Adsorption and Adsorption isotherm Questions in English

(A) The froth flotation process used for the concentration of sulphide ores is based on the principle of adsorption.
In this process,the ore particles are preferentially wetted by the oil (froth collector),which leads to their adsorption on the surface of the froth,while the gangue particles are wetted by water.

(C) The decomposition of gaseous $H_2$ on a metal surface like tungsten is a heterogeneous catalytic reaction.
At high pressures,the metal surface is completely covered with gas molecules.
Since the rate of reaction depends on the number of active sites available on the surface,and the surface is already saturated,the rate becomes independent of the concentration of the reactant.
Therefore,it follows $0^{th}$ order kinetics.

(D) The Freundlich adsorption isotherm is given by the equation $\frac{x}{m} = k \cdot p^{1/n}$,where $n > 1$.
As the pressure $p$ increases,the value of $p^{1/n}$ increases,but since $1/n < 1$,the rate of increase of $\frac{x}{m}$ decreases as $p$ increases.
Therefore,the value of $\frac{x}{m}$ does not increase rapidly with an increase in $p$ because $n > 1$.

(D) Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent. It is a specific process,irreversible,and generally slow. It is highly dependent on temperature,as it typically increases initially with temperature due to the activation energy requirement and then decreases.

(C) Physical adsorption (physisorption) is characterized by weak van der Waals forces between the adsorbate and the adsorbent.
Because these forces are weak,the enthalpy of adsorption is low,typically ranging from $-20 \, kJ \, mol^{-1}$ to $-40 \, kJ \, mol^{-1}$.
In contrast,chemical adsorption (chemisorption) involves strong chemical bonds and has a much higher enthalpy of adsorption,typically ranging from $-80 \, kJ \, mol^{-1}$ to $-400 \, kJ \, mol^{-1}$.

(C) $1$. Adsorption is a surface phenomenon involving the release of energy,making it an exothermic process $(\Delta H < 0)$.
$2$. Physisorption involves weak van der Waals forces and is reversible. It decreases with an increase in temperature.
$3$. Chemisorption involves strong chemical bonds and is often irreversible. It initially increases with temperature due to the requirement of activation energy.
$4$. Therefore,the statement that 'Physisorption increases with an increase in temperature' is incorrect,as physisorption typically decreases as temperature rises.

(C) $(i)$ Physical adsorption is a reversible process; it can be reversed by changing temperature or pressure.
$(ii)$ The enthalpy of physical adsorption is low,typically $20-40 \ kJ \ mol^{-1}$,whereas chemical adsorption has a higher enthalpy of $80-240 \ kJ \ mol^{-1}$.
$(iii)$ It occurs on all solid adsorbents to some extent and does not require activation energy.
$(iv)$ It decreases with an increase in temperature.

(C) Adsorption is a surface phenomenon where gas molecules are trapped on the surface of a solid.
When gas molecules are adsorbed,their freedom of movement is restricted,which leads to a decrease in the entropy of the system $(\Delta S < 0)$.
For a process to be spontaneous,the Gibbs free energy change $(\Delta G)$ must be negative $(\Delta G = \Delta H - T\Delta S < 0)$.
Since $\Delta S$ is negative,the term $-T\Delta S$ becomes positive.
For $\Delta G$ to be negative,$\Delta H$ must be negative,which means the process is exothermic.

(D) The extent of adsorption of a gas on a solid surface depends on the ease of liquefaction of the gas.
Easily liquefiable gases have higher critical temperatures $(T_c)$ and are adsorbed more strongly.
The critical temperatures of the given gases are:
$H_2$ $(33 \ K)$,$CO$ $(134 \ K)$,$CO_2$ $(304 \ K)$,and $NH_3$ $(405 \ K)$.
Since $NH_3$ has the highest critical temperature,it is the most easily liquefiable and will be adsorbed to the greatest extent on the surface of charcoal.

(C) Chromatography is a technique used for the separation of components of a mixture. It is based on the principle of differential adsorption of the components of a mixture on a stationary phase. Therefore,the process involved is adsorption.

(D) The $As_2S_3$ sol is a negatively charged sol.
During its preparation,$As_2S_3$ is formed by the reaction: $As_2O_3 + 3H_2S \rightarrow As_2S_3 + 3H_2O$.
The particles of $As_2S_3$ preferentially adsorb $S^{2-}$ ions from the excess $H_2S$ present in the medium,which gives the sol a negative charge.

