Colloids and Surfaces MCQ Quiz in मल्याळम - Objective Question with Answer for Colloids and Surfaces - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is \(\rm \frac{1}{\sqrt {K_{eq}}}\)

Explanation:-

In case of diatomic molecule,

\(\theta=\frac{(k P)^{1 / 2}}{1+(k p)^{1 / 2}}\)

\(\frac{1}{\theta}=\frac{1+(\mathrm{kp})^{1 / 2}}{(\mathrm{kp})^{1 / 2}}\)

\(\frac{1}{\theta}=\frac{1}{(\mathrm{k} \cdot \mathrm{P})^{1 / 2}}+1\)

\(\text { or, } \frac{1}{\theta}=\frac{1}{\mathrm{k}^{1 / 2} \cdot \mathrm{P}^{1 / 2}}+1\)

comparing it with straight line equation, 

y = mx + c 

\(\text { the plot of } \frac{1}{\theta} \mathrm{v} / \mathrm{s} \frac{1}{\mathrm{p}^{1 / 2}} \text { gives a slope of } \frac{1}{\mathrm{k}^{1 / 2}}\)

Conclusion:-

So,  the slope of is \(\rm \frac{1}{\sqrt {K_{eq}}}\)

Concept:-

• Adsorption: Adsorption is a surface process where molecules or atoms (adsorbate) accumulate on the surface of a material (adsorbent). The process can be described via rate equations, where ka (adsorption rate constant) and kb (desorption rate constant) feature prominently for the proportion of molecules bound to the surface at a given time versus those released back into the gaseous phase.
• Langmuir Isotherm Model: The key concept in the Langmuir adsorption isotherm model is that the surface of the adsorbent is uniform and each site on the surface is equivalent, i.e., each site has an equal chance of adsorbing a molecule.
  The model is formulated as:

         θ = kaP / (1 + kaP)

Explanation:-

The correct relation between K and P at a fixed temperature is :

\(K\times P = \frac{\theta}{1-\theta}\)

\(K = \frac{1}{P}(\frac{\theta}{1-\theta})\)

From above equation \(K \propto p^{-1}\)

Conclusion:-

So, For a fixed fractional coverage, the correct relation between K and P at a fixed temperature is \(K \propto p^{-1}\)

The correct answer ø = Kp/(1+Kp)

Concept:-

Adsorption isotherms: it illustrate the connection between the degree of adsorption (denoted as θ) and the concentration or pressure of the adsorbate (denoted as p) at a specific temperature. The Langmuir isotherm model is commonly applied to represent this relationship, especially when describing the adsorption of a monolayer of molecules on a surface.

Equilibrium Constant (Kp): The equilibrium constant (Kp) in the Langmuir isotherm equation reveals the balance between the rate of adsorption and the rate of desorption. Kp relies on factors such as the adsorption energy and the number of available sites on the surface. A higher Kp indicates a more significant adsorption tendency.

Fractional Surface Coverage (θ): The symbol θ represents the fractional surface coverage, signifying the proportion of available adsorption sites that are occupied by adsorbate molecules. It ranges from 0 (indicating no adsorption) to 1 (indicating complete coverage by a monolayer). The Langmuir isotherm equation predicts how θ changes with variations in p.

Explanation:-

The Langmuir isotherm is a fundamental model used to describe the adsorption of molecules onto a surface, such as in the case of gas adsorption on a solid surface. The correct form of the Langmuir isotherm equation is as follows:

ø = Kp/(1+Kp)

θ represents the fractional surface coverage or the fraction of sites occupied by adsorbed molecules.
Kp is the equilibrium constant, which is related to the adsorption energy and the number of available sites.
p is the pressure of the gas or the concentration of the adsorbate in the gas phase.

Concept:-

• Surface tension is the force exerted per unit length on the surface of a liquid due to the cohesive nature of its molecules. It is the tendency of the surface of a liquid to minimize its surface area and form a shape that has the minimum surface area possible.

• The SI unit of surface tension is Newtons per meter (N/m) or Joules per square meter (J/m²).

Explanation:-

• The surface tension of a liquid is influenced by the intermolecular forces between its molecules, which determine the cohesive forces that hold the molecules together. A decrease in surface tension indicates a decrease in the strength of the cohesive forces between the molecules, while an increase in surface tension indicates an increase in the strength of the cohesive forces.
• Given that the surface tension of the solution of A is smaller than that of pure water, it implies that the intermolecular forces between the molecules of A are weaker than those of water. This can occur if A is a surfactant or a solute that disrupts the hydrogen bonding network of water molecules at the surface.
• On the other hand, the fact that the surface tension of the solution of B is greater than that of pure water suggests that the intermolecular forces between the molecules of B are stronger than those of water. This can occur if B is a polar compound that can form hydrogen bonds with water molecules or if B has a larger molecular size that contributes to stronger van der Waals forces between the molecules.
• Therefore, one can infer that the surface concentration of A is larger than that of B, since the weaker intermolecular forces of A compared to water would require a higher concentration of A at the surface to compensate for the loss of cohesive forces. Hence, the correct option is (3) - the surface concentration of A is larger than that of B.

