Answer

(c) Homogeneous catalysis does not occur at the interface of phases as in case of homogeneous catalysis reactant and catalyst have same phase and their distribution is uniform throughout.

• (a) Crystallisation occurs at the interface of phases because it involves the formation of solid crystals from a liquid or gas phase, where the solid-liquid or solid-gas interface is crucial.
• (b) Heterogeneous catalysis occurs at the interface of phases because the catalyst is in a different phase (usually solid) than the reactants (usually liquid or gas), and the reaction takes place on the surface of the catalyst.
• (d) Corrosion occurs at the interface of phases because it involves the interaction between a metal (solid phase) and its environment (often a liquid or gas phase), leading to the deterioration of the metal.

Answer

(b) As we know that, at equilibrium $\Delta G=0$

$$ \begin{aligned} \Delta H-T \Delta S & =0 \\ \Delta H & =T \Delta S \end{aligned} $$

Hence, at equilibrium enthalpy change is equal to product of temperature and entropy change.

• (a) $\Delta H>0$: This option is incorrect because at equilibrium, the Gibbs free energy change ($\Delta G$) is zero, which implies that $\Delta H = T \Delta S$. If $\Delta H > 0$, it would not necessarily satisfy the condition $\Delta G = 0$ unless $\Delta S$ is also positive and appropriately scaled by temperature, which is not guaranteed by this option alone.

• (c) $\Delta H>T \Delta S$: This option is incorrect because if $\Delta H > T \Delta S$, then $\Delta G = \Delta H - T \Delta S$ would be greater than zero, indicating that the system is not at equilibrium. At equilibrium, $\Delta G$ must be zero.

• (d) $\Delta H<T \Delta S$: This option is incorrect because if $\Delta H < T \Delta S$, then $\Delta G = \Delta H - T \Delta S$ would be less than zero, indicating that the process is spontaneous and not at equilibrium. At equilibrium, $\Delta G$ must be zero.

Answer

(d) Gas-gas interface can not be obtained as they are completely miscible in nature.

e.g., air is a mixture of various gases such as, $O_2, N_2, CO_2$ etc.

• Liquid-liquid interface can be obtained because immiscible liquids, such as oil and water, can form distinct layers when combined.
• Solid-liquid interface can be obtained because a solid can be in contact with a liquid, such as ice in water.
• Liquid-gas interface can be obtained because a liquid can be in contact with a gas, such as water in contact with air.

Answer

(c) Sorption stands for both absorption and adsorption. We can understand this by using following figures

• (a) Absorption: This option is incorrect because absorption refers specifically to the process where a substance is taken up into the volume of another substance, such as a liquid or solid, rather than just adhering to its surface.

• (b) Adsorption: This option is incorrect because adsorption refers specifically to the process where a substance adheres to the surface of another substance, rather than being taken up into its volume.

• (d) Desorption: This option is incorrect because desorption is the process by which a substance is released from or through a surface, which is the opposite of both absorption and adsorption.

Answer

(b) Extent of physisorption of a gas increases with decrease in temperature. Because in physisorption particles are held to the surface by weak van der Waals’ force of attraction hence on increasing temperature they get desorbed easily.

• (a) Increase in temperature: Physisorption is an exothermic process, meaning it releases heat. Increasing the temperature provides energy to the adsorbed molecules, causing them to desorb from the surface, thus decreasing the extent of physisorption.

• (c) Decrease in surface area of adsorbent: The extent of physisorption is directly proportional to the surface area of the adsorbent. A decrease in surface area reduces the number of available sites for adsorption, thereby decreasing the extent of physisorption.

• (d) Decrease in strength of van der Waals’ forces: Physisorption relies on weak van der Waals’ forces to hold the gas molecules on the surface of the adsorbent. A decrease in the strength of these forces would result in weaker interactions between the gas molecules and the surface, thus reducing the extent of physisorption.

Answer

(a) Extent of adsorption of adsorbate from solution phase increases with increase in amount of adsorbate in solution. As amount of adsorbate in the solution increases interaction of adsorbate with adsorbent increases which lead to increase in extent of adsorption.

(b) Decrease in surface area of adsorbent: A decrease in the surface area of the adsorbent reduces the available sites for adsorption, thereby decreasing the extent of adsorption.

(c) Increase in temperature of solution: An increase in temperature generally decreases the extent of adsorption because adsorption is usually an exothermic process, and higher temperatures favor desorption.

(d) Decrease in amount of adsorbate in solution: A decrease in the amount of adsorbate in the solution reduces the interaction between the adsorbate and the adsorbent, leading to a lower extent of adsorption.

Answer

(a) For phenomenon of adsorption $\Delta H<0$, i.e., enthalpy change during phenomenon of adsorption is negative because during adsorption, there is always a decrease in residual forces of the surface which lead to decrease in surface energy which appears as heat.

So, adsorption is an exothermic process and $\Delta H<0$.

• (b) $\Delta G<0$: This option is correct for adsorption. The Gibbs free energy change ($\Delta G$) must be negative for a spontaneous process, and adsorption is a spontaneous process.

• (c) $\Delta S<0$: This option is correct for adsorption. The entropy change ($\Delta S$) is negative because the randomness of the system decreases when molecules adhere to the surface.

• (d) $\Delta H<0$: This option is correct for adsorption. The enthalpy change ($\Delta H$) is negative because adsorption is an exothermic process, releasing heat.

Answer

(d) Physisorption is a process in which adsorbate get adsorbed on the adsorbent surface by weak van der Waals’ force of attraction. On increasing temperature the interaction between adsorbate and adsorbent becomes weak and adsorbate particles get desorbed.

