NARAYANA Class Xii Chem Chemical Kinetics

S T U D Y M A T E R I A L PHASE - I CHEMICAL KINETICS & NUCLEAR CHEMISTRY IIT-JEE

CHEMISTRY

NARAYANA INSTITUTE OF CORRESPONDENCE COURSES

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C O N T E N T S

CONTENTS

CHEMICAL KINETICS & NUCLEAR CHEMISTRY 1.

Theory

2.

Solved Problems (i)

Subjective Type Problems

(ii)

Single Choice Problems

(iii) Multiple Choice Problems (iv) Miscellaneous Problems

3.

•

Comprehension Type Problems

•

Matching Type Problems

•

Assertion-Reason Type Problems

Assignments (i)

Subjective Questions

(ii)

Single Choice Questions

(iii) Multiple Choice Questions (iv) Miscellaneous Questions

(v) 4.

•

Comprehension Type Questions

•

Matching Type Questions

•

Assertion-Reason Type Questions

Problems Asked in IIT-JEE

Answers

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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Chemistry : Chemical Kinetics & Nuclear Chemistry

INSTITUTE OF CORRESPONDENCE COURSES

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 ∆x is known as average rate, where ∆x is the change in concentration in ∆t ∆t 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.

The quantity

1.2

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

 ∆[A]  d[A]   = dt ∆t → 0  ∆ t 

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




Similarly we can prove that − So, rate of reaction = −

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

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 m1A + m2 B  → n1C + n 2 D Rate of reaction = −

1 d[A] 1 d[B] 1 d[C] 1 d[D] =− =+ =+ m1 dt m 2 dt n1 dt n 2 dt

Illustration 1 : Dinitrogen pentaoxide decomposes as follows: d [ N 2O5 ] 1 N 2O5  → 2 NO2 + O2 . If, − = K ′[ N 2O5 ] 2 dt d [ NO2 ] = K ′′[ N 2O5 ] dt d [O2 ] = K ′′′[ N 2O5 ] dt Derive a relation in, K ′, K '' and K ''' Solution :

d[N 2O5 ] 1 d[NO 2 ] d[O 2 ] =+ = +2 dt 2 dt dt 1 On substituting values, K′[N 2O5 ] = K′′[N 2O5 ] = 2K′′′[N 2O5 ] 2 ′ ′′ ′ ′′ or 2K = K = K

For the given change −

EXERCISE 1 1.

For the reaction; 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), the rate of reaction in terms of d[NH3 ] , then write rate expression in terms of O2, NO and disappearance of NH3 is − dt 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 × 10–3 M in 10 second. Calculate: a)

rate of appearance of B,

b) rate of the reaction, c)

rate of disappearance of A 3

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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. NO 2(g) + CO(g)  → NO (g) + CO 2(g) [net chemical equation]

…(i)

Suppose this chemical reaction takes place in two steps NO 2(g) + NO 2(g)  → NO3(g) + NO (g) [Elementary reaction]

…(ii)

NO3(g) + CO (g)  → NO 2(g) + CO 2(g) [Elementary reaction]

…(iii)

Thus net reaction (i) is obtained by the combination of two elementary reaction (ii) and (iii) and NO3 is a reaction intermediate as it is produced and consumed during the course of reaction. 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 + D as we have seen −

1 d[A] d[C] = ∝ [A]m [B]n 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,

d[C] = k[A]m [B]n 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 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320



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 it’s 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  →R The differential rate law is written as d[P] 1 d[Q] Rate = − =− = k[P]m [Q]n 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 dC = kCn dt 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. k=

dC / dt Cn

Units of k: (concentration)1–ntime–1 For a zero order reaction (n = 0) Units of k = mol/L/s For a first order reaction (n = 1) Units of k = time–1 For a second order reaction (n = 2) Units of k = (mol/l)–1time–1 = l/mol/s


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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 litre–1sec–1)

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)

Solution :

Calculate the rate constant

d) Determine the reaction rate when conc. of Cl2 and NO are 0.2 M and 0.4 M respectively. 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 × 10–3 = k[0.05]m [0.05]n

…(2)

3 × 10–3 = k [0.15]m [0.05]n

…(3)

9 × 10–3 = 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., k =

d) Further,

r 1× 10−3 = [Cl2 ][NO]2 [0.05]1[0.05]2

= 8 litre2 mol–2 sec–1

r = k[Cl2]1 [NO]2 = 8 [0.2]1 [0.4]2 = 0.256 mol litre–1 sec–1


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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

fast

NOBr2

slow ii) NOBr2 + NO  → 2NOBr

Suggest the rate law expression 3.

The thermal decomposition of N2O5 occurs in the following steps: Step I:

slow N2O5  → NO2 + NO3

Step II

fast N2O5 + NO3 → 3NO2 + 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 [A]0 , which is the concentration at t = 0 and [A ]t , the [A ]t

concentration at time t

∫

[A ]0

t

−

d[A] = k ∫ dt dt 0


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ln

[A]0 [A]0 2.303 = kt or k = log [A]t t [A ]t A→

t=0

a

t=t

a–x

k= 3.2

Products

2.303 a log t a−x

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. k= at

2.303 a log t a−x

t = t 1/ 2 ; x = k=

a 2

2.303 a log t1/ 2 a/2

t1/ 2 =

0.6932 k

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

2.303 a log t a−x

at

t = t 718 ; x=

at

k=

7a 8

2.303 a log 7a t 718 a− 8

t 718 =

2.303 2.303 log 8 = × 3 log 2 k k

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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.

d[A] = k[A]n , it can be shown that dt

t1/ 2 =

2n −1 − 1 k(n − 1)a 0n −1

i.e.t1/ 2 ∝

1

(n ≥ 2)

a n0 −1

Expression for first order reaction is 2.303 a log t a−x 2.303 a 2.303 t= log = [log a − log(a − x)] or K a−x k 2.303 2.303 t= log a − log(a − x) k k Comparing this equation with Y = mx + C k=

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

2.303 log a and a slope k

2.303 k

2.303 log a K

2.303 K

time

slope = −

log(a − x) 3.5

EXAMPLES OF FIRST ORDER REACTION

3.5.1 Decomposition of H2O2 2H 2O2  → 2H 2O + O2 H 2O2  → H 2O + O [slow i.e. rate determining step] O + O  → O 2 [fast] 9 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828319

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So

Rate ∝ [H 2 O2 ]1

or

R = k[H 2 O2 ] 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 KMnO 4 solution in presence of dil. H 2SO4 2KMnO 4 + 3H 2SO4  → K 2SO4 + 2MnSO 4 + 3H 2O + 5O [H 2O2 + O  → H 2 O + O2 ] × 5 –––––––––––––––––––––––––––––––––––––––––––––––– 2KMnO 4 + 3H 2SO 4 + 5H 2O 2  → K 2SO4 + 2MnSO 4 + 8H 2O + 5O 2 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 V0 and Vt be the volumes of KMnO 4 used at zero time and after t time respectively. then

V0 ∝ a

[a = initial conc. of H 2O2 ]

Vt ∝ a − x

[a –x = conc. of H 2O2 after t time]

V0 a = Vt a − x k=

V 2.303 log 0 t Vt

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 VolumeV/(ml)

0

10

20

30

40

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 2.303 a k1 = log t a-x The volume of KMnO4 used, evidently corresponds to the undecomposed hydrogen peroxide.


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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 2.303 25 log = 0.022287 min-1 = 0.000314 s-1 when t = 10 min. k1 = 10 20.0 2 .303 25 log = 0.023230 min -1 = 0.0003871 s -1 when t = 20 min. k1 = 10 15.7 2.303 25 when t = 30 min. k1 = log = 0.02369 min -1 = 0.0003948 s -1 30 12.5 2.303 25 log = 0.023897 min -1 = 0.0003983 s -1 when t = 40 min. k1 = 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 NH 4 NO 2  → N 2 + 2H 2O Rate ∝ [NH 4 NO2 ]1 Rate = K[NH 4 NO2 ] order = 1 Kinetics of this reaction may be studied by collecting the nitrogen evolved and measuring its volume after regular intervals of time Let Vt and V∞ be the volumes of N 2 liberated after t time and at the end of reaction. NH 4 NO 2  → N 2 + 2H 2O Initial conc. Conc. after t time

a a–x

0 x

Conc. at t = ∞

0

a

Vt ∝ x V∞ ∝ a or

V∞ − Vt ∝ a − x V∞ a = V∞ − Vt a − x k=

2.303 V∞ log t V∞ − Vt


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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

C6H5N2Cl → C6H5Cl + N2

Solution : 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. 2.303 162 log For t = 20 min, k1 = = 0.0032min −1 20 162 − 10 2.303 162 log = 0.0033min −1 For t = 50 min, k1 = 50 162 − 25 2.303 162 log For t = 70 min, k1 = = 0.0032min −1 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 O

O Cl

N

H

CH3

N

CH3

rearrangement  → isomerism

N-chloro-N-phenylacetamide

Cl N-(4-chlorophenyl)acetamide

Rate ∝ [N-chloroacetinalide]1 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 I2 against standard sodium thiosulphate solution. N-chloroacetinalide reacts with KI and liberates I2 but p-chloroacetinalide does not. I2 + 2Na 2S2O3  → Na 2S4O6 + 2NaI 12 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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N-chloroacetinalide  → P − chloroacetinalide a 0 a–x x

t=0 t=t

Let V0 and Vt be the volumes of sodium thiosulphate used at t = 0 and t = t respectively, then V0 ∝ a Vt ∝ a − x V0 a = Vt a − x k= Illustration

V 2.303 log 0 t Vt

5 : Acetochloracetanilide (N-chloroacetanilide) is converted into pchloroacetanilide, 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 k=

2.303 a log t 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 k=

(ii)

When

t = 2 hours, (a – x) = 22.5 k=

(iii) When

2.303 45 log = 0.3411 1 32 2.303 45 log = 0.3468 t 22.5

t = 6 hours, (a – x) = 5.7


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k=

∴

2.303 45 log = 0.3443 6 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 k=

2.303 a log t a−x

0.300 =

2.303 45 log t (45 − x)

45 – x = 16.03 ∴

x = 28.97

Hence fraction of N-chloroacetanilide reacted in 3 hours = 3.6

28.97 = 0.643 45

PSEUDO FIRST ORDER REACTION If molecularity of a reaction is 2 but order is 1 then the reaction is known as pseudo firstorder reaction or pseudo unimolecular reaction. Hydrolysis of tertiary butyl bromide : (CH3 )3 CBr + OH −  →(CH3 )3 OH + Br − This reaction takes place in two steps. CH3

i)

CH3 slow Br  → H 3C

H3C

CH3

CH3

ii)

+ Br -

CH3

CH3 fast H3C + OH - →

H3C CH3

OH CH3

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Order = 1 So the reaction is pseudo unimolecular 3.6.1 Hydrolysis of Ester catalysed by an acid +

H CH3COOC 2H 5 + H 2O  → CH3COOH + C6 H5OH

Rate ∝ [CH 3COOC 2 H 5 ][H 2O] Since water is in large excess so its concentration almost remains constant Rate ∝ [CH 3COOC 2 H 5 ]1 Rate = k[CH 3COOC 2 H 5 ] 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 V0 , Vt and 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 ( V0 ) corresponds to the concentration of catalyst and volume of NaOH used after t time (Vt ) corresponds to the concentration of catalyst and acetic acid formed. +

H CH3COOC 2H 5 + H 2O  → CH3COOH + C2 H5OH

t=0

a

0

t=t

a–x

x

t=∞

0

a

Vt − V0 ∝ x

[x = conc. of CH3 COOH formed after t time]

V∞ − V0 ∝ a (V∞ − V0 ) − (Vt − V0 ) ∝ a − x V∞ − Vt ∝ a − x V∞ − V0 a = V∞ − Vt a − x k=

V − V0 2.303 log ∞ t V∞ − Vt 15

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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 30°C. 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. k1 =

