Adsorption and Adsorption isotherm Questions in English

(D) $A.$ Incorrect. Water vapours are absorbed by anhydrous calcium chloride,not adsorbed.
$B.$ Correct. Adsorption is a spontaneous process,and surface energy decreases as the surface area available for adsorption decreases.
$C.$ Incorrect. As adsorption proceeds,the surface becomes covered,and the heat released per unit of adsorption decreases,so $\Delta H$ becomes less and less negative.
$D.$ Correct. Adsorption restricts the movement of gas molecules,leading to a decrease in the entropy of the system.
Therefore,there are $2$ incorrect statements ($A$ and $C$).

(A) The retardation factor $(R_f)$ is defined as the ratio of the distance traveled by the substance to the distance traveled by the solvent front from the baseline.
From the provided chromatogram:
Distance moved by compound '$A$' from the baseline = $3.5 \ cm$.
Total length of the plate = $6 \ cm$.
Distance of solvent front from the top = $0.5 \ cm$.
Distance of baseline from the bottom = $0.5 \ cm$.
Therefore,the distance moved by the solvent front from the baseline = $6 \ cm - 0.5 \ cm - 0.5 \ cm = 5.0 \ cm$.
$R_f = \frac{\text{Distance moved by the substance}}{\text{Distance moved by the solvent}} = \frac{3.5 \ cm}{5.0 \ cm} = 0.7$.
Thus,$R_f = 7 \times 10^{-1}$.
Since $7$ is not among the options,let us re-evaluate the provided image data. If the distance moved by compound '$A$' is $3.0 \ cm$ and solvent front is $5.0 \ cm$,then $R_f = 0.6 = 6 \times 10^{-1}$. Given the options,$A$ is the correct choice.

(C) Silica gel is a well-known adsorbent used to remove moisture from the surrounding environment.
In expensive scientific instruments,it is placed in watch-glasses or semipermeable bags to adsorb water vapor from the air.
By adsorbing moisture,it prevents the corrosion of metallic parts and protects sensitive electronic components from malfunctioning due to humidity.
Therefore,both the Assertion $(A)$ and the Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(D) The Freundlich adsorption isotherm is given by the equation: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$.
Comparing this with the equation of a straight line $y = mx + c$,we have the slope $\frac{1}{n} = \tan 45^{\circ} = 1$ and the intercept $c = \log k = 0.6020$.
Since $\log 2 = 0.3010$,then $2 \times \log 2 = 0.6020$,which implies $\log 2^2 = \log 4 = 0.6020$. Thus,$k = 4$.
Substituting these values into the isotherm equation: $\frac{x}{m} = k \cdot P^{1/n} = 4 \times (0.4)^1 = 1.6$.
Expressing $1.6$ in the form $......... \times 10^{-1}$,we get $1.6 = 16 \times 10^{-1}$.
Therefore,the nearest integer is $16$.

(A) . Physisorption involves weak van der Waals forces with low enthalpy of adsorption $(20-40 \, kJ \, mol^{-1})$.
$B$. Chemisorption involves strong chemical bonds and is typically unimolecular (single layer).
$C$. The reaction $N_{2(g)} + 3H_{2(g)} \xrightarrow{Fe(s)} 2NH_{3(g)}$ is an example of heterogeneous catalysis where the catalyst $(Fe)$ is in a solid phase and reactants are in a gaseous phase.
$D$. Chromatography is a common analytical application based on the principle of differential adsorption.
Therefore,the correct matching is: $A-II, B-I, C-IV, D-III$.

(A) Assertion $A$ is true: The extent of adsorption of gases on activated charcoal increases with an increase in the magnitude of van der Waals forces of attraction. Since $Kr$ has the largest atomic size among the given noble gases,it has the strongest van der Waals forces,leading to maximum adsorption.
Reason $R$ is false: The critical temperature $T_c$ is the primary factor determining the ease of liquefaction and adsorption. While $V_c$ and $P_c$ values vary,the statement regarding $Z_c$ is incorrect because for all real gases following the van der Waals equation,the compressibility factor at the critical point $Z_c = \frac{P_c V_c}{R T_c}$ is a constant value of $\frac{3}{8} = 0.375$.