• High pressure: High pressure is a favorable condition for physical adsorption because it increases the number of adsorbate molecules in the vicinity of the adsorbent surface, thereby enhancing the adsorption process.

• Negative $\Delta H$: A negative enthalpy change ($\Delta H$) indicates that the adsorption process is exothermic, which is a characteristic of physical adsorption. This makes it a favorable condition.

• Higher critical temperature of adsorbate: A higher critical temperature of the adsorbate implies stronger intermolecular forces, which enhances the likelihood of adsorption occurring. Therefore, it is a favorable condition for physical adsorption.

Answer

(b) On increasing temperature physisorption changes to chemisorption. As temperature increases, energy of activation of adsorbate particles increases which lead to formation of chemical bond between adsorbate and adsorbent.

Hence, physisorption transform into chemisorption.

• (a) Decrease in temperature: Physical adsorption (physisorption) typically decreases with a decrease in temperature because it is an exothermic process. Lower temperatures favor physisorption, not chemisorption.

• (c) Increase in surface area of adsorbent: Increasing the surface area of the adsorbent generally increases the extent of adsorption (both physisorption and chemisorption) but does not directly cause a transition from physisorption to chemisorption.

• (d) Decrease in surface area of adsorbent: Decreasing the surface area of the adsorbent would reduce the overall adsorption capacity and does not influence the transition from physisorption to chemisorption.

Answer

(a) In physisorption adsorbent does not show specificity for any particular gas because involved van der Waals’ forces are universal. It means extent of van der Waals’ interaction between adsorbate and adsorbent is constant for all gases.

(b) Gases involved behave like ideal gases: This is incorrect because the behavior of gases as ideal gases does not influence the specificity of adsorption. Ideal gas behavior pertains to the relationship between pressure, volume, and temperature of gases, not to the nature of the adsorption process.

(c) Enthalpy of adsorption is low: This is incorrect because the low enthalpy of adsorption in physisorption indicates weak interactions between the adsorbate and adsorbent, but it does not explain the lack of specificity. The universality of van der Waals’ forces is the key reason for the non-specificity.

(d) It is a reversible process: This is incorrect because the reversibility of the adsorption process does not determine the specificity of the adsorbent for different gases. Reversibility simply means that the adsorbed gas can be easily desorbed, but it does not affect the nature of the interaction between the adsorbent and different gases.

Answer

(b) Absorption means penetration of adsorbate molecules into the bulk of the adsorbent e.g., water on calcium chloride. When water is spread over calcium chloride, water get penetrate into bulk of the calcium chloride.

• (a) Water on silica gel: This is an example of adsorption, not absorption. In adsorption, water molecules adhere to the surface of silica gel rather than penetrating into its bulk.

• (c) Hydrogen on finely divided nickel: This is also an example of adsorption. Hydrogen molecules adhere to the surface of the finely divided nickel rather than being absorbed into its bulk.

• (d) Oxygen on metal surface: This is another example of adsorption. Oxygen molecules adhere to the surface of the metal rather than penetrating into its bulk.

Thinking Process

This problem includes concept of extent of conjugation and critical temperature. Extent of adsorption is directly related to critical temperature of gases.

Answer

(d) Lesser the value of critical temperature of gases lesser will be the extent of adsorption. Here $H_2$ has lowest value of critical temperature, i.e., 33.

Hence, hydrogen gas shows least adsorption on a definite amount of charcoal.

• (a) $CO_2$: $CO_2$ has a critical temperature of 304 K, which is significantly higher than that of $H_2$. Therefore, it will have a higher extent of adsorption compared to $H_2$.

• (b) $SO_2$: $SO_2$ has a critical temperature of 630 K, which is much higher than that of $H_2$. This means $SO_2$ will have a much higher extent of adsorption compared to $H_2$.

• (c) $CH_4$: $CH_4$ has a critical temperature of 190 K, which is also higher than that of $H_2$. Thus, $CH_4$ will have a higher extent of adsorption compared to $H_2$.

Answer

(c) (i) Reaction in which catalyst is in different phase than other (reactants and products) is known as heterogeneous catalysis.

(ii) $2 SO_2(g) \xrightarrow{Pt(s)} 2 SO_3(g)$

Here, reactant $SO_2$ and product $SO_3$ are in gaseous phase while platinum is in solid phase. So, this reaction represents a heterogeneous catalysis.

(iii) $N_2(g)+3 H_2(g) \xrightarrow{Fe(s)} 2 NH_3(g)$

Similarly, here $N_2$ and $H_2$ reactants are in gaseous phase while $NH_3$ is in solid phase. Whereas in other reactions catalyst is in same phase with reactant(s) and product(s).

• In reaction (i), the catalyst ( NO(g) ) is in the same phase (gaseous) as the reactants ( SO_2(g) ) and ( O_2(g) ), so it does not represent heterogeneous catalysis.
• In reaction (iv), the catalyst ( HCl(l) ) is in the same phase (liquid) as the reactants ( CH_3COOCH_3(l) ) and ( H_2O(l) ), so it does not represent heterogeneous catalysis.

Answer

(b) Associated colloid At high concentration of soap in water, soap particles present in the solution get associated around and lead to formation of associated colloid.

• (a) Molecular colloid: Molecular colloids are formed by the dispersion of individual molecules in a medium. Soap in water does not form molecular colloids because soap molecules tend to aggregate into micelles rather than remaining as individual molecules.

• (c) Macromolecular colloid: Macromolecular colloids are formed by large molecules, such as polymers, dispersed in a medium. Soap molecules are not large enough to be considered macromolecules; they form micelles instead of dispersing as individual large molecules.

