Adsorption and Adsorption isotherm Questions in English

(A) The removal of coloring matter from a sugar solution by charcoal is an example of $Adsorption$.
In this process,the colored molecules accumulate on the surface of the charcoal particles rather than being absorbed into the bulk of the solid.
Therefore,it is a surface phenomenon known as $Adsorption$.

(A) Activated charcoal acts as an adsorbent. When it is added to an acetic acid solution,the acetic acid molecules are adsorbed onto the surface of the charcoal. This process reduces the amount of acetic acid present in the bulk solution,thereby decreasing the concentration of the solution.

(B) The spontaneity of a process is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$.
For a process to be spontaneous,$\Delta G$ must be negative.
During the adsorption of a gas on a solid surface,the gas molecules become restricted,leading to a decrease in randomness,so $\Delta S < 0$ (negative).
Substituting this into the equation: $\Delta G = \Delta H - T(\text{negative}) = \Delta H + T \Delta S$.
For $\Delta G$ to be negative,$\Delta H$ must be sufficiently negative to overcome the positive $T \Delta S$ term.
Therefore,$\Delta H$ must be highly negative.

(A) The decomposition reaction is $PH_3 \xrightarrow{W} P + \frac{3}{2} H_2$.
According to the Langmuir adsorption isotherm,the fraction of surface coverage $\theta$ is given by $\theta = \frac{kP}{1 + kP}$.
At low pressure,$kP \ll 1$,so $\theta \approx kP$.
Since the rate of reaction is proportional to the surface coverage $(\text{Rate} \propto \theta)$,at low pressure,the rate becomes proportional to the partial pressure of $PH_3$ $(\text{Rate} \propto P_{PH_3})$,which corresponds to a first-order reaction.

(D) Adsorption is a spontaneous process that occurs with the release of energy and a decrease in the entropy of the substance.
For a spontaneous process,the Gibbs free energy change,$\Delta G$,must be negative.
The relationship is given by $\Delta G = \Delta H - T \Delta S$.
Since the process is exothermic,the enthalpy change,$\Delta H$,is negative.
As the molecules get adsorbed on the surface,their randomness decreases,meaning the entropy change,$\Delta S$,is also negative.

(A) In Freundlich adsorption isotherm,the equation is given by $\frac{x}{m} = k p^{1/n}$.
Here,$x$ is the mass of the adsorbate,$m$ is the mass of the adsorbent,$p$ is the pressure,and $k$ and $n$ are constants that depend on the nature of the adsorbent and adsorbate at a particular temperature.
The value of $n$ is always greater than $1$,which implies that $n > 1$.
Therefore,the value of the exponent $1/n$ must lie between $0$ and $1$ in all cases.

(D) The relation $x/m = p \times T$ is incorrect for the adsorption process.
Adsorption is typically described by the Freundlich or Langmuir isotherms,where the extent of adsorption $(x/m)$ is a function of pressure $(p)$ at constant temperature $(T)$,or a function of temperature $(T)$ at constant pressure $(p)$.
The relationship $p = f(T)$ at constant $x/m$ represents isosteres.
The expression $x/m = p \times T$ does not represent any standard adsorption isotherm or thermodynamic relation.

(A) The main assumptions of the Langmuir adsorption isotherm are:
$(i)$ The surface of the solid is homogeneous,meaning all adsorption sites are equivalent in their ability to adsorb the particles.
$(ii)$ Adsorption is limited to the formation of a unimolecular layer (monolayer).
$(iii)$ There are no lateral interactions between the adsorbed molecules.
$(iv)$ The heat of adsorption is constant and independent of the surface coverage.

(D) According to the Freundlich adsorption isotherm,the relation is given by $x/m = K p^{1/n}$.
Taking logarithm on both sides,we get $\log(x/m) = \log K + \frac{1}{n} \log p$.
Comparing this with the equation 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 slope of the plot is $1/n$.

(A) According to Langmuir's model of adsorption,the rate of adsorption is proportional to the pressure of the gas $(p)$ and the number of vacant sites on the surface.
The mass of gas striking a given area of surface is directly proportional to the pressure of the gas $(p)$ because the frequency of molecular collisions with the surface increases linearly with pressure.
Therefore,the correct statement is that the mass of gas striking a given area of surface is proportional to the pressure of the gas.

