Vk logt V

= V

Where V0, Vt and V represent the volumes of alkali used at the commencement of the reaction, after time t and at the end of the reaction respectively, Hence

V - V0 = 21.05 - 9.62 = 11.43

Time V - Vt 0 1t

303 gt V V

2. V Vlo k =

75 min 21.05 - 12.10 = 8.95 -12.303 11.43log 0.003259 min75 8.95

=

119 min 21.05 - 13.10 = 7.95 -12.303 11.43log 0.003264 min119 7.95

=

83 min 21.05 - 14.75 = 6.30 -12.303 11.43log 0.003254 min183 6.30

= A constant value of k shows that hydrolysis of ethyl acetate is a first order reaction

3.6.2 Inversion of Cane Sugar

H12 22 11 2 6 12 6 6 12 6Cane sugar glu cose fructose

C H O H O C H O C H O++ +

12 22 11 2Rate [C O O ][H O]Since water is solvent its concentration is constant.

So Rate = K[ 12 22 11C H O ]

Molecularity = 2

Order = 1 So the reaction is pseudo unimolecular


16



If some inert liquid is used as solvent then concentration of water can be changed and rate changes on changing the concentration of water. The reaction then becomes second order. Rate of this reaction is also proportional to the concentration of catalyst H+. But concentration of catalyst remains constant during the reaction. If we change concentration of water as well as that of H+ then reaction becomes 3rd order.

12 22 11 2Rate [C H O ][H O][H ]+

But under ordinary conditions and are constant so, 2H O H+

12 22 11Rate [C H O ]The initial solution of cane sugar is destrorotatory but the product mixture is laevorotatory. Since the reaction involves inversion in the sign of rotation, ths is known as inversion of cane sugar.

H12 22 11 2 6 12 6 6 12 6Dextrorotatory Dextrorotatory LaevorotatoryC H O H O C H O C H O

++ +

laevorotatory

Specific rotation +52.5 92

Cane sugar

Glucose

Fructose

dextro rotatory

dextro rotatory

Laevo rotatory

Prism

Kinetics of this reaction can be studied by taking cane sugar solution in a polarimeter and measuring the angles of rotation after regular intervals of time.

Let and be the angles of rotation at 0 time, after t time and at the end of reaction. 0r , rt r

r tr

0r

90-90

180

0

So, (i) 0 tr r x (ii) 0r r a


17



0 0 t(r r ) (r r ) a x or r r t a x

0t

r r ar r a x

=

0t

r r2.303k logt r

= r

Illustration 7 :The optical rotations of sucrose in 0.5 N HCl at 35C at various time intervals are given below. Show that the reaction is of first order :

Time (minutes) 0 10 20 30 40 Rotation (degree) +32.4 +28.8 +25.5 +22.4 +19.6 -11.1

Solution : The inversion of sucrose will be first order reaction if the above data confirm to

the equation , k t

r rr rt

102 303=

. log

Where r0, rt and r represent optical rotations at the commencement of the reaction after time t and at the completion of the reaction respectively

n the case a0 = r0 - r = +32.4 - (-11.1) = +43.5

The value of k at different times is calculated as follows :

Time rt rt - r k

10 min +28.8 39.9 2.303 43.5log10 39.9

= 0.008625 min-1

20 min +25.5 36.6 2. = 0.008625 min303 43.5log20 36.6

-1

0 min +22.4 33.5 2. = 0.008694 min303 43.5log30 33.5

-1

40 min +19.6 30.7 2. = 0.008717 min303 43.5log40 30.7

-1

The constancy of k1 indicates that the inversion of sucrose is a first - order reaction.


18



EXERCISE 3 : 1. From the data obtained for decomposition of H2O2, show that decomposition of a first

order reaction.

Time in minutes 0 10 20 30 40

Volume of KMnO4 (ml) 25.0 20.0 15.7 1.5 9.6

2. From the following data for the decomposition of ammonium nitrite in aqueous solution, show that the reaction is a first order.

Time in minutes 10 15 20 15 Volume of N2(ml) 6.25 9.00 11.4 13.65 35.05

3. A certain volume of ethyl acetate was added to a flask containing HCl (as catalyst)./ 5 m l of reaction mixture was withdrawn at different intervals of time and titrated against a standard alkali. The following data were obtained

Time in minutes 0 75 119 183 Volume of NaOH(ml) 9.62 12.1 13.1 12.75 21.05

Show that hydrolysis of ester is a first order reaction.

4. ACTIVATION ENERGY (E) For a chemical reaction to take place reactant molecules must make collisions among themselves. Actually a fraction of collisions are responsible for the formation of products i.e. not all collisions are effective enough to give products. The collisions which converts the reactants into products are known as effective collisions. Effective collisions between the molecules which have energies equal to or above a certain minimum value. This minimum value of energy which must be possessed by the molecules in order to make effective collision is called threshold energy. Now most of the times, the reactants do not possess threshold energy. So in order to make effective collisions an additional energy is needed. This additional energy which is absorbed by the molecules so that they achieve threshold energy is called as activation energy.

bEaE

Pote

ntia

l ene

rgy

Reaction co-ordinate

Reactants

Products

Threshold energy

HbE

aE

Reaction co-ordinate

ReactantsProducts

HProducts

Threshold energy


19



= activation energy of forward reaction aE

= activation energy of backward reaction bE

= Heat of reaction H

b

b

= aH E E = a bH E E Since Since E E aE E> a b< So, > i.e. reaction is endothermic So, i.e. reaction is exothermic H 0 H 0 k2

Let the rate of formation of B is x% rate of decomposition of A.

[ ] [ ]d B d Ax%dt dt

+ =

k1 [A]t = x % k0 [A]t or, k1 = x% k0

( ) ( )1/ 2 1/ 21 overalln 2 x n 2

t 100 t= l l

or, ( ) ( )1/ 2 overall1/ 2 1 100 tt x=

where (t1/2)1 is the half life of A, while it is converting into B and (t1/2)overall is the half life of A.