2.303 V − V0 log ∞ t V∞ − Vt

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

2.303 V − V0 log ∞ = k1 t V∞ − Vt

75 min

21.05 - 12.10 = 8.95

2.303 11.43 log = 0.003259 min -1 75 8.95

119 min

21.05 - 13.10 = 7.95

2.303 11.43 log = 0.003264 min -1 119 7.95

83 min

21.05 - 14.75 = 6.30

2.303 11.43 log = 0.003254 min -1 183 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 +

H C12 H 22O11 + H 2O  → C6 H12O6 + C6 H12O6 Cane sugar

glu cose

fructose

Rate ∝ [C12 O22 O11 ][H 2 O] Since water is solvent its concentration is constant. So

Rate = K[C12 H22O11 ] Molecularity = 2 Order = 1

So the reaction is pseudo unimolecular


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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. Rate ∝ [C12 H 22 O11 ][H 2 O][H + ] But under ordinary conditions H 2O and H + are constant so, Rate ∝ [C12 H 22 O11 ] 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. +

H C12 H 22O11 + H 2O  → C6H12O6 + C6 H12O6 Dextrorotatory

Dextrorotatory

Laevorotatory

−92°

Specific rotation +52.5° laevorotatory

Fructose Laevo rotatory Glucose Prism

dextro rotatory Cane sugar dextro rotatory

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 r0 , rt and r∞ be the angles of rotation at 0 time, after t time and at the end of reaction. 0° r∞

rt r0

-90°

90°

180°

So,

r0 − rt ∝ x

…(i)

r0 − r∞ ∝ a

…(ii)


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(r0 − r∞ ) − (r0 − rt ) ∝ a − x or rt − r∞ ∝ a − x r0 − r∞ a = rt − r∞ a − x k=

r −r 2.303 log 0 ∞ t rt − r∞

Illustration 7 :The optical rotations of sucrose in 0.5 N HCl at 35°C at various time intervals are given below. Show that the reaction is of first order : Time (minutes) Rotation (degree) Solution :

0

10

20

30

40

∞

+32.4

+28.8

+25.5

+22.4

+19.6

-11.1

The inversion of sucrose will be first order reaction if the above data confirm to 2.303 r −r log 0 ∞ the equation , k 1 = t rt − r∞ 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 43.5 log = 0.008625 min-1 10 39.9

20 min

+25.5

36.6

2.303 43.5 log = 0.008625 min-1 20 36.6

0 min

+22.4

33.5

2.303 43.5 log = 0.008694 min-1 30 33.5

40 min

+19.6

30.7

2.303 43.5 log = 0.008717 min-1 40 30.7

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


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EXERCISE 3 : 1.

From the data obtained for decomposition of H2O2, show that decomposition of a first order reaction. Time in minutes Volume of KMnO4 (ml)

2.

10

20

30

40

25.0

20.0

15.7

1.5

9.6

From the following data for the decomposition of ammonium nitrite in aqueous solution, show that the reaction is a first order. Time in minutes Volume of N2(ml)

3.

0

10

15

20

15

∞

6.25

9.00

11.4

13.65

35.05

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 0

75

119

183

∞

9.62

12.1

13.1

12.75

21.05

Time in minutes Volume of NaOH(ml)

Show that hydrolysis of ester is a first order reaction.

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. Threshold energy Potential energy

4.

Ea

Ea

Eb

Eb Products

∆H Reactants

Reactants

∆H

Products Products

Threshold energy

Reaction co-ordinate

Reaction co-ordinate


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E a = activation energy of forward reaction E b = activation energy of backward reaction ∆H = Heat of reaction ∆H = E a − E b Since

Ea > E b

So, ∆H > 0 i.e. reaction is endothermic 4.1

∆H = E a − E b Since

Ea < Eb

So, ∆H < 0 i.e. reaction is exothermic

FACTORS INFLUENCING THE RATE OF A REACTION (i) Concentration of the reactants: More the concentration of the reactants more will be the rate. (ii) Temperature: The rate of a reaction increases with an increase in temperature. (iii) Nature of reactants: Physical and chemical state of the reactants also affects the rate of reaction. (iv) Effect of Catalyst: A reaction proceeds much faster in the presence of a catalyst. (v) Effect of radiations: Photochemical reactions are faster than thermal ones because in the former all the energy of photons is completely used in exciting molecules. While in latter the energy is distributed into various forms like translational, rotational and vibrational

4.2

EFFECT OF TEMPERATURE ON REACTION RATE The increase of temperature generally increases rate and therefore rate constant of a reaction. k Temperature coefficient = T +10 kT k T = rate constant t°C k T +10 = rate constant at T + 10°C Normally T and T+ 10 are taken as 25°C and 35°C. The value of temperature coefficient at room temperature lies between 2 and 3. Temperature coefficient is only an approximate method of indicating influence of temperature on reaction rate. The exact variation of rate constant with temperature is given by Arrhenius equation. d nk E = dT RT 2 E → Energy of activation T → Temperature is Kelvin scale 20 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

…(i)

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R → Gas constant E

∫ d nk = ∫ RT

2

dT. {E is supposed to be constant]

nK slope = −

E 1 nk =  −  + C R T or

E R

1/ T

E nk = − +C RT

…(iv)

on comparing this equation with Y = mx + C 4.3

ANOTHER FORM OF ARRHENIUS EQUATION k = Ae− E / RT A → frequency factor e − E / RT → fraction of molecules which have energy equal to or greater than activation energy. Let k1 is rate constant at temperature T1 and k 2 is rate constant at temperature T2 . nk1 = −

E +C RT1

nk 2 = −

E +C RT2

E1 1   −  R  T1 T2  k E T −T n 2 = × 2 1 k1 R T1T2 nk 2 − nk1 =

or

log

k2 E T −T = × 2 1 k1 2.303R T1T2

Illustration 8: The activation energy of a non-catalysed reaction at 37°C is 200 kcal/ mol and the activation energy of the same reaction when catalysed decreased to only 6.0 kcal/mol. Calculate the ratio of rate constants of the two reactions. Solution :

We know that: k = Ae − Ea / RT Let k = rate constant for non-catalysed reaction and kc rate constant for catalysed reaction. Ea be the activation energy of non-catalysed reaction and Eac be the energy of activation of catalysed reaction. 1.

k = Ae − Ea / RT

2.

k c = Ae − Ea / RT


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⇒

1 (E ac − E a ) k = e RT kc

⇒

log

k 1 = (Eac − Ec ) k c 2.303RT

⇒

log

k 1 = (6 × 103 − 200 × 103 ) k c 2.303 × 2 × 310

⇒

log

k = −9.8 kc

⇒

k = 1.56 ×10 −10 kc

or

kc = 6.3 × 109 k

Illustration 9 : A first order reaction A → B requires activation energy of 70 kJ/mol. When 20% solution of A was kept at 25°C for 20 minutes, 25% decomposition too place. What will be the percent decomposition in the same time in a 30% solution maintained at 40°C? Assume that the activation energy remains constant in this range of temperature. Solution :

Note: it does not matter whether you take 20%, 3%, 40% or 70% of A. At 25°C, 20% of A decomposes 25% C0 Ct

⇒

kt = 2.303log

⇒

k(40) = 2.303log

⇒

k(at 25°C) = 0.0143min -1

⇒

100 75

log

k 40° Ea  T2 − T1  =   k 25° 2.303R  T1T2 

log

k 40° 70 × 103  40 − 25  = 0.0143 2.303 × 8.31  298 × 313 

k (at 40°C)=0.055min -1

Now calculate the % decomposition at 40°C using first order kinetics kt = 2.303log

C0 Ct 100 100 − x

⇒

0.055 × 40 = 2.303log

⇒

x = 67.1 ≡ 67.1% decomposition of A at 40°C.


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4.4

ACTIVATION ENERGY AND TEMPERATURE DEPENDENCY The temperature dependency of reactions is determined by activation energy and temperature level given below is a graph showing the temperature dependency of the reaction rate. Following conclusions are drawn from this graph. (i) From Arrhenius law a plot of ln k vs

1 gives a straight line, with large slope for large T

E and small slope for small E. (ii) Reactions with high activation energy are far more temperature sensitive then reactions with low activation energy. High E

200 Low E

100 50 nK

slope = −

20

E R

10 5

∆T = 1000 for doubling rate 2000°K

1000°K 1/ T

∆T = 87 for doubling rate 463 °K 376° K

iii) A given reaction is much more temperature sensitive at low temperatures then at high temperatures. EXERCISE 4 : 1.

(i) Which reaction will have the greater temperature dependence for the rate constant, one with small value of Ea or one with large value of Ea (ii) For a reaction, rate constant is given as lnK(min–1) = 31.33 –

11067 . Evaluate, K,E T

and A for the chemical reaction at 25°C. 2.

For a particular first order reaction at 27°C, the concentration of reactant is reduced to one half of its initial value after 5000s. At 37°C, the concentration is halved after 1000s. Calculate (a) the rate constant of the reaction at 27°C, (b) the time required for the concentration to be reduced to one quarter of its value at 37°C, (c) the activation energy of the reaction.

3.

The Arrhenius equations for the rate constant of decomposition of methyl nitrite and ethyl nitrate are  −152300Jmol−1   −157700Jmol−1  −1 14 k1 (s −1 ) = 1013 exp  and k (s ) = 10 exp    2 RT RT     respectively. Find the temperature at which the rate constants are equal. 23 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828319

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4.5

EFFECT OF CATALYST ON RATE OF REACTION A catalyst is a substance that takes part in a reaction. It increases the rate of reaction by providing a new pathway. Now let us consider a reaction i.e. hydrolysis of an ester catalysed by an acid. +

H Ester + H 2 O  → acid + alcohol

The rate of the reaction is given as Rate = k[ester][H + ] Since concentration of H + in the beginning and at the end is same so. Potential energy

Rate = k′[Ester] where

k′ = k[H + ]

Alternative path provided by the catalyst.

Ea

Eb Reactants

∆H Products

k′ is known as pseudo or effective rate constant Reaction co-ordinate

Now according to the definition of the catalyst are substances which increases the rate of reaction by their presence but their presence is not required for the reaction to occur. Now according to this definition, absence of H + should not let the reaction to happen, therefore the rate of reaction in the absence of H + should be, Rate = K[ester] If [H + ] < 1 M then rate of reaction would becomes less than the rate which we had in its absence. In order to avoid above controversies, we are required to go through the functioning of catalyst. There are two types of catalyst. a)

Reactant catalyst

b) Non-reactant catalyst.

Former increases the rate of reaction by existing in the rate law. Therefore, in the absence of reactant catalyst the reaction rate will be zero i.e. no reaction will take place. The non-reacting catalyst on the other hand do not exist in the rate expression. Changing the concentration of non-reacting catalyst has no effect on the rate. These catalyst increase the rate of reaction by either decreasing activation energy or by increasing Arrhneius constant.

5.

KINETICS OF SOME COMPLEX FIRST ORDER REACTIONS

5.1

PARALLEL REACTIONS First order parallel chain reaction: If A decomposes into B and C parallely following two paths with first order rate constants k1 and k2 respectively, then it is known as first order parallel chain reaction. Let the overall rate constant of A is k0 by which decay into B and C.


