Chapter Chemical Kinetic - Kaysons Education · Consider a binary electrolyte AB which dissociates...

Kaysons Educations Chemical Kinetic

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Day – 1

Chemical Kinetics

For any hypothetical reaction,

A + B ⇋ C + D

There are following three main aspects related to the study of reactions:

(i):- Whether the reaction at all takes place or not, i.e., spontaneity of a reaction (as discussed in

terms of Gibbs energy change in thermodynamic).

(ii):- If it spontaneous, then to what extent the reaction takes place before equilibrium is attained

(as studied under Chemical Equilibrium).

(iii):- Rates of speeds of reactions, which the branch of thermodynamics is unable to answer.

The branch of chemistry which deals with the study of the speeds or the rates of chemical

reactions, the factors affecting the rates of the reactions and the mechanism by which the reactions

proceed is known as Chemical Kinetics.

(i):- For prediction of reaction rates and knowing the reasons & factors affecting the rate.

(ii):- Prediction of mechanism of reactions

Classification of reactions

(i):- Very fast reactions, i.e., which take place instantaneously (time taken is generally 10–12

to

10–16

second), e.g., ionic reactions like

(ii):- Very slow reactions, i.e., which may take days or months, e.g., rusting of iron.

(iii):- Reactions which are neither very slow nor very fast but take place at moderate speeds.

Chemical Kinetic Chapter

1


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Rate of Reaction

For a reaction,

R → P

(i):- Average

(ii):- Instantaneous

Rate of Reaction

In general, for the reaction

The rate can be expressed in any one of the following ways:

For example,

Rate of decomposition (disappearance) of N2O5

Similarly, for the reaction,

Rate of reaction


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Illustration

For the decomposition of dinitrogen pentoxide at 200oC,

If the initial pressure is 114 mm and after 25 minutes of the reaction, total pressure of the gaseous

mixture is 133 mm, calculate the average rate of reaction in atm min–1

.

Solution

(a):- If p is the decrease in pressure of N2O5 in 25 minutes, then

Initial pressure 114 mm 0 0

After 25 min 114 – p p

i.e., Decrease in pressure in 25 min = 38 mm

= 0.002 atm min–1

.

Law of Mass Action (Guldberg and Waage, 1864)

“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.”

For any reaction,

Rate Law Expression

k = Rate Constant = Velocity constant.

α:- Order of reaction with respect to A.

β:- Order of reaction with respect to B.

Rate Constant or Velocity Constant


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The rate constant or velocity constant of a reaction is equal to rate of the reaction when

concentration of each of the reactants is unity.

Order of Reaction

The sum of the exponents (powers) to which the concentration terms in the rate law equation are

raised to express the observed rate of the reaction is called order of the reaction.

For Example:-

For the reaction,

aA + bB → cC + dD

R = k[A]a [B]

b

Order of Reaction = a + b.

Illustration

The rate for a reaction is found to be

How would the rate of reaction change when

(i):- Concentration of H+ is doubled.

(ii):- Concentration of I– is halved.

(iii):- Concentration of each of are tripled?

Solution

Suppose initially the concentrations are

And [H+] = c mol L

–1

∴ Rate = kabc2

(i):- New [H+] = 2c

∴ New Rate = k ab (2c)2 = 4 kabc2

= 4 times.

(ii):-

i.e., Rate of reaction is halved.

New rate = k (3a) (3b) (3c)2

= 81 k abc2

= 81 times.


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Integration of rate Expressions

1. Zero – order Reaction

A → P

At time t = 0, [A]0 0

At time t = t, [A] [P]

Where k0 is the rate constant. Rearranging,

If at t = 0, the initial concentration is [A]0 and the concentration at t = t, is [A], then, integration

yields

This is called the integrated rate equation for zero – order reaction.

