ROUTERA


Chapter 5 Surface Chemistry

Class 12th Chemistry Chapter Hots


1. Explain the phenomenon of adsorption. Discuss the differences between physical and chemical adsorption with suitable examples.

Answer: Adsorption is the process in which molecules of a substance (adsorbate) adhere to the surface of a solid or liquid (adsorbent) due to physical or chemical forces. This phenomenon is crucial in various fields such as catalysis, environmental science, and material science.

Physical Adsorption (Physisorption):

  • Involves weak van der Waals forces (physical forces) between the adsorbate and the surface of the adsorbent.
  • It is generally reversible.
  • The amount of adsorption increases with an increase in temperature and pressure.
  • The adsorbed layer is usually thin (monolayer).
  • Example: Adsorption of gases like nitrogen on activated charcoal.

Chemical Adsorption (Chemisorption):

  • Involves the formation of chemical bonds between the adsorbate and the surface of the adsorbent.
  • It is usually irreversible and occurs at higher temperatures.
  • The amount of adsorption decreases with an increase in temperature.
  • A monolayer of adsorbate is formed, and the process is often highly specific.
  • Example: Hydrogen gas adsorbed on platinum surface.

Difference:

  • Forces: Physisorption involves weak van der Waals forces, while chemisorption involves stronger covalent or ionic bonds.
  • Reversibility: Physisorption is reversible, whereas chemisorption is irreversible.
  • Temperature Dependence: Physisorption increases with temperature, but chemisorption decreases.

2. Discuss the Langmuir Adsorption Isotherm. Derive the expression for the fractional coverage of the adsorbent.

Answer: The Langmuir Adsorption Isotherm describes adsorption on a surface with a limited number of adsorption sites, where each site can hold only one molecule (monolayer adsorption). It assumes that the adsorbate molecules form a monolayer without interaction between adsorbed molecules.

The equation for the Langmuir Adsorption Isotherm is:

1θ=1K+1KP\frac{1}{\theta} = \frac{1}{K} + \frac{1}{K \cdot P}

Where:

  • θ\theta is the fraction of the surface covered by the adsorbate,
  • KK is the adsorption equilibrium constant,
  • PP is the pressure of the adsorbate gas.

Rearranging:

θ=KP1+KP\theta = \frac{K \cdot P}{1 + K \cdot P}

This expression shows that as the pressure increases, the coverage θ\theta approaches 1, meaning that the surface becomes completely covered with the adsorbate.

Fractional Coverage: Fractional coverage (θ\theta ) is the ratio of the number of sites occupied by adsorbate molecules to the total number of sites available.


3. Explain the Freundlich Adsorption Isotherm and its applicability. Derive the mathematical expression for the isotherm.

Answer: The Freundlich Adsorption Isotherm is an empirical equation that describes adsorption on heterogeneous surfaces. It assumes that the adsorption sites have unequal energies, and adsorption decreases with the increase in concentration.

The Freundlich adsorption isotherm is given by:

x/m=kP1/nx/m = k P^{1/n}

Where:

  • x/mx/m is the amount of adsorbate per unit mass of adsorbent,
  • PP is the pressure of the adsorbate gas,
  • kk and nn are constants that depend on the system.

Taking the logarithm of both sides:

log(xm)=logk+1nlogP\log \left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log P

This is a straight-line equation where:

  • The slope is 1n\frac{1}{n} ,
  • The intercept is logk\log k .

Applicability:

  • The Freundlich isotherm is applicable to systems where the adsorbent has a heterogeneous surface (different types of adsorption sites).
  • It is commonly used when the adsorption capacity does not follow the Langmuir model (i.e., for multilayer adsorption).

4. What are the factors affecting the adsorption of gases on solid surfaces?

Answer: The adsorption of gases on solid surfaces is influenced by several factors:

  1. Nature of the Adsorbent: The surface area, porosity, and surface energy of the adsorbent affect its adsorption capacity. For example, activated charcoal has a high surface area, making it a good adsorbent.

  2. Nature of the Adsorbate: The type of gas being adsorbed plays a role. Polar molecules are more likely to be adsorbed due to stronger dipole interactions with the adsorbent surface.

  3. Pressure: At low pressures, the amount of gas adsorbed increases with pressure. However, at high pressures, the adsorption reaches a saturation point, and the surface becomes completely covered.

  4. Temperature: Adsorption generally decreases with an increase in temperature. This is because adsorption is an exothermic process, and higher temperatures provide more energy to the adsorbate molecules, overcoming the forces binding them to the surface.