Similarly, halflife of A, i.e. (t1/2)2 while it is converting into C will be

( ) ( )( )1/ 2 overall

1/ 2 2

100 tt

100 x=

(c) For the first order parallel chain reaction

1k

A

2k

mB

nC

[ ][ ]t 1

2t

B mkC mk

=

and koverall = k1 + k2


26



5.2 CONSECUTIVE OR SEQUENTIAL REACTIONS Consider the following consecutive reaction

1 2K KA B CB will exist in the sequential reaction provided that the rate of formation of B must be higher than is rates of decay; which is governed by its rate constant of formation and deformation. The stage of the reaction where the rate of formation of B is equal to its rate of decay is known as steady state.

Now from the sequential reaction, it is very clear that

[ ] [ ] [ ] k,t1 t td A k A A A edt = = 0 (i) [ ] [ ] [ ]1 2t td B k A k Bdt+ = or [ ] [ ] [ ] [ ] 1k t1 1t t 0d B k B k A k A edt + = = (ii) Solving first degree differential equation (ii) with integrating factor , then we get concentration of B at any time t.

2k te

[ ] [ ] 1 21 k t k t0t2 1

k AB e

k k = e

(b) Time at which concentration of B is maximum

22 1 1

1 kt nk k k

= l

(c) Maximum concentration of B

[ ] [ ]x

2max 0

1

kB Ak

= where 2

1 2

kxk k

=

(d) Time at which [ ]d Bdt

is maximum

21 2

2 kt lk k k

= 1n

(e) At steady state,

k1 [A]t = k2 [B]t (f) If k1 < < k2, then sequential reaction converts into

1kA C


27



[ ] [ ]1 td A k Adt = Illustration 10 : 227Ac has a half-life of 22.0 years with respect to radioactive decay. The decay

follows two parallel, paths, one leading to 222Th and the other to 223Fr. The percentage yields of these two daughter nuclides are 2.0 and 98.0 respectively. What are the decay constants () for each of the separate paths?

Solution : The rate constant of the decay is

k = 1

2

693 0.693t 2

=0. 2

If k1 and k2 are the rate constants of the reactions leading to 222Th and Fr223, respectively we have

k1 + k2 =0.693

22;

12

k 2k 9

=8

On solving for k1 and k2, we get

k2 = 0.03087 y1

k1 = 0.00063 y1

EXERCISE 5 : 1. An organic compound A decomposes following two parallel first order mechanisms.

5 111

2

k 1 and k 1.3 10 sk 9

= = AB

C Calculate the concentration ratio of C to A, if an experiment is allowed to start with only

A for one hour.

2. For a consecutive first order reaction A B the values of K1 2K K C, 1 and K2 are 45 sec1 and 15 sec1 respectively. If the reaction is carried out with pure A at a concentration of 1.0 mol dm3,

a) How much time till be required for the concentration of B to reach a minimum.

b) What will be the maximum concentration of B. c) What will be the composition of the reacting system after a time interval of 10

minute.


28



5.3 REACTIONS INVOLVING OPPOSING OR REVERSIBLE REACTIONS Such reactions results in equilibrium. In other words the reactant changes to product and vice versa. Say we have an opposing reaction in which both forward and backward reactions are first order, viz., (k1 and k2 are rate constant of forward and backward reaction)

Say initial conc. of A and B are a and b mol L1 respectively. If after time t, x moles/L of A change into B, then conc. of A and B will be (a x) and (b + x) respectively. The net rate of the reaction would be given as :

Rate = k1 (a x) k2 (b + x) (i) [ both processes occur simultaneously] When equilibrium is reached, the net rate is zero

Thus, k1 (a xe) = k2 (b + xe) (e = equilibrium)

Hence, (b + xe) 2

1

kk= (a xe) or b =

2

1

kk (a xe) xe

Substituting value of b in eqn. (i)

Rate of reaction,

+ xx)xa(kkk)xa(k

dt ee21

21=dx

on solving, we get Rate = (k1 + k2) (xe x)

After rearranging and integrating the equation, we get an equation similar to first order reaction as shown below.

or

or

dt)kk(xx

dx21

e+=

0

x t

1 2x 0e

dx (k k )dtx x

= + log t)kk(

xxxx

21e

0e +=

A Bk1k2

Or xx

xxlog

t1)kk(

e

0e21

=+

The equation is similar to first order reaction except that the measured rate constant is now the sum of the forward and the reverse rate constants.


29



5.4 PRESSURE CHANGE METHOD This method is used for gaseous reactions.

As reaction proceeds there is change in pressure.

For a reaction, A(g) B(g) + C(g) Initial pressure at t = 0 P0 0 0

Pressure at time (t) (P0 x) x x

(Here x is no. of moles of A which change to produce)

Thus, total pressure (Pt) at time (t) = P0 x + x + x

or Pt = P0 + x, x = Pt P0

a x = P0 (Pt P0) = 2P0 Pt

Thus, 00 t

P2.303k logt (2P

= P )

6. METHODS OF DETERMINING ORDER

(i) Integration Method : In this method, the data is substituted into integrated rate equations for different order reactions. The equation which gives a constant value of K decides the order of reaction. Suppose we have to check whether a reaction is following first order kinetics or not then we will calculate values of K by using expression.

0t

1 [A]K lnt [A]

= . If for different values of t and [ K comes constant then order of reaction will be 1 other wise we will switchover to higher order expressions.

tA] ,

Integrated Expressions for various reactions

Zero order 0xt

=k

First order 12.303 aog

t (a k l x)

Second order 21 xk t a(a x) =

Third order 3 2 21 x(2a x)k t a (a x ) =


30



(ii) Graphical Method: In this method the data are plotted according to different integrated rate equations so as to yield a straight line suppose log (ax) vs t is a straight line then order of reaction will be 1.

(a x) Zero order

t

log(a x) First order

t

Second order

t

1(a x)

Third order

t

21

(a x)

(iii) Half life Method: As we have seen for nth order reaction

1/ 2 n 11t

a

Let 2

t be the half life at initial concentration a and be the half life at intial concentration

1/1 1 1/ 22t

2a .

Then 1/ 21 n 1

1

1ta

1/ 22 n 1

2

1ta

1/ 21/ 2

n 11 2

2 1

t at a

=

So by knowing half life at two different initial concentrations, order of reaction can be calculated.