Half life Period

Half – life period (t1/2) is the time in which half of the substance has reacted. This implies that

When [A] = [A]0/2, t = t1/2. Substituting these value in equation, we get

Thus, half life period of a zero order reaction is directly proportional to initial concentration, i.e.,

t1/2 ∝ [A]0. Hence, a plot of t1/2 versus [A]0 will be a straight line passing through the origin and

slope = 1/2k

Kaysons Education Chemical Equilibrium

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Day – 1

Chemical Equilibrium

Type of Reaction

1. Irreversible Reaction

The chemical reactions which move in one direction, i.e., forward direction only are called

irreversible reactions.

For example

Thermal decomposition of potassium chlorate,

2. Reversible Reactions

The chemical reactions which take place in both directions under similar conditions are called

reversible reactions.

For example

Chemical Equilibrium Chapter

2


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Chemical Equilibrium

At equilibrium,

Reaction does not go to completion.

Reaction appears to have stopped.

Equilibrium can be attained from both sides.

△G = 0

Heterogeneous Reaction

The reversible reaction in which one or more than one phase are present.

For example:-

Homogeneous Reaction

The reversible reaction in which only one phase is present.


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For example:-

Physical Equilibrium

If the opposing processes involve only physical changes, the equilibrium called physical

equilibrium.

1. Solid Liquid Equilibrium

Rate of meting of ice = Rate of freezing of water.

The equilibrium is represented as

2. Liquid Vapour Equilibrium

Rate of evaporation = Rate of condensation.

The equilibrium is represented

3. Solid Vapour Equilibrium

4. Solid Solution Equilibrium

In this case, at equilibrium,

5. Gas Solution Equilibrium

Henry’s Law

The mass of a gas dissolved in a given mass of solvent at any temperature is directly proportional

to the pressure of the gas above the solvent.

Nature of gas

Nature of liquid

Temperature


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Self Efforts

Assertion and Reason

1. Assertion:- For a reaction at equilibrium, the free energy for the reaction is minimum.

Reason:- The free energy for both reactants and products decreases and become equal.

(a) Both Assertion and Reason are true and Reason is correct explanation of Assertion.

(b) Both Assertion and Reason are true and Reason is not correct explanation of Assertion.

(c) Assertion is true, Reason is false.

(d) Reason is true, Assertion is false.

2. Assertion:- Chemical equilibrium represents a state of a reversible reaction in which measurable

properties of the system (pressure, concentration, colour etc) become constant under the given set of

conditions.

Reason:- The chemical equilibrium is an apparent state of rest in which two opposing reactions are

proceeding at the same rate.

(a) Both Assertion and Reason are true and Reason is correct explanation of Assertion.

(b) Both Assertion and Reason are true and Reason is not correct explanation of Assertion.

(c) Assertion is true, Reason is false.

(d) Reason is true, Assertion is false.

Passage

Melting of ice is an endothermic process and take place with decrease in volume because density of ice is

less that of water. Therefore at constant volume there is decrease in pressure. Thus increasing of pressure

on ice ⇌ water system at a constant temperature will shift the equilibrium towards right i.e., it will cause

the ice to melt. Hence in order to retain ice in equilibrium with water at the higher pressure will lower the

melting point or freezing point of ice.

3. The boiling of water at 60oC is

(a) Spontaneous reaction

(b) Non-spontaneous reaction

(c) Chemical reaction

(d) None of these

4. A liquid is in equilibrium with its vapour in a sealed container at constant temperature. Which of the

following is correct for the above equilibrium?

(a) The vapour pressure will fall due to sudden increase in volume.


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(b) The valour pressure remains constant.

(c) The vapour pressure will decrease with decreasing the volume.

(d) All option are correct.

5. The yield of water is maximum at

(a) High temperature and high pressure.

(b) Low temperature and low pressure.

(c) High temperature and low pressure.

(d) Low temperature and high pressure.

Answer

1. Both Assertion and Reason are true and Reason is correct explanation of Assertion.

2. Both Assertion and Reason are true and Reason is correct explanation of Assertion.

3. Non-spontaneous reaction. 4. The vapour pressure will fall due to sudden increase in

volume.