Explanation:-

• Correct option is (3). 

Concept:

→ The order of a surface catalyzed unimolecular reaction can depend on the pressure of the reactant. At very low pressures of the reactant, the reaction is typically first-order, while at very high pressures, the reaction is typically zero-order.

→ This is because the rate of a surface catalyzed reaction depends on the concentration of the reactant molecules adsorbed onto the surface of the catalyst.

→ At very low pressures of the reactant, there are few reactant molecules adsorbed onto the catalyst surface, and the rate of the reaction is limited by the rate at which reactant molecules adsorb onto the surface.

→ In this case, the rate of the reaction is proportional to the concentration of the reactant, and the reaction is first-order.

→ At very high pressures of the reactant, the catalyst surface becomes saturated with reactant molecules, and the rate of the reaction becomes limited by the rate at which the reactant molecules react with each other on the catalyst surface.

→ In this case, the rate of the reaction is independent of the concentration of the reactant, and the reaction is zero-order.

Conclusion: Therefore, the order of a surface catalyzed unimolecular reaction, at very low and very high pressures of the reactant, would be 1 and 0, respectively.

Concept:

The height to which a liquid rises in a capillary tube can be calculated using the following equation:

\( h=\frac{2Tcos\Theta }{\rho gr}\),

where:

• h is the height to which the liquid rises
• T is the surface tension of the liquid
• θ is the angle of contact between the liquid and the capillary tube
• ρ is the density of the liquid
• g is the acceleration due to gravity
• r is the radius of the capillary tube.

Explanation:

We can rearrange this equation to solve for the surface tension:

\(T=\frac{\rho grh}{2cos\theta } \)

Substituting the given values, we get:

\(T=\frac{1.1\times9.81\times5.0\times0.2}{2cos0^{o} }\)

Note that the angle of contact between the liquid and the capillary tube is assumed to be 0° because the liquid wets the capillary completely.

Converting the units to Nm⁻¹, we get:

T ≈ 0.054 Nm⁻¹.

Conclusion:
Therefore, the surface tension of the liquid is closest to 0.05 Nm⁻¹.

Concept:

The surface area occupied = Total number of molecules combining for complete monolayer coverage × Area of 1 molecule

At STP,

1 mole gas = 22.4 L

22.4 L = 6.022 × 10²³ molecules

22.4 L = N_A

No. of molecules in 1 L = \(\frac{N_A}{22.4}\)

No. of molecules in V_m = \(\frac{N_A}{22.4} × V_m\)

where V_m = volume corresponding to total a complete monolayer coverage.

Explanation:

∵ The volume of N₂ at STP required to form a monolayer on a porous solid surface is = 22.4 cm³ g^-1

→ Area occupied by one molecule of nitrogen gas molecule = 16.2 Å²

= 16.2 × 10^-16 cm²

we know that (1 L = 10³ cm³)

∴ 22400 cm³ of N₂ at STP contains N_A molecules of N₂

∴ 22.4 cm³g^-1 of N₂ contains

= \(\frac{22.4 × N_A}{22400}\)

= 6.022 × 10²⁰ molecules of N₂

∴ Surface area occupied by 6.022 × 10²⁰ molecules of N₂,

= 6.022 × 10²⁰ × 16.2 × 10^-16 cm²

= 97.55 × 10⁴

= 9.755 × 10⁵ cm² g^-1

Conclusion:-
So, the option '2' is correct.

Explanation:-

Micelles are the clusters of aggregated particles that are formed due to the combination of colloids in the solution.

Important Points

1.  The process of formation of micelles is called micellization
2. Micellization is a temperature-dependent process therefore it depends on the rates of enthalpy and entropy.
3. Enthalpy Change is very low in micellization and can be given by Δ H_m = − nRT² (∂ ln CMC / ∂T)_p 
4. Entropy Change in Micellization is estimated to be very high.  

​THERMODYNAMIC CHANGES -    ΔG = RT  ln(CMC)

Concept:

BET EQUATION 

• We can use the BET equation for determining the specific surface area from the adsorption data. Also this equation is used to give the volume of gas needed for the formation of a monolayer on the surface of the sample. 

