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Kaysons Educations Chemical Kinetic
Page 1
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
Kaysons Educations Chemical Kinetic
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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
Kaysons Educations Chemical Kinetic
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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 kabc
2
= 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
Kaysons Education Chemical Equilibrium
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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.
Kaysons Education Chemical Equilibrium
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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
Kaysons Education Chemical Equilibrium
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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.
Kaysons Education Chemical Equilibrium
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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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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
Kaysons Education Ionic Equilibrium
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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
Kaysons Education Ionic Equilibrium
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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–
Kaysons Education Ionic Equilibrium
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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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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
Kaysons Education Electrolysis and Electrolytic Conductance
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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:-
Kaysons Education Electrolysis and Electrolytic Conductance
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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.
Kaysons Education Electrolysis and Electrolytic Conductance
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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,
Kaysons Education Electrolysis and Electrolytic Conductance
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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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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
Kaysons Education Ideal Solutions
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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
Kaysons Education Ideal Solutions
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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.
Kaysons Education Ideal Solutions
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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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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
Kaysons Education Surface Chemistry
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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:
Kaysons Education Surface Chemistry
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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.
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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.