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k1

B

k2

C

A

∵

+

∵

+

∵

−

∵

−

d [ B] dt d [ C] dt d [A ] dt d [A] dt

= k1 [ A ]t

…(i)

= k 2 [ A ]t

…(ii)

= k 0 [ A ]t

…(iii)

=+

d [ B] dt

+

d [ C] dt

…(iv)

Putting equations (i), (ii) and (iii) in equation (iv), we get k 0 [ A ]t = k1 [ A ]t + k 2 [ A ]t

or,

k 0 = k1 + k 2

From equation (i) and (ii), we get d [ B] dt

or,

= k1 [ A ]t

[ B]t = ∫ d [ B] = k1 ∫ [ A ]t dt

Similarly, [ C]t = ∫ d [ C] = k 2 ∫ [ A ]t dt

…(v) …(vi)

Divide equation (v) by equation (vi), we find

[ B ]t [ C ]t

=

k2 (at any time t) k1

∵

[ A ]0 = [ A]t + [ B]t + [ C]t

or

[ A ]0 − [ A ]t = [ B]t + [C]t

∴

( k1 + k 2 ) t = 2.303log

…(vii)

(viii)

[ A]0 [ A ]t

If k1 and k2 , t and [A]0 is known, then [A]t can be calculated then by solving equations (vii) and (viii), the concentration of B and C at any time i.e., [B]t and [C]t can be calculated. (b) Half lives of A in different path of parallel chain reaction. ∵ for parallel chain reaction


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k1

B

k2

C

A

k0 = k1 + k2 or

k0 > k1 and k0 > k2

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

d [A]

∴

+

⇒

k1 [A]t = x % × k0 [A]t

or,

k1 = x% × k0

⇒

x ln2 l n2 = × ( t1/ 2 )1 100 ( t1/ 2 )overall

or,

( t1/ 2 )1 =

dt

= x% × −

dt

100 × ( t1/ 2 )overall 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, half–life of A, i.e. (t1/2)2 while it is converting into C will be

( t1/ 2 )2 =

100 × ( t1/ 2 )overall

(100 − x )

(c) For the first order parallel chain reaction k1

mB

k2

nC

A

[ B ]t [ C ]t and

=

mk1 mk 2

koverall = k1 + k2


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5.2

CONSECUTIVE OR SEQUENTIAL REACTIONS Consider the following consecutive reaction K1 K2 A  → B  →C

B 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 − +

or

d [A] dt d [ B]

dt

…(i)

= k1 [ A ]t − k 2 [ B]t

dt

d [ B]

= k1 [ A ]t ⇒ [ A ]t = A 0 e − k,t

+ k [ B]t = k1 [ A ]t = k1 [ A ]0 e − k1 t

…(ii)

Solving first degree differential equation (ii) with integrating factor e k 2 t , then we get concentration of B at any time t.

[ B ]t =

k1 [ A ]0

e − k1 t − e − k 2 t   k 2 − k1 

(b) Time at which concentration of B is maximum t=

1 k ln 2 k 2 − k1 k1

(c) Maximum concentration of B

[ B]max (d) Time at which t=

x

k  k2 = [ A ]0  2  where x = k1 − k 2  k1 

d [ B] dt

is maximum

2 k ln 2 k1 − k 2 k1

(e) At steady state, k1 [A]t = k2 [B]t (f) If k1 < < k2, then sequential reaction converts into k1 A  →C


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∴

−

d [A ] dt

= k1 [ A ]t

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 (λ) for each of the separate paths? Solution :

The rate constant of the decay is k=

0.693 0.693 = t 12 22

If k1 and k2 are the rate constants of the reactions leading to respectively we have 0.693 k1 + k2 = 22 k1 2 = k 2 98

222

Th and Fr223,

On solving for k1 and k2, we get k2 = 0.03087 y−1 k1 = 0.00063 y−1 EXERCISE 5 : 1.

An organic compound A decomposes following two parallel first order mechanisms. B k1 1 = and k1 = 1.3 × 10−5 s−1 k2 9

A C

Calculate the concentration ratio of C to A, if an experiment is allowed to start with only A for one hour. 2.

K1 K2 For a consecutive first order reaction A  → B  →C, the values of K1 and K2 are –1 –1 45 sec and 15 sec respectively. If the reaction is carried out with pure A at a concentration of 1.0 mol dm–3,

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.


;

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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 L–1 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) =

k1 (a – xe) k2

or

b=

k1 (a – xe) –xe k2

Substituting value of b in eqn. (i) Rate of reaction,

k  dx = k 1 (a − x ) − k 2  1 (a − x e ) − x e + x  dt k2 

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

dx = (k 1 + k 2 )dt xe − x

∫

x

x0

log or Or

k

A k1 B 2

t dx = ∫ (k1 + k 2 )dt 0 xe − x

xe − x0 = (k1 + k 2 )t xe − x

x − x0 1 (k 1 + k 2 ) = log e t xe − x

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.


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5.4

PRESSURE CHANGE METHOD This method is used for gaseous reactions. As reaction proceeds there is change in pressure. →

B(g) + C(g)

For a reaction,

A(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,

6.

k=

2.303 P0 log t (2P0 − Pt )

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. 1 [A]0 K = ln . If for different values of t and [A]t , K comes constant then order of t [A]t reaction will be 1 other wise we will switchover to higher order expressions. Integrated Expressions for various reactions x t

Zero order

k0 =

First order

k1

Second order

1 x  k2 =   t  a(a − x) 

Third order

1  x(2a − x)  k3 =  2  t  a (a − x 2 ) 

2.303 a log t (a − x)


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(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 (a–x) vs t is a straight line then order of reaction will be 1.

Zero order

(a − x)

First order

log(a − x)

t

t

1 (a − x)2

1 (a − x)

Second order t

Third order t

(iii) Half life Method: As we have seen for nth order reaction t1/ 2 ∝

1 a n −1

Let t11/ 2 be the half life at initial concentration a1 and t 21/ 2 be the half life at intial concentration a 2 . Then

t11/ 2 ∝ t 21/ 2 ∝ t11/ 2 t 21/ 2

1 a1n −1 1 a n2 −1

a  = 2   a1 

n −1

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 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828319

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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 :— α -particles

Sl. No.

β-particles

γ -rays.

1.

2 units +ve charge & 4 Unit –ve charge with no Electromagnetic waves units mass. mass of short wavelength.

2.

Represented as

2

He 4 or Represented as

–1

eo

Represented as 0 γ 0

He ++

7.3

3.

They have high ionizing Ionizing power is less than Ionizing power is very power less that of α particle

4.

These have penetrating power

5.

Velocity is of the order of Velocity is of the order of Velocity is same as that of velocity of light 2 × 109 cm/s 2.8 × 1010 cm/s ( 3 × 1010 m/s).

6.

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

low Penetrating power is 100 Penetrating power is 100 times that of α particle times that of β particle.

When a β particle is There is no change in emitted, atomic number atomic number or atomic increases by one unit and mass. atomic mass remains unchanged

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.


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(i) About 160 stable nuclides have an even number of protons and an even number of neutrons, for example, 126 C (ii) About 50 stable nuclides have an even number of protons and an odd number of neutrons, for example, 25 12 Mg . (iii) About 50 stable unclides have an odd number of protons and an even number of neutrons, for example, 199 F . (iv) Only 4 stable nuclides have an odd number of protons and an odd number of neutrons. They are 12 H, 36 Li, 105 B, and 147 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 20826 Pb ; the uranium-235 series, in 20827 Pb ; and the thorium-232 series, in lead (Z = 82);

208 82

208 82

Pb . All these terminating includes have the magic number 82 in

Pb is doubly magic, with 82 protons and 126 neutrons.

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 27 39 40 Al, 19 K, and 20 Ca . Nuclides with Z > 20 require a large number this group are 42 He, 186 O, 13 of neutrons than protons to moderate the effect of increasing proton repulsions. For 133 209 example, the N/Z ratio in 56 26 Fe is 30/26 = 1.15; in 55 Cs it is 78/55 = 1.42, and in 83 Bi , 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. 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 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828319

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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, 74 30

Solution :

118 50

Sn,

234 91

Pa, 54 25 Mn,

Zn ?

118 50

Sn : This nuclide has 50 protons and (118 – 50) = 68 neutrons. This is an eveneven combination, the most common for stable nuclides. Also, the neutron : proton ratio of 68:50 lies within the belt of stability so 11508 Sn is non-radioactive. 234 91

Pa : Atomic number 91 exceeds the limit for the naturally occurring stable nuclides (Z > 83).

234 91

Pa is radioactive.

54 25

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


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74 30

Zn : 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 the belt and is radioactive. 7.4

74 30

Zn lies above

MODE OF DECAY Alpha-particle emission : An alpha (a ) particle has the same composition as a helium nucleus : two protons and two neutrons. Thus an a 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 4 2 He . We can represent a -particle emission with a nuclear equation, as in the radioactive decay of uranium-238. Sum of mass numbers :

238 238 92 U

¾ ¾®

Sum of atomic numbers : 92

234 + 4 = 238 234 + 24 He 90 Th 90

+ 2

= 92

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 (β - ) 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 -01 e . 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 0

n ¾ ¾®

1 1

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 : Sum of mass numbers :

234 234 90 Th Sum of atomic numbers : 90

¾ ¾®

234 + 234 + 91 Pa 91 +

0

= 234

0 -1

e -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.


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Gama-ray emission : Gamma (γ) 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 alphaparticle emission, 23% of the thorium-230 nuclei formed are in an excited state : 230m 90 Th . These nuclei then emit energy as gamma rays. Sum of mass numbers :

230 234m 90

Th

230 + 0 = 230 230 + g 90 Th 90 + 0 = 90

¾ ¾®

Sum of atomic numbers : 90

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 1

p ¾ ¾®

1 0

n + 10e

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. Sum of mass numbers :

26 Al 13

26 13

Sum of atomic numbers :

¾ ¾®

26 + Mg + 12 +

26 12

0 = 26 e 1 = 13

0 1

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

e + 11P ¾ ¾®

1 0

n

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 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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Sum of mass numbers :

125 125 + 53 I Sum of atomic numbers : 53

0 0 -1 e -1

¾ ¾®

125 Te 52

125 52

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 Solution :

84 40

Zr to undergo ?

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 : Electron capture :

84 40

84 40

Zr ¾ ¾®

84 39

Y + 10e

Zr + -01e ¾ ¾®

84 39

Y

Notice that in each case, the neutron:proton ration increases from 44/40 in

84 40

Zr to

45/39 in 84 39 Y . 7.5

NUCLEAR FISSION Splitting of a nucleus into lighter nucleus is known as nuclear fission. U 235 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. Ba140 + 36 Kr 93 + 30 n1 144 + 38 Sr 90 + 20 n1 54 Xe 144 + 37 Rb 90 + 20 n1 55 Cs 56

92

U 235 + 0 n1

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, that’s 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 supply–mostly 11 H —will last for billions of years.


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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 1

H + 13H  → 42He + 10n

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 fusion–a 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.

KINETICS OF RADIOACTIVE DISINTEGRATION Rate of decay of disintegration : Rate of disintegration of radioactive isotope is proportional to it’s number of atoms. Rate of decay ∝ N –

dN ∝ N. dt

or

–

dN ∝N dt

–

dN = λN dt

Time

Rate of decay of A

N  → number of atoms of radioactive isotope. No. of atoms of A

8.

Time

… (i)

λ is known as decay constant or disintegration constant. If

dt = 1 second.