Initial Rate Method

In this method concentration of one of the reactants is varied by a known factor and its effect on the initial rate of the reaction is studied for example if the rate doubles on doubling the concentration of a reactant then order with respect to that reactant will be 1. If rate becomes four times on doubling the concentration then order with respect to that reactant will be two. This way by changing the concentration of all the reactants one by one and keeping the concentrations of rest all reactants constant we can find order with respect to each of reacting species.


31



7. NUCLEAR CHEMISTRY

7.1 INTRODUCTION In chemical reactions, atoms of the reactants combine by a rearrangement of extra nuclear electrons but nuclei of the atoms remains unchanged. In a nuclear reaction on the other hand, it is the nucleus of the atom which is involved.

7.2 RADIOACTIVITY This can be defined as the spontaneous emission of certain kinds of radiations by some elements and the elements emitting such radiation are called radioactive elements. Radiation emitted by radioactive substances are of three types :

Sl. No.

-particles -particles -rays.

1. 2 units +ve charge & 4 units mass.

Unit ve charge with no mass

Electromagnetic waves of short wavelength.

2. Represented as 2 or

4HeHe++

Represented as o1e Represented as 0

0

3. They have high ionizing power

Ionizing power is less than that of particle

Ionizing power is very less

4. These have low penetrating power

Penetrating power is 100 times that of particle

Penetrating power is 100 times that of particle.

5. Velocity is of the order of cm/s 92 10

Velocity is of the order of cm/s 102.8 10

Velocity is same as that of velocity of light ( 3 1 m/s). 100

6. When an -particle is emitted atomic number decreases by 2 units and atomic mass decreases by 4 units

When a particle is emitted, atomic number increases by one unit and atomic mass remains unchanged

There is no change in atomic number or atomic mass.

7.3 STABILITY OF NUCLEI At some point, you may have wondered, If positive charges repel one another, how is it possible for protons to be packed so closely in the nuclei of atoms ? The answer is that there are attractive nuclear forces that are much stronger than electrostatic forces. The strengths of these forces are closely related to the numbers of protons and neutrons in a nucleus. We begin with some observations about the naturally occurring stable nuclides.


32



(i) About 160 stable nuclides have an even number of protons and an even number of neutrons, for example, 12 6 C

(ii) About 50 stable nuclides have an even number of protons and an odd number of neutrons, for example, . 2512 Mg

(iii) About 50 stable unclides have an odd number of protons and an even number of neutrons, for example, 19 . 9 F

(iv) Only 4 stable nuclides have an odd number of protons and an odd number of neutrons. They are 1 3 .

2 6 10 145 7H, Li, B, and N

One theoretical approach to nuclear stability is the nuclear shell theory. In simplest terms, this theory proposes that the protons and the neutrons each exist in shells within the nucleus. This is much like the existence of electrons in shells outside the nucleus. The similarity extends to the special stability associated with the closing of shells, similar to what is seen with the noble gases in electrons configuration. In the nuclear shell theory, a special stability is associated with nuclei that have any of the following numbers as numbers of protons or neutrons.

2, 8, 20, 28, 50, 82, 126 These numbers are called magic numbers because scientists recognized their significance in relation to nuclear stability before they had developed a theory to explain them. One observation consistent with these magic numbers is the fact that alpha particles are very stable; they have 2 protons and 2 neutrons and are doubly magic. Another observation is that tin (Z = 50) has ten naturally occurring stable nuclides, more than any other element. The magic number of protons (50) seems to allow for a greater variation in the number of neutrons in the tin nucleus than in others. Also, the uranium-238 radioactive decay series terminates in the nuclide 20 ; the uranium-235 series, in 20 ; and the

thorium-232 series, in . All these terminating includes have the magic number 82 in

lead (Z = 82); is doubly magic, with 82 protons and 126 neutrons.

682 Pb

782 Pb

20882 Pb

20882 Pb

A crucial factor in the stability of a nucleus is the ratio of the neutron number (N) to the proton number (Z). Some nuclides of the lightest elements have an N/Z ratio of 1, and as a group these nuclides have an average ratio slightly greater than 1. Examples of nuclides in this group are 4 12 . Nuclides with Z > 20 require a large number of neutrons than protons to moderate the effect of increasing proton repulsions. For example, the N/Z ratio in 5626 is 30/26 = 1.15; in

13 it is 78/55 = 1.42, and in 20 , 126/83 = 1.52. For nuclides with Z > 83, the protons repulsions are too large to be overcome by proton-neutron interactions, and the nuclides are all radioactive.

6 27 39 408 13 19 20He, O, Al, K, and Ca

Fe 355 Cs9

83 Bi

The general pattern for nuclear stability in terms of neutron and proton numbers is shown in figure, a graph of neutron number (N) versus proton number (Z). All of the naturally occurring stable nuclides are indicated by dots within the belt labeled the belt of stability. However, other nuclides in this belt that are not shown are radioactive. Also all nuclides


33



falling outside the belt are radioactive, and their mode of decay is one that brings the nuclides formed in the decay process into the belt. Nuclides above the belt decay by beta emission, and those below the belt by positron emission and electron capture. Many of the nuclides in the upper right corner decay by alpha emission. If the numbers 114 and 184 are also magic numbers, as some scientists think they are, there might be a small island of stability centered on the nuclide with Z = 114 and N = 184. This nuclide may someday be synthesized, and there have even been unsuccessful attempts to find element 114 in natural sources.

Illustration 11 :Which of the following would you expect to be radioactive, 11

?

8 23450 91Sn, Pa,

5425 Mn,

7430 Zn

Solution : : This nuclide has 50 protons and (118 50) = 68 neutrons. This is an even-even combination, the most common for stable nuclides. Also, the neutron : proton ratio of 68:50 lies within the belt of stability so 11 is non-radioactive.

11850 Sn

850 Sn

23 : Atomic number 91 exceeds the limit for the naturally occurring stable

nuclides (Z > 83). is radioactive.

491 Pa

23491 Pa

5425 : This nuclide has 25 protons and (54 25) = 29 neutrons. This is an odd-odd combination found only in four stable nuclides of low atomic numbers. We should expect that it is radioactive.

