Surface Chemistry

The complete reaction,

$ \mathrm{CO}(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \xrightarrow{\mathrm{Cu}} \underset{\substack{\text { Formaldetryde } \\ \mathrm{}\mathrm{~}}}{\mathrm{HCHO(g)}} $

Catalyst $X=\mathrm{Cu}$

$ \mathrm{CO}(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \xrightarrow{\mathrm{Ni}} \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g}) $

Catalyst $Y=\mathrm{Ni}$

Statement $A$ and $D$ are correct while $B$ and $C$ are incorrect. The correct form of $B$ and $C$ are $B$ as follow.

(B) Zeta potential is related to stability of colloidal particles

(C) Brownian motion becomes slower if the viscosity is high. Higher viscosity resists the random motion of particles.

1. Sol: A colloidal system where a solid is dispersed in a liquid.

   • Examples: Paint, ink, blood, mud.

   • From List-II, Paint (III) is a solid dispersed in a liquid.

   • So, A - III.

2. Foam: A colloidal system where a gas is dispersed in a liquid.

   • Examples: Whipped cream, soap lather, shaving cream.

   • From List-II, Whipped cream (II) is gas (air) dispersed in cream (liquid).

   • So, B - II.

3. Gel: A colloidal system where a liquid is dispersed in a solid. Gels are typically semi-solid.

   • Examples: Jelly, cheese, butter, gelatin.

   • From List-II, Butter (IV) is primarily water (liquid) dispersed in solid fat.

   • So, C - IV.

4. Aerosol: A colloidal system where a solid or liquid is dispersed in a gas.

   • Examples: Fog, mist, cloud (liquid in gas); smoke (solid in gas).

   • From List-II, Cloud (I) is fine droplets of water (liquid) dispersed in air (gas).

   • So, D - I.

Combining these matches:

• A - III

• B - II

• C - IV

• D - I

Most effective coagulating agent for $\mathrm{Sb}_2 \mathrm{~S}_3$ sol is $\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$. This is because $\mathrm{Sb}_2 \mathrm{~S}_3$ forms a negative charge sol, and according to the Hardy-Schulze rule, the coagulating power increases with oppositely charged ions.

Freundlich adsorption isotherm equation:

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

where

• $ \frac{x}{m} $ = mass of gas adsorbed per unit mass of adsorbent,

• $ p $ = equilibrium pressure,

• $ k $ and $ n $ = empirical constants,

• $ n > 1 $.

Or, taking logarithms,

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

Option A:

$ \frac{x}{m} = k\, p^{1/n} \quad (n > 1) $

✅ Correct.

Option B:

Extent of adsorption of gas is more at high temperature than at low temperature.

Adsorption is an exothermic process.

Hence, increasing temperature decreases adsorption (Le Chatelier’s principle).

❌ Incorrect statement.

Option C:

$ \frac{1}{n} \text{ represents the slope of the isotherm (in log–log plot)} $

Since $\log\frac{x}{m} = \log k + \frac{1}{n}\log p$,

✅ Correct.

Option D:

$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$ holds good over a limited range of pressures.

At very high or very low pressures, Freundlich isotherm deviates.

✅ Correct.

✅ Final Answer:

Option B — Extent of adsorption of gas is more at high temperature than at low temperature — is not correct about Freundlich adsorption isotherm.

Option A:

Brownian movement and Tyndall effect are shown by colloidal systems.

✅ Correct — both Brownian movement and the Tyndall effect are characteristic properties of colloids.

Option B:

Hardy-Schulze rule is related with coagulation.

✅ Correct — the Hardy–Schulze rule relates the coagulating power of ions to their charge.

Option C:

Gold number is a measure of the protection power of a lyophilic colloid.

✅ Correct — the gold number represents the minimum amount of protective (lyophilic) colloid required to prevent coagulation of a standard gold sol.

Option D:

Aerosol is a colloidal system in which gas is dispersed in liquid.

❌ Incorrect — an aerosol is a colloid in which a solid or liquid is dispersed in a gas, not the other way around.