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dN =λ N Decay constant ( λ ) : Thus decay constant ( λ ) is equal to the fraction of radioactive isotope disintegrating per second. From equation (i) Then

–

dN = λ dt N ∫ – log N = λt + C

∫–

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

–log N = λ t – log N 0 N0 N 1 N λ = ln 0 t N λt = log

N 2.303 log 0 t N All radioactive decay follows first order kinetics. λ=

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 t1/ 2 = 1600 years). Rate of decay (activity ) = λ × No. of atoms in 1 g Ra. =

.6932 1× 6.023 × 1023 × =3.7 × 1010 dps 1600 × 365 × 24 × 60 × 60 223

= 3.7 × 1010 Becque rrel 1 curie

= 3.7 × 1010 dps = 3.7 × 1010 Becquerrel

1 Rutherford = 106 dps. Initial activity ∝ Initial amount Activity after `t’ time ∝ amount left undisintegrated after t time. λ=

2.303 initial activity log t activity after t time


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as we have seen – ln N = λt – ln N 0 ln

N = – λt N0

N = e – λt N0 or

N = No e – λt

Half life period (t1/ 2 ) : It is the time after which activity of a radioactive isotope reduces to half of it’s initial value. When

t = t1/ 2 , N = N 0 / 2 λ=

N 2.303 log 0 t N

t1/ 2 =

N 2.303 log 0 N0 / 2 λ

t1/ 2 =

0.6932 λ

Unit of λ = time –1 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 = K=

1 λ

t 1 = 1/ 2 λ 0.6932

K = 1.44 t1/ 2 8.2

RADIOACTIVE EQUILIBRIUM 92

–α –β –β U 238  → 90Th 234  → 91Pa 234  → 92 U 234

Suppose a radioactive isotope. A is producing another radioactive substance B, which in turn is producing C. A  → B  →C Rate of decay of A ∝ N1 [ N1 is no. of atoms of A]. 40 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320



Rate of decay of A = λ1N1 [ λ1 is decay constant of A] Rate of decay of B ∝ N 2 [N 2 is no. of atoms of B] = λ 2 N 2 [λ 2 is decay constant of B] 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 λ1 N1 = λ 2 N 2 λ1 N 2 0.6932 / T1 = = λ 2 N1 0.6932 / T2 λ1 N 2 T1 = = λ 2 N1 T2 1 K λ1 K 2 N T = = 2 = 1 N1 T2 λ 2 K1

λ=

but so 8.3

CARBON DATING 14 7 atomspheric nitrogen

N

6

+ 0 n1  → 6C14 + 1H1 cos mic rays

C14 is radiocarbon ( β -emitter) 6

C14  → 7 N14 + –1β0

Carbon dating is based upon the fact that radio carbon –14 is produced in the atmosphere by the action of cosmic rays C14 is a β emitter with a half life of 5730 years. It’s rate of formation has been constant so the atmosphere has an equilibrium concentration of C14 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 C14 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. N 2.303 λ= log 0 t N λN 0 2.303 λ= log t λN 41 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828319

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8.4

λ=

2.303 Initial activity log t activity after t time

λ=

2.303 C / C in fresh sample log 14 12 t C14 / C12 in old sample

MINERAL DATING (GEOCHRONOLOGY). Determination of age of rocks, earth etc. is known as mineral dating N 2.303 λ= log 0 t N 2.303 P+D λ= log t 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 :

K=

0.693 0.693 = T1 / 2 5770

= 1.201 × 10–4 year–1 K=

N 2.303 log 0 t Nt

1.201 × 10–4 = 4.002 = ∴

N 2.303 log 0 11540 Nt

N0 Nt

Nt 1 (Remaining fraction) = N0 4.002

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.


Chemistry : Chemical Kinetics and Nuclear Chemistry


SECTION - I SUBJECTIVE TYPE PROBLEMS Problem 1:

Solution :

The half life of a chemical reaction at a particular concentration is 100 min. When the concentration of the reactant is doubled, the half life becomes 50 min, then what is the order of the reaction ? (t1/ 2 )1  a 2  =  (t1/ 2 )2  a1 

n −1

n −1

100  2  ⇒ =  50  1  ⇒ 2 = 2(n–1) ⇒ =n–1 ⇒ n=2 Problem 2:

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

Solution :

Let the dextrorotatory isomer A and laevoratatory is B θ0 = initial observed rotation θt = observed rotation at time t A

!!" #!!

t=0

θ0

⇒

θ0 – θ

B 0 –θ

From question, θ0 = 100, θt = 60 ⇒

60 = 100 – 2θ

∴

θ = 20 = θ0 – θ = 100 – 20 At 27°C,r

∴

Keq. = K1 =

[B] 20 1 = = [A] 80 4

Similarly, at 127°C. Keq. = K2 =

[B] 10 1 = = [A] 90 9

since 2.30 log =

1/ 4 ∆H  1 1  = −  1/ 9 R  300 400  43

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9 ∆H  100  = 4 R  300 × 400 

⇒

2.303 log

⇒

2.303 [log 9 – log 4] =

⇒

2.303 [2 log 3 – 2 log 2] =

⇒

2.303 (2 × 0.477 – 2 × 0.301)

⇒

∆H = 2.303 [0.954 – 0.602] × 8.314 × 1200

∆H 1 × R 1200 ∆H 1 × R 1200

∆H = 8.188 kJ Problem 3:

A mixture of two substances X & Y produces Z simultaneously through parallel chain reaction as under

I X Z Y I The rate constant for the first path is 1.2 × 10 –2 s –1 and of second path is 4 × 10–3 s–1. After 40 seconds, the mole percentage of Z is found to be 36. Find out its 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 = 100 Initial 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 C

Now. a–p, k1=1.2×10-2s-1

X C p+q Y -3 -1

100–a–q, k2=4×10 s

Now, ⇒

p 1.2 × 10−2 3 = = q 4 × 10−3 1

p = 3q Again, p + q = 36 4q = 36

∴

q=9

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∴

p = 27

a Now k1 × 40 = 2.303 log a − 27 1.2 × 10−2 × 40 a = log 2.303 a − 27 ⇒ ⇒ ⇒

0.208 = log

a a − 27

a = 1.6 a − 27 a = 72

Now, for 60 seconds, let the amount of “X” converted into Z is p′ and the amount of Y converted into Z is q′.

1.2 × 10−2 × 60 72 = log 2.303 72 − p' ⇒

72 0.313 = log 72 − p′ 72 2.050 = 72 − p′

p′ = 36.88 q′ = 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 10–3% molecule to cross over the energy barrier? Given that, ln10–5 = –11.515.

Solution :

Fraction of molecules crossing over the energy barrier = e− Ea / RT

Problem 5 :

10−3 = e− Ea / RT 100

⇒

=

⇒

10–5 = e− Ea / RT

⇒

– 11.515 = −

⇒

Ea = 11.515 × 8.314 × 298 Joule

=

28.53 KJ

Ea RT

The reaction given below, involving the gases is observed to be first order with rate constant 7.48 × 10−3 sec−1.Calculate the time required for the total pressure in a system containing A at an initial pressure of 0.1 atm to rise to 0.145 atm and also find the total pressure after 100 sec. 2A(g) → 4B(g)

+ C(g) 45


NARAYANA



Solution :

→

2A(g)

4B(g)

+

C(g)

0

Po Po − P′

0

2P′

P′/2

Ptotal =Po − P′ + 2P′ + P′/2 = Po + P′ =

2 (0.145 − 0.1) =0.03 atm 3

k =

Po 2.303 log t Po − P '

t=

3P ' 2

2.303  0.1  log   −3 7.48 × 10  0.07 

t = 47.7 sec Also, k =

 0.1  2.303 log   t  Po − P ' 

7.48 × 10−3=

2.303  0.1  log   100  0.1 − P ' 

0.1 − P′ = 0.047 P′ = 0.053 Ptotal = 0.1 + Problem 6 :

3 (0.053) ≈ 0.180 atm. 2

For the reaction : C2H5I + OH− → C2H5OH + I−

the rate constant was found to have a value of 5.03 × 10−2 mol−1 dm3 s−1 at 289 K and 6.71 mol−1 dm3 s−1 at 333 K. What is the rate constant at 305 K. Solution :

k1 = 5.03 × 10−2 mol−1 dm3 s−1 at T1 = 289 K k2 = 6.71 mol−1 dm3 s−1 at T2 = 333 K Ea  333 − 289   6.71  log    −2  =  5.03 × 10  2.303 × 8.314  333 × 289 

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

Ea  1 1  k2 log k = 2.303R  T − T  1 2   1 k2 88914 1     1 = − log    −2   5.03 × 10  2.303 × 8.314  298 305 

On solving we get, k2 = 0.35 mol−1 dm3 s−1 46 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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Problem 7 :

The decomposition of Cl2O7 at 400 K in the gas phase to Cl2 and O2 is of I order. After 55 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. Cl2O7

Solution :

 →

Cl2

7 O 2 2

+

Mole at t = 0

a

0

0

Mole 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=

2.303 0.062 log10 t 0.044

K = 6.23 × 10–3 sec–1. (b) Let at t = 100 sec, (a – x) ∝ P ∴6.23 × 10–3 =

2.303 0.062 log10 100 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 life of A is 8 days, what is the half life of B?

Solution :

Let the initial number of nuclide of X be x0 Let the initial number of nuclide of Y be y0 Let the number of nuclide of X after time t be xt Let the number of nuclide of Y after time t be yt ∴

xt = ½ y0; yt = ¼ x0 x0 = 1.414 y0

Now,

xt = x0 e −λ1t = ½ y0

…(1)

yt = y0 e −λ2 t = ¼ x0

….(2)

Dividing equation. (1) by (2)

2y0 x0 e(λ 2 −λ1 )t = x0 y0 2= Since

x02 (λ 2 −λ1 )t e y 02 x0 =

2 y0

x02 = 2y02


NARAYANA



∴

2 = 2e(λ 2 −λ1 )t

∴

e(λ 2 −λ1 )t = 1

Since t is not zero, hence λ2 = λ1. Therefore the half life of the other radioactive element is also 8 days. Problem 9 :

On analysis a sample of Uranium was found to contain 0.277 g of 82Pb206 and 1.667 g of 92U238. The half life period of 92U238 is 4.51×109 years. If all the lead were assumed to have come from decay of 92U238, what is the age of the earth?

Solution :

92U

238

82Pb

1.667 mole 238

= 1.667 g =

206

= 0.227g =

0.227 mole 206

∵ All the lead have come from decay of U. ∴ Moles of Pb formed =

0.277 206

∴ moles of U decayed =

0.277 206

∴ Total moles of Uranium = Also N for U238 =

1.667 0.277 + , ie N0238 206

1.667 238

1.667 0.277 + 9 238 206 N 2.803 2.303 × 4.51 × 10 0 log ∵ for U238 t = = log 1.667 k N 0.693 238 ∴

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 be 26° and that after completion of reaction was 10° at 27°C. Find the half - time of the reaction assuming it to follow pseudo first order kinetics. The observed rotation per mole of A,B and C are 60°, 50° and –80°. A plot of logarithm of rate constant of the above reaction vs T–1 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

+

H 2O

→

2B

+

C

a

excess

0

0

(a-x)

const.

x

2x

0

const.

2a

a




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

x = 0.1 K=

2.303 a 2.303 0.5 2.303 log = log = t (a − x) 40 0.5 − 0.1 40 log 5/4

∴

K = 5.57 × 10–3 min–1

∴

t1/2 =

0.693 0.693 = = 124.4 min K 0.00557

From Arrhenius equation, K = Ae–Ea/RT LogK =

−Ea + log A; Intercept = log A 2.303RT

WhenT = 27 + 273 = 300K , log K = log (5.57 × 10–3) = –2.254 Putting the value in above equation – 2.254 = ∴

− Ea 1 × + 15.046 2.303 × 8.314 300

Ea = 99.37 kJ /mol

In order that t1/2 of reaction may decrease from 124.4 min, rate constant will have to increase 4 time. Log (4 × 5.57 × 10–3) = ∴

– 1.652 – 15.046 =

∴

T = 310.80 K

99370 1 × + 15.046 2.303 × 8.314 T

99370 1 × 2.303 × 8.314 T


NARAYANA



SECTION - II SINGLE CHOICE PROBLEMS Problem 1:

!!" nB with time is presented in the figure, the value of n The progress of the reaction A #!! is (a) 2

(b) 3

(c) 4

(d) 5 B

Con. ol/litre

0.6

0.3 A

1

5

3

7

Time/hour

Solution :

A

!!" #!!

nB

t=0

0.06

0

(initial conc. in mol L–1)

At equi.

0.6 – x

nx

(conc. at equilibrium)

From the graph 0.6 – x = 0.3 Also from the graph nx = 0.6 ∴n=2 ∴ (a) Problem 2 :

Solution :

For the elementary reaction, 2NO + H2 → N2O + H-2O; the half life time is 19.2 sec at 820°C 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 are 600 mm of Hg and 20 mm of Hg respectively would be (a) 19.2 sec

(b) 10 sec

(c) 830 sec

(d) 415 sec

According to Ostwald isolation method of order determination, the concentration terms of reactants present in excess remain virtually unchanged and thus can be merged with the rate constant. In this case, the rate of the reaction depends only on the concentration o f that reactant 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 may consider the rate of reaction to be independent of PNO and hence reaction will follow the rate law.

dp(N 2O) dt

= k ′PH 2 ; where k ′k(PNO ) 2 = Pseudo first order reaction rate constant.