• (d) Lyophilic colloid: Lyophilic colloids are formed by substances that have a strong affinity for the solvent, leading to stable colloidal dispersions. While soap can form stable dispersions, the key characteristic at high concentrations is the formation of micelles (associated colloids), not just the affinity for the solvent.

Answer

(b) Aqueous solution of soap above critical micelle concentration lead to formation of colloidal solution. Tyndall effect is a characteristic of colloidal solution in which colloidal particles show a coloured appearance when sunlight is passes through it and seen from the perpendicular side.

• (a) Aqueous solution of soap below critical micelle concentration: Below the critical micelle concentration, soap molecules are not aggregated into micelles and thus do not form a colloidal solution. Without the formation of colloidal particles, the Tyndall effect cannot be observed.

• (c) Aqueous solution of sodium chloride: Sodium chloride in water forms a true solution where the solute particles are ions that are too small to scatter light. Therefore, it does not exhibit the Tyndall effect.

• (d) Aqueous solution of sugar: Sugar dissolves in water to form a true solution with molecules that are too small to scatter light. As a result, it does not show the Tyndall effect.

Answer

(c) Lyophobic sol can be protected by addition of lyophilic sol. As lyophobic sols are readily precipitated on addition of small amount of electrolytes or shaking, or heating hence they are made stable by adding lyophillic sol which stabilises the lyophobic sols.

• (a) By addition of oppositely charged sol: Adding an oppositely charged sol can lead to coagulation or precipitation of the lyophobic sol due to neutralization of charges, rather than protecting it.

• (b) By addition of an electrolyte: Adding an electrolyte to a lyophobic sol typically causes coagulation or precipitation because the added ions neutralize the charges on the colloidal particles, leading to their aggregation.

• (d) By boiling: Boiling a lyophobic sol can cause the particles to collide more frequently and with greater energy, which can lead to coagulation or precipitation rather than stabilization.

Answer

(d) Freshly prepared precipitate sometimes gets converted to colloidal solution by peptisation. Peptisation is a process in which by addition of a suitable peptising agent precipitate gets converted into colloidal solution.

• (a) Coagulation: Coagulation is the process where colloidal particles aggregate to form larger particles, leading to the formation of a precipitate. It is the opposite of peptisation and does not convert a precipitate into a colloidal solution.

• (b) Electrolysis: Electrolysis involves the use of an electric current to drive a non-spontaneous chemical reaction. It is not typically used to convert a precipitate into a colloidal solution.

• (c) Diffusion: Diffusion is the movement of particles from an area of higher concentration to an area of lower concentration. It does not involve the conversion of a precipitate into a colloidal solution.

Thinking Process

This problem includes concept of Hardy-Schulze law. According to which higher charge on oppositely charge ion of electrolyte decide the coagulating power of colloid.

Answer

(b) According to Hardy-Schulze law, greater the charge on anion greater will be its coagulating power.

Here, $PO_4^{3-}$ have highest charge. Hence, $PO_4^{3-}$ have highest coagulating power.

• (a) $Na_2 S$: The anion $S^{2-}$ has a charge of 2. According to the Hardy-Schulze law, the coagulating power is directly proportional to the charge on the ion. Since $S^{2-}$ has a lower charge compared to $PO_4^{3-}$, its coagulating power is less.

• (c) $Na_2 SO_4$: The anion $SO_4^{2-}$ has a charge of 2. Similar to $S^{2-}$, the charge is lower than that of $PO_4^{3-}$, resulting in a lower coagulating power.

• (d) $NaCl$: The anion $Cl^{-}$ has a charge of 1. This is the lowest charge among the given options, making its coagulating power the least according to the Hardy-Schulze law.

Answer

(d) Sol is a colloidal system in which solid substance is a dispersed phase and a liquid is a dispersion medium. e.g., paint, cell fluids etc.

In paints solid colouring particles are dissolved in liquid dispersion medium.

• (a) Solid sol: A solid sol is a colloidal system where both the dispersed phase and the dispersion medium are solids. This does not match the given condition of having a liquid as the dispersion medium.

• (b) Gel: A gel is a colloidal system where a liquid is dispersed in a solid. This is the opposite of the given condition, where a solid should be the dispersed phase and a liquid the dispersion medium.

• (c) Emulsion: An emulsion is a colloidal system where both the dispersed phase and the dispersion medium are liquids. This does not match the given condition of having a solid as the dispersed phase.

Answer

(d) The value of colligative properties of colloidal solution are of small order in comparison to those of true solution of same concentration because of colloidal particles are comparatively less in number.

This is due to slight large size of colloidal particles in comparison to particles present in true solution. Size of colloidal particles is in between $1 nm$ to $1000 nm$.

• (a) The enormous surface area of colloidal particles does not directly affect the colligative properties. Colligative properties depend on the number of particles in the solution, not their surface area.

• (b) The fact that colloidal particles remain suspended in the dispersion medium does not influence the colligative properties. Colligative properties are determined by the number of solute particles, not their state of suspension.

• (c) Forming lyophilic colloids does not impact the colligative properties. Lyophilic colloids are characterized by their affinity for the solvent, but this does not change the number of particles in the solution, which is what affects colligative properties.

Answer

(b) Correct sequence is $I \rightarrow III \rightarrow Ii \rightarrow I V \rightarrow V$

Each transformation denotes a meaningful process as follows

I $\rightarrow$ adsorption of $A$ and $B$ on surface

III $\rightarrow$ II interaction between $A$ and $B$ to form intermediate

• Option (a): The sequence $I \rightarrow II \rightarrow II \rightarrow IV \rightarrow V$ is incorrect because it repeats step II twice, which is not a logical progression in the mechanism of catalysis. Each step should represent a distinct stage in the process.

