PHYSICAL CHEMISTRY�Unit III

Chemical kinetics

Syllabus

• Mechanism of Complex reaction, The equilibrium approximation, Steady state approximation, Collision and Encounter, third order reaction.
• Effect of temperature on reaction rate, Effect of catalyst, the Arrhenius Equation. The theories of reaction rate, The Lindeman theory of uni-molecular reaction,
• Kinetics of Complex reaction, opposing or reversible reaction, consecutive reactions, Chain reaction, branch chain reaction, Activated complex theory of bi-molecular reaction. �

Chemical Kinetics

Chemical kinetics or kinetics is the subject of physical chemistry that explain following

The chemical process,

Rates of reaction (R)

Rearrangement of atom,

Various variable affect on rate of reaction etc.

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Reaction

• Chemical change is followed by several steps is called complex reaction.
• The reaction are uni, bi or tri-molecular as molecularity.
• Third Order reaction:

Rate of reaction is depend upon on third power of reactant’s Conc. is define as third order reaction.

d[x]/dt α [A]³

d[x]/dt = K3[A]³

Third order reaction

3A --🡪Product ( A is same reactant)

A + B + C ---🡪 Product

according to rate law the rate of reaction is given by following eq.

d[x]/dt = K3[A][B][C] or

d[x]/dt = K3[A]³ … … (1)

according to rate law after time is extend, product is formed and conc. C of reactant is decrease.

Third order reaction and Rate constant

When [X] is product conc. than [a-x] is conc. of reactant, where ‘a’ is initial conc. of reactant and ‘[x]’ is conc. of product.

now

d[x]/dt α [A]³ where ‘A’ is 3 mole/lit

d[x]/dt = K3[a-x]³ (2)

=> d[x]/[a-x]³ = K3dt (3)

upon integrate above Eq.(3)

Third order reaction�and rate constant(K)

ʃ(d[x]/[a-x]³ ) = ʃ K3dt ʃ1/xⁿ dx=xⁿ⁺¹ /n-1

½ (a-x)² = K3.t + C …. (4)

now when t = 0 than x = 0

therefore C = ½ (a)² …. (5)

Value C is apply in eq. (4)

½ (a-x)² = K3.t + ½ (a)²

simplify and rearrange above eq. we gets,

K3 = x(2a-x)/2t a²(a-x)² …. (6)

Third order reaction and rate constant

Eq. (6) is rate constant of third order reaction.

( when conc. [A]=[B]=[C] = a mole/lit )

Half life time(t1/2) of third order reaction :

The time of reaction to complete 50% of reaction is called half life time of reaction.

According to half life time definition when

(t =t1/2 ) , x = a/2 this value is apply in eq. (6)

Half life time(t1/2)

K3 = x(2a-x)/2t a²(a-x)²

= a/2 (2a – a/2)

2t a²(a-a/2)²

t1/2 = 1/K3.(a)2 …. (7)

Characteristic of third order reaction:

1. Third order reaction rate is proportional to third power of initial conc. of reactant.

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Characteristic of third order reaction

(2) half life time of third order reaction is inversely proportional to second power of reactant conc.

t1/2 = 1/K3.(a)2

(3) Unit of K3 is min^-1.(mole/lit) ^-2

MECHENISM OF COMPLEX REACTION

• Reactions are discussed in two different way
• macroscopic and microscopic

Macroscopic is study of amt react, rate of reaction and amt of product formed.

Microscopic is study of molecular energy, collision, Encounter of atom or molecule.

Reaction mechanism is sequence of reaction process and steps, which describe way of reactant to product.

Mechanism of Complex Reactions:

• Mechanism of Complex Reactions:� Gaseous reactions are classified as�Uni-molecular and bi-molecular elementary�reactions. Reaction which occur in numbers of steps is known as complex reaction. Each step has its own molecularity. Two approximation are used to determining the mechanism of a complex�reaction. (1)The equilibrium approximation�and (2) the steady-state approximation �

Mechanism of Complex Reactions:

(1)The equilibrium approximation and

(2) the steady-state approximation � The equilibrium approximation:

Consider a reaction which reactant R gives product P through the formation�of number of steps,

R -🡪 [I1] -🡪 [I2]-🡪 [I3] -🡪 P�

The equilibrium approximation:

• The whole reaction reactions sequence may be described in terms of a single reaction co-ordinate composing the co- ordinates of the individual steps.
• The intermediates are unstable, stable at shown point in graph.