Half-life time of first order reaction is independent of initial concentration. Hence the required half-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 expressed in mole ml–1 the relation between their rate constants k′1 and k′2 will be (b) k′2 × 103 = k′2 (a) k′1 = k′2 (c) k′2 × 10–3 = k′1 (d) k′1 = 10k′2

Solution :

Since kα[a]1− n For first order reaction,

k1α[a]1−1 k1α1

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

k2 ∝

1 a

k2 ∝

1 1 = -3 −3 a mole dm a × 10 mole ml-1 1

k ′2 ∝

a mole ml

−1

∴ k 2 = k ′2 × 103 = k1′ Problem 4 :

Solution :

∴ (c) B For the first order reaction 3A → B concentration varies with time as shown in the adjacent graph. The half-life of the Conc. reaction would be A (In M) (a) 2 minutes (b) 4 minutes 2 4 6 8 (c) 8 minutes Time (in minutes) (d) 16 minutes From graph it follows that at time 4 minute from the start of the reaction [A] = [B] 3A → B t=0 a t = 4 min a – x x/3

x 3a ⇒x= 3 4 3a a ∴ a−x=a− = 4 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) ∴

a−x=


NARAYANA



Problem 5 :

The inversion of cane sugar proceeds with half−life of 500 minute at pH 5 for any concentration of sugar. However if pH = 6, the half−life changes to 50 minute. The rate law expression for the sugar inversion can be written as (a) r = K[sugar]2[H]6 r = K[sugar]0[H+]6

(c) Solution :

(b) r = K[sugar]1[H]0 (d) r = K[sugar]0[H+]1

Since t1/2 does not depends upon the sugar concentration means it is first order respect to sugar concentration. t1/2 ∝ [sugar]1. t1/2 × an-1 = k

( t1/ 2 )1 ( t1/ 2 )2

=

[H + ]11− n [H + ]12− n 1− n

500  10−5  =  50  10−6 

10 = (10)1-n Hence n = 0 ∴ (b) Problem 6 :

Two substances A and B are present such that [A0] = 4[B0] and half−life of A is 5 minute and that of B is 15 minute. If they start decaying at the same time following first order kinetics how much time later will the concentration of both of them would be same. (a) 15 minute

(b) 10 minute

(c) 5 minute

(d) 12 minute n

Solution :

1 1 Amount of A left in n1 haves =   [A 0 ] 2 n

1 2 Amount of B left n2 halves =   [B0 ] 2 At the end, according to the question [A 0 ] [B0 ] = n 2n1 22

⇒

4 1 = n , [[A 0 ] = 4[B0 ]] n1 2 22

∴

2n1 = 4 ⇒ 2n1 − n 2 = (2) 2 ∴ n1 − n 2 = 2 n2 2

∴

n 2 = (n1 − 2)

…(1)

Also t = n1 × t1/ 2(A) t = n 2 × t1/ 2(B) (Let concentration both become equal after time t)


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∴

n1 × t1/ 2(A) n 2 × t1/ 2(B)

=1⇒

n1 × 5 n =1⇒ 1 = 3 n 2 × 15 n2

…(2)

For equation (1) and (2) n1 = 3, n2 = 1 t = 3 × 5 = 15 minute ∴ (a) Problem 7 :

Solution :

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 the rate of reaction relative to the initial rate is (a) 1/48

(b) 1/24

(c) 9/16

(d) 1/6

R1 = K[A][B] = K[0.6][0.80]2

After reaction A

+

2B

 →

C

+

D

0.6 – 0.2

0.8 – 0.4

0.2

0.2

0.4

0.4

0.2

0.2

R 2 K(0.4)(0.4) 2 1 = = R1 6 (0.6)(0.8) 2 ∴ (d) Problem 8 :

Solution :

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 4 mole L–1 of their respective reactants. Number of mole per litre of A and R remaining unreacted 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

In the case of first order reaction t1/2 will remain constant independent of initial concentration so. –1 –1 20 min 2 mole L–1 20 min 20 min 4 mole L–1  →  → 1 mole L  → 0.5 moile L .

That is, after 60 minutes 0.5 mole L–1 of A will be left unreacted. In the case of second order reaction t1/2 is inversely proportional to initial concentration of reactant 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. –1 –1 20 min 40 min 4 mole L–1  → 2 mole L  → 1 mole L .

That is, after 60 minutes, the concentration of R remaining unreacted will be 1 mole L–1. Note that t1/2 a = 20 × 4 = 40 × 2 = 80 mole L–1 min, a constant. ∴ (b)


NARAYANA



Problem 9 :

The thermal decomposition of acetaldehyde : CH3CHO → CH4 + CO, has rate constant

d[CH3CHOO] will dt change if concentration of acetaldehyde is doubled keeping the temperature constant? of 1.8 × 10–3 mole–1/2L1/2 min–1 at a given temperature. How would -

Solution :

(a) will increase by 2.828 times

(b) will increase by 11.313 times

(c) will not change

(d) will increase by 4 times

Unit of the rate constant order.

mole –1/2L1/2.min1/2

suggests that the reaction obeys kinetics of 1.5

Rate = k [CH3CHO]3/2 or

Rate moleL−1 min −1 = k= = Mole–1/2, L1/2.min–1 [CH3CHO]3/ 2 (mole L-1 )3/ 2

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

∴ (a) Problem 10 : The reaction ; 2O3 → 3O2, is assigned the following mechanism. I. O3 → O2 + O (slow) II. O3 + O → 2O2 The rate law of if the reaction will, therefore be (b) r α [O3]2 [O2]–1 (a) r α [O3]2[O2] (c) r α [O3] (d) r α [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 =

K II K C [O3 ]2 [O 2 ]

∴ (b)


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SECTION - III MULTIPLE CHOICE PROBLEMS Problem 1:

Arrhenius equation may be represented as

(a) ln

A Ea = k RT

(b)

(c) log A = log k + Solution :

Ea 2.303RT

d ln k Ea = dt RT  Ea  k =  RT  A

(d) log  −

(a, c) (a) is correct as explained below : k = Ae–Ea/RT ⇒

(b) is not correct

−

k = e− Ea / RT A

k −Ea A Ea = ⇒ ln = A RT k RT

d d d  Ea  (ln A) − (ln k) =   dT dt dt  RT 

d −Ea (ln k) = dT RT 2

⇒

ln

[ln A does not change with T]

d Ea (ln k) = dT RT 2

(c) is correct. k = Ae–Ea/RT ln k = ln A – Ea/RT ⇒ ln A = ln k + Ea/RT ⇒ log A = log k +

Ea 2.303RT

Hence (d) is not correct it follows from explanation of (a), (b) and (c). Problem 2 :

A positive catalyst (a) increase the average kinetic energy of the reacting molecules (b) decreases the activation energy (c) alters the reaction mechanism (d) increases collision frequency

Solution :

(b, c) (b) is correct because the catalyst lowers down the activation energy by altering the mechanism. (c) is correct because it provides alternative short-cut path or mechanism of the reaction. (a) is not correct because K.E. ∝ T. (d) is wrong because collision frequency is increased by temperature and pressure.


NARAYANA



Problem 3:

For first order reaction (a) the degree of dissociation is equal to (1 – e–kt) (b) a plot of reciprocal of conc. of reactant vs time gives a straight line (c) t75% = 3t1/2 (d) The pre-exponential factor in Arrhenius equation has the dimension of time T–1.

Solution :

(a, d) (a) is correct α =

[A 0 ] − [A] [A 0 ] − [A 0 ]e− kt = = 1 − e− kt [A 0 ] [A 0 ]

(d) is correct, k = Ae–Ea/RT Units of pre-exponential factor A = units of k = mol L–1 s–1 Dimension of time in A is T–1, i.e., s–1. (b) is not correct because log (conc.) vs time is straight line. (c) is not correct because t75% = 2t1./2 Problem 4 :

For the reaction H2(g) + Br2(g) → 2HBr (g) Which of the following statements are wrong

Solution :

(a) order is 2, molecularity is 2

(b) order is 3/2, molecularity is 2

(c) order is 1, molecularity is 2

(d) order is 2, molecularity is 1

(a, c, d) (a) is correct because it is a wrong statement as order cannot be determined theoretically. (c) is correct because it is a wrong statement as order is 1.5 and not 1. (d) is correct becasue it is a wrong statement as mostly order is not greater than molecularity. (b) is not correct as it is a correct statement because experimentally, order of this reaction is found to be 3/2 and it involves two molecules. ∴ molecularity is 2.

Problem 5 :

Which of the following statements are correct. (a) A plot of log kp vs 1/T is linear (b) A plot of log A vs time is linear for 1st order reaction (c) A plot of log P vs 1/T is linear at constant volume (d) A plot of P vs 1/V is linear at constant temperature

Solution :

(a, b, d) (a) is correct because ln k = ln A − (b) is correct because k =

Ea RT

[A ] 2.303 log 0 t [A]

1 at constant temperature. V (c) is not correct because P ∝ T . (d) is correct because P ∝


NARAYANA



Problem 6 :

Which of the following is a first order reaction (a) decomposition of NH4NO2 to N2 and H2O (b) catalytic decomposition of ammonia (c) decomposition of H2O2 in aqueous solution (d) all of these

Solution :

(a, c) (a) is correct as it is observed that the rate of reaction is directly proportional to the concentration of NH4NO2 (The amount of NH4NO2 reacted is known by measuring the volume of N2 formed). (c) is correct as it is observed that the rate of reaction is directly proportional to the concentration of NO2. (b) is a 2nd order reaction. It follows that (d) cannot be correct.

Problem 7 :

Solution :

Which of the following will emit positron (a)

13 7N

(b)

30 15 P

(c)

3 1H

(d)

14 6C

(a, b) 13 → 136 C + (a) is correct ∵ n p < 1 7 N 

0 +1 e

(positron)

n (b) is correct ∵ p = 1 but still it is radioactive isotope or phosphorus and emits positron so as to increase n/p ratio. 30 15 P

 → 30 14 Si +

0 +1 e

While (c) and (d) are incorrect. Problem 8 :

Which of the following statements is/are correct? (a) Nuclear fusion involves chain reaction (b) Nuclear fusion forms products which are radioactive (c) Nuclear fusion takes place at high temperature (106 K) (d) Nuclear fusion produces more energy than nuclear fission.

Solution :

(b, c, d) (b) is correct because nuclear fission forms the product which are radioactive. (c) is correct because high temperature of the order of 106 K is required so as to overcome repulsion between the nuclei to be fused together. (d) is correct because in fusion reaction energy released is more than fission reaction e.g., sun is source of energy due to fusion reaction between isotopes of hydrogen. (a) is not correct because nuclear fusion does not involve chain reaction.


NARAYANA



Problem 9 :

Solution :

If n/p is less than 1, the nucleide can (a) emit positron

(b) K-capture

(c) emit β-particle

(d) emit α-particle

(a, b) (a) is correct because n/p increases during positron emission because proton gets converted into neutron. (b) k-capture means capture of electrons which will convert proton into neutron ∴ increasing n/p ratio.

1 0 → 01 n 1 p + −1 e 

(c) β-particle emission leads to decrease in n/p ratio. (d) α-particle emission leads to increase in n/p ratio but n/p should not be less than 1 in parent nucleides. e.g., Problem 10 :

Solution :

238 → 234 92 U  90Th

n + 24 He; for p

238 92 U

n is 1.58; for p

234 90Th

is 1.60.

Which of the following are used for measuring radioactivity? (a) Gieger Muller counter

(b) Wilson cloud chamber

(c) Scintillation counter

(d) none of these

(a, b, c) Geiger Muller Counter, Wilson Cloud Chamber, Scintillation counter are used for measuing radioactivity ∴ (a), (b), (c) are correct.


