Adsorption and Adsorption isotherm Questions in English

(C) The Freundlich adsorption isotherm equation is given by $\frac{x}{m} = kP^{1/n}$,where $x$ is the mass of the adsorbate,$m$ is the mass of the adsorbent,$P$ is the pressure,and $k$ and $n$ are constants.
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,plotting $\log(\frac{x}{m})$ versus $\log P$ yields a straight line.

(C) Chemisorption is an exothermic process that requires activation energy.
Initially,as temperature increases,the rate of adsorption increases because more molecules gain the necessary activation energy to form chemical bonds with the surface.
However,as the temperature continues to rise,the exothermic nature of the process causes the equilibrium to shift towards desorption,leading to a decrease in the amount of gas adsorbed $(x/m)$.
Therefore,the graph of $x/m$ versus $T$ for chemisorption shows an initial increase followed by a decrease,resulting in a peak.

(B) In column chromatography,the compound that is more strongly adsorbed on the stationary phase moves slower through the column.
Given that the adsorption of $I > II$,compound $I$ is more strongly adsorbed than compound $II$.
Therefore,compound $I$ moves slower and travels a shorter distance,while compound $II$ moves faster and travels a longer distance.
The $R_f$ value is defined as the ratio of the distance traveled by the substance to the distance traveled by the solvent front.
Since compound $II$ travels a greater distance,it has a higher $R_f$ value than compound $I$.

(B) According to the Freundlich adsorption isotherm,the relationship is given by:
$\frac{x}{m} = kP^{\frac{1}{n}}$
Taking the logarithm on both sides:
$\log_{10} \frac{x}{m} = \log_{10} (kP^{\frac{1}{n}})$
$\log_{10} \frac{x}{m} = \frac{1}{n} \log_{10} P + \log_{10} k$
This equation follows the linear form $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log P$,the slope $m = \frac{1}{n}$,and the intercept $c = \log k$.
Therefore,the slope of the plot is $\frac{1}{n}$.

(D) Physical adsorption (physisorption) is characterized by weak van der Waals forces between the adsorbate and the adsorbent.
It is a reversible process that is favored by low temperatures and high pressures.
It increases with an increase in the surface area of the adsorbent.
Unlike chemisorption,physical adsorption is multi-layered in nature.
Therefore,the statement that 'Unilayer adsorption occurs' is incorrect,as it is a property of chemisorption.

(C) According to the Freundlich adsorption isotherm:
$\log \frac{x}{m} = \log k + \frac{1}{n} \log P$
Comparing this with the equation of a straight line $y = mx + c$,the slope is $\frac{1}{n}$.
Given that the slope is $0.5$,we have $\frac{1}{n} = 0.5 = \frac{1}{2}$.
Substituting this into the adsorption equation $\frac{x}{m} = k \cdot P^{1/n}$,we get:
$\frac{x}{m} = k \cdot P^{1/2}$
This implies that the adsorption is proportional to the square root of pressure $(P^{1/2})$.

(D) The characteristics given suggest that this is physical adsorption (physisorption).
Physical adsorption is characterized by weak van der Waals forces,reversibility,and the formation of multimolecular layers.
It is an exothermic process that typically decreases with an increase in temperature.
The enthalpy of adsorption is low (usually $20-40 \ kJ \ mol^{-1}$).
Since it involves weak forces and no chemical bond formation,the energy of activation required is very low.

(C) Adsorption is a surface phenomenon where molecules of a substance accumulate on the surface of a solid or liquid.
During adsorption,the residual forces on the surface are satisfied by the incoming molecules,leading to a decrease in surface energy.
Therefore,the statement that residual forces increase is incorrect.
For adsorption,$\Delta H < 0$ (exothermic),$\Delta S < 0$ (decrease in randomness),and $\Delta G < 0$ (spontaneous process).

(C) The Freundlich adsorption isotherm is given by the equation: $x/m = k \cdot p^{1/n}$
Taking the logarithm on both sides: $\log(x/m) = \log k + (1/n) \log p$
This equation is of the form $y = mx + c,$ where $y = \log(x/m),$ $x = \log p,$ slope $m = 1/n,$ and intercept $c = \log k.$
Therefore,plotting $\log(x/m)$ against $\log p$ yields a straight line.

(C) Adsorption is a surface phenomenon that involves the release of energy.
Both physisorption (physical adsorption) and chemisorption (chemical adsorption) are exothermic processes because they involve the formation of bonds (van der Waals forces in physisorption and chemical bonds in chemisorption) which leads to a decrease in the enthalpy of the system $(\Delta H < 0)$.

