Chemical Kinetics-RC

October 13, 2022 admin Off Uncategorized

INTRODUCTION

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

Different reactions occur at different rate. In fact a chemical reaction involves redistribution of bonds –– breaking of bond(s) in the reactant molecule(s) and making of bonds in the product molecule(s). The rate of a chemical reaction actually depends upon the strength of the bond(s) and number of bonds to be broken during the reaction. It takes longer time for the reactant molecules to acquire higher amount of energy which they do by collision. Hence reactions involving strong bond – breaking occur at relatively slower rate while those involving weak
bond – breaking occur at relatively faster rate. On the basis of rate, reactions are classified as.

Instantaneous or extremely fast reactions i.e. reactions with half-life of the order of fraction of second.
Extremely slow reactions i.e. reactions with half-life of the order of years.
Reactions of moderate or measurable rate.

Ionic reactions are instantaneous. If a drop of silver nitrate solution is added to a solution of the chloride salt of any metal or solution of HCl, a white precipitate of silver chloride appears within twinkling of eye. This is because of the fact that in aqueous solution an ionic compound exists as its constituent ions. No bond needs to be broken during the reaction. Hence reaction takes no time to complete. The half life period of an ionic reaction is of the order of 10–10 s.

Na+ + Cl– + Ag+ + ⎯→ AgCl↓ + Na+ +

Free radicals being very unstable (reactive) due to the presence of unpaired electron, reactions involving free radicals also occur instantaneously. Thus, the reactions, are instantaneous.

∙CH3 + Cl2 ⎯→ CH3Cl + ∙Cl

∙CH3 + ∙CH3 ⎯→ H3C – CH3

Some molecular reactions involving reactant(s) containing odd electron completes within a fraction of second. The speed of such reactions is attributable to the tendency of the odd electron molecule (paramagnetic in nature) to transform into stable spin-paired molecule (diamagnetic) by dimerisation. An example of such reaction is the dimerisation of nitrogen dioxide into nitrogen tetraoxide as mentioned below.

∙NO2 + ∙NO2 ⎯→ N2O4

There are some molecular reactions which are known to be extremely slow. Their half-lives are of the order of several years. Some examples of the type of reactions are as given below:

4 ⎯→ 2Fe2O3.xH2O

[Cr(H2O)6]3+ + I– ⎯→ [Cr(H2O)5I]2+ + H2O

Note that the first reaction given above is called “rusting of iron”. The second one is not ionic reaction as it appears at the first sight. Here in this reaction it is the co-ordinate bond between central metal ion i.e. Cr3+ (acceptor) and water molecule (donor) that is broken and covalent bond between Cr3+ and I– that is formed. The half-life of this reaction is in years.

Most molecular reactions especially organic reactions occur at measurable rate. The half-life of such reactions are of the order of minutes, hours, days. Examples of such reactions are numerous. Some of these are given below.

CH3COOC2H5 + H2O CH3COOH + C2H5OH

+ H2O

H2O2 (aq) H2O +

2N2O5 4NO2 +

NH4NO2 (aq) 2H2O +

In Chemical Kinetics we deal with the rates of only those reactions which occur with measurable rate i.e. which are neither too fast nor too slow. The rates of fast reactions are also determined using lasers.

RATE OF REACTION

The rate of a reaction means the speed with which the reaction takes place. This is expressed either in terms of decrease in the concentration of a reactant per unit time or increase in the concentration of a product per unit time.

Rate of reaction

Or,

The term means is the amount of time elapsed. For example, a car driver starts his journey at 9.00 AM with odometer reading x miles. At 11.00 AM, he reaches his destination. The odometer reading at destination is y miles. The rate of his travel can be calculated as

The above example indicates that the car has been driven with uniform rate but actually it has been driven sometimes faster and sometimes slower depending upon the condition of road. Thus, the overall rate is an average rate and the rate at which the car was moving at any instan

t, called instantaneous rate. The rate measured over a long time interval is called average rate and the rate measured for an infinitesimally small time interval is called instantaneous rate.

In general, for any reaction of the type

Average rate of reaction

Where [A] signifies the molar conc. of (A) and stands for the change in molar concentration of A. The negative sign placed before a reaction rate symbol signifies a decrease in concentration of the reactant with increase of time and a positive sign before the rate symbol signifies that the concentration of product increases with increase of time.

Average rate of reaction

The rate measured over a long time interval is called average rate. The rate of reaction (average rate) is defined as the change of concentration of any one of the reactants (or products) per unit time.

Average rate of reaction

Consider the reaction between CO and

This equation shows that one mole of CO reacts with one mole of one mole each of are formed. The average rate of reaction can be expressed either by decrease in conc. of reactant or by the increase in conc. of any one of products

Thus,

For the reaction,

When 2 moles of decomposed, one mole of and 2 moles of is formed. The rate of increase in the conc. of therefore is half that of the disappearance of the conc. of and increase in conc. of is the same of the disappearance of the conc. of in the same time interval.

In general, for a reaction,

The rate is expressed as:

Instantaneous rate

With the progress of reaction the conc. of reactants decreases while that of product increases. According to law of mass action the rate of reaction decreases moment to moment as shown by graph of rate vs. time.

Rate varies from moment to moment so rate of reaction has to be specified at a given instant of time called instantaneous rate

Where dC is the infinitesimal change in conc. during infinitesimal time interval dt after time t i.e. between t and t + dt.

Consider a reaction,To know the rate of reaction at any time t, a tangent is drawn to curve at the point corresponding to that time and it is extended on either side so as to cut the axes, say at the point A and B. Then

Thus the rate of reaction at time 10 minutes

Units of the rate of reaction

Since concentration is usually expressed in moles / time and time is taken in seconds or minutes, the unit of the rate of reaction is moles or or moles

Illustration 1. Why we prefer instantaneous rate of reaction over average rate of reaction?

Solution: The rate of reaction decreases continuously with time except for a zero order reaction. Therefore, average rate of reaction has no significance for the reaction. But instantaneous rate of reaction for a given instant of time does not change with time.

Illustration 2. Define rate of a reaction.

Solution: Rate of a reaction may be defined as the change in any one of the reactants or products per unit time.

Illustration 3. Define specific rate constant.

Solution: It is defined as the rate of a chemical reaction when the concentration of each reactant appearing in the rate equation is taken as unity.

Illustration 4. The following reaction was carried out in water. .

The initial concentration of I− was 0.50 and concentration after 10 minutes was 0.46 mole lit−1. Calculate the rate of disappearance of I− and rate of appearance of iodine.

Solution:

Rate of disappearance of I− =

= 0.004

Rate of appearance of iodine = of I−)

Illustration 5. The rate of formation of nitric oxide (NO) in the following reaction is

Find the rate of disappearance of oxygen.

Solution: The rate of above reaction in terms of oxygen and NO are

∴ Rate of disappearance of oxygen =

= =

Exercise 1.

For the hypothetical reaction:

write the rate equation in term of the disappearance of B2 and formation of AB2.

Exercise 2.

In the reaction

Express the rate of disappearance of Br− in terms of formation of Br2.

Exercise 3.

The reaction, takes place in a closed flask. It is found that concentration of NO2 increases by in
5 seconds. Calculate the rate of the reaction and the rate of change of concentration of N2O5.

Exercise 4.

At 250°C, the half life for decomposition of N2O5 is 5.7 hr and is independent of initial pressure of N2O5. The specific rate constant is

(A) 0.693/5.7 (B) 0.693×3.7

(C) 5.7 /0.693 (D) none

LAW OF MASS ACTION

“At a given temperature, the rate of a reaction at a particular instant is proportional to the product of the active masses of the reactants at that instant raised to powers which are numerically equal to the numbers of their respective molecules in the stoichiometric equation describing the reaction”.

Active mass = molar concentration of the substance

where W = mass of substance, M = molecular mass in grams.

V = volume in litres

Consider a simple reaction

If CA is the molar concentration or active mass of A at a particular instant, then

Where K is a proportionality constant or rate constant.

If CA = 1 then

Rate

Let us consider a general reaction

If [A] = [B] = 1 mole / lit, then

Rate = K

Rate of reaction at unit concentration of reactant is called rate constant.

The value of rate constant depends on :

(i) Nature of reactant

(ii) Temperature

(iii) Catalyst

MOLECULARITY

A chemical reaction that take place in one and only one step i.e., all that occurs in a single step is called elementary reaction while a chemical reaction occurring in the sequence of two or more steps is called complicated reaction. The sequence of steps through which a complicated reaction takes place is called reaction mechanism. Each step in a mechanism is an elementary step reaction.

The molecularity of an elementary reaction is defined as the minimum number of molecules, atoms or ions of the reactants required for the reaction to occur and is equal to the sum of stoichiometric coefficient of the reactants in the chemical equation of the reaction. Thus, the molecularity of some elementary reactions are as mentioned below :

Elementary reactions Molecularity

Reaction with molecularity equal to one, two, three etc; are called unimolecular, biomolecular, trimolecular etc. respectively.

A complicated reaction has no molecularity of its own but molecularity of each of the steps (elementary reactions) involved in the mechanism.

For example; consider the reaction; which is complicated reaction and takes place in the sequence of following three steps:

(i) (fast and reversible)

(ii) (slow)

(iii) (fast)

The molecularity of each step in the mechanism is two, so that we say that the reaction takes in the sequence of three steps each of which is bimolecular. There is another way also. According to which molecularity of a complicated reaction is taken to be equal to the molecularity of the slowest step i.e. rate determining step (r.d.s) in the mechanism.

For example, the reaction

is said to be unimolecular nucleophilic substitution. Since the reaction occurs in the sequence of the following three steps and the slowest step i.e. r.d.s. is unimolecular.

(i)
(ii)
(iii)

Reactions of higher molecularity (molecularity > 3) are rare. This is because a reaction takes place by collision between reactant molecules and as number of a reactant molecules i.e. molecularity increases the chance of their coming together and colliding simultaneously decreases.

ORDER OF REACTION

The mathematical expression showing the dependence of rate on the concentration of reactant is known as rate law or rate expression of the reaction and sum of the indices (powers) of the concentration terms appearing in the rate law as observed experimentally is called order of reaction. To understand what is order of reaction, consider the reaction :

Kinetic experiment carried out at upon this reaction has shown following rate data.

Experiment Number
1
2
3

From the experiment number 1 and 2, it is evident that rate increases 4 fold when conc. of NO is doubled keeping the conc. of constant i.e. is constant again from experiment number 2 and 3, it is evident that when concentration of is doubled keeping the conc. of NO constant, the rate is just doubled i.e.

is constant

Order of reaction with respect to NO is 2 and with respect to The overall order of reaction is 2 + 1 = 3. This order of reaction suggest that the reaction is complicated and it does not occur in single step. In order to explain this reaction following mechanism has been proposed.

(i)

(ii)

(iii)

Rate of overall reaction = Rate of step II

where K = Rate constant of step II

being intermediate for the overall reaction, its concentration has to be evaluated in terms of the concentration of reactant and this can be done by applying law of mass action upon the equilibrium of step I. Thus,

where equilibrium constant of step I, putting this value of concentration of in the above rate expression, we get

or Rate of reaction

Rate of reaction

Where is another constant, rate constant of overall reaction.