NARAYANA



MISCELLANEOUS PROBLEMS SECTION - IV COMPREHENSION TYPE PROBLEMS Write up – I Nucleus is considered to resemble a liquid drop in which nucleons are present having very rich density i.e. 130 trillion tonnes m–3 which is about trillion times the density of water. It has high surface tension of about 1.24 × 1018 times, the surface tension of water. The high surface tension keeps the nucleons together against the force of repulsion. Problem 1 :

The nuclear volume is (a) directly proportional to mass number (c) inversely proportional to atomic number

(b) inversely proportional to mass number (d) independent of mass number

Solution :

Ans. (a)

Problem 2 :

4 3 πr r α[A]1/ 3 , ∴ V ∝ |A|1/3 where A is mass number. 3 The radius of nucleus of 12C is equal to (b) 4 × 10–15 m (a) 3.5 × 10–15 m (d) 6.0 × 10–15 m (c) 5 × 10–15 m

Solution :

Ans. (a)

The nuclear volume =

r = 1.4 × 10–15 (12)1/3 = 1.4 × 10–15 × 2.3 = 3.2 × 10–15 m Problem 3 :

Solution :

Radius of nucleus is directly proportional to (a)

[A] 2

(b) [A] 1/3

(c)

[A] 3

(d) [A]

Ans. (b) r α [A]1/3 where ‘A’ is mass number

Write-up II For a particular first order reaction at 27°C, the concentration of reactant is reduced to one half of its initial value after 5000s. At 37°C, the concentration is halved after 1000s. Problem 4 :

Solution :

The constant of the reaction at 27°C is : (a)

1.386 × 10–4 s–1

(b) 6.93 × 10–4 s–1

(c)

11.231 × 10–4 s–1

(d) 9.232 × 10–4 s–1

Ans. (a) (T50)300 K = 5000s k1 (3000 k) =

0.693 0.693 = 1.386 × 10-4 s–1 = (T50 )300 k 5000 59


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Problem 5 :

Solution :

What is the time required for the concentration to be reduced to one quarter of its value at 37°C? (a)

1000 s

(b) 500 s

(c)

2000 s

(d) 1500 s

Ans. (c) (T50)310 k = 1000 s k2 (310 k) =

0.693 = 6.93 × 10−4 s −1 1000

T75 = 2 × T50 for first order reaction, (when concentration is reduced to one quarter i.e.. 75% reaction) Hence, T75 (310 k) = 2000 s Problem 6 :

Solution :

The activation energy of the reaction is : (a)

124.46 kJ/mol

(b) 650 kJ/mol

(c)

229.12 kJ/mol

(d) 693.15 kJ/mol

Ans. (a)

Ea =

2.303 RT1T2 k log10 2 = 124.46 kJ/mol T2 − T1 k1 MATCHING TYPE PROBLEM

7.

Column (I)

Column (II)

(a)

9 4 → 126 C + 01 n 4 Be + 2 He 

(b)

24 12 Mg

(c)

2 1D

(d)

NaCl BrO3− + 3I −  → Br − + I 2 + IO3−

1 + 42 He  → 27 14 Si + 0 n

+ 31T  → 24 He + 01 n + 17.6 MeV

(a) → (r), This reaction is the transmutation of 94 Be to

Sol.

(p)

Hydrogen bomb

(q)

Radiolysis in Redox systems

(r)

Artificial transmutation

(s)

Induced radio-activity

12 6C

by using 42 He as projectile.

(b) → (r), (s); This reaction is an artificial transmutation leading to the formation of unstable nuclei 27 14 Si

which gives stable 27 14 Si

 →

27 13 Al

27 0 13 Al + +1 e

(c) → (p) ; This is the reaction taking place in a hydrogen bomb. (d) → (q); This is radiolysis where a substance is exposed to high energy radiation such as gamma rays. It involves two stages High energy → A + e− Primary stage : A  Radiation




Secondary stage : A+ + e– → A M + e – → M–

or

A+ + M– → Å + M Here NaCl is the one which induces this reaction which is redox reaction. ASSERTION-REASON TYPE PROBLEMS The question given below consist of an ASSERTION and the REASON. Use the following key for the appropriate answers (a)

If both Assertion and Reason are correct and Reason is the correct explanation for Assertion

(b)

If both Assertion and Reason are correct and Reason is not the correct explanation for Assertion

(c)

If Assertion is correct but Reason is not correct.

(d)

If Assertion is incorrect but Reason is correct.

8. Assertion : Photochemical reactions are zero order reactions. Reason :

Photochemical reactions involve free radicals and they are very fast reactions.

Solution :

Photochemical reactions are extremely fast, therefore, lead to completion. Their rate is not affected by concentration of reactant. Ans. (a)

9. Assertion : 2FeCl3 + SnCl2 → 2FeCl2 + SnCl4 is a 3rd order reaction. Reason :

The rate constant for third order reaction has unit L2 mol–2 s–1.

Solution :

rate = k [FeCl3]2 [SnCl2]1 rate = k[A]3 ⇒ mol L–1 s–1 = k [mol L–1]3 ⇒ k = L2 mol–2 s–1 Ans. (b)

10.Assertion : β-particles are emitted by nucleus Reason :

In nucleus, 01 n  → 11 p +

Solution :

A is correct, R is correct and ‘R’ is correct explanation of ‘A’

0 −1 e

Ans. (a)


NARAYANA



ASSIGNMENTS SECTION - I SUBJECTIVE QUESTIONS

LEVEL - I 1.

The isomerism of cyclopropane to propene is of first order with k = 8.25 × 10–4 s–1 at 500°C. If cyclopropane is enclosed in a heated tube at 500°C. What will be the minimum time needed to yield at least 10% propene in resultant gas sample.

2.

For a reaction, A → B + C, it was found that the end of 10 minutes from the start the total optical rotation of the system ws 50° and when the reaction is complete, it was 100°. Assuming that only B and C are optically active and dextro rotatory. What would be the rate constant of this first order reaction.

3.

At a given instant thee are 25% undecayed radioactive nuclei in a sample. After 10 seconds the number of undecayed nuclei reduces to 12.5%. What is the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number.

4.

Naturally occurring potassium consist of 0.01% K40, which has a half life of 1.28 × 109 yr. What would be the activity of 1.0 g sample of KCl.

5.

A catalyst lowers the activation energy of a reaction from 20 kJ mole–1 to 10 kJ mole–1. What is the temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27°C.

6.

For the first order reaction A(g) → 2B(g) + C(g), the initial pressure is PA = 90 mm Hg, the pressure after 10 minutes is found to be 180 mm Hg. What is the rate constant of the reaction ?

7.

The activity of a radioactive sample reduced from 20 Ci to 1.25 Ci in 2000 years. What are the half life of the sample and its decay constant.

8.

k1 k2 In the sequential chain reaction A  → B  → C at the steady state, the half life of B is 30 seconds, then what will be the half life of A? Given that at any time, the number of atoms of A is twice the number of atoms of B.

9.

What is the percentage fraction of the molecule crossing over the energy barrier at 27ºC. Given that Ea(b) = 80 cal, ∆H = 20 cal and antilog (0.072) = 1.181.

10.

At 278°C the half life period for the first order thermal decomposition of ethylene oxide is 363 min and the energy of activation of the reaction in 52,00 cal/mole. From these data estimate the time required for ethylene oxide to be 75% decomposed at 450°C.


NARAYANA



LEVEL - II 1.

For the following reaction, the rate law has been determined to be rate = k[A][B]2 with k = 2 × 10−6 mol−2 L2 sec−1 2A + B + C → A2B + C For this reaction the initial conc of the various species. [A] = 0.1 mol L−1; [B] = 0.2 mol L−1; [C] = 0.8 mol L−1. Determine the rate after 0.04 mol L−1 of A has reacted.

2.

For the reaction process A + B ——→ products, the rate is first order w.r.t A and second order w.r.t B. If 1 mol each of A and B were introduced into a 1 L vessel, and the initial rate were 1 × 10-2 (mol/ L)sec-1. Calculate the rate when half the reactants have been turned into products.

3.

10g atoms of an α-active radioisotope are disintegrating in a sealed container. In one hour helium gas collected at STP is 11.2 litre. Calculate the half-life of the radioisotope.

4.

The gas phase decomposition of dimethyl ether follows first-order kinetics.

→ CH 4(g) + H 2(g) + CO(g) CH3OCH3(g)  The reaction is carried out at constant volume in a container at 500°C and has a half-life of 14.5 minutes. Initially only dimethyl ether is present at a pressure of 0.40 atmosphere. What is the total pressure of the system after 12 minutes? Assume ideal gas behaviour. 5.

In the Arrhneius equation for a certain reaction, the value of A and Ea (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?

6.

The reaction 2NO(g) + 2H2 → N2(g) + 2H2O(g) has been studied at 904°C Sl. No.

Initial Concentration

Rate of appearance of N2 (mol litre–1 sec–1

NO

H2

1.

0.42M

0.122M

0.136

2.

0.21M

0.122M

0.0339

3.

0.21M

0.244M

0.0678

4.

0.105M

0.488M

0.0339

Determine the reaction order 7.

The decomposition of N2O5 according to the equation 2N2O2(g) + O2(g) is a first order reaction. After 30 min from start of decomposition in a closed vessel the total pressure developed is found to be 284.5 mm Hg. On complete decomposition, the total pressure is 584.5 mm Hg. Calculate the rate constant of the reaction.

8.

A certain physiologically important first order reaction has an activation energy equal to 45.0 kJ/mol at normal body temperature (37°C). Without a catalyst, the rate constant for the reaction is 5.0 × 10–4s– 1 . To be effective in the human body, where the reaction is catalysed by an enzyme, the rate constant must be at least 2.0 × 10–2s–1. If the activation energy is the only factor affected by the presence of the enzyme, by how much must the enzyme lower the activation energy of the reaction to achieve the desired rate?


NARAYANA



1 → 2NO 2 + O 2 is 2.4 hours at 30°C. What time would The half-life period for the reaction, N 2 O5  2 be required to reduce 5 × 1010 molecules of N2O5 to 108 molecules?

9.

10.

For a certain reaction, it takes 5 minutes for the initial concentration of 0.5 mole/L to become 0.25 mole/L and another 5 minutes to become 0.125 mole/L.What is the order and specific rate constant of the reaction?

LEVEL - III 1.

Two I order reactions having same reactant concentration proceed at 25°C at the same rate. The temperature coefficient of the rate of the first reaction is 2 and that of second reaction is 3. Find the also of the rates of these reactions at 75°C.

2.

In a certain reaction Bm+ 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 agent which reacts only with Bn+ and B(n+4)+. In this process, it converts Bn+ to B(n–2)+ and B(n+4)+ to B(n-1)+. At t = 0, the volume of reagent consumed is 25 mL and at t = 10 min, the volume used is 32 mL. Calculate the rate constant of the conversion of Bn+ to B(n+4)+ assuming it to be a first order reaction.

3.

For the reaction 2NO + H2 → N2O + H2O the value of −

dp was found to be 1.50 torr s–1 for a pressure of 359 torr of NO and 0.25 torr s–1 for dt

a pressure of 152 torr, the pressure of H2 being constant. On the other hand, when the pressure of NO was kept constant, −

dp was 1.60 torr s–1 for a hydrogen pressure of 289 torr and 0.79 torr s–1 for a dt

pressure of 1.47 torr. Determine the order of the reaction. 4.

Calculate the value of rate constant and time required for solution to become optically inactive. Time/min

0.0

36.8

46.0

68

∞

Rotation of Polarized light /degree

24.1 21.4 12.4

10.0

5.5

–10.7

7.2

5.

Decomposition of H2O2 is a first order reaction. A solution of H2O2 labelled as 20 volumes was left open. Due to this some H2O-2 decomposed. To determine the new volume strength after 6 hours, 10 Ml of this solution was diluted to 100 Ml. 10 Ml of this diluted solution was titrated against 25 Ml of 0.025 M KmnO4 acidified solution. Calculate the rate constant for decomposition of H2O2.

6.