(C) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = kP^{1/n}$.
Taking logarithm on both sides: $\log(\frac{x}{m}) = \log k + \frac{1}{n} \log P$.
This is in the form of a straight line equation $y = mx + c$,where the slope is $\frac{1}{n}$.
From the given plot,the slope is calculated as: $\text{Slope} = \frac{\text{change in } y}{\text{change in } x} = \frac{2}{4} = \frac{1}{2}$.
Therefore,$\frac{1}{n} = \frac{1}{2}$.
Substituting this back into the original equation,we get: $\frac{x}{m} \propto P^{1/2}$.

(A) The extent of adsorption of a gas on a solid surface is directly related to its ease of liquefaction.
Easier liquefaction occurs for gases with higher critical temperatures.
Therefore,the smaller the value of the critical temperature of a gas,the lesser is the extent of its adsorption.
Comparing the given values,the critical temperature of $H_2$ $(33 \ K)$ is the lowest among all the gases listed.
Thus,$H_2$ is the least adsorbed gas.

(A) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k P^{1/n}$
Taking the logarithm on both sides,we get: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$
This equation represents a straight line of the form $y = mx + c$,where the slope is $\frac{1}{n}$.
From the given graph,the slope is calculated as: $\text{Slope} = \frac{\text{rise}}{\text{run}} = \frac{2}{3}$.
Therefore,$\frac{1}{n} = \frac{2}{3}$.
Substituting this back into the original equation,we get: $\frac{x}{m} = k P^{2/3}$.
Thus,$\frac{x}{m}$ is proportional to $p^{2/3}$.

(A) Calculate the number of moles of surfactant: $n = \frac{M \times V \, (mL)}{1000} = \frac{10^{-3} \times 10}{1000} = 10^{-8} \, mol$.
Calculate the number of molecules: $N = n \times N_A = 10^{-8} \times 6 \times 10^{23} = 6 \times 10^{15} \, \text{molecules}$.
Area covered by one molecule $(a^2)$: $a^2 = \frac{\text{Total Area}}{\text{Number of molecules}} = \frac{0.24 \, cm^2}{6 \times 10^{15}} = 0.04 \times 10^{-15} \, cm^2 = 4 \times 10^{-17} \, cm^2$.
Calculate edge length $(a)$: $a = \sqrt{4 \times 10^{-17} \, cm^2} = 2 \times 10^{-8.5} \, cm \approx 6.32 \times 10^{-9} \, cm$.
Re-evaluating based on standard approximation: If $N_A = 6 \times 10^{23}$, $a^2 = \frac{0.24}{10^{-5} \times 6 \times 10^{23}} = 4 \times 10^{-20} \, cm^2$.
$a = \sqrt{4 \times 10^{-20}} = 2 \times 10^{-10} \, cm = 2 \times 10^{-12} \, m = 2 \, pm$.

(A) The Freundlich adsorption isotherm is given by $\frac{x}{m} = k p^{1/n}$.
According to the given equation $\frac{x}{m} = k p^{0.5}$,the extent of adsorption $\frac{x}{m}$ is directly proportional to the square root of pressure $p$.
Therefore,an increase in pressure $p$ leads to an increase in the extent of adsorption.
Physical adsorption is an exothermic process,which means it releases heat.
According to Le Chatelier's principle,for an exothermic process,a decrease in temperature $T$ favors the forward reaction,thereby increasing the extent of adsorption.
Thus,adsorption increases with an increase in $p$ and a decrease in $T$.

(C) The $R_f$ (retention factor) value is defined as the ratio of the distance traveled by the solute to the distance traveled by the solvent front.
Since the solute cannot travel further than the solvent front,the $R_f$ value is always $\le 1$.
$A$ higher $R_f$ value indicates that the substance has a greater affinity for the mobile phase and a lower affinity for the stationary phase (i.e.,lower adsorption).
Therefore,the statement "Higher $R_f$ value means higher adsorption" is incorrect.

(D) Physisorption is directly related to the ease of liquefaction of a gas.
Easily liquefiable gases have higher critical temperatures and stronger van der Waals forces of attraction.
Among the given options,$H_2O$ has the highest critical temperature and is the most easily liquefiable due to hydrogen bonding.
Therefore,$H_2O$ has the maximum enthalpy of physisorption.

