GSEB Class 12 Chemistry Solutions Chapter 5 Surface Chemistry

GSEB Class 12 Chemistry Surface Chemistry InText Questions and Answers

 

Question 1. Write any two characteristics of chemisorption?
Answer:
(i) Chemisorption demonstrates high specificity.
(ii) This process involves the formation of chemical compounds, making it inherently irreversible in nature.
In simple words: Chemisorption is very particular about what it sticks to, and once a chemical bond forms, it's hard to undo.

🎯 Exam Tip: When describing characteristics of chemisorption, remember to highlight its specificity and irreversible nature, often linked to chemical bond formation. These are key distinguishing features.

 

Question 2. Why does physisorption decrease with the increase of temperature?
Answer:
Physisorption is generally a reversible process and is exothermic, meaning it releases heat. The equilibrium can be represented as:
Solid + Gas \( \rightleftharpoons \) Gas/Solid + Heat
Since physisorption is an exothermic process, according to Le Chatelier's principle, an increase in temperature will shift the equilibrium to the left, favoring desorption and thus decreasing the extent of physisorption.
In simple words: Physisorption gives off heat, so if you add more heat (increase temperature), the adsorbed gas will prefer to leave the surface, reducing adsorption.

🎯 Exam Tip: Always relate the temperature effect on physisorption to its exothermic nature and Le Chatelier's principle. This demonstrates a deeper understanding of chemical equilibria.

 

Question 3. Why are powdered substances more effective adsorbents than their crystalline forms?
Answer:
Powdered substances possess a significantly larger surface area compared to their crystalline forms. A greater surface area directly correlates with an increased number of available sites for adsorption, leading to higher adsorption efficiency.
In simple words: Crushing a substance into a powder creates much more surface for gases or liquids to stick to, making it a better adsorbent.

🎯 Exam Tip: The key concept here is surface area. Remember that finely divided materials expose more surface, which is crucial for efficient adsorption processes.

 

Question 4. Why is it necessary to remove CO when ammonia is obtained by Haber's process?
Answer:
Carbon monoxide (CO) acts as a catalytic poison in the Haber's process. Its presence can reduce or destroy the activity of the catalyst used for ammonia synthesis, making its removal essential for efficient production.
In simple words: In the Haber's process, CO acts like a "poison" for the catalyst, making it stop working. So, we must get rid of it.

🎯 Exam Tip: For industrial processes like Haber's, understanding the role of impurities and catalytic poisons is vital. Always explain *why* a substance needs to be removed in terms of its effect on the catalyst.

 

Question 5. Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
Answer:
The hydrolysis of esters, such as ethyl acetate, is initially slow but accelerates over time because it is an autocatalytic reaction. This reaction is catalyzed by an acid (\( \text{H}^+ \)). One of the products formed during the hydrolysis, acetic acid (\( \text{CH}_3\text{COOH} \)), acts as a catalyst for the reaction. As the reaction proceeds, more acetic acid is produced, which in turn increases the reaction rate.
\( \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \longrightarrow \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \)
In simple words: Ester breakdown starts slow because it needs an acid catalyst. The reaction itself makes an acid (acetic acid), which then speeds up the reaction, so it gets faster later on.

🎯 Exam Tip: Focus on explaining autocatalysis in terms of product formation. Identifying the catalytic species (acetic acid) is crucial for a complete answer.

 

Question 6. What is the process of catalysis?
Answer:
In catalysis, after the products are formed on the surface of the catalyst, they must detach from it. This detachment process is known as desorption. Desorption is crucial because it frees up the active sites on the catalyst's surface, making them available for new reactant molecules to adsorb and undergo further reactions.
In simple words: Catalysis involves reactants sticking to a surface, reacting, forming products, and then these products leaving the surface. This "leaving" part, called desorption, is important because it makes space for new reactions.

🎯 Exam Tip: When discussing catalysis, beyond just stating that a catalyst speeds up a reaction, it's important to mention the role of desorption in regenerating the catalyst surface for continuous activity.

 

Question 7. Why is it essential to wash the precipitate with water before estimating it quantitatively?
Answer:
Before quantitative estimation, it is crucial to wash the precipitate with water to remove any impurities that may have adsorbed onto its surface. These adsorbed impurities can stick to the precipitate and lead to inaccurate measurements if not thoroughly removed.
In simple words: We wash the solid (precipitate) with water to get rid of any unwanted stuff stuck to its surface. If we don't, our measurements won't be correct.

🎯 Exam Tip: Emphasize the removal of *adsorbed* impurities to ensure the purity of the precipitate, which is critical for accurate quantitative analysis.

GSEB Class 12 Chemistry Surface Chemistry Text Book Questions and Answers

 

Question 1. Distinguish between the meaning of the terms adsorption and absorption. Give one example of each?
Answer:
The distinction between adsorption and absorption can be summarized as follows:

Absorption	Adsorption
1. This phenomenon involves the uniform distribution of gas or liquid particles throughout the entire bulk of a solid material.	1. This phenomenon describes the accumulation of gas or liquid particles on the surface of a solid, resulting in a higher concentration there compared to the bulk.
2. The concentration of the substance remains consistent throughout the material.	2. The concentration of the substance on the adsorbent's surface is higher than within its bulk.
3. Absorption typically occurs at a uniform rate.	3. Adsorption is rapid initially and then gradually slows down.

Example of Absorption: Water absorbed by a sponge.
Example of Adsorption: Ammonia gas adsorbed on charcoal.
In simple words: Absorption is when a substance soaks *into* something evenly, like a sponge absorbing water. Adsorption is when a substance only sticks *to the surface* of something, like gas molecules sticking to a charcoal surface.

🎯 Exam Tip: When distinguishing between absorption and adsorption, clearly state whether the process involves bulk penetration (absorption) or surface concentration (adsorption) and provide a suitable example for each.

 

Question 2. What is the difference between physisorption and chemisorption?
Answer:
The key differences between physisorption and chemisorption are outlined below:

Physisorption	Chemisorption
1. Adsorbate molecules are held on the surface by weak van der Waals' forces.	1. Adsorbate molecules are held on the surface by strong chemical bonds.
2. The enthalpy of adsorption is quite low (typically in the range of 20-40 kJ mol\(^{-1}\)).	2. The enthalpy of adsorption is high (typically in the range of 40-400 kJ mol\(^{-1}\)).
3. This process is generally reversible.	3. This process is irreversible.
4. It is not highly specific in nature.	4. It is highly specific in nature.
5. Forms multimolecular layers on the surface.	5. Forms a monomolecular layer on the surface.
6. Occurs predominantly at low temperatures and diminishes with increasing temperature.	6. Takes place at high temperatures.
7. It typically does not require activation energy.	7. It usually requires activation energy.

In simple words: Physisorption is like weak magnets holding things, happening at low temperatures and easily reversible, forming many layers. Chemisorption is like strong glue, needing higher temperatures to start, forming only one layer, and hard to reverse.

🎯 Exam Tip: For this comparison, focus on the nature of forces (van der Waals vs. chemical bonds), enthalpy values, reversibility, and specificity. These are the most critical distinguishing points.

 

Question 3. Give reason why a finely divided substance is more effective as an adsorbent?
Answer:
A finely divided substance provides a significantly larger surface area compared to the same mass of the substance in a coarser or crystalline form. The extent of adsorption is directly proportional to the available surface area of the adsorbent. Therefore, substances with a greater exposed surface are more effective adsorbents.
In simple words: When a substance is ground into a fine powder, it exposes much more surface. More surface means more places for molecules to stick, making it a better adsorbent.

🎯 Exam Tip: The direct relationship between increased surface area and enhanced adsorption capacity is the fundamental reason. Always link "finely divided" to "larger surface area."