(C) Adsorption is a spontaneous process,so the Gibbs free energy change,$\Delta G$,must be negative $(\Delta G < 0)$.
Since adsorption is an exothermic process,the enthalpy change,$\Delta H$,is negative $(\Delta H < 0)$.
As gas molecules are adsorbed on the solid surface,their randomness decreases,so the entropy change,$\Delta S$,is negative $(\Delta S < 0)$.

(B) Physical adsorption (physisorption) is an exothermic process,which means $\Delta H < 0$.
According to Le Chatelier's principle,for an exothermic process,an increase in temperature will shift the equilibrium in the backward direction,leading to a decrease in the extent of adsorption.
Therefore,the statement that adsorption increases with an increase in temperature is incorrect for physical adsorption.

(D) $(i)$ At a constant temperature,the extent of adsorption $(x/m)$ is proportional to pressure up to a certain limit,but deviations occur at very high pressures.
$(ii)$ Since adsorption is an exothermic process,an increase in temperature decreases the extent of adsorption.
$(iii)$ Chemisorption is unimolecular,whereas physisorption can form multimolecular layers on the adsorbent.
$(iv)$ As the particle size of the adsorbent decreases,the surface area increases,which leads to an increase in the extent of adsorption. Therefore,the statement that particle size does not affect adsorption is incorrect.

(B) Peptization is the process of converting a precipitate into a colloidal sol by shaking it with an electrolyte in the presence of a small amount of a dispersion medium.
In the case of $Fe(OH)_3$ precipitate,when $FeCl_3$ is added,the common $Fe^{3+}$ ions are preferentially adsorbed on the surface of the $Fe(OH)_3$ particles.
This results in the formation of positively charged colloidal particles of $Fe(OH)_3 \cdot Fe^{3+}$.
Therefore,the charge on the colloidal particles is due to the adsorption of $Fe^{3+}$ ions.

(B) Adsorption is an exothermic process,which means heat is released during the process. Therefore,the enthalpy change $\Delta H$ is negative $(\Delta H < 0)$.
During adsorption,the gas molecules are trapped on the solid surface,which leads to a decrease in the randomness of the molecules. Therefore,the entropy change $\Delta S$ is negative $(\Delta S < 0)$.

(A) The Langmuir adsorption isotherm equation is given by $\frac{x}{m} = \frac{aP}{1 + bP}$.
At high pressure,the term $bP$ is much greater than $1$ $(bP >> 1)$.
Therefore,the equation simplifies to $\frac{x}{m} = \frac{aP}{bP} = \frac{a}{b}$.
Since $a$ and $b$ are constants,the amount of gas adsorbed $(\frac{x}{m})$ becomes independent of pressure and reaches a constant limiting value.

(B) When $AgNO_3$ is added in excess to $KI$ solution,the $AgI$ precipitate is formed.
Since $AgNO_3$ is in excess,the $Ag^{+}$ ions are present in the dispersion medium.
The $AgI$ particles preferentially adsorb the common ion from the dispersion medium.
Thus,$AgI$ particles adsorb $Ag^{+}$ ions from the solution,resulting in the formation of positively charged colloidal particles: $[AgI]Ag^{+}$.

(D) According to the preferential adsorption theory,the colloidal particles of a sol preferentially adsorb the common ion present in the dispersion medium.
In the case of $SnCl_4$ sol prepared in the presence of excess $HCl$,the common ion present is the chloride ion $(Cl^-)$.
Therefore,the colloidal particles will preferentially adsorb $Cl^-$ ions from the medium.

(B) Physical adsorption is an exothermic process,meaning the enthalpy of adsorption $(\Delta H_{ads})$ is always negative,typically ranging from $20-40 \ kJ/mol$.
Option $B$ states that the enthalpy is positive,which is incorrect.

(D) Adsorption is a spontaneous process,so the change in Gibbs free energy $(\Delta G)$ is negative $(\Delta G < 0)$.
Since the particles of the adsorbate are trapped on the surface of the adsorbent,their randomness decreases,so the change in entropy $(\Delta S)$ is negative $(\Delta S < 0)$.
Adsorption is an exothermic process,so the change in enthalpy $(\Delta H)$ is negative $(\Delta H < 0)$.
Therefore,none of the thermodynamic parameters $(\Delta G, \Delta S, \Delta H)$ are zero during the process of adsorption.

(D) The extent of adsorption of a gas on a solid adsorbent depends on the ease of liquefaction of the gas.
Easily liquefiable gases have higher critical temperatures $(T_c)$ and are adsorbed to a greater extent.
The critical temperatures of the given gases are:
$NH_3$ $(405.5 \ K)$ > $CO_2$ $(304.1 \ K)$ > $CH_4$ $(190.6 \ K)$ > $H_2$ $(33.2 \ K)$.
Since the extent of adsorption follows the order of critical temperatures,the correct order of volume adsorbed is $NH_3 > CO_2 > CH_4 > H_2$.