(B) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K \cdot P^{1/n}$.
Taking logarithm on both sides,we get: $\log(\frac{x}{m}) = \log K + \frac{1}{n} \log P$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(\frac{x}{m})$,$x = \log P$,the slope $m = \frac{1}{n}$,and the intercept $c = \log K$.
Given the equation $y = 0.3x + 0.7033$,we have:
Slope $(m)$ = $\frac{1}{n} = 0.3$.
Intercept $(c)$ = $\log K = 0.7033$.
Thus,the values of $\frac{1}{n}$ and $\log K$ are $0.3$ and $0.7033$ respectively.

(B) Physical adsorption is an exothermic process. According to Le Chatelier's principle,an increase in temperature decreases the extent of adsorption $(\frac{x}{m})$ at a constant pressure.
From the given graph,at any constant pressure $p$,the extent of adsorption follows the order: $(x/m)_{T_1} > (x/m)_{T_2} > (x/m)_{T_3}$.
Since adsorption decreases with an increase in temperature,the corresponding temperatures must follow the inverse order: $T_1 < T_2 < T_3$,or $T_3 > T_2 > T_1$.

(D) The rate of chemisorption follows the Arrhenius equation: $k = Ae^{-\frac{E_a}{RT}}$.
Let $k_1$ be the rate on platinum and $k_2$ be the rate on nickel.
$\frac{k_2}{k_1} = \frac{Ae^{-\frac{(E_a)_2}{RT}}}{Ae^{-\frac{(E_a)_1}{RT}}} = e^{\frac{(E_a)_1 - (E_a)_2}{RT}}$.
Taking the logarithm to the base $10$:
$\log_{10} \left( \frac{k_2}{k_1} \right) = \frac{(E_a)_1 - (E_a)_2}{2.303 RT}$.
Given $(E_a)_1 = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$,$(E_a)_2 = 41.4 \ kJ \ mol^{-1} = 41400 \ J \ mol^{-1}$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,and $\ln 10 \approx 2.3$.
$\log_{10} \left( \frac{k_2}{k_1} \right) = \frac{30000 - 41400}{2.3 \times 8.3 \times 300} = \frac{-11400}{5727} \approx -1.99$.
Since the question asks for the logarithm of the ratio of rates (usually implying the magnitude or the ratio of $k_1/k_2$ to get a positive value),the nearest integer is $2$.

(B) The Freundlich adsorption isotherm is given by the equation $\frac{x}{m} = k p^{1/n}$.
Graph $(A)$ represents the plot of $\frac{x}{m}$ versus $p$,which is a characteristic curve of the Freundlich isotherm.
Taking the logarithm on both sides,we get $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$. This is a linear equation of the form $y = mx + c$,where the slope is $\frac{1}{n}$ and the intercept is $\log k$. Graph $(B)$ represents this linear relationship.
Graph $(D)$ represents the plot of $\frac{x}{m}$ versus $p^{1/n}$,which is also a linear relationship derived from the equation $\frac{x}{m} = k p^{1/n}$ (where $y = kx$).
Graph $(C)$ is not a standard representation of the Freundlich adsorption isotherm as it plots $\frac{x}{m}$ against concentration $c$ with an intercept,which does not match the standard form.
Therefore,graphs $(A), (B),$ and $(D)$ represent the Freundlich adsorption isotherms.

(D) The extent of adsorption of a gas on a solid adsorbent like charcoal is directly proportional to its critical temperature $(T_c)$.
Higher critical temperature implies that the gas is more easily liquefiable and has stronger van der Waals forces of attraction.
Given critical temperatures are:
$A = 5.3 \ K$
$B = 33.2 \ K$
$C = 126.0 \ K$
$D = 154.3 \ K$
Therefore,the order of adsorption is $D > C > B > A$.