 

Question 4. What are the factors which influence the adsorption of a gas on a solid?
Answer:
Several factors influence the adsorption of a gas on a solid surface:
(i) Nature of the gas
(ii) Surface area of the adsorbent
(iii) Pressure
(iv) Temperature
(v) Activity of the adsorbent
In simple words: How much gas sticks to a solid depends on the type of gas, how much surface the solid has, the gas pressure, the temperature, and how "active" the solid's surface is.

🎯 Exam Tip: Listing these five factors is essential. For a more comprehensive answer, be prepared to briefly explain how each factor influences adsorption (e.g., higher surface area leads to more adsorption).

 

Question 5. What is an adsorption isotherm? Describe Freundlich adsorption isotherm?
Answer:
An adsorption isotherm is a graph that illustrates the relationship between the extent of adsorption of a gas per unit mass of adsorbent (\( \frac{x}{m} \)) and the pressure (P) of the gas, while maintaining a constant temperature.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र अधिशोषण समतापी वक्र को दर्शाता है, जिसमें गैस की अधिशोषित मात्रा (\( \frac{x}{m} \)) को दाब (P) के साथ प्लॉट किया गया है। वक्र दर्शाता है कि दाब बढ़ने पर अधिशोषण की मात्रा बढ़ती है, लेकिन एक निश्चित दाब (संतृप्ति दाब, Ps) के बाद यह स्थिर हो जाती है।
The Freundlich adsorption isotherm is an empirical equation that describes the quantitative relationship between the extent of adsorption and pressure. It is expressed as:
\( \frac{x}{m} = \text{KP}^{1/n} \)
Where:
\( x \) = mass of the adsorbate
\( m \) = mass of the adsorbent
\( P \) = pressure of the gas
\( K \) and \( n \) = constants that depend on the nature of the adsorbent, adsorbate, and temperature.
This isotherm suggests that at low pressures, the extent of adsorption is directly proportional to pressure (\( \frac{x}{m} \propto \text{P} \)). At high pressures, adsorption becomes independent of pressure (\( \frac{x}{m} \) is constant), implying that \( \text{P}^{1/n} \) approaches a constant value where \( 1/n = 0 \).
In simple words: An adsorption isotherm shows how much gas sticks to a solid at different pressures when the temperature stays the same. The Freundlich equation is a formula that tries to explain this relationship, showing that adsorption usually increases with pressure but then levels off.

🎯 Exam Tip: When defining adsorption isotherm, emphasize constant temperature. For Freundlich isotherm, state the equation, explain its terms (x, m, K, n), and briefly describe the pressure dependence at low and high pressure limits.

 

Question 6. What do you understand by activation of adsorbent? How is it achieved?
Answer:
Activation of an adsorbent refers to the process of enhancing its adsorption power. This can be achieved through several methods:
(i) **Increasing the surface area:** By grinding the adsorbent into a finer powder or creating a porous structure.
(ii) **Making the surface rough:** A rougher surface provides more active sites for adsorption.
(iii) **Removing previously adsorbed gases:** Heating or creating a vacuum can desorb existing gases, freeing up active sites for new adsorption.
In simple words: Activating an adsorbent means making it better at sticking things to its surface. We can do this by making its surface bigger, rougher, or by cleaning off anything already stuck to it.

🎯 Exam Tip: Remember that activation aims to increase the number and accessibility of active sites on the adsorbent. Mentioning surface area, roughness, and removal of old adsorbates are key points.

 

Question 7. What role does adsorption play in heterogeneous catalysis?
Answer:
In heterogeneous catalysis, adsorption is a fundamental step. The theory posits that gaseous reactants are adsorbed onto the surface of a solid catalyst, often forming a unimolecular layer. These adsorbed molecules then interact with the surface atoms to form an activated complex at specific locations known as active centers. Subsequently, this activated complex decomposes to form the product. After formation, the products desorb from the catalyst surface, making the active centers available again for fresh reactant molecules. This cyclic process of adsorption, reaction, and desorption is central to how heterogeneous catalysts function.
For example, in the Contact process, \( \text{SO}_2 \) and \( \text{O}_2 \) gases adsorb on the surface of a platinum (Pt) catalyst to form \( \text{SO}_3 \).
The number of active sites on a catalyst's surface can be increased by subdividing the catalyst into finer particles or by making its surface rougher. This theory successfully explains why a catalyst's mass and composition remain unchanged at the reaction's conclusion, but it falls short in explaining the actions of catalytic promoters and poisons.

**Nature of solid catalysts:**
Solid catalysts include substances like metals, metal alloys, metal oxides, and metal sulfides. Their catalytic action involves two important aspects:
(a) **Activity:** This refers to the catalyst's ability to accelerate a chemical reaction. A prominent example is the reaction of \( \text{H}_2 \) and \( \text{O}_2 \) to form \( \text{H}_2\text{O} \) on a platinum surface, which occurs with explosive violence. In the absence of a catalyst, a pure hydrogen and oxygen mixture does not react at all.
(b) **Selectivity:** This is the catalyst's ability to direct a reaction towards a specific product. For instance, when acetylene (\( \text{CH}\equiv\text{CH} \)) reacts with \( \text{H}_2 \):
\( \text{CH}\equiv\text{CH} + \text{H}_2 \xrightarrow{\text{Pt}} \text{CH}_3\text{CH}_3 \text{(Ethane)} \)
\( \text{CH}\equiv\text{CH} + \text{H}_2 \xrightarrow{\text{Lindlar's catalyst}} \text{CH}_2=\text{CH}_2 \text{(Ethene)} \)
Similarly, \( \text{H}_2 \) and \( \text{CO} \) can yield different products depending on the catalyst used:
\( \text{CO}_{(\text{g})} + 3\text{H}_{2(\text{g})} \xrightarrow{\text{Ni}} \text{CH}_{4(\text{g})} + \text{H}_2\text{O}_{(\text{g})} \)
\( \text{CO}_{(\text{g})} + 2\text{H}_{2(\text{g})} \xrightarrow{\text{Cu/ZrO}+\text{Cr}_2\text{O}_3} \text{CH}_3\text{OH}_{(\text{g})} \)
\( \text{CO}_{(\text{g})} + \text{H}_{2(\text{g})} \xrightarrow{\text{Cu}} \text{HCHO}_{(\text{g})} \)
Since heterogeneous catalysts are generally solids and their function is a surface phenomenon, heterogeneous catalysis is sometimes referred to as surface catalysis.
In simple words: Adsorption is key in how solid catalysts work. Reactant gases stick to specific spots on the catalyst surface, react, turn into products, and then leave, making space for more reactants. Catalysts can also be active (speed up reactions greatly) and selective (guide reactions to make specific products).

🎯 Exam Tip: When discussing heterogeneous catalysis, explain the three-step process: adsorption, surface reaction (forming activated complex), and desorption. Provide examples for both activity and selectivity, demonstrating how different catalysts yield different products.

 

Question 8. Why is adsorption always exothermic?
Answer:
Adsorption is invariably an exothermic process. This can be understood thermodynamically: when gas molecules adsorb onto a surface, their freedom of movement (randomness) decreases, leading to a reduction in entropy (\( \Delta S \) is negative). For a process to be spontaneous, the Gibbs free energy change (\( \Delta G \)) must be negative. The relationship is given by the equation: \( \Delta G = \Delta H - T\Delta S \). Since \( \Delta S \) is negative, \( -T\Delta S \) becomes positive. For \( \Delta G \) to be negative, \( \Delta H \) must be sufficiently negative (i.e., the process must be exothermic) to outweigh the positive \( -T\Delta S \) term. Thus, adsorption is an exothermic process.
In simple words: When gas sticks to a surface, it becomes more orderly, so its "messiness" (entropy) goes down. For this to happen naturally, the process must release heat (be exothermic) to make the overall energy change favorable.

🎯 Exam Tip: To explain why adsorption is exothermic, always refer to the decrease in entropy (\( \Delta S < 0 \)) of the adsorbed species and the spontaneity condition (\( \Delta G < 0 \)), using the Gibbs-Helmholtz equation \( \Delta G = \Delta H - T\Delta S \).