(C) Physisorption is an exothermic process because the formation of weak van der Waal's forces between the adsorbate and adsorbent releases energy.
Therefore,the enthalpy of adsorption $(\Delta H_{adsorption})$ is always negative,typically ranging between $-20 \text{ to } -40 \text{ kJ/mol}$.
Option $C$ states that the enthalpy is positive,which is incorrect.

(D) The Freundlich adsorption isotherm is mathematically represented as $x/m = kP^{1/n}$ where $n > 1$.
At low pressure,$1/n \approx 1$,so $x/m \propto p^1$.
At high pressure,$1/n \approx 0$,so $x/m \propto p^0$.
At intermediate pressure,$x/m \propto p^{1/n}$.
Therefore,all the given relations are correct for different ranges of pressure.

(C) The initial amount of acetic acid in $50 \, mL$ of $0.06 \, N$ solution is calculated as:
$w_1 = \frac{N \times M_{wt} \times V}{1000} = \frac{0.06 \times 60 \times 50}{1000} = 0.18 \, g = 180 \, mg$.
The final amount of acetic acid in the filtrate after adsorption is:
$w_2 = \frac{0.042 \times 60 \times 50}{1000} = 0.126 \, g = 126 \, mg$.
The amount of acetic acid adsorbed by $3 \, g$ of charcoal is:
$w_{ads} = 180 \, mg - 126 \, mg = 54 \, mg$.
The amount of acetic acid adsorbed per gram of charcoal is:
$\frac{54 \, mg}{3 \, g} = 18 \, mg/g$.

(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: $\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log p$.
Comparing this with the linear equation $y = mx + c$,where $y = \log(x/m)$,$x = \log p$,the slope is $1/n$ and the intercept is $\log k$.
Therefore,only $1/n$ appears as the slope.

(C) Physical adsorption (physisorption) involves weak van der Waals forces and does not require significant activation energy.
Conversely,chemisorption involves strong chemical bonds and typically requires high activation energy.
Therefore,the statement that physical adsorption requires high activation energy is false.

(A) $1$. Physical adsorption (physisorption) is exothermic and decreases with an increase in temperature. Chemical adsorption (chemisorption) often increases initially with temperature due to activation energy requirements.
$2$. Both physical and chemical adsorption increase with an increase in pressure as more gas molecules are forced onto the surface.
$3$. Physisorption is reversible and forms multilayers,whereas chemisorption is irreversible and forms a monolayer.
$4$. Therefore,the correct statement is that both increase with an increase in pressure.

(A) Adsorption is multilayer in the case of physical adsorption (physisorption) due to weak van der Waals forces.
In contrast,chemical adsorption (chemisorption) is unilayer because it involves the formation of strong chemical bonds between the adsorbate and adsorbent molecules.

(D) According to the Freundlich adsorption isotherm,$\frac{x}{m} = K P^{1/n}$.
Taking logarithm on both sides,we get $\log \left( \frac{x}{m} \right) = \log K + \frac{1}{n} \log P$.
The plot of $\log \left( \frac{x}{m} \right)$ versus $\log P$ is a straight line with slope $= \frac{1}{n}$ and intercept $= \log K$.
Given that the line is inclined at an angle of $45^{\circ}$,the slope is $\frac{1}{n} = \tan 45^{\circ} = 1$.
Thus,$n = 1$.
Given $P = 0.5 \ atm$ and $K = 10$,substituting these values into the equation:
$\frac{x}{m} = K \times P^{1/n} = 10 \times (0.5)^{1} = 5$.
Therefore,the amount of solute adsorbed per gram of adsorbent is $5 \ g$.

(A) Gas masks contain activated charcoal,which acts as a highly effective adsorbent.
Poisonous gases from the atmosphere get adsorbed onto the surface of the activated charcoal,thereby purifying the air.
This process is based on the principle of adsorption.

(D) The Freundlich adsorption isotherm is given by $\frac{x}{m} = K(P)^{1/n}$.
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 $y = mx + c$,the slope is $\frac{1}{n}$.
Given that the line is inclined at an angle of $45^{\circ}$,the slope is $\tan 45^{\circ} = 1$.
Therefore,$\frac{1}{n} = 1$,which implies $n = 1$.
Substituting the values $K = 10$,$P = 0.5 \ atm$,and $n = 1$ into the isotherm equation:
$\frac{x}{m} = 10 \times (0.5)^{1} = 5$.
Since $m = 1 \ g$,the amount of solute adsorbed $x$ is $5 \ g$.