5. High temperature and high pressure.

Kaysons Education Ionic Equilibrium

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Day – 1

Electrolytes

The compounds which give ions either in molten state or in solution are called electrolytes.

(i) Strong electrolytes

These electrolytes are almost completely ionized when dissolved in a polar medium like water.

(ii) Weak electrolytes

These are not completely ionized when dissolved in a polar solvent and they behave as poor

conductors of electricity.

For example:-

CH3COOH,H3PO4, H3BO3,

Degree of ionization ‘α’ may be defined as a fraction of total number of molecules of an

electrolyte which dissociate into ions.

Ionic Equilibrium Chapter

3


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Ostwald’s Dilution Law

Consider a binary electrolyte AB which dissociates into A+ and B

– ions and the equilibrium state is

represented by the equation.

Initially t = 0 C 0 0

At equilibrium C(1- α) Cα Cα

So, dissociation constant may be given as:

For very weak electrolytes,

α < < < 1, (1 – α) ≈ 1

∴ K = Cα2,

From equation (ii), it is clear that degree of ionization increases on dilution.

Thus, degree of dissociation of a weak electrolyte is proportional to the square root of dilution.

Common Ion Effect

Let AB be the weak electrolyte. Considering its dissociation,

And applying law of mass action, we have

The degree of dissociation of an electrolyte (weak) is suppressed by the addition of another

electrolyte (strong) containing a common ion. This is termed as common ion effect.

Illustration

A 0.001M solution of acetic acid is 1.34% ionized (degree of dissociation = 0.0134) at 289K.

What is the ionization constant of acetic acid?

Solution

Initial conc. C mol L–1

0 0

At equilibrium C(1- α) Cα Cα


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Illustration

A solution contains 0.1M H2S and 0.3M HCl. Calculate the concentration of S2–

and HS–

ions in

solution. Given Ka1 and Ka2for H2S are 10–7

and 1.3 × 10–13

respectively.

Solution

Further HS– = H

+ + S

2–

Multiplying both equations

Due to common ion, the ionisation of H2S is suppressed and the [H+] in solution is due to the

presence of 0.3M HCl.

Putting the value of [S2–

] in equation (ii),

Classical Concept of Acids and Bases

Acid

Acid is a substance which in aqueous solution have the following property.

Has a sour test.

React with active metal to give H2.

Conduct electricity.

Turns blue litmus red.

React with base and loose its property.

Base

Base is a substance which in aqueous solution gives the following property.

Has a bitter test.

Have soapy touch.

Conduct electricity.

Turns red litmus blue.

React with acids and loose its property.

Arrhenius Concept of Acids and Base

An acid is defined as a substance which contains hydrogen and which when dissolved into water

gives hydrogen ions (H+).


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Such acids are called strong acids.

Such acids are called weak acids.

A base is defined as a substance which contains hydroxyl groups and which when dissolved in

water gives hydroxide ions (OH–).

These are called strong bases.

Substances like NH4OH, Ca(OH)2, Mg(OH)2, Al(OH)3 etc. Dissociate to a small extent as follows:

These are called weak bases.

Arrhenius described neutralization as the process in which hydrogen ions and hydroxide ions

combine to from unionized molecules of water.

Nature of H+ ions and OH–


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Limitations

Inability to explain acidic and basic character of certain substances.

Inability to explain the reaction between an acid and base in absence of water.

Bronsted-Lowry Concept of Acid and Base.

An acid is defined as a substance which has the tendency to give a proton (H+).

A base is defined as a substance which has tendency to accept a proton.

In other words, an acid is a proton donor whereas a base is a proton acceptor.

Water acts both as an acid as well as a base and hence is called amphoteric or amphiprotic.

Kaysons Education Electrolysis and Electrolytic Conductance

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Day – 1

Electrical Conductors

Metallic Electrolytic

(NaCl solution)

(i):- Metallic

(ii):- Electrolytic

Distinction between metallic and electrolytic conduction

Metallic Conduction Electrolytic conduction

1. Electric current flows by movement of

electrons.