  5. Surface Area of Adsorbent: A larger surface area allows for more adsorption sites, leading to higher adsorption capacity.

  6. Size and Shape of Molecules: Larger molecules may be more easily adsorbed because they can fit into smaller pores of the adsorbent.


5. Derive the BET (Brunauer-Emmett-Teller) Adsorption Isotherm for multilayer adsorption.

Answer: The BET adsorption isotherm describes multilayer adsorption, which is particularly important when the surface area of the adsorbent is large enough to allow for multiple layers of adsorbate molecules.

The BET equation is derived by assuming that:

  • Adsorption occurs in multiple layers.
  • The adsorbate molecules in the second and higher layers are adsorbed by the molecules in the first layer, and so on.

The BET equation is given by:

1[V(P0P)]=c1Vmc+PVmc(P0P)\frac{1}{[V(P_0 - P)]} = \frac{c - 1}{V_m c} + \frac{P}{V_m c(P_0 - P)}

Where:

  • VV is the volume of adsorbate adsorbed at pressure PP ,
  • P0P_0 is the saturation pressure of the adsorbate,
  • VmV_m is the volume of gas adsorbed when the surface is covered by a monolayer,
  • cc is the BET constant, related to the enthalpy of adsorption.

This equation is linearized and can be used to calculate the surface area of the adsorbent, typically from a plot of 1V(P0P)\frac{1}{V(P_0 - P)} versus PP0\frac{P}{P_0} .


6. What is the role of adsorption in heterogeneous catalysis? Discuss with examples.

Answer: Heterogeneous catalysis involves a catalyst in a different phase than the reactants, typically a solid catalyst and gaseous or liquid reactants. Adsorption plays a crucial role in the catalytic process, as reactant molecules must first adsorb onto the surface of the catalyst to undergo chemical reactions.

Steps in heterogeneous catalysis:

  1. Adsorption of reactants: Reactant molecules are adsorbed onto the catalyst surface, where they are activated.
  2. Activation and reaction: Adsorbed molecules react with one another, breaking bonds and forming new products.
  3. Desorption of products: Once the products are formed, they desorb from the catalyst surface, freeing the surface for new reactant molecules.

Examples:

  • The Haber process for ammonia synthesis: Nitrogen and hydrogen gases are adsorbed onto the surface of an iron catalyst, where they react to form ammonia.
  • The Catalytic converter in car engines: Nitrogen oxides and carbon monoxide are adsorbed onto platinum, palladium, or rhodium catalysts and undergo reduction to form less harmful gases.

7. Explain the concept of chemisorption with respect to surface chemistry. How does it differ from physisorption?

Answer: Chemisorption refers to the type of adsorption where a chemical bond is formed between the adsorbate (the substance being adsorbed) and the adsorbent (the surface on which adsorption occurs). It is typically more specific and involves stronger forces compared to physisorption.

Characteristics of Chemisorption:

  • Involves the formation of strong covalent or ionic bonds between the adsorbate and the surface.
  • It is usually irreversible and occurs at higher temperatures.
  • It is specific to particular reactants and adsorbents.
  • The adsorbed layer is usually a monolayer, and the rate of chemisorption decreases as the surface becomes covered.

Difference from Physisorption:

  • Forces: Chemisorption involves strong covalent or ionic bonds, while physisorption involves weak van der Waals forces.
  • Reversibility: Chemisorption is often irreversible, while physisorption is usually reversible.
  • Temperature Dependence: Chemisorption decreases with increasing temperature, while physisorption increases with temperature.

8. Discuss the concept of colloids and classify them based on the state of dispersion medium and dispersed phase.

Answer: Colloids are heterogeneous mixtures where the size of the dispersed particles is in the range of 1 to 1000 nm. These particles are dispersed throughout the medium and do not settle over time.

Classification of Colloids:

  • Based on the state of the dispersion medium and dispersed phase:
    1. Solids in liquid: Examples are paints, inks, and colloidal gold.
    2. Liquids in liquid: Examples include emulsions such as milk.
    3. Gases in liquids: Examples are whipped cream and froths.
    4. Solids in solids: Examples include alloys and some types of rubber.
    5. Liquids in solids: Examples include gels like agar-agar and jelly.
    6. Gases in solids: Examples include pumice and sponge.

Colloids exhibit properties such as Brownian motion, Tyndall effect, and electrophoresis, which help in distinguishing them from true solutions.