(C) Chemisorption (chemical adsorption) involves the formation of strong chemical bonds between the adsorbate and the adsorbent.
$1$. It is highly specific in nature.
$2$. It is generally irreversible.
$3$. The enthalpy of adsorption $(\Delta H)$ is high,typically in the range of $80-240 \, kJ/mol$,not $540 \, kJ/mol$.
$4$. Like physisorption,it increases with an increase in the surface area of the adsorbent.
Therefore,the statement that $\Delta H$ is of the order of $540 \, kJ/mol$ is incorrect.

(D) The Freundlich adsorption isotherm equation is $\frac{x}{m} = k C^{1/n}$.
Taking the logarithm on both sides,we get: $\log (\frac{x}{m}) = \log k + \frac{1}{n} \log C$.
This is a linear equation of the form $y = mx + c$,where $y = \log (\frac{x}{m})$,$x = \log C$,slope $m = \frac{1}{n}$,and intercept $c = \log k$.
The given graph is a horizontal line parallel to the $\log C$ axis,which means the slope of the line is zero.
Therefore,$\frac{1}{n} = 0$.

(C) According to the Freundlich adsorption isotherm,$\log (\frac{x}{m}) = \log K + \frac{1}{n} \log P$.
From the given graph,the slope is $\tan(45^{\circ}) = 1$,so $\frac{1}{n} = 1$.
The intercept on the y-axis is $\log K = 0.3$. Since $\log 2 \approx 0.301$,we have $K = 2$.
Substituting these values into the Freundlich equation: $\frac{x}{m} = K \cdot P^{1/n} = 2 \cdot P^1$.
Given $m = 50 \ gm$ and $P = 2 \ atm$,we get $\frac{x}{50} = 2 \cdot 2 = 4$.
Therefore,$x = 50 \cdot 4 = 200 \ gm$.

(B) Adsorption is a spontaneous process,so $\Delta G < 0$.
It is an exothermic process,so the enthalpy change $\Delta H < 0$.
Since gas molecules are adsorbed on the solid surface,their freedom of movement decreases,leading to a decrease in entropy,so $\Delta S_{\text{system}} < 0$.
For a spontaneous process,the total entropy change $\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surrounding}} > 0$.
Since $\Delta S_{\text{system}}$ is negative,$\Delta S_{\text{surrounding}}$ must be positive and greater in magnitude than $|\Delta S_{\text{system}}|$ to make the process spontaneous. Thus,$\Delta S_{\text{surrounding}} > 0$ is the correct parameter.

(A) The froth floatation process is used for the concentration of sulphide ores.
In this process,the 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 molecules (like pine oil or xanthates) adsorb onto the surface of the ore particles,making them hydrophobic and allowing them to attach to air bubbles and rise to the surface as froth.

(D) According to the Freundlich adsorption isotherm equation:
$\log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log P$
Comparing this with the equation of a straight line $y = mx + c$,the slope is $\frac{1}{n}$.
Given that the line is inclined at an angle of $45^o$,the slope is:
$\frac{1}{n} = \tan 45^o = 1$
Given $k = 10$ and $P = 0.5 \ atm$,we substitute these values into the adsorption isotherm equation:
$\frac{x}{m} = k \times P^{1/n}$
$\frac{x}{m} = 10 \times (0.5)^1$
$\frac{x}{m} = 5$
Since we are calculating the amount of solute adsorbed per gram of adsorbent $(m = 1 \ g)$,the amount of solute $x$ is $5 \ g$.

(B) The Freundlich adsorption isotherm equation is given by $\log (x/m) = \log k + (1/n) \log P$.
Comparing this with the equation of a straight line $y = mx + c$,we have the slope $1/n = \tan 45^o = 1$ and the intercept $\log k = 0.3010$.
Calculating $k$,we get $k = \text{antilog}(0.3010) = 2$.
Substituting the values into the adsorption equation: $(x/m) = k \cdot P^{1/n} = 2 \times (0.3)^1 = 0.6$.

(B) is the incorrect statement.
Chemical adsorption (chemisorption) typically increases with an increase in temperature initially because it requires activation energy,and it increases with an increase in pressure.
Physical adsorption (physisorption) is reversible,whereas chemical adsorption is generally irreversible.
Activation energy for chemical adsorption is significantly higher than that of physical adsorption.