(B) According to the Freundlich adsorption isotherm,the relationship is given by $x/m = K P^{1/n}$.
Here,$x/m$ is the amount of gas adsorbed per unit mass of the adsorbent at pressure $P$.
$K$ and $n$ are constants that depend on the nature of the gas and the adsorbent.
The value of $1/n$ ranges between $0$ and $1$ (i.e.,$0 \le 1/n \le 1$),where $n > 1$.
At low pressure,$1/n = 1$,so $x/m \propto P^1$.
At high pressure,$1/n = 0$,so $x/m \propto P^0$ (independent of pressure).
Thus,the general expression is $x/m = K P^{1/n}$ with $0 \le 1/n \le 1$.

(C) Physical adsorption (physisorption) is not specific in nature because it arises due to weak van der Waals forces between the adsorbent and the adsorbate. It can occur between any gas and solid. Chemical adsorption (chemisorption) is highly specific as it involves the formation of chemical bonds. Therefore,the statement in option $C$ is incorrect.

(C) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k \times p^{\frac{1}{n}}$.
Taking the logarithm on both sides,we get: $\log(\frac{x}{m}) = \log(k) + \frac{1}{n} \log(p)$.
This equation represents a straight line of the form $y = mx + c$,where the slope is $\frac{1}{n}$.
If the slope is zero,then $\frac{1}{n} = 0$.
Substituting this into the original equation: $\frac{x}{m} = k \times p^0 = k$.
Since $\frac{x}{m}$ is constant and does not depend on $p$,the extent of adsorption is independent of the pressure of the gas.

(B) Physical adsorption is a spontaneous process,which implies that the change in Gibbs free energy,$\Delta G$,must be negative $(\Delta G < 0)$.
During adsorption,the gas molecules are trapped on the surface of the adsorbent,leading to a decrease in randomness,so $\Delta S$ is negative $(\Delta S < 0)$.
Adsorption is an exothermic process,meaning heat is released,so the enthalpy change,$\Delta H$,is negative $(\Delta H < 0)$.
Therefore,the correct statement is $\Delta G = -ve$.

(C) Physical adsorption is a multilayered process due to weak van der Waals forces,so statement $A$ is correct.
Chemical adsorption involves the formation of strong chemical bonds,making it irreversible,so statement $B$ is correct.
Physical adsorption is an exothermic process; therefore,it decreases with an increase in temperature. Statement $C$ is incorrect because it does not increase with temperature.
Langmuir adsorption isotherm is primarily used to explain chemical adsorption (chemisorption),so statement $D$ is correct.
Thus,the incorrect statement is $C$.

(A) The froth floatation process is used for the concentration of sulphide ores.
In this process,the sulphide ore particles are preferentially wetted by the oil (frother),while the gangue particles are wetted by water.
This selective wetting is based on the principle of adsorption,where the collector (like pine oil or xanthates) gets adsorbed on the surface of the sulphide ore particles,making them hydrophobic and allowing them to attach to air bubbles and rise to the surface as froth.

(A) The extent of adsorption of a gas on a solid adsorbent like activated charcoal depends on the ease of liquefaction of the gas.
Easily liquefiable gases (which have higher critical temperatures) are adsorbed more strongly.
Among the given options,$CO_2$ has the highest critical temperature $(304.1 \ K)$ compared to $CH_4$ $(190.6 \ K)$,$N_2$ $(126.2 \ K)$,and $H_2$ $(33.2 \ K)$.
Therefore,$CO_2$ is adsorbed to the maximum extent.

(C) Given the Langmuir adsorption isotherm equation: $\frac{x}{m} = \frac{ap}{1 + bp}$.
Taking the reciprocal of both sides,we get: $\frac{m}{x} = \frac{1 + bp}{ap}$.
This can be rewritten as: $\frac{m}{x} = \frac{1}{ap} + \frac{bp}{ap} = \left( \frac{1}{a} \right) \left( \frac{1}{p} \right) + \frac{b}{a}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \frac{m}{x}$ and $x = \frac{1}{p}$:
The slope $(m)$ is $\frac{1}{a}$ and the intercept $(c)$ is $\frac{b}{a}$.

(A) Physical adsorption (physisorption) is characterized by the formation of multimolecular layers on the surface of an adsorbent.
As the pressure $(P)$ of the gas increases,the extent of adsorption $(x/m)$ increases.
At low pressures,the adsorption is proportional to the pressure.
As the pressure increases further,the rate of increase of adsorption decreases,leading to a characteristic curve that approaches a limiting value at the saturation pressure $(P_s)$.
This behavior is described by the Freundlich adsorption isotherm,which shows a smooth,continuous curve that levels off as it approaches $P_s$,representing the saturation of the surface sites.
Option $A$ correctly depicts this standard behavior of physical adsorption.