Mn


34



7430 : This nuclide has 30 protons and (74 30) = 44 neutrons. So Z = 30, the upper limit of the belt of stability is at about N = 40. The nuclide 7430 lies above the belt and is radioactive.

ZnZn

7.4 MODE OF DECAY Alpha-particle emission : An alpha ( particle has the same composition as a helium nucleus : two protons and two neutrons. Thus an particle has a mass of 4 u and a charge of 2+. Because they carry a positive charge, a particles are deflected in electric and magnetic fields. The penetrating power of alpha particles through matter is low; the particles can generally be stopped by a sheet of paper. The symbol for an alpha particles is

. We can represent a -particle emission with a nuclear equation, as in the radioactive decay of uranium-238.

)aa

42 He

238 234 492 90 2

238 234 4 238U Th He

92 90 2 92

+ =

++ =

Sum of mass numbers :

Sum of atomic numbers :

When a nucleus emits an alpha particles, its atomic number decreases by 2 and its mass number decreases by 4. The new nuclide is that of a different element than the decaying nuclide.

Beta-particles emission : Beta particles are electrons. Like all electrons, they have very little mass and carry a charge of 1. Beta particles are deflected in electric and magnetic fields, but in the opposite direction from alpha particles. Beta particles are more penetrating than a particles. They can pass through aluminum foil 2 to 3 mm thick. The symbol for a beta particle is .

-( )

01e-

Although an atomic nucleus contains the protons and neutrons that make up an alpha particle, a nucleus does not contain electrons. Instead, a neutron in converted to a proton and an electron, represented in the following nuclear equation.

1 10 1n p - + 01e

Because the atomic number represents the positive charge on a particle, the neutron has an atomic number of 0 (no charge). The electron has the equivalent of an atomic number of 1; it carries the same charge as a proton, but negative in sign. An example of a radioactive decay that produces beta particles is shown below :

234 234 090 91 1

234 234 0 234Th Pa e

90 91 1 90-

+ =

++ - =



When a nucleus emits a beta particle, its atomic number increases by 1 and its mass number is unchanged. The new nuclide is that of a different element than the decaying nuclide.


35



Gama-ray emission : Gamma rays are a highly penetrating form of electromagnetic radiation. They consist of photons, and thus they are not particles of matter. They are emitted by energetic nuclei as a means of reaching a lower energy state. In a nuclear equation for gama-ray emission, we represent the energetic nucleus by affixing the symbol m to its mass number. For example, in the radioactive decay of uranium-238 by alpha-particle emission, 23% of the thorium-230 nuclei formed are in an excited state : 23 . These nuclei then emit energy as gamma rays.

()

0m90 Th

234m 23090 90

230 230 0 230Th Th

90 90 0 90

+ =

+ g+ =



When a nucleus emits a gamma ray, both its atomic number and mass number remain constant. The new and old nuclide are of the same element. As we would expect for a form of electromagnetic radiation, gamma rays are unaffected by electric and magnetic fields.

Positron emission : Positrons are particles having the same mass as electrons but carrying a charge of 1+. They are sometimes called positive electrons and referred to as b particles. Their penetrating power through matter is very limited because when a positron comes into contact with an electron, the two particles annihilate each other and are converted to two gamma rays. Positrons are formed in the nucleus through the conversion of a proton to a neutron and a positron.

+

1 11 0p n + 01e

n

Positrons are most commonly emitted in the radioactive decay of certain nuclides of the lighter elements. The radioactive decay of alumium-26 is 82% by positron emission.

26 26 013 12 1

26 26 0 26Al Mg e

13 12 1 13

+ =

++ =



When a decaying nucleus emits a positron, its atomic number decreases by 1 and its mass number is unchanged. The new nuclide is that of a different element from the decaying nuclide. Electron capture : Electron capture (EC) is a process in which the nucleus absorbs an electron from an inner electronic shell, usually the first or second. An X ray is released when an electron from a higher quantum level drops to fill the level vacated by the captured electron. Once inside the nucleus, the captured electron. Once inside the nucleus, the captured electron combines with a proton to form a neutron.

0 1 11 1 0e P- +

Nuclear equations for electron capture usually show the captured electrons as a reactant. Iodine-125, used in medicine to diagnose pancreatic function and intestinal fat absorption, decays by electron capture.


36



125 0 12553 1 52

125 0 125I e

53 1 52-

+ -


Sum of atomic numbers :Te

The result of electron capture is the same as positron emission. The atomic number of the nucleus decreases by 1, and the mass number is unchanged. The new nuclide is that of a different element than the decaying nuclide.

Illustration 12 :What kind of radioactive decay would you expect the nuclide 84 to undergo ? 40 Zr

Solution : When we check the belt of stability at Z = 40, we see that a nuclide with N = 44 lies below the belt. This confirms that the nuclide is radioactive. We would expect a decay that moves the neutron:proton ratio closer to the belt. This means converting a proton to a neutron. The atomic number goes down by one, and the mass number remains the same. These changes are achieved either by positron emission or electron capture.

Positron emission : 8440 84 039 1Zr Y e +

Electron capture : 8440 0 841 39Zr e Y-+

Notice that in each case, the neutron:proton ration increases from 44/40 in to 45/39 in 8439 .

8440 Zr

Y

7.5 NUCLEAR FISSION Splitting of a nucleus into lighter nucleus is known as nuclear fission. U isotope is unstable and when it is hited by slow moving neutrons, it splits into a number of fragments, each of which has mass smaller that that of Uranium with the evolution of large amount of energy. This process is called nuclear fission.

235

235 192 0U n+

140 93 156 36 0

144 90 154 38 0

144 90 155 37 0

Ba Kr 3 nXe Sr 2 nCs Rb 2 n

+ ++ ++ +

It has been observed that during nuclear fission, the sum of the masses of the products formed is slightly less than the masses of target species and bombarding neutrons. This loss is known as mass defect. This loss in mass is converted to energy. Loss in lamu produces 931.48 Mev energy.

