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.

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} $

Bredig's arc method is suitable for the preparation of colloidal solutions of metals like gold, silver, platinum etc. The charge of sol particles will be negative. Hence, $X$ is metal sol and $q$ is negative.

Sodium lauryl sulphate is an anionic surfactant used in cosmetics and pharma ceuticals.

When sodium lauryl sulphate is dissolved in water is an appreciable amount, ionic micelle formation takes place.

The correct match is

$ \mathrm{A} \rightarrow \mathrm{III}, \mathrm{~B} \rightarrow \mathrm{I}, \mathrm{C} \rightarrow \mathrm{IV}, \mathrm{D} \rightarrow \mathrm{II} . $

• When ferric chloride is added to NaOH solution, a negatively charged sol is obtained due to adsorption of $\mathrm{OH}^{-}$ions.



• Milk is an emulsion in which oil is mixed with water (act as dispersion medium).

• The protecting power of a lyophilic solution is expressed in terms of gold number.

• Colloidal antimony is used in curing kala-azar.

Adsorption is an exothermic process, it decreases with increase in temperature and varies inversely proportional to temperature that is $\frac{x}{m} \propto \frac{1}{T}$. So, according to graph $\frac{x}{m}$ is more for $T_4$. Therefore, the corerct order is $T_4>T_3>T_2$.

Tyndall effect is scattering of a beam of light by a colloidal medium containing suspended particle. True solution doesn't show tyndall effect. Therefore, it is used to distinguish between true and colloidal solution. It is possible when the dispersed medium and disperssed phase differ much in their refraction indices. This effect is seen when size of colloidal particle is smaller than wavelength of light used. Thus, correct statements are given in I and II only.

Let's analyze the factors step by step:

High surface area (I)

More surface area means more sites available for the adsorbate molecules to stick to, which favors adsorption.

Low temperatures (II)

Physical adsorption is generally an exothermic process. At lower temperatures, the kinetic energy of the adsorbate molecules is reduced, allowing the weak van der Waals forces to effectively hold them onto the surface.

High pressures (V)

Increasing the pressure increases the number of gas molecules hitting the surface, which enhances the likelihood of adsorption.

On the other hand:

High temperatures (III) actually decrease physisorption because the higher kinetic energy of the molecules makes it harder for the weak forces to capture them.

Low pressures (IV) reduce the number of collisions with the surface, thereby not favoring adsorption.

Thus, the factors that favor physical adsorption are I, II, and V.

The correct answer is Option B: I, II and V only.

The colloidal particles when subjected to electric current/potential start moving towards electrode. This process is called as electrophoresis.

According to Freundlich adsorption isotherm,

$ \begin{aligned} \frac{x}{m} & =k p^{1 / n} \\ \frac{10}{500} & =k(10)^{1 / n} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \frac{14}{500} & =k(20)^{1 / n} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \frac{10}{14} & =\left(\frac{10}{20}\right)^{1 / n} \\ 0.71 & =(0.5)^{1 / n} \\ \log (0.71) & =\log (0.5)^{1 / n} \\ -0.14 & =1 / n \log (0.5) \\ & =1 / n(-0.3) \\ \frac{1}{n} & =\frac{0.14}{0.3}=0.46 .=0.5 \end{aligned} $

Moles of HCl added to coagulate 200 mL of colloid

$ \begin{aligned} & =\frac{\text { mass of } \mathrm{HCl} \text { taken }}{\text { molecular mass of } \mathrm{HCl}} \\ & =\frac{0.73}{(1+35.5)}=0.02 \mathrm{~mol} \\ & =20 \mathrm{~m} \mathrm{~mol} \end{aligned} $

Now, we know

Flocculation value is the minimum number of moles of electrolyte per litre required to coagulate a colloid.

200 mL of sol required 20 mmol of HCl

So, 1000 mL of sol required

$ =\frac{20}{200} \times 1000=100 \mathrm{~m} \mathrm{~mol} $

Thus, the flocculation value of HCl for the colloid is 100 .

