Surfaces, Interfaces and Colloids Questions and Answers

Q: If θ is the fraction of the surface covered at PA (pressure of the adsorbate A), then :

The correct relation is θ = Vads at PA / Vmax.The fractional coverage is the volume of gas adsorbed at the given pressure divided by the volume needed for a complete monolayer. It lies between 0 and 1, so Vmax must be in the denominator; the inverted ratio would exceed 1.The rate statements are the wrong way round. In the Langmuir model, adsorption needs a vacant site and a colliding molecule, so its rate is proportional to (1 - θ)PA; desorption depends only on the occupied sites, so its rate is proportional to θ, with no pressure term.

Q: When a drop of liquid at the tip of a capillary is balanced by surface tensional forces, its weight is equal to :

The weight of the drop equals 2πrγ.The drop hangs from the tip of the capillary, held by surface tension acting along the circumference of the tip. Surface tension γ is a force per unit length, and the length of the contact line is the circumference 2πr. At the moment of detachment the upward force just balances the weight (Tate's law):\[ mg = 2\pi r\gamma \]This is the basis of the drop-weight (stalagmometer) method. πr2γ uses an area where a length is needed, and the other two forms are not dimensionally a force.

Q: The pressure difference across a curved interface can be written as :

The keyed form is ΔP = r[1/R1 + 1/R2], where the factor printed as r stands for the surface tension γ.The Young-Laplace equation gives the pressure difference across a curved interface with principal radii of curvature R1 and R2:\[ \Delta P = \gamma (\frac{1}{R_{1}} + \frac{1}{R_{2}}) \]For a sphere, R1 = R2 = R and this becomes 2γ/R. The curvatures add, which rules out the forms with a minus sign, and the surface tension multiplies the curvature rather than dividing it, which rules out the forms with 1/r. Read with the printed r as γ, the first option is the Young-Laplace equation.

Q: According to Langmuir theory of adsorption which of the following is correct ?

According to Langmuir, the heat of adsorption is independent of coverage.The Langmuir model assumes a uniform surface of equivalent sites, a monolayer, and no interactions between adsorbed molecules. Because every site is the same and neighbours do not affect each other, the heat of adsorption is the same whether the surface is nearly empty or nearly full.A heat of adsorption that changes with coverage (A) and interacting adsorbates (C) are the deviations that the Temkin and Freundlich treatments allow for. Dissociative adsorption (D) can be handled within the Langmuir framework, so 'always non-dissociative' is not part of the theory.

Q: According to Langmuir model of adsorption, there is an equilibrium between adsorption and desorption as given below :\[ \mathrm{A_{(g)} + S_{(s)} \rightleftharpoons A{-}S_{(s)}} \]where A & S represent the adsorbate and the solid surface sites and A-S is the adsorbed species.The adsorption coefficient K is defined as ka/kd where ka and kd are the rate constants for adsorption and desorption respectively.Which of the following is correct ?(I) \(\mathrm{K = \frac{[A{-}S]}{[A][S]}}\)(II) \(\mathrm{K = \frac{[A][S]}{[A{-}S]}}\)(III) A large value of K indicates strong adsorption(IV) A large value of K indicates strong desorption

The correct statements are (I) and (III).At equilibrium the rates of adsorption and desorption are equal:\[ \begin{aligned} k_{a}[\mathrm{A}][\mathrm{S}] &= k_{d}[\mathrm{A{-}S}] \\ K = \frac{k_{a}}{k_{d}} &= \frac{[\mathrm{A{-}S}]}{[\mathrm{A}][\mathrm{S}]} \end{aligned} \]So (I) is right and (II), its reciprocal, is wrong. A large K means adsorption is fast compared with desorption and most of the adsorbate stays bound, which is strong adsorption (III), not strong desorption (IV).

Q: For a solid/liquid interface the following equation is applicable :\( \dfrac{\gamma_{SV} - \gamma_{SL}}{\gamma_{LV}} = \mathrm{cos}\,\theta \)where θ is the contact angle, γSV, γSL and γLV are the surface tension of solid, interfacial tension of S/L and surface tension of liquid.If a metal pan has to be made non-sticking for cooking appliance :

The pan should be coated with a low energy substance.From Young's equation, food or water sticks least when it wets the surface least, which means a large contact angle and a small cosθ. That requires a small γSV, a low-energy solid surface. This is why non-stick pans are coated with PTFE (Teflon), whose surface energy is very low.A high-energy coating would raise γSV and increase wetting and sticking. Surfactants in the water lower γLV and γSL, which improves wetting, the opposite of what is wanted.