Electric current flows by the movement of ions.

2. No chemical change occurs. Ions are oxidized or reduced at the electrodes.

3. It does not involve the transfer of any matter. It involves transfer of matter in the form of ions.

4. Ohm’s law followed. Ohm’s law followed.

5. Resistance increases with increase of

temperature.

Resistance decreases with increase of temperature.

6. Faraday’s law is not followed. Faraday’s law is not followed.

Electrolysis and

Electrolytic Conductance

Chapter

4


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Electrolysis

The process of chemical decomposition of an electrolyte by passage of electric current through its

solution is called electrolysis

Examples:-

(i):- Electrolysis of molten sodium chloride.

At cathode:-

At anode:-

Overall reaction:-

(ii):- Electrolysis of acidulated water.

At cathode:

(Reduction, Primary change)

(Secondary change)

At anode:

(Reduction, Primary change)

(Secondary change)

Over all reaction:

At cathode At anode

(iii):- Electrolysis of an aqueous solution of sodium chloride.

(almost completely ionized)

(only slightly ionized)

At cathode:-


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At anode:-

Faraday’s Laws

(1) Faraday’s First Law of Electrolysis

The amount of chemical reactions and hence the mass of any substance deposited or liberated at

any electrode id directly proportional to the quantity of the electricity passed through the

electrolyte (solution or melt). If W gram of the substance is deposited on passing Q columns of

electricity, then

Where Z is constant of proportionality and is called electrochemical equivalent of the substance

deposited.

Electrochemical equivalent

Electrochemical equivalent of a substance may be defined as the mass of the substance deposited

when a current of one ampere is passed for one second, i.e., a quantity of electricity equal to one

coulomb is passed.

(2) Faraday’s Second Law of Electrolysis

When the same quantity of electricity is passed through solutions of different electrolytes

`connected in series, the weights of the substances produced at the electrodes are directly

proportional to their equivalent weights.

Quantitative aspects of Electrolysis

One electron produces one sodium atom.

Two moles of electrons produce one mole of Cl2

This quantity of

electricity is called one faraday.

One faradays (i.e, 96500 coulombs) of electricity deposits one gram equivalent of the substance.


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Transport number or Transference number

The current flowing through an electrolytic solution is carried by the ions. The fraction of the

current carried by an ion is called its transport number or transference number.

Transport number of cation,

Transport number of anion,

Evidently,

Illustration

An electric current of 100 ampere is passed through a molten liquid of sodium chloride for 5

hours. Calculate the volume of chlorine gas liberated at the electrode at NTP.

Solution

The reaction taking place at anode is:

1 mole

The amount of chlorine liberated by passing 100 × 5 × 60 × 60 coulomb of electric charge

Electroplating

The process of coating an inferior metal with a superior metal by electrolysis is known as

electroplating.

Illustration

How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a

metal surface of 80 cm2 with 0.005 mm thick layer? Density of silver is 10.5 g/cm

3.

Solution

Mass of silver to be deposited = volume × density

= Area × thickness × density

Given: Area = 80 cm2 , thickness = 0.0005 cm and density = 10.5 g/ cm

3

Mass of silver to be deposited = 80 × 0.0005 × 10.5 = 0.42 g

Applying to silver E = Z × 96500

Let the current be passed for t seconds.

We know that,

W = Z× I × t

So,


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or

Illustration

A current of 1.70 ampere is passed through 300mL of 0.160 M solution of zinc sulphate for 230

second with a current efficiency 0f 90 per cent. Find out the molarity of Zn2+

ions after the

deposition of zinc. Assume the volume of the solution to remains constant during electrolysis.

Solution

Amount of actual charge passed

No. of moles of Zn deposited by passing 351.9 coulomb of the charge

Molarity of Zn2+

ions after deposition of zinc

Kaysons Education Ideal Solutions

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Day – 1

Ideal Solution

A homogeneous mixture whose composition can be varied within certain limits is termed a true

solution.