In general, if rate law of a reaction represented by the equation.

is experimentally found to be as follows :

Then order w.r.t. A = m, order w.r.t. B = n

Overall order = m + n

It may be noted that ‘m’ may or may not be equal to a and similarly ‘n’ may or may not be equal to b, m and n are experimental values, which really depends upon reaction mechanism and experimental condition, may not be predicted by just seeing the chemical equation of the reaction. An example of this is as follows:

(i)

Order of reaction is 1.

(ii)

Order of reaction is 2.

Illustration 6. For the reaction , the rate constant is 1.26 × 10−3 L mol−1s−1. What is the order of the reaction?

Solution: The units of rate constant is L mol−1s−1 or (mol L−1)−1s−1+. Equate this with general expression of (mol L−1)1−ns−1.

∴ −1 = 1 − n or n = 2

The order of reaction = 2

Illustration 7. For a reaction,

The experimental rate law is. Propose the mechanism of the reaction.

Solution: The rate law is since rate law is proportional to single power of NO2 and single power of F2, it implies that only one molecule of NO2 and one molecule of F2 are involved in the slow step. Thus the various steps are:

Illustration 8. The reaction, is experimentally found to have a rate given as. Suggest a mechanism consistent with the data.

Solution:

Exercise 5.

The gas – phase decomposition of acetaldehyde.

follows the rate law

If pressure is in atmosphere and the time is in minutes. Find the units for (i) rate of reaction (ii) rate constant.

Exercise 7.

(a) The reaction

Proceeds by the following mechanism

(i)

(ii)

what is the rate law expression?

(b) For the reaction

The experimental rate law is propose the mechanism of the reaction.

Pseudo first order reaction

Reaction whose actual order is different from that expected using rate law expression are called pseudo – order reactions, eg.

Expected rate law :

, Expected order = 1 + 1 = 2

Actual rate law :

, Actual order = 1

Water is taken in excess; therefore, its concentration may be taken constant. This reaction is therefore, pseudo first order. Similarly, the acid catalysed hydrolysis of ester, viz,

follows first order kinetics.

It is also a pseudo – first order reaction.

Illustration 9. If the decomposition of nitrogen (V) oxide

Following a first order kinetics

(i) Calculate the rate constant for 0.04 M solution, if the instantaneous rate is .

(ii) Also calculate the rate of reaction when the concentration of N2O5 is 1.20 M.

(iii) What concentration of N2O5 would give a rate of?

Solution: (i) As the given rate is of the first order, therefore, rate =

(ii) Now if the concentration of N2O5 is 1.2 M, then

rate =

(iii) To obtain concentration of N2O5 when the rate is

or 0.7 M

Illustration 10. The experimental data for the reaction

Experiment	[A] mole/lit	[B2] mol/lit	Initial rate
1.	0.50	0.50
2.	0.50	1.00
3.	1.00	1.00

Write the rate law equation.

Solution: Let the rate law equation be

Now,

…(i)

…(ii)

….(iii)

Divide (ii) by (i)

Now divide (iii) by (ii)

2x = 1

x = 0

∴ Rate law =

Exercise 8.

Three experiments were carried out for the reaction between Cl2 and NO

, the following data were obtained.

Experiment	Initial concentration		Initial rate
Experiment	[Cl2]	[NO]	Initial rate
1.	*0.02*	*0.01*
2.	*0.02*	*0.03*
3.	*0.04*	*0.03*

Determine:

(i) the orders with respect to Cl2 & NO

(ii) the rate law.

(iii) the rate constant

Difference between order and molecularity

(i) Order is an experimental property while molecularity is the theoretical property.

(ii) Order concerns with kinetics (rate law) while molecularity concerns with mechanism.

(iii) Order may be any number, fraction, integral or even zero whereas molecularity is always an integer expecting zero.

REACTIONS OF VARIOUS ORDERS

(i) Zero order reactions

A reaction is said to be zero order if its rate is independent of the conc. of the reactants, consider the general reactions:

If it is reac. of zero order

or d[A] = -K dt

Integrating both sides, we get

[A] = -Kt + I ………………….. (i)

where I is a constant of integration

At t = 0,

Substituting this value of I in equation (i), we get

Some important characteristics of reaction of zero order.

(i) Any reaction of zero order must obey equation (ii). As it is a equation of straight line
(y = mx + c), the plot of [A] versus t will be a straight line with slope and intercept on the conc. axis as shown in figure

(ii) Half reaction period:

Half life period is the time in which half of the substance has reacted.

When substituting those values in equation (iii), we get

Unit of K:

Examples of zero order reaction

(i) Photochemical reaction between hydrogen and chlorine

(ii) Decomposition of N2O on hot platinum surface :

(iii) Decomposition of NH3 in presence of molybdenum or tungsten

Illustration 11. Write the integrated rate equation for a zero order reaction.

Solution:

where k = rate constant

[R]0 = initial concentration of the reactant

[R] = concentration of the reactant at time, t.

Exercise 9.

The following graph is a plot of the rate of reaction versus concentration of the reactant. What is the order of the reaction?

First order reaction

A reaction is said to be of the first order if the rate of the reaction depends upon one conc. term only.

Consider the reaction

Let ‘a’ be the conc. of A at the start and after time t, the conc. becomes (a – x), i.e., x has been changed into products. The rate of reaction after time ‘t’ is given by the expression

Upon integration of above equation,

where c is integration constant

when t = 0, x = 0

Putting the value of ‘c’,

If the initial concentration is and the concentration after time t is [A], then putting and (a – x) = [A], equation becomes

This equation can be written in the exponential form as

Illustration 12. Give any one example of

(i) zero order reaction

(ii) first order reaction

Solution: (i)

r = k[NH3]°

(ii)

r = k[NH4NO2]

Illustration 13. For decomposition of N2O5 in CCl4 solution at 320 K.

Show that the reaction is of first order and also calculate the rate constant:

Time in min

∞

Vol. of O2

Evolved (in ml)

6.30

8.95

11.40

13.50

34.75

Solution: If the reaction is of first order, it must obey the equation

In the above reaction, NO2 remains in solution and oxygen is liberated and collected at different intervals of time.

Therefore

, Vα α a

Substituting these values in the first order equation.

Time t = 10

Time t = 15

Time t = 20

Time t = 25

Since the value of K comes out to be content the reaction is therefore is of first order.

Illustration 14. At 375 K, a gaseous reaction

is observed to be of first order. On starting with pure A it was found that at the end of 10 minutes the total pressure of the system was 176 mm of mercury and after a long time, when dissociation of A was complete, it was 270 mm, from these data calculate

(i) the initial pressure of A.

(ii) the pressure of A at the end of 10 minutes

(iii) the rate constant

Solution: Suppose initial pressure of A = P mm

Decrease in the pressure of A after time t = p mm

A ⎯⎯⎯⎯→ 2B + C Total pressure

Initial pressure P 0 0 P

Pressure after

time (t) P − p 2p p P + 2p

Final pressure 0 2P P 3P

(i) Final pressure = 270 mm

∴ 3P = 270 or P = 90 mm

(ii) Pressure after 10 minutes = 176 mm

∴ P + 2p = 176

90 + 2p = 176 or p = 43 mm

∴ Pressure of A after 10 min = P – p

= 90 − 43 = 47 mm

(iii) a α p x α p

∴ K =

Exercise 10.

Write the rate equation and order of a reaction when

(i) [NH3] is very slow.

(ii) [NH3] concentration is very high.

Exercise 11.

For a chemical reaction X ⎯⎯→ Y, the rate of reaction increases by a factor of 1.837, when the concentration of X is increased by 1.5 times. The order of the reaction with respect t X is

(A) 1 (B) 1.5

(C) 2 (D) –1

Exercise 12.

50% of the amount of a radioactive substance decomposes in 5 years. The time required for the decomposition of 99.9% of the substance will be

(A) 10 years (B) between 10 and 40 years

(C) less than 10 years (D) between 49 and 50 years

Some important characteristics of first order reaction

(a) A change in conc. unit will not change the numerical value of K. Let the new unit be n times the first one.

(b) Graphical method:

Comparing it with y = mx + c

Slope

The time taken for any fraction of the reaction to be completed is independent of the initial concentration.

when the half reaction is completed

Half – Life of a nth Order Reaction:

To find out the t½ for a nth order reaction where n ≠1.

∴ ⇒ ⇒

⇒ ⇒

⇒ ⇒ ( order n≠1)

Therefore for a nth order reaction, the half life is inversely related to the initial concentration raised to the power of (n–1).

Note:

It can be noted that for a zero order reaction t1/2 =.

Illustration 15. Define half-life of a reaction.

Solution: The half-life of a reaction is defined as the time required for the reaction to be 50% complete.

It is also defined as the time during which the concentration of reactant is reduced to one half of its initial concentration.

Exercise 13.

The following graph is a plot of t1/2 and concentration. What is the order of the reaction?

Examples of first order reaction

(i) Decomposition of H2O2 in aqueous solution

(ii)

(iii)

Unit of rate constant

where n = order of reaction

Illustration 16. The t1/2 of a first order reaction is 60 minutes. What percentage will be left after 240 minutes?

Solution: No. of half life =

∴ Amount left [A] =

= 0.0625 of [A]0

= 6.25 %

Illustration 17. A first order reaction has a specific reaction rate of 10−3 s−1. How much time will it take for 10 g of the reactant to reduce to 2.5 g?

Solution: Concentration after ‘n’ half life [A] =

or n = 2

∴ time required to reduce to 2.5 g

= 693 × 2 ⇒ 1386 sec

Exercise 14.

A first order reaction is 75% completed in 60 minutes. Find t0.5 of the reaction.

SOME COMPLEX FIRST ORDER REACTIONS

Parallel Reactions:

In such reactions (mostly organic) a single reactant gives two products B and C with different rate constants. If we assume that both of them are first order, we get.

… (1)

… (2)

and …(3)

Let us assume that in a time interval, dt, x moles / lit of B was produced and y moles / lit of C was produced.

∴ and

∴ = . We can also see that from (2) and (3),

∴. This means that irrespective of how much time is elapsed, the ratio of concentration of B to that of C from the start (assuming no B and C in the beginning) is a constant equal to k1/k2.

Exercise 15.

A substrate undergoes a first order decomposition. The decomposition follows two parallel first order reactions as

The % distribution of C and B are

(A) 80% & 20% (B) 25% & 75%

(C) 20% & 80% (D) 75% & 25%

Sequential Reactions:

ABC. In this A decomposes to B which in turn decomposes to C.

∴ = k1[A] …(1)

= k1 [A] –k2[B] …(2)

= k2 [B] …(3)

∴tmax = … (4)

Substituting equation (4) in equation (3)

Bmax =

FACTORS AFFECTING THE RATE OF A CHEMICAL REACTION

The various factors which influences the rate of reaction are

(i) concentration of reactants

(ii) temperature of reactants

(iii) nature of reacting substances

(iv) presence of catalyst

(v) exposure to radiations

Concentration of the reactants

Greater is the concentration of reactants, more will be the chances of collisions between the reacting particles, consequently, larger is the rate of the reaction. For gaseous reactants as the concentration is related to the pressure, therefore, greater is the pressure more will be number of molecules per unit volume and consequently, greater will be the rate of encounters between the molecules.