Q: Which of the following is correct ?(I) CMC of an anionic surfactant is higher than that of a non-ionic surfactant of the same chain length(II) CMC of an anionic surfactant is lower than that of a non-ionic surfactant of the same chain length(III) CMC of an ionic surfactant decreases in presence of small quantity of added electrolyte(IV) CMC of an ionic surfactant increases in presence of small quantity of added electrolyte

The correct statements are (I) and (III).A surfactant forms micelles when the gain from hiding its hydrocarbon tails outweighs the cost of crowding its head groups together. Ionic head groups repel one another electrostatically, which makes micellization harder, so an anionic surfactant needs a higher concentration to form micelles than a non-ionic surfactant with the same tail. Statement (I) is true and (II) false; for C12 chains the CMC of sodium dodecyl sulfate is about 8 mM, while non-ionic C12 surfactants lie near 0.1 mM.Added electrolyte screens the repulsion between ionic head groups (and its counter-ions bind to the micelle surface), so micelles form more easily and the CMC of an ionic surfactant decreases. Statement (III) is true and (IV) false. The options that include (II) or (IV) contain one of the false statements.

Q: Which of the following are directly related to surface tension phenomena ?(I) Spherical shapes of droplets in the absence of external forces(II) A sail boat in water(III) A needle floating in water(IV) Water rising in the xylem of plants

The phenomena directly related to surface tension are (I), (III) and (IV).(I) A free droplet takes a spherical shape because surface tension minimizes its surface area, and a sphere has the least area for a given volume.(III) A steel needle laid gently on water floats, supported by the surface film, although steel is far denser than water.(IV) The rise of water in the narrow xylem vessels of plants is treated as capillary action, which depends on surface tension together with the adhesion of water to the vessel walls.(II) A sailing boat floats by buoyancy (Archimedes' principle) and moves by wind on its sails, neither of which involves surface tension. Every option containing (II) is therefore wrong.

Q: For the adsorption of N2 on activated carbon at 77 K :

For this adsorption ΔH < 0 and ΔS < 0.Nitrogen adsorbs readily on activated carbon at 77 K, so the process is spontaneous and ΔG < 0. Gas molecules that settle on the surface lose most of their freedom to move, so the entropy falls: ΔS < 0. With ΔG = ΔH − TΔS, a negative ΔG and a negative ΔS are possible only if ΔH is negative:\[ \begin{aligned} &\Delta G = \Delta H - T\Delta S \lt 0,\ \Delta S \lt 0 \\ &\Rightarrow \Delta H \lt 0 \end{aligned} \]Adsorption is therefore exothermic; this physical adsorption releases heat comparable to the heat of condensation of nitrogen.ΔH = 0 or ΔH > 0 combined with ΔS < 0 would make ΔG positive and the adsorption non-spontaneous, and ΔS = 0 ignores the loss of freedom.

Q: According to Langmuir model :(I) All adsorption sites are equivalent(II) Adsorption and desorption are cooperative processes(III) Probability of adsorption is not affected by occupancy state of adsorption site(IV) Only monolayer adsorption

The Langmuir assumptions here are (I) and (IV).The Langmuir model treats the surface as a set of identical, equivalent sites (I), each holding at most one molecule, so adsorption stops at a monolayer (IV). It also assumes no interaction between adsorbed molecules, so the chance of adsorbing on a site does not depend on whether neighbouring sites are filled.(II) is wrong: in the model adsorption and desorption are independent, non-cooperative events whose rates balance at equilibrium. (III) as worded is also wrong: the probability of adsorption on a site does depend on its own occupancy, since an occupied site cannot take another molecule; the rate of adsorption is proportional to the fraction of empty sites, (1 − θ). So every option containing (II) or (III) is wrong.

11.

If θ is the fraction of the surface covered at P_A (pressure of the adsorbate A), then :

A.rate of adsorption is proportional to \(\theta \times P_{A}\)
B.rate of desorption is proportional to \((1 - \theta) \times P_{A}\)
C.\(\theta = \dfrac{V_{ads}\ \text{at}\ P_{A}}{V_{max}}\)
D.\(\theta = \dfrac{V_{max}}{V_{ads}\ \text{at}\ P_{A}}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The correct relation is θ = V_ads at P_A / V_max.

The fractional coverage is the volume of gas adsorbed at the given pressure divided by the volume needed for a complete monolayer. It lies between 0 and 1, so V_max must be in the denominator; the inverted ratio would exceed 1.

The rate statements are the wrong way round. In the Langmuir model, adsorption needs a vacant site and a colliding molecule, so its rate is proportional to (1 - θ)P_A; desorption depends only on the occupied sites, so its rate is proportional to θ, with no pressure term.

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12.

When a drop of liquid at the tip of a capillary is balanced by surface tensional forces, its weight is equal to :

A.\(\pi r^{2}\gamma\)
B.\(2\pi r\gamma\)
C.\(\gamma/2\pi r\)
D.\(\frac{4}{3}\pi r^{3}\gamma\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The weight of the drop equals 2πrγ.