(B) Initial moles of acetic acid $= M \times V = 0.6 \ M \times 0.1 \ L = 0.06 \ mol$.
Final moles of acetic acid $= 0.5 \ M \times 0.1 \ L = 0.05 \ mol$.
Moles adsorbed $= 0.06 - 0.05 = 0.01 \ mol$.
Mass of acetic acid adsorbed $= 0.01 \ mol \times 60 \ g/mol = 0.6 \ g$.
Mass of carbon used $= 2 \ g$.
Amount adsorbed per gram of carbon $= \frac{0.6 \ g}{2 \ g} = 0.3 \ g/g$.

(B) Physisorption is an exothermic process. According to Le Chatelier's principle,for an exothermic process,the extent of adsorption decreases with an increase in temperature. Therefore,the statement that it increases with an increase in temperature is incorrect.

(D) The Freundlich adsorption isotherm is given by the equation $\frac{x}{m} = kP^{1/n}$,which represents the relationship between the amount of adsorbate and pressure at constant temperature.
The Langmuir adsorption isotherm is given by $\frac{x}{m} = \frac{bP}{1+aP}$,which also relates the amount of adsorbate to pressure at constant temperature.
Adsorption isotherms are plots of $\frac{x}{m}$ versus $P$ at constant $T$,and adsorption isobars are plots of $\frac{x}{m}$ versus $T$ at constant $P$.
Adsorption isosteres are plots of $P$ versus $T$ at constant $\frac{x}{m}$.
The relation $\frac{x}{m} = P \times T$ does not represent any standard adsorption isotherm,isobar,or isostere,as it does not maintain either $P$ or $T$ as a constant variable.
Therefore,option $D$ is not related to the adsorption process.

(A) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K \cdot P^{\frac{1}{n}}$
Taking the logarithm on both sides,we get: $\log \left( \frac{x}{m} \right) = \log K + \frac{1}{n} \log P$
Comparing this equation with the straight-line equation $y = mx + c$,where $y = \log \left( \frac{x}{m} \right)$,$x = \log P$,$m = \frac{1}{n}$,and $c = \log K$.
Thus,the slope of the plot is equal to $\frac{1}{n}$.

(C) In chemical adsorption (chemisorption),a chemical bond is formed between the adsorbate and the adsorbent.
$1$. The enthalpy of adsorption is high,typically ranging from $80-240 \ kJ \ mol^{-1}$,not less than $20 \ kJ \ mol^{-1}$.
$2$. Chemical bonds are formed,not Van der Waals forces (which are characteristic of physisorption).
$3$. Since chemical bonds are formed,the process is specific and usually results in the formation of a unimolecular layer.
$4$. Adsorption is a spontaneous process where the surface area decreases,leading to a decrease in entropy $(\Delta S < 0)$.
Therefore,the correct statement is that a unimolecular layer is formed.

(D) The extent of adsorption of gases 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 to a greater extent.
Among the noble gases,the atomic size increases from $He$ to $Xe$,which leads to an increase in the magnitude of van der Waals forces.
Therefore,$Xe$ has the highest critical temperature and is most easily liquefied,resulting in the maximum extent of adsorption.

(B) Physisorption is an exothermic process,which can be represented as: Adsorbate + Adsorbent $\rightleftharpoons$ Adsorption Complex + Heat.
According to Le Chatelier's principle,for an exothermic process,an increase in temperature shifts the equilibrium in the backward direction.
Therefore,the extent of physical adsorption decreases with an increase in temperature.

(B) Soap molecules are surfactants,which consist of a long hydrophobic hydrocarbon chain and a hydrophilic polar head group.
When soap is added to water,these molecules accumulate at the surface,with the hydrophobic tails pointing away from the water.
This arrangement disrupts the cohesive forces between water molecules at the surface,thereby reducing the surface tension of the water.
This reduction in surface tension is what allows soap to act as a cleaning agent by enabling water to penetrate into fabrics and emulsify dirt.

(C) The extent of adsorption of a gas on a solid adsorbent like charcoal is directly proportional to its critical temperature $(T_c)$.
This is because gases with higher critical temperatures are more easily liquefiable and have stronger van der Waals forces of attraction,leading to greater adsorption.
Comparing the given critical temperatures:
$A = 304 \ K$
$B = 630 \ K$
$C = 200 \ K$
$D = 540 \ K$
Since gas $C$ has the lowest critical temperature $(200 \ K)$,it will show the least adsorption on a definite amount of charcoal.