Constituents

Solution can not be separated by filtration, settling or centrifugal action.

Binary Solution

When the solution is composed of only two chemical substances, it is termed a binary solution.

Types of Solution

S.No. Solute Solvent Types of Solution Examples

SOLID SOLUTIONS (Solid Solvent)

1.

2.

3.

Solid

Liquid

Gas

Solid

Solid

Solid

Solid in Solid

Liquid in Solid

Gas in Solid

Alloys (brass, German silver, bronze,

22 caeat gold etc.).

Hydrated salts, Amalgam of Hg with

Na.

Dissolved gases in minerals or H2 in

Pd.

LIQUID SOLUTIONS (Liquid Solvent)

4.

5.

Solid

Liquid

Liquid

Liquid

Solid in Liquid

Liquid in Liquid

Salt or glucose or sugar or urea

solution in water.

Methanol or ethanol in water.

Ideal Solutions Chapter

5


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6.

Gas

Liquid

Gas in Liquid

Aerated drinks, O2 water.

GASEOUS SOLUTIONS (Gaseous Solvent)

7.

8.

9.

Solid

Liquid

Gas

Gas

Gas

Gas

Solid in Gas

Liquid in Gas

Gas in Gas

Iodine vapurs in air, camphor in N2

gas.

Humidity in air, chloroform mixed

with N2 gas.

Air (O2 + N2).

Out of the various types of solutions listed in the table, the most significant types of solutions are

those which are in liquid phase, i.e., liquid solutions. Hence, we shall confine ourselves to the

study of solutions of solids, liquids or gases in liquids.

Concentration Units

The concentration of a solute is the amount of solute dissolved in a given quantity of solvent or

solution. The quantity of solvent or solution can be expressed in terms of volume or in terms of

mass or molar mass. Thus there are several ways of expressing the concentration of a solution.

(a) Molarity (M):- Moles of solute present in one litre solution.

(b) Molality (m):- Moles of solute present in one kilogram of solvent.

(c) Normality (N):- Number of equivalents present in one litre solution.

(d) Mole fraction:- The mole fraction of a component substance A(XA) in a solution is defined as the

moles of component substance divided by the total moles of solutions.

Mass percent:- The mass percent of a component A in solution is defined as


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Part per million (PPM):- It is defined as the parts of given component in one million parts of solution.

Mathematically.

Concentration in gram per litre is also termed as strength of the solution. Let w g of the solute be

present in V litre of solution, then

Methods of Expressing the Concentration of a Solution

(i) Mass percentage of per cent by mass

It is defined as the amount of solute in gram present in 100 gram of the solution.

Thus,

Mass percentage of solute = Mass fraction × 100

10 % solution of sugar means that 10 gram of sugar is present in 100 gram of the solution, i.e., 10

gram of sugar has been dissolved in 90 gram of water.

(ii) Per cent by volume

It is defined as the volume of solute in mL present in 100 mL solution.

(iii) Per cent mass by volume

It is defined as the mass of solute present in 100 mL of solution.

(iv) Strength or concentration (Gram per litre)

It is defined as the amount of the solute in gram present in one litre of the solution.

Concentration in gram per litre is also termed as strength of the solution. Let w g of the solute be

present inV litre of solution, then

Note:- V is not the volume of the solvent. V is actually the final volume after dissolving a definite quantity

of solute in the solvent.


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Illustration

If 20 ml of ethanol (density = 0.7893 gm/ml) is mixed with 40 ml of water (density = 0.9971

gm/ml) at 25oC, the final solution has density of 0.9571 gm/ml. Calculate the percentage change in

total volume of mixing. Also calculate the molality of alcohol in the final solution.