The inversion of the sugar proceeds with constant half life of 500 minute at Ph = 5 for any concentration of sugar. However, if Ph = 6, the half life changes to 50 minute. Derive the rate law for inversion of cane sugar.

7.

Bi-cyclohexane was found to undergo two parallel first order rearrangements. At 730 K, the first order rate constant for the formation of cyclohexane was measured as 1.26 × 10–4 s–1 and for the formation of methyl cyclopentene the rate constant was 3.8 × 10–5s–1. What was the % distribution of the rearrangement products.




8.

If a reaction A → Products, the concentrations of reactant A are C0, Ac0, a2C0, a3C0.. after time interval 0, t, 2t, 3t, …. Where a is constant. Given 0 < a < 1. Show that the reaction is of I order. Also calculate the reaction in K, a and t.

9.

For a homogeneous gaseous phase reaction 2A → 3B + C, the initial pressure was P0 while pressure at time ‘t’ was P. Find the pressure after time 2t. Assume first order reaction.

10.

The decomposition of a compound A, at temperature T according to the equation.

2P(g)  → 4Q(g ) + R (g ) + S(l) is the first order reaction. After 30 minutes 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 minutes, 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.


NARAYANA



SECTION - II SINGLE CHOICE QUESTIONS 1.

The instantaneous rate of disappearance of the MnO −4 ion in the following reaction is 4.56 × 10–3 Ms–1.

2MnO −4 + 10I − + 16H +  → 2Mn 2 + + 5I 2 + 8H 2 O The rate of appearance of I2 is

2.

(a) 1.14 × 10–3 M s–1

(b) 5.7 × 10–3 M s–2

(c) 4.56 × 10–4 M s–1

(d) 1.14 × 10–2 M s–1

The rate constant for a zero-order reaction is (a) k =

C0 2t

(c) k = ln 3.

4.

5.

6.

7.

C0 − C t 2t

(b) k =

C0 − C t t

C0 (d) k = C t

The rate expression for the reaction A(g) + B(g) → C(g) is rate = kC2A C1/B 2 . What changes in the initial concentrations of A and B will cause the rate of reaction to increase by a factor of eight? (a) CA × 2; CB × 2

(b) CA × 2; CB+ × 4

(c) CA × 1; CB × 4

(d) CA × 4; CB × 1

In an exothermic reaction X → Y, the activation energy is 100kJ mol–1 of X. The enthalpy of the reaction is –140 kJ mol–1. The activation energy of the reverse reaction Y → X is (a) 40 kJ mol–1

(b) 340 kJ mol–1

(c) 240 kJ mol–1

(d) 100 kJ mol–1

The rate constant, the activation energy and the pre-exponential factor of a chemical reaction at 25°C are 8.0 × 10–4 s–1, 112 kJ mol–1 and 4 × 1015 s–1 respectively. The value of the rate constant as T → ∞ is (a) 8 × 1016s–1

(b) 4 × 104s–1

(c) 4 × 1015s–1

(d) 112 × 1012s–1

Two first order reactions have half-lives in the ratio 3:2. Calculate the ratio of time intervals t1 : t2; t1 is the time period for 25% completion of the first reaction and t2 for 75% completion of the second reaction. (a) 0.311:1

(b) 0.420:1

(c) 0.273:1

(d) 0.119:1

Which of the following rate law has an overall order of 0.5 for the reactions involving substances x,y,z? (a) Rate = k(Cx) (Cy) (Cz) (b) Rate = k (Cx)0.5 (Cy)0.5 (Cz)0.5

R ( Cx )( Cz )

o

(c) Rate = k(Cx)1.5 (Cy)–1(Cz)0

(d) Rate =

(Cy )2


NARAYANA



The reaction A(g) + 2B(g) → C(g) + D(g) is an elementary process. In an experiment, the initial partial pressure of A & B are PA = 0.60 and PB = 0.80 atm. When PC = 0.2 atm the rate of reaction relative to the initial rate is

8.

(a) 1/48

(b) 1/24

(c) 9/16

(d) 1/6

For a second order reaction of the type rate = k [A]2, the plot of [A]t versus t is linear with a

9.

(a) positive slope and zero intercept (b) positive slope and non zero intercept (c) negative slope and zero intercept (d) negative slope and non zero intercept 10.

11.

12.

If concentration are measured in mole/lit and time in minutes, the unit for the rate constant of a 3rd order reaction are (a) mol lit–1 min–1

(b) lit2 mol–2min–1

(c) lit.mol–1 min–1

(d) min–1

Radioactivity is affected by (a) Temperature

(b) Pressure

(c) State of chemical combination

(d) None

No. of β – particles emitted during the change a X b →c Yd is (a)

a −d (b) c +  +b  2 

a −d 4

b−d (c) c +  −a  2 

13.

For two atoms

Z1 X

a −d (d) c +  −b  2  A1

and

Z2

X A 2 where Z1 ≠ Z2 and A1 ≠ A 2 , A1 – 2Z1 = A2 – 2Z2, the species

would be called (a) Isotonic (c) Isoters 14.

15.

16.

(b) Isodiaphers (d) none

Which statement is true about N/P ratio (a) It increases by β –emission

(b) It increases by α –emission

(c) It increases by γ –emission

(d) None of these

How many α and β particles should be eliminated so that an isodipher is formed (a) n α , n β

(b) n α , (n + 1) β

(c) n α

(d) n β

+ 0 n1 →92 U 236 → Pr oducts + x 0 n1 + 3.2 × 10−11 J . The energy released when 1 gm of finally undergoes fission is 92 U

235

(a) 12.75 × 108 kJ

(b) 16.4 × 107 kJ

(c) 8.2 × 107 kJ

(d) 6.5 × 106 kJ

92 U

235


NARAYANA



17.

18.

At radioactive equilibrium the ratio between two atoms of radioactive elements A and B are 3×109 : 1. If t1/2 of A is 1010 years. What is t1/2 of B (a) 30 years

(b) 0.3 years

(c) 3.3 years

(d) none

In the sequence of following nuclear reactions 92 X

238

−α −β −β − nα Y  → → Z  → L  →84 M 218 the value of n is

(a) 3

(b) 4

(c) 5 19.

(d) 6 1

4

The mass of 1H is 1.0076 amu and that of 2He is 4.0015 amu the energy released in the reaction

41 H1 →2 He 4 + 20 e1 is nearly

20.

(a) 9 meV

(b) 27 meV

(c) 36 meV

(d) 54 meV

Co–60 has t1/2 = 5.3 years the time taken for 7/8 of the original sample of disintegrate will be (a) 4.6 years

(b) 9.2 years

(c) 10.6 years

(d) 15.9 years


NARAYANA



SECTION - III MULTIPLE CHOICE QUESTIONS 1.

2.

Molecularity of a reaction can’t be (a) zero

(b) (–) ve

(c) fractional

(d) greater than 3 (genrally

st

For a 1 order reaction (a) dx/dt = k(a–x) (b) K =

2.302 a log t a−x

(c) t 2 − t1 =

a − x1 2.302 log k a − x2

(d) Plot between t and log(a–x) will be straight line 3.

4.

5.

6.

For a 1st order reaction (a) t 3/ 4 = 2t1/ 2

(b) t 7 / 8 = 3t1/ 2

(c) t 99% = 2t 90%

(d) t 90% =

10 t 50% 3

Conditions when K =a (a) Ea = 0

(b) T = 0

(c) T = ∞ Which are temperature independent?

(d) all

(a) A

(b) Ea

(c) R

(d) K

Which is true statement (a) lightest radioactive isotope is tritium (b) nuclides having at no. > 82 are radioactive (c) highest degree of radioactive is shown by uranium (d) all of these are true

7.

8.

Primary emission is (a) α –emission

(b) β –emission

(c) γ –emission

(d) all

A radioactive element present in VIII group of the periodic table. If it emits one α –particle, the new position of the nuclide will be (a) VI B

(b) VIII

(c) VII B

(d) IB 69


NARAYANA



Which is true about γ emission

9.

(a) γ rays are the most penetrating (b) emission of γ rays leads to the formation of a new element (c) γ –emission is known as secondary emission (d) all of these 10.

Which is true about decay constant ( λ ) (a) unit of λ is time –1 (b) value of λ is always less than 1 (c) λ is independent of temperature (d) λ is defined as the ratio of no. of atoms disintegrating per unit time to the total no. of atom present at that time


NARAYANA



MISCELLANEOUS QUESTIONS SECTION - IV COMPREHENSION TYPE QUESTIONS Write-up I In a parallel reaction reactant cobalt which is a transition metal decays by parallel pathways to form 2 products B and C.

β - emission (k1 ) 58 27 Co

B

β + emission (k2 ) . C

The ratio of concentration of [B] : [C] are 2 : 98 and the half life of Co is 22 years. 1.

2.

3.

The decay constant for (B) should be (a) 63 × 10–5 y–1

(b) 126 × 10–5 y–1

(c) 863 × 10–3 y–1

(d) None of these

The product B will be (a) Copper

(b) Manganese

(c) Nickel

(d) Zinc

The product C will be (a) Copper

(b) Manganese

(c) Nickel

(d) Iron

Write-up II Ozone is prepared in laboratory by passing silent electric discharge through pure and dry oxygen in an apparatus known as ozoniser. This conversion from oxygen to ozone is a reversible and endothermic reaction. When oxygen is subjected to an ordinary electric discharge, most of the O3 produced will get decomposed. When any insulating material such as glass, is inserted in the space between the two electrodes and high current density is applied, silent electric discharge passes on between the two electrodes. By this process no spark is produced and much less heat is generated, and as a result the decomposition of the produced ozone is much retarded. The decomposition of ozone is believed to occur by the following mechanism: Kf !!! " O3 #!! ! O2 + O K

(fast step)

K1 O3 + O  → 2O2

(slow)

b


NARAYANA



4.

5.

6.

Molecularity of reaction is defined by (a) slow step

(b) reversible step

(c) from overall reaction

(d) fast step

When the concentration of O2 is increased, for the same concentration of ozone, its rate (a) increases

(b) decreases

(c) remains the same

(d) cannot be answered

When the concentration of O3 is increased, for the same concentration of oxygen, its rate (a) increases

(b) decreases

(c) remains the same

(d) cannot be answered

MULTIPLE MATCHING TYPE QUESTIONS Match the following 7.

Column – I (a) Decomposition of H2O2

Column – II (p) 10t1/2

k 308 K (b)

k 298 K

(q) 1st order

(c) Arrhenius equation

(r) Temperature coefficient

(d) t 99.9%

E a  T2 − T1  k2 (s) log k = 2.303 R  T T  1  1 2 

8.

Column – I

Column – II

(a) Rate of reaction

(p) Can be fractional

(b) rate constant

(q) Will be a whole number

(c) Order of reaction

(r) Decreases with increase in concentration

(d) Molecularity

(s) Increases with increase in temperature


NARAYANA



ASSERTION-REASON TYPE QUESTIONS The question given below consist of an ASSERTION and the REASON. Use the following key for the appropriate answers

9.

10.

11.

12.

13.

14.

15.

(a)

If both Assertion and Reason are correct and Reason is the correct explanation for Assertion

(b)

If both Assertion and Reason are correct and Reason is not the correct explanation for Assertion

(c)

If Assertion is correct but Reason is not correct.

(d)

If Assertion is incorrect but Reason is correct.

Assertion :

In zero order reaction, the conc. vs time graph is a straight line.

Reason :

The rate of change of conc. per unit time in zero order reaction remains constant.

Assertion :

Photochemical reactions always occur in the presence of light.

Reason :

Photochemical reactions do not require activation energy.

Assertion :

Many of photochemical changes have positive sign of , yet they are spontaneous.

Reason :

The activation energy in photochemical reactions is provided by light energy.

Assertion :

The order of a reaction can have a fractional value.

Reason :

The molecularity of a reaction can have a fractional value.

Assertion :

For a zero order reaction, rate of reaction is independent of conc. of reactants.

Reason :

For a zero order reaction, reaction proceeds at a constant rate which is equal to rate constant of the reaction.