Effect of temperature on reaction rate

The rates of almost all reactions increase with the increase in temperature. In most of the cases the rate of the reaction becomes almost double for every 10° rise of temperature. This is also expressed in the terms of Temperature co-efficient. Which is the ratio of rate constant of the reaction at two temperatures differing by 10°. The two temperature generally selected are 298K and 308 K. Thus

Explanation of the effect of temperature

According to collision theory of reaction rates.

Rate of a reaction = f × z

Where f = no. of effective collision

z = frequency of collision

Thus, the increase in rate is due to either of the above factors is f or z or due to both of these.

It may be observed that the increase in the total number of collision per unit volume per unit time (collision frequency) is not so much responsible for the higher reaction rate as is the increase in the fraction of effective collisions. Let us, for eg, calculate the increase in collision frequency when temperature increases from 298 to 308 K. As we know that collision frequency is directly proportional to the square root of absolute temperature, therefore, the rate of collision frequencies at these temperatures follows as

From the above ratio it is clear that there is a insignificant increase in the collision frequency. Hence it can not explain the observed increase in the ratio of the reaction with increase in temperature.

Let us now consider the effect of increase in temperature on the number of effective collisions.

Now, as we know that the rise in temperature increases the kinetic energy of the molecules. therefore the energy distribution curve gets flattened and shifts towards higher energy region. A close revels examination of the curves in the graph clearly reveals that the fraction of molecules possessing higher kinetic energy i.e. energy greater than threshold energy, as indicated by shaded portion becomes almost double and therefore, the rate of reaction almost doubles for 10° rise in temperature. Thus, increase in the rate of reaction with increase in temperature is mainly due to increase in no. of collisions which are energetically effective.

Note:

The reaction rate dependence on temperature can also explained by Vant Hoff’s equation.

Nature of reacting substance

The nature of reacting substances affect, the rates significantly. For e.g. the oxidation of ferrous (Fe+2) by KMnO4 in acidic medium is practically instaneous. On the other hand, oxidation of oxalate ions by KMnO4 in acidic medium is comparatively much slower.

Catalyst

A catalyst is a substance, which increases the rate of a reaction without itself being consumed at the end of the reaction, and the phenomenon is called catalysis. There are some catalysts which decrease the rate of reaction and such catalysts are called negative catalyst. Obviously, the catalyst accelerating the rate will be positive catalyst. However, the term positive is seldom used and catalyst itself implies positive catalyst.

Catalyst are generally foreign substances but sometimes one of the product formed may act as a catalyst and such catalyst is called “auto catalyst” and the phenomenon is called auto catalysis. Thermal decomposition of KClO3 is found to be accelerated by the presence of MnO2. Here MnO2 (foreign substance) acts as a catalyst.

2KClO3 + [MnO2] ⎯→ 2KCl + 3O2↑ + [MnO2]

MnO2 can be received in the same composition and mass at the end of the reaction. In the permanganate titration of oxalic acid in the presence of bench H2SO4 (acid medium), it is found that the titration in the beginning there is slow discharge of the colour of permanganate solution but after sometime the discharge of the colour become faster. This is due to the formation of MnSO4 during the reaction which acts as a catalyst for the same reaction. Thus, MnSO4 is an “auto catalyst” for this reaction. This is an example of auto catalyst.

2KMnO4 + 3H2SO4 + 5H2C2O4 ⎯→ K2SO4 + 8H2O + 10CO2 + 2MnSO4

General characteristics of catalyst

A catalyst does not initiate the reaction. It simply fastens it.
Only a small amount of catalyst can catalyse the reaction.
A catalyst does not alter the position of equilibrium i.e. magnitude of equilibrium constant and hence ΔG0. It simply lowers the time needed to attain equilibrium. This means if a reversible reaction in absence of catalyst completes to go to the extent of 75% till attainment of equilibrium, and this state of equilibrium is attained in 20 minutes then in presence of a catalyst also the reaction will go to 75% of completion before the attainment of equilibrium but the time needed for this will be less than 20 minutes.
A catalyst drives the reaction through a different route for which energy barrier is of shortest height and hence Ea is of lower magnitude. That is, the function of the catalyst is to lower down the activation.

Ea = Energy of activation in absence of catalyst.

E′a = Energy of activation in presence of catalyst.

Ea – E′a = lowering of activation energy by catalyst.

If k and kcat be the rate constant of a reaction at a given temperature T, and Ea and E′a are the activation energies of the reaction in absence and presence of catalyst, respectively, the

Since Ea > Ea′ so kcat > k. the ratio gives the number of times the rate of reaction will increase by the use of catalyst at a given temperature and this depends upon Ea –. Greater the value of Ea –, more number of times kcat is greater than k.

The rate of reaction in the presence of catalyst at any temperature T1 may be made equal to the rate of reaction in absence of catalyst but for this sake we will have to raise the temperature. Let this temperature be T2 then

Illustration 18.

Let k1:k2 = 1 :15. Calculate the ratio, at the end of one hour assuming that k1 = x hr–1

Solution: =

∴ = (k1+k2) dt

Integrating with in the required limits, we get

= (k1 + k2 )t

∴ ln = (k1+k2) t

Since = =

∴ ln = 16x

Illustration 19. A hydrogenation reaction is carried out at 500°K. If the same reaction is carried out in the presence of catalyst at the same rate, the temperature required is 400°K. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by 20 KJ/mole.

Solution: By Arhenius equation,

Let Ea of the reaction in absence of catalyst = xKJ/mole

∴ Ea for the reaction in presence of catalyst = (x − 20) KJ/mole

By equation of (i) and (ii)

∴ x = 100 KJ/mole

Exercise 16.

(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) The rate of a particular reaction doubles when temperature changes from 27°C to 37°C. Calculate energy of activation of such a reaction.

COLLISION THEORY OF REACTION RATES:

According to collision theory, a reaction takes place because the molecules collide with each other. The number of collision that takes place per second per unit volume of the reaction mix is called collision frequency. At ordinary temperature and pressure the value of collision frequency is so high (in a gaseous reaction) that if all the collisions were effective, the reaction should be completed in fraction of seconds. However, in actual practice, that is not so. This is explained on the basis of two factors.

(i) Energy factor

For a collisions to be effective, the colliding molecules must have energy more than a particular value. The minimum energy which the colliding molecules must have in order that the collision between them may be effective is called threshold energy. Thus at ordinary temperature and pressure, most of the molecules may not posses energy equal to or greater than threshold value.

A collision between high energy molecules overcomes the repulsion and bring the formation of an unstable molecule cluster, called the activated complex.

Thus, every chemical reaction whether exothermic or endothermic has an energy barrier which has to be overcome before reactants can be transformed into products. If the reactant molecules have sufficient energy, they can reach the peak of the energy barrier after collision and then they change into products. If the activation energy for a reaction is low, the fraction of effective collisions will be large and reaction will be fast. On the other hand, if the activation energy is high, then fraction of effective collisions will be small and the reaction will be slow. When temperature is increased, the number of active molecules increases, i.e., the number of effective collisions will increase and the rate of reaction will increase.

Activation energy Ea = E(activated complex) – E(ground state)

ΔH = activation energy of forward reaction − activation energy of backward reaction.

(ii) Orientation factor

In some cases it is found that even if a larger number of colliding molecules have energy more than threshold value, still the reaction is slow. This is because of improper orientation of the colliding molecules at the time of collisions.

Rate of reaction is directly proportional to the number of effective collisions.

Rate =

= z × f

According to kinetic theory of gases, the fraction of molecules having energy more than a particular value, E at temperature T is given by

Rate =

As rate of reaction is directly related to rate constant K, we can also write

K =

Arrhenius Equation

The variation of equilibrium constant of a reaction with temperature is described by Van’t Hoff equation of thermodynamics which is as follows:

If k1 and k2 be the rate constants of forward reaction and backward reaction, respectively then Kp = k1/k2. Further,
ΔH = Ea1 – Ea2 .Putting these in the above equation we get,

Splitting into two parts

(For FR)

(For BR)

where Z is constant

Arrhenius sets Z equal to zero and without specifying FR and BR, he gave the following equation called Arrhenius equation.

…(i)

From this equation it is evident that rate of change of logarithm of rate constant with temperature depends upon the magnitude of energy of activation of the reaction. Higher the Ea smaller the rate of change of logarithm of rate constant with temperature. That is, rate of the reaction with low Ea increases slowly with temperature while rate of the reaction with high Ea increases rapidly with temperature. It is also evident that rate of increase of logarithm of rate constant will go on decreasing with increase of temperature.

Integrating Equation 4 assuming Ea to be constant we get,

lnk = ..(ii)

or k = …(iii)

Equation (iii) is integrated form of Arrhenius equation. The constant A called pre-exponential factor is the frequency factor since it is somewhat related with collision frequency. It is a constant for a given reaction. From Equation (iii) it is evident that as T → ∞, k → A. Thus, the constant A is the rate constant of reaction at infinity temperature. The rate constant goes on increasing with temperature.

So, when T approaches infinity, k will be maximum. That is to say, A is the maximum rate constant of a reaction.

It is also to be noted that the exponential term i.e. e–Ea/RT measures the fraction of total number of molecules in the activated state or fraction of the total number of effective collisions. If nEa and n be the number of molecules of reactant in the activated state and the total number of molecules of the reactant present in the reaction vessel respectively, then

Equation (ii) may also be put as

logk = + logA …(iv)

Since and logA both are constants for a given reaction. So from equation (iv) it is evident that a plot of log k vs. will be a straight line of the slope equal to and intercept equal to logA as shown below.

tanθ = – tan(180 – θ) = –

∴ Ea =

logA = OA

Thus, from this plot Ea and A both can be determined accurately.

If k1 be the rate constant of a reaction at two different temperature T1 and T2 respectively then from equation (iv), we may write

log k1 = ⋅ and log k2 = ⋅

Subtracting former from the latter we get

= … (v)

With the help of this equation it is possible to calculate Ea of a reaction provided, rate constants of reaction at two different temperatures are known. Alternatively one can calculate rate constant of a reaction at a given temperature provided that rate constant of the reaction at some other temperature and also Ea of the reaction is known.

Exercise 17.

The activation energies of two reactions are Ea1 and Fa2 with Ea1 > Ea2. If the temperature of the reaction system is increased from T1 to T2, predict which of the following alternative is correct?

(A) (B)

(C) (D)

Exercise 18.

A catalyst lower the activation energy of the forward reaction by 20 kJ mol-1. It also changes the activation energy of the backward reaction by an amount

(A) equal to that of forward reaction

(B) equal to twice that of the forward reaction

(C) Which is determined only by the average energy of products

(D) Which is determined by the average energy of products relative to that of reactants

RADIOACTIVITY

Henri Becquerel in 1896 discovered that rays emitted from Uranium compounds had the properties of affecting photographic plates, ionising air and penetrating through substances. The spontaneous emission of such radiations is known as radioactivity and the elements emitting these radiations are called radioactive elements.