• Option (c): The sequence $I \rightarrow II \rightarrow II \rightarrow V \rightarrow IV$ is incorrect because it repeats step II twice and places step V before step IV. In the correct sequence, step IV (formation of the product) should come before step V (desorption of the product).

• Option (d): The sequence $I \rightarrow II \rightarrow III \rightarrow V \rightarrow IV$ is incorrect because it places step III after step II, which disrupts the logical order of the mechanism. Step III (interaction between A and B to form an intermediate) should come immediately after step I (adsorption of A and B on the surface).

Answer

(c) River water is a colloidal solution of clay and sea. Water contains various electrolytes. When river water comes in contact with sea water, then the electrolytes present in sea water coagulate the suspended colloidal particles which ultimately settle down at the point of contact.

• Emulsification: Emulsification is the process of mixing two immiscible liquids, such as oil and water, to form a stable mixture. This process is not involved in the formation of a delta, which is related to the settling of solid particles from a colloidal solution.

• Colloid formation: Colloid formation refers to the creation of a colloidal system where fine particles are dispersed in a continuous medium. While river water is a colloidal solution, the formation of a delta is due to the coagulation of these colloidal particles, not their initial formation.

• Peptisation: Peptisation is the process of converting a precipitate into a colloidal solution by adding a suitable electrolyte. This process is the reverse of coagulation and does not contribute to the formation of a delta, which requires the settling of particles rather than their dispersion.

Answer

(c) According to the Freundlich adsorption isotherm

$$ \frac{x}{m}=k p^{\frac{1}{n}} $$

Taking log on both side $\underset{\substack{\uparrow \\ \\ \mathrm{Y}}}{\log \frac{x}{m}}=\underset{\substack{\uparrow \\ \\ \mathrm{= m}}}{(\frac{1}{n})} \underset{\substack{\uparrow \\ \\ x}}{\log p}+\underset{\substack{\uparrow \\ \\ \mathrm{+ C}}}{\log k}$

On comparing it with equation of straight line and drawing the graph $\log \frac{x}{m}$ versus $\log p$, we get a straight line with intercept $\log k$ and slope of the straight line gives the value of $\frac{1}{n}$.

• Option (a): The curve in option (a) does not represent a straight line when plotted as (\log \frac{x}{m}) versus (\log p). According to the Freundlich adsorption isotherm, the plot should be a straight line, indicating a linear relationship between (\log \frac{x}{m}) and (\log p).

• Option (b): The curve in option (b) also does not represent a straight line when plotted as (\log \frac{x}{m}) versus (\log p). The Freundlich adsorption isotherm requires a linear plot, which is not the case here.

• Option (d): The curve in option (d) does not follow the linear relationship required by the Freundlich adsorption isotherm when plotted as (\log \frac{x}{m}) versus (\log p). The Freundlich isotherm necessitates a straight line, which is not observed in this option.

Answer

(d) Absorption of ionic species from solution is not responsible for the presence of electric charge on the sol particles. Charge on the sol particles is due to

(i) electrons capture by sol particles during electro dispersion of metal.

(ii) preferential adsorption of ionic species from solution.

(iii) formation of Helmholtz electrical double layer.

• (a) Electron capture by sol particles: This process is responsible for the presence of electric charge on the sol particles because during electro dispersion of metal, sol particles can capture electrons, leading to the development of a charge.

• (b) Adsorption of ionic species from solution: This process is responsible for the presence of electric charge on the sol particles because sol particles can preferentially adsorb ions from the solution, which contributes to their charge.

• (c) Formation of Helmholtz electrical double layer: This process is responsible for the presence of electric charge on the sol particles because the formation of an electrical double layer around the sol particles results in the development of a charge.

Answer

(b) Above figure represent adsorption of yellowish brown colour of raw sugar by animal charcoal.

Here, aqueous solution of raw sugar is filtered by using animal charcoal. Yellowish brown colour of raw sugar is adsorbed and filterate is colourless which gives white colour on cystallisation. Hence, this phenomenon is adsorption.

• Absorption: Absorption refers to the process where a substance is taken up into the volume of another substance, such as a liquid being absorbed by a sponge. In the given figure, the process involves the surface phenomenon where the yellowish brown color of raw sugar is removed by animal charcoal, which is characteristic of adsorption, not absorption.

• Coagulation: Coagulation is the process where particles in a liquid clump together to form a solid or semi-solid mass. This process is typically used in the context of blood clotting or in water treatment to remove suspended particles. The figure does not depict the clumping of particles but rather the removal of color through surface interaction, which is not coagulation.

• Emulsification: Emulsification is the process of mixing two immiscible liquids, such as oil and water, to form a stable mixture. This process involves the dispersion of one liquid into another in the form of tiny droplets. The figure does not show the mixing of two immiscible liquids but rather the removal of color from a solution, which is not emulsification.

Thinking Process

This problem is based on concept of micelle formation and CMC (critical micelle concentration).

Answer

$(b, c)$

Micelle formation Some substances at low concentration behaves as a normal electrolytes but at higher concentration exhibit colloidal behaviour due to formation of micelles.

CMC (Critical Micelle Concentration) The concentration above which it behaves as a micelle known as critical micelle concentration (CMC).

e.g., soap solution in aqueous solution above particular concentration (called CMC $=10^{-4}-10^{3} mol L^{-1}$ ) forms soap micelles.

On dilution soap solution behaves as a normal electrolyte and after adding excess of water intermolecular force of attraction between the soap particles decreases and soap solution micelles may revert to individual ions.