9. What is the role of surface energy in the formation of colloids? Discuss its significance.

Answer: Surface energy refers to the excess energy at the surface of a material compared to its bulk. It plays a crucial role in the formation of colloids because particles of the dispersed phase must overcome their surface energy to interact with the dispersion medium.

Significance:

  • Formation of colloids: For a colloidal particle to remain suspended in a dispersion medium, the surface energy of the particles must be minimized. Colloidal particles are stabilized by the presence of surfactants or stabilizers that reduce surface energy.
  • Tendency of particles to aggregate: High surface energy encourages particles to aggregate to minimize their surface area, which leads to the formation of larger particles. By reducing surface energy, colloidal stability can be achieved, preventing aggregation.
  • Wettability and interaction: Surface energy determines how particles interact with the dispersion medium. For example, hydrophobic particles exhibit high surface energy when immersed in water, whereas hydrophilic particles have lower surface energy and are better dispersed.

10. Explain the concept of "Gibbs Adsorption Isotherm" and how it relates to surface tension.

Answer: The Gibbs Adsorption Isotherm explains how the adsorption of a substance at the surface of a liquid alters its surface tension. It relates the change in surface tension to the concentration of the adsorbate at the surface. This is particularly useful for understanding the behavior of surfactants and other substances that adsorb at liquid interfaces.

The Gibbs Adsorption Isotherm is given by:

dγdlnC=ΔGRT\frac{d\gamma}{d \ln C} = -\frac{\Delta G}{RT}

Where:

  • γ\gamma is the surface tension,
  • CC is the concentration of the adsorbate in the bulk,
  • RR is the gas constant,
  • TT is the temperature, and
  • ΔG\Delta G is the change in Gibbs free energy.

This equation describes how the surface tension changes with the concentration of the adsorbate. Surfactants, when adsorbed on the surface, decrease the surface tension of water and can be used in applications such as detergents, emulsions, and foams.


11. Explain the term "emulsions". What are the different types of emulsions, and how are they stabilized?

Answer: An emulsion is a colloidal system in which one liquid is dispersed as droplets in another immiscible liquid. Emulsions are commonly used in the food, pharmaceutical, and cosmetic industries.

Types of Emulsions:

  1. Oil in Water (O/W): In this type, oil droplets are dispersed in water. Example: Milk, mayonaisse.
  2. Water in Oil (W/O): Here, water droplets are dispersed in oil. Example: Butter, cold creams.

Stabilization of Emulsions:

  • Surfactants (Emulsifiers): These are molecules with hydrophilic and hydrophobic parts. Surfactants reduce the surface tension between the two immiscible liquids, thus stabilizing the emulsion by preventing the droplets from coalescing. Examples: lecithin, soap, and detergent.
  • Mechanical agitation: This helps break large droplets into smaller ones, creating a more stable emulsion.
  • Change in temperature: Temperature can influence the viscosity of the continuous phase and help in stabilizing emulsions.

12. Explain the phenomenon of "Tyndall effect" and its application in colloidal systems.

Answer: The Tyndall effect refers to the scattering of light by colloidal particles in a dispersion. When a beam of light passes through a colloidal system, the particles in the colloid scatter the light in different directions, making the path of the light visible. This effect helps distinguish colloidal solutions from true solutions, which do not scatter light.

Applications:

  1. Identification of colloids: The Tyndall effect is used to identify colloidal particles. If a beam of light passes through a solution and is scattered, the solution is colloidal.
  2. Characterization of colloidal particles: The intensity of the scattered light is related to the size and concentration of colloidal particles.
  3. Meteorological applications: The Tyndall effect explains why the sky appears blue and why fog scatters light.

13. What are the factors affecting the stability of colloids? Explain the concept of coagulation and flocculation.

Answer: The stability of colloids depends on several factors that prevent the particles from aggregating and settling out of the dispersion medium. These factors include:

  1. Electrostatic repulsion: In many colloids, the particles carry charges that cause them to repel each other, preventing aggregation.
  2. Steric repulsion: The presence of large organic molecules (such as polymers or surfactants) around colloidal particles can create a physical barrier, preventing particles from coming close enough to aggregate.
  3. Solvent quality: The type of solvent and its interactions with the dispersed phase also affect stability.

Coagulation: This refers to the process by which the colloidal particles aggregate and settle out of the dispersion. It can be caused by the addition of electrolytes or other destabilizing agents that neutralize the surface charge of colloidal particles.