Rutherford and Soddy (1903) gave an ingenious interpretation of the radioactive processes and the origin of α and β emissions on the basis of neutron-proton theory.

(i) The atomic nucleii of radioactive elements are unstable.

(ii) The nucleii of radioactive elements are undergoing a process of disintegration forming atoms of new elements called daughter elements which are distinct in physical and chemical properties from parent elements.

(iii) α and β particles are ejected from the nucleus.

(iv) γ-rays are secondary products of atomic disintegration.

Radioactivity is found to be of three types α, β and γ radiations.

	α particles	β particles	γ ray
1.	It carries two units positive charge and four units mass.	It carries unit negative charge with no mass.	They are electromagnetic waves with very short wave length.
2.	It is represented as helium nucleus. or He ions (He++)	It is represented as	It is represented as
3.	They have high ionising power.	Ionisation power is less than that of α particle.	Ionisation power is low.
4.	The velocity of α particle is 2 × 109 cm s−1.	Velocity is less than that of light. It varies from 2.36 to 2.83 × 1010 cm s−1	Velocity is the same as that of light. viz. 3 × 1010 cm s−1.
5.	They have low penetrating power.	Penetration power is 100 times that of α particle	Penetration power is 100 times that of β particles.
6.	The range i.e. the distance upto which radioactive properties can be shown is very small, 8 − 12 cm.	Range is more than that of α particle.	Range is more.
7.	They cause luminescence on zinc sulphide screen.	Very little effect on zinc sulphide screen.	Very little effect on zinc sulphide screen.
8.	When an α particle is emitted, atomic number decreases by 2 units and mass number by 4 units.	When a β particle comes out, atomic no increases by one unit, mass number remains unchanged.	There is no change in atomic number or mass number when a γ-ray is emitted.
9.	Deflected in magnetic field.	Deflected to a greater extent in magnetic field.	Not deflected in magnetic field.

TYPES OF RADIOACTIVE DECAY

Types of decay	Radiation	Equivalent process	Nuclear change		Usual Nuclear condition
Types of decay	Radiation	Equivalent process	At Numbers	Mass Number	Usual Nuclear condition
(i) Alpha emission (α)		⎯	– 2	– 4	Z > 83
(ii) Beta emission (β)			+1	0	too large
(iii) Positron emission (β+)			– 1	0	too small
(iv) Electron capture	X-rays		––	––	too small
(v) Gamma emission (γ)	γ	⎯	0	0	Excited nucleus

Illustration 20. Define radioactivity. Who discovered natural radioactivity?

Solution: Radioacitvity is a phenomenon of spontaneous emission of invisible radiations by certain substances. Natural radioactivity was discovered by Henri Becquerel.

Exercise 19.

How does the temperature changes the radioactivity of an element?

CAUSE OF RADIOACTIVITY

Except in the case of ordinary hydrogen all other nuclide contain both neutrons and protons. A look at the stable nuclides show that the ratio N/P (neutron/proton) in them is either equal to 1 or more than 1. The ratio is ≈ 1 in all the light stable nuclides up to calcium and thereafter the ratio is greater than 1 for heavy nuclides.

The stable nuclides lie within the shaded area which is called the region or zone of stability. All those nuclides falling outside this zone are invariably radioactive and unstable in nature. Nuclides falling above the stability zone have an excess of neutrons while those lying below have more protons. Both of these cause instability. These nuclides attain stability by making adjustment in the N/P ratio.

Illustration 21. How can we increase the N/Z ratio is

Solution: The N/Z ratio can be increased by

(i) Emission of β+

(ii) K – capture

THEORY OF RADIOACTIVE DISINTEGRATION

Rutherford and Soddy, in 1903, postulated that radioactivity is a nuclear phenomenon and all the radioactive changes are taking place in the nucleus of the atom. They presented an interpretation of the radioactive process and the origin of radiations in the form of a theory known as theory of radioactive disintegration.

The disintegration process may proceed in one of the following two ways:

(a) α-particle emission: When an α-particle is emitted from the nucleus of an atom of the parent element, the nucleus of the new element, called daughter element, possess atomic mass or atomic mass number less by four units and nuclear charge or atomic number less by 2 units, because α-particle has mass of 4 units and nuclear charge of two units.

Parent element Daughter element

Atomic mass W W – 4

Atomic number Z Z – 2

e.g. ⎯→

⎯→

(b) β-particle emission: β particle is merely an electron which has negligible mass. Whenever a β-particle is emitted from the nucleus of a radioactive atom, the nucleus of the new element formed, possess the same atomic mass but nuclear charge is increased by one unit over the parent element. β – particle emission is due to the result of decay of neutron into proton and electron.

⎯→ +

The electron produced escapes as a β-particles leaving proton in the nucleus.

Parent element Daughter element

Atomic mass W W

Atomic number Z Z + 1

e.g. ⎯→ +

⎯→

Illustration 22. Arrange the alpha (α), beta (β) and gama (γ) radiations in the increasing order of the properties indicated:

(i) penetrating power

(ii) ionizing power

(iii) extent of deflection in a magnetic field.

Solution: (i) Increasing penetrating power order is α < β < γ.

(ii) Increasing ionizing power order is γ < β < α.

(iii) Increasing extent of deflection in a magnetic field is γ < α < β.

GROUP DISPLACEMENT LAW

According to this law, “When an α – particle is emitted the daughter element has atomic number
2 units less than that of the parent element. It is consequently displaced two places to the left in the periodic table. When a β-particle is emitted, the daughter element has an atomic number 1 unit higher than that of the parent element. It is consequently displaced one place to the right in the periodic table.

Counting of number of α and β particles in a radioactive transformation

Parent element Daughter element

⎯→

Number of α – particles = =

Number of β-particles: Let ‘x’ β-particles and ‘y’ α – particles be emitted

Atomic number of parent element – 2y + x = Atomic number of daughter element

Z1 – 2y + x = Z2

∴x = (Z2 – Z1 + 2y)

Illustration 23. Which particle emission produces isobars?

Solution: Emission of β− particle produces isobars.

Illustration 24. Complete the nuclear reaction:

(i)

(ii)

Solution: (i)

(ii)

Illustration 25. Calculate the number of α and β particles emitted when 92U238 changes into radioactive 82Pb206

Solution: 92U238 ⎯⎯→ 82Pb206+x (2He4) + y(-1e0)

Equating mass number in both sides

238 = 206 + 4x + y

x =

Equating atomic number on both sides:

92 = 82 + 2x + y (-1) = 82+2×8+ y(-1)

∴y =6

∴ No of α-particles = 8

No of β particles =6

Illustration 26. The atomic mass to the is 232.18 and its atomic number is 90. In terms of its radioactive disintegration six α and four β – particles are emitted. What is the mass number and atomic number of product?

Solution: 90Th232 ⎯⎯⎯→ AZm + 6α + 4β

Equating mass number in both sides

232= m + 6×4 + 4×0 (Mass no of Th = 232)

∴ m =208

Equating, at no, in both sides: 90 = A + 2 × 6 + 4 × (-1) ∴ A = 82

∴ Z has atom no = 82

Mass no = 208

RATE OF DISINTEGRATION AND HALF-LIFE PERIOD

The radioactive decay of the different radioactive substances differ widely. The rate of disintegration of a given substance depends upon the nature of disintegrating substance and its total amount. The law of radioactive disintegration may be defined as the quantity of radioactive substance which disappears in unit time is directly proportional to the amount of radioactive substances present or yet not decayed.

Rutherford introduced a constant known as half – life period. It is defined as “time during which half the amount of a given sample of the radioactive substance disintegrates”.

Half life periods vary from billions of years for some radio isotopes to a fraction of a second.

Half life period is represented as t1/2. Let the initial amount of a radioactive substance be No. After one half life period (t1/2) it becomes = No/2. After two half life periods (2t1/2) it becomes = No/4 and after n half life periods (nt1/2) it becomes = N0. Thus, for the total disintegration of a radioactive substance an infinite time will be required.

Amount of radioactive substance left after n half life periods N = and total time T = n × t1/2. Where n is a whole number.

Illustration 27. A radioisotope has t1/2 = 5 years. After a given amount of decays for 15 years, what fraction of the original isotope remains?

Solution: t1/2 = 5 year

t = 15 year ∴ n =

Let original amount be No and amount after time t be N

∴ N =

Hence (⅛)th of the original amount left.

Illustration 28. The half life period of is 140 days. In how many days 1 gm of this isotope is reduced to 0.25 gm?

Solution: Original quantity (No) = 1 gm

Final quantity of isotope = 0.25 gm

As N =

∴ ∴ n = 2

n = ∴ n×t1/2 = 2×140= 280 days.

Disintegration constant: A chemical reaction whose rate varies directly to the concentration of one molecular species only, is termed as first order reaction. Radioactive disintegration is similar to such a chemical reaction as one radioactive species changes into other. This can be represented as.

A ⎯→ B

Suppose the number of atoms of a radioactive substance present at the start of observation, i.e., when t = 0, is No and after time t the number of atoms remaining unchanged is N. At this instant of very small number of atoms dN disintegrate in a small time dt; the rate of change of A into B is given. The negative sign indicates the number of atoms decreases as time increases.

Since rate of disintegration or change is proportional to the number of atoms present at that time, the relation becomes.

= λ.N …(i)

‘λ‘ is called the disintegration constant or decay constant.

Evidently, = λ.dt …(ii)

If dt = 1 second, λ = … (iii)

Thus, λ may be defined as the fraction of the total number of atoms which disintegrates per second at any time.

Integrating equation (ii),
or, – log N = λt + C ……… (iv)

where C is the integration constant.

When t = 0, N = N0

Putting values in equation (iv)

– log N0 = C

Putting the value of C in equation (iv)

– log N = λt – log N0

or, log No – log N = λt

or, log = λt

or, 2.303 log10 = λt

or, λ = … (v)

This equation is called kinetic equation and is obeyed by first order reactions.

Relationship between half life period and radioactive disintegration constant

When t = t1/2 N =

Putting the values in equation (v)

λ = =

So, λ = [ log102 = 0.3010]

or, t1/2 =

Thus, half life period of a given radioactive substance does not depend on the initial amount of a radioactive substance but depends only on the disintegration constant of the radioactive element.

Illustration 29. What mass of C – 14 isotope will have an activity equal to one curie? Given that the half life period of C – 14 is 5730 years.

Solution: Rate of disinitegration

= 1 curie =

∴ Amount of C – 14 in grams

Illustration 30. The half life period of is 60 days. What percent of original radioactivity would be present after 180 days?

Solution: t1/.2 = 60 days, t = 180 days

Applying the formula N =

We get N =

Exercise 20.

(i) The half – life of tritium is 12.4 years. Calculate its decay constant in S−1.

(ii) The half – life period of a radioactive element is 27.96 days. Calculate the time taken by a given sample to reduce to 1/8th of its activity.

Average life Period (T)

Since total decay period of any element is infinity, it is meaningless to use the term total decay period for radioelement. Thus the term average life is used which is determined by the following relation.