Tyndall effect is light scattering by particles in colloid. The particles size is smaller than that of suspension but greater than solution that means Tyndall effect is shown only by colloidal solutions. Out of given options, clouds, milk and suspension are colloidal solutions. So, they can show Tyndall effect. But sugar solution is a pure solution so, it does not show Tyndall effect.

$1 \mathrm{~cm}=10^7 \mathrm{~nm}$

Edge length of original cube $=10^7 \mathrm{~nm}$

Volume of original cube $(V)=\left(10^7\right)^3 \mathrm{~nm}$

Edge length of new cube $=1 \mathrm{~nm}$

Volume of new cube $=(1)^3 \mathrm{~nm}$

If the number of new cubes formed is $n$, then volume of total new cubes will be $\left(V_1\right)=n \times(1)^3 \mathrm{~nm}$

But $V=V_1$

$ \therefore\left(10^7\right)^3=n \times(1)^3 \therefore n=10^{21} $

The surface area of the original cube

$ (S)=6 \times\left(10^7\right)^2 \mathrm{~nm}^2=6 \times a^2 $

The surface area of the total cubes formed after crushing the original cube

$ \begin{aligned} & =\left(S_1\right)=6 \times 10^{21} \times(1)^2 \mathrm{~nm}^2 \\ \frac{S_1}{S} & =\frac{6 \times 10^{21} \times(1)^2}{6 \times\left(10^7\right)^2}=10^7 \end{aligned} $

I. Esterification reaction is catalysed by mineral acid such as HCl (l).

II. Conversion of $\mathrm{SO}_2$ into $\mathrm{SO}_3$ is an important reaction in manufacture of sulphuric acid which is catalysed by $\mathrm{Pt}(s)$.

III. In lead chamber process conversion of $\mathrm{SO}_2$ into $\mathrm{SO}_3$ by reaction with $\mathrm{O}_2$ is catalysed by $\mathrm{NO}(\mathrm{g})$.

IV. In Haber's process ammonia is manufactured by reacting $\mathrm{N}_2$ and $\mathrm{H}_2$ in presence of catalysed $\mathrm{Fe}(\mathrm{s})$.

According to Hardy Schulze rule, greater the valency of the coagulating ion, greater is its coagulating power. $\mathrm{As}_2 \mathrm{~S}_3$ is a negatively charged sol. For coagulation, oppositely charged sol is needed. Thus, out of the given, $\mathrm{FeCl}_3\left(\mathrm{Fe}^{3+}\right)$ is most effective for causing coagulation of $\mathrm{As}_2 \mathrm{~S}_3$ sol.

Gold number is the amount (in milligrams) of aprotective colloid, which is just sufficient to prevent the coagulation of 10 mL of standard hydro gold sol, when 1 mL of $10 \% \mathrm{NaCl}$ solution is added to it.

$\because 10$ cc gold sol $=1$ cc of $10 \% ~ \mathrm{NaCl} \therefore 100 \mathrm{~mL}$ of gold sol = l cc of gelatin Therefore, gold number of gelatin is

$ \frac{1 \times 10}{100}=0.01 $

Hence, option (d) is the correct answer.

Adsorption processes are exothermic ( $\Delta H<0$ ) and degree of randomness of adsorbate particles decreases, i.e. $\Delta S<0$.

According to Gibbs' equation, $\Delta G=\Delta H-T \Delta S$

at lower temperature adsorption becomes spontaneous, i.e. $\Delta G<0$

In chemisorption, adsorption increases with temperature. Because the plot between $x / m$ and $T$ (temperature) is as follows. where, $x / m=$ mass fraction of chemisorption.

Thus, option (c) is the correct answer.

The lyophilic colloids differ in their protective power. The protective power is measured in terms of gold number. It may be noted that smaller the value of the gold number, greater will be the protecting power of protective colloids. Therefore, the reciprocal of gold number is a measure of the protective power of the colloid, gold number of gelatin, haemoglobin and sodium acetate are $5 \times 10^{-3}, 5 \times 10^{-2}$ and $7 \times 10^{-1}$, respectively, hence, the protective actions will be in the order : Haemoglobin > gelatin >sodium acetate