Rate of Reaction,Order of Reaction,Molecularity of Reaction, Collision Theory of Reaction Rate

Rate of Reaction:



Rate of change of extent of reaction is the rate of reaction.


Rate of reaction is positive for product and negative for reactant.


For reaction aA →bB
Rate =1/b(Δ[B]/ Δ t)  = -1/a (Δ [A]/ Δt)


It goes on decreasing as the reaction progress due to decrease in the concentration(s) of the reactant(s).










Unit of rate of reaction : mol L-1 s-1


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 a chemical change, reactants and products are involved. As the chemical reaction proceeds, the concentration of the reactants decreases, i.e., products are produced.


The rate of reaction (average rate) is defined as the change of concentration of any one of its reactants (or products) per unit time.





Order of Reaction





For reaction aA + bB + ….. → cC+ ….



R ∝[A]m[B]n or R = k[A]m[B]n….

Where m and n may or may not be equal to a & b.

m is order of reaction with respect to A and n is the order of reaction with respect to B.

m + n +… is the overall order of the reaction.

Elementary Reaction:



It is the reaction which completes in a single step.


A reaction may involve more than one elementary reactions or steps also.


Overall rate of reaction depends on the slowest elementary step and thus it is known as rate determining step.





Molecularity of Reaction:



Number of molecules taking part in an elementary step is known as its molecularity.


Order of an elementary reaction is always equal to its molecularity.


Elementary reactions with molecularity greater than three are not known because collisions in which more than three particles come together simultaneously are rare.








Chemical Reaction







Molecularity





PCl5  →  PCl3 + Cl2   

Unimolecular


2HI  →  H2 + I2 

Bimolecular


2SO2 + O2  →  2SO3

Trimolecular


NO + O3  →  NO2 + O2

Bimolecular


2CO + O2  →  2CO2

Trimolecular


2FeCl3 + SnCl2 → SnCl2 + 2FeCl2

Trimolecular

 






Differential and Integrated Rate Laws:

Zero Order Reactions:




For Reaction: A → Product

[A]0-[A]t  = k0t

Where,

[A]0 = Initial concentration of A

[A]t = Concentration of A at time t.  

k0  =  Rate constant for zero order reaction.

Half Life:



t1/2 = [A]0/2k

Unit of rate constant = mol dm-3s-1


Examples: 


 Enzyme catalyzed reactions are zero order with respect to substrate concentration.


 Decomposition of gases on the surface of metallic catalysts like decomposition of HI on gold surface.





First Order Reactions:




A → Product

(Δ [A] /A) = -k1Δt

 or k1=( 2.303/ t)log ([A]0 / [A]t) 

Half Life:

t1/2 = 0.693/k1

Half life is independent of the initial concentration of the reactant for a first order reaction.

Units of k1 =  s-1

Examples:

N2O5 →  2NO2 + 1/2O2

Br2 → 2Br

2HNO3 → 2NO + H2O

 H2O2→ H2O + 1/2O2 



Pseudo First Order Reactions:


These are the reactions in which more than one species is involved in the rate determining step but still the order of reaction is one.

Examples:


Acid hydrolysis of ester: CH3COOEt + H3O+ →CH3COOH + EtOH 


Inversion of cane sugar:



  


Decomposition of benzenediazonium halides C6H5N=NCl +H2O → C6H5OH +N2 +HCl





Half – Life of a nth Order Reaction:


kt1/2 =  (2n-1-1)/(n-1)[A0]n-1

Where, n = order of reaction ≠1



Parallel  Reactions:


The reactions in which a substance reacts or decomposes in more than one way are called parallel or side reactions.





If we assume that both  of them are first order, we get.



k1 = fractional yield of B × kav

k2 = fractional yield of C × kav

If k1 >  k2 then

A → B main and

A → C is side reaction

Let after a definite interval x mol/litre of B and y mol/litre of C are formed.



i.e



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.









Sequential Reactions:


This reaction is defined as that reaction which proceeds from reactants to final products through one or more intermediate stages. The overall reaction is a result of several successive or consecutive steps.

A → B → C and so on



…....(i)

…......(ii)

….......(iii)

Integrating equation (i), we get






   

   

 





Arrhenius Equation:


k = A exp(-Ea/RT)

Where, k = Rate constant

A = pre-exponential factor

Ea = Activation energy



      




Temperature Coefficient:


The temperature coefficient of a chemical reaction is defined as the ratio of the specific reaction rates of a reaction at two temperature differing by 10oC.

μ = Temperature coefficient= k(r+10)/kt

Let temperature coefficient of a reaction be ' μ ' when temperature is raised from T1to T2; then the ratio of rate constants or rate may be calculated as







Its value lies generally between 2 and 3.