Average life (T) =

Relation between average life and half – life

Average life(T) of an element is the inverse of its decay constant, i.e.,

T =

Substituting the value of λ =

Average life (T) = 1.44 × Half life (t1/2)

Illustration 31. Prove that time required for 99.9% decay of radioactive species is almost ten times of its half life period.

Solution: t = log

No = 100, N= 100-=99.9 =0.1

t99.9%= …(1)

t1/2 = …(2)

UNITS OF RADIOACTIVITY

The standard unit of radioactivity is curie (C) which is defined as that amount of any radioactive material which gives 3.7 × 1010 disintegrations per second (dps). i.e., 1C = 3.7 × 1010 dps.

Now a – days, the unit curie is replaced by Rutherford (rd) which is defined as the amount of a radioactive substance which undergoes 106 dps; i.e., 1 rd = 106 dps.

1 C = 3.7 × 1010 dps = 37 × 103 rd

1 mC = 3.7 × 107 dps = 37 rd

However, in SI system the unit of radioactivity is becquerel (Bq)

1Bq = 1 disintegration per second

106 Bq = 1 rd

3.7 × 1010 Bq = 1 C

Illustration 32. Define one curie?

Solution: One curie (Ci) is defined as the amount of radioactive isotope that gives
3.7 × 1010 distintegrations per second.

Radioactive Equilibrium

According to the theory of radioactive disintegration, the product of a radioactive disintegration may also be radioactive and the rate of disintegration of the daughter element depends upon the amount of it present at different time. When steady state is reached,

or, λ1N1 = λ2N2

Where λ1 and λ2 are the disintegration constants for the decay of parent and daughter radioactive species and N1 and N2 are the number of atoms disintegrating at a given time.

Thus,

Activity: Activity is the rate at which the radioactive substance decays

Activity = λ×N =

Specific Activity: Activity of the unit mass of radio isotope.

Illustration 33. 90Sr shows β– activity and its half- life period is 28 years. What is the activity of sample containing 1 gm of 90Sr?

Solution: Activity = λN

Parallel Path: Let the radioactive element ‘A’ decay to B and C in two parallel paths.

Decay constant of ‘A’ = Decay constant of B + Decay constant of C

λA = (λB+ λC)

(Frication of B) =

(Frication of C) =

Maximum yield of daughter element

Let A ⎯⎯⎯⎯⎯⎯→ B (daughter element)

If λA and λB are decay constants of ‘A’ and ‘B’. Maximum activity time of daughter element can be calculated as:

tmax

Number of atoms of daughter element B after time t:

[NB] =

Nuclear Fission: It is a nuclear reaction in which a heavy nucleus splits into lighter nuclei and energy in released.

⇒ Californium – 252 decays both by alpha emission (97%) and by spontaneous fission (3%). During spontaneous fission, the nucleus splits into two stabler nuclei plus several neutrons.

⎯→ + +

⇒ When a neutron strikes nucleus, the nucleus splits into roughly equal parts, giving off several neutrons

⎯→ + +

or, + +

⇒ If the neutrons from each nuclear fission are absorbed by other uranium – 235 nuclei, these nuclei split and release even more neutrons. Thus a chain reaction can occur.

Spallation: Bombardment with high energy charged particles can break some target nuclei into several smaller nuclei with the emission of a large number of nucleus (10 to 20 or more). These are called spallation reaction.

+ ⎯→ +

Nuclear fusion: In this type of nuclear reaction, certain light nuclei may fuse together with the liberation of tremendous amount of energy. To achieve this, the colliding nuclei must possess enough kinetic energy to overcome the initial repulsion between the positively charged cores. This energy may be made available by raising the temperature of the reacting system to several million degress. Such reactions are therefore also known as thermonuclear reactions.

⎯→Mev

The energy of a fusion process is due to mass defect (converted to Binding Energy). The high temperature required to initiate such reactions may be attainted initially through a fission process.

Artificial Radioactivity: Bombardment of stable elements with high energy α-particles, protons, neutrons, deutrons or γ-rays produce radioactive nuclide. These radionuclide do not occur naturally and may be called man – made or artificial. The radioactivity exhibited by these artificial radio nuclide is referred to as artificial radioactivity or as induced radioactivity.

+ ⎯→ +

Binding Energy and Packing fraction

The nearest integer to the mass of a nuclide is called the mass number of that isotope. The difference between the actual isotopic mass and mass number is termal as mass defect of the nuclide.

Thus Δm (mass defect) = isotopic mass – mass number

Aston introduced a term “Packing fraction” for each nuclide to compare their mass excess, it was defined as:

Packing fraction (f)=

Binding energy (B) = Δm × c2

Binding energy of the nucleus is equal to the energy required to split the nucleus into its component nucleons. Nuclear binding energy is generally expressed in MeV (mega – electron volt).

Since 1 a.m.u. = 1.66 × 10–27 Kg

∴ (B) for one a.m.u. = 1.66 × 10–27 × (3×108)2J

= 931 × 106 eV = 931 MeV

∴ (B) = ΔM (in a.m.u.) × 931 MeV

Binding energy per nucleon

Applications of Radioactivity

(i) Radio Carbon Dating: The application of the C – 14 dating is based on the fundamental assumption that the intensity of cosmic ray and hence of 14C in the atmosphere has been remaining constant over many thousands of years. This gives the initial activity of C – 14 corresponding to the time when the plant or animal died and assimilation of radio active carbon ceased to continue.

+ ⎯→ +

⎯→ +

Age =

or N = No

where y = t/t1/2

(ii) Rock Dating: It is based on the kinetics of radioactive decay. It is assumed that no lead was initially in the sample and the whole of it came from the uranium.

Initial radioactivity = [U] = N0 in terms of gm atoms

Final radioactivity = [U] = N

We have assumed that due to high value of tx/2 of uranium its amount remains unchanged.

Thus t =

Where y =, hence t is calculated.

Illustration 34. The amount isotope in a pice of wood is found to be one fifth of that present in fresh piece of wood. Calculate the age of wood (t1/2 = 5577 year.

Solution: t = ⇒ t =

Given N = ∴ t =

Illustration 35. A sample of uranium mineral was hound to contain 206Pb and 238U in the ratio of 0.008:1. Estimate the age of the mineral (half-life of 238U = 4.51×109 yeas)

Solution: We know that t = log

⇒ t = =

= 5.945×109 years

ANSWER TO EXERCISES

Exercise 1:

Exercise 2:

Rate of disappearance of Br−

Exercise 3:

Exercise 4:

(A)

Exercise 5:

(i) Rate of a reaction = atm (min)−1

(ii) Rate constant = atm−1/2 min−1

Exercise 6:

In the absence of Hbr, the rate equation is reduced to

and the overall order is

Exercise 7:

(a)

(b)

Exercise 8:

Exercise 9:

As rate is independent of the concentration the order of the reaction is zero.

Exercise 10:

(i) At very low concentration of NH3, 1 + k2[NH3] ≈ 1.

∴ Rate = k1 [NH3] and order of reaction = 1

(ii) At very high concentration of NH3,

1 + k2[NH3] ≈ k2[NH3]

∴ order of reaction = 0

Exercise 11:

(B)

Exercise 12:

(B)

Exercise 13:

The order of the reaction is one, as t1/2 is independent of the concentration of the reactant.

Exercise 14:

30 minutes

Exercise 15:

(A)

Exercise 16:

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

(ii)

Exercise 17:

(B)

Exercise 18:

(A)

Exercise 19:

Radioactivity of an element is independent of temperature.

Exercise 20:

(i)

(ii) 83.9 days

MISCELLANEOUS EXERCISES

Exercise 1: The plot of log[R] versus time is a straight line with slope equals to
−2.303 × 10−2 s−1. What is the rate constant of this reaction?

Exercise 2: The reaction is first order with respect to A, second order with respect to B and some order with respect to C. The overall order of the reaction is three half. Write the rate law and find the order of a reaction with respect to C.

Exercise 3: Define activation energy. What are its units?

Exercise 4: What is the value of the rate constant when T approaches infinity in the Arrhenious equation?

Exercise 5: Define collision frequency factor.

Exercise 6: Explain what do you understand by transition state.

Exercise 7: What do you understand by the term ‘activated complex’?

Exercise 8: Define molecularity.

Exercise 9: The pressure of a gas decomposing at the surface of a solid catalyst has been measured at different times and the results are given below:

t/s : 0 100 200 300

p/Pascal :

Determine the order of reaction, its rate constant and half-life period.

Exercise 10: The rate of decomposition of H2O2 at a particular temperature is measured by titrating the solution with acidic KMnO4 solution.

Following results were obtained:

Time in minutes : 0 10 20

Volume of KMnO4 in cc : 22.8 13.8 8.3

Show that is a first reaction. Calculate k for the reaction.

ANSWER TO MISCELLANEOUS EXERCISES

Exercise 1: As the plot of log[R] versus time is a straight line the slope is equal to

∴ k = −2.303 × slope

k = 2.303 × (− 2.303 × 10−2)

k = 5.30 × 10−2 s−1

Exercise 2: The rate law is r =

where x = order w.r.t, C.

The given is 1 = 2 + x = 3/2

∴

So, the order w.r.t, C = and the rate law is

Exercise 3: The activation energy is defined as the excess energy (over and above the average energy of the reactants) required by the reactants to undergo chemical reactions. It is expressed in J mol−1.

Exercise 4: and

at T ∞, k = A

Exercise 5: The collision frequency factor is defied as the number of collisions between the reactant molecules A and B per unit volume per unit time divided by the NA and NB, the number of molecular per unit volume of A and B respectively. It is denoted by ZAB.

Exercise 6: When colliding molecules possesses the kinetic energy equal to Ea, the atomic configuration of species formed at this stage is different from the reactant as well as the products. This stage is called activated state or transition state.

Exercise 7: Activated complex refers to the specific configuration of a transition state which is supposed to be in equilibrium with the reactant molecules and has all the attributes of a normal molecule except that one of the vibrational degree of freedom along the reaction is converted into transitional degree of freedom.

Exercise 8: The molecularity of a reaction is defined as the number of reacting molecules which colloide simultaneously to bring about a chemical reaction.

Exercise 9: The rate of reaction between different time interval is

*Interval*	*Rate*
0 – 100 s
100 – 200 s
200 – 300 s

As rate remains constant therefore, reaction is of zero order.

Exercise 10: To show that the reaction is first order, we calculate k at different interval of time using

If k comes out to be same then it proves that order is one.

Volume of KMnO4 required at any stage corresponding to the amount of H2O2 ∝ [R]0 and Vt ∝ [R].

At 10 min

At 20 min

As the value of k is almost same, the reaction is of first order.

SOLVED PROBLEMS

Subjective:

Board Type Questions

Prob 1. Differentiate between the rate of reaction and the rate constant.

*Sol.*	Rate of reaction		Rate constant
	(i)	It is defined as the change in concentration of the reactant or product with time, each divided by its stoichimetric coefficient.	(i)	It is defined as the rate of chemical reaction when the concentration of each reactant appearing in the rate equation is taken as unity.
	(ii)	It always has a unit of (conc)/time.	(ii)	Its units depends on the order of a reaction. For nth order reaction the units of rate constant = (conc)1−n time−1
	(iii)	It depends on the initial concentration of the reactants.	(iii)	It is independent of the initial concentration of the reactants.