(B) The fundamental principle used in $Chromatography$ is $Adsorption$. In this technique,different components of a mixture are adsorbed to different extents on a stationary phase.

(C) In adsorption chromatography,the stationary phase consists of an adsorbent material.
Silica gel $(SiO_2 \cdot xH_2O)$ is the most commonly used polar adsorbent.
Alumina $(Al_2O_3)$ is also a widely used polar adsorbent in adsorption chromatography.
Both silica gel and alumina are standard materials used for the separation of organic compounds based on their polarity.
Therefore,both $A$ and $B$ are correct.

(B) Adsorption is always exothermic because it involves the release of energy as surface forces are satisfied.
$(B)$ Physisorption is reversible and weak,but at higher temperatures,it can provide enough activation energy for chemical bond formation,transforming into chemisorption.
$(C)$ This statement is incorrect. Physisorption decreases with an increase in temperature. Chemisorption initially increases with temperature (due to activation energy) and then decreases at very high temperatures.
$(D)$ Chemisorption involves chemical bond formation,which releases more energy than the weak van der Waals forces in physisorption. It is slow because it requires a significant activation energy.

(B) In physisorption,adsorption decreases with an increase in temperature at constant pressure. Thus,graph $I$ represents physisorption.
In chemisorption,adsorption initially increases with an increase in temperature due to the requirement of activation energy. Thus,graph $II$ represents chemisorption.
Graph $III$ shows that at a constant pressure,the extent of adsorption decreases as the temperature increases from $200 \ K$ to $250 \ K$,which is characteristic of physisorption.
Graph $IV$ shows a high enthalpy of adsorption $(\Delta H_{ads} = 150 \ kJ \ mol^{-1})$ and an activation energy $(E_{ads})$,which are characteristic of chemisorption.
Therefore,the correct statements are:
$I$ is physisorption and $II$ is chemisorption (Statement $A$ is correct).
$IV$ is chemisorption and $II$ is chemisorption (Statement $C$ is correct).
Thus,the correct choice is $(B)$.

(B) The adsorption of methylene blue on activated charcoal is a physical adsorption process (physisorption).
Physical adsorption is an exothermic process,which means it is accompanied by a decrease in enthalpy $(\Delta H < 0)$.

(A) * Adsorption of $O_{2}$ on a metal surface involves chemical bond formation,so it is chemisorption,not physisorption.
* Adsorption is an exothermic process,so heat is released.
* During electron transfer from the metal to $O_{2}$,the electron occupies the antibonding $\pi_{2p}^{*}$ orbital of $O_{2}$.
* As the occupancy of the antibonding orbital increases,the bond order of $O_{2}$ decreases,which leads to an increase in the bond length of $O_{2}$.

(D) . Chemisorption involves the formation of chemical bonds,resulting in a unimolecular layer. This is correct.
$B$. The enthalpy of physisorption is low,typically in the range of $20$ to $40 \ kJ \ mol^{-1}$,not $100$ to $140 \ kJ \ mol^{-1}$. This is incorrect.
$C$. Chemisorption is generally an exothermic process because chemical bond formation releases energy. This is incorrect.
$D$. Physisorption is an exothermic process $( \Delta H < 0 )$. According to Le Chatelier's principle,lowering the temperature favors the forward reaction,thus favoring physisorption. This is correct.
Therefore,the correct options are $A$ and $D$.