(B) The extent of adsorption of a gas on a solid surface depends on the ease of liquefaction of the gas.
Easily liquefiable gases like $SO_2, NH_3, HCl$ have higher critical temperatures and are adsorbed to a greater extent compared to permanent gases like $N_2, O_2, H_2$.
Since $SO_2$ is more easily liquefiable than $N_2, O_2,$ and $H_2$,it is adsorbed to a greater extent on activated charcoal.

(A) Physical adsorption is an exothermic process. According to Le Chatelier's principle,it is favored at low temperatures. Therefore,the rate of physical adsorption increases with a decrease in temperature.

(D) In the process of adsorption,the substance that gets accumulated on the surface of another substance is called the $Adsorbate$.
Here,acetic acid is the substance that accumulates on the surface of the activated charcoal.
Therefore,acetic acid is the $Adsorbate$ and activated charcoal is the $Adsorbent$.

(D) When $FeCl_3$ is added to hot water,a positively charged ferric hydroxide sol is formed due to the preferential adsorption of $Fe^{+3}$ ions from the dispersion medium.

(A) According to the Freundlich adsorption isotherm,the relationship between the extent of adsorption $(x/m)$ and pressure $(P)$ is given by $x/m = k \cdot P^{1/n}$,where $n > 1$.
$1$. At low pressure,$1/n \approx 1$,so $x/m = k \cdot P^1$,which means it is directly proportional to pressure.
$2$. At intermediate pressure,$x/m = k \cdot P^{1/n}$,where $0 < 1/n < 1$.
$3$. At high pressure,$1/n \approx 0$,so $x/m = k \cdot P^0 = \text{constant}$,meaning it becomes independent of pressure.
Comparing these with the given statements,all three relationships are correct representations of the Freundlich isotherm under the specified conditions.

(C) The term 'sorption' is used when both adsorption and absorption occur simultaneously in a process.
Adsorption is a surface phenomenon,while absorption is a bulk phenomenon.
When both processes take place together,the phenomenon is collectively called 'sorption'.

(B) The statement in option $B$ is incorrect because both physisorption and chemisorption can occur at various pressures depending on the nature of the adsorbent and adsorbate. Physisorption is generally favored at low temperatures and high pressures,while chemisorption often requires activation energy and can occur at higher temperatures.

(A) The Freundlich adsorption isotherm is given by $\frac{x}{m} = k p^{1/n}$.
Taking logarithm on both sides: $\log(x/m) = \log k + \frac{1}{n} \log p$.
This is in the form of a straight line equation $y = mx + c$,where $y = \log(x/m)$,$m = \text{slope} = \frac{1}{n}$,and $c = \text{intercept} = \log k$.
Given: $\text{Angle} = 45^o$,so $\text{slope} = \tan(45^o) = 1$. Thus,$\frac{1}{n} = 1$.
Given: $\text{Intercept} = \log k = 0.3010$. Since $\log(2) \approx 0.3010$,we have $k = 2$.
The equation becomes $\log(x/m) = 0.3010 + 1 \times \log(0.2)$.
$\log(x/m) = \log(2) + \log(0.2) = \log(2 \times 0.2) = \log(0.4)$.
Therefore,$\frac{x}{m} = 0.4$.

(A) Physisorption is an exothermic process,which means it releases heat.
According to Le Chatelier's principle,for an exothermic process,a decrease in temperature favors the forward reaction.
Therefore,the rate and extent of physisorption increase as the temperature decreases.

(B) Physical adsorption,also known as physisorption,occurs when the adsorbate molecules are held to the surface of the adsorbent by weak intermolecular forces. These forces are known as $Van \ der \ Waals$ forces. Unlike chemisorption,no chemical bonds are formed in this process.

(A) In $Physisorption$ (physical adsorption),the adsorbate molecules are held to the surface by weak $Van \text{ } der \text{ } Waals$ forces. Due to these weak forces,multiple layers of adsorbate can accumulate on the adsorbent surface,leading to the formation of a multimolecular layer. In contrast,$Chemisorption$ involves strong chemical bonds and is typically limited to the formation of a unimolecular layer.

(D) Adsorption is a surface phenomenon. The extent of adsorption depends on the surface area of the adsorbent. Smaller particle size leads to a larger surface area,which increases the extent of adsorption. Therefore,the statement that the particle size of the adsorbent does not affect the value of adsorption is incorrect.