Flocculation: This is the process in which colloidal particles aggregate into loose, fluffy clusters (flocs) rather than completely coagulating into a solid mass. Flocculation is a milder process than coagulation and is used in water treatment and other industrial applications.


14. Explain the concept of "electrophoresis" and its application in the study of colloids.

Answer: Electrophoresis is the motion of charged particles under the influence of an electric field. In colloidal systems, electrophoresis is used to study the nature and properties of colloidal particles by observing their movement in an electric field.

Applications:

  • Determining the charge of colloidal particles: By applying an electric field, colloidal particles can move toward either the anode or cathode, depending on their charge. This helps determine whether the particles are positively or negatively charged.
  • Purification of colloidal systems: Electrophoresis can be used to separate particles based on their charge and size.
  • Characterization of colloidal particles: The rate of movement provides information about the size and shape of the particles, as smaller particles move faster than larger ones.

15. What is the importance of the surface area to volume ratio in surface chemistry?

Answer: The surface area to volume ratio is an important factor in surface chemistry, especially for reactions occurring at surfaces, such as adsorption and catalysis. The larger the surface area, the greater the number of active sites available for reactions.

Importance:

  1. Enhanced reactivity: A higher surface area means more molecules can interact with the surface, leading to an increase in reaction rate.
  2. Increased adsorption: Adsorption is a surface phenomenon, so a higher surface area allows more molecules to be adsorbed, which is important in processes like catalysis and purification.
  3. Catalyst efficiency: Catalysts are often more efficient when they have a high surface area (e.g., powdered catalysts like activated carbon or platinum in catalytic converters).

16. What is a catalyst? Explain the role of catalysts in surface chemistry with suitable examples.

Answer: A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. Catalysts work by providing an alternative reaction pathway with a lower activation energy.

Role of Catalysts in Surface Chemistry:

  1. Adsorption of reactants: In heterogeneous catalysis, reactant molecules are adsorbed onto the surface of the catalyst, where they are activated and can undergo reactions.
  2. Formation of intermediate complexes: The adsorbed reactant molecules may form unstable intermediates that decompose to form the products.
  3. Desorption of products: After the reaction, the products are desorbed from the catalyst surface, freeing the active sites for new reactant molecules.

Example:

  • Haber Process: In the synthesis of ammonia, nitrogen and hydrogen gases are adsorbed onto the surface of iron, where they react to form ammonia.
  • Catalytic Converters: In car exhaust systems, platinum, palladium, and rhodium catalysts adsorb harmful gases like carbon monoxide and nitrogen oxides, promoting their conversion into less harmful substances.

17. Explain the process of "dialysis" and its application in the purification of colloids.

Answer: Dialysis is the process of separating small solute particles from colloidal particles by using a semipermeable membrane. The small particles can pass through the membrane, while the larger colloidal particles are retained.

Applications:

  1. Purification of colloids: Dialysis is used to remove impurities or unwanted solutes from colloidal dispersions, such as removing ions or small molecules.
  2. Medical applications: Dialysis is used in kidney dialysis to remove waste products from the blood of patients whose kidneys are not functioning properly.
  3. Desalting of proteins: Dialysis is used in biochemical research to remove salts and small molecules from proteins and other biological macromolecules.

18. Discuss the phenomenon of "adsorption" and the factors affecting it. Explain the Langmuir adsorption isotherm.

Answer: Adsorption is the process by which molecules from a gas or liquid phase adhere to the surface of a solid or liquid. This phenomenon plays a significant role in surface chemistry, especially in processes like catalysis, chromatography, and purification.

Factors Affecting Adsorption:

  1. Nature of adsorbent: The adsorption capacity depends on the surface area and chemical properties of the adsorbent.
  2. Nature of adsorbate: The size, polarity, and chemical nature of the molecules being adsorbed influence the extent of adsorption.
  3. Temperature: Adsorption is generally exothermic, and an increase in temperature typically decreases the extent of adsorption.
  4. Pressure: For gases, the adsorption increases with increasing pressure as more molecules are available to interact with the surface.

Langmuir Adsorption Isotherm: The Langmuir adsorption isotherm describes the adsorption of molecules onto a solid surface. It assumes that:

  • Adsorption occurs at specific sites on the surface.
  • Each site can hold only one molecule (monolayer adsorption).
  • The rate of adsorption is proportional to the number of available sites.