R -🡪 [I1] -🡪 [I2]-🡪 [I3] -🡪 P

Ko, K1, K2 ……. Different Equilibrium constant

The steady-state approximation :

This approximation is used' to determine�mechanism of a reaction in which the lowest rate-determining step does not exits.

R -🡪 [I1] -🡪 [I2]-🡪 [I3] -🡪 P �The rate of formation of an intermediate�is equal to the rate of decomposition,�so that = = = 0

�

=

=

= 0

=

=

= 0

Collisions and Encounters :�

• Reactions occurring in liquids and solids, a�molecule ‘A’ is surrounding by a cage of�Solvent molecules. Molecule. ‘A' bumps into�the walls of the cage. This is known as cage effect, or Frank Robinovich effect.
• Molecule A' is undergoing diffusion�from one solvent cage to another cage where it may encounter another B and react with it.

Collisions and Encounters :�

• For gaseous reactions the word collision' is used, where as condensed phases (liquids. and solids) the term 'encounter' is used. Because of the cage effect, there is no clear distinction between uni, bi and tri- molecular reactions in condensed phases.
• Investigation of the mechanism of chemical reactions is an which is both fascinating and frustrating.

Collisions and Encounters :�

• There is no last word about a reaction mechanism. If a particular mechanism is proved, it is always possible to give a more�complicated mechanism that is equally�Consistent with the facts.
• Thus we must accept the simplest mechanism that is consistent with all of the evidence that is available at a given time.�

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Decomposition Reaction

To derive the rate equation for the reaction, rate- determining step is used.

This rate- determining step is the slowest steps among all the steps. It is assume that all steps are in equilibrium.

• Rate(dx/dt) :

At constant temp. and pressure(1atm)

differential of Conc. Change in respect of unit time(t) is called rate of reaction.

Reaction of decomposition of O3(g)

• The gaseous decomposition of ozone

2O3(g) --🡪 3O2(g)

According to a rate equation, Decomposition rate of Ozone is given by following equation

Rate (r) = - d[O3]/dt =K [O3(g)]² /[O2]

shows that following mechanism is consists

Step 1 : O3 <=> O2 + O (fast reaction)

Step 2 : O + O3 🡺 2O2 (slow reaction)

Reaction of decomposition of O3(g)

Solution : From the slow rate determining step, (RDS)

Rate (r) = - d[O3]/dt =K1[O] [O3] …(1)

from the fast equilibrium step,

K= [O2][O]/[O3]

therefore,

[O] = K[O3]/[O2] …. (2)

Reaction of decomposition of O3(g)

By substituting equation (2) in (1)

r = K1K[O3]²/[O2]

= K[O3]²/[O2]

For the thermal decomposition of ozone to oxygen , the following mechanism is suggested :

Step 1 : O3 <=> O2 + O

Reaction of decomposition of O3(g)

Step 2 : O + O3 🡺 2O2

Use steady state approximation and other suitable approximation to account for the observed rate law,

r = - K[O3]²/[O2]

Solution :

Ozone decomposes in steps 1 and 2 and is formed in the reverse of step 1.

Reaction of decomposition of O3(g)

Hence,

r = d[O3]/dt

= - K [O3]+K-1[O2][O]-K2[O3][O]…… (1)

Using steady state approximation for the oxygen atom , which is the intermediate having very short existence time.

d[O]/dt = K1 [O3]-K-1[O2][O]-K2[O3][O] = 0

Therefore

K1 [O3]={K-1[O2]+K2[O3]}[O]

Reaction of decomposition of O3(g)

Therefore , [O]= K1[O3]/K-1[O2]+K2[O3]….-(2)

therefore , the rate of decomposition of ozone is given by following eq.

r = - K[O3]²/[O2]

Exa : Decomposition N2O5 is given by following reaction N2O5(g)🡪2NO2+1/2 O2(g)

N2O5 gas is decomposed in following sequence of steps :

Decomposition reaction of N2O5(g)

Step 1:N2O5 <=> NO2 + NO3 (fast equilibrium)

Step 2: NO2+NO3🡪NO2+NO+O2 (slow)

Step 3: NO+NO3🡪2NO2 (fast)

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Factor affecting on rate of reaction:

(1) Temperature & Pressure

(2) Concentration

(3) Catalyst

Temp Effect on rection rate

  we are know that increase in temperature is remarkable change of the reaction rate.� The ratio of two rate constants of a reaction at two different temp, differing from 10 o C is called  the temperature co-efficient of the reaction.�    

Temp Effect on rection rate

The temperatures usually selected for  this Experimental study is  25 and 35 C.� Temp Co- efficient = K35/K25�The Value of the temperature co-efficient for most of the reactions is near about 2 and in Some cases it is close to 3 also.�

Temp Effect on rection rate

On the basis of the collision theory, we are explain why there is a large difference of reaction rates upon a small change of�temperature.