Solution

Mass of ethanol = v × d

W = 20 × 0.7893 gm = 15.786 gm

Mass of water = v × d

= 40 × 0.9971 gm = 39.884 gm

Total volume = 60 ml

Total mass = 15.786 + 39.884 = 55.67 gm

Let the volume of solution = x ml

Change in volume = 60 – 58.165 = 1.865 ml

= 3.05 %.

= 8.604 m.

Illustration

What would be the molality of a solution made by mixing equal volumes of 30.0% by mass of

H2SO4 (density 1.218g cm–3

) and 70% by mass of H2SO4 (density 1.610g cm–3

)?

Solution

Let 100 mL of one solution be mixed with 100 mL of the other solution.

Mass of 100 mL of 30% H2SO4 = 100 × 1.218 = 121.8g

Mass of H2SO4 in 121.8g of 30% H2SO4

Mass of water = (121.8 – 36.54) = 85.26g

Mass of 100 mL of 70% H2SO4 = 100 × 1.61 = 161.0g

Mass of H2SO4 in 161.0g of 70% H2SO4

Mass of water = (161.0 – 112) = 48.30g

Total H2SO4 in solution = 36.54 + 112.7 = 149.24g

Total mass of water in solution = (85.26 + 48.30)

= 11.4 m.


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Illustration

Calculate the mole fraction of ethylene glycol (C2H6O2) and water in a solution containing 20% of

C2H6O2 by mass.

Solution

20% of C2H6O2 by mass means that 20g of C2H6O2 are present in 100g of the solution, i.e.,

Mass of solute (C2H6O2) = 20 g

Mass of solvent (H2O) = 100 – 20g = 80 g

Molar mass of C2H6O2 = 62 g mol–1

Molar mass of H2O = 18g mol–1

∴ Mole fraction of C2H6O2 in the solution

Mole fraction of H2O in the solution

= 1 – 0.068 = 0.932.

Illustration

What volume of 95% sulphuric acid (density = 1.85 g/cm3) and what mass of water must be taken

to prepare 100 cm3 of 15% solution of sulphuric acid (density = 1.10g/cm

3)?

Solution

Step – I:-

Calculation of H2SO4 present in 100cm3 of 15% H2SO4 (d = 1.10 g/cm

3).

15% H2SO4 means that 15g H2SO4 are present in 100g of the solution

= 90.91 cm3

90.91 cm3 of solution contain 15g of H2SO4

∴ 100 cm3 of solution will contain H2SO4

Step – II:-

Calculation of volume of 95% H2SO4 (d = 1.85 g/cm3) containing 16.5 of H2SO4.

95% H2SO4 means that 95 g of H2SO4 are present in 100g of the solution.

But volume of 100 g of this solution

Thus, 95g H2SO4 are present in 54.05 cm3

∴ 16.5g H2SO4 will be present in

Thus, 95g H2SO4 to be taken = 9.4 cm3

Kaysons Education Surface Chemistry

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Day – 1

Surface Chemistry

Surface chemistry is that branch of chemistry which deals with the study of the phenomena

occurring at the surface of interface, i.e., at the boundary separating two bulk phases.

Adsorption is due to the fact that the surface particles of the adsorbent are in different sate than the

particles inside the bulk. Inside the bulk all the forces acting between the particles are mutually

balanced but on the surface the particles are not surrounded by atoms or molecules of their kind

on all sides and hence they possess unbalanced or residual attracting the adsorbate particles on its

surface.

Terms

The phenomenon of attracting and retaining the molecules of a substance on the surface of a liquid

or a solid resulting into a higher concentration of the molecules on the surface is called

adsorption. The substance thus adsorbed on the surface is called the adsorbate and the substance

on which it is adsorded is called adsorbent. The reverse process, i.e., removal of the adsorbed

substance from the surface is called desorption (which can be brought about by heating or

reducing the pressure). The adsorption of gases on the surface of metals is called occlusion.

Free Energy Change During Adsorption

Adsorption-an Exothermic Process

When adsorption takes place, the residual forces on the surface of the adsorbent decrease. In other

words, surface energy decreases. This appears in the form of heat called heat of adsorption. Hence,

adsorption is invariably an exothermic process, i.e., ∆Hadsorption is always negative.