 

Question 9. How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?
Answer:
Colloidal solutions are classified into eight types based on the physical states (solid, liquid, or gas) of their dispersed phase and dispersion medium, as summarized in the table below:

Sl. No.	Dispersed Phase	Dispersion Medium	Name	Examples
1.	Solid	Solid	Solid sol	Some colored glasses
2.	Solid	Liquid	Sol	Some paints, muddy water
3.	Solid	Gas	Aerosol	Smoke, dust
4.	Liquid	Solid	Gel	Cheese, butter, jellies
5.	Liquid	Liquid	Emulsion	Milk, hair cream
6.	Liquid	Gas	Aerosol	Fog, mist, cloud
7.	Gas	Solid	Solid foam	Pumice stone, foam rubber
8.	Gas	Liquid	Foam	Froth, whipped cream

*Note: Gas in gas mixtures are homogeneous and do not form colloidal systems.
In simple words: Colloidal solutions are grouped by whether the dispersed substance and the medium it's spread in are solid, liquid, or gas. This gives us eight main types, each with its own name and examples like fog (liquid in gas) or jelly (liquid in solid).

🎯 Exam Tip: Remember that gas-in-gas systems are true solutions, not colloids. Be able to recall at least 3-4 common types and their examples, such as sol, gel, emulsion, and aerosol.

 

Question 10. Discuss the effect of pressure and temperature on the adsorption of gases on solids?
Answer:
Adsorption occurs due to the unbalanced attractive forces present on the surface of an adsorbent. While forces within the bulk of the adsorbent are balanced, surface particles experience net attractive forces, leading to adsorption. The force of adsorption increases when a solid is pulverized (increasing surface area).

During adsorption, new bonds form between the adsorbent and adsorbate. This process is generally exothermic (\( \Delta H \) is negative). The enthalpy change when one mole of an adsorbate is adsorbed on an adsorbent surface is termed the heat of adsorption or enthalpy of adsorption. Furthermore, when a gas is adsorbed, the movement of its molecules is restricted, resulting in a decrease in entropy (\( \Delta S \) is negative).

For adsorption to be spontaneous, \( \Delta G \) must be negative. From the equation \( \Delta G = \Delta H - T\Delta S \), if \( \Delta H \) is negative and \( \Delta S \) is negative (making \( -T\Delta S \) positive), \( \Delta G \) will be negative only when \( \Delta H \) is sufficiently negative, particularly at low temperatures. As adsorption proceeds, \( \Delta H \) becomes less negative, eventually equaling \( T\Delta S \), at which point \( \Delta G \) becomes zero, indicating equilibrium.

**Effect of Pressure:**
As pressure increases, the magnitude of adsorption generally increases. This relationship, showing the variation of adsorption (\( \frac{x}{m} \)) with pressure (P) at a constant temperature, is depicted graphically by an adsorption isotherm. Here, \( x \) represents the amount of gas adsorbed by mass \( m \) of the adsorbent at pressure \( P \).

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र अधिशोषण समतापी वक्र को दर्शाता है, जिसमें गैस की अधिशोषित मात्रा (\( \frac{x}{m} \)) को दाब (P) के साथ प्लॉट किया गया है। वक्र दर्शाता है कि दाब बढ़ने पर अधिशोषण की मात्रा बढ़ती है, लेकिन एक निश्चित दाब (संतृप्ति दाब, Ps) के बाद यह स्थिर हो जाती है।
At a specific pressure, \( \text{P}_{\text{s}} \) (saturation pressure or equilibrium pressure), \( \frac{x}{m} \) reaches a maximum value and remains constant. Beyond \( \text{P}_{\text{s}} \), further increases in pressure have no effect on adsorption.

**Freundlich Adsorption Isotherm:**
This empirical equation describes the variation of adsorption amount per unit mass with pressure:
\( \frac{x}{m} = \text{KP}^{1/n} \)
... (5.1)
where 'k' and 'n' are constants dependent on the nature of the adsorbent, adsorbate, and temperature. Taking logarithms on both sides yields:
\( \log \frac{x}{m} = \log \text{k} + \frac{1}{n} \log \text{P} \)
... (5.2)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र Freundlich अधिशोषण समतापी का लॉगरिदमिक प्लॉट दर्शाता है, जिसमें \( \log \frac{x}{m} \) को \( \log \text{P} \) के विरुद्ध प्लॉट किया गया है। यह एक सीधी रेखा है जिसका ढलान \( 1/n \) है और y-अक्ष पर अवरोधन (intercept) \( \log \text{k} \) है। यह वक्र Freundlich समीकरण के सत्यापन में मदद करता है।
Equations 5.1 and 5.2 are different representations of the Freundlich adsorption isotherm. A graph of \( \log \frac{x}{m} \) versus \( \log \text{P} \) yields a straight line with an intercept of \( \log \text{k} \) on the \( \log \frac{x}{m} \) axis and a slope of \( \frac{1}{n} \).
The factor \( \frac{1}{n} \) typically ranges between 0 and 1.
At high pressure, \( \frac{1}{n} = 0 \), so \( \frac{x}{m} = \text{KP}^0 = \text{k} \) (a constant), meaning adsorption is independent of pressure.
At low pressure, \( \frac{1}{n} = 1 \), so \( \frac{x}{m} = \text{KP} \), indicating direct proportionality between adsorption and pressure.
The primary drawback of the Freundlich adsorption isotherm is its failure to accurately predict adsorption at high pressures.
In simple words: Adsorption always releases heat (exothermic), and it also makes molecules less free (decreases entropy). For it to happen naturally, the heat released must be enough to balance the decrease in freedom. As for pressure, generally, more pressure means more adsorption, but only up to a point, called the saturation pressure. Temperature works in reverse: higher temperatures usually mean less adsorption because it's an exothermic process.

🎯 Exam Tip: For temperature, connect it to the exothermic nature and spontaneity via Gibbs Free Energy. For pressure, explain the general trend (increase with P, then saturation) and introduce the Freundlich isotherm with its equation and logarithmic form, noting its limitations at high pressure.

 

Question 11. What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Answer:
**Lyophilic Sols:**
Lyophilic sols are colloidal solutions where there is a strong attractive interaction (affinity) between the dispersed phase and the dispersion medium. These sols are very stable because the dispersed phase particles are well-solvated (surrounded by a layer of the dispersion medium), which stabilizes them. Lyophilic sols are considered reversible sols; if the dispersed phase is separated from the medium (e.g., by evaporation), the sol can be readily re-formed by simply remixing the components.
Example: Starch sol, albumin, gelatin.

**Lyophobic Sols:**
In contrast, lyophobic sols (hydrophobic sols when the medium is water) have little or no affinity between the dispersed phase and the dispersion medium. These substances do not spontaneously form colloidal solutions upon simple mixing. They require special preparation methods and are less stable than lyophilic sols. If the solid particles in a lyophobic sol coagulate or precipitate, they cannot be easily converted back into a sol by remixing. Therefore, they are called irreversible sols.
Example: Colloidal metals (like gold sol), colloidal hydroxides.

**Why hydrophobic sols are easily coagulated:**
The stability of hydrophobic sols primarily depends on the presence of an electrical charge on their colloidal particles. These particles typically acquire a charge by adsorbing specific ions from the dispersion medium. If this charge is neutralized or removed (for example, by adding an electrolyte), the particles can come closer, aggregate, and precipitate out, a process known as coagulation. Lyophilic sols, however, have an additional stabilizing factor of solvation, making them much more resistant to coagulation.
In simple words: Lyophilic sols "love" their liquid medium, so they mix easily and are stable, like starch in water. Lyophobic sols "hate" their liquid medium, so they're hard to make and are unstable, like gold in water. Hydrophobic (water-hating) sols are easily coagulated because their stability mostly comes from their surface charge, and if that charge is removed, they clump together and settle out.