According to this theory, when �Chemical reaction occurs, there must be�collisions between the reactant molecules.�

Temp Effect on rection rate

However most of the collisions taking place between the molecules are successfully effective. Only those collisions are successful, which associated with a amt of certain energy called Threshold energy.� Arrhenius used the Maxwell's distribution� law of molecular energies to explain the� temperature dependence of reaction rates.

Temp Effect on reaction Rate

As the temperature increases from T1 to T2 , the energy distribution changes. There are more molecules on the high side of kinetic energy. The number of molecules whose energies are equal to or greater than the threshold energy E temperature T1 is represented by the shaded area ‘efcd’ and at temperature T2 by the shaded area ‘abcd’. The shaded area ‘abcd’ is nearly twice the shaded

Temp Effect of Reaction Rate

• Area ‘efcd’. This means number of molecules having energy equal to or greater than the threshold energy becomes nearly,

Temp Effect of Reaction Rate

• Double even with a small increase of temperature from T1 to T2 . The number of effective collisions and hence the rate of reaction becomes double with a small increase in temperature.
• The colliding molecules must possess certain minimum energy to give products. However most of the molecules have much less kinetic energy than this minimum energy .

Temp Effect of Reaction rate

• The excess energy that the reactant molecules having energy less than the threshold energy must acquire in order to react to give products is known as Activation energy.

Thus,

Activation Energy = Threshold energy – Energy

actually possessed by molecules.

Temp Effect of Reaction Rate

• The an energy Barrier between reactants enthalpy and products enthalpy. Reactants must cross this barrier to give products. The magnitude of energy barrier is called threshold energy.

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Effect of Catalyst

• A catalyst is a substance that can increase the rate of a reaction but which itself remains unchanged in amount and chemical composition at the end of the reaction. When catalyst is added , a new reaction path with a lower energy barrier is provided. This path is denoted by dotted lines in the following figure. Since the energy barrier is reduced in magnitude , a large number of molecules of

Effect of Catalyst

• the reactants can cross it. This increases the rate of the reaction.

A catalyst does not alter the position of equilibrium in a reversible reaction. Equilibrium is attained quickly just by speeding up both the forward and the backward reactions.

Effect of Catalyst

The Arrhenius Equation

Arrhenius proposed the following equation to calculate the activation energy of a reaction having rate constant K at temperature T.

K = A . e^-Ea/RT (1)

where Ea is known as Arrhenius activation energy and ‘A’ is known as Arrhenius pre-exponential factor . ‘A’ has the same units as the rate constant K.

Hence ‘A’ is also known as collision frequency factor.

The Arrhenius Equation

Taking logarithms of equation(1)

ln K = - Ea/RT + ln A -- -- (2)

If we plot a graph of lnK against (1/ T), it will be a straight line with slope = - Ea/R and intercept = ln A.

Upon differentiating equation (2) with respect to temperature

The Arrhenius Equation

dln K/dt = Ea/RT² --- {d xⁿ /dx = x^n-1 }

(3)

Upon integrating equation (3) between temperatures T1 and T2 when the corresponding rate constants K1 and K2 respectively, and assuming that Ea is constant over this temperature range.

ln (K2/K1) = Ea/R [T2-T1/T1T2] (4)

This equation is known as Arrhenius equation.

The Arrhenius Equation

Thus, from rate constants at two different temperatures, activation energy can be determined.

Ex: Calculate the activation energy of a reaction whose rate constant is tripled by a 10^o c rise in temperature in the vicinity of 27^o c.

Solution : The given temperature is 27^oC

= 300K.

The Arrhenius Equation

In the vicinity of this temperature, the two temperatures are 295 and 305 K.