(D) Adsorption is a spontaneous process,so $\Delta G < 0$.
It is an exothermic process,so $\Delta H < 0$.
Since the gas molecules are adsorbed on the solid surface,their freedom of movement decreases,leading to a decrease in entropy,so $\Delta S < 0$.
Therefore,none of the given thermodynamic quantities $(\Delta H, \Delta S, \Delta G)$ are positive for the adsorption process.

(C) The Freundlich adsorption isotherm is given by the equation $\frac{x}{m} = kP^{1/n}$.
Taking logarithms on both sides,we get $\log(\frac{x}{m}) = \log k + \frac{1}{n} \log P$.
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 $\log(x/m)$ vs $\log P$ gives a straight line.

(B) During adsorption,the negative value of $\Delta H$ goes on decreasing as the process proceeds.
This is because the number of available active adsorption sites on the surface decreases,leading to a reduction in the heat released (exothermicity) as the surface becomes covered.

(A) In column chromatography,the separation of components depends on their differential adsorption on the stationary phase.
The components that are most readily adsorbed are retained near the top of the column.
The components that are less strongly adsorbed move further down the column to various distances.

(B) 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 $(T_c)$ have stronger intermolecular forces of attraction,making them easier to liquefy.
Easily liquefiable gases are adsorbed to a greater extent on the surface of a solid.
Therefore,a gas with a higher critical temperature $(T_c)$ will be adsorbed to a greater extent.

(D) Physical adsorption decreases with an increase in temperature because an increase in temperature leads to an increase in kinetic energy of molecules and hence,their desorption.
Since physical adsorption is an exothermic process,it occurs more readily at lower temperatures and decreases with an increase in temperature (Le-Chatelier's Principle).
If the adsorption is a spontaneous phenomenon,it should be exothermic,at least for adsorption from the gaseous phase.
$A$ molecule in the gas phase will have an entropy higher than in the adsorbed state,then the $\Delta S$ should be negative. The only way to have a $\Delta G$ negative (spontaneous) is to have a $\Delta H$ negative (exothermic process).
Therefore,the process will be spontaneous $(\Delta G < 0)$ if the temperature is low.

(C) Physisorption is caused by weak van der Waals forces,so it requires very low activation energy. In contrast,chemisorption involves the formation of chemical bonds,which requires a significant amount of activation energy. Therefore,the statement that activation energy is high for physisorption and low for chemisorption is incorrect.

(B) The froth flotation process is based on the difference in the wetting properties of the ore and the gangue particles with water and oil. The sulphide ore particles are preferentially wetted by oil (hydrophobic),while the gangue particles are wetted by water (hydrophilic). This selective wetting is a surface phenomenon known as adsorption.

(D) Physical adsorption is characterized by weak $Vander \text{ } Waal's$ forces between the adsorbate and the adsorbent.
Due to these weak forces,multiple layers of adsorbate molecules can accumulate on the surface of the adsorbent,leading to the formation of multimolecular layers.
Since $Vander \text{ } Waal's$ adsorption is another name for physical adsorption,and the $Freundlich$ adsorption isotherm describes the behavior of physical adsorption,all these terms refer to the same phenomenon.
Therefore,the correct answer is $All \text{ } of \text{ } these$.

(D) The amount of gas adsorbed by a solid depends on the nature of the gas.
Easily liquefiable gases,which have higher critical temperatures,are adsorbed more readily because they possess stronger van der Waals forces.
The critical temperatures of these gases are in the order: $NH_3 (405.5 \ K) > CO_2 (304.1 \ K) > CH_4 (190.6 \ K) > H_2 (33.2 \ K)$.
Therefore,the volume of gases adsorbed by $1 \ g$ of charcoal at $288 \ K$ follows the same order: $NH_3 > CO_2 > CH_4 > H_2$.

(B) Initial moles of acetic acid: $(n_{CH_3COOH})_{\text{initial}} = 500 \ mL \times 0.5 \ M = 250 \ mmol = 0.25 \ mol$.
Final moles of acetic acid: $(n_{CH_3COOH})_{\text{final}} = 500 \ mL \times 0.3 \ M = 150 \ mmol = 0.15 \ mol$.
Moles adsorbed on charcoal: $\Delta n = 0.25 \ mol - 0.15 \ mol = 0.1 \ mol$.
Number of molecules adsorbed: $N = \Delta n \times N_A = 0.1 \times 6 \times 10^{23} = 6 \times 10^{22} \text{ molecules}$.