Entropy Change During Adsorption and Adsorption Equilibrium

Surface Chemistry Chapter

6


Page 187

As the molecules of the adsorbate are held on the surface of the solid adsorbent, entropy decreases,

i.e., ∆S is also negative. As ∆G = ∆H – T∆S, hence for the process of adsorption to occur, ∆G

must be negative. As ∆S is negative, ∆G can be negative only if ∆H is negative and ∆H > T∆S in

magnitude. This is true in the beginning. However, as the adsorption proceeds, ∆H keeps on

decreasing and T∆S keeps on increasing till ultimately ∆H becomes equal to T∆S so that ∆G = 0.

This state is called adsorption equilibrium.

In a number of cases, both adsorption and adsorption take place simultaneously. For such

processes, simply the term sorption is used.

Adsorption Absorption

1. It is a surface phenomenon, i.e., it occurs

only at the surface of the adsorbent.

1. It is a bulk phenomenon, ie., occurs

throughout the body of the material.

2. In this phenomenon, the concentration on

the surface of adsorbent is different from that

in the bulk.

2. In this phenomenon, the concentration is

same throughout the material.

3. Its rate is high in the beginning and then

decreases till equilibrium is attained.

3. Its rate remains same throughout the

process.

When the concentration of the adsorbate is more on the surface of the adsorbent than in the bulk,

it called positive adsorption. If the concentration of the adsorbate increases in the bulk after

adsorption, it is called negative adsorption.

Factors Affecting Adsorption of Gases by Solids

Almost all solids absorb gases to some extent. However, the exact amount of a gas absorbed

depends upon a number of factors, as briefly explained below:

(i) Nature and Surface area of the Adsorbent

It is observed that the same gas is absorbed to different extents by different solids at the same

temperature. Further as may be expected the greater the surface area of the adsorbent, greater is

the volume of the gas adsorbed. It is for this reason that substance like charcoal and silica gel are

excellent adsorbents because they have highly porous structures and hence large surface areas.

The surface area per gram of the adsorbent is called specific surface area.

(ii) Nature of the gas being adsorbed

A gas which is more easily liquefiable or is more soluble in water is more readily adsorbed.

(iii) Temperature

Studying the adsorption of any particular gas by some particular adsorbent, it is observed that the

adsorption decreases with increase of temperature and vice versa. For example, one gram of

charcoal absorbs about 10 mL of N2 at 273K, 20 mK at 244K and 45mL at 195K. The decrease of

absorption with increase of temperature may be explained as follows:


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Like any other equilibrium, adsorption is a process involving a true equilibrium. The two opposing

processes involved are condensation (i.e., adsorption) of the gas molecules on the surface of the

soild and evaporation (i.e., desorption) of condensation (or adsorpation) is exothermic so that the

equilibrium may be represented as:

Applying Le Chatelier‟s principle, it can be seen that increase of temperature decreases the

adsorption and vice versa.

The amount of heat evolved when one mole of the gas is absorbed on the adsorbent is called the

heat of adsorption.

(iv) Pressure

At constant temperature, the absorption of a gas increases with increase of pressure. It is observed

that at low temperature, the absorption of a gas increases very rapidly as the pressure is increased

from small values.

(v) Activation of the solid adsorbent

It means increasing the adsorbing power of an adsorbent. This is usually done by increasing the

surface area (or the specific area) of the adsorbent which can be achieved in any of the following

ways:

(a):- By making the surface of the adsorbent rough.

(b):- By subdividing the adsorbent into smaller pieces or grains.

(c):- By removing the gases already adsorbed, e.g., charcoal is activated by heating in superheated

steam or in vacuum at a temperature between 623 to 1273K.