• (a) Micelle formation by soap in aqueous solution is possible at all temperatures

  This statement is incorrect because micelle formation is not possible at all temperatures. Micelle formation typically occurs above a certain temperature known as the Krafft temperature. Below this temperature, the solubility of the soap molecules is too low to form micelles.

• (d) Soap solution behaves as a normal strong electrolyte at all concentrations

  This statement is incorrect because soap solutions do not behave as normal strong electrolytes at all concentrations. At concentrations above the critical micelle concentration (CMC), soap molecules aggregate to form micelles, and the solution exhibits colloidal behavior rather than behaving as a strong electrolyte.

Answer

$(a, b)$

(a) Same reactants may give different products by using different catalysts as different catalysts have different specific functions to mold the reaction towards specific product.

e.g., starting with $H_2$ and $CO$, and using different catalysts, we get different products.

(i) $CO(g)+3 H_2 \xrightarrow{Ni} CH_4(g)+H_2 O(g)$

(ii) $CO(g)+2 H_2(g) \xrightarrow[Cr_2 O_3]{\stackrel{Cu / ZnO}{\longrightarrow}} CH_3 OH(g)$

(iii) $CO(g)+H_2(g) \xrightarrow{Cu} HCHO(g)$

(b) Catalyst does not change $\Delta H$ of reaction as $\Delta H$ of reaction is difference between enthalpy of reactants and products. So, it does not change during catalysed reaction.

• (c) Catalyst is required in large quantities to catalyse reactions: This statement is incorrect because catalysts are typically required in very small quantities relative to the reactants. They function by providing an alternative reaction pathway with a lower activation energy, and they are not consumed in the reaction, allowing them to be used repeatedly.

• (d) Catalytic activity of a solid catalyst does not depend upon the strength of chemisorption: This statement is incorrect because the catalytic activity of a solid catalyst often depends on the strength of chemisorption. Chemisorption involves the formation of strong chemical bonds between the catalyst surface and the reactants, which is crucial for the activation of the reactants and the subsequent reaction. If the chemisorption is too weak, the reactants may not be adequately activated; if it is too strong, the reactants may become too tightly bound to the catalyst surface, hindering the reaction.

Thinking Process

To solve this problem follow steps given below.

Write Freundlich equation and transform it into different form depending upon value of $n$ then choose the correct answer.

Answer

$(a, c)$

According to Freundlich adsorption isotherm

$$ \frac{x}{m} \propto p^{\frac{1}{n}} \Rightarrow \frac{x}{m}=k p^{\frac{1}{n}} $$

(a) At $\frac{1}{n}=0$ this equation becomes $\frac{x}{m}=k p^{0}=k$

Extent of adsorption is independent of pressure.

(c) At $x=0, \quad \frac{x}{m}=k p^{\frac{1}{0}}=k p^{\infty}$

Hence, $\frac{x}{m}$ vs $p$ graph can be plotted as

• (b) When $\frac{1}{n}=0$, the adsorption is directly proportional to pressure

  This option is incorrect because when $\frac{1}{n}=0$, the equation $\frac{x}{m}=k p^{0}=k$ indicates that the extent of adsorption is independent of pressure, not directly proportional to it.

• (d) When $n=0$, plot of $\frac{X}{m} vs p$ is a curve

  This option is incorrect because when $n=0$, the term $\frac{1}{n}$ becomes undefined (since division by zero is not possible). Therefore, the scenario where $n=0$ is not physically meaningful in the context of the Freundlich adsorption isotherm, and no valid plot can be derived from it.

Answer

$(b, c)$

$H_2$ gas is adsorbed on activated charcoal to a very little extent in comparison to easily liquefiable gases due to

(i) very low van der Waals’ forces and

(ii) very low critical temperature equal to $33 K$.

• (a) very strong van der Waals’ interaction: This option is incorrect because if the van der Waals’ interactions were very strong, $H_2$ gas would be adsorbed to a greater extent, not to a very little extent.

• (d) very high critical temperature: This option is incorrect because a high critical temperature would indicate that the gas is easily liquefiable, which would result in a higher extent of adsorption on activated charcoal.

Answer

$(b, d)$

Presence of equal and similar charges on colloidal particles provides stability to colloids as repulsive forces between charge particles having same charge prevent them from colliding when they come closer to each other.

• (a) Mixing two oppositely charged sols neutralises their charges and stabilises the colloid: This statement is incorrect because mixing two oppositely charged sols would neutralize their charges, leading to coagulation or precipitation rather than stabilization. The neutralization of charges removes the repulsive forces that keep the colloidal particles apart, causing them to aggregate and settle out of the dispersion medium.

• (c) Any amount of dispersed liquid can be added to emulsion without destabilising it: This statement is incorrect because adding an excessive amount of dispersed liquid to an emulsion can lead to phase separation and destabilization. Emulsions have a limited capacity to hold the dispersed phase, and exceeding this capacity can cause the emulsion to break, leading to the separation of the dispersed and continuous phases.

Answer

$(b, d)$

Emulsions are liquid-liquid colloidal system. They can’t be broken by adding more amount of dispersion medium and adding emulsifying agent as on adding more amount of dispersion medium they become dilute and on adding emulsifying agent they get stabilises.

• Heating: Heating can increase the kinetic energy of the particles in the emulsion, potentially causing the dispersed phase to coalesce and separate from the continuous phase, thus breaking the emulsion.
• Freezing: Freezing can cause the dispersed phase to solidify and separate from the continuous phase, leading to the breakdown of the emulsion.