K = A e^-Ea/RT

K305/K295 = e^-Ea/RT305 /e^-Ea/RT295 = 3

Taking logs , ln3 = - Ea/R [1/305-1/295]

= Ea/R[10/(305x295)]

Ea = (ln3)(8.314Jk^- mol^-)(305K)(295K)/10K

= 82182 J mol^-1 = 88.182 kJmol^-1

The Arrhenius Equation

Ex : The activation energy of a non- catalyzed reaction at 37^oC is 83.68 kJmol^-1 and the activation energy of the same reaction catalyzed by an enzyme is 25.10 kJmol^-1.

Calculate the ratio of the rate of constants of the enzyme – catalyzed and the non –catalyzed reactions.

Solution : Arrhenius equation K = A e^-Ea/RT

The Arrhenius Equation

• Let us and Kx be the rate constants of the enzyme catalyzed and non- catalyzed reactions, respectively.

Assuming that the Arrhenius pre – exponential factor A is the same in the both cases,

(Ke/K) = e^-Ea/RT (Enzyme catalyzed)

e^-Ea/RT (Non – catalyzed )

The Arrhenius Equation

= exp (83.68-25.10/ RT)

= exp (58.58 kJmol^-

8.314 x10^-3 kJK^- mol^- x 310 K )

= e ^22.728

ln ke/ kn = 22.728

hence

ke/kn = 10¹⁰

Thus, the enzyme – catalyzed reaction is about 10 billion times faster than the non-catalyzed reaction.

THEORY OF REACTION

• Lindeman Theory of Uni-molecular reaction
• Activated Complex theory of bi molecular reaction

�Lindeman Theory of Uni-molecular reaction�

• A bimolecular reaction can be easily understood on the basis of collision theory, If we consider unimolecular reaction

A 🡪P, then� the molecule ‘A’ acquires the necessary�activation energy by colliding with another�molecule. This means the reaction is second�order. Actually unimoleculer reactions are�first order.

Lindeman Theory of Uni-molecular reaction

• To understand such reactions the theory was proposed by F. A. Lindeman.� According to this theory a unimolecular�reaction

A🡪 P

occurs through the following mechanism.� A + A 🡨 🡪 A* + A� A* --- P (Product)

Lindeman Theory of Uni-molecular reaction

• Here A is energized molecule which is�capable to rearrange or decompose. In other�words, the vibrational energy of A' is more�then threshold energy for the overall reaction� A* is simply a molecule in a higher�vibrational energy level and not an activated complex.
• In the first step the energized with another molecule. The kinetic into the vibrational energy of the first.�

Lindeman Theory of Uni-molecular reaction

• Second molecule may be any molecule including product, or a foreign molecule present in the system. This molecule may not present in the overall reaction

A🡪P.

The rate constant for all the energization step is K1. A* may be de- energized back to A in the reverse step with rate constant K^-1. In second step with rate constant K2 , A* give product by decomposition or rearrangement.

Lindeman Theory of Uni-molecular reaction

• In this theory there is existence of a time lag between the energization step�and decomposition step. During this time�lag, A* can be de-energized back to 'A?�whenever such short-lived reactive species�is formed in a reaction, steady-state�approximation is used. According to this�approximation, rate of formation of such�species is equal to its rate of decomposition.

Lindeman Theory of Uni-molecular reaction

• In this theory there is existence of a time lag between the energization step�and decomposition step. During this time� A* can be de-energized back to ‘A’ ?�whenever such short-lived reactive species�is formed in a reaction, steady-state�approximation is used. According to this�approximation, rate of formation of such�species is equal to its rate of decomposition.

Lindeman Theory of Uni-molecular reaction

• Here A* is such short lived species

Rate of formation of [A*] =

rate of decomposition

K1 [A] ² = K^-1[A][A*]+K2[A*]

K1 [A] ² - K^-1[A][A*]-K2[A*]=0

therefore ,

[A*] = K1[A] ² / K-1[A]+K2 ----- (1) �

Lindeman Theory of Uni-molecular reaction

The rate of reaction is given by

r = -d[A]/dt = K2 [A*] ------- (2)

upon substituting equation (1) in equation (2)

r = K1K2[A] ² / K-1 [A]+K2 …. (3)

Lindeman Theory of Uni-molecular reaction

This rate law has no definite order.

Let us� Consider two limiting cases.