(D) Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent. This process requires an activation energy to initiate the bond formation.
Initially,as the temperature $T$ increases,more gas molecules gain sufficient energy to overcome the activation energy barrier,leading to an increase in the extent of adsorption $(x/m)$.
However,at higher temperatures,the thermal energy becomes sufficient to break the chemical bonds formed,leading to desorption.
Therefore,the adsorption isobar for chemisorption shows an initial increase followed by a decrease in $x/m$ as temperature $T$ increases.
This corresponds to the plot shown in option $D$.

(B) An adsorption isobar is a plot of the extent of adsorption $(x/m)$ versus temperature $(T)$ at a constant pressure.
For chemisorption,the process involves the formation of chemical bonds,which requires an initial activation energy.
Therefore,as the temperature increases,the rate of adsorption initially increases due to the overcoming of the activation energy barrier.
However,at higher temperatures,the desorption process becomes dominant due to the breaking of chemical bonds,causing the extent of adsorption to decrease.
Thus,the plot of $x/m$ versus $T$ for chemisorption shows an initial increase followed by a decrease,which corresponds to the graph in option $B$.

(C) Graph $(I)$ shows the variation of the amount of gas adsorbed with temperature for physical adsorption (physiosorption). In physiosorption,adsorption decreases with an increase in temperature because it is an exothermic process.
Graph $(II)$ shows the variation for chemical adsorption (chemisorption). Chemisorption initially increases with temperature due to the activation energy requirement and then decreases at higher temperatures.
Graph $(III)$ represents the Freundlich adsorption isotherm,where the amount of gas adsorbed increases with pressure. For physical adsorption,adsorption decreases as temperature increases. Since the curve for $T_1$ is below the curve for $T_2$ at the same pressure,it implies $T_1 > T_2$.

(B) The extent of adsorption of a gas on a solid surface depends on the ease of liquefaction of the gas.
Easily liquefiable gases are adsorbed more readily.
The ease of liquefaction is directly related to the critical temperature $(T_c)$ of the gas.
Higher the critical temperature of a gas,greater is the magnitude of intermolecular forces of attraction,and thus,it is more easily adsorbed on a solid surface.
Therefore,the gas with the highest critical temperature will be adsorbed to a greater extent.

(B) According to the Freundlich adsorption isotherm,the extent of adsorption $(x/m)$ is given by $x/m = kP^{1/n}$.
At high pressure,the value of $1/n$ approaches $0$,making $x/m$ constant.
Therefore,the rate of physical adsorption becomes independent of pressure at high pressure.

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

(D) The $As_2S_3$ sol is a negatively charged sol.
This negative charge arises due to the preferential adsorption of common ions,specifically $S^{2-}$ ions,on the surface of the $As_2S_3$ particles from the dispersion medium.

(B) For any spontaneous adsorption process,the Gibbs free energy change,$ \Delta G $,must be negative $( \Delta G < 0 )$.
Adsorption is an exothermic process,meaning heat is released,so the enthalpy change,$ \Delta H $,is negative $( \Delta H < 0 )$.
Since gas molecules are restricted to the surface upon adsorption,the randomness of the system decreases,leading to a negative entropy change,$ \Delta S < 0 $.

(B) Absorption is a bulk phenomenon where a substance is uniformly distributed throughout the body of a solid or liquid.
Adsorption is a surface phenomenon where particles accumulate only at the surface of a solid or liquid rather than in the bulk.
Therefore,absorption is a bulk phenomenon and adsorption is a surface phenomenon.

(A) Adsorption is multilayer in the case of physical adsorption (physisorption) due to weak van der Waals forces.
In contrast,chemisorption is unilayer because it involves the formation of strong chemical bonds between the adsorbate and adsorbent molecules,which limits the adsorption to a single layer.

(B) Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent.
$1$. It is highly specific in nature.
$2$. It is generally irreversible.
$3$. It is a unimolecular process,meaning it forms a monolayer,not a multilayer.
$4$. It is dependent on temperature; typically,it increases with an increase in temperature initially due to activation energy requirements.
Therefore,the statement 'independent of temperature' is not applicable to chemisorption.