The drop hangs from the tip of the capillary, held by surface tension acting along the circumference of the tip. Surface tension γ is a force per unit length, and the length of the contact line is the circumference 2πr. At the moment of detachment the upward force just balances the weight (Tate's law):

\[ mg = 2\pi r\gamma \]

This is the basis of the drop-weight (stalagmometer) method. πr²γ uses an area where a length is needed, and the other two forms are not dimensionally a force.

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13.

The pressure difference across a curved interface can be written as :

A.\(\Delta P = r\left[\frac{1}{R_{1}} + \frac{1}{R_{2}}\right]\)
B.\(\Delta P = r\left[\frac{1}{R_{1}} - \frac{1}{R_{2}}\right]\)
C.\(\Delta P = \frac{1}{r}\left[\frac{1}{R_{1}} + \frac{1}{R_{2}}\right]\)
D.\(\Delta P = \frac{1}{r}\left[\frac{1}{R_{1}} - \frac{1}{R_{2}}\right]\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The keyed form is ΔP = r[1/R₁ + 1/R₂], where the factor printed as r stands for the surface tension γ.

The Young-Laplace equation gives the pressure difference across a curved interface with principal radii of curvature R₁ and R₂:

\[ \Delta P = \gamma (\frac{1}{R_{1}} + \frac{1}{R_{2}}) \]

For a sphere, R₁ = R₂ = R and this becomes 2γ/R. The curvatures add, which rules out the forms with a minus sign, and the surface tension multiplies the curvature rather than dividing it, which rules out the forms with 1/r. Read with the printed r as γ, the first option is the Young-Laplace equation.

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14.

According to Langmuir theory of adsorption which of the following is correct ?

A.Heat of adsorption changes with coverage.
B.Heat of adsorption is independent of coverage.
C.The adsorbed species interact with each other.
D.Adsorption is always non-dissociative.

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

According to Langmuir, the heat of adsorption is independent of coverage.

The Langmuir model assumes a uniform surface of equivalent sites, a monolayer, and no interactions between adsorbed molecules. Because every site is the same and neighbours do not affect each other, the heat of adsorption is the same whether the surface is nearly empty or nearly full.

A heat of adsorption that changes with coverage (A) and interacting adsorbates (C) are the deviations that the Temkin and Freundlich treatments allow for. Dissociative adsorption (D) can be handled within the Langmuir framework, so 'always non-dissociative' is not part of the theory.

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15.

According to Langmuir model of adsorption, there is an equilibrium between adsorption and desorption as given below :

\[ \mathrm{A_{(g)} + S_{(s)} \rightleftharpoons A{-}S_{(s)}} \]

where A & S represent the adsorbate and the solid surface sites and A-S is the adsorbed species.

The adsorption coefficient K is defined as k_a/k_d where k_a and k_d are the rate constants for adsorption and desorption respectively.

Which of the following is correct ?

(I) \(\mathrm{K = \frac{[A{-}S]}{[A][S]}}\)
(II) \(\mathrm{K = \frac{[A][S]}{[A{-}S]}}\)
(III) A large value of K indicates strong adsorption
(IV) A large value of K indicates strong desorption

A.(I) and (III)
B.(II) and (III)
C.(I) and (IV)
D.(I) and (II)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The correct statements are (I) and (III).

At equilibrium the rates of adsorption and desorption are equal:

\[ \begin{aligned} k_{a}[\mathrm{A}][\mathrm{S}] &= k_{d}[\mathrm{A{-}S}] \\ K = \frac{k_{a}}{k_{d}} &= \frac{[\mathrm{A{-}S}]}{[\mathrm{A}][\mathrm{S}]} \end{aligned} \]

So (I) is right and (II), its reciprocal, is wrong. A large K means adsorption is fast compared with desorption and most of the adsorbate stays bound, which is strong adsorption (III), not strong desorption (IV).

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16.

For a solid/liquid interface the following equation is applicable :

\( \dfrac{\gamma_{SV} - \gamma_{SL}}{\gamma_{LV}} = \mathrm{cos}\,\theta \)

where θ is the contact angle, γ_SV, γ_SL and γ_LV are the surface tension of solid, interfacial tension of S/L and surface tension of liquid.

If a metal pan has to be made non-sticking for cooking appliance :

A.Edible surfactants are to be added to the cooking water.
B.The metal has to be coated with high energy substance.
C.The metal has to be coated with a low energy substance.
D.Edible surfactants that can adsorb at S/L interface can be added.

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The pan should be coated with a low energy substance.