If k-1[A] >>k, then

r = K1K2[A] / K-1

This is the rate equation for a first order �reaction. In a gaseous reaction, this is the�high-pressure limit. At very high pressure,�[A] is very large so that k-1, [A] >> k,�If k2 >>k..[A] then� r = k1 [A] ²

Lindeman Theory of Uni-molecular reaction

This is the rate equation for a second order�reaction. This is the low pressure limit.� The experimental rate is defined as� r = k uni [A]�where Kuni is the rate constant for�unimolecular reaction. If we compare�equation (3) and equation (4)� Kuni = K1K2 [A]/ K-1 [A]+K2

= K1K2 / K-1 + K2 / [A]

Lindeman Theory of Uni-molecular reaction

Ex : Consider the following Lindeman mechanism for the uni-molecular decomposition of a molecule A in the presence of a species M :

A + M 🡪 A* + M (activation )

A* + M🡪 A + M (deactivation)

A* 🡪 P (decomposition)

Using the steady state approximation , derive the rate law for the formation of the product.

Lindeman Theory of Uni-molecular reaction

Solution :

r = -d[A]/dt = d[P]/dt = K2 [A*]…….. (1)

Applying steady –state approximation ,

rate of formation of [A*] = rate of its

disappearance

K1 [A][M] = K-1 [A*][M]+K2[A*]

Therefore, [A*] = K1 [A][M]/K-1[M]+K2 .. (2)

Upon substituting equation (2) in equation (1),

r = K1K2 [M][A]/K-1[M]+K2

Activated Complex Theory �of Bi-Molecular Reactions

As a result of the development of quantum mechanics, another theoretical approach to chemical reaction rates is developed which gives deeper understanding of the reaction process.

It is known as the absolute reaction rate theory or the transition state theory. This theory is also known as activated complex theory .

This theory was developed by Eyeing , Polyani and Evans.

Activated Complex Theory �of Bi-Molecular Reactions

According to this theory two molecules A2 and B2 form activated complex then decomposes to give the product AB.

A2 + B2 <=> [A2B2] ^# 🡪 2AB

or A-A A-----A A B

+ < = > 🡪 +

B-B B------B A B

(Reactants) (Activated (Product)

complex)

Activated Complex Theory �of Bi-Molecular Reactions

• Generally, activated complex can be treated as a chemical species in equilibrium with the reactants which then decomposes into products.
• However , according to thermodynamics , it is a special molecule in which one vibrational degree of freedom has been converted to a translational degree of freedom along the reaction co-ordinate.
• The reaction co-ordinate could be the bond length which changes in going from the reactants to products.

Activated Complex Theory of � Bi-Molecular Reactions

• The reaction co-ordinate is a measure of the progress of a reaction. Activated complex is unstable because it is at the maximum of the potential energy barrier separating the products from the reactants.
• The difference between the energy of activated complex and the energy of the reactants is the activation energy Ea.

Activated Complex Theory of �Bi-Molecular Reactions

Activated Complex Theory of �Bi-Molecular Reactions

• Consider a simple bimolecular reaction,

A + B 🡨🡪 (AB) ^# 🡪 Products

Where (AB) ^# is the activated complex and K ^# is the equilibrium constant between the reactants and activated complex.

In (AB) ^# one of the Vibrational degrees of freedom has become a translational degree of freedom.

Activated Complex Theory of �Bi-Molecular Reactions

• From classical mechanics the energy of vibration is given by RT/NA.
• Now R/NA = KB Boltzmann constant.
• Therefore ,

Energy = KB.T

Now, from the quantum mechanics Vibrational energy is hν, where h is planks constant and is Vibrational frequency.

Activated Complex Theory of �Bi-Molecular Reactions

• Thus, hν = RT/ NA
• There fore, ν = RT/NA.h

= KBT/h

The vibrational frequency is the rate at which the activated complex molecules move across the energy barrier . Thus , the rate constant can be identified with ν.

Activated Complex Theory of�Bi-Molecular Reactions

The Rate of reaction is

-d[A] /dt = KK2 [(AB) ^#]

= KKBT/h [(AB) ^#] - - - (1)

where the factor K is known as the transition co-efficient. K is a measure of the probability that a molecule , once it crosses the barrier , will keep on going ahead and not return.

Activated Complex Theory of �Bi-Molecular Reactions

• The value of K is taken equal to unity.

Now ,according to equilibrium constant

K ^# = [(AB) ^#] / [A][B]

therefore ,

[(AB) ^#] = K ^# [A][B] --- (2)

Upon substituting equation (2) in (1).