7.6 NUCLEAR FUSION What would you thing of constructing one giant nuclear reactor out in space that would transmit abundant energy to Earth almost forever ? Well, thats exactly what we earthlings have in our sun, 92 million miles away. The sun is powered by the fusion of atomic nuclei, and its fuel supplymostly 11 will last for billions of years. H


37



On Earth, scientists have unleashed the extraordinary energy of uncontrolled fusion reactions in hydrogen bombs. In a hydrogen bomb, nuclear fusion is initiated by the fission reaction in a fission (atomic) bomb. However, such a totally uncontrolled fusion reaction cannot be used for practical purposes. Control of fusion reactions as energy sources is probably still decades away.

Scientists face a daunting challenge in developing a fusion energy source. The most promising nuclear reaction is the deuterium-tritium reaction.

2 3 4 11 1 2 0H H He n+ +Before they will fuse, however, the nuclei of deuterium and tritium must be forced extremely close together. And because the positively charged nuclei repel one another so strongly, close approach requires that the nuclei have enormously high thermal energies. At the required temperature, gases are completely ionized into a mixture of atomic nuclei and electrons known as plasma. A temperature of over 40,000,000 K is necessary to initiate self-sustaining fusiona nuclear reaction that releases more energy than it takes to get it started. Another requirement is that the plasma be confined at an enormously high density long enough for the fusion to plasma be confined at an enormously high density long enough for the fusion to occur. Moreover, this confinement must be done without the plasma contacting the walls of the reactor, where it would immediately lose heat and thus its capability to fuse. One method is to confine the plasma in a magnetic field.

8. KINETICS OF RADIOACTIVE DISINTEGRATION Rate of decay of disintegration : Rate of disintegration of radioactive isotope is proportional to its number of atoms.

Rate of decay N dN

dtN.

N number of atoms of radioactive isotope.

Time

No.

of a

tom

s of

A

Time

Rat

e of

dec

ay o

f A

dN Ndt

or dNdt

= N (i) is known as decay constant or disintegration constant. If dt = 1 second.


38



Then dNN

= Decay constant ( ) : Thus decay constant ( ) is equal to the fraction of radioactive isotope disintegrating per second.

From equation (i)

dN dN

= t log N t C= +where C is integration constant when t = 0; N = No. (initial number of atoms) log N = C 0So log N = t log 0N 0Nt log

N =

0N1 lnt N

=

0N2.303 logt N

= All radioactive decay follows first order kinetics.

8.1 ACTIVITY AND UNITS OF ACTIVITY The unit of radioactivity is measured as the rate at which it changes into daughter nucleus. It has been derived on the scale of disintegration of radium. Let us consider 1 g of Ra (at. Wt. 223 and years). 1/ 2t 1600= Rate of decay (activity ) No. of atoms in 1 g Ra. =

2310.6932 1 6.023 10 3.7 10 dps

1600 365 24 60 60 223 = =

= Becque rrel 103.7 10 1 curie = 3. dps = Becquerrel 107 10 103.7 10 1 Rutherford = 10 dps. 6

Initial activity Initial amount Activity after `t time amount left undisintegrated after t time. 2.303 initial activitylog

t activity after t time =


39



as we have seen

0 ln N t ln N=

0

Nln tN

=

t0

N eN

=

or tN No e =Half life period (t : It is the time after which activity of a radioactive isotope reduces to half of its initial value.

1/ 2 )

When 1/ 2 0 / 2t t , N N= =

0N2.303 logt N

=

01/ 20 / 2

N2.303t logN

=

1/ 20.6932t =

Unit of 1time =Specific activity : This is defined as activity per unit mass of the sample. Average life (K) : The reciprocal of decay constant is defined as average life of radioactive isotope.

Average life 1= 1/ 2t1K

0.6932= =

1/ 2K 1.44 t=8.2 RADIOACTIVE EQUILIBRIUM

238 234 234 23492 90 91 92U Th Pa U

Suppose a radioactive isotope. A is producing another radioactive substance B, which in turn is producing C.

A B C1Rate of decay of A is no. of atoms of A]. 1N [ N


40



Rate of decay of A = is decay constant of A] 1 1 1N [ 2Rate of decay of B is no. of atoms of B] 2N [N

= is decay constant of B] 2 2 2N [ at radio active equilibrium Rate of formation of B = Rate of decay of B

But rate of formation of B = Rate of decay of A So at equilibrium = 1 1 2 2N N 1 2

2 1

N 0.6932 / TN 0.6932 / T

= =1

2

1 22 1

N TN T

= =1

2

but 1K

=

so 1 2 22 1 1

K NK N T

= = =1

2

T

1H

1

8.3 CARBON DATING 14 1 14 17 0 6

atomspheric cos micnitrogen rays

N n C+ +

146 C is radiocarbon ( -emitter) 14 14 06 7C N +Carbon dating is based upon the fact that radio carbon 14 is produced in the atmosphere by the action of cosmic rays C is a emitter with a half life of 5730 years. Its rate of formation has been constant so the atmosphere has an equilibrium concentration of C corresponding to nearly 16 disintegration per minute per gram of carbon. All living plants and animals contains this equilibrium concentration but when a plant or animal dies no further exchange between it and environment takes place and C content start decreasing as a result of radio active decay. After 5730 years there will remain only 8 disintegration per minute per gram of carbon. By measuring the activity of a fresh piece and the old piece, the age of old piece may be determined.

14 14

14

0N2.303 logt N

=

0N2.303 logt N

=


41



2.303 Initial activitylogt activity after t time

=

14 1214 12

C / C in fresh sample2.303 logt C / C in old sample

=

8.4 MINERAL DATING (GEOCHRONOLOGY). Determination of age of rocks, earth etc. is known as mineral dating

0N2.303 logt N

=

2.303 P Dlogt P

+ = P = no. of atoms of parent element D = no. of atoms of daughter element.

Illustration 13 :Radioactive decay is a first order process. Radioactive carbon wood sample decays with a half life of 5770 years. What is the rate constant in (years)1 for the decay? What fraction would remain after 11540 years?

