Adsorption and Adsorption isotherm Questions in English

(C) Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent,which requires an activation energy. Therefore,initially,the rate of adsorption increases with an increase in temperature. The provided graph shows the variation of the extent of adsorption $(x/m)$ with temperature $(T)$ at a constant pressure $(p)$,where the adsorption first increases and then decreases,which is characteristic of chemisorption.

(C) Adsorption is a spontaneous process,so the change in Gibbs free energy must be negative,i.e.,$ \Delta G < 0 $.
Adsorption is an exothermic process,which means heat is released,so the enthalpy change is negative,i.e.,$ \Delta H < 0 $.
As gas molecules are adsorbed on the surface of a solid,their freedom of movement decreases,leading to a decrease in entropy,i.e.,$ \Delta S < 0 $.
According to the Gibbs-Helmholtz equation,$ \Delta G = \Delta H - T \Delta S $,for the process to be spontaneous at a given temperature,$ \Delta G $ must be negative.

(A) Adsorption is a spontaneous process where gas molecules are trapped on a solid surface,leading to a decrease in the randomness of the system. Thus,$\Delta S < 0$.
Since the process is spontaneous,the Gibbs free energy change must be negative $(\Delta G < 0)$.
From the equation $\Delta G = \Delta H - T\Delta S$,for $\Delta G$ to be negative when $\Delta S$ is negative,$\Delta H$ must be negative.
Therefore,the process is exothermic $(\Delta H < 0)$.

$(A)$ van der Waals' forces are responsible for physisorption, not chemisorption.
Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent.
Therefore, Assertion $(A)$ is false.
Chemisorption is an activated process and generally requires high activation energy.
Thus, high temperature is favourable for chemisorption as it helps in overcoming the activation energy barrier.
Therefore, Reason $(R)$ is true.
Hence, the correct option is $(A)$ is false, but $(R)$ is true.

(C) Physical adsorption (physisorption) is characterized by weak van der Waals forces between the adsorbate and the adsorbent.
Because these forces are weak,the activation energy required for physical adsorption is very low,not high.
Therefore,the statement that 'Activation energy of physical adsorption is very high' is incorrect.
In contrast,chemical adsorption (chemisorption) typically requires a high activation energy due to the formation of chemical bonds.

(B) The Freundlich adsorption isotherm is given by the equation $\frac{x}{m} = kp^{1/n}$.
Physical adsorption is an exothermic process,which means that the extent of adsorption decreases with an increase in temperature.
Therefore,the statement that the extent of adsorption of gas is more at high temperature than at low temperature is incorrect.
Option $B$ is the correct answer.

(C) The Freundlich adsorption isotherm equation is given by: $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log p$,slope $= \frac{1}{n}$,and intercept $= \log k$.
Given: Slope $(\frac{1}{n}) = 0.5$ and Intercept $(\log k) = 1.0$.
Pressure $(p) = 100 \ atm$.
Substituting the values into the equation: $\log \frac{x}{m} = 1.0 + 0.5 \times \log(100)$.
Since $\log(100) = 2$,we get: $\log \frac{x}{m} = 1.0 + 0.5 \times 2 = 1.0 + 1.0 = 2.0$.
Therefore,$\frac{x}{m} = 10^2 = 100$.

(B) Adsorption is an exothermic process. According to Le Chatelier's principle,for physical adsorption,the extent of adsorption $(x / m)$ decreases with an increase in temperature at a constant pressure.
From the given graph,at a constant pressure,the extent of adsorption follows the order: $T_2 > T_3 > T_4$.
Since the extent of adsorption is inversely proportional to temperature for physisorption,the correct order of temperatures is $T_4 > T_3 > T_2$.

