(1) Enthalpy of adsorption ($$\Delta$$H_ads) :

Enthalpy change when 1 mole of adsorbate is adsorbed on the surface of adsorbent, is known as enthalpy of adsorption.

$$\to$$ Sign of enthalpy of adsorption :-

$$\Delta$$H_ads = $$-$$ve, [Exothermic]

Whenever adsorbate get's adsorbed on the surface of adsorbent then a new bond formed so energy get's released. That is why exothermic.

Exception : Adsorption of H₂O on the surface glass is endothermic.

(2) Entropy of adsorption : ($$\Delta$$S_ads)

Entropy change when 1 mole of adsorbate is adsorbed on the surface of adsorbent, is known as entropy of adsorption.

$$\to$$ Sign of entropy of adsorption :-

$$\Delta$$S_ads = $$-$$Ve

As free motion of molecules decreases when adsorbate get's adsorber on the surface of adsorbent.

(3) Gibb's energy change of adsorption : ($$\Delta$$G_ads)

Gibb's energy change when 1 mol of adsorbate is adsorbed on the surface of adsorbent is known as Gibb's energy change of adsorption.

Sign of Gibb's energy change of adsorption :-

We know, when

$$\Delta$$G < 0 $$\Rightarrow$$ Process spontaneous

$$\Delta$$G > 0 $$\Rightarrow$$ Process non-spontaneous

$$\Delta$$G = 0 $$\Rightarrow$$ Process Equilibrium

To keep $$\Delta$$G negative, magnitude of $$\Delta$$H should be high and magnitude of ($$-$$ T$$\Delta$$S) is low. It is possible when T is low as for 1 mole of adsorbate value of $$\Delta$$H and $$\Delta$$S is fixed so only change in T can make value of $$-$$T$$\Delta$$S high or low. If magnitude of $$-$$T$$\Delta$$S is greater than magnitude of $$\Delta$$H then $$\Delta$$G becomes positive process becomes non-spontaneous.

In case of adsorption of gases on solid surface, gases which can be more easily liquified can be more easily adsorbed. Gases having high critical temperature are easily liquified because Vanderwall's force of attraction is maximum at critical temperature (T_C).

T_C = $$\frac{8a}{27R_b}$$

Whose T_C (critical temperature) is more that gas has more "a" (Vanderwall constant for force of attraction)

$$\therefore$$ Higher the value of "a" then more easily will be the liquification of the gas.

Note :

(1) Value of 'a' for H₂ $$ \simeq $$ 0

(2) Value of 'a' for N₂ < Value of O₂ as molar mass of N₂ = 28 and molar mass of O₂ = 32.

(3) As for non polar gases we find value of "a" based on molar mass. Whose molar mass is more will have more "a" in case of non polar gases.

(4) For polar gases there is no rule to compare value of "a". You have to remember the order.

Order : SO₂ > NH₃ > CO₂ > CO > O₂ > H₂

Graph of $$\frac{x}{m}$$ vs T in Physisorption :

Graph of $$\frac{x}{m}$$ vs T in Chemisorption :

Here, $$\frac{x}{m}$$ = Extent of adsorption

x = mass of adsorbate

m = mass of adsorbent

Experimentally found that, for both physisorption and Chemisorption

Extent of adsorption $$\propto$$ Pressure

$$\Rightarrow\frac{x}{m}\propto p$$

In 1909 he studied extent of adsorption vs pressure experimentally. And he found that

$${x \over m} \propto {(p)^{{1 \over n}}}$$ [Experimentally found]

$$ \Rightarrow $$ $${x \over m} = k{p^{{1 \over n}}}$$

Where, k and n are constant which depends on nature of gas, solid and temperature.

where, $$n \ge 1$$

$$ \Rightarrow 0 < {1 \over n} \le 1$$

Most of the time $$\frac{1}{n}$$ is in between 0.1 to 0.5.

[Note : It is obeyed when adsorbate forms single or monolayer on the surface of adsorbent. That means it is applicable only for physisorption]

We have,

$${x \over m} = k{p^{{1 \over n}}}$$

Taking log both sides we get,

$$\log \left( {{x \over m}} \right) = \log \left( {k{p^{{1 \over n}}}} \right)$$

$$ \Rightarrow \log \left( {{x \over m}} \right) = \log k + {1 \over n}\log p$$

Comparing with equation $$y = mx + c$$, we get

$$y = \log \left( {{x \over m}} \right)$$

$$m = {1 \over n}$$

$$x = \log p$$

$$c = \log k$$

$$\tan \theta = m = {1 \over n}$$

He found those following conditions

$$\bullet$$ $${x \over m} \propto p$$ when n = 1 ; it happens at low pressure.

$$\bullet$$ $${x \over m} \propto {p^{{1 \over n}}}$$ when n > 1 ; at moderate pressure.

$$\bullet$$ $${x \over m} \propto {p^0}$$ (At high pressure)

$$\therefore$$ Freundlich adsorption isotherm fails at high pressure.

In case of Freundlich adsorption isotherm $${x \over m}$$ vs p graph :

At low pressure it is straight line.

And at high pressure, $${x \over m}$$ independent of p.

After the failure of Freundlich adsorption isotherm at high pressure Langmuir introduce his new adsorption isotherm at 1916.

He made some assumption in his adsorption isotherm $$\to$$

Assumptions :

(1) Adsorption of gas on solid surface only occur till whole of surface of solid is covered by uniform monolayer of gas molecule.

(2) Gass behaving ideally during adsorption.

(3) Already adsorbed gas molecules will not interact with each other.

(4) There occurs two opposite type of processes $$\to$$

(a) adsorption or condensation of gas molecules on free surface.

(b) desorption or evaporation of adsorbed molecules from surface.

We found that

$$ \Rightarrow \theta = {{KP} \over {1 + KP}}$$ [Let $${{{K_a}} \over {{K_d}}} = K$$]

We know extent of adsorption $$\left( {{x \over m}} \right)$$ means mass of adsorbate adsorbed per unit gram of adsorbent.

And $$\theta$$ means fraction of surface covered by molecule, it means how much surface of adsorbent is adsorbed by the adsorbate.

So, when $$\theta$$ is more then extent of adsorption $$\left( {{x \over m}} \right)$$ is more.

$$\therefore$$ Extent of adsorption $$\left( {{x \over m}} \right)$$ is directly proportion to $$\theta$$.

$$\frac{x}{m}\propto \theta$$

$$ \Rightarrow {x \over m} = K'\theta $$

$$ \Rightarrow {x \over m} = K' \times {{KP} \over {1 + KP}}$$

$$ \Rightarrow {x \over m} = {{K'KP} \over {1 + KP}}$$

$$ \Rightarrow {x \over m} = {{aP} \over {1 + bP}}$$

Let $$K'K = a$$ (New constant)

$$K = b$$ (Another constant)

Also $${1 \over {{x \over m}}} = {{1 + bP} \over {aP}} = {1 \over {aP}} + {b \over a}$$

$$ \Rightarrow {1 \over {\left( {{x \over m}} \right)}} = {b \over a} + \left( {{1 \over a}} \right){1 \over P}$$

Graph between $${1 \over {\left( {{x \over m}} \right)}}$$ and $${1 \over P}$$ :