🎯 Exam Tip: For lyophilic vs. lyophobic, highlight the affinity (or lack thereof) for the medium, reversibility, and stability. When explaining coagulation of hydrophobic sols, emphasize their primary dependence on surface charge for stability.

 

Question 12. What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Answer:
(i) **Multimolecular colloids:** In this type, the colloidal particles are aggregates of many atoms or small molecules (with diameters typically less than 1 nm) held together by weak van der Waals' forces. These aggregates collectively form particles within the colloidal size range.
Example: Gold sol (containing clusters of gold atoms), sulfur sol (containing about a thousand \( \text{S}_8 \) molecules).

(ii) **Macromolecular colloids:** Here, the dispersed phase consists of large molecules (macromolecules), typically polymers, whose sizes are already within the colloidal range. These big molecules themselves act as the colloidal particles, having molecular masses ranging from thousands to millions.
Example: Polyethylene, polystyrene, nylon, rubber, starch, proteins.

(iii) **Associated colloids (Micelles):** These are distinct from both multimolecular and macromolecular colloids. Associated colloids are substances that behave as normal electrolytes at low concentrations but exhibit colloidal properties at higher concentrations. Above a certain critical micelle concentration (CMC) and Kraft temperature, their molecules aggregate to form micelles. Micelles are dynamic structures that can revert back to individual ions upon dilution.
Example: Soaps and detergents.

**Mechanism of micelle formation (for soaps):**
Soaps, like sodium stearate (\( \text{CH}_3(\text{CH}_2)_{16}\text{COONa} \)), dissociate in water to give ions:
\( \text{R-COONa} \longrightarrow \text{R-COO}^- \text{(Anion)} + \text{Na}^+ \text{(Cation)} \)
The anion has two parts: a long hydrocarbon chain (R, hydrophobic/water-repelling 'tail') and a carboxyl group ( \( \text{COO}^- \), hydrophilic/water-dissolving 'head'). In solution, the hydrophobic tails aggregate inwards to avoid water, while the hydrophilic heads remain exposed on the surface, forming a charged spherical aggregate called a micelle. A micelle can contain many ions (e.g., up to 100). Similarly, in detergents (e.g., \( \text{CH}_3(\text{CH}_2)_{11}\text{SO}_4^-\text{Na}^+ \)), the polar group is \( -\text{SO}_4^- \) and the non-polar group is the long hydrocarbon chain, with a similar micelle formation mechanism.
In simple words: Multimolecular colloids are small groups of many tiny molecules stuck together (like sulfur particles). Macromolecular colloids are just very big molecules themselves that are already colloid-sized (like proteins). Associated colloids (micelles) are special molecules, like soap, that act normal when dilute but clump together into bigger, colloid-sized structures called micelles when concentrated.

🎯 Exam Tip: Clearly define each type based on particle composition/size. For associated colloids, explain the critical micelle concentration (CMC) and Kraft temperature, and briefly describe the hydrophilic/hydrophobic parts in micelle formation.

 

Question 13. What are enzymes? Write in brief the mechanism of enzyme catalysis?
Answer:
Enzymes are biological catalysts, which are complex protein molecules produced by living cells. They significantly accelerate specific biochemical reactions within living organisms.

**Mechanism of Enzyme Catalysis:**
The mechanism of enzyme action typically involves a series of steps:
(a) **Binding of the enzyme to the substrate:** The substrate (reactant) binds to the enzyme's specific active site, forming an enzyme-substrate complex.
\( \text{E (Enzyme)} + \text{S (Substrate)} \longrightarrow \text{[ES] (Enzyme-Substrate Complex)} \)
(b) **Product formation within the enzyme-substrate complex:** The substrate undergoes reaction within the active site, converting into the product while still bound to the enzyme.
\( \text{[ES] (Enzyme-Substrate Complex)} \longrightarrow \text{EP (Enzyme-Product Complex)} \)
(c) **Release of products from the enzyme:** The product detaches from the enzyme's active site, leaving the enzyme free to catalyze further reactions.
\( \text{EP (Enzyme-Product Complex)} \longrightarrow \text{E (Enzyme)} + \text{P (Product)} \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एंजाइम उत्प्रेरण की प्रक्रिया को दर्शाता है। इसमें, एंजाइम (E) पर एक सक्रिय साइट होती है जहाँ सबस्ट्रेट (S) फिट होता है, जिससे एंजाइम-सबस्ट्रेट कॉम्प्लेक्स [ES] बनता है। इस कॉम्प्लेक्स में, सबस्ट्रेट उत्पाद (P) में परिवर्तित होता है, जिससे एंजाइम-उत्पाद कॉम्प्लेक्स EP बनता है। अंत में, उत्पाद एंजाइम से अलग हो जाता है, जिससे एंजाइम मुक्त हो जाता है और एक नए सबस्ट्रेट के साथ प्रतिक्रिया करने के लिए तैयार हो जाता है।
Examples:
(a) The enzyme maltase catalyzes the hydrolysis of maltose into glucose.
(b) The enzyme pepsin converts proteins into amino acids.
In simple words: Enzymes are biological helpers that speed up body reactions. They work like a lock and key: the enzyme (lock) has an active site where the substrate (key) fits, forming a complex. The enzyme then helps change the substrate into a product, which is released, leaving the enzyme ready for another reaction.

🎯 Exam Tip: When explaining enzyme catalysis, describe the "lock and key" mechanism involving active sites, enzyme-substrate complex formation, and product release. Using the E+S -> ES -> EP -> E+P sequence is highly effective.

 

Question 14. How are colloids classified on the basis of
1. Physical states of components
2. Nature of dispersion medium and
3. Interaction between dispersed phase and dispersion medium?
Answer:
Colloids are classified based on the following criteria:
1. **Physical States of Components:** This classification considers whether the dispersed phase and the dispersion medium are solid, liquid, or gas. There are eight possible combinations, excluding gas-in-gas (which forms a true solution), leading to types like sols (solid in liquid), gels (liquid in solid), emulsions (liquid in liquid), and aerosols (solid or liquid in gas). (Refer to Question 9 for a detailed table).
2. **Nature of Dispersion Medium:** Colloids can also be classified based on the chemical nature of the dispersion medium. For example:
* **Hydrosols (Aqueous sols):** When water is the dispersion medium.
* **Alcosols:** When alcohol is the dispersion medium.
* **Benosols:** When benzene is the dispersion medium.
3. **Interaction Between Dispersed Phase and Dispersion Medium:** This classifies colloids based on the affinity between the two phases:
* **Lyophilic Sols (Solvent-loving):** Exhibit a strong affinity between the dispersed phase and the dispersion medium, forming stable and reversible colloids (e.g., starch in water).
* **Lyophobic Sols (Solvent-hating):** Have little or no affinity between the dispersed phase and the dispersion medium, requiring special preparation and being less stable and irreversible (e.g., gold sol in water).
In simple words: Colloids are sorted in three ways: first, by if their parts are solid, liquid, or gas; second, by what the main liquid or gas is (like water for hydrosols); and third, by how much the dispersed bits "like" or "dislike" the surrounding medium (solvent-loving or solvent-hating).

🎯 Exam Tip: Ensure you list all three classification bases. For each, give a brief explanation or example. For interaction, clearly distinguish between lyophilic and lyophobic sols and their characteristics.

 

Question 15. Explain what is observed when
1. A beam of light is passed through a colloidal sol.
2. An electrolyte, NaCl is added to hydrated ferric oxide sol.
3. Electric current is passed through a colloidal sol
Answer:
Here's what is observed in each scenario:
1. **A beam of light is passed through a colloidal sol:** The path of the light beam becomes visible, and the light is scattered. This phenomenon is known as the **Tyndall effect**.
2. **An electrolyte, NaCl is added to hydrated ferric oxide sol:** The colloidal particles of hydrated ferric oxide precipitate or coagulate, forming a solid mass. This process is called **coagulation**.
3. **Electric current is passed through a colloidal sol:** The colloidal particles move towards either the anode (positive electrode) or the cathode (negative electrode), depending on the charge of the colloidal particles. This movement is called **electrophoresis**.
In simple words: 1. When light shines through a colloid, its path becomes visible (Tyndall effect). 2. Adding salt to a colloid can make the tiny particles clump together and settle out (coagulation). 3. Running electricity through a colloid makes the charged particles move towards one of the electrical poles (electrophoresis).