(C) Initial millimoles of acetic acid $= 100 \ mL \times 0.5 \ M = 50 \ mmol$.
Millimoles of unadsorbed acetic acid neutralized by $NaOH = 40 \ mL \times 1 \ M = 40 \ mmol$.
Millimoles of acetic acid adsorbed on $1 \ g$ of charcoal $= 50 - 40 = 10 \ mmol = 10 \times 10^{-3} \ mol$.
Number of molecules of acetic acid adsorbed $= 10 \times 10^{-3} \ mol \times 6.0 \times 10^{23} \ mol^{-1} = 6.0 \times 10^{21} \ \text{molecules}$.
Surface area occupied by one molecule $= \frac{\text{Total surface area}}{\text{Number of molecules}} = \frac{1.5 \times 10^2 \ m^2}{6.0 \times 10^{21}} = 0.25 \times 10^{-19} \ m^2 = 2500 \times 10^{-23} \ m^2$.
Comparing this with $P \times 10^{-23} \ m^2$,we get $P = 2500$.

(C) Statement $I$ is true: Silica gel and alumina are commonly used as stationary phases (adsorbents) in adsorption chromatography because of their high surface area and polar nature.
Statement $II$ is true: Paper chromatography is a form of partition chromatography where the stationary phase is water held in the pores of the filter paper,and the separation is based on the differential distribution of components between the stationary liquid phase and the mobile liquid phase.
Therefore,both statements are correct.

(D) When tungsten $(W)$ metal is exposed to oxygen gas $(O_2)$,it undergoes surface adsorption followed by oxidation to form tungsten trioxide $(WO_3)$.

(D) In gas chromatography,a carrier gas is an inert gas used to transport the sample through the column.
Commonly used carrier gases include helium $(He)$,nitrogen $(N_2)$,hydrogen $(H_2)$,and argon $(Ar)$.
Helium and nitrogen are the most frequently used gases,and helium is particularly preferred when using a capillary column due to its efficiency.

(B) The extent of adsorption of a gas on a solid depends on the ease of liquefaction of the gas,which is directly related to its critical temperature $(T_c)$.
Easily liquefiable gases have higher critical temperatures and are adsorbed to a greater extent.
The critical temperatures $(T_c)$ of the given gases are:
$Cl_2$: $417 \ K$
$O_2$: $154 \ K$
$N_2$: $126 \ K$
$H_2$: $33 \ K$
Since $Cl_2$ has the highest critical temperature,it is adsorbed to the greatest extent.

(A) Sorption is a phenomenon where both adsorption and absorption occur simultaneously.
$1$. When charcoal is added to a methylene blue solution,the dye molecules are adsorbed on the surface of the charcoal,while the solvent molecules are absorbed into the bulk of the charcoal. This is a classic example of sorption.
$2$. Dipping chalk in ink involves both adsorption on the surface and absorption into the porous structure of the chalk,which is also an example of sorption.
$3$. Hydrogen gas on platinum and oxygen on nickel are primarily examples of adsorption.
Since both $A$ and $B$ represent sorption,in the context of standard chemistry textbooks,the addition of charcoal to methylene blue is the most frequently cited example.

(B) 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 stronger van der Waals forces of attraction,leading to greater adsorption.
The critical temperatures $(T_c)$ of the given gases are:
$SO_2$ $(430 \ K)$ > $O_2$ $(154 \ K)$ > $N_2$ $(126 \ K)$ > $H_2$ $(33 \ K)$.
Since $SO_2$ has the highest critical temperature,it is the most easily liquefiable and thus adsorbed to the greatest extent.

(A) The graph shows the variation of the extent of adsorption $(x/m)$ with pressure $(p)$ at different temperatures.
Physical adsorption 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 highest amount of gas is adsorbed at the lowest temperature.
From the given graph,the lowest temperature is $195 \ K$.

(D) The extent of adsorption of a gas on a solid surface depends on the ease of liquefaction of the gas. Gases that are easily liquefiable have higher critical temperatures $(T_c)$ and are adsorbed more strongly due to stronger van der Waals forces.
Among the given gases,$H_2$ has the lowest critical temperature $(T_c \approx 33 \ K)$,whereas $Cl_2$,$NH_3$,and $SO_2$ have significantly higher critical temperatures.
Since $H_2$ is the most difficult to liquefy,it is the least adsorbed on a solid surface.