Solution : 5770

693.0T

693.0K2/1==

= 1.201 104 year1

t

0

NN

logt303.2K =

1.201 104 t

0

NN

log11540

303.2=

4.002 =t

0

NN

0

t

NN (Remaining fraction) =

002.41

EXERCISE 6 : 1. The bones of a prehistoric bison were found to have a 14C activity of 2.80 dis/min. g

carbon. Approximately how long ago did the animal live? (t1/2 = 5730 years) 14C activity of fresh sample = 15.3 dis/min/gC.

2. Potassium contains 9.310 atom % 39K, having mas 38.9637 u; 0.0118 atom % 40K, which has mass of 40.0 u and is radioactive with t1/2 = 1.3 109 y and 6.88 atom % 41K having a mass of 40.96184u. Calculate the specific activity of naturally occurring potassium.

3. A mixture of 239Pu and 240Pu has a specific activity of 6 109 dis/s/g, The half lives of the isotopes are 2.44 104y and 6.08 103y specific activity of naturally occurring potassium.


42

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Chemistry : Chemical Kinetics and Nuclear ChemistryINSTITUTE OF CORRESPONDENCE COURSESNARAYANA

SECTION - ISUBJECTIVE TYPE PROBLEMS

Problem 1: The half life of a chemical reaction at a particular concentration is 100 min. Whenthe concentration of the reactant is doubled, the half life becomes 50 min, then what isthe order of the reaction ?

Solution :n 1

1/ 2 1 2

1/ 2 2 1

(t ) a(t ) a

=

n 1100 250 1

= 2 = 2(n1)

= n 1 n = 2

Problem 2: An optically active drug has one chiral center and only dextrorotatory isomer iseffective. Moreover it becomes ineffective when its optical activity reduces to 60% ofthe original in the equilibrium mixture at 27C and at one 127C, its optical activityreduces upto 80%. Then find out H of the reaction assuming that the specific rotationof the isomers remains constant during this temperature range. [log 3 = .477]

Solution : Let the dextrorotatory isomer A and laevoratatory is B0 = initial observed rotationt = observed rotation at time t

A !!"#!! B

t = 0 0 0 0 From question,

0 = 100, t = 60 60 = 100 2 = 20 = 0 = 100 20 At 27C,r

Keq. = K1 =[B] 20 1[A] 80 4

= =

Similarly, at 127C.

Keq. = K2 =[B] 10 1[A] 90 9

= =

since

2.30 log =1/ 4 H 1 11/ 9 R 300 400

=

Chemistry : Chemical Kinetics and Nuclear Chemistry


INSTITUTE OF CORRESPONDENCE COURSESNARAYANA

2.303 log 9 H 1004 R 300 400

=

2.303 [log 9 log 4] = H 1

R 1200

2.303 [2 log 3 2 log 2] = H 1

R 1200

2.303 (2 0.477 2 0.301) H = 2.303 [0.954 0.602] 8.314 1200

H = 8.188 kJProblem 3: A mixture of two substances X & Y produces Z simultaneously through parallel chain

reaction as under

X

Y

Z

I

I

The rate constant for the first path is 1.2 102 s1 and of second path is4 103 s1. After 40 seconds, the mole percentage of Z is found to be 36. Find outits mole-percentage after 60 seconds. Given that antilog (0.208) = 1.6 and Antilog(0.303) = 2.50.

Solution : Let the total moles of mixture = 100Initial moles of X = a Initial moles of Y = 100 a

Out of a moles of X, p moles converts into C and q moles of Y converts into CNow.

C p+q

Xap, k =1.210 s1

-2 -1

Y100aq, k =410 s2

-3 -1

Now, 2

3p 1.2 10 3q 14 10

= =

p = 3qAgain, p + q = 364q = 36

q = 9



p = 27

Now k1 40 = 2.303 log a

a 27

21.2 10 40 alog2.303 a 27

=

0.208 = alog

a 27

a

a 27 = 1.6

a = 72Now, for 60 seconds, let the amount of X converted into Z is p and the amount of Yconverted into Z is q.

21.2 10 60 72log

2.303 72 p'

=

0.313 = 72log

72 p

2.050 = 72

72 p

p = 36.88q = 12.29

mole percentage of Z is = p + q = 49.17%Problem 4: What is the amount of activation energy in kilo joules at 298 K for 103% molecule to

cross over the energy barrier? Given that, ln105 = 11.515.

Solution : Fraction of molecules crossing over the energy barrier = aE / RTe

= a3

E / RT10 e100

=

105 = aE / RTe

11.515 = aERT

Ea = 11.515 8.314 298 Joule= 28.53 KJ

Problem 5 : The reaction given below, involving the gases is observed to be first order with rateconstant 7.48 103 sec1.Calculate the time required for the total pressure in asystem containing A at an initial pressure of 0.1 atm to rise to 0.145 atm and also findthe total pressure after 100 sec.

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




Solution : 2A(g) 4B(g) + C(g)Po 0 0Po P 2P P/2

Ptotal =Po P + 2P + P/2 = Po + 3P '2

P = 23 (0.145 0.1) =0.03 atm

k = o

o

P2.303 logt P P '

t = 32.303 0.1log

0.077.48 10

t = 47.7 sec

Also, k = o

2.303 0.1logt P P '

7.48 103= 2.303 0.1log100 0.1 P '

0.1 P = 0.047P = 0.053

Ptotal = 0.1 + 32

(0.053) 0.180 atm.

Problem 6 : For the reaction :C2H5I + OH

C2H5OH + I

the rate constant was found to have a value of 5.03 102 mol1 dm3 s1 at 289 K and6.71 mol1 dm3 s1 at 333 K. What is the rate constant at 305 K.

Solution : k1 = 5.03 102 mol1 dm3 s1 at T1 = 289 Kk2 = 6.71 mol

1 dm3 s1 at T2 = 333 K

log 26.71

5.03 10

=aE 333 289

2.303 8.314 333 289

On solving we get, Ea = 88.914 kJThe rate constsnt at 305 K may be determined from the relation:

log a2

1 1 2

Ek 1 1k 2.303R T T

=

log 2 2k 88914 1 1

2.303 8.314 298 3055.03 10

= On solving we get, k2 = 0.35 mol

1 dm3 s1



Problem 7 : The decomposition of Cl2O7 at 400 K in the gas phase to Cl2 and O2 is of I order. After55 sec at 400 K, the pressure of Cl2O7 falls from 0.062 to 0.044 atm. Calculate :(a) The rate constant.(b) Pressure of Cl2O7 after 100 sec.