  Collision Theory of Reaction Rate


A chemical reaction takes place due to collision among reactant molecules.


The number of collisions taking place per second per unit volume of the reaction mixture is known as collision frequency (Z).


The value of collision frequency is very high, of the order of 1025 to 1028 in case of binary collisions.


Every collision does not bring a chemical change.


The collisions that actually produce the products are effective collisions.


The effective collisions which bring chemical change are few in comparison to the form a product are ineffective elastic collisions, i.e., molecules just collide and


disperse in different directions with different velocities.


For a collision to be effective, the following two barriers are to be cleared.




Energy Barrier


Orientation Barrier









Radioactivity:


All radioactive decay follow 1st order kinetics

For radioactive decay A ->B

-(dNA/dt) =l NA

Where, l =  decay constant of reaction

NA  = number of nuclei of the radioactive substance at the time when rate is calculated.

Arrhenius equation is not valid for radioactive decay.

Integrated Rate Law: Nt = Noe-lt

Half Life:  t1/2= 0.693/λ

Average life time: Life time of a single isolated nucleus, tav= 1/λ

Activity: Rate of decay

A = dNA/dt, Also, At = Aoe-lt

Specific Activity: activity per unit mass of the sample.



Units: dps or Becquerre

UPPGT PRAVAKTA EXAM 2016 SOLUTION ANSWER KEY

Silicones are substances that are being very widely used in cosmetics. That's because it's more or less the common basic ingredient that any formulator probably uses in every one of their creams, skin serums, shampoos, hair care products… These substances make the formula much softer, contributing to a better spread



Zeise's salt, potassium trichloro(ethene)platinate(II), is thechemical compound with the formulaK[PtCl3(C2H4)]

An alkelene having the molecular formula C9H18 on ozonolysis gives 2,2-dimethyl propanal and 2-butanone. The alkene is

Answer

2, 2, 4-trimethyl-3-hexene

An alkelene having the molecular formula C9H18 on ozonolysis gives 2,2-dimethyl propanal and 2-butanone. The alkene is

Answer

2, 2, 4-trimethyl-3-hexene

Abnormal colligative properties and Van't Hoff Factor

Since colligative properties depend upon the number of particles of the solute, in some cases where the solute associates or dissociates in solution, abnormal results for molecular masses are obtained.

Van't Hoff Factor :

Van't Hoff, in order to account for all abnormal cases introduced a factor i known as the Van't Hoff factor, such that



Association :

There are many organic solutes which in non-aqueous solutions undergo association, that is, two or more molecules of the solute associate to form a bigger molecule. Thus, the number of effective molecules decreases and, consequently the osmotic pressure, the elevation of boiling point or depression of freezing point, is less than that calculated on the basis of a single molecule. Two examples are : acetic acid in benzene and chloroacetic acid in naphthalene.



Association of Acetic acid in benzene through hydrogen bonding

Degree of Association :

The fraction of the total number of molecules which combine to form bigger molecule.

Consider one mole of solute dissolved in a given volume of solvent. Suppose n simple molecules combine to form an associated molecule,

i.e. nA  (A)n

Let a be the degree of association, then,

The number of unassociated moles = 1-a

The number of associated moles = a/n

Total number of effective moles = 1-a+a/n



i = 1- a (1–1/n)

Obviously, i < 1

Refer to the following video for Van’t Hoff Factor





Example 1: 


Question: 

Acetic acid (CH3COOH) associates in benzene to form double molecules. 1.65 g of acetic acid when dissolved in 100g of benzene raised the boiling point by 0.36°C. Calculate the Van't Hoff Factor and the degree of association of acetic acid in benzene (Molal elevation constant of benzene is 2.57).

Solution: 

Normal molar mass of acetic acid = 60

Observed molar mass of acetic acid.







0.508 = 1–(1–1/n) = 1 – (1–1/2) = 1–1/2

/2 = 1– 0.508 = 0.492

  = 2  0.492 = 0.984

Thus acetic acid is 98.4% associated in benzene.




Dissociation

Inorganic acids, bases and salts in aqueous solutions undergo dissociation, that is, the molecules break down into positively and negatively charged ions. In such cases, the number of effective particles increases and, therefore, osmotic pressure, elevation of boiling point and depression of freezing point are much higher than those calculated on the basis of an undissociated single molecule.

Degree of Dissociation

Degree of dissociation means the fraction of the total number of molecules which dissociates in the solution, that is, breaks into simpler molecules or ions. Consider one mole of an univalent electrolyte like potassium chloride dissolved in a given volume of water. Let a be its degree of dissociation.

Then the number of moles of KCI left undissociated will be 1-a. At the same time, a moles of K+ ions and a moles of Cl-ions will be produced, as shown below.