(D) Adsorption is a spontaneous process where gas molecules are trapped on a solid surface,leading to a decrease in randomness,so $\Delta S < 0$.
Since it is an exothermic process,heat is released,so $\Delta H < 0$.
For a spontaneous process,the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ must be negative.
Since $\Delta S$ is negative,$-T\Delta S$ becomes positive.
Thus,$(A)$,$(B)$,and $(C)$ are all correct.

(A) Physical adsorption (physisorption) is characterized by the formation of multi-molecular layers on the surface of the adsorbent.
Therefore,the statement that it is usually mono-molecular layered is incorrect.

(C) $1$. Physical adsorption (physisorption) is multilayered due to weak van der Waals forces.
$2$. Chemical adsorption (chemisorption) is irreversible because it involves the formation of strong chemical bonds.
$3$. Physical adsorption is an exothermic process; therefore,it decreases with an increase in temperature. It does not increase first.
$4$. Langmuir adsorption isotherm is primarily used to describe chemisorption,where a monolayer is formed on the surface.

(B) In chromatography,the stationary phase acts as an adsorbent. Common adsorbents used in column chromatography include alumina $(Al_2O_3)$ and silica gel $(SiO_2 \cdot xH_2O)$. Therefore,alumina is the correct choice among the given options.

(C) Graph $I$ shows that the amount adsorbed decreases with an increase in temperature at constant pressure,which is characteristic of physisorption.
Graph $II$ shows that the amount adsorbed increases with an increase in temperature,which is characteristic of chemisorption (as it requires activation energy).
Graph $III$ shows the adsorption isotherm at different temperatures,where adsorption decreases as temperature increases $(200 \ K > 250 \ K)$,which is characteristic of physisorption.
Therefore,$I$ is physisorption,$II$ is chemisorption,and $III$ is physisorption.
Comparing this with the given options,statement $C$ is correct.

(D) The correct order is $Xe > Kr > Ar > Ne$ for polarisation.
As the size of the noble gas atom increases down the group,the valence electrons are held less tightly by the nucleus,making the electron cloud more easily polarisable.
Therefore,the polarisability increases as $He < Ne < Ar < Kr < Xe$.
Option $A$,$B$,and $C$ are incorrect because for noble gases,adsorption tendency,critical temperature,and boiling point all increase with increasing atomic size $(He < Ne < Ar < Kr < Xe)$.

(B) Chemisorption involves the formation of strong chemical bonds between the adsorbate and the adsorbent,making it a highly specific process.
It is generally irreversible in nature because the chemical bonds formed are strong.
It typically results in the formation of a mono-molecular layer on the surface.
Since it involves chemical bond formation,it is usually exothermic,meaning the heat of adsorption is negative (or high in magnitude).
Therefore,the statement that it is reversible is not applicable to chemisorption.

(D) Gas masks are used to protect against toxic gases. Wood charcoal is a form of activated carbon which has a large surface area and a porous structure. Due to this,it acts as an adsorbent and effectively adsorbs toxic gases from the air,making it safe for breathing.

(C) $Van \ der \ Waals$ adsorption (physical adsorption) is an exothermic process. According to $Le \ Chatelier's \ Principle$,low temperature favors exothermic processes. Furthermore,physical adsorption involves the accumulation of gas molecules on the surface,which is favored by high pressure. Therefore,low temperature and high pressure are the favorable conditions for $Van \ der \ Waals$ adsorption.

(D) The Freundlich adsorption isotherm is given by the equation $(x/m) = k \cdot p^{1/n}$,where $n > 1$.
At low pressure,the value of $1/n$ is approximately $1$,so $(x/m) \propto p^1$.
At high pressure,the value of $1/n$ approaches $0$,so $(x/m) \propto p^0$ (independent of pressure).
At intermediate pressure,$(x/m) \propto p^{1/n}$.
Therefore,the relationship holds true across different pressure ranges.

(C) Physical adsorption (physisorption) is caused by weak van der Waals forces between the adsorbate and the adsorbent.
It is not specific in nature because any gas can be adsorbed on any solid to some extent.
It is reversible in nature and involves low enthalpy of adsorption.

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

(C) Physical adsorption,also known as physisorption,occurs when adsorbate molecules are held on the surface of an adsorbent by weak intermolecular forces known as $Van \ der \ Waals$ forces. These forces are non-specific and reversible in nature.