When gas is dispersed in a liquid, the system is called foam (e.g., whipped cream).

✅ Answer: Option D

The statement given in option (c) is incorrect regarding adsorption theory of heterogeneous catalysis. The correct form is,

In heterogeneous catalysis, product molecules will not permanently bonded to catalyst surface.

They are intially absorbed, but then desorbed after the reaction occur allowing the catalyst to be reused.

$ \begin{aligned} &\text { The correct match is }\\ &\text { A-I, B-II, C-IV, D-III } \end{aligned} $

The statement given in $I$ is incorrect. It's correct form is, If concentration < CMC : No micelles form hence not stable. Since, $10^{-7} \ll 5 \times 10^{-4}$, the concentration is below CMC, so micelles will not form and thus they can't be stable.

Let's match each sol with its appropriate method of preparation:

(A) As₂S₃ (Arsenic sulfide sol)

Arsenic sulfide sol is typically prepared by passing hydrogen sulfide (H₂S) gas through a solution of arsenious oxide (As₂O₃). This reaction is a classic example of double decomposition.

As₂O₃ + 3H₂S → As₂S₃ (sol) + 3H₂O

So, (A) matches with (IV) Double decomposition.

(B) Au (Gold sol)

Gold sol, a metallic sol, is commonly prepared by Bredig's arc method. This method involves striking an electric arc between two gold electrodes immersed in a dispersion medium (like water), which vaporizes the gold, and the vapor then condenses to form colloidal particles.

So, (B) matches with (I) Bredig's arc method.

(C) S (Sulfur sol)

Sulfur sol can be prepared by the oxidation of hydrogen sulfide (H₂S) with an oxidizing agent such as sulfur dioxide (SO₂).

2H₂S + SO₂ → 3S (sol) + 2H₂O

So, (C) matches with (II) Oxidation.

(D) Fe(OH)₃ (Ferric hydroxide sol)

Ferric hydroxide sol is prepared by the hydrolysis of ferric chloride (FeCl₃). When a small amount of ferric chloride solution is added to hot water, it undergoes hydrolysis to form colloidal particles of ferric hydroxide.

FeCl₃ + 3H₂O → Fe(OH)₃ (sol) + 3HCl

So, (D) matches with (III) Hydrolysis.

Combining these matches:

A - IV

B - I

C - II

D - III

This corresponds to Option C.

The final answer is $\boxed{\text{C}}$

$(i) \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \xrightarrow[200 \mathrm{~atm}]{\mathrm{Fe}, 773 \mathrm{~K}} 2 \mathrm{NH}_3(g): $

Haber's process

$ X=\mathrm{Fe}(\text { iron }) $

Water gas shift reaction $Y=$ Iron chromate

$ \begin{aligned} &\text { Steam reforming of methane }\\ &Z=\text { Nickel } \end{aligned} $

The correct answer is:

✅ Option B — Eosin dye

Explanation:

• Argentometric titrations are titrations involving silver nitrate (AgNO₃) as the titrant, typically used for the determination of halide ions such as Cl⁻, Br⁻, and I⁻.

• One common method is the Fajans method, where an adsorption indicator, such as fluorescein or eosin, is used.

• Before the equivalence point, halide ions are present in excess, and the precipitate surface (AgX) is negatively charged.

• After the equivalence point, excess Ag⁺ ions make the surface positively charged, allowing the negatively charged dye ions (like eosin) to adsorb on the surface and cause a visible color change — signaling the end point.

Hence, Eosin dye (an adsorption indicator) is used in argentometric titrations by Fajans’ method.