Solution : Cl2O7 Cl2 +7

22 O

Mole at t = 0 a 0 0Mole at t = 55 sec. (a x) x 7x/2(a) Since Pressure of Cl2O7 is given and therefore,

a 0.062(a x) 0.044

K = 102.303 0.062log

t 0.044K = 6.23 103 sec1.

(b) Let at t = 100 sec, (a x) P

6.23 103 = 102.303 0.062log100 P

P = 0.033 atm.Problem 8: A radioactive element A says to B I am half of what you were when you are one

fourth of what I was. Moreover I was 1.414 time than what you were. If the half lifeof A is 8 days, what is the half life of B?

Solution : Let the initial number of nuclide of X be x0Let the initial number of nuclide of Y be y0Let the number of nuclide of X after time t be xtLet the number of nuclide of Y after time t be yt xt = y0; yt = x0

x0 = 1.414 y0Now, xt = 1

t0e

x = y0 (1)

yt = y0 2te = x0 .(2)

Dividing equation. (1) by (2)

( )2 1 t0 0

0 0

2y ey

=

x

x

( )2 12

t020

2 ey

=

x

Since x0 = 2 y02 20 02y=x




2 = ( )2 1 t2e

( )2 1 te = 1Since t is not zero, hence 2 = 1. Therefore the half life of the other radioactive element isalso 8 days.

Problem 9 : On analysis a sample of Uranium was found to contain 0.277 g of 82Pb206 and 1.667 gof 92U238. The half life period of 92U238 is 4.51109 years. If all the lead were assumedto have come from decay of 92U238, what is the age of the earth?

Solution : 92U238 = 1.667 g = 1.667238 mole

82Pb206 = 0.227g =

0.227206 mole

All the lead have come from decay of U.

Moles of Pb formed = 0.277206

moles of U decayed = 0.277206

Total moles of Uranium = 1.667238 +

0.277206 , ie N0-

Also N for U238 = 1.667238

for U238 t = 0N2.803 log

k N =

92.303 4.51 100.693

log

1.667 0.277238 206

1.667238

+

t = 1.143 109 yrs .Problem 10 : An optically active compound A is hydrolysed by dilute acid to give

two optically compounds B and C according to the following chemical equation,A + H2O 2B + C. The angle of rotation after 40 minutes was observed to be26 and that after completion of reaction was 10 at 27C. Find the half - time of thereaction assuming it to follow pseudo first order kinetics. The observed rotation permole of A,B and C are 60, 50 and 80. A plot of logarithm of rate constant of theabove reaction vs T1 give straight line with intercept equal 15.046 on log K axis.Calculate at what temperature half - time of the reaction will be 31.1 min.

Solution : A + H2O 2B + Ca excess 0 0(a-x) const. x 2x0 const. 2a a



After completion of reaction 50 29 + (80) a = 10, a = 0.5After 40 minute. 60(a-x) + 50 2x 80 x = 26 x = 0.1

K = 2.303 a 2.303 0.5 2.303log log

t (a x) 40 0.5 0.1 40= =

log 5/4

K = 5.57 103 min1

t1/2 = 0.693

K = 0.693

0.00557 = 124.4 min

From Arrhenius equation, K = AeEa/RT

LogK = Ea

2.303RT

+ log A; Intercept = log A

WhenT = 27 + 273 = 300K , log K = log (5.57 103) = 2.254Putting the value in above equation

2.254 = Ea

2.303 8.314

1300 + 15.046

Ea = 99.37 kJ /molIn order that t1/2 of reaction may decrease from 124.4 min, rate constant will have toincrease 4 time.

Log (4 5.57 103) = 99370

2.303 8.314 1T + 15.046

1.652 15.046 = 99370

2.303 8.314 1T

T = 310.80 K




SECTION - IISINGLE CHOICE PROBLEMS

Problem 1: The progress of the reaction A !!"#!! nB with time is presented in the figure, the value of nis(a) 2 (b) 3(c) 4 (d) 5

1 3 5 7

0.3

Con

. ol/l

itre

Time/hour

A

B0.6

Solution : A !!"#!! nB

t = 0 0.06 0 (initial conc. in mol L1)At equi. 0.6 x nx (conc. at equilibrium)From the graph0.6 x = 0.3Also from the graphnx = 0.6 n = 2 (a)

Problem 2 : For the elementary reaction, 2NO + H2 N2O + H-2O; the half life time is 19.2 sec at820C when partial pressure of NO and H2 are 600 mm of Hg and 10 mm of Hg respectively.The value of half time at the same temperature when partial pressure of NO and H-2 are600 mm of Hg and 20 mm of Hg respectively would be(a) 19.2 sec (b) 10 sec(c) 830 sec (d) 415 sec

Solution : According to Ostwald isolation method of order determination, the concentration terms ofreactants present in excess remain virtually unchanged and thus can be merged with therate constant. In this case, the rate of the reaction depends only on the concentration o f thatreactant which is not present in excess. The reaction is known to be pseudo first order.The partial pressure of NO is very large in comparison to that of hydrogen. Thus, we mayconsider the rate of reaction to be independent of PNO and hence reaction will follow therate law.

2

2

(N O) 2H NO

dpk P ; where k k(P )

dt = = Pseudo first order reaction rate constant.