(D) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = kp^{1/n}$
For the first condition:
$\frac{10}{500} = k(10)^{1/n} \quad ...(i)$
For the second condition:
$\frac{14}{500} = k(20)^{1/n} \quad ...(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{14/500}{10/500} = \frac{k(20)^{1/n}}{k(10)^{1/n}}$
$\frac{14}{10} = (\frac{20}{10})^{1/n}$
$1.4 = 2^{1/n}$
Taking logarithm on both sides:
$\log(1.4) = \frac{1}{n} \log(2)$
$0.1461 = \frac{1}{n} \times 0.3010$
$\frac{1}{n} = \frac{0.1461}{0.3010} \approx 0.485$
Rounding to the nearest option,the slope $\frac{1}{n}$ is approximately $0.5$.

(C) Chemisorption requires high activation energy,so it is referred to as activated adsorption.
In chemisorption,the extent of adsorption $(x/m)$ first increases and then decreases with an increase in temperature.
When the adsorption isobar is plotted,the graph initially increases because the supplied heat helps molecules overcome the activation energy barrier required for chemisorption.
However,at higher temperatures,the extent of adsorption decreases due to the exothermic nature of the adsorption process at equilibrium.
This behavior is represented by a bell-shaped curve.
Therefore,Option $C$ is correct.

(B) According to the Freundlich adsorption isotherm: $\frac{x}{m} = kP^{1/n}$.
Taking logarithm on both sides: $\log (x/m) = \log k + \frac{1}{n} \log P$.
This equation represents a straight line $y = mx + c$,where the slope is $\frac{1}{n}$ and the intercept is $\log k$.
Given slope $= \tan 45^{\circ} = 1$,so $\frac{1}{n} = 1$.
Given intercept $= \log k = 0.3$. Since $\log 2 \approx 0.301$,we have $k = 2$.
At $P = 1 \ atm$,the amount adsorbed $\frac{x}{m} = kP^{1/n} = 2 \times (1)^1 = 2$.

(D) Physical adsorption (physisorption) is an exothermic process. According to Le Chatelier's principle,an increase in temperature decreases the extent of adsorption at a constant pressure.
From the given graph,at a constant pressure $p$,the extent of adsorption $\frac{x}{m}$ follows the order: $T_2 > T_3 > T_1$.
Since $\frac{x}{m}$ is inversely proportional to temperature for physisorption,the temperature order must be the reverse of the adsorption extent order.
Therefore,the correct order of temperatures is $T_1 < T_3 < T_2$.

(B) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K p^{\frac{1}{n}}$.
Here,$\frac{x}{m}$ represents the mass of adsorbate per unit mass of adsorbent,$p$ is the pressure,and $K$ and $n$ are constants depending on the nature of the adsorbent and adsorbate at a particular temperature.
Given that the value of $\frac{1}{n} = 1$.
Substituting this value into the equation,we get: $\frac{x}{m} = K p^{1} = K p$.
Therefore,the correct option is $B$.

(B) For the Freundlich adsorption isotherm equation:
$\frac{x}{m} = k p^{\frac{1}{n}}$
Taking the logarithm on both sides:
$\log \frac{x}{m} = \log (k p^{\frac{1}{n}})$
$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$
Comparing this with the linear equation $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log p$,slope $m = \frac{1}{n}$,and intercept $c = \log k$.
Thus,the intercept of the plot is $\log k$.

(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 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 a graph between $\log \frac{x}{m}$ and $\log p$ yields a straight line.

(C) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k p^{1/n}$.
Taking the logarithm on both sides,we get: $\log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log p$.
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 $\log (x / m)$ on the $y$-axis and $\log p$ on the $x$-axis yields a straight line with a positive slope of $1/n$ and an intercept of $\log k$ on the $y$-axis.

(B) The Freundlich adsorption isotherm is given by the empirical relationship:
$\frac{x}{m} = K p^{1/n}$
where:
$x$ is the mass of the adsorbate,
$m$ is the mass of the adsorbent,
$p$ is the pressure,
$K$ and $n$ are constants that depend on the nature of the adsorbent and adsorbate at a particular temperature.