Let’s recall the Freundlich adsorption isotherm:

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

Or, taking logarithms on both sides:

$ \log (x/m) = \log k + \frac{1}{n} \log p $

Given data

• Slope = 1 ⇒ $\frac{1}{n} = 1 \Rightarrow n = 1$

• $k = 0.1$

• $p = 2$ atm

• $\log 2 = 0.30$

Calculation

$ x/m = k \, p^{1/n} = 0.1 \times 2^{1} = 0.1 \times 2 = 0.2 $

✅ Extent of adsorption = 0.2

Correct option: (C) 0.2

(a) Colloidal antimony → Used in treatment of Kala-azar

→ (III)

(b) Argyrol (colloidal silver) → Used as an eye lotion

→ (I)

(c) Colloidal gold → Used for intramuscular injections (tonic for nervous disorders)

→ (II)

(d) Milk of magnesia → Used for stomach disorders (antacid/laxative)

→ (IV)

✅ Correct match:

(a) – (III), (b) – (I), (c) – (II), (d) – (IV)

Hence, the correct answer is:

Option A: A-III, B-I, C-II, D-IV

Given:

Freundlich adsorption isotherm:

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

Taking log on both sides:

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

Given from the question:

$ \text{slope} = \frac{1}{n} = 3 $

$ \text{intercept} = \log k = 0.30 $

We are asked to find $\frac{x}{m}$ at $ p = 2 \text{ atm} $, given $\log 2 = 0.3$.

Step 1: Write the equation

$ \log \frac{x}{m} = 0.30 + 3 \log p $

Step 2: Substitute $ \log p = \log 2 = 0.3 $

$ \log \frac{x}{m} = 0.30 + 3(0.3) $

$ \log \frac{x}{m} = 0.30 + 0.90 = 1.20 $

Step 3: Find $\frac{x}{m}$

$ \frac{x}{m} = 10^{1.20} $

We know $10^{1.2} = 10 \times 10^{0.2} \approx 10 \times 1.58 = 15.8 \approx 16$

✅ Final Answer:

$ \boxed{\frac{x}{m} = 16} $

Correct Option: C) 16

Using Freundlich adsorption isotherm equation,

$ \begin{aligned} & \frac{x}{m}=k p^{1 / n} \\ & p=2, n=2 \Rightarrow \frac{x}{m}=k(2)^{1 / 2} \\ & \frac{x}{m}=k \sqrt{2} \text { or } \frac{x}{m}=1.414 \mathrm{k} \end{aligned} $

When both the dispersed phase and the dispersion medium are liquid, the colloid is known as an emulsion.

✅ Correct Answer: Option B — Emulsion

Explanation:

• Sol: Solid dispersed in liquid

• Gel: Liquid dispersed in solid

• Foam: Gas dispersed in liquid

• Aerosol: Solid or liquid dispersed in gas

• Emulsion: Liquid dispersed in liquid (e.g. milk, mayonnaise)

Hence, the type of colloid is emulsion.

Let's recall the Freundlich adsorption isotherm formula:

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

Taking logarithms on both sides:

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

✅ Correct option: Option B

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

The incorrect property of physical adsorption is,

It increases with increase of temperature. The correct form is, It decreases with increase of temperature.

The correct order is,

$ \mathrm{II}>\mathrm{III}>\mathrm{IV}>\mathrm{I} $

Identify the counter-ions in each electrolyte

$\mathrm{KCl}: \mathrm{Cl}^{-}, \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]:\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$,

$ \mathrm{K}_3 \mathrm{PO}_4: \mathrm{PO}_4^{3-}, \mathrm{K}_2 \mathrm{SO}_4: \mathrm{SO}_4^{2-} $

Charge $-1,-4,-3,-2$

Now applying Hardy-Schulze rule

The higher the charge of counter ions, the greater is precepitals power

$ \mathrm{Fe}\left(\mathrm{CN}_6\right)^{4-}>\mathrm{PO}_4^{3-}>\mathrm{SO}_4^{2-}>\mathrm{Cl}^{-} $

(A) Kieselguhr (Diatomaceous earth)

→ Used for filtration (it is a fine, porous, siliceous material).

✅ So, (A) → (III) Filtration plants

(B) Silica gel

→ Used in chromatography as an adsorbent.

✅ So, (B) → (I) Chromatographic material

(C) ZSM-5 (a type of zeolite catalyst)

→ Used to convert alcohol directly into gasoline (in petrochemical industry).