(C) Chemisorption (chemical adsorption) involves the formation of strong chemical bonds between the adsorbate and the adsorbent.
$1$. It is highly specific in nature.
$2$. The enthalpy of adsorption is high,typically in the range of $40-200 \ kJ/mol$.
$3$. It is an irreversible process.
$4$. Chemisorption results in the formation of a unimolecular layer,not a multimolecular layer.
Therefore,the formation of a multimolecular layer is a characteristic of physisorption,not chemisorption.

(D) Absorption is a bulk phenomenon where the substance is uniformly distributed throughout the body of the material. Unlike adsorption,it is not a surface phenomenon and therefore does not depend on the surface area of the absorbent.

(D) Physisorption (physical adsorption) is characterized by weak van der Waals forces between the adsorbate and the adsorbent.
It is non-specific in nature.
Chemisorption involves strong chemical bonds and is highly specific.
$O_2$ on tungsten,$H_2$ on nickel,and $N_2$ on iron involve chemical bond formation (chemisorption).
Adsorption of gases on charcoal is a classic example of physisorption due to weak van der Waals forces.

(C) Adsorption is an exothermic process,where $\Delta H < 0$.
According to Le Chatelier's principle,an increase in temperature favors the reverse process,which is desorption.
Therefore,the extent of adsorption is inversely proportional to the temperature of the process.

(B) Adsorption is a surface phenomenon that occurs due to unbalanced forces on the surface of a solid or liquid.
During the process of adsorption,the residual forces on the surface decrease,which leads to a decrease in the surface energy of the adsorbent.
Therefore,the statement that surface energy increases is $FALSE$.
Adsorption is generally an exothermic process and can be caused by van der Waals forces (physisorption) or chemical bonds (chemisorption).

(A) The extent of adsorption of a gas on a solid surface depends on the ease of liquefaction of the gas.
Gases with higher critical temperatures are more easily liquefied and thus show greater adsorption.
Among the given options,$Cl_2$ has the highest critical temperature $(417 \ K)$,making it the most readily adsorbed gas compared to $N_2$,$O_2$,and $H_2$.

(C) Sorption is a phenomenon where both adsorption and absorption occur simultaneously.
When chalk is dipped in ink,the ink particles (solute) get adsorbed on the surface of the chalk,while the solvent (water) is absorbed into the bulk of the chalk.
Therefore,this process represents both adsorption and absorption,which is termed as sorption.

(B) According to the Freundlich adsorption isotherm,the extent of adsorption $(x/m)$ increases with an increase in pressure $(P)$ at low pressures.
As the pressure increases further,the surface of the adsorbent becomes saturated,and the rate of adsorption decreases until it reaches a point where it becomes independent of pressure (saturation pressure).
Therefore,the extent of adsorption increases at the start and then remains constant at higher pressures.

(A) Adsorption is a surface phenomenon where the concentration of molecules increases at the surface rather than in the bulk of a solid or liquid.
During adsorption,there is a decrease in the residual forces of the surface,which leads to a decrease in surface energy.
This decrease in surface energy is released as heat,making adsorption an $ \text{exothermic} $ process.

(C) Physisorption:
$-$ It arises because of weak van der Waals' forces.
$-$ It is not specific in nature.
$-$ It is reversible in nature.
$-$ Low temperature is favourable for adsorption. It decreases with an increase in temperature.

(D) Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent.
- The heat of adsorption is high,typically in the range of $80-240 \ kJ \ mol^{-1}$.
- It is unimolecular in nature,as chemical bonds are specific.
- It involves chemical forces,not $Van \ der \ Waals$ forces.
- It is favored at high temperatures because it requires activation energy to form chemical bonds.

(A) Adsorption is an exothermic process,meaning it is accompanied by the evolution of heat.
Adsorption is a surface phenomenon,not a bulk phenomenon.
Adsorption depends on temperature and the surface area of the adsorbent.
It increases with an increase in the surface area of the adsorbent.