BET equation is given below, 

\(\frac{{\rm{z}}}{{\left( {{\rm{1 - z}}} \right)\,{\rm{V}}}}\,\,{\rm{ = }}\,\frac{{\left( {{\rm{c - 1}}} \right)\,{\rm{z}}}}{{{\rm{c}}\,\,\mathop {\rm{V}}\nolimits_{{\rm{mono}}} }}\,{\rm{ + }}\,\frac{{\rm{1}}}{{{\rm{c}}\,\mathop {\rm{V}}\nolimits_{{\rm{mono}}} }}\)

V= Volume of the gas adsorbed at pressure P. 

V_Mono= amount of gas corresponding to one monolayer. 

z = p/p⁰, ratio of pressure of gas and pressure at saturation. 

c = constant. 

a plot of \(\dfrac{z}{(1-z){V}}\)  versus z gives a slope \(\,\frac{{\left( {{\rm{c - 1}}} \right)\,}}{{{\rm{c}}\,\,\mathop {\rm{V}}\nolimits_{{\rm{mono}}} }}\,\)  and intercept equal to \(\,\frac{1}{{{\rm{c}}\,\,\mathop {\rm{V}}\nolimits_{{\rm{mono}}} }}\,\)  at z=0 .

Explanation:

Given, Adsorption of N₂ on TiO₂ at 75K. 

Here the intercept, \(\,\frac{1}{{{\rm{c}}\,\,\mathop {\rm{V}}\nolimits_{{\rm{mono}}} }}\,\)= 4.0 × 10-6 mm-3 

Slope, \(\,\frac{{c - 1}}{{{\rm{c}}\,\,\mathop {\rm{V}}\nolimits_{{\rm{mono}}} }}\,\) =  1.0 × 10-3 mm-3.

first, we have to calculate the value of constant "c" from the given slope.

\(\begin{array}{l} \,\,\frac{{c - 1}}{{{\rm{c}}\,\,\mathop {\rm{V}}\nolimits_{{\rm{mono}}} }}\,\, = \,1.0 \times {10^{ - 3}}\,m{m^{ - 3}}\\ \left( {c - 1} \right) \times \frac{1}{{{\rm{c}}\,\,\mathop {\rm{V}}\nolimits_{{\rm{mono}}} }}\, = \,1.0 \times {10^{ - 3}}\,m{m^{ - 3}} \end{array}\)

Given, 

\(\frac{1}{{{\rm{c}}\,\,\mathop {\rm{V}}\nolimits_{{\rm{mono}}} }}\,\, = \,4.0 \times {10^{ - 6}}m{m^{ - 3}}\)

\(\begin{array}{c} \left( {c - 1} \right) \times \left( {4.0 \times {{10}^{ - 6}}\,m{m^{ - 3}}} \right)\, = \,1.0 \times {10^{ - 3}}\,m{m^{ - 3}}\\ c = \,251 \end{array}\)

Further, 

\(\begin{array}{c} \frac{1}{{c\,\mathop V\nolimits_{mono} }}\, = \,\,4.0 \times {10^{ - 6}}m{m^{ - 3}}\\ \frac{1}{{\mathop V\nolimits_{mono} }}\,\, = c \times \left( {4.0 \times {{10}^{ - 6}}m{m^{ - 3}}} \right)\\ c = \,\,251\\ \frac{1}{{\mathop V\nolimits_{mono} }}\, = \,251 \times \,\left( {4.0 \times {{10}^{ - 6}}m{m^{ - 3}}} \right) \end{array}\)

\(\mathop V\nolimits_{mono} \, = \,\frac{1}{{251 \times \left( {4.0 \times {{10}^{ - 6}}\,m{m^{ - 3}}} \right)}}\)

\(\,\therefore \,\,\,\mathop V\nolimits_{mono} = \,996\,m{m^3}\)

The volume corresponding to the monolayer coverage is 996mm³.

Hint

To measure the BET surface area we often use Nitrogen. 

Concept:

Freundlich Adsorption Isotherm

• The Freundlich adsorption isotherm is an empirical relationship that describes the adsorption of gases or solutes onto surfaces. It is applicable when adsorption occurs on heterogeneous surfaces and the adsorbed molecules do not interact significantly with one another.
• The Freundlich isotherm is typically expressed as:

  x/m = k * p^1/n

  where x/m is the amount of adsorbate per unit mass of adsorbent, p is the pressure (or concentration), k is the Freundlich constant, and n is a constant that reflects the adsorption intensity.

Explanation:

• The correct expression for the Freundlich adsorption isotherm is:

  x/m = k * p^1/n

  This formula indicates that the amount of adsorbate (x/m) is proportional to the pressure raised to the power of 1/n, where n is greater than 1, and k is a constant that depends on the system.
• Other options do not correctly represent the Freundlich adsorption isotherm.

Hence, the correct answer is x/m = k * p^1/n.