Types of Adsorption

When a gas is held (adsorbed) on the surface of a solid by van der Waals forces (which are weak

intermolecular forces of attraction) without resulting into the formation of any chemical bond


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between the adsorbate and the adsorbent, it is called „physical adsorption‟ or „van der Walls

adsorption‟ or „physisorption‟.

When a gas is held on to the surface of a solid by forces similar to those of a chemical bond

(which may be covalent or ionic in nature), the type of adsorption is called chemical adsorption or

chemisorptions. This type of adsorption results into the formation of what is called a „surface

compound‟.

Distinction between Physical Adsorption and Chemisorption

Physical Adsorption Chemisorption

1. The forces operating in these cases are weak

van der Waals forces.

1. The forces operating in these cases are

similar to those of a chemical bond.

2. The heats of adsorption are low, viz., about

20-40 kJ mol–1

.

2. The heats of adsorption are high, viz., about

40-400kJ mol–1

.

3. No compound formation takes place in these

cases.

3. Surface compounds are formed.

4. The process is reversible, i.e., desorption of

the gas occurs by increasing the temperature or

decreasing the pressure.

4. The process is irreversible. Efforts to free

the absorbed gas give some definite

compound.

5. It does not require any a activation energy. 5. It requires activation energy.

6. This type of adsorption usually takes place at

low temperature and decreases with increase of

temperature.

6. This type of adsorption first increases with

increase of temperature. The effect is called

activated adsorption.

7. It is not specific in nature, i.e., all gases are

adsorbed on all solids to some extent.

8. It is specific in nature and occurs only when

there is some possibility of compound

formation between the gas being absorbed and

the solid absorbent.

8. The amount of the gas adsorbed is related to

the ease of liquefaction of the gas.

8. There is no such correlation.

9. It forms multimolecular layer. 9. It forms unimolecular layer.

10. It increases with increase in the surface area

of the adsorbent.

10. It also increases with increase in the

surface area of the adsorbent.

11. It increases with of pressure. 11. It also increases with increase of pressure.

Mechanism of Chemisorption

Taking the example of adsorption of H2 on the surface of platinum, as shown in the figure below,

hydrogen molecules are first attracted towards the surface by weak van der Waals forces and

hence absorbed on it due to the presence of unbalanced attractive forces or free valencies on the

surface of the solid (physical adsorption). The adsorbed molecules then dissociate into atoms

which are then chemisorbed and hence held strongly.


Page 190

When the solid adsorbent is broken into pieces, the number of unbalanced forces (free valencies)

increases and hence adsorption increases. For example,

Adsorption of nitrogen on Iron

This is an example where the gas is physiosorbed at low temperature and chemisorbed at high

temperature. At 83K, nitrogen is physiosorbed on iron surface as N2 which decreases with increase

of temperature and at room temperature there is almost no adsorption. At 773K and above, there is

chemisorptions of nitrogen as N atoms.

Freundlich Adsorption Isotherm

When the adsorbent and the absorbate are enclosed in a closed vessel, after an initial decrease in

the pressure of the gas, gas pressure as well as the amount of gas absorbed reach constant values.

This is because after the equilibrium is attained, rate of adsorption becomes equal to the rate of

desorption.

The extent of adsorption is usually expressed as x/m, where m is the mass of the adsorbent and x

is the mass of the adsorbate when adsorption equilibrium is reached.

A graph between the amount of the gas adsorbed per gram of the adsorbent (x/m) and the

equilibrium pressure of the adsorbate at constant temperature is called the adsorption isotherm.

At s value of Ps of equilibrium pressure, x/m reaches its maximum value and then it remains

constant even though the pressure P is increased. This is the saturation state and Ps is the saturation

pressure.

Chemical Kinetic Ch-01.pdfChemical Equlibrium Ch-02.pdfIonic Equilibrium Ch-03.pdfElectrolysis and Electrolytic Conductance Ch-04.pdfIdeal Solution Ch-05.pdfSurface Chemistry Ch-06.pdf

Chapter Chemical Kinetic - Kaysons Education · Consider a binary electrolyte AB which dissociates...

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