✅ So, (C) → (IV) To convert alcohol directly into gasoline

(D) Hydrated zeolites

→ Used in softening of hard water (by ion exchange).

✅ So, (D) → (II) Softening of hard water

Now, putting it all together:

(A)–III, (B)–I, (C)–IV, (D)–II

✅ Correct Answer: Option D

In the catalytic reaction of hydrogenation of vegetable oil, both reactants are in different phase.

Statement I: Gold sol is prepared by Bredig's arc method.

• True.

  Bredig’s arc method is indeed used for preparing metal sols like gold, platinum, and silver sols.

  In this method, an electric arc is struck between metal electrodes under water containing a stabilizer. Metal vapors formed are condensed to give colloidal particles (the sol).

Statement II: Bredig's arc method involves only dispersion but not condensation.

• False.

  Bredig's arc method is actually a combination of dispersion and condensation methods:

• The metal is dispersed (broken up) by vaporization (dispersion).

• The vaporized metal atoms condense to form colloidal particles (condensation).

Hence, both processes are involved.

✅ Correct answer: Option C

Statement-I is correct, but Statement-II is not correct.

Among the given options, I and III are correct. The correct form of II is Starch is an example of macromolecular colloids.

Commonly used adsorbents in adsorption chromatography are silica gel and alumina.

Principle used in chromotography is adsorption.

Promoters are those substance which do not themselves acts a catalyst but increases the activity of catalyst. Iron ( Fe ) is used as catalyst in Haber's process. If small amount of Mo is mixed with iron then catalytic power of Fe increases.

Catalytic poison are substance which decreases the activity of catalyst.

CO is used as catalytic poison in Haber's process.

Given, 10 mL of 0.5 M NaCl

Molarity

$ \begin{aligned} & =\frac{\text { Number of moles of solute }(\mathrm{in} \mathrm{~mL})}{\text { Volume of solution }} \\\\ 0.5 & =\frac{x}{10} \\\\ x & =5 \text { millimoles } \end{aligned} $

Flocculating value is given by

$ =\frac{\text { Millimoles of electrolytes }}{\text { Volume of solution (in L) }} $

Volume of solution is given is 1 L So, Flocculating value $ =\frac{5}{1}=5 \mathrm{~m} \mathrm{~mol} $

According to Freundlich adsorption isotherm,

$\frac{x}{m}=k p^{1 / n}\quad \ldots \text { (i) }$

Taking $\log$ on both side,

$ \begin{aligned} & \log \left(\frac{x}{m}\right)=\log k+\frac{1}{n} \log p \ldots \text { (ii) } \\\\ & \text { Slope }=\frac{1}{n} \quad \ldots \text { (iii) } \\\\ & \text { From graph slope }=\frac{3}{6}=\frac{1}{2} \\\\ & \therefore \quad \frac{x}{m}=k p^{1 / 2} \text { or } \frac{x}{m} \propto p^{1 / 2} \end{aligned} $

According to Freundlich adsorption isotherm,

$ \frac{x}{m}=k p^{1 / n}\quad \ldots \text { (i) } $

Taking $\log$ on both side,

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

Intercept, $\log k=1$

So,$\quad\quad k=10\quad \ldots \text { (ii) } $

Put the values of $k=10, n=0.5\quad \text { (given as slope) }$

$ \begin{aligned} \frac{x}{m} & =10 \times(100)^{0.5}[\text { Also } p=100 \mathrm{~atm}] \\ \Rightarrow \quad \frac{x}{m} & =100 \end{aligned} $

Option A

• Statement I is true. Adsorption (both physisorption and chemisorption) releases heat, so it is an exothermic process.

• Statement II is also true. When you add charcoal to a closed vessel containing gases, some molecules stick to the charcoal surface, decreasing the number of gas‐phase molecules and hence lowering the pressure below the original value $p$.

Adsorption by charcoal is stronger for more easily liquefiable gases (i.e. those with higher critical temperatures), because stronger van der Waals forces enhance physisorption. Hence, at the same pressure (and temperature):

Amount adsorbed ∝ (critical temperature)

The gas with the lowest critical temperature (190 K) is A, so it will be adsorbed the least.