Half-life time of first order reaction is independent of initial concentration. Hence the requiredhalf-life time will be 19.2 sec. (a)

Problem 3 : Two reactions one of first order and other of second order have same values of rate constants(k1 = k2) when concentration are expressed in mole dm

3. If the concentrations are expressedin mole ml1 the relation between their rate constants k1 and k2 will be(a) k1 = k2 (b) k2 10

3 = k2(c) k2 10

3 = k1 (d) k1 = 10k2Solution : Since 1 nk [a]

For first order reaction,1 1

1k [a]

1k 1So k is constant and independent of unit of a, Therefore k1 = k1For second order reaction

21ka

2 -3 3 -11 1k

a mole dm a 10 mole ml =

121k

a mole ml

32 2 1k k 10 k = =

(c)Problem 4 : For the first order reaction 3A B

concentration varies with time as shownin the adjacent graph. The half-life of thereaction would be(a) 2 minutes(b) 4 minutes

Conc.(In M)

Time (in minutes)

2 4 6 8

A

B

(c) 8 minutes(d) 16 minutes

Solution : From graph it follows that at time 4 minute from the start of the reaction[A] = [B]

3A Bt = 0 at = 4 min a x x/3

x 3aa x x3 4

= =

3a aa x a4 4

= =

i.e., in 4 minutes concentration of A has reduced to 25% of the initial value therefore,t3/4 = 4 min. As t3/4 = 2t1/2 for a first order reaction, so half-life of this reaction = 2 min (a)




Problem 5 : The inversion of cane sugar proceeds with halflife of 500 minute at pH 5 for anyconcentration of sugar. However if pH = 6, the halflife changes to 50 minute. The rate lawexpression for the sugar inversion 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]0[H+]1

Solution : Since t1/2 does not depends upon the sugar concentration means it is first order respect tosugar concentration. t1/2 [sugar]

1.t1/2 a

n-1 = k

( )( )

1 n1/ 2 1 1

1 n1/ 2 22

t [H ]t [H ]

+

+ =

1 n5

6500 1050 10

=

10 = (10)1-n

Hence n = 0(b)

Problem 6 : Two substances A and B are present such that [A0] = 4[B0] and halflife of A is 5 minuteand that of B is 15 minute. If they start decaying at the same time following first orderkinetics how much time later will the concentration of both of them would be same.(a) 15 minute (b) 10 minute(c) 5 minute (d) 12 minute

Solution : Amount of A left in n1 haves = 1n

01 [A ]2

Amount of B left n2 halves = 2n

01 [B ]2

At the end, according to the question

1 2

00

n n[B ][A ]

2 2=

[ ]1 2 0 0n n

4 1 , [A ] 4[B ]2 2

= =

1

1 2

2

nn n 2

1 2n2 4 2 (2) n n 22

= = =

2 1n (n 2)= (1)

Also 1 1/ 2(A)t n t= 2 1/ 2(B)t n t=

(Let concentration both become equal after time t)



1 1/ 2(A) 1 1

2 1/ 2(B) 2 2

n t n 5 n1 1 3n t n 15 n

= = = (2)

For equation (1) and (2)n1 = 3, n2 = 1t = 3 5 = 15 minute

(a)Problem 7 : The reaction A(g) + 2B(g) C(g) + D(g) is an elementary process. In an experiment, the

initial partial pressure of A and B are PA = 0.60 and PB = 0.80 atm. When PC = 0.2 atm therate of reaction relative to the initial rate is(a) 1/48 (b) 1/24(c) 9/16 (d) 1/6

Solution : 1R K[A][B]= 2K[0.6][0.80]=

After reaction

A + 2B C + D

0.6 0.2 0.8 0.4 0.2 0.20.4 0.4 0.2 0.2

22

21

R K(0.4)(0.4) 1R 6(0.6)(0.8)

= =

(d)Problem 8 : A first order reaction : A Products and a second order reaction :

2R Products both have half - time of 20 minutes when they are carried out taking 4mole L1 of their respective reactants. Number of mole per litre of A and R remainingunreacted after 60 minutes from the short of the reaction, respectively will be.(a) 1 and 0.5 (b) 0.5 and negligible(c) 0.5 and 1 (d) 1 and 0.25

Solution : In the case of first order reaction t1/2 will remain constant independent of initial concentrationso.

4 mole L1 20 min 2 mole L1 20 min 1 mole L1 20 min 0.5 moile L1.

That is, after 60 minutes 0.5 mole L1 of A will be left unreacted.In the case of second order reaction t1/2 is inversely proportional to initial concentration ofreactant i.e., t1/2 will go on doubling as concentration of reactant will go on getting half.That is, t1/2 a will be constant, so.

4 mole L1 20 min 2 mole L1 40 min 1 mole L1.

That is, after 60 minutes, the concentration of R remaining unreacted will be 1 mole L1.Note that t1/2 a = 20 4 = 40 2 = 80 mole L

1 min, a constant. (b)




Problem 9 : The thermal decomposition of acetaldehyde : CH3CHO CH4 + CO, has rate constant

of 1.8 103 mole1/2L1/2 min1 at a given temperature. How would 3d[CH CHOO]-

dt will

change if concentration of acetaldehyde is doubled keeping the temperature constant?(a) will increase by 2.828 times (b) will increase by 11.313 times(c) will not change (d) will increase by 4 times

Solution : Unit of the rate constant mole 1/2L1/2.min1/2 suggests that the reaction obeys kinetics of 1.5order.

Rate = k [CH3CHO]3/2

or k =1 1

3/ 2 -1 3/ 23

Rate moleL min[CH CHO] (mole L )

= = Mole1/2, L1/2.min1

So, by doubling the concentration of acetaldehyde the rate will increase 21.5 i.e., 2.828times. (a)

Problem 10 : The reaction ; 2O3 3O2, is assigned the following mechanism.I. O3 O2 + OII. O3 + O 2O2 (slow)The rate law of if the reaction will, therefore be(a) r [O3]

2[O2] (b) r [O3]2 [O2]

1

(c) r [O3] (d) r [O3] [O2]2

Solution : Step II, being r.d.s.Rate of overall reaction = rate of Step II = KII [O3][O]Putting the value of [O] from the equilibrium of Step I,

Rate = 2

II C 3

2

K K [O ][O ]

(b)



SECTION - IIIMULTIPLE CHOICE PROBLEMS

Problem 1: Arrhenius equation may be represented as

(a)A Ealnk RT= (b)

d ln k Eadt RT

=

(c)Ealog A log k

2.303RT= + (d)

Ea klogRT A

=

Solution : (a, c)(a) is correct as e

NARAYANA Class Xii Chem Chemical Kinetics

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