The catalytic poisons decrease the activity of the catalyst because they are preferentially adsorbed on the surface of catalyst.

As excess of $$KI$$  has been added, $${I^ - }$$  ions are adsorbed on $$AgI$$  forming a fixed layer ( and giving it a negative charge ). It then attracts the counter ions $$\left( {{K^ + }} \right)$$  from the medium forming a second layer ( diffused layer ).

$$S{b_2}{S_3}$$  is a negatively charged sol. The most effective coagulating agent will be electrolyte with highest positive charge on the cation, $$viz.,A{l_2}{\left( {S{O_4}} \right)_3}.$$    The order is $$A{l^{3 + }} > C{a^{2 + }} > N{a^ + } > NH_4^ + $$

According to Hardy-Schulze rule "The amount of electrolyte required to coagulate a fixed amount of a sol depends upon the sign of charge and valency of the flocculating ion."
Thus, the coagulating power vary in the order.
$$A{l^{3 + }} > Z{n^{2 + }} > N{a^ + }$$

$$\eqalign{ & {\text{Mass of acetic acid adsorbed by}}\,\,2\,g\,\,{\text{charcoal}} \cr & = 100 \times {10^{ - 3}} \times \left( {0.6 - 0.5} \right) \times 60 \cr & \Rightarrow 0.6;\frac{x}{m} = \frac{{0.6}}{2} \Rightarrow 0.3 \cr} $$

This is an example of heterogeneous catalysis as the catalyst is present in different phase than that of the reactants.
\[4N{{H}_{3\left( g \right)}}+5{{O}_{2\left( g \right)}}\xrightarrow{P{{t}_{\left( s \right)}}}\]       \[4N{{O}_{\left( g \right)}}+6{{H}_{2}}{{O}_{\left( g \right)}}\]

van der Waals' adsorption i.e., physisorption increases at low temperature and high pressure.

The lyophobic sols are less stable than lyophilic sols. The lyophilic sols are thus used to protect the lyophobic sols. This property of lyophilic sols is known as protective action of lyophilic sols which is represented by gold number.

$$\eqalign{ & \frac{x}{m} = K{P^{\frac{1}{n}}} \cr & {\text{log}}\,\frac{x}{m} = {\text{log}}\,K + \frac{1}{n} \times {\text{log}}\,P \cr & {\text{Slope}} = \frac{1}{n} \cr} $$

According to Langmuir's Model of adsorption of a gas on a solid surface the mass of gas adsorbed $$(x)$$ per gram of the adsorbent $$(m)$$ is directly proportional to the pressure of the gas $$(p)$$ at constant temperature i.e.
$$\frac{{\text{x}}}{{\text{m}}} \propto {\text{p}}$$