(C) Physical adsorption,also known as physisorption,occurs when the adsorbate is held on the surface of the adsorbent by weak intermolecular forces.
These forces are known as $Van \ der \ Waals$ forces.
Unlike chemisorption,which involves strong chemical bonds (ionic or covalent),physisorption is reversible and involves low enthalpy of adsorption.

(C) Chemisorption involves the formation of strong chemical bonds between the adsorbate and the adsorbent.
Due to the nature of chemical bonding,it is specific and typically results in the formation of a uni-molecular layer on the surface of the adsorbent.
In contrast,physisorption involves weak Van der Waals forces and can lead to multi-molecular layers.
The enthalpy of chemisorption is high,typically ranging from $80$ to $240 \ kJ \ mol^{-1}$,whereas physisorption has low enthalpy values $(20-40 \ kJ \ mol^{-1})$.

(D) Adsorption is a spontaneous process,which implies that the change in Gibbs free energy,$\Delta G$,must be negative $(\Delta G < 0)$.
Since adsorption involves the accumulation of molecules on a surface,the randomness of the system decreases,meaning the change in entropy,$\Delta S_{\text{system}}$,is negative $(\Delta S < 0)$.
Adsorption is an exothermic process,which means heat is released,so the change in enthalpy,$\Delta H$,is negative $(\Delta H < 0)$.
Therefore,all the given values are less than zero.

(D) The Freundlich adsorption isotherm is given by the equation $\frac{x}{m} = k \cdot p^{1/n}$ (where $n > 1$).
At low pressure,the exponent $\frac{1}{n} \approx 1$,so $\frac{x}{m} \propto p^1$.
At high pressure,the exponent $\frac{1}{n} \approx 0$,so $\frac{x}{m} \propto p^0$.
In the intermediate range,$\frac{x}{m} \propto p^{1/n}$.
Since all these relations hold true for different pressure ranges,option $D$ is the correct answer.

(B) $1$. Physisorption is caused by weak van der Waals forces and is reversible in nature.
$2$. Chemisorption involves strong chemical bonds and is irreversible.
$3$. Physisorption is multi-layered,whereas chemisorption is unimolecular (forms a single layer).
$4$. The enthalpy of adsorption for chemisorption $(80-240 \ kJ \ mol^{-1})$ is much higher than that for physisorption $(20-40 \ kJ \ mol^{-1})$.
$5$. Therefore,the statement that chemisorption forms multimolecular layers is incorrect.

(A) Adsorption is a spontaneous process where gas molecules are trapped on the surface of a solid.
According to the Gibbs free energy equation,$\Delta G = \Delta H - T\Delta S$.
For a spontaneous process,$\Delta G$ must be negative.
During adsorption,the freedom of movement of gas molecules decreases,leading to a decrease in entropy $(\Delta S < 0)$.
Since $-T\Delta S$ becomes positive,for $\Delta G$ to be negative,the enthalpy change $(\Delta H)$ must be negative (exothermic).

(D) The extent of adsorption of a gas on a solid surface (adsorbent) is influenced by several factors:
$1$. Nature of the gas: Easily liquefiable gases (like $NH_3$,$HCl$,$CO_2$) are adsorbed more readily than permanent gases (like $H_2$,$O_2$,$N_2$).
$2$. Pressure of the gas: According to the Freundlich adsorption isotherm,the extent of adsorption $(x/m)$ increases with an increase in pressure.
$3$. Temperature of the system: Adsorption is generally an exothermic process; therefore,according to Le Chatelier's principle,the extent of adsorption decreases with an increase in temperature.
Thus,all the given factors affect the extent of adsorption.

(B) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k \cdot p^{1/n}$ (where $n > 1$).
Taking the logarithm on both sides,we get: $\log(\frac{x}{m}) = \log(k) + \frac{1}{n} \log(p)$.
This equation is in the form of a straight line $y = mx + c$,where $y = \log(\frac{x}{m})$,$x = \log(p)$,slope $m = \frac{1}{n}$,and intercept $c = \log(k)$.
Therefore,the plot of $\log(\frac{x}{m})$ vs $\log(p)$ gives a straight line.

(A) In chemical adsorption (chemisorption),the adsorbate molecules are held to the surface of the adsorbent by strong chemical bonds.
Because these bonds are specific and require direct interaction with the surface atoms,chemisorption is limited to the formation of a single layer (unimolecular layer).
Therefore,the correct answer is $1$.

(C) In chemical adsorption (chemisorption),the adsorbate molecules are held to the surface of the adsorbent by strong chemical bonds.
Since these bonds are specific and require direct contact with the surface,the adsorption is limited to a single layer of molecules.
Therefore,chemical adsorption is always unimolecular.