Answer: Option D.

The gold number is defined as the minimum mass (in milligrams) of a protecting colloid required to prevent the coagulation of 10 mL of gold sol when 1 mL of 10% sodium chloride solution is added.

Looking at the data:

• Experiment 3: 25 mg A – Not prevented

• Experiment 4: 32 mg A – Not prevented

• Experiment 5: 33 mg A – Prevented

• Experiment 2: 35 mg A – Prevented

• Experiment 1: 40 mg A – Prevented

Since 32 mg does not prevent coagulation and 33 mg does, the minimum effective mass is 33 mg.

Thus, the gold number of A is:

$\text{Gold Number} = 33 \text{ mg}$

The correct option is Option B: 33.

The minimum concentration of electrolyte in millimoles required to cause coagulation of one litre of colloidal solution is called coagulation value from the experimental data, minimum concentration of $\mathrm{Cl}^{-}$when coagulation just started is $7 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$. We need to convert it into millimoles. $7 \times 10^{-5} \mathrm{~mol} / \mathrm{L}=7 \times 10^{-2}$ millimoles $/ \mathrm{L}$

Let's analyze each pair step by step:

Colloidal antimony has historically been used in the treatment of Kala-azar (a form of visceral leishmaniasis). Thus, it matches with:

$\text{Colloidal antimony} \rightarrow \text{Kalaazar (A)}$

Milk of magnesia is a suspension of magnesium hydroxide commonly used as an antacid and laxative. It helps in relieving stomach disorders. Therefore:

$\text{Milk of magnesia} \rightarrow \text{Stomach disorder (D)}$

Silver sol is a colloidal silver solution known for its antimicrobial properties and has been used topically, for example as an eye lotion.

$\text{Silver sol} \rightarrow \text{Eye lotion (C)}$

Gold sol (colloidal gold) was used in the past in the treatment of rheumatoid arthritis and similar conditions via intramuscular injections.

$\text{Gold sol} \rightarrow \text{Intermuscular injection (B)}$

Putting it all together, the correct matching is:

I → A

II → C

III → D

IV → B

Thus, the correct answer is Option B.

Colloidal sols can be prepared using the method of peptisation. This process involves converting a precipitate into a colloidal sol by shaking it with a dispersion medium.

Chemiosorption is a form of adsorption in which the adsorbed material is held together by chemical bonds.

It has high specificity and occur only when there is a chemical bond between adsorbent and adsorbate. It is proportional to surface area.

Enthalpy of adsorption in chemiosorption is in the range of $80-240 \mathrm{~kJ} / \mathrm{mol}$

$\mathrm{As}_2 \mathrm{~S}_3$ sol is a negatively charged sol.

A sol is a colloidal suspension made out of tiny solid particle in continuous liquid medium.

Heterogeneous catalysis occurs when the catalyst is in a different phase than the reactants and products. For example, in the reaction:

$ A(g) + B(g) \xrightarrow{C(s)} D(g) $

The catalyst $ C $ is in the solid phase, while both the reactants $ A $ and $ B $, and the product $ D $, are in the gaseous phase. This difference in phase between the catalyst and the other substances involved characterizes heterogeneous catalysis.

Let's analyze each pairing:

Aerosol: An aerosol is a dispersion of fine solid particles or liquid droplets in a gas. Smoke is a common example, as it consists of tiny particles suspended in air.

  • Match: Aerosol – Smoke (IV)

Foam: A foam is a mass of small bubbles formed in or on a liquid. Soap lather is a familiar example of a foam.

  • Match: Foam – Soap lather (II)

Emulsion: An emulsion is a mixture of two or more liquids that are normally immiscible. Milk is an emulsion where fat droplets are dispersed in water.

  • Match: Emulsion – Milk (I)

Gel: A gel is a semi-solid system in which a liquid is dispersed in a solid network. Cheese is an example, as it has a gel-like structure formed by protein networks.