🎯 Exam Tip: For each observation, state the correct phenomenon (Tyndall effect, coagulation, electrophoresis) and provide a concise explanation of *what* happens to the light or the colloidal particles.

 

Question 16. What are emulsions? What are their different types? Give example of each type?
Answer:
Emulsions are colloidal solutions formed by two immiscible liquids, where one liquid acts as the dispersed phase and the other as the dispersion medium. These are primarily of two types:
(a) **Oil-in-Water (O/W) emulsions:** In this type, oil is the dispersed phase (present as droplets) and water is the dispersion medium.
Example: Milk (fat droplets in water), vanishing cream, hair cream.
(b) **Water-in-Oil (W/O) emulsions:** In this type, water is the dispersed phase (present as droplets) and oil is the dispersion medium.
Example: Butter (water droplets in fat), cod liver oil, cold cream.
In simple words: Emulsions are mixtures of two liquids that normally don't mix, like oil and water. They come in two kinds: oil mixed into water (like milk) or water mixed into oil (like butter).

🎯 Exam Tip: Define emulsions as mixtures of immiscible liquids. Clearly distinguish between O/W and W/O types by identifying the dispersed phase and dispersion medium, and provide a common example for each.

 

Question 17. What is demulsification? Name two demulsifiers?
Answer:
Demulsification is the process of breaking an emulsion into its constituent immiscible liquids. This separation can be achieved by various physical or chemical methods, such as heating, freezing, centrifugation, or adding specific chemical agents.
Two types of effective demulsifiers are:
1. Dehydrating agents (which remove the dispersion medium).
2. Electrolytes (which can neutralize the charge on the dispersed phase, causing coagulation).
In simple words: Demulsification is the act of separating an emulsion back into its two original liquids. Things like dehydrating agents or electrolytes can act as demulsifiers, helping to break the emulsion apart.

🎯 Exam Tip: Define demulsification as breaking an emulsion. For demulsifiers, mention general categories like dehydrating agents or electrolytes, rather than specific chemical names, to show a broader understanding.

 

Question 18. Explain how the cleansing action of soap is due to emulsification and micelle formation. Comment?
Answer:
The cleansing action of soap is fundamentally based on its ability to form micelles and emulsify oily or greasy dirt. Soap molecules, like sodium stearate, consist of a long hydrocarbon chain (hydrophobic 'tail') and a polar carboxylate group (hydrophilic 'head').

When soap is added to water, its molecules aggregate to form micelles above the Critical Micelle Concentration (CMC). In these micelles, the hydrophobic tails point inwards, forming a non-polar core, while the hydrophilic heads face outwards into the water. When dirty clothes with oily or greasy stains are introduced, the hydrophobic tails of the soap molecules penetrate the oil or grease layer, while the hydrophilic heads remain in the water.

This interaction traps the oil/grease particles within the micelle structure, forming an emulsion where the oil/grease is suspended in water. The formation of these micelles around the dirt particles reduces the interfacial tension between the oil and water. The negatively charged surface of the micelle (due to the hydrophilic heads) also prevents the re-aggregation of the dirt particles. Consequently, the oily dirt particles are displaced from the fabric into the water, forming a stable emulsion that can be easily washed away, thus cleaning the cloth.
In simple words: Soap cleans by forming tiny balls called micelles. The oily part of the soap traps the grease and oil from dirt, while the watery part of the soap stays in the water. This lifts the dirt off the clothes and keeps it suspended in the water, letting it be rinsed away.

🎯 Exam Tip: When explaining soap's cleansing action, clearly describe micelle formation, specifying the hydrophobic tail trapping grease and the hydrophilic head interacting with water. Emphasize the role of emulsification in lifting and suspending dirt.

 

Question 19. Give four examples of heterogeneous catalysis?
Answer:
Heterogeneous catalysis occurs when the catalyst exists in a different physical phase from that of the reactants. Here are four examples:
(i) **Manufacture of Sulfuric Acid (Contact Process):** Involves the oxidation of sulfur dioxide using a solid catalyst like vanadium pentoxide (\( \text{V}_2\text{O}_5 \)) or platinum (Pt).
\( 2\text{SO}_{2(\text{g})} + \text{O}_{2(\text{g})} \xrightarrow{\text{V}_2\text{O}_{5(\text{s})}} 2\text{SO}_{3(\text{g})} \)
(ii) **Haber Process for Ammonia Synthesis:** Nitrogen and hydrogen gases react in the presence of solid iron (Fe) as a catalyst.
\( \text{N}_{2(\text{g})} + 3\text{H}_{2(\text{g})} \xrightarrow{\text{Fe}_{(\text{s})}} 2\text{NH}_{3(\text{g})} \)
(iii) **Oxidation of Ammonia (Ostwald's Process):** Ammonia is oxidized to nitric oxide (NO) using a solid platinum (Pt) catalyst.
\( 4\text{NH}_{3(\text{g})} + 5\text{O}_{2(\text{g})} \xrightarrow{\text{Pt}_{(\text{s})}} 4\text{NO}_{(\text{g})} + 6\text{H}_2\text{O}_{(\text{g})} \)
(iv) **Hydrogenation of Vegetable Oils:** Vegetable oils are converted into solid vegetable ghee (vanaspati) by reaction with hydrogen gas in the presence of a solid nickel (Ni) catalyst.
\( \text{Vegetable oil}_{(\text{l})} + \text{H}_{2(\text{g})} \xrightarrow{\text{Ni}_{(\text{s})}} \text{Vegetable ghee}_{(\text{s})} \)
In simple words: Heterogeneous catalysis is when the catalyst is in a different state (like solid) than the chemicals reacting (like gases). Examples include making ammonia with iron, sulfuric acid with vanadium pentoxide, nitric acid with platinum, and turning vegetable oil into ghee with nickel.

🎯 Exam Tip: For each example, identify the reactants, products, and the specific heterogeneous catalyst involved. Writing the balanced chemical equations with the catalyst noted above the arrow is crucial.

 

Question 20. What do you mean by activity and selectivity of catalysts?
Answer:
(a) **Activity of a catalyst:** The activity of a catalyst refers to its ability to increase the rate of a particular chemical reaction. This capability is strongly influenced by chemisorption; the adsorption of reactants on the catalyst surface should be neither too strong nor too weak, but just strong enough to make the catalyst active and facilitate the reaction. A highly active catalyst can dramatically speed up a reaction that would otherwise be extremely slow or not occur at all.

(b) **Selectivity of the catalyst:** The selectivity of a catalyst is its power to direct a reaction to yield a specific desired product from multiple possible products. This means that by choosing different catalysts for the same reactants, one can obtain different products. For example, the reaction between hydrogen (\( \text{H}_2 \)) and carbon monoxide (CO) can yield different products depending on the catalyst:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दर्शाता है कि विभिन्न उत्प्रेरकों का उपयोग करके एक ही अभिकारकों (एथाइन और हाइड्रोजन) से विभिन्न उत्पाद (एथेन या एथीन) कैसे प्राप्त किए जा सकते हैं, जो उत्प्रेरक की चयनात्मकता को उजागर करता है।
\( \text{CH}\equiv\text{CH} + \text{H}_2 \xrightarrow{\text{Pt}} \text{CH}_3\text{CH}_3 \text{(Ethane)} \)
\( \text{CH}\equiv\text{CH} + \text{H}_2 \xrightarrow{\text{Lindlar's catalyst}} \text{CH}_2=\text{CH}_2 \text{(Ethene)} \)
Similarly, for the reaction of \( \text{H}_2 \) and \( \text{CO} \):
\( \text{CO}_{(\text{g})} + 3\text{H}_{2(\text{g})} \xrightarrow{\text{Ni}} \text{CH}_{4(\text{g})} + \text{H}_2\text{O}_{(\text{g})} \)
\( \text{CO}_{(\text{g})} + 2\text{H}_{2(\text{g})} \xrightarrow{\text{Cu/ZrO}+\text{Cr}_2\text{O}_3} \text{CH}_3\text{OH}_{(\text{g})} \)
\( \text{CO}_{(\text{g})} + \text{H}_{2(\text{g})} \xrightarrow{\text{Cu}} \text{HCHO}_{(\text{g})} \)
In simple words: Catalyst activity means how well a catalyst speeds up a reaction. Catalyst selectivity means how well a catalyst directs a reaction to make a specific product, rather than other possible products, from the same starting materials.