From Young's equation, food or water sticks least when it wets the surface least, which means a large contact angle and a small cosθ. That requires a small γ_SV, a low-energy solid surface. This is why non-stick pans are coated with PTFE (Teflon), whose surface energy is very low.

A high-energy coating would raise γ_SV and increase wetting and sticking. Surfactants in the water lower γ_LV and γ_SL, which improves wetting, the opposite of what is wanted.

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17.

Which of the following is correct ?

(I) CMC of an anionic surfactant is higher than that of a non-ionic surfactant of the same chain length
(II) CMC of an anionic surfactant is lower than that of a non-ionic surfactant of the same chain length
(III) CMC of an ionic surfactant decreases in presence of small quantity of added electrolyte
(IV) CMC of an ionic surfactant increases in presence of small quantity of added electrolyte

A.(I) and (II)
B.(I) and (III)
C.(II) and (III)
D.(III) and (IV)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The correct statements are (I) and (III).

A surfactant forms micelles when the gain from hiding its hydrocarbon tails outweighs the cost of crowding its head groups together. Ionic head groups repel one another electrostatically, which makes micellization harder, so an anionic surfactant needs a higher concentration to form micelles than a non-ionic surfactant with the same tail. Statement (I) is true and (II) false; for C₁₂ chains the CMC of sodium dodecyl sulfate is about 8 mM, while non-ionic C₁₂ surfactants lie near 0.1 mM.

Added electrolyte screens the repulsion between ionic head groups (and its counter-ions bind to the micelle surface), so micelles form more easily and the CMC of an ionic surfactant decreases. Statement (III) is true and (IV) false. The options that include (II) or (IV) contain one of the false statements.

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18.

Which of the following are directly related to surface tension phenomena ?

(I) Spherical shapes of droplets in the absence of external forces
(II) A sail boat in water
(III) A needle floating in water
(IV) Water rising in the xylem of plants

A.(I) and (II)
B.(I), (II) and (III)
C.(I), (III) and (IV)
D.(II), (III) and (IV)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The phenomena directly related to surface tension are (I), (III) and (IV).

(I) A free droplet takes a spherical shape because surface tension minimizes its surface area, and a sphere has the least area for a given volume.
(III) A steel needle laid gently on water floats, supported by the surface film, although steel is far denser than water.
(IV) The rise of water in the narrow xylem vessels of plants is treated as capillary action, which depends on surface tension together with the adhesion of water to the vessel walls.

(II) A sailing boat floats by buoyancy (Archimedes' principle) and moves by wind on its sails, neither of which involves surface tension. Every option containing (II) is therefore wrong.

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19.

For the adsorption of N₂ on activated carbon at 77 K :

A.\(\Delta H = 0;\ \Delta S \lt 0\)
B.\(\Delta H \lt 0;\ \Delta S \lt 0\)
C.\(\Delta H \gt 0;\ \Delta S \lt 0\)
D.\(\Delta H = 0;\ \Delta S = 0\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

For this adsorption ΔH < 0 and ΔS < 0.

Nitrogen adsorbs readily on activated carbon at 77 K, so the process is spontaneous and ΔG < 0. Gas molecules that settle on the surface lose most of their freedom to move, so the entropy falls: ΔS < 0. With ΔG = ΔH − TΔS, a negative ΔG and a negative ΔS are possible only if ΔH is negative:

\[ \begin{aligned} &\Delta G = \Delta H - T\Delta S \lt 0,\ \Delta S \lt 0 \\ &\Rightarrow \Delta H \lt 0 \end{aligned} \]

Adsorption is therefore exothermic; this physical adsorption releases heat comparable to the heat of condensation of nitrogen.

ΔH = 0 or ΔH > 0 combined with ΔS < 0 would make ΔG positive and the adsorption non-spontaneous, and ΔS = 0 ignores the loss of freedom.

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20.

According to Langmuir model :

(I) All adsorption sites are equivalent
(II) Adsorption and desorption are cooperative processes
(III) Probability of adsorption is not affected by occupancy state of adsorption site
(IV) Only monolayer adsorption

A.(I), (II) and (III)
B.(I) and (IV)
C.(II) and (III)
D.(II) and (IV)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The Langmuir assumptions here are (I) and (IV).

The Langmuir model treats the surface as a set of identical, equivalent sites (I), each holding at most one molecule, so adsorption stops at a monolayer (IV). It also assumes no interaction between adsorbed molecules, so the chance of adsorbing on a site does not depend on whether neighbouring sites are filled.

(II) is wrong: in the model adsorption and desorption are independent, non-cooperative events whose rates balance at equilibrium. (III) as worded is also wrong: the probability of adsorption on a site does depend on its own occupancy, since an occupied site cannot take another molecule; the rate of adsorption is proportional to the fraction of empty sites, (1 − θ). So every option containing (II) or (III) is wrong.

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