Prob 2. Define order of a reaction. Can it be a fractional value? If yes then give an example of a fractional order reaction.

Sol. The order of a reaction is defined as the sum of the exponents to which each concentration terms is raised in the experimently derived rate equation. For example in the reaction

A + B + C ⎯→ Products and the rate law expression is

Then, overall order of reaction = x + y + z.

Order of a reaction can be fractional value. An example of a fractional reaction is gas-phase decomposition of CH3CHO.

and

Prob 3. Differentiate between order and molecularity of a reaction.

*Sol.*	Order of a reaction		Molecularity of a reaction
	(i)	It is sum of the exponents to which the concentration terms in the rate law expression are raised to express observe rate of reaction	(i)	In simple reactions, it is equal to the number of molecules of the reactants while in complex reactions, it is number of molecules involved in rate determining step.
	(ii)	It need not be whole number −it is can be factional or zero also.	(ii)	Always a whole number.
	(iii)	It can only be experimentally determined.	(iii)	It can be obtained by adding the molecules of the slowest step.
	(iv)	It is for the overall reaction and no separate steps are written to obtain it.	(iv)	Overall molecularity of a complex reaction has no significance; only slowest step is significant.
	(v)	It does not give an ideal of the mechanism of the reaction.	(v)	It explains the mechanism of the reaction.

Prob 4. How would you compare chemical reactions and the nuclear reaction?

*Sol.*	Chemical Reactions		Nuclear Reactions
	(i)	This involves making and breaking of bonds in order to rearrange the atoms.	(i)	This involves the conversion of one element into another.
	(ii)	Only electrons in the atomic orbitals are involved in the making and breaking of bonds.	(ii)	If involves protons, neutrons, electrons, and other elementary particles.
	(iii)	A involves the absorption of or release of relatively small amounts release energy.	(iii)	There reactions are accompanied by absorption or release of tremendous amounts of energy.
	(iv)	Rates of reaction are influences by temperature, pressure, concentrations and catalyst.	(iv)	Rates of reaction normally are not affected by temperature, pressure, and catalyst.

Prob 5. Calculate the rate constant of a reaction at 293 K when the energy of activation is
103 KJ mol−1 and the rate constant at 273 K is.

Sol.

T1 = 273°K, T2 = 293°K

Prob 6. The rate law for the reaction,

at 200°C is found to be rate =,

(a)How would the rate change if [Cl2O] is reduced to one third of its original value?

(b)How should the [Cl2O] be charged in order to double the rate?

(c)How would the rate change if [Cl2O] is raised to three fold of its original value?

Sol. (a) Rate equation for the reaction

Let new rate be r′; so

(b) In order to have the rate = 2r, let the concentration of Cl2O be x.

so 2r = Kx2 …(i)

we known that r = …(ii)

Dividing equation (i) and (ii)

⇒

x2 = or x =

Prob 7. Explain in brief the collision theory of reaction rates.

Sol. According to this theory, the reaction takes places as a result of the collision between the reactant molecules. The number of collisions per unit volume per unit time is known as collision frequency ZAB. But all collisions are not effective. The collision which actually converts the reactants into products are called effective collisions. These collisions takes the molecules to the top of the energy barrier and finally results in the formation of products.

There are two conditions for the effective collision:

(i) Energy barrier: The reactant molecules must posses the minimum energy known as threshold energy before they could react and form products.

(ii) Orientation barrier: The reactant molecules must be properly oriented in order to have an effective collision.

The rate constant is given by then

Where ZAB = collision frequency

P = orientation factor and = fraction of total collision which are effective.

Prob 8. What is the difference between

Sol. The symbol represents the electron in or from an atomic orbital. The symbol represents an electron, that although physically identical to any other electron, comes from a nucleus and not form an atomic orbital.

The emission of involves the conversion of a neutron into a proton.

Prob 9. How are the radioactive decay series distinguished? Which one of the decay series is not natural but artificial?

Sol. A radioactive decay series is a sequence of nuclear reactions that ultimately result in the formation of stable isotope.

There are four decay series distinguished by whether the mass number are

(i) divided by four (4n series) or

(ii) divided by four with remainder equals to one (4n + 1)

(iii) divided by four with remainder equals to two (4n + 2)

(iv) divided by four with remainder equals to three (4n + 3)

The (4n + 1) series in not natural but artificial series starts with and ends with .

Prob 10. Explain the following:

(i) Mass defect (ii) Binding energy

Sol. (i) Mass defect is the difference between the actual mass of an isotope of an element and the sum of the masses of protons, neutrons and electrons present in it.

(ii) Binding energy is the energy required to hold the nucleus together. Binding energy of a nucleus is generally quoted as energy in million volts (MeV) per nucleons.

The binding energy per nucleon is a measure of the stability of the nucleus. The greater the binding energy per nucleon, the more stable is the nucleons.

IIT Level Questions

Prob 11. Decomposition of N2O5 is expressed by the equation,

If during a certain time interval, the rate of decomposition of N2O5 is, what will be the rates of formation of NO2 and O2 during the same interval?

Sol. The rate expression for the decomposition of N2O5 is

and

Prob 12. The rate of reaction triples when temperature changes from 20°C to 50°C. Calculate energy of activation for the reaction

Sol. The Arrhenius equation is

Given:

and T2 = 50 + 273 = 323°K

Substituting the given values in the Arhenius equation,

Ea = 28.81 KJ mole−1

Prob 13. In Arhenius equation for a certain reaction, the value of A and Ea (activation energy) are and respectively. At what temperature, the reaction will have specific rate constant

Sol. According to Arhenius equation

or logeK =

or log10K =

T =

Prob 14. The energy of activation for a reaction is 100 KJ mol−1. Presence of a catalyst lowers the energy of activation by 75%. What will be effect on rate of reaction at 20°C, other things being equal?

Sol. The Arhenius equation is

K =

In absence of catalyst,

In presence of catalyst, K2 =

So, or

⇒

As the things being equal in presence or absence of a catalyst,

Prob 15. The experimental data for the reaction

2A + B2 ⎯⎯→ 2AB is as follows:

*Expt. No.*	*[A] mole lit*−1	*[B] mole lit*−1	*Rate of reaction*
1	0.50	0.50
2	0.50	1.00
3	1.00	1.00

Write the most probable equation for the rate of reaction giving reason for you answer.

Sol.

from these three equation

α = 0, β = 1

Rate = K[B2]

Prob 16. Thermal decomposition of a compound is of the first order. If 50% of a sample of the compound is decomposed in 120 minutes. How long will it take for 90% of compound to decompose?

Sol. Half life of reaction = 120 min.

We known that

Applying first order reaction equation,

if a = 100, x = 90 or (a − x) = 10,

so t =

Prob 17. Catalytic decomposition of nitrous oxide by gold at 900°C at an initial pressure of
200 mm was 50% in 53 minutes and 73% in 100 minutes.

(a) what is order of reaction?

(b) how much it will decompose in 100 minutes at the same temperature but an initial pressure of 600 mm?

Sol. (a) Using first order kinetics equation and substituting given values,

In first case:

In second case:

K is same in both cases, so reaction is of first order.

(b) In first order reaction, the time required for the completion of same fraction is independent of initial concentration, the percentage decomposition in
100 minutes when the initial pressure is 600 mm will also be 73%.

Prob 18. The gas phase decomposition

follows the first order rate law. At a given temperature the rate constant of the reaction is.The initial pressure of N2O5 is 0.1 atm.

(i) Calculate the time of decomposition of N2O5 so that the total pressure becomes 0.15 atm.

(ii) What will be total pressure after 150 seconds?

Sol. (i) The decomposition reaction is represented by

Total pressure = p0 − 2x + 4x + x = p0 + 3x = 0.15 atm.

or 3x = 0.15 – 0.10 = 0.05

Pressure of N2O5 after t seconds,

Now applying first order rate law

(ii) After 150 second,

P0 − 2x = P = 0.0324

2x = (0.1 − 0.0324) = 0.06757⇒ x = 0.0337

Total pressure = P0 + 3x = 0.1 + 0.1012 = 0.201 atm

Prob 19. At a certain temperature the half life for the catalytic decomposition of ammonia were found as follows:

Pressure (Pascals) 6667 13333 26666

Half life period (hrs) 3.52 1.92 1.0

Calculate order of reaction.

Sol. , where n is order of reaction.

From the given data,

= (2)n−1

n = 2

Thus order of reaction is 2.

Prob 20. The reaction is a first order reaction with a half-life of
3.5 × 104 s at 320 K. What percentage of SO2Cl2 would be decomposed on heating at 320 K for 920 minutes?

Sol.

t = 90 min = 90 × 60 = 5400 s

If [R]0 = 100 and [R] = conc. of SO2Cl2 at time 90 min then

Percentage of SO2Cl2 decomposed = [R]0 − [R]

= = 11.2%

Prob 21. 88Ra226 emits α– particle to produce radon gas 86Rn222 half-life period of both are respectively 1620 years and 3.8 days calculate

(a) Volume of Radon in equilibrium with 0.226 g radium at 27°C and

(b) Decay constant λ of Ra

Sol. (a) No of atoms in 0.226 g Ra =

or, NRa =

∴ N atoms are present in 222 g of Rn

∴ wt of Rn gas at eq =

∴ at NTP 222 g Rn occupy 22400 cc volume

∴ Volume of

Now,

P1 = 1 at m T1=273°C

P2 = 1 atm T2 = 273+27=300 K

V1 =

V2 = = V2 = 1.581×10-3 cc

(b) Decay constant of Rn = (t1/2 = 3.8 days)

Prob 22. For the following 1st order parallel chain reaction

given that K1 = 5×10-5 s-1 & K2 = 6×10-6 s-1 then find out half life of A.

Sol. For parallel chain reaction

Koverall = K1 + K2

= 5×10-5 + 6×10-6 s-1 = 56×10-6 s-1

∴ t1/2 =

= =1.242×104 Sec

t1/2 = 1.242×104 Sec

Prob 23. (half life = 12.8 h) decays by β emission (38%), β+ emission (19%) and e– capture (43%). Calculate partial half – lives for each of the decay process.

Sol.

Let the rate constant of the above emission process be k1, k2 & k3 respectively and the overall rate constant be K. Then

K = K1+ K2 + K3 =

Also, K1 = 0.38 K = 0.38× h-1

t1 = = 33.68 h

Similarly t2 = = = 67.36 h

t3 = = = 29.76 h

Where t1, t2 & t3 are the partial half lives for respective emissions.

Prob 24. A sample of river water was found to contain 8 × 10–8 tritium atoms, per atom of ordinary hydrogen. The half life of radioactive tritium is 12.3 years.
[log 95.186=1.978]

(a) What will be the ratio of the tritium to normal hydrogen after 49 years, if the sample is stored in a place where additional tritium atoms cannot be formed?

(b) How many tritium atoms would 10 gram of such sample contain 40 years after initial sampling?