🎯 Exam Tip: Define activity in terms of reaction rate enhancement and mention the role of optimal chemisorption. For selectivity, define it as directing product formation and provide clear chemical equation examples to illustrate how different catalysts lead to different products from the same reactants.

 

Question 21. Describe some features of catalysis by zeolites?
Answer:
Zeolites are aluminosilicates that possess a distinctive three-dimensional network structure. They are highly porous materials characterized by a variety of cavity sizes. The catalytic behavior of zeolites is highly dependent on the size and shape of these cavities, which typically range from 260-740 pm. Only reactant molecules with a suitable shape and size can enter these specific pores, get adsorbed, and undergo reaction. This unique characteristic leads to zeolites being known as shape-selective catalysts.
In simple words: Zeolites are like tiny chemical sieves with specific pore sizes. Only molecules that are the right size and shape can fit into these pores to react, which is why they are called "shape-selective" catalysts.

🎯 Exam Tip: Key features to mention include their aluminosilicate composition, three-dimensional porous structure, specific cavity sizes (range 260-740 pm), and the resulting property of shape-selectivity.

 

Question 22. What is shape-selective catalysis?
Answer:
Shape-selective catalysis is a catalytic process that is highly specific and depends crucially on the pore structure of the catalyst and the size and shape of both the reactant and product molecules. Catalysts like zeolites exhibit this property. Their pore sizes, typically ranging from 260-740 pm, act as molecular sieves. Only molecules small enough to enter these pores can undergo reaction, while larger molecules are excluded. Similarly, if the product molecules are too large to diffuse out of the pores, the reaction can be hindered.
In simple words: Shape-selective catalysis is when a catalyst's tiny holes (pores) only allow molecules of a specific size and shape to react. It's like a molecular filter, ensuring only certain reactions can happen.

🎯 Exam Tip: When defining shape-selective catalysis, always link it to the catalyst's pore structure (like zeolites) and the critical role of reactant/product molecule size and shape in allowing or preventing access to the active sites.

 

Question 23. Explain the following terms:
(i) Electrophoresis
(ii) Coagulation
(iii) Dialysis
(iv) Tyndall effect.
Answer:(i) Electrophoresis describes the movement of colloidal particles within an electric field. These particles gain a positive or negative charge, leading to the formation of a Helmholtz double layer and exhibiting distinct electrical properties. Observing their migration direction helps determine the charge on the colloidal particles. (ii) Coagulation occurs when a colloidal solution, like ferric hydroxide, precipitates upon adding an electrolyte such as sodium chloride. This process, also known as flocculation, involves the precipitation of a colloidal solution through the introduction of a suitable electrolyte. (iii) Dialysis is a method used to eliminate dissolved substances from a colloidal solution by utilizing diffusion across a semi-permeable membrane. (iv) The Tyndall effect is observed when a beam of light passes through a colloidal solution, illuminating its path due to the scattering of light by the colloidal particles.
In simple words: These terms describe different behaviors of tiny particles suspended in a liquid. Electrophoresis is how charged particles move in an electric field. Coagulation is when these particles clump together and settle. Dialysis is filtering out small dissolved things from the suspension. The Tyndall effect is when you can see a light beam passing through the suspension because the particles scatter the light.

🎯 Exam Tip: Focus on defining each term clearly and concisely. For coagulation, remember to mention the role of electrolytes.

 

Question 24. Give four uses of emulsions?
Answer:Emulsions have several practical applications:
(i) The cleaning efficiency of soaps and detergents relies on the principle of emulsification.
(ii) They are utilized in pharmaceutical products, for instance, cod liver oil.
(iii) Various disinfectants, such as phenyl, Dettol, and Lysol, form oil-in-water emulsions when diluted with water.
(iv) The froth flotation method for mineral ore concentration employs emulsification principles.
(v) The digestive process of fats within the intestine also involves emulsification.
In simple words: Emulsions are useful for cleaning with soaps, making medicines like cod liver oil, diluting disinfectants, concentrating ores, and are essential for fat digestion in our bodies.

🎯 Exam Tip: Provide distinct examples for each use to demonstrate a broad understanding of emulsion applications.

 

Question 25. What are micelles? Give an example of a miceller system.
Answer:Micelles are aggregates of particles formed by certain substances, known as associated colloids, which behave as typical electrolytes at low concentrations but exhibit colloidal properties at higher concentrations. Their formation requires specific conditions, namely a temperature above the Kraft temperature and a concentration exceeding the critical micelle concentration (CMC). Upon dilution, these colloids disassociate back into individual ions, with micelles often comprising a hundred or more molecules. The mechanism of micelle formation involves surface-active agents like soaps and synthetic detergents. When these substances dissolve in water, they ionize. For example, sodium stearate (\( \text{CH}_3(\text{CH}_2)_{16}\text{COO}^-\text{Na}^+ \)) dissociates as
\( \text{R COONa} \implies \text{RCOO}^- + \text{Na}^+ \)
The anion has two distinct parts: a water-soluble (hydrophilic) carboxyl group, known as the head, and a water-repelling (hydrophobic) hydrocarbon chain, referred to as the tail. In an aqueous solution, the hydrophobic tails cluster together, while the hydrophilic heads position themselves on the micelle's surface, rendering it charged. A single micelle can be composed of approximately 100 or more such ions.
In simple words: Micelles are tiny clusters formed by special molecules, like soap, when their concentration is high enough in water. These molecules have a water-loving "head" and a water-hating "tail." The tails hide inside, away from water, and the heads face outwards, making the cluster stable and able to trap oily dirt.

🎯 Exam Tip: Define micelles, explain their formation conditions (Kraft temperature, CMC), and describe the dual nature of surfactant molecules (hydrophilic head, hydrophobic tail) with a relevant example like soap.

 

Question 26. Explain the terms with suitable examples:
(i) Alcosol
(ii) Aerosol
(iii) Hydrosol.
Answer:Colloidal systems are classified based on the physical states of the dispersed phase and the dispersion medium.
(i) **Alcosol:** This is a type of colloidal system where the dispersion medium is alcohol. * *Example:* Colloidal solutions of certain substances (e.g., cellulose nitrate) in alcohol.
(ii) **Aerosol:** An aerosol is a colloidal system in which the dispersion medium is a gas, with either a solid or a liquid as the dispersed phase. * *Examples:* Smoke (solid dispersed in gas), fog (liquid dispersed in gas).
(iii) **Hydrosol:** A hydrosol is a colloidal system where water serves as the dispersion medium. * *Examples:* Muddy water (solid dispersed in liquid water), milk (liquid fat dispersed in liquid water).
In simple words: These terms describe different types of colloids based on what they're made of. An alcosol has alcohol as its main liquid. An aerosol has tiny solid or liquid particles floating in a gas, like mist. A hydrosol has tiny particles suspended in water, like muddy water.

🎯 Exam Tip: Clearly define each type of sol by specifying its dispersed phase and dispersion medium. Provide distinct and common examples for better understanding.