  • Match: Gel – Cheese (III)

This matching corresponds to Option D:

A-IV, B-II, C-I, D-III.

The gold number is defined as the minimum mass (in mg) of a protecting colloid that is required to prevent the coagulation of 10 mL of gold sol when 1 mL of 10% NaCl solution is added.

Looking at the provided data:

• Experiment 1: 24 mg → Coagulation not prevented

• Experiment 2: 23 mg → Coagulation not prevented

• Experiment 3: 26 mg → Coagulation prevented

• Experiment 4: 27 mg → Coagulation prevented

• Experiment 5: 25 mg → Coagulation prevented

Since the smallest mass at which coagulation is prevented is 25 mg (from Experiment 5), the gold number of substance $X$ is:

$\text{Gold Number} = 25 \text{ mg}$

Thus, the correct option is:

Option D: 25

The correct answer is Option D – Gold sol.

Here’s a brief explanation:

• In pharmaceutical dosage forms, a "sol" is a colloidal dispersion where tiny solid particles are dispersed in a liquid medium. For an injection, the formulation must be sterile, non-irritant, and biocompatible.

• Gold sol (a colloidal gold suspension) has been used historically for its anti-inflammatory properties in the treatment of rheumatoid arthritis. This form of chrysotherapy involves intramuscular injections.

• The other options are not used as intramuscular injections:

  – Antimony sol is not typically administered by injection in its colloidal form.

  – Silver sol is used more for topical applications (like in wound dressings) rather than injections.

  – The emulsion of milk of magnesia is formulated for oral use as an antacid and laxative, not for injection.

Thus, gold sol is the only sol among the options that is used as an intramuscular injection.

In coagulation of colloidal sols, the effectiveness of a coagulating ion depends mainly on its charge. A higher charge on the ion neutralizes the charge on the sol particles more efficiently, requiring a lower amount to induce coagulation. This observation is encapsulated in the Schulze-Hardy rule.

Here’s the reasoning using the given options:

For a positively charged sol, the anion with the highest negative charge will have the maximum coagulating power.

The given ions and their charges are:

$\mathrm{Cl}^-$ has a charge of -1.

$\mathrm{SO}_4^{2-}$ has a charge of -2.

$\mathrm{PO}_4^{3-}$ has a charge of -3.

$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ has a charge of -4.

Among these, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ has the highest negative charge.

Therefore, in the coagulation of a positively charged sol, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ will have the maximum coagulating power.

The correct answer is: Option D.

Let's evaluate each statement:

Statement I: "Easily liquefiable gases are readily adsorbed."

Gases that are easily liquefiable typically have higher critical temperatures, indicating stronger intermolecular forces.

These stronger forces favor adsorption onto surfaces, making such gases more likely to be adsorbed.

Statement II: "Adsorption enthalpy for physisorption is less compared to adsorption enthalpy for chemisorption."

Physisorption involves weak van der Waals forces and usually has an enthalpy change in the range of about 20–40 kJ/mol.

In contrast, chemisorption involves the formation of chemical bonds, leading to a much higher enthalpy change.

Since both statements accurately reflect the principles of adsorption:

The correct answer is Option A: Both Statement I and Statement II are correct.

According to Freundlich, there is an empirical relationship between the amount of gas adsorbed by a unit mass of solid adsorbent and the pressure at a given temperature:

$ \frac{x}{m} = k \cdot p^{1/n} \quad \text{(for $n > 1$)} $

Here, $x$ represents the mass of gas adsorbed, $m$ is the mass of the adsorbent, and $p$ is the pressure. The constants $k$ and $n$ are specific to the system being examined.

By taking the logarithm of both sides of the equation, we have:

$ \log \frac{x}{m} = \log k + \frac{1}{n} \log p \quad \ldots \text{(i)} $

To verify the validity of the Freundlich isotherm, you can plot $\log \frac{x}{m}$ on the y-axis and $\log p$ on the x-axis. If the plot forms a straight line, the isotherm is valid. If not, it indicates a lack of alignment with the Freundlich equation.