Sol. (a) 49 years is exactly four half life’s

After 49 years,

Number of tritium atom = tritium initial

= × 8 × 10–8 = 5 × 10–19

∴ Ratio = =

(b) =

NH atom = mol × Av.NO = 6.7 × 1023 atoms

∴ NO. of tritium atoms =8 × 10–18 × 6.7 × 1023 = 5 × 106

K = log

= log = log

On solving,

After 40 years, number of tritium atom left = .106 × 5 × 106

= .530 × 106 = 5.3 × 105 atom

Prob 25. The half life of radon is 3.8 days. Calculate the time in which its of the amount will be left behind. (Log 2= 0.3010)

Sol.

per day = 0.182 per day

Let the initial amount (N0) be a then Nt = a/20

t =

Objective:

Prob 1. For the reaction 4A + B ⎯⎯→ 2C + 2D, which of the following statement is not correct?

(A) the rate of disappearance of B is one fourth of the rate of disappearance of A

(B) the rate of formation of C is one – half of the rate of consumption of A.

(C) the rate of appearance of D is half of the rate of disappearance of B.

(D) the rate of formation of C and D are equal.

Sol. The rate of appearance of D is double the rate of disappearance of B.

∴ (C)

Prob 2. When concentration of reactant in reaction A ⎯→ B is increased by 8 times, the rate increases only 2 times. The order of reaction would be

(A) 2 (B) 1/3

(C) 4 (D) ½

Sol. (B)

(i) r = kan (ii) 2r = k(8a)n dividing (ii) by (i), 2 = 8n or 23n = 21, n = 1/3

Prob 3. For the following first order reaction A ⎯→ products, which one of the following is correct plot of log[A] versus time?

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

Sol. Plot of log(A) vs time is linear with (−ve) slope.

Prob 4. A catalyst

(A) increases in average kinetic energy of reacting molecules.

(B) increases the activation energy.

(C) alters the reaction mechanism.

(D) increases the frequency of collisions of reacting species.

Sol. A catalyst alters the reaction mechanism.

Prob 5. The rate constant of a reaction increases with increases of temperature because

(A) the activation energy increases

(B) the population of activated molecules increases

(C) the activation energy decreases

(D) the population of activated molecules decreases

Sol. With increase in temperature, population of activated molecules increases.

∴ (B)

Prob 6. In the formation of sulphur trioxide by contact process,, the rate of reaction was measured as . The rate of reaction expressed in terms of SO3 will be

(A) (B)

(C) (D)

Sol.

∴ (B)

Prob 7. The rate of gaseous reaction is equal to K[A][B]. The volume of the reaction vessel containing these gases is suddenly reduced to one – fourth of the initial volume. The rate of reaction would be

(A) 1/16 (B) 16/1

(C) 1/8 (D) 8/1

Sol. Initially rate = K[A][B]

When volume of the vessel is reduced to one – fourth, concentrations becomes
4 times, hence new rate = i.e. it becomes 16 times.

Prob 8. If initial concentration is reduced to 1/4th in a zero order reaction, the time taken for half the reaction to complete

(A) remains same (B) becomes 4 times

(C) becomes one – fourth (D) doubles

Sol. For zero order reactions, t1/2 =

When [A0] is reduced to 1/4th, t1/2 will become 1/4th.

Prob 9. The half – life of a reaction is halved as the initial concentration of the reactant is doubled. The order of reaction is

(A) 0.5 (B) 1

(C) 2 (D) 0

Sol.

or 2 = 2n−1

or (n − 1) = 1 or n = 2

Prob 10. A first order reaction is half – completed in 45 minutes. How long does it need for 99.9% of the reaction to be completed?

(A) 20 hours (B) 10 hours

(C) 7.5 hours (D) 5 hours

Sol.

Or = 448 min ≈ 7.5 hours

Prob 11. For a first order reaction with half life of 150 seconds, the time taken for the concentration of the reactant to fall from M/10 to M/100 will be approximately

(A) 1500 sec (B) 500 sec

(C) 900 sec (D) 600 sec

Sol.

≈ 500 sec

Prob 12. The half life of a first order reaction is 10 minute. If initial amount is 0.08 mole/lit and concentration at some instant is 0.01 mole/lit then t =

(A) 10 min (B) 30 min

(C) 20 min (D) 40 min

Sol. 0.08 mole/lit to 0.01 mole/lit−1 involves 3 half lives.

Prob 13. The rate of chemical reaction doubles for every 10°C rise in temperature. If the temperature is increased by 60°C, the rate of reaction increased by about

(A) 20 times (B) 32 times

(C) 64 times (D) 128 times

Sol. 10° increases has been made 6 times, the rate will increase 26 = 64 times.

Prob 14. The rate constant of a zero order reaction is 0.2 mole dm−3 hr−1. If the concentration of the reactant after 30 minutes is 0.05 mole dm−3, then its initial concentration would be

(A) (B)

(C) (D)

Sol. For zero order reaction

= 0.1 mole dm−3

Now conc. = 0.05 mole dm−3, hence initial conc.

Prob 15. In a particular reaction the time required to complete half of the reaction was found to increase 9 times when the initial concentration of reactant was reduced to one third. What is the order of reaction?

(A) zero (B) first

(C) second (D) third

Sol.

Prob 16. The rate of reaction is doubled for every 10°C rise in temperature. The increase in rate as result of increase in temperature from 10°C to 100°C is

(A) 112 (B) 512

(C) 400 (D) 256

Sol. ==2. For increases of temperature from 10°C to 90°C rate will increase by (2)9 time i.e 512 times.

∴ (B)

Prob 17. For a first order reaction A ⎯⎯→ Products, the rate of reaction, at [A] = 0.2 M is 1.0×10-2 mol lt-1. The half life period for the reaction is:

(A) 823 sec (B) 440 sec

(C) 416 sec (D) 14 sec

Sol. r = K [A] ∴ K =

t1/2 = min = 832 sec

∴ (A)

Prob 18. Two substance A and B are present [A] = 4 [B]. The 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 the concentration of both of them would be same

(A) 15 minute (B) 10 minute

(C) 5 minute (D) 12 minute

Sol. Amount of A in n1 halves =

Amount of B in n2 halves =

If = When A decays to n1 halves and B decays to n2 halves

= 4

4 =

∴n2 = (n1-2) …(1)

Now t = n1×t1/2(A) = n2×t2(B) = ⇒

∴ …(2)

by (1) & (2) n1= 3, n2 =1 ∴ t=3×5 = 15 minute

∴ (A)

Prob 19. The chemical reaction 203 ⎯⎯→ 302 proceeds as following:

O3 O2 + O (fast)

O + O3 ⎯→ 2O2 (slow)

The rate expression should be

(A) r= KI [O3]2 (B) r= KI (O3)2 [O2]-1

(C) r = KI [O3] [O2] (D) unpredictable

Sol. Kc = ⇒

Rate= K [O] [O3]

∴Rate =

∴ (B)

Prob 20. The number of α and β particles emitted in nuclear reaction
90Th236 ⎯→ 83Br212 are respectively

(A) 4, 1 (B) 3, 7

(C) 8, 1 (D) 4, 7

Sol. 90Th232 ⎯→ 83Bi212 + x α + yβ.

228 – 212=4 x

∴ x = 4

90 = 83+2x –y = 83+8-y ∴ y=1

∴ (A)

Fill in the blanks

Prob 21. A plot of versus time is a straight line, the order of a reaction is……………and slope of the line equals to………………….

Sol. One, 2.303/K

Prob 22. An activated complex is formed at the ………………..of the potential energy barriers graph.

Sol. Top

True & False

Prob 23. Rate of a reaction depends only on the number of collision per unit volume per unit time.

Sol. False

Prob 24. The minimum amount of energy which the colliding molecules must possess is known as energy of activation.

Sol. False

Prob 25. When an α-particle emits from a radioactive element, its mass number decreases by two unit and atomic number decreases by four unit.(False)

Sol. False

ASSIGNMENT PROBLEMS

Subjective:

Level- 0

The rate of decomposition of dimethyl ether to form CH4, H2 and CO in a closed vessel is expressed as,

If the pressure measured in bar and time in minutes, then what are the units of rate and rate constant?

A reaction is first order in A and second order in B.

(i) write differential rate equation. (ii) how is the rate affected when

(a) concentration of B is tripled (b) concentration of A and B is doubled?

Nitric oxide, NO reacts with oxygen to produce nitrogen dioxide:

What is the predicted rate law, if the mechanism is

and

For a homogenous decomposition of N2O5 into NO2 and O2.

Find out the order with respect to [N2O5]

Can a reaction have zero activation energy?

State any one condition under which a bimolecular reaction may be kinetically of first order?

What do you understand by the rate of a reaction? How is it expressed? How is the rate of reaction determined?

What do you understand by order of a reaction? How does rate law differ from law of mass action? Give two examples each of the reactions of (i) zero order (ii) first order.

List the main points of difference between order and molecularity of a reaction. What do you mean by the mechanism of a reaction. Explain taking at least two examples.

Derive the equation for rate constant for a first order reaction. What would be the units of the first order rate constant if the concentration is expressed in moles per litre and time in seconds?
Write short notes on (i) fast reactions (ii) photochemical reactions.

Write the rate law for a first order reaction and justify the statement that half life of such a reaction is independent of the initial concentration of the reactants?

The reaction is feasible. How is that hydrogen and oxygen mixture allowed to stand at room temperature shows no formation of water at all?

What are pseudounimolecular reactions? Explain the help of suitable example.

What is activation energy? How is the rate constant of a reaction related to its activation energy?

Level – I

For the following reaction the rate law has been determined to be R = K[A]2[B] where K =.

For this reaction the initial concentration of various species.

Determine the rate after 0.4 mole lit−1 of A has recited.

For the reaction 2A + B ⎯→ C, the rate of formation of C is 0.25 mole lit−1 hr−1. What is the rate of disappearance of A and B?

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

For a chemical reaction the energy of activation is 85 KJ mol−1. If the frequency factor is, what is rate constant at 400°K?

Calculate the ratio of the catalysed and uncatalysed rate constant at 20°C if the energy of activiation of a catalysed reaction is 20 KJ mol−1 and for the uncatalysed reaction is
75 KJ mol−1.

For a first order reaction, the rate constant is 0.1 sec−1. How much time will it take to reduce the concentration from initial value of 0.6 mol−1 to 0.06 mol lit−1?

For the reaction, 2NO + 2H2 ⎯→ N2 + 2H2O the following kinetic data were obtained.

Expt. No.	[A] mole lit−1	[B] mole lit−1	Rate of reaction
1	0.12	0.12
2	0.12	0.24
3	0.24	0.24

Determine the rate law and rate constant.

In a reaction A ⎯→ B + C, the following data were obtained:

t in seconds 0 900 1800

conc. of A 50.8 19.7 7.62

Prove that it is a first order reaction.

A first order reaction is 20% completed in 10 minutes. Calculate

(i) the specific rate constant of the reaction.

(ii) the time taken for the reaction to go to 75% completion

The half life of a first order reaction is 10 seconds. Calculate the time for completion of 99.9% of the reaction.

Level – II

The first order reaction has K = at 200°C. If the reaction is allowed to run for 10 hours at the same temperature, what percentage of the initial concentration would have changed into the product? What is the half life period of this reaction?