KCl  K+ + Cl-

1-           

Thus, the total number of moles after dissociation = 1-++ = 1+

Hence, i = (1+)/1

Since, as already, mentioned, osmotic pressure, vapour pressure lowering, boiling point elevation or freezing point depressions vary inversely as the molecular weight of the solute, it follows that

i = 1+  = 1+ (2–1)

In general, i = 1+ (n–1) a, Where, n = number of particles ( ions) formed after dissociation

From the above formula, it is clear that i > 1

Knowing, the observed molar mass and the Van't Hoff factor, i, the degree of dissociation, a can be easily calculated.

Now, if we include Van’t Hoff factor in the formulae for colligative properties we obtain the normal results.



Note: The value of i is taken as one when solute is non electrolyte.



Example 2:


Question: 

The freezing point depression constant for HgCl2 is 34.3 km–1. For a solution of 0.849 of mercurous chloride (empirical formula HgCl) in 50 g of HgCl­2, the freezing point depression is 1.24. What is the molecular weight of mercurous chloride in this solution? What is its molecular formula?

Solution: 

Tf = Kf m
Molality of mercurous chloride solution = T/Kf = 0.036

M (molecular weight of mercurous chloride) = (0.849/0.036)(100/50) = 471.67

Empirical formula wt = 235.5

Molecular formula of mercurous chloride = (HgCl)2 or Hg2Cl2




Van’t Hoff Theory of Dilute Solutions  

Van’t Hoff realized that an analogy exists between gases and solutions provided osmotic pressure of solutions is used in place of ordinary gas pressure. He showed that for dilute solutions of non-electrolysis the following laws hold good.  


Boyle-van’t Hoff law:  



The osmotic pressure (P or α) of a solution is directly proportional to its concentration (C) when the temperature is kept constant. The concentration of the solution containing one gram mole in V litres is equal to 1/V (C = 1/V)

Thus P ∝ C                 (when temperature is constant)  

or P ∝ 1/V

or PV = constant            

or     πV = constant  

Van’t Hoff presumed that the osmotic pressure is due to the bombardment of solute molecules against the semipermeable membrane as the gas pressure is due to hits recorded by gas molecules against the walls of its container.  


Pressure-Temperature law (Gay-Lussac-van’t Hoff law):  



Concentration remaining same; the osmotic pressure of a dilute solution is directly proportional to its absolute temperature (T), i.e.,  

P ∝ T  

or P/T = constant      
or           
μ/T constant  

Combining the two laws, i.e., when concentration and temperature both are changing, the osmotic pressure will be given by:  

P ∝ CT  

or P = kCT  

or P = k.1/V.T           (since C = 1/V)

or PV = ST or πV = ST  

S is called molar solution constant.  

Here V is the volume of solution containing one gram mole of the solute. The value of 5 comes out to 0.082 lit atm K−1 mol−1 which is in agreement with the value of R, the molar gas constant. In case the solution contains n gram moles in V litres, the general equation would become  PV = nST  or     πV = ST  


Third law:  



Equimolecular solutions of different solutes exert equal osmotic pressure under identical conditions of temperature. Such solutions which have the same osmotic pressure are termed isotonic or iso-osmotic. When two isotonic solutions are separated by a semipermeable membrane, no flow of solvent molecules is observed on either side.  

The law is similar to Avogadro’s hypothesis. It can be stated as, “Equal volumes of dilute solutions of different solutes, having the same temperature and osmotic pressure, contain equal number of molecules.”  

For solution I,             PV = n1ST  

For solution II,            PV = n2ST  

Thus, n1 must be equal to n2 when P, V and T are same.  

The analogy of dilute solutions with gases is this perfect.  

This led van’t Hoff to suggest that a solute in dissolved state (i.e., in solution) behaves as a gas and the osmotic pressure of the solution is equal to the pressure which the solute would exert if it were a gas at the same temperature and occupying the same volume as that of the solution. 



Question 1: Total number of molecules which combine to form bigger molecule is called

a. Van't hoff factor 

b. degree of dissociation 

c. degree of association

d. dissociation constant

Question 2: According to Boyle-van’t Hoff law

a. πV = constant  

b. i = 1+  = 1+ (2–1)

c. Tf = Kf m

d. πV = nRT 

Question 3: For the reaction nA  (A)n, i = 

a. 1- a (1–1/n)

b. a (1–1/n)

c. 1- (1–1/n)

d. 1- n (1–1/a)

Question 4: For association of molecules in solution

a. i = 0

b. i = 1

c. i > 1

d. i < 1





Q.1

Q.2

Q.3

Q.4


c

a

a

d




collected by:-NKGupta