-d[A]/dt = KBT/h K ^#[A][B]

Thus, the rate constant K2 may be given as ,

K2 = (KBT/h)K ^# ------ (3)

Activated Complex Theory of �Bi-Molecular Reactions

The equilibrium constant can be expressed in terms of (ΔG^o) ^# , called the standard Gibbs free energy of activation.

(ΔG^o) ^# = - RTlnK ^# and

(ΔG^o) ^# = (ΔH ^o) ^#-T(ΔS ^o) ^#…… (4)

(ΔG^o) ^# = - RTlnK ^#

lnK ^# = (ΔG^o) ^# / RT

• Therefore, K ^# = e ^{–(ΔGo)#/RT}
• Therefore , K ^# = e ^(ΔSo)#/RT e ^{–(ΔHo)#/RT} ….. (5)

Activated Complex Theory �of Bi-Molecular Reactions

• Upon substituting equation (5) in equation (3),
• K2 = (KBT/Ћ) e ^(ΔSo)#/RT e ^{–(ΔHo)#/RT} …… (6)
• This equation is known as Eyering equation. Here,(ΔS ^o) ^# is the standard entropy of activation and (ΔH ^o) ^# is the standard enthalpy of activation. The Eyering equation can be applied to solutions also.
• Taking logarithms of both sides of equation (6)

Activated Complex Theory �of Bi-Molecular Reactions

i.e. K2 = (KBT/h) e ^(ΔSo)#/RT e ^{–(ΔHo)#/RT}

lnK2 = ln KBT/ h + (ΔS ^o) ^# /R - (ΔH ^o) ^# /RT

Upon differentiating with respect to T

d(lnK)/dT =(ΔH ^o) ^#/ RT²+1/T

[d xⁿ /dx = nx^n-1 ]

[d lnT /dx = 1/T]

therefore,

d(lnK2)/dT = (ΔH ^o) ^# + RT / RT² ….. (7)

Activated Complex Theory �of Bi-Molecular Reactions

Now , Arrhenius equation ,

K2 = A.e^-Ea/RT

Taking logarithms of both sides ,

lnK2 = ln A – Ea/RT

Upon differentiating with respect to T ,

d(lnK2)/dT = Ea/RT² ………… (8)

From equations (7) and (8),

Ea = (ΔH ^o) ^# + RT

• Therefore , (ΔH ^o) ^# = Ea-RT ………………. (9)

Chain reaction

• Reaction which are initiate by free radical or ions is called chain reaction, these are extend rapid with three different step
• (a) Initiation
• (b) Chain Propagation
• (c) Chain termination
• Reaction: Br2(g) is reacted with H2(g)

Mechanism of reaction

(i) Br2(g) --K1-🡪 2Br˚ [Chain initiation]

(ii) Br˚ + H2(g) --K2-🡪 HBr + H˚

[Chain propagation]

(iii) H˚+ Br2(g) K3-🡪 HBr + Br˚ ‘’

(iv) HBr + H˚ K4 --🡪 Br˚ + H2(g) ‘’

(V) 2Br˚ --K5-🡪 Br2(g) [ Chain termination]

Mechanism of reaction

Here K1, K2, …are rate constant of different steps.

In the chain reaction, the intermediate is consumed, products are formed and the intermediate is regenerated. This regeneration allows the cycle to be repeated again and again.

Mechanism of reaction

• Step:(1) Chain initiation :

Bromine molecule collides with any molecule and acquires energy to give bromine atoms(free Radical), Reactive bromine atoms are formed.

• Step: (2),(3) Chain propagation :
• Here Br˚ converting H2 and Br2 into HBr and regenerating Br˚ and H˚

Mechanism of reaction

Step 4 : Chain retardation :

Here , the chain is inhibited by the destruction of product , HBr , thereby decreasing the rate of its formation.

Step 5 : Chain termination.

This reactive species Br˚ and H˚, which are responsible for propagating the chain are known as chain carriers.

Mechanism of reaction

The rate law for this reaction in the temperature range 500-1500 K is

r = d[HBr]/dt = K2 [Br2 ] ^½

This type of reactions were studied by Rice – Herzfeld.

Mechanism of [Br2(g) + H2(g)]

The reaction Mechanism of

H2(g) + Br2(g) -🡪 2HBr is given below.