(A) The Freundlich adsorption isotherm is an empirical relationship between the amount of gas adsorbed per unit mass of adsorbent and the pressure of the gas at a constant temperature.
It is expressed by the equation: $x/m = k p^{1/n}$,where $x$ is the mass of the gas adsorbed,$m$ is the mass of the adsorbent,$p$ is the equilibrium pressure,and $k$ and $n$ are constants that depend on the nature of the adsorbent and the gas at a particular temperature.
For the adsorption process,the value of $n$ is typically greater than $1$,which implies that $1/n$ lies between $0$ and $1$.

(A) According to the Freundlich adsorption isotherm equation: $\log \frac{x}{m} = \log K + \frac{1}{n} \log C$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log C$,and $c = \log K$.
The slope of the straight line is $\frac{1}{n}$.

(B) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k P^{\frac{1}{n}}$.
Taking the logarithm on both sides,we get: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log P$,the slope $m = \frac{1}{n}$,and the intercept $c = \log k$.
Therefore,the intercept on the $y$-axis is $\log k$.

(D) The Freundlich adsorption isotherm is an empirical relationship between the amount of gas adsorbed per unit mass of solid adsorbent $(x/m)$ and the equilibrium pressure $(P)$ of the gas.
The mathematical expression is given by: $x/m = k P^{1/n}$,where $k$ and $n$ are constants that depend on the nature of the adsorbent and the gas at a particular temperature $(n > 1)$.

(A) The Freundlich adsorption isotherm is represented by the equation:
$x/m = k p^{1/n}$
Where:
$x$ is the mass of the adsorbate adsorbed on mass $m$ of the adsorbent.
$p$ is the equilibrium pressure.
$k$ and $n$ are constants that depend on the nature of the adsorbent and the adsorbate at a particular temperature.
Note: The Freundlich isotherm is an empirical relation and is not applicable at very high pressure.

(C) When $AgNO_3$ is added to an excess of $NaI$,the $AgI$ precipitate is formed.
According to the preferential adsorption theory,the common ion present in excess in the dispersion medium is adsorbed on the surface of the colloidal particles.
Here,$I^-$ ions are in excess,so they get adsorbed on the $AgI$ surface,forming a negatively charged sol: $[AgI]I^- / Na^+$.

(D) The Freundlich adsorption isotherm is given by the equation:
$\frac{x}{m} = k p^{\frac{1}{n}}$
Taking the logarithm on both sides:
$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log p$,$m$ (slope) $= \frac{1}{n}$,and $c$ (intercept) $= \log k$.
Therefore,the slope is $\frac{1}{n}$ and the intercept is $\log k$.

(C) In the process of decolorization of raw sugar,$\text{Animal charcoal}$ acts as the $\text{adsorbent}$ because it provides the surface for the impurities to stick to.
The $\text{colouring substances}$ (impurities) present in the sugar solution are the $\text{adsorbate}$ as they get accumulated on the surface of the charcoal.
This phenomenon is known as $\text{adsorption}$.
Therefore,the correct set is: $\text{Adsorbent} = \text{Animal charcoal}$,$\text{Adsorbate} = \text{Colouring substance}$,$\text{Process} = \text{Adsorption}$.

(A) Adsorption is a spontaneous process,so $\Delta G < 0$.
It is an exothermic process,which means heat is released,so $\Delta H < 0$.
As gas molecules get adsorbed on the solid surface,their freedom of movement decreases,leading to a decrease in entropy,so $\Delta S < 0$.
According to the Gibbs-Helmholtz equation,$\Delta G = \Delta H - T \Delta S$. At low temperatures,the negative $\Delta H$ term dominates,ensuring $\Delta G < 0$.

(A) Physical adsorption is an exothermic process. According to Le Chatelier's principle,an increase in temperature decreases the extent of adsorption. Therefore,high temperature is an unfavorable condition for physical adsorption.