 

Question 27. Comment on the statement that "colloid is not a substance but a state of substance"?
Answer:The assertion that "colloid is not a substance but a state of substance" is accurate. This is because "colloidal" describes a specific physical state, not an inherent property of a particular material. Any substance can be prepared in a colloidal state if its particles are appropriately sized (typically between 1 nm and 1000 nm) and dispersed using suitable methods. Therefore, the term refers to a condition of matter rather than a distinct chemical substance itself.
In simple words: The statement is true because being a colloid isn't about *what* a substance is, but *how* it exists. Any material can become a colloid if its particles are small enough (1-1000 nm) and spread out in another substance.

🎯 Exam Tip: Emphasize the particle size range (1-1000 nm) as the critical factor defining the colloidal state, irrespective of the substance's chemical identity.

GSEB Class 12 Chemistry Surface Chemistry Additional Important Questions And Answers

 

Question 1. What are Lyophilic and Lyophobic sols? Give an example of each type. Which one of these two types of sols is easily coagulated and why?
Answer:(i) **Lyophilic Sols:** These colloidal solutions are formed when substances like gum, gelatin, or starch are mixed directly with a suitable liquid dispersion medium. They are reversible; if the dispersed phase and medium are separated (e.g., by evaporation), the sol can be reformed by simply remixing the components. * *Examples:* Sols of gum, starch, gelatin.
(ii) **Lyophobic Sols:** These sols do not form simply by mixing substances like metals and their sulfides with a dispersion medium. They require special preparation methods and are irreversible. * *Examples:* Sols of metals (e.g., gold sol), metal sulfides.
**Coagulation:** Lyophobic sols are much more easily coagulated than lyophilic sols. The stability of lyophilic sols relies on two factors: the presence of a charge on colloidal particles and their solvation (interaction with the dispersion medium). In contrast, lyophobic sols depend primarily on the presence of charge for their stability. If this charge on hydrophobic sols is neutralized (e.g., by adding electrolytes), the particles aggregate and precipitate, making them prone to coagulation.
In simple words: Lyophilic sols (like starch in water) are stable because their particles like the water and are often covered by it, making them reversible. Lyophobic sols (like gold in water) are unstable and easily clump together (coagulate) because their particles don't like water and rely only on electric charge for stability. Remove the charge, and they precipitate.

🎯 Exam Tip: Differentiate between lyophilic and lyophobic sols based on their affinity for the dispersion medium, reversibility, and method of preparation. Crucially, explain *why* lyophobic sols coagulate easily, linking it to their dependence on charge for stability.

 

Question 2. Adsorption is found to be a surface phenomenon?
(i) What do you mean by physical and chemical adsorption?
(ii) Explain the term enthalpy of adsorption?
(iii) List out the important differences between physical adsorption (physisorption) and chemical adsorption (chemisorption)?
Answer:(i) **Physical Adsorption (Physisorption) vs. Chemical Adsorption (Chemisorption):** * **Physical adsorption** occurs when adsorbate molecules are held onto the adsorbent's surface by weak physical forces, such as van der Waals forces. * **Chemical adsorption** involves the formation of strong chemical bonds between the adsorbate molecules and the adsorbent surface.
(ii) **Enthalpy of Adsorption:** This term refers to the amount of heat energy released when one mole of an adsorbate is adsorbed onto the surface of an adsorbent. * For physical adsorption, the enthalpy typically ranges from 20-40 kJ mol^-1. * For chemical adsorption, the enthalpy is significantly higher, ranging from 40-400 kJ mol^-1.
(iii) Here are the key distinctions between physical adsorption (physisorption) and chemical adsorption (chemisorption):

Physisorption	Chemisorption
1. Adsorbate molecules are held by weak van der Waals forces on the surface.	1. Adsorbate molecules are attached to the surface via strong chemical bonds.
2. The enthalpy of adsorption is quite low (typically 20-40 kJ mol-1).	2. The enthalpy of adsorption is considerably high (typically 40-400 kJ mol-1).
3. This process is generally reversible.	3. This process is inherently irreversible.
4. It exhibits a non-specific nature.	4. It is highly specific in its nature.
5. It results in the formation of multimolecular layers.	5. It leads to the formation of monomolecular layers.
6. It typically occurs at low temperatures and decreases as temperature rises.	6. It occurs predominantly at high temperatures.
7. It generally does not necessitate activation energy.	7. It usually requires activation energy.

In simple words: Physical adsorption is weak, reversible, non-specific, forms multiple layers, happens at low temps, and needs no activation energy. Chemisorption is strong, irreversible, specific, forms one layer, happens at high temps, and needs activation energy. The heat released (enthalpy) is low for physical adsorption and high for chemical adsorption.

🎯 Exam Tip: When listing differences, ensure a clear comparison for each point. Key differentiating factors include the nature of forces, enthalpy values, reversibility, specificity, and layer formation.

 

Question 3. When a small amount of silica gel is placed in a vessel containing moist air, the moisture is adsorbed by silica gel.
(i) What do you mean by adsorption?
(ii) What are the factors which will affect the adsorption of gases on solids?
(iii) Discuss the influence of pressure on adsorption of a gas on a solid.
Answer:(i) **Adsorption:** This refers to the process where molecules of a substance (adsorbate) are selectively attracted and held onto the surface of a liquid or solid (adsorbent), leading to a higher concentration of these molecules on the surface compared to the bulk.
(ii) **Factors Affecting Adsorption of Gases on Solids:** Several factors influence the extent of gas adsorption on a solid surface: * The inherent nature of the gas. * The characteristics of the adsorbent material. * The surrounding temperature. * The applied pressure of the gas. * The activation state of the adsorbent surface.
(iii) **Influence of Pressure on Gas Adsorption on Solids:** At a constant temperature, increasing the pressure of a gas generally leads to an increase in its adsorption onto a solid surface. A graph illustrating the extent of adsorption (x/m) against pressure (P) at a constant temperature is known as an adsorption isotherm. This isotherm typically shows that adsorption increases with pressure until it reaches a maximum value, termed the saturation pressure (Ps). Beyond this saturation pressure, further increases in pressure have no additional effect, and the amount of gas adsorbed remains constant.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक अधिशोषण समतापी वक्र (adsorption isotherm) को दर्शाता है, जहाँ गैस की अधिशोषित मात्रा (\(x/m\)) को दाब (P) के विरुद्ध आलेखित किया गया है। वक्र दर्शाता है कि दाब बढ़ने पर अधिशोषण बढ़ता है और एक निश्चित दाब, संतृप्ति दाब (Ps) पर अधिकतम मान पर पहुँच जाता है, जिसके बाद दाब बढ़ाने पर अधिशोषण की मात्रा स्थिर रहती है।
In simple words: Adsorption is when gas or liquid molecules stick to a surface. The amount that sticks depends on the type of gas, the type of surface, the temperature, the pressure, and how "active" the surface is. For a gas on a solid, more pressure usually means more gas sticks to the surface, but there's a limit; after a certain "saturation pressure," adding more pressure won't make more gas stick.

🎯 Exam Tip: Define adsorption clearly, distinguishing it from absorption. List all key factors influencing gas adsorption. Explain the concept of adsorption isotherm and saturation pressure with the general trend.