(i) Br2(g) --K1-🡪 2Br˚ [Chain initiation]

(ii) Br˚ + H2(g) --K2-🡪 HBr + H˚

[Chain propagation]

(iii) H˚+ Br2(g) K3-🡪 HBr + Br˚ ‘’

(iv) HBr + H˚ K4 --🡪 Br˚ + H2(g) ‘’

(V) 2Br˚ --K5-🡪 Br2(g) [ Chain termination]

Mechanism of [Br2(g) + H2(g)]

From above mechanism and apply steady-state approximation for [H˚] and [Br˚], we derive the rate law expression for the formation of (HBr).

• Now, here(HBr) is formed in steps 2 and 3 and consumed in step: (4) according to steady- state approximation

d[HBr]/dt = K2[Br][H2]+K3[H][Br]-K4[H][HBr] (1)

d[H]/dt = K2 [Br][H2] – K3 [H][Br2]–K4[H][HBr]= 0 ….. (2)

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Mechanism of reaction

d[Br]/dt = 2K1 [Br2] – K2[Br][H2] + K3 [H][Br2]+

K4 [H][HBr] – 2K5 [Br] ² = 0

(3) Adding equations (2) and (3),

2K1[Br2] – 2K5[Br] ² = 0

Therefore , [Br] = (K1/K2) ^½ [Br] ^½ … (4)

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Mechanism of reaction

From equation (2)

[H] = K2 [Br][H2]/ K2 [Br2] + K4[HBr]… (5)

Upon substituting equation (4) into (5),

[H] = K2 (K1/K2) ^½ [Br] ^½ [H2] …(6)

K3 [Br2]+K4[HBr]

Upon substituting equations (4) and (6) into (1)

Mechanism of reaction

d[HBr]/dt = K2 (K1/K2) ^½[Br] ^½[H2] + K2K3 (K1/K5) ^½

[Br] ^½[H] / K3 [Br2]+K4[HBr] – K4K2

(K1/K5) ^½[Br] ^½ [H2][HBr] / K3 [Br2] + K4 [HBr]

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Mechanism of reaction

= 2K2(K1/K5) ^½[Br] ^3/2[H2] +

K3 [Br2] + K4 [HBr]

K4K2(K1/K5) ^½ [Br2] ^½ [H2][HBr]

K3 [Br2] + K4 [HBr]

- K4K2(K1/K5) ^½ [Br] ^½ [H2][HBr] ..(7)

K3 [Br2] + K4 [HBr]

Mechanism of reaction

d[HBr] = 2K2(K1/K5) ^½[Br] ^½ [H2] … (8)

dt K3 [Br2] + K4 [HBr]

Dividing , the numerator and denominator by [Br2] ,

d[HBr] / dt = 2K2 (K1/K2) ^½ [Br2] ^½ [H2]

K3 + K4 [HBR] / [Br2]

= 2K2 (K1/K5) ^½ [Br2] ^½ [H2]

1 + K4 / K3 [HBr] / [Br2]

Mechanism of reaction

k4/k3 [HBr]/[Br2] < 1

Therefore , r = d[HBr] / dt = K [H2][Br2] ^½

Where K= 2K2 (K1/K5) ^½ is the observed rate law.

Important Question

SQ Ans. the following question

(1) Explain the terms

(a)Collision (b)Encounters (c)Cage effect

(d)Threshold Energy (e) Catalyst

(f) Complex reaction (g) Temp Co-Efficient

(2) Explain the rate constant when (T= α) in

Arrhenius Eq.

(3) Give Arrhenius Eq. ? Why ‘A’ is define as

collision frequency ?

(4) Explain the steady- state approximation law.

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LQ

(1) Explain the reaction between

H2(g) & Br2(g)

Using steady state approximation.

(2)Prove that the rate of (H2(g) & Br2(g)) is

Rate(r) = d[HBr]/dt =K [H2][Br2] ^1/2

(3) Explain brief of Activated Complex theory

of Complex reaction ( absolute rate theory)

LQ

(5) Explain the Uni-Molecular theory of reaction

rate ( Lindeman theory)

(6) Explain the reaction rate of Arrhenius theory.

(7) Explain the rate of decomposition of

Ozone(O3) , The rate(r) = - K[O3]²/[O2]

Cal.

Ex: 1 Calculate the activation energy of a reaction whose rate is tripled by rise temp in the vicinity of 27^oC

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