The Langmuir equation is:

1θ=1K+1bC\frac{1}{\theta} = \frac{1}{K} + \frac{1}{bC}

Where:

  • θ\theta is the fraction of the surface occupied by adsorbate molecules,
  • KK is the adsorption constant,
  • CC is the concentration of the adsorbate, and
  • bb is a constant related to adsorption.

19. What are the differences between physisorption and chemisorption? Explain with examples.

Answer: Physisorption and chemisorption are two types of adsorption processes that differ in the nature of the forces involved and the strength of adsorption.

Physisorption:

  • Occurs due to van der Waals forces (weak forces).
  • It is a reversible process.
  • It usually occurs at low temperatures and is independent of surface area.
  • Example: Adsorption of gases like nitrogen, oxygen, and carbon dioxide on charcoal.

Chemisorption:

  • Occurs due to the formation of chemical bonds (strong bonds).
  • It is an irreversible process.
  • It generally occurs at higher temperatures.
  • Example: Adsorption of hydrogen on a metal surface during hydrogenation reactions.

20. What is the significance of the surface area in catalysis?

Answer: In catalysis, the surface area plays a crucial role because it determines the number of active sites available for the reactants to adsorb and undergo the reaction. Catalysts with larger surface areas are more efficient because they provide more active sites for the reactants to interact with.

Significance:

  • Enhanced reaction rate: A larger surface area increases the number of reactant molecules that can interact with the catalyst, thus speeding up the reaction.
  • Increased efficiency: Catalysts with large surface areas (such as powdered catalysts) are more efficient in industrial processes like the Haber process for ammonia synthesis.

21. How do colloids behave differently from true solutions and suspensions?

Answer: Colloids, true solutions, and suspensions are all heterogeneous mixtures, but they differ in their particle size, stability, and behavior.

Colloids:

  • Particle size: 1-1000 nm.
  • The particles do not settle over time.
  • They exhibit the Tyndall effect (scattering of light).

True solutions:

  • Particle size: <1 nm (molecular level).
  • The particles do not scatter light.
  • They are stable and do not settle.

Suspensions:

  • Particle size: >1000 nm.
  • The particles settle over time due to gravity.
  • They do not exhibit the Tyndall effect.

22. Explain the role of adsorption in the process of catalysis.

Answer: In catalysis, adsorption plays a crucial role because it brings the reactants close to the catalytic surface, thereby increasing the rate of the reaction. Adsorption helps in the following ways:

  1. Activation of reactants: Adsorption of reactant molecules on the catalyst surface weakens the bonds within the molecules, making them easier to react.
  2. Formation of intermediate complexes: Adsorbed reactants can form unstable intermediate complexes on the catalyst surface, which decompose to form the desired products.
  3. Increasing collision frequency: Adsorption increases the effective concentration of reactants on the catalyst surface, increasing the frequency of collisions between molecules.

23. What are surfactants, and how do they stabilize colloidal systems?

Answer: Surfactants are molecules that have both hydrophilic (water-attracting) and hydrophobic (water-repelling) parts. These molecules adsorb at the surface of a liquid, such as water, and reduce surface tension.

Stabilization of Colloidal Systems:

  • Surfactants help stabilize colloidal systems by reducing the surface tension and preventing aggregation of colloidal particles.
  • They form a barrier around the colloidal particles, preventing them from coalescing and thereby stabilizing the colloid.

24. What is the role of electrolytes in the coagulation of colloids?

Answer: Electrolytes play a significant role in coagulating colloidal systems by neutralizing the charges on the colloidal particles, leading to their aggregation.

  • Coagulation occurs when the electrostatic repulsion between particles is overcome by the addition of ions.
  • The critical coagulation concentration (CCC) is the concentration of electrolyte required to neutralize the charge and cause the colloidal particles to aggregate.
  • For example, adding sodium chloride to a negatively charged colloidal solution reduces the repulsive forces, causing the particles to aggregate and settle out.

25. Explain the difference between "sol" and "gel" in colloidal systems.

Answer: In colloidal systems, a sol is a colloidal dispersion of solid particles in a liquid, while a gel is a semi-solid system where the dispersed phase forms a network within the dispersion medium.

Sol:

  • It is a fluid-like colloidal system.
  • The dispersed phase is solid particles suspended in a liquid.
  • Example: Paint, blood.

Gel:

  • It is a more rigid or semi-solid colloidal system.
  • The dispersed phase forms a network structure within the liquid.
  • Example: Jelly, agar-agar.