The catalytic decomposition of H2O2 was studied by titrating it at different intervals with KMnO4 and the following data were obtained:

t in seconds 0 600 1200

volume of KMnO4 (ml) 22.8 13.8 8.3

Calculate the velocity constant for the reaction assuming it to be first order reaction.

The decomposition of is first order reaction with a rate constant at 318°K. If AB has 26664.5 Pa pressure at the initial stage, what be the partial pressure of AB after half an hour?

Temperature coefficient μ = of a reaction is 1.82. Calculate the energy of activation in calories. (R = 1.987 cal degree−1 mole−1)

Two reactions (i) A ⎯→ products, (ii) B ⎯→ products, follows first order kinetics. The rate of the reaction (i) is doubled when the temperature is raised from 300°K to 310°K. The half life for this reaction at 310°K is 30 minutes. At the same temperature B decomposes twice as fast as A. if the energy of activation for the reaction (ii) is half that of reaction (i), calculate the rate constant of the reaction (ii) at 300°K.

The inversion of cane sugar is studied by recording the angle of rotation of a plane of polarised light. In a typical experiment at 18°C, 20 ml of a 15% cane sugar solution was mixed with 2 ml of conc. HCl and the angle of rotation was recorded with passage of time and the following data were collected:

Time (min.) 0 5 10 15 ∞

Angle of rotation 17.2 16.7 16.2 15.7 −6.0

Show that the inversion of cane sugar follows the first order kinetic and also determine the rate constant of inversion.

The reaction,

Sucrose ⎯→ glucose + fructose

Takes place at 308 K in 0.5 N HCl. At time zero the initial rotation of mixture is 32.4°. After 10 minutes the total rotation is 28.8°. If the rotation of sucrose per mole is 85°, that of glucose is 74° and of fructose is −86.04°, calculate the rate constant of the reaction. The volume of the solution is maintained at 1 lit.

A radioactive element A says to B, “I am half of what you were and you are one fourth of what I was. More over, I was 1.414 times than what you were”. If the half life of A is 8 days, what is half life of B?

For a first order reaction A ⎯→ B the dependence of rate constant K on temperature is given by log10K =, where T is in Kelvin.

A 20 % solution of A by mass decomposes to the extent of 25% in 20 minutes at 298K. What will be the percent dissociation of A in the same time at 340°K, if we start with its 30% solution?

The half life period of is 140 days. In how many days 1 gm of this isotope is reduced to 0.25 gms?

Objective:

Level – I

Which of the following is true for order of reaction?

(A) it is equal to the sum of exponents of the molar concentrations of the reactants in the rate equation

(B) it is always a whole number

(D) it can be determined theoretically

The rate of reaction A + B + C ⎯→ products is given by rate =.

The order of reaction is

(A) 1 (B) 3

The rate of reaction 2X + Y ⎯→ products is given by , if X is present in large excess, the order of the reaction is

(A) zero (B) two

The rate of a reaction at different times is found as follows

Time (in minutes)	Rate (in mole L−1S−1)
0
10
20
30

The order of reaction is

(A) zero (B) one

In Arhenius equation, the fraction of effective collision is given by

(A) (B) A

The plot of log k versus is linear with a slope of

(A) Ea/R (B) −Ea/R

In the reaction 2A + B ⎯→ products, the order w.r.t. A is found to be one and w.r.t. B is equal to 2. Concentration of A is doubled and that of B is halved, the rate of reaction
will be

(A) doubled (B) halved

The rate constant of a reaction is mole dm−3S−1. The order of the reactions is

(A) zero (B) 1

A gaseous reaction,

Shows increase in pressure from 100 mm to 120 mm in 5 minutes. The rate of disappearance of A2 is

(A) 4 mm min−1 (B) 8 mm min−1

A first order reaction is half – completed in 45 minutes. How long does it need for 99.9% of the reaction to be completed?

(A) 20 hours (B) 10 hours

For the first order reaction, half life is 14 sec. The time required for the initial concentration to reduce to 1/8th of its value is

(A) 28 s (B) 42 s

The rate constant for the reaction. If the rate is, then concentration of N2O5 in mole L−1 is

(A) 1.4 (B) 1.2

In the first order reaction the concentration of the reactants is reduced to 25% in one hour. The half life period for the reaction

(A) 2 hrs (B) 4 hrs

The half life period of a first order reaction is 10 minute. If initial amount is 0.08 mole/lit and concentration at some instant is 0.01 mole/lit, then t is equal to

(A) 10 min (B) 30 min

For a first order reaction T3/4 is 1200 seconds. The specific rate constants is (in sec−1)

(A) 10−3 (B) 10−2

is a stable isotope is expected to disintegrate by

(A) α – emission (B) β – emission

The number of α and β particles lost when changes to are

(A) 8α,6β (B) 6α,8β

Of the following nuclides the one most likely to be radioactive is:

(A) (B)

In the nuclear reaction

⎯⎯→

The mass loss is nearly 0.02 amu. Hence the energy released (in units of million Kcal /mol) in the process is approx.

(A) 425 (B) 220

If half-life period is 100 years, average life is nearly

(A) 100 years (B) 70 years

Level – II

Which of the following rate law has an overall order of 0.5 for reaction involving substances X, Y and Z?

(A) rate = K(Cx) (Cy)(Cz) (B) rate = K(Cx)0.5 (Cy)0.5 (Cz)0.5

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 = 0.40 and PB = 0.60 atm. When PC = 0.2 atm and the rate of reaction relative to the intial rate is

(A) 1/48 (B) 1/24

If concentration are measured in mole/lit and time in minutes, the unit for the rate constant of a nth order reaction are

(A) (B)

In a reaction the threshold energy is equal to

(A) average energy of the reactants

(B) activation energy

(D) activation energy − average energy of the reactants

For a reaction A + B ⎯→ C + D, ΔH = −30 KJ mole−1 the activation energy for the forward reaction is 65 KJ mol−1. The activation energy for the backward reaction is

(A) 35 KJ mole−1 (B) 95 KJ mole−1

For a reaction A + 2B ⎯→ C + D, the following data were obtained


[A]	[B]
0.1	0.1
0.3	0.2
0.3	0.4
0.4	0.1

The correct rate law expression will be

(A) rate = k[A] [B] (B) rate = k[A] [B]2

For a first order reaction with half life of 150 seconds, the time taken for the concentration of the reactant to fall from M/10 to M/100 will be approximately

(A) 1500 s (B) 500 s

The rate constant of a reaction at temperature 200° K is 10 times less than the rate constant at 400°K. What is the activation energy (Ea) of the reaction?

(A) 1842.4 R (B) 921.2R

If a is the initial concentration and K is the rate constant of a zero order reaction, the time for the reaction to go to competition will be

(A) K/a (B) a/K

The rate constant of a zero order reaction is 0.2 mole dm−3 hr−1. If the concentration of the reactant after 30 minutes is 0.05 mole dm−3, then its initial concentration would be

(A) (B)

For a zero order reaction, the plot of concentration vs time is linear with

(A) +ve slope and zero intercept (B) −ve slope and zero intercept

For the reaction A ⎯→ B, when concentration of A is made 1.5 times, the rate of reaction becomes 1.837 times. The order of reaction is

(A) 1 (B) 1.5

What is the order of a reaction which has a rate expression rate = K[A]3/2 [B]−1?

(A) 3/2 (B) 1/2

For a hypothetical reaction A + B ⎯→ C + D, the rate = K[A]1/2 [B]3/2. On doubling the concentration of A and B the rate will be

(A) 4 time (B) 2 times

For a reaction pA + qB ⎯→ Products, the rate law expression is r = K[A]m[B]n then

(A) (p + q) ≠ (m + n) (B) (p + q) =(m + n)

The half -life a first order reaction is 24 hours. If we start with 10M initial conc. of the reactant then conc. after 96 hours will be

(A) 6.25 M (B) 1.25 M

During a particular reaction 20% of the reactant decompose in 1 hour, 40% in 2 hours, 60% in 3 hours and so on. The unit of rate constant

(A) hr–1 (B) lit mole–1 hr–1

For a reaction it is observed that. What is the order of the reaction?

(A) 0 (B) 1

⎯⎯→ is written as

(A) (B)

Which of the following nuclear reactions in an Illustration of nuclear fusion

(A) ⎯⎯→

(B) ⎯⎯→

(D) ⎯⎯→

ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level – O

In terms of concentrations, units of rate = mole lit−1min−1

Unit of K =

= L1.2 mole-1/2min1

In terms of pressure, unit of rate = bar min−1.

Unit of K =

(ii) rate = Kab2

if [B] is tripled,

Rate = Ka(3b)2 = 9Kab2 = 9 times

If both [A] and [B] are doubled,

Rate = K(2a)(2b)2 = 8Kab2 = 8 times

Rate depends on the slow step.

for first reaction if equilibrium constant is Kc

Putting the value of [NO3] in equation (i)

Rate = K.Kc[NO]2[O2]

K.Kc = K′

Rate = K′[NO]2[O2]

The overall order of reaction is 3.

Order is 1.
If Ea = 0, then according to Arhenius equation, K =

i.e. rate constant = collision frequency.

This means every collision results into a chemical reaction which can not be true. Hence
Ea ≠ 0.

A bimolecular reaction may become kinetically of first order if one of the reactant is present in excess.

Level – I

4.
6. 23 seconds

99.7 sec

Level – II

t1/2 = 128.3 hrs, 5.27% 2.

12981.7 Pa 4. 10.924 Kcal/mole

0.03265 min−1 6.

0.0102 min−1 8. 8 days

100%

Original quantity of the isotope (N0) = 1 gm

Final quantity of the isotope N = 0.25 g

We know that

n = 2

Time taken = T = n × t1/2

= 2 × 140 = 280 days

Objective:

Level – I

A 2. D 3. C
A 5. C 6. D
B 8. A 9. B
C 11. B 12. D
C 14. B 15. A
D 16. A 18. D
A 20. C

Level – II

C 2. D 3. B
C 5. B 6. B
B 8. B 9. B
B 11. D 12. B
B 14. A 15. C
B 17. C 18. B
C 20. C

About The Author

admin

Editable Study Material for JEE mains, JEE advanced, Foundation, CBSE. Online Tuition, Online Tutors, Home Tutors, Coaching, Online Classes by Study Innovations IIT JEE study material, NEET Study Material, JEE mains Study Material, JEE advanced study material, AIIMS Study Material, IIT JEE Foundation study material, NEET Foundation study material, CBSE Study Material, Test Series, Question Bank, ICSE Study Material, School Exams study material, board exams study material, XII board exams Study Material, X board exams Study Material, Study Material, JEE mains, JEE advanced, Video Lectures, Study Innovations, online tuition, home tuition, online tutors, coaching & tutorials for English, Mathematics, Science, Physics, Chemistry, Biology, soft copy study material, editable study material, study material provider,customised study material,customised study material provider for coaching institutes,study material for coaching classes, how to make study material for coaching institute, jee neet study material,

Half – Life of a nth Order Reaction:

SOME COMPLEX FIRST ORDER REACTIONS

About The Author

admin

Quick Links