 

Question 4. A graph showing the adsorption isotherm according to Freundlich isotherm is given.
(a) How are the extent of adsorption and pressure related?
(b) Draw a plot of \( \log \left( \frac{x}{m} \right) \), to \( \log P \) and explain how you obtain the value of \( k \) and \( \frac{1}{n} \).
Answer:(a) **Relationship between Extent of Adsorption and Pressure:** According to the Freundlich adsorption isotherm, the relationship between the extent of adsorption \( \left( \frac{x}{m} \right) \) and pressure \( P \) is expressed as \( \frac{x}{m} = kP^{1/n} \), where \( k \) and \( n \) are empirical constants dependent on the adsorbent, adsorbate, and temperature. * At low pressure, the relationship simplifies to \( \frac{x}{m} = kP \). * At high pressure, \( \frac{x}{m} = kP^0 = k \), indicating adsorption becomes independent of pressure (saturation). * At moderate pressure, the general form \( \frac{x}{m} = kP^{1/n} \) applies. Taking the logarithm of the Freundlich equation yields: \( \log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log P \).
(b) **Plot of \( \log \left( \frac{x}{m} \right) \) vs. \( \log P \) and Determination of \( k \) and \( \frac{1}{n} \):**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक Freundlich अधिशोषण समतापी वक्र के लघुगणकीय रूप को दर्शाता है। इसमें \( \log(x/m) \) को \( \log P \) के विरुद्ध आलेखित किया गया है, जिसके परिणामस्वरूप एक सीधी रेखा प्राप्त होती है। इस सीधी रेखा का अंतःखंड (intercept) \( \log k \) के बराबर होता है, और इसकी ढलान (slope) \( 1/n \) के बराबर होती है, जो Freundlich समीकरण के स्थिरांकों को निर्धारित करने में मदद करता है। A plot of \( \log \left( \frac{x}{m} \right) \) on the y-axis against \( \log P \) on the x-axis, based on the linearized Freundlich equation \( \log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log P \), yields a straight line. * The **intercept** of this straight line on the \( \log \left( \frac{x}{m} \right) \) axis directly corresponds to \( \log k \), from which the constant \( k \) can be determined. * The **slope** of this straight line provides the value of \( \frac{1}{n} \).
In simple words: The amount of gas sticking to a surface depends on pressure, following the Freundlich equation. At low pressure, it increases with pressure. At very high pressure, it becomes constant. If you plot the logarithm of the amount adsorbed against the logarithm of pressure, you get a straight line. The point where this line crosses the y-axis gives you a value related to 'k', and the steepness of the line (slope) gives you '1/n', which are constants in the Freundlich equation.

🎯 Exam Tip: State the Freundlich equation and its linearized form. Emphasize that a plot of \( \log(x/m) \) versus \( \log P \) yields a straight line, and clearly explain how to derive the constants \( k \) and \( 1/n \) from its intercept and slope, respectively.

 

Question 7. It is found that when Litmus is shaken with animal charcoal adsorption takes place
(i) Find out the adsorbent and adsorbate in this process.
(ii) Explain the terms adsorbent and adsorbate.
(iii) Name the process of removing adsorbate from the adsorbent.
(iv) Explain the term “sorption'.
Answer:(i) In this process, **Litmus is the adsorbate**, and **animal charcoal is the adsorbent**.
(ii) **Adsorbent:** The material that provides the surface for adsorption to occur. **Adsorbate:** The substance whose molecules are adsorbed onto the surface of the adsorbent.
(iii) The process of removing an adsorbed substance from the surface of the adsorbent is called **desorption**.
(iv) **Sorption:** This term is used to describe the phenomenon where both adsorption and absorption occur simultaneously.
In simple words: Charcoal is the surface (adsorbent), and litmus is what sticks to it (adsorbate). The adsorbent is the material that captures other substances on its surface, and the adsorbate is the substance that gets captured. Removing the stuck substance from the surface is called desorption. Sorption is a general term used when both sticking to the surface (adsorption) and soaking into the bulk (absorption) happen at the same time.

🎯 Exam Tip: Clearly distinguish between adsorbent and adsorbate. Ensure you know the correct term for removing an adsorbed substance (desorption) and for combined processes (sorption).

 

Question 6. The teacher asked you to prepare ferric hydroxide sol by adding small amount of FeCl3 to water. But you get a precipitate of Fe(OH)3.
(i) Suggest a method to convert Fe(OH)3 precipitate to Fe(OH)3 sol.
(ii) What is the role of the added substance?
(iii) What happens when NaCl is added to Fe(OH)3 sol?
(iv) Give reason.
Answer:(i) **Method for converting precipitate to sol:** To transform the \( \text{Fe(OH)}_3 \) precipitate into a colloidal sol, a small amount of ferric chloride (\( \text{FeCl}_3 \)) solution can be added to the freshly prepared precipitate. This process is known as peptization.
(ii) **Role of the added substance:** The \( \text{FeCl}_3 \) solution acts as a **peptizing agent**. It helps to convert the precipitate into a colloidal solution by promoting the dispersion of the precipitate into colloidal particles.
(iii) **Effect of adding NaCl to \( \text{Fe(OH)}_3 \) sol:** When sodium chloride (\( \text{NaCl} \)) is introduced to a ferric hydroxide colloidal solution, the sol undergoes **coagulation** (also termed flocculation). This is the process where colloidal particles aggregate and precipitate out of the solution.
(iv) **Reason for coagulation:** The chloride ions (\( \text{Cl}^- \)) from the sodium chloride act as an electrolyte. Since ferric hydroxide sol is positively charged, these negatively charged chloride ions neutralize the positive charge on the colloidal particles, leading to their aggregation and subsequent precipitation.
In simple words: To turn the \( \text{Fe(OH)}_3 \) solid into a colloid, you can add a tiny bit of ferric chloride solution, which breaks up the solid into tiny particles. The ferric chloride solution helps spread the solid particles into the liquid, acting like a dispersal agent. Adding salt (NaCl) to the \( \text{Fe(OH)}_3 \) colloid will make it clump together and settle out. This happens because the salt's negative ions cancel out the positive charge on the colloid particles, making them stick to each other.

🎯 Exam Tip: For converting precipitate to sol, mention peptization and the role of the peptizing agent. For coagulation, explain how electrolytes (specifically, the oppositely charged ions) neutralize the colloidal particles, leading to precipitation.

 

Question 7. Explain the observations which are likely to be made when
(i) A beam of light is allowed to pass through a colloidal solution.
(ii) An electric current is allowed to pass through a colloidal solution.
(iii) Explain the chemistry behind each case.
Answer:(i) **Observation with a beam of light:** When a beam of light is directed through a colloidal solution, the path of the light becomes distinctly visible and illuminated. This phenomenon is known as the **Tyndall effect**.
(ii) **Observation with an electric current:** Upon passing an electric current through a colloidal solution, the colloidal particles are observed to migrate towards either the cathode (negative electrode) or the anode (positive electrode).
(iii) **Chemistry behind the observations:** * **Tyndall Effect (for light beam):** The visible path of light results from the scattering of light by the relatively large colloidal particles. These particles are big enough to scatter light but too small to be seen individually, making the light path apparent. * **Electrophoresis (for electric current):** The migration of colloidal particles in an electric field is termed **electrophoresis**. Colloidal particles acquire an electric charge, either positive or negative, which leads to the formation of a Helmholtz double layer around them. Because they are charged, these particles move towards the oppositely charged electrode, demonstrating their electrical properties and allowing the determination of their charge.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक कोलाइडल घोल के माध्यम से प्रकाश किरण के मार्ग को दर्शाता है, जिसे टिंडल प्रभाव कहते हैं। इसमें एक शंकु आकार का प्रकाशित पथ दिखाई देता है जहाँ प्रकाश कोलाइडल कणों द्वारा प्रकीर्णित होता है। यह दर्शाता है कि एक बीकर में रखे कोलाइडल विलयन में प्रकाश की किरण डालने पर, प्रकाश का मार्ग स्पष्ट रूप से दृश्यमान हो जाता है।
In simple words: When light shines through a colloid, you can see the light beam itself because the tiny particles scatter the light. This is the Tyndall effect. When electricity passes through a colloid, the charged particles move towards the positive or negative electrode. The light beam becomes visible because colloid particles are large enough to scatter light. The particles move in electricity because they have an electric charge, and charged particles move towards the opposite charge. This movement is called electrophoresis.

🎯 Exam Tip: Define the Tyndall effect and electrophoresis clearly. For each, explain the underlying principle: light scattering by colloidal particles for the Tyndall effect, and the movement of charged colloidal particles in an electric field for electrophoresis.