Assertion :

To separate U-235 from the more abundant, all the uranium converted into UF6

Reason :

UF6 is one of the few compounds that exist in gaseous state under ordinary conditions which helps in separation of U-235 from U-238.

Assertion :

Nucleide

Reason :

Nucleide having odd number of protons and neutrons are generally unstable.

30 13 Al

is less stable than

40 20 Ca


NARAYANA



SECTION - V (PROBLEMS ASKED IN IIT-JEE) A.

Only one option is correct (Objective Questions)

1.

A positron is emited from

2.

3.

23 11 Na

. The ratio of the atomic mass and atomic number of the resulting

nuclide is (2007) (a) 22/10 (b) 22/11 (c) 23/10 (d) 23/12 Consider a reaction aG + bH → Products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is (2007) (a) 0 (b) 1 (c) 2 (d) 3 23

Na is the more stable isotope of Na. Find out the process by which

decay: (a) β– emission (c) β+ emission 4.

235 92 U

(b) α emission (d) K electron capture

6.

94 38 Sr

from the absorption of a slow

, followed by nuclear fission is :

(a) 0 (c) 1 5.

can undergo radioactive (2003)

and The number of neutrons accompanying the formation of 139 54 Xe neutron by

24 11 Na

(1999) (b) 2 (d) 3

The rate constant for the reaction, 2N 2 O5 → 4NO 2 + O 2 is 3.0 × 10 –5 sec –1. If the rate is (2000) 2.40 ×10–5 mol/L/sec., then the concentration of N2O5 (in mol/L) is : (a) 1.4 (b) 1.2 (c) 0.04 (d) 0.8 If l is the intensity of absorbed light and C is the concentration of AB for the photochemical process. * AB + hν → AB* , the rate of formation of AB is directly proportional to: (a) C (b) I

(c) I 2 7.

(2001)

(d) C.I

Consider the chemical reaction,

N 2 (g ) + 3H 2 (g ) → 2NH3 ( g ) The rate of this reaction can be expressed in terms of time derivatives of concentration of N2 (g), H2 (g) or NH3 (g). Identify the correct relationship amongst the rate expressions: (2002) (a) Rate = − (c) Rate = −

d [N 2 ] dt d [N 2 ] dt

=− =

1 d [H 2 ] 1 d [ NH3 ] = 3 dt 2 dt

1 d [H 2 ] 1 d [ NH3 ] = 3 dt 2 dt

(b) Rate = − (d) Rate = −

d [N2 ] dt d [N2 ] dt

= −3 =−

d [H 2 ] dt

d [H 2 ] dt

=2

=

d [ NH3 ] dt

d [ NH 3 ] dt


NARAYANA



In a first order reaction the concentration of reactant decreases from 800 mol/dm3 in 2×104 sec. The (2003) rate constant of reaction in sec–1 is :

8.

(a) 2 × 104

(b) 3.45 × 10–5

(c) 1.386 × 10à

(d) 2 × 10–4

(A) follows first order reaction. (A) → product

9.

Concentration of A, changes from 0.1 M to 0.025 M in 40 minutes. Find the rate of reaction of A when concentration of A is 0.01 M: (2004)

10.

(a) 3.47 × 10–4 M min –1

(b) 3.47 × 10–5 M min –1

(c) 1.73 × 10–4 M min –1

(d) 1.73 × 10–5 M min –1

Which of the following statement for order of reaction is not correct?

(2005)

(a) Order can be determined experimentally (b) Order of reaction is equal to sum of the power of concentration terms in differential rate law (c) It is not affected with stoichiometric coefficient of the reactants (d) Order can not be fractional

B.

More than one options are correct (Objective Question)

1.

The nuclear reactions accompanied with emission of neutron (s) are (a)

30 + 42 He →15 P

30 →14 Si + e10 Decrease in atomic number is observed during : (a) alpha emission (c) positron emission

(c) 2.

17 13 Al 30 15 P

(b)

12 1 6 C +1

H →14 7 N

(d)

241 96 Am

244 + 42 He →97 Bk + e10

(1988)

(1988) (b) beta emission (d) electron capture

SUBJECTIVE 1.

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the pre-exponential factor for the reaction is 3.56 × 109 s –1 , calculate its rate constant at 318 K and also the energy of activation. [IIT 1997]

2.

The rate constant of a reaction is 1.5 × 107 sec –1 at 50°C and 4.5 × 107 sec –1 at 100°C. Evaluate the Arrhenius parameters A and E a .

3.

4.

[IIT 1998]

A first order reaction A  → B requires activation energy of 70 kJ mol –1 . When a 20% solution of A was kept at 25°C for 20 minute, 25% decomposition took place. What will be the per cent decomposition in the same time in a 30% solution maintained at 40°C ? Assume that activiation energy remains constant in this range of temperature. [IIT 1993] A 1st order reaction is 50% complete in 30 minute at 27°C and in 10 minute at 47°C. Calculate the : (a) Rate constant for reaction at 27°C and 47°C (b) Energy of activation for the reaction. (c) Energy of activation for the reverse reaction if heat of reaction is –50 kJ mol –1 .

[IIT 1988]


NARAYANA



5.

At 380°C, the half life period for the first order decomposition of H 2O 2 is 360 min. The energy of activation of the reaction is 200 kJ mol –1 . Calculate the time required for 75% decomposition at 450°C. [IIT 1995] A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in presence of a catalyst at the same rate, the temperature required is 400 K. Calculate the activation energy of the

6.

[IIT 2000] reaction if the catalyst lowers the activation energy barrier by 20 kJ mol –1 . The rate constant for the first order decomposition of a certain reaction is given by the equation,

7.

ln K(sec –1 ) = 14.34 –

1.25 × 104 T

Calculate : (a) The energy of activation (b) The rate constant at 500 K. (c) At what temperature will its half life period be256 minute?

[IIT 1997]

Two reactions (I) A  → Products (II) B  → Products follow first kinetics. The rate of the reaction (I) is doubled when temperature is raised from 300 K to 310 K. The half life for this reaction at 310 K is 30 minute. 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 reaction (II) at 300 K. [IIT 1992] From the following data from the reaction between A and B,

8.

9.

[A] mol litre–1

[B] mol litre–1

2.5 × 10 –4

Initial rate mol litre–1 Sec–1 300 K

320 K

3.0 × 10 –5

5.0 × 10 –4

2.0 × 10 –3

5.0 × 10 –4

6.0 × 10 –5

4.0 × 10 –3

—

1.0 × 10 –3

6.0 × 10 –5

1.6 × 10 –2

—

Calculate : (i) The order of reaction with respect to A and with respect to B. (ii) The rate constant at 300 K. (iii) The energy of activation. (iv) The pre-exponential factor. 10.

[IIT 1994]

In the Arrhenius equation for a certain reaction, the values of A and E a (energy of activation) are 4 × 1013 sec –1 and 98.6 kJ mol –1 respectively. If the reaction is of first order, at what temperature will its half life period be 10 minute? [IIT 1990]


NARAYANA



ANSWERS EXERCISE - 1 4 d[O 2 ] d[NO] 4 d[H 2 O] = = 5 dt dt 6 dt

1.

−

2.

1.2 × 10–2 atm min–1

3.

(a)

5 × 10–4 mol litre–1sec–1

(c)

2.5 × 10–4 mol litre–1sec–1

(b)

1.25 × 10–4 mol litre–1 sec–1

(b)

r2 = 8r1

EXERCISE - 2 1

2

K[A] [B] [C]

0

1.

(a)

2.

rate = K1[NO]2[Br2]

3.

(a)

2O3(g) → 3O2(g);

(b)

intermediate is O(g);

(c)

first step is unimolecular second is bimolecular.

EXERCISE - 3 Do your self

EXERCISE - 4 1. 2.

(i)

Greater the temperature dependence for reaction with larger value of Ea.

(ii)

A = 6.73 × 1011 sec–1 K = 5.10 × 105sec–1 Ea = 92.011 Kj /mol

(a)

1.386 × 10–4s–1

(b) 2000s (c) 124.46kJ mol 3.

T = 282 K

EXERCISE - 5 1.

0.537

2.

(a)

132 sec

(b)

0.58 mol dm–3

(c) [A] = 0, [B] = 0.12 mol dm–3, [C] = 0.88 mol dm–3

EXERCISE - 6 1.

14,000 years

2.

Specific activity = 30.69 dis.g–1s–1

3.

239

Pu = 45.1%,

240

Pu = 54.9%


NARAYANA



SECTION - I (Subjective Questions) LEVEL - I 1.

127.7 s

2.

0.069 min–1

3.

40 sec

4.

4.37 × 108 yr–1

5.

327°C

6.

1.15 × 10–3 sec–1

7.

500 yrs and 1.386 × 10–3 yrs–1

8.

60 sec

9.

84.7%

10.

236.48 min

LEVEL - II 1.

3.38 × 10–9 mol L–1 sec–1

2.

1.25 × 10–3

3.

13.51 hours

4.

0.7488 atm

5.

311.35K

7.

K1 = 5.2 × 10–3 min–1

8.

10kJ/mol

9.

21.5 hours

2.

2.07 × 10–2 min–1

4.

k = 0.0119, t = 105.4 min

6.

K [Sugar]1 [H+]0

10.

–1

First; 0.138 minute .

LEVEL - III 1.

7.5937

3.

rate = k[NO]2 [H2] –1

5.

0.022 hr

7.

0.768, 0.232

8.

K comes constant and thus, it is I order reaction

K=

2.303 1 log t a (2P 0 − P)2 P0

9.

PT = 2P 0 −

10.

Pt = 268.11 mm Hg, t7/8 = 43.76 min


NARAYANA



SECTION - II (Single Choice Questions) 1. 3. 5. 7. 9. 11. 13. 15. 17. 19.

(d) (b) (c) (c) (b) (d) (b) (c) (c) (b)

2. 4. 6. 8. 10. 12. 14. 16. 18. 20.

(b) (c) (a) (d) (b) (c) (b) (c) (b) (d)

SECTION - III (Multiple Choice Questions) 1. 3. 5. 7. 9.

(a, b, c, d) (a, b, c, d) (a, b, c) (a, b) (a, c)

2. 4. 6. 8. 10.

(a, b, c, d) (a, c) (a, b) (a, b, c) (a, b, c, d)

SECTION - IV (Comprehension Type Questions) 1. 3. 5.

(a) (d) (b)

2. 4. 6.

(c) (a) (a)

(Multiple Matching Type Questions) 7. 8.

(a)

(q)

(b)

(r)

(c)

(s)

(d)

(p)

(a)

(r, s)

(b)

(s)

(c)

(p, q)

(d)

(q)


NARAYANA



(Assertion Reason Type Questions) 9. 11. 13. 15.

(a) (b) (b) (a)

10. 12. 14.

(c) (c) (a)

SECTION - V (Problems asked in IIT-JEE) A.

Only one option is correct (Objective Questions)

B.

1.

(c)

2.

(d)

3.

(a)

4.

(d)

5.

(d)

6.

(d)

7.

(a)

8.

(c)

9.

(a)

10.

(d)

More than one option are correct (Objective Questions) 1.

(a, d)

2.

(a, c, d)

6.

100 kJ mol –1

SUBJECTIVE 1.

K318 = 9.22 × 10–4 sec–1 Ea = 18.3 kcal mol–1

2.

Ea = 2.2 × 104 J mol–1 A = 5.42 × 1010

3.

67.21%

4.

(a) K1 at 27°C = 2.31 × 10–2 min–1 K2 at 47°C = 6.93 × 10–2 min–1 (b) Ea = 43.85 kJ mol–1 (c) 93.85 kJ mol–1

5.

20.39 min

7.

24.83 k cal mol–1, 2.35 × 10–5 sec–1; T = 513 K

8.

0.0327 min–1

9.

(i) 2 and 1 (ii) 2.66 × 108 litre2 mol–2 sec –1 (iii) 55.33 kJ (iv) 1.140 × 1018

10.

311.35 K


NARAYANA Class Xii Chem Chemical Kinetics

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