Answer

$(a, d)$

The droplets present in emulsion has negative charge. It can be precipitated by adding electrolyte such as $KCl, NaCl$ etc. Since, glucose and urea do not produce ions on dissolving in water.

Hence, they are non-electrolyte and do not precipitate the negatively charged emulsion.

• Glucose: Glucose does not produce ions when dissolved in water. It is a non-electrolyte and therefore cannot precipitate the negatively charged emulsion.

• Urea: Urea also does not produce ions when dissolved in water. It is a non-electrolyte and thus cannot precipitate the negatively charged emulsion.

Answer

(c, $d$)

Lyophilic colloids (liquid loving colloids) which are also known as reversible colloid can’t be coagulated easily. The stability of these colloids are due to (i) charge on colloidal particles and (ii) solvation of colloidal particles.

• Lyophobic colloids (liquid hating colloids) can be coagulated easily because they are less stable due to the absence of solvation and rely primarily on the charge for stability.
• Irreversible colloids cannot be re-dispersed once coagulated, making them easier to coagulate compared to reversible colloids.

Answer

$(a, c)$

Lyophobic sol is unstable in nature when lyophilic sol is added to lyophobic sol then lyophobic sol is protected because a film of lyophilic sol is formed over lyophobic sol.

Hence, (a) and (c) are correct.

• Option (b) is incorrect because the lyophilic sol does not need protection; it is already stable due to its affinity for the solvent.
• Option (d) is incorrect because it is the lyophilic sol that forms a protective film over the lyophobic sol, not the other way around.

Thinking Process

This problem is based on concept of electroosmosis. It can be solved by knowing the exact meaning of electroosmosis

Answer

$(b, c)$

The movement of colloidal particles under an applied electric potential is called electrophoresis. When this movement of particles is prevented by some suitable means, it is observed that the dispersion medium begins to move in an electric field. This phenomenon is termed as electroosmosis.

• (a) Reverse osmosis takes place: Reverse osmosis is a process where water molecules move through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration by applying pressure. This process is unrelated to the movement of colloidal particles or the dispersion medium in an electric field.

• (d) Dispersion medium becomes stationary: When electrophoresis is prevented, the dispersion medium does not remain stationary. Instead, it begins to move under the influence of the electric field, which is the basis of electroosmosis.

Answer

$(a, b)$

In a reaction catalyst changes physically and qualitatively as it is unaltered during the reaction and remain quantitatively intact after completion of reaction and chemically does not change.

• Option (c) chemically is incorrect because a catalyst does not undergo any permanent chemical change during the reaction; it remains chemically the same before and after the reaction.
• Option (d) quantitatively is incorrect because the amount of catalyst remains unchanged throughout the reaction; it is not consumed or altered in quantity.

Answer

$(a, d)$

When a chalk stick is dipped in ink absorption as well as absorption both occurs. Adsorption of coloured substance and absorption of solvent takes place.

• Option (b) Adsorption of solvent: This option is incorrect because adsorption typically involves the adhesion of molecules from a gas, liquid, or dissolved solid to a surface. In the case of a chalk stick dipped in ink, the solvent (usually water) is absorbed into the porous structure of the chalk rather than just adhering to its surface.

• Option (c) Absorption and adsorption both of solvent: This option is incorrect because while both absorption and adsorption occur, the adsorption is primarily of the colored substance (the dye in the ink) rather than the solvent itself. The solvent is mainly absorbed into the chalk.

Answer

It facilitates the adsorption of desired species. If surface is covered by the gases of air then it will not be available for adsorption of desired gases. So, it is very important to have clean surface in surface studies i.e., study of surface chemistry.

Answer

Chemisorption referred to as activated adsorption as it involves chemical bond formation between reactant and adsorbent molecules. Formation of chemical bond requires high activation energy. So, it is activated on increasing temperature.

Answer

At lower concentration, soap behaves as a normal solution of electrolyte in water. However, after a certain concentration, called critical micelle concentration, colloidal solution is formed due to aggregation of colloidal particles.

$CMC$ for soap solution is $10^{-4}$ to $10^{-3} mol L^{-1}$.

Answer

Gold sol is a solvent repelling sol i.e., a lyophobic sol and unstable in nature. Addition of gelatin stabilises the gold sol because gelatin forms lyophilic sol and act as a protective colloid.

Answer

As we know artificial rainfall occurs when oppositely charged clouds meets. Since, clouds are colloidal in nature and carry charge. Spray of silver iodide, an electrolyte from aeroplane results in coagulation of colloidal water particles leading to rain. Sometimes electrified sand is also used for this purpose.

Answer

Emulsifying agent is added to emulsion to stabilise the emulsion. Emulsifying agent form a layer between suspended particles and the medium and hence stabilises the emulsion. Ice cream (emulsion) is stabilised by emulsifying agent like gelatin.

Answer

$4 $% solution of nitrocellulose in a mixture of alcohol and ether is called collodion.

Answer

We add alum to purify water as alum coagulates the colloidal impurities present in water, so that these impurities get settle down and remove by decantation or filtration. Thus, water gets purified by adding alum to water.

Answer

When electric potential is applied to colloidal solution, the colloidal particles move towards one or other electrode. Positively charged particles move towards the cathode while negatively charged particles move towards the anode.

The movement of colloidal particles under an applied electric potential is called electrophoresis. When electrophoresis is prevented by some means, then the dispersion medium begins to move in an electric field. This phenomenon is termed as electroosmosis.

Answer

Brownian movement may be defined as continuous zig-zag movement of colloidal particles in a colloidal sol. A state of continuous zig-zag motion of colloidal particles appears to be in view due to unbalanced bombardment of the particles of dispersed phase by molecules of dispersion medium. This Brownian movement stabilises the sol.

Answer

Positively charged ions coagulate the negatively charged sol and negatively charged ion coagulate the positively charged sol. Positively charged sol of hydrated ferric oxide is formed when $FeCl_3$ is added in excess of hot water. On adding excess of $NaCl$ to this sol, negatively charged $Cl^{-}$ions coagulate the positively charged sol of hydrated ferric oxide.

Answer

Emulsifying agents stabilise the emulsion by forming an interfacial layer between suspended particles and the dispersion medium. e.g., gelatin is added to ice cream to stabilise it.

Answer

Some medicines are more effective in colloidal form because they have large surface area so easily assimilated in the body.

Answer

Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tanin (negatively charged), a colloid, it results in mutual coagulation and gets harden. Thus, leather get hardened after tanning.

Answer

In Cottrell precipitator, smoke particles (charged) are passed through a chamber containing plates with charge opposite to the smoke particles, smoke particles lose their charge on the plates and get precipitated.

Answer

To distinguish between dispersed phase and dispersion medium we increase the concentration of any one dispersion medium or dispersed phase then notice the change.

When dispersion medium is added to an emulsion, it gets diluted to any extent. But on adding dispersed phase it forms a separate layer, if added in excess.

Answer

Minimum quantity of an electrolyte required to cause precipitation of a sol is called its coagulating value. Greater the charge on flocculating ion and smaller is the amount of electrolyte required for precipitation, higher is the coagulating power of coagulating ion (Hardy-Schulze Law).

Phosphate ion bear -3 charge while chloride ion carries only -1 charge and due to high charge phosphate ion has high coagulating power than that of chloride ion.

Answer

Blood is a colloidal sol. When we rub the injured part with moist alum then coagulation of blood takes place. Hence, main reason is coagulation, which stops the bleeding.

Answer

Charge on sol is decided by adsorption of ions present in medium. Adsorption of positively charged $Fe^{3+}$ ions takes place by the sol of hydrated ferric oxide. Thus, $Fe(OH)_3$ colloid has positive charge when prepared by adding $Fe(OH)_3$ to hot water.

Answer

Behaviour of physisorption and chemisorption on increase in temperature can be explained on the basis of nature of forces present to bind their particles. Physisorption involves weak van der Waals’ forces which weakens with increase in temperature.

Chemisorption involves formation of chemical bond which requires activation energy hence, it is favoured by rise in temperature.

Answer

Traces of electrolyte which stabilises the colloids is removed completely making the colloid unstable. So, coagulation occurs on prolonged dialysis.

Answer

White coloured precipitate of silver halide becomes coloured in the presence of dye eosin because dye eosin (coloured) gets adsorbed on the surface of silver halide precipitate.

Answer

Role of activated charcoal in gas mask can be explained on the basis of adsorption. Activated charcoal adsorbs various poisonous gases on its surface present in coal mines.

Answer

Formation of delta at the meeting place of sea and river water is due to coagulation. River water (colloid of sea water + clay) has many dissolved electrolytes. The place where river meets sea is the site for coagulation. Deposition of coagulated clay results in delta formation.

Answer

Adsorption of $H_2$ on finely divided nickel (physisorption) involves weak van der Waals’ forces. When temperature is increased, hydrogen molecules dissociate into hydrogen atoms, form chemical bonds with the metal atoms at the surface (chemisorption).

Answer

Desorption is important for a substance to act as a good catalyst so that after the reaction, the products formed on the surface separate out (desorbed) to create free surface again for other reactant molecules to approach the surface and react.

If desorption does not occur then other reactants are left with no space on the catalysts surface for adsorption and reaction will stop.

Answer

Diffusion of gas molecules occur at the surface of catalyst (solid) followed by adsorption. In the same way, the product formed diffuse from the surface of the catalyst leaving the surface free for more reactant molecules to get adsorbed and undergo reaction.

Answer

When gaseous molecules come in contact with the surface of a solid catalyst, a weak chemical bond is formed between the surface (catalyst) molecules and reactant (gas) molecules. Thus, concentration of reactant molecules increases at the surface.

The rate of reaction increases by adsorption of different molecules side by side facilitating the chemical reaction. Adsorption, being exothermic also help in increasing the rate of reaction (chemisorption increases with rise in temperature).

Answer

The optimum temperature range for enzymatic activity is $298-310 K$, i.e., enzymes are inactive beyond this temperature range (high or low).

Thus, during fever (temperature $>310 K$ ) the activity of enzymes may be affected.

Answer

A. $\rightarrow(2)$

B. $\rightarrow(3)$

C. $\rightarrow(4)$

D. $\rightarrow(1)$

A. When sulphur vapours passed through cold water it leads to formation of molecular colloids.

B. When soap is mixed with water above critical micelle concentration it lead to formation of associated colloids.

C. White of egg whipped with water is an example of macromolecular colloids in which high molecular mass proteneous molecule acts as a colloidal particle.

D. Soap mixed with water below critical micelle concentration is known as normal electrolyte solution.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow(2)$

D. $\rightarrow(1)$

A. Dispersion medium moves in an electric field is known as electroosmosis.

B. Solvent molecules pass through semipermeable membrane towards solvent side is known as reverse-osmosis.

C.Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charge electrodes is known as electrophoresis.

D. Solvent molecules pass through semipermeable membranes towards solution side is known as osmosis.

Answer

A. $\rightarrow(2)$

B. $\rightarrow$ (3)

C. $\rightarrow(4)$

D. $\rightarrow(1)$

A. Lyophobic colloid (solvent hating colloid) are readily protected by small amount of electrolyte. These colloids are also stabilised by addition of lyophilic colloids which makes a protective layer around lyophobic sol. Hence, lyophilic sol are known as protective colloid.

B. Liquid-liquid colloid is also known as emulsion if they are partially miscible or immiscible liquids.

C. When $FeCl_3$ is added to hot water it lead to the formation of positively charged colloid.

D. When $NaOH$ is added to $FeCl_3$ it lead to the formation of negatively charged colloid.

Answer

A. $\rightarrow(2)$

B. $\rightarrow(3)$

C. $\rightarrow(4)$

D. $\rightarrow(1)$

Colloids are classified on the basis of types of dispersed phase and dispersion medium.

Answer

A. $\rightarrow$ (4) $\quad$

B. $\rightarrow$ (3) $\quad$

C. $\rightarrow$ (1) $\quad$

D. $\rightarrow(2)$

A. Purification of colloid can be done by dialysis in which ions/particles are removed from solution through semipermeable membrane.

B. Peptisation is a process in which when small quantity of electrolyte (peptising agent) is added to precipitate. It leads to formation of colloidal solution.

C. The process of removing of oily or greasy dirt from the cloth is done by emulsification.

D. Process of setting of colloidal particle is called coagulation. Electrophoresis is a process in which on applying electric potential to the electrodes dipped in sol, the oppositely charged particles of colloidal solution move towards oppositely charged electrodes, get discharged and precipitated.

Answer

A. $\rightarrow(4) \quad$

B. $\rightarrow(3) \quad$

C. $\rightarrow(1) \quad$

D. $\rightarrow(2)$

A. Butter is an example of dispersion of liquid in solid.

B. Pumice stone is an example of dispersion of gas in solid in which gas bubbles are pearced within solid particles.

C. Milk is a dispersion of liquid in liquid in which fats and protein are dissolved in milk.

D. Paint is an example of solid in liquid.

Answer

(c) Assertion is correct, but reason is wrong.

An ordinary filter paper impregnated with collodion solution stops the flow of colloidal particles because pore size of the filter paper becomes smaller than the size of colloidal particles.

Answer

(b) Assertion and reason both are correct, but reason does not explain assertion.

Colloidal solutions show colligative properties as colloidal particles have large size so colloidal particles have small value of colligative properties because number of particles are small in comparison to normal solution.

Answer

(e) Assertion is incorrect, but reason is correct.

Colloidal solutions show Brownian motion and this Brownian motion is responsible for stability of sols.

Answer

(a) Assertion and reason both are correct and reason is correct explanation of assertion. According to Hardy-Schulze law greater the charge/valency on flocculating ion added greater will be its power to cause precipitation.

Coagulating power $\propto$ Valency of flocculating ion

Answer

(a) Assertion and reason both are correct and reason is correct explanation of assertion. Detergents with low CMC are more economic to use as cleansing action of detergents involves the formation of micelles. These are formed when the concentration of detergent becomes equal to $CMC$. If $CMC$ has lower value then it will form $CMC$ easily and readily.

Answer

Catalyst is used to increase rate of reaction. Heterogeneous catalyst is used to increase rate of reaction in which catalyst is not in phase with reactants and products.

Role of adsorption in heterogeneous catalysis are

(i) Diffusion of reactants to the surface of the catalyst.

(ii) Reactants are adsorbed on the catalyst surface.

(iii) Occurrence of chemical reaction at catalyst surface.

(iv) Desorption.

(v) Diffusion of reaction products away from the catalysts surface.

e,g.,

Answer

There are various applications of adsorption in chemical analysis.

Some of which are as follows

(i) In thin layer chromatography The lower adsorbing particles comes out readily while another having higher adsorbing tendency comes out later. On this basis compounds are separated or analysed.

(ii) In adsorption indicators Surface of certain precipitate has an ability to adsorb some dyes to produce a characteristic colour e.g., AgX adsorbs eosin dye and thereby producing a characteristic colour at the end point.

(iii) In qualitative analysis Specific material has specific adsorption tendency so particular ion can be identified very easily.

(iv) In the separation of inert gases Different inert gases are adsorbed to different extents at different temperatures on coconut charcoal. This forms the basis of their separation from a mixture.

Answer

In froth floatation process sulphide ore is shaken with pine oil and water, the ore particles are adsorbed on froth that floats and the gangue particles settle down in tank. Thus, role of adsorption in froth floatation process can be understood as following processes.

(i) Adsorption of pine oil on sulphide ore particles.

(ii) Formation of emulsion takes place.

(iii) Froth is formed along with ore particles.

(iv) Mechanism of the functioning of shape selective catalysis.

As sulphide are extracted using froth floatation method therefore, only sulphide ore particles will show these type of adsorbing tendency.

Answer

The catalytic reaction which depends upon pore structure of catalyst and the size of the reactant and product molecules is known as shape selective catalysts. Zeolites are good shape selective catalyst because of honey comb like structure.

(i) They are microporous aluminosilicates with $Al-O-Si$ framework and general formula $M _{x / n}[(AlO _2) _{x}(SiO_2) _{y}] m H_2 O$

(ii) The reactions taking place in zeolites depend upon the size and shape of the reactant and product molecules as well as upon the pores and cavities of the zeolites.

(iii) Zeolites are widely used as catalysts in petrochemical industries for cracking of hydrocarbonds and isomerisation. They are also used for removing permanent hardness of water.

(iv) e.g., ZSM-5 is a catalyst used in petroleum industry

$$ \text { Alcohols } \underset{\text { Dehydration }}{\stackrel{Z S M-5}{\longrightarrow}} \text { Gasoline(petrol